HMMT 二月 2011 · 冲刺赛 · 第 29 题
HMMT February 2011 — Guts Round — Problem 29
题目详情
- [ 16 ] Let ABC be a triangle such that AB = AC = 182 and BC = 140. Let X lie on AC such that 1 CX = 130. Let the line through X perpendicular to BX at X meet AB at X . Define X , X , . . . , 1 1 1 1 2 2 3 as follows: for n odd and n ≥ 1, let X be the intersection of AB with the perpendicular to X X n +1 n − 1 n through X ; for n even and n ≥ 2, let X be the intersection of AC with the perpendicular to n n +1 X X through X . Find BX + X X + X X + . . . . n − 1 n n 1 1 2 2 3
解析
- [ 16 ] Let ABC be a triangle such that AB = AC = 182 and BC = 140. Let X lie on AC such that 1 CX = 130. Let the line through X perpendicular to BX at X meet AB at X . Define X , X , . . . , 1 1 1 1 2 2 3 as follows: for n odd and n ≥ 1, let X be the intersection of AB with the perpendicular to X X n +1 n − 1 n through X ; for n even and n ≥ 2, let X be the intersection of AC with the perpendicular to n n +1 X X through X . Find BX + X X + X X + . . . . n − 1 n n 1 1 2 2 3 1106 Answer: 5 Let M and N denote the perpendiculars from X and A to BC , respectively. Since triangle ABC 1 is isosceles, we have M is the midpoint of BC . Moreover, since AM is parallel to X N , we have 1 N C M C X N 70 5 1 = ⇔ = = , so N C = 50. Moreover, since X N ⊥ BC , we find X C = 120 by 1 1 X C AC 130 182 13 1 the Pythagorean Theorem. Also, BN = BC − N C = 140 − 50 = 90, so by the Pythagorean Theorem, X B = 150 . 1 We want to compute X X = X B tan( ∠ ABX ). We have 2 1 1 1 ( ) ( ) 12 4 − 1 + tan( ∠ ABC ) tan( ∠ X BC ) 1 5 3 ( ) ( ) tan( ∠ ABX ) = tan( ∠ ABC − ∠ X BC ) = = 1 1 12 4 tan( ∠ ABC ) − tan( ∠ X BC ) 1 + 1 5 3 16 16 15 = = . 63 63 15 16 65 Hence X X = 150 · , and by the Pythagorean Theorem again, X B = 150 · . 2 1 2 63 63 AX n Next, notice that is constant for every nonnegative integer n (where we let B = X ). Indeed, 0 AX n +2 since X X is parallel to X X for each n , the dilation taking X to X for some n also n n +1 n +2 n +3 n n +2 takes X to X for all k . k k +2 Guts Round X X AX n +2 n +3 n Since 4 AX X ∼ 4 AX X with ratio for each even n , we can compute that = n +2 n +3 n n +1 AX X X n +2 n n +1 65 150 · AX n +2 63 = 1 − for every nonnegative integer n . Notice we use all three sides of the above similar AX 182 n triangles. We now split our desired sum into two geometric series, one with the even terms and one with the odd terms, to obtain 16 150 · 150 63 BX + X X + . . . = ( BX + X X + . . . ) + ( X X + X X + . . . ) = + 1 1 2 1 2 3 1 2 3 4 65 65 150 · 150 · 63 63 182 182 79 · 150 1106 63 = = . 65 150 · 63 5 182