# Heron Triangles and Elliptic Curves

## Inhaltsverzeichnis

### A tale of two triangles

The following question arose in a group of teachers and mathematicians working for the Focus on Math project. If two triangles have the same area and the same perimeter, are they necessarily congruent? It turns out that the answer is no. For example, the triangle with sides 3, 4, and 5 has the same area and perimeter as the triangle with sides 41/15, 101/21, and 156/35.

Indeed, the perimeter is

$\frac{41}{15} + \frac{101}{21} + \frac{156}{35} = \frac{287 + 505 + 468}{105} = \frac{1260}{105} = 12 = 3 + 4 + 5.$

The triangle with sides 3, 4, and 5, shown on the right, has area $\frac{1}{2} 4 \cdot 3 = 6$. To find the area of the triangle on the left, we use Heron's formula, which states that the area $A$ of a triangle with side lengths $a$, $b$, and $c$, is given by

$A = \frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} = \sqrt{s(s-a)(s-b)(s-c)},$

where $s = \frac12(a+b+c)$ is the semiperimeter of the triangle. A quick calculation using this formula shows that the area of the triangle on the left is also 6.

##### Note

For a description of how the teachers and mathematicians in Focus on Math worked on this question see Steven Rosenberg, Michael Spillane, and Daniel~B. Wulf, Delving deeper: Heron triangles and moduli spaces, Mathematics Teacher 101 (2008), no.~9, 656.

### The space of triangles

Figure 1:The three angles $\alpha$, $\beta$, and $\gamma$ determine a triangle.

How do we find examples like this? The secret is to find the right way of representing the space of all triangles. There are many possible ways to do this. For example, we could think of the space of triangles as the subspace of all triples $(a,b,c) \in \mathbb{R}^3$ corresponding to the three sides of the triangle. Not every point in $\mathbb{R}^3$ corresponds to a triangle; for example, all the coordinates must be positive. Can you think of other restrictions?

There's another way of putting coordinates on the space of triangles using angles instead of lengths. Every triangle has an inscribed circle, and the radius $r$ of the circle has a simple relationship with the area $A$ and semiperimeter $s$, namely

$A = rs. \qquad\qquad (1)$

To see why this is true, drop perpendiculars from the center of the circle to the sides of the triangle, as in Figure 1. These perpendiculars form the altitudes of 3 smaller triangles with bases on the sides of of the big triangle and and vertices at the center of the inscribed circle. Adding up the areas of these triangles we get equation (1).

Equation (1) tells us that if two triangles have the same area and same semiperimeter, then the radii of their inscribed circles are also the same. So if we are looking for two such triangles we will find them in the space of all triangles inscribed around a fixed circle. Instead of using lengths to parameterize this space, we will use the angles formed by the three radii at the center of incircle, as in in Figure 1.

### A curve of triangles with constant area and perimeter

Figure 2: the curve of triangles

Inside this space we can find curves corresponding to a whole family of triangles with the same values of $A$ and $s$.

First, we express $s$ in terms of the angles $\alpha$, $\beta$, and $\gamma$ and the radius $r$ of the incircle, as follows. The radii and the lines from the vertices to the incenter break the triangle into six right triangles. Because the lines from the vertices to the center bisect the angles of the big triangle, these right triangles occur in congruent pairs. Taking one base length from each pair and adding , we get

$s=r (\tan \frac{\alpha}{2} + \tan \frac{\beta}{2} + \tan \frac{\gamma}{2}). \qquad\qquad (2)$

Equations (1) and (2) together tell us that if the area $A$ and semiperimeter $s$ are constant, then so is the sum of the tangents:

$\tan \frac{\alpha}{2} + \tan \frac{\beta}{2} + \tan \frac{\gamma}{2} = \frac{s^2}{A}. \qquad \qquad (3)$

Second, we translate this condition into an equation defining a curve in the plane. Let $x = \tan (\alpha/2)$, $y = \tan(\beta/2)$, and $z = \tan(\gamma/2)$. Since $\alpha+\beta+\gamma = 2\pi$, we have

$\frac{\gamma}{2} = {\pi} - \frac{\alpha}{2} - \frac{\beta}{2},$

so

$z = \tan \left( \frac{\gamma}{2} \right)= \tan\left({\pi} - \frac{\alpha}{2} - \frac{\beta}{2}\right) = - \tan\left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = - \frac{x + y}{1-xy}.$

Then, if $k$ is the constant $s^2/A$, equation (3) becomes for fixed $k$, the equation

$x + y - \frac{x + y}{1-xy} = k,$

which we rewrite as

$x^2y+xy^2=kxy - k.$

Every triangle with area $A$ and semiperimeter $s$ determines a point on this curve, and every point on the curve in a certain region of the plane corresponds to a triangle. The region corresponds to angles that actually work in Figure 1, namely angles satisfying $\alpha + \beta + \gamma = 2\pi$ and $0 < \alpha, \beta, \gamma < \pi$, which corresponds to the region $x > 0$, $y > 0$ and $xy > 1$ (since $z > 0$).

Figure 2 shows this curve for $k = 6$, the value corresponding to the triangle with sides 3, 4, and 5. Every point on the component of this curve in the positive quadrant corresponds to a triangle; the side lengths of the triangle are $a = x+ y$, $b = y + z$, and $c = z + x$. In particular, the points $(1,2)$, $(2,1)$, $(2, 3)$, $(3, 2)$, $(1, 3)$, and $(3, 1)$ all correspond to the triangle with sides $3$, $4$, and $5$, with the sides taken in different orders.

### Finding points on the curve

Because the curve in Figure 2 is defined by an equation of degree 3, we can find points on it using the secant method. Two points on the curve determine a secant which cuts the curve in one more point; finding the point amounts to solving a cubic equation in $x$, two of whose roots are already known. Since we already have 6 points on the curve, there are lots of possibilities for secants, and generating more points generates more possibilities. In fact, the curve has infinitely many points with rational coordinates. The two-secant procedure illustrated in Figure 2 leads to the point $(54/35, 25/21)$ (marked with a circle), which corresponds to the triangle with sides $41/15$, $101/21$, and $156/35$.

The secant procedure works for any cubic curve in the plane; such curves are called elliptic curves (not because the curves are themselves ellipses, but because they arise in the study of a certain class of complex functions called elliptic functions). The secant procedure allows one to define a group structure on the set of rational points on a elliptic curves (that is, points whose coordinates are rational numbers).

### The mathematical moral

The study of elliptic curves is a central area of research in number theory, with applications to the cryptographic schemes behind secure financial transactions on the web. Elliptic curves played a central role in the proof of Fermat's Last Theorem.

The story described in this article shows the remarkable unity of mathematics, starting as it does in high school and ending in research. Along the way we encountered a fundamental idea in modern mathematics: the idea of solving a problem about a particular type of object (triangles with area 6 and perimeter 12, for example) by situating the object in a more general space (the space of all triangles) and finding the right way of parameterizing that space.