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South African Mathematical Olympiad Third Round

Geometry Difficulty 5.7 AIME, harder Prove it South Africa

Let $ABIH$ , $BDEC$ , and $ACFG$ be arbitrary rectangles constructed (externally) on the sides of triangle $ABC$ . Choose point $S$ outside rectangle $ABIH$ (on the opposite side as triangle $ABC$ ) such that $\angle SHI = \angle FAC$ and $\angle HIS = \angle EBC$ . Prove that the lines $FI$ , $EH$ , and $CS$ are concurrent.

Solution

Let $T = EH \cap FI$ . Let $K$ and $L$ be the feet of the perpendiculars from $A$ to $FI$ and from $B$ to $EH$ , respectively. Then $K$ and $L$ are on the circumcircle $\Gamma$ of $ABIH$ . Note that $K$ is also on the circumcircle of $ACFG$ and $L$ is also on the circumcircle of $BDEC$ . Let $CK$ extended intersect $\Gamma$ in $M$ , and let $CL$ extended intersect $\Gamma$ in $N$ . Then the lines $HM$ and $IN$ intersect in $S$ . $[\angle EBC = \angle ELC = \angle HLN = \angle HIN]$ . But $\angle HIS = \angle EBC$ , so $S$ is on $IN$ . Similarly, $S$ is on $HM$ .

Consider the self-crossing hexagon $IKMHLN$ inscribed in the circle $\Gamma$ . By Pascal's Theorem, the points $IK \cap HL = T$ , $KM \cap LN = C$ , and $MH \cap NI = S$ are collinear. This shows that $T$ is a point that belongs to all three lines $FI$ , $EH$ and $CS$ , i.e., these lines are concurrent.

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