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

(Redirected page to 2010 AMC 12A Problems/Problem 14)
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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 ==
 
 
 
<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 <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=17|ab=A}}
 
{{AMC12 box|year=2010|num-b=13|num-a=15|ab=A}}
 
 
 
[[Category:Introductory Geometry Problems]]
 
{{MAA Notice}}
 

Latest revision as of 13:26, 26 May 2020