Difference between revisions of "2010 AIME II Problems/Problem 12"
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Perimeter <math>=2a+14c=2(223)+14(15)=\boxed{676}</math> | Perimeter <math>=2a+14c=2(223)+14(15)=\boxed{676}</math> | ||
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== See also == | == See also == | ||
{{AIME box|year=2010|num-b=11|num-a=13|n=II}} | {{AIME box|year=2010|num-b=11|num-a=13|n=II}} |
Revision as of 20:28, 3 April 2010
Problem 12
Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is . Find the minimum possible value of their common perimeter.
Solution
Let the first triangle has side lengths , , ,
and the second triangle has side lengths , , ,
where .
Equal perimeter:
Equal Area:
$
Since and are integer, the minimum occurs when , , and
Perimeter
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |