HMMT 二月 2013 · 几何 · 第 4 题
HMMT February 2013 — Geometry — Problem 4
题目详情
- 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 r 1 of line O O with ω . If AB = O A , find all possible values of . 1 2 2 1 r 2 ◦
解析
- 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 r 1 of line O O with ω . If AB = O A , find all possible values of . 1 2 2 1 r 2 √ √ − 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 2 Case 2: Similar to case 1, let us only consider the triangle ABO . AB = AO = O O = r because 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 . 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 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 ◦