2015 AMC 10A Problem 25

Question

Let

S

be a square of side length

1.

Two points are chosen independently at random on the sides of

S.

The probability that the straight-line distance between the points is at least

\dfrac{1}{2}

is

\dfrac{a-b\pi}{c},

where

a,

b,

and

c

are positive integers with

\gcd(a,b,c)=1.

What is

a+b+c?

59

60

61

62

63

Condition on whether the second point is on the same, adjacent, or opposite side

The adjacent-side case removes a quarter circle of radius

\frac{1}{2}

Fix one of the two points. The second point is on the same side with probability

\frac14

, on an adjacent side with probability

\frac12

, and on the opposite side with probability

\frac14

.

On the same side, two coordinates

a,b\in[0,1]

are at distance at least

\frac12

when

|a-b|\ge\frac12

. This region consists of two right triangles with total area

\frac14

.

On adjacent sides, the distance has the form

\sqrt{a^2+b^2}

. The failing region is a quarter circle of radius

\frac12

, so the success probability is

1-\frac{\pi}{16}

.

On opposite sides, the distance is always at least

1

, so the success probability is

1

. Therefore the desired probability is

\begin{aligned} &\frac14\cdot\frac14 \\ &\quad {}+\frac12\left(1-\frac{\pi}{16}\right) \\ &\quad {}+\frac14=\frac{26-\pi}{32}. \end{aligned}

Hence

a+b+c=26+1+32=59

.

Thus, A is the correct answer.

Problem 25 in other years