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Problem 1987

National Olympiad, first round

Geometry Difficulty 6.8 Prove it Cono Sur Mathematical Olympiad · Argentina

Let $ABC$ be an acute triangle. Denote by $D, E, F$ the midpoints of sides $BC, CA, AB$ respectively. The circle with diameter $AB$ intersects lines $AB$ and $AC$ again at $P$ and $Q$ respectively. The line through $P$ parallel to $BC$ meets line $DE$ at $R$ , the line through $Q$ parallel to $BC$ meets line $DF$ at $S$ . The circumcircle of $DPR$ meets $AB$ again at $X$ , the circumcircle of $DQS$ meets $AC$ again at $Y$ , and those two circles meet again at $Z$ . Prove that $Z$ is the midpoint of $XY$ .

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

Since $D$ and $E$ are midpoints we know that $DE \parallel AB$ , and by definition $PR \parallel BC$ , hence $PRDB$ is a parallelogram. Analogously, $QSDC$ is a parallelogram. Thus $PR = BD = DC = SQ$ .

Since $AD$ is a diameter, we know that $AB \perp PD$ and $AC \perp QD$ . Then, because of the parallel lines, $PD \perp DR$ and $QD \perp DS$ . Hence $PR$ and $SQ$ , which have the same length, are diameters of the circumcircles of $DPR$ and $DQS$ respectively.

In the cyclic quadrilateral $PXRD$ , we have $\angle DPX = 90^\circ$ , so $DX$ is a diameter of the circumcircle of $DPR$ . Likewise, $DY$ is a diameter of the circumcircle of $DQS$ .

Since $DX$ is a diameter, we have that $XZ \perp ZD$ , and since $DY$ is a diameter we have that $YZ \perp ZD$ . Therefore $X, Z, Y$ are collinear. But we also know that both dashed circles have the same diameter, so $DX = DY$ . Therefore $DZ$ is an altitude of the isosceles triangle $DXY$ ; this implies that $Z$ is the midpoint of $XY$ , and we are done.

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