HMMT 二月 2011 · CALCGEO 赛 · 第 20 题

英文解析

20. Let ω and ω be two circles that intersect at points A and B . Let line l be tangent to ω at P and to
    1 2 1
    ω at Q so that A is closer to P Q than B . Let points R and S lie along rays P A and QA , respectively,
    so that P Q = AR = AS and R and S are on opposite sides of A as P and Q . Let O be the circumcenter 2
    of triangle ASR , and let C and D be the midpoints of major arcs AP and AQ , respectively. If ∠ AP Qis 45 degrees and ∠ AQP is 30 degrees, determine ∠ COD in degrees.
    Answer: 142 . 5
    RAPQ
    ′
    SCBT
    We use directed angles throughout the solution.TD
    Let T denote the point such that ∠ T CD = 1 / 2 ∠ AP Q and ∠ T DC = 1 / 2 ∠ AQP . We claim that T isthe circumcenter of triangle SAR .
    Calculus & Geometry Individual Test
    Since CP = CA , QP = RA , and ∠ CP Q = ∠ CP A + ∠ AP Q = ∠ CP A + ∠ ACP = ∠ CAR , we have
    ∼
    4 CP Q 4 CAR . By spiral similarity, we have 4 CP A ∼ 4 CQR .
    ′ ′ ′=
    Let T denote the reflection of T across CD . Since ∠ T CT = ∠ AP Q = ∠ ACP , we have 4 T CT ∼
    ′
    4 ACP ∼ 4 RCQ . Again, by spiral similarity centered at C , we have 4 CT R ∼ 4 CT Q . But
    ′ ′ ′ ′
    ∼
    CT = CT , so 4 CT R 4 CT Q and T R = T Q . Similarly, 4 DT T ∼ 4 DAQ , and spiral similarity
    ′ ′=
    ∼
    centered at D shows that 4 DT A 4 DT Q . Thus T A = T Q = T R .
    ′=
    We similarly have T A = T P = T S , so T is indeed the circumcenter. Therefore, we have ∠ COD =
    ° °
    45 30
    ° °
    ∠ CT D = 180 − − = 142 . 5 .
    2 2
    Calculus & Geometry Individual Test