Triangle $ABC$ in which $AB <BC$, is inscribed in a circle $\omega$ and circumscribed about a circle $\gamma$ with center $I$. The line $\ell$ parallel to $AC$, touches the circle $\gamma$ and intersects the arcs $BAC$ and $BCA$ at points $P$ and $Q$, respectively. It is known that $PQ = 2BI$. Prove that $AP + 2PB = CP$.
Problem
Source: 2016 239 J5
Tags: geometry
mofumofu
21.06.2021 09:53
[asy][asy]
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[/asy][/asy]
Fun problem, but I can't believe how this is the first question of the day for a Grade 8-9 contest...
First we show that $BMIP$ cyclic. By poncelet porism the other tangents from $P,Q$ to the incircle meet at some point $H$ on $(ABC)$, so $GI=GP=GQ$; then $\triangle GBI\cong \triangle GMP$ so $GM=GB$, thus $BMIP$ is an isosceles trapezium. Now by Ptolemy's on $PGCA$, we have $PG\cdot AC=AG(CP-AP)$. Thus we want to show
\begin{align*}
PG\cdot AC=AG\cdot 2BP&\Leftrightarrow GI\cdot 2AN = AG\cdot 2IN (\text{since} BP=IM=IN, GI=GP)\\
&\Leftrightarrow\frac{GI}{IN}=\frac{AG}{AN}\\
\end{align*}
which is true since $\triangle DIN\sim\triangle DGI$ so $\frac{GI}{IN}=\frac{DG}{DI}=\frac{DG}{DA}=\frac{AG}{AN}$ and we are done.
Quick commentaryOnce again despite the short solution I found this quite difficult; the condition and result to prove just looks so weird. I had admittedly the help of GGB to hypothesize that $BPIM$ cyclic. Upon first look it should be quite clear that there is most likely some kind of Ptolemy used somewhere; I spent some time playing around with the isosceles trapezium $PQCA$ but got nowhere. The perpendicular bisector of $BC$ should be quite apparent, as it provides symmetry for $PQ$ and $AC$ (and bring in more familiar points like midpoint of arcs). It took me awhile before I noticed $G$ and its importance, but once I saw the poncelet(how is this in a grade 8-9 contest?!) the rest of the problem was quite smooth.
jayme
21.06.2021 14:59
Dear Mathlinkers, why it is known that PQ = 2.BI? or it is simply a condition? Sincerely Jean-Louis