BR = 2XY wanted, AR = BP, AQ = AB ,tangent to circumcircle related

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BR = 2XY wanted, AR = BP, AQ = AB ,tangent to circumcircle related

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Source: Switzerland - Swiss ΜΟ 2017 p5

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#1 h Jul 14, 2020, 11:17 PM

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Let $ABC$ be a triangle with $AC> AB$ . Let $P$ be the intersection of $BC$ and the tangent through $A$ around the triangle $ABC$ . Let $Q$ be the point on the straight line $AC$ , so that $AQ = AB$ and $A$ is between $C$ and $Q$ . Let $X$ and $Y$ be the center of $BQ$ and $AP$ . Let $R$ be the point on $AP$ so that $AR = BP$ and $R$ is between $A$ and $P$ . Show that $BR = 2XY$ .

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#2 h Jul 15, 2020, 1:42 PM

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Let $D=BQ\cap AP$ , $\angle BAC=\alpha$ , $\angle ABC=\beta$ , $\angle ACB=\gamma$ , $\ell$ the bisector of $\angle BAC$ , $J=\ell\cap BC$

Fact 1: $Y$ is the midpoint of $RD$

Proof: since $QX=XB$ and $AQ=AB$ , we have $AX\perp BQ$ . $\angle AQB=\angle ABQ=\frac{\alpha}{2}$ $\Rightarrow$ $BD\parallel \ell$ .

$\triangle PAJ$ is isosceles because $\angle PAJ=\angle PAB+\angle BAJ=\gamma+\frac{\alpha}{2}$ and $\angle AJP=\angle AJB=\pi-\angle BAJ-\angle ABJ=\pi-\beta-\frac{\alpha}{2}=\gamma+\frac{\alpha}{2}$ . Then $AP=PJ$

It follows that $BDAJ$ is an isosceles trapezoid, with $AD=BJ$ .

Since $AP=PJ$ , $AR=PB$ , then $PR=BJ=AD$ which is equivalent to $RY=YD$ as $PY=AY$ $_{\blacksquare}$

Let $H$ be the midpoint of $QR$ and $G$ be the midpoint of $BR$ . From fact 1 it follows that $H,Y,G$ are collinear and $HY\parallel BQ\parallel \ell$ , so $HY\perp AX$

Fact 2: $AH\parallel BC$

Proof: $AQDJ$ is a parallelogram because $\angle DQA=\angle DBA=\angle DJA=\frac{\alpha}{2}$ (the last equality is due to the fact that $ADBJ$ is an isosceles trapezoid)

Let $L=HY\cap BC$ . $YL=\frac{1}{2}AJ$ as $PY=AY$ . $HY=frac{1}{2}DQ=\frac{1}{2}AJ=YL$

Since $HY=YL$ and $PY=AY$ , we have that $PHAL$ is a parallelogram. Therefore $AH\parallel BC$ $_{\blacksquare}$

Fact 3: $H$ and $Y$ are symmetric wrt $AX$

Proof: $AHLJ$ is a parallelogram since $AH\parallel BC$ and $HL\parallel AJ$ . Since $PY=AY$ , we have $PL=LJ$

Then $AH=LJ=AY$ . Since $AH=AY$ and $HY\perp AX$ , we get that $H$ and $Y$ are symmetrical wrt $AX$ $_{\blacksquare}$

As a result, $XY=HX=GB=\frac{1}{2}BR$ ( $HX=GB$ because $H,X,G$ are the midpoints of $RQ,QB,BR$ )

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#3 h Jul 15, 2020, 6:05 PM

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Let $S$ be the intersection of segments $BE$ , $AP$ . Since $AC>AB\implies \angle ABC>\angle ACB$ and $AX\perp BE$ we have $\angle XAB=\frac{\angle ABC+\angle ACB}{2}>\angle ACB=\angle SAB$ which proves $S$ belongs to the segment $BX$ . $AE=AB\implies \angle ABS=\frac{\angle BAC}{2}\implies \angle ASB=180^\circ-\frac{\angle BAC}{2}-\angle ACB=\frac{\angle BAC}{2}+\angle ABC=\angle SBC$ Therefore $AR=PB=PS$ and $Y$ is the midpoint of segment $RS$ . Let $T$ be the midpoint of segment $BR$ . We will prove $XY=BT$ . Denote $G$ as intersection of lines $YT,BC$ . Segment $YT$ is a midline in triangle $BSR$ so $GY\parallel BS$ and $BG=SY$ . Because $\angle TGB=180^\circ-\angle GBS= \angle YSX$ we need only $SX=TG$ to get $\triangle TGB\equiv\triangle XSY$ . Let's focus on this equality.
Menelaos in $\triangle PGY$ with line $B,T,R$ gives $1=\frac{PB}{BG}\cdot\frac{TG}{TY}\cdot\frac{RY}{PR}\iff 1=\frac{PB}{PR}\cdot\frac{TG}{TY}\iff 2TG=\frac{PR\cdot BS}{PB}$ By sine theorem $\frac{PC}{AP}=\frac{\sin\angle ABC}{\sin\angle ACB}=\frac{AC}{AB}.$ Menelaos in $\triangle BQC$ with line $P,S,A$ yields $1=\frac{CP}{BP}\cdot\frac{BS}{QS}\cdot\frac{QA}{AC}\iff 1=\frac{CP}{BP}\cdot\frac{BS}{BS+2SX}\cdot\frac{AB}{AC}\iff 1=\frac{AP}{BP}\cdot\frac{BS}{BS+2SX}\iff$ $\frac{2SX}{BS}=\frac{AP-PS}{PS}\iff 2SX=\frac{BS\cdot AS}{PS}$ .
Since $PR=AS$ and $PS=PB$ we have finished. QED
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#4 h Oct 26, 2023, 3:20 PM • 3 Y

Y by ismayilzadei1387, Yunis019, FriIzi

Take a point B' on extension of BY such that BY=YB'.Then ABPB' is parallelogram and triangles AQB' and ABR are congruent.Rest is easy.

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ismayilzadei1387

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#5 h Oct 28, 2023, 7:02 PM • 1 Y

Y by Ismayil_Orucov

Ismayil_Orucov wrote:

Take a point B' on extension of BY such that BY=YB'.Then ABPB' is parallelogram and triangles AQB' and ABR are congruent.Rest is easy.

Thanks for Elegant Solution

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