2009 AMC 10A Problem 12

The diagonal $BD$ splits the shape into two triangles. In any triangle, the two shorter sides together must be longer than the third side. Use this in both triangles.

1. Triangle $BCD$ has sides $17$, $5$ and $BD$. The side $17$ is long, so the other two together must beat it:
$$5 + BD > 17, \qquad BD > 12.$$
2. Triangle $ABD$ has sides $5$, $9$ and $BD$. Here $BD$ must be less than the other two together:
$$BD < 5 + 9 = 14.$$
3. So $BD$ is more than $12$ and less than $14$. The only whole number that fits is $13$.
4. Check both triangles with $BD = 13$: $5 + 13 = 18 > 17$ and $5 + 9 = 14 > 13$. Both are real triangles.

The answer is $\boxed{\textbf{(C)}\ 13}$.

A side shared by two triangles gets limits from both. A lower limit comes from the triangle with a long side that the shared side must help cover ($BD$ must be more than $17 - 5$), and an upper limit from the other triangle ($BD$ must be less than $5 + 9$). Squeezing a whole number between the two limits is a common AMC pattern.

The diagonal $BD$ is a side of two triangles, and each gives a triangle inequality.

In $\triangle BCD$ with sides $17$, $5$ and $BD$: the longest side must be less than the sum of the other two, so $17 < 5 + BD$, giving $BD > 12$.

In $\triangle ABD$ with sides $5$, $9$ and $BD$: $BD < 5 + 9 = 14$.

So $12 < BD < 14$, and the only integer in that range is $13$. (Both triangles then exist: $5 + 13 > 17$ and $5 + 9 > 13$.)

The answer is $\boxed{\textbf{(C)}\ 13}$.