PUMaC 2013 · 几何（A 组） · 第 6 题

PUMaC 2013 — Geometry (Division A) — Problem 6

[ 6 ] On a circle, points A, B, C, D lie counterclockwise in this order. Let the orthocenters of ABC, BCD, CDA, DAB be H, I, J, K respectively. Let HI = 2 , IJ = 3 , JK = 4 , KH = 5. 2 Find the value of 13( BD ) . ◦ ◦ ◦

解析

[ 6 ] On a circle, points A, B, C, D lie counterclockwise in this order. Let the orthocenters of ABC, BCD, CDA, DAB be H, I, J, K respectively. Let HI = 2 , IJ = 3 , JK = 4 , KH = 5. 2 Find the value of 13( BD ) . ! ! ! ! Solution We repeatedly take advantage of the fact that OH = OA + OB + OC for a triangle ABC with circumcenter O , orthocenter H . (This is a crucial lemma to a synthetic proof of the Euler Line and is well-known.) ! ! ! ! ! ! ! ! ! ! Using this, we have OH = OA + OB + OC and OI = OB + OC + OD . Thus we have HI = AD . ! ! ! ! ! ! Likewise we have IJ = DC , JK = CB , KH = BA . Thus the given length equations is actually just AB = 5 , BC = 4 , CD = 3 , DA = 2. From this one can use many ways to prove that s (2 ⇤ 5 + 4 ⇤ 3)(2 ⇤ 4 + 3 ⇤ 5) BD = . (2 ⇤ 5 + 4 ⇤ 3) (This is also well-known.)