# 2011 AMC 12A Problems/Problem 22

The following problem is from both the 2011 AMC 12A #22 and 2011 AMC 10A #25, so both problems redirect to this page.

## Problem

Let $R$ be a unit 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?

$\textbf{(A)}\ 1500 \qquad \textbf{(B)}\ 1560 \qquad \textbf{(C)}\ 2320 \qquad \textbf{(D)}\ 2480 \qquad \textbf{(E)}\ 2500$

## Solution 1

There must be four rays emanating from $X$ that intersect the four corners of the square region. Depending on the location of $X$, the number of rays distributed among these four triangular sectors will vary. We start by finding the corner-most point that is $100$-ray partitional (let this point be the bottom-left-most point).

We first draw the four rays that intersect the vertices. At this point, the triangular sectors with bases as the sides of the square that the point is closest to both do not have rays dividing their areas. Therefore, their heights are equivalent since their areas are equal. The remaining $96$ rays are divided among the other two triangular sectors, each sector with $48$ rays, thus dividing these two sectors into $49$ triangles of equal areas.

Let the distance from this corner point to the closest side be $a$ and the side of the square be $s$. From this, we get the equation $\frac{a\times s}{2}=\frac{(s-a)\times s}{2}\times\frac1{49}$. Solve for $a$ to get $a=\frac s{50}$. Therefore, point $X$ is $\frac1{50}$ of the side length away from the two sides it is closest to. By moving $X$ $\frac s{50}$ to the right, we also move one ray from the right sector to the left sector, which determines another $100$-ray partitional point. We can continue moving $X$ right and up to derive the set of points that are $100$-ray partitional.

In the end, we get a square grid of points each $\frac s{50}$ apart from one another. Since this grid ranges from a distance of $\frac s{50}$ from one side to $\frac{49s}{50}$ from the same side, we have a $49\times49$ grid, a total of $2401$ $100$-ray partitional points. To find the overlap from the $60$-ray partitional, we must find the distance from the corner-most $60$-ray partitional point to the sides closest to it. Since the $100$-ray partitional points form a $49\times49$ grid, each point $\frac s{50}$ apart from each other, we can deduce that the $60$-ray partitional points form a $29\times29$ grid, each point $\frac s{30}$ apart from each other. To find the overlap points, we must find the common divisors of $30$ and $50$ which are $1, 2, 5,$ and $10$. Therefore, the overlapping points will form grids with points $s$, $\frac s{2}$, $\frac s{5}$, and $\frac s{10}$ away from each other respectively. Since the grid with points $\frac s{10}$ away from each other includes the other points, we can disregard the other grids. The total overlapping set of points is a $9\times9$ grid, which has $81$ points. Subtract $81$ from $2401$ to get $2401-81=\boxed{\textbf{(C)}\ 2320}$.

## Solution 2

Position the square region $R$ so that the bottom-left corner of the square is at the origin. Then define $s$ to be the sidelength of $R$ and $X$ to be the point $(rs, qs)$, where $0.

There must be four rays emanating from $X$ that intersect the four corners of $R$. The areas of the four triangles formed by these rays are then $A_1=\frac{qs\times s}{2}$, $A_2=\frac{(s-rs)\times s}{2}$, $A_3=\frac{(s-qs)\times s}{2}$, and $A_4=\frac{rs\times s}{2}$.

If a point is $n$-ray partitional, then there exist positive integers $a, b, c, d$ such that $a+b+c+d=n$ and $\frac{A_1}{a}=\frac{A_2}{b}=\frac{A_3}{c}=\frac{A_4}{d}$. Substituting in our formulas for $A_1$, $A_2$, $A_3$, and $A_4$ and canceling equal terms, we get $$\frac{q}{a}=\frac{1-r}{b}=\frac{1-q}{c}=\frac{r}{d}.$$

Taking $\frac{q}{a}=\frac{1-q}{c}$ and solving for $q$, we get $q=\frac{a}{a+c}$, and taking $\frac{1-r}{b}=\frac{r}{d}$ and solving for $r$, we get $r=\frac{d}{b+d}$. Finally, from $\frac{q}{a}=\frac{r}{d}$, we have $qd=ar$ $\Leftrightarrow$ $\frac{ad}{a+c}=\frac{ad}{b+d}$ $\Leftrightarrow$ $a+c=b+d$.

So for a point $X$ to be $100$-ray partitional, $a+b+c+d=100$, so $a+c=b+d=50$. $X$ must then be of the form $\left(\frac{d}{50}s, \frac{a}{50}s\right)$. Since $X$ is in the interior of $R$, $a$ and $d$ can be any positive integer from $1$ to $49$ (with $b$ and $c$ just equaling $50-d$ and $50-a$, respectively). Thus, there are $49\times 49=2401$ points that are $100$-ray partitional.

However, the problem asks for points that are not only $100$-ray partitional but also not $60$-ray partitional. Points that are $60$-ray partitional are of the form $\left(\frac{m}{30}s, \frac{n}{30}s\right)$, where $m$ and $n$ are also positive integers. We count the number of points $\left(\frac{d}{50}s, \frac{a}{50}s\right)$ that can also be written in this form. For a given $d$, $\frac{d}{50}=\frac{m}{30}$ if and only if $m=\frac{3}{5}d$, and likewise with $a$ and $n$. We can then see that a point is both $100$-ray partitional and $60$-ray partitional if and only if $a$ and $d$ are both divisible by $5$. There are $9$ integers between $1$ and $49$ that are divisible by $5$, so out of our $2401$ points that are $100$-ray partitional, $9\times 9=81$ are also $60$-ray partitional.

Our answer then is just $2401-81=\boxed{\textbf{(C)}\ 2320}$.

## Solution 3

For the sake of simplicity, let $R$ be a $60 \times 60$ square and set the bottom-left point as the origin. Then, $R$ has vertices: $$(0,0), (60,0), (60,60), (0,60).$$

Now, let a point in the square have coordinates $(x, y).$

In order for the point to be $100-$ ray partitional, we must be able to make $100$ triangles with area $60^2/100 = 36.$ For it to not be $60$-ray partitional, we cannot make $60$ triangles with area $60^2/60 = 60.$

When we draw such a triangle, it's base will always be on one of the sides of the square. If it is on the bottom side, then the height must be $y$. So, the base of each triangle must be $\frac{72}{y}.$ So, there will be $\frac{60}{\frac{72}{y}} = \frac{60y}{72}$ total triangles on that side.

If the triangle is on the right side of the square, then the height has to be $60 - x.$ So, the base will be $\frac{72}{60-x}$ and there will be $\frac{60 (60-x)}{72}$ triangles.

If the triangle is on the top, the height will be $60-y$, the base will be $\frac{72}{60-y}$ and there will be $\frac{60 (60-y)}{72}$ triangles along this side.

The triangles on the left will have height $x$, base $\frac{72}{x}$ and $\frac{60x}{72}$ such triangles must exist.

Simplifying, we get $\frac{5y}{6}, \frac{5 (60-x)}{6}, \frac{5(60-y)}{6}, \frac{5x}{6}$ triangles on the respective sides of the square.

All of these numbers must be integers. Let $x = 60a$ and $y = 60b$ where $0 < a, b < 1.$

Then, the amounts become: $$50a, 50b, 50 - 50a, 50 - 50b.$$

As long as $50a$ and $50b$ are integers, $50 - 50a$ and $50 - 50b$ will also be integers.

For this to happen, $a$ and $b$ must be of the form $\frac{x}{50}$ where $1 \le x \le 49.$ So, we have a total of $49^2 = 2401$ points that are $100$-ray partitional.

Now, we must calculate the number of $100$- ray partitional points that are also $60$-ray partitional.

Using a method similar to before, if a point is $60$-ray partitional, then we must be able to make $30a, 30b, 30 - 30a, 30 -30b$ triangles on the different sides.

So, $30a$ and $30b$ must be integers. This means $a$ and $b$ have to be of the form $\frac{y}{30}$ where $0

If a point is both $100$-ray partitional and $60$-ray partitional, then it can be written as $$\frac{x}{50} = \frac{y}{30}$$. Note that whenever $x$ is divisible by $5$, a $y$ will always exist to satisfy the above equation.

So, $x$ has to be in the range $(0, 50)$ and must be divisible by $5$. There are $9$ possibilities, namely $5, 10, 15, 20, 25, 30, 35, 40, 45.$

The $x$-coordinate of the point and the $y$-coordinate of the point can each use any of these $9$ possibilities, giving $9^2 = 81$ numbers that are both $100$-ray partitional and $60$-ray partitional.

Overall, we are left with $2401 - 81 = \boxed{2320}$ solutions.