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

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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]].
 
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 ==
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== Solution 1==
 
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>.
 
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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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>.
 
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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== Solution 2==
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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
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<center>
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<cmath>\sqrt{s(8x)(8x)(s-16x)}=\sqrt{s(7x)(7x)(s-14x)}</cmath>
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<cmath>8\sqrt{s-16x}=7\sqrt{s-14x}</cmath>
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<cmath>64s-1024x=49s-686x</cmath>
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<cmath>15s=338x</cmath>
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</center>
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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>.
  
 
== See also ==
 
== See also ==

Revision as of 12:42, 1 April 2013

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 1

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 $2a+14c=2(233)+14(15)=\boxed{676}$.

Solution 2

Let $s$ be the semiperimeter of the two triangles. Also, let the base of the longer triangle be $16x$ and the base of the shorter triangle be $14x$ for some arbitrary factor $x$. Then, the dimensions of the two triangles must be $s-8x,s-8x,16x$ and $s-7x,s-7x,14x$. By Heron's Formula, we have

\[\sqrt{s(8x)(8x)(s-16x)}=\sqrt{s(7x)(7x)(s-14x)}\] \[8\sqrt{s-16x}=7\sqrt{s-14x}\] \[64s-1024x=49s-686x\] \[15s=338x\]

Since $15$ and $338$ are coprime, to minimize, we must have $s=338$ and $x=15$. However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by $2$, which gives us a final answer of $\boxed{676}$.

See also

2010 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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All AIME Problems and Solutions
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