Let $ABCD$ be a rectangle. Points $E$ and $F$ are on diagonal $AC$ such that $F$ lies between $A$ and $E$; and $E$ lies between $C$ and $F$. The circumcircle of triangle $BEF$ intersects $AB$ and $BC$ at $G$ and $H$ respectively, and the circumcircle of triangle $DEF$ intersects $AD$ and $CD$ at $I$ and $J$ respectively. Prove that the lines $GJ, IH$ and $AC$ concur at a point.
Concurrency is equivalent to $\frac{AG}{CJ} = \frac{CH}{AI}$ which is equivalent to $\triangle AGI\sim\triangle CJH$. By power of point $HDJB$ and $GBDI$ are concyclic so
\[
\triangle AGI\sim\triangle ADB = \triangle CBD\sim\triangle CJH
\]and we are done.
This is probably one of the easiest olympiad papers I have seen in recent history. While every olympiad is getting harder and harder, INAMO seems to be getting worse and easier.
Using phantom point, let $AC$ intersect $HI$ and $GJ$ at $K$ and $L$ respectively. We will prove that $K = L$
By POP, we got
\[CJ \times CD = CF \times CE = CH \times CB \]\[AI \times AD = AE \times AF = AG \times AB \]
Notice that $\triangle AKI \sim \triangle CKH \ \ ; \ \ \triangle ALG \sim \triangle CLJ$, thus
\[\frac{AK}{CK} = \frac{AI}{CH} \ \ ; \ \ \frac{AL}{CL} = \frac{AG}{CJ}\]
From there we can conclude that
\[\frac{AK}{CK} = \frac{AL}{CL}\]
Because both $K$ and $L$ are on $AC$ therefore, $K = L \ \ \blacksquare$