South-Afrika, 2011

Geometry Difficulty 7.0 National Olympiad, round 2 Prove it South Africa

Given a non-isosceles triangle $ABC$ , let $D$ , $E$ and $F$ denote the midpoints of the sides $BC$ , $CA$ and $AB$ respectively. The circumcircle of triangle $BCF$ and the line $BE$ meet again at $P$ , and the circumcircle of $ABE$ and the line $AD$ meet again in $Q$ . Finally, the lines $DP$ and $FQ$ meet at $R$ . Prove that the centroid $G$ of the triangle $ABC$ lies on the circle $PQR$ .

Solution

We will use the following lemma:

Lemma 1. Let $AD$ be a median in triangle $ABC$ . Then $\cot \angle BAD = 2 \cot A + \cot B$ and $\cot \angle ADC = \frac{1}{2}(\cot B - \cot C)$ .

Proof. Let $CC_1$ and $DD_1$ be the perpendiculars from $C$ and $D$ to $AB$ . Using signed lengths, we write
$\begin{align*} \cot BAD &= \frac{AD_1}{DD_1} \\ &= \frac{\frac{1}{2}(AC_1 + AB)}{\frac{1}{2}CC_1} \\ &= \frac{CC_1 \cot A + CC_1(\cot A + \cot B)}{CC_1} \\ &= 2 \cot A + \cot B. \end{align*}$
Similarly, denoting by $A_1$ the projection of $A$ onto $BC$ , we get
$\cot ADC = \frac{DA_1}{AA_1}$
$\begin{aligned} &= \frac{\frac{1}{2}BC - A_1C}{AA_1} \\ &= \frac{\frac{1}{2}(AA_1 \cot B + AA_1 \cot C) - AA_1 \cot C}{AA_1} \\ &= \frac{1}{2}(\cot B - \cot C). \end{aligned}$
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Turning to the given problem, by the lemma we get
$\begin{aligned} \cot BPD &= 2 \cot BPC + \cot PBC \\ &= 2 \cot BFC + \cot PBC \quad (\text{from circle } BFPC) \\ &= 2 \cdot \frac{1}{2}(\cot A - \cot B) + 2 \cot B + \cot C \\ &= \cot A + \cot B + \cot C. \end{aligned}$
Similarly, $\cot GQF = \cot A + \cot B + \cot C$ , so $\angle GPR = \angle GQF$ and $GPRQ$ is cyclic.

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