HMMT 二月 2004 · 冲刺赛 · 第 41 题

英文解析

41. A tetrahedron has all its faces triangles with sides 13, 14, 15. What is its volume?
    √
    Solution: 42 55
    Let ABC be a triangle with AB = 13 , BC = 14 , CA = 15. Let AD, BE be altitudes.
    Then BD = 5 , CD = 9. (If you don’t already know this, it can be deduced from the
    2 2 2 2 2 2 2 2
    Pythagorean Theorem: CD − BD = ( CD + AD ) − ( BD + AD ) = AC − AB = 56,
    while CD + BD = BC = 14, giving CD − BD = 56 / 14 = 4, and now solve the linear
    √
    2 2
    system.) Also, AD = AB − BD = 12. Similar reasoning gives AE = 33 / 5,
    EC = 42 / 5.
    GIEA
    C F D BH
    Now let F be the point on BC such that CF = BD = 5, and let G be on AC suchthat CG = AE = 33 / 5. Imagine placing face ABC flat on the table, and letting Xbe a point in space with CX = 13 , BX = 14. By mentally rotating triangle BCXabout line BC , we can see that X lies on the plane perpendicular to BC through F .14
    In particular, this holds if X is the fourth vertex of our tetrahedron ABCX . Similarly,
    X lies on the plane perpendicular to AC through G . Let the mutual intersection ofthese two planes and plane ABC be H . Then XH is the altitude of the tetrahedron.
    To find XH , extend F H to meet AC at I . Then 4 CF I ∼ 4 CDA , a 3-4-5 triangle,
    so F I = CF · 4 / 3 = 20 / 3, and CI = CF · 5 / 3 = 25 / 3. Then IG = CI − CG = 26 / 15,
    and HI = IG · 5 / 4 = 13 / 6. This leads to HF = F I − HI = 9 / 2, and finally
    √
    √ √
    2 2 2 2
    XH = XF − HF = AD − HF = 3 55 / 2.
    Now XABC is a tetrahedron whose base 4 ABC has area AD · BC/ 2 = 12 · 14 / 2 = 84,
    √ √ √
    and whose height XH is 3 55 / 2, so its volume is (84)(3 55 / 2) / 3 = 42 55.
    2 2