Question

Let $R$ be a square region and $n \ge 4$ 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?

$1500$

$1560$

$2320$

$2480$

$2500$

Solution:

For even $n = 2m,$ the $n$ -ray partitional points are exactly $\left(\tfrac{i}{m}, \tfrac{j}{m}\right)$ with $1 \le i, j \le m - 1,$ giving $(m-1)^2$ points.

For $n = 100$ ( $m = 50$ ) there are $49^2 = 2401$ points. A point is both $100$ - and $60$ -ray partitional iff its coordinates are multiples of $\tfrac{1}{10},$ i.e. it is $20$ -ray partitional, giving $9^2 = 81$ points.

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

Thus, the correct answer is C.

2011 AMC 12A Problem 22

Solution:

Problem 22 in Other Years