Difference between revisions of "2010 AMC 12A Problems/Problem 14"
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== See also == | == See also == | ||
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{{AMC12 box|year=2010|num-b=13|num-a=15|ab=A}} | {{AMC12 box|year=2010|num-b=13|num-a=15|ab=A}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:46, 18 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 
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 
.
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
