HMMT 十一月 2018 · 冲刺赛 · 第 29 题

HMMT November 2018 — Guts Round — Problem 29

[ 15 ] An isosceles right triangle ABC has area 1. Points D, E, F are chosen on BC, CA, AB respectively such that DEF is also an isosceles right triangle. Find the smallest possible area of DEF .

解析

[ 15 ] An isosceles right triangle ABC has area 1. Points D, E, F are chosen on BC, CA, AB respectively such that DEF is also an isosceles right triangle. Find the smallest possible area of DEF . Proposed by: Yuan Yao 1 Answer: 5 Without loss of generality, suppose that AB is the hypotenuse. If F is the right angle, then F must be the midpoint of AB . To prove this, let X and Y be the feet from F to BC and AC . Since \ XF Y = \ DF E = 90 , we have \ XF D = \ Y F E so XF = DF cos \ XF D = EF cos \ Y F E = Y F. Hence F is equidistant from AC and BC so it is the midpoint of AB . Then the minimum area is achieved by minimizing DF ; this occurs when DF is perpendicular to BC . The triangle DEF then 1 becomes the medial triangle of ABC , so its area is . 4 If F is not the right angle, without loss of generality, let the right angle be D . Place this triangle in p p the complex plane such that C is the origin, B = 2, and A = 2 i . Now since D is on the real axis and E is on the imaginary axis, D = x and E = yi , and we can obtain F by a 90 degree counterclockwise rotation of D around E : this evaluates to F = y + ( x + y ) i . For F p p to be on AB , the only constraint is to have y + ( x + y ) = 2 = ) x = 2 2 y . To minimize the area, we minimize p p 2 2 2 2 2 2 DE x + y ( 2 2 y ) + y 5 y 4 2 y + 2 = = = 2 2 2 2 p p 1 2 2 which has a minimum of at y = . Since this is between 0 and 2, this is indeed a valid 5 5 configuration. 1 Finally, we take the smallest area of . 5