HMMT 二月 2013 · 冲刺赛 · 第 24 题

英文解析

24. [ 14 ] Given a point p and a line segment l , let d ( p, l ) be the distance between them. Let A , B , and Cbe points in the plane such that AB = 6, BC = 8, AC = 10. What is the area of the region in the
    ( x, y )-plane formed by the ordered pairs ( x, y ) such that there exists a point P inside triangle ABCwith d ( P, AB ) + x = d ( P, BC ) + y = d ( P, AC )?
    Answer: . Place ABC in the coordinate plane so that A = (0 , 6) , B = (0 , 0) , C = (8 , 0).288
    Consider a point P = ( a, b ) inside triangle ABC . Clearly, d ( P, AB ) = a, d ( P, BC ) = b . Now, we see 5
    Guts Round
    6 · 8
    that the area of triangle ABC is = 24, but may also be computed by summing the areas of triangles
    6 · a 2
    P AB, P BC, P CA . The area of triangle P AB is = 3 a , and similarly the area of triangle P BC is
    24 − 3 a − 4 b 2
    4 b . Thus, it follows easily that d ( P, CA ) = . Now, we have
    ( )5
    24 8 4 24 3 9
    ( x, y ) = − a − b, − a − b .
    5 5 b 5 5 5
    The desired region is the set of ( x, y ) obtained by those ( a, b ) subject to the constraints a ≥ 0 , b ≥
    0 , 6 a + 8 b ≤ 48.
    Consequently, our region is the triangle whose vertices are obtained by evaluating ( x, y ) at the vertices
    ( a, b ) of the triangle. To see this, let f ( a, b ) output the corresponding ( x, y ) according to the above.
    Then, we can write every point P in ABC as P = m (0 , 0) + n (0 , 6) + p (8 , 0) for some m + n + p = 1.
    24 24
    Then, f ( P ) = mf (0 , 0) + nf (0 , 6) + pf (8 , 0) = m ( , ) + n ( − 8 , 0) + p (0 , − 6), so f ( P ) ranges over
    5 5
    the triangle with those three vertices.
    ( )
    24 24
    Therefore, we need the area of the triangle with vertices , , (0 , − 6) , ( − 8 , 0), which is easily
    5 5
    computed (for example, using determinants) to be .288 5