[Geometry]
Problem: Let $T=T N Y W R$. A regular $n$-gon has exactly $T$ more diagonals than a regular $(n-1)$-gon. Compute the value of $n$.

Solution: Using the formula $D(n)=\frac{n(n-3)}{2}$ twice yields $D(n)-D(n-1)=\frac{n^{2}-3 n}{2}-\frac{n^{2}-5 n+4}{2}=\frac{2 n-4}{2}=n-2$. So $T=n-2$, thus $n=T+2$, and with $T=17, n=19$.

Final answer: 19