Problem Solving and Recreational Mathematics

Paul Yiu

Department of Mathematics Florida Atlantic University

Summer 2012

Chapters 16–33

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 Contents

16 The golden ratio 601 16.1 Divisionofasegmentinthegoldenratio ...... 601 16.2 Theregularpentagon ...... 603 16.3 Construction of 36◦, 54◦, and 72◦ angles ...... 604 16.4 Themostnon-isoscelestriangle ...... 608

17 Medians and angle Bisectors 609 17.1 Apollonius’Theorem ...... 609 17.2 Anglebisectortheorem ...... 611 17.3 Theanglebisectors ...... 612 17.4 Steiner-LehmusTheorem ...... 613

18 Dissections 615 18.1 Dissection of the 6 6 square...... 615 18.2 Dissectionof a 7 ×7 squareintorectangles...... 617 18.3 Dissectarectangletoformasquare× ...... 619 18.4 Dissectionofasquareintothreesimilarparts ...... 620

19 Pythagorean triangles 701 19.1 PrimitivePythagoreantriples ...... 701 19.1.1 Rationalangles ...... 702 19.1.2 Some basic properties of primitive Pythagorean triples...... 702 19.2 A Pythagorean trianglewithan inscribedsquare . . . . 704 19.3 When are x2 px q bothfactorable? ...... 705 19.4 Dissection of− a square± into Pythagorean triangles . . . . 705

20 Integer triangles with a 60◦ or 120◦ angle 707 20.1 Integer triangles with a 60◦ angle ...... 707 iv CONTENTS

20.2 Integer triangles with a 120◦ angle ...... 710

21 Triangles with centroid on incircle 713 21.1 Construction ...... 714 21.2 Integertriangleswithcentroidontheincircle ...... 715

22 The area of a triangle 801 22.1 Heron’sformulafortheareaofatriangle ...... 801 22.2 Herontriangles...... 803 22.2.1 TheperimeterofaHerontriangleiseven . . . . 803 22.2.2 The area of a Heron triangle is divisible by 6 . . 803 22.2.3 Heron triangles with sides < 100 ...... 804 22.3 Heron triangles with sides in arithmetic progression . . 805 22.4 IndecomposableHerontriangles ...... 807 22.5 Herontriangleaslatticetriangle...... 809

23 Heron triangles 811 23.1 Herontriangleswithareaequaltoperimeter ...... 811 23.2 Herontriangleswithintegerinradii ...... 812 23.3 Division of a triangle into two subtriangles with equal incircles ...... 813 23.4 Inradiiinarithmeticprogression...... 817 23.5 Herontriangleswithintegermedians ...... 818 23.6 Herontriangleswithsquareareas ...... 819

24 TriangleswithsidesandonealtitudeinA.P. 821 24.1 Newton’ssolution ...... 821 24.2 Thegeneralcase ...... 822

25 The Pell Equation 901 25.1 The equation x2 dy2 =1 ...... 901 25.2 The equation x2 − dy2 = 1 ...... 903 25.3 The equation x2 − dy2 = c− ...... 903 − 26 Figurate numbers 907 26.1 Whichtriangularnumbersaresquares?...... 907 26.2 Pentagonalnumbers ...... 909 26.3 Almostsquaretriangularnumbers...... 911 26.3.1 Excessivesquaretriangularnumbers ...... 911 26.3.2 Deﬁcientsquaretriangularnumbers ...... 912 CONTENTS v

27 Special integer triangles 915 27.1 AlmostisoscelesPythagoreantriangles ...... 915 27.1.1 The generators of the almost isosceles Pythagorean triangles...... 916 27.2 Integer triangles (a, a +1, b) with a 120◦ angle . . . . . 917

28 Heron triangles 1001 28.1 Herontriangleswithconsecutivesides ...... 1001 28.2 Heron triangles with two consecutivesquare sides . . . 1002

29 Squares as sums of consecutive squares 1005 29.1 Sumofsquaresofnaturalnumbers ...... 1005 29.2 Sumsofconsecutivesquares: oddnumbercase . . . . . 1008 29.3 Sumsofconsecutivesquares: evennumbercase . . . . 1010 29.4 Sumsofpowersofconsecutiveintegers ...... 1012

30 Lucas’ problem 1013 30.1 Solution of n(n + 1)(2n +1)=6m2 for even n . . . . 1013 30.2 The Pell equation x2 3y2 =1 revisited ...... 1014 30.3 Solution of n(n + 1)(2−n +1)=6m2 for odd n . . . . .1015

31 Some geometry problems 1101

32 Basic geometric constructions 1109 32.1 Somebasicconstructionprinciples ...... 1109 32.2 Geometricmean ...... 1110 32.3 Harmonicmean ...... 1111 32.4 A.M G.M. H.M...... 1112 ≥ ≥ 33 Construction of a triangle from three given points 1115 33.1 Someexamples ...... 1115 33.2 Wernick’sconstructionproblems ...... 1117 Chapter 16

The golden ratio

16.1 Division of a segment in the golden ratio

Given a segment AB, a point P in the segment is said to divide it in 2 AP √5+1 the golden ratio if AP = P B AB. Equivalently, P B = 2 . We shall denote this golden ratio by ϕ· . It is the positive root of the quadratic equation x2 = x +1.

A P B A P B

Construction 16.1 (Division of a segment in the golden ratio). Given a segment AB, (1) draw a right triangle ABM with BM perpendicular to AB and half in length, (2) mark a point Q on the hypotenuse AM such that MQ = MB, (3) mark a point P on the segment AB such that AP = AQ. Then P divides AB into the golden ratio. 602 The golden ratio

Suppose P B has unit length. The length ϕ of AP satisﬁes

ϕ2 = ϕ +1. This equation can be rearranged as

1 2 5 ϕ = . − 2 4 Since ϕ> 1, we have 1 ϕ = √5+1 . 2 Note that AP ϕ 1 2 √5 1 = = = = − . AB ϕ +1 ϕ √5+1 2 This explains the construction above. 16.2 The regular pentagon 603

16.2 The regular pentagon

Consider a regular pentagon ACBDE. It is clear that the ﬁve diagonals all have equal lengths. Note that (1) ∠ACB = 108◦, (2) triangle CAB is isosceles, and (3) ∠CAB = ∠CBA = (180◦ 108◦) 2=36◦. − ÷ In fact, each diagonal makes a 36◦ angle with one side, and a 72◦ angle with another. C

A B P

E D

It follows that (4) triangle PBC is isosceles with ∠PBC = ∠P CB = 36◦, (5) ∠BPC = 180◦ 2 36◦ = 108◦, and (6) triangles CAB and− PBC× are similar. Note that triangle ACP is also isosceles since (7) ∠ACP = ∠AP C = 72◦. This means that AP = AC. Now, from the similarity of CAB and PBC, we have AB : AC = BC : P B. In other words AB AP = AP P B, or AP 2 = AB P B. This means that P divides AB in· the golden· ratio. · 604 The golden ratio

16.3 Construction of 36◦, 54◦, and 72◦ angles

Angles of sizes 36◦, 54◦, and 72◦ can be easily constructed from a segment divided in the golden ratio.

36 36 36◦ ◦ B A P ϕ

ϕ cos36◦ = 2 .

54◦ 72◦

36◦ B A P 1

54◦ 72◦ B A P ϕ D

1 cos72◦ = 2ϕ . 16.3 Construction of 36◦, 54◦, and 72◦ angles 605

Exercise

1. Three equal segments A1B1, A2B2, A3B3 are positioned in such a way that the endpoints B2, B3 are the midpoints of A1B1, A2B2 respectively, while the endpoints A1, A2, A3 are on a line perpendicular to A1B1. Show that A2 divides A1A3 in the golden ratio.

B2 B3

A1 A2 A3

2. Given an equilateral triangle ABC, erect a square BCDE externally on the side BC. Construct the circle, center C, passing through E, to intersect the line AB at F . Show that B divides AF in the golden ratio.

A B F 3. Given a segment AB, erect a square on it, and an adjacent one with base BC. If D is the vertex above A, construct the bisector of angle ADC to intersect AB at P . Calculate the ratio AP : P B.

A P B C 606 The golden ratio

4. Let D and E be the midpoints of the sides AB and AC of an equilateral triangle ABC. If the line DE intersects the circumcircle of ABC at F , calculate the ratio DE : EF . A

D E F

B C

5. Given a segment AB with midpoint M, let C be an intersection of the circles A(M) and B(A), and D the intersection of C(A) and A(C) inside B(A). Prove that the line CD divides AB in the golden ratio.

D A B M P

6. The three small circles are congruent. Show that each of the ratios OA T X ZT AB , XY , TO . is equal to the golden ratio.

Z X Y

T B O A 16.3 Construction of 36◦, 54◦, and 72◦ angles 607

7. 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? 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 com- paring the lengths of a pair of sides: a b b c a c , , , , , . b a c b 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 16.1. A number η is the ratio of nonisoscelity of a triangle if and only if ϕ 1 < η 1. − ≤ a 1 b 1 1 Proof. First note that if b = r < 1, then r = a > 1. Since 2 (r + r ) > 1, 1 it follows that r is closer to 1 than r . If a b c are the lengths of the three sides of a triangle, the ratio of nonisoscelity≤ ≤ is a b η = max , . b c Since a + b>c, we have a a + b c 1 η +1 +1= > . ≥ b b b ≥ η Therefore, η2 + η > 1. Since the roots of x2 + 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− t2 + t> 1 and 1 1∈ − t +1 > t . The numbers t 1 t form the sides of a triangle with ratio of nonisoscelity equal to t≤. ≤ Chapter 17

Medians and angle Bisectors

17.1 Apollonius’ Theorem

Theorem 17.1. Given triangle ABC, let D be the midpoint of BC. The length of the median AD is given by AB2 + AC2 = 2(AD2 + BD2).

B D C Proof. Applying the law of cosines to triangles ABD and ACD, and noting that cos ADB = cos ADC, we have − AB2 = AD2 + BD2 2AD BD cos ADB; − · · AC2 = AD2 + CD2 2AD CD cos ADC, − · · = AD2 + BD2 +2AD BD cos ADB. · · The result follows by adding the ﬁrst and the third lines.

If ma denotes the length of the median on the side BC, 1 m2 = (2b2 +2c2 a2). a 4 − 610 Medians and angle Bisectors

Example 17.1. Suppose the medians BE and CF of triangle ABC are perpendicular. This means that BG2 + CG2 = BC2, where G is the 4 2 4 2 centroid of the triangle. In terms of the lengths, we have 9 mb + 9 mc = 2 2 2 2 2 2 2 2 2 2 2 a ; 4(mb + mc)=9a ; (2c +2a b )+(2a +2b c )=9a ; b2 + c2 =5a2. − − This relation is enough to describe, given points B and C, the locus of A for which the medians BE and CF of triangle ABC are perpendicular. Here, however, is a very easy construction: From b2 + c2 =5a2, we have m2 = 1 (2b2 +2c2 a2) = 9 a2; m = 3 a. The locus of A is a 4 − 4 a 2 the circle with center at the midpoint of BC, and radius 3 BC. 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. Calcu- late the lengths of its medians. 3. Triangle ABC has sidelengths a = 17, b = 13, c = 7. Show that the medians are in the proportions of the sides.

7 13

B 17 C 17.2 Angle bisector theorem 611

17.2 Angle bisector theorem

Theorem 17.2 (Angle bisector theorem). The bisectors of an angle of a triangle divide its opposite side in the ratio of the remaining sides. If AX and AX′ respectively the internal and external bisectors of angle BAC, then BX : XC = c : b and BX′ : X′C = c : b. − Z

Z′ b c

B X C X′

Proof. Construct lines through C parallel to the bisectors AX and AX′ to intersect the line AB at Z and Z′. (1) Note that ∠AZC = ∠BAX = ∠XAC = ∠ACZ. This means AZ = AC. Clearly, BX : XC = BA : AZ = BA : AC = c : b. (2) Similarly, AZ′ = AC, and BX′ : X′C = BA : AZ′ = BA : AC = c : b. − − 612 Medians and angle Bisectors

17.3 The angle bisectors

The length of the internal bisector of angle A is 2bc A t = cos , a b + c 2

ta′ ta

Q B P C and that of the external bisector of angle A is 2bc A t′ = sin . a b c 2 | − | Example 17.2. The triangle (a, b, c) = (125, 154, 169) is a Heron triangle with rational angle bisectors. Its area is 9240. 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. 3 4. 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.

1tb = a = 6. 2ta = b = 24. 3 975 26208 12600 ta = 7 , tb = 253 , tc = 209 . 17.4 Steiner-Lehmus Theorem 613

17.4 Steiner-Lehmus Theorem

Theorem 17.3. A triangle is isosceles if it has two equal angle bisectors.

F Y Z E

B C

Proof. Suppose the bisectors BE = CF , but triangle ABC not isosceles. We may assume ∠B < ∠C. Construct parallels to BC through E and F to intersect AB and AC at Z and Y respectively. (1) In the isosceles triangles ZBE and Y CF with equal bases BE and CF , ∠ZBE < ∠Y CF = EZ EC. This clearly implies YF

tb′ C

B a

tc′

Z 614 Medians and angle Bisectors

1 (1) From tb′ = a, we have 2 (π β)=2γ; β +4γ = π. 1 − (2) From tc′ = a, we have 2 (π γ)= π 2(π β)=2β π; 4β+γ =3π. 11π π − −π − − From these, β = 15 , γ = 15 , and α = 5 .

Exercise

1. The bisector ta and the external bisector tb′ of triangle ABC satisfy 4 ta = tb′ = c. Calculate the angles of the triangle.

tb′ ta X

A c B 2. AD and BE are angle bisectors of triangle ABC, with D on BC and E on AC. Suppose that AD = AB and BE = BC. Determine the angles of the triangle. Show that if CF is the external bisector of angle C, then CF = CA. 5 A

B D C

4 2π 7π 2π Answer: α = 15 , β = 15 , γ = 5 . 5 2π 6π 5π Answer: α = 13 ,β = 13 , γ = 13 . Chapter 18

Dissections

18.1 Dissection of the 6 6 square × Since 1+2+3+4+5+6+7+8=62, it is reasonable to look for a dissection of a 6 6 square into eight pieces with areas 1, 2, ..., 8 square units. ×

1. 5 cuts: Construct a line through A which completes the dissection.

2. 4 straight cuts: Construct a line through B which completes the dissection.

B 616 Dissections

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

C 18.2 Dissection of a 7 7 square into rectangles 617 × 18.2 Dissection of a 7 7 square into rectangles × Dissect a square into seven rectangles of different shapes but all having the same area. 1

R7 R2 R3

Suppose each side of the square has length 7, and for i = 1, 2 ..., 7, the rectangle Ri has (horizontal) length ai and (vertical) height bi. Set- ting a1 = r, we compute easily in succession, b1, b2, a2, a3, b3, b4, a4, a5, b5, using aibi =7. Now, a =7 a a , 6 − 2 − 3 b =7 b b , 6 − 1 − 5 a =7 a a a , 7 − 2 − 3 − 5 b =7 b b b . 7 − 1 − 4 − 6

rectangle ai bi R 7 1 r r r 7(r 1) R − 2 r 1 r R − 7 3 7 r 7 r 7−r 7(6− r) R − 4 6−r 7 r 5(7r −7 r2) 7(6 r)(−r 1)3 R − − − − 5 (6 r)(r 1) 5(7r 7 r2 −r(r 2)− 7(r 1)(29−r −35 4r2 ) R − − − − 6 r 1 5r(7r 7 r2) (7 r)(6 r)(28− 27r+4r2 ) 7(6− r−) R − − − − 7 5 r 5(7 r)(7r 7 r2) − − − −

1Problem 2567, Journal of Recreational Math., . A dissection into seven rectangles was given by Hess and Ibstedt. Markus G¨otz showed that no solution with n rectangles exists for 2 n 6. ≤ ≤ 618 Dissections

The relations a6b6 =7 and a7b7 =7 both reduce to (2r 7)(2r2 14r +15)= 0. − − 7 7 √19 Since r = 2 or −2 would make a7, b7 negative, we must have r = 7+√19 2 . The dimensions of the rectangles are as follows.

rectangle ai bi R 7+√19 77 √19 1 2 −15 R 7(8+√19) 8 √19 2 15 −3 R 7(7+√19) 7 √19 3 15 −2 8+√19 7(8 √19) R − 4 3 15 R 5 21 5 3 5 5(1+√19) 7( 1+√19) R − 6 6 15 5( 1+√19) 7(1+√19) R − 7 6 15 18.3Dissectarectangletoformasquare 619

18.3 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 −

√ab a √ab This is clear since − b √b x =(a √ab) = √a(√a √b) = √ab b. − · √ab − · √a −

a √ab −

√ab 620 Dissections

18.4 Dissection of a square into three similar parts

Dissect a square into three similar parts, no two congruent.

2 D 1 x C

x2 x +1 −

A B x2 +1 Solution. x2 +1= x(x2 x + 1); x3 2x2 + x 1=0. − − − Exercise 1. Show that the area of the largest rectangle is x4. Solution. Thisareais

(x2 + 1)(x2 x + 1) − = x4 x3 + x2 + x2 x +1 − − = x4 x3 +2x2 x +1 − − = x4. 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

Q P Y

A x X B

2. Let P and Q be points on the sides AB and BC of rectangle ABCD. D C

T1 T2

A P B (a) Show that if the areas of triangles AP D, BP Q and CDQ are equal, then P and Q divide the sides in the golden ratio. (b) If, in addition, DP = P Q, show that the rectangle is golden, AB ∠ i.e., BC = ϕ, and DP Q is a right angle. Chapter 19

Pythagorean triangles

19.1 Primitive Pythagorean triples

A Pythagorean triangle is one whose sidelengths are integers. An easy way to construct Pythagorean triangles is to take two distinct positive integers m > n and form

(a, b, c)=(2mn, m2 n2, m2 + n2). − Then, a2 + b2 = c2. We call such a triple (a, b, c) a Pythagorean triple. The Pythagorean triangle has perimeter p = 2m(m + n) and area A = mn(m2 n2). −

2 n + 2 2mn m

A C m2 n2 − A 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. 702 Pythagorean triangles

19.1.1 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 m2 n2 2mn A m2+n2 m2+−n2 m2 n2 m2 n2 2mn m2−n2 B m2+−n2 m2+n2 2mn−

A B More basic than these is the fact that tan 2 and tan 2 are rational: A n B m n tan = , tan = − . 2 m 2 m + n

This is easily seen from the following diagram showning the incircle of the right triangle, which has r =(m n)n.

− B

n ) n + m ( (m + n)n

) n I r − m ( m r (m − n)n

A C m(m − n) (m − n)n

19.1.2 Some basic properties of primitive Pythagorean triples 1. Exactly one leg is even. 2. Exactly one leg is divisible by 3. 3. Exactly one side is divisible by 5. 4. The area is divisible by 6. 19.1 Primitive Pythagorean triples 703

Appendix: 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, 145 10, 1 99, 20, 101 10, 3 91, 60, 109 10, 7 51, 140, 149 10, 9 19, 180, 181 11, 2 117, 44, 125 11, 4 105, 88, 137 11, 6 85, 132, 157 11, 8 57, 176, 185 11, 10 21, 220, 221 12, 1 143, 24, 145 12, 5 119, 120, 169 12, 7 95, 168, 193 12, 11 23, 264, 265 13, 2 165, 52, 173 13, 4 153, 104, 185 13, 6 133, 156, 205 13, 8 105, 208, 233 13, 10 69, 260, 269 13, 12 25, 312, 313 14, 1 195, 28, 197 14, 3 187, 84, 205 14, 5 171, 140, 221 14, 9 115, 252, 277 14, 11 75, 308, 317 14, 13 27, 364, 365 15, 2 221, 60, 229 15, 4 209, 120, 241 15, 8 161, 240, 289 15, 14 29, 420, 421 16, 1 255, 32, 257 16, 3 247, 96, 265 16, 5 231, 160, 281 16, 7 207, 224, 305 16, 9 175, 288, 337 16, 11 135, 352, 377 16, 13 87, 416, 425 16, 15 31, 480, 481 17, 2 285, 68, 293 17, 4 273, 136, 305 17, 6 253, 204, 325 17, 8 225, 272, 353 17, 10 189, 340, 389 17, 12 145, 408, 433 17, 14 93, 476, 485 17, 16 33, 544, 545 18, 1 323, 36, 325 18, 5 299, 180, 349 18, 7 275, 252, 373 18, 11 203, 396, 445 18, 13 155, 468, 493 18, 17 35, 612, 613 19, 2 357, 76, 365 19, 4 345, 152, 377 19, 6 325, 228, 397 19, 8 297, 304, 425 19, 10 261, 380, 461 19, 12 217, 456, 505 19, 14 165, 532, 557 19, 16 105, 608, 617 19, 18 37, 684, 685 20, 1 399, 40, 401 20, 3 391, 120, 409 20, 7 351, 280, 449 20, 9 319, 360, 481 20, 11 279, 440, 521 20, 13 231, 520, 569 20, 17 111, 680, 689 20, 19 39, 760, 761 21, 2 437, 84, 445 21, 4 425, 168, 457 21, 8 377, 336, 505 21, 10 341, 420, 541 21, 16 185, 672, 697 21, 20 41, 840, 841 22, 1 483, 44, 485 22, 3 475, 132, 493 22, 5 459, 220, 509 22, 7 435, 308, 533 22, 9 403, 396, 565 22, 13 315, 572, 653 22, 15 259, 660, 709 22, 17 195, 748, 773 22, 19 123, 836, 845 22, 21 43, 924, 925 23, 2 525, 92, 533 23, 4 513, 184, 545 23, 6 493, 276, 565 23, 8 465, 368, 593 23, 10 429, 460, 629 23, 12 385, 552, 673 23, 14 333, 644, 725 23, 16 273, 736, 785 23, 18 205, 828, 853 23, 20 129, 920, 929 24, 1 575, 48, 577 24, 5 551, 240, 601 24, 7 527, 336, 625 24, 11 455, 528, 697 24, 13 407, 624, 745 24, 17 287, 816, 865 24, 19 215, 912, 937 25, 2 621, 100, 629 25, 4 609, 200, 641 25, 6 589, 300, 661 25, 8 561, 400, 689 25, 12 481, 600, 769 25, 14 429, 700, 821 25, 16 369, 800, 881 25, 18 301, 900, 949 26, 1 675, 52, 677 26, 3 667, 156, 685 26, 5 651, 260, 701 26, 7 627, 364, 725 26, 9 595, 468, 757 26, 11 555, 572, 797 26, 15 451, 780, 901 26, 17 387, 884, 965 27, 2 725, 108, 733 27, 4 713, 216, 745 27, 8 665, 432, 793 27, 10 629, 540, 829 27, 14 533, 756, 925 27, 16 473, 864, 985 28, 1 783, 56, 785 28, 3 775, 168, 793 28, 5 759, 280, 809 28, 9 703, 504, 865 28, 11 663, 616, 905 28, 13 615, 728, 953 29, 2 837, 116, 845 29, 4 825, 232, 857 29, 6 805, 348, 877 29, 8 777, 464, 905 29, 10 741, 580, 941 29, 12 697, 696, 985 30, 1 899, 60, 901 30, 7 851, 420, 949 31, 2 957, 124, 965 31, 4 945, 248, 977 31, 6 925, 372, 997 704 Pythagorean triangles

19.2 A Pythagorean triangle with an inscribed square

How many matches of equal lengths are required to make up the following conﬁguration?

Suppose the shapeof the right triangleis givenby a primitive Pythagorean triple (a, b, c). The lengthof a side of the square must be a commonmul- tiple 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? 19.3 When are x2 px q bothfactorable? 705 − ± 19.3 When are x2 px q both factorable? − ± For integers p and q, the quadratic polynomials x2 px+q and x2 px q both factorable (over integers) if and only if p2 4−q and p2 +4q are− both− squares. Thus, p and q are respectively the hypotenuse− and area of a Pythagorean triangle.

p2 +4q =(a + b)2, p2 4q =(b c)2. − −

b a

p p b b − a

b − a q

a b c = p q x2 − px + q x2 + px − q 3 4 5 6 (x − 2)(x − 3) (x − 1)(x + 6) 5 12 13 30 (x − 3)(x − 10) (x − 2)(x + 15) 8 15 17 60 (x − 5)(x − 12) (x − 3)(x + 20) The roots of x2 px+q are s a and s b, while thoseof x2+px q are s c and s. Here,− s is the semiperimeter− − of the Pythagorean triangle.− − − 19.4 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. 255 105

136 289

375 360 424 224

360 706 Pythagorean triangles

Exercise 1. A 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 A, 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?

A 60 B

P 60 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 3333 3. ··· 7. 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 Chapter 20 60 Integer triangles with a ◦ or 120 ◦ angle

20.1 Integer triangles with a 60◦ angle

If triangle ABC has C = 60◦, then

c2 = a2 ab + b2. (20.1) −

Integer triangles with a 60◦ angle therefore correspond to rational points in the ﬁrst quadrant on the curve

x2 xy + y2 =1. (20.2) − Note that the curve contains the point P = ( 1, 1). By 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 − − 2t 1 t(2 t) (x, y)= − , − , t2 t +1 t2 t +1 − − 1 which is in the ﬁrst quadrant if 2 < t 2. By symmetry, we may 1 ≤ simply take 2 < t 1 to avoid repetition. q ≤ Putting t = p for relatively prime integers p, q, and clearing denomi- 708 Integertriangleswitha 60◦ or 120◦ angle

P nators, we obtain

a =p(2q p), − b =q(2p q), − c =p2 pq + q2, − 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) 20.1 Integer triangles with a 60◦ angle 709

Exercise 1. A 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.

b The answer to this problem is given by

a + b √a2 ab + b2 x = − − . 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. Another example is a = 16, b = 21 with x =3.

2. 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.

1Answer: a, b, c, with gcd(p + q, 6) = 3. 710 Integertriangleswitha 60◦ or 120◦ angle

20.2 Integer triangles with a 120◦ angle

If triangle ABC has C = 120◦, then

c2 = a2 + ab + b2. (20.3)

Integer triangles with a 120◦ angle therefore correspond to rational points in the ﬁrst quadrant on the curve

x2 + xy + y2 =1. (20.4)

Note that the curve contains the point Q = ( 1, 0). By 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 − 1 t2 t(2 + t) Q(t)= − , , t2 + t +1 t2 + t +1 which is in the ﬁrst quadrant if 0

a =p2 q2, − b =q(2p + q), c =p2 + pq + q2, with 0

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) 712 Integertriangleswitha 60◦ or 120◦ angle

Exercise

1. An 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 = a2 + ab + b2. a2 1 From this, b = 2 −a . For no positive integer a is b positive. There is no such triangle.−

2. Give an example of an integer triangle with 120◦ whose adjacent sides are consecutive integers.

3. In triangle ABC, α = 120◦. AX is the bisector of angle A. Show 1 1 1 that t = b + c .

c b t

B X C

4. In the diagram below, ABX, BCY , and CDZ are equilateral tri- ∠ 1 1 1 2 angles. Suppose XYZ = 120◦. Show that b = a + c .

A a B b C c D

2Hint: Extend ZY to intersect AB at T . Show that T C = a. Chapter 21

Triangles with centroid on incircle

G′

I G

B X D C

Suppose the centroid G of triangle ABC lies on the incircle. The median AD intersects the incircle at another point G′. If AG′ = k AD, then by the intersecting chords theorem, we have · 2 (i) AY = AG′ AG, and 2 · (ii) DX = DG DG′. 2 · 1 Clearly, AG = 3 ma, DG = 3 ma. If AG′ = k ma, then DG′ = (1 k)ma. · 1 − a a 1 1 Since AY = 2 (b+c a) and DX = 2 XB = 2 2 (c+a b)= 2 (b c), these lead to − − − − −

1 2 (b + c a)2 = m k m , 4 − 3 a · · a 1 1 (b c)2 = m (1 k)m . 4 − 3 a · − a 714 Triangles with centroid on incircle

Eliminating k, we have

2 4m2 = 3((b + c a)2 + 2(b c)2). · a − − 2 2 2 2 From Apollonius’ theorem, 4ma = 2b +2c a . Simplifying, we obtain − 5(a2 + b2 + c2) 6(bc + ca + ab)=0. −

Theorem 21.1. The centroid of triangle (a, b, c) lies on its incircle if and only if 5(a2 + b2 + c2) 6(bc + ca + ab)=0. − 21.1 Construction

Let A0B0C0 be an equilateral triangle with center I0, and C the image h 1 of the incircle under the homothety I0, 2 . For every point P on the circle C, let X, Y , Z be the pedals on the sides of A0B0C0. The triangle with sides PX, PY , PZ has its centroid on its incircle. 21.2Integertriangleswithcentroidontheincircle 715

21.2 Integer triangles with centroid on the incircle

a b By putting x = c and y = c , we have

5x2 6xy +5y2 6x 6y +5=0. − − − 3 3 √ 1 √ This is an ellipse with center 2 , 2 and semiaxes 2 and 2 2. It contains the rational point Q(1, 2). The line through (1, 2) with slope 2+t 2+t 2 t has equation y 2= 2 t (x 1), or (t+2)x+(t 2)y (3t 2)=0. − − − − − − −

It intersects the conic at

1 2t +2t2 2 2t + t2 − , − . 1+ t2 1+ t2

q Writing t = p for relatively prime integers p and q, we have

a = p2 2pq +2q2, − b = 2p2 2pq + q2, − c = p2 + q2.

Here are some examples, with relatively prime integers p and q satisfying q

p q a b c 2 1 2 5 5 3 2 5 10 13 4 3 10 17 25 5 3 13 29 34 5 4 17 26 41 6 5 26 37 61 7 4 25 58 65 7 5 29 53 74 7 6 37 50 85 8 5 34 73 89 8 7 50 65 113 9 5 41 97 106 9 7 53 85 130 9 8 65 82 145 10 7 58 109 149 10 9 82 101 181

Exercise 1. Find the (shape of the) triangle which has a median trisected by the incircle. Chapter 22

The area of a triangle

22.1 Heron’s formula for the area of a triangle

Theorem 22.1. The area of a triangle of sidelengths a, b, c is given by

= s(s a)(s b)(s c), △ − − − 1 p where s = 2 (a + b + c).

I ra

A C Y Y ′ s − b s − c s − a Proof. Consider the incircle and the excircle on the opposite side of A. From the similarity of triangles AIZ and AI′Z′, r s a = − . ra s

From the similarity of triangles CIY and I′CY ′, r r =(s b)(s c). · a − − 802 The area of a triangle

From these, (s a)(s b)(s c) r = − − − , r s and the area of the triangle is

= rs = s(s a)(s b)(s c). △ − − − p

Exercise 1. Prove that 1 2 = (2a2b2 +2b2c2 +2c2a2 a4 b4 c4). △ 16 − − − 22.2 Heron triangles 803

22.2 Heron triangles

A Heron triangle is an integer triangle whose area is also an integer.

22.2.1 The perimeter of a Heron triangle is even Proposition 22.2. 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 a4, b4, c4 is congruent to 0 or 1, according as the number is even or odd. To render in (??) the sum 2a2b2 +2b2c2 +2c2a2 a4 b4 c4 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.

22.2.2 The area of a Heron triangle is divisible by 6 Proposition 22.3. 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 −1+1+( 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. 804 The area of a triangle

22.2.3 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) 22.3 Heron triangles with sides in arithmetic progression 805

22.3 Heron triangles with sides in arithmetic progression

We write s a = u, s b = v, and s c = w. − − − a, b, c are in A.P. if and only if u, v, w are in A.P. Let u = v d and w = v + d. Then we require 3v2(v d)(v + d) to be a square.− This means v2 d2 =3t2 for some integer−t. − Proposition 22.4. Let d be a squarefree integer. If gcd(x, y, z)=1 and x2+dy2 = z2, then there are integers m and n satisfying gcd(dm, n)=1 such that (i)

x = m2 dn2, y =2mn, z = m2 + dn2 − if m and dn are of different parity, or (ii) m2 dn2 m2 + n2 x = − , y = mn, z = , 2 2 if m and dn are both odd.

For the equation v2 = d2 +3t2, we take v = m2 +3n2, d = m2 3n2, and obtain u =6n2, v = m2 +3n2, w =2m2, leading to −

a = 3(m2 + n2), b = 2(m2 +3n2), c = m2 +9n2,

for m, n of different parity and gcd(m, 3n)=1.

m n a d a a + d area − 12 15 14 13 84 1 4 51 38 25 456 2 1 15 26 37 156 2 5 87 74 61 2220 3 2 39 62 85 1116 3 4 75 86 97 3096 4 1 51 98 145 1176 4 5 123 146 169 8760 5 2 87 158 229 4740 5 4 123 182 241 10920

If m and n are both odd, we obtain 806 The area of a triangle

m n a d a a + d area − 113 4 5 6 1 5 39 28 17 210 3 1 15 28 41 126 3 5 51 52 53 1170 5 1 39 76 113 570 22.4 Indecomposable Heron triangles 807

22.4 Indecomposable Heron triangles

A Heron triangle can be constructed by joining two integer right triangles along a common leg. Beginning 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). ∪

13 15 15 12 12 13

5 5 9 4

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 840 168 840 420 840 280 = , = , = , 25 5 34 17 39 13 none an integer. 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 10. 3. It is knownthat the smallest acute Heron triangle also has its longest sides two consecutive integers. Identify this triangle. 22.5 Heron triangle as lattice triangle 809

22.5 Heron triangle as lattice triangle

A 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. A decomposable Heron triangle can clearly be realized as a lattice triangle. It turns out that this is also true of the indecomposable one. Theorem 22.5. Every Heron triangle can be realized as a lattice triangle. Example 22.1. The indecomposable Heron triangle (25, 34, 39; 420) as a lattice triangle

(15, 36)

39 (30, 16)

(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. 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. 1 1 1 1 (i) uv + uw + vw = 4 ; 4(v+w) (ii) u = vw 4 . We may− assume w v u. From (i) w2 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). Also, 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) 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 = n2. u + v + w

n2(u+v) If we assume u v w, then (i) w = uv n2 , ≤ ≤ − (ii) n u √3n, and ≤ ≤ √ 2 2 (iii) v n(n+ n +u ) . ≤ u Therefore, the number T (n) of integer triangles with inradius n is ﬁnite. Here are the beginning values of T (n):

n 123 4 5 6 7 8 910 T (n) 1 5 13 18 15 45 24 45 51 52 ··· ···

Exercise 1. Find all Heron triangles with inradius 1. 2. Given an integer n, how many Pythagorean triangles are there with inradius n? 1 2 Answer. 2 d(2n ). 3 3. Find all Heron triangles with inradius 2 . 23.3 Division of a triangle into two subtriangles with equal incircles 813

23.3 Division of a triangle into two subtriangles with equal incircles

Given a triangle ABC, to locate a point P on the side BC so that the incircles of triangles ABP and ACP have equal radii.

b b

b b B P C

Suppose BP : PC = k : 1 k, and denote the length of AP by x. By Stewart’s Theorem, −

x2 = kb2 + (1 k)c2 k(1 k)a2. − − − Equating the inradii of the triangles ABP and ACP , we have 2k 2(1 k) △ = − △ . c + x + ka b + x + (1 k)a − This latter equation can be rewritten as c + x + ka b + x + (1 k)a = − , (23.1) k 1 k − or c + x b + x = , (23.2) k 1 k − from which x + c k = . 2x + b + c

Now substitution into (1) gives x2(2x + b + c)2 = (2x + b + c)[(x + c)b2 +(x + b)c2] (x + b)(x + c)a2. − 814 Heron triangles

Rearranging, we have (x + b)(x + c)a2 = (2x + b + c)[(x + c)b2 +(x + b)c2 x2[(x + b)+(x + c)]] = (2x + b + c)[(x + b)(c2 x2)+(x−+ c)(b2 x2)] = (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 4x2]. − − From this, 1 1 x2 = ((b + c)2 a2)= (b + c + a)(b + c a)= s(s a). 4 − 4 − −

Let the excircle on the side BC touch AC at Y . Construct a semicircle on AY as diameter, and the perpendicular from the incenter I to AC to intersect this semicircle at Q. P is the point on BC such that AP has the same length as AQ. Z

b Q

A b

b I b b

b b b b b C B P

b Y

K. W. Lau (Solution to Problem 1097, b Crux Math., 13 (1987) 135– 136) has proved an interesting formula which leads to a simple construction of the point P . If the angle between the median AD and the angle bisector AX is θ, then −−→AD −−→AX = m w cos θ = s(s a). · a · a · − 23.3 Division of a triangle into two subtriangles with equal incircles 815

Proof. 1 −−→AD = −→AB + −−→BC, 2 c −−→AX = −→AB + −−→BC; b + c 1 c c −−→AD −−→AX = AB 2 + + −→AB −−→BC + BC 2 · | | 2 b + c · 2(b + c)| | 1 c b2 a2 c2 ca2 = c2 + + − − + 2 b + c · 2 2(b + c) 4c2(b + c)+(b +3c)(b2 a2 c2)+2ca2 = − − 4(b + c) (a + b + c)(b + c a) = − 4 = s(s a). −

b A

b Y

b b b b b B X P D C

This means if the perpendicular from X to AD is extended to intersect the circle with diameter AD at a point Y , then AY = s(s a). Now, the circle A(Y ) intersects the side BC at two points, one of which− is the required point P . 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 B = 9. AP = 10. The two small triangles have the same inradius 2. 816 Heron triangles

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

B P C 23.4 Inradii in arithmetic progression 817

23.4 Inradii in arithmetic progression

(1) Triangle ABC below has sides 13, 14, 15. The three inradii are 2, 3, 4. C

4 3 2 B A D (2) The following four inradii are in arithmetic progression. What is the shape of the large triangle? 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. But it is not a Heron triangle. It has an area . Buchholz 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 mb = and mc = . The third one, however, has irrational length: ma = . 23.6 Heron triangles with square areas 819

23.6 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

4 2 ak = 20k +4k +1, 6 4 2 bk = 8k 4k 2k +1, 6 − 4 − 2 ck = 8k +8k + 10k .

Here, ak, bk

A4 + B4 + C4 = D4, with gcd(A, B, C, D)=1. The triangle with a = B4 + C4, b = C4 + A4, c = A4 + B4

is a Heron triangle with area (ABCD)2. The ﬁrst example which Elkies constructed is

A =2682440, B =15365639, C =18796760, D =20615673.

1American Mathematical Monthly, 98 (1991). 2On A4 + B4 + C4 = D4, Math. Comp., 51 (1988) 825 – 835. Chapter 24

Triangles with sides and one altitude in A.P.

24.1 Newton’s solution

Problem 29 of Isaac Newton’s Lectures on Algebra ([Whiteside, pp.234 – 237]) studies triangles whose sides and one altitude are in arithmetic progression. Newton considered a triangle ABC with an altitude DC. Clearly, DC is shorter than AC and BC. Setting AC = a, BC = x, DC = 2x a, and AB =2a x, he obtained − − 16x4 80ax3 + 144a2x2 10a3x + 25a4 =0. ( ) − − † “Divide this equation by 2x a and there will result 8x3 36ax2 + 54a2x 25a3 =0”. Newton did− not solve this equation nor− did he give any numerical− example. Actually, ( ) can be rewritten as † (2x 3a)3 +2a3 =0, − a √3 so that x = 2 (3 2), the other two roots being complex. By 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 , , . 822 Triangles with sides and one altitude in A.P.

24.2 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 A.P. in some other order. Note that the altitude in question is either the ﬁrst or the second terms of the A.P. (in increasing order). Assuming 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 3 2 = (1+2x)2(1+4x). △ 16 1 On the other hand, = 2 1 (1 + kx) for k = 1, 2, 3. These lead to the equations △ · ·

for k =1: 48x3 + 56x2 + 16x 1=0, • − for k =2: 48x3 + 44x2 +8x 1=0, • − for k =3: 48x3 + 24x 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 48x3 +56x2 +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 A.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 A.P. Instead of constructing an equation in x, we seek one such triangle with sides 15, 17+2z, 18+3z, and the altitude 16 + z on the 24.2 The general case 823 shortest side. By considering the area of the triangle in two different ways, we obtain the cubic equation

z3 120z +16=0. ( ) − ∗ This can be solved by writing z = 4√10 sin θ for an angle θ. Using the trigonometric identity sin 3θ = 3 sin θ 4 sin3 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. Chapter 25

The Pell Equation

25.1 The equation x2 dy2 =1 − Let d be a fixed integer. We consider the Pell equation x2 dy2 = 1. Clearly, if d is negative or is a (positive) square integer, then the− equation has only finitely many solutions. Theorem 25.1. Let d be a nonsquare, positive integer. The totality of positive solutions of the Pell equation x2 dy2 = 1 form an infinite − sequence (xn,yn) defined recursively by

xn+1 = axn + dbyn, yn+1 = bxn + ayn; x1 = a, y1 = b, where (x1,y1)=(a, b) is the fundamental solution, with a, b positive and b smallest possible.

Examples 1. The fundamental solution of the Pell equation x2 2y2 = 1 is (3,2). This generates an inﬁnite sequence of nonnegative− solutions (xn,yn) deﬁned by

xn+1 =3xn +4yn, yn+1 =2xn +3yn; x0 =1, y0 =0. The beginning terms are

n 12345 6 7 8 9 10 ... xn 3 17 99 577 3363 19601 114243 665857 3880899 22619537 ... yn 2 12 70 408 2378 13860 80782 470832 2744210 15994428 ... 902 The Pell Equation

2. Fundamental solution (a, b) of x2 dy2 =1 for d< 100: −

dab dab da b 232 321 59 4 652 783 83 1 10 19 6 11 10 3 12 7 2 13 649 180 14 15 4 15 4 1 17 33 8 18 17 4 19 170 39 20 9 2 21 55 12 22 197 42 23 24 5 24 5 1 26 51 10 27 26 5 28 127 24 29 9801 1820 30 11 2 31 1520 273 32 17 3 33 23 4 34 35 6 35 6 1 37 73 12 38 37 6 39 25 4 40 19 3 41 2049 320 42 13 2 43 3482 531 44 199 30 45 161 24 46 24335 3588 47 48 7 48 7 1 50 99 14 51 50 7 52 649 90 53 66249 9100 54 485 66 55 89 12 56 15 2 57 151 20 58 19603 2574 59 530 69 60 31 4 61 1766319049 226153980 62 63 8 63 8 1 65 129 16 66 65 8 67 48842 5967 68 33 4 69 7775 936 70 251 30 71 3480 413 72 17 2 73 2281249 267000 74 3699 430 75 26 3 76 57799 6630 77 351 40 78 53 6 79 80 9 80 9 1 82 163 18 83 82 9 84 55 6 85 285769 30996 86 10405 1122 87 28 3 88 197 21 89 500001 53000 90 19 2 91 1574 165 92 1151 120 93 12151 1260 94 2143295 221064 95 39 4 96 49 5 97 62809633 6377352 98 99 10 99 10 1

3. Pell’s equations whose fundamental solutions are very large:

d a b 421 3879474045914926879468217167061449 189073995951839020880499780706260 541 3707453360023867028800645599667005001 159395869721270110077187138775196900 601 38902815462492318420311478049 1586878942101888360258625080 613 464018873584078278910994299849 18741545784831997880308784340 661 16421658242965910275055840472270471049 638728478116949861246791167518480580 673 4765506835465395993032041249 183696788896587421699032600 769 535781868388881310859702308423201 19320788325040337217824455505160 919 4481603010937119451551263720 147834442396536759781499589 937 480644425002415999597113107233 15701968936415353889062192632 949 609622436806639069525576201 19789181711517243032971740 991 379516400906811930638014896080 12055735790331359447442538767

4. The equation x2 4729494y2 = 1 arises from the famous Cattle problem of Archimedes,− and has smallest positive solution

x = 109931986732829734979866232821433543901088049, y = 50549485234315033074477819735540408986340. 25.2 The equation x2 dy2 = 1 903 − −

Exercise 1. Solve the Pell equations (a) x2 +3y2 = 1; (b) x2 4y2 = 1 for integer solutions. 1 −

2. Find the 10 smallest nonnegative solutions of the Pell equation x2 3y2 =1. 2 −

3. If (a, b) is the fundamental solution of the Pell equation x2 dy2 = − 1, generating the inﬁnite sequence of nonnegative solutions (x0,y0)= (1, 0), (x1,y1)=(a, b), (x2,y2),..., (xn,yn), ...,then

xn+1 =2axn xn 1; yn+1 =2ayn yn 1. − − − −

25.2 The equation x2 dy2 = 1 − − The negative Pell’s equation

x2 dy2 = 1, − − may or may not have solutions. 3 Here are the smallest positive solution (a, b) of x2 dy2 = 1 for the ﬁrst 24 values of d: − −

dab dab dab 211 521 1031 13 18 5 17 4 1 26 5 1 29 70 13 37 6 1 41 32 5 50 7 1 53 182 25 58 99 13 61 29718 3805 65 8 1 73 1068 125 74 43 5 82 9 1 85 378 41 89 500 53 97 5604 569 101 10 1

25.3 The equation x2 dy2 = c − Theorem 25.2. Let c> 1 be a positive integer.

1(a). (x,y) = ( 1, 0); (b). (x,y) = ( 1, 0). 2 ± ±

n 12345 6 7 8 9 10 11 ... xn 2 7 26 97 362 1351 5042 18817 70226 262087 978122 ... yn 1 4 15 56 209 780 2911 10864 40545 151316 564719 ...

3This depends on the parity of the length of the period of continued fraction expansion of √d. 904 The Pell Equation

(a) If the equation x2 dy2 = c is solvable, it must have a fundamental solution (u, v) in the− range

1 b 0 < u (a + 1)c, 0 v √c. | | ≤ r2 ≤ ≤ 2(a + 1) ·

Every solution appears in a doubly inﬁnite sequencep (xn,yn)

un+1 = aun + dbvn, vn+1 = bun + avn, u1 = u, v1 = v, for some (u, v) in the range above. (b) Same conclusion for the equation x2 dy2 = c, except that it must have a solution (u, v) in the range − −

1 b 0 u (a 1)c, 0 < v √c. ≤| | ≤ 2 − ≤ 2(a 1) · r − Example 25.1. The fundamental solution of xp2 8y2 = 1 is (a, b) = (3, 1). − Consider the equation x2 8y2 = 9. We need only search for solutions (u, v) satisfying 0 v − 1 3= 3 . With v =0, we have ≤ ≤ √2(3+1) · 2√2 u = 3. x2± 8y2 = 7. There are two inﬁnite sequences of positive solutions: − −

(x, y)=( 1, 1), (5, 2), (31, 11), (181, 64), (1055, 373), (6149, 2174), (35839, 12671),... − and (x, y)=(1, 1), (11, 4), (65, 23), (379, 134), (2209, 781), (12875, 4552),....

Example 25.2. Consider the equation x2 23y2 =4 11 23. It is easy to see that x and y must be both even, and− 23 divides ·x. With· x = 46h, y =2k, we have 23h2 k2 = 11, or k2 23h2 = 11. The fundamental solution of x2 23y2−= 1 being (a, b)− = (24, 5)−, we need only ﬁnd y in the range 1− h 2 It is now easy to see that only h = 2 gives ≤ ≤ k = 9. From this we obtain (x1,y1) = (92, 18). The other solutions are generated recursively by

xn+1 = 24xn + 115yn, yn+1 =5xn + 24yn, x1 = 92, y1 = 18. 25.3 The equation x2 dy2 = c 905 −

Here are the ﬁrst 5 solutions. n 123 4 5 ... xn 92 4278 205252 9847818 472490012 ... yn 18 892 42798 2053412 98520978 ··· Chapter 26

Figurate numbers

26.1 Which triangular numbers are squares ?

The triangular numbers are the 1 T =1+2+3+ + m = m(m + 1). m ··· 2 The ﬁrst few of these are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... .

2 Suppose the triangular number Tm is also a square, i.e., Tm = n for 1 2 2 2 some integer n. 2 m(m +1) = n , (2m + 1) =8n +1. We make use of the fundamental solution of the Pell equation x2 8y2 = 1, namely, (3, 1) to generate the solutions: −

2m +1 3 8 2m +1 k+1 = k , nk 1 3 nk +1 or m 3m +4n +1 k+1 = k k . nk 2mk +3nk +1 +1 908 Figurate numbers

Here are some beginning values: k 0123 4 5 . . . mk 1 8 49 288 1681 9800 . . . nk 1 6 35 204 1189 6930 . . .

These yield the square triangular numbers

2 2 2 2 T1 =1 , T8 =6 , T49 = 35 , T288 = 204 , ...

Exercise 1. Find all positive integers m > n satisfying 1+2+ +(n 1)=(n +1)+(n +2)+ + m. ··· − ···

2 Solution. Tn 1 = Tm Tn = Tm = Tn 1 + Tn = n . − − ⇒ − 26.2 Pentagonal numbers 909

26.2 Pentagonal numbers

The pentagonal numbers are the sums 1 P :=1+4+7+ + (3m 2) = m(3m 1). m ··· − 2 −

Exercise 1. Which pentagonal numbers are squares? 2 2 2 2 2 Solution. If Pm = n , 3m m =2n , (6m 1) = 24n +1. We make use of the fundamental− solution of x2 − 24y2 = 1, namely, (a, b)=(5, 1) to generate the solutions −

6m 1 5 24 6m 1 k+1 − = k − . nk 1 5 nk +1 This, however, does not always yield integer values for mk+1. The solutions of the Pell equation x2 24y2 = 1, when arranged in − ascending order, have xk 1 (mod 6) only for odd values of k. Therefore, we modify the≡ recursion − as follows:

6m 1 5 24 2 6m 1 49 240 6m 1 k+1 − = k − = k − . nk 1 5 nk 10 49 nk +1 This gives

m 49m + 40n 8 k+1 = k k − . nk+1 60mk + 49nk 10 − 910 Figurate numbers

Here are some beginning values: k 01 2 3 . . . mk 1 81 7921 776161 . . . nk 1 99 9701 950599 . . .

2. Which pentagonal numbers are triangular numbers? 2 2 Solution. If Pm = Tn, 3m m = n + n. Completing squares leads to , − (6m 1)2 = 3(2n + 1)2 2. (26.1) − − Now, the Pell equation x2 3y2 = 2 has smallest solution (1, 1). We make use the fundamental− solution− of x2 3y2 = 1, namely, (a, b)=(2, 1) to generate solutions of (26.1): −

2 6m 1 2 3 6m 1 7 12 6m 1 k+1 − = k − = k − . 2nk +1 1 2 2nk +1 4 7 2nk +1 +1 This gives m 7m +4n +1 k+1 = k k . nk 12mk +7nk +1 +1 Here are some beginning values: k 012 3 4 . . . mk 1 12 165 2296 31977 . . . nk 1 20 285 3976 55385 . . . 26.3 Almost square triangular numbers 911

26.3 Almost square triangular numbers

There are two types of almost square triangular numbers, deﬁcient and excessive.

26.3.1 Excessive square triangular numbers

2 If the triangular number Tm = n +1 for some integer n, we call it an excessive square triangular number. This is equivalent to

(2m + 1)2 8n2 =9. − Its solutions can be found from those of x2 8y2 =9: −

mk+1 = 3mk +4nk +1,

nk+1 = 2mk +3nk +1, with initial entries given by (2m0 + 1, n0) = (3, 0), i.e., (m0, n0) = (1, 0). The beginning values are k 012 3 4 . . . mk 1 4 25 148 865 . . . nk 0 3 18 105 612 . . . This gives the excessive triangular numbers

2 T1 = 0 +1, 2 T4 = 3 +1, 2 T25 = 18 +1, 2 T148 = 105 +1, . . 912 Figurate numbers

26.3.2 Deﬁcient square triangular numbers

2 If the triangular number Tm = n 1 for some integer n, we call it a deﬁcient square triangular number.− This is equivalent to

(2m + 1)2 8n2 = 7. − − Its solutions can be found from the same recurrence relations, but with initial entries given by (2m0 +1, n0)=(1, 1) or ( 1, 1), i.e., (m0, n0)= (0, 1) or ( 1, 1). These yield two distinct sequences− of deﬁcient triangular numbers:−

T = 12 1, 0 − T = 42 1, 5 − T = 232 1, 32 − T = 1342 1, 189 − . . and

T = 22 1, 2 − T = 112 1, 15 − T = 642 1, 90 − T = 3732 1, 527 − . . 26.3 Almost square triangular numbers 913

Exercise 1. A developer wants to build a community in which the n (approxi- mately 100) homes are arranged along a circle, numbered consecu- tively from 1, 2,... n, and are separated by the club house, which is not numbered. He wants the house numbers on one side of club house adding up to the same sum as the house numbers on the other sides. Between which two houses should he build the club house? How many houses are there altogether? Solution. Suppose houses 1, 2,..., m are on one side of the club house. 1+2 + n = 2(1+2+ + m); n(n +1)=2m(m +1); 4n2 +4n = 2(4···m2 +4m); (2n+1)···2 = 2(2m+1)2 1; (2n+1)2 2(2m + 1)2 = 1. From the solutions of x2 2y2−= 1, we have− (2n +1, 2m +1)− = (239, 169). This gives (n,− m) = (119− , 84). 2. Later the developer finds out that government law requires the club house also to be numbered. If he wants to maintain equal house number sums on both sides, he finds that he has to build signifi- cantly fewer homes. How many homes should he build, and what is the number of the club house? 3. Find five positive integers n for which both 2n +1 and 3n +1 are squares. 4. The voting population of your county is about one million. Each voter is assigned a registration number from 1, 2, 3,..., N, the exact number of voters. When a county official typed in her own name, her registration number M appeared on the computer screen. Her assistant, being an excellent student from your mathematics class, remarked that the probability that the registration numbers of two more names she would randomly enter are both less than M 1 is exactly 2 . And she was correct. The assistant’s father was very proud. He copied the numbers, and related the story to you. Now, give him more reason to be proud by telling the exact population N of the county, and the registration number M of his daugh- ter’s superior, (which he can easily verify). Answer. : Voting population 803762, first number 568346. Chapter 27

Special integer triangles

27.1 Almost isosceles Pythagorean triangles

An almost isosceles right triangle is one whose shorter sides differ by 1. Let a and a + 1 be these two sides and y the hypotenuse. Then y2 = a2 +(a + 1)2 = 2a2 +2a +1. From this, 2y2 = (2a + 1)2 + 1. The equation With x = 2a + 1, this reduces to the Pell equation x2 2y2 = 1, which we know has solutions, with the of this equations − − are (xk,yk) given recursively by smallest positive one (1, 1), and the equation x2 2y2 = 1 has fundamental solution (3, 2). It follows that the solutions− are given recursively by

xk+1 = 3xk +4yk, yk+1 = 2xk +3yk, x0 =1, y0 =1.

If we write xk =2ak +1, these become

ak+1 = 3ak +2yk +1, yk+1 = 4ak +3yk +2, a0 =0, y0 =1.

The beginning values of ak and yk are as follows.

k 1234 5 6 . . . ak 3 20 119 696 4059 23660 . . . yk 5 29 169 985 5741 33461 . . . 916 Special integer triangles

32 +42 = 52, 202 + 212 = 292, 1192 + 1202 = 1692, 6962 + 6972 = 9852, . .

27.1.1 The generators of the almost isosceles Pythagorean triangles Consider the almost isosceles Pythagorean triangles along with their generators:

k 1 2 3 4 5 6 . . .

ak 3 20 119 696 4059 23660 . . . yk 5 29 169 985 5741 33461 . . . mk 1 2 5 12 29 70 . . . nk 2 5 12 29 70 169 . . .

The generators form a sequence

1, 2, 5, 12, 29, 70, 169,... which satisﬁes the second order recurrence

mk+2 =2mk+1 + mk, m1 =1, m2 =2. Note that the hypotenuses of the triangles are among the generators. 27.2 Integer triangles (a,a +1,b) with a 120◦ angle 917

27.2 Integer triangles (a, a +1,b) with a 120◦ angle

If (a, a +1, b) is an integer triangle with a 120◦ angle, the consecutive sides must be the shorter ones. This means that b2 = a2 + a(a +1)+(a + 1)2 =3a2 +3a +1. Completing squares leads to the Pell equation (2b)2 3(2a + 1)2 =1. (27.1) − Since the solutions of x2 3y2 =1, when arranged in ascending order, have x even and y odd only− in alternate terms, the solutions of (27.1) are given recursively by

2b 2 3 2 2b 7 12 2b k+1 = k = k , 2ak +1 1 2 2ak +1 4 7 2ak +1 +1 or a 7a +4b +3 k+1 = k k . bk 12ak +7bk +6 +1 The beginning values are k 012 3 4 . . . ak 0 7 104 1455 20272 . . . bk 1 13 181 2521 35113 . . .

Note that the values of bk are all sums of two consecutive squares: 1= 02 +12, 13= 22 +32, 181= 92 + 102, 2521 = 352 + 362, 35113 = 1322 + 1332, . . This observation can be justiﬁed by rewriting (27.1) in the form (2b 1)(2b +1) = 3(2a + 1)2. − Since 2b 1 and 2b +1 are consecutive odd numbers, they are relatively prime. Therefore,− either 918 Special integer triangles

(i) 2b +1= m2, 2b 1=3n2, or (ii) 2b +1=3m2, 2−b 1= n2. In (i), m2 3n2 =2−, an impossibility. We must have (ii), and − n2 +1 n 1 2 n +1 2 b = = − + , 2 2 2 n 1 n+1 the sum of the squares of the consecutive numbers −2 and 2 .

Exercise

2 2 1. If we write bk = ck +(ck +1) , the numbers ck are given recursively by c =4c c +1, c =0, c =2. k+2 k+1 − k 0 1

2. Prove that there is no integer triangle with two sides which are consecutive integers and have a 60◦ angle between them.

3. Find all integer triangles with a 60◦ and two sides which are consecutive integers. Answer. (3, 7, 8) and (5, 7, 8). Chapter 28

Heron triangles

28.1 Heron triangles with consecutive sides

If (b 1, b, b +1, ) is a Heron triangle, then b must be an even integer. We write− b =2m.△ Then s =3m, and 2 =3m2(m 1)(m + 1). This requires m2 1=3k2 for an integer n△, and =3mn−. The solutions of m2 3n2 =1− can be arranged in a sequence△ − m 2 3 m m 2 k+1 = k , 1 = . nk 1 2 nk n 1 +1 1 From these, we obtain the side lengths and the area. The middle sides form a sequence (bk) given by b =4b b , b =2, b =4. k+2 k+1 − k 0 1 The areas of the triangles form a sequence = 14 , =0, =6. △k+2 △k+1 −△k △0 △1

k b Heron triangle k △k 0 2 0 (1, 2, 3, 0) 1 4 6 (3, 4, 5, 6) 2 14 84 (13, 14, 15, 84) 3 4 5 1002 Heron triangles

28.2 Heron triangles with two consecutive square sides

Construct an inﬁnite family of Heron triangles each with two sides the squares of consecutive integers. Here is one example: the triangle (25, 36, 29) has area 360. Solution. Let a = v2 and b =(v +1)2. Since the perimeter of a Heron triangle must be an even number, the third side must be odd. Write c =2u 1. With these, −

s = u + v + v2, s a = u + v, − s b = u v 1, − − − s c = v2 u + v +1. − −

To obtain a Heron triangle, we ﬁnd a relation between u and v which makes one or more of these expressions squares. By putting u =3v, we have c =6v 1 and −

s = v(v + 4), s a = 4v, − s b = 2v 1, − − s c = (v 1)2. − −

This is a Heron triangle if and only if 4v2(v 1)2(v + 4)(2v 1) is a square. Equivalently, (v + 4)(2v 1) is a square.− We take 2v −1= x2 and v +4= y2 for integers x and−y. These must satisfy x2 2−y2 = 9. − − Beginning with the smallest solution (x1,y1) = (3, 3), we obtain an inﬁnite sequence of solutions

x 3 4 x n+1 = = n . yn 2 3 yn +1 28.2 Heron triangles with two consecutive square sides 1003

nxyv a b c ∆ 133 5 25 36 29 360 2 21 15 221 48841 49284 1325 30630600 3 123 87 7565 57229225 57244356 45389 1224657967320 . .

The parameter v can be generated recursively by v = 34v v + 56, v =5, v = 221. n+2 n+1 − n 1 2 This gives a Heron triangle (v2, (v + 1)2, 6v 1) with area 2v(v 1)xy. − − Exercise Make use of this recurrence to ﬁnd one more such triangle. Answer. v = 257045;(a, b, c) = (66072132025, 66072646116, 1542269), ∆ = 48036763841709240. Chapter 29

Squares as sums of consecutive squares

29.1 Sum of squares of natural numbers

Theorem 29.1. 1 12 +22 +32 + + n2 = n(n + 1)(2n + 1). ··· 6 Proof. Let T =1+2+3 + n = 1 n(n + 1) and n ··· 2 S =12 +22 +32 + + n2. n ···

23 = 13 + 3 12 + 3 1 +1 33 = 23 + 3 · 22 + 3 · 2 +1 43 = 33 + 3 · 32 + 3 · 3 +1 . · · . n3 = (n 1)3 + 3(n 1)2 + 3(n 1) + 1 (n + 1)3 = −n3 + 3 −n2 + 3 −n + 1 · · Combining these equations, we have 3 3 (n + 1) =1 +3Sn +3Tn + n.

Since Tn is known, we have 1 3 1 S = (n + 1)3 n 1 n(n + 1) = n(n + 1)(2n + 1). n 3 − − − 2 6 1006 Squares as sums of consecutive squares

Exercise 1. Find 12 +32 +52 + + (2n 1)2. ··· − 2. Find n so that n2 +(n + 1)2 is a square. 3. Show that

32 +42 =52, 102 + 112 + 122 =132 + 142, 212 + 222 + 232 + 242 =252 + 262 + 272, 362 + 372 + 382 + 392 + 402 =412 + 422 + 432 + 442, . .

4. Find all integers n so that the mean and the standard deviation of n consecutive integers are both integers. Solution. If the mean of n consecutive integers is an integer, n must be odd. We may therefore assume the numbers to be m, (m 1), ..., 1, 0, 1, ..., m 1, m. The standard deviation− − − − − 1 of these number is 3 m(m + 1). For this to be an integer, we 1 2 2 2 must have 3 m(m +1)q = k for some integer k. m + m = 3k ; n2 = (2m + 1)2 = 12k2 +1. The smallest positive solution of the Pell equation n2 12k2 = 1 being (7, 2), the solutions of this − equation are given by (ni,ki), where

ni+1 = 7ni + 24ki, ki+1 = 2ni +7ki, n0 =1, k0 =0. The beginning values of n and k are i 123 4 5 . . . ni 7 97 1351 18817 262087 . . . ki 2 28 390 5432 75658 . . .

5. Find the ﬁrst two instances when the average of the squares of n consecutive squares is equal to n2. Describe how to generate all such identities. 29.1 Sum of squares of natural numbers 1007

6. Find all sequences of 11 consecutive positive integers, the sum of whose squares is the square of an integer. Answer.

182 + 192 + + 282 = 772, 382 + 392 + ··· + 482 = 1432, 4562 + 4572 +··· + 4662 = 15292, 8542 + 8552 + ··· + 8642 = 28492, 91922 + 91932 +··· + 92022 = 305032, 171322 + 171332 +··· + 171422 = 568372, . ··· .

7. Find all natural numbers n> 1 such that (12 +22 + + n2)((n + 1)2 +(n + 2)2 + +(n + n)2) ··· ··· is a perfect square. 1008 Squares as sums of consecutive squares

29.2 Sums of consecutive squares: odd number case

Suppose the sum of the squares of 2k +1 consecutive positive integers is a square. If the integers are b, b 1,...,b k. We require ± ± 1 (2k + 1)b2 + k(k + 1)(2k +1) = a2 3 for an integer a. From this we obtain the equation 1 a2 (2k + 1)b2 = k(k + 1)(2k + 1). (E ) − 3 k

1. Suppose 2k +1 is a square. Show that (Ek) has solution only when k = 6m(m + ǫ) for some integers m > 1, and ǫ = 1. In each case, the number of solutions is ﬁnite. ±

Number of solutions of (Ek) when 2k +1 is a square

2k +1 25 49 121 169 289 361 529 625 841 961 . . . 011 2 7 3 5 3 310 . . . 2. Find the unique sequence of 49 (respectively 121) consecutive positive integers whose squares sum to a square.

3. Find the two sequences of 169 consecutive squares whose sums are squares.

4. Suppose 2k +1 is not a square. If k +1 is divisible 9=32 or by any prime of the form 4k +3 7, then the equation (Ek) has no solution. Verify that for the following≥ values of k < 50, the equation (Ek) has no solution: 29.2 Sums of consecutive squares: odd number case 1009

k = 6, 8, 10, 13, 17, 18, 20, 21, 22, 26, 27, 30, 32, 34, 35, 37, 40, 41, 42, 44, 45, 46, 48,...

1 k(k+1) − 3 5. Suppose p =2k+1 is a prime. If the Legendre symbol p = 1, then the equation (Ek) has no solution. Verify that for the fol- − lowing values of k < 50, the equation (Ek) has no solution: 1, 2, 3, 8, 9, 14, 15, 20, 21, 26, 33, 39, 44.

6. For k 50, it remains to consider (Ek) for the following values of k: ≤ 5, 7, 11, 16, 19, 23, 25, 28, 29, 31, 36, 38, 43, 47, 49.

Among these, only for k = 5, 11, 16, 23, 29 are the equations (Ek) solvable. 7. Work out 5 sequences of 23 consecutive integers whose squares add up to a square in each case. Answer: 72 +82 + + 292 = 922; 8812 + 8822 + ···+ 9032 = 42782; 427872 + 427882 + ···+ 428092 = 2052522; 20534012 + 20534022 + ···+ 20534232 = 98478182; ··· ········· 2 2 8. Consider the equation (E36) : u 73v = 12 37 73. This equation does in fact have solutions (u, v−) = (4088,· 478)· , (23360, 2734). The fundamental solution of the Pell equation x2 73y2 =1 being (a, b) = (2281249, 267000), we obtain two sequences− of solutions of (E73): Answer:

(4088, 478), (18642443912, 2181933022), (85056113063608088, 9955065049008478),... (23360, 2734), (106578370640, 12474054766), (486263602888235360, 56912849921762734),...

This means, for example, the sum of the squares of the 73 numbers with center 478 (respectively 2734) is equal to the square of 4088 (respectively 23360). 1010 Squares as sums of consecutive squares

29.3 Sums of consecutive squares: even number case

Suppose the sum of the squares of the 2k consecutive numbers

b k +1, b k +2,...,b,...,b + k 1, b + k, − − − is equal to a2. This means

2 2 2k 2 (2a) 2k(2b + 1) = (4k 1). (E′ ) − 3 − k Note that the numbers 2k, 4k2 1 are relatively prime. −

1. Show that the equation (Ek′ ) has no solution if 2k is a square.

2. Suppose 2k is not a square. Show that if 2k +1 is divisible by 9, or by any prime of the form 4k +1, then the equation (Ek′ ) has no solution.

3. For k 50, the equation (Ek′ ) has no solution for the following values≤ of k:

k = 3, 4, 5, 9, 11, 13, 15, 17, 21, 23, 24, 27, 29, 31, 33, 35, 38, 39, 40, 41, 45, 47, 49.

4. Let k be a prime. The equation (Ek′ ) can be written as 4k2 1 (2b + 1)2 2ky2 = − . − − 3

By considering Legendre symbols, the equation (Ek′ ) has no solution for the following values of k 50: ≤ k =5, 7, 17, 19, 29, 31, 41, 43.

5. Excluding square values of 2k < 100, the equation (Ek′ ) has solutions only for k =1, 12, 37, 44.

6. Show that (34, 0), (38, 3), (50, 7) are solutionsof (E”12). Construct from them three inﬁnite sequences of expressions of the sum of 24 consecutive squares as a square. 29.3 Sums of consecutive squares: even number case 1011

7. The equation (E37′ ) has solutions (185, 2), (2257,261), and (2849, 330). From these we construct three inﬁnite sequences of expressions of the sum of 74 consecutive squares as a square. Answer:

2252 + 2262 + + 2982 = 22572; 2942 + 2952 + ··· + 3672 = 28492; 130962 + 130972 + ···+ 131792 = 7638652. ···

8. The equation (E44′ ) has solutions (242, 4) and (2222,235). From these we obtain two inﬁnite sequences of expressions of the sum of 88 consecutive squares as a square.

1922 + 1932 + + 2792 = 22222; 59252 + 59262 +··· 60122 = 559902. ··· 1012 Squares as sums of consecutive squares

29.4 Sums of powers of consecutive integers

Consider the sum of the k-powers of the ﬁrst natural numbers: S (n):=1k +2k + + nk. k ···

Theorem 29.2 (Bernoulli). Sk(n) is a polynomial in n of degree k +1 without constant term. It can be obtained recursively as

Sk+1(n)= (k + 1)Sk(n)dn + cn, Z where c is determined by the condition that the sum of the coefﬁcients is 1.

Examples

1 (1) S2(n)= 6 n(n + 1)(2n + 1). (2) S (n)=13 +23 + + n3 = 1 n2(n + 1)2. 3 ··· 4 (3) S (n)= 1 n5 + 1 n4 + 1 n3 1 n. 4 5 2 3 − 30 Exercise Show that 1k +2k + + nk =(1+2+ + n)p ··· ··· for all integers n if and only if (k,p)=(1, 1) or (3, 2). 1

1Math. Mag. 82:1 (2009) 64, 67. Chapter 30

Lucas’ problem

A square pyramid of cannon balls contains a square number of cannon balls only when it has 24 cannon balls along its base. 1 Theorem 30.1 (Lucas).

12 +22 + + n2 = m2 = (n, m)=(1, 1) or (24, 70). ··· ⇒ Equivalently, these are the only positive integer solutions of

n(n + 1)(2n +1)=6m2.

The elementary solution of Lucas’ problem presented here is based on Anglin [?].

30.1 Solution of n(n + 1)(2n +1)=6m2 for even n

We make use of three nontrivial results.

(A) (Fermat) The area of a Pythagorean triangle cannot be a square.

(B) There are no positive integers x such that 2x4 +1 is a square.

Note that the numbers n, n +1 and 2n +1 are mutually relatively prime.

1E. Lucas, Question 1180, Nouvelles Annales de Math´ematiques, ser. 2, 14 (1875) 336. 1014 Lucas’ problem

If n is even, then n +1 and 2n +1 are both odd. Each of them is of the form k2 or 3k2. In any case, n +1 2 and 2n +1 2 (mod 3). This means that n 0 (mod 3), and there6≡ are integers p,6≡q, r such that ≡ n =6p2, n +1= q2, 2n +1= r2. From these, 6p2 = r2 q2 =(r q)(r + q). Since q and r are both odd, r p and r + q are both− even. It− follows that p is even, and − r q r + q p 2 − =6 . 2 · 2 2 There are two possibilities. r q r+q 2 2 (a) One of −2 and 2 is of the form 3x and the other 2y for some integers x and y. In this case, q = (3x2 2y2) and p =2xy. Since ± − (3x2 2y2)2 = q2 =6p2 +1=24x2y2 +1 − = (3x2 6y2)2 = 32y4 +1 =2(2y)4 +1, ⇒ − we must have y = 0 by (B). Correspondingly, x = 0, p = 0, and n cannot be a positive integer. r q r+q 2 2 (b) One of −2 and 2 is of the form 6x and the other y for some integers x and y. Then q = (6x2 y2) and p =2xy. Since ± − (6x2 y2)2 = q2 =6p2 +1=24x2y2 +1 − = (6x2 3y2)2 =8y4 +1, ⇒ − we must have y =0 or y =1 by (C). Correspondingly, x =0 or x =1. Hence, the only positive even integer value of n =6p2 = 6(2xy)2 = 24.

30.2 The Pell equation x2 3y2 =1 revisited − The positive solutions of the Pell equation x2 3y2 =1 form a sequence − (xk,yk).

1. xh+k = xhxk +3yhyk, yh+k = xhyk + xkyh.

2. For h k, xh k = xhxk 3yhyk, yh k = xhyk + xkyh. ≥ − − − − 3. x =4x x , y =4y y . k+2 k+1 − k k+2 k+1 − k 4. x =2x2 1=6y2 +1, y =2x y . 2k k − k 2k k k 30.3 Solution of n(n + 1)(2n +1)=6m2 for odd n 1015

5. Let h,k,r be nonnegative integers such that 2hr k is nonnegative. Then − r x2hr k ( 1) xk (mod xh). ± ≡ −

6. If k is even, then xk is odd, not divisible by 5, and 5 is a quadratic residue of xk if and only if k is divisible by 3.

7. If k is even, then xk is odd, and 2 is a quadratic residue of xk if and only if k is divisible by 4. −

2 8. xk =4z +3 for an integer z only if xk =7.

30.3 Solution of n(n + 1)(2n +1)=6m2 for odd n

Suppose n is odd. Then each of n and 2n +1 is of the form k2 or 3k2. This means that n 2 (mod 3). Since n +1 is even, it is of the form 2k2 or 6k2. This means6≡ that n +1 1. Therefore, n 1 (mod 3), and 2n +1 is divisible by 3. There are6≡ nonnegative integers≡ p, q, r such that n = p2, n +1=2q2, 2n +1=3r2. From these, we have 6r2 +1=4n+3=4p2 +3. Note that 3r2 4q2 = 1. It follows that − − (6r2 + 1)2 3(4qr)2 = 12r2(3r2 +1 4q2)+1=1. − − Therefore, (6r2 +1, 4qr) is a solution of the Pell equation x2 3y2 =1, 2 2 − and 6r +1= xk for some k. By (8), we must have 6r +1=7, r =1. Hence, the only positive odd integer value of n is given by 2n +1=3, i.e., n =1. Chapter 31

Some geometry problems

1. AX and BY are angle bisectors of triangle ABC, with X on BC and Y on AC. Suppose AX = AC and BY = AB. (a) Determine the angles of triangle ABC.

(b) If CZ′ is the external bisector of angle C, with Z′ on the exten- sion of BA, show that CZ′ = AC.

Z′

B X C A B Solution. (a) 2 +B = C, 2 +C = A. Together with A+B+C = π, we have 6π 2π 5π A = , B = , C = . 13 13 13

(b) Half of the external angle of C = 1 π 5π = 4π = 2B. 2 − 13 13 Therefore, CZ′ = BC. 1102 Some geometry problems

3 3

1 1 5 5 B C X 1103

1 ◦ tan7 = √6 √3+ √2 2=(√2 1)(√3 √2), 2 − − − − 1 ◦ tan37 = √6+ √3 √2 2=(√2+1)(√3 √2), 2 − − − 1 ◦ tan52 = √6 √3 √2+2=(√2 1)(√3+ √2), 2 − − − 1 ◦ tan82 = √6+ √3+ √2+2=(√2+1)(√3+ √2). 2

Solution. Consider a square ABCD with sidelength 2 and an equilateral triangle XCD with X in the interior of the square. Let M be the midpoint of AB.

D C

30◦

Y 45◦

X T

A M B

Since triangle DAX is isosceles with ∠ADX = 30◦, ∠DAX = 75◦, and ∠XAM = 15◦. In the right triangle XAM,

AM = 1, MX = 2 √3, − AX = 12 + (2 √3)2 = 8 4√3= 8 2√12 = √6 √2. − − − − q q q 1104 Some geometry problems

If AT bisects angle XAM, by the angle bisector theorem AM T M = XM · AM + AX 1 = (2 √3) − · 1+ √6 √2 1 − = (2 + √3)(1 + √6 √2) − 1 = 2+ √2+ √3+ √6 1 = (√2+1)(√3+ √2) = (√2 1)(√3 √2). − − From the right triangle XAM, we have

1 ◦ T M tan7 = =(√2 1)(√3 √2). 2 AM − −

(2) Applying the law of tangents to triangle XCY in which ∠X = 45◦ and ∠C = 30◦, we have

X+C tan 2 CY + XY X C = . tan − CY XY 2 − Now, XY = AX = √2(√3 1), CY = 2 2(2 √3) = 2(√3 1). This means − − − − tan37 1 ◦ √2+1 2 = , tan7 1 ◦ √2 1 2 − and 1 ◦ tan37 =(√2+1)(√3 1). 2 − 1105

3. Let D be the midpoint of BC of triangle ABC. Suppose that ∠BAM = ∠C and ∠DAC = 15◦. Calculate angle C.

15◦

B D C

Answer.: 30◦.

B D C

Solution. Since ∠DAB = ∠ACD, the line AB is tangent to the circle ACD. Let O be the center of the circle, and P , Q the projections of A, D on OC. Since ∠DOQ = 2∠DAC = 30◦, DQ = 1 OD = 1 OA. Therefore, BP = AO. Since OB is 2 · 2 · a diameter of the circle through O, A, B, P , ∠AOC = 90◦. It follows that ∠ACB = ∠OCD ∠OCA = 75◦ 45◦ = 30◦. − − 1106 Some geometry problems

4. Calculate the length of AC.

C 1 1

30◦ D B

Answer. x = √3 2. Solution. Let AB = x. If AE is the bisector of angle A, then 1 AE = 1+x . AD is the bisector angle BAE.

C 1 1

30◦ D B

√3 x

1 x

2 1 2 √3 (x + 2)(x3 2) (1 + x)2 = 1+ + = − =0. x x ⇒ x2 ! 1107

5. In a right - angled triangle, establish the existence of a unique interior point with the property that the line through the point perpendicular to any side cuts off a triangle of the same area.

A A A

P P Y Z2 P

C B C B C B X X1

Solution. (1) Since the three triangles Z1BX, AZ2Y and X1BZ are similar, they are congruent if their areas are equal. Therefore, BX = BZ and triangles BPX and BPZ are congruent. It follows that P lies on the bisector of angle B.

A A A

M M Z

P P Y Z2 P

C B C B C B X X1

(2) Note that Z1B = AZ2. From this, AZ1 = Z2B. Extend CP to intersect AB at M. We have

AZ1 : AM = CP : CM = BZ2 : BM = Z2B : MB. Therefore, AM = MB and CM is the median on the hypotenuse. (3) The point P is the intersection of the bisector of angle B (smaller acute angle) and the median on the hypotenuse.

Exercise

(a) Calculate the coordinates of P in terms of a, b, c. Here, c2 = a2 + b2. a2 ab Answer. 2a+c , 2a+c . a+c 2 (b) Show that the common area of the triangles is 2a+c of the given right triangle. Chapter 32

Basic geometric constructions

32.1 Some basic construction principles

Theorem 32.1 (Perpendicular bisector locus). Given two distinct A and B on a plane, a point P is equidistant from A and B if and only if P lies on the perpendicular bisector of the segment AB. Theorem 32.2 (Angle bisector locus). A point P is equidistant from two given intersecting lines if and only if it lies on the bisector of an angle between the two lines. Note that two intersecting lines have two angle bisectors. Theorem 32.3. If two circles are tangent to each other, the line joining their centers passes through the point of tangency. The distance between their centers is the sum (respectively difference) of their radii if the tangency is external (respectively internal). 1110 Basic geometric constructions

32.2 Geometric mean

We present two ruler-and-compass constructions of the geometric means of two quantities given as lengths of segments. These are based on Eu- clid’s proof of the Pythagorean theorem. Construction 32.1. Given two segments of length a < b, mark three points P , A, B on a line such that P A = a, P B = b, and A, B are on the same side of P . Describe a semicircle with P B as diameter, and let the perpendicular through A intersect the semicircle at Q. Then P Q2 = P A P B, so that the length of P Q is the geometric mean of a and b. ·

A P B

Construction 32.2. Given two segments of length a, b, mark three points A, P , B on a line (P between A and B) such that P A = a, P B = b. Describe a semicircle with AB as diameter, and let the perpendicular through P intersect the semicircle at Q. Then P Q2 = P A P B, so that the length of P Q is the geometric mean of a and b. ·

P A B 32.3 Harmonic mean 1111

32.3 Harmonic mean

Let ABCD be a trapezoid with AB//CD. If the diagonals AC and BD intersect at K, and the line through K parallel to AB intersect AD and BC at P and Q respectively, then P Q is the harmonic mean of AB and CD: 2 1 1 = + . P Q AB CD

A b B

K P Q harmonic mean

D a C

Another construction

harmonic mean

a 1112 Basic geometric constructions

32.4 A.M G.M. H.M. ≥ ≥ A trapezoid with parallel sides a and b is given. Here is a construction of the geometric mean √ab as a parallel to the bases, making use of the arithmetic and harmonic means. As is well known, the parallel through 2ab the intersection of the diagonals gives the harmonic mean a+b . Let H be the endpoint of this parallel on the side AB of the trapezoid. Construct the perpendicular to AB at H to intersect the circle with diameter AB at P . Bisect the right angle AP B and let the bisector intersect AB at G.

B b

G M

A a

Then the parallel through G to the bases has length √ab. Proof. Note that AP B is a right angle, and AP 2 : BP 2 = AH : BH = a : b. Since PG bisects angle AP B, AG : BG = AP : P B = √a : √b. It follows that the parallel through G has length

a √b + b √a √ab(√a + √b) · · = = √ab. √b + √a √a + √b 32.4 A.M G.M. H.M. 1113 ≥ ≥

Exercise 1. Given triangle ABC, construct the equilateral triangles BCX, CAY and ABZ externally on the sides of the triangle. Join AX, BY , CZ. What can you say about the intersections, lengths, and direc- tions of these lines (segments)?

2. Show that the 90◦ angle of a right triangle is bisected by the line joining it to the center of the square on the hypotenuse. 3. Make a sketch to show that for two given positive quantities a and b, a + b 2ab √ab . 2 ≥ ≥ a + b

4. Construct the following diagram.

B A

C D

5. Construct the following diagram. B A

C D 1114 Basic geometric constructions

6. Two congruent circles of radii a have their centers on each other. Consider the circle tangent to one of them internally, the other externally, and the line joining their centers. It is known that this √3 circle has radius 4 a. Construct the circle.

7. An 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, the common inradius is (√3 √2)a. Construct the partition. − Chapter 33

Construction of a triangle from three given points

33.1 Some examples

A A A

Hb Tb Mc M b Tc O

G I Hc H

B Ma C B Ta C B Ha C

Solve the construction problems of a triangle ABC from the given data in each case.

A1 : B,C,H B1 : B,C,I C1 : A, I, Mb A2 : A, B, Ma B2 : A, Ma, Mb C2 : A, H, Mb A3 : A, H, Ma B3 : A,O,H C3 : A, B, Ta A4 : A,O,G B4 : A, I, Tb C4 : A, G, Mb A5 : O, Mb, Mc B5 : A, Mb, Mc C5 : A,G,Hb

A1 A is the orthocenter of triangle BCH.

A2 Extend BMa to C such that BMa = MaC. 1116 Construction of a triangle from three given points

A3 Construct the parallel to AH through Ma, and the point O on this 1 MaO HA line such that the vector = 2 . Construct the circle O(A). This is the circumcircle of the· triangle. Construct the perpendicular to OMa at Ma to intersect the circumcircle at B and C.

A4 Construct the circumcircle O(A). Extend AG to Ma such that AG : GMa =2:1. Join OMa. Construct the perpendicular to OMa at Ma to intersect the circumcircle at B and C.

A5 O is the orthocenter of MaMbMc. Therefore, Ma can be constructed as the orthocenter of triangle OMbMc. Now, construct lines through Ma, Mb, Mc parallel respectively to MbMc, McMa, MaMb. These lines bound the required triangle ABC. B1 Let X be the perpendicular foot of I on BC. Construct the circle I(X), then the tangents from B and C to the circle. These tangents intersect at A. 1

B2 Trisect the segment AMa at G. Join Mb to G and extend it to B such that MbG : GB =1:2. Extend BMa to C such that BMa = MaC. B3 Construct the circumcircle O(A). Construct a line through O parallel to AH, and on this line mark a point Ma such that the vector 1 OMa AH = 2 . The perpendicular to OMa at Ma intersects the circumcircle· at B and C.

B4 Construct the perpendicular foot Y of I on the line ATb and the circle I(Y ). This is the incircle of the triangle. Construct A(Y ) to intersect the incircle again at Z. The lines AZ and ITb intersect at B. Construct the circle B(Y ) to intersect the incircle again at X. The lines BX and ATb intersect at C.

B5 Extend AMc to B such that AMc = McB and AMb to C such that AMb = MbC.

C2 Extend AMb to C such that AMb = MbC. B is the orthocenter of triangle HAC.

C4 Extend AMb to C such that AMb = MbC. Extend MbG to B such that GB =2 M G. · b 1These tangents can be constructed easily as follows. Let the circles B(X) and C(X) intersect the circle I(X) again at Z and Y respectively. The lines BZ and CY are the tangents, and their intersection is A. 33.2 Wernick’s construction problems 1117

33.2 Wernick’s construction problems

1. A,B,O L 36. A,Mb,Tc S 71.O,G,H R 106. Ma,Hb,Tc U 2. A,B,Ma S 37. A,Mb,I S 72. O,G,Ta U 107. Ma,Hb,I U 3. A,B,Mc R 38. A,G,Ha L 73.O,G,I U 108. Ma,H,Ta U 4. A,B,G S 39. A,G,Hb S 74.O,Ha,Hb U 109. Ma,H,Tb 5.A,B,Ha L 40. A,G,H S 75.O,Ha,H S 110. Ma,H,I 6.A,B,Hc L 41. A,G,Ta S 76.O,Ha,Ta S 111. Ma,Ta,Tb 7.A,B,H S 42. A,G,Tb U 77.O,Ha,Tb 112. Ma,Ta,I S 8. A,B,Ta S 43. A,G,I S 78.O,Ha,I 113. Ma,Tb,Tc 9. A,B,Tc L 44.A,Ha,Hb S 79.O,H,Ta U 114. Ma,Tb,I U 10.A,B,I S 45.A,Ha,H L 80.O,H,I U 115. G,Ha,Hb U 11. A,O,Ma S 46.A,Ha,Ta L 81.O,Ta,Tb 116. G,Ha,H S 12. A,O,Mb L 47.A,Ha,Tb S 82.O,Ta,I S 117. G,Ha,Ta S 13. A,O,G S 48.A,Ha,I S 83. Ma, Mb, Mc S 118. G,Ha,Tb 14.A,O,Ha S 49.A,Hb,Hc S 84. Ma, Mb, G S 119. G,Ha,I 15.A,O,Hb S 50.A,Hb,H L 85. Ma, Mb,Ha S 120. G,H,Ta U 16.A,O,H S 51.A,Hb,Ta S 86. Ma, Mb,Hc S 121. G,H,I U 17. A,O,Ta S 52.A,Hb,Tb L 87. Ma, Mb,H S 122. G,Ta,Tb 18. A,O,Tb S 53.A,Hb,Tc S 88. Ma, Mb,Ta U 123. G,TA,I 19.A,O,I S 54.A,Hb,I S 89. Ma, Mb,Tc U 124.Ha,Hb,Hc S 20. A,MA, Mb S 55. A,H,Ta S 90. Ma, Mb,I S 125.Ha,Hb,H S 21. A,Ma, G R 56. A,H,Tb U 91. Ma,G,Ha L 126.Ha,Hb,Ta S 22. A,Ma,Ha L 57.A,H,I S 92. Ma,G,Hb S 127.Ha,Hb,Tc 23. A,Ma,Hb S 58. A,Ta,Tb S 93. Ma,G,H S 128.Ha,Hb,I 24. A,Ma,H S 59. A,Ta,I L 94. Ma,G,Ta S 129.Ha,H,I L 25. A,Ma,Ta S 60. A,Tb,Tc S 95. Ma,G,Tb U 130.Ha,H,Tb U 26. A,Ma,Tb U 61. A,Tb,I S 96. Ma,G,I S 131.Ha,H,I S 27. A,Ma,I S 62.O,Ma, Mb S 97. Ma,Ha,Hb S 132.Ha,Ta,Tb 28. A,Mb, Mc S 63.O,Ma, G S 98. Ma,Ha,H L 133.Ha,Ta,I S 29. A,Mb, G S 64.O,Ma,Ha L 99. Ma,Ha,Ta L 134.Ha,Tb,Tc 30. A,Mb,Ha L 65.O,Ma,Hb S 100. Ma,Ha,Tb U 135.Ha,Tb,I 31. A,Mb,Hb L 66.O,Ma,H S 101. Ma,Ha,I 136. H,Ta,Tb 32. A,Mb,Hc L 67.O,Ma,Ta L 102. Ma,Hb,Hc L 137. H,Ta,I 33. A,Mb,H S 68.O,Ma,Tb U 103. Ma,Hb,H S 138.Ta,Tb,Tc U 34. A,Mb,Ta S 69.O,Ma,I S 104. Ma,Hb,Ta S 139.Ta,Tb,I S 35. A,Mb,Tb L 70. O,G,Ha S 105. Ma,Hb,Tb S

R Redundant. Given the location of two of the points of the triple, the location of the third point is determined. L Locus Restricted. Given the location of two points, the third must lie on a certain locus. S Solvable. Known ruler and compass solutions exist for these triples. U Unsolvable with ruler and compass.