Baltic Way 2023 Shortlist

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 $Q$.

$\square$