HMMT 二月 2011 · ALGGEO 赛 · 第 20 题
HMMT February 2011 — ALGGEO Round — Problem 20
题目详情
- 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, 2 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 of triangle ASR , and let C and D be the midpoints of major arcs AP and AQ , respectively. If ∠ AP Q is 45 degrees and ∠ AQP is 30 degrees, determine ∠ COD in degrees.
解析
- 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, 2 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 of triangle ASR , and let C and D be the midpoints of major arcs AP and AQ , respectively. If ∠ AP Q is 45 degrees and ∠ AQP is 30 degrees, determine ∠ COD in degrees. Answer: 142 . 5 Q P A R ′ T B C S D T We use directed angles throughout the solution. Let T denote the point such that ∠ T CD = 1 / 2 ∠ AP Q and ∠ T DC = 1 / 2 ∠ AQP . We claim that T is the circumcenter of triangle SAR . 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 1 http://en.wikipedia.org/wiki/Balanced_ternary Algebra & Geometry Individual Test