Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.3 National olympiad Prove it Belarus

Point $P$ inside an acute-angled triangle $A_1A_2A_3$ is chosen so that its projections $P_1, P_2, P_3$ onto the sides $A_1A_2$ , $A_2A_3$ , $A_3A_1$ respectively lie on the sides of the triangle.
Prove that for points $X_1$ , $X_2$ , $X_3$ on the sides $A_1A_2$ , $A_2A_3$ , $A_3A_1$ respectively, $\max \left\{ \frac{X_1X_2}{P_1P_2}, \frac{X_2X_3}{P_2P_3}, \frac{X_3X_1}{P_3P_1} \right\} \ge 1$ if

a) $X_1$ , $X_2$ , $X_3$ are the midpoints of the corresponding sides;
b) $X_1$ , $X_2$ , $X_3$ are the feet of the corresponding altitudes;
c) $X_1$ , $X_2$ , $X_3$ are arbitrary points on the corresponding sides.

(IMO-2010 Shortlist, Problem G3, modified)

Solution

a), b) (Solution of M. Mankevich, A. Nekrashevich, A. Tanana.) The statement from a) and b) simply follow from the general

Lemma. If there is a point $X$ such that $X_1$ , $X_2$ , $X_3$ are the orthogonal projections of $X$ onto the sides $A_2A_3$ , $A_3A_1$ , $A_1A_2$ respectively, then the statement of the problem is valid.

Indeed, $P$ belongs to one of the quadrilaterals $A_1X_2XX_3$ , $A_2X_3XX_1$ , $A_3X_1XX_2$ ; say, $P$ belongs to $A_2X_1XX_3$ . Then
$P_1P_3 = A_2P \sin \angle A_2 \le A_2X \sin \angle A_2,$
i.e. $X_1X_3 / P_1P_3 \ge 1$ . Thus the lemma is proved.

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