4. Let $H$ be the orthocenter of $\triangle A B C$ and let $A_{1}, B_{1}, C_{1}$ be the feet of the altitudes dropped from $A, B, C$ respectively. Let $\frac{\overline{A H}}{\overline{H A_{1}}}+\frac{\overline{B H}}{\overline{H B_{1}}}+\frac{\overline{C H}}{\overline{H C_{1}}}=2008$. Calculate the product $\frac{\overline{A H}}{\overline{H A_{1}}} \cdot \frac{\overline{B H}}{\overline{H B_{1}}} \cdot \frac{\overline{C H}}{\overline{H C_{1}}}$.
Solution. We have,

$$
\frac{\overline{A H}}{\overline{H A_{1}}}+1=\frac{\overline{A H}}{\overline{H A_{1}}}+\frac{\overline{H A_{1}}}{\overline{H A_{1}}}=\frac{\overline{A A_{1}}}{\overline{H A_{1}}}=\frac{P_{\Delta A B C}}{P_{\triangle H B C}}.
$$

Similarly,

$$
\frac{\overline{B H}}{\overline{H B_{1}}}+1=\frac{P_{\triangle A B C}}{P_{\triangle H C A}} \text { and } \frac{\overline{C H}}{\overline{H C_{1}}}+1=\frac{P_{\triangle A B C}}{P_{\triangle H A B}}.
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_ea01241e897ed9d0948dg-07.jpg?height=279&width=432&top_left_y=795&top_left_x=1012)

Let $x=P_{\triangle H B C}, y=P_{\triangle H C A}, z=P_{\triangle H A B}$, then $P_{\triangle A B C}=x+y+z$. With these notations and the previous transformations, the equation $\frac{\overline{A H}}{\overline{H A_{1}}}+\frac{\overline{B H}}{\overline{H B}}+\frac{\overline{C H}}{\overline{H C_{1}}}=2008$ becomes

$$
\frac{x+y+z}{x}-1+\frac{x+y+z}{y}-1+\frac{x+y+z}{z}-1=2008,
$$

which is equivalent to

$$
(x+y+z)(y z+x z+x y)=2011 x y z
$$

The product that needs to be calculated is

$$
\begin{aligned}
\left(\frac{x+y+z}{x}-1\right) \cdot\left(\frac{x+y+z}{y}-1\right) \cdot\left(\frac{x+y+z}{z}-1\right) & =\frac{y+z}{x} \cdot \frac{x+z}{y} \cdot \frac{x+y}{z}=\frac{(y+z) \cdot(x+z) \cdot(x+y)}{x y z} \\
& =\frac{(x+y+z) \cdot(y z+x z+x y)-x y z}{x y z}=\frac{2011 x y z-x y z}{x y z}=2010.
\end{aligned}
$$

## IV year