Saudi Arabian Mathematical Competitions

Let $A$ and $B$ be two neighboring stations. We have $\widehat{A O B}=\frac{\pi}{5}$, hence $A B=2 R \sin \frac{\pi}{10}$, where $R=22236+3960=26196$. We will prove that $\sin \frac{\pi}{10}=\frac{\sqrt{5}-1}{4}$.

Since $\sin \pi=0$, then $\sin \left(\frac{2 \pi}{5}+\frac{3 \pi}{5}\right)=0$. We have:
$$\begin{gathered} \sin \frac{2 \pi}{5} \cos \frac{3 \pi}{5}+\cos \frac{2 \pi}{5} \sin \frac{3 \pi}{5}=0 \Rightarrow \\ 2 \sin \frac{\pi}{5} \cos \frac{\pi}{5}\left(4 \cos ^{3} \frac{\pi}{5}-3 \cos \frac{\pi}{5}\right) \\ +\left(2 \cos ^{2} \frac{\pi}{5}-1\right)\left(3 \sin \frac{\pi}{5}-4 \sin ^{3} \frac{\pi}{5}\right)=0 \end{gathered}$$
We may divide by $\sin \frac{\pi}{5}$ and we get
$$\begin{gathered} 2 \cos \frac{\pi}{5}\left(4 \cos ^{3} \frac{\pi}{5}-3 \cos \frac{\pi}{5}\right) \\ +\left(2 \cos ^{2} \frac{\pi}{5}-1\right)\left(-1+4 \cos ^2 \frac{\pi}{5}\right)=0 \end{gathered}$$
Denote $y=\cos \frac{\pi}{5}$. Then, our equation becomes
$$16 y^{4}-12 y^{2}+1=0$$
with the solutions given by $y^{2}=\frac{3 \pm \sqrt{5}}{8}$. Since $\frac{\pi}{6}<\frac{\pi}{5}<\frac{\pi}{4}$, then $\cos \frac{\pi}{5} \in\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}\right)$, so that
$$y^{2}=\frac{3+\sqrt{5}}{8} \text{ and } \cos \frac{\pi}{5}=\sqrt{\frac{3+\sqrt{5}}{8}}=\frac{\sqrt{5}+1}{4}$$
Now, $\sin \frac{\pi}{10}=\sqrt{\frac{1-\cos \frac{\pi}{5}}{2}}=\frac{\sqrt{5}-1}{4}$. It follows
$$A B=26196 \cdot \frac{\sqrt{5}-1}{2}=-13098+13098 \sqrt{5}$$