HMMT 二月 2007 · 冲刺赛 · 第 36 题
HMMT February 2007 — Guts Round — Problem 36
题目详情
- [ 25 ] The Marathon. Let ω denote the incircle of triangle ABC . The segments BC, CA , and AB are tangent to ω at D , E , and F , respectively. Point P lies on EF such that segment P D is perpendicular to BC . The line AP intersects BC at Q . The circles ω and ω pass through B and C , respectively, 1 2 and are tangent to AQ at Q ; the former meets AB again at X , and the latter meets AC again at Y . The line XY intersects BC at Z . Given that AB = 15 , BC = 14, and CA = 13, find b XZ · Y Z c . 5
解析
- [ 25 ] The Marathon. Let ω denote the incircle of triangle ABC . The segments BC, CA , and AB are tangent to ω at D , E , and F , respectively. Point P lies on EF such that segment P D is perpendicular to BC . The line AP intersects BC at Q . The circles ω and ω pass through B and C , respectively, 1 2 and are tangent to AQ at Q ; the former meets AB again at X , and the latter meets AC again at Y . The line XY intersects BC at Z . Given that AB = 15 , BC = 14, and CA = 13, find b XZ · Y Z c . ′ ′ ′ Answer: 101 . Construct D diametrically opposed to D , so that ∠ DF D and ∠ DED are right, and ′ ′ ′ note that P lies on DD . By standard angle chasing, m ∠ F DD = β (half angle B ) and m ∠ D DE = γ . ′ ◦ ′ ◦ ′ ′ Thus, m ∠ DD F = 90 − β and m ∠ ED D = 90 − γ . Then by the law of sines, DE : ED : D F : ′ ′ F D = cos( γ ) : sin( γ ) : sin( β ) : sin( γ ). Now using 4 DEP ∼ 4 F D P and 4 DF P ∼ 4 ED P , we have ′ EP ED · ED sin( γ ) cos( γ ) c = = = . ′ P F F D · F D sin( β ) sin( β ) b ′ ′ Let the dilation centered at A sending E to C map P and F to P and F , respectively. Note that ′ ′ ′ ′ AF = AC as AE and AF are equal tangents, and CP : P F = EP : P F = c : b by similarity. Then by Menelaus’ theorem, ′ ′ BQ CP F A BQ c b 1 = = , ′ ′ QC P F AB QC b c ( ) 2 1 2 2 2 so that BQ = QC and AQ is actually a median. So, AQ = 2 b + 2 c − a = 148 . Now by Power 4 2 of a Point, AB · AX = AQ = AC · AY, so AX = 148 / 15 and AY = 148 / 13. Moreover, BXCY is cyclic as 4 ABC ∼ 4 AY X. Thus, XZ · Y Z = BZ · CZ , and it suffices to compute BZ/CZ . Menelaus once more gives BZ CY AX 1 = , ZC Y A XB 13 whence, BZ/CZ = ( AY /AX )( BX/CY ) = (15 / 13)((77 · 13) / (21 · 15)) = 11 / 3. We write CZ = 3 d and BZ = 11 d . Because AX < AB and AY < AC , Z does not lie on segment BC . Given the configuration 1617 2 information, BC = 8 d = 14, so d = 7 / 4, and finally b BZ · CZ c = b 33 d c = b c = 101 . 16 14