Problem (Geometry):
Given a regular 16-gon, extend three of its sides to form a triangle none of whose vertices lie on the 16-gon itself. Compute the number of noncongruent triangles that can be formed in this manner.

Solution:
Label the sides of the polygon, in order, $s_{0}, s_{1}, \ldots, s_{15}$. First note that two sides of the polygon intersect at a vertex if and only if the sides are adjacent. So the sides chosen must be nonconsecutive. Second, if nonparallel sides $s_{i}$ and $s_{j}$ are extended, the angle of intersection is determined by $|i-j|$, as are the lengths of the extended portions of the segments. In other words, the spacing of the extended sides completely determines the shape of the triangle. So the problem reduces to selecting appropriate spacings, that is, finding integers $a, b, c \geq 2$ whose sum is 16 . However, diametrically opposite sides are parallel, so (for example) the sides $s_{3}$ and $s_{11}$ cannot both be used. Thus none of $a, b, c$ may equal 8 . Taking $s_{0}$ as the first side, the second side would be $s_{0+a}=s_{a}$, and the third side would be $s_{a+b}$, with $c$ sides between $s_{a+b}$ and $s_{0}$. To eliminate reflections and rotations, specify additionally that $a \geq b \geq c$. The allowable partitions are in the table below.

| $a$ | $b$ | $c$ | triangle |
| :---: | :---: | :---: | :---: |
| 12 | 2 | 2 | $s_{0} s_{12} s_{14}$ |
| 11 | 3 | 2 | $s_{0} s_{11} s_{14}$ |
| 10 | 4 | 2 | $s_{0} s_{10} s_{14}$ |
| 10 | 3 | 3 | $s_{0} s_{10} s_{13}$ |
| 9 | 5 | 2 | $s_{0} s_{9} s_{14}$ |
| 9 | 4 | 3 | $s_{0} s_{9} s_{13}$ |
| 7 | 7 | 2 | $s_{0} s_{7} s_{14}$ |
| 7 | 6 | 3 | $s_{0} s_{7} s_{13}$ |
| 7 | 5 | 4 | $s_{0} s_{7} s_{12}$ |
| 6 | 6 | 4 | $s_{0} s_{6} s_{12}$ |
| 6 | 5 | 5 | $s_{0} s_{6} s_{11}$ |

Thus there are $\mathbf{1 1}$ distinct such triangles.

Final answer: 11