2010 AIME II Problem 2

Question

A point

P

is chosen at random in the interior of a unit square

S.

Let

d(P)

denote the distance from

P

to the closest side of

S.

The probability that

\frac{1}{5} \le d(P) \le \frac{1}{3}

is equal to

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

The set of points at distance at least

t

from every side is a concentric square of side

1 - 2t

The favorable region lies inside the square of side

\frac{3}{5}

and outside the square of side

\frac{1}{3};

subtract areas

The points with

d(P) \ge t

form a concentric square of side

1 - 2t.

So

d(P) \ge \frac{1}{5}

puts

P

inside the concentric square of side

\frac{3}{5},

and

d(P) \le \frac{1}{3}

keeps

P

outside the open concentric square of side

\frac{1}{3}.

Since the unit square has area

1,

the probability is the area between those two squares:

\begin{aligned} &\left(\frac{3}{5}\right)^2 - \left(\frac{1}{3}\right)^2 \\ &= \frac{9}{25} - \frac{1}{9} \\ &= \frac{81 - 25}{225} = \frac{56}{225}. \end{aligned}

Thus

m + n = 56 + 225 = 281.

Problem 2 in other years