2017 AIME I Problem 8

Attempt Problem 8 of the 2017 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AIME I solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

8.

Two real numbers aa and bb are chosen independently and uniformly at random from the interval (0,75).(0, 75). Let OO and PP be two points in the plane with OP=200.OP = 200. Let QQ and RR be points on the same side of line OPOP such that the degree measures of POQ\angle POQ and POR\angle POR are aa and b,b, respectively, and OQP\angle OQP and ORP\angle ORP are both right angles. The probability that QR100QR \le 100 is equal to mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Answer: 41
Concepts:geometric probabilityinscribed anglechordtrigonometry
Difficulty rating: 2920
Solution:

Since OQP=ORP=90,\angle OQP = \angle ORP = 90^\circ, both QQ and RR lie on the circle with diameter OP,\overline{OP}, whose radius is 100.100. The angle QOR=ab\angle QOR = |a - b| is an inscribed angle in this circle, so the chord satisfies QR=2100sinab.QR = 2 \cdot 100 \cdot \sin|a - b|. Because ab<75,|a - b| \lt 75^\circ, the condition QR100,QR \le 100, i.e. sinab12,\sin|a - b| \le \frac{1}{2}, is equivalent to ab30.|a - b| \le 30.

In the 75×7575 \times 75 square of equally likely pairs (a,b),(a, b), the region ab>30|a - b| \gt 30 consists of two right triangles with legs 7530=45,75 - 30 = 45, so the probability is 1452752=1925=1625.1 - \frac{45^2}{75^2} = 1 - \frac{9}{25} = \frac{16}{25}.

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

← Problem 7#7
Full Exam

Problem 8 in Other Years