IMO 2005
- Raiyan Jamil
- Posts:138
- Joined:Fri Mar 29, 2013 3:49 pm
Let $ABCD$ be a given convex quadrilateral with sides $BC$ and $AD$ equal in length and not parallel. Let $E$ and $F$ be interior points of the sides $BC$ and $AD$ respectively such that $BE = DF$. The lines $AC$ and $BD$ meet at $P$, the lines $BD$ and $EF$ meet at $Q$, the lines $EF$ and $AC$ meet at $R$. Consider all the triangles $PQR$ as $E$ and $F$ vary. Show that the circumcircles of these triangles have a common point other than $P$.
A smile is the best way to get through a tough situation, even if it's a fake smile.
- Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
Re: IMO 2005
We let $K=BC \cap AD$ and $T$ be the centre of the spiral similarity that sends $CB$ to $AD$. Now since $BE/BC=DF/DA$, $T$ is also the centre of spiral similarity that sends $EB$ to $FD$. So $KETF$ is cyclic. Now, $T$ is also the centre of spiral similarity that sends $CE$ to $AF$ and thus $CA$ to $EF$, and so $\angle TCA=\angle TEF$. Therefore $CERT$ is cyclic. Similarly, $DFQT$ is cyclic.
Now, using directed angles, $\angle TRP=\angle TRC=\angle TEC=\angle TEK=\angle TFK=\angle TFD=\angle TQD=\angle TQP$. So, $PRQT$ is cyclic. So $T$ is the intended point through which all the circles pass, as it depends only on $ABCD$.
Now, using directed angles, $\angle TRP=\angle TRC=\angle TEC=\angle TEK=\angle TFK=\angle TFD=\angle TQD=\angle TQP$. So, $PRQT$ is cyclic. So $T$ is the intended point through which all the circles pass, as it depends only on $ABCD$.
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Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.