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HMMT 二月 1998 · 几何 · 第 8 题

HMMT February 1998 — Geometry — Problem 8

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

Question Eight . [6 points]
It is not possible to construct a segment of length π using astraightedge, compass, and a given segment of length 1. Thefollowing construction, given in 1685 by Adam Kochansky, yields asegment whose length agrees with π to five decimal places:
Construct a circle of radius 1 and call its center O . Construct adiameter AB of this circle and a line l tangent to the circle at A .
Next, draw a circle with radius 1 centered at A , and call one of theintersections with the original circle C . Now from C draw an arc ofradius 1 intersecting the circle around A at D , where D lies outsideof the circle centered at O . Draw OD and let E be its point ofintersection with l . Construct H on AE such that A is between
H and E , and HE =3.
The distance between B and H is then close to π ; calculate its exactvalue.

解析

英文解析

  1. Since OD ⊥ AC and 4 AOC is equilateral, we have AOD = 30 . So AE = , and
    √3


    √ ( )
    1 402
    2 2 2

    BH = AB + AH = 2 + 3 − = − 2 3 ≈ 3 . 141533339.
    233