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[几何] 内接圆与外接圆半径之比

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tommywong posted 2014-5-5 22:11 |Read mode
mathchina.com/cgi-bin/topic.cgi?forum=5&t … ic=20129&show=25
设r与R分别为三角形ABC的内接圆与外接圆半径,而且∠A为三角形ABC中最大的角,M为BC的中点,X为外接圆在三角形ABC的B和C二点上的切线的交点,证明:

$\displaystyle\frac{r}{R}\ge\frac{AM}{AX}$
现充已死,エロ当立。
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