Mediterranean Mathematics Competition

The triangle $A^{\prime }{B}^{\prime }{C}^{\prime }$ is the pedal triangle of the symmetrical point of the incenter $I$ of $ABC$ with respect to the circumcenter of $ABC$. So, the relation between the areas $S=[ABC]$ and $S^{\prime }=[A^{\prime }{B}^{\prime }{C}^{\prime }]$ is given by $$S^{\prime }=S\cdot \frac{R^2-{\overline{OI}}^2}{4R^2}=\frac{r}{2R}.$$ In the triangle $A^{\prime }{B}^{\prime }{C}^{\prime }$, as $A{B}^{\prime }=s-c=r\mathrm{ctg}\frac{C}{2}$, $A{C}^{\prime }=s-b=r\cdot \mathrm{ctg}\frac{B}{2}$ and $$\begin{array}{rl}{\overline{B^{\prime }{C}^{\prime }}}^2& =a^{\prime 2}=r^2\left[{\left(\mathrm{ctg}\frac{B}{2}+\mathrm{ctg}\frac{A}{2}\right)}^2-4\mathrm{ctg}\frac{A}{2}\mathrm{ctg}\frac{B}{2}\mathrm{ctg}\frac{C}{2}\right]=\\ & =r^2\frac{{\cos }^2\frac{A}{2}}{{\sin }^2\frac{B}{2}{\sin }^2\frac{C}{2}}\left(1-\sin B\sin C\right)=a^2\left(1-\frac{bc}{4R^2}\right)\end{array}$$ but as $bc=2R{h}_a$, we get $$a^{\prime 2}=\frac{a^2}{2R}(2R-h_a)$$

and as we have $$S^{\prime 2}=\frac{a^{\prime 2}{b}^{\prime 2}{c}^{\prime 2}}{16R^{\prime 2}}=\frac{a^2{b}^2{c}^2}{16R^2}\frac{1}{8R{R}^{\prime 2}}(2R-h_a)(2R-h_b)(2R-h_c),$$ the relation searched holds.