AAMICABLE HERON TRIANGLES
IWAN PRATON AND NART SHALQINIA
BSTRACT . A Heron triangle is a triangle whose side lengths and areaare integers. Two Heron triangles are amicable if the perimeter of oneis the area of the other. We show, using elementary techniques, thatthere is only one pair of amicable Heron triangles. I NTRODUCTION
A Heron triangle is a triangle whose side lengths and area are integers.They are named for the Greek mathematician Heron (or Hero) of Alexan-dria, who is usually credited with inventing the formula for the area A ofa triangle in terms of its side lengths a , b , c : A = (cid:112) s ( s − a )( s − b )( s − c );here s is the semiperimeter ( a + b + c ).Heron triangles form a popular topic (e.g., [1], [2], [6], [7]), and newfacts about them are still being discovered. For example, Hirakawa andMatsumura [5] showed recently that there is a unique pair (up to scal-ing) of right and isosceles Heron triangles with the same perimeter andthe same area. Surprisingly, the proof uses sophisticated tools from thetheory of hyperelliptic curves. The result has been featured in a Number-phile video [4].The video focuses mostly on equable Heron triangles (called Super-Hero triangles in the video), i.e., Heron triangles where the perimeter isequal to the area. This seems analogous to perfect numbers . Recall that aperfect number is a positive integer whose aliquot sum is equal to itself.(The aliquot sum of n is the sum of the divisors of n , excluding n .) Inequable Heron triangles, the perimeter and area play the roles of n andits aliquot sum.Venturing beyond a single number, recall that a pair of positive inte-gers n and m form an amicable pair if the aliquot sum of n is equal to m and the aliquot sum of m is equal to n . Analogously, we define twoHeron triangles H and H to be amicable if the area of H is equal to theperimeter of H and the perimeter of H is equal to the area of H . a r X i v : . [ m a t h . M G ] J a n micable Heron triangles exist: the triangles with side lengths (3, 25, 26)and (9, 12, 15) form an example. They are an unusual looking pair. (SeeFigure 1.) F IGURE
1. A unique pair of trianglesSomewhat surprisingly, there is no other example.
Theorem.
There is only one pair of amicable Heron triangles: the (3, 25, 26) and (9, 12, 15) triangles.
In contrast to [5], we use completely elementary methods to prove thisresult. P
ROOFS
We begin by establishing our notation. Suppose we have a Heron tri-angle with side lengths a , b , c . Its semiperimeter s = ( a + b + c )/2 is aninteger. We define x = s − a , y = s − b , z = s − c , and set x ≤ y ≤ z . ThenHeron’s formula for the area of the triangle becomes (cid:112) sx y z . Lemma 1.
Suppose H is one of a pair of amicable Heron triangles withperimeter p and area A. Then A divides p .Proof. For any Heron triangle,2 A p = sx y z s = x y z is an integer. Apply this result to the partner triangle of H . Since theperimeter of H is equal to the area of its partner and the area of H isequal to the perimeter of its partner, we get the result of the lemma. (cid:3) We now take care of the case where both amicable triangles are equable.It is well-known that there are only five equable Heron triangles [3]: trian-gles with side lengths (5, 12, 13), (6, 8, 10), (6, 25, 29), (7, 15, 20), and (9, 10, 17). heir perimeters (and thus their areas) are all different, so none of themform an amicable pair. Equable triangles are not amicable.We conclude that if we have an amicable pair, then for one of the tri-angles the perimeter is larger than the area. These triangles are long andskinny, similar to the second triangle in Figure 1. There are not manysuch triangles. From now on, let H denote a triangle of this kind. Lemma 2.
For H as above, x + y + z ) > x y z . ( ∗ ) Proof.
This is a simple consequence of the perimeter of H being largerthan the area of H . (cid:3) These two lemmas are our main tool for cutting down the possible val-ues of x , y , and z . In fact, they suffice to show that there are only finitelymany. Lemma 3.
Let H be as above. Then there are only a finite number of x , y , zvalues that satisfy lemmas 1 and 2.Proof. We first show that x ≤
3. If x ≥
4, then by Lemma 2,4( z + z + z ) ≥ x + y + z ) > x y z ≥ · · z = z ,a contradiction, so x ≤ y ≤
9. If y ≥
10, then4(3 + z + z ) ≥ x + y + z ) > x y z ≥ z =⇒ + z > z =⇒ > z ,which is a contradiction since z ≥ y ≥
9. We conclude that there are onlyfinitely many values of x and y .We now tackle z . Lemma 1 states that 2 p / A is an integer, which means8 s (cid:112) sx y z ∈ (cid:78) =⇒ s sx y z ∈ (cid:78) =⇒ x + y + z ) x y z ∈ (cid:78) =⇒ x + y + z ) z ∈ (cid:78) .Let c = x + y . Then 64( z + c ) / z = z + zc + c + c / z ) is an integer,which implies that 64 c / z is an integer. So z must be a divisor of 64( x + y ) , and since there are only finitely many values of 64( x + y ) , there areonly finitely many values of z also. (cid:3) We now need to investigate only a finite number of cases. It turns outthat the possibilities can be cut down considerably by using the require-ment that the area of H is an integer. We provide an example here; othercases are similar.Suppose x = y =
4. The 2 p / A calculation in Lemma 3 showsthat 4 z divides 64( z + ; we conclude that z is a divisor of 2 · . Since z ≥
4, there are 18 possibilities for z . The area of H is (cid:112) z ( z + he 18 possible value of z , only z = x = y =
4, and z = x , y , z , we come up withjust four cases that satisfy all the requirements mentioned above: • x = y = z =
4, producing H with side lengths 5, 5, 8. • x = y = z =
3, producing H with side lengths 3, 4, 5. • x = y = z =
24, producing H with side lengths 3, 25, 26. • x = y = z = H with side lengths 3, 865, 866.The first two cases are easy to eliminate. The first case produces a tri-angle of area 12 and perimeter 18. Its amicable partner, if it exists, musthave semiperimeter 6. This yields three possibilities: x = y = z = x = y = z = x = y = z =
2, none of which yields an area of 18.The second case produces a triangle of area 6, but it is impossible to havea partner triangle of perimeter 6, the only possibility being an equilateraltriangle with side-length 2 that has an irrational area.If H is the fourth triangle listed above then its perimeter is 1734 and itsarea is 1224. Thus its partner triangle has perimeter 1224 and area 1734.Therefore for the partner triangle, we have x y z = /612 = x , y , and z are all odd. This contradictsthat the semiperimeter of the partner triangle is 612. Therefore H has nopartner triangle.The third case produces the amicable pair mentioned in the Theorem.This concludes the proof of the Theorem: there is a unique pair of ami-cable Heron triangles. R EFERENCES [1] J. Carlson, Determination of Heronian triangles. Fibonacci Quart. 8 (1970), no. 5,499–506, 551.[2] Wm F. Cheney, Heronian Triangles. Amer. Math. Monthly 36 (1929), no. 1, 22–28.[3] L. Dickson,
History of the Theory of Numbers, Volume Il . Dover Publications (2005).[4] B. Haran, Superhero triangles, [5] Y. Hirakawa and H. Matsumura, A unique pair of triangles. J. Number Theory, 194(2019), 297–302.[6] R. Nelsen, Almost Equilateral Heronian Triangles. Math. Mag. 93 (2020), no. 5, 378–379.[7] P. Yiu, Construction of indecomposable Heronian triangles. Rocky Mountain J.Math. 28 (1998), no. 3, 1189–1202.F
RANKLIN & M
ARSHALL C OLLEGE
Email address : [email protected], [email protected]@fandm.edu, [email protected]