2010 USAJMO Problems/Problem 4

Problem

A triangle is called a parabolic triangle if its vertices lie on a parabola $y = x^2$ . Prove that for every nonnegative integer $n$ , there is an odd number $m$ and a parabolic triangle with vertices at three distinct points with integer coordinates with area $(2^nm)^2$ .

Solution

Let the vertices of the triangle be $(a, a^2), (b, b^2), (c, c^2)$ . The area of the triangle is the absolute value of $A$ in the equation:

$A = \frac{1}{2}\det\left\vert \begin{array}{c c} b-a & c - a\\ b^2 - a^2 & c^2 - a^2 \end{array}\right\vert = \frac{(b-a)(c-a)(c-b)}{2}$

If we choose $a < b < c$ , $A > 0$ and gives the actual area. Furthermore, we clearly see that the area does not change when we subtract the same constant value from each of $a$ , $b$ and $c$ . Thus, all possible areas can be obtained with $a = 0$ , in which case $A = \frac{1}{2}bc(c-b)$ .

If a particular choice of $b$ and $c$ gives an area $A = (2^nm)^2$ , with $n$ a positive integer and $m$ a positive odd integer, then setting $b' = 4b$ , $c' = 4c$ gives an area $A' = 4^3 A = 8^2 A = (2^{n+3}m)^2$ .

Therefore, if we can find solutions for $n = 0$ , $n = 1$ and $n = 2$ , all other solutions can be generated by repeated multiplication of $b$ and $c$ by a factor of $4$ .

Setting $b=1$ and $c=2$ , we get $A=1 = (2^0\cdot1)^2$ , which yields the $n=0$ case.

Setting $b=1$ and $c=9$ , we get $A = 9\cdot4 = (2^1\cdot3)^2$ , which yields the $n=1$ case.

Setting $b=1$ and $c=5$ , we get $A=1\cdot2\cdot5 = 10$ . Multiplying these values of $b$ and $c$ by $10$ , we get $b'=10$ , $c'=50$ , $A'=10\cdot20\cdot50 = 100^2 = (2^2\cdot5^2)^2$ , which yields the $n=2$ case. This completes the construction.

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2010 USAJMO Problems/Problem 4

Problem

Solution