Shortlist JBMO, 2008

Geometry Difficulty 6.3 National Olympiad Prove it JBMO

Problem:

Consider $ABC$ an acute-angled triangle with $AB \neq AC$ . Denote by $M$ the midpoint of $BC$ , by $D, E$ the feet of the altitudes from $B, C$ respectively and let $P$ be the intersection point of the lines $DE$ and $BC$ . The perpendicular from $M$ to $AC$ meets the perpendicular from $C$ to $BC$ at point $R$ . Prove that lines $PR$ and $AM$ are perpendicular.

Solution

Solution:

Let $F$ be the foot of the altitude from $A$ and let $S$ be the intersection point of $AM$ and $RC$ . As $PC$ is an altitude of the triangle $PRS$ , the claim is equivalent to $RM \perp PS$ , since the latter implies that $M$ is the orthocenter of $PRS$ . Due to $RM \perp AC$ , we need to prove that $AC \parallel PS$ , in other words
$\frac{MC}{MP} = \frac{MA}{MS}$
Notice that $AF \parallel CS$ , so $\frac{MA}{MS} = \frac{MF}{MC}$ . Now the claim is reduced to proving $MC^2 = MF \cdot MP$ , a well-known result considering that $AF$ is the polar line of $P$ with respect to circle of radius $MC$ centered at $M$ .

The "elementary proof" on the latter result may be obtained as follows: $\frac{PB}{PC} = \frac{FB}{FC}$ , using, for instance, Menelaus and Ceva theorems with respect to $ABC$ . Cross-multiplying one gets $(PM - x)(FM + x) = (x - FM)(PM + x)$
- $x$ stands for the length of $MC$ - and then $PM \cdot FM = x^2$ .

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