AMC10 2011 A

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 $49^2-9^2=40\cdot 58=2320$.