Question

Consider the set of all triangles $OPQ$ where $O$ is the origin and $P$ and $Q$ are distinct points in the plane with nonnegative integer coordinates $(x, y)$ such that $41x + y = 2009.$ Find the number of such distinct triangles whose area is a positive integer.

Answer

Solution:

The points on the line with nonnegative integer coordinates are $P_i = (i,\, 2009 - 41i)$ for $i = 0, 1, \ldots, 49$ — fifty points in all. For $P = P_i$ and $Q = P_j,$ the shoelace formula gives $[OPQ] = \frac{1}{2}\left|\,i(2009 - 41j) - j(2009 - 41i)\,\right| = \frac{2009}{2}\,|i - j|.$

This is automatically positive for distinct points, and since $2009$ is odd, it is an integer exactly when $i - j$ is even, that is, when $i$ and $j$ have the same parity. There are $25$ even and $25$ odd indices, so the number of triangles is $\binom{25}{2} + \binom{25}{2} = 300 + 300 = 600.$

2009 AIME I Problem 11

Solution:

Problem 11 in Other Years