Difference between revisions of "2010 AMC 10A Problems/Problem 16"

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== Problem ==
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#REDIRECT [[2010_AMC_12A_Problems/Problem_14]]
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?
 
 
 
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37</math>
 
 
 
== Solution ==
 
By the [[Angle Bisector Theorem]], we know that <math>\frac{AB}{3} = \frac{BC}{8}</math>. If we use the lowest possible integer values for AB and BC (the measures of AD and DC, respectively), then <math>AB + BC = AD + DC = AC</math>, contradicting the [[Triangle Inequality]]. If we use the next lowest values (<math>AB = 6</math> and <math>BC = 16</math>), the Triangle Inequality is satisfied. Therefore, our answer is <math>6 + 16 + 3 + 8 = \boxed{33}</math>, or choice <math>\textbf{(B)}</math>.
 
 
 
== See also ==
 
{{AMC10 box|year=2010|num-b=15|num-a=16|ab=A}}
 
 
 
[[Category:Introductory Geometry Problems]]
 

Latest revision as of 12:26, 26 May 2020