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

Revision as of 14:13, 22 December 2022

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 have side lengths $a$ , $a$ , $14c$ , and the second triangle have 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:

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

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 $\boxed{676}$ .

Solution 3

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 49: / Line 49: @@
 ~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 ==

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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