HMMT 二月 2012 · 几何 · 第 9 题
HMMT February 2012 — Geometry — Problem 9
题目详情
- Let O, O , O , O , O be points such that O , O, O and O , O, O are collinear in that order, OO = 1 2 3 4 1 3 2 4 1 √ ◦ 1 , OO = 2 , OO = 2 , OO = 2, and ∡ O OO = 45 . Let ω , ω , ω , ω be the circles with respective 2 3 4 1 2 1 2 3 4 centers O , O , O , O that go through O . Let A be the intersection of ω and ω , B be the intersection 1 2 3 4 1 2 of ω and ω , C be the intersection of ω and ω , and D be the intersection of ω and ω , with A, B, C, D 2 3 3 4 4 1 all distinct from O . What is the largest possible area of a convex quadrilateral P P P P such that P 1 2 3 4 i lies on O and that A, B, C, D all lie on its perimeter? i 2 2 2
解析
- Let O, O , O , O , O be points such that O , O, O and O , O, O are collinear in that order, OO = 1 2 3 4 1 3 2 4 1 √ ◦ 1 , OO = 2 , OO = 2 , OO = 2, and ∡ O OO = 45 . Let ω , ω , ω , ω be the circles with respective 2 3 4 1 2 1 2 3 4 centers O , O , O , O that go through O . Let A be the intersection of ω and ω , B be the intersection 1 2 3 4 1 2 of ω and ω , C be the intersection of ω and ω , and D be the intersection of ω and ω , with A, B, C, D 2 3 3 4 4 1 all distinct from O . What is the largest possible area of a convex quadrilateral P P P P such that P 1 2 3 4 i lies on O and that A, B, C, D all lie on its perimeter? i √ Answer: 8 + 4 2 We first maximize the area of triangle P OP , noting that the sum of the area of 1 2 P OP and the three other analogous triangles is the area of P P P P . Note that if A 6 = P , P , without 1 2 1 2 3 4 1 2 ◦ ◦ loss of generality say ∠ OAP < 90 . Then, ∠ OO P = 2 ∠ OAP , and since ∠ OAP = 180 − ∠ OAP > 1 1 1 1 2 1 ◦ 90 , we see that ∠ OO P = 2 ∠ OAP as well, and it follows that OO P ∼ OO P . This is a spiral 2 2 1 1 1 2 2 similarity, so OO O ∼ OP P , and in particular ∠ P OP = ∠ O OO , which is fixed. By the sine 1 2 1 2 1 2 1 2 area formula, to maximize OP · OP , which is bounded above by the diameters 2( OO ) , 2( OO ). In 1 2 1 2 a similar way, we want P , P to be diametrically opposite O , O in their respective circles. 3 4 3 4 When we take these P , we indeed have A ∈ P P and similarly for B, C, D , since ∠ OAP = ∠ OAP = i 1 2 1 2 ◦ 90 . To finish, the area of the quadrilateral is the sum of the areas of the four triangles, which is √ √ √ √ 1 2 2 · · 2 · (1 · 2 + 2 · 2 + 2 · 2 + 2 · 1) = 8 + 4 2 . 2 2 2 2 2