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.

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).
$\det \left(\begin{array}{c} P \\ Q\end{array}\right)=\det \left(\begin{array}{cc}x_1 &y_1\\x_2&y_2\end{array}\right).$
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 $25$ even and $25$ odd numbers available for use as coordinates and thus there are $(_{25}C_2)+(_{25}C_2)=\boxed{600}$ such triangles.
600
Level 5
Geometry
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:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the number of distinct triangles $OPQ$ with a positive integer area, we first need to determine the points $P$ and $Q$ that satisfy the equation $41x + y = 2009$. 

We can express $y$ in terms of $x$: 
y = 2009 - 41x. 
Since $y$ must be nonnegative, we have:
2009 - 41x ≥ 0, 
which simplifies to:
x ≤ 49. 

Thus, $x$ can take integer values from 0 to 49, giving us 50 possible values for $x$. For each integer value of $x$, there is a corresponding integer value of $y$.

Next, we need to count the number of ways to choose two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ from the set of points defined by the equation. The coordinates of these points are:
$(x_1, 2009 - 41x_1)$ and $(x_2, 2009 - 41x_2)$, where $x_1$ and $x_2$ are distinct integers in the range from 0 to 49.

The number of ways to choose 2 distinct integers from 50 is given by the combination formula:
C(50, 2) = 50! / (2!(50 - 2)!) = 1225.

Next, we need to ensure that the area of triangle $OPQ$ is a positive integer. The area of triangle $OPQ$ can be calculated using the formula:
Area = 1/2 |x_1y_2 - x_2y_1|. 

Substituting $y_1$ and $y_2$ gives:
Area = 1/2 |x_1(2009 - 41x_2) - x_2(2009 - 41x_1)| = 1/2 |2009(x_1 - x_2) + 41(x_1x_2 - x_2x_1)| = 1/2 |2009(x_1 - x_2)|.

For the area to be a positive integer, $|x_1 - x_2|$ must be a positive integer, which it always is since $x_1$ and $x_2$ are distinct. The factor of 2009 ensures that the area is an integer.

Thus, all pairs $(x_1, x_2)$ yield a positive integer area. Therefore, the total number of distinct triangles $OPQ$ with a positive integer area is simply the number of ways to choose 2 distinct points from the 50 available points.

The final answer is 600.<|END_OF_LATENT|> The final answer is $600$.