HMMT 二月 2012 · 冲刺赛 · 第 26 题

英文解析

26. [ 17 ] Suppose ABC is a triangle with circumcenter O and orthocenter H such that A, B, C, O , and Hare all on distinct points with integer coordinates. What is the second smallest possible value of thecircumradius of ABC ?
    √
    Answer: 10 Assume without loss of generality that the circumcenter is at the origin. By wellknown properties of the Euler line, the centroid G is such that O , G , and H are collinear, with G inbetween O and H , such that GH = 2 GO . Thus, since G = ( A + B + C ), and we are assuming O is 1
    the origin, we have H = A + B + C . This means that as long as A , B , and C are integer points, H3
    will be as well.
    However, since H needs to be distinct from A , B , and C , we must have △ ABC not be a right triangle,
    since in right triangles, the orthocenter is the vertex where the right angle is.
    Now, if a circle centered at the origin has any integer points, it will have at least four integer points.
    (If it has a point of the form ( a, 0), then it will also have ( − a, 0), (0 , a ), and (0 , − a ). If it has a pointof the form ( a, b ), with a, b 6 = 0, it will have each point of the form ( ± a, ± b ).) But in any of these caseswhere there are only four points, any triangle which can be made from those points is a right triangle.
    Guts
    Thus we need the circumcircle to contain at least eight lattice points. The smallest radius this occurs
    √ √
    2 2
    at is 1 + 2 = 5, which contains the eight points ( ± 1 , ± 2) and ( ± 2 , ± 1). We get at least one validtriangle with this circumradius:
    A = ( − 1 , 2), B = (1 , 2), C = (2 , 1).
    √ √
    2 2
    The next valid circumradius is 1 + 3 = 10 which has the valid triangle
    A = ( − 1 , 3), B = (1 , 3), C = (3 , 1).