Difference between revisions of "2010 AMC 12A Problems/Problem 14"

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{{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #14]] and [[2010 AMC 10A Problems|2010 AMC 10A #16]]}}
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== Problem ==
 
== Problem ==
 
Nondegenerate <math>\triangle ABC</math> has integer side lengths, <math>\overline{BD}</math> is an angle bisector, <math>AD = 3</math>, and <math>DC=8</math>. What is the smallest possible value of the perimeter?
 
Nondegenerate <math>\triangle ABC</math> has integer side lengths, <math>\overline{BD}</math> is an angle bisector, <math>AD = 3</math>, and <math>DC=8</math>. What is the smallest possible value of the perimeter?

Revision as of 20:19, 26 May 2020

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 $\frac{AB}{3} = \frac{BC}{8}$. 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 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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 12 Problems and Solutions

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