HMMT 十一月 2024 · 冲刺赛 · 第 21 题

HMMT November 2024 — Guts Round — Problem 21

[11] Two points are chosen independently and uniformly at random from the interior of the X-pentomino shown below. Compute the probability that the line segment between these two points lies entirely within the X-pentomino. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2024, November 09, 2024 — GUTS ROUND Organization Team Team ID# 3 3 3

解析

[11] Two points are chosen independently and uniformly at random from the interior of the X- pentomino shown below. Compute the probability that the line segment between these two points lies entirely within the X-pentomino. Proposed by: Benjamin Shimabukuro 21 Answer: = 0 . 84 25 Solution: Here are the cases in which the line segment is guaranteed to lie entirely within the X- pentomino: • At least one point lies in center square. • The two points lie in the same square. • The two points are in opposite edge squares (north and south, or east and west). Thus, the only case left to consider is the one where one point is in the north square and the other point is in the west square (and the other 3 analogous cases). By symmetry, the segment will lie outside the X-pentomino half of the time in this case, as illustrated by the red and blue segments in the following diagram. 4 2 8 The probability that two points are in non-opposite edge squares is · = . 5 5 25 Half the time, the segment will like outside the X-pentomino. So, the line segment will lie outside the 1 8 4 X-pentomino with probability · = . Thus, the probability that the line segment lies within the 2 25 25 4 21 X-pentomino is 1 − = . 25 25 3 3 3