HMMT 二月 2006 · 冲刺赛 · 第 38 题
HMMT February 2006 — Guts Round — Problem 38
题目详情
- [15] Suppose ABC is a triangle with incircle ω , and ω is tangent to BC and CA at D and E respectively. The bisectors of ∠ A and ∠ B intersect line DE at F and G respectively, such that BF = 1 and F G = GA = 6. Compute the radius of ω .
解析
- Suppose ABC is a triangle with incircle ω , and ω is tangent to BC and CA at D and E respectively. The bisectors of ∠ A and ∠ B intersect line DE at F and G respectively, such that BF = 1 and F G = GA = 6. Compute the radius of ω . √ 2 5 Answer: 5 1 1 1 Solution: Let α, β, γ denote the measures of ∠ A, ∠ B, ∠ C, respectively. We have 2 2 2 ◦ ◦ ◦ ◦ m ∠ CEF = 90 − γ, m ∠ F EA = 90 + γ, m ∠ AF G = m ∠ AF E = 180 − α − (90 + γ ) = β = m ∠ ABG , so ABF G is cyclic. Now AG = GF implies that BG bisects ∠ ABF . Since BG by definition bisects ∠ ABC , we see that F must lie on BC . Hence, F = D . If I denotes the incenter of triangle ABC , then ID is perpendicular to BC , but since A, I, F are collinear, we have that AD ⊥ BC . Hence, ABC is isoceles with AB = AC . Furthermore, BC = 2 BF = 2. Moreover, since ABF G is cyclic, ∠ BGA is a right angle. ′ ′ ′ Construct F on minor arc GF such that BF = 6 and F G = 1, and let AB = x . By 13 √ ′ ′ 2 the Pythagorean theorem, AF = BG = x − 36, so that Ptolemy applied to ABF G 2 yields x − 36 = x + 36. We have ( x − 9)( x + 8) = 0. Since x is a length we find x = 9. Now we have AB = AC = 9. Pythagoras applied to triangle ABD now yields √ √ √ √ 1 2 2 AD = 9 − 1 = 4 5, which enables us to compute [ ABC ] = · 2 · 4 5 = 4 5. 2 Since the area of a triangle is also equal to its semiperimeter times its inradius, we √ √ 2 5 have 4 5 = 10 r or r = . 5 REMARK. In fact, ABF G is always a cyclic quadrilateral for which AB plays a di- ameter. That is, we could have proven this fact without using F G = GA .