EGMO 2018 P5

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Ananya Promi
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EGMO 2018 P5

Unread post by Ananya Promi » Thu Apr 19, 2018 10:21 pm

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, $\angle{ABP}=\angle{QBC}$

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Ananya Promi
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Re: EGMO 2018 P5

Unread post by Ananya Promi » Thu Apr 19, 2018 10:25 pm

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 $MM'=0$ :shock:
By similarity we get $NB^2=NM.NL=NQ.NP$
So, $\triangle NBQ \sim -\triangle NPB$
$\measuredangle NBQ = \measuredangle BPN$
$\measuredangle NBA+ \measuredangle ABQ = \measuredangle BCN + \measuredangle PBC$
$\measuredangle ABQ = \measuredangle PBC$
$\measuredangle ABP = \measuredangle QBC$

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Atonu Roy Chowdhury
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Re: EGMO 2018 P5

Unread post by Atonu Roy Chowdhury » Fri Apr 20, 2018 12:20 pm

Okay, I copied it from my aops post.
Let $X$ be the midpoint of the arc $AB$ not containing $C$. Also let $Y$ and $Z$ be the tangency point of $\Omega$ with $AB$ and $\Gamma$ respectively. $D$ is the feet of angle bisector of $\angle ACB$. $I$ denotes the incenter of $\triangle ABC$

Lemma 1: $X,Y,Z$ collinear.
Proof: Let $O_1$ and $O_2$ denotes the circumcenters of $\Omega$ and $\Gamma$ respectively. We know that, $Z, O_1, O_2$ collinear.
Also, $YO_1 \perp AB$ and $XO_2 \perp AB$ implies $YO_1 || XO_2$. Now, homothety gives us $X,Y,Z$ collinear. $\square$

Lemma 2: $Y,Z,C,D$ cyclic.
Proof: $\angle YZC = \angle XZC = \angle XAC = \angle A + \frac {\angle C}{2}$
$\angle YDX = \angle A + \frac {\angle C}{2}$ $\square$

Lemma 3: $XB$ is tangent to $\bigodot PQB$
Proof: $\angle XBD = \angle DCB = \frac {\angle C}{2} \Rightarrow XB$ is tangent to $\bigodot BDC \Rightarrow XD.XC=XB^2$
$XP.XQ = XY.XZ = XD.XC = XB^2$ $\square$

Lemma 4: $BI$ bisects $\angle QBP$
Proof: Lemma 3 gives us $$XP.XQ=XB^2=XI^2$$
$$\Rightarrow \frac{XP}{XI}= \frac {XI}{XQ}$$
$$\Rightarrow \frac{XP-XI}{XI}= \frac {XI-XQ}{XQ}$$
$$\Rightarrow \frac{PI}{XI}= \frac {QI}{XQ}$$
$$\Rightarrow \frac{PI}{QI}= \frac {XI}{XQ} = \frac{XP}{XI} = \frac{XP}{XB}$$
$$\triangle XPB \sim \triangle BPQ \Rightarrow \frac{XP}{XB}=\frac{BP}{BQ}$$
$$\therefore \frac{PI}{QI} = \frac{BP}{BQ} \square$$

Now our problem statement follows immediately.
$Q.E.D$
This was freedom. Losing all hope was freedom.

thczarif
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Location: Dhaka,Bangladesh

Re: EGMO 2018 P5

Unread post by thczarif » Sat Jul 27, 2019 12:02 pm

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

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