Solution: Let $(I_a)$ be the A-excircle of triangle $ABC. (I_a)$ touches $BC$ at $L$. $\frac{BI}{BI_a}=\cot \angle BII_a=\cot (90^o-\angle C/2)=\tan C/2=\frac{IY}{CY}=\frac{IY}{BR}.$ On the other side, $\angle RBI_a=\angle 180^o-\angle C+90^o-\angle B/2=\angle BIY$. So $\Delta BIY\sim \Delta I_aBR$, which follows that $RI_a\perp BY. (1)$ But $BL=BR$ hence $B$ lies on the radical axis of $R$ and $(I_a) (2)$ From $(1)$ and $(2)$ we obtain $GB$ is the radical axis of $R$ and $(I_a)$. Therefore $GR^2=GI_a^2-R^2_a$. Similarly we are done.

Attachments:

Nice solution livetolove! I have another solution that basically uses Stewart all over the place (plus the fact that $ZS=BC$ and $CS=BZ$, together with the known results $BZ=\frac{c+a-b}{2}$ and $CY=\frac{a+b-c}{2}$), resulting after some boring calculations in an expression for $GS^2$ symmetric in $b,c$, hence equal to $GR^2$ by symmetry in variables. No extra "added value" in my solution, so I will not post some boring calculations...

livetolove212 wrote: From $(1)$ and $(2)$ we obtain $GB$ is the radical axis of $R$ and $(I_a)$. We can prove this in an even simpler way: $BX_a=BR$ and $YY_a=YR$, where $X_a=(I_a)\cap BC$, $Y_a=(I_a)\cap CA$, and $Z_a=(I_a)\cap AB$, so $BY$ is the radical axis of $(I_a)$ and $R$.

Ok, let me endorse the dark side. We use barycentric coordinates. (This was done in 30 minutes by hand, if anyone wonders about feasibility). Let $a = y+z, b=x+z, c=x+y$, side lengths. Of course $ABC$ the the reference triangle. We'll generally used normalized coordinates here. Of course $Y = \left(\frac{s-c}{b}, 0, \frac{s-a}{b} \right)$ and $Z = \left( \frac{s-b}{c}, \frac{s-a}{c}, 0 \right)$. Let $t = xy+yz+xz$. Then the Gergonne point (un-normalized) $G = \text{cyc}\left( \frac{1}{x} \right) = \text{cyc} \left(\frac{yz}{t} \right)$. Now for $R, S$. Clearly, $R = B+Y-C = \left(\frac{z}{x+z}, 1, -\frac{z}{x+z} \right)$ and $S = \left( \frac{y}{x+y}, 1, -\frac{y}{x+y} \right)$. Now for the final computation, we find $|RG|^2$ and show that this quantity is symmetric in $y, z$ to finish. Well, a simple computation shows that $\vec{RG} = \left( \frac{xz^2}{(x+z)t}, \frac{y(x+z)}{t}, -\frac{zt+xy(x+z)}{(x+z)t} \right)$. By the barycentric distance formula, \[|RG|^2 = (y+z)^2 \cdot \frac{y(x+z)}{t} \cdot \frac{zt+xy(x+z)}{(x+z)t} + (x+z)^2 \cdot \frac{xz^2}{(x+z)t} \cdot \frac{zt+xy(x+z)}{(x+z)t} - (x+y)^2 \cdot \frac{xz^2}{(x+z)t} \cdot \frac{y(x+z)}{t} \] \[ = \frac{(y+z)^2yzt+\left( (y+z)^2 xy^2(x+z) + xz^3t+x^2yz^2(x+z) - (x+y)^2 xyz^2 \right)}{t^2} \] \[ = \frac{(y+z)^2yzt + x (x y^4+2 x y^3 z-x y^2 z^2+2 x y z^3+x z^4+y^4 z+y^3 z^2+y^2 z^3+y z^4)}{t^2}, \] which is clearly symmetric in $y, z$.

Mine is a normal approach but a hand-paining bashing,but still not intolerable,as I had no other option.Apply Menelaus theorem to get each of the ratios $\frac{GC}{GZ}$ and $\frac{GB}{GY}$.Then apply Stewart in $\triangle{ZCS}$ and $\triangle{YBR}$.(In fact,finding out one of the lengths will do,by the symmetric nature of the expression.

let $A-$excircle$(W)$ touch $AB,AC,BC$ at $M,N,P$. $YR=BC=YN,BM=BP=CY$ so $BY$ is radius exist of point$R$ and circle $W$ so $GR$ is equal to length of tangent from $G$ to $W$. similar we find $GS$ is equal to length of tangent from $G$ to $W$. so we are done

Let the center of the excircle be Ia,and let it touch the side BC at X1,and let X be the point where the incircle touches BC and let the excircle touches AB and AC at points X3 and X2,respectively.Now,let k1 be the circumcircle of X3X1R and k2 be the circumcircle of X1X2S.Now,the centers of k1 and k2 are B and C,respectively.Now,it is easy to show that RIa is perpendicular to BY(Lemma:In a convex quadritedral ABCD diagonals are perpendicular iff AB^2+CD^2=BC^2+AD^2,this lemma aplyed on the quadritedral RBIaY kills it) and similary CZ is perpendicular to SIa.Let k3 be the circle with center G and radius GR.Clearly,IaX1 is the radical axis of k1 and k2 and IaR is the radical axis of k1 and k3,so we have that IaS is the radical axis of k3 and k2 and S lies on k2,so it lies on k3 and we are finished.

Very beautiful problem! The configuration is quite rich here and it admits an elegant proof Let us denote by $\omega_a$ the excircle opposite to the vertex $A$ in triangle $ABC$; $V,W$ as the touching points of the circle $\omega_a$ with the lines $AC,AB$, extended beyond $C,B$; respectively. Let the incircle $\omega$ touch the side $BC$ at point $X$ and the circle $\omega_a$ touch the side $BC$ at the point $U$. Let $a,b,c$ denote the sides of the triangle $ABC$ in the usual way and $s$ be the semi-perimeter. Notice that $BR=CY=CX=BU$ since the points $X,U$ are known to be symmetric wrt the midpoint of $BC$ and $YR=BC=(s-(s-a))=AW-AY=YW$. Therefore, the line $BY$ is the radical axis of the circles $R,\omega$ where we assume $R$ to be a "point circle". Similarly, $CZ$ is the radical axis of the circles $S,\omega_a$ where $S$ is also a point circle. The intersection point $G$ of the lines $BY$ and $CZ$ is thus, the radical center of the circles $R,S,\omega_a$ and we conclude that $GR=GS$.

Question: what is the motivation for adding the $A$-excenter and $A$-excircle?

Can anybody tell why on earth $GR \neq GS$ in my picture ? [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(14.213303567689671cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 5.115636810410432, xmax = 19.328940378100103, ymin = -0.713488980527826, ymax = 5.611693345167297; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen ffqqff = rgb(1.,0.,1.); draw((11.587164375684239,4.003326457516075)--(9.277525367805563,1.9786429116483892)--(11.542171407998303,1.7911722129569367)--cycle, linewidth(2.) + zzttqq); /* draw figures */ draw((11.587164375684239,4.003326457516075)--(9.277525367805563,1.9786429116483892), linewidth(2.) + zzttqq); draw((9.277525367805563,1.9786429116483892)--(11.542171407998303,1.7911722129569367), linewidth(2.) + zzttqq); draw((11.542171407998303,1.7911722129569367)--(11.587164375684239,4.003326457516075), linewidth(2.) + zzttqq); draw((xmin, 0.8766233766233832*xmin-6.154252702986418)--(xmax, 0.8766233766233832*xmax-6.154252702986418), linewidth(2.) + dotted + blue); /* line */ draw((xmin, 49.166666666680925*xmin-565.698922013791)--(xmax, 49.166666666680925*xmax-565.698922013791), linewidth(2.) + dotted + blue); /* line */ draw((xmin, -0.08278145695364258*xmin + 2.746649978519712)--(xmax, -0.08278145695364258*xmax + 2.746649978519712), linewidth(2.) + dotted + blue); /* line */ draw((xmin, 0.3298117400621979*xmin-1.0811938733787465)--(xmax, 0.3298117400621979*xmax-1.0811938733787465), linewidth(2.) + dotted + blue); /* line */ draw((xmin, 0.9020263512574868*xmin-8.620170547788263)--(xmax, 0.9020263512574868*xmax-8.620170547788263), linewidth(2.) + dotted + blue); /* line */ draw(circle((10.892588441380406,2.511309674254309), 0.6640924854657327), linewidth(2.) + blue); draw(circle((9.84443121263534,0.25976581915050795), 1.666248340309526), linewidth(2.) + blue); draw((11.587164375684239,4.003326457516075)--(10.837801299206813,1.8494809967972916), linewidth(2.) + ffqqff); draw((9.277525367805563,1.9786429116483892)--(11.556543610473554,2.4978055013236053), linewidth(2.) + ffqqff); draw((11.542171407998303,1.7911722129569367)--(10.45482151829024,3.0106882383719773), linewidth(2.) + ffqqff); draw((11.556543610473554,2.4978055013236053)--(13.821189650666295,2.310334802632153), linewidth(2.) + ffqqff); draw((13.821189650666295,2.310334802632153)--(11.542171407998303,1.7911722129569367), linewidth(2.) + ffqqff); draw((10.45482151829024,3.0106882383719773)--(8.190175478097501,3.19815893706343), linewidth(2.) + ffqqff); draw((8.190175478097501,3.19815893706343)--(9.277525367805563,1.9786429116483892), linewidth(2.) + ffqqff); draw(circle((11.020913489997902,2.375788434830691), 2.9477738948162835), linewidth(2.) + ffqqff); /* dots and labels */ dot((11.587164375684239,4.003326457516075),dotstyle); dot((9.277525367805563,1.9786429116483892),dotstyle); dot((11.542171407998303,1.7911722129569367),dotstyle); dot((9.84443121263534,0.25976581915050795),linewidth(4.pt) + dotstyle); dot((10.892588441380406,2.511309674254309),linewidth(4.pt) + dotstyle); dot((10.837801299206813,1.8494809967972916),linewidth(4.pt) + dotstyle); dot((9.981895476597057,1.9203341278080333),linewidth(4.pt) + dotstyle); dot((10.45482151829024,3.0106882383719773),linewidth(4.pt) + dotstyle); dot((11.556543610473554,2.4978055013236053),linewidth(4.pt) + dotstyle); dot((9.866173443047902,2.4946655750101834),linewidth(4.pt) + dotstyle); dot((11.549357509235929,2.144488857140271),linewidth(4.pt) + dotstyle); dot((8.190175478097501,3.19815893706343),dotstyle); dot((13.821189650666295,2.310334802632153),dotstyle); dot((11.020913489997902,2.375788434830691),linewidth(4.pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] ?

A quick solution using well-known results: Let $M$ be the midpoint of $YZ$, $X$ be where the incircle meets $BC$, $N$ be the midpoint of the $X$-altitude of $\triangle XYZ$, $I,I_A$ be the incenter and $A$-excenter of $\triangle ABC$, and $H$ be the orthocenter of $\triangle BIC$. We know $M, N, G$ are collinear on $\ell$, the X-Schwatt line of $\triangle XYZ$, and by 2009 Iran TST Day 3 #3 $\ell$ passes through $H$. Since $RYSZ$ is a parallelogram, $RS$ has midpoint $M$, hence we need only to show $\ell$ is the perpendicular bisector of $RS$, or equivalently that $HR=HS$. Reflect $H,R,S$ over the midpoint of $BC$, so $H$ is mapped to $I_A$. Clearly $R$ is mapped to a point $R'$ on $\overline{AC}$ with $CR' = BR = CY$, i.e. $R'$ is the reflection of $Y$ over $C$, and similarly $S'$ is the reflection of $Z$ over $B$. We'd like to show $I_AR'=I_AS'$,or equivalently that the lengths of the tangents from $R',S'$ to the $A$-excircle are equal, but it's trivial to compute that $AR' = s+b-c, AS' = s+c-b$, hence both tangents have length $|b-c|$, as desired.

April wrote: Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram. Prove that $GR=GS$. Proposed by Hossein Karke Abadi, Iran Probably the same as the above? Still I'm so amazed I'll post anyway Let $J$ be the $A$-excenter, $U,V$ be the projections of $J$ on $\overline{AB}, \overline{AC}$, $I$ be the incenter and $H$ the orthocenter of $\triangle BIC$. Let $M,N$ be midpoints of $\overline{BC}, \overline{YZ}$ respectively. Observe $YR=ZS$ hence $N$ is the midpoint of $RS$. Now Iran TST 2009/3 and Schwatt line tell us $\overline{HN}$ passes through $G$. Thus, it is enough to show $HR=HS$. Reflect about $M$; then we wish $JR'=JS'$. Note that $R',S'$ lie on $\overline{AC}, \overline{AB}$ and $US'=VR'=|AB-AC|$ hence we're done. $\blacksquare$

My solution (posting almost 2 years later ) : Draw in the $A-$excircle of $ABC$, say $\odot (I_a)$. Let it touch $AB, AC$ at $L,M$. $BL = CY = BR$. We also have $YR = BC = YM$. So the radical center of $\odot (R,0), \odot (S,0), \odot (I_a)$ is $BY \cap CZ$, which is $G$, and thus $GR = GS$.

wu2481632 wrote: Question: what is the motivation for adding the $A$-excenter and $A$-excircle? Anyone?

My solution(complex bash!!): 1.Set the incircle as the unit circle. 2.By complex intersection formula,compute G.(feasible) 3.Compute R and S using the fact that if ABCD is a parallelogram a+d=b+c holds. 4.Obtain the absolute values and prove that they are equal. It took me 30 mins. Drunken_Master wrote: wu2481632 wrote: Question: what is the motivation for adding the $A$-excenter and $A$-excircle? Anyone? I think it is kind of common.Have u observed it in the pernicious Russian olympaid Perimeter problem?(EGMO chapter 2 last example) It is indeed kind of a short trick like reflecting the orthocenter.

Drunken_Master wrote: wu2481632 wrote: Question: what is the motivation for adding the $A$-excenter and $A$-excircle? Anyone? Pretty much the only ways you can handle parallelograms are either lengths, parallel sides, or midpoints. Lengths go well together with the incircle if there is also the excircle, mainly due to isotomic points and equal tangents. So it is natural to bring in the excircle touchpoints. This finishes the problem instantly.

Electron_Madnesss wrote: Lastly $G$ is,of course, the gergonne point of $\triangle ABC \implies G = (s-a : s-b : s-c)$. $G=((s-b)(s-c):(s-c)(s-a):(s-a)(s-b))$. The nagel point has coordinates $(s-a:s-b:s-c)$.

Thanks. Anyways the solution had a fault

I have drawn a diagram.

Attachments:

Let the $A-\text{excircle}$ $\omega$ meet $AC,$ $AB$ by $E,F$ respectively. Since $|BR|=|BF|,\text{ } |YR|=|BC|=|YE|$ line $BY$ is the radical axis of $R,\omega,$ and analogously for $CZ.$ Therefore $G$ is the radical center of $R,S,\omega$ and the conclusion follows.

We employ complex numbers. Let the incircle be the unit circle with $1$ being the tangency point of the unit circle and $BC$. We'll use $y,z$ as variables. Using the intersection of tangents formula we compute that: \[b=\frac{2z}{z+1}\quad c=\frac{2y}{y+1}\quad r=y+b-c=\frac{y^2 z + y^2 + y z + 2 z- y }{(y + 1) (z + 1)}\quad s=c+z-b=\frac{y z^2 + z^2+ y z + 2 y - z}{(y + 1) (z + 1)}\]Thus we only need to compute $g$. As $G=BY\cap CZ\Longrightarrow \frac{g-y}{b-y}=\overline{\left(\frac{g-y}{b-y}\right)}$ and $\frac{g-z}{c-z}=\overline{\left(\frac{g-z}{c-z}\right)}$. After solving these two equations we get \begin{align} g&=\frac{((\overline{b}-\overline{y})y-(b-y)\overline{y})(c-z)-((\overline{c}-\overline{z})z-(c-z)\overline{z})(b-y)}{(c-z)(\overline{b}-\overline{y})-(b-y)(\overline{c}-\overline{z})}\nonumber \\ &=\frac{\left((\frac{2}{z+1}-\frac{1}{y})y-(\frac{2z}{z+1} - y)\frac{1}{y}\right) (\frac{2y}{y + 1} - z) - \left((\frac{2}{y+1} - \frac{1}{z}) z - (\frac{2y}{y + 1} - z)\frac{1}{z}\right) (\frac{2z}{z + 1} - y)}{(\frac{2y}{y+1} - z) (\frac{2}{z+1} - \frac{1}{y}) - (\frac{2z}{z+1} - y) (\frac{2}{y+1} - \frac{1}{z})}\nonumber \\ &=2\times\frac{\left(\frac{y}{z+1}-\frac{z}{y(z+1)}\right) (\frac{2y}{y + 1} - z) - \left(\frac{z}{y+1}-\frac{y}{z(y + 1)}\right) (\frac{2z}{z + 1} - y)}{(\frac{2y}{y+1} - z) (\frac{2}{z+1} - \frac{1}{y}) - (\frac{2z}{z+1} - y) (\frac{2}{y+1} - \frac{1}{z})}\nonumber \\ &=\frac{2 (-y^2z^2 + y^2 z - y^2 + y z^2 + y z-z^2)}{6 y z-y^2z- z-y^2-z^2 -y z^2-y} \nonumber \\ \nonumber \end{align}Now we compute $|g-r|$ and $|g-s|$: \begin{align} |g-r|&=\left|\frac{2 (-y^2z^2 + y^2 z - y^2 + y z^2 + y z-z^2)}{6 y z-y^2z- z-y^2-z^2 -y z^2-y}-\frac{y^2 z + y^2 + y z - y + 2 z}{(y + 1) (z + 1)}\right|\nonumber \\ &=\frac{|y|\cdot|y - z|\cdot |yz - 3z + y + 1|\cdot |y z + y + z - 3|}{|y+1|\cdot |z+1|\cdot |6 y z-y^2z- z-y^2-z^2 -y z^2-y|} \nonumber\\ \nonumber \end{align}\begin{align} |g-s|&=\left| \frac{2 (-y^2z^2 + y^2 z - y^2 + y z^2 + y z)}{6 y z-y^2z- z-y^2-z^2 -y z^2-y} - \frac{y z^2 + y z + 2 y + z^2 - z}{(y + 1) (z + 1)} \right| \nonumber\\ &=\frac{|z|\cdot |z-y| \cdot |yz-3y + z + 1|\cdot |y z + y + z - 3|}{|y+1|\cdot |z+1|\cdot |6 y z-y^2z- z-y^2-z^2 -y z^2-y|}\nonumber \\ \nonumber \end{align}As $|y|=1=|z|\Longrightarrow |g-s|=|g-r|\iff |yz-3z+y+1|=|yz-3y+z+1|$, which is the case: \[|yz-3z+y+1|=(yz-3z+y+1)\left(\frac{1-3y+z+yz}{yz}\right)=\frac{yz-3z+y+1}{yz}\cdot (1-3y+z+yz)=|yz-3y+z+1|\]

One of the best synthetic problems in existence. [asy][asy] // ISL 2009 G3 size(7cm); defaultpen(0.8); defaultpen(fontsize(9pt)); dotfactor*=0.6; pen purplefill,pinkfill,lightbluedraw,bluedraw,purpledraw,pinkdraw; purplefill = RGB(238,204,255); pinkfill = RGB(255,204,255); lightbluedraw = RGB(85,187,255); bluedraw = RGB(0,102,255); purpledraw = RGB(170,34,255); pinkdraw = RGB(255,17,255); pair A,B,C,I,X,Y,Z,R,S,XX,P,Q,G; A = dir(115); B = dir(210); C = dir(330); I = incenter(A,B,C); X = foot(I,B,C); Y = foot(I,C,A); Z = foot(I,A,B); R = B+Y-C; S = C+Z-B; XX = B+C-X; P = intersectionpoints(circle(C,distance(C,XX)),C--2C-A)[0]; Q = intersectionpoints(circle(B,distance(B,XX)),B--2B-A)[0]; G = extension(B,Y,C,Z); filldraw(circumcircle(X,Y,Z),pinkfill,pinkdraw); filldraw(circumcircle(XX,P,Q),purplefill,purpledraw); draw(11B-10A--A--11C-10A,bluedraw); draw(B--C,bluedraw); draw(B--Y,purpledraw); draw(C--Z,purpledraw); draw(B--R--Y,lightbluedraw); draw(C--S--Z,lightbluedraw); dot("$A$",A,dir(100)); dot("$B$",B,dir(200)); dot("$C$",C,dir(20)); dot("$X$",X,1.5*dir(280)); dot("$Y$",Y,dir(50)); dot("$Z$",Z,dir(150)); dot("$R$",R,dir(140)); dot("$S$",S,dir(30)); dot("$X'$",XX,dir(270)); dot("$P$",P,dir(50)); dot("$Q$",Q,dir(150)); dot("$G$",G,dir(270)); clip((1.7,1.5)--(-2,1.5)--(-2,-2)--(1.7,-2)--cycle); [/asy][/asy] Let the $A$-excircle $\Omega$ touch $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ at $X'$, $P$, and $Q$, respectively, and let the incircle touch $\overline{BC}$ at $X$. Since $BR=CY=CX=BX'=BQ$ and $YR=BC=BX+XC=CX'+CY=PY$, $\overline{BY}$ is the radical axis of $\Omega$ and the circle $\omega_R$ of radius $0$ centered at $R$. Similarly, $\overline{CZ}$ is the radical axis of $\Omega$ and the circle $\omega_S$ of radius $0$ centered at $S$. Thus, $G$ is the radical center of $\Omega$, $\omega_R$, and $\omega_S$, so $GR=GS$. $\square$

Draw the $A$-excircle $\gamma$, which meets $BC$ and $AC$ and $F$ and $H$, respectively. $RB=YC=BF$ and $RY=BC=YH$. Thus, $BY$ is the radical axis of $(R)$ and $\gamma$. Similarly, $CZ$ is the radical axis of $(S)$ and $\gamma$. The perpendicular bisector of $RS$ is the radical axis of $R$ and $S$ and must concur with $CZ$ and $SR$. Thus, $GR=GS$.

Wow. Apply barycentric coordinates on $\triangle ABC$. We have $Y=(s-c:0:s-a), Z=(s-b:s-a:0)$. Also, since $G$ is the Gergonne point, we have $G=((s-b)(s-c):(s-a)(s-c):(s-a)(s-b))$. By Parallelogram Lemma we now get $$R=\left(\frac{s-c}{b}, 1, \frac{s-a-b}{b}\right), S=\left(\frac{s-b}{c}, \frac{s-a-c}{c}, 1\right).$$Distance formula immediately finishes.

I think this is problem is one of the best synthetic problems on olympiad geometry. ISL 2009 G3 wrote: Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram. Prove that $GR=GS$. Let $\omega$ be the $A$-excircle, and let $W, A_2$, and $A_1$ be the intersections of $\omega$ with $BC$, $AC$, $AB$, respectively. Also, define $Z$ as the intersection of the incircle with $BC$. By the given conditions of the problem we have $BR=CY=CZ=BW=BA_1$, similarly $CS=CA_2$. Furthermore, we have $XA_1=XB+BA_1=BZ+BW=BZ+CZ=BC=XS$, similarly $YA_2=YR$. Consider the circles $\Omega_1$, $\Omega_2$ with radius $0$, and centers $R$, and $S$, respectively. The radical axis of $\Omega_1$, and $\omega$ is $BY$, and the radical axis of $\Omega_2$, and $\omega$ is $CX$. By the Radical Axis Theorem, the radical axis of $\omega$, $\Omega_1$, and $\Omega_2$ is $BY\cap CX=G\implies Pow_{G}(\Omega_1)= Pow_{G}(\Omega_2)\implies GS^2=GR^2\implies GS=GR.\blacksquare$

Employ barycentric coordinates on $\triangle ABC$. Note that $Y=((s-c)/b,0,(s-a)/b)$ and $Z=((s-b)/c,(s-a)/c,0)$ Then, by the parallelograms, we have $$R=((s-c)/b,1,(s-a)/b-1),S=((s-b)/c,(s-a)/c-1,1).$$Note that it is well known that $G$ is the Gergonne point, so $$G=((s-b)(s-c):(s-a)(s-c):(s-a)(s-b)).$$Using the distance formula directly is too complicated, so we will use a Ravi substitution first: let $$x=s-a,y=s-b,z=s-c.$$Then, $$R=(z/(x+z),1,-z/(x+z)),S=(y/(x+y),-y/(x+y),1).$$Let $k=xy+xz+yz$. Most importantly, we can now write a homogenized form for $G$ that is not as messy: $$G=(yz/k,xz/k,xy/k).$$We can then compute $$\overrightarrow{GR}=(\frac{-xz^2}{k(x+z)}, \frac{-yx-yz}{k}, \frac{x^2y+xyz+kz}{k(x+z)}).$$Using the distance formula, $GR^2$ becomes an expression where swapping the variables $y$ and $z$ does not change the expression. Therefore, if we swap the labels of poins $B$ and $C$ in the original problem, then $R$ and $S$ will swap, but since $GR^2$ does not change, $GR=GS$, so we are done.

This one deserves a post. Let $\omega_R$ and $\omega_S$ denote the circles with radius $0$ centered at $R$ and $S,$ respectively. Let $\Omega$ be the $A$-excircle, and denote its tangents to $AB,BC,CA$ by $M,N,D,$ respectively. Define $X$ the intouch point on $BC.$ Observe $BR=CY=CX=BD=BM$ and $YR=BC=BX+CX=CN+CY=NY.$ Since $BR^2=BD^2$ and $YR^2=YN^2,$ BY is the radical axis of $\omega_R$ and $\Omega.$ Similarly we find $CD^2=CS^2$ and $ZM^2=ZS^2,$ implying $CZ$ is the radical axis of $\omega_S$ and $\Omega.$ Thus $G$ is the radical center of $\Omega, \omega_R, \omega_S,$ which finishes.

WOWIE! Construct $\omega_A$ the A-excircle. If the incircle and this excircle touches $BC$ at $D,D'$ respectively, then $BD'=CD=CY=RB$ and if the excircle touches $AC$ at $E$ then $YE=CY+AE=AD+CD=BC=RY.$ Thud $BY$ is the radical axis of $\omega_A$ and the circle with radius $0$ centered at $R$. Similarly, $CZ$ is radical axis of $\omega_A$ and $S$. Thus we find by radical center $GR^2=GS^2$ as desired

Beautiful Consider the $A$-excircle, $\omega_{a}$. We claim that the radical center of $\omega_{a}$ and a circle with radius $0$ centered at $R$ and $S$ is $G$. Note that this would finish the problem because $POW_{S}(G) = POW_{R}(G) = RG^{2} = GS^{2}$. Note that \[POW_{R}(B) = BR^{2} = POW_{\omega_{a}}(B)\]and \[POW_{R}(Y) = RY^{2} = BC^{2} = POW_{\omega_{a}}(Y)\]so $BY$ is the radical axis of $R$ and $\omega_{a}$. We can apply similar reasoning to $S$ and $\omega_{a}$ to find that $CZ$ is the radical axis of $S$ and $\omega_{a}$. Because $G = BY\cap CZ$, we must have that $G$ is the radical center, finishing the proof.

Construct the A-excircle $w_a$, and point circles at R and S (R) and (S), as well as extouch points on AB and AC F and E. Now, the following are well known (but is very easily length chased): BR=CY=BF;YR=BC=YE; whence BY is radax of (R),$w_a$; similarly for CZ. Then G is the radical center of them, whence GR=GS.

what. Let $\omega_A$ be the $A$-excircle, and denote by $D$, $E$, $F$ its tangency points with $BC, AC, AB$. Also, let the incircle be tangent to $\overline{BC}$ at $X$, and $\omega_R, \omega_S$ be the circles centered at $R$ and $S$ with radius $0$. Since $$\text{Pow}_{\omega_R}(B) = BR^2 = CY^2 = CX^2 = BD^2 = \text{Pow}_{\omega_A}(B)$$and $$\text{Pow}_{\omega_R}(Y) = YR^2 = BC^2 = (BX + XC)^2 = (CD + CX)^2 = (CE + CY)^2 = YE^2 = \text{Pow}_{\omega_A}(Y),$$we find that line $BY$ is the radical axis of $\omega_R$ and $\omega_A$. Similarly, $CZ$ is the radical axis of $\omega_S$ and $\omega_A$, and the conclusion follows by radical center lemma on $\omega_R, \omega_S$, and $\omega_A$.

Let the intersections of the $A$-excircle with $BC$, $AC$, and $AB$ be $D'$, $E$, and $F$, and let the intersection of the incircle with $BC$ be $D$. $\newline$ We will prove that $G$ is the radical center of the circles of radius zero at $R$ and $S$, and the $A$-excircle(which we will denote as $w_A$). $\newline$ This proves that $GR = GS$, due to properties of the radical center$(GR^2 - 0^2 = GS^2 - 0^2)$. Note that \[\text{Pow}_R(C) = CS^2 = BZ^2 = BD^2 = CD'^2 = CE^2 = \text{Pow}_{w_A}(C)\]And \[\text{Pow}_S(Z) = SZ^2 = BC^2 = (BD + DC)^2 = (BD' + BD)^2 = (ZB + BF)^2 = ZF^2 = \text{Pow}_{w_A}(Z)\]So $CZ$ is the radical axis of $S$ and $w_A$. By symmetry, $BY$ is the radical axis of $R$ and $w_A$, so $G$ is the radical center of the three circles, so we are done.

April wrote: Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram. Prove that $GR=GS$. Proposed by Hossein Karke Abadi, Iran $BR=YC=BA'$ and $YR=BC=YB'$ so $BY$ is the radical axes of $(I_A),R$ Similar $CZ$ is the radical axes of $(I_A),S$ So $G$ is the center of the three circles $(I_A),S,R$ which gives that $GR=GS$

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