Problem 1865

Solution:

Let $R$ be the circumradius of $\triangle ABC$ and $\Delta$ its area. We have $OD = R \cos A$ and $DC = \frac{a}{2}$, so
$$[ODC] = \frac{1}{2} \cdot OD \cdot DC = \frac{1}{2} \cdot R \cos A \cdot R \sin A = \frac{1}{2} R^2 \sin A \cos A$$
Again $HE = 2R \cos C \cos A$ and $EA = c \cos A$. Hence
$$[HEA] = \frac{1}{2} \cdot HE \cdot EA = \frac{1}{2} \cdot 2R \cos C \cos A \cdot c \cos A = 2R^2 \sin C \cos C \cos^2 A$$
Further
$$[GFB] = \frac{\Delta}{6} = \frac{1}{6} \cdot 2R^2 \sin A \sin B \sin C = \frac{1}{3} R^2 \sin A \sin B \sin C$$
Equating (1) and (2) we get $\tan A = 4 \sin C \cos C$. And equating (1) and (3), and using this relation we get
$$\begin{aligned} 3 \cos A & = 2 \sin B \sin C = 2 \sin (C + A) \sin C \\ & = 2(\sin C + \cos C \tan A) \sin C \cos A \\ & = 2 \sin^2 C (1 + 4 \cos^2 C) \cos A \end{aligned}$$
Since $\cos A \neq 0$ we get $3 = 2t(-4t + 5)$ where $t = \sin^2 C$. This implies $(4t - 3)(2t - 1) = 0$ and therefore, since $\sin C > 0$, we get $\sin C = \sqrt{3}/2$ or $\sin C = 1/\sqrt{2}$. Because $\triangle ABC$ 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 $\pi/4$.