IMO 2011 Problem 6

[Moon]

Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let $\ell$ be a tangent line to $\Gamma$, and let $\ell_a, \ell_b$ and $\ell_c$ be the lines obtained by reflecting $\ell$ in the lines $BC$, $CA$ and $AB$, respectively.
Show that the circumcircle of the triangle determined by the lines $\ell_a, \ell_b$ and $\ell_c$ is tangent to the circle $\Gamma$.

[Tahmid Hasan]

Let ABC be an acute triangle with circumcircle $\Gamma$. Let $\ell$ be a tangent line to $\Gamma$ and let $\ell_a,\ell_b,\ell_c$ be the lines obtained by reflecting $\ell$ in the lines $BC, CA ,AB$, respectively. Show that the circumcircle of the triangle determined by the lines $\ell_a,\ell_b,\ell_c$ is tangent to the circle $\Gamma$.
solution:let $R$ be the tangent point.let$F=\ell \cap AB,E=\ell \cap AC,D=\ell \cap BC$
$X=l_c \cap l_b,Y=l_a \cap l_c,Z=l_a \cap l_b$.let the circumcircle of $\triangle XYZ$ be $\delta$
by reflection we can say $BF,BD,AE$ bisects $\angle XFE,\angle FDZ,\angle XEF$ respectively.
by locus in $\triangle XFE,\triangle YFD,\triangle DEZ$
we can say $XA,YB,ZC$ bisects $\angle YXZ,\angle XYZ,\angle XZY$,so $A$ is the incentre and $B,C$ the excenter of $\triangle XFE,\triangle YFD,\triangle DEZ$ respectively
let $I$ be the incentre of $\triangle XYZ$.
Lemma:if $I$ is the incenter and $J$ is the $A$ excenter of $\triangle ABC$,$\angle BIC=90^\circ+\frac{1}{2}A$ and $\angle BJC=90^\circ-\frac{1}{2}A$.
applying this on $\triangle XFE,\triangle DFY$ we get $\angle YXZ=180^\circ-2 \angle A,\angle XYZ=180^\circ-2 \angle B$
so $\angle XZY=180^\circ-2 \angle C$.
applying the lemma again on $\triangle XYZ$ we get $\angle XIZ=180^\circ- \angle B$.hence $A,B,C,I$ are concyclic.
let $R_a,R_b,R_c$ be the reflection of $R$ in $BC,CA,AB$ respectively.
theorem(1):in $\triangle ABC$ the internal and external bisector of $\angle A$ and perpendicular bisector of $BC$ intersect on the circumcircle.
by reflection we get $AR=AR_c,AR=AR_b$,hence $AR_b=AR_c$,so $A$ is on the perpendicular bisector of $R_bR_c$ and since $XA$ bisects $\angle R_bXR_c$,applying theorem (1),we get $X,R_c,A,R_b$ are concyclic(let the circle be $\alpha$).
in the same manner we get $Y,R_c,B,R_a$(let the circle be $\beta$) and $Z,R_a,R_b,C$ (let the circle be $\gamma$)concyclic.
claim1:$R_a,R_b,R_c$ are collinear.
$\angle ZXY+\angle XR_cR_a=\angle ZXY+\angle XYZ+\angle R_bR_aZ=\angle R_bZR_a+\angle R_bR_aZ$
so in $\triangle XR_cR_b,\triangle ZR_aR_b$ we can say,$\angle XR_bR_c=\angle ZR_bR_a$
now $180^\circ=\angle XR_bR_a+\angle ZR_bR_a=\angle XR_bR_a+\angle XR_bR_c$,so $R_aR_bR_c$ are collinear.
let $S$ be the second intersection of $\alpha,\beta$.
claim2:$S$ lies on $\gamma$ too.now $\angle R_bSR_a=\angle R_cSR_a-R_cSR_b$
$=180^\circ-\angle XYZ-\angle ZXY=2 \angle B-(180^\circ-2 \angle A)=180^\circ-2 \angle C$.
so claim2 is verified.
claim3:$S$ is on $\Gamma$.
$\angle ASB=\angle ASR_c+\angle BSR_c=\angle RXY+\angle XYR=\angle ARB$
claim3 is also verified.
claim4:$S$ lies on $\delta$.
now $\angle XSY=\angle XSR_c+\angle YSR_c=\angle XR_bR_c+\angle YR_aR_c$
$=\angle ZR_bR_a+\angle ZR_aR_b=\angle XZY$.
so claim4 is also verified.
hence $S$ is a common point of $\Gamma,\delta$.
now our task is to prove that $S$ is the tangent point.
claim5:the intersection point of $AR_b$ and $SZ$ (let the point be $P$) lies on $\Gamma$.
$\angle ARF=\angle ARC$(alternate segment angle)
the reflection of $\ell$ and $AR$ in $AC$ are $XZ$ and $AR_b$.
so we get $\angle XR_bA=\angle ARF$ and $\angle ACR_b=\angle ACR$
hence $\angle XR_bA=\angle ACR_b$ or,$\angle ACR_b=\angle PR_bZ$.....(1)
in $\gamma$,$\angle SCR_b=\angle SZR_b$.....(2)
adding (1),(2) we get $\angle ACS=\angle PR_bZ+\angle SZR_b=\angle APS$
so claim5 is also verified.
now in $\alpha$,$\angle SAP=\angle SXZ$
so from alternate segment theorem we can say,$\Gamma$ and $\delta$ are tangent at $S$.
finally solved it after sticking with this problem 7 days at a stretch. (「(だ)ってばよ!」)

[Tahmid Hasan]

not much advanced geometry was needed to solve this problem yet it was quite challenging