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International Mathematical Arhimede Contest

Geometry Difficulty 5.7 AIME, harder Prove it Spain

Let $ABCD$ be a cyclic quadrilateral inscribed in a circle $\gamma$ . Let $X$ be a point on the extension of $AC$ such that $XB$ and $XD$ are tangent to $\gamma$ . The tangent at $C$ intersects $XD$ at $Q$ . Let $E$ be the intersection of $AQ$ with $\gamma$ distinct from $A$ . Prove that lines $AD$ , $BE$ , and $CQ$ are concurrent.

Solution

Suppose that $AD$ intersects $CQ$ at $Y$ and $AD$ intersects $BE$ at $Z$ . Since $\triangle XAD \sim \triangle XDC$ , then we have $\frac{AD}{DC} = \frac{XA}{XD} = \frac{XA}{XB} = \frac{AB}{BC}$

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from which follows $AB \cdot DC = BC \cdot AD$ . By Ptolemy's theorem,
$AB \cdot DC = BC \cdot AD = \frac{CA \cdot DB}{2} \\ \Rightarrow \frac{DB}{AB} = \frac{2 DC}{CA}$
Likewise,
$CA \cdot ED = CE \cdot AD = \frac{AE \cdot DC}{2} \\ \Rightarrow \frac{DC}{CA} = \frac{2 ED}{AE}$
Since $\triangle YDC \sim \triangle YCA$ then we have $\frac{YD}{YC} = \frac{DC}{CA} = \frac{YC}{YA}$ .

Thus, taking into account the preceding, we have
$\frac{YD}{YA} = \frac{YD \cdot YA}{YA^2} = \left(\frac{YC}{YA}\right)^2 = \left(\frac{DC}{CA}\right)^2 = \left(\frac{2ED}{AE}\right)^2$
Using the similar triangles $ABZ$ and $EDZ$ , we get $\frac{ZD}{ZB} = \frac{ED}{AB}$ . Likewise, using the similar triangles $DBZ$ and $EAZ$ we have $\frac{ZA}{ZB} = \frac{EA}{DB}$ . Thus,
$\frac{ZD}{ZA} = \frac{ED \cdot DB}{EA \cdot AB} = \left(\frac{2ED}{AE}\right)^2$
Finally, we have $Y = Z$ on account of the preceding and we are done.

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