HMMT 二月 2025 · 几何 · 第 3 题

HMMT February 2025 — Geometry — Problem 3

Point P lies inside square ABCD such that the areas of △ P AB , △ P BC , △ P CD , and △ P DA are 1, 2, 3, and 4, in some order. Compute P A · P B · P C · P D .

解析

Point P lies inside square ABCD such that the areas of △ P AB , △ P BC , △ P CD , and △ P DA are 1, 2, 3, and 4, in some order. Compute P A · P B · P C · P D . Proposed by: Karthik Venkata Vedula √ Answer: 8 10 Solution: A D P B C Let h , h , h , and h be the lengths of the altitudes from P to sides AB , BC , CD , and DA , 1 2 3 4 respectively. Then, the problem statement implies that { h , h , h , h } = { x, 2 x, 3 x, 4 x } for some x . 1 2 3 4 Furthermore, the area of the square is 1 + 2 + 3 + 4 = 10, so we have √ h + h = 10 = h + h . 1 3 2 4 Hence either ( { h , h } = { x, 4 x } and { h , h } = { 2 x, 3 x } ), or ( { h , h } = { x, 4 x } and { h , h } = 1 3 2 4 2 4 1 3 √ { 2 x, 3 x } ). In any case, we get 5 x = 10 from above, and finish via the Pythagorean theorem: p p p p 2 2 2 2 2 2 2 2 P A · P B · P C · P D = x + (2 x ) · (2 x ) + (4 x ) · (4 x ) + (3 x ) · (3 x ) + x √ 4 = 50 10 x 2 √ 2 = (50 10) · 5 √ = 8 10 .