2010 AMC 12A Problems/Problem 14
- 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 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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