HMMT 二月 2011 · ALGGEO 赛 · 第 16 题
HMMT February 2011 — ALGGEO Round — Problem 16
题目详情
- Let ABCD be a quadrilateral inscribed in the unit circle such that ∠ BAD is 30 degrees. Let m denote the minimum value of CP + P Q + CQ , where P and Q may be any points lying along rays AB and AD , respectively. Determine the maximum value of m . 2 π 2 π
解析
- Let ABCD be a quadrilateral inscribed in the unit circle such that ∠ BAD is 30 degrees. Let m denote the minimum value of CP + P Q + CQ , where P and Q may be any points lying along rays AB and AD , respectively. Determine the maximum value of m . Answer: 2 Algebra & Geometry Individual Test G A H R E Q F P B D C For a fixed quadrilateral ABCD as described, we first show that m , the minimum possible length of CP + P Q + QC , equals the length of AC . Reflect B , C , and P across line AD to points E , F , and R , respectively, and then reflect D and F across AE to points G and H , respectively. These two ◦ reflections combine to give a 60 rotation around A , so triangle ACH is equilateral. It also follows ◦ that RH is a 60 rotation of P C around A , so, in particular, these segments have the same length. Because QR = QP by reflection, CP + P Q + QC = CQ + QR + RH. The latter is the length of a broken path CQRH from C to H , and by the “shortest path is a straight line” principle, this total length is at least as long as CH = CA . (More directly, this follows from the triangle inequality: ( CQ + QR )+ RH ≥ CR + RH ≥ CH ). Therefore, the lower bound m ≥ AC indeed holds. To see that this is actually an equality, note that choosing Q as the intersection of segment CH with ray AD , and choosing P so that its reflection R is the intersection of CH with ray AE , aligns path CQRH with segment CH , thus obtaining the desired minimum m = AC . We may conclude that the largest possible value of m is the largest possible length of AC , namely 2: the length of a diameter of the circle. 2 π 2 π