HMMT 二月 2026 · 冲刺赛 · 第 17 题
HMMT February 2026 — Guts Round — Problem 17
题目详情
- [11] A point P is selected uniformly at random on one of the straight edges of a quarter circle, and another point Q is chosen independently and uniformly at random on the other straight edge. Compute the probability there exists a point A on the arc of the quarter circle such that ∠ P AQ is obtuse.
解析
- [11] A point P is selected uniformly at random on one of the straight edges of a quarter circle, and another point Q is chosen independently and uniformly at random on the other straight edge. Compute the probability there exists a point A on the arc of the quarter circle such that ∠ P AQ is obtuse. Proposed by: Rohan Bodke π Answer: 1 − 4 Solution: Q M O P Let M be the midpoint of P Q and O be the center of the quarter circle. Denote ω by the circle with diameter P Q . The region of points X for which ∠ P XQ is obtuse is the interior of ω . Thus, the point A exists if and only if ω intersects the arc of the quarter circle at two points, which only happens when OM plus the radius of ω is greater than the radius of the quarter circle. Because ω passes through M , the radius of ω is equal to OM . Thus, we need to find the probability that 2 · OM is greater than the radius of the quarter circle. To compute this probability, set the coordinate O to (0 , 0) , ray OP to x-axis and OQ to y-axis. WLOG the radius of the quarter circle is 2 . Then, P = ( x, 0) and Q = (0 , y ) where x and y are random variable independently sampled uniformly from [0 , 2] . Hence, the distribution of M = ( x/ 2 , y/ 2) is uniform on the red unit square as shown in the image. We need to find the probability that 2 OM > 2 , which happens when M is outside the circle of radius 1 centered at O . This is exactly the portion of the red unit square that does not lie on the unit circle centered at O , which is 1 − π/ 4 π = 1 − . 1 4 ©2026 HMMT