# 2009 AIME I Problems/Problem 11

## Problem

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.

## Solution

Let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$

We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem).

$\det \left({\matrix {P \above Q}}\right)=\det \left({\matrix {x_1 \above x_2}\matrix {y_1 \above y_2}\right).$ (Error compiling LaTeX. ! Package amsmath Error: Old form `\matrix' should be \begin{matrix}.)

Since the triangle has half the area of the parallelogram, we just need the determinant to be even.

The determinant is

$$(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))$$

Since $2009$ is not even, $((x_1)-(x_2))$ must be even, thus the two $x$'s must be of the same parity. Also note that the maximum value for $x$ is $49$ and the minimum is $0$. There are then $25$ even and $25$ odd numbers and thus there are $(_{25}C_2)+(_{25}C_2)=\boxed{600}$ such triangles.