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

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== Solution 2(Trick) ==
 
== Solution 2(Trick) ==
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We find that <math>\frac{AB}{BC}=\frac{3}{8}</math> by the [[Angle Bisector Theorem]] so we let the lengths be <math>3n</math> and <math>8n</math>, respectively where <math>n</math> is a positive integer. Also since <math>AD=3</math> and <math>BC=8</math>, we notice that the perimeter of the triangle is the sum of these, namely <math>3n+8n+3+8=11n+11.</math> This can be factored into <math>11(n+1)</math> and so the sum must be a multiple of <math>11</math>. The only answer choice which is a multiple of <math>11</math> is <math>\textbf{(B)}</math>.
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We find that <math>\frac{AB}{BC}=\frac{3}{8}</math> by the [[Angle Bisector Theorem]] so we let the lengths be <math>3n</math> and <math>8n</math>, respectively where <math>n</math> is a positive integer. Also since <math>AD=3</math> and <math>BC=8</math>, we notice that the perimeter of the triangle is the sum of these, namely <math>3n+8n+3+8=11n+11.</math> This can be factored into <math>11(n+1)</math> and so the sum must be a multiple of <math>11</math>. The only answer choice which is a multiple of <math>11</math> is <math>\boxed{\textbf{(B)}11}</math>. ~mathboy282
  
 
==Video Solution==
 
==Video Solution==

Revision as of 21:04, 4 December 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 $\frac{AB}{BC} = \frac{3}{8}$. 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 $\textbf{(B)}$.

Solution 2(Trick)

We find that $\frac{AB}{BC}=\frac{3}{8}$ by the Angle Bisector Theorem so we let the lengths be $3n$ and $8n$, respectively where $n$ is a positive integer. Also since $AD=3$ and $BC=8$, we notice that the perimeter of the triangle is the sum of these, namely $3n+8n+3+8=11n+11.$ This can be factored into $11(n+1)$ and so the sum must be a multiple of $11$. The only answer choice which is a multiple of $11$ is $\boxed{\textbf{(B)}11}$. ~mathboy282

Video Solution

https://youtu.be/rsURe5Xh-j0

~IceMatrix

See also

2010 AMC 12A (Problems • Answer Key • Resources)
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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