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Difference between revisions of "2010 AMC 10A Problems/Problem 16"
m (→‎Solution)
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== Solution ==
 
== 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>.
 
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>.
  

Revision as of 12:28, 13 August 2017

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 (Problems • Answer Key • Resources)
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

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