2011 AMC 10A Problem 25

Let the square have corners $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$, so its area is $1$. Let the point be $X = (p, q)$ inside it. Each of the $n$ triangles must have area $\frac{1}{n}$.

1. Every corner is on a ray. Each triangle has corner $X$, and its opposite side runs along the edge of the square. A triangle's side is straight, so it cannot bend around a corner of the square. So there must be a ray through each of the four corners.
2. The four big triangles. The four corner rays split the square into four triangles, one on each side of the square. Each has base $1$ (a side of the square), and its height is the distance from $X$ to that side:
- bottom side: height $q$, area $\frac{q}{2}$;
- top side: height $1 - q$, area $\frac{1 - q}{2}$;
- left side: height $p$, area $\frac{p}{2}$;
- right side: height $1 - p$, area $\frac{1 - p}{2}$.
3. The other rays cut these into equal pieces. Say the bottom triangle is cut into $k$ small triangles, each of area $\frac{1}{n}$. Then $\frac{q}{2} = \frac{k}{n}$, so $q = \frac{2k}{n}$. So $q$ must be a multiple of $\frac{2}{n}$. The same reasoning applies to $1 - q$, $p$ and $1 - p$.
4. And this is enough. If $p$ and $q$ are multiples of $\frac{2}{n}$ between $0$ and $1$, divide each side of the square into equal pieces of the right length. All small triangles on one side have the same height and the same base, so they all have area $\frac{1}{n}$. So the $n$-ray partitional points are exactly those with
$$p \text{ and } q \text{ each one of } \frac{2}{n}, \frac{4}{n}, \ldots, 1 - \frac{2}{n}.$$
5. Check with the smallest case. For $n = 4$, the only choice is $\frac{2}{4} = \frac12$, so the only point is the center. That makes sense: four equal triangles, one on each side.
6. Count for $n = 100$. Each coordinate is one of $\frac{2}{100} = \frac{1}{50}, \frac{2}{50}, \ldots, \frac{49}{50}$: $49$ choices. So there are $49 \times 49 = 2401$ points.
7. Count for $n = 60$. Each coordinate is one of $\frac{1}{30}, \frac{2}{30}, \ldots, \frac{29}{30}$.
8. Points on both lists. A coordinate is on both lists when it is a multiple of $\frac{1}{50}$ and of $\frac{1}{30}$. Those are the multiples of $\frac{1}{10}$: $\frac{1}{10}, \frac{2}{10}, \ldots, \frac{9}{10}$, which is $9$ values. (For example, $\frac{3}{10} = \frac{15}{50} = \frac{9}{30}$.) So $9 \times 9 = 81$ points are both $100$-ray and $60$-ray partitional.
9. Subtract: $2401 - 81 = 2320$.

The answer is $\boxed{\textbf{(C)}\ 2320}$.

The whole problem opens up once you see that the rays have to go through the corners. After that, triangles that share the top point $X$ and have bases on the same side have equal area exactly when their bases are equal. So the condition becomes a grid of allowed points, and the rest is counting, with the two grids overlapping on the coarser grid of tenths.

The count of $n$-ray partitional points is $\left(\frac{n}{2} - 1\right)^2$. For $n = 100$ that is $49^2$, and the overlap with $n = 60$ is $9^2$. Use the difference of squares to finish quickly:
$$49^2 - 9^2 = (49 - 9)(49 + 9) = 40 \times 58 = 2320.$$