Points $A$ and $P$ are marked in the plane not lying on the line $\ell$. For all right triangles $ABC$ with hypotenuse on $\ell$, show that the circumcircle of triangle $BPC$ passes through a fixed point other than $P$.
Problem
Source: Saint Petersburg MO 2016
Tags: geometry, circumcircle, Power of Point, coaxial circles
anantmudgal09
11.12.2016 21:25
Let $D$ be on $\ell$ with $AD \perp \ell$. Then, $DA^2=DB\cdot DC$ is fixed, so $D$ has equal powers in all circles $(BCP)$, whence, they are coaxial. Their other common point is on the line $PD$.
Aiscrim
11.12.2016 21:47
Inverting with pole $A$, we have to prove that for a fixed point $P$ and any diameter $XY$ in a fixed circle $\mathcal{C}(O,R)$, the circle $(PXY)$ goes through a fixed point. Letting $\{Q\}=PO\cap (PXY)$, we have $OP\cdot OQ=OX\cdot OY=R^2$, so $Q$ is fixed.
nikolapavlovic
21.03.2017 15:39
Let $Q$ be the foot of perpendicular from $A$ to $\ell$.It's well known that $AQ^2=BQ\cdot CQ$ $\implies$ the other fixed point is the image of $P$ in inversion in $Q$ with radius $AQ$ Edit:pretty much the same solution as in #3 but i didn't saw it before it was to late.
khanhnx
07.01.2023 18:03
Let $D$ be orthogonal projection of $A$ on $\ell,$ $Q$ be a point on $PD$ such as $\overline{DP} \cdot \overline{DQ} = - AD^2$. Then $\overline{DP} \cdot \overline{DQ} = - AD^2 = \overline{DB} \cdot \overline{DC}$ or $Q \in (BCP)$