Problem 12.2. In triangle $A B C$ with orthocenter $H$ one has that

$$
A H \cdot B H \cdot C H=3 \text { and } A H^{2}+B H^{2}+C H^{2}=7
$$

Find:

a) the circumradius of $\triangle A B C$;

b) the sides of $\triangle A B C$ with maximum possible area.

Oleg Mushkarov, Nikolai Nikolov
12.2. a) If $\triangle A B C$ is acute, then by the Law of cosines for $\triangle A H B$ we get that

$$
A B^{2}=A H^{2}+B H^{2}-2 A H \cdot B H \cos (\pi-\gamma)
$$

Since $A B=2 R \sin \gamma$ and $C H=2 R \cos \gamma$ (by the Extended Law of sines), we obtain $A B^{2}+C H^{2}=4 R^{2}$. Therefore

$$
4 R^{2}=A H^{2}+B H^{2}+C H^{2}+\frac{A H \cdot B H \cdot C H}{R}
$$

Then $4 R^{3}=7 R+3$, i.e. $(R+1)(2 R+1)(2 R-3)=0$, whence $R=\frac{3}{2}$.

If $\triangle A B C$ is obtuse, then we get analogously that

$$
4 R^{2}=A H^{2}+B H^{2}+C H^{2}-\frac{A H \cdot B H \cdot C H}{R}
$$

and therefore $4 R^{3}=7 R-3$, i.e. $(R-1)(2 R-1)(2 R+3)=0$. Since $3=A H \cdot B H \cdot C H<(2 R)^{3}$, we conclude that $R=1$.

The existence of $\triangle A B C$ with $R=\frac{3}{2}$ and $R=1$ follows from b).

b) Denote by $S$ the area of $\triangle A B C$. Since $S=\frac{A B \cdot B C . C A}{4 R}$, we have

$$
S^{2}=\frac{\left(4 R^{2}-A H^{2}\right)\left(4 R^{2}-B H^{2}\right)\left(4 R^{2}-C H^{2}\right)}{16 R^{2}}
$$

Setting $x=A H^{2}, y=B H^{2}, z=C H^{2}$ and $t=4 R^{2}$, we get

$$
S^{2}=\frac{t^{3}-7 t^{2}+t(x y+y z+z x)-9}{4 t}
$$

Without loss of generality we may assume that $x \geq y \geq z$. Then $x \geq \frac{7}{3}$ and therefore

$$
x y+y z+z x=\frac{9}{x}+x(7-x)=15-\frac{(x-3)^{2}(x-1)}{x} \leq 15
$$

where the equality is attained if $x=3$. Hence

$$
S^{2} \leq \frac{t^{3}-7 t^{2}+15 t-9}{4 t}
$$

Since $R=\frac{3}{2}$ or $R=1$, we conclude that $S_{\max }=\sqrt{8}$ and it is achieved for an acute $\triangle A B C$ with $R=\frac{3}{2}, A H=B H=\sqrt{3}$ and $C H=1$. The sides of this triangle are $\sqrt{6}, \sqrt{6}$ and $\sqrt{8}$.