Nondegenerate ${\triangle ABC}$ has integer side lengths, ${\overline{BD}}$ is an angle bisector, $AD = 3$, and $DC=8$. What is the smallest possible value of the perimeter? $\textbf{(A)}\ 30 \qquad$ $ \textbf{(B)}\ 33 \qquad$ $ \textbf{(C)}\ 35 \qquad$ $ \textbf{(D)}\ 36 \qquad$ $ \textbf{(E)}\ 37$
Solution
By angle bisector theorem (theorem 8), $AB:CB=AD:CD=3:8$. Since $AB,BC$ are integers and $\gcd(3,8)=1$, $AB=3x, BC=8x$ for some positive integer $x$.
The third side is $AC=AD+DC=11$. The triangle inequalities are $3x+11>8x$ (so $x<\frac{11}5$) and $3x+8x>11$ (so $x>1$); the third one, $8x+11>3x$, always holds. So $x=2$.
Nondegenerate $\triangle ABC$ has integer side lengths, $\overline{BD}$ is an angle bisector, $AD = 4$, and $DC=9$. What is the smallest possible value of the perimeter?