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Selection Examinations for the IMO, 2013

Geometry Difficulty 8.6 Shortlist Prove it Slovenia

Let $A$ , $B$ , $C$ and $D$ be points on a circle $\Gamma$ such that the lines $AB$ and $CD$ intersect at the point $T$ , where $A$ lies between $B$ and $T$ and $D$ lies between $C$ and $T$ . Let the line through $D$ that is parallel to the line $AB$ intersect the circle $\Gamma$ again at the point $E$ , and let the line $ET$ intersect $\Gamma$ again at the point $F$ . Let $G$ denote the point of intersection of the lines $CF$ and $AB$ . Let $X$ be the midpoint of the line segment $AB$ and let $Y$ be the point that we obtain by reflecting the point $T$ through the point $G$ . Prove that the points $X$ , $Y$ , $C$ and $D$ are concyclic.

Solution

The points $C$ , $D$ , $E$ and $F$ are concyclic and the lines $AB$ and $DE$ are parallel, so we have $\angle GTF = \angle DEF = \angle DCF$ . Hence, triangles $\Delta TGF$ and $\Delta CGT$ are similar and they have an equal side ratio. We notice that
$\frac{|GF|}{|GT|} = \frac{|GT|}{|GC|} \text{ or } |GT|^2 = |GF||GC|.$
According to the power-of-a-point theorem for the point $G$ and the circle $\Gamma$ we also have $|GF||GC| = |GA||GB|$ . If we take into account that $|GA| = |TA| - |GT|$ and $|GB| = |TB| - |GT|$ , we get
$|GT|^2 = (|TA| - |GT|)(|TB| - |GT|) = |TA||TB| - |GT|(|TA| + |TB|) + |GT|^2,$
which means that $|TA||TB| = |GT|(|TA| + |TB|)$ .
Because $X$ is the midpoint of the line segment $AB$ , we have $|TX| = \frac{|TA|+|TB|}{2}$ , and from the definition of the point $Y$ we get $|TY| = 2|GT|$ . Summing it all up, we derive
$|TX||TY| = |GT|(|TA| + |TB|) = |TA||TB|.$
The power-of-a-point theorem for the point $T$ and the circle $\Gamma$ now says that $|TA||TB|$ is equal to $|TD||TC|$ . We now derive $|TX||TY| = |TD||TC|$ . The points $X$ , $Y$ , $C$ and $D$ are thus concyclic.

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