HMMT 二月 2024 · 冲刺赛 · 第 23 题

HMMT February 2024 — Guts Round — Problem 23

[12] Let ℓ and m be two non-coplanar lines in space, and let P be a point on ℓ . Let P be the point on 1 2 m closest to P , P be the point on ℓ closest to P , P be the point on m closest to P , and P be the 1 3 2 4 3 5 point on ℓ closest to P . Given that P P = 5 , P P = 3 , and P P = 2 , compute P P . 4 1 2 2 3 3 4 4 5 2 2

解析

[12] Let ℓ and m be two non-coplanar lines in space, and let P be a point on ℓ . Let P be the point 1 2 on m closest to P , P be the point on ℓ closest to P , P be the point on m closest to P , and P be 1 3 2 4 3 5 the point on ℓ closest to P . Given that P P = 5 , P P = 3 , and P P = 2 , compute P P . 4 1 2 2 3 3 4 4 5 Proposed by: Luke Robitaille √ 39 Answer: 4 Solution: The figure below shows the situation of the problem when projected appropriately, which will be explained later. ℓ P 1 √ 25 P 3 − √ h 2 √ 9 − 4 h 2 − P 5 m h √ 2 a 2 − P 2 h 2 P 4 Let a be the answer. By taking the z -axis to be the cross product of these two lines, we can let the lines be on the planes z = 0 and z = h , respectively. Then, by projecting onto the xy -plane, we get the above √ √ √ 2 2 2 diagram. The projected lengths of the first four segments are 25 − h , 9 − h , and 4 − h , and √ 2 2 2 a − h . By similar triangles, these lengths must form a geometric progression. Therefore, 25 − h , 2 2 2 2 2 9 − h , 4 − h , a − h is a geometric progression. By taking consecutive differences, 16 , 5 , 4 − a is a √ 25 39 2 geometric progression. Hence, 4 − a = = ⇒ a = . 16 4 2 2