Difference between revisions of "2010 AIME II Problems/Problem 12"
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Since <math>15</math> and <math>338</math> are coprime, to minimize, we must have <math>s=338</math> and <math>x=15</math>. However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by <math>2</math>, which gives us a final answer of <math>\boxed{676}</math>. | Since <math>15</math> and <math>338</math> are coprime, to minimize, we must have <math>s=338</math> and <math>x=15</math>. However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by <math>2</math>, which gives us a final answer of <math>\boxed{676}</math>. | ||
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+ | == Solution 3== | ||
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+ | Let the first triangle have sides <math>16n,a,a</math>, so the second has sides <math>14n,a+n,a+n</math>. The height of the first triangle is <math>\frac{7}{8}</math> the height of the second triangle. Therefore, we have <cmath>a^2-64n^2=\frac{49}{64}((a+n)^2-49n^2).</cmath> Multiplying this, we get <cmath>64a^2-4096n^2=49a^2+98an-2352n^2,</cmath> which simplifies to <cmath>15a^2-98an-1744n^2=0.</cmath> Solving this for <math>a</math>, we get <math>a=n\cdot\frac{218}{15}</math>, so <math>n=15</math> and <math>a=218</math> and the perimeter is <math>15\cdot16+218+218=\boxed{676}</math>. | ||
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+ | ~john0512 | ||
== See also == | == See also == |
Revision as of 21:24, 25 November 2021
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
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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