6. Circles $W_{1}$ and $W_{2}$ with equal radii are given. Let $P$,$Q$ be the intersection of the circles. points $B$ and $C$ are on $W_{1}$ and $W_{2}$ such that they are inside $W_{2}$ and $W_{1}$ respectively. Points $X$,$Y$ $\neq$ $P$ are on $W_{1}$ and $W_{2}$ respectively, such that $\angle{BPQ}=\angle{BYQ}$ and $\angle{CPQ}=\angle{CXQ}$.Denote by $S$ as the other intersection of $(YPB)$ and $(XPC)$. Prove that $QS,BC,XY$ are concurrent.
Problem
Source: 2023 Iran MO
Tags: geometry
v4913
17.05.2023 23:08
Note that $X, Y$ are the intersections of $\omega_1, \omega_2$ with the lines through $B, C$ perpendicular to $PQ$. Thus, $(XPC), (YPB)$ are $\omega_1, \omega_2$ translated by $QC, QB$ in the direction of $QC, QB$ respectively, which means that they both pass through the point which forms a parallelogram with $Q, B, C$. Thus, it suffices to show that $XY$ passes through the midpoint of $BC$, which is obvious since $CX = BY$. $\square$
sami1618
22.12.2024 18:56
Let \( M \) be the midpoint of \( XY \). We claim \( M \) is the desired intersection point. It suffices to show that \( BYCX \) and \( QYSX \) are both parallelograms. Claim: Point \( B \) is the orthocenter of triangle \( PQY \). Similarly, point \( C \) is the orthocenter of triangle \( PQX \). Proof: Let \( B' \) be the reflection of \( B \) about \( PQ \). Notice that \( B' \) lies on \( W_2 \). Since \[ \angle B'YQ = \angle B'PQ = \angle BPQ = \angle BYQ, \]points \( B \), \( B' \), and \( Y \) are collinear. Because \( BB' \perp PQ \), we have \( BY \perp PQ \). From the equal angles, we also have \( BP \perp QY \), proving the claim. Claim: Quadrilateral \( BYCX \) is a parallelogram. Proof: Let \( O_1 \) and \( O_2 \) be the centers of \( W_1 \) and \( W_2 \), respectively. Let \( N \) be the midpoint of \( O_1O_2 \). Notice \( BY \parallel CX \) and \[ BY = 2NO_2 = 2NO_1 = CX. \] Claim: Quadrilateral \( QYSX \) is a parallelogram. Proof: Let \( O_3 \) and \( O_4 \) be the centers of \( (YPB) \) and \( (XPC) \), respectively. By the first claim, \( W_1 \), \( W_2 \), \( (YPB) \), and \( (XPC) \) all have equal radii. Thus, \( PO_1QO_2 \), \( PO_4XO_1 \), \( PO_3SO_4 \), and \( PO_2YO_3 \) are parallelograms (specifically rhombi). We have \[ \overrightarrow{Q} = \overrightarrow{O_1} + \overrightarrow{O_2} - \overrightarrow{P}, \quad \overrightarrow{X} = \overrightarrow{O_4} + \overrightarrow{O_1} - \overrightarrow{P}, \]\[ \overrightarrow{S} = \overrightarrow{O_3} + \overrightarrow{O_4} - \overrightarrow{P}, \quad \overrightarrow{Y} = \overrightarrow{O_2} + \overrightarrow{O_3} - \overrightarrow{P}. \]Thus, $ \overrightarrow{Q} + \overrightarrow{S} = \overrightarrow{X} + \overrightarrow{Y}, $ as desired. (This is known as the "AT" lemma. There is also a simple proof using congruent triangles.)
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