1998 AIME Problem 9

Question

Two mathematicians take a morning coffee break each day. They arrive at the cafeteria independently, at random times between

9

a.m. and

10

a.m., and stay for exactly

m

minutes. The probability that either one arrives while the other is in the cafeteria is

40\%,

and

m = a - b\sqrt{c},

where

a,

b,

and

c

are positive integers, and

c

is not divisible by the square of any prime. Find

a + b + c.

Answer

Plot the two arrival times as a point in a

60 \times 60

square; they meet exactly when

|x - y| \lt m

The complement consists of two right triangles that fit together into a square of side

60 - m,

so

(60 - m)^2 = \frac{3}{5} \cdot 3600

Let the arrival times be

x

and

y

minutes after

9

a.m., so

(x, y)

is uniform in a

60 \times 60

square. The two people meet exactly when

|x - y| \lt m.

The non-meeting region

|x - y| \ge m

consists of two right triangles with legs

60 - m,

with total area

(60 - m)^2.

Meeting with probability

40\%

means

(60 - m)^2 = 0.6 \cdot 3600 = 2160,

so

60 - m = \sqrt{2160} = 12\sqrt{15}.

Thus

m = 60 - 12\sqrt{15},

and

a + b + c = 60 + 12 + 15 = 87.

Problem 9 in other years