PUMaC 2009 · 几何（A 组） · 第 8 题
PUMaC 2009 — Geometry (Division A) — Problem 8
题目详情
- Consider 4 ABC and a point M in its interior so that ∠ M AB = 10 , ∠ M BA = 20 , ∠ M CA = ◦ ◦ 30 and ∠ M AC = 40 . What is ∠ M BC ? 2
解析
- Consider 4 ABC and a point M in its interior so that ∠ M AB = 10 , ∠ M BA = 20 , ∠ M CA = ◦ ◦ 30 and ∠ M AC = 40 . What is ∠ M BC ? Solution. 60. Solution 1: ◦ ◦ Choose point X on BC such that ∠ XAM = 10 and ∠ XAC = 30 . Choose point Y on AC ◦ such that ∠ Y BM = 20 and Y 6 = A . Let the intersection of AX and BY be Z . By construction, ∠ ABM = ∠ ZBM and ∠ BAM = ∠ ZAM , therefore M is the incenter of ◦ 1 ◦ Triangle ABZ . Hence ∠ BM Z = 90 + ∠ BAZ = 100 = ∠ BM C . This shows that points 2 M , Z , C are collinear. ◦ ◦ Since ZM bisects ∠ AZB , we have ∠ AZY = ∠ CZY = 60 , also ∠ ZAC = ∠ ZCA = 30 , ∼ hence 4 AZY 4 CZY . Therefore BY is the perpendicular bisector of AC = ⇒ ∠ CBY = = ◦ ◦ ∠ ABY = 40 = ⇒ ∠ M BC = ∠ M BZ + ∠ CBY = 60 4 Solution 2: Set up rectangluar coordinates s.t. M = (0, 0), A = (0, 1). Then ◦ BA : y = tan 80 x + 1 ◦ CA : y = − tan 50 x + 1 ◦ BM : y = tan 60 x + 1 ◦ CM : y = − tan 20 x + 1 ◦ ◦ ◦ ◦ ◦ The y-coordinates of B and C are thus tan 20 / (tan 20 − tan 50 ) and tan 60 / (tan 60 − ◦ tan 80 ) respectively. We conjecture that these two y-coordinates are equal. To prove this, notice that ◦ ◦ tan 20 tan 60 ◦ ◦ ◦ ◦ = ⇐⇒ tan 20 tan 80 = tan 60 tan 50 ◦ ◦ ◦ ◦ tan 20 − tan 50 tan 60 − tan 80 ◦ ◦ ◦ ◦ ⇐⇒ tan 20 tan 40 tan 80 = tan 60 . Last equality is true by the triple angle formula ◦ ◦ tan x × tan(60 − x ) × tan(60 + x ) = tan(3 x ) . ◦ ◦ Hence AM ⊥ BC , ∠ M BC = 90 − ∠ M AB − ∠ M BA = 60 . 5