2011 AMC 10A Problem 25

An $n$-ray point lies on an $(\frac{n}{2}-1)\times(\frac{n}{2}-1)$ grid

Subtract the overlap of the $49\times49$ and $29\times29$ grids

Scale the square to have side length $1,$ and write $X=(u,v),$ where $u$ and $v$ are its distances from the left and bottom sides. Every corner must be joined to $X$; otherwise one of the regions containing that corner would not be a triangle.

Each of the $n$ triangles has area $\frac{1}{n}.$ A triangle whose base lies on the bottom side has height $v,$ so its base has length $\frac{2}{nv}.$ Therefore the number of triangles along the bottom side is $\frac{nv}{2},$ which must be a positive integer. Applying the same argument to all four sides shows that $$\begin{gathered} \dfrac{nu}{2},\quad\dfrac{n(1-u)}{2},\\ \dfrac{nv}{2},\quad\dfrac{n(1-v)}{2} \end{gathered}$$ are positive integers. Conversely, whenever these four numbers are integers, subdividing each side into the indicated number of equal bases and joining the division points to $X$ produces the required triangles.

For $n=100,$ this says $u=\frac{i}{50}$ and $v=\frac{j}{50}$ for $i,j\in\{1,2,\ldots,49\}.$ Hence the $100$-ray points form a $49\times49$ grid. Similarly, the $60$-ray points have coordinates $u=\frac{i}{30}$ and $v=\frac{j}{30}$ with $i,j\in\{1,2,\ldots,29\}.$

A coordinate belongs to both grids exactly when $\frac{i}{50}=\frac{j}{30},$ or $3i=5j.$ Thus the common coordinates are $\frac{1}{10},\frac{2}{10},\ldots,\frac{9}{10},$ giving a $9\times9$ overlap. The requested number is $$49^2-9^2=2401-81=2320.$$

Thus, C is the correct answer.