Difference between revisions of "2010 AMC 12A Problems/Problem 14"
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| − | By the [[Angle Bisector Theorem]], we know that <math>\frac{AB}{BC} = \frac{3}{8}</math>. If we use the lowest possible integer values for <math>AB</math> and <math>BC</math> (the | + | By the [[Angle Bisector Theorem]], we know that <math>\frac{AB}{BC} = \frac{3}{8}</math>. If we use the lowest possible integer values for <math>AB</math> and <math>BC</math> (the lengths of <math>AD</math> and <math>DC</math>, 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>. |
== Solution 2(Trick) == | == Solution 2(Trick) == | ||
Revision as of 19:29, 4 July 2021
- 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 
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 
and 
(the lengths of 
and 
, 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 
.
Solution 2(Trick)
We find that 
by the Angle Bisector Theorem so we let the lengths be 
and 
, respectively where 
is a positive integer. Also since 
and 
, we notice that the perimeter of the triangle is the sum of these, namely 
This can be factored into 
and so the sum must be a multiple of 
. The only answer choice which is a multiple of is 
. ~mathboy282
Video Solution by the Beauty of Math
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 | ||
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
