HMMT 二月 2011 · ALGGEO 赛 · 第 12 题

英文解析

12. Let ABCDEF be a convex equilateral hexagon such that lines BC , AD , and EF are parallel. Let
    H be the orthocenter of triangle ABD . If the smallest interior angle of the hexagon is 4 degrees,
    determine the smallest angle of the triangle HAD in degrees.
    Answer: 3
    Algebra & Geometry Individual Test
    ′A
    F BA
    ′H
    E CB
    ′
    F BADD
    E CH
    Note that ABCD and DEF A are isosceles trapezoids, so ∠ BAD = ∠ CDA and ∠ F AD = ∠ EDA . In d
    °
    order for the hexagon to be convex, the angles at B , C , E , and F have to be obtuse, so ∠ A = ∠ D = 4 .
    Letting s be a side length of the hexagon, AD = AB cos ∠ BAD + BC + CD cos ∠ CDA = s (1 +
    2 cos ∠ BAD ), so ∠ BAD is uniquely determined by AD . Since the same equation holds for trapezoid
    ° ° ° °
    DEF A , it follows that ∠ BAD = ∠ F AD = ∠ CDA = ∠ EDA = 2 . Then ∠ BCD = 180 − 2 = 178 .
    ° ° °
    Since 4 BCD is isosceles, ∠ CDB = 1 and ∠ BDA = 1 . (One may also note that ∠ BDA = 1 byobserving that equal lengths AB and BC must intercept equal arcs on the circumcircle of isoscelestrapezoid ABCD ).
    ′ ′ ′
    Let A , B , and D be the feet of the perpendiculars from A , B , and D to BD , DA , and AB , respectively.
    Angle chasing yields
    ′ ′ ° ′ ′ ° ′ ′
    ∠ AHD = ∠ AHB + ∠ DHB = (90 − ∠ A AB ) + (90 − ∠ D DB )
    ° ° °
    = ∠ BDA + ∠ BAD = 1 + 2 = 3
    ° ′ °
    ∠ HAD = 90 − ∠ AHB = 89
    ° ′ °
    ∠ HDA = 90 − ∠ DHB = 88
    °
    Hence the smallest angle in 4 HAD is 3 .
    It is faster, however, to draw the circumcircle of DEF A , and to note that since H is the orthocenterof triangle ABD , B is the orthocenter of triangle HAD . Then since F is the reflection of B across
    ° ° °
    AD , quadrilateral HAF D is cyclic, so ∠ AHD = ∠ ADF + ∠ DAF = 1 + 2 = 3 , as desired.
    Algebra & Geometry Individual Test