In an acute-angled triangle $A B C$, the altitude $A D$, the median $B E$, and the bisector $C F$ intersect at point $O$. Find $\angle C$, if $O E=2 O C$.

#
Apply the Law of Cosines to triangle EOC.

Let $A C=2, \angle C=2 \gamma$. Then $C E=1, C D=2 \cos 2 \gamma, O C=2 \cos 2 \gamma \sec \gamma, O E=4 \cos 2 \gamma \sec \gamma$.

By the Law of Cosines, $E O^{2}=C O^{2}+C E^{2}-2 C O \cdot C E \cos \gamma$, or $16 \cos ^{2} 2 \gamma \sec ^{2} \gamma=4 \cos ^{2} 2 \gamma \sec ^{2} \gamma+1-4 \cos 2 \gamma \Leftrightarrow$

$12 \cos ^{2} 2 \gamma=\cos ^{2} \gamma-4 \cos 2 \gamma \cos ^{2} \gamma \Leftrightarrow 24 \cos ^{2} 2 \gamma=(1+\cos 2 \gamma)(1-4 \cos 2 \gamma)$.

From this, $\cos 2 \gamma=1 / 7$.

## Answer

$\arccos 1 / 7 ;$