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Difference between revisions of "2010 AIME II Problems/Problem 12"

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== Problem 12 ==
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== 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.
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Two non[[congruent]] integer-sided [[isosceles triangle]]s 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]].
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== Solution ==
 
== Solution ==
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Let the first triangle has side lengths <math>a</math>, <math>a</math>, <math>14c</math>, and the second triangle has side lengths <math>b</math>, <math>b</math>, <math>16c</math>, where <math>a, b, 2c \in \mathbb{Z}</math>.
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Let the first triangle has side lengths <math>a</math>, <math>a</math>, <math>14c</math>,
 
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and the second triangle has side lengths <math>b</math>, <math>b</math>, <math>16c</math>,
 
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where <math>a, b, 2c \in \mathbb{Z}</math>.
 
  
 
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Since <math>a</math> and <math>b</math> are integer, the minimum occurs when <math>a=233</math>, <math>b=218</math>, and <math>c=15</math>
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Since <math>a</math> and <math>b</math> are integer, the minimum occurs when <math>a=233</math>, <math>b=218</math>, and <math>c=15</math>. Hence, the perimeter is <math>2a+14c=2(233)+14(15)=\boxed{676}</math>.
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Perimeter <math>=2a+14c=2(233)+14(15)=\boxed{676}</math>
 
  
 
== 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}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 12:00, 6 April 2010

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 $8: 7$. Find the minimum possible value of their common perimeter.

Solution

Let the first triangle has side lengths $a$, $a$, $14c$, and the second triangle has side lengths $b$, $b$, $16c$, where $a, b, 2c \in \mathbb{Z}$.


Equal perimeter:

$\begin{array}{ccc} 2a+14c&=&2b+16c\\ a+7c&=&b+8c\\ c&=&a-b\\ \end{array}$


Equal Area:

$\begin{array}{cccc} 7c(\sqrt{a^2-(7c)^2})&=&8c(\sqrt{b^2-(8c)^2})&{}\\ 7(\sqrt{(a+7c)(a-7c)})&=&8(\sqrt{b+8c)(b-8c)})&{}\\ 7(\sqrt{(a-7c)})&=&8(\sqrt{(b-8c)})&\text{(Note that} a+7c=b+8c)\\ 49a-343c&=&64b-512c&{}\\ 49a+169c&=&64b&{}\\ 49a+169(a-b)&=&64b&\text{(Note that} c=a-b)\\ 218a&=&233b&{}\\ \end{array}$

Since $a$ and $b$ are integer, the minimum occurs when $a=233$, $b=218$, and $c=15$. Hence, the perimeter is $2a+14c=2(233)+14(15)=\boxed{676}$.

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
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