2010 AMC 10A Problem 16

This is a "smallest possible" question, so find the smallest choice that really makes a triangle.

1. The angle bisector rule. When $BD$ cuts angle $B$ in half, it divides side $AC$ in the same ratio as the two sides that meet at $B$:
$$\frac{AB}{BC} = \frac{AD}{DC} = \frac{3}{8}.$$
2. Possible side pairs. The sides are whole numbers in the ratio $3 : 8$. Since $3$ and $8$ share no common factor, $AB$ and $BC$ must be $3$ and $8$, or $6$ and $16$, or $9$ and $24$, and so on.
3. The third side. $AC = AD + DC = 3 + 8 = 11$.
4. Try the smallest pair: $3$, $8$, $11$. In a real triangle, the two shorter sides must add to more than the longest side. Here $3 + 8 = 11$, exactly equal, so the three points would lie on one line. The problem says the triangle is nondegenerate (not flat), so this fails.
5. Try the next pair: $6$, $16$, $11$. The longest side is $16$, and $6 + 11 = 17$, which is more than $16$. This triangle exists.
6. Its perimeter is $6 + 16 + 11 = 33$. Every larger pair gives a bigger perimeter, so $33$ is the smallest.

The answer is $\boxed{\textbf{(B)}\ 33}$.

The angle bisector rule turns a line inside the triangle into a ratio of side lengths. Whole-number sides in a ratio like $3 : 8$ must be $3$ and $8$ times the same whole number. Whenever you look for the smallest triangle, check the triangle inequality: the first candidate is often flat.