2010 AIME II Problems/Problem 12
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 .
Solution 3
Let the first triangle have sides , so the second has sides . The height of the first triangle is the height of the second triangle. Therefore, we have Multiplying this, we get which simplifies to Solving this for , we get , so and and the perimeter is .
~john0512
Note
We use and instead of and to ensure that the triangle has integral side lengths. Plugging and directly into Heron's gives , but for this to be true, the second triangle would have side lengths of , which is impossible.
~jd9
See also
Video Solution: https://www.youtube.com/watch?v=IUxOyPH8b4o
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