CMO 2012

[Phlembac Adib Hasan]

In the triangle $ABC$, $\angle A$ is biggest. On the circumcircle of $\triangle ABC$, let $D$ be the midpoint of $\displaystyle \widehat{ABC}$ and $E$ be the midpoint of $\displaystyle \widehat{ACB}$. The circle $c_1$ passes through $A,B$ and is tangent to $AC$ at $A$, the circle $c_2$ passes through $A,E$ and is tangent $AD$ at $A$. $c_1$ and $c_2$ intersect at $A$ and $P$. Prove that $AP$ bisects $\angle BAC$.

[Tahmid Hasan]

Dude, search before posting
viewtopic.php?f=25&t=2326&p=11671&hilit ... 012#p11671

[photon]

let , $c_2$ intersects $AB$ at $X$.
$\angle PAD=\angle AEP$ , $\angle PAC=\angle PBA$ ;
$\angle PAD-\angle PAC=\angle AEP-\angle PBA\Rightarrow \frac{1}{2}\angle B=(\angle AXP)-(\angle AXP-\angle BPX) \Rightarrow \angle BPX= \frac{1}{2}\angle B$
$AD$ intersects $c_1$ at $D_1$.$\angle EAD_1=180^o-A-\frac{1}{2}\angle B -\frac{1}{2}\angle C$
$\Rightarrow \angle EXA=\frac{1}{2}(\angle B+\angle C)$
$\Rightarrow \angle EBX+\angle BEX=\frac{1}{2}(\angle B+\angle C)$ ; $\text{Because }\angle EBX=\angle ECA=\frac{1}{2}\angle C$
$\therefore \angle BEX=\frac{1}{2}\angle B$
finally , $\displaystyle \frac{sin\angle BEX}{BX}=\frac{sin\angle BPX}{BX} \Rightarrow \frac{sin\angle BXE}{BE}=\frac{sin\angle BXP}{BP}$
$\displaystyle \Rightarrow \frac{sin\angle BXE}{sin\angle BXP}=\frac{BE}{BP} \Rightarrow \frac{sin\angle APE}{sin\angle AEP}=\frac{BE}{BP} \Rightarrow \frac{AE}{AP}=\frac{BE}{BP}$ ;
$AP=BP ; \Rightarrow \angle PAB=\angle PBA=\angle PAC$ , $AP$ bisects that angle.

there might be typo ,my "preview" didn't show latex well

[Nadim Ul Abrar]

I also solved this in Tahmid's way .. But at first I accidentally proved something else :p

In the $\triangle ABC$ , $\angle A$ is biggest. On the circumcircle of $\triangle ABC$ , let $D$ be the midpoint of arc$ABC$ and $E$ be the midpoint of arc$ACB$ . The circle $c_1$ passes through $A,D$ and is tangent to $AC$ at $A$ , the circle $c_2$ passes through $A,E$ and is tangent $AB$ at $A$ . $c_1$ and $c_2$ intersect at $A$ and $P$ . Prove that $AP$ bisects $\angle BAC$ .

[*Mahi*]

In the $\triangle ABC$ , $\angle A$ is biggest. On the circumcircle of $\triangle ABC$ , let $D$ be the midpoint of arc$ABC$ and $E$ be the midpoint of arc$ACB$ . The circle $c_1$ passes through $A,D$ and is tangent to $AC$ at $A$ , the circle $c_2$ passes through $A,E$ and is tangent $AB$ at $A$ . $c_1$ and $c_2$ intersect at $A$ and $P$ . Prove that $AP$ bisects $\angle BAC$ .

Let $F = $ center of $c_1$ and $G =$ center of $c_2$, renaming $P$ to $H$.
As $\angle HAC = \angle ADH = \frac 12 \angle AFH$ and similarly $\angle HAB = \angle AEH = \frac 12 \angle AGH$, if we can prove that $c_1, c_2$ have equal radius, then we are done, as $AH$ is the common chord.

Now, if $I = AG \cap c_2$ other than $A$, then $\angle AIE = 90^\circ - \angle EAG = \angle BAE = \angle ABE$, so diameter of $c_2 =$ $\frac {AE} {\sin \angle AIE} = \frac {AE} {\sin \angle ABE} = 2R$, and similarly $c_1$ has the same diameter, which implies all of $c_1, c_2$ and circumcircle of $\triangle ABC$ has the same radius, which proves our claim.

[Nadim Ul Abrar]

if we can prove that $c_1, c_2$ have equal radius, then we are done.

yesss. I also used this idea .

[SANZEED]

My solution is similar to Mahi vai's one.

[Tahmid Hasan]

In the $\triangle ABC$ , $\angle A$ is biggest. On the circumcircle of $\triangle ABC$ , let $D$ be the midpoint of arc$ABC$ and $E$ be the midpoint of arc$ACB$ . The circle $c_1$ passes through $A,D$ and is tangent to $AC$ at $A$ , the circle $c_2$ passes through $A,E$ and is tangent $AB$ at $A$ . $c_1$ and $c_2$ intersect at $A$ and $P$ . Prove that $AP$ bisects $\angle BAC$ .

Let $F = $ center of $c_1$ and $G =$ center of $c_2$, renaming $P$ to $H$.
As $\angle HAC = \angle ADH = \frac 12 \angle AFH$ and similarly $\angle HAB = \angle AEH = \frac 12 \angle AGH$, if we can prove that $c_1, c_2$ have equal radius, then we are done, as $AH$ is the common chord.

I have a little different approach from there.
Let $r_1,r_2,r$ be the radius of $c_1,c_2,\odot ABC$ respectively.
$2r_1=\frac{AD}{\sin ABD}=2r=\frac{AE}{\sin ACE}=2r_2$.
so we are done.

[Tahmid Hasan]

Oops, made a little mistake before; fixed it now . The new solution is similar to Mahi's now.