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### IMO 2007 Problem 4

Posted: **Wed Jan 09, 2013 10:14 am**

by **Phlembac Adib Hasan**

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**

by **zadid xcalibured**

### Re: IMO 2007 Problem 4

Posted: **Thu Jan 10, 2013 8:14 am**

by **Phlembac Adib Hasan**

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**

by **Nadim Ul Abrar**

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**

by **Ananya Promi**

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, $$[RQL]=\frac{1}{2}QL*RY=\frac{1}{8}abtan\frac{1}{2}C$$