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Geometry Difficulty 8.8 Shortlist Prove it Baltic Way

Let $ABCDEF$ be a cyclic and convex hexagon. A point $P$ is called admissible if it does not lie on the circumcircle of $ABCDEF$ or on any of the lines $AD$ , $BE$ and $CF$ . An admissible point is called fantastic if $\odot(ADP)$ , $\odot(BEP)$ and $\odot(CFP)$ intersect in exactly two points. Prove that if there exists a fantastic point, then all admissible points are fantastic.

Solution

The problem follows readily from the following claim:

Claim: An admissible point $P$ is fantastic if and only if the lines $AD$ , $BE$ and $CF$ all intersect in a point.

Proof. Let $\Gamma$ be the circumcircle of $ABCDEF$ . Start by supposing that $P$ is fantastic. Let the other intersection of $\odot(ADP)$ , $\odot(BEP)$ and $\odot(CFP)$ be $Q$ . Since the radical axis of $\Gamma$ and of $\odot(ADP)$ and $\odot(BEP)$ intersect, we have that the lines $AD$ , $BE$ and $PQ$ intersect. Likewise we also have that $AD$ , $CF$ and $PQ$ intersect. Therefore the lines $AD$ , $BE$ and $CF$ all intersect in a point.

Suppose now that $AD$ , $BE$ and $CF$ all intersect in a point $S$ . Let $PS$ intersect $\odot(ADP)$ again in $Q$ . Now we have that
$|SC| \cdot |SF| = |SB| \cdot |SE| = |SA| \cdot |SD| = |SP| \cdot |SQ|$
Hence $\odot(ADP)$ , $\odot(BEP)$ and $\odot(CFP)$ intersect in $P$ and $Q$ .

$\Box$

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.