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Macedonian Mathematical Competitions

Geometry Difficulty 6.5 National Olympiad Prove it North Macedonia

Let $H$ be the orthocenter of the triangle $\triangle ABC$ and $A_1$ , $B_1$ , $C_1$ be the feet of the altitudes from $A$ , $B$ , $C$ respectively.
Find $\frac{\overline{AH}}{HA_1} \cdot \frac{\overline{BH}}{HB_1} \cdot \frac{\overline{CH}}{HC_1}$ if $\frac{\overline{AH}}{HA_1} + \frac{\overline{BH}}{HB_1} + \frac{\overline{CH}}{HC_1} = 2008$ .

Solution

We have
$\frac{\overline{AH}}{HA_1} + 1 = \frac{\overline{AH}}{HA_1} + \frac{\overline{HA_1}}{HA_1} = \frac{\overline{AA_1}}{HA_1} = \frac{P_{\triangle ABC}}{P_{\triangle HBC}}.$
Similarly, $\frac{\overline{BH}}{HB_1} + 1 = \frac{P_{\triangle HBC}}{P_{\triangle HCA}}$ and $\frac{\overline{CH}}{HC_1} + 1 = \frac{P_{\triangle HBC}}{P_{\triangle HAB}}$ . If $x = P_{\triangle HBC}$ , $y = P_{\triangle HCA}$ , $z = P_{\triangle HAB}$ , then

$P_{\triangle ABC} = x + y + z$ . Now $\frac{\overline{AH}}{HA_1} + \frac{\overline{BH}}{HB_1} + \frac{\overline{CH}}{HC_1} = 2008$ is equivalent to
$\frac{x+y+z}{x} - 1 + \frac{x+y+z}{y} - 1 + \frac{x+y+z}{z} - 1 = 2008,$
from where $(x+y+z)(yz+xz+xy) = 2011xyz$ .
Finally
$\begin{aligned} & \left(\frac{x+y+z}{x}-1\right) \cdot \left(\frac{x+y+z}{y}-1\right) \cdot \left(\frac{x+y+z}{z}-1\right) = \frac{y+z}{x} \cdot \frac{x+z}{y} \cdot \frac{x+y}{z} = \frac{(y+z)(x+z)(x+y)}{xyz} = \\ & = \frac{(x+y+z)(yz+xz+xy)-xyz}{xyz} = \frac{2011xyz-xyz}{xyz} = 2010 \end{aligned}$

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