HMMT 二月 2012 · 几何 · 第 9 题

英文解析

9. 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 ilies 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 · (1 · 2 + 2 · 2 + 2 · 2 + 2 · 1) = 8 + 4 2 .2
   2 2
   2 2 2