Given triangle $ABC$. Let $BPCQ$ be a parallelogram ($P$ is not on $BC$). Let $U$ be the intersection of $CA$ and $BP$, $V$ be the intersection of $AB$ and $CP$, $X$ be the intersection of $CA$ and the circumcircle of triangle $ABQ$ distinct from $A$, and $Y$ be the intersection of $AB$ and the circumcircle of triangle $ACQ$ distinct from $A$. Prove that $\overline{BU} = \overline{CV}$ if and only if the lines $AQ$, $BX$, and $CY$ are concurrent. Proposed by Li4.
Problem
Source: 2024 Taiwan TST Round 2 Mock P1
Tags: geometry, Please refran from saying easy
math_comb01
29.03.2024 11:13
Sketch: Notice that the second condition is equivalent to $\measuredangle AQB = \measuredangle AQC$ now $BU=CV$ follows from a few applications of sin rule, similarly the other direction also follows from applying sin rule a few times .
sami1618
29.07.2024 02:00
Let $ABA'C$ be a parallelogram. We show that both conditions are equivalent to $A'P$ bisecting $\angle BPC$. Part 1: $AQ$, $BX$, and $CY$ concur iff $A'P$ bisects $\angle BPC$ If $AQ$, $BX$, and $CY$ concur at $Z$ then $BZ\cdot ZX=AZ\cdot ZQ=CZ\cdot ZY$ so $BCXY$ is cyclic, if $BCXY$ is cyclic then $AQ$, $BX$, and $CY$ concur by the Radical Axes Theorem. Then $BCXY$ cyclic iff $\angle BXA=\angle CYA$ iff $\angle BQA=\angle CQA$ iff $AQ$ bisects $\angle BQC$ iff $A'P$ bisects $\angle BPC$. Part 2: $BU=CV$ iff $A'P$ bisects $\angle BPC$ As $[A'BU]=[A'CV]$ we have that $BU=CV$ iff $d(A',BU)=d(A',CV)$ iff $A'P$ bisects $\angle BPC$.
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