Let $R$ be a square region and $n \ge4$ an integer. A point $X$ in the interior of $R$ is called n-ray partitional if there are n rays emanating from $X$ that divide $R$ into n triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?
设 $R$ 是一个正方形区域，$n \ge4$ 是一个整数。$R$ 内部的点 $X$ 称为 n-射线分割点，如果从 $X$ 发出 n 条射线，将 $R$ 分成 n 个面积相等的三角形。有多少点是 100-射线分割点但不是 60-射线分割点？
Answer (C): Assume without loss of generality that $R$ is bounded by the square with vertices $A=(0,0)$, $B=(1,0)$, $C=(1,1)$, and $D=(0,1)$, and let $X=(x,y)$ be $n$-ray partitional. Because the $n$ rays partition $R$ into triangles, they must include the rays from $X$ to $A$, $B$, $C$, and $D$. Let the number of rays intersecting the interiors of $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$ be $n_1,n_2,n_3,$ and $n_4$, respectively. Because $\triangle ABX\cup\triangle CDX$ has the same area as $\triangle BCX\cup\triangle DAX$, it follows that $n_1+n_3=n_2+n_4=\frac{n}{2}-2$, so $n$ is even. Furthermore, the $n_1+1$ triangles with one side on $\overline{AB}$ have equal area, so each has area $\frac{1}{2}\cdot\frac{1}{n_1+1}\cdot y$. Similarly, the triangles with sides on $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$ have areas $\frac{1}{2}\cdot\frac{1}{n_2+1}\cdot(1-x)$, $\frac{1}{2}\cdot\frac{1}{n_3+1}\cdot(1-y)$, and $\frac{1}{2}\cdot\frac{1}{n_4+1}\cdot x$, respectively. Setting these expressions equal to each other gives
\[
x=\frac{n_4+1}{n_2+n_4+2}=\frac{2(n_4+1)}{n}
\quad\text{and}\quad
y=\frac{n_1+1}{n_1+n_3+2}=\frac{2(n_1+1)}{n}.
\]
Thus an $n$-ray partitional point must have the form $X=\left(\frac{2a}{n},\frac{2b}{n}\right)$ with $1\le a<\frac{n}{2}$ and $1\le b<\frac{n}{2}$. Conversely, if $X$ has this form, $R$ is partitioned into $n$ triangles of equal area by the rays from $X$ that partition $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA}$ into $b,$ $\frac{n}{2}-a,$ $\frac{n}{2}-b,$ and $a$ congruent segments, respectively.
Assume $X$ is 100-ray partitional. If $X$ is also 60-ray partitional, then $X=\left(\frac{a}{50},\frac{b}{50}\right)=\left(\frac{c}{30},\frac{d}{30}\right)$ for some integers $1\le a,b\le 49$ and $1\le c,d\le 29$. Thus $30a=50c$ and $30b=50d$; that is, both $a$ and $b$ are multiples of 5. Conversely, if $a$ and $b$ are multiples of 5, then
\[
X=\left(\frac{a}{50},\frac{b}{50}\right)=\left(\frac{3a/5}{30},\frac{3b/5}{30}\right)
\]
is 60-ray partitional. Because there are exactly 9 multiples of 5 between 1 and 49, the required number of points $X$ is equal to $49^2-9^2=40\cdot 58=2320$.
答案（C）：不失一般性，设 $R$ 为以 $A=(0,0)$、$B=(1,0)$、$C=(1,1)$、$D=(0,1)$ 为顶点的正方形区域，令 $X=(x,y)$ 为 $n$ 射线可分点。由于这 $n$ 条射线把 $R$ 分成若干三角形，它们必然包含从 $X$ 指向 $A,B,C,D$ 的射线。设与 $\overline{AB}$、$\overline{BC}$、$\overline{CD}$、$\overline{DA}$ 内部相交的射线条数分别为 $n_1,n_2,n_3,n_4$。因为 $\triangle ABX\cup\triangle CDX$ 的面积与 $\triangle BCX\cup\triangle DAX$ 的面积相同，可得
\[
n_1+n_3=n_2+n_4=\frac{n}{2}-2,
\]
因此 $n$ 为偶数。又因为在 $\overline{AB}$ 上作为一边的 $n_1+1$ 个三角形面积相等，所以每个面积为 $\frac{1}{2}\cdot\frac{1}{n_1+1}\cdot y$。类似地，在 $\overline{BC}$、$\overline{CD}$、$\overline{DA}$ 上作为一边的三角形面积分别为 $\frac{1}{2}\cdot\frac{1}{n_2+1}\cdot(1-x)$、$\frac{1}{2}\cdot\frac{1}{n_3+1}\cdot(1-y)$、$\frac{1}{2}\cdot\frac{1}{n_4+1}\cdot x$。令这些表达式两两相等，得到
\[
x=\frac{n_4+1}{n_2+n_4+2}=\frac{2(n_4+1)}{n}
\quad\text{且}\quad
y=\frac{n_1+1}{n_1+n_3+2}=\frac{2(n_1+1)}{n}.
\]
因此，$n$ 射线可分点必须形如
\[
X=\left(\frac{2a}{n},\frac{2b}{n}\right),
\]
其中 $1\le a<\frac{n}{2}$ 且 $1\le b<\frac{n}{2}$。反之，若 $X$ 具有此形式，则从 $X$ 出发的射线可将 $R$ 分割成 $n$ 个面积相等的三角形：这些射线把 $\overline{AB},\overline{BC},\overline{CD},\overline{DA}$ 分别分成 $b,\ \frac{n}{2}-a,\ \frac{n}{2}-b,\ a$ 段全等线段。
设 $X$ 是 100 射线可分点。若 $X$ 同时也是 60 射线可分点，则
\[
X=\left(\frac{a}{50},\frac{b}{50}\right)=\left(\frac{c}{30},\frac{d}{30}\right),
\]
其中整数满足 $1\le a,b\le 49$ 且 $1\le c,d\le 29$。于是 $30a=50c$ 且 $30b=50d$，即 $a,b$ 都是 5 的倍数。反之，若 $a,b$ 都是 5 的倍数，则
\[
X=\left(\frac{a}{50},\frac{b}{50}\right)=\left(\frac{3a/5}{30},\frac{3b/5}{30}\right)
\]
为 60 射线可分点。由于在 1 到 49 之间恰有 9 个 5 的倍数，所以所求点 $X$ 的个数为
\[
49^2-9^2=40\cdot 58=2320.
\]