HMMT 二月 2010 · 几何 · 第 10 题
HMMT February 2010 — Geometry — Problem 10
题目详情
- [ 8 ] Circles ω and ω intersect at points A and B . Segment P Q is tangent to ω at P and to ω at Q , 1 2 1 2 and A is closer to P Q than B . Point X is on ω such that P X ‖ QB , and point Y is on ω such that 1 2 ◦ ◦ QY ‖ P B . Given that ∠ AP Q = 30 and ∠ P QA = 15 , find the ratio AX/AY .
解析
- [ 8 ] Circles ω and ω intersect at points A and B . Segment P Q is tangent to ω at P and to ω at Q , 1 2 1 2 and A is closer to P Q than B . Point X is on ω such that P X ‖ QB , and point Y is on ω such that 1 2 ◦ ◦ QY ‖ P B . Given that ∠ AP Q = 30 and ∠ P QA = 15 , find the ratio AX/AY . √ Answer: 2 − 3 P X ω C 1 M A Q B ω 2 Y Let C be the fourth vertex of parallelogram AP CQ . The midpoint M of P Q is the intersection of the 3 diagonals of this parallelogram. Because M has equal power with respect to the two circles ω and 1 ← → ← → 4 ω , it lies on AB , the circles’ radical axis . Therefore, C lies on AB as well. 2 Using a series of parallel lines and inscribed arcs, we have: ∠ AP C = ∠ AP Q + ∠ CP Q = ∠ AP Q + ∠ P QA = ∠ ABP + ∠ QBA = ∠ P BQ = ∠ XP B, where the last equality follows from the fact that P X ‖ QB . 3 http://en.wikipedia.org/wiki/Power_of_a_point 4 http://en.wikipedia.org/wiki/Radical_axis Geometry Subject Test ◦ We also know that ∠ BXP = 180 − ∠ P AB = ∠ CAP , so triangles BXP and CAP are similar. By 5 the spiral similarity theorem , triangles BP C and XP A are similar, too. By analogous reasoning, triangles BQC and Y QA are similar. Then we have: 2 AX AX/BC AP/CP AP = = = 2 AY AY /BC AQ/CQ AQ where the last inequality holds because AP CQ is a parallelogram. Using the Law of Sines, the last √ 2 ◦ sin 15 expression equals = 2 − 3. 2 ◦ sin 30 5 This theorem states that if 4 P AB and 4 P XY are similar and oriented the same way, then 4 P AX and 4 P BY are similar too. It is true because the first similarity implies that AP/BP = XP/Y P and ∠ AP B = ∠ XP Y , which proves the second similarity. Geometry Subject Test