IMO 2023 P2

IMO 2023 P2

Well, IMO 2023 Day 1 problems are out and I thought of trying the geometry problem which was P2. 

Problem: Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.

Well, here's my proof, but I would rather call this my rough work tbh. There are comments in the end!

Proof

Define $A'$ as the antipode of $A$. And redefine $P=A'D\cap (ABC)$. Define $L=SP\cap (PDB)$. 

Claim1: $L-B-E$ collinear
Proof: Note that $$\angle SCA=\angle SCB-\angle ACB=90-A/2-C.$$

So $$\angle SPA=90-A/2-C\implies \angle SPA'=90-(90-A/2-C)=A/2+C\implies \angle LBD=A/2+C\implies \angle LBD=A/2+C.$$

By angle chase, $$\angle DBC=90-A/2,\angle EBC=90-C.$$ So $E-B-L$ collinear.

Claim2: $LD||BC$
Proof: We know that $$\angle EBC=90-C,\angle ELD=\angle BLD=\angle BPD=\angle BPA'=\angle BAO=90-C.$$

So $LD||BC$.

Define $F$ as midpoint of arc $BC$ not containing $A$.

Now define $X=PP\cap AF$. Define $O$ as the circumcenter of $(ABC)$. 

Claim3: $PXOF$ is cyclic
Proof: Let $\angle PAB=\theta$. So $\angle PSB=\theta$ and $\angle PSF=\theta+A/2$. So $\angle POF=2\theta+A$. 

And $\angle PXF=\angle PAX+\angle APX$. But $\angle PAX=\theta+A/2$. And $$\angle APX=\angle APS+\angle SPX=\angle ACS+\angle SPX=90-A/2-C+\angle SPX.$$

So to show cyclicity, we need to show that $$2\theta+A=\angle PXF=90+\theta-C+\angle SPX.$$
Or enough to show $$\angle SPX=\theta+A-90+C.$$

But $$\angle SPX=\angle SPB-\angle XPB$$ 

$$=\angle SAB-\angle XPD-\angle DPB= 90+A/2-\angle DBP-\angle DLB$$

$$=90+A/2-\angle DLP-(90-C)$$

$$=C+A/2-\angle DLP$$

as $SF||AD,BC||LD, AD\perp BC$ we get $$C+A/2-\angle DLP=C+A/2-(90-A/2-\theta)=C+A-90+\theta.$$

And so we get $PXOF$ is cyclic.

Claim4: $XA=XP$
Proof: Note that $$\angle XPA=\angle PXF-\angle PAX=2\theta+A-\theta-A/2=\theta+A/2$$. So $\angle XPA=\angle XAP$. And hence the claim.

But $$XA^2=XP^2=XD\cdot XB.$$ So $XA$ is tangent to $(DBA)$. Hence $$\angle ABD=\angle DAS=\angle DAF=\angle AFS=\angle ABS.$$

And hence we get $B-X-S$ collinear.

And we are done!

This was my proof when I first tried the problem. I won't call this a "proof" but rather "raw progress which somehow manage to finish the problem". I am someone who writes every single observation on paper, and this was how my mind was working tbh. 

As you can see, claim 3 is not needed and we can directly show $XA=XP$ but I still added it. 

My rough diagram

As soon as I made the diagram, I conjecture $L-P-S$ must be collinear. But then point $P$ was still weird you know. To show that "the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$", we had to introduce the arc midpoint of $BC$ and hence $F$ was introduced. Point $O$ was introduced because I was using a compass and had to mark point $O$ and then realised $AO,PD$ concur at the circle. And then the rest conjectures followed very intuitively. 

Additionally, my initial idea for claim 4 was to show $OX||PA'$ and that would finish it. I believe there is an angle chase to it, but I didn't do it as I was lazy.

I personally think the problem is a G3/G4 level. There are basically three claims for this problem, showing $L-P-S$, $D-P-A'$ and $XP=XA$ and then you use Power of point to finish it.

Well, it was a cute problem! I hope you guys tried it too! :). What was your solution to this problem?

~Sunaina