HMMT 十一月 2013 · 冲刺赛 · 第 28 题

HMMT November 2013 — Guts Round — Problem 28

[ 15 ] Triangle ABC has AB = 4, BC = 3, and a right angle at B . Circles ω and ω of equal radii are 1 2 drawn such that ω is tangent to AB and AC , ω is tangent to BC and AC , and ω is tangent to ω . 1 2 1 2 Find the radius of ω . 1 ◦

解析

[ 15 ] Triangle ABC has AB = 4, BC = 3, and a right angle at B . Circles ω and ω of equal radii are 1 2 drawn such that ω is tangent to AB and AC , ω is tangent to BC and AC , and ω is tangent to ω . 1 2 1 2 Find the radius of ω . 1 5 Answer: Solution 1. Denote by r the common radius of ω , ω , and let O , O be the centers 1 2 1 2 7 of ω and ω respectively. Suppose ω hits AC at B for i = 1 , 2, so that O O = B B = 2 r . 1 2 i i 1 2 1 2 Extend angle bisector AO to hit BC at P . By the angle bisector theorem and triangle similarity 1 r BP 3 r 4 4 AB O ∼ 4 ABP , we deduce = = . Similarly, = , so 1 1 AB AB 4+5 CB 3+5 1 2 5 = AC = AB + B B + B C = 3 r + 2 r + 2 r = 7 r, 1 1 2 2 5 or r = . 7 1 Solution 2. Use the same notation as in the previous solution, and let α = ∠ A . By constructing 2 right triangles with hypotenuses AO , O O , and O C and legs parallel to AB and BC , we obtain 1 1 2 2 4 = AB = r cot α + 2 r cos ∠ A + r. 4 1+ 1+cos 2 α 4 5 But cot α = = = 3 and cos ∠ A = , so the above equation simplifies to 3 sin 2 α 5 5 8 28 r 4 = r (3 + + 1) = , 5 5 5 or r = . 7 ◦