HMMT 二月 2017 · 几何 · 第 2 题

HMMT February 2017 — Geometry — Problem 2

Let ABC be a triangle with AB = 13 , BC = 14, and CA = 15. Let be a line passing through two sides of triangle ABC . Line cuts triangle ABC into two figures, a triangle and a quadrilateral, that have equal perimeter. What is the maximum possible area of the triangle?

解析

Let ABC be a triangle with AB = 13 , BC = 14, and CA = 15. Let be a line passing through two sides of triangle ABC . Line cuts triangle ABC into two figures, a triangle and a quadrilateral, that have equal perimeter. What is the maximum possible area of the triangle? Proposed by: Sam Korsky 1323 Answer: 26 There are three cases: ` intersects AB, AC , ` intersects AB, BC , and intersects AC, BC . These cases are essentially identical, so let intersect segment AB at M and segment AC at N . Then the condition is equivalent to AM + M N + AN = M B + BC + CN + M N AM + AN = M B + CN + 15 but AN + CN = 14 and AM + BM = 13, so that BM + CN = 27 − AM − AN = AM − AN − 15 implying that AM + AN = 21. Now let ∠ BAC = θ for convenience, so that 1 [ AM N ] = AM · AN · sin θ 2 21 12 which is maximized when AM = AN = . Further we can easily calculate sin θ = (e.g. by LOC); 2 13 note that this is why the area is maximized in this case (we want to maximize sin θ , which is equivalent to maximizing θ , so θ should be opposite the largest side). Our answer is thus 1 21 21 12 1323 · · · = 2 2 2 13 26 Alternatively we could also calculate AM AN [ AM N ] = [ ABC ] · · AB AC 21 21 2 2 = 84 · · 13 14 which gives the same answer.