EGMO 2018 P5

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Ananya Promi
Posts: 36
Joined: Sun Jan 10, 2016 4:07 pm

EGMO 2018 P5

Let $\tau$ be the circumcircle of triangle $ABC$. A circle $\Omega$ is tangent to the line segment $AB$ and is tangent to $\tau$ at a point lying on the same side of the line $AB$ as $C$. The angle bisector of $\angle{BCA}$ intersects $\Omega$ at two different points $P$ and $Q$.
Prove that, $\angle{ABP}=\angle{QBC}$

Ananya Promi
Posts: 36
Joined: Sun Jan 10, 2016 4:07 pm

Re: EGMO 2018 P5

Let $N$ be the intersection point of $(ABC)$ and the angle bisector of $\angle{ACB}$ other than $C$
Suppose $\Omega$ touches $\tau$ at $L$ and $AB$ at $M$
Let $M'=NL \cap AB$ and $XY$ be the common tangent of $\tau$ and $\Omega$
Define $K=XY \cap AB$
$KM=KL$
$\measuredangle KLM' = \measuredangle KLN = \measuredangle KLB + \measuredangle BLN = \measuredangle LAB + \measuredangle BAN = \measuredangle LNB + \measuredangle NBA = \measuredangle M'NB + \measuredangle NBM' = \measuredangle NM'B = \measuredangle LM'B = \measuredangle LM'K$
So, we get $KM' = KL$ which means $MM'=0$
By similarity we get $NB^2=NM.NL=NQ.NP$
So, $\triangle NBQ \sim -\triangle NPB$
$\measuredangle NBQ = \measuredangle BPN$
$\measuredangle NBA+ \measuredangle ABQ = \measuredangle BCN + \measuredangle PBC$
$\measuredangle ABQ = \measuredangle PBC$
$\measuredangle ABP = \measuredangle QBC$

Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm

Re: EGMO 2018 P5

Okay, I copied it from my aops post.
This was freedom. Losing all hope was freedom.

thczarif
Posts: 17
Joined: Mon Sep 25, 2017 11:27 pm

Re: EGMO 2018 P5

Let $D$ be the second intersection of $\tau$ and the bisector $\angle BCA$.Let $E,F$ be the tangency point
of $\Omega$ with $AB$ and $\tau$ respectively.

$Lemma$ 1: $DE \cdot DF=DB^2$

$Proof$ : $$\angle DFB= \frac{\angle C}{2}= \angle DBE$$.
So, $\triangle DFB$ is similar to $\triangle DBE$
$$\rightarrow DE \cdot DF=DB^2$$

$Lemma$ 2: $DP \cdot DQ=DB^2$

$Proof$ : By, POP we get, $$P_\Omega(D)=DP \cdot DQ=DE \cdot DF=DB^2$$

$Lemma$ 3: $BI$ is the angle bisector of $\angle PBQ$ where $I$ is the incentre of $\triangle ABC$

$Proof$ : From $Lemma$ $3$ we get $$DP \cdot DQ=DB^2$$
And, $P,Q,D$ collinear. So, $D$ is the centre of the Appolonian Circle of $\triangle BPQ$.
Let, the Appolonian Circle $(D,DB)$ intersect $PQ$ at $I$. So, $BI$ is the angle bisector of $\angle PBQ$.
And from $Fact$ $5$ we get $I$ is the incentre of $\triangle ABC$.

Now, $$\angle PBI=\angle QBI$$
$$\rightarrow \angle PBI + \angle ABI=\angle QBI + \angle CBI$$
$$\rightarrow \angle ABP= \angle QBC$$