# 2010 AMC 12A Problems/Problem 14

The following problem is from both the 2010 AMC 12A #14 and 2010 AMC 10A #16, so both problems redirect to this page.

## Problem

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 the Angle Bisector Theorem, we know that $\frac{AB}{BC} = \frac{3}{8}$. If we use the lowest possible integer values for $AB$ and $BC$ (the lengths of $AD$ and $DC$, respectively), then $AB + BC = AD + DC = AC$, contradicting the Triangle Inequality. If we use the next lowest values ( $AB = 6$ and $BC = 16$), the Triangle Inequality is satisfied. Therefore, our answer is $6 + 16 + 3 + 8 = \boxed{33}$, or choice $\textbf{(B)}$.

## Solution 2(Trick)

We find that $\frac{AB}{BC}=\frac{3}{8}$ by the Angle Bisector Theorem so we let the lengths be $3n$ and $8n$, respectively where $n$ is a positive integer. Also since $AD=3$ and $BC=8$, we notice that the perimeter of the triangle is the sum of these, namely $3n+8n+3+8=11n+11.$ This can be factored into $11(n+1)$ and so the sum must be a multiple of $11$. The only answer choice which is a multiple of $11$ is $\boxed{\textbf{(B)} 33}$. ~mathboy282

## Video Solution by the Beauty of Math

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