2010 AMC 10A Problems/Problem 16

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

$[asy] pair A,B,C,D; C=(0,0); B=(4,0); A=(3,1); D=(2,0.666); draw(A--B--C--cycle); draw(B--D); label("$A$",A,N); label("$B$",B,S); label("$C$",C,S); label("$D$",D,NW); label("$3$",A--D,N); label("$8$",C--D,N); [/asy]$

By the Angle Bisector Theorem, we know that $\frac{AB}{3} = \frac{BC}{8}$ . If we use the lowest possible integer values for AB and BC (the measures 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 $\textbf{(B)}$ .

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2010 AMC 10A Problems/Problem 16

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