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

Latest revision as of 11:54, 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

@@ Line 1: / Line 1: @@
-== Problem 12 ==
+== 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 <math>8: 7</math>. Find the minimum possible value of their common perimeter.
+== Solution 1==
+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
-== Solution ==
+<center>
+<cmath>\sqrt{s(8x)(8x)(s-16x)}=\sqrt{s(7x)(7x)(s-14x)}</cmath>
+<cmath>8\sqrt{s-16x}=7\sqrt{s-14x}</cmath>
+<cmath>64s-1024x=49s-686x</cmath>
+<cmath>15s=338x</cmath>
+</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>.
-Let the first triangle has side lengths <math>a</math>, <math>a</math>, <math>14c</math>,
+== Solution 2==
-and the second triangle has side lengths <math>b</math>, <math>b</math>, <math>16c</math>,
+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>.
-where <math>a, b, 2c \in \mathbb{Z}</math>.
+~john0512
-<br/>
-Equal perimeter: <math>2a+14c=2b+16c \rightarrow a+7c=b+8c \rightarrow c=a-b</math>
-<center>
+== Note ==
-<math>\begin{array}{ccc}
-a+14c&=&2b+16c\\
-a+7c&=&b+8c\\
-c&=&a-b\\
-\end{array}</math>
-</center>
-<br/>
-Equal Area:
+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.
-<center>
-<math>\begin{array}{ccc}
-c(\sqrt{a^2-(7c)^2})&=&8c(\sqrt{b^2-(8c)^2})&{}\\
-(\sqrt{(a+7c)(a-7c)})&=&8(\sqrt{b+8c)(b-8c)})&{}\\
-(\sqrt{(a-7c)})&=&8(\sqrt(b-8c)})&\text{---(Note that} a+7c=b+8c)\\
-a-343c&=&64b-512c&{}\\
-a+169c&=&64b&{}\\
-a+169(a-b)&=&64b&\text{---(Note that} c=a-b)\\
-a&=&233b&{}\\
-\end{array}</math>
-</center>
-Since <math>a</math> and <math>b</math> are integer, the minimum occurs when <math>a=223</math>, <math>b-218</math>, and <math>c=15</math>
-Perimeter <math>=2a+14c=2(223)+14(15)=\boxed{676}</math>
+~jd9
 == See also ==
+Video Solution: https://www.youtube.com/watch?v=IUxOyPH8b4o
 {{AIME box|year=2010|num-b=11|num-a=13|n=II}}
+[[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

2010 AIME II (Problems • Answer Key • Resources)
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