The Area of a Triangle

Chapter 22

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 construc- tion 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.