Question

Let $R$ be a square region and $n \geq 4$ 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:

Let us first find all the points that are $100$ -ray and $60$ -ray partitional.

First, consider an extreme $100$ -ray partitional point. Let this be the point in the top-left corner.

Note that we must draw rays through the vertices of the square, since otherwise we will end up with non-triangular regions.

Since this is the topmost and leftmost point, we have that the areas of the top triangular region and the left triangle region must be minimized.

This means that they do not have any rays going through them, which also means that their areas must be the same.

Then we have that the distance from the point to the top and left side must be the same. We now have $96$ rays to split among the other $2$ regions.

Since the other two regions also have the same area, we will have to have $48$ rays in each region. This means that those two regions are split into $49$ equal regions.

Let $x$ be the distance between the point and the top of the square. We then have that $\dfrac{1 \cdot x}{2} = \dfrac{(1 - x) \cdot 1}{2} \cdot \dfrac{1}{49}.$

Simplifying gives $49x = 1 - x$ $x = \dfrac{1}{50}.$ Now, if we move the point right to the next $100$ -partitional point, we have that a ray from the right region gets moved to the left region.

Doing the same analysis again would tell us that the point is $\dfrac{2}{50}$ away from the left side and $\dfrac{1}{50}$ from the top side.

Repeating this process, moving the point right and down, gets us that all the $100$ -partitional points form a $49 \times 49$ grid, with each point $\dfrac{1}{50}$ away from adjacent points.

Similarly, we can find the $60$ -partitional points form a $29 \times 29$ grid where the points are $\dfrac{1}{30}$ apart.

We now have to find the overlap between these two grids. Note that the gcd of $60$ and $100$ is $10.$ This means that all the points that are on both grids themselves form another grid that is $9 \times 9$ and $\dfrac{1}{10}$ apart.

This means that there are $9^2 = 81$ points that are on both grids. Then there are $49^2 - 81 = 2401 - 81 = 2320$ points that are $100$ -partitional and not $60$ -partitional.

Thus, C is the correct answer.

2011 AMC 10A Problem 25

Solution:

Problem 25 in Other Years