Subfield: Geometry


Question: Let $A B C$ be a triangle. The incircle of $A B C$ touches the sides $A B$ and $A C$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $B Y$ and $C Z$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $B C Y R$ and $B C S Z$ are parallelograms.

Prove that $G R=G S$.

Solution:
Denote by $k$ the incircle and by $k_{a}$ the excircle opposite to $A$ of triangle $A B C$. Let $k$ and $k_{a}$ touch the side $B C$ at the points $X$ and $T$, respectively, let $k_{a}$ touch the lines $A B$ and $A C$ 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

$$
\begin{gathered}
Z P=Z B+B P=X B+B T=B X+C X=Z S \text { and } \\
C Q=C T=B X=B Z=C S .
\end{gathered}
$$

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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 $Z C$, which is the radical axis of $S$ and $k_{a}$. Similar arguments yield that $B Y$ is the radical axis of $R$ and $k_{a}$. So the point of intersection of $Z C$ and $B Y$, which is $G$ by definition, is the radical center of $R, S$ and $k_{a}$, from which the claim $G R=G S$ follows immediately.
Denote $x=A Z=A Y, y=B Z=B X, z=C X=C Y, p=Z G, q=G C$. Several lengthy calculations (MENELAOs' theorem in triangle $A Z C$, law of Cosines in triangles $A B C$ and $A Z C$ and STEWART's theorem in triangle $Z C S$ ) give four equations for $p, q, \cos \alpha$



and $G S$ in terms of $x, y$, and $z$ that can be resolved for $G S$. The result is symmetric in $y$ and $z$, so $G R=G S$. More in detail this means:

The line $B Y$ intersects the sides of triangle $A Z C$, so MEnELaOS' theorem yields $\frac{p}{q} \cdot \frac{z}{x} \cdot \frac{x+y}{y}=1$, hence

$$
\frac{p}{q}=\frac{x y}{y z+z x}
\tag{1}
$$

Since we only want to show that the term for $G S$ is symmetric in $y$ and $z$, we abbreviate terms that are symmetric in $y$ and $z$ by capital letters, starting with $N=x y+y z+z x$. So (1) implies

$$
\frac{p}{p+q}=\frac{x y}{x y+y z+z x}=\frac{x y}{N} \quad \text { and } \quad \frac{q}{p+q}=\frac{y z+z x}{x y+y z+z x}=\frac{y z+z x}{N}
\tag{2}
$$

Now the law of Cosines in triangle $A B C$ yields

$$
\cos \alpha=\frac{(x+y)^{2}+(x+z)^{2}-(y+z)^{2}}{2(x+y)(x+z)}=\frac{2 x^{2}+2 x y+2 x z-2 y z}{2(x+y)(x+z)}=1-\frac{2 y z}{(x+y)(x+z)}
$$

We use this result to apply the law of Cosines in triangle $A Z C$ :

$$
\begin{aligned}
(p+q)^{2} & =x^{2}+(x+z)^{2}-2 x(x+z) \cos \alpha \\
& =x^{2}+(x+z)^{2}-2 x(x+z) \cdot\left(1-\frac{2 y z}{(x+y)(x+z)}\right) \\
& =z^{2}+\frac{4 x y z}{x+y} .
\end{aligned}
\tag{3}
$$

Now in triangle $Z C S$ the segment $G S$ is a cevian, so with STEWART's theorem we have $p y^{2}+q(y+z)^{2}=(p+q)\left(G S^{2}+p q\right)$, 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

$$
\begin{aligned}
G S^{2} & =\frac{x y}{N} y^{2}+\frac{y z+z x}{N}(y+z)^{2}-\frac{x y}{N} \cdot \frac{y z+z x}{N} \cdot\left(z^{2}+\frac{4 x y z}{x+y}\right) \\
& =\frac{x y^{3}}{N}+\underbrace{\frac{y z(y+z)^{2}}{N}}_{M_{1}}+\frac{z x(y+z)^{2}}{N}-\frac{x y z^{3}(x+y)}{N^{2}}-\underbrace{\frac{4 x^{2} y^{2} z^{2}}{N^{2}}}_{M_{2}} \\
& =\frac{x y^{3}+z x(y+z)^{2}}{N}-\frac{x y z^{3}(x+y)}{N^{2}}+M_{1}-M_{2} \\
& =\underbrace{\frac{x\left(y^{3}+y^{2} z+y z^{2}+z^{3}\right)}{N}}_{M_{3}}+\frac{x y z^{2} N}{N^{2}}-\frac{x y z^{3}(x+y)}{N^{2}}+M_{1}-M_{2} \\
& =\frac{x^{2} y^{2} z^{2}+x y^{2} z^{3}+x^{2} y z^{3}-x^{2} y z^{3}-x y^{2} z^{3}}{N^{2}}+M_{1}-M_{2}+M_{3} \\
& =\frac{x^{2} y^{2} z^{2}}{N^{2}}+M_{1}-M_{2}+M_{3},
\end{aligned}
$$

a term that is symmetric in $y$ and $z$, indeed.