HMMT 二月 2010 · 几何 · 第 10 题

英文解析

10. [ 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
    C 1ωXP
    BQAMω
    Y2
    Let C be the fourth vertex of parallelogram AP CQ . The midpoint M of P Q is the intersection of the diagonals of this parallelogram. Because M has equal power with respect to the two circles ω and 3
    ← → ← →1
    ω , it lies on AB , the circles’ radical axis . Therefore, C lies on AB as well.4
    Using a series of parallel lines and inscribed arcs, we have:2
    ∠ 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 .
    http://en.wikipedia.org/wiki/Power_of_a_point 3
    http://en.wikipedia.org/wiki/Radical_axis 4
    Geometry Subject Test
    °
    We also know that ∠ BXP = 180 − ∠ P AB = ∠ CAP , so triangles BXP and CAP are similar. Bythe spiral similarity theorem , triangles BP C and XP A are similar, too.5
    By analogous reasoning, triangles BQC and Y QA are similar. Then we have:
    AX AX/BC AP/CP AP2 = = =
    AY AY /BC AQ/CQ AQ2
    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
    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 5
    too. It is true because the first similarity implies that AP/BP = XP/Y P and ∠ AP B = ∠ XP Y , which proves the secondsimilarity.
    Geometry Subject Test