Harvard-MIT November Tournament

Solution:

Answer: $\frac{169}{23}$

We first show that quadrilateral $G_{1} G_{2} G_{4} G_{3}$ is cyclic. Note that by the law of cosines,
$$\cos \angle G_{2} G_{1} G_{3}=\frac{7^{2}+15^{2}-13^{2}}{2 \cdot 7 \cdot 15}=\frac{1}{2}$$
so $\angle G_{2} G_{1} G_{3}=60^{\circ}$. However, we know that $\angle G_{3} G_{1} G_{4}=30^{\circ}$, so $G_{1} G_{4}$ is an angle bisector. Now, let $G_{1} G_{4}$ intersect the circumcircle of triangle $G_{1} G_{2} G_{3}$ at $X$. Then, the minor arcs $\widehat{G_{2} X}$ and $\widehat{G_{3} X}$ are subtended by the equal angles $\angle G_{2} G_{1} X$ and $\angle G_{3} G_{1} X$, implying that $G_{2} X=G_{3} X$, i.e. $X$ is on the perpendicular bisector of $G_{2} G_{3}$, $l$. Similarly, since $G_{4} G_{2}=G_{4} G_{3}$, $G_{4}$ lies on $l$. However, since $l$ and $G_{1} G_{4}$ are distinct (in particular, $G_{1}$ lies on $G_{1} G_{4}$ but not $l$), we in fact have $X=G_{4}$, so $G_{1} G_{2} G_{4} G_{3}$ is cyclic.

We now have $G_{5} G_{2} G_{4} \sim G_{5} G_{3} G_{1}$ since $G_{1} G_{2} G_{4} G_{3}$ is cyclic. Now, we have $\angle G_{4} G_{3} G_{2}=\angle G_{4} G_{1} G_{2}=30^{\circ}$, and we may now compute $G_{2} G_{4}=G_{4} G_{3}=13 / \sqrt{3}$. Let $G_{5} G_{2}=x$ and $G_{5} G_{4}=y$. Now, from $G_{5} G_{4} G_{2} \sim G_{5} G_{1} G_{3}$, we have:
$$\frac{x}{y+13 / \sqrt{3}}=\frac{13 / \sqrt{3}}{15}=\frac{y}{x+7}$$
Equating the first and second expressions and cross-multiplying, we get
$$y+\frac{13 \sqrt{3}}{3}=\frac{15 \sqrt{3} x}{13} .$$
Now, equating the first and third expressions and substituting gives
$$\left(\frac{15 \sqrt{3} x}{13}-\frac{13 \sqrt{3}}{3}\right)\left(\frac{15 \sqrt{3} x}{13}\right)=x(x+7)$$
Upon dividing both sides by $x$, we obtain a linear equation from which we can solve to get $x=169 / 23$.