9. Point $Y$ lies on line segment $X Z$ such that $X Y=5$ and $Y Z=3$. Point $G$ lies on line $X Z$ such that there exists a triangle $A B C$ with centroid $G$ such that $X$ lies on line $B C, Y$ lies on line $A C$, and $Z$ lies on line $A B$. Compute the largest possible value of $X G$.
Answer:
$\frac{20}{3}$

Solution: The key claim is that we must have $\frac{1}{X G}+\frac{1}{Y G}+\frac{1}{Z G}=0$ (in directed lengths).
We present three proofs of this fact.
Proof 1: By a suitable affine transformation, we can assume without loss of generality that $A B C$ is equilateral. Now perform an inversion about $G$ with radius $G A=G B=G C$. Then the images of $X, Y, Z$ (call them $\left.X^{\prime}, Y^{\prime}, Z^{\prime}\right)$ lie on $(G B C),(G A C),(G A B)$, so they are the feet of the perpendiculars from $A_{1}, B_{1}, C_{1}$ to line $X Y Z$, where $A_{1}, B_{1}, C_{1}$ are the respective antipodes of $G$ on $(G B C),(G A C),(G A B)$. But now $A_{1} B_{1} C_{1}$ is an equilateral triangle with medial triangle $A B C$, so its centroid is $G$. Now the centroid of (degenerate) triangle $X^{\prime} Y^{\prime} Z^{\prime}$ is the foot of the perpendicular of the centroid of $A_{1} B_{1} C_{1}$ onto the line, so it is $G$. Thus $X^{\prime} G+Y^{\prime} G+Z^{\prime} G=0$, which yields the desired claim.

Proof 2: Let $W$ be the point on line $X Y Z$ such that $W G=2 G X$ (in directed lengths). Now note that $(Y, Z ; G, W)$ is a harmonic bundle, since projecting it through $A$ onto $B C$ gives $\left(B, C ; M_{B C}, \infty_{B C}\right)$. By harmonic bundle properties, this yields that $\frac{1}{Y G}+\frac{1}{Z G}=\frac{2}{W G}$ (in directed lengths), which gives the desired.

Proof 3: Let $P \neq G$ be an arbitrary point on the line $X Y Z$. Now, in directed lengths and signed areas, $\frac{G P}{G X}=\frac{[G B P]}{[G B X]}=\frac{[G C P]}{[G C X]}$, so $\frac{G P}{G X}=\frac{[G B P]-[G C P]}{[G B X]-[G C X]}=\frac{[G B P]-[G C P]}{[G B C]}=\frac{3([G B P]-[G C P])}{[A B C]}$. Writing analogous equations for $\frac{G P}{G Y}$ and $\frac{G P}{G Z}$ and summing yields $\frac{G P}{G X}+\frac{G P}{G Y}+\frac{G P}{G Z}=0$, giving the desired.
With this lemma, we may now set $X G=g$ and know that
$$
\frac{1}{g}+\frac{1}{g-5}+\frac{1}{g-8}=0
$$

Solving the quadratic gives the solutions $g=2$ and $g=\frac{20}{3}$; the latter hence gives the maximum (it is not difficult to construct an example for which $X G$ is indeed 20/3).