1 Solution of Final. Dr. Franz Rothe December 25, Figure 1: Dissection proof of the Pythagorean theorem in a special case
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1 Math 3181 Dr. Franz Rothe December 25, 2012 Name: 1 Solution of Final Figure 1: Dissection proof of the Pythagorean theorem in a special case 10 Problem 1. Given is a right triangle ABC with angle α = 60. Squares are erected on its three sides. As done in the figure on page 1, one can construct dissections of them which prove the Theorem of Pythagoras. What kind of quadrilateral is ALKB. Put a = 2 3 and determine b and the segment lengths x = LE and y = BL. Answer. The quadrilateral ABLK is a parallelogram. Its opposite sides are congruent. Hence y = AK = AC + CK = AC + LE = b + x. Secondly, a = BE = BL + LE = y + x. Given values are a = 2 3, hence b = 2 and c = 4. x + y = a and y x = b imply x = a b = 3 1 and y = a + b =
2 Figure 2: One of many possible constructions of the parallel. 10 Problem 2. Given is a line l through two points A and B and a point P not lying on the line. One of many possible constructions of the parallel to line l through the point P is shown in the figure on page 2. Explain how the point Q is obtained. What does one know about the lines AB and P Q. Explain how the point R is obtained. What does one know about the lines P R and P Q. Explain why the lines P R and AB are parallel. Answer. The point Q is the second intersection point of the two circles around A and B through point P. The lines AB and P Q are perpendicular. Point R is the intersection of the circle around A with the line AQ. By Thales Theorem, the lines P R and P Q are perpendicular. Since both lines P R and AB are perpendicular to line P Q, they are parallel. 2
3 10 Problem 3 (Another triangle). Construct a triangle with angles of 30, 45 and 105, using only compass and straightedge, but no protractor. Describe your construction. Figure 3: Construction of a triangle with angles of 45, 30 and 105. Answer. On an arbitrary segment OB, an equilateral triangle OBD is erected. As described in Euclid I.1, D is an intersection point of a circle around B through O with a circle around O through B. At point O, we erect the perpendicular onto this diameter. Let E be the intersection of the perpendicular with the circle around O, lying on the same side of AB as point D. Let A be the second endpoint of diameter AOB. The rays AD and BE intersect in point C. We thus get a triangle ABC with the angles 30, 45 and 105 at its vertices A, B and C. The angle BAD = 30 since triangle OBD is equilateral and ADB = 90. The angle ABC = 45 is obtained from the right isosceles triangle OBE. Finally, one calculates the third angle BCA = = 105 by the angle sum of triangle ABC. 3
4 10 Problem 4 (Tangents to a circle). Given is a circle C with center O, and a point P outside of C. Construct the tangents from point P to the circle C. Actually do and describe the construction! Answer. Figure 4: Tangents to a circle Construction 1 (Tangents to a circle). One begins by constructing a second circle T with diameter OP. (I call this circle the Thales circle over the segment OP ). The Thales circle intersects the given circle in!! two points T and S. The lines P T and P S are the two tangents from P to circle C. Validity of the Construction. By Thales theorem, the angle!! P T O is a right angle, because it is an angle in the semicircle over diameter OP. Since point T lies on the circle C, too, the segment OT is a radius of that circle. By Euclid III. 16, The line perpendicular to a diameter is!! tangent to a circle. Since T P is perpendicular to the radius OT, and hence to a diameter, it is a tangent of circle C. 4
5 10 Problem 5. Construct a right triangle with projections p = 3 and q = 4 of the legs onto the hypothenuse. Use Thales theorem. Describe your construction. Measure the lengths of the two legs of your triangle. Figure 5: Construction of a right triangle with projections p = 3, q = 4. Answer. Adjacent to each other on one line, we draw segments with the given lengths AF = q = 4 and F B = p = 3. We erect the perpendicular on line AB at point F and draw a semicircle with diameter AB. The semicircle and the perpendicular intersect at point C. The triangle ABC is a right triangle with hypothenuse AB, and the projections q = AF and F B = p have the lengths as required. I measure the two legs as a = 4.6 and b = Problem 6. What are the lengths of the two legs of the triangle from the last problem. Use the leg theorem to calculate exact root expressions. Answer. The leg theorem gives the squares a 2 = (p + q)p = 21 and b 2 = (p + q)q = 28. Hence the lengths of the legs are a = 21 and b = 28 5
6 Figure 6: Constructing a circle through a point and touching two lines. 10 Problem 7. Given are two intersecting lines l and m and a point A different from their intersection point. We have to construct a circle through the point A touching both lines. Describe the construction given in the figure on page 6. Answer. We draw the angular bisector of the the two given lines, bisecting the angle α in the interior of which the given point A lies. We choose any point O on the bisector and draw a circle around it touching both line l and line m. We draw the ray emanating from the intersection point Z of the two given lines and pointing into the interior of the angle α. This ray intersects the circle in two points C and D. The parallel to line DO through point A intersects the bisector in point O. The circle around O through point A does touch the two given lines and solves the problem. A second solution is obtained by drawing the parallel to line CO through point A. This line intersects the bisector in the center O 2. The circle around O 2 through 6
7 point A does touch the two given lines and yields the second solution of the problem. 10 Problem 8. We explain that there are two pairs of similar triangles ZAO ZDO and ZF O ZF O and prove from the resulting proportions that the circle around O through point A touches the two given lines l and m. Complete the following reasoning: Reason for the construction. The two triangles ZAO ZDO are equiangular because the lines AO and DO are!! parallel by construction. Because the lines F O and F O are both perpendicular to the same!! line m and hence parallel, the triangles ZF O ZF O are!! equiangular. Since equiangular triangles are similar, we get the proportions Division yields ZO AO = ZO DO and ZO F O = ZO F O AO F O = DO F O = 1 Indeed, the latter quotient equals 1 since both points D and F lie on a!! circle around O. Hence the former quotient equals one, too. Hence both points A and F lie on a circle C around O. This means that the constructed circle C touches the line!! m. Since the centers O and O lie on the!! angle bisector of the two given lines l and m, the circle O touches the line l, too. 7
8 Figure 7: How to trisect a segment but not an angle. 10 Problem 9. A right triangle with the angles 30, 60, 90, the hypothenuse has twice the length of the shorter leg. Give any convincing reason you want for this fact. Take two equilateral triangles with the common side AB. Together they give the rhombus ACBF as shown in the figure on page 8. We draw the long diagonal CF. Let M and N be the midpoints of segments AC and AF. The segment BM intersects the long diagonal CF in point D. The segment BN intersects the long diagonal CF in point E. We see and explain the following facts: At vertex B we get the angles CBM = 30, MBN = 60 and NBF = 30. The points D and E trisect the long diagonal CF. 8
9 Figure 8: How to trisect a segment but not an angle. 10 Problem 10. Complete the following reasoning: The different angles at vertex B. The ray BM bisects the angle!! CBA. But this is the angle of an!! equilateral triangle, and hence measures!! 60. Hence the bisected angle CBM measures 30. Similarly, we see that the most left angle NBF measures!! 30. The middle angle MBN is obtained by addition of two bisected angles of 30 and hence measures!! Problem 11. Complete the following reasoning: The trisection of the long diagonal CF. The ray BM bisects the angle!! CBA. Hence the triangles BMA = BMC are congruent. The point M is indeed the midpoint of side AC. The angle BMA is congruent to its supplement BMC and hence a!! right angle. The angles in the small triangle MCD are!! 90, 30 and 60, at the vertices M, C and D, respectively. The small triangles MCD = OBD, because of the vertical angles at D, the right angles at M and O and the congruent sides MC = OB. The small triangles OBD = OBE, because of the vertical angles at D, the right angles at O, the congruent angles at B and the common side!! OB. 9
10 For any right triangle with the angles 30, 60, 90, the hypothenuse has twice the length of the!! shorter leg. Hence CD = BD = 2 DO = DE Similarly we see that EF = DE. Hence points D and E trisect the long diagonal CF. 10
11 10 Problem 12. The angle of 120 can be constructed, but not trisected with ruler and compass. Complete the following argument: Proof. The construction of an angle of 120 was done in a previous problem. To investigate the trisection, we use the trigonometric formula 4 cos 3 α 3 cos α = cos 3α with 3α = 120 and obtain 4 cos cos 40 = cos 120 = 1 2 Hence k = 2 cos 40 is a root of the cubic equation x 3 3x + 1 = 0 This cubic polynomial has no rational root, as shown in the Lemma below. Since it has degree only!! three, this fact already implies that the polynomial cannot at all be factored over the rational numbers. Hence the polynomial is irreducible. For an!! irreducible polynomial, the field extension generated by root has dimension given by the!! degree of this polynomial. Hence we have obtained [Q(cos 40 ) : Q] = 3 We know that, for every constructible number k, the dimension [Q(k) : Q] is a!! power of two. Since 3 is not a power of two, we conclude that cos 40 is not a constructible number. Lemma 1. The polynomial x 3 3x + 1 has no rational root. Proof. Suppose towards a!! contradiction that x = p/q with p and q nonzero relatively prime integers is a rational root of x 3 3x + 1. By plugging the fraction into this polynomial, we obtain both p 3 = 3pq 2 q 3 = (3p q 2 )q and q 3 = p 3 + 3pq 2 = p( p 2 + 3q 2 ). The first relation implies that q divides p 3. Since p and q are relatively prime, the integer q can only be ±1. The second relation q 3 = p( p 2 + 3q 2 ) implies that p!! divides q 3. Since p and q are relatively prime, this implies p = ±1. Hence x = p/q is equal to!! ±1. But neither one of these two numbers is a root of x 3 3x + 1 = 0. This contradiction shows that there is no rational root of the polynomial x 3 3x+1. 11
12 10 Problem 13. In the coordinate plane are given: the circle with radius 3 and center (0, 0) and the line through points P = ( 1, 0) and Q = (0, 1). 1. What are the equations for the circle and the line. 2. What are the coordinates (x, y) for the intersection points of the circle and the line. 3. Give the exact root expressions for both coordinates of one intersection point. For the construction of which regular polygon could this be a useful first step? Answer. The equations for the circle and the line are x 2 + y 2 = 9 and y = x + 1 To get the coordinates (x, y) of the intersection points of the circle and the line, we plug y = x + 1 into the equation of the circle and get x 2 + (x + 1) 2 = 9 2x 2 + 2x 8 = 0 x 2 + x 4 = 0 The two intersection points have coordinates ( 1 17, 1 ) 17 and 2 2 x 1,2 = 1 ± 17 2 ( , 1 + ) These values could be used as a first step in a construction of a regular 17-gon. 12
13 10 Problem 14. The cubic polynomial x 3 2 is irreducible over the rational numbers, since 3 2 is irrational. One can use the Eisenstein criterium to check that the polynomial x 5 2 is irreducible over the rational numbers. Which ones of the roots 2, 3 2, 4 2, 5 2 are in the constructible field, which ones are not. Answer. The numbers!! 2 and!! 4 2 = 2 are constructible. The dimension of an extension generated by one root of an irreducible polynomial is given by its degree: [Q( 3 2) : Q] = 3 and [Q( 5 2) : Q] = 5 We have shown that the!! dimension [Q(k) : Q] is a!! power of two for every constructible number k. Since neither 3 nor 5 is a power of two, we conclude that neither 3 2 nor 5 2 are constructible. 13
14 Figure 9: Trisection with Nicolson s angle. Construction 2 (Trisection with Nicolson s angle (1883)). Nicolson s angle consists of a strip of two parallels with distance AB to which an extra square of congruent side BC is attached at point B. Thus B becomes the midpoint of segment AC. The points A, B and C are marked on the Nicolson angle. To trisect any given angle P OR, we first draw the parallel to side OP at distance AB which cuts the interior of the given angle. Next we place the Nicolson angle in such a way that on the parallel to side OP lies point A; on the other side OR lies point C; The vertex O lies on the edge of the Nicolson angle with extension through point B. The rays OA and OB trisect the given angle P OR. 14
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