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 $m + n = 16 + 25 = 41.$

2017 AIME I Problem 8

Answer: 41

Difficulty rating: 2920

Small Hint:

Big Hint:

Solution:

Problem 8 in Other Years