Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.5 National olympiad Prove it Czech-Polish-Slovak Mathematical Match

Let $ABC$ be a triangle. Line $\ell$ is parallel to $BC$ and it respectively intersects side $AB$ at point $D$ , side $AC$ at point $E$ , and the circumcircle of the triangle $ABC$ at points $F$ and $G$ , where points $F, D, E, G$ lie in this order on $\ell$ . The circumcircles of triangles $FEB$ and $DGC$ intersect at points $P$ and $Q$ . Prove that points $A, P, Q$ are collinear.

(Slovakia)

Solution

Let $\omega_B, \omega_C$ be the circumcircles of triangles $FEB$ and $DGC$ respectively. Since $PQ$ is the radical axis of $\omega_B$ and $\omega_C$ , it is sufficient to prove that the powers of $A$ with respect to $\omega_B$ and $\omega_C$ are equal. If we denote by $X$ the second intersection of $\omega_B$ and $AC$ , and by $Y$ the second intersection of $\omega_C$ and $AB$ , then this is equivalent to
$AX \cdot AE = AY \cdot AD.$
Lines $DE$ and $BC$ are parallel, yielding $\frac{AB}{AD} = \frac{AC}{AE}$ , so the above is equivalent to
$AX \cdot AC = AY \cdot AB.$
This, in turn, is equivalent to $B, Y, X, C$ being concyclic. From angles in $\omega_B$ and $\omega_C$ we have
$\angle BYC = \angle DYC = \angle DGC \quad \text{and} \quad \angle BXC = \angle BXE = \angle BFE.$
On the other hand, trapezoid $BFGC$ is inscribed in the circumcircle of $ABC$ , so it is isosceles, hence $\angle BFE = \angle DGC$ . We conclude that $\angle BYC = \angle BXC$ , so $B, Y, X, C$ are indeed concyclic and we are done. □

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