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### IMO 2007 Problem 4

Posted: Wed Jan 09, 2013 10:14 am
In $\triangle ABC$, the internal bisector of $\angle C$ meets the circumcircle again at $R$, the perpendicular bisector of $BC$ at $P$ and the perpendicular bisector of $AC$ at $Q$. The mid point of $B$ is $K$ and the mid point of $AC$ is $L$. Prove that $\triangle RPK$ and $\triangle RQL$ have the same area.

### Re: IMO 2007 Problem 4

Posted: Wed Jan 09, 2013 7:49 pm

### Re: IMO 2007 Problem 4

Posted: Thu Jan 10, 2013 8:14 am
My Proof
Let $O$ be the circumcenter of $ABC$. I proved $OP=OQ$ by angle chasing. So $a^2-OP^2=a^2-OQ^2$ ($a=$ radius of $\bigcirc ABC$). Hence $RP\cdot PC=RQ\cdot QC$. By the similarity of $\triangle LCQ$ and $\triangle PKC$, the fact follows immediately.

### Re: IMO 2007 Problem 4

Posted: Fri Jan 11, 2013 10:45 pm
My proof :

$\displaystyle \frac{1}{2}. RC .\frac{1}{2}BC.sin(\frac{c}{2})-\frac{1}{2} .(\frac{1}{2}BC)^2.tan(\frac{c}{2})=\frac{1}{2}. RC. \frac{1}{2}AC.sin(\frac{c}{2})-\frac{1}{2} .(\frac{1}{2}AC)^2.tan(\frac{c}{2})$

$\longleftrightarrow \displaystyle RC.cos(\frac{c}{2})=\frac{AB+AC}{2}$ if $AB \neq AC$ ortherwise $AB=AC$

Which is true indeed (trust me )

### Re: IMO 2007 Problem 4

Posted: Fri Jun 02, 2017 1:59 pm
Let's draw $RX$ perpendicular on $PK$ which intersects thevcircumcirle of triangle $ABC$ at $M$
$RX$ is parallel to $BC$
So, $BRMC$ is a trapizoid.
So, $BR$=$MC$
Again, $BR$=$AR$ is known as $CR$ in the angle bisector of angle $ACB$
So, $AR$=$MC$ and it makes $ARCM$ cyclic trapizoid
Then $MR$=$b$
$XK$ bisects both $BC$ and also $RM$
So, $RX$=$$\frac{1}{2}b$$
Similarly we can prove $RY$=$$\frac{1}{2}a$$
And draw $RY$ perpendicular on $QL$
Here in the right angled triangle $PKC$,
$PK$=$$\frac{1}{2}atan\frac{1}{2}C$$
Similarly in triangle $LCQ$,
$QL$=$$\frac{1}{2}btan\frac{1}{2}C$$
So, $$[RPK]=\frac{1}{2}PK*RX=\frac{1}{8}abtan\frac{1}{2}C$$
Again, $$[RQL]=\frac{1}{2}QL*RY=\frac{1}{8}abtan\frac{1}{2}C$$