Quelltext der Seite Heron Triangles and Elliptic Curves

===A tale of two triangles=== 
[[Image:Heron_triangle.png|left|frame]][[Image:Triangle_3_4_5.png|frame]]

The following question arose in a group of teachers and mathematicians
working for the [http://www.focusonmath.org 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

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

The triangle with sides 3, 4, and 5, shown on the right, has area <math>\frac{1}{2} 4
\cdot 3 = 6</math>. To find the area of the  triangle on the left, we use
[http://en.wikipedia.org/wiki/Heron%27s_formula Heron's formula], which states that the area <math>A</math> of a triangle with
side lengths <math>a</math>, <math>b</math>, and <math>c</math>, is given by 

<math>
    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)},
</math>

where <math>s = \frac12(a+b+c)</math> is the semiperimeter of the triangle.
A
[http://www74.wolframalpha.com/input/?i=A+%3D+sqrt%286*%286-41%2F15%29*%286-101%2F21%29*%286-156%2F35%29%29 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 the
this question see Steven Rosenberg, Michael Spillane, and Daniel~B. Wulf,
<i>Delving deeper: Heron triangles and moduli spaces</i>,
Mathematics Teacher <b>101</b> (2008), no.~9, 656.

===The space of triangles===

[[Image:Heron_Triangles_Elliptic_Curves_Figure_2.png|frame|Figure 1:The three angles <math>\alpha</math>, <math>\beta</math>, and <math>\gamma</math> 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 <math>(a,b,c) \in \mathbb{R}^3</math> corresponding
to the three sides of the triangle. Not every point in <math>\mathbb{R}^3</math>
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 <math>r</math> of the circle has a simple relationship with
the area <math>A</math> and semiperimeter <math>s</math>, namely

<math>
  A = rs. \qquad\qquad (1)
</math>

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===


[[Image:Heron_Triangles_Elliptic_Curves_Figure_3.png|frame|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 <math>A</math> and <math>s</math>. 

First, we express <math>s</math> in terms of the angles <math>\alpha</math>, <math>\beta</math>, and
<math>\gamma</math> and the radius <math>r</math>
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 

<math>
        s=r (\tan \frac{\alpha}{2} + \tan \frac{\beta}{2} + \tan
        \frac{\gamma}{2}). \qquad\qquad (2)
</math>



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

<math>
\tan \frac{\alpha}{2} + \tan \frac{\beta}{2} + \tan
\frac{\gamma}{2} = \frac{s^2}{A}. \qquad \qquad (3)
</math>

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

<math>
  \frac{\gamma}{2} = {\pi} - \frac{\alpha}{2} - \frac{\beta}{2}, 
</math>

so 

<math>
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}. 
</math>

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

<math>
  x + y - \frac{x + y}{1-xy} = k,
</math>

which we rewrite as 

<math>
  x^2y+xy^2=kxy - k.
</math>

Every triangle with area <math>A</math> and semiperimeter <math>s</math> 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 
<math>\alpha + \beta + \gamma = 2\pi</math> and <math>0 < \alpha, \beta, \gamma < \pi</math>, which
corresponds to the region <math>x > 0</math>, <math>y > 0</math> and <math>xy > 1</math> (since <math>z > 0</math>).

Figure 2 shows this curve for <math>k = 6</math>, 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 <math>a =
x+ y</math>, <math>b = y + z</math>, and <math>c = z + x</math>. In particular, the points
<math>(1,2)</math>, <math>(2,1)</math>, <math>(2, 3)</math>, <math>(3, 2)</math>, <math>(1, 3)</math>, 
and <math>(3, 1)</math> all correspond to the triangle with sides <math>3</math>, <math>4</math>, and
<math>5</math>, 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 <math>x</math>, 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 <math>(54/35, 25/21)</math>
(marked with a circle), which corresponds to the triangle with sides
<math>41/15</math>, <math>101/21</math>, and <math>156/35</math>.

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.