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Czech and Slovak Mathematical Olympiad

Geometry Difficulty 6.1 National Olympiad Prove it Czech Republic

In the triangle $ABC$ , let us denote $M, N, P$ the midpoints of the sides $BC, CA, AB$ respectively and let $G$ be the centroid of $ABC$ . Let the circumcircle of $BGP$ intersects the line $MP$ at a point $K$ different from $P$ , and let the circumcircle of $CGN$ intersects the line $MN$ at a point $L$ different from $N$ . Prove $|\angle BAK| = |\angle CAL|$ .

Solution

Obviously, $MP$ intersects the median $BN$ between points $B$ and $G$ , so the point $K$ lies on the ray $PM$ and $BKGP$ is cyclic. Similarly, the point $L$ lies on the ray $NM$ and $CLGN$ is cyclic. Due to $MP \parallel CA$ and $MN \parallel BA$ we have
$|\angle BPK| = |\angle BPM| = |\angle BAC| = |\angle MNC| = |\angle LNC|,$
while the two cyclic quadrilaterals imply
$|\angle BKP| = |\angle BGP| = |\angle NGC| = |\angle NLC|.$

We see that triangles BPK and CNL are similar according to the condition AA. By the condition SAS also triangles ABK and ACL are similar since
(i) $|\angle ABK| = |\angle PBK| = |\angle NCL| = |\angle ACL|$ ,
(ii) $\frac{|AB|}{|BK|} = 2 \cdot \frac{|PB|}{|BK|} = 2 \cdot \frac{|NC|}{|CL|} = \frac{|AC|}{|CL|}.$
Thus, the equality $|\angle BAK| = |\angle CAL|$ is proved.

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