Problem (Geometry):
Compute the smallest possible value of $n$ such that two diagonals of a regular $n$-gon intersect at an angle of 159 degrees.

Solution:
Let the vertices of the polygon be $A_{0}, A_{1}, \ldots, A_{n-1}$. Considering the polygon as inscribed in a circle, the angle between diagonals $\overline{A_{0} A_{i}}$ and $\overline{A_{0} A_{j}}$ is $\frac{1}{2} \cdot\left(\frac{360^{\circ}}{n}\right) \cdot|j-i|=\left(\frac{180|j-i|}{n}\right)^{\circ}$. The diagonal $\overline{A_{k} A_{k+j}}$ can be considered as the rotation of $\overline{A_{0} A_{j}}$ through $k / n$ of a circle, or $\left(\frac{360 k}{n}\right)^{\circ}$. So the diagonals $A_{0} A_{i}$ and $A_{k} A_{k+j}$ intersect at a combined angle of $\left(\frac{180|j-i|}{n}\right)^{\circ}+\left(\frac{360 k}{n}\right)^{\circ}$. Without loss of generality, assume $i<j$ (otherwise relabel vertices in the opposite order, with $A_{k}$ becoming $A_{0}$ ). Then the desired number is the least $n$ such that

$$
\left(\frac{180(j-i)}{n}\right)+\frac{360 k}{n}=159
$$

Multiply both sides of the equation by $n$ and factor to obtain $180(j-i+2 k)=159 n$, thus $60(j-i+2 k)=53 n$. Because 53 and 60 are relatively prime and $(j-i+2 k)$ is an integer, it follows that $60 \mid n$. So the smallest possible value is $n=\mathbf{6 0}$; one set of values that satisfies the equation is $i=1, j=54, k=0$.

Final answer: 60