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HMMT 十一月 2013 · 冲刺赛 · 第 29 题

HMMT November 2013 — Guts Round — Problem 29

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 15 ] Let 4 XY Z be a right triangle with ∠ XY Z = 90 . Suppose there exists an infinite sequence ofequilateral triangles X Y T , X Y T , . . . such that X = X, Y = Y , X lies on the segment XZ for all
    0 0 0 1 1 1 0 0 ii ≥ 0, Y lies on the segment Y Z for all i ≥ 0, X Y is perpendicular to Y Z for all i ≥ 0, T and Y arei i i iseparated by line XZ for all i ≥ 0, and X lies on segment Y T for i ≥ 1.
    i i − 1 i − 1
    Let P denote the union of the equilateral triangles. If the area of P is equal to the area of XY Z , find
    .XY
    Y Z
    2 2 2
解析

英文解析

  1. [ 15 ] Let 4 XY Z be a right triangle with ∠ XY Z = 90 . Suppose there exists an infinite sequence ofequilateral triangles X Y T , X Y T , . . . such that X = X, Y = Y , X lies on the segment XZ for all
    0 0 0 1 1 1 0 0 ii ≥ 0, Y lies on the segment Y Z for all i ≥ 0, X Y is perpendicular to Y Z for all i ≥ 0, T and Y arei i i iseparated by line XZ for all i ≥ 0, and X lies on segment Y T for i ≥ 1.
    i i − 1 i − 1
    Let P denote the union of the equilateral triangles. If the area of P is equal to the area of XY Z , find
    .XY
    Y Z
    Answer: 1 For any region R , let [ R ] denote its area.
    2 4
    Let a = XY , b = Y Z , ra = X Y . Then [ P ] = XYTXY T (1+r+r+⋅⋅⋅1 + r + r + · · ·), [ XY Z ] = XYYXXY Y X ($1 +
    1 1 0 1 1
    √ √
    2 4 2
    r + r + · · ·$), Y Y = ra 3, and b = ra 3(1 + r + r + · · · ) (although we can also get this by similartriangles).1
    √ √
    √
    a 3 1 3 − 1 XY a 1 − r 2
    √
    Hence = ( ra + a )( ra 3), or 2 r ( r + 1) = 1 = ⇒ r = . Thus = = = 1.
    4 2 2 Y Z br 3
    2 2 2