HMMT 二月 2003 · 几何 · 第 10 题

HMMT February 2003 — Geometry — Problem 10

Convex quadrilateral M AT H is given with HM/M T = 3 / 4, and AT M = M AT = ◦ 6 AHM = 60 . N is the midpoint of M A , and O is a point on T H such that lines M T, AH, N O are concurrent. Find the ratio HO/OT . 2

解析

Convex quadrilateral M AT H is given with HM/M T = 3 / 4, and AT M = M AT = ◦ 6 AHM = 60 . N is the midpoint of M A , and O is a point on T H such that lines M T, AH, N O are concurrent. Find the ratio HO/OT . Solution: 9 / 16 6 6 Triangle M AT is equilateral, so HM/AT = HM/M T = 3 / 4. Also, AHM = AT M , so the quadrilateral is cyclic. Now, let P be the intersection of M T, AH, N O . Extend M H and N O to intersect at point Q . Then by Menelaus’s theorem, applied to triangle 4 AHM and line QN P , we have HQ M N AP · · = 1 , QM N A P H while applying the same theorem to triangle T HM and line QP O gives HQ M P T O · · = 1 . QM P T OH Combining gives HO/OT = ( M P/P T ) · ( AN/N M ) · ( HP/P A ) = ( M P/P A ) · ( HP/P T ) (because AN/N M = 1). But since M AT H is cyclic, 4 AP T ∼ 4 M P H , so M P/P A = 2 HP/P T = HM/AT = 3 / 4, and the answer is (3 / 4) = 9 / 16. (See figure.) Q T H O P A M N 5