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 $\textbf{(B)}$.

See also

2010 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS