Two $n$-sided polygons intersect at 80 points on their boundaries. What is the minimum value of $n$?
The boundary of two $n$-sided polygons can intersect in at most $n^2$ points. Therefore, 80 intersection points can only occur if $n$ is at least 9.

We will show that the two polygons cannot even be nonagons, so the value of $n$ must be at least 10. Suppose, for instance, that the boundaries of two nonagons intersect in 80 points. Then one of the polygons must have a side that intersects all nine sides of the other, otherwise the number of intersection points could be at most 72.

Suppose that $PQ$ is such a side, that is, $PQ$ intersects every side of the $A_1A_2 \ldots A_9$ polygon. $A_1$ and $A_2$ are on different sides of $PQ$, as are $A_2$ and $A_3$, so $A_1$ and $A_3$ must be on the same side. Continuing this reasoning, we find that $A_1$ and $A_9$ are on the same side, so $PQ$ cannot intersect the side $A_9A_1$. The contradiction thus implies that $n > 9$.

Decagons, however, can be constructed in the desired manner. In the diagram, one of the polygons has 5 "spikes," and the other has 4. Two "spikes" intersect in 4 points, so the total number of intersection points is $5 \cdot 4 \cdot 4 = 80$.

Therefore, the value of $n$ is at least 10.