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Saudi Arabian Mathematical Competitions

Geometry Difficulty 6.7 National Olympiad Prove it Saudi Arabia

Let $ABCD$ be a quadrilateral inscribed a circle ( $O$ ). Assume that $AB$ and $CD$ intersect at $E$ , $AC$ and $BD$ intersect at $K$ , and $O$ does not belong to the line $KE$ . Let $G$ and $H$ be the midpoints of $AB$ and $CD$ respectively. Let ( $I$ ) be the circumcircle of the triangle $GKH$ . Let ( $I$ ) and $(O)$ intersect at $M, N$ such that $MGHN$ is convex quadrilateral. Let $P$ be the intersection of $MG$ and $HN$ , $Q$ be the intersection of $MN$ and $GH$ .
1. Prove that $IK$ and $OE$ are parallel.
2. Prove that $PK$ is perpendicular to $IQ$ .

Solution

Denote $F$ as the intersection of $AD$ , $BC$ . Suppose that $FK$ intersects $AB$ , $CD$ at $S$ , $T$ respectively. Since the harmonic points of complete quadrilateral, we have $(E, T, D, C) = -1$ . Then $ET \cdot EH = ED \cdot EC$ . Similarly, $ES \cdot EG = EA \cdot EB$ . But $EA \cdot EB = ED \cdot EC$ then $ES \cdot EG = ET \cdot EH$ which implies that $G, H, T, S$ are concyclic.

Denote $Q'$ as the midpoint of $FK$ then $Q'$ , $G$ , $H$ belong to the Gauss line of the complete quadrilateral $AFBKCD$ . We shall prove that $Q'$ belongs to the radical axis of $(O), (I)$ .
It is easy to see that $EK$ is the antipole of $F$ then denote $EK \cap (O) = \{X, Y\}$ then $FX$ , $FY$ are the tangent lines of $(O)$ . Let $U, V$ be midpoints of $FX$ , $FY$ .
We consider the power of point to circle $(O)$ and degenerate circle $F$ .
$\mathscr{P}_{U/(O)} = UX^2, \quad \mathscr{P}_{U/(F)} = UF^2$
but $UX = UF$ then $\mathscr{P}_{U/(O)} = \mathscr{P}_{U/(F)}$ . This implies that $U$ belongs to the radical axis of $(O), (F)$ . Similarly with the point $V$ then $UV$ is the radical axis of $(O), (F)$ . But $Q' \in UV$ which is the midline of triangle $FXY$ , then
$\mathscr{P}_{Q'/(F)} = \mathscr{P}_{Q'/(O)} \Leftrightarrow \mathscr{P}_{Q'/(O)} = Q'F^2 = Q'K^2 = Q'S \cdot Q'T = Q'G \cdot Q'H = \mathscr{P}_{Q'/(I)}.$
Then $Q'$ belongs to the radical axis of $(I), (O)$ which means $Q' \in MN$ or $Q' \equiv Q$ .
Thus $QK^2 = \mathscr{P}_{Q/(I)}$ implies that $QK$ is the tangent line of $(I)$ . Hence, $QK \perp KI$ .
Using the Brocard's theorem, we have $K$ is the orthocenter of triangle $OEF$ then $EO \perp FK$ . Combining all these results, we get $EO \parallel IK$ .

2) Two lines pass through $Q$ and intersect $(I)$ at $M, N$ and $G, H$ then $P$ is the intersection of $MG$ , $NH$ which means $P$ belongs to the antipole of $Q$ .
Note that $QK$ is tangent to $(I)$ which means $K$ also belong to the antipole of the pole $Q$ respect to circle $(I)$ .
Then $PK$ is the antipole of $Q$ respect to the circle $(I)$ . This implies that $PK \perp IQ$ .

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