HMMT 二月 2013 · 团队赛 · 第 3 题

HMMT February 2013 — Team Round — Problem 3

[ 15 ] Let ABC be a triangle with circumcenter O such that AC = 7. Suppose that the circumcircle of AOC is tangent to BC at C and intersects the line AB at A and F . Let F O intersect BC at E . Compute BE .

解析

[ 15 ] Let ABC be a triangle with circumcenter O such that AC = 7. Suppose that the circumcircle of AOC is tangent to BC at C and intersects the line AB at A and F . Let F O intersect BC at E . Compute BE . 7 Answer: EB = O is the circumcenter of △ ABC = ⇒ AO = CO = ⇒ ∠ OCA = ∠ OAC . 2 Because AC is an inscribed arc of circumcircle △ AOC , ∠ OCA = ∠ OF A . Furthermore BC is tangent to circumcircle △ AOC , so ∠ OAC = ∠ OCB . However, again using the fact that O is the circumcenter of △ ABC , ∠ OCB = ∠ OBC . We now have that CO bisects ∠ ACB , so it follows that triangle CA = CB . Also, by AA similarity 2 2 we have EOB ∼ EBF . Thus, EB = EO · EF = EC by the similarity and power of a point, so EB = BC/ 2 = AC/ 2 = 7 / 2. Team Round