2015 AMC 10A Problem 25

Where the two points are depends on which sides they land on, so split into cases, solve each one, and then combine.

Step 1: how likely each case is.

1. Each point lands on one of the $4$ sides, each equally likely. Fix the side of the first point. The second point is on the same side with chance $\frac{1}{4}$, on one of the $2$ next-door sides with chance $\frac{2}{4} = \frac{1}{2}$, and on the opposite side with chance $\frac{1}{4}$.

Step 2: the chance of success in each case. In each case, describe the two points by two numbers between $0$ and $1$ (how far along its side each point is). All pairs are equally likely, so a chance is an area inside a $1 \times 1$ square.

2. Same side. Positions $u$ and $v$ on one side; the distance is the gap between them. We need the gap to be at least $\frac{1}{2}$. Examples: $u = 0.1$, $v = 0.7$ works; $u = 0.3$, $v = 0.6$ does not. In the $1 \times 1$ square of pairs, the good pairs fill two corner triangles, each with legs $\frac{1}{2}$ and area $\frac{1}{8}$. So the chance is $\frac{1}{8} + \frac{1}{8} = \frac{1}{4}$.
3. Next-door sides. Put the shared corner at $(0, 0)$, one point at $(u, 0)$ and the other at $(0, v)$. The distance is $\sqrt{u^2 + v^2}$. It is less than $\frac{1}{2}$ when the pair lies inside a quarter circle of radius $\frac{1}{2}$ around the corner, with area
$$\frac{1}{4} \cdot \pi \cdot \left(\frac{1}{2}\right)^2 = \frac{\pi}{16}.$$
So the chance of success is $1 - \frac{\pi}{16}$.
4. Opposite sides. The two points are at least $1$ apart (the width of the square), which is more than $\frac{1}{2}$. The chance of success is $1$.

Step 3: combine. Multiply each case's chance by its success chance and add:
$$\frac{1}{4} \cdot \frac{1}{4} + \frac{1}{2}\left(1 - \frac{\pi}{16}\right) + \frac{1}{4} \cdot 1 = \frac{2}{32} + \frac{16}{32} - \frac{\pi}{32} + \frac{8}{32} = \frac{26 - \pi}{32}.$$

So $a = 26$, $b = 1$, $c = 32$, which share no common factor, and $a + b + c = 26 + 1 + 32 = 59$.

The answer is $\boxed{\textbf{(A)}\ 59}$.

When the situation comes in different kinds (here, which pair of sides), handle each kind separately and weight it by how often it happens. Inside each kind, two positions between $0$ and $1$ make a square of possibilities, and the chance is an area. The $\pi$ in the answer form was a hint that one case involves a circle: the points near a corner.

Find the chance of failure (distance less than $\frac{1}{2}$) and subtract from $1$. Same side fails with chance $\frac{3}{4}$, contributing $\frac{1}{4} \cdot \frac{3}{4} = \frac{3}{16}$. Next-door sides contribute $\frac{1}{2} \cdot \frac{\pi}{16} = \frac{\pi}{32}$. Opposite sides never fail. So the failure chance is $\frac{6 + \pi}{32}$, and the success chance is $1 - \frac{6 + \pi}{32} = \frac{26 - \pi}{32}$.

A uniformly random point on the boundary amounts to picking one of the four sides (each with probability $\frac14$) and then a uniform position along it. For two points, the sides are the same with probability $\frac14$, adjacent with probability $\frac12$, and opposite with probability $\frac14$.

Same side. Positions $u, v \in [0,1]$; we need $|u - v| \ge \frac12$. In the unit square of $(u, v)$ this region is two right triangles with legs $\frac12$, total area $2 \cdot \frac18 = \frac14$.

Adjacent sides. Put the shared corner at the origin, so the points are $(u, 0)$ and $(0, v)$ and the distance is $\sqrt{u^2 + v^2}$. The failing region $u^2 + v^2 < \frac14$ is a quarter disc of radius $\frac12$, area $\frac14 \cdot \pi \cdot \frac14 = \frac{\pi}{16}$. Success probability $1 - \frac{\pi}{16}$.

Opposite sides. The distance is at least the side length $1 > \frac12$, so success probability $1$.

Combine with the weights:
$$\frac14 \cdot \frac14 + \frac12\left(1 - \frac{\pi}{16}\right) + \frac14 \cdot 1 = \frac{1}{16} + \frac12 + \frac14 - \frac{\pi}{32} = \frac{13}{16} - \frac{\pi}{32} = \frac{26 - \pi}{32}.$$
Here $a = 26$, $b = 1$, $c = 32$ with $\gcd(26, 1, 32) = 1$, so $a + b + c = 59$.

The answer is $\boxed{\textbf{(A)}\ 59}$.