HMMT 二月 2008 · 冲刺赛 · 第 24 题

HMMT February 2008 — Guts Round — Problem 24

[ 10 ] Suppose that ABC is an isosceles triangle with AB = AC . Let P be the point on side AC so that AP = 2 CP . Given that BP = 1, determine the maximum possible area of ABC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 11 HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

解析

[ 10 ] Suppose that ABC is an isosceles triangle with AB = AC . Let P be the point on side AC so that AP = 2 CP . Given that BP = 1, determine the maximum possible area of ABC . 9 Answer: Let Q be the point on AB so that AQ = 2 BQ , and let X be the intersection of BP 10 and CQ . The key observation that, as we will show, BX and CX are fixed lengths, and the ratio of areas [ ABC ] / [ BCX ] is constant. So, to maximize [ ABC ], it is equivalent to maximize [ BCX ]. Using Menelaus’ theorem on ABP , we have BX · P C · AQ = 1 . XP · CA · QB Since P C/CA = 1 / 3 and AQ/QB = 2, we get BX/XP = 3 / 2. It follows that BX = 3 / 5. By symmetry, CX = 3 / 5. Also, we have 5 [ ABC ] = 3[ BP C ] = 3 · [ BXC ] = 5[ BXC ] . 3 6 ◦ Note that [ BXC ] is maximized when ∠ BXC = 90 (one can check that this configuration is indeed ( ) 2 1 1 3 9 possible). Thus, the maximum value of [ BXC ] is BX · CX = = . It follows that the 2 2 5 50 9 maximum value of [ ABC ] is . 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 11 HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND