2011 AMC 12A Problem 22

The $n$-ray partitional points form a grid: for $n = 2m,$ they are the interior points $\left(\tfrac{i}{m}, \tfrac{j}{m}\right),$ an $(m-1) \times (m-1)$ array

A point is both $100$- and $60$-ray partitional exactly when it is $20$-ray partitional

Scale the square to $[0,1]^2$ and write $X=(x,y).$ The rays must include those through the four vertices. Every small triangle has area $\frac{1}{n}.$ The triangles whose bases partition the bottom side together have area $\frac{y}{2},$ so their number is $\frac{ny}{2}.$ Similarly, the numbers along the top, left, and right sides are $\frac{n(1-y)}{2},$ $\frac{nx}{2},$ and $\frac{n(1-x)}{2}.$

These four numbers must be positive integers. Hence $n$ is even and $$\begin{gathered} X=\left(\dfrac{2i}{n},\dfrac{2j}{n}\right), \\ 1\le i,j\le\dfrac n2-1. \end{gathered}$$ Conversely, partitioning each side into the indicated number of equal segments and joining the division points to $X$ produces $n$ equal-area triangles. Thus these are exactly the partitional points.

For $n=100,$ the points are $(\frac{i}{50},\frac{j}{50})$ with $1\le i,j\le49,$ giving $49^2=2401.$ Such a point is also $60$-ray partitional exactly when $\frac{i}{50}=\frac{c}{30}$ and $\frac{j}{50}=\frac{d}{30}$ for integers $c,d.$ Thus $i$ and $j$ must both be multiples of $5.$ There are $9$ choices for each, so the overlap has $9^2=81$ points.

So the count is $2401 - 81 = 2320.$

Thus, the correct answer is C.