Let $ABC$ be an isosceles triangle with $AB = AC$. Points $P$ and $Q$ inside $\triangle ABC$ are such that $\angle BPC = \frac{3}{2} \angle BAC$, $BP = AQ$ and $AP = CQ$. Prove that $AP = PQ$. Proposed by Fedir Yudin
Problem
Source: Ukraine MO 2021 8.4
Tags: geometry
khina
02.05.2021 21:25
think in apollonian circles!
.
Note that the locus of all possible $P$ is just the apollonius circle preserving $\frac{OB}{AB}$. We thus use this definition exclusively.
Let $X$ be the center of this apollonian circle. First, let's tie in the fact that $AOB$ is isoceles. Note that $\angle{XBO} = \angle{BAX} = \angle{OBA}$ (as $XB$ is tangent to the circumcircle of $OBA$), so $OB$ bisects $\angle{ABX}$. Thus $\frac{AB}{AO} = \frac{XP}{XO}$.
Let's now use this ratio. Note that $\frac{PQ}{PO} = \frac{AB}{AO} = \frac{XP}{XO}$. However, $XP$ is also tangent to the circumcircle of $OPA$, so we also find that $\frac{AP}{PO} = \frac{XP}{XO}$. This thus indicates that $AP = PQ$, and so we are done!
zuss77
20.05.2021 01:57
With $O$ - circumcenter, we have $ABP\mapsto CAQ$ is a rotation around $O$. So with $ABC$ fixed and $P$ moving $PQ/PO$ is fixed. When $P\in AO$ we see that $BP$ bisects $\angle ABO$. So $P$ moves on $B$-Apollonian circle of $AO$. Hence $AP/PO$ is also fixed. We need to check one case. Take $P\equiv B$ and Q.E.D.
Math4Life2020
15.09.2022 20:30
Sorry, but what is O the circumcenter of?