HMMT 二月 2019 · 几何 · 第 5 题

HMMT February 2019 — Geometry — Problem 5

Isosceles triangle ABC with AB = AC is inscribed in a unit circle Ω with center O . Point D is the √ reflection of C across AB . Given that DO = 3, find the area of triangle ABC .

解析

Isosceles triangle ABC with AB = AC is inscribed in a unit circle Ω with center O . Point D is the √ reflection of C across AB . Given that DO = 3, find the area of triangle ABC . Proposed by: Lillian Zhang √ √ 2+1 2 − 1 Answer: OR 2 2 Solution 1. Observe that 1 1 1 ◦ ◦ ∠ DBO = ∠ DBA + ∠ ABO = ∠ CBA + ∠ BAO = ( ∠ CBA + ∠ BCA ) + ( ∠ BAC ) = (180 ) = 90 . 2 2 2 √ ◦ Thus BC = BD = 2 by the Pythagorean Theorem on 4 DBO . Then ∠ BOC = 90 , and the distance √ 2 from O to BC is . Depending on whether A is on the same side of BC as O , the height from A to 2 √ √ √ √ √ 2 2 2 2 ± 1 BC is either 1 + or 1 − , so the area is ( 2 · (1 ± )) / 2 = . 2 2 2 2 Solution 2. One can observe that ∠ DBA = ∠ CBA = ∠ ACB by property of reflection and ABC being isosceles, hence DB is tangent to Ω and Power of a Point (and reflection property) gives BC = √ √ 2 2 BD = OD − OB = 2. Proceed as in Solution 1. Note. It was intended, but not specified in the problem statement that triangle ABC is acute, so we accepted either of the two possible answers.