HMMT 二月 2012 · 几何 · 第 10 题

英文解析

10. Let C denote the set of points ( x, y ) ∈ R such that x + y ≤ 1. A sequence A = ( x , y ) | i ≥ 0 ofi i ipoints in R is ‘centric’ if it satisfies the following properties:2
    • A = ( x , y ) = (0 , 0), A = ( x , y ) = (1 , 0).
    0 0 0 1 1 1
    • For all n ≥ 0, the circumcenter of triangle A A A lies in C .
    n n +1 n +2
    2 2
    Let K be the maximum value of x + y over all centric sequences. Find all points ( x, y ) such
    2012 2012
    2 2
    that x + y = K and there exists a centric sequence such that A = ( x, y ).
    2012
    Geometry Test
    √ √
    Answer: ( − 1006 , 1006 3) , ( − 1006 , − 1006 3) Consider any triple of points △ A A A withn n +1 n +2
    circumcenter P . By the Triangle Inequality we have A P ≤ A A + A P ≤ A A +1. Since P is then n n n 0 0 n n 0 ncircumcenter, we have P A = P A . Finally we have A a ≤ P A +1 = A P +1 ≤ A A +2.
    n n n n +2 n +2 0 n n +2 n n n 0
    √
    √
    2 2
    2 2
    Therefore x + y ≤ x + y + 2.
    n +2 n +2 n n
    It is also clear that equality occurs if and only if A , A , P , A are all collinear and P lies on then 0 n n +2 nunit circle.
    √
    2 2 2
    From the above inequality it is clear that x + y ≤ 2012. So the maximum value of K is 2012 .
    2012 2012
    Now we must find all points A that conforms to the conditions of the equality case. P must lie on
    2 0
    the unit circle, so it lies on the intersection of the unit circle with the perpendicular bisector of A A ,
    0 1
    ( )
    √
    1 1 3
    which is the line x = . Thus P must be one of , ± . From now on we assume that we take the
    2 2 20
    positive root, as the negative root just reflects all successive points about the x -axis.
    ( )
    √
    √
    1 3
    If P = , then A , P , A must be colinear, so A = (1 , 3).
    0 0 0 2 2
    2 2
    Then since we must have A , P , A , A colinear and P on the unit circle, it follows that
    0 2 n 2 n 2 n +2 2 n
    ( )
    √
    √
    n 1 3 n +1
    P = ( − 1) , . Then by induction we have A = ( − 1) ( n, n 3). To fill out the rest of the
    2 n 2 n
    2 2
    ( )
    √
    √
    n n +1 1 3
    sequence, we may take A = ( − 1) ( n + 1 , − n 3) and P = ( − 1) , − so that we get
    2 n +1 2 n +1
    2 2
    the needed properties.
    Therefore the answer is
    √ √
    A ∈ { ( − 1006 , 1006 3) , ( − 1006 , − 1006 3) }
    2012
    .
    Geometry Test