IMO Problem Shortlist

Denote by $k$ the incircle and by $k_a$ the excircle opposite to $A$ of triangle $ABC$. Let $k$ and $k_a$ touch the side $BC$ at the points $X$ and $T$, respectively, let $k_a$ touch the lines $AB$ and $AC$ at the points $P$ and $Q$, respectively. We use several times the fact that opposing sides of a parallelogram are of equal length, that points of contact of the excircle and incircle to a side of a triangle lie symmetric with respect to the midpoint of this side and that segments on two tangents to a circle defined by the points of contact and their point of intersection have the same length. So we conclude $$\begin{array}{c}ZP=ZB+BP=XB+BT=BX+CX=ZS\text{ and }\\ CQ=CT=BX=BZ=CS.\end{array}$$ So for each of the points $Z,C$, their distances to $S$ equal the length of a tangent segment from this point to $k_a$. It is well-known, that all points with this property lie on the line $ZC$, which is the radical axis of $S$ and $k_a$. Similar arguments yield that $BY$ is the radical axis of $R$ and $k_a$. So the point of intersection of $ZC$ and $BY$, which is $G$ by definition, is the radical center of $R,S$ and $k_a$, from which the claim $GR=GS$ follows immediately.

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Alternative solution.

Denote $x=AZ=AY$, $y=BZ=BX$, $z=CX=CY$, $p=ZG$, $q=GC$. Several lengthy calculations (Menelaos' theorem in triangle $AZC$, law of Cosines in triangles $ABC$ and $AZC$ and Stewart's theorem in triangle $ZCS$) give four equations for $p,q,\cos \alpha$ and $GS$ in terms of $x,y$, and $z$ that can be resolved for $GS$. The result is symmetric in $y$ and $z$, so $GR=GS$. More in detail this means: The line $BY$ intersects the sides of triangle $AZC$, so Menelaos' theorem yields $\frac{p}{q}\cdot \frac{z}{x}\cdot \frac{x+y}{y}=1$, hence $$\begin{array}{c}\frac{p}{q}=\frac{xy}{yz+zx}.\end{array}\text{\hspace{1em}(1)}$$ Since we only want to show that the term for $GS$ is symmetric in $y$ and $z$, we abbreviate terms that are symmetric in $y$ and $z$ by capital letters, starting with $N=xy+yz+zx$. So (1) implies $$\begin{array}{c}\frac{p}{p+q}=\frac{xy}{xy+yz+zx}=\frac{xy}{N}\text{ and }\frac{q}{p+q}=\frac{yz+zx}{xy+yz+zx}=\frac{yz+zx}{N}.\end{array}\text{\hspace{1em}(2)}$$ Now the law of Cosines in triangle $ABC$ yields $$\cos \alpha =\frac{(x+y{)}^2+(x+z{)}^2-(y+z{)}^2}{2(x+y)(x+z)}=\frac{2x^2+2xy+2xz-2yz}{2(x+y)(x+z)}=1-\frac{2yz}{(x+y)(x+z)}.$$ We use this result to apply the law of Cosines in triangle $AZC$: $$\begin{array}{rl}(p+q{)}^2& =x^2+(x+z{)}^2-2x(x+z)\cos \alpha \\ & =x^2+(x+z{)}^2-2x(x+z)\cdot \left(1-\frac{2yz}{(x+y)(x+z)}\right)\\ & =z^2+\frac{4xyz}{x+y}\end{array}\text{\hspace{1em}(3)}$$ Now in triangle $ZCS$ the segment $GS$ is a cevian, so with Stewart's theorem we have $p{y}^2+q(y+z{)}^2=(p+q)\left(G{S}^2+pq\right)$, hence $$G{S}^2=\frac{p}{p+q}\cdot y^2+\frac{q}{p+q}\cdot (y+z{)}^2-\frac{p}{p+q}\cdot \frac{q}{p+q}\cdot (p+q{)}^2.$$ Replacing the $p$'s and $q$'s herein by (2) and (3) yields $$\begin{array}{rl}G{S}^2& =\frac{xy}{N}{y}^2+\frac{yz+zx}{N}(y+z{)}^2-\frac{xy}{N}\cdot \frac{yz+zx}{N}\cdot \left(z^2+\frac{4xyz}{x+y}\right)\\ & =\frac{x{y}^3}{N}+\underset{M_1}{\underbrace{\frac{yz(y+z{)}^2}{N}}}+\frac{zx(y+z{)}^2}{N}-\frac{xy{z}^3(x+y)}{N^2}-\underset{M_2}{\underbrace{\frac{4x^2{y}^2{z}^2}{N^2}}}\\ & =\frac{x{y}^3+zx(y+z{)}^2}{N}-\frac{xy{z}^3(x+y)}{N^2}+M_1-M_2\\ & =\underset{M_3}{\underbrace{\frac{x\left(y^3+y^{2}z+y{z}^2+z^3\right)}{N}}}+\frac{xy{z}^{2}N}{N^2}-\frac{xy{z}^3(x+y)}{N^2}+M_1-M_2\\ & =\frac{x^2{y}^2{z}^2+x{y}^2{z}^3+x^{2}y{z}^3-x^{2}y{z}^3-x{y}^2{z}^3}{N^2}+M_1-M_2+M_3\\ & =\frac{x^2{y}^2{z}^2}{N^2}+M_1-M_2+M_3,\end{array}$$ a term that is symmetric in $y$ and $z$, indeed.