2017 AIME I Problem 8

Question

Two real numbers

a

and

b

are chosen independently and uniformly at random from the interval

(0, 75).

Let

O

and

P

be two points in the plane with

OP = 200.

Let

Q

and

R

be points on the same side of line

OP

such that the degree measures of

\angle POQ

and

\angle POR

are

a

and

b,

respectively, and

\angle OQP

and

\angle ORP

are both right angles. The probability that

QR \le 100

is equal to

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

Answer: 41

Concepts:geometric probability inscribed angle chord trigonometry

Difficulty rating: 2920

Small hint:

The right angles at

Q

and

R

place both points on the circle with diameter

\overline{OP},

which has radius

100

Big hint:

The inscribed angle

\angle QOR = |a - b|

gives

QR = 200\sin|a-b|,

so the condition is

|a - b| \le 30;

find that region’s area in the

75 \times 75

square

Solution:

Since

\angle OQP = \angle ORP = 90^\circ,

both

Q

and

R

lie on the circle with diameter

\overline{OP},

whose radius is

100.

The angle

\angle QOR = |a - b|

is an inscribed angle in this circle, so the chord satisfies

QR = 2 \cdot 100 \cdot \sin|a - b|.

Because

|a - b| \lt 75^\circ,

the condition

QR \le 100,

i.e.

\sin|a - b| \le \frac{1}{2},

is equivalent to

|a - b| \le 30.

In the

75 \times 75

square of equally likely pairs

(a, b),

the region

|a - b| \gt 30

consists of two right triangles with legs

75 - 30 = 45,

so the probability is

1 - \frac{45^2}{75^2} = 1 - \frac{9}{25} = \frac{16}{25}.

Therefore

m + n = 16 + 25 = 41.