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试题:△ABC与△KMT(顶点的字母均依逆时针方向标注)为同一平面上的正三角形,且有。证明,线段CM与AT相垂直,并且CM:AT=3~(1/2)。(第19届全俄数学奥林匹克第三轮竞赛试题) 本文应用复数知识给出试题的简单证明: 如图,AB∩TK=O,以点O作为原点,过点O平行于(?)的直线作为实轴建立复平面O-xy。 ∴AKBT为平行四边形。设B(a,b),T(c,d),则B,T关于O点的对称点A(-a.-b),K(-c,-d)。
Question: △ABC and △KMT (the vertices are marked counterclockwise) are regular triangles on the same plane, and there are. It is proved that the line segment CM is perpendicular to AT and CM:AT=3~(1/2). (The 19th All-Russian Mathematical Olympiad in the third round of the contest) This paper uses the plural knowledge to give a simple proof of the test: As shown in the figure, AB ∩ TK = 0, with the point O as the origin, the point O is parallel to the (?) The complex plane O-xy is established as a real axis. ∴ AKBT is a parallelogram. Let B(a,b),T(c,d), then B, T point about O point of symmetry A(-a.-b), K(-c,-d).