
Problem 5. In an acute triangle $A B C, O$ is the circumcenter, $H$ is the orthocenter and $G$ is the centroid. Let $O D$ be perpendicular to $B C$ and $H E$ be perpendicular to $C A$, with $D$ on $B C$ and $E$ on $C A$. Let $F$ be the midpoint of $A B$. Suppose the areas of triangles $O D C, H E A$ and $G F B$ are equal. Find all the possible values of $\widehat{C}$.


Solution. Let $R$ be the circumradius of $\triangle A B C$ and $\Delta$ its area. We have $O D=R \cos A$ and $D C=\frac{a}{2}$, so

$$
[O D C]=\frac{1}{2} \cdot O D \cdot D C=\frac{1}{2} \cdot R \cos A \cdot R \sin A=\frac{1}{2} R^{2} \sin A \cos A
$$

Again $H E=2 R \cos C \cos A$ and $E A=c \cos A$. Hence

$$
[H E A]=\frac{1}{2} \cdot H E \cdot E A=\frac{1}{2} \cdot 2 R \cos C \cos A \cdot c \cos A=2 R^{2} \sin C \cos C \cos ^{2} A
$$

Further

$$
[G F B]=\frac{\Delta}{6}=\frac{1}{6} \cdot 2 R^{2} \sin A \sin B \sin C=\frac{1}{3} R^{2} \sin A \sin B \sin C
$$

Equating (1) and (2) we get $\tan A=4 \sin C \cos C$. And equating (1) and (3), and using this relation we get

$$
\begin{aligned}
3 \cos A & =2 \sin B \sin C=2 \sin (C+A) \sin C \\
& =2(\sin C+\cos C \tan A) \sin C \cos A \\
& =2 \sin ^{2} C\left(1+4 \cos ^{2} C\right) \cos A
\end{aligned}
$$

Since $\cos A \neq 0$ we get $3=2 t(-4 t+5)$ where $t=\sin ^{2} C$. This implies $(4 t-3)(2 t-1)=0$ and therefore, since $\sin C>0$, we get $\sin C=\sqrt{3} / 2$ or $\sin C=1 / \sqrt{2}$. Because $\triangle A B C$ is acute, it follows that $\widehat{C}=\pi / 3$ or $\pi / 4$.

We observe that the given conditions are satisfied in an equilateral triangle, so $\widehat{C}=\pi / 3$ is a possibility. Also, the conditions are satisfied in a triangle where $\widehat{C}=\pi / 4, \widehat{A}=\tan ^{-1} 2$ and $\widehat{B}=\tan ^{-1} 3$. Therefore $\widehat{C}=\pi / 4$ is also a possibility.

Thus the two possible values of $\widehat{C}$ are $\pi / 3$ and $\pi / 4$.
