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Problem 1597

National Olympiad, first round

Geometry Difficulty 6.1 Prove it Brazilian Math Olympiad · Brazil

Let $ABC$ be a triangle, $M$ be the midpoint of side $AC$ and $N$ be the midpoint of side $AB$ . Let $r$ and $s$ be the reflections of lines $BM$ and $CN$ across line $BC$ , respectively. Lines $r$ and $s$ meet line $MN$ at points $D$ and $E$ , respectively. The circumcircles of $BDM$ and $CEN$ meet at points $X$ and $Y$ , lines $BE$ and $CD$ meet at $Z$ and lines $r$ and $s$ meet at $W$ . Prove that lines $XY$ , $WZ$ and $BC$ meet at a single point.

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

First, notice that $MN \parallel BC$ , so $\frac{WD}{DB} = \frac{EW}{EC}$ . Applying Ceva's theorem to triangle $BCW$ and cevians $WP$ , $CD$ and $BE$ we have $\frac{WD}{DB} \cdot \frac{BP}{PC} \cdot \frac{CE}{EW} = 1 \iff BP = PC$ , so $WZ$ meets $BC$ at its midpoint $P$ .

Now it remains to prove that $P$ lies on line $XY$ , which is the radical axis of the two circles. But, because of the reflection and the fact that $MN \parallel BC$ ,

$\angle MBC = \angle CBW = \angle MDB$ , and thus $BC$ is tangent to the circumcircle of $BDM$ . The power of $P$ with respect to this circle is $PB^2 = BC^2/4$ . Analogously, the power of $P$ with respect to the circumcircle of $CEN$ is $BC^2/4$ , and thus $P$ lies on the radical axis of these circles, and we are done.

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