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Taiwan IMO Selection Camp, 2023

Geometry Difficulty 6.7 National Olympiad Prove it Taiwan

Using the three sides of triangle $ABC$ as sides, construct equilateral triangles $ABC_1, BCA_1, CAB_1$ outward from $ABC$ , respectively. Let point $P$ be the other intersection point of the circumcircle of $ABC_1$ and the circumcircle of $CAB_1$ ( $P \neq A$ ). Find a point $Q$ on the circumcircle of $CAB_1$ such that $PQ$ is parallel to $BA_1$ . Find a point $R$ on the circumcircle of $ABC_1$ such that $PR$ is parallel to $CA_1$ .
Prove that the line joining the centroid of triangle $ABC$ and the centroid of triangle $PQR$ is parallel to line $BC$ .

Let $ABC$ be a triangle. Let $ABC_1, BCA_1, CAB_1$ be three equilateral triangles that do not overlap with $ABC$ . Let $P$ be the intersection of the circumcircles of triangles $ABC_1$ and $CAB_1$ ( $P \neq A$ ). Let $Q$ be the point on the circumcircle of triangle $CAB_1$ so that $PQ$ is parallel to $BA_1$ . Let $R$ be the point on the circumcircle of triangle $ABC_1$ so that $PR$ is parallel to $CA_1$ .
Show that the line connecting the centroid of triangle $ABC$ and the centroid of triangle $PQR$ is parallel to $BC$ .

Solution

Brute-force trigonometric method. Orient the lines $PA$ , $PB$ , $PC$ so that the angle from $PA$ to $PB$ , from $PB$ to $PC$ , and from $PC$ to $PA$ are all $\frac{2\pi}{3}$ . Also orient the lines $PQ$ , $PR$ so that the angle between $\overrightarrow{BC}$ and $\overrightarrow{PQ}$ , and between $\overrightarrow{PR}$ and $\overrightarrow{BC}$ , are also $\frac{2\pi}{3}$ . Let $\angle(\overrightarrow{BC}, PA) = \alpha$ , $\angle(\overrightarrow{BC}, PB) = \beta$ , $\angle(\overrightarrow{BC}, PC) = \gamma$ . It is easy to see that $\gamma = \beta + \frac{2\pi}{3} = \alpha - \frac{2\pi}{3}$ . The problem is equivalent to proving the identity:
$PQ \sin \frac{2\pi}{3} - PR \sin \frac{2\pi}{3} = PA \sin \alpha + PB \sin \beta + PC \sin \gamma. \quad (*)$

The left-hand side of (*) can be reduced to $\frac{PQ - PR}{2}$ . By Ptolemy's theorem, we know that
$PQ \sin \angle(\mathrm{PC}, \mathrm{PA}) + \mathrm{PC} \sin \angle(\mathrm{PA}, \mathrm{PQ}) + \mathrm{PA} \sin \angle(\mathrm{PQ}, \mathrm{PC}) = 0.$
By the orientation given at the start of the proof, we have $\angle(PC,PA) = \frac{2\pi}{3}$ . The second angle in the above expression can be written as $\angle(PA,PQ) = \frac{2\pi}{3} - \alpha = -\gamma$ , and the third angle can be written as $\angle(PQ,PC) = \angle(\vec{BC},PB) = \beta$ . Therefore
$\frac{1}{2}PQ = PC \sin \gamma - PA \sin \beta.$
Similarly we obtain
$-\frac{1}{2}PR = PB \sin \beta - PA \sin \gamma.$

$- \sin \beta - \sin \gamma = \sin \alpha,$
that is, $\sin \alpha + \sin \beta + \sin \gamma = 0$ . This identity can be proved using the identity $(1+\omega+\omega^2)z=0$ , where $z$ is an arbitrary complex number and $\omega$ is a primitive cube root of unity. $\square$

Alternative proof. Let $F$ be on $A_1BC$ with $PF$ parallel to $BC$ . According to the remark above, it suffices to show that $FQR$ and $ABC$ share the same centroid.
Lemma. Let $X, Y, Z$ be moving points on three separate circles with the same angular speed. Then the centroid of $XYZ$ also moves in a circle with the same angular speed, and the center of its locus is the centroid of the centers of the loci of $X, Y, Z$ .
Proof of lemma. This is obvious by, say, complex coordinates. □
Now if $X, Y, Z$ are on $A_1BC, AB_1C$ , and $ABC_1$ , respectively, with $X, Y, Z$ starting at $B, C, A$ , then after a while they arrive at $C, A, B$ . Therefore the locus of the centroid of $XYZ$ visits the centroid of $ABC$ at least twice in each cycle, showing that in fact the centroid of $XYZ$ is always the same, which has to be the centroid of $ABC$ . Now when $X$ travels to $F$ , it is easy to chase the angle and show that $Y$ travels to $Q$ and $Z$ travels to $R$ . □

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.