Square-like triangles Ralph H. Buchholz August 2004 Abstract We demonstrate that there are infinitely many triangles with the property that their area is the square of the tierce-semi-perimeter. These turn out to correspond to rational points on a rank one elliptic curve.

Keywords : Heron triangle, elliptic curve.

1

Introduction

If one considers the set of rational-sided rectangles, then a simple calculation reveals that the square is the only type which has the property that its area is equal to the square of half its semiperimeter. Clearly, if a rectangle has sides a and b then its semiperimeter is s = a + b and ab =

 s 2 2

implies a = b.

Now since the perimeter-to-area ratio is larger for triangles than it is for squares it seems appropriate to adjust the condition to ask if we can find solutions to Area(a, b, c) =

 s 2 3

.

(1)

An example of such a triangle is (a, b, c) = (9, 10, 17) with area 36 and semiperimeter 18 so that 36 = (18/3)2 .

2

Transformation to an elliptic curve

If we let ∆(a, b, c) denote the area of a triangle with sides a, b, c then we immediately recall what Hero of Alexandria taught us about 2000 years ago, namely, p ∆(a, b, c) = s(s − a)(s − b)(s − c) 1

where s is the semiperimeter of the triangle. Using this we can rewrite equation (1) as  s 4 s(s − a)(s − b)(s − c) = 3 or equivalently, since s 6= 0, 34 (s − a)(s − b)(s − c) = s3 . We find it useful to consider this as a homogeneous curve, F = 0, where F is defined by F = 81(−a + b + c)(a − b + c)(a + b − c) − (a + b + c)3 .

(2)

It was instructive (at least for the author) to follow the path provided by Knapp [2] to convert this into an elliptic curve in Weierstrass normal form. The first step is to find a rational flex. All flexes are determined by finding the solutions of the equation Resultant(F, |H(F )|) = 0, where H(F ) is the Hessian matrix of second partial derivatives given by   Faa Fab Fac H(F ) =  Fba Fbb Fbc  . Fca Fcb Fcc Thus we calculate |H(F )| = −11(a3 + b3 + c3 ) + 13(a(b2 + c2 ) + b(c2 + a2 ) + c(a2 + b2 )) − 36abc 26 39 and hence obtain Resultant(F, |H(F )|) = 232 331 bc(b + c)p6 (b, c) where p6 = 169c6 − 897c5 b + 2094c4 b2 − 2705c3 b3 + 2094c2 b4 − 897cb5 + 169b6 . We could pause here to consider the solutions to p6 = 0 but it is a trivial exercise to show that there are no non-trivial rational ones. So we just ignore it and conclude that F and |H(F )| have a common zero, and hence a point of inflexion when b + c = 0, namely at (a, b, c) = (0, 1, −1). The next step is to move this point to the point at infinity in P2 (C), namely (0, 1, 0). This is accomplished through the judicious choice of a matrix φ1 with determinant 1 such that φ1 (0, 1, −1) = (0, 1, 0). Clearly,   0 1 1 φ1 = 0 1 0 1 0 0 2

does the job nicely. The transformed curve, which Knapp calls F φ1 , is given by F φ1 = F (φ−1 1 (a, b, c)) = F (c, b, a − b) = −82(a3 + c3 ) + 78ac(a + c) + 324b(b − a)(c − a). The tangent line at the point at infinity is given by  φ1   φ1   φ1  ∂F ∂F ∂F a+ b+ c=0 ∂a [0,1,0] ∂b [0,1,0] ∂c [0,1,0] or 4a − 4c = 0. We can twist this line to  a  φ−1 2 = 0 c

c = 0 by setting  0 p 1 0 0 q

where a and c are solutions to the tangent equation and p and q are chosen so that φ−1 2 is non-singular. We choose   1 0 1 −1 0 φ2 = 0 1 1 0 −1 so that (F φ1 )φ2 = F φ1 (φ−1 2 (a, b, c)) = F φ1 (a + c, b, a − c) = 8(−81b2 c + 81abc + 81bc2 − 81ac2 − a3 ). Finally, we simply need to ensure that the coefficients of b2 c and a3 in (F φ1 )φ2 are the same magnitude and opposite sign. If we set   −t 0 0 −1 φ3 =  0 t 0 0 0 1 where t = Cbbc /Caaa = −648/ − 8 = 81 then we get ((F φ1 )φ2 )φ3 = (F φ1 )φ2 (φ−1 3 (a, b, c)) = (F φ1 )φ2 (−81a, 81b, c) = 23 38 (−81abc + ac2 − 81b2 c + 81a3 + bc2 ). Thus we have, if we dehomogenise at c, the elliptic curve E : y 2 + (x − 1/81)y = x3 + x/81

3

and the complete transformation, Φ : E 7→ F , is given by (a, b, c) := Φ(x, y) −1 −1 = φ−1 1 (φ2 (φ3 (x, y, 1))) = (−81x − 1, 81y, −81x − 81y + 1).

Now using the Magma package one easily finds that the group of rational points Z E(Q) is isomorphic to the group 3Z ⊕ Z and in fact E(Q) = hT, P i where T = (0, 1/81) and P = (−1/36, 17/648). Using the mapping Φ we find that Φ(P ) = [10/8, 17/8, 9/8] which when scaled becomes the example mentioned in the introduction. Similarly, we can enumerate as many such triangles as we wish by simply considering odd multiples of the infinite order point, P , as shown in Table 1. Note the rather large sizes of the sides of the second and n 1 3 5

7

sides [10,17,9] [219306474,103852745,142121521] [15416595835563213110186, 25794834030387295378353, 36477747308373903790345] [59730389701899336405908520804697131111784778, 78154024353491180473910733247001559447683401, 29575567589570974106298998432091015725306385]

Table 1: Square-like triangles corresponding to odd multiples of P subsequent trinagles.

3

Generalisations

If we vary the fraction of the semiperimeter so that we ask for solutions to the equation  s 2 ∆(a, b, c) = (3) n then, by following the analogous transformation path as that in the previous section we find that, this corresponds to the family of elliptic curves  4  4 1 1 3 En : y + xy − y=x + x. n n 2

Again using Magma to determine the group En (Q) leads to the ranks shown in Table 2. An alternate generalisation is to consider cyclic quadrilaterals with the 4

n 1 2 3 4 5 6 7

rank 0 0 1 1 1 0 1

torsion Z/3Z Z/3Z Z/3Z Z/3Z Z/3Z Z/3Z Z/3Z

Table 2: Rank of elliptic curves En (Q) property that Area(a, b, c, d) =

 s 2 2

which leads to consideration of the equation 16(−a + b + c + d)(a − b + c + d)(a + b − c + d)(a + b + c − d) = (a + b + c + d)4 . (4) This quartic surface in four variables is not amenable to the techniques used for the triangular problem and so at present the best we can do is to attempt to find a solution through computational means. Unfortunately, the only rational d satisfying equation 4 that have turned up, with a, b, c inside the box a ≤ b ≤ c ≤ 500, were those in which a = b = c = d. One could, of course, consider other cyclic n-gons (see [1]) or even higher dimensional variants—but we leave these for the future.

4

References 1. Ralph H. Buchholz and James A. MacDougall, Cyclic polygons with rational sides and area, submitted. 2. Knapp, Elliptic Curves 3. http://www.hyperdictionary.com/

5

Square-like triangles

rational points on a rank one elliptic curve. ... The first step is to find a rational flex. All flexes .... Table 1: Square-like triangles corresponding to odd multiples of P.

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