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

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== Problem ==
 
== Problem ==
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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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==
Let the first triangle have side lengths <math>a</math>, <math>a</math>, <math>14c</math>, and the second triangle have side lengths <math>b</math>, <math>b</math>, <math>16c</math>, where <math>a, b, 2c \in \mathbb{Z}</math>.
 
 
<br/>
 
Equal perimeter:
 
 
<center>
 
<math>\begin{array}{ccc}
 
2a+14c&=&2b+16c\\
 
a+7c&=&b+8c\\
 
c&=&a-b\\
 
\end{array}</math>
 
</center>
 
<br/>
 
 
Equal Area:
 
 
<center>
 
<math>\begin{array}{cccl}
 
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}</math>
 
</center>
 
 
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>.
 
 
== Solution 2==
 
 
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>.
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== Solution 2==
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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
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== Note ==
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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.
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~jd9
  
 
== See also ==
 
== See also ==
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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

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 $s$ 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 $\boxed{676}$.

Solution 2

Let the first triangle have sides $16n,a,a$, so the second has sides $14n,a+n,a+n$. The height of the first triangle is $\frac{7}{8}$ the height of the second triangle. Therefore, we have \[a^2-64n^2=\frac{49}{64}((a+n)^2-49n^2).\] Multiplying this, we get \[64a^2-4096n^2=49a^2+98an-2352n^2,\] which simplifies to \[15a^2-98an-1744n^2=0.\] Solving this for $a$, we get $a=n\cdot\frac{218}{15}$, so $n=15$ and $a=218$ and the perimeter is $15\cdot16+218+218=\boxed{676}$.

~john0512


Note

We use $16x$ and $14x$ instead of $8x$ and $7x$ to ensure that the triangle has integral side lengths. Plugging $8x$ and $7x$ directly into Heron's gives $s=338$, but for this to be true, the second triangle would have side lengths of $\frac{223}{2}$, which is impossible.

~jd9

See also

Video Solution: https://www.youtube.com/watch?v=IUxOyPH8b4o

2010 AIME II (ProblemsAnswer KeyResources)
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Problem 13
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