HMMT 十一月 2013 · 冲刺赛 · 第 28 题

英文解析

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 .
    Answer: Solution 1. Denote by r the common radius of ω , ω , and let O , O be the centers 51
    1 2 1 2
    of ω and ω respectively. Suppose ω hits AC at B for i = 1 , 2, so that O O = B B = 2 r .7
    1 2 i i 1 2 1 2
    Extend angle bisector AO to hit BC at P . By the angle bisector theorem and triangle similarityr BP 3 r 41
    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
    or r = .5
    Solution 2. Use the same notation as in the previous solution, and let α = ∠ A . By constructing 17
    right triangles with hypotenuses AO , O O , and O C and legs parallel to AB and BC , we obtain 2
    1 1 2 2
    4 = AB = r cot α + 2 r cos ∠ A + r.
    1+4
    1+cos 2 α 4
    But cot α = = = 3 and cos ∠ A = , so the above equation simplifies to 5
    sin 2 α 53
    8 28 r 5
    4 = r (3 + + 1) = ,
    5 5
    or r = .5
    °7