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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 4.5 AIME Prove it Soviet Union

Problem:

$ABC$ is an acute-angled triangle. The angle bisector $AD$ , the median $BM$ and the altitude $CH$ are concurrent. Prove that angle $A$ is more than $45$ degrees.

Solution

Solution:

We use Ceva's theorem. Since $AD$ , $BM$ , $CH$ are concurrent, we have $(BD/DC) \cdot (CM/MA) \cdot (AH/BH) = 1$ . But $CM = MA$ and since $AD$ is the angle bisector $BD/DC = AB/AC$ , so $(AB/AC) \cdot (AH/BH) = 1$ . Hence $AH/AC = BH/AB < 1$ . So angle $HAC >$ angle $HCA$ . But angle $AHC = 90^\circ$ , so angle $A > 45^\circ$ .

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