Difference between revisions of "2010 AMC 10A Problems/Problem 16"
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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 13:28, 13 August 2017
Problem
Nondegenerate has integer side lengths,
is an angle bisector,
, and
. What is the smallest possible value of the perimeter?
Solution
By the Angle Bisector Theorem, we know that . If we use the lowest possible integer values for AB and BC (the measures of AD and DC, respectively), then
, contradicting the Triangle Inequality. If we use the next lowest values (
and
), the Triangle Inequality is satisfied. Therefore, our answer is
, or choice
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.