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Problem 1645

National Olympiad, first round

Geometry Difficulty 6.2 Prove it Saudi Arabian Mathematical Competitions · Saudi Arabia · 2015

Let $ABC$ be a triangle and $G$ its centroid. Let $G_{a}$ , $G_{b}$ and $G_{c}$ be the orthogonal projections of $G$ on sides $BC$ , $CA$ , respectively $AB$ . If $S_{a}$ , $S_{b}$ and $S_{c}$ are the symmetrical points of $G_{a}$ , $G_{b}$ , respectively $G_{c}$ with respect to $G$ , prove that $AS_{a}$ , $BS_{b}$ and $CS_{c}$ are concurrent.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $H_{a}$ be the foot of altitude from $A$ , $M_{a}$ the midpoint of side $BC$ , $A'$ the intersection point of line $AS_{a}$ and side $BC$ and $A''$ the intersection point of the parallel line to $BC$ passing through $S_{a}$ with $AM_{a}$ .

Because $GG_{a}$ and $AH_{a}$ are parallel, we have
$\frac{H_{a}M_{a}}{G_{a}M_{a}} = \frac{AM_{a}}{GM_{a}} = 3$
Because $S_{a}A''$ and $BC$ are parallel, we have
$\frac{G_{a}M_{a}}{S_{a}A''} = \frac{G_{a}G}{S_{a}G} = 1$
We also have
$\frac{S_{a}A''}{A'M_{a}} = \frac{AA''}{AM_{a}} = 1 - \frac{M_{a}A''}{AM_{a}} = 1 - 2\frac{M_{a}G}{AM_{a}} = \frac{1}{3}$
Multiplying these three relations we deduce that $A'M_{a} = M_{a}H_{a}$ , which means that $AS_{a}$ and $AH_{a}$ are isotomic conjugate. Similarly, $BS_{b}$ and $BH_{b}$ are isotomic conjugate and $CS_{c}$ and $CH_{c}$ are isotomic conjugate. Since the altitudes of a triangle are concurrent, so are $AS_{a}$ , $BS_{b}$ and $CS_{c}$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.