Difference between revisions of "2010 AIME II Problems/Problem 12"
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== Problem == | == Problem == | ||
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+ | 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 <math>8: 7</math>. Find the minimum possible value of their common perimeter. | ||
== Solution 1== | == Solution 1== | ||
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Let <math>s</math> be the semiperimeter of the two triangles. Also, let the base of the longer triangle be <math>16x</math> and the base of the shorter triangle be <math>14x</math> for some arbitrary factor <math>x</math>. Then, the dimensions of the two triangles must be <math>s-8x,s-8x,16x</math> and <math>s-7x,s-7x,14x</math>. By Heron's Formula, we have | Let <math>s</math> be the semiperimeter of the two triangles. Also, let the base of the longer triangle be <math>16x</math> and the base of the shorter triangle be <math>14x</math> for some arbitrary factor <math>x</math>. Then, the dimensions of the two triangles must be <math>s-8x,s-8x,16x</math> and <math>s-7x,s-7x,14x</math>. By Heron's Formula, we have | ||
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</center> | </center> | ||
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>. | ||
+ | |||
+ | == Solution 2== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ~john0512 | ||
+ | |||
+ | |||
+ | == Note == | ||
+ | |||
+ | We use <math>16x</math> and <math>14x</math> instead of <math>8x</math> and <math>7x</math> to ensure that the triangle has integral side lengths. Plugging <math>8x</math> and <math>7x</math> directly into Heron's gives <math>s=338</math>, but for this to be true, the second triangle would have side lengths of <math>\frac{223}{2}</math>, which is impossible. | ||
+ | |||
+ | ~jd9 | ||
== 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}} |
Latest revision as of 12:54, 21 November 2023
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 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 2
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
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 |
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