Let $ABC$ be an acute-angled scalene triangle. Let $P$ be a point on the extension of $AB$ past $B$, and $Q$ a point on the extension of $AC$ past $C$ such that $BPQC$ is a cyclic quadrilateral. Let $N$ be the foot of the perpendicular from $A$ to $BC$. If $NP = NQ$ then prove that $N$ is also the centre of the circumcircle of $APQ$.
Problem
Source: 2015 NZOMC Camp Selections p7
Tags: geometry, circumcircle, cyclic quadrilateral, Circumcenter
L567
10.01.2021 12:58
Hasn't this one already been posted by you? I remember seeing it...
parmenides51
10.01.2021 13:30
I always post problems with sources, trying to complete the whole geometry problem set from each contest, probably a problem was not that original after all
Aurn0b
11.01.2021 09:08
$\textbf{Solution :}$
Let the circumcenter of $\triangle APQ$ be $O$.
$\textbf{Claim:}$ $O$ lies on the line $AN$.
Proof : Let $X$ be the foot of perpendicular from $A$ to $PQ$.
$\angle ANB= \angle AXQ = 90^{\circ}$
$\angle ABN= \angle AQX $ (As $(BPQC)$ is cyclic )
$\Rightarrow \angle PAN = \angle BAN = \angle XAQ$
Now,
$\angle OAP = \angle QAX = \angle NAP $
As $N$ and $O$ lies on the same side of $AB$, we can say that $A,N,O$ are collinear.(Proved)
So, $OP=OQ$, and we also know that $NP=NQ$, as $AN$ cannot be the perpendicular on $PQ$(as traingle ABC is a scalene triangle, hence also is triangle APQ), its easy to see that $O$ and $N $ are the same point.
THUS, $N$ is the circumcenter of $\triangle APQ. \blacksquare$