HMMT 二月 2013 · 几何 · 第 4 题

英文解析

4. Let ω and ω be circles with centers O and O , respectively, and radii r and r , respectively. Suppose
   1 2 1 2 1 2
   that O is on ω . Let A be one of the intersections of ω and ω , and B be one of the two intersections
   2 1 1 2
   1 rof line O O with ω . If AB = O A , find all possible values of .
   1 2 2 1
   2 r
   √ √
   − 1+ 5 1+ 5
   Answer: , There are two configurations to this problem, namely, B in between the
   2 2
   segment O O and B on the ray O O passing through the side of O Case 1: Let us only consider
   1 2 1 2 2
   the triangle ABO . AB = AO = O O = r because of the hypothesis and AO and O O are radii
   2 1 1 2 1 1 1 2
   of w . O B = O A = r because they are both radii of w .
   1 2 2 2 2
   Then by the isosceles triangles, ∠ AO B = ∠ ABO = ∠ ABO = ∠ O AB . Thus can establish that
   1 1 2 2
   △ ABO ∼ △ O AB .
   1 2
   Thus,
   r r
   2 1
   r r − r=
   1 2 1
   2 2
   r − r + r r = 0
   1 2
   1 2
   By straightforward quadratic equation computation and discarding the negative solution,
   √
   r − 1 + 5 =1
   r 2
   Case 2: Similar to case 1, let us only consider the triangle ABO . AB = AO = O O = r because 2
   1 1 1 2 1
   of the hypothesis and AO and O O are radii of w . O B = O A = r because they are both radii of
   1 1 2 1 2 2 2
   w .
   Then by the isosceles triangles, ∠ AO B = ∠ ABO = ∠ ABO = ∠ O AB . Thus can establish that 2
   1 1 2 2
   △ ABO ∼ △ O AB .
   1 2
   Now,
   r r
   2 1
   r r + r=
   1 2 1
   2 2
   r − r − r r = 0
   1 2
   1 2
   By straightforward quadratic equation computation and discarding the negative solution,
   √
   r 1 + 5 =1
   r 2
   °2