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| − | == Problem ==
| + | #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?
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| − | <math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37</math>
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| − | == Solution ==
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| − | <asy>
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| − | pair A,B,C,D;
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| − | C=(0,0);
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| − | B=(4,0);
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| − | A=(3,1);
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| − | D=(2,0.666);
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| − | draw(A--B--C--cycle);
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| − | draw(B--D);
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| − | label("$A$",A,N);
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| − | label("$B$",B,S);
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| − | label("$C$",C,S);
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| − | label("$D$",D,NW);
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| − | label("$3$",A--D,N);
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| − | label("$8$",C--D,N);
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| − | </asy>
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| − | 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>.
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| − | == See also ==
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| − | {{AMC10 box|year=2010|num-b=15|num-a=17|ab=A}}
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| − | [[Category:Introductory Geometry Problems]]
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| − | {{MAA Notice}}
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