Problem - 3138
Let $H$ be the orthocenter of acute $\triangle{ABC}$. Show that $$a\cdot BH\cdot CH + b\cdot CH\cdot AH+c\cdot AH\cdot BH=abc$$
where $a=BC, b=CA,$ and $c=AB$.
Let $S$ be the area of this triangle and $R$ be its circumradius, then
\begin{align*}
abc &=4RS = 4R(S_{\triangle{BHC}}+S_{\triangle{CHA}}+S_{\triangle{AHB}})\\
&=4R\Big(\frac{1}{2}\cdot BH\cdot CH\cdot\sin(\pi - A) +\frac{1}{2}\cdot CH\cdot AH\cdot\sin(\pi - B)+\frac{1}{2}\cdot AH\cdot BH\cdot\sin(\pi - C)\Large)\\
&=2R\sin A \cdot BH\cdot CH + 2R\sin B \cdot CH \cdot AH + 2R\sin C \cdot AH\cdot BH\\
&=a\cdot BH\cdot CH + b\cdot CH\cdot AH+c\cdot AH\cdot BH
\end{align*}