Let $ABCD$ be a cyclic quadrilateral with circumcircle $\Omega$ and let diagonals $AC$ and $BD$ intersect at $X$. Suppose that $AEFB$ is inscribed in a circumcircle of triangle $ABX$ such that $EF$ and $AB$ are parallel. $FX$ meets the circumcircle of triangle $CDX$ again at $G$. Let $EX$ meets $AB$ at $P$, and $XG$ meets $CD$ at $Q$. Denote by $S$ the intersection of the perpendicular bisector of $\overline{EG}$ and $\Omega$ such that $S$ is closer to $A$ than $B$. Prove that line through $S$ parallel to $PQ$ is tangent to $\Omega$.
Problem
Source: KJMO 2021 P3
Tags: geometry, cyclic quadrilateral, circumcircle, perpendicular bisector, tangent
sarjinius
30.11.2021 17:40
Let $O$, $O_1$ and $O_2$ be the circumcenters of $\Omega, (ABX)$ and $(CDX)$ respectively, let $I, M$ and $N$ be the midpoints of $XO, XF$ and $XG$ respectively. Note that $OO_1 \perp AB$, since $\angle ABX + \angle DXO_2 = \angle ACD + (90^{\circ} - \angle ACD) = 90^{\circ}$, we have $XO_2 \perp AB$, and $OO_1 \parallel XO_2$. Similarly, we can show $OO_2 \parallel O_1X \perp CD$. Thus, $OO_1XO_2$ is a parallelogram, which implies $I$ is also the midpoint of $O_1O_2$. Since $O_1M \parallel O_2N \perp MN$, $IM = IN$. Hence, $OF = 2IM = 2IN = OG$. Also, note that $OO_1$ is the perpendicular bisector of $AB$, the perpendicular bisector of $EF$ must pass through $O_1$, and $AB \parallel EF$. Thus, $OO_1$ is also the perpendicular bisector of $EF$, and $OE = OF$. Hence, the line tangent to $\Omega$ at $S$ is parallel to $EG$. It suffices to show that $EG \parallel PQ$. Let $FX$ intersect $AB$ at $R$. By angle chasing, $\angle AFQ = \angle AFX = \angle ABX = \angle ABD = \angle ACD = \angle ACQ$, thus $A, F, C, Q$ are concyclic. Similarly, $\angle CGR = \angle CGX = \angle CDX = \angle CDB = \angle CAB = \angle CAR$, thus $A, R, C, G$ are concyclic. By power of a point, $XF \cdot XQ = XA \cdot XC = XR \cdot XG \implies \frac{XR}{XF} = \frac{XQ}{XG}$. Since $PR \parallel EF$, $\frac{XR}{XF} = \frac{XP}{XE}$. Hence, $\frac{XP}{XE} = \frac{XQ}{XG} \iff EG \parallel PQ$, which solves the problem.
genius09
24.12.2021 17:35
Since $AB \parallel EF$, $\angle BXF = \angle AXF = \angle CXG$. It is easy to show $\triangle XBP \sim XCQ$ and $\triangle XEB \sim \triangle XGC$. It means $\frac{XE}{XP} = \frac{XG}{XQ}$, so $EG \parallel PQ$. Also, because $PA\cdot PB=PE \cdot PX$ and $QC \cdot QD = QG \cdot QX$, $PQ$ is a radical axis of $\Omega$ and $(XEG)$. Hence, if we denote $O, O_1$ as its center respectively, $OO_1 \perp PQ$. It means $OO_1$ is the perpendicular bisector of $EG$, then $SO \perp PQ$. Therefore the tangent line passing through $S$ is parallel to $PQ$.
Mahdi_Mashayekhi
11.05.2022 11:45
Let $O$ and $O'$ be circumcircles of $ABCD$ and $EGX$. Note that $PA.PB = PE.PX$ and $QD.QC = QG.QX$ so $PQ$ is Radical Axis of $ABCD$ and $EGX$ so $OO' \perp EG$. Note that we need to prove $OO' \perp PQ$ so we need to prove $PQ || EG$. Note that $\angle XBP = \angle XCQ$ and $\angle CXQ = \angle CXG = \angle AXF = \angle EXB = \angle BXP$ so $BXP$ and $CXQ$ are similar also Note that $\angle XEB = \angle XAB = \angle CAB = \angle BDC = \angle XDC = \angle XGC$ and $\angle BXE = \angle CXG$ so $BXE$ and $CXG$ are similar so $\frac{PX}{QX} = \frac{BX}{CX} = \frac{EX}{GX} \implies EG || PQ$. we're Done.