Problem Solving and Recreational Mathematics

Transcription

1 Problem Solving and Recreational Mathematics Paul Yiu Department of Mathematics Florida tlantic University Summer 2012 Chapters July 20 Monday 6/25 7/2 7/9 7/16 7/23 7/30 Wednesday 6/27 *** 7/11 7/18 7/25 8/1 Friday 6/29 7/6 7/13 7/20 7/27 8/3

2 Contents 16 The golden ratio Division of a segment in the golden ratio The regular pentagon Construction of 36, 54, and 72 angles The most non-isosceles triangle Medians and angle isectors pollonius Theorem ngle bisector theorem The angle bisectors Steiner-Lehmus Theorem Dissections Dissection of the 6 6 square Dissection of a 7 7 square into rectangles Dissect a rectangle to form a square Dissection of a square into three similar parts Pythagorean triangles Primitive Pythagorean triples Rational angles Some basic properties of primitive Pythagorean triples Pythagorean triangle with an inscribed square When are x 2 px ± q both factorable? Dissection of a square into Pythagorean triangles Integer triangles with a 60 or 120 angle Integer triangles with a 60 angle

3 iv CONTENTS 20.2 Integer triangles with a 120 angle Triangles with centroid on incircle Construction Integer triangles with centroid on the incircle The area of a triangle Heron s formula for the area of a triangle Heron triangles The perimeter of a Heron triangle is even The area of a Heron triangle is divisible by Heron triangles with sides < Heron triangles with sides in arithmetic progression Indecomposable Heron triangles Heron triangle as lattice triangle Heron triangles Heron triangles with area equal to perimeter Heron triangles with integer inradii Division of a triangle into two subtriangles with equal incircles Inradii in arithmetic progression Heron triangles with integer medians Heron triangles with square areas Triangles with sides and one altitude in.p Newton s solution The general case The Pell Equation The equation x 2 dy 2 = The equation x 2 dy 2 = The equation x 2 dy 2 = c Figurate numbers Which triangular numbers are squares? Pentagonal numbers lmost square triangular numbers Excessive square triangular numbers Deficient square triangular numbers

4 CONTENTS v 27 Special integer triangles lmost isosceles Pythagorean triangles The generators of the almost isosceles Pythagorean triangles Integer triangles (a, a + 1, b) with a 120 angle Heron triangles Heron triangles with consecutive sides Heron triangles with two consecutive square sides Squares as sums of consecutive squares Sum of squares of natural numbers Sums of consecutive squares: odd number case Sums of consecutive squares: even number case Sums of powers of consecutive integers Lucas problem Solution of n(n + 1)(2n + 1) = 6m 2 for even n The Pell equation x 2 3y 2 = 1 revisited Solution of n(n + 1)(2n + 1) = 6m 2 for odd n Some geometry problems asic geometric constructions Some basic construction principles Geometric mean Harmonic mean M G.M. H.M Construction of a triangle from three given points Some examples Wernick s construction problems

5 Chapter 16 The golden ratio 16.1 Division of a segment in the golden ratio Given a segment, a point P in the segment is said to divide it in the golden ratio if P 2 = P. Equivalently, P = 5+1. We P 2 shall denote this golden ratio by ϕ. It is the positive root of the quadratic equation x 2 = x + 1. M Q P P Construction 16.1 (Division of a segment in the golden ratio). Given a segment, (1) draw a right triangle M with M perpendicular to and half in length, (2) mark a point Q on the hypotenuse M such that MQ = M, (3) mark a point P on the segment such that P = Q. Then P divides into the golden ratio.

6 602 The golden ratio Suppose P has unit length. The length ϕ of P satisfies ϕ 2 = ϕ + 1. This equation can be rearranged as ( ϕ 1 ) 2 = Since ϕ > 1, we have ϕ = 1 2 ( ). Note that P = This explains the construction above. ϕ ϕ + 1 = 1 ϕ = =

7 16.2 The regular pentagon The regular pentagon Consider a regular pentagon CDE. It is clear that the five diagonals all have equal lengths. Note that (1) C = 108, (2) triangle C is isosceles, and (3) C = C = ( ) 2 = 36. In fact, each diagonal makes a 36 angle with one side, and a 72 angle with another. C P E D It follows that (4) triangle PC is isosceles with PC = PC = 36, (5) PC = = 108, and (6) triangles C and PC are similar. Note that triangle CP is also isosceles since (7) CP = PC = 72. This means that P = C. Now, from the similarity of C and PC, we have : C = C : P. In other words P = P P, or P 2 = P. This means that P divides in the golden ratio.

8 604 The golden ratio 16.3 Construction of 36, 54, and 72 angles ngles of sizes 36, 54, and 72 can be easily constructed from a segment divided in the golden ratio. C P ϕ cos 36 = ϕ 2. C C P 1 54 D 72 ϕ P cos 72 = 1 2ϕ.

9 16.3 Construction of 36, 54, and 72 angles 605 Exercise 1. Three equal segments 1 1, 2 2, 3 3 are positioned in such a way that the endpoints 2, 3 are the midpoints of 1 1, 2 2 respectively, while the endpoints 1, 2, 3 are on a line perpendicular to 1 1. Show that 2 divides 1 3 in the golden ratio Given an equilateral triangle C, erect a square CDE externally on the side C. Construct the circle, center C, passing through E, to intersect the line at F. Show that divides F in the golden ratio. D C E 3. Given a segment, erect a square on it, and an adjacent one with base C. If D is the vertex above, construct the bisector of angle DC to intersect at P. Calculate the ratio P : P. F D P C

10 606 The golden ratio 4. Let D and E be the midpoints of the sides and C of an equilateral triangle C. If the line DE intersects the circumcircle of C at F, calculate the ratio DE : EF. D E F O C 5. Given a segment with midpoint M, let C be an intersection of the circles (M) and (), and D the intersection of C() and (C) inside (). Prove that the line CD divides in the golden ratio. C D M P 6. The three small circles are congruent. Show that each of the ratios O, TX, ZT. is equal to the golden ratio. XY TO Z X Y T O

11 16.3 Construction of 36, 54, and 72 angles Which of the two equilateral triangles inscribed in a regular pentagon has larger area? 8. Which of the two squares inscribed in a regular pentagon has larger area?

12 608 The golden ratio 16.4 The most non-isosceles triangle Given a triangle with sides a, b, c, there are six ratios obtained by comparing the lengths of a pair of sides: a b, b a, b c, c b, a c, c a. The one which is closest to 1 is called the ratio of nonisoscelity of the triangle. The following theorem shows that there is no most non-isosceles triangle. Theorem number η is the ratio of nonisoscelity of a triangle if and only if ϕ 1 < η 1. Proof. First note that if a = r < 1, then 1 = b > 1. Since 1(r+ 1) > 1, b r a 2 r it follows that r is closer to 1 than 1. r If a b c are the lengths of the three sides of a triangle, the ratio of nonisoscelity is ( a η = max b, b ). c Since a + b > c, we have η + 1 a b + 1 = a + b b > c b 1 η. Therefore, η 2 + η > 1. Since the roots of x 2 + x 1 are ϕ 1 > 0 and ϕ < 0, we must have η > ϕ 1. Therefore, η (ϕ 1, 1]. Conversely, for each number t (ϕ 1, 1], we have t 2 + t > 1 and t+1 > 1. The numbers t 1 1 form the sides of a triangle with ratio t t of nonisoscelity equal to t.

13 Chapter 17 Medians and angle isectors 17.1 pollonius Theorem Theorem Given triangle C, let D be the midpoint of C. The length of the median D is given by 2 + C 2 = 2(D 2 + D 2 ). D C Proof. pplying the law of cosines to triangles D and CD, and noting that cos D = cos DC, we have 2 = D 2 + D 2 2D D cosd; C 2 = D 2 + CD 2 2D CD cosdc, = D 2 + D 2 + 2D D cos D. The result follows by adding the first and the third lines. If m a denotes the length of the median on the side C, m 2 a = 1 4 (2b2 + 2c 2 a 2 ).

14 610 Medians and angle isectors Example Suppose the medians E and CF of triangle C are perpendicular. This means that G 2 + CG 2 = C 2, where G is the centroid of the triangle. In terms of the lengths, we have 4 9 m2 b m2 c = a 2 ; 4(m 2 b + m2 c ) = 9a2 ; (2c 2 + 2a 2 b 2 ) + (2a 2 + 2b 2 c 2 ) = 9a 2 ; b 2 + c 2 = 5a 2. This relation is enough to describe, given points and C, the locus of for which the medians E and CF of triangle C are perpendicular. Here, however, is a very easy construction: From b 2 + c 2 = 5a 2, we have m 2 a = 1 4 (2b2 + 2c 2 a 2 ) = 9 4 a2 ; m a = 3 a. The locus of is 2 the circle with center at the midpoint of C, and radius 3 C. 2 Exercise 1. The triangle with sides 7, 8, 9 has one median equal to a side. Which median and which side are these? 2. The lengths of the sides of a triangle are 136, 170, and 174. Calculate the lengths of its medians. 3. Triangle C has sidelengths a = 17, b = 13, c = 7. Show that the medians are in the proportions of the sides C

15 17.2 ngle bisector theorem ngle bisector theorem Theorem 17.2 (ngle bisector theorem). The bisectors of an angle of a triangle divide its opposite side in the ratio of the remaining sides. If X and X respectively the internal and external bisectors of angle C, then X : XC = c : b and X : X C = c : b. Z c Z b X C X Proof. Construct lines through C parallel to the bisectors X and X to intersect the line at Z and Z. (1) Note that ZC = X = XC = CZ. This means Z = C. Clearly, X : XC = : Z = : C = c : b. (2) Similarly, Z = C, and X : X C = : Z = : C = c : b.

16 612 Medians and angle isectors 17.3 The angle bisectors The length of the internal bisector of angle is t a = 2bc b + c cos 2, t a t a Q P C and that of the external bisector of angle is t a = 2bc b c sin 2. Example The triangle (a, b, c) = (125, 154, 169) is a Heron triangle with rational angle bisectors. Its area is The lengths of the bisectors are,, and. Exercise 1. The triangle with sides 6, 7, 8 has one angle bisector equal to a side. Which angle bisector and which side are these? 1 2. The triangle with sides 17, 24, 27 has one angle bisector equal to a side. Which angle bisector and which side are these? 2 3. The triangle with sides 84, 125, 169 has three rational angle bisectors. Calculate the lengths of these bisectors Prove that the bisector of the right angle of a Pythagorean triangle cannot have rational length. 5. Give an example of a Pythagorean triangle with a rational bisector. 6. Prove that the bisectors of the acute angles of a Pythagorean triangle cannot be both rational. 1 t b = a = 6. 2 t a = b = t a = 975 7, t b = , tc =

17 17.4 Steiner-Lehmus Theorem Steiner-Lehmus Theorem Theorem triangle is isosceles if it has two equal angle bisectors. F Z Y E C Proof. Suppose the bisectors E = CF, but triangle C not isosceles. We may assume < C. Construct parallels to C through E and F to intersect and C at Z and Y respectively. (1) In the isosceles triangles ZE and Y CF with equal bases E and CF, ZE < Y CF = EZ < Y F. (2) Y = F = C < = E. Therefore, Y C F C C EC C = Y + 1 < E + 1 = C, and Y C > EC. This clearly implies C Y C EC EC Y F < EZ, contradicting (1) above. Example The analogue of the Steiner-Lehmus theorem does not hold for external angle bisectors. It is very easy to construct one such triangle with the extra condition t b = t c = a. Y t b C a t c Z

18 614 Medians and angle isectors (1) From t b = a, we have 1 (π β) = 2γ; β + 4γ = π. 2 (2) From t c = a, we have 1 (π γ) = π 2(π β) = 2β π; 4β+γ = 3π. 2 From these, β = 11π, γ = π, and α = π Exercise 1. The bisector t a and the external bisector t b of triangle C satisfy t a = t 4 b = c. Calculate the angles of the triangle. Y C t b ta X c 2. D and E are angle bisectors of triangle C, with D on C and E on C. Suppose that D = and E = C. Determine the angles of the triangle. Show that if CF is the external bisector of angle C, then CF = C. 5 E D C 4 nswer: α = 2π 15, β = 7π 15, γ = 2π 5. 5 nswer: α = 2π 13, β = 6π 13, γ = 5π 13.

19 Chapter 18 Dissections 18.1 Dissection of the 6 6 square Since = 6 2, it is reasonable to look for a dissection of a 6 6 square into eight pieces with areas 1, 2,..., 8 square units cuts: Construct a line through which completes the dissection straight cuts: Construct a line through which completes the dissection.

20 616 Dissections 3. 5 straight cuts: Construct two lines through C to complete the dissection. C

21 18.2 Dissection of a 7 7 square into rectangles Dissection of a 7 7 square into rectangles Dissect a square into seven rectangles of different shapes but all having the same area. 1 R 4 R 5 R 7 R 2 R 3 R 6 R 1 Suppose each side of the square has length 7, and for i = 1, 2..., 7, the rectangle R i has (horizontal) length a i and (vertical) height b i. Setting a 1 = r, we compute easily in succession, b 1, b 2, a 2, a 3, b 3, b 4, a 4, a 5, b 5, using a i b i = 7. Now, a 6 =7 a 2 a 3, b 6 =7 b 1 b 5, a 7 =7 a 2 a 3 a 5, b 7 =7 b 1 b 4 b 6. rectangle a i b i R 1 r 7 R 2 r r 1 r 7(r 1) r R 3 7 r 7 7 r R 4 6 r 5(7r 7 r R 2 ) 5 (6 r)(r 1) r(r 2) R 6 r 1 (7 r)(6 r)(28 27r+4r R 2 ) 7 5 r 7 r 7(6 r) 7 r 7(6 r)(r 1)3 5(7r 7 r 2 7(r 1)(29r 35 4r 2 ) 5r(7r 7 r 2 ) 7(6 r) 5(7 r)(7r 7 r 2 ) 1 Problem 2567, Journal of Recreational Math.,. dissection into seven rectangles was given by Hess and Ibstedt. Markus Götz showed that no solution with n rectangles exists for 2 n 6.

22 618 Dissections The relations a 6 b 6 = 7 and a 7 b 7 = 7 both reduce to (2r 7)(2r 2 14r + 15) = 0. Since r = 7 2 or would make a 7, b 7 negative, we must have r = The dimensions of the rectangles are as follows. rectangle a i b i R R 2 7(8+ 19) 15 7(7+ 19) R R R 5 3 R 6 5(1+ 19) 6 R 7 5( 1+ 19) (8 19) ( 1+ 19) 15 7(1+ 19) 15

23 18.3 Dissect a rectangle to form a square Dissect a rectangle to form a square Dissect a given rectangle by two straight cuts into three pieces that can be reassembled into a square. If this is possible, the lengths of the cuts must be as follows, and it is enough to verify that x = ab b. a ab ab b x This is clear since x = (a ab) ab a ab b = a( a b b) = ab b. ab a a ab b ab

24 620 Dissections 18.4 Dissection of a square into three similar parts Dissect a square into three similar parts, no two congruent. D 1 x 2 C x x 2 x + 1 x Solution. x = x(x 2 x + 1); x 3 2x 2 + x 1 = 0. Exercise 1. Show that the area of the largest rectangle is x 4. Solution. This area is (x 2 + 1)(x 2 x + 1) = x 4 x 3 + x 2 + x 2 x + 1 = x 4 x 3 + 2x 2 x + 1 = x 4.

25 18.4 Dissection of a square into three similar parts 621 Exercise 1. Show how the square should be dissected so that it can be reassembled into a rectangle. D C x Q Y P x X 2. Let P and Q be points on the sides and C of rectangle CD. D C T 3 Q T 1 T 2 P (a) Show that if the areas of triangles PD, PQ and CDQ are equal, then P and Q divide the sides in the golden ratio. (b) If, in addition, DP = PQ, show that the rectangle is golden, i.e., = ϕ, and DPQ is a right angle. C

26 Chapter 19 Pythagorean triangles 19.1 Primitive Pythagorean triples Pythagorean triangle is one whose sidelengths are integers. n easy way to construct Pythagorean triangles is to take two distinct positive integers m > n and form (a, b, c) = (2mn, m 2 n 2, m 2 + n 2 ). Then, a 2 + b 2 = c 2. We call such a triple (a, b, c) a Pythagorean triple. The Pythagorean triangle has perimeter p = 2m(m + n) and area = mn(m 2 n 2 ). m 2 + n 2 2mn m 2 n 2 C Pythagorean triple (a, b, c) is primitive if a, b, c do not have a common divisor (greater than 1). Every primitive Pythagorean triple is constructed by choosing m, n to be relatively prime and of opposite parity.

27 702 Pythagorean triangles Rational angles The (acute) angles of a primitive Pythagorean triangle are called rational angles, since their trigonometric ratios are all rational. sin cos tan 2mn m 2 n 2 m 2 +n 2 m 2 +n 2 m2 n 2 m 2 +n 2 2mn m 2 n 2 2mn m 2 n 2 m 2 +n 2 2mn More basic than these is the fact that tan 2 and tan 2 are rational: tan 2 = n m, tan 2 = m n m + n. This is easily seen from the following diagram showning the incircle of the right triangle, which has r = (m n)n. m(m n) (m + n)n I r r (m + n)n (m n)n m(m n) (m n)n C Some basic properties of primitive Pythagorean triples 1. Exactly one leg is even. 2. Exactly one leg is divisible by Exactly one side is divisible by The area is divisible by 6.

28 19.1 Primitive Pythagorean triples 703 ppendix: Primitive Pythagorean triples < 1000 m, n a, b, c m, n a, b, c m, n a, b, c m, n a, b, c 2, 1 3, 4, 5 3,2 5,12, 13 4, 1 15, 8, 17 4,3 7,24, 25 5, 2 21,20, 29 5,4 9,40, 41 6, 1 35, 12, 37 6,5 11, 60, 61 7, 2 45,28, 53 7,4 33, 56, 65 7, 6 13, 84, 85 8,1 63, 16, 65 8, 3 55,48, 73 8,5 39, 80, 89 8, 7 15, 112, 113 9,2 77, 36, 85 9, 4 65,72, 97 9,8 17, 144, , 1 99, 20, , 3 91, 60, , 7 51, 140, , 9 19, 180, , 2 117, 44, , 4 105, 88, , 6 85, 132, , 8 57, 176, , 10 21, 220, , 1 143, 24, , 5 119, 120, , 7 95, 168, , 11 23, 264, , 2 165, 52, , 4 153, 104, , 6 133, 156, , 8 105, 208, , 10 69, 260, , 12 25, 312, , 1 195, 28, , 3 187, 84, , 5 171, 140, , 9 115, 252, ,11 75, 308, , 13 27, 364, , 2 221, 60, , 4 209, 120, , 8 161, 240, , 14 29, 420, , 1 255, 32, , 3 247, 96, , 5 231, 160, , 7 207, 224, , 9 175, 288, , , 352, ,13 87, 416, , 15 31, 480, , 2 285, 68, , 4 273, 136, , 6 253, 204, , 8 225, 272, , , 340, , , 408, ,14 93, 476, , 16 33, 544, , 1 323, 36, , 5 299, 180, , 7 275, 252, , , 396, , , 468, , 17 35, 612, , 2 357, 76, , 4 345, 152, , 6 325, 228, , 8 297, 304, , , 380, , , 456, , , 532, , , 608, ,18 37, 684, , 1 399, 40, , 3 391, 120, , 7 351, 280, , 9 319, 360, , , 440, , , 520, , , 680, ,19 39, 760, , 2 437, 84, , 4 425, 168, , 8 377, 336, , , 420, , , 672, , 20 41, 840, , 1 483, 44, , 3 475, 132, , 5 459, 220, , 7 435, 308, , 9 403, 396, , , 572, , , 660, , , 748, , , 836, ,21 43, 924, , 2 525, 92, , 4 513, 184, , 6 493, 276, , 8 465, 368, , , 460, , , 552, , , 644, , , 736, , , 828, , , 920, , 1 575, 48, , 5 551, 240, , 7 527, 336, , , 528, , , 624, , , 816, , , 912, , 2 621, 100, , 4 609, 200, , 6 589, 300, , 8 561, 400, , , 600, , , 700, , , 800, , , 900, , 1 675, 52, , 3 667, 156, , 5 651, 260, , 7 627, 364, , 9 595, 468, , , 572, , , 780, , , 884, , 2 725, 108, , 4 713, 216, , 8 665, 432, , , 540, , , 756, , , 864, , 1 783, 56, , 3 775, 168, , 5 759, 280, , 9 703, 504, , , 616, , , 728, , 2 837, 116, , 4 825, 232, , 6 805, 348, , 8 777, 464, , , 580, , , 696, , 1 899, 60, , 7 851, 420, , 2 957, 124, , 4 945, 248, , 6 925, 372, 997

29 704 Pythagorean triangles 19.2 Pythagorean triangle with an inscribed square How many matches of equal lengths are required to make up the following configuration? Suppose the shape of the right triangle is given by a primitive Pythagorean triple (a, b, c). The length of a side of the square must be a common multiple of a and b. The least possible value is the product ab. There is one such configuration consisting of (i) two Pythagorean triangles obtained by magnifying (a, b, c) a and b times, (ii) a square of side ab. The total number of matches is (a + b)(a + b + c) + 2ab = (a + b + c)c + 4ab. The smallest one is realized by taking (a, b, c) = (3, 4, 5). It requires 108 matches. How many matches are required in the next smallest configuration?

30 19.3 When are x 2 px ± q both factorable? When are x 2 px ± q both factorable? For integers p and q, the quadratic polynomials x 2 px+q and x 2 px q both factorable (over integers) if and only if p 2 4q and p 2 +4q are both squares. Thus, p and q are respectively the hypotenuse and area of a Pythagorean triangle. p 2 + 4q = (a + b) 2, p 2 4q = (b c) 2. b a p b p b a q b a a q a b c = p q x 2 px + q x 2 + px q (x 2)(x 3) (x 1)(x + 6) (x 3)(x 10) (x 2)(x + 15) (x 5)(x 12) (x 3)(x + 20) The roots of x 2 px+q are s a and s b, while those of x 2 +px q are s c and s. Here, s is the semiperimeter of the Pythagorean triangle Dissection of a square into Pythagorean triangles Here is the smallest square which can be dissected into three Pythagorean triangles and one with integer sides and integer area

31 706 Pythagorean triangles Exercise 1. man has a square field, 60 feet by 60 feet, with other property adjoining the highway. He put up a straight fence in the line of 3 trees, at, P, Q. If the distance between P and Q is 91 feet, and that from P to C is an exact number of feet, what is this distance? P? 91 Q D C 2. Give an example of a primitive Pythagorean triangle in which the hypotenuse is a square. 3. Give an example of a primitive Pythagorean triangle in which the even leg is a square. 4. Find the smallest Pythagorean triangle whose perimeter is a square (number). 5. Find the shortest perimeter common to two different primitive Pythagorean triangles. 6. Show that there are an infinite number of Pythagorean triangles whose hypotenuse is an integer of the form For each natural number n, how many Pythagorean triangles are there such that the area is n times the perimeter? How many of these are primitive? 8. Find the least number of toothpicks (of equal size) needed to form

32 Chapter 20 Integer triangles with a 60 or 120 angle 20.1 Integer triangles with a 60 angle If triangle C has C = 60, then c 2 = a 2 ab + b 2. (20.1) Integer triangles with a 60 angle therefore correspond to rational points in the first quadrant on the curve x 2 xy + y 2 = 1. (20.2) Note that the curve contains the point P = ( 1, 1). y passing a line of rational slope t through P to intersect the curve again, we obtain a parametrization of the rational points. Now, such a line has equation y = 1 + t(x + 1). Solving this simultaneously with (20.2) we obtain (x, y) = ( 1, 1) = P, and (x, y) = ( ) 2t 1 t 2 t + 1, t(2 t), t 2 t + 1 which is in the first quadrant if 1 < t 2. y symmetry, we may 2 simply take 1 < t 1 to avoid repetition. 2 Putting t = q for relatively prime integers p, q, and clearing denomi- p

33 708 Integer triangles with a 60 or 120 angle t P nators, we obtain a =p(2q p), b =q(2p q), c =p 2 pq + q 2, with p < q p. Dividing by gcd(a, b) = gcd(p + q, 3), we obtain the 2 primitive integer triangles with a 60 angle: p q (a, b, c) 1 1 (1, 1,1) 3 2 (3, 8,7) 4 3 (8, 15, 13) 5 3 (5, 21, 19) 5 4 (5, 8,7) 6 5 (24, 35, 31) 7 4 (7, 40, 37) 7 5 (7, 15, 13) 7 6 (35, 48, 43) 8 5 (16, 55, 49) 8 7 (16, 21, 19) 9 5 (9, 65, 61) 9 7 (45, 77, 67) 9 8 (63, 80, 73) 10 7 (40, 91, 79) 10 9 (80, 99, 91)

34 20.1 Integer triangles with a 60 angle 709 Exercise 1. standard calculus exercise asks to cut equal squares of dimension x from the four corners of a rectangle of length a and breadth b so that the box obtained by folding along the creases has a greatest capacity. a x The answer to this problem is given by b x = a + b a 2 ab + b 2. 6 How should one choose relatively prime integers a and b so that the resulting x is an integer? 1 For example, when a = 5, b = 8, x = 1. nother example is a = 16, b = 21 with x = Prove that there is no integer triangle with a 60 angle whose adjacent sides are consecutive integers. Proof. The 60 angle being strictly between the other two angles, its opposite side has length strictly between two consecutive integers. It cannot be an integer. 3. Find all integer triangles with a 60 angle and two sides consecutive integers. 1 nswer: a, b, c, with gcd(p + q, 6) = 3.

35 710 Integer triangles with a 60 or 120 angle 20.2 Integer triangles with a 120 angle If triangle C has C = 120, then c 2 = a 2 + ab + b 2. (20.3) Integer triangles with a 120 angle therefore correspond to rational points in the first quadrant on the curve x 2 + xy + y 2 = 1. (20.4) t Q Note that the curve contains the point Q = ( 1, 0). y passing a line of rational slope t through P to intersect the curve again, we obtain a parametrization of the rational points. Now, such a line has equation y = t(x+1). Solving this simultaneously with (20.2) we obtain (x, y) = ( 1, 0) = Q, and Q(t) = ( ) 1 t 2 t 2 + t + 1, t(2 + t), t 2 + t + 1 which is in the first quadrant if 0 < t < 1. It is easy to check that Q(t) and Q ( 1 t 1+2t) are symmetric about the line y = x. To avoid repetition we may restrict to 0 < t < Putting t = q p for relatively prime integers p, q satisfying q < p,

36 20.2 Integer triangles with a 120 angle 711 and clearing denominators, we obtain a =p 2 q 2, b =q(2p + q), c =p 2 + pq + q 2, with 0 < q < p. Dividing by gcd(a, b) = gcd(p q, 3), we obtain the primitive integer triangles with a 120 : p q (a, b, c) 3 1 (8,7, 13) 4 1 (5,3, 7) 5 1 (24, 11, 31) 6 1 (35, 13, 43) 7 1 (16, 5, 19) 7 2 (45, 32, 67) 8 1 (63, 17, 73) 9 1 (80, 19, 91) 9 2 (77, 40, 103) 10 1 (33, 7, 37) 10 3 (91, 69, 139)

37 712 Integer triangles with a 60 or 120 angle Exercise 1. n integer triangle has a 120 angle. Show that the two longer sides cannot differ by 1. Solution. If the sides are a, b, b + 1, then (b + 1) 2 = a 2 + ab + b 2. From this, b = a2 1. For no positive integer a is b positive. There 2 a is no such triangle. 2. Give an example of an integer triangle with 120 whose adjacent sides are consecutive integers. 3. In triangle C, α = 120. X is the bisector of angle. Show that 1 t = 1 b + 1 c. c t b X C 4. In the diagram below, X, CY, and CDZ are equilateral triangles. Suppose XY Z = 120. Show that 1 b = 1 a + 1 c. 2 Z X Y a b C c D 2 Hint: Extend ZY to intersect at T. Show that TC = a.

38 Chapter 21 Triangles with centroid on incircle Y G I G X D C Suppose the centroid G of triangle C lies on the incircle. The median D intersects the incircle at another point G. If G = k D, then by the intersecting chords theorem, we have (i) Y 2 = G G, and (ii) DX 2 = DG DG. Clearly, G = 2 3 m a, DG = 1 3 m a. If G = k m a, then DG = (1 k)m a. Since Y = 1 2 (b+c a) and DX = a 2 X = a (c+a b) = 1 2 (b c), these lead to 1 4 (b + c a)2 = 2 3 m a k m a, 1 4 (b c)2 = 1 3 m a (1 k)m a.

39 714 Triangles with centroid on incircle Eliminating k, we have 2 4m 2 a = 3((b + c a) 2 + 2(b c) 2 ). From pollonius theorem, 4m 2 a = 2b 2 + 2c 2 a 2. Simplifying, we obtain 5(a 2 + b 2 + c 2 ) 6(bc + ca + ab) = 0. Theorem The centroid of triangle (a, b, c) lies on its incircle if and only if 5(a 2 + b 2 + c 2 ) 6(bc + ca + ab) = Construction Let 0 0 C 0 be an equilateral triangle with center I 0, and C the image of the incircle under the homothety h ( I 0, 1 2). For every point P on the circle C, let X, Y, Z be the pedals on the sides of 0 0 C 0. The triangle with sides PX, PY, PZ has its centroid on its incircle.

40 21.2 Integer triangles with centroid on the incircle Integer triangles with centroid on the incircle y putting x = a and y = b, we have c c 5x 2 6xy + 5y 2 6x 6y + 5 = 0. This is an ellipse with center ( 3, ) and semiaxes 2 and It contains the rational point Q(1, 2). The line through (1, 2) with slope 2+t 2+t has equation y 2 = (x 1), or (t+2)x+(t 2)y (3t 2) = 0. 2 t 2 t Q t It intersects the conic at ( 1 2t + 2t t 2, ) 2 2t + t t 2 Writing t = q p for relatively prime integers p and q, we have a = p 2 2pq + 2q 2, b = 2p 2 2pq + q 2, c = p 2 + q 2. Here are some examples, with relatively prime integers p and q satisfying q < p < 2q:

41 716 Triangles with centroid on incircle p q a b c Exercise 1. Find the (shape of the) triangle which has a median trisected by the incircle.

42 Chapter 22 The area of a triangle 22.1 Heron s formula for the area of a triangle Theorem The area of a triangle of sidelengths a, b, c is given by where s = 1 (a + b + c). 2 = s(s a)(s b)(s c), Ia r a I r Y s b C s c Y s a Proof. Consider the incircle and the excircle on the opposite side of. From the similarity of triangles IZ and I Z, r = s a. r a s From the similarity of triangles CIY and I CY, r r a = (s b)(s c).

43 802 The area of a triangle From these, (s a)(s b)(s c) r =, s and the area of the triangle is = rs = s(s a)(s b)(s c). Exercise 1. Prove that 2 = 1 16 (2a2 b 2 + 2b 2 c 2 + 2c 2 a 2 a 4 b 4 c 4 ).

44 22.2 Heron triangles Heron triangles Heron triangle is an integer triangle whose area is also an integer The perimeter of a Heron triangle is even Proposition The semiperimeter of a Heron triangle is an integer. Proof. It is enough to consider primitive Heron triangles, those whose sides are relatively prime. Note that modulo 16, each of a 4, b 4, c 4 is congruent to 0 or 1, according as the number is even or odd. To render in (??) the sum 2a 2 b 2 + 2b 2 c 2 + 2c 2 a 2 a 4 b 4 c 4 0 modulo 16, exactly two of a, b, c must be odd. It follows that the perimeter of a Heron triangle must be an even number The area of a Heron triangle is divisible by 6 Proposition The area of a Heron triangle is a multiple of 6. Proof. Since a, b, c are not all odd nor all even, and s is an integer, at least one of s a, s b, s c is even. This means that is even. We claim that at least one of s, s a, s b, s c must be a multiple of 3. If not, then modulo 3, these numbers are +1 or 1. Since s = (s a)+(s b)+(s c), modulo 3, this must be either ( 1) or 1 1+( 1)+( 1). In each case the product s(s a)(s b)(s c) 1 (mod 3) cannot be a square. This justifies the claim that one of s, s a, s b, s c, hence, must be a multiple of 3. Exercise 1. Prove that if a triangle with integer sides has its centroid on the incircle, the area cannot be an integer.

45 804 The area of a triangle Heron triangles with sides < 100 (a, b, c, ) (a, b, c, ) (a, b, c, ) (a, b, c, ) (a, b, c, ) (3, 4, 5, 6) (5, 5,6, 12) (5, 5, 8, 12) (5, 12, 13, 30) (10, 13, 13, 60) (4, 13, 15, 24) (13, 14, 15, 84) (9, 10, 17, 36) (8, 15, 17, 60) (16, 17, 17, 120) (11, 13, 20, 66) (7, 15, 20, 42) (10, 17, 21, 84) (13, 20, 21, 126) (13, 13, 24, 60) (12, 17, 25, 90) (7, 24, 25, 84) (14, 25, 25, 168) (3, 25, 26, 36) (17, 25, 26, 204) (17, 25, 28, 210) (20, 21, 29, 210) (6, 25, 29, 60) (17, 17, 30, 120) (11, 25, 30, 132) (5, 29, 30, 72) (8, 29, 35, 84) (15, 34, 35, 252) (25, 29, 36, 360) (19, 20, 37, 114) (15, 26, 37, 156) (13, 30, 37, 180) (12, 35, 37, 210) (24, 37, 37, 420) (16, 25, 39, 120) (17, 28, 39, 210) (25, 34, 39, 420) (10, 35, 39, 168) (29, 29, 40, 420) (13, 37, 40, 240) (25, 39, 40, 468) (15, 28, 41, 126) (9, 40, 41, 180) (17, 40, 41, 336) (18, 41, 41, 360) (29, 29, 42, 420) (15, 37, 44, 264) (17, 39, 44, 330) (13, 40, 45, 252) (25, 25, 48, 168) (29, 35, 48, 504) (21, 41, 50, 420) (39, 41, 50, 780) (26, 35, 51, 420) (20, 37, 51, 306) (25, 38, 51, 456) (13, 40, 51, 156) (27, 29, 52, 270) (25, 33, 52, 330) (37, 39, 52, 720) (15, 41, 52, 234) (5, 51, 52, 126) (25, 51, 52, 624) (24, 35, 53, 336) (28, 45, 53, 630) (4, 51, 53, 90) (51, 52, 53, 1170) (26, 51, 55, 660) (20, 53, 55, 528) (25, 39, 56, 420) (53, 53, 56, 1260) (33, 41, 58, 660) (41, 51, 58, 1020) (17, 55, 60, 462) (15, 52, 61, 336) (11, 60, 61, 330) (22, 61, 61, 660) (25, 52, 63, 630) (33, 34, 65, 264) (20, 51, 65, 408) (12, 55, 65, 198) (33, 56, 65, 924) (14, 61, 65, 420) (36, 61, 65, 1080) (16, 63, 65, 504) (32, 65, 65, 1008) (35, 53, 66, 924) (65, 65, 66, 1848) (21, 61, 68, 630) (43, 61, 68, 1290) (7, 65, 68, 210) (29, 65, 68, 936) (57, 65, 68, 1710) (29, 52, 69, 690) (37, 37, 70, 420) (9, 65, 70, 252) (41, 50, 73, 984) (26, 51, 73, 420) (35, 52, 73, 840) (48, 55, 73, 1320) (19, 60, 73, 456) (50, 69, 73, 1656) (25, 51, 74, 300) (25, 63, 74, 756) (35, 44, 75, 462) (29, 52, 75, 546) (32, 53, 75, 720) (34, 61, 75, 1020) (56, 61, 75, 1680) (13, 68, 75, 390) (52, 73, 75, 1800) (40, 51, 77, 924) (25, 74, 77, 924) (68, 75, 77, 2310) (41, 41, 80, 360) (17, 65, 80, 288) (9, 73, 80, 216) (39, 55, 82, 924) (35, 65, 82, 1092) (33, 58, 85, 660) (29, 60, 85, 522) (39, 62, 85, 1116) (41, 66, 85, 1320) (36, 77, 85, 1386) (13, 84, 85, 546) (41, 84, 85, 1680) (26, 85, 85, 1092) (72, 85, 85, 2772) (34, 55, 87, 396) (52, 61, 87, 1560) (38, 65, 87, 1140) (44, 65, 87, 1386) (31, 68, 87, 930) (61, 74, 87, 2220) (65, 76, 87, 2394) (53, 75, 88, 1980) (65, 87, 88, 2640) (41, 50, 89, 420) (28, 65, 89, 546) (39, 80, 89, 1560) (21, 82, 89, 840) (57, 82, 89, 2280) (78, 89, 89, 3120) (53, 53, 90, 1260) (17, 89, 90, 756) (37, 72, 91, 1260) (60, 73, 91, 2184) (26, 75, 91, 840) (22, 85, 91, 924) (48, 85, 91, 2016) (29, 75, 92, 966) (39, 85, 92, 1656) (34, 65, 93, 744) (39, 58, 95, 456) (41, 60, 95, 798) (68, 87, 95, 2850) (73, 73, 96, 2640) (37, 91, 96, 1680) (51, 52, 97, 840) (65, 72, 97, 2340) (26, 73, 97, 420) (44, 75, 97, 1584) (35, 78, 97, 1260) (75, 86, 97, 3096) (11, 90, 97, 396) (78, 95, 97, 3420)

46 22.3 Heron triangles with sides in arithmetic progression Heron triangles with sides in arithmetic progression We write s a = u, s b = v, and s c = w. a, b, c are in.p. if and only if u, v, w are in.p. Let u = v d and w = v + d. Then we require 3v 2 (v d)(v + d) to be a square. This means v 2 d 2 = 3t 2 for some integer t. Proposition Let d be a squarefree integer. If gcd(x, y, z) = 1 and x 2 +dy 2 = z 2, then there are integers m and n satisfying gcd(dm, n) = 1 such that (i) x = m 2 dn 2, y = 2mn, z = m 2 + dn 2 if m and dn are of different parity, or (ii) x = m2 dn 2, y = mn, z = m2 + n 2, 2 2 if m and dn are both odd. For the equation v 2 = d 2 +3t 2, we take v = m 2 +3n 2, d = m 2 3n 2, and obtain u = 6n 2, v = m 2 + 3n 2, w = 2m 2, leading to a = 3(m 2 + n 2 ), b = 2(m 2 + 3n 2 ), c = m 2 + 9n 2, for m, n of different parity and gcd(m, 3n) = 1. m n a d a a + d area If m and n are both odd, we obtain

47 806 The area of a triangle m n a d a a + d area

48 22.4 Indecomposable Heron triangles Indecomposable Heron triangles Heron triangle can be constructed by joining two integer right triangles along a common leg. eginning with two primitive Pythagorean triangles, by suitably magnifying by integer factors, we make two integer right triangles with a common leg. Joining them along the common leg, we obtain a Heron triangle. For example, (13, 14, 15; 84) = (12, 5, 13; 30) (12, 9, 15; 54) We may also excise (5, 12, 13) from (9, 12, 15), yielding (13, 4, 15; 24) = (12, 9, 15; 54) \ (12, 5, 13; 30). Does every Heron triangle arise in this way? We say that a Heron triangle is decomposable if it can be obtained by joining two Pythagorean triangles along a common side, or by excising a Pythagorean triangle from a larger one. Clearly, a Heron triangle is decomposable if and only if it has an integer height (which is not a side of the triangle). The first example of an indecomposable Heron triangle was obtained by Fitch Cheney. The Heron triangle (25, 34, 39; 420) is not decomposable because it does not have an integer height. Its heights are none an integer = 168 5, = , = ,

49 808 The area of a triangle Exercise 1. Find all shapes of Heron triangles that can be obtained by joining integer multiples of (3, 4, 5) and (5, 12, 13). 2. Find two indecomposable Heron triangles each having its longest sides two consecutive integers, and the shortest side not more than It is known that the smallest acute Heron triangle also has its longest sides two consecutive integers. Identify this triangle.

50 22.5 Heron triangle as lattice triangle Heron triangle as lattice triangle Heron triangle is decomposable if it can be decomposable into (the union or difference of) two Pythagorean triangles. It is decomposable if one or more heights are integers. decomposable Heron triangle can clearly be realized as a lattice triangle. It turns out that this is also true of the indecomposable one. Theorem Every Heron triangle can be realized as a lattice triangle. Example The indecomposable Heron triangle (25, 34, 39; 420) as a lattice triangle (15, 36) (30, 16) 34 (0, 0) Exercise 1. The triangle (5, 29, 30; 72) is the smallest Heron triangle indecomposable into two Pythagorean triangles. Realize it as a lattice triangle, with one vertex at the origin. 2. The triangle (15, 34, 35; 252) is the smallest acute Heron triangle indecomposable into two Pythagorean triangles. Realize it as a lattice triangle, with one vertex at the origin.

51 Chapter 23 Heron triangles 23.1 Heron triangles with area equal to perimeter Suppose the area of an integer triangle (a, b, c) is numerically equal to its perimeter. Write w = s a, v = s b, u = s c. Note that s = u + v + w. We require s(s a)(s b)(s c) = 4s 2. Equivalently, uvw = 4(u + v + w). There are at least two ways of rewriting this. (i) 1 uv + 1 uw + 1 vw = 1 4 ; (ii) u = 4(v+w) vw 4. We may assume w v u. From (i) w 2 vw 12 and we must have w 3. If w = 3, then v = 3 or 4. In neither case can u be an integer according to (ii). If w = 2, then uv = 2(2 + u + v), (u 2)(v 2) = 8, (v 2) 2 8; v = 3 or 4. Therefore, (u, v, w) = (10, 3, 2) or (6, 4, 2). If w = 1, then v 5 by (ii). lso, uv = 4(1 + u + v), (u 4)(v 4) = 20, (v 4) 2 20; v 8. Therefore, v = 5, 6, 7 or 8. Since (v, w) = (7, 1) does not give an integer u, we only have (u, v, w) = (24, 5, 1), (14, 6, 1), (9, 8, 1). Summary There are only five integer triangles with area equal to perimeter: (u, v, w) (1, 5, 24) (1, 6, 14) (1, 8, 9) (2, 3, 10) (2, 4, 6) (a, b, c) (6, 25, 29) (7, 15, 20) (9, 10, 17) (5, 12, 13) (6, 8, 10)

52 812 Heron triangles 23.2 Heron triangles with integer inradii Suppose an integer triangle (a, b, c) has inradius n, an integer. Let w = s a, v = s b, u = s c. We have uvw u + v + w = n2. If we assume u v w, then (i) w = n2 (u+v) uv n 2, (ii) n u 3n, and (iii) v n(n+ n 2 +u 2 ) u. Therefore, the number T(n) of integer triangles with inradius n is finite. Here are the beginning values of T(n): n T(n) Exercise 1. Find all Heron triangles with inradius Given an integer n, how many Pythagorean triangles are there with inradius n? nswer. 1 2 d(2n2 ). 3. Find all Heron triangles with inradius 3 2.

53 23.3 Division of a triangle into two subtriangles with equal incircles Division of a triangle into two subtriangles with equal incircles Given a triangle C, to locate a point P on the side C so that the incircles of triangles P and CP have equal radii. P C Suppose P : PC = k : 1 k, and denote the length of P by x. y Stewart s Theorem, x 2 = kb 2 + (1 k)c 2 k(1 k)a 2. Equating the inradii of the triangles P and CP, we have 2k c + x + ka = This latter equation can be rewritten as or from which c + x + ka k = c + x k k = Now substitution into (1) gives 2(1 k) b + x + (1 k)a. b + x + (1 k)a, (23.1) 1 k = b + x 1 k, (23.2) x + c 2x + b + c. x 2 (2x+b+c) 2 = (2x+b+c)[(x+c)b 2 +(x+b)c 2 ] (x+b)(x+c)a 2.

54 814 Heron triangles Rearranging, we have (x + b)(x + c)a 2 = (2x + b + c)[(x + c)b 2 + (x + b)c 2 x 2 [(x + b) + (x + c)]] = (2x + b + c)[(x + b)(c 2 x 2 ) + (x + c)(b 2 x 2 )] = (2x + b + c)(x + b)(x + c)[(c x) + (b x)] = (2x + b + c)(x + b)(x + c)[(b + c) 2x] = (x + b)(x + c)[(b + c) 2 4x 2 ]. From this, x 2 = 1 4 ((b + c)2 a 2 ) = 1 (b + c + a)(b + c a) = s(s a). 4 Let the excircle on the side C touch C at Y. Construct a semicircle on Y as diameter, and the perpendicular from the incenter I to C to intersect this semicircle at Q. P is the point on C such that P has the same length as Q. Z Q I P C Y K. W. Lau (Solution to Problem 1097, Crux Math., 13 (1987) ) has proved an interesting formula which leads to a simple construction of the point P. If the angle between the median D and the angle bisector X is θ, then D X = m a w a cosθ = s(s a).

55 23.3 Division of a triangle into two subtriangles with equal incircles 815 Proof. D D = X = C, 2 c C; b + ( c c X = 2 + = c 2 + ( c b + c b + c ) ) C + b2 a 2 c 2 2 c 2(b + c) C 2 + ca2 2(b + c) = 4c2 (b + c) + (b + 3c)(b 2 a 2 c 2 ) + 2ca 2 4(b + c) (a + b + c)(b + c a) = 4 = s(s a). Y X P D C This means if the perpendicular from X to D is extended to intersect the circle with diameter D at a point Y, then Y = s(s a). Now, the circle (Y ) intersects the side C at two points, one of which is the required point P. Here are two examples of Heron triangles with subdivision into two Heron triangles with equal inradii. (1) (a, b, c) = (15, 8, 17); CP = 6, P = 9. P = 10. The two small triangles have the same inradius 2.

56 816 Heron triangles C P (2) (a, b, c) = (51, 20, 65); P = 33, CP = 18. P = 34. The two small triangles have the same inradius 4. P C

57 23.4 Inradii in arithmetic progression Inradii in arithmetic progression (1) Triangle C below has sides 13, 14, 15. The three inradii are 2, 3, 4. C 2 D 4 (2) The following four inradii are in arithmetic progression. What is the shape of the large triangle? 3

58 818 Heron triangles 23.5 Heron triangles with integer medians It is an unsolved problem to find Heron triangles with integer medians. The triangle (a, b, c) = (136, 170, 174) has three integer medians. ut it is not a Heron triangle. It has an area. uchholz and Rathbun have found an infinite set of Heron triangles with two integer medians. Here is the first one. Let a = 52, b = 102, c = 146. This is a Heron triangle with area and two integer medians m b = and m c =. The third one, however, has irrational length: m a =.

59 23.6 Heron triangles with square areas Heron triangles with square areas Fermat has shown that there does not exist a Pythagorean triangle whose area is a perfect square. However, the triangle with sides 9, 10, 17 has area 36. In fact, there infinitely many primitive Heron triangles whose areas are perfect squares. Here is one family constructed by C. R. Maderer. 1 For each positive integer k, define a k = 20k 4 + 4k 2 + 1, b k = 8k 6 4k 4 2k 2 + 1, c k = 8k 6 + 8k k 2. Here, a k, b k < c k < a k + b k. The triangle with sides (a k, b k, c k ) has area ((2k)(2k 2 1)(2k 2 + 1)) 2. N. D. Elkies has constructed 2 an infinite sequence of solutions of the Diophantine equation C 4 = D 4, with gcd(,, C, D) = 1. The triangle with a = 4 + C 4, b = C 4 + 4, c = is a Heron triangle with area (CD) 2. The first example which Elkies constructed is = , = , C = , D = merican Mathematical Monthly, 98 (1991). 2 On C 4 = D 4, Math. Comp., 51 (1988)

60 Chapter 24 Triangles with sides and one altitude in.p Newton s solution Problem 29 of Isaac Newton s Lectures on lgebra ([Whiteside, pp ]) studies triangles whose sides and one altitude are in arithmetic progression. Newton considered a triangle C with an altitude DC. Clearly, DC is shorter than C and C. Setting C = a, C = x, DC = 2x a, and = 2a x, he obtained 16x 4 80ax a 2 x 2 10a 3 x + 25a 4 = 0. ( ) Divide this equation by 2x a and there will result 8x 3 36ax a 2 x 25a 3 = 0. Newton did not solve this equation nor did he give any numerical example. ctually, ( ) can be rewritten as (2x 3a) 3 + 2a 3 = 0, so that x = a 2 (3 3 2), the other two roots being complex. y taking a = 2, we may assume the sides of the triangles to be,,, and the altitude on the longest side to be. The angles of the triangles are,,.

61 822 Triangles with sides and one altitude in.p The general case Recalling the Heron triangle with sides 13, 14, 15 with altitude 12 on the side 14, we realize that these lengths can be in.p. in some other order. Note that the altitude in question is either the first or the second terms of the.p. (in increasing order). ssuming unit length for this altitude, and x > 0 for the common difference, we have either 1. the three sides of the triangles are 1 + x, 1 + 2x, and 1 + 3x, or 2. the sides of the triangles are 1 x, 1+x, and 1+2x, and the altitude on the shortest side is 1. In (1), the area of the triangle, by the Heron formula, is given by 2 = 3 16 (1 + 2x)2 (1 + 4x). On the other hand, = 1 2 the equations 1 (1 + kx) for k = 1, 2, 3. These lead to for k = 1: 48x x x 1 = 0, for k = 2: 48x x 2 + 8x 1 = 0, for k = 3: 48x x 1 = 0. The case k = 3 has been dealt with in Newton s solution. For k = 2, the polynomial factors as so that we have x =. This leads to the Heron triangle with sides 13, 14, 15, and altitude 12 on the side 14the triangles are,,. For k = 1, it is easy to see, using elementary calculus, that the polynomial 48x 3 +56x 2 +16x 1 has exactly one real root, which is positive. This gives a similarity class of triangle with the three sides and the altitude on the shortest side in.p. More detailed calculation shows that the angles of such triangles are,,. Now we consider (2), when the altitude in question is the second term of the.p. Instead of constructing an equation in x, we seek one such triangle with sides 15, z, z, and the altitude 16 + z on the

62 24.2 The general case 823 shortest side. y considering the area of the triangle in two different ways, we obtain the cubic equation z 3 120z + 16 = 0. ( ) This can be solved by writing z = 4 10sin θ for an angle θ. Using the trigonometric identity sin 3θ = 3 sinθ 4 sin 3 θ, we reduce this to sin 3θ = so that the positive roots of ( ) are the two numbers z =,. We obtain two similarity classes of triangles, respectively with angles and,,,,,. There are altogether five similarity classes of triangles whose three sides and one altitude, in some order, are in arithmetic progression.

63 Chapter 25 The Pell Equation 25.1 The equation x 2 dy 2 = 1 Let d be a fixed integer. We consider the Pell equation x 2 dy 2 = 1. Clearly, if d is negative or is a (positive) square integer, then the equation has only finitely many solutions. Theorem Let d be a nonsquare, positive integer. The totality of positive solutions of the Pell equation x 2 dy 2 = 1 form an infinite sequence (x n, y n ) defined recursively by x n+1 = ax n + dby n, y n+1 = bx n + ay n ; x 1 = a, y 1 = b, where (x 1, y 1 ) = (a, b) is the fundamental solution, with a, b positive and b smallest possible. Examples 1. The fundamental solution of the Pell equation x 2 2y 2 = 1 is (3,2). This generates an infinite sequence of nonnegative solutions (x n, y n ) defined by x n+1 = 3x n +4y n, y n+1 = 2x n +3y n ; x 0 = 1, y 0 = 0. The beginning terms are n x n y n

64 902 The Pell Equation 2. Fundamental solution (a, b) of x 2 dy 2 = 1 for d < 100: d a b d a b d a b Pell s equations whose fundamental solutions are very large: d a b The equation x y 2 = 1 arises from the famous Cattle problem of rchimedes, and has smallest positive solution x = , y =