Difference between revisions of "2010 AMC 12A Problems/Problem 14"
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+ | {{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #14]] and [[2010 AMC 10A Problems|2010 AMC 10A #16]]}} | ||
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== Problem == | == Problem == | ||
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? | 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? |
Revision as of 19:19, 26 May 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 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 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.