Difference between revisions of "2010 AIME II Problems/Problem 12"
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== See also == | == See also == | ||
+ | Video Solution: https://www.youtube.com/watch?v=IUxOyPH8b4o | ||
{{AIME box|year=2010|num-b=11|num-a=13|n=II}} | {{AIME box|year=2010|num-b=11|num-a=13|n=II}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 09:41, 26 June 2021
Contents
Problem
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 1
Let the first triangle have side lengths , , , and the second triangle have side lengths , , , where .
Equal perimeter:
Equal Area:
Since and are integer, the minimum occurs when , , and . Hence, the perimeter is .
Solution 2
Let be the semiperimeter of the two triangles. Also, let the base of the longer triangle be and the base of the shorter triangle be for some arbitrary factor . Then, the dimensions of the two triangles must be and . By Heron's Formula, we have
Since and are coprime, to minimize, we must have and . However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by , which gives us a final answer of .
See also
Video Solution: https://www.youtube.com/watch?v=IUxOyPH8b4o
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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