Revised Edition: 2016 ISBN All rights reserved.

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2 Revised Edition: 2016 ISBN All rights reserved. Published by: Library Press 48 West 48 Street, Suite 1116, New York, NY 10036, United States

3 Table of Contents Chapter 1 - Introduction to Triangle Chapter 2 - Altitude and Angle Bisector Theorem Chapter 3 - Centroid and Ceva's Theorem Chapter 4 - Fermat Point and Heron's Formula Chapter 5 - Incircle & Excircles of a Triangle and Inertia Tensor of Triangle Chapter 6 - Law of Cosines and Law of Sines Chapter 7 - Equilateral Triangle and Heronian Triangle Chapter 8 - Bell Number Chapter 9 - Binomial Coefficient Chapter 10 - Boustrophedon Transform and Eulerian Number Chapter 11 - Gilbreath's Conjecture and Lah Number Chapter 12 - Leibniz Harmonic Triangle and Narayana Number Chapter 13 - Pascal Matrix Chapter 14 - Pascal's Pyramid Chapter 15 - Pascal's Simplex Chapter 16 - Pascal's Triangle

4 Chapter 1 Introduction to Triangle Triangle A triangle Edges and vertices 3 Schläfli symbol {3} (for equilateral) Area Internal angle (degrees) various methods; see below 60 (for equilateral) A triangle is one of the basic shapes of geometry: a polygon with three corners or vertices and three sides or edges which are line segments. A triangle with vertices A, B, and C is denoted ABC. In Euclidean geometry any three non-collinear points determine a unique triangle and a unique plane (i.e. a two-dimensional Euclidean space).

5 Types of triangles Euler diagram of types of triangles, using the definition that isosceles triangles have at least 2 equal sides, i.e. equilateral triangles are isosceles. By relative lengths of sides Triangles can be classified according to the relative lengths of their sides: In an equilateral triangle all sides have the same length. An equilateral triangle is also a regular polygon with all angles measuring 60. In an isosceles triangle, two sides are equal in length. An isosceles triangle also has two angles of the same measure; namely, the angles opposite to the two sides of the same length; this fact is the content of the Isosceles triangle theorem. Some mathematicians define an isosceles triangle to have exactly two equal sides, whereas others define an isosceles triangle as one with at least two equal sides. The latter definition would make all equilateral triangles isosceles triangles. The Right Triangle, which appears in the Tetrakis square tiling, is isosceles. In a scalene triangle, all sides are unequal. The three angles are also all different in measure. Some (but not all) scalene triangles are also right triangles. Equilateral Isosceles Scalene In diagrams representing triangles (and other geometric figures), "tick" marks along the sides are used to denote sides of equal lengths-- the equilateral triangle has tick marks on all 3 sides, the isosceles on 2 sides. The scalene has single, double, and triple tick marks, indicating that no sides are equal. Similarly, arcs on the inside of the vertices are used to indicate equal angles. The equilateral triangle indicates all 3 angles are equal; the

6 isosceles shows 2 identical angles. The scalene indicates by 1, 2, and 3 arcs that no angles are equal. By internal angles Triangles can also be classified according to their internal angles, measured here in degrees. A right triangle (or right-angled triangle, formerly called a rectangled triangle) has one of its interior angles measuring 90 (a right angle). The side opposite to the right angle is the hypotenuse; it is the longest side of the right triangle. The other two sides are called the legs or catheti (singular: cathetus) of the triangle. Right triangles obey the Pythagorean theorem: the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse: a 2 + b 2 = c 2, where a and b are the lengths of the legs and c is the length of the hypotenuse. Special right triangles are right triangles with additional properties that make calculations involving them easier. One of the two most famous is the right triangle, where = 5 2. In this situation, 3, 4, and 5 are a Pythagorean Triple. The other one is an isosceles triangle that has 2 angles that each measure 45 degrees. Triangles that do not have an angle that measures 90 are called oblique triangles. A triangle that has all interior angles measuring less than 90 is an acute triangle or acute-angled triangle. A triangle that has one angle that measures more than 90 is an obtuse triangle or obtuse-angled triangle. A "triangle" with an interior angle of 180 (and collinear vertices) is degenerate. A triangle that has two angles with the same measure also has two sides with the same length, and therefore it is an isosceles triangle. It follows that in a triangle where all angles have the same measure, all three sides have the same length, and such a triangle is therefore equilateral. Right Obtuse Acute Oblique

7 Basic facts Triangles are assumed to be two-dimensional plane figures, unless the context provides otherwise. In rigorous treatments, a triangle is therefore called a 2-simplex. Elementary facts about triangles were presented by Euclid in books 1 4 of his Elements, around 300 BC. A triangle, showing exterior angle d The measures of the interior angles of the triangle always add up to 180 degrees (same color to point out they are equal). The measures of the interior angles of a triangle in Euclidean space always add up to 180 degrees. This allows determination of the measure of the third angle of any triangle given the measure of two angles. An exterior angle of a triangle is an angle that is a linear pair (and hence supplementary) to an interior angle. The measure of an exterior angle of a

8 triangle is equal to the sum of the measures of the two interior angles that are not adjacent to it; this is the exterior angle theorem. The sum of the measures of the three exterior angles (one for each vertex) of any triangle is 360 degrees. The sum of the lengths of any two sides of a triangle always exceeds the length of the third side, a principle known as the triangle inequality. Since the vertices of a triangle are assumed to be non-collinear, it is not possible for the sum of the length of two sides be equal to the length of the third side. Two triangles are said to be similar if every angle of one triangle has the same measure as the corresponding angle in the other triangle. The corresponding sides of similar triangles have lengths that are in the same proportion, and this property is also sufficient to establish similarity. A few basic theorems about similar triangles: If two corresponding internal angles of two triangles have the same measure, the triangles are similar. If two corresponding sides of two triangles are in proportion, and their included angles have the same measure, then the triangles are similar. (The included angle for any two sides of a polygon is the internal angle between those two sides.) If three corresponding sides of two triangles are in proportion, then the triangles are similar. Two triangles that are congruent have exactly the same size and shape: all pairs of corresponding interior angles are equal in measure, and all pairs of corresponding sides have the same length. (This is a total of six equalities, but three are often sufficient to prove congruence.) Some sufficient conditions for a pair of triangles to be congruent are: SAS Postulate: Two sides in a triangle have the same length as two sides in the other triangle, and the included angles have the same measure. ASA: Two interior angles and the included side in a triangle have the same measure and length, respectively, as those in the other triangle. (The included side for a pair of angles is the side that is common to them.) SSS: Each side of a triangle has the same length as a corresponding side of the other triangle. AAS: Two angles and a corresponding (non-included) side in a triangle have the same measure and length, respectively, as those in the other triangle. Hypotenuse-Leg (HL) Theorem: The hypotenuse and a leg in a right triangle have the same length as those in another right triangle. This is also called RHS (rightangle, hypotenuse, side). Hypotenuse-Angle Theorem: The hypotenuse and an acute angle in one right triangle have the same length and measure, respectively, as those in the other right triangle. This is just a particular case of the AAS theorem.

9 An important case: Side-Side-Angle (or Angle-Side-Side) condition: If two sides and a corresponding non-included angle of a triangle have the same length and measure, respectively, as those in another triangle, then this is not sufficient to prove congruence; but if the angle given is opposite to the longer side of the two sides, then the triangles are congruent. The Hypotenuse-Leg Theorem is a particular case of this criterion. The Side-Side-Angle condition does not by itself guarantee that the triangles are congruent because one triangle could be obtuse-angled and the other acuteangled. Using right triangles and the concept of similarity, the trigonometric functions sine and cosine can be defined. These are functions of an angle which are investigated in trigonometry. The Pythagorean theorem A central theorem is the Pythagorean theorem, which states in any right triangle, the square of the length of the hypotenuse equals the sum of the squares of the lengths of the two other sides. If the hypotenuse has length c, and the legs have lengths a and b, then the theorem states that The converse is true: if the lengths of the sides of a triangle satisfy the above equation, then the triangle has a right angle opposite side c. Some other facts about right triangles: The acute angles of a right triangle are complementary.

10 If the legs of a right triangle have the same length, then the angles opposite those legs have the same measure. Since these angles are complementary, it follows that each measures 45 degrees. By the Pythagorean theorem, the length of the hypotenuse is the length of a leg times 2. In a right triangle with acute angles measuring 30 and 60 degrees, the hypotenuse is twice the length of the shorter side, and the longer side is equal to the length of the shorter side times 3 : For all triangles, angles and sides are related by the law of cosines and law of sines (also called the cosine rule and sine rule). Points, lines, and circles associated with a triangle There are hundreds of different constructions that find a special point associated with (and often inside) a triangle, satisfying some unique property. Often they are constructed by finding three lines associated in a symmetrical way with the three sides (or vertices) and then proving that the three lines meet in a single point: an important tool for proving the existence of these is Ceva's theorem, which gives a criterion for determining when three such lines are concurrent. Similarly, lines associated with a triangle are often constructed by proving that three symmetrically constructed points are collinear: here Menelaus' theorem gives a useful general criterion. In this section just a few of the most commonly encountered constructions are explained. The circumcenter is the center of a circle passing through the three vertices of the triangle. A perpendicular bisector of a side of a triangle is a straight line passing through the midpoint of the side and being perpendicular to it, i.e. forming a right angle with it. The three perpendicular bisectors meet in a single point, the triangle's circumcenter; this point is the center of the circumcircle, the circle passing through all three vertices. The diameter of

11 this circle, called the circumdiameter, can be found from the law of sines stated above. The circumcircle's radius is called the circumradius. Thales' theorem implies that if the circumcenter is located on one side of the triangle, then the opposite angle is a right one. If the circumcenter is located inside the triangle, then the triangle is acute; if the circumcenter is located outside the triangle, then the triangle is obtuse. The intersection of the altitudes is the orthocenter An altitude of a triangle is a straight line through a vertex and perpendicular to (i.e. forming a right angle with) the opposite side. This opposite side is called the base of the altitude, and the point where the altitude intersects the base (or its extension) is called the foot of the altitude. The length of the altitude is the distance between the base and the vertex. The three altitudes intersect in a single point, called the orthocenter of the triangle. The orthocenter lies inside the triangle if and only if the triangle is acute. The intersection of the angle bisectors is the center of the incircle An angle bisector of a triangle is a straight line through a vertex which cuts the corresponding angle in half. The three angle bisectors intersect in a single point, the incenter, the center of the triangle's incircle. The incircle is the circle which lies inside the triangle and touches all three sides. Its radius is called the inradius. There are three other important circles, the excircles; they lie outside the triangle and touch one side as well as the

12 extensions of the other two. The centers of the in- and excircles form an orthocentric system. The intersection of the medians is the centroid A median of a triangle is a straight line through a vertex and the midpoint of the opposite side, and divides the triangle into two equal areas. The three medians intersect in a single point, the triangle's centroid or geometric barycenter. The centroid of a rigid triangular object (cut out of a thin sheet of uniform density) is also its center of mass: the object can be balanced on its centroid in a uniform gravitational field. The centroid cuts every median in the ratio 2:1, i.e. the distance between a vertex and the centroid is twice the distance between the centroid and the midpoint of the opposite side. Nine-point circle demonstrates a symmetry where six points lie on the edge of the triangle The midpoints of the three sides and the feet of the three altitudes all lie on a single circle, the triangle's nine-point circle. The remaining three points for which it is named are the midpoints of the portion of altitude between the vertices and the orthocenter. The radius of the nine-point circle is half that of the circumcircle. It touches the incircle (at the Feuerbach point) and the three excircles.

13 Euler's line is a straight line through the centroid (orange), orthocenter (blue), circumcenter (green) and center of the nine-point circle (red). The centroid (yellow), orthocenter (blue), circumcenter (green) and center of the ninepoint circle (red point) all lie on a single line, known as Euler's line (red line). The center of the nine-point circle lies at the midpoint between the orthocenter and the circumcenter, and the distance between the centroid and the circumcenter is half that between the centroid and the orthocenter. The center of the incircle is not in general located on Euler's line. If one reflects a median at the angle bisector that passes through the same vertex, one obtains a symmedian. The three symmedians intersect in a single point, the symmedian point of the triangle.

14 Computing the area of a triangle The area of a triangle can be demonstrated as half of the area of a paralellogram which has the same base length and height. Calculating the area of a triangle is an elementary problem encountered often in many different situations. The best known and simplest formula is: selection of frequently used formulae for the area of a triangle. where b is the length of the base of the triangle, and h is the height or altitude of the triangle. The term 'base' denotes any side, and 'height' denotes the length of a perpendicular from the vertex opposite the side onto the line containing the side itself. In 499 CE Aryabhata, a great mathematician-astronomer from the classical age of Indian mathematics and Indian astronomy, used this method in the Aryabhatiya (section 2.6). Although simple, this formula is only useful if the height can be readily found. For example, the surveyor of a triangular field measures the length of each side, and can find the area from his results without having to construct a 'height'. Various methods may be used in practice, depending on what is known about the triangle. The following is a Using vectors The area of a parallelogram embedded in a three-dimensional Euclidean space can be calculated using vectors. Let vectors AB and AC point respectively from A to B and from A to C. The area of parallelogram ABDC is then which is the magnitude of the cross product of vectors AB and AC. The area of triangle ABC is half of this,.

15 The area of triangle ABC can also be expressed in terms of dot products as follows: In two-dimensional Euclidean space, expressing vector AB as a free vector in Cartesian space equal to (x 1,y 1 ) and AC as (x 2,y 2 ), this can be rewritten as: Applying trigonometry to find the altitude h Using trigonometry The height of a triangle can be found through the application of trigonometry. Knowing SAS: Using the labels in the image on the left, the altitude is h = a sin γ. Substituting this in the formula Area = ½bh derived above, the area of the triangle can be expressed as: (where α is the interior angle at A, β is the interior angle at B, γ is the interior angle at C and c is the line AB). Furthermore, since sin α = sin (π - α) = sin (β + γ), and similarly for the other two angles: Knowing AAS:

16 and analogously if the known side is a or c. Knowing ASA: and analogously if the known side is b or c. Using coordinates If vertex A is located at the origin (0, 0) of a Cartesian coordinate system and the coordinates of the other two vertices are given by B = (x B, y B ) and C = (x C, y C ), then the area can be computed as ½ times the absolute value of the determinant For three general vertices, the equation is: In three dimensions, the area of a general triangle {A = (x A, y A, z A ), B = (x B, y B, z B ) and C = (x C, y C, z C )} is the Pythagorean sum of the areas of the respective projections on the three principal planes (i.e. x = 0, y = 0 and z = 0): Using line integrals The area within any closed curve, such as a triangle, is given by the line integral around the curve of the algebraic or signed distance of a point on the curve from an arbitrary oriented straight line L. Points to the right of L as oriented are taken to be at negative distance from L, while the weight for the integral is taken to be the component of arc length parallel to L rather than arc length itself.

17 This method is well suited to computation of the area of an arbitrary polygon. Taking L to be the x-axis, the line integral between consecutive vertices (x i,y i ) and (x i+1,y i+1 ) is given by the base times the mean height, namely (x i+1 x i )(y i + y i+1 )/2. The sign of the area is an overall indicator of the direction of traversal, with negative area indicating counterclockwise traversal. The area of a triangle then falls out as the case of a polygon with three sides. While the line integral method has in common with other coordinate-based methods the arbitrary choice of a coordinate system, unlike the others it makes no arbitrary choice of vertex of the triangle as origin or of side as base. Furthermore the choice of coordinate system defined by L commits to only two degrees of freedom rather than the usual three, since the weight is a local distance (e.g. x i+1 x i in the above) whence the method does not require choosing an axis normal to L. When working in polar coordinates it is not necessary to convert to cartesian coordinates to use line integration, since the line integral between consecutive vertices (r i,θ i ) and (r i+1,θ i+1 ) of a polygon is given directly by r i r i+1 sin(θ i+1 θ i )/2. This is valid for all values of θ, with some decrease in numerical accuracy when θ is many orders of magnitude greater than π. With this formulation negative area indicates clockwise traversal, which should be kept in mind when mixing polar and cartesian coordinates. Just as the choice of y-axis (x = 0) is immaterial for line integration in cartesian coordinates, so is the choice of zero heading (θ = 0) immaterial here. Using Heron's formula The shape of the triangle is determined by the lengths of the sides alone. Therefore the area can also be derived from the lengths of the sides. By Heron's formula: where is the semiperimeter, or half of the triangle's perimeter. Three equivalent ways of writing Heron's formula are

18 Formulas mimicking Heron's formula Three formulas have the same structure as Heron's formula but are expressed in terms of different variables. First, denoting the medians from sides a, b, and c respectively as m a,m b, and m c and their semi-sum (m a + m b + m c ) / 2 as σ, we have Next, denoting the altitudes from sides a, b, and c respectively as h a, h b, and h c,and denoting the semi-sum of the reciprocals of the altitudes as we have And denoting the semi-sum of the angles' sines as where D is the diameter of the circumcircle: Using Pick's Theorem The theorem states:, we have where I is the number of internal lattice points and B is the number of lattice points lying on the border of the polygon. Other area formulas Numerous other area formulas exist, such as where r is the inradius, and s is the semiperimeter;

19 for circumdiameter D; and for angle 90. Denoting the radius of the inscribed circle as r and the radii of the excircles as r 1, r 2, and r 3, the area can be expressed as In 1885, Baker gave a collection of over a hundred distinct area formulas for the triangle (although the reader should be advised that a few of them are incorrect). These include #9, #39a, #39b, #42, and #49: for circumradius (radius of the circumcircle) R, and Upper bound on the area The area of any triangle with perimeter p is less than or equal to holding if and only if the triangle is equilateral. with equality Computing the sides and angles There are various standard methods for calculating the length of a side or the size of an angle. Certain methods are suited to calculating values in a right-angled triangle; more complex methods may be required in other situations.

20 Trigonometric ratios in right triangles The hypotenuse is the side opposite the right angle, or defined as the longest side of a right-angled triangle, in this case h. A right triangle always includes a 90 (π/2 radians) angle, here with label C. Angles A and B may vary. Trigonometric functions specify the relationships among side lengths and interior angles of a right triangle. In right triangles, the trigonometric ratios of sine, cosine and tangent can be used to find unknown angles and the lengths of unknown sides. The sides of the triangle are known as follows: The opposite side is the side opposite to the angle we are interested in, in this case a. The adjacent side is the side that is in contact with the angle we are interested in and the right angle, hence its name. In this case the adjacent side is b. Sine, cosine and tangent The sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse. In our case Note that this ratio does not depend on the particular right triangle chosen, as long as it contains the angle A, since all those triangles are similar.

21 The cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse. In our case The tangent of an angle is the ratio of the length of the opposite side to the length of the adjacent side. In our case The acronym "SOH-CAH-TOA" is a useful mnemonic for these ratios. Inverse functions The inverse trigonometric functions can be used to calculate the internal angles for a right angled triangle with the length of any two sides. Arcsin can be used to calculate an angle from the length of the opposite side and the length of the hypotenuse Arccos can be used to calculate an angle from the length of the adjacent side and the length of the hypontenuse. Arctan can be used to calculate an angle from the length of the opposite side and the length of the adjacent side. In introductory geometry and trigonometry courses, the notation sin 1, cos 1, etc., are often used in place of arcsin, arccos, etc. However, the arcsin, arccos, etc., notation is standard in higher mathematics where trigonometric functions are commonly raised to powers, as this avoids confusion between multiplicative inverse and compositional inverse.

22 The sine, cosine and tangent rules A triangle with sides of length a, b and c and angles of α, β and γ respectively The law of sines, or sine rule, states that the ratio of the length of a side to the sine of its corresponding opposite angle is constant, that is For a triangle with length of sides a, b, c and angles of α, β, γ respectively, given two This ratio is equal to the diameter of the circumscribed circle of the given triangle. Another interpretation of this theorem is that every triangle with angles α, β and γ is similar to a triangle with side lengths equal to sinα, sinβ and sinγ. This triangle can be constructed by first constructing a circle of diameter 1, and inscribing in it two of the angles of the triangle. The length of the sides of that triangle will be sinα, sinβ and sinγ. The side whose length is sinα is opposite to the angle whose measure is α, etc. The law of cosines, or cosine rule, connects the length of an unknown side of a triangle to the length of the other sides and the angle opposite to the unknown side. As per the law: known lengths of a triangle a and b, and the angle between the two known sides γ (or the angle opposite to the unknown side c), to calculate the third side c, the following formula can be used: If the lengths of all three sides of any triangle are known the three angles can be calculated:

23 The law of tangents or tangent rule, is less known than the other two. It states that: It is not used very often, but can be used to find a side or an angle when you know two sides and an angle or two angles and a side. Further formulas for general Euclidean triangles The following formulas are also true for all Euclidean triangles: and and equivalently for m b and m c, relating the medians and the sides;, for semiperimeter s, where the bisector length is measured from the vertex to where it meets the opposite side; and the following formulas involving the circumradius R and the inradius r : in terms of the altitudes, and,

24 . Suppose two adjacent but non-overlapping triangles share the same side of length f and share the same circumcircle, so that the side of length f is a chord of the circumcircle and the triangles have side lengths (a, b, f) and (c, d, f), with the two triangles together forming a cyclic quadrilateral with side lengths in sequence (a, b, c, d). Then :84 Let M be the centroid of a triangle with vertices A, B, and C, and let P be any interior point. Then the distances between the points are related by :174 Let p a, p b, and p c be the distances from the centroid to the sides of lengths a, b, and c. Then :173 and The product of two sides of a triangle equals the altitude to the third side times the diameter of the circumcircle. Carnot's Theorem states that the sum of the distances from the circumcenter to the three sides equals the sum of the circumradius and the inradius. Here a segment's length is considered to be negative if and only if the segment lies entirely outside the triangle. Euler's theorem states that the distance d between the circumcenter and the incenter is given by. d 2 = R(R 2r) where R is the circumradius and r is the inradius. Thus for all triangles R 2r. If we denote that the orthocenter divides one altitude into segments of lengths u and v, another altitude into segment lengths w and x, and the third altitude into segment lengths y and z, then uv = wx = yz.

25 The distance from a side to the circumcenter equals half the distance from the opposite vertex to the orthocenter. The sum of the squares of the distances from the vertices to the orthocenter plus the sum of the squares of the sides equals twelve times the square of the circumradius. Morley's trisector theorem The Morley triangle, resulting from the trisection of each interior angle Morley's trisector theorem states that in any triangle, the three points of intersection of the adjacent angle trisectors form an equilateral triangle, called the Morley triangle. Figures inscribed in a triangle As discussed above, every triangle has a unique inscribed circle (incircle) that is interior to the triangle and tangent to all three sides. Every triangle has a unique Steiner inellipse which is interior to the triangle and tangent at the midpoints of the sides. Marden's theorem shows how to find the foci of this ellipse. Every triangle has three inscribed squares (squares in its interior such that all four of a square's vertices lie on a side of the triangle, so two of them lie on the same side and hence one side of the square coincides with part of a side of the triangle). However, in the case of a right triangle two of the squares coincide and have a vertex at the triangle's right angle, so a right triangle has only two distinct inscribed squares. Within a given triangle, a longer common side is associated with a smaller inscribed square. If an inscribed square has side of length s and the triangle has a side of length x, part of which side coincides with a side of the square, then s, x, and the triangle's area A are related according to

26 The largest possible ratio of the area of the inscribed square to the area of the triangle is 1/2, which occurs when x 2 = 2A. Non-planar triangles A non-planar triangle is a triangle which is not contained in a (flat) plane. Some examples of non-planar triangles in non-euclidean geometries are spherical triangles in spherical geometry and hyperbolic triangles in hyperbolic geometry. While the measures of the internal angles in planar triangles always sum to 180, a hyperbolic triangle has measures of angles that sum to less than 180, and a spherical triangle has measures of angles that sum to more than 180. A hyperbolic triangle can be obtained by drawing on a negatively curved surface, such as a saddle surface, and a spherical triangle can be obtained by drawing on a positively curved surface such as a sphere. Thus, if one draws a giant triangle on the surface of the Earth, one will find that the sum of the measures of its angles is greater than 180 ; in fact it will be between 180 and 540. In particular it is possible to draw a triangle on a sphere such that the measure of each of its internal angles is equal to 90, adding up to a total of 270. Specifically, on a sphere the sum of the angles of a triangle is 180 (1+4f), where f is the fraction of the sphere's area which is enclosed by the triangle. For example, suppose that we draw a triangle on the Earth's surface with vertices at the North Pole, at a point on the equator at 0 longitude, and a point on the equator at 90 West longitude. The great circle line between the latter two points is the equator, and the great circle line between either of those points and the North Pole is a line of longitude; so there are right angles at the two points on the equator. Moreover, the angle at the North Pole is also 90 because the other two vertices differ by 90 of longitude. So the sum of the angles in this triangle is =270. The triangle encloses 1/4 of the northern hemisphere (90 /360 as viewed from the North Pole) and therefore 1/8 of the Earth's surface, so in the formula f = 1/8; thus the formula correctly gives the sum of the triangle's angles as 270. From the above angle sum formula we can also see that the Earth's surface is locally flat: If we draw an arbitrarily small triangle in the neighborhood of one point on the Earth's surface, the fraction f of the Earth's surface which is enclosed by the triangle will be arbitrarily close to zero. In this case the angle sum formula simplifies to 180, which we know is what Euclidean geometry tells us for triangles on a flat surface.

27 Chapter 2 Altitude and Angle Bisector Theorem Altitude Orthocenter In geometry, an altitude of a triangle is a straight line through a vertex and perpendicular to (i.e. forming a right angle with) the opposite side or an extension of the opposite side. The intersection between the (extended) side and the altitude is called the foot of the altitude. This opposite side is called the base of the altitude. The length of the altitude is the distance between the base and the vertex. Altitudes can be used to compute the area of a triangle: one half of the product of an altitude's length and its base's length equals the triangle's area, as well as being related to the sides of the triangle through trigonometric functions. In an isosceles triangle (a triangle with two congruent sides), the altitude having the incongruent side as its base will have the midpoint of that side as its foot. Also the altitude having the incongruent side as its base will form the angle bisector of the vertex. In a right triangle, the altitude with the hypotenuse as base divides the hypotenuse into two lengths p and q. If we denote the length of the altitude by h, we then have the relation h 2 = pq.

28 The orthocenter The three altitudes intersect in a single point, called the orthocenter of the triangle. The orthocenter lies inside the triangle (and consequently the feet of the altitudes all fall on the triangle) if and only if the triangle is not obtuse (i.e. does not have an angle greater than a right angle). The orthocenter, along with the centroid, circumcenter and center of the nine-point circle all lie on a single line, known as the Euler line. The center of the nine-point circle lies at the midpoint between the orthocenter and the circumcenter, and the distance between the centroid and the circumcenter is half that between the centroid and the orthocenter. The orthocenter is also equidistant from each of the sides of the triangle. The isogonal conjugate and also the complement of the orthocenter is the circumcenter. Four points in the plane such that one of them is the orthocenter of the triangle formed by the other three are called an orthocentric system or orthocentric quadrangle. Let A, B, C denote the angles of the reference triangle, and let a = BC, b = CA, c = AB be the sidelengths. The orthocenter has trilinear coordinates sec A : sec B : sec C and barycentric coordinates ((a 2 + b 2 c 2 )(a 2 b 2 + c 2 ):(a 2 + b 2 c 2 )( a 2 + b 2 + c 2 ):(a 2 b 2 + c 2 )( a 2 + b 2 + c 2 )).

29 Orthic triangle Triangle abc is the orthic triangle of triangle ABC If the triangle ABC is oblique (not right-angled), the points of intersection of the altitudes with the sides of the triangles form another triangle, A'B'C', called the orthic triangle or altitude triangle. It is the pedal triangle of the orthocenter of the original triangle. Also, the incenter (that is, the center for the inscribed circle) of the orthic triangle is the orthocenter of the original triangle. The orthic triangle is closely related to the tangential triangle, constructed as follows: let L A be the line tangent to the circumcircle of triangle ABC at vertex A, and define L B and L C analogously. Let A" = L B L C, B" = L C L A, C" = L C L A. The tangential triangle, A"B"C", is homothetic to the orthic triangle. The orthic triangle provides the solution to Fagnano's problem which in 1775 asked for the minimum perimeter triangle inscribed in a given acute-angle triangle. The orthic triangle of an acute triangle gives a triangular light route. Trilinear coordinates for the vertices of the orthic triangle are given by

30 A' = 0 : sec B : sec C B' = sec A : 0 : sec C C' = sec A : sec B : 0 Trilinear coordinates for the vertices of the tangential triangle are given by A" = a : b : c B" = a : b : c C" = a : b : c Some additional altitude theorems Equilateral triangle theorem Inradius theorems For any point P within an equilateral triangle, the sum of the perpendiculars to the three sides is equal to the altitude of the triangle. Consider an arbitrary triangle with sides a, b, c and with corresponding altitudes α, β, η. The altitudes and incircle radius r are related by Let c, h, s be the sides of 3 squares associated with the right triangle; the square on the hypotenuse, and the triangle's 2 inscribed squares respectively. The sides of these squares (c>h>s) and the incircle radius r are related by a similar formula: The symphonic theorem * In the case of the right triangle, the sides of the 3 squares c, h, s are related to each other by the symphonic theorem, as are the 3 altitudes α, β, η. The symphonic theorem states that triples (c 2,h 2,s 2 ) and (α 2,β 2,η 2 ) are harmonic, and that triples are Pythagorean: and

31 Area theorem Denoting the altitudes from sides a, b, and c respectively as h a, h b, and h c,and denoting the semi-sum of the reciprocals of the altitudes as we have Angle bisector theorem In this diagram, BD:DC = AB:AC In geometry, the angle bisector theorem relates the length of the side opposite one angle of a triangle to the lengths of the other two sides of the triangle. Consider a triangle ABC. Let the angle bisector of angle A intersect side BC at a point D. The angle bisector theorem states that the ratio of the length of the line segment BD to the length of segment DC is equal to the ratio of the length of side AB to the length of side AC The generalized angle bisector theorem states that if D lies on BC, then This reduces to the previous version if AD is the bisector of BAC.

32 Proof Let B 1 be the base of altitude in the triangle ABD through B and let C 1 be the base of altitude in the triangle ACD through C. Then, DB 1 B and DC 1 C are right, while the angles B 1 DB and C 1 DC are congruent if D lies on the segment BC and they are identical otherwise, so the triangles DB 1 B and DC 1 C are similar (AAA), which implies that The angle bisector theorem is commonly used when you have angle bisectors and side lengths.

33 Chapter 3 Centroid and Ceva's Theorem Centroid Centroid of a triangle In geometry, the centroid, geometric center, or barycenter of a plane figure or twodimensional shape X is the intersection of all straight lines that divide X into two parts of equal moment about the line. Informally, it is the "average" (arithmetic mean) of all points of X. The definition extends to any object X in n-dimensional space: its centroid is the intersection of all hyperplanes that divide X into two parts of equal moment. In physics, the word centroid means the geometric center of the object's shape, as above, but barycenter may also mean its physical center of mass or the center of gravity, depending on the context. Informally, the center of mass (and center of gravity in a uniform gravitational field) is the average of all points, weighted by the local density or specific weight. If a physical object has uniform density, then its center of mass is the same as the centroid of its shape. In geography, the centroid of a region of the Earth's surface, projected radially onto said surface, is known as its geographical center. Properties The geometric centroid of a convex object always lies in the object. A non-convex object might have a centroid that is outside the figure itself. The centroid of a ring or a bowl, for example, lies in the object's central void.

34 If the centroid is defined, it is a fixed point of all isometries in its symmetry group. In particular, the geometric centroid of an object lies in the intersection of all its hyperplanes of symmetry. The centroid of many figures (regular polygon, regular polyhedron, cylinder, rectangle, rhombus, circle, sphere, ellipse, ellipsoid, superellipse, superellipsoid, etc.) can be determined by this principle alone. In particular, the centroid of a parallelogram is the meeting point of its two diagonals. This is not true for other quadrilaterals. For the same reason, the centroid of an object with translational symmetry is undefined (or lies outside the enclosing space), because a translation has no fixed point. Locating the centroid Plumb line method The centroid of a uniform two-dimensional lamina, such as (a) below, may be determined, experimentally, by using a plumbline and a pin to find the center of mass of a thin body of uniform density having the same shape. The body is held by the pin inserted at a point near the body's perimeter, in such a way that it can freely rotate around the pin; and the plumb line is dropped from the pin (b). The position of the plumbline is traced on the body. The experiment is repeated with the pin inserted at a different point of the object. The intersection of the two lines is the centroid of the figure (c). (a)

35 (b) (c) This method can be extended (in theory) to concave shapes where the centroid lies outside the shape, and to solids (of uniform density), but the positions of the plumb lines need to be recorded by means other than drawing. Of a finite set of points The centroid of a finite set of k points in is

36 By geometric decomposition The centroid of a plane figure X can be computed by dividing it into a finite number of simpler figures, computing the centroid C i and area A i of each part, and then computing Holes in the figure X, overlaps between the parts, or parts that extend outside the figure can all be handled using negative areas A i. Namely, the measures A i should be taken with positive and negative signs in such a way that the sum of the signs of A i for all parts that enclose a given point p is 1 if p belongs to X, and 0 otherwise. For example, the figure below (a) is easily divided into a square and a triangle, both with positive area; and a circular hole, with negative area (b). (a) (b) (c)

37 The centroid of each part can be found in any list of centroids of simple shapes (c). Then the centroid of the figure is the weighted average of the three points. The horizontal position of the centroid, from the left edge of the figure is The same formula holds for any three-dimensional objects, except that each A i should be the volume of X i, rather than its area. It also holds for any subset of, for any dimension d, with the areas replaced by the d-dimensional measures of the parts. By integral formula The centroid of a subset X of can also be computed by the integral where the integral is taken over the whole space, and g is the characteristic function of the subset, which is 1 inside X and 0 outside it. Note that the denominator is simply the measure of the set X (However, this formula cannot be applied if the set X has zero measure, or if either integral diverges.) Another formula for the centroid is where C k is the kth coordinate of C, and S k (z) is the measure of the intersection of X with the hyperplane defined by the equation x k = z. Again, the denominator is simply the measure of X. For a plane figure, in particular, the barycenter coordinates are where A is the area of the figure X; S y (x) is the length of the intersection of X with the vertical line at abscissa x; and S x (y) is the analogous quantity for the swapped axes.

38 Of an L-shaped object This is a method of determining the centroid of an L-shaped object. 1. Divide the shape into two rectangles, as shown in fig 2. Find the centroids of these two rectangles by drawing the diagonals. Draw a line joining the centroids. The centroid of the shape must lie on this line AB. 2. Divide the shape into two other rectangles, as shown in fig 3. Find the centroids of these two rectangles by drawing the diagonals. Draw a line joining the centroids. The centroid of the L-shape must lie on this line CD. 3. As the centroid of the shape must lie along AB and also along CD, it is obvious that it is at the intersection of these two lines, at O. The point O might not lie inside the L-shaped object. Of triangle and tetrahedron The centroid of a triangle is the point of intersection of its medians (the lines joining each vertex with the midpoint of the opposite side). The centroid divides each of the medians in the ratio 2:1, which is to say it is located ⅓ of the perpendicular distance between each side and the opposing point (see figures above). Its Cartesian coordinates are the means of the coordinates of the three vertices. That is, if the three vertices are a = (x a,y a ), b = (x b,y b ), and c = (x c,y c ), then the centroid is The

39 centroid is therefore at in barycentric coordinates. The centroid is also the physical center of mass if the triangle is made from a uniform sheet of material; or if all the mass is concentrated at the three vertices, and evenly divided among them. On the other hand, if the mass is distributed along the triangle's perimeter, with uniform linear density, the center of mass may not coincide with the geometric centroid. The area of the triangle is 1.5 times the length of any side times the perpendicular distance from the side to the centroid. Similar results hold for a tetrahedron: its centroid is the intersection of all line segments that connect each vertex to the centroid of the opposite face. These line segments are divided by the centroid in the ratio 3:1. The result generalizes to any n-dimensional simplex in the obvious way. If the set of vertices of a simplex is considering the vertices as vectors, the centroid is, then The geometric centroid coincides with the center of mass if the mass is uniformly distributed over the whole simplex, or concentrated at the vertices as n equal masses. The isogonal conjugate of a triangle's centroid is its symmedian point. Centroid of polygon The centroid of a non-self-intersecting closed polygon defined by n vertices (x 0,y 0 ), (x 1,y 1 ),..., (x n 1,y n 1 ) is the point (C x, C y ), where and where A is the polygon's signed area,

40 In these formulas, the vertices are assumed to be numbered in order of their occurrence along the polygon's perimeter, and the vertex (x n, y n ) is assumed to be the same as (x 0, y 0 ). Note that if the points are numbered in clockwise order the area A, computed as above, will have a negative sign; but the centroid coordinates will be correct even in this case. Centroid of cone or pyramid The centroid of a cone or pyramid is located on the line segment that connects the apex to the centroid of the base. For a solid cone or pyramid, the centroid is 1/4 the distance from the base to the apex. For a cone or pyramid that is just a shell (hollow) with no base, the centroid is 1/3 the distance from the base plane to the apex. Ceva's theorem Ceva's theorem, case 1: the three lines are concurrent at a point O inside ABC

41 Ceva's theorem, case 2: the three lines are concurrent at a point O outside ABC Ceva's theorem is a theorem in elementary geometry. Given a triangle ABC, and points D, E, and F that lie on lines BC, CA, and AB respectively, the theorem states that lines AD, BE and CF are concurrent if and only if where AF indicates the directed distance between A and F (i.e. distance in one direction along a line is counted as positive, and in the other direction is counted as negative). There is also an equivalent trigonometric form of Ceva's Theorem, that is, AD,BE,CF concur if and only if The theorem was proved by Giovanni Ceva in his 1678 work De lineis rectis, but it was also proven much earlier by Yusuf Al-Mu'taman ibn Hűd, an eleventh-century king of Zaragoza.

42 Associated with the figures are several terms derived from Ceva's name: cevian (the lines AD, BE, CF are the cevians of O), cevian triangle (the triangle DEF is the cevian triangle of O); cevian nest, anticevian triangle, Ceva conjugate. (Ceva is pronounced Chay'va; cevian is pronounced chev'ian.) Proof of the theorem Suppose AD, BE and CF intersect at a point O. Because and have the same height, we have Similarly, From this it follows that Similarly, and Multiplying these three equations gives as required. Conversely, suppose that the points D, E and F satisfy the above equality. Let AD and BE intersect at O, and let CO intersect AB at F'. By the direction we have just proven,

43 Comparing with the above equality, we obtain Adding 1 to both sides and using AF' + F'B = AF + FB = AB (case 1), or subtracting both sides from 1 and using F'B AF' = FB AF = AB (case 2) we obtain Thus F'B = FB, so that F and F' coincide (recalling that the distances are directed). Therefore AD, BE and CF = CF' intersect at O, and both implications are proven. For the trigonometric form of the theorem, one approach is to view the three cevians, concurrent at point O, as partitioning the triangle into three smaller triangles:,, and. Applying the law of sines to each triangle we get: When the three equations are multiplied, the right side will equal 1. The six sines on the left side, when rearranged, will yield the expression given in the theorem. Generalizations The theorem can be generalized to higher dimensional simplexes using barycentric coordinates. Define a cevian of an n-simplex as a ray from each vertex to a point on the opposite (n-1)-face (facet). Then the cevians are concurrent if and only if a mass distribution can be assigned to the vertices such that each cevian intersects the opposite facet at its center of mass. Moreover, the intersection point of the cevians is the center of mass of the simplex. Routh's theorem gives the area of the triangle formed by three cevians in the case that they are not concurrent. Ceva's theorem can be obtained from it by setting the area equal to zero and solving. The analogue of the theorem for general polygons in the plane has been known since the early nineteenth century (Grünbaum & Shephard 1995, p. 266). The theorem has also been generalized to triangles on other surfaces of constant curvature (Masal'tsev 1994).

44 Chapter 4 Fermat Point and Heron's Formula Fermat point Fig 1. Construction for the first isogonic center, X(13) In geometry the Fermat point of a triangle, also called Torricelli point, is the solution to the problem of finding a point such that the total distance from the three vertices of the triangle to the point is the minimum possible. It is so named because this problem is first raised by Fermat in a private letter. The Fermat point gives solution to the Steiner tree problem for three points.

45 Construction In order to locate the Fermat point of a triangle with largest angle at most 120 : 1. Construct three regular triangles out of the three sides of the given triangle. 2. For each new vertex of the regular triangle, draw a line from it to the opposite triangle's vertex. 3. These three lines intersect at the Fermat point. When a triangle has an angle greater than 120 the Fermat point is sited at the obtuse angled vertex. The Fermat point has a near-identical twin called the first isogonic center or X(13) and it is important not to confuse the two. Their relationship is explored below. Geometry Fig 2. Geometry of the Fermat point

46 Given any Euclidean triangle ABC and an arbitrary point P let d(p) = PA+PB+PC. The aim of this section is to identify a point P 0 such that d(p 0 ) < d(p) for all P P 0. If such a point exists then it will be the Fermat point. In what follows Δ will denote the points inside the triangle and will be taken to include its boundary Ω. A key result that will be used is the dogleg rule which asserts that if a triangle and a polygon have one side in common and the rest of the triangle lies inside the polygon then the triangle has a shorter perimeter than the polygon. [ If AB is the common side extend AC to cut the polygon at X. Then by the triangle inequality the polygon perimeter > AB+AX+XB = AB+AC+CX+XB AB+AC+BC. ] Let P be any point outside Δ. Associate each vertex with its remote zone; that is, the halfplane beyond the (extended) opposite side. These 3 zones cover the entire plane except for Δ itself and P clearly lies in either one or two of them. If P is in two (say the B and C zones intersection) then setting P' = A implies d(p') = d(a) < d(p) by the dogleg rule. Alternatively if P is in only one zone, say the A-zone, then d(p') < d(p) where P' is the intersection of AP and BC. So for every point P outside Δ there exists a point P' in Ω such that d(p') < d(p). Case 1. The triangle has an angle 120 o. Without loss of generality suppose that the angle at A is 120 o. Construct the equilateral triangle AFB and for any point P in Δ (except A itself) construct Q so that the triangle AQP is equilateral and has the orientation shown. Then the triangle ABP is a 60 o rotation of the triangle AFQ about A so these two triangles are congruent and it follows that d(p) = CP+PQ+QF which is simply the length of the path CPQF. As P is constrained to lie within ABC, by the dogleg rule the length of this path exceeds AC+AF = d(a). Therefore d(a) < d(p) for all P є Δ, P A. Now allow P to range outside Δ. From above a point P' є Ω exists such that d(p') < d(p) and as d(a) d (P') it follows that d(a) < d(p) for all P outside Δ. Thus d(a) < d(p) for all P A which means that A is the Fermat point of Δ. In other words the Fermat point lies at the obtuse angled vertex. Case 2. The triangle has no angle 120 o. Let P be any point inside Δ and construct the equilateral triangle CPQ. Then CQD is a 60 o rotation of CPB about C so d(p) = PA+PB+PC = AP+PQ+QD which is simply the length of the path APQD. Let P 0 be the point where AD and CF intersect. This point is commonly called the first isogonic center. By the angular restriction P 0 lies inside Δ moreover BCF is a 60 o rotation of BDA about B so Q 0 must lie somewhere on AD. Since CDB = 60 o it follows that Q 0 lies between P 0 and D which means AP 0 Q 0 D is a straight line so d(p 0 ) = AD. Moreover if P P 0 then either P or Q won't lie on AD which means d(p 0 ) = AD < d(p). Now allow P to range outside Δ. From above a point P' є Ω exists such that d(p') < d(p) and as d(p 0 ) d(p') it follows that d(p 0 ) < d(p) for all P outside Δ. That means P 0 is the Fermat point of Δ. In other words the Fermat point is coincident with the first isogonic center.

47 Concurrency Fig 3. Geometry of the first isogonic center Here is a proof using properties of concyclic points to show that the three red lines in Fig 1 are concurrent and cut one another at angles of 60. In Fig 3 suppose RC and BQ intersect at F, and two lines, AF and FP, are drawn. We aim to prove that the points A,F,P are collinear. The triangles RAC and BAQ are congruent because the second is a 60 rotation of the first about A. Hence ARF = ABF and AQF = ACF. By converse of angle in the same segment, ARBF and AFCQ are both concyclic. Thus AFB = AFC = BFC = 120. Because BFC and BPC add up to 180, BPCF is also concyclic. Hence BFP = BCP = 60. Because BFP + BFA = 180, AFP is a straight line. Q.E.D. This proof only applies in Case 2 since if BAC > 120 A lies inside the circumcircle of BPC which switches the relative positions of A and F. However it is easily modified to cover Case 1. Then AFB = AFC = 60 hence BFC = AFB + AFC = 120 which means BPCF is concyclic so BFP = BCP = 60 = BFA. Therefore A lies on FP.

48 Clearly any 3 lines perpendicular to the red ones in Fig 1, in particular those joining the centres of the circles, must also cut at angles of 60 and thereby form an equilateral triangle. This curiosity is known as Napoleon's Theorem. Another proof Another approach to find a point within the triangle, from where sum of the distances to the vertices of triangle is minimum, is to use one of the optimization (mathematics) methods. In particular, method of the lagrange multipliers and the law of cosines. We draw lines from the point within the triangle to its vertices and call them X, Y and Z. Also, let the lengths of these lines be x, y, and z, respectively. Let the acute angle between X and Y be α, Y and Z be β. Then the angle between X and Z is (2π α β). Using the method of lagrange multipiers we have to find the minimum of the lagrangian, which is expressed as: x + y + z + λ 1 (x 2 + y 2 2xy cos(α) a 2 ) + λ 2 (y 2 + z 2 2yz cos(β) b 2 ) + λ 3 (z 2 + x 2 2zx cos(α + β) c 2 ). where a, b and c are the lengths of the sides of the triangle. Calculating the partial derivatives δ/δx, δ/δy, δ/δz, δ/δα, δ/δβ gives a system of 5 equations: δ/δx: 1 + λ 1 (2x 2y cos(α)) + λ 3 (2x 2z cos(α + β)) = 0 δ/δy: 1 + λ 1 (2y 2x cos(α)) + λ 2 (2y 2z cos(β)) = 0 δ/δz: 1 + λ 2 (2z 2y cos(β)) + λ 3 (2z 2x cos(α + β)) = 0 δ/δα: λ 1 y sin(α) + λ 3 z sin(α + β) = 0 δ/δβ: λ 2 y sin(β) + λ 3 x sin(α + β) = 0 After some algebraic manipulations equations for α and β separate from the rest of the parameters, giving: sin(α) = sin(β) sin(α + β) = sin(β) that gives: α = β = 120 o Q.E.D. Note: if one of the vertices of triangle has angle not less than 120 o, then the Fermat point is at that vertex.

49 Properties When the largest angle of the triangle is not larger than 120, X(13) is the Fermat point. The angles subtended by the sides of the triangle at X(13) are all equal to 120 (Case 2), or 60, 60, 120 (Case 1). The circumcircles of the three regular triangles in the construction intersect at X(13). Trilinear coordinates for the first isogonic center, X(13): csc(a + π/3) : csc(b + π/3) : csc(c + π/3), or, equivalently, sec(a π/6) : sec(b π/6) : sec(c π/6). Trilinear coordinates for the second isogonic center, X(14): csc(a π/3) : csc(b π/3) : csc(c π/3), or, equivalently, sec(a + π/6) : sec(b + π/6) : sec(c + π/6). Trilinear coordinates for the Fermat point: 1 u + uvw sec(a π/6) : 1 v + uvw sec(b π/6) : 1 w + uvw sec(c π/6) where u, v, w respectively denote the Boolean variables (A<120 ), (B<120 ), (C<120 ). The isogonal conjugate of X(13) is the first isodynamic point, X(15): sin(a + π/3) : sin(b + π/3) : sin(c + π/3). The isogonal conjugate of X(14) is the second isodynamic point, X(16): sin(a π/3) : sin(b π/3) : sin(c π/3). The following triangles are equilateral: antipedal triangle of X(13) antipedal triangle of X(14) pedal triangle of X(15) pedal triangle of X(16) circumcevian triangle of X(15) circumcevian triangle of X(16) The lines X(13)X(15) and X(14)X(16) are parallel to the Euler line. The three lines meet at the Euler infinity point, X(30). The points X(13), X(14), circumcenter, nine-point center lie on a Lester circle.

50 Aliases X(13) and X(14) are also known as the first Fermat point and the second Fermat point respectively. Alternatives are the positive Fermat point and the negative Fermat point. However all these different names can be confusing and are perhaps best avoided. The problem is that much of the literature blurs the distinction between the Fermat point and the first Fermat point whereas it is only in Case 2 above that they are actually the same. History This question was proposed by Fermat, as a challenge to Evangelista Torricelli. He solved the problem in a similar way to Fermat's, albeit using intersection of the circumcircles of the three regular triangles instead. His pupil, Viviani, published the solution in Heron's formula A triangle with sides a, b, and c In geometry, Heron's (or Hero's) formula, named after Heron of Alexandria, states that the area A of a triangle whose sides have lengths a, b, and c is where s is the semiperimeter of the triangle: Heron's formula can also be written as:

51 Heron's formula is distinguished from other formulas for the area of a triangle, such as half the base times the height or half the modulus of a cross product of two sides, by requiring no arbitrary choice of side as base or vertex as origin. History A formula equivalent to Heron's namely: The formula is credited to Heron (or Hero) of Alexandria, and a proof can be found in his book, Metrica, written c. A.D. 60. It has been suggested that Archimedes knew the formula, and since Metrica is a collection of the mathematical knowledge available in the ancient world, it is possible that it predates the reference given in the work. was discovered by the Chinese independently of the Greeks. It was published in Shushu Jiuzhang ( Mathematical Treatise in Nine Sections ), written by Qin Jiushao and published in A.D Proof A modern proof, which uses algebra and is quite unlike the one provided by Heron (in his book Metrica), follows. Let a, b, c be the sides of the triangle and A, B, C the angles opposite those sides. We have by the law of cosines. From this proof get the algebraic statement: The altitude of the triangle on base a has length b sin(c), and it follows

52 The difference of two squares factorization was used in two different steps. Proof using the Pythagorean theorem Triangle with altitude h cutting base c into d + (c d) Heron's original proof made use of cyclic quadrilaterals, while other arguments appeal to trigonometry as above, or to the incenter and one excircle of the triangle. The following argument reduces Heron's formula directly to the Pythagorean theorem using only elementary means. In the form 4A 2 = 4s(s a)(s b)(s c), Heron's formula reduces on the left to (ch) 2, or (cb) 2 (cd) 2 using b 2 d 2 = h 2 by the Pythagorean theorem, and on the right to

53 (s(s a) + (s b)(s c)) 2 (s(s a) (s b)(s c)) 2 via the identity (p + q) 2 (p q) 2 = 4pq. It therefore suffices to show and The former follows immediately by substituting (a + b + c)/2 for s and simplifying. Doing this for the latter reduces s(s a)(s b)(s c) only as far as (b 2 + c 2 a 2 )/2. But if we replace b 2 by d 2 + h 2 and a 2 by (c d) 2 + h 2, both by Pythagoras, simplification then produces cd as required. Numerical stability Heron's formula as given above is numerically unstable for triangles with a very small angle. A stable alternative involves arranging the lengths of the sides so that: a b c and computing The parentheses in the above formula are required in order to prevent numerical instability in the evaluation. Generalizations Heron's formula is a special case of Brahmagupta's formula for the area of a cyclic quadrilateral. Heron's formula and Brahmagupta's formula are both special cases of Bretschneider's formula for the area of a quadrilateral. Heron's formula can be obtained from Brahmagupta's formula or Bretschneider's formula by setting one of the sides of the quadrilateral to zero. Heron's formula is also a special case of the formula of the area of the trapezoid based only on its sides. Heron's formula is obtained by setting the smaller parallel side to zero. Expressing Heron's formula with a Cayley Menger determinant in terms of the squares of the distances between the three given vertices,

54 illustrates its similarity to Tartaglia's formula for the volume of a three-simplex. Another generalization of Heron's formula to polygons inscribed in a circle was discovered by David P. Robbins. Heron-looking formula for tetrahedra If U, V, W, u, v, w are lengths of edges of the tetrahedron (first three form a triangle; u opposite to U and so on), then where

55 Chapter 5 Incircle & Excircles of a Triangle and Inertia Tensor of Triangle Incircle & excircles of a triangle A triangle (black) with incircle (blue), incentre (I), excircles (orange), excentres (J A,J B,J C ), internal angle bisectors (red) and external angle bisectors (green) In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches (is tangent to) the three sides. The center of the incircle is called the triangle's incenter.

56 An excircle or escribed circle of the triangle is a circle lying outside the triangle, tangent to one of its sides and tangent to the extensions of the other two. Every triangle has three distinct excircles, each tangent to one of the triangle's sides. The center of the incircle can be found as the intersection of the three internal angle bis - ectors. The center of an excircle is the intersection of the internal bisector of one angle and the external bisectors of the other two. Because the internal bisector of an angle is perpendicular to its external bisector, it follows that the center of the incircle together with the three excircle centers form an orthocentric system. Relation to area of the triangle The radii of the in- and excircles are closely related to the area of the triangle. Let A be the triangle's area and let a, b and c, be the lengths of its sides. By Heron's formula, the area of the triangle is where is the semiperimeter. The radius of the incircle (also known as the inradius, r) is Thus, the area, A, of a triangle may be found by multiplying the inradius by the semiperimeter: A = r s. The excircle at side a has radius Similarly the radii of the excircles at sides b and c are respectively and

57 From these formulas we see that the excircles are always larger than the incircle and that the largest excircle is the one attached to the longest side. A triangle, ΔABC, with incircle (blue), incenter (blue, I), contact triangle (red, ΔT a T b T c ) and Gergonne point (green, Ge) Nine-point circle and Feuerbach point The circle tangent to all three of the excircles as well as the incircle is known as the ninepoint circle. The point where the nine-point circle touches the incircle is known as the Feuerbach point. Gergonne triangle and point The Gergonne triangle of ABC is denoted by the vertices T A, T B and T C that are the three points where the incircle touches the reference triangle ABC and where T A is opposite of A, etc. This triangle T A T B T C is also known as the contact triangle or intouch triangle of ABC. The incircle of ABC is the circumcircle of T A T B T C. The three lines AT A, BT B and CT C intersect in a single point, the triangle's Gergonne point Ge - X(7). Interestingly, the Gergonne point of a triangle is the symmedian point of its Gergonne triangle.

58 The touchpoints of the excircle with segments BC,CA,AC are the vertices of the extouch triangle. The points of intersection of the interior angle bisectors of ABC with the segments BC,CA,AB are the vertices of the incentral triangle. Nagel triangle and point The Nagel triangle of ABC is denoted by the vertices X A, X B and X C that are the three points where the excircles touches the reference triangle ABC and where X A is opposite of A, etc. This triangle X A X B X C is also known as the extouch triangle of ABC. The circumcircle of the extouch triangle X A X B X C is called the Mandart circle. The three lines AX A, BX B and CX C intersect in a single point, the triangle's Nagel point Na - X(8). Trilinear coordinates for the vertices of the intouch triangle are given by Trilinear coordinates for the vertices of the extouch triangle are given by Trilinear coordinates for the vertices of the incentral triangle are given by Trilinear coordinates for the vertices of the excentral triangle are given by Trilinear coordinates for the Gergonne point are given by

59 , or, equivalently, by the Law of Sines, Trilinear coordinates for the Nagel point are given by. or, equivalently, by the Law of Sines, It is the isotomic conjugate of the Gergonne point. Coordinates of the incenter The Cartesian coordinates of the incenter are a weighted average of the coordinates of the three vertices. (The weights are positive so the incenter lies inside the triangle as stated above.) If the three vertices are located at (x a,y a ), (x b,y b ), and (x c,y c ), and the opposite sides of the triangle have lengths a, b, and c, then the incenter is at., where Trilinear coordinates for the incenter are given by Barycentric coordinates for the incenter are given by Equations for four circles Let x : y : z be a variable point in trilinear coordinates, and let u = cos 2 (A/2), v = cos 2 (B/2), w = cos 2 (C/2). The four circles described above are given by these equations:

60 Incircle: A-excircle: B-excircle: C-excircle: Other incircle properties Suppose the tangency points of the incircle divide the sides into lengths of x and y, y and z, and z and x. Then where r is the incircle's radius. If the altitudes from sides of lengths a, b, and c are h a, h b, and h c and the inradius is r, then The incircle radius is no larger than half the circumcircle radius (Euler's triangle inequality). The distance between the circumcenter and the incenter is incircle radius and R is the circumcircle radius. where r is the The product of the incircle radius and the circumcircle radius of a triangle with sides a, b, and c is

61 Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter (the center of its incircle). Inertia tensor of triangle The inertia tensor of a triangle (like the inertia tensor of any body) can be expressed in terms of covariance of the body: where covariance is defined as area integral over the triangle: Covariance for a triangle in three-dimensional space, assuming that mass is equally distributed over the surface with unit density, is where represents 3 3 matrix containing triangle vertex coordinates in the rows, is twice the area of the triangle, Substitution of triangle covariance in definition of inertia tensor gives eventually A proof of the formula The proof given here follows the steps from the chapter.

62 Covariance of a canonical triangle Let's compute covariance of the right triangle with the vertices (0,0,0), (1,0,0), (0,1,0). Following the definition of covariance we receive The rest components of C are zero because the triangle is in z = 0. As a result Covariance of the triangle with a vertex in the origin Consider a linear operator that maps the canonical triangle in the triangle,,. The first two columns of contain and respectively, while the third column is arbitrary. The target triangle is equal to the triangle in question (in particular their areas are equal), but shifted with its zero vertex in the origin. Covariance of the triangle in question The last thing remaining to be done is to conceive how covariance is changed with the translation of all points on vector.

63 where is the centroid of dashed triangle. It's easy to check now that all coefficients in before is and before is. This can be expressed in matrix form with as above.

64 Chapter 6 Law of Cosines and Law of Sines Law of cosines Figure 1 A triangle. The angles α, β, and γ are respectively opposite the sides a, b, and c. In trigonometry, the law of cosines (also known as the cosine formula or cosine rule) is a statement about a general triangle that relates the lengths of its sides to the cosine of one of its angles. Using notation as in Fig. 1, the law of cosines states that where γ denotes the angle contained between sides of lengths a and b and opposite the side of length c. The law of cosines generalizes the Pythagorean theorem, which holds only for right triangles: if the angle γ is a right angle (of measure 90 or π/2 radians), then cos(γ) = 0, and thus the law of cosines reduces to The law of cosines is useful for computing the third side of a triangle when two sides and their enclosed angle are known, and in computing the angles of a triangle if all three sides are known. By changing which legs of the triangle play the roles of a, b, and c in the original formula, one discovers that the following two formulas also state the law of cosines:

65 History Fig. 2 Obtuse triangle ABC with perpendicular BH Though the notion of cosine was not yet developed in his time, Euclid's Elements, dating back to the 3rd century BC, contains an early geometric theorem equivalent to the law of cosines. The case of obtuse triangle and acute triangle (corresponding to the two cases of negative or positive cosine) are treated separately, in Propositions 12 and 13 of Book 2. Trigonometric functions and algebra (in particular negative numbers) being absent in Euclid's time, the statement has a more geometric flavor: Proposition 12 In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle. Euclid's Elements, translation by Thomas L. Heath. Using notation as in Fig. 2, Euclid's statement can be represented by the formula This formula may be transformed into the law of cosines by noting that CH = (CB) cos(π γ) = (CB) cos(γ). Proposition 13 contains an entirely analogous statement for acute triangles. It was not until the development of modern trigonometry in the Middle Ages by Muslim mathematicians, especially the discovery of the cosine, that the general law of cosines was formulated. The Persian astronomer and mathematician al-battani generalized Euclid's result to spherical geometry at the beginning of the 10th century, which permitted him to calculate the angular distances between stars. In the 15th century, al-kashi in Samarqand computed trigonometric tables to great accuracy and provided the first explicit statement of the law of cosines in a form suitable for triangulation. In France, the law of cosines is still referred to as the theorem of Al-Kashi.

66 The theorem was popularised in the Western world by François Viète in the 16th century. At the beginning of the 19th century, modern algebraic notation allowed the law of cosines to be written in its current symbolic form. Applications Fig. 3 Applications of the law of cosines: unknown side and unknown angle The theorem is used in triangulation, for solving a triangle, i.e., to find (see Figure 3) the third side of a triangle if one knows two sides and the angle between them: the angles of a triangle if one knows the three sides: the third side of a triangle if one knows two sides and an angle opposite to one of them (one may also use the Pythagorean theorem to do this if it is a right triangle): These formulas produce high round-off errors in floating point calculations if the triangle is very acute, i.e., if c is small relative to a and b or γ is small compared to 1. It is even possible to obtain a result slightly greater than one for the cosine of an angle. The third formula shown is the result of solving for a the quadratic equation a 2 2 ab cos γ + b 2 c 2 = 0. This equation can have 2, 1, or 0 positive solutions corresponding

67 to the number of possible triangles given the data. It will have two positive solutions if b sin(γ) < c < b, only one positive solution if c b or c = b sin(γ), and no solution if c < b sin(γ). These different cases are also explained by the Side-Side-Angle congruence ambiguity. Proofs Using the distance formula Consider a triangle with sides of length a, b, c, where γ is the measurement of the angle opposite the side of length c. We can place this triangle on the coordinate system by plotting By the distance formula, we have Now, we just work with that equation: An advantage of this proof is that it does not require the consideration of different cases for when the triangle is acute vs. obtuse.

68 Using trigonometry Fig. 4 An acute triangle with perpendicular Drop the perpendicular onto the side c to get (see Fig. 4) (This is still true if α or β is obtuse, in which case the perpendicular falls outside the triangle.) Multiply through by c to get By considering the other perpendiculars obtain Adding the latter two equations gives Subtracting the first equation from the last one we have which simplifies to

69 This proof uses trigonometry in that it treats the cosines of the various angles as quantities in their own right. It uses the fact that the cosine of an angle expresses the relation between the two sides enclosing that angle in any right triangle. Other proofs (below) are more geometric in that they treat an expression such as a cos(γ) merely as a label for the length of a certain line segment. Many proofs deal with the cases of obtuse and acute angles γ separately. Using the Pythagorean theorem Fig. 5 Obtuse triangle ABC with height BH Case of an obtuse angle. Euclid proves this theorem by applying the Pythagorean theorem to each of the two right triangles in Fig. 5. Using d to denote the line segment CH and h for the height BH, triangle AHB gives us and triangle CHB gives us Expanding the first equation gives us Substituting the second equation into this, the following can be obtained This is Euclid's Proposition 12 from Book 2 of the Elements. To transform it into the modern form of the law of cosines, note that

70 Case of an acute angle. Euclid's proof of his Proposition 13 proceeds along the same lines as his proof of Proposition 12: he applies the Pythagorean theorem to both right triangles formed by dropping the perpendicular onto one of the sides enclosing the angle γ and uses the binomial theorem to simplify. Fig. 6 A short proof using trigonometry for the case of an acute angle Another proof in the acute case. Using a little more trigonometry, the law of cosines by applying can be deduced by using the Pythagorean theorem only once. In fact, by using the right triangle on the left hand side of Fig. 6 it can be shown that: using the trigonometric identity Remark. This proof needs a slight modification if b < a cos(γ). In this case, the right triangle to which the Pythagorean theorem is applied moves outside the triangle ABC. The only effect this has on the calculation is that the quantity b a cos(γ) is replaced by a cos(γ) b. As this quantity enters the calculation only through its square, the rest of the proof is unaffected. Note. This problem only occurs when β is obtuse, and may be avoided by reflecting the triangle about the bisector of γ. Observation. Referring to Fig 6 it's worth noting that if the angle opposite side a is α then:

71 This is useful for direct calculation of a second angle when two sides and an included angle are given. Using Ptolemy's theorem Proof of law of cosines using Ptolemy's theorem Referring to the diagram, triangle ABC with sides AB = c, BC = a and AC = b is drawn inside its circumcircle as shown. Triangle ABD is constructed congruent to triangle ABC with AD = BC and BD = AC. Perpendiculars from D and C meet base AB at E and F respectively. Then: Now the law of cosines is rendered by a straightforward application of Ptolemy's theorem to cyclic quadrilateral ABCD:

72 Plainly if angle B is 90 degrees, then ABCD is a rectangle and application of Ptolemy's theorem yields Pythagoras' theorem: By comparing areas One can also prove the law of cosines by calculating areas. The change of sign as the angle γ becomes obtuse makes a case distinction necessary. Recall that a 2, b 2, and c 2 are the areas of the squares with sides a, b, and c, respectively; if γ is acute, then ab cos(γ) is the area of the parallelogram with sides a and b forming an angle of ; if γ is obtuse, and so cos(γ) is negative, then ab cos(γ) is the area of the parallelogram with sides a' and b forming an angle of. Fig. 7a Proof of the law of cosines for acute angle γ by "cutting and pasting" Acute case. Figure 7a shows a heptagon cut into smaller pieces (in two different ways) to yield a proof of the law of cosines. The various pieces are in pink, the areas a 2, b 2 on the left and the areas 2ab cos(γ) and c 2 on the right; in blue, the triangle ABC, on the left and on the right;

73 in grey, auxiliary triangles, all congruent to ABC, an equal number (namely 2) both on the left and on the right. The equality of areas on the left and on the right gives Fig. 7b Proof of the law of cosines for obtuse angle γ by "cutting and pasting" Obtuse case. Figure 7b cuts a hexagon in two different ways into smaller pieces, yielding a proof of the law of cosines in the case that the angle γ is obtuse. We have in pink, the areas a 2, b 2, and 2ab cos(γ) on the left and c 2 on the right; in blue, the triangle ABC twice, on the left, as well as on the right. The equality of areas on the left and on the right gives The rigorous proof will have to include proofs that various shapes are congruent and therefore have equal area. This will use the theory of congruent triangles. Using geometry of the circle Using the geometry of the circle, it is possible to give a more geometric proof than using the Pythagorean theorem alone. Algebraic manipulations (in particular the binomial theorem) are avoided.

74 Case of acute angle γ, where a > 2b cos(γ). Drop the perpendicular from A onto a = BC, creating a line segment of length b cos(γ). Duplicate the right triangle to form the isosceles triangle ACP. Construct the circle with center A and radius b, and its tangent h = BH through B. The tangent h forms a right angle with the radius b (Euclid's Ele- ments: Book 3, Proposition 18), so the yellow triangle in Figure 8 is right. Apply the Fig. 8a The triangle ABC (pink), an auxiliary circle (light blue) and an auxiliary right triangle (yellow) Pythagorean theorem to obtain Then use the tangent secant theorem (Euclid's Elements: Book 3, Proposition 36), which says that the square on the tangent through a point B outside the circle is equal to the product of the two lines segments (from B) created by any secant of the circle through B. In the present case: BH 2 = BC BP, or Substituting into the previous equation gives the law of cosines:

75 Note that h 2 is the power of the point B with respect to the circle. The use of the Pythagorean theorem and the tangent secant theorem can be replaced by a single application of the power of a point theorem. Fig. 8b The triangle ABC (pink), an auxiliary circle (light blue) and two auxiliary right triangles (yellow) Case of acute angle γ, where a < 2b cos γ. Drop the perpendicular from A onto a = BC, creating a line segment of length b cos(γ). Duplicate the right triangle to form the isosceles triangle ACP. Construct the circle with center A and radius b, and a chord through B perpendicular to c = AB, half of which is h = BH. Apply the Pythagorean theorem to obtain Now use the chord theorem (Euclid's Elements: Book 3, Proposition 35), which says that if two chords intersect, the product of the two line segments obtained on one chord is

76 equal to the product of the two line segments obtained on the other chord. In the present case: BH 2 = BC BP, or Substituting into the previous equation gives the law of cosines: Note that the power of the point B with respect to the circle has the negative value h 2. Fig. 9 Proof of the law of cosines using the power of a point theorem Case of obtuse angle γ. This proof uses the power of a point theorem directly, without the auxiliary triangles obtained by constructing a tangent or a chord. Construct a circle with center B and radius a (see Figure 9), which intersects the secant through A and C in C and K. The power of the point A with respect to the circle is equal to both AB 2 BC 2 and AC AK. Therefore, which is the law of cosines. Using algebraic measures for line segments (allowing negative numbers as lengths of segments) the case of obtuse angle (CK > 0) and acute angle (CK < 0) can be treated simultaneously.

77 Vector formulation The law of cosines is equivalent to the formula in the theory of vectors, which expresses the dot product of two vectors in terms of their respective lengths and the angle they enclose. Fig. 10 Vector triangle Proof of equivalence. Referring to Figure 10, note that and so we may calculate: The law of cosines formulated in this notation states: which is equivalent to the above formula from the theory of vectors.

78 Isosceles case When a = b, i.e., when the triangle is isosceles with the two sides incident to the angle γ equal, the law of cosines simplifies significantly. Namely, because a 2 + b 2 = 2a 2 = 2ab, the law of cosines becomes or Analog for tetrahedra An analogous statement begins by taking tetrahedron. Denote the dihedral angles by Law of cosines in non-euclidean geometry to be the areas of the four faces of a etc. Then Spherical triangle solved by the law of cosines A version of the law of cosines also holds in non-euclidean geometry. In spherical geometry, a triangle is defined by three points u, v, and w on the unit sphere, and the arcs of great circles connecting those points. If these great circles make angles A, B, and C with opposite sides a, b, c then the spherical law of cosines asserts that each of the following relationships hold:

79 In hyperbolic geometry, a pair of equations are collectively known as the hyperbolic law of cosines. The first is where sinh and cosh are the hyperbolic sine and cosine, and the second is Like in Euclidean geometry, one can use the law of cosines to determine the angles A, B, C from the knowledge of the sides a, b, c. However, unlike Euclidean geometry, the reverse is also possible in each of the models of non-euclidean geometry: the angles A, B, C determine the sides a, b, c. Law of sines A triangle In trigonometry, the law of sines (also known as the sine law, sine formula, or sine rule) is an equation relating the lengths of the sides of an arbitrary triangle to the sines of its angles. According to the law, where a, b, and c are the lengths of the sides of a triangle, and A, B, and C are the opposite angles (see the figure above). Sometimes the law is stated using the reciprocal of this equation:

80 The law of sines can be used to compute the remaining sides of a triangle when two angles and a side are known a technique known as triangulation. It can also be used when two sides and one of the non-enclosed angles are known. In some such cases, the formula gives two possible values for the enclosed angle, leading to an ambiguous case. The law of sines is one of two trigonometric equations commonly applied to find lengths and angles in a general triangle, the other being the law of cosines. Examples The following are examples of how to solve a problem using the law of sines: Given: side a = 20, side c = 24, and angle C = 40 Using the law of sines, we conclude that Or another example of how to solve a problem using the law of sines: If two sides of the triangle are equal to R and the length of the third side, the chord, is given as 100 feet and the angle C opposite the chord is given in degrees, then and

81 Numeric problems Like the law of cosines, although the law of sines is mathematically true, it has problems for numeric use. Much precision may be lost if an arcsine is computed when the sine of an angle is close to one. Some applications The sine law can be used to prove the angle sum identity for sine when α and β are each between 0 and 90 degrees. To prove this, make an arbitrary triangle with sides a, b, and c with corresponding arbitrary angles A, B and C. Draw a perpendicular to c from angle C. This will split the angle C into two different angles, α and β, that are less than 90 degrees, where we choose to have α to be on the same side as A and β be on the same side as B. Use the sine law identity that relates side c and side a. Solve this equation for the sine of C. Notice that the perpendicular makes two right angles triangles, also note that sin(a) = cos(α), sin(b) = cos(β) and that c = a sin(β) + b sin(α). After making these substitutions you should have sin(c) =sin(α + β) = sin (β)cos(α) + (b/a)sin(α)cos(α). Now apply the sine law identity that relates sides b and a and make the substitutions noted before. Now substitute this expression for (b/a) into the original equation for sin(α + β) and you will have the angle sum identity for α and β in terms of sine. The only thing that was used in the proof that was not a definition was the sine law. Thus the sine law is equivalent to the angle sum identity when the angles sum is between 0 and 180 degrees and when each individual angle is between 0 and 90 degrees. The sine law along with the prosthaphaeresis and shift identities can be used to prove the law of tangents and Mollweide's formulas (Dresdin 2009, Plane Trigonometry pg ). The ambiguous case When using the law of sines to solve triangles, there exists an ambiguous case where two separate triangles can be constructed (i.e., there are two different possible solutions to the triangle).

82 Given a general triangle ABC, the following conditions would need to be fulfilled for the case to be ambiguous: The only information known about the triangle is the angle A and the sides a and b, where the angle A is not the included angle of the two sides (in the above image, the angle C is the included angle). The angle A is acute (i.e., A < 90 ). The side a is shorter than the side b (i.e., a < b). The side a is longer than the altitude of a right angled triangle with angle A and hypotenuse b (i.e., a > b sin A). Given all of the above premises are true, the angle B may be acute or obtuse; meaning, one of the following is true: OR Relation to the circumcircle In the identity the common value of the three fractions is actually the diameter of the triangle's circumcircle. It can be shown that this quantity is equal to

83 where S is the area of the triangle and s is the semiperimeter The second equality above is essentially Heron's formula. Spherical case In the spherical case, the formula is: Here, α, β, and γ are the angles at the center of the sphere subtended by the three arcs of the spherical surface triangle a, b, and c, respectively. A, B, and C are the surface angles opposite their respective arcs. Hyperbolic case In hyperbolic geometry when the curvature is 1, the law of sines becomes In the special case when B is a right angle, one gets which is the analog of the formula in Euclidean geometry expressing the sine of an angle as the opposite side divided by the hypotenuse. History The spherical law of sines was discovered in the 10th century. It is variously attributed to al-khujandi, Abul Wafa Bozjani, Nasir al-din al-tusi and Abu Nasr Mansur.

84 Al-Jayyani's The book of unknown arcs of a sphere in the 11th century introduced the general law of sines. The plane law of sines was later described in the 13th century by Nasīr al-dīn al-tūsī. In his On the Sector Figure, he stated the law of sines for plane and spherical triangles, and provided proofs for this law. Derivation Make a triangle with the sides a, b, and c, and angles A, B, and C. Draw the altitude from vertex C to the side across c; by definition it divides the original triangle into two right angle triangles. Mark the length of this line h. It can be observed that: Therefore and Doing the same thing with the line drawn between vertex A and side a will yield: Determine an angle For 2nd angle:

85 for 3rd angle: C = 180 A B, Determine a side A law of sines for tetrahedra A tetrahedron structure with vertices O, A, B, C. The product of the sines of OAB, OBC, OCA appears on one side of the identity, and the product of the sines of OAC, OCB, OBA on the other. A corollary of the law of sines as stated above is that in a tetrahedron with vertices O, A, B, C, we have One may view the two sides of this identity as corresponding to clockwise and counterclockwise orientations of the surface. Putting any of the four vertices in the role of O yields four such identities, but in a sense at most three of them are independent: If the "clockwise" sides of three of them are multiplied and the product is inferred to be equal to the product of the "counterclockwise" sides of the same three identities, and then common factors are cancelled from both sides,

86 the result is the fourth identity. One reason to be interested in this "independence" relation is this: It is widely known that three angles are the angles of some triangle if and only if their sum is a half-circle. What condition on 12 angles is necessary and sufficient for them to be the 12 angles of some tetrahedron? Clearly the sum of the angles of any side of the tetrahedron must be a half-circle. Since there are four such triangles, there are four such constraints on sums of angles, and the number of degrees of freedom is thereby reduced from 12 to 8. The four relations given by this sines law further reduce the number of degrees of freedom, not from 8 down to 4, but only from 8 down to 5, since the fourth constraint is not independent of the first three. Thus the space of all shapes of tetrahedra is 5-dimensional.

87 Chapter 7 Equilateral Triangle and Heronian Triangle Equilateral triangle Equilateral triangle An equilateral triangle is a regular polygon. Family 2-simplex Edges and vertices 3 Schläfli symbols {3} Coxeter Dynkin diagrams Symmetry group Dihedral (D 3 ) Internal angle (degrees) 60

88 In geometry, an equilateral triangle is a triangle in which all three sides are equal. In traditional or Euclidean geometry, equilateral triangles are also equiangular; that is, all three internal angles are also congruent to each other and are each 60. They are regular polygons, and can therefore also be referred to as regular triangles. In an equilateral triangle all sides have the same length. An equilateral triangle is also a regular polygon with all angles measuring 60. Properties Assuming the lengths of the sides of the equilateral triangle are, we can determine using the Pythagorean theorem that: The area is The perimeter is The radius of the circumscribed circle is The radius of the inscribed circle is The geometric center of the triangle is the center of the circumscribed and inscribed circles And the altitude or height is. For any point P in the plane, with distances p, q, and t from the vertices A, B, and C respectively, 3(p 4 + q 4 + t 4 + a 4 ) = (p 2 + q 2 + t 2 + a 2 ) 2. For any interior point P in an equilateral triangle, with distances d, e, and f from the sides, d+e+f = the altitude of the triangle, independent of the location of P. For any point P on the inscribed circle of an equilateral triangle, with distances p, q, and t from the vertices, 4(p 2 + q 2 + t 2 ) = 5a 2

89 and 16(p 4 + q 4 + t 4 ) = 11a 4. For any point P on the minor arc BC of the circumcircle, with distances p, q, and t from A, B, and C respectively, and p = q + t q 2 + qt + t 2 = a 2 ; moreover, if point D on side BC divides PA into segments PD and DA with DA having length z and PD having length y, then which also equals if t q; and The triangle of largest area of all those inscribed in a given circle is equilateral; and the triangle of smallest area of all those circumscribed around a given circle is equilateral. An equilateral triangle is the most symmetrical triangle, having 3 lines of reflection and rotational symmetry of order 3 about its center. Its symmetry group is the dihedral group of order 6 D 3.

90 A regular tetrahedron is made of four equilateral triangles Equilateral triangles are found in many other geometric constructs. The intersection of circles whose centers are a radius width apart is a pair of equilateral arches, each of which can be inscribed with an equilateral triangle. They form faces of regular and uniform polyhedra. Three of the five Platonic solids are composed of equilateral triangles. In particular, the regular tetrahedron has four equilateral triangles for faces and can be considered the three dimensional analogue of the shape. The plane can be tiled using equilateral triangles giving the triangular tiling. A result finding an equilateral triangle associated to any triangle is Morley's trisector theorem.

91 Geometric construction Construction of equilateral triangle with compass and straightedge An equilateral triangle is easily constructed using a compass. Draw a straight line, and place the point of the compass on one end of the line, and swing an arc from that point to the other point of the line segment. Repeat with the other side of the line. Finally, connect the point where the two arcs intersect with each end of the line segment Alternate method: Draw a circle with radius r, place the point of the compass on the circle and draw another circle with the same radius. The two circles will intersect in two points. An equilateral triangle can be constructed by taking the two centers of the circles and either of the points of intersection. The proof that the resulting figure is an equilateral triangle is the first proposition in Book I of Euclid's Elements. Heronian triangle In geometry, a Heronian triangle is a triangle whose sidelengths and area are all rational numbers. It is named after Hero of Alexandria. Any such rational triangle can be scaled up to a corresponding triangle with integer sides and area, and often the term Heronian triangle is used to refer to the latter.

92 Properties Any triangle whose sidelengths are a Pythagorean triple is Heronian, as the sidelengths of such a triangle are integers, and its area (being a right-angled triangle) is just half of the product of the two sides at the right angle. A triangle with sidelengths c, e and b + d, and height a An example of a Heronian triangle which is not right-angled is the one with sidelengths 5, 5, and 6, whose area is 12. This triangle is obtained by joining two copies of the rightangled triangle with sides 3, 4, and 5 by the side of length 4. This approach works in general, as illustrated in the picture to the right. One takes a Pythagorean triple (a, b, c), with c being largest, then another one (a, d, e), with e being largest, constructs the triangles with these sidelengths, and joins them together by the side of length a, to obtain a triangle with integer sidelengths c, e, and b + d, with rational area (one half times the base times the height). An interesting question to ask is whether all Heronian triangles can be obtained by joining together two right-angled triangles described in this procedure. The answer is no. If one takes the Heronian triangle with sidelengths 0.5, 0.5, and 0.6, which is just the triangle described above shrunk 10 times, it clearly cannot be decomposed into two triangles with integer sidelengths. Nor for example can a 5, 29, 30 triangle with area 72, since none of its altitudes are integers. However, if one allows for Pythagorean triples with rational entries, not necessarily integers, then the answer is affirmative. Note that a triple with rational entries is just a scaled version of a triple with integer entries. Theorem Given a Heronian triangle, one can split it into two right-angled triangles, whose sidelengths form Pythagorean triples with rational entries.

93 Proof of the theorem Consider again the illustration to the right, where this time it is known that c, e, b + d, and the triangle area A are rational. We can assume that the notation was chosen so that the sidelength b + d is the largest, as then the perpendicular onto this side from the opposite vertex falls inside this segment. To show that the triples (a, b, c) and (a, d, e) are Pythagorean, one must prove that a, b, and d are rational. Since the triangle area is one can solve for a to find which is rational, as both A and b + d are rational. Left is to show that b and d are rational. From the Pythagorean theorem applied to the two right-angled triangles, one has and One can subtract these two, to find or or The right-hand side is rational, because by assumption, c, e, and b + d are rational. Then, b d is rational. This, together with b + d being rational implies by adding these up that b is rational, and then d must be rational too. Q.E.D.

94 Exact formula for Heronian triangles The following formulas generate all Heronian triangles: where m,n, and k are rational numbers, and where a, b and c are now the side lengths. Examples The list of fundamental integer Heronian triangles, sorted by area and, if this is the same, by perimeter, starts as in the following table. Fundamental means that the greatest common divisor of the three side lengths equals 1. Area Perimeter length b+d length e length c

95

96 Almost-equilateral Heronian triangles A Heronian triangle is a triangle with rational sides, area and inradius. Since the area of an equilateral triangle with rational sides is an irrational number, no equilateral triangle is Heronian. However, there is a unique sequence of Heronian triangles that are "almost equilateral" because the three sides, expressed as integers, are of the form n 1, n, n + 1. The first few examples of these almost-equilateral triangles are set forth in the following table. Side length Area Inradius n 1 n n Subsequent values of n can be found by multiplying the last known value by 4, then subtracting the next to the last one (52 = , 194 = , etc), as expressed in This sequence (sequence A in OEIS) can also be generated from the solutions to the Pell's equation x² 3y² = 1, which can in turn be derived from the regular continued fraction expansion for 3.

97 Chapter 8 Bell Number In combinatorics, the nth Bell number, named after Eric Temple Bell, is the number of partitions of a set with n members, or equivalently, the number of equivalence relations on it. Starting with B 0 = B 1 = 1, the first few Bell numbers are: 1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, , (sequence A in OEIS). Partitions of a set The traditional Japanese symbols for the chapters of the Tale of Genji are based on the 52 ways of partitioning five elements.

98 In general, B n is the number of partitions of a set of size n. A partition of a set S is defined as a set of nonempty, pairwise disjoint subsets of S whose union is S. For example, B 3 = 5 because the 3-element set {a, b, c} can be partitioned in 5 distinct ways: { {a}, {b}, {c} } { {a}, {b, c} } { {b}, {a, c} } { {c}, {a, b} } { {a, b, c} }. B 0 is 1 because there is exactly one partition of the empty set. Every member of the empty set is a nonempty set (that is vacuously true), and their union is the empty set. Therefore, the empty set is the only partition of itself. Note that, as suggested by the set notation above, we consider neither the order of the partitions nor the order of elements within each partition. This means the following partitionings are all considered identical: Another view of Bell numbers { {b}, {a, c} } { {a, c}, {b} } { {b}, {c, a} } { {c, a}, {b} }. Bell numbers can also be viewed as the number of distinct possible ways of putting n distinguishable balls into one or more indistinguishable boxes. For example, let us suppose n is 3. We have three balls, which we will label a, b, and c, and three boxes. If the boxes can not be distinguished from each other, there are five ways of putting the balls in the boxes: All three balls go in to one box. Since the boxes are anonymous, this is only considered one combination. a goes in to one box; b and c go in to another box. b goes in to one box; a and c go in to another box. c goes in to one box; a and b go in to another box. Each ball goes in to its own box. Properties of Bell numbers The Bell numbers satisfy this recursion formula:

99 They also satisfy "Dobinski's formula": = the nth moment of a Poisson distribution with expected value 1. And they satisfy "Touchard's congruence": If p is any prime number then or, generalizing Each Bell number is a sum of Stirling numbers of the second kind The Stirling number is the number of ways to partition a set of cardinality n into exactly k nonempty subsets. More generally, the Bell numbers satisfy the following recurrence: The nth Bell number is also the sum of the coefficients in the polynomial that expresses the nth moment of any probability distribution as a function of the first n cumulants; this way of enumerating partitions is not as coarse as that given by the Stirling numbers. The exponential generating function of the Bell numbers is Asymptotic limit and bounds Several asymptotic formulae for the Bell numbers are known. One such is

100 Here where W is the Lambert W function. (Lovász, 1993) In (Berend, D. and Tassa, T., 2010), the following bounds were established: moreover, if then for all, where and Triangle scheme for calculating Bell numbers The triangular array whose right-hand diagonal sequence consists of Bell numbers The Bell numbers can easily be calculated by creating the so-called Bell triangle, also called Aitken's array or the Peirce triangle: 1. Start with the number one. Put this on a row by itself.

101 2. Start a new row with the rightmost element from the previous row as the leftmost number 3. Determine the numbers not on the left column by taking the sum of the number to the left and the number above the number to the left (the number diagonally up and left of the number we are calculating) 4. Repeat step three until there is a new row with one more number than the previous row 5. The number on the left hand side of a given row is the Bell number for that row. For example, the first row is made by placing one by itself. The next (second) row is made by taking the rightmost number from the previous row (1), and placing it on a new row. We now have a structure like this: 1 1 ''x'' The value x here is determined by adding the number to the left of x (one) and the number above the number to the left of x (also one) y The value y is determined by copying over the number from the right of the previous row. Since the number on the right hand side of the previous row has a value of 2, y is given a value of two ''x'' Again, since x is not the leftmost element of a given row, its value is determined by taking the sum of the number to x's left (three) and the number above the number to x's left (two). The sum is five. Here is the first five rows of this triangle: The fifth row is calculated thus: Take 15 from the previous row = = = 37

102 = 52 Computer program The following is example code in the Ruby programming language that prints out all the Bell numbers from the 1st to the 300th inclusive (the limits can be adjusted) #!/usr/bin/env ruby def print_bell_numbers(start, finish) # Initialize the Bell triangle as a two-dimensional array triangle = Array[Array] # Make sure "start" is less than "finish", and both numbers are at least 1 (finish, start = start, finish) if finish < start start = 1 if start < 1 finish = 1 if finish < 1 element row 1] 1.upto(finish-1) do row_num end # Set the first element of the current row to be the last # of the previous row current_row = [triangle[row_num-1][row_num-1]] # Calculate the rest of the elements in this row, then add the # to the Bell triangle 1.upto(row_num) do col_num sum = triangle[row_num-1][col_num-1] + current_row[col_num- end current_row.push(sum) triangle[row_num] = current_row end # Print out the Bell numbers start.upto(finish) do num puts triangle[num-1][0] end # Adjust the limits here print_bell_numbers(1, 300) And here an equivalent version written in Python def bell_numbers(start, stop): ## Swap start and stop if start > stop if stop < start: start, stop = stop, start

103 if start < 1: start = 1 if stop < 1: stop = 1 t = [] ## Initialize the triangle as a twodimensional array c = 1 ## Bell numbers count while c <= stop: if c >= start: yield t[-1][0] ## Yield the Bell number of the previous row row = [t[-1][-1]] ## Initialize a new row for b in t[-1]: row.append(row[-1] + b) ## Populate the new row c += 1 ## We have found another Bell number t.append(row) ## Append the row to the triangle for b in bell_numbers(1, 300): print b Prime Bell numbers The number in the nth row and kth column is the number of partitions of {1,..., n} such that n is not together in one class with any of the elements k, k + 1,..., n 1. For example, there are 7 partitions of {1,..., 4} such that 4 is not together in one class with either of the elements 2, 3, and there are 10 partitions of {1,..., 4} such that 4 is not together in one class with element 3. The difference is due to 3 partitions of {1,..., 4} such that 4 is together in one class with element 2, but not with element 3. This corresponds to the fact that there are 3 partitions of {1,..., 3} such that 3 is not together in one class with element 2: for counting partitions two elements which are always in one class can be treated as just one element. The 3 appears in the previous row of the table. The first few Bell numbers that are primes are: 2, 5, 877, , , corresponding to the indices 2, 3, 7, 13, 42 and 55 (sequence A in OEIS). The next prime is B 2841, which is approximately As of 2006, it is the largest known prime Bell number. Phil Carmody showed it was a probable prime in After 17 months of computation with Marcel Martin's ECPP program Primo, Ignacio Larrosa Cañestro proved it to be prime in He ruled out any other possible primes below B 6000, later extended to B by Eric Weisstein.

104 Chapter 9 Binomial Coefficient The binomial coefficients can be arranged to form Pascal's triangle. In mathematics, binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. They are indexed by two nonnegative integers; the binomial coefficient indexed by n and k is usually written, and it is the coefficient of the x k term in the polynomial expansion of the binomial power (1 + x) n. Arranging binomial coefficients into rows for successive values of n, and in which k ranges from 0 to n, gives a triangular array called Pascal's triangle. This family of numbers also arises in many other areas than algebra, notably in combinatorics. For any set containing n elements, the number of distinct k-element subsets of it that can be formed (the k-combinations of its elements) is given by the binomial

105 coefficient. Therefore is often read as "n choose k". The properties of binomial coefficients have led to extending the meaning of the symbol beyond the basic case where n and k are nonnegative integers with k n; such expressions are then still called binomial coefficients. The notation was introduced by Andreas von Ettingshausen in 1826, although the numbers were already known centuries before that. The earliest known detailed discussion of binomial coefficients is in a tenth-century commentary, due to Halayudha, on an ancient Hindu classic, Pingala's chandaḥśāstra. In about 1150, the Hindu mathematician Bhaskaracharya gave a very clear exposition of binomial coefficients in his book Lilavati. Alternative notations include C(n, k), n C k, n C k,,, in all of which the C stands for combinations or choices. Definition and interpretations For natural numbers (taken to include 0) n and k, the binomial coefficient can be defined as the coefficient of the monomial X k in the expansion of (1 + X) n. The same coefficient also occurs (if k n) in the binomial formula (valid for any elements x,y of a commutative ring), which explains the name "binomial coefficient". Another occurrence of this number is in combinatorics, where it gives the number of ways, disregarding order, that k objects can be chosen from among n objects; more formally, the number of k-element subsets (or k-combinations) of an n-element set. This number can be seen as equal to the one of the first definition, independently of any of the formulas below to compute it: if in each of the n factors of the power (1 + X) n one temporarily labels the term X with an index i (running from 1 to n), then each subset of k indices gives after expansion a contribution X k, and the coefficient of that monomial in the result will be the number of such subsets. This shows in particular that is a natural number for any natural numbers n and k. There are many other combinatorial interpretations of binomial coefficients (counting problems for which the answer is given by a binomial coefficient expression), for instance the number of words formed of n bits (digits 0 or 1) whose sum is k is given by, while the number of ways to write where every a i is a nonnegative integer is given by

106 . Most of these interpretations are easily seen to be equivalent to counting k- combinations. Computing the value of binomial coefficients Several methods exist to compute the value of power or counting k-combinations. without actually expanding a binomial Recursive formula One has a recursive formula for binomial coefficients with initial values The formula follows either from tracing the contributions to X k in (1 + X) n 1 (1 + X), or by counting k-combinations of {1, 2,..., n} that contain n and that do not contain n separately. It follows easily that when k > n, and for all n, so the recursion can stop when reaching such cases. This recursive formula then allows the construction of Pascal's triangle. Multiplicative formula A more efficient method to compute individual binomial coefficients is given by the formula where the numerator of the first fraction is expressed as a falling factorial power. This formula is easiest to understand for the combinatorial interpretation of binomial coefficients. The numerator gives the number of ways to select a sequence of k distinct objects, retaining the order of selection, from a set of n objects. The denominator counts

107 the number of distinct sequences that define the same k-combination when order is disregarded. Factorial formula Finally there is a formula using factorials that is easy to remember: where n! denotes the factorial of n. This formula follows from the multiplicative formula above by multiplying numerator and denominator by (n k)!; as a consequence it involves many factors common to numerator and denominator. It is less practical for explicit computation unless common factors are first canceled (in particular since factorial values grow very rapidly). The formula does exhibit a symmetry that is less evident from the multiplicative formula (though it is from the definitions) Generalization and connection to the binomial series The multiplicative formula allows the definition of binomial coefficients to be extended by replacing n by an arbitrary number α (negative, real, complex) or even an element of any commutative ring in which all positive integers are invertible: (1) With this definition one has a generalization of the binomial formula (with one of the variables set to 1), which justifies still calling the binomial coefficients: (2) This formula is valid for all complex numbers α and X with X < 1. It can also be interpreted as an identity of formal power series in X, where it actually can serve as definition of arbitrary powers of series with constant coefficient equal to 1; the point is that with this definition all identities hold that one expects for exponentiation, notably

108 If α is a nonnegative integer n, then all terms with k > n are zero, and the infinite series becomes a finite sum, thereby recovering the binomial formula. However for other values of α, including negative integers and rational numbers, the series is really infinite. Pascal's triangle 1000 th row of Pascal's triangle, arranged vertically, with grey-scale representations of decimal digits of the coefficients, right-aligned. The left boundary of the image corresponds roughly to the graph of the logarithm of the binomial coefficients, and illustrates that they form a log-concave sequence.

109 Pascal's rule is the important recurrence relation (3) which can be used to prove by mathematical induction that is a natural number for all n and k, (equivalent to the statement that k! divides the product of k consecutive integers), a fact that is not immediately obvious from formula (1). Pascal's rule also gives rise to Pascal's triangle: 0: 1 1: 1 1 2: : : : : : : Row number n contains the numbers for k = 0,,n. It is constructed by starting with ones at the outside and then always adding two adjacent numbers and writing the sum directly underneath. This method allows the quick calculation of binomial coefficients without the need for fractions or multiplications. For instance, by looking at row number 5 of the triangle, one can quickly read off that (x + y) 5 = 1 x x 4 y + 10 x 3 y x 2 y x y y 5. The differences between elements on other diagonals are the elements in the previous diagonal, as a consequence of the recurrence relation (3) above. Combinatorics and statistics Binomial coefficients are of importance in combinatorics, because they provide ready formulas for certain frequent counting problems: There are ways to choose k elements from a set of n elements. There are ways to choose k elements from a set of n if repetitions are allowed.

110 There are strings containing k ones and n zeros. There are strings consisting of k ones and n zeros such that no two ones are adjacent. The Catalan numbers are The binomial distribution in statistics is The formula for a Bézier curve. Binomial coefficients as polynomials can be simplified and defined as a poly- For any nonnegative integer k, the expression nomial divided by k!: This presents a polynomial in t with rational coefficients. As such, it can be evaluated at any real or complex number t to define binomial coefficients with such first arguments. These "generalized binomial coefficients" appear in Newton's generalized binomial theorem. For each k, the polynomial can be characterized as the unique degree k polynomial p(t) satisfying p(0) = p(1) =... = p(k 1) = 0 and p(k) = 1. Its coefficients are expressible in terms of Stirling numbers of the first kind, by definition of the latter: The derivative of can be calculated by logarithmic differentiation:

111 Binomial coefficients as a basis for the space of polynomials Over any field containing Q, each polynomial p(t) of degree at most d is uniquely expressible as a linear combination of the sequence p(0), p(1),, p(k). Explicitly,. The coefficient a k is the k th difference (3.5) Integer-valued polynomials Each polynomial is integer-valued: it takes integer values at integer inputs. (One way to prove this is by induction on k, using Pascal's identity.) Therefore any integer linear combination of binomial coefficient polynomials is integer-valued too. Conversely, (3.5) shows that any integer-valued polynomial is an integer linear combination of these binomial coefficient polynomials. More generally, for any subring R of a characteristic 0 field K, a polynomial in K[t] takes values in R at all integers if and only if it is an R-linear combination of binomial coefficient polynomials. Example The integer-valued polynomial 3t(3t + 1)/2 can be rewritten as Identities involving binomial coefficients The factorial formula facilitates relating nearby binomial coefficients. For instance, if k is a positive integer and n is arbitrary, then and, with a little more work,

112 Powers of -1 A special binomial coefficient is ; it equals powers of -1: Series involving binomial coefficients The formula (5) is obtained from (2) using x = 1. This is equivalent to saying that the elements in one row of Pascal's triangle always add up to two raised to an integer power. A combinatorial interpretation of this fact involving double counting is given by counting subsets of size 0, size 1, size 2, and so on up to size n of a set S of n elements. Since we count the number of subsets of size i for 0 i n, this sum must be equal to the number of subsets of S, which is known to be 2 n. That is, Equation 5 is a statement that the power set for a finite set with n elements has size 2 n. The formulas (6a) and (6b) follow from (2), after differentiating with respect to x (twice in the latter) and then substituting x = 1. The Chu-Vandermonde identity, which holds for any complex-values m and n and any non-negative integer k, is (7a)

113 and can be found by examination of the coefficient of x k in the expansion of (1 + x) m (1 + x) n m = (1 + x) n using equation (2). When m = 1, equation (7a) reduces to equation (3). A similar looking formula, which applies for any integers j, k, and n satisfying 0 j k n, is (7b) and can be found by examination of the coefficient of x n + 1 in the expansion of using j = k, equation (7b) gives From expansion (7a) using n = 2m, k = m, and (1), one finds When (8) Let F(n) denote the n th Fibonacci number. We obtain a formula about the diagonals of Pascal's triangle (9) This can be proved by induction using (3) or by Zeckendorf's representation (Just note that the lhs gives the number of subsets of {F(2),...,F(n)} without consecutive members, which also form all the numbers below F(n+1)). Also using (3) and induction, one can show that (10) Again by (3) and induction, one can show that for k = 0,..., n 1

114 (11) as well as (12) which is itself a special case of the result from the theory of finite differences that for any polynomial P(x) of degree less than n, (13a) Differentiating (2) k times and setting x = 1 yields this for taking linear combinations of these. When P(x) is of degree less than or equal to n, where a n is the coefficient of degree n in P(x). More generally for (13b),, when 0 k < n, and the general case follows by (13b) (13c) where m and d are complex numbers. This follows immediately applying (13b) to the polynomial Q(x):=P(m + dx) instead of P(x), and observing that Q(x) has still degree less than or equal to n, and that its coefficient of degree n is d n a n. The infinite series (14)

115 is convergent for k 2. This formula is used in the analysis of the German tank problem. It is equivalent to the formula for the finite sum which is proved for M>m by induction on M. Using (8) one can derive Identities with combinatorial proofs and Many identities involving binomial coefficients can be proved by combinatorial means. For example, the following identity for nonnegative integers (which reduces to (6) when q = 1): (15) (16) (16b) can be given a double counting proof as follows. The left side counts the number of ways of selecting a subset of [n] of at least q elements, and marking q elements among those selected. The right side counts the same parameter, because there are ways of choosing a set of q marks and they occur in all subsets that additionally contain some subset of the remaining elements, of which there are 2 n q. The recursion formula

116 where both sides count the number of k-element subsets of {1, 2,..., n} with the right hand side first grouping them into those that contain element n and those that do not. The identity (8) also has a combinatorial proof. The identity reads Suppose you have 2n empty squares arranged in a row and you want to mark (select) n of them. There are ways to do this. On the other hand, you may select your n squares by selecting k squares from among the first n and n k squares from the remaining n squares. This gives Now apply (4) to get the result. Sum of coefficients row The number of k-combinations for all k,, is the sum of the nth row (counting from 0) of the binomial coefficients. These combinations are enumerated by the 1 digits of the set of base 2 numbers counting from 0 to 2 n 1, where each digit position is an item from the set of n. Continuous identities Certain trigonometric integrals have values expressible in terms of binomial coefficients: For and (17) (19) (18)

117 These can be proved by using Euler's formula to convert trigonometric functions to complex exponentials, expanding using the binomial theorem, and integrating term by term. Generating functions Ordinary generating functions For a fixed n, the ordinary generating function of the sequence is: For a fixed k, the ordinary generating function of the sequence is: The bivariate generating function of the binomial coefficients is: Another bivariate generating function of the binomial coefficients, which is symmetric, is: Exponential generating function The exponential bivariate generating function of the binomial coefficients is:

118 Divisibility properties In 1852, Kummer proved that if m and n are nonnegative integers and p is a prime number, then the largest power of p dividing equals p c, where c is the number of carries when m and n are added in base p. Equivalently, the exponent of a prime p in equals the number of nonnegative integers j such that the fractional part of k/p j is greater than the fractional part of n/p j. It can be deduced from this that n/gcd(n,k). is divisible by A somewhat surprising result by David Singmaster (1974) is that any integer divides almost all binomial coefficients. More precisely, fix an integer d and let f(n) denote the number of binomial coefficients with n < N such that d divides. Then Since the number of binomial coefficients with n < N is N(N+1) / 2, this implies that the density of binomial coefficients divisible by d goes to 1. Another fact: An integer n 2 is prime if and only if all the intermediate binomial coefficients are divisible by n. Proof: When p is prime, p divides for all 0 < k < p because it is a natural number and the numerator has a prime factor p but the denominator does not have a prime factor p. When n is composite, let p be the smallest prime factor of n and let k = n/p. Then 0 < p < n and

119 otherwise the numerator k(n 1)(n 2)... (n p+1) has to be divisible by n = k p, this can only be the case when (n 1)(n 2)... (n p+1) is divisible by p. But n is divisible by p, so p does not divide n 1, n 2,..., n p+1 and because p is prime, we know that p does not divide (n 1)(n 2)... (n p+1) and so the numerator cannot be divisible by n. Bounds and asymptotic formulas The following bounds for hold: Stirling's approximation yields the bounds: and, in general, m 2 and n 1, and the approximation as The infinite product formula (cf. Gamma function, alternative definition) for yields the asymptotic formulas as. This asymptotic behaviour is contained in the approximation as well. (Here H k is the k th harmonic number and γ is the Euler Mascheroni constant).

120 The sum of binomial coefficients can be bounded by a term exponential in n and the binary entropy of the largest n / k that occurs. More precisely, for and, it holds where is the binary entropy of ε. A simple and rough upper bound for the sum of binomial coefficients is given by the formula below (not difficult to prove) Generalizations Generalization to multinomials Binomial coefficients can be generalized to multinomial coefficients. They are defined to be the number: where While the binomial coefficients represent the coefficients of (x+y) n, the multinomial coefficients represent the coefficients of the polynomial The case r = 2 gives binomial coefficients:

121 The combinatorial interpretation of multinomial coefficients is distribution of n distinguishable elements over r (distinguishable) containers, each containing exactly k i elements, where i is the index of the container. Multinomial coefficients have many properties similar to these of binomial coefficients, for example the recurrence relation: and symmetry: where (σ i ) is a permutation of (1,2,...,r). Generalization to negative integers If, then Taylor series extends to all n. Using Stirling numbers of the first kind the series expansion around any arbitrarily chosen point z 0 is Binomial coefficient with n=1/2 The definition of the binomial coefficients can be extended to the case where n is real and k is integer. In particular, the following identity holds for any non-negative integer k :

122 This shows up when expanding series : into a power series using the Newton binomial Identity for the product of binomial coefficients One can express the product of binomial coefficients as a linear combination of binomial coefficients: Partial Fraction Decomposition where the connection coefficients are multinomial coefficients. In terms of labelled combinatorial objects, the connection coefficients represent the number of ways to assign m+n-k labels to a pair of labelled combinatorial objects of weight m and n res pectively that have had their first k labels identified, or glued together to get a new labelled combinatorial object of weight m+n-k. (That is, to separate the labels into three portions to apply to the glued part, the unglued part of the first object, and the unglued part of the second object.) In this regard, binomial coefficients are to exponential generating series what falling factorials are to ordinary generating series. The partial fraction decomposition of the inverse is given by and

123 Newton's binomial series Newton's binomial series, named after Sir Isaac Newton, is one of the simplest Newton series: The identity can be obtained by showing that both sides satisfy the differential equation (1+z) f'(z) = α f(z). The radius of convergence of this series is 1. An alternative expression is where the identity is applied. Two real or complex valued arguments The binomial coefficient is generalized to two real or complex valued arguments using the gamma function or beta function via This definition inherits these following additional properties from Γ: moreover,

124 The resulting function has been little-studied, apparently first being graphed in (Fowler 1996). Notably, many binomial identities fail: but for n positive (so n negative). The behavior is quite complex, and markedly different in various octants (that is, with respect to the x and y axes and the line y = x), with the behavior for negative x having singularities at negative integer values and a checkerboard of positive and negative regions: in the octant it is a smoothly interpolated form of the usual binomial, with a ridge ("Pascal's ridge"). in the octant and in the quadrant the function is close to zero. in the quadrant the function is alternatingly very large positive and negative on the parallelograms with vertices ( n,m + 1),( n,m),( n 1,m 1),( n 1,m) in the octant 0 > x > y the behavior is again alternatingly very large positive and negative, but on a square grid. in the octant 1 > y > x + 1 it is close to zero, except for near the singularities. Generalization to q-series The binomial coefficient has a q-analog generalization known as the Gaussian binomial coefficient. Generalization to infinite cardinals The definition of the binomial coefficient can be generalized to infinite cardinals by defining: where A is some set with cardinality α. One can show that the generalized binomial coefficient is well-defined, in the sense that no matter what set we choose to represent the cardinal number α, will remain the same. For finite cardinals, this definition coincides with the standard definition of the binomial coefficient. Assuming the Axiom of Choice, one can show that α. for any infinite cardinal

125 Binomial coefficient in programming languages The notation is convenient in handwriting but inconvenient for typewriters and computer terminals. Many programming languages do not offer a standard subroutine for computing the binomial coefficient, but for example the J programming language uses the exclamation mark: k! n. Naive implementations of the factorial formula, such as the following snippet in Python: def binomialcoefficient(n, k): from math import factorial return factorial(n) / (factorial(k) * factorial(n - k)) are very slow and are uselessly calculating factorials of very high numbers (in languages as C or Java they suffer from overflow errors because of this reason). A direct implementation of the multiplicative formula works well: def binomialcoefficient(n, k): if k > n - k: # take advantage of symmetry k = n - k c = 1 for i in range(k): c = c * (n - i) c = c / (i + 1) return c The example mentioned above can be also written in functional style. The following Scheme example uses recursive definition Rational arithmetic can be easily avoided using integer division The following implementation uses all these ideas (define (binomial n k) ;; Helper function to compute C(n,k) via forward recursion (define (binomial-iter n k i prev) (if (>= i k) prev (binomial-iter n k (+ i 1) (/ (* (- n i) prev) (+ i 1))))) ;; Use symmetry property C(n,k)=C(n, n-k)

126 (if (< k (- n k)) (binomial-iter n k 0 1) (binomial-iter n (- n k) 0 1))) Another way to compute the binomial coefficient when using large numbers is to recognize that lnγ(n) is a special function that is easily computed and is standard in some programming languages such as using log_gamma in Maxima, LogGamma in Mathematica, or gammaln in MATLAB. Roundoff error may cause the returned value to not be an integer.

127 Chapter 10 Boustrophedon Transform and Eulerian Number Boustrophedon transform In mathematics, the boustrophedon transform is a procedure which maps one sequence to another. The transformed sequence is computed by filling a triangular array in boustrophedon (zig-zag) manner. Definition The boustrophedon transform: Start with the original sequence (in blue), then add numbers as indicated by the arrows, and finally read off the transformed sequence on the other side (in red). Given a sequence, the boustrophedon transform yields another sequence,, which is constructed by filling up a triangle as pictured on the right. Number the rows in the triangle starting from 0, and fill the rows consecutively. Let k denote the number of the row currently being filled. If k is odd, then put the number a k on the right end of the row and fill the row from the right to the left, with every entry being the sum of the number to the right and the number to the upper right. If k is even, then put the number a k on the left end and fill the row from the left to the right, with every entry being the sum of the number to the left and the number to the upper left.

128 The numbers b k forming the transformed sequence can then be found on the left end of odd-numbered rows and on the right end of even-numbered rows, that is, opposite to the numbers a k. Recurrence relation A more formal definition uses a recurrence relation. Define the numbers T k,n (with k n 0) by Then the transformed sequence is defined by b n = T n,n. The exponential generating function In the case a 0 = 1, a n = 0 (n > 0), the resulting triangle is called the Seidel Entringer Arnold Triangle and the numbers T k,n are called Entringer numbers (sequence A in OEIS). In this case the numbers in the transformed sequence b n are called the Euler up/down numbers. This is sequence A on the On-Line Encyclopedia of Integer Sequences. These enumerate the number of alternating permutations on n letters and are related to the Euler numbers and the Bernoulli numbers. The exponential generating function of a sequence (a n ) is defined by The exponential generating function of the boustrophedon transform (b n ) is related to that of the original sequence (a n ) by The exponential generating function of the unit sequence is 1, so that of the up/down numbers is sec x + tan x.

129 Eulerian number In combinatorics the Eulerian number A(n, m), is the number of permutations of the numbers 1 to n in which exactly m elements are greater than the previous element (permutations with m "ascents"). They are the coefficients of the Eulerian polynomials: This polynomial appears as the numerator in an expression for the generating function of the sequence. Other notations for A(n, m) are E(n, m) and History In 1755 Leonhard Euler investigated in his book Institutiones calculi differentialis polynomials α 1 (x) = 1, α 2 (x) = x + 1, α 3 (x) = x 2 + 4x + 1, etc. These polynomials are a shifted form of what are now called the Eulerian polynomials A n (x).

130 Basic properties For a given value of n > 0, the index m in A(n, m) can take values from 0 to n 1. For fixed n there is a single permutation which has 0 ascents; this is the falling permutation (n, n 1, n 2,..., 1). There is also a single permutation which has n 1 ascents; this is the rising permutation (1, 2, 3,..., n). Therefore A(n, 0) and A(n, n 1) are 1 for all values of n. Reversing a permutation with m ascents creates another permutation in which there are n m 1 ascents. Therefore A(n, m) = A(n, n m 1). Values of A(n, m) can be calculated "by hand" for small values of n and m. For example n m Permutations A(n, m) 1 0 (1) A(1,0) = (2, 1) A(2,0) = 1 1 (1, 2) A(2,1) = 1 0 (3, 2, 1) A(3,0) = 1 1 (1, 3, 2) (2, 1, 3) (2, 3, 1) (3, 1, 2) A(3,1) = 4 2 (1, 2, 3) A(3,2) = 1 For larger values of n, A(n, m) can be calculated using the recursion formula For example A(n,m) = (n m)a(n 1,m 1) + (m + 1)A(n 1,m). Values of A(n, m) (sequence A in OEIS) for 0 n 9 are: n \ m

131 The above triangular array is called the Euler triangle or Euler's triangle, and it shares some common characteristics with Pascal's triangle. The sum of row n is the factorial n!. Closed-form expression A closed-form expression for A(n, m) is Summation properties It is clear from the combinatoric definition that the sum of the Eulerian numbers for a fixed value of n is the total number of permutations of the numbers 1 to n, so Other summation properties of the Eulerian numbers are: The alternating sum of the Eulerian numbers for a fixed value of n is related to the Bernoulli number B n+1 where B n is the n th Bernoulli number. Identities The Eulerian numbers are involved in the generating function for the sequence of n th powers

132 Worpitzky's identity expresses x n as the linear combination of Eulerian numbers with binomial coefficients: Another interesting identity is given by the following manipulation: we have that the terms on the right side are positive, so we may So for switch the sum. The terms on the left make a geometric series, and we know that converges. After all of that, we use the above identity to finish the manipulation: Finally, for we get Notice that the sum on the right-hand side is the sum of the Eulerian polynomials shown at the top of this page. Eulerian numbers of the second kind The permutations of the multiset which have the property that for each k, all the numbers appearing between the two occurrences of k in the permutation are greater than k are counted by the double factorial number (2n-1)!!. The

133 Eulerian number of the second kind, denoted counts the number of all such permutations that have exactly m ascents. For instance, for n=3 there are 15 such permutations, 1 with no ascents, 8 with a single ascent, and 6 with two ascents: The Eulerian numbers of the second kind follows directly from the above definition: satisfy the recurrence relation, that with initial condition for n=0, expressed in Iverson bracket notation:. Correspondingly, the Eulerian polynomial of second kind, here denoted P n (no standard notation exists for them) are and the above recurrence relations are translated into a recurrence relation for the sequence P n (x): with initial condition P 0 (x) = 1. The latter recurrence may be written in a somehow more compact form by means of an integrating factor: so that the rational function

134 u n (x): = (x 1) 2n P n (x) satisfies a simple autonomous recurrence: whence one obtains the Eulerian polynomials as P n (x)=(1-x) 2n u n (x), and the Eulerian numbers of the second kind as their coefficients. Here are some values of the second order Eulerian numbers (sequence A in OEIS): n \ m The sum of the n-th row, which is also the value P n (1), is then (2n-1)!!.

135 Chapter 11 Gilbreath's Conjecture and Lah Number Gilbreath's conjecture Motivating arithmetic Gilbreath's conjecture is a hypothesis, or a conjecture, in number theory regarding the sequences generated by applying the forward difference operator to consecutive prime numbers and leaving the results unsigned, and then repeating this process on consecutive terms in the resulting sequence, and so forth. The statement is named after mathematician Norman L. Gilbreath who, in 1958, presented it to the mathematical community after observing the pattern by chance while doing arithmetic on a napkin. In 1878, eighty years before Gilbreath's discovery, François Proth had, however, published the same observations along with an attempted proof, which was later shown to be false. Gilbreath observed a pattern while playing with the ordered sequence of prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,... Computing the absolute value of the difference between term n+1 and term n in this sequence yields the sequence 1, 2, 2, 4, 2, 4, 2, 4, 6, 2,... If the same calculation is done for the terms in this new sequence, and the sequence that is the outcome of this process, and again ad infinitum for each sequence that is the output of such a calculation, the first five sequences in this list are given by 1, 0, 2, 2, 2, 2, 2, 2, 4,... 1, 2, 0, 0, 0, 0, 0, 2,... 1, 2, 0, 0, 0, 0, 2,... 1, 2, 0, 0, 0, 2,... 1, 2, 0, 0, 2,... What Gilbreath and François Proth before him noticed is that the first term in each series of differences appears to be 1.

136 The conjecture Stating Gilbreath's observation formally is significantly easier to do after devising a notation for the sequences in the previous section. Toward this end, let {p n } denote the ordered sequence of prime numbers p n, and define each term in the sequence {d n } by d n = p n + 1 p n, where n is positive. Also, for each integer k greater than 1, let the terms in by be given. Gilbreath's conjecture states that every term in the sequence for positive k is 1. Verification and attempted proofs As of 2011, no valid proof of the conjecture has been published. As mentioned in the introduction, François Proth released what he believed to be a proof of the statement that was later shown to be flawed. Andrew Odlyzko verified that is 1 for in 1993, but the conjecture remains an open problem. Lah number In mathematics, Lah numbers, discovered by Ivo Lah in 1955, are coefficients expressing rising factorials in terms of falling factorials. Unsigned Lah numbers have an interesting meaning in combinatorics: they count the number of ways a set of n elements can be partitioned into k nonempty linearly ordered subsets. Lah numbers are related to Stirling numbers. Unsigned Lah numbers: Signed Lah numbers:

137 L(n, 1) is always n!; using the interpretation above, the only partition of {1, 2, 3} into 1 set can be ordered in 6 ways: {(1, 2, 3)}, {(1, 3, 2)}, {(2, 1, 3)}, {(2, 3, 1)}, {(3, 1, 2)} or {(3, 2, 1)} L(3, 2) corresponds to the 6 partitions with two ordered parts: {(1), (2, 3)}, {(1), (3, 2)}, {(2), (1, 3)}, {(2), (3, 1)}, {(3), (1, 2)} or {(3), (2, 1)} L(n, n) is always 1: partitioning {1, 2, 3} into 3 non-empty subsets results in subsets of length 1. {(1), (2), (3)} Paraphrasing Karamata-Knuth notation for Stirling numbers, it was proposed to use the following alternative notation for Lah numbers: Rising and falling factorials Let x (n) represent the rising factorial represent the falling factorial. and let (x) n Then and For example, Compare the third row of the table of values. Identities and relations

138 with s(n,k) the Stirling numbers of the first kind, the Stirling numbers of the second kind and with the conventions L(0,0) = 1 and L(n,k) = 0 if k > n. L(n,1) = n! L(n,2) = (n 1)n! / 2 L(n,3) = (n 2)(n 1)n! / 12 L(n,n 1) = n(n 1) L(n,n) = 1 Table of values Below is a table of values for the Lah numbers:

139 Chapter 12 Leibniz Harmonic Triangle and Narayana Number Leibniz harmonic triangle The Leibniz harmonic triangle is a triangular arrangement of fractions in which the outermost diagonals consist of the reciprocals of the row numbers and each inner cell is the absolute value of the cell above minus the cell to the left. To put it algebraically, (where r is the number of the row, starting from 1, and c is the column number, never more than r) and L(r, c) = L(r - 1, c - 1) L(r, c - 1). The first eight rows are: The denominators are listed in (sequence A in OEIS), while the numerators, which are all 1s, are listed in A Whereas each entry in Pascal's triangle is the sum of the two entries in the above row, each entry in the Leibniz triangle is the sum of the two entries in the row below it. For example, in the 5 th row, the entry (1/30) is the sum of the two (1/60)s in the 6 th row.

140 Just as Pascal's triangle can be computed by using binomial coefficients, so can Leibniz's:. Furthermore, the entries of this triangle can be computed from Pascal's, "the terms in each row are the initial term divided by the corresponding Pascal triangle entries." (Wells, 1986) This triangle can be used to obtain examples for the Erdős Straus conjecture when n is divisible by 4. If one takes the denominators of the nth row and adds them, then the result will equal n2 n 1. For example, for the 3rd row, we have = 12 = It is worth noting that Narayana number In combinatorics, the Narayana numbers N(n, k), n = 1, 2, 3..., 1 k n, form a triangular array of natural numbers, called Narayana triangle, that occur in various counting problems. They are named for T.V. Narayana ( ), a mathematician from India. The Narayana numbers can be expressed in terms of binomial coefficients: An example of a counting problem whose solution can be given in terms of the Narayana numbers N(n, k), is the number of expressions containing n pairs of parentheses which are correctly matched and which contain k distinct nestings. For instance, N(4, 2) = 6 as with four pairs of parentheses six sequences can be created which each contain two times the sub-pattern '()': ()((())) (())(()) (()(())) ((()())) ((())()) ((()))() From this example it should be obvious that N(n, 1) = 1, since the only way to get a single sub-pattern '()' is to have all the opening parentheses in the first n positions, followed by all the closing parentheses. Also N(n, n) = 1, as distinct nestings can be achieved only by the repetitive pattern ()()()... (). More generally, it can be shown that the Narayana triangle is symmetric: N(n, k) = N(n, n k + 1).

141 The first eight rows of the Narayana triangle read: k = n = (sequence A in OEIS) The sum of the rows in this triangle equal the Catalan numbers: To illustrate this relationship, the Narayana numbers also count the number of paths from (0, 0) to (2n, 0), with steps only northeast and southeast, not straying below the x-axis, with k peaks. The following figures represent the Narayana numbers N(4, k): N(4, k) N(4, 1) = 1 path with 1 peak: Paths N(4, 2) = 6 paths with 2 peaks: N(4, 3) = 6 paths with 3 peaks: N(4, 4) = 1 path with 4 peaks: The sum of N(4, k) is , or 14, which is the same as Catalan number C 4. This sum coincides with the interpretation of Catalan numbers as the number of monotonic paths along the edges of an n n grid that do not pass above the diagonal.

142 Partitions In the study of partitions, we see that in a set containing n elements, we may partition that set in B n different ways, where B n is the n th Bell number. Furthermore, the number of ways to partition a set into exactly k partitions we use the Stirling numbers S(n,k). Both of these concepts are a bit off-topic, but a necessary foundation for understanding the use of the Narayana numbers. In both of the above two notions crossing partitions are accounted for. To reject the crossing partitions and count only the non-crossing partitions, we may use the Catalan numbers to count the non-crossing partitions of all n elements of the set, C n. To count the non-crossing partitions in which the set is partitioned in exactly k ways, we use the Narayana number N(n,k).

143 Chapter 13 Pascal Matrix In mathematics, particularly matrix theory and combinatorics, the Pascal matrix is an infinite matrix containing the binomial coefficients as its elements. There are 3 ways this can be achieved - either as an upper-triangular matrix, a lower-triangular matrix, or as a symmetric matrix. The 5 5 truncations of these are shown below. Upper triangular: lower triangular: symmetric: These matrices have the pleasing relationship S n = L n U n. From this it is easily seen that all three matrices have determinant 1, as the determinant of a triangular matrix is simply the product of its diagonal elements, which are all 1 for both L n and U n. In other words, matrices S n, L n, and U n are unimodular, with L n and U n having trace n. The elements of the symmetric Pascal matrix are the binomial coefficients, i.e. In other words,

144 Thus the trace of S n is given by with the first few terms given by the sequence 1, 3, 9, 29, 99, 351, 1275,... (Sloane's A006134). Construction The Pascal matrix can actually be constructed by taking the matrix exponential of a special subdiagonal or superdiagonal matrix. The example below constructs a 7-by-7 Pascal matrix, but the method works for any desired n n Pascal matrices. (Note that dots in the following matrices represent zero elements.) It is important to note that you cannot simply assume exp(a)exp(b) = exp(a + B), for A and B n n matrices. Such an identity only holds when AB = BA (i.e. the matrices A and B commute). In the construction of symmetric Pascal matrices like that above, the sub- and superdiagonal matrices do not commute, so the (perhaps) tempting simplification involving the addition of the matrices cannot be made. A useful property of the sub- and superdiagonal matrices used in the construction is that both are nilpotent; that is, when raised to a sufficiently high integer power, they degenerate into the zero matrix. As the n n generalised shift matrices we are using become zero when raised to power n, when calculating the matrix exponential we need only consider the first n + 1 terms of the infinite series to obtain an exact result. Variants Interesting variants can be obtained by obvious modification of the matrix-logarithm PL_7 and then application of the matrix exponential.

145 The first example below uses the squares of the values of the log-matrix and constructs a 7-by-7 "Laguerre"- matrix (or matrix of coefficients of Laguerre-polynomials) The Laguerre-matrix is actually used with some other scaling and/or the scheme of alternating signs. (Literature about generalizations to higher powers is not found yet) The second example below uses the products v(v + 1) of the values of the log-matrix and constructs a 7-by-7 "Lah"- matrix (or matrix of coefficients of Lah numbers) Using v(v 1) instead provides a diagonal shifting to bottom-right. The third example below uses the square of the original PL 7 -matrix, divided by 2, in other words: the first-order binomials (binomial(k,2) ) in the second subdiagonal and constructs a matrix, which occurs in context of the derivatives and integrals of the Gaussian error-function: If this matrix is inverted (using, for instance, the negative matrix-logarithm), then this matrix has alternating signs and gives the coefficients of the derivatives (and by extension) the integrals of the Gauss' - error-function. (Literature about generalizations to higher powers is not found yet.)

146 Chapter 14 Pascal's Pyramid In mathematics, Pascal's Pyramid is a three-dimensional arrangement of the trinomial numbers, which are the coefficients of the trinomial expansion and the trinomial distribution. Pascal's Pyramid is the three-dimensional analog of the two-dimensional Pascal's Triangle, which contains the binomial numbers and relates to the binomial expansion and the binomial distribution. The binomial and trinomial numbers, coefficients, expansions, and distributions are subsets of the multinomial constructs with the same names. Pascal's Pyramid is more precisely called "Pascal's Tetrahedron", since it has four triangular surfaces. (The pyramids of ancient Egypt had five surfaces: a square base and four triangular sides.) Structure of the Tetrahedron Because the Tetrahedron is a three-dimensional object it is difficult to display it on a piece of paper or a computer screen. So the Tetrahedron is divided into a number of levels, or floors, or slices, or layers. The top layer (the apex) is labeled "Layer 0". Other layers can be thought of as overhead views of the Tetrahedron with the previous layers removed. The first six layers are as follows: Layer 0 1 Layer Layer Layer

147 Layer Layer Overview of the Tetrahedron The layers of the Tetrahedron have been deliberately displayed with the point down so that the Tetrahedron is not confused with Pascal's Triangle. There is three-way symmetry of the numbers in each layer. The number of terms in the n th Layer is the n th triangular number: (n + 1) (n + 2) / 2. The sum of the values of the numbers in the n th Layer is 3 n. Each number in any layer is the sum of the three adjacent numbers in the layer above. Each number in any layer is a simple whole number ratio of the adjacent numbers in the same layer. Each number in any layer is a coefficient of the Trinomial Distribution and the Trinomial Expansion. This non-linear arrangement makes it easier to: o display the Trinomial Expansion in a coherent way; o compute the coefficients of the Trinomial Distribution; o calculate the numbers of any Tetrahedron layer. The numbers along the three edges of the n th Layer are the numbers of the n th Line of Pascal's Triangle. And almost all the properties listed above have parallels with Pascal's Triangle and Multinomial Coefficients. Trinomial Expansion connection The numbers of the Tetrahedron are derived from Trinomial Expansion. The n th Layer is the detached coefficient matrix (no variables or exponents) of a trinomial expression (e.g.: A + B + C) raised to the n th power. The trinomial is expanded by repeatedly multiplying the trinomial by itself: (A + B + C) 1 (A + B + C) n = (A + B + C) n+1

148 Each term in the first expression is multiplied by each term in the second expression; and then the coefficients of like terms (same variables and exponents) are added together. Here is the expansion of (A + B + C) 4 : 1A 4 B 0 C 0 + 4A 3 B 0 C 1 + 6A 2 B 0 C 2 + 4A 1 B 0 C 3 + 1A 0 B 0 C 4 + 4A 3 B 1 C A 2 B 1 C A 1 B 1 C 2 + 4A 0 B 1 C 3 + 6A 2 B 2 C A 1 B 2 C 1 + 6A 0 B 2 C 2 + 4A 1 B 3 C 0 + 4A 0 B 3 C 1 + 1A 0 B 4 C 0 Writing the expansion in this non-linear way shows the expansion in a more understandable way. It also makes the connection with the Tetrahedron obvious the coe-fficients here match those of Layer 4. All the implicit coefficients, variables, and exponents, which are normally not written, are also shown to illustrate another relationship with the Tetrahedron. (Usually, "1A" is "A"; "B 1 " is "B"; and "C 0 " is "1"; etc.) The exponents of each term sum to the Layer number (n), or 4, in this case. More significantly, the value of the coefficients of each term can be computed directly from the exponents. The formula is: (x + y + z)! / (x! y! z!), where x, y, z are the exponents of A, B, C, respectively, and "!" means factorial (e.g.: n! = n). The exponent formulas for the 4 th Layer are: The exponents of each expansion term can be clearly seen and these formulae simplify to the expansion coefficients and the Tetrahedron coefficients of Layer 4. Trinomial Distribution connection The numbers of the Tetrahedron can also be found in the Trinomial Distribution. This is a discrete probability distribution used to determine the chance some combination of events occurs given three possible outcomes the number of ways the events could occur is multiplied by the probabilities that they would occur. The formula for the Trinomial Distribution is: [ n! / ( x! y! z!) ] [ (P A ) x (P B ) y (P C ) z ]

149 where x, y, z are the number of times each of the three outcomes does occur; n is the number of trials and equals the sum of x+y+z; and P A, P B, P C are the probabilities that each of the three events could occur. For example, in a three-way election, the candidates got these votes: A, 16%; B, 30%; C, 54%. What is the chance that a randomly-selected four-person focus group would contain the following voters: 1 for A, 1 for B, 2 for C? The answer is: [ 4! / ( 1! 1! 2!) ] [ (16%) 1 (30%) 1 (54%) 2 ] = = 17% The number 12 is the coefficient of this probability and it is number of combinations that can fill this "112" focus group. There are 15 different arrangements of four-person focus groups that can be selected. Expressions for all 15 of these coefficients are: The numerator of these fractions (above the line) is the same for all expressions. It is the sample size a four-person group and indicates that the coefficients of these arrangements can be found on Layer 4 of the Tetrahedron. The three numbers of the denominator (below the line) are the number of the focus group members that voted for A, B, C, respectively. Shorthand is normally used to express combinatorial functions in the following "choose" format (which is read as "4 choose 4, 0, 0", etc.).

150 But the value of these expression is still equal to the coefficients of the 4 th Layer of the Tetrahedron. And they can be generalized to any Layer by changing the sample size (n). This notation makes an easy way to express the sum of all the coefficients of Layer n: = 3 n. Addition of coefficients between layers The numbers on every layer (n) of the Tetrahedron are the sum of the three adjacent numbers in the layer (n 1) "above" it. This relationship is rather difficult to see without intermingling the layers. Below are italic Layer 3 numbers interleaved among bold Layer 4 numbers: The relationship is illustrated by the lower, central number 12 of the 4 th Layer. It is "surrounded" by three numbers of the 3 rd Layer: 6 to the "north", 3 to the "southwest", 3 to the "southeast". (The numbers along the edge have only two adjacent numbers in the layer "above" and the three corner numbers have only one adjacent number in the layer above, which is why they are always "1". The missing numbers can be assumed as "0", so there is no loss of generality.) This relationship between adjacent layers is not a magical coincidence. Rather, it comes about through the two-step Trinomial Expansion process. Continuing with this example, in Step 1, each term of (A + B + C) 3 is multiplied by each term of (A + B + C) 1. Only three of these multiplications are of interest in this example: Layer 3 term Multiply by Product term 6A 1 B 1 C 1 1B 1 6A 1 B 2 C 1 3A 1 B 2 C 0 1C 1 3A 1 B 2 C 1 3A 0 B 2 C 1 1A 1 3A 1 B 2 C 1 (The multiplication of like variables causes the addition of exponents; e.g.: D 1 D 2 = D 3.)

151 Then, in Step 2, the summation of like terms (same variables and exponents) results in: 12A 1 B 2 C 1, which is the term of (A + B + C) 4 ; while 12 is the coefficient of the 4 th Layer of the Tetrahedron. Symbolically, the additive relation can be expressed as: C(x,y,z) = C(x 1,y,z) + C(x,y 1,z) + C(x,y,z 1) where C(x,y,z) is the coefficient of the term with exponents x, y, z and x+y+z = n is the layer of the Tetrahedron. This relationship will work only if the Trinomial Expansion is laid out in the non-linear fashion as it is portrayed in the section on the "Trinomial Expansion connection". Ratio between coefficients of same layer On each layer of the Tetrahedron, the numbers are simple whole number ratios of the adjacent numbers. This relationship is illustrated for horizontally adjacent pairs on the 4 th Layer by the following: 1 <1:4> 4 <2:3> 6 <3:2> 4 <4:1> 1 4 <1:3> 12 <2:2> 12 <3:1> 4 6 <1:2> 12 <2:1> 6 4 <1:1> 4 1 Because the Tetrahedron has three-way symmetry, the ratio relation also holds for diagonal pairs (in both directions), as well as for the horizontal pairs shown. The ratios are controlled by the exponents of the corresponding adjacent terms of the Trinomial Expansion. For example, one ratio in the illustration above is: 4 <1:3> 12 The corresponding terms of the Trinomial Expansion are: 4A 3 B 1 C 0 and 12A 2 B 1 C 1 The following rules apply to the coefficients of all adjacent pairs of terms of the Trinomial Expansion: The exponent of one of the variables remains unchanged (B in this case) and can be ignored. For the other two variables, one exponent increases by 1 and one exponent decreases by 1. o The exponents of A are 3 and 2 (the larger being in the left term).

152 o The exponents of C are 0 and 1 (the larger being in the right term). The coefficients and larger exponents are related: o 4 3 = 12 1 o 4 / 12 = 1 / 3 These equations yield the ratio: "1:3". The rules are the same for all horizontal and diagonal pairs. The variables A, B, C will change. This ratio relationship provides another (somewhat cumbersome) way to calculate Tetrahedron coefficients: The coefficient of the adjacent term equals the coefficient of the current term multiplied by the current-term exponent of the decreasing variable divided by the adjacent-term exponent of the increasing variable. For x = 0: C(x,y,z 1) = C(x,y 1,z) z / y C(x,y 1,z) = C(x,y,z 1) y / z For y = 0: C(x 1,y,z) = C(x,y,z 1) x / z C(x,y,z 1) = C(x 1,y,z) z / x For z = 0: C(x,y 1,z) = C(x 1,y,z) y / x C(x 1,y,z) = C(x,y 1,z) x / y The ratio of the adjacent coefficients may be a little clearer when expressed symbolically. Each term can have up to six adjacent terms: where C(x,y,z) is the coefficient and x, y, z are the exponents. In the days before pocket calculators and personal computers, this approach was used as a school-boy short-cut to write out Binomial Expansions without tedious algebraic expansions or clumsy factorial computations. This relationship will work only if the Trinomial Expansion is laid out in the non-linear fashion as it is portrayed in the section on the "Trinomial Expansion connection". Relationship with Pascal's Triangle It is well known that the numbers along the three outside edges of the n th Layer of the Tetrahedron are the same numbers as the n th Line of Pascal's Triangle. However, the connection is actually much more extensive than just one row of numbers. This relationship is best illustrated by comparing Pascal's Triangle down to Line 4 with Layer 4 of the Tetrahedron. Pascal's Triangle

153 Tetrahedron Layer Multiplying the numbers of each line of Pascal's Triangle down to the n th Line by the numbers of the n th Line generates the n th Layer of the Tetrahedron. In the following example, the lines of Pascal's Triangle are in italic font and the rows of the Tetrahedron are in bold font. 1 1 = = = = = The multipliers ( ) compose Line 4 of Pascal's Triangle. This relationship demonstrates the fastest and easiest way to compute the numbers for any layer of the Tetrahedron without computing factorials, which quickly become huge numbers. (Extended precision calculators become very slow beyond Tetrahedron Layer 200.) If the coefficients of Pascal's Triangle are labeled C(i,j) and the coefficients of the Tetrahedron are labeled C(n,i,j), where n is the layer of the Tetrahedron, i is the row, and j is the column, then the relation can be expressed symbolically as: C(i,j) C(n,j) = C(n,i,j) i = 0 to n, j = 0 to i

154 [It is important to understand that i, j, n are not exponents here, just sequential labeling indexes.] Parallels to Pascal's Triangle and Multinomial Coefficients This table summarizes the properties of the Trinomial Expansion and the Trinomial Distribution, and it compares them to the Binomial and Multinomial Expansions and Distributions: Type of polynomial bi-nomial tri-nomial multi-nomial Order of polynomial 2 3 m Example of polynomial A+B A+B+C A+B+C+...+M Geometric structure (1) triangle tetrahedron m-simplex Element structure line layer group Symmetry of element 2-way 3-way m-way Number of terms per (n+1) (n+2) / n+1 element 2 Sum of values per element 2 n 3 n m n Example of term A x B y A x B y C z A x B y C z...m m Sum of exponents, all terms Coefficient equation (2) Sum of coefficients "above" Ratio of adjacent coefficients n n n n! / (x! y!) n! / (x! y! z!) 2 3 m 2 6 m (m 1) (n+1) (n+2)... (n+m 1) / (m 1) n! / (x 1! x 2! x 3!... x m!) (1) A simplex is the simplest linear geometric form that exists in any dimension. Tetrahedrons and triangles are examples in 3 and 2 dimensions, respectively.(2) The formula for the binomial coefficient is usually expressed as: n! / (x! (n x)!); where n x = y.

155 Other properties Exponential construction Arbitrary layer n can be obtained in a single step using the following formula: where b is the radix and d is the number of digits of any of the central multinomial coefficients, that is then wrapping the digits of its result by d(n+1), spacing by d and removing leading zeros. This method generalised to arbitrary dimension can be used to obtain slices of any Pascal's simplex. Examples For radix b = 10, n = 5, d = 2: = = ~ ~ ~ wrapped by d(n+1) spaced by d leading zeros removed For radix b = 10, n = 20, d = 9: Pascal's pyramid layer #20.

156 Sum of coefficients of a layer by rows Summing the numbers in each row of a layer n of Pascal's pyramid gives where b is the radix and d is the number of digits of the sum of the 'central' row (the one with the greatest sum). For radix b = 10: 1 ~ 1 \ 1 ~ 1 \ 1 ~ 1 \ 1 ~ 1 \ 1 ~ \ 1 ~ 2 \ 2 \ 2 ~ 4 \ 3 \ 3 ~ 06 \ 4 \ 4 ~ \ 2 \ 1 ~ 4 \ 3 \ 6 \ 3 ~ 12 \ 6 \12 \ 6 ~ \ 3 \ 3 \ 1 ~ 08 \ 4 \12 \12 \ 4 ~ \ 4 \ 6 \ 4 \ 1 ~ Sum of coefficients of a layer by columns Summing the numbers in each column of a layer n of Pascal's pyramid gives where b is the radix and d is the number of digits of the sum of the 'central' column (the one with the greatest sum). For radix b = 10:

157 Usage While studying genetics, it's common to use the Pascal's pyramid to find out the proportion between different genotypes on the same crossing. This is done by checking the line that is equivalent to the number of phenotypes (genotypes + 1). That line will be the proportion.

158 Chapter 15 Pascal's Simplex In mathematics, Pascal's simplex is a generalisation of Pascal's triangle into arbitrary number of dimensions, based on the multinomial theorem. Induction of Pascal's simplices Each Pascal's m-simplex is a semi-infinite object, which consists of a semi-infinite series (n 0) of finite (m 1)-simplices, where m is the number of terms of a polynomial and n is a power the polynomial is raised to. Let be a semi-infinite Pascal's m-simplex and its n th component, a finite (m 1)-simplex with edge length n. Pascal's 1-simplex (a point) is the coefficient of multinomial expansion of a polynomial with 1 term raised to the power of n: Arrangement of : which equals 1 for all n. Pascal's 2-simplex is known as Pascal's triangle.

159 (a line) consists of the coefficients of binomial expansion of a polynomial with 2 terms raised to the power of n: Arrangement of : Pascal's 3-simplex is known as Pascal's tetrahedron. (a triangle) consists of the coefficients of trinomial expansion of a polynomial with 3 terms raised to the power of n: Arrangement of : Pascal's m-simplex consists of the coefficients of multinomial expansion of a polynomial with m terms raised to the power of n:

160 Components of Pascal's simplices Inheritance of components is numericaly equal to each (m 1)-face (there is m + 1 of them) of, or: From this follows, that the whole is (m + 1)-times included in, or: Example Equality of sub-faces Conversely, is (m + 1)-times bounded by, or: From this follows, that for given n, all i-faces are numericaly equal in n th components of all Pascal's (m > i)-simplices, or: Example

161 The 3 rd component (2-simplex) of Pascal's 3-simplex is bounded by 3 equal 1-faces (lines). Each 1-face (line) is bounded by 2 equal 0-faces (vertices): 2-simplex 1-faces of 2-simplex 0-faces of 1-face Also, for all m and all n: Number of coefficients For the n th component ((m 1)-simplex) of Pascal's m-simplex, the number of the coefficients of multinomial expansion it consists of is given by: that is, either by a sum of the number of coefficients of an (n 1) th component ((m 1)- simplex) of Pascal's m-simplex with the number of coefficients of an n th component ((m 2)-simplex) of Pascal's (m 1)-simplex, or by a number of all possible partitions of an n th power among m exponents. Example Number of coefficients of n th component ((m 1)-simplex) of Pascal's m-simplex m-simplex n th component n = 0 n = 1 n = 2 n = 3 n = 4 n = 5 1-simplex 0-simplex simplex 1-simplex simplex 2-simplex simplex 3-simplex simplex 4-simplex simplex 5-simplex

162 Chapter 16 Pascal's Triangle The first six rows of Pascal's triangle In mathematics, Pascal's triangle is a triangular array of the binomial coefficients in a triangle. It is named after the French mathematician Blaise Pascal in much of the Western world, although other mathematicians studied it centuries before him in India, Persia, China, Germany, and Italy. The rows of Pascal's triangle are conventionally enumerated starting with row n = 0 at the top. The entries in each row are numbered from the left beginning with k = 0 and are usually staggered relative to the numbers in the adjacent rows. A simple construction of the triangle proceeds in the following manner. On row 0, write only the number 1. Then, to construct the elements of following rows, add the number directly above and to the left with the number directly above and to the right to find the new value. If either the number

163 to the right or left is not present, substitute a zero in its place. For example, the first number in the first row is = 1, whereas the numbers 1 and 3 in the third row are added to produce the number 4 in the fourth row. This construction is related to the binomial coefficients by Pascal's rule, which states that if then for any nonnegative integer n and any integer k between 0 and n. Pascal's triangle has higher dimensional generalizations. The three-dimensional version is called Pascal's pyramid or Pascal's tetrahedron, while the general versions are called Pascal's simplices. Each number in the triangle is the sum of the two directly above it.

164 History Yang Hui (Pascal's) triangle, as depicted by the Chinese using rod numerals.

165 Blaise Pascal's version of the triangle The set of numbers that form Pascal's triangle were well known before Pascal. However, Pascal developed many applications of it and was the first one to organize all the information together in his treatise, Traité du triangle arithmétique (1653). The numbers originally arose from Hindu studies of combinatorics and binomial numbers and the Greeks' study of figurate numbers. The earliest explicit depictions of a triangle of binomial coefficients occur in the 10th century in commentaries on the Chandas Shastra, an Ancient Indian book on Sanskrit prosody written by Pingala between the 5th and 2nd century BC. While Pingala's work only survives in fragments, the commentator Halayudha, around 975, used the triangle to explain obscure references to Meru-prastaara, the "Staircase of Mount Meru". It was also realised that the shallow diagonals of the triangle sum to the Fibonacci numbers. At around the same time, it was discussed in Persia (Iran) by the Persian mathematician, Al-Karaji ( ). It was later repeated by the Persian poet-astronomer-mathematician Omar Khayyám ( ); thus the triangle is referred to as the Khayyam triangle in Iran. Several theorems related to the triangle were known, including the

166 binomial theorem. Khayyam used a method of finding nth roots based on the binomial expansion, and therefore on the binomial coefficients. In 13th century, Yang Hui ( ) presented the arithmetic triangle that is the same as Pascal's triangle. Pascal's triangle is called Yang Hui's triangle in China. The "Yang Hui's triangle" was known in China in the early 11th century by the Chinese mathematician Jia Xian ( ). Petrus Apianus ( ) published the triangle on the frontispiece of his book on business calculations in the 16th century. This is the first record of the triangle in Europe. In Italy, it is referred to as Tartaglia's triangle, named for the Italian algebraist Niccolò Fontana Tartaglia ( ). Tartaglia is credited with the general formula for solving cubic polynomials, (which may in fact be from Scipione del Ferro but was published by Gerolamo Cardano 1545). Binomial expansions Pascal's Traité du triangle arithmétique (Treatise on Arithmetical Triangle) was publicshed posthumously in In this, Pascal collected several results then known about the triangle, and employed them to solve problems in probability theory. The triangle was later named after Pascal by Pierre Raymond de Montmort (1708) who called it "Table de M. Pascal pour les combinaisons" (French: Table of Mr. Pascal for combinations) and Abraham de Moivre (1730) who called it "Triangulum Arithmeticum PASCALIANUM" (Latin: Pascal's Arithmetic Triangle), which became the modern Western name. Pascal's triangle determines the coefficients which arise in binomial expansions. For an example, consider the expansion (x + y) 2 = x 2 + 2xy + y 2 = 1x 2 y 0 + 2x 1 y 1 + 1x 0 y 2. Notice the coefficients are the numbers in row two of Pascal's triangle: 1, 2, 1. In general, when a binomial like x + y is raised to a positive integer power we have: (x + y) n = a 0 x n + a 1 x n 1 y + a 2 x n 2 y a n 1 xy n 1 + a n y n, where the coefficients a i in this expansion are precisely the numbers on row n of Pascal's triangle. In other words, This is the binomial theorem.

167 Notice that the entire right diagonal of Pascal's triangle corresponds to the coefficient of y n in these binomial expansions, while the next diagonal corresponds to the coefficient of xy n 1 and so on. To see how the binomial theorem relates to the simple construction of Pascal's triangle, consider the problem of calculating the coefficients of the expansion of (x + 1) n+1 in terms of the corresponding coefficients of (x + 1) n (setting y = 1 for simplicity). Suppose then that Now The two summations can be reorganized as follows: (because of how raising a polynomial to a power works, a 0 = a n = 1). We now have an expression for the polynomial (x + 1) n+1 in terms of the coefficients of (x + 1) n (these are the a i s), which is what we need if we want to express a line in terms of the line above it. Recall that all the terms in a diagonal going from the upper-left to the lower-right correspond to the same power of x, and that the a-terms are the coefficients of the polynomial (x + 1) n, and we are determining the coefficients of (x + 1) n+1. Now, for any given i not 0 or n + 1, the coefficient of the x i term in the polynomial (x + 1) n+1 is

168 equal to a i (the figure above and to the left of the figure to be determined, since it is on the same diagonal) + a i 1 (the figure to the immediate right of the first figure). This is indeed the simple rule for constructing Pascal's triangle row-by-row. It is not difficult to turn this argument into a proof (by mathematical induction) of the binomial theorem. Since (a + b) n = b n (a/b + 1) n, the coefficients are identical in the expansion of the general case. An interesting consequence of the binomial theorem is obtained by setting both variables x and y equal to one. In this case, we know that (1 + 1) n = 2 n, and so Combinations In other words, the sum of the entries in the nth row of Pascal's triangle is the nth power of 2. A second useful application of Pascal's triangle is in the calculation of combinations. For example, the number combinations of n things taken k at a time (called n choose k) can be found by the equation But this is also the formula for a cell of Pascal's triangle. Rather than performing the calculation, one can simply look up the appropriate entry in the triangle. For example, suppose a basketball team has 10 players and wants to know how many ways there are of selecting 8. Provided we have the first row and the first entry in a row numbered 0, the answer is entry 8 in row 10: 45. That is, the solution of 10 choose 8 is 45. Relation to binomial distribution and convolutions When divided by 2 n, the nth row of Pascal's triangle becomes the binomial distribution in the symmetric case where p = 1/2. By the central limit theorem, this distribution approaches the normal distribution as n increases. This can also be seen by applying Stirling's formula to the factorials involved in the formula for combinations. This is related to the operation of discrete convolution in two ways. First, polynomial multiplication exactly corresponds to discrete convolution, so that repeatedly convolving the sequence {..., 0, 0, 1, 1, 0, 0,...} with itself corresponds to taking powers of 1 + x, and hence to generating the rows of the triangle. Second, repeatedly convolving the distribution function for a random variable with itself corresponds to calculating the distribution function for a sum of n independent copies of that variable; this is exactly the

169 situation to which the central limit theorem applies, and hence leads to the normal distribution in the limit. Patterns and properties Pascal's triangle has many properties and contains many patterns of numbers. Rows 1000 th row of Pascal's triangle, arranged vertically, with grey-scale representations of decimal digits of the coefficients right-aligned. The left boundary of the image corresponds roughly to the graph of the logarithm of the binomial coefficients, and illustrates that they form a log-concave sequence.

170 When adding all the digits in a single row, each successive row has twice the value of the row preceding it. For example, row 1 has a value of 1, row 2 has a value of 2, row 3 has a value of 4, and so forth. The value of a row, if each entry is considered a decimal place (and numbers larger than 9 carried over accordingly) is a power of 11 ( 11 n, for row n). Thus, in row two, '1,2,1' becomes 11 2, while '1,5,10,10,5,1' in row six becomes (after carrying) 161,051, which is This property is explained by setting x = '10' in the binomial expansion of (x + 1) row=n, and adjusting values to the decimal system. But x can be chosen to allow rows to represent values in any base - such as base 3; ['1,2,1'] = 4 2 (16), ['1,3,3,1'] = 4 3 (64) - or base 9; = 10 2 (100), = 10 3 (1000) and ['1,5,10,10,5,1'] = 10 5 (100,000). In particular, for x = 1 place value remains constant (1 place =1). Thus entries can simply be added in interpreting the value of a row. The sum of the elements of row m is equal to 2 m 1. For example, the sum of the elements of row 5 is = 16, which is equal to 2 4 = 16. This follows from the binomial theorem proved above, applied to (1 + 1) m 1. If rows are numbered starting with n = 0, the sum of the elements in the row is simply 2 n, so row 0 adds to 2 0 = 1, row 1 adds to 2 1 = 2, etc. Some of the numbers in Pascal's triangle correlate to numbers in Lozanić's triangle. The sum of the squares of the elements of row m equals the middle element of row (2m 1). For example, = 70. In general form: Another interesting pattern is that on any row m, where m is odd, the middle term minus the term two spots to the left equals a Catalan number, specifically the (m + 1)/2 Catalan number. For example: on row 5, 6 1 = 5, which is the 3 rd Catalan number, and (5 + 1)/2 = 3. Another interesting property of Pascal's triangle is that in rows where the second number (immediately following '1') is prime, all the terms in that row except the 1s are multiples of that prime. Diagonals The diagonals of Pascal's triangle contain the figurate numbers of simplices: The diagonals going along the left and right edges contain only 1's. The diagonals next to the edge diagonals contain the natural numbers in order. Moving inwards, the next pair of diagonals contain the triangular numbers in order. The next pair of diagonals contain the tetrahedral numbers in order, and the next pair give pentatope numbers.

171 The symmetry of the triangle implies that the n th d-dimensional number is equal to the d th n-dimensional number. An alternative formula that does not involve recursion is as follows: where n (d) is the rising factorial. The geometric meaning of a function P d is: P d (1) = 1 for all d. Construct a d-dimensional triangle (a 3-dimensional triangle is a tetrahedron) by placing additional dots below an initial dot, corresponding to P d (1) = 1. Place these dots in a manner analogous to the placement of numbers in Pascal's triangle. To find P d (x), have a total of x dots composing the target shape. P d (x) then equals the total number of dots in the shape. A 0-dimensional triangle is a point and a 1-dimensional triangle is simply a line, and therefore P 0 (x) = 1 and P 1 (x) = x, which is the sequence of natural numbers. The number of dots in each layer corresponds to P d 1 (x). Calculating an individual row or diagonal by itself This algorithm is an alternative to the standard method of calculating individual cells with factorials. Starting at the left, the first cell's value v 0 is 1. For each cell after, the value is determined by multiplying the value to its left by a slowly changing fraction: where r = row + 1, starting with 0 at the top, and c = the column, starting with 0 on the left. For example, to calculate row 5, r = 6. The first value is 1. The next value is 1 5/1 = 5. The numerator decreases by one, and the denominator increases by one with each step. So 5 4/2 = 10. Then 10 3/3 = 10. Then 10 2/4 = 5. Then 5 1/5 = 1. Notice that the last cell always equals 1, the final multiplication is included for completeness of the series. A similar pattern exists on a downward diagonal. Starting with the one and the natural number in the next cell, form a fraction. To determine the next cell, increase the numerator and denominator each by one, and then multiply the previous result by the fraction.

172 For example, the row starting with 1 and 7 form a fraction of 7/1. The next cell is 7 8/2 = 28. The next cell is 28 9/3 = 84. (Note that for any individual row it is only necessary to calculate half (rounded up) the terms in the row due to symmetry.) Overall patterns and properties Sierpinski triangle The pattern obtained by coloring only the odd numbers in Pascal's triangle closely resembles the fractal called the Sierpinski triangle. This resemblance becomes more and more accurate as more rows are considered; in the limit, as the number of rows approaches infinity, the resulting pattern is the Sierpinski triangle, assuming a fixed perimeter. More generally, numbers could be colored differently according to whether or not they are multiples of 3, 4, etc.; this results in other similar patterns. Imagine each number in the triangle is a node in a grid which is connected to the adjacent numbers above and below it. Now for any node in the grid, count the number of paths there are in the grid (without backtracking) which connect this node to the top node (1) of the triangle. The answer is the Pascal number

173 associated to that node. The interpretation of the number in Pascal's Triangle as the number of paths to that number from the tip means that on a Plinko game board shaped like a triangle, the probability of winning prizes nearer the center will be higher than winning prizes on the edges. One property of the triangle is revealed if the rows are left-justified. In the triangle below, the diagonal coloured bands sum to successive Fibonacci numbers

174 Construction as matrix exponential Binomial matrix as matrix exponential. All the dots represent 0. Number of elements of polytopes Due to its simple construction by factorials, a very basic representation of Pascal's triangle in terms of the matrix exponential can be given: Pascal's triangle is the exponential of the matrix which has the sequence 1, 2, 3, 4, on its subdiagonal and zero everywhere else. Pascal's triangle can be used as a lookup table for the number of elements (such as edges and corners) within a polytope (such as a triangle, a tetrahedron, a square and a cube). Let's begin by considering the 3rd line of Pascal's triangle, with values 1, 3, 3, 1. A 2- dimensional triangle has one 2-dimensional element (itself), three 1-dimensional elements (lines, or edges), and three 0-dimensional elements (vertices, or corners). The meaning of the final number (1) is more difficult to explain (but see below). Continuing with our example, a tetrahedron has one 3-dimensional element (itself), four 2 dimensional elements (faces), six 1-dimensional elements (edges), and four 0-dimen-sional elements (vertices). Adding the final 1 again, these values correspond to the 4th row of the triangle (1, 4, 6, 4, 1). Line 1 corresponds to a point, and Line 2 corresponds to a line segment (dyad). This pattern continues to arbitrarily high-dimensioned hypertetrahedrons (known as simplices). To understand why this pattern exists, one must first understand that the process of building an n-simplex from an (n 1)-simplex consists of simply adding a new vertex to the latter, positioned such that this new vertex lies outside of the space of the original simplex, and connecting it to all original vertices. As an example, consider the case of building a tetrahedron from a triangle, the latter of whose elements are enumerated by row 3 of Pascal's triangle: 1 face, 3 edges, and 3 vertices (the meaning of the final 1 will be explained shortly). To build a tetrahedron from a triangle, we position a new vertex above the plane of the triangle and connect this vertex to all three vertices of the original triangle.

175 The number of a given dimensional element in the tetrahedron is now the sum of two numbers: first the number of that element found in the original triangle, plus the number of new elements, each of which is built upon elements of one fewer dimension from the original triangle. Thus, in the tetrahedron, the number of cells (polyhedral elements) is 0 (the original triangle possesses none) + 1 (built upon the single face of the original triangle) = 1; the number of faces is 1 (the original triangle itself) + 3 (the new faces, each built upon an edge of the original triangle) = 4; the number of edges is 3 (from the original triangle) + 3 (the new edges, each built upon a vertex of the original triangle) = 6; the number of new vertices is 3 (from the original triangle) + 1 (the new vertex that was added to create the tetrahedron from the triangle) = 4. This process of summing the number of elements of a given dimension to those of one fewer dimension to arrive at the number of the former found in the next higher simplex is equivalent to the process of summing two adjacent numbers in a row of Pascal's triangle to yield the number below. Thus, the meaning of the final number (1) in a row of Pascal's triangle becomes understood as representing the new vertex that is to be added to the simplex represented by that row to yield the next higher simplex represented by the next row. This new vertex is joined to every element in the original simplex to yield a new element of one higher dimension in the new simplex, and this is the origin of the pattern found to be identical to that seen in Pascal's triangle. A similar pattern is observed relating to squares, as opposed to triangles. To find the pattern, one must construct an analog to Pascal's triangle, whose entries are the coefficients of (x + 2) Row Number, instead of (x + 1) Row Number. There are a couple ways to do this. The simpler is to begin with Row 0 = 1 and Row 1 = 1, 2. Proceed to construct the analog triangles according to the following rule: That is, choose a pair of numbers according to the rules of Pascal's triangle, but double the one on the left before adding. This results in: The other way of manufacturing this triangle is to start with Pascal's triangle and multiply each entry by 2 k, where k is the position in the row of the given number. For example, the 2nd value in row 4 of Pascal's triangle is 6 (the slope of 1s corresponds to the zeroth entry in each row). To get the value that resides in the corresponding position in the analog triangle, multiply 6 by 2 Position Number = = 6 4 = 24. Now that the analog triangle has been constructed, the number of elements of any dimension that compose an

176 arbitrarily dimensioned cube (called a hypercube) can be read from the table in a way analogous to Pascal's triangle. For example, the number of 2-dimensional elements in a 2- dimensional cube (a square) is one, the number of 1-dimensional elements (sides, or lines) is 4, and the number of 0-dimensional elements (points, or vertices) is 4. This matches the 2nd row of the table (1, 4, 4). A cube has 1 cube, 6 faces, 12 edges, and 8 vertices, which corresponds to the next line of the analog triangle (1, 6, 12, 8). This pattern continues indefinitely. To understand why this pattern exists, first recognize that the construction of an n-cube from an (n 1)-cube is done by simply duplicating the original figure and displacing it some distance (for a regular n-cube, the edge length) orthogonal to the space of the original figure, then connecting each vertex of the new figure to its corresponding vertex of the original. This initial duplication process is the reason why, to enumerate the dimensional elements of an n-cube, one must double the first of a pair of numbers in a row of this analog of Pascal's triangle before summing to yield the number below. The initial doubling thus yields the number of "original" elements to be found in the next higher n- cube and, as before, new elements are built upon those of one fewer dimension (edges upon vertices, faces upon edges, etc.). Again, the last number of a row represents the number of new vertices to be added to generate the next higher n-cube. Fourier transform of sin(x) n+1 /x In this triangle, the sum of the elements of row m is equal to 3 m 1. Again, to use the elements of row 5 as an example: = 81, which is equal to 3 4 = 81. As stated previously, the coefficients of (x + 1) n are the nth row of the triangle. Now the coefficients of (x 1) n are the same, except that the sign alternates from +1 to 1 and back again. After suitable normalization, the same pattern of numbers occurs in the Fourier transform of sin(x) n+1 /x. More precisely: if n is even, take the real part of the transform, and if n is odd, take imaginary part. Then the result is a step function, whose values (suitably normalized) are given by the nth row of the triangle with alternating signs. For example, the values of the step function that results from compose the 4th row of the triangle, with alternating signs. This is a generalization of the following basic result (often used in electrical engineering): is the boxcar function. The corresponding row of the triangle is row 0, which consists of just the number 1.

177 If n is congruent to 2 or to 3 mod 4, then the signs start with 1. In fact, the sequence of the (normalized) first terms corresponds to the powers of i, which cycle around the intersection of the axes with the unit circle in the complex plane: Elementary cellular automaton The pattern produced by an elementary cellular automaton using rule 60 is exactly Pascal's triangle of binomial coefficients reduced modulo 2 (black cells correspond to odd binomial coefficients). Rule 102 also produces this pattern when trailing zeros are omitted. Rule 90 produces the same pattern but with an empty cell separating each entry in the rows. Extensions Pascal's Triangle can be extended to negative row numbers. First write the triangle in the following form: m = 0 m = 1 m = 2 m = 3 m = 4 m = 5... n = n = n = n = n = Next, extend the column of 1s upwards: m = 0 m = 1 m = 2 m = 3 m = 4 m = 5... n = n = n = n = n = n = n = n = n = Now the rule:

178 can be rearranged to: which allows calculation of the other entries for negative rows: m = 0 m = 1 m = 2 m = 3 m = 4 m = 5... n = n = n = n = n = n = n = n = n = This extension preserves the property that the values in the mth column viewed as a function of n are fit by an order m polynomial, namely This extension also preserves the property that the values in the nth row correspond to the coefficients of (1 + x) n :. For example: Another option for extending Pascal's triangle to negative rows comes from extending the other line of 1s:

179 m = -4 m = -3 m = -2 m = -1 m = 0 m = 1 m = 2 m = 3 m = 4 m = 5... n = n = n = n = n = n = n = n = n = Applying the same rule as before leads to m = -4 m = -3 m = -2 m = -1 m = 0 m = 1 m = 2 m = 3 m = 4 m = 5... n = n = n = n = n = n = n = n = n = Note that this extension also has the properties that just as we have,

180 Also, just as summing along the lower-left to upper-right diagonals of the Pascal matrix yields the Fibonacci numbers, this second type of extension still sums to the Fibonacci numbers for negative index. Either of these extensions can be reached if we define and take certain limits of the Gamma function, Γ(z).