Survey of Geometry. Paul Yiu. Department of Mathematics Florida Atlantic University. Spring 2007

Transcription

1 Survey of Geometry Paul Yiu Department of Mathematics Florida tlantic University Spring 2007

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3 ontents 1 The circumcircle and the incircle The law of cosines and its applications The circumcircle and the law of sines The incircle and excircles Heron s formula for the area of a triangle Exercise The shoemaker s knife The shoemaker s knife ircles in the shoemaker s knife rchimedean circles in the shoemaker s knife Introduction to Triangle Geometry Preliminaries oordinatization of points on a line enters of similitude of two circles Tangent circles Harmonic division Homothety The power of a point with respect to a circle Menelaus and eva theorems Menelaus and eva Theorems Desargues Theorem The incircle and the Gergonne point The excircles and the Nagel point The nine-point circle The Euler triangle as a midway triangle The orthic triangle as a pedal triangle

4 iv ONTENTS The nine-point circle The OI-line The homothetic center of the intouch and excentral triangles The centers of similitude of the circumcircle and the incircle Reflection of I in O Orthocenter of intouch triangle entroids of the excentral and intouch triangles Mixtilinear incircles Euler s formula and Steiner s porism Euler s formula Steiner s porism Homogeneous arycentric oordinates arycentric coordinates evian and traces Homogeneous barycentric coordinates of classical triangle centers The centroid The incenter Gergonne point The Nagel point The orthocenter rea and barycentric coordinates arycentric coordinates of the circumcenter The inferior and superior triangles The nine-point center Isotomic conjugates ongruent-parallelians point Reflections and isogonal conjuates Reflection triangle and isogonal conjugate Simson line theorem The circumcenter and the orthocenter Gergonne and Nagel points Reflections of the Euler line Isogonal conjugate of an infinite point

5 hapter 1 The circumcircle and the incircle 1.1 The law of cosines and its applications Given a triangle, we denote by a, b, c the lengths of the sides,, respectively. Theorem 1.1 (The law of cosines). c 2 = a 2 + b 2 2ab cos. Theorem 1.2 (Stewart). If X is a point on the side (or its extension) such that X : X = λ : µ, then X 2 = λb2 + µc 2 λ + µ λµa2 (λ + µ). 2 Proof. Use the cosine formula to compute the cosines of the angles X and X, and note that cos = cos X. orollary 1.3 (pollonius theorem). The length m a of the median D is given by m 2 a = 1 4 (2b2 +2c 2 a 2 ). orollary 1.4. The length w a of the (internal) bisector of angle is given by ( ( ) ) 2 a wa 2 = bc 1. b + c Proof. pply Stewart s Theorem with λ = c and µ = b. Remark. If we write s = 1 2 (a + b + c), then w2 a = 4bcs(s a) (b+c) 2.

6 2 The circumcircle and the incircle Exercise Show that the (4,5,6) triangle has one angle equal to twice of another. 2. If =2, show that c 2 =(a + b)b. 3. Find a simple relation between the sum of the areas of the three squares S 1, S 2, S 3, and that of the squares T 1, T 2, T 3. T 3 S 2 S 1 T 2 S 3 T 1 4. is a triangle with a =12, b + c =18, and cos = 7. alculate the 38 lengths of b and c, 1 and show that a 3 = b 3 + c The lengths of the sides of a triangle are 136, 170, and 174. alculate the lengths of its medians (a) m b = m c if and only if b = c. (b) m 2 a + m2 b + m2 c = 3 4 (a2 + b 2 + c 2 ). (c) If m a : m b : m c = a : b : c, show that the triangle is equilateral. 7. Suppose m b : m c = c : b. Show that either (i) b = c, or 1 MM E688, P.. Pizá. Here, b =9 5, and c = nswers: 158, 131, 127.

7 1.2 The circumcircle and the law of sines 3 (ii) the quadrilateral EGF is cyclic. Show that the triangle is equilateral if both (i) and (ii) hold. [Hint: Show that b 2 m 2 b c2 m 2 c = 1(c b)(c + 4 b)(b2 + c 2 2a 2 )]. 8. The lengths of the sides of a triangle are 84, 125, 169. alculate the lengths of its internal bisectors The circumcircle and the law of sines The perpendicular bisectors of the three sides of a triangle are concurrent at the circumcenter of the triangle. This is the center of the circumcircle, the circle passing through the three vertices of the triangle. F O E O D D Theorem 1.5 (The law of sines). Let R denote the circumradius of a triangle with sides a, b, c opposite to the angles,, respectively. a sin = b sin = c sin =2R. Since the area of a triangle is given by = 1 bc sin, the circumradius can 2 be written as R = abc 4. 3 nswers: 975 7, ,

8 4 The circumcircle and the incircle Exercise Where is the circumcenter of a right triangle? What is its circumradius? 4 2. (The orthocenter) Given triangle, construct lines through parallel to, through parallel to, and through parallel to. These three lines bound a triangle (so that lies on, on, and on respectively). Show that the altitudes of triangle are the perpendicular bisectors of the sides of. Deduce that the three altitudes of a triangle always intersect at a point (called its orthocenter). 3. Let H be the orthocenter of triangle. Show that (i) is the orthocenter of triangle H; (ii) the triangles H, H, H and have the same circumradius. 4. The internal bisectors of angles and intersect the circumcircle of at and. (i) Show that if =, then =. (ii) If =, does it follow that =? The incircle and excircles The internal angle bisectors of a triangle are concurrent at the incenter of the triangle. This is the center of the incircle, the circle tangent to the three sides of the triangle. If the incircle touches the sides, and respectively at X, Y, and Z, Y = Z = s a, X = Z = s b, X = Y = s c. Denote by r the inradius of the triangle. r = 2 a + b + c = s. 4 The midpoint of the hypotenuse; half of the length of the hypotenuse. 5 (ii) No. = if and only if = or = π 3.

9 1.3 The incircle and excircles 5 Y Z I X The internal bisector of each angle and the external bisectors of the remaining two angles are concurrent at an excenter of the triangle. n excircle can be constructed with this as center, tangent to the lines containing the three sides of the triangle. I b I c X Y Z I a The exradii of a triangle with sides a, b, c are given by r a = s a, r b = s b, r c = s c. The areas of the triangles I a, I a, and I a are 1ar 2 a, 1br 2 a, and 1cr 2 a respectively. Since we have from which r a = s a. = I a + I a + I a, = 1 2 r a( a + b + c) =r a (s a),

10 6 The circumcircle and the incircle Exercise Show that the inradius of a right triangle (of hypotenuse c) iss c. 2. square of side a is partitioned into 4 congruent right triangles and a small square, all with equal inradii r. alculate r. 3. The incenter of a right triangle is equidistant from the midpoint of the hypotenuse and the vertex of the right angle. Show that the triangle contains a 30 angle. I 4. The incircle of triangle touches the sides and at Y and Z respectively. Suppose Y = Z. Show that the triangle is isosceles. 5. line parallel to hypotenuse of a right triangle passes through the incenter I. The segments included between I and the sides and have lengths 3 and 4. alculate the area of the triangle. I

11 1.3 The incircle and excircles 7 6. If the incenter is equidistant from the three excenters, show that the triangle is equilateral. 7. The circle I intersects the sides, at E and F respectively. Show that EF is tangent to the incircle of. 6 Z I Y F X E 8. (a) The triangle is isosceles and the three small circles have equal radii. Suppose the large circle has radius R. Find the radius of the small circles. 7 (b) The large circle has radius R. The four small circles have equal radii. alculate this common radius. 8 6 Hint: Show that IF bisects angle F E. 7 Let θ be the semi-vertical angle of the isosceles triangle. The inradius of the triangle is = 2R sin θ(1 sin θ). If this is equal to R 2 (1 sin θ), then sin θ = 1 4. From 2R sin θ cos 2 θ 1+sin θ thi,s the inradius is 3 8 R. 8 Let θ be the smaller acute angle of one of the right triangles. The inradius of the right triangle 2R sin θ cos θ is 1+sin θ+cos θ. If this is equal to R 5 2 (1 sin θ), then 5sinθ cos θ =1. From this, sin θ = 13, and the radius is 4 13 R.

12 8 The circumcircle and the incircle 9. n equilateral triangle of side 2a is partitioned symmetrically into a quadrilateral, an isosceles triangle, and two other congruent triangles. If the inradii of the quadrilateral and the isosceles triangle are equal, find this inradius Heron s formula for the area of a triangle onsider a triangle with area. Denote by r the inradius, and r a the radius of the excircle on the side of triangle. It is convenient to introduce the semiperimeter s = 1 (a + b + c). 2 I a I r a r Y Y = rs. 9 ( 3 2)a.

13 1.4 Heron s formula for the area of a triangle 9 From the similarity of triangles IZ and I Z, r = s a. r a s From the similarity of triangles IY and I Y, r r a =(s b)(s c). From these, This latter is the famous Heron formula Exercise 1.4 r = (s a)(s b)(s c), s = s(s a)(s b)(s c). 1. The altitudes a triangle are 12, 15 and 20. What is the area of the triangle? Find the inradius and the exradii of the (13,14,15) triangle. 3. If one of the ex-radii of a triangle is equal to its semiperimeter, then the triangle contains a right angle r a + 1 r b + 1 r c = 1 r. 5. r a r b r c = rs Show that (i) r a + r b + r c = s3 +(ab+bc+ca)s ; (ii) (s a)(s b)(s c) = s 3 +(ab + bc + ca)s abc. Deduce that r a + r b + r c =4R + r. 10 = 150. The lengths of the sides are 25, 20 and 15.

14 10 The circumcircle and the incircle 7. The length of each side of the square is 6a, and the radius of each of the top and bottom circles is a. alculate the radii of the other two circles (a) D is a square of unit side. P is a point on so that the incircle of triangle P and the circle tangent to the lines P, P and D have equal radii. Show that the length of P satisfies the equation 2x 3 2x 2 +2x 1=0. D D P y Q x (b) D is a square of unit side. Q is a point on so that the incircle of triangle Q and the circle tangent to Q, Q, D touch each other at a point on Q. Show that the radii x and y of the circles satisfy the equations y = x(3 6x +2x2 ), 1 2x 2 Deduce that x is the root of x + y =1. 4x 3 12x 2 +8x 1=0. 11 a and 3 4 a.

15 1.4 Heron s formula for the area of a triangle 11 ppendix: Synthetic proofs of Steiner - Lehmus Theorem First proof. 12 Suppose β<γin triangle. We show that the bisector M is longer than the bisector N. hoose a point L on M such that NL = 1 β. Then, N, L, are 2 concyclic since NL = NL. Note that N = β< 1 (β + γ) = L, 2 and both are acute angles. Since smaller chords of a circle subtend smaller acute angles, we have N < L. It follows that N < M. G N L M N M Second proof. 13 Suppose the bisectors M and N in triangle are equal. We shall show that β = γ. If not, assume β < γ. ompare the triangles M and N. These have two pairs of equal sides with included angles M = 1 2 β< 1 2 γ = N, both of which are acute. Their opposite sides therefore satisfy the relation M < N. omplete the parallelogram MGN, and consider the triangle NG. This is isosceles since N = M = NG. Note that GN = 1 2 β + GM, 12 Gilbert - McDonnell, merican Mathematical Monthly, vol. 70 (1963) M. Descube, 1880.

16 12 The circumcircle and the incircle GN = 1 2 γ + GM. Since β<γ, we conclude that GM > GM. From this, M > GM = N. This contradicts the relation M < N obtained above.

17 hapter 2 The shoemaker s knife 2.1 The shoemaker s knife Let P be a point on a segment. The region bounded by the three semicircles (on the same side of ) with diameters, P and Pis called a shoemaker s knife. Suppose the smaller semicircles have radii a and b respectively. Let Q be the intersection of the largest semicircle with the perpendicular through P to. This perpendicular is an internal common tangent of the smaller semicircles. Q H U R V K O 1 O P O 2 O 1 O P O 2 Exercise 1. Show that the area of the shoemaker s knife is πab. 2. Let UV be the external common tangent of the smaller semicircles, and R the intersection of PQand UV. Show that (i) UV = PQ; (ii) UR = PR = VR = QR. Hence, with R as center, a circle can be drawn passing through P, Q, U, V.

18 14 The shoemaker s knife 3. Show that the circle through U, P, Q, V has the same area as the shoemaker s knife. 2.2 ircles in the shoemaker s knife Theorem 2.1 (rchimedes). The two circles each tangent to P, the largest semicircle and one of the smaller semicircles have equal radii t, given by t = ab a + b. Q O 1 O P O 2 O 1 O P O 2 Proof. onsider the circle tangent to the semicircles O(a+b), O 1 (a), and the line PQ. Denote by t the radius of this circle. alculating in two ways the height of the center of this circle above the line, wehave (a + b t) 2 (a b t) 2 =(a + t) 2 (a t) 2. From this, t = ab a + b. The symmetry of this expression in a and b means that the circle tangent to O(a + b), O 2 (b), and PQhas the same radius t. Theorem 2.2 (rchimedes). The circle tangent to each of the three semicircles has radius given by ab(a + b) ρ = a 2 + ab + b. 2

19 2.2 ircles in the shoemaker s knife 15 X Y O 1 O P O 2 Proof. Let OO 2 = θ. y the cosine formula, we have Eliminating θ, we have (a + ρ) 2 = (a + b ρ) 2 + b 2 +2b(a + b ρ)cosθ, (b + ρ) 2 = (a + b ρ) 2 + a 2 2a(a + b ρ)cosθ. a(a + ρ) 2 + b(b + ρ) 2 =(a + b)(a + b ρ) 2 + ab 2 + ba 2. The coefficients of ρ 2 on both sides are clearly the same. This is a linear equation in ρ: a 3 + b 3 +2(a 2 + b 2 )ρ =(a + b) 3 + ab(a + b) 2(a + b) 2 ρ, from which 4(a 2 + ab + b 2 )ρ =(a + b) 3 + ab(a + b) (a 3 + b 3 )=4ab(a + b), and ρ is as above. Theorem 2.3 (Leon ankoff). If the incircle (ρ) of the shoemaker s knife touches the smaller semicircles at X and Y, then the circle through the points P, X, Y has the same radius as the rchimedean circles. Z X Y O 1 O P O 2

20 16 The shoemaker s knife Proof. The circle through P, X, Y is clearly the incircle of the triangle O 1 O 2, since X = Y = ρ, O 1 X = O 1 P = a, O 2 Y = O 2 P = b. The semiperimeter of the triangle O 1 O 2 is a + b + ρ =(a + b)+ ab(a + b) (a + b)3 = a 2 + ab + b2 a 2 + ab + b. 2 The inradius of the triangle is given by abρ a + b + ρ = ab ab(a + b) = ab (a + b) 3 a + b. This is the same as t, the common radius of rchimedes twin circles. onstruction of incircle of shoemaker s knife Let Q 1 and Q 2 be the highest points of the semicircles O 1 (a) and O 2 (b) respectively. The intersection of O 1 Q 2 and O 2 Q 1 is a point 3 above P, and 3 P = ab = t. This gives a very easy construction of ankoff s circle in Theorem 2.3 above. From this, we obtain the points X and Y. The center of the a+b incircle of the shoemaker s knife is the intersection of the lines O 1 X and O 2 Y. The incircle of the shoemaker s knife is the circle (X). It touches the largest semicircle of the shoemaker at Z, the intersection of O with this semicircle. Z Q 1 X Q 2 3 Y O 1 O P O 2 Note that 3 (P ) is the ankoff circle, which has the same radius as the rchimedean circles.

21 2.3 rchimedean circles in the shoemaker s knife 17 Exercise 1. Show that the area of triangle O 1 O 2 is ab(a+b)2 a 2 +ab+b Show that the center of the incircle of the shoemaker s knife is at a distance 2ρ from the line. 3. Show that the area of the shoemaker s knife to that of the heart (bounded by semicircles O 1 (a), O 2 (b) and the lower semicircle O(a + b))isasρ to a + b. Z Q 1 Q 2 O 1 O P O 2 X Y O 1 O P O 2 4. Show that the points of contact of the incircle (ρ) with the semicircles can be located as follows: Y, Z are the intersections with Q 1 (), and X, Z are the intersections with Q 2 (). 2.3 rchimedean circles in the shoemaker s knife Let UV be the external common tangent of the semicircles O 1 (a) and O 2 (b), which extends to a chord HK of the semicircle O(a+b). Let 4 be the intersection of O 1 V and O 2 U. Since O 1 U = a, O 2 V = b, and O 1 P : PO 2 = a : b, 4 P = ab a+b = t. This means that the circle 4(t) passes through P and touches the common tangent HK of the semicircles at N. Let M be the midpoint of the chord HK. Since O and P are symmetric (isotomic conjugates) with respect to O 1 O 2, OM + PN = O 1 U + O 2 V = a + b.

22 18 The shoemaker s knife H U 5 M N V K 4 O 1 O P O 2 it follows that (a + b) QM = PN =2t. From this, the circle tangent to HK and the minor arc HK of O(a + b) has radius t. This circle touches the minor arc at the point Q. Theorem 2.4 (Thomas Schoch). The incircle of the curvilinear triangle bounded by the semicircle O(a + b) and the circles (2a) and (2b) has radius t = ab a+b. S O 1 O P O 2 Proof. Denote this circle by S(x). SO 1 O 2. y pollonius theorem, Note that SO is a median of the triangle (2a + x) 2 +(2b + x) 2 =2[(a + b) 2 +(a + b x) 2 ]. From this, x = ab a + b = t. Exercise 1. The circles ( 1 ) and ( 1) are each tangent to the outer semicircle of the shoemaker s knife, and to OQ 1 at Q 1 ; similarly for the circles ( 2 ) and ab ). Show that they have equal radii t = ( 2 a+b.

23 2.3 rchimedean circles in the shoemaker s knife Q 1 1 Q 2 2 O 1 O P O 2 2. We call the semicircle with diameter O 1 O 2 the midway semicircle of the shoemaker s knife. Show that the circle tangent to the line PQ and with center at the intersection of (O 1 ) and the midway semicircle has radius t = ab. a+b Q O 1 O P O 2 3. Show that the radius of the circle tangent to the midway semicircle, the outer semicircle, and with center on the line PQhas radius t = ab a+b. Q O 1 O P O 2

24 20 The shoemaker s knife

25 hapter 3 Introduction to Triangle Geometry 3.1 Preliminaries oordinatization of points on a line Let and be two fixed points on a line L. Every point X on L can be coordinatized in one of several ways: (1) the ratio of division t = X, X (2) the absolute barycentric coordinates: an expression of X as a convex combination of and : X =(1 t) + t, which expresses for an arbitrary point P outside the line L, the vector PX as a combination of the vectors P and P. (3) the homogeneous barycentric coordinates: the proportion X : X, which are masses at and so that the resulting system (of two particles) has balance point at X. P X

26 22 Introduction to Triangle Geometry enters of similitude of two circles onsider two circles O(R) and I(r), whose centers O and I are at a distance d apart. nimate a point X on O(R) and construct a ray through I oppositely parallel to the ray OX to intersect the circle I(r) at a point Y. You will find that the line XY always intersects the line OI at the same point T. This we call the internal center of similitude, or simply the insimilicenter, of the two circles. It divides the segment OI in the ratio OT : TI = R : r. The absolute barycentric coordinates of P with respect to OI are T = R I + r O. R + r X O Y T T I Y If, on the other hand, we construct a ray through I directly parallel to the ray OX to intersect the circle I(r) at Y, the line XY always intersects OI at another point T. This is the external center of similitude, or simply the exsimilicenter, of the two circles. It divides the segment OI in the ratio OT : T I = R : r, and has absolute barycentric coordinates T = R I r O. R r Tangent circles If two circles are tangent to each other, the line joining their centers passes through the point of tangency, which is a center of similitude of the circles.

27 3.1 Preliminaries 23 O T I O I T Harmonic division Two points X and Y are said to divide two other points and harmonically if X X = Y Y. They are harmonic conjugates of each other with respect to the segment. Examples 1. For two given circles, the two centers of similitude divide the centers harmonically. 2. Given triangle, let the internal bisector of angle intersect at X. The harmonic conjugate of X in is the intersection of with the external bisector of angle. 3. Let and be distinct points. If M is the midpoint of the segment, it is not possible to find a finite point N on the line so that M, N divide, harmonically. This is because N = M = 1, requiring N M N = N = N, and = N N =0, a contradiction. We shall agree to say that if M and N divide, harmonically, then N is the infinite point of the line. Exercise 1. If X, Y divide, harmonically, then, divide X, Y harmonically.

28 24 Introduction to Triangle Geometry X X 2. Given a point X on the line, construct its harmonic associate with respect to the segment. Distinguish between two cases when X divides internally and externally The centers and of two circles (a) and (b) are at a distance d apart. The line intersect the circles at and respectively, so that, are between,. 4. Given two fixed points and and a positive constant k 1, the locus of the points P for which P : P = k is a circle. 1 Make use of the notion of centers of similitude of two circles.

29 3.1 Preliminaries Homothety Given a point T and a nonzero constant k, the similarity transformation h(t,k) which carries a point X to the point X on the line TX satisfying TX : TX = k :1is called the homothety with center T and ratio k. Explicitly, h(t,k)(p )=(1 k)t + kp. ny two circles are homothetic. Let P and Q be the internal and external centers of similitude of two circles O(R) and I(r). oth the homotheties h(q, r ) R and h(p, r ) transform the circle O(R) into I(r). R The power of a point with respect to a circle The power of a point P with respect to a circle = O(R) is the quantity (P ):=OP 2 R 2. This is positive, zero, or negative according as P is outside, on, or inside the circle. If it is positive, it is the square of the length of a tangent from P to the circle. Theorem (Intersecting chords) If a line L through P intersects a circle at two points X and Y, the product PX PY (of signed lengths) is equal to the power of P with respect to the circle. T O X Y P T

30 26 Introduction to Triangle Geometry 3.2 Menelaus and eva theorems Menelaus and eva Theorems onsider a triangle with points X, Y, Z on the side lines,, respectively. Theorem 3.1 (Menelaus). The points X, Y, Z are collinear if and only if X X Y Y Z Z = 1. Y Z X Theorem 3.2 (eva). The lines X, Y, Z are concurrent if and only if X X Y Y Z Z = Desargues Theorem s a simple illustration of the use of the Menelaus and eva theorems, we prove the following rmdesargues Theorem: Given three circles, the exsimilicenters of the three pairs of circles are collinear. Likewise, the three lines each joining the insimilicenter of a pair of circles to the center of the remaining circle are concurrent. We prove the second statement only. Given three circles (r 1 ), (r 2 ) and (r 3 ), the insimilicenters X of () and (), Y of (), (), and Z of (), () are the points which divide,, in the ratios X X = r 2 r 3, Y Y = r 3 r 1, Z Z = r 1 r 2.

31 3.2 Menelaus and eva theorems 27 Z P Y X X P Y Y Z X Z It is clear that the product of these three ratio is +1, and it follows from the eva theorem that X, Y, Z are concurrent.

32 28 Introduction to Triangle Geometry The incircle and the Gergonne point The incircle is tangent to each of the three sides,, (without extension). Its center, the incenter I, is the intersection of the bisectors of the three angles. The inradius r is related to the area by S =(a + b + c)r. Y Z G e I X If the incircle is tangent to the sides at X, at Y, and at Z, then Y = Z = b + c a 2, Z = X = c + a b 2, X = Y = a + b c. 2 These expressions are usually simplified by introducing the semiperimeter s = (a + b + c): 1 2 Y = Z = s a, Z = X = s b, X = Y = s c. lso, r = s. It follows easily from the eva theorem that X, Y, Z are concurrent. The point of concurrency G e is called the Gergonne point of triangle. Triangle XY Z is called the intouch triangle of. learly, X = + 2, Y = + 2, Z = +. 2

33 3.2 Menelaus and eva theorems 29 It is always acute angled, and YZ =2r cos 2, ZX =2r cos 2, XY =2r cos 2. Exercise 1. Given three points,, not on the same line, construct three circles, with centers at,,, mutually tangent to each other externally. X Y Z 2. onstruct the three circles each passing through the Gergonne point and tangent to two sides of triangle. The 6 points of tangency lie on a circle Two circles are orthogonal to each other if their tangents at an intersection are perpendicular to each other. Given three points,, not on a line, construct three circles with these as centers and orthogonal to each other. (1) onstruct the tangents from to the circle (b), and the circle tangent to these two lines and to (a) internally. (2) onstruct the tangents from to the circle (a), and the circle tangent to these two lines and to (b) internally. (3) The two circles in (1) and (2) are congruent. r. 2 This is called the dams circle. It is concentric with the incircle, and has radius (4R+r) 2 +s 2 4R+r

34 30 Introduction to Triangle Geometry G e I 4. Given a point Z on a line segment, construct a right-angled triangle whose incircle touches the hypotenuse at Z Let be a triangle with incenter I. (1a) onstruct a tangent to the incircle at the point diametrically opposite to its point of contact with the side. Let this tangent intersect at Y 1 and at Z 1. (1b) Same in part (a), for the side, and let the tangent intersect at Z 2 and at X 2. (1c) Same in part (a), for the side, and let the tangent intersect at X 3 and at Y 3. (2) Note that Y 3 = Z 2. onstruct the circle tangent to and at Y 3 and Z 2. How does this circle intersect the circumcircle of triangle? 6. The incircle of touches the sides,, at D, E, F respectively. X is a point inside such that the incircle of X touches at D also, and touches X and X at Y and Z respectively. 3 P. Yiu, G. Leversha, and T. Seimiya, Problem 2415 and solution, rux Math. 25 (1999) 110; 26 (2000)

35 3.2 Menelaus and eva theorems 31 (1) The four points E, F, Z, Y are concyclic. 4 (2) What is the locus of the center of the circle EFZY? 5 7. Given triangle, construct a circle tangent to at Y and at Z such that the line YZpasses through the centroid G. Show that YG: GZ = c : b. G Y Z The excircles and the Nagel point Let X, Y, Z be the points of tangency of the excircles (I a ), (I b ), (I c ) with the corresponding sides of triangle. The lines X, Y, Z are concurrent. The common point N a is called the Nagel point of triangle. Exercise 1. onstruct the tritangent circles of a triangle. (1) Join each excenter to the midpoint of the corresponding side of. These three lines intersect at a point M i. (This is called the Mittenpunkt of the triangle). 4 International Mathematical Olympiad IMO 1996.

36 32 Introduction to Triangle Geometry I b I c Z Z I Y N a Y X X I a (2) Join each excenter to the point of tangency of the incircle with the corresponding side. These three lines are concurrent at another point T. (3) The lines M i and T are symmetric with respect to the bisector of angle ; so are the lines M i, T and M i, T (with respect to the bisectors of angles and ). 2. onstruct the excircles of a triangle. (1) Let D, E, F be the midpoints of the sides,,. onstruct the incenter S p of triangle DEF, 6 and the tangents from S to each of the three excircles. (2) The 6 points of tangency are on a circle, which is orthogonal to each of the excircles. 6 This is called the Spieker point of triangle.

37 3.3 The nine-point circle 33 I b I c Z Y Z T M i I Y X X I a 3. Let D, E, F be the midpoints of the sides,,, and let the incircle touch these sides at X, Y, Z respectively. The lines through X parallel to ID, through Y to IE and through Z to IF are concurrent The nine-point circle The Euler triangle as a midway triangle Let P be a point in the plane of triangle. The midpoints of the segments P, P, P form the midway triangle of P. It is the image of under the homothety h(p, 1 ). The midway triangle of the orthocenter H is called the Euler 2 7 rux Math. Problem The reflection of the Nagel point N a in the incenter. This is X 145 of ET.

38 34 Introduction to Triangle Geometry I b I c Z S p Y X I a triangle. The circumcenter of the midway triangle of P is the midpoint of OP. In particular, the circumcenter of the Euler triangle is the midpoint of OH, which is the same as N, the circumcenter of the medial triangle. (See Exercise??.). The medial triangle and the Euler triangle have the same circumcircle The orthic triangle as a pedal triangle The pedals of a point are the intersections of the sidelines with the corresponding perpendiculars through P. They form the pedal triangle of P. The pedal triangle of the orthocenter H is called the orthic triangle of. The pedal X of the orthocenter H on the side is also the pedal of on the same line, and can be regarded as the reflection of in the line EF. It follows that EXF = EF = EDF,

39 3.3 The nine-point circle 35 F Y E Z P I X D O O P since EDF is a parallelogram. From this, the point X lies on the circumcircle of the medial triangle DEF; similarly for the pedals Y and Z of H on the other two sides and.

40 36 Introduction to Triangle Geometry Y Z P X The nine-point circle From 3.3.1, above, the medial triangle, the Euler triangle, and the orthic triangle all have the same circumcircle. This is called the nine-point circle of triangle. Its center N, the midpoint of OH, is called the nine-point center of triangle. Exercise 1. Show that (i) the incenter is the orthocenter of the excentral triangle (ii) the circumcircle is the nine-point circle of the excentral triangle, (iii) the circumcenter of the excentral triangle is the reflection of I in O. 2. Let P be a point on the circumcircle. What is the locus of the midpoint of HP? Why? 3. If the midpoints of P, P, P are all on the nine-point circle, must P be the orthocenter of triangle? 8 8 P. Yiu and J. Young, Problem 2437 and solution, rux Math. 25 (1999) 173; 26 (2000) 192.

41 3.3 The nine-point circle 37 Y F E N O Z H X D 4. Let be a triangle and P a point. The perpendiculars at P to P, P, P intersect,, respectively at,,. (1),, are collinear. 9 (2) The nine-point circles of the (right-angled) triangles P, P, P are concurrent at P and another point P. Equivalently, their centers are collinear (Triangles with nine-point center on the circumcircle) egin with a circle, center O and a point N on it, and construct a family of triangles with (O) as circumcircle and N as nine-point center. (1) onstruct the nine-point circle, which has center N, and passes through 9. Gibert, Hyacinthos 1158, 8/5/ P. Hatzipolakis, Hyacinthos 3166, 6/27/01. The three midpoints of,, are collinear. The three nine-point circles intersect at P and its reflection in this line.

42 38 Introduction to Triangle Geometry I b I c I 0 I I a P P

43 3.4 The OI-line 39 the midpoint M of ON. (2) nimate a point D on the minor arc of the nine-point circle inside the circumcircle. (3) onstruct the chord of the circumcircle with D as midpoint. (This is simply the perpendicular to OD at D). (4) Let X be the point on the nine-point circle antipodal to D. omplete the parallelogram ODX (by translating the vector DO to X). The point lies on the circumcircle and the triangle has nine-point center N on the circumcircle. Here is a curious property of triangles constructed in this way: let,, be the reflections of,, in their own opposite sides. The reflection triangle degenerates, i.e., the three points,, are collinear The OI-line The homothetic center of the intouch and excentral triangles The OI-line of a triangle is the line joining the circumcenter and the incenter. We consider several interesting triangle centers on this line, which arise from the homothety of the intouch and excentral triangles. These triangles are homothetic since their corresponding sides are parallel, being perpendicular to the same angle bisector of the reference triangle. The ratio of homothetic is clearly 2R. Their circumcenters are r I =2O I and I. The homothetic center is the point T such that 2O I = h ( T, 2R r ) (I) = ( 1 2R r ) T + 2R r I. This gives (2R + r)i 2r O T =. 2R r It is the point which divides OI externally in the ratio 2R + r : 2r O. ottema, Hoofdstukken uit de Elementaire Meetkunde, hapter The point T is X 57 in ET.

44 40 Introduction to Triangle Geometry I b I c Y Z T I X The centers of similitude of the circumcircle and the incircle These are the points T + and T which divide the segment OI harmonically in the ratio of the circumradius and the inradius. 13 I a T + = 1 (r O + R I), R + r T = 1 ( r O + R I). R r 13 T + and T are respectively X 55 and X 56 in ET.

45 3.4 The OI-line 41 M Y Z T I T + O X Reflection of I in O M The reflection I of I in O is the circumcenter of the excentral triangle. It is the intersections of the perpendiculars from the excenters to the sidelines. 14 The midpoint M of the arc is also the midpoint of II a. Since IM and I I a are parallel, I I a =2R. Similarly, I I b = P c =2R. This shows that the rmexcentral triangle has circumcenter I =2O I and circumradius 2R. Since I is the orthocenter of I a I b I c, its follows that O, the midpoint of I and I, is the nine-point center of the excentral triangle. In other words, the circumcircle of triangle is the nine-point circle of the excentral triangle. part from, the second intersection of I b I c with the circumcircle of is the midpoint M of I b I c. From this we deduce the following interesting formula: 15 r a + r b + r c =4R + r. 14 I appears as X 40 in ET. 15 Proof. r a +r b +r c =2MD+(I I a I X )=2R+2OD+I I a I X =4R+IX =4R+r.

46 42 Introduction to Triangle Geometry I b M I c I O I X c D X b M I a Orthocenter of intouch triangle The OI-line is the Euler line of the excentral triangle, since O and I are the nine-point center and orthocenter respectively. The corresponding sides of the rmintouch triangle and the rmexcentral triangle are parallel, being perpendicular to the respective angle bisectors. Their Euler lines are parallel. Since the intouch triangle has circumcenter I, its Euler line is actually the line OI, which therefore contains its orthocenter. This is the point which divides OI in the ratio R + r : r. 16 It also lies on the line joining the Gergonne and Nagel points. 16 This is the point X 65 in ET.

47 3.4 The OI-line 43 Y Z H G e I O Na X entroids of the excentral and intouch triangles The centroid of the excentral triangle is the point which divides OI in the ratio 1 :4. 17 r From the homothety h(t, ), it is easy to see that the centroid of the 2R intouch triangle is the point which divides OI in the ratio 3R + r : r. 18 Exercises 1. an any of the centers of similitude of (O) and (I) lie outside triangle? 17 This is the point which divides the segment I I in to the ratio 1:2.ItisX 165 of ET. 18 The centroid of the intouch triangle is X 354 of ET.

48 44 Introduction to Triangle Geometry I b I c Y Z T I O X 2. Show that the distance between the circumcenter and the Nagel point is R 2r Identify the midpoint of the Nagel point and the delongchamps point as a point on the OI-line onstruct the orthic triangle of the intouch triangle. This is homothetic to 19 Feuerbach s theorem. 20 Reflection of I in O. This is because the pedal triangle of X 40 is the cevian triangle of the Nagel point, and the reflections of the pedals of the Nagel point in the respective traces form the pedals of the de Longchamps point. I a

49 3.4 The OI-line 45. Identify the homothetic center. 21 Y Z H I O X 5. onstruct the external common tangent of each pair of excircles. These three external common tangents bound a triangle. (i) This triangle is perpsective with. Identify the perspector. 22 (ii) Identify the incenter of this triangle Extend and by length a and join the two points by a line. Similarly define the two other lines. The three lines bound a triangle with perspector X Let H be the orthocenter of the intouch triangle XY Z, and X, Y, Z its pedals on the sides YZ, ZX, XY respectively. Identify the common point of the three lines X, Y, Z as a point on the OI-line. Homothetic center of intouch and excentral triangles. 21 T. 22 Orthocenter of intouch triangle. 23 Reflection of I in O. 24 2/18/03.

50 46 Introduction to Triangle Geometry 8. Let P be the point which divides OI in the ratio OP : PI = R :2r. There is a circle, center P, radius Rr, which is tangent to three congruent circles R+2r of the same radius, each tangent to two sides of the triangle. onstruct these circles There are three circles each tangent internally to the circumcircle at a vertex, and externally to the incircle. It is known that the three lines joining the points of tangency of each circle with (O) and (I) pass through the internal center T + of similitude of (O) and (I). onstruct these three circles. 26 I T + O 10. Let T + be the insimilicenter of (O) and (I), with pedals Y and Z on and respectively. If Y and Z are the pedals of Y and Z on, calculate the length of Y Z Let P be the centroid of the excentral triangle, with pedals X, Y, Z on the 25. P. Hatzipolakis, Hyacinthos message 793, pril 18, P. Hatzipolakis and P. Yiu, Triads of circles, preprint. 27.P. Hatzipolakis and P. Yiu, Pedal triangles and their shadows, Forum Geom., 1 (2001)

51 3.4 The OI-line 47 Y Z I T + O Z X Y sides,, respectively. Show that 28 Y + Z = Z + X = X + Y = 1 (a + b + c) Mixtilinear incircles mixtilinear incircle of triangle is one that is tangent to two sides of the triangle and to the circumcircle internally. Denote by the point of tangency of the mixtilinear incircle K(ρ) in angle with the circumcircle. The center K clearly lies on the bisector of angle, and K : KI = ρ : (ρ r). In terms of barycentric coordinates, K = 1 ( (ρ r) + ρi). r 28 The projections of O and I on the side are the midpoint D of, and the point of tangency D of the incircle with this side. learly, D = a 2 and D = 1 2 (c + a b). It follows that X = D D D = 4 3 D 1 3 D = 1 (3a + b c). 6 Similarly, Z = 1 6 (3c + b a), and X + Z = 1 3 (a + b + c). similar calculation shows that Y + Z = X + Y = 1 3 (a + b + c).

52 48 Introduction to Triangle Geometry lso, since the circumcircle O( ) and the mixtilinear incircle K( ) touch each other at,wehaveok : K = R ρ : ρ, where R is the circumradius. From this, K = 1 R (ρo +(R ρ) ). K I O omparing these two equations, we obtain, by rearranging terms, RI ro R r = R(ρ r) + r(r ρ). ρ(r r) We note some interesting consequences of this formula. First of all, it gives the intersection of the lines joining and OI. Note that the point on the line OI represented by the left hand side is T, the external center of similitude of the circumcircle and the incircle. This leads to a simple construction of the mixtilinear incircle. Given a triangle, let P be the external center of similitude of the circumcircle (O) and incircle (I). Extend P to intersect the circumcircle at. The intersection of I and O is the center K of the mixtilinear incircle in angle. The other two mixtilinear incircles can be constructed similarly. 3.5 Euler s formula and Steiner s porism

53 3.5 Euler s formula and Steiner s porism 49 I O T K Euler s formula M The distance between the circumcenter and the incenter of a triangle is given by OI 2 = R 2 2Rr. Let the bisector of angle intersect the circumcircle at M. onstruct the circle M() to intersect this bisector at a point I. This is the incenter since I = 1 2 IM = 1 2 M = 1 2, and for the same reason I = 1. Note that 2 (1) IM = M = M =2Rsin, 2 (2) I = r sin 2, and (3) by the theorem of intersecting chords, OI 2 R 2 = the power of I with respect to the circumcircle = I IM = 2Rr Steiner s porism onstruct the circumcircle (O) and the incircle (I) of triangle. nimate a point on the circumcircle, and construct the tangents from to the incircle (I). Extend these tangents to intersect the circumcircle again at and. The lines is always tangent to the incircle. This is the famous theorem on Steiner

54 50 Introduction to Triangle Geometry I O M porism: if two given circles are the circumcircle and incircle of one triangle, then they are the circumcircle and incircle of a continuous family of poristic triangles. Exercises 1. r 1 R. When does equality hold? 2 2. Suppose OI = d. Show that there is a right-angled triangle whose sides are d, r and R r. Which one of these is the hypotenuse? 3. Given a point I inside a circle O(R), construct a circle I(r) so that O(R) and I(r) are the circumcircle and incircle of a (family of poristic) triangle(s). 4. Given the circumcenter, incenter, and one vertex of a triangle, construct the triangle. 5. onstruct an animation picture of a triangle whose circumcenter lies on the incircle Hint: OI = r.

55 3.5 Euler s formula and Steiner s porism 51 Y Z I O X 6. What is the locus of the centroids of the poristic triangles with the same circumcircle and incircle of triangle? How about the orthocenter? 7. Let be a poristic triangle with the same circumcircle and incircle of triangle, and let the sides of,, touch the incircle at X, Y, Z. (i) What is the locus of the centroid of XY Z? (ii) What is the locus of the orthocenter of XY Z? (iii) What can you say about the Euler line of the triangle XY Z?

56 52 Introduction to Triangle Geometry

57 hapter 4 Homogeneous arycentric oordinates 4.1 arycentric coordinates The notion of barycentric coordinates dates back to Möbius. In a given triangle, every point P is coordinatized by a triple of numbers (x : y : z) in such a way that the system of masses x at, y at, and z at will have its balance point at P. mass y at and a mass z at will balance at the point X = y+z y+z on the line. mass x at and a mass y + z at X will balance at the point P = x +(y + z)x x +(y + z) = x + y + z. x + y + z We say that with reference to triangle, the point P has x + y + z (i) absolute barycentric coordinate and x + y + z (ii) homogeneous barycentric coordinates (x : y : z). 4.2 evian and traces ecause of the fundamental importance of the eva theorem in triangle geometry, we shall follow traditions and call the three lines joining a point P to the vertices of the reference triangle the cevians of P. The intersections X, Y, Z of these cevians with the side lines are called the traces of P. The coordinates of the

58 54 Homogeneous arycentric oordinates traces can be very easily written down: X =(0:y : z), Y =(x :0:z), Z =(x : y :0). X P Z Y Theorem 4.1 (eva theorem). Three points X, Y, Z on,, respectively are the traces of a point if and only if they have coordinates of the form for some x, y, z. X = 0 : y : z, Y = x : 0 : z, Z = x : y : 0, 4.3 Homogeneous barycentric coordinates of classical triangle centers The centroid The midpoint points of the sides have coordinates X = (0 : 1 : 1), Y = (1 : 0 : 1), Z = (1 : 1 : 0). The centroid G has coordinates (1:1:1). 1 1 The centroid appears in Kimberling s Encyclopedia of Triangle centers (ET, available at as the point X 2.

59 4.3 Homogeneous barycentric coordinates of classical triangle centers 55 X G Z The incenter Y X I Z The traces of the incenter have coordinates Y X = (0 : b : c), Y = (a :0:c), Z = (a : b :0). The incenter I has coordinates I =(a : b : c). 2 2 The incenter appears in ET as the point X 1.

60 56 Homogeneous arycentric oordinates Gergonne point s a s a Z X G e I s c s b s b Y s c The points of tangency of the incircle with the side lines are These can be reorganized as X = 0 : s c : s b, Y = s c : 0 : s a, Z = s b : s a : X = 0 : : s b Y = 1 1 : 0 : s a Z = 1 1 : : 0. s a s b, s c, s c It follows that X, Y, Z intersect at a point with coordinates ( ) 1 s a : 1 s b : 1. s c This is called the Gergonne point G e of triangle. 3 3 The Gergonne point appears in ET as the point X 7.

61 4.3 Homogeneous barycentric coordinates of classical triangle centers The Nagel point The points of tangency of the excircles with the corresponding sides have coordinates X = (0 : s b : s c), Y = (s a :0:s c), Z = (s a : s b :0). These are the traces of the point with coordinates (s a : s b : s c). This is the Nagel point N a of triangle. 4 I c I a s b Z s c s a Na Y s a s c X s b I b 4 The Nagel point appears in ET as the point X 8.

62 58 Homogeneous arycentric oordinates The orthocenter The trace X of the orthocenter H divides in the ratio X : X = c cos : b cos. Z X H In homogeneous barycentric coordinates, Y X =(b cos : c cos ) = Similarly, the other two traces are ( a Y = cos :0: ( a Z = cos : The orthocenter is the point 5 H = ( a cos : ( 0: b cos : c ), cos ). b cos :0 b cos : ) c. cos ) c. cos y the law of cosines, these coordinates can be rewritten as ( ) 1 H = b 2 + c 2 a : 1 2 c 2 + a 2 b : 1. 2 a 2 + b 2 c 2 5 The orthocenter appears in ET as the point X 4.

63 4.4 rea and barycentric coordinates rea and barycentric coordinates Theorem 4.2. If in homogeneous barycentric coordinates with reference to triangle, P =(x : y : z), then P : P : P = x : y : z. Theorem 4.3. If for i =1, 2, 3, P i = x 1 +y 1 +z i (in absolute barycentric coordinates), then the area of the oriented triangle P 1 P 2 P 3 is x 1 y 1 z 1 P 1 P 2 P 3 = x 2 y 2 z 2 x 3 y 3 z 3. Theorem 4.4 (Routh theorem). If X, Y, Z are points on the lines,, respectively such that X : X = λ :1, Y : Y= µ :1, Z : Z = µ :1, then the cevian lines X, Y, Z bound a triangle with area (λµν 1) 2 (µν + µ +1)(νλ + ν +1)(λµ + λ +1). 1 Z ν P 1 Y R Q µ λ X 1 Proof. In homogeneous barycentric coordinates with reference to triangle, X =(0:1:λ), Y =(µ :0:1), Z =(1:ν :0). Those of P, Q, R can be worked out easily: P = Y Z Q = Z X R = X Y Y =(µ :0:1) Z =(1:ν :0) X =(0:1:λ) Z =(1:ν :0) X =(0:1:λ) Y =(µ :0:1) P =(µ : µν :1) Q =(1:ν : νλ) R =(λµ :1:λ)

64 60 Homogeneous arycentric oordinates This means that the absolute barycentric coordinates of P, Q, R are 1 P = (µ + µν + ), µν + µ +1 1 Q = ( + ν + νλ), νλ + ν +1 1 R = (λµ + + λ). λµ + λ +1 From these, µ µν 1 1 ν νλ λµ 1 λ rea(pqr)= (µν + µ +1)(νλ + ν +1)(λµ + λ +1) (λµν 1) 2 = (µν + µ +1)(νλ + ν +1)(λµ + λ +1). Example 4.1. Given a triangle, X, Y, Z are points on the side lines specified by the ratios of divisions X : X =2:1, Y : Y=5:3, Z : Z =3:2. The lines X, Y, Z bound a triangle PQR. Suppose triangle has area. Find the area of triangle PQR. Z 2 3 P 3 Y R 2 X 1 We make use of homogeneous barycentric coordinates with respect to. X =(0:1:2), Y =(5:0:3), Z =(2:3:0). Q 5

65 4.4 rea and barycentric coordinates 61 Those of P, Q, R can be worked out easily: P = Y Z Q = Z X R = X Y Y =(5:0:3) Z =(2:3:0) X =(0:1:2) Z =(2:3:0) X =(0:1:2) Y =(5:0:3) P = (10 : 15 : 6) Q =(2:3:6) R =(10:3:6) This means that the absolute barycentric coordinates of X, Y, Z are P = ( ), Q = ( ), R = ( ) The area of triangle PQR = = arycentric coordinates of the circumcenter onsider the circumcenter O of triangle. O Since O =2, the area of triangle O is 1 2 O O sin O = 1 2 R2 sin 2.

66 62 Homogeneous arycentric oordinates Similarly, the areas of triangles O and O are respectively 1 2 R2 sin 2 and 1 2 R2 sin 2. It follows that the circumcenter O has homogeneous barycentric coordinates O : O : O = 1 2 R2 sin 2 : 1 2 R2 sin 2 : 1 2 R2 sin 2 =sin2 :sin2 :sin2 =a cos : b cos : c cos =a b2 + c 2 a 2 : b c2 + a 2 b 2 : c a2 + b 2 c 2 2bc 2ca 2ab =a 2 (b 2 + c 2 a 2 ):b 2 (c 2 + a 2 b 2 ):c 2 (a 2 + b 2 c 2 ). 4.5 The inferior and superior triangles The triangle whose vertices are the midpoints of the sides of triangle is called the inferior triangle of. Its vertices are the points =(0:1:1), =(1:0:1), =(1:1:0). This triangle is homothetic to at the centroid G, with ratio 1. Equivalently, 2 we say that it is the image of under the homothety h(g, 1 ). This means 2 that h(g, 1 ) is a one-to-one correspondence of points on the plane such that 2 P and P have the same homogeneous barycentric coordinates with reference to and whenever X, X and G are collinear and GP GP = 1 2. We call P the inferior of P. More explicitly, P divides PG in the ratio PP : P G =3: 1, so that P = 1 (3G P ). Suppose P has homogeneous barycentric 2 coordinates (x : y : z) with reference to. It is the point P = x+yz. 6 x+y+z 6 The conversion from homogeneous barycentric coordinates to absolute barycentric coordinates is called normalization.

67 4.5 The inferior and superior triangles 63 Thus, P = 1 2 (3G P ) = 1 ( ) x + yz x + y + z = 1 ( ) (x + y + z)( + + ) (x + y + z) 2 x + y + z (y + z) +(z + x) +(x + y) =. 2(x + y + z) It follows that the homogenous coordinates of P are (y + z : z + x : x + y). The superior triangle of is its image under the homothety h(g, 2). Its vertices are the points =( 1 :1:1), =(1: 1 :1), =(1:1: 1) The nine-point center The nine-point circle is the circumcircle of the inferior triangle. Its center N is the inferior of the circumcenter. From N = 3G O and O = 1 (3G H), we obtain 2 2 N = 1 7 (O + H). The nine-point center N is the midpoint of OH. 2 Exercise 1. The Nagel point N a lies on the line joining the incenter to the centroid; it divides IG in the ratio IN a : N a G =3: Find the coordinates of the circumcenter O by using the fact that OG : GH =1:2. 3. The superior of P is the point P GP on the line GP for which = 2. GP Show that if P =(x : y : z), then P =( x+y +z : x y +z : x+y z). 4. Find the coordinates of the superior of the incenter. 5. Find the coordinates of the inferior of the incenter. Show that it is the centroid of the perimeter of triangle. 7 The nine-point center appears in ET as the point X 5.