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

Revision as of 20:31, 3 April 2010

Problem 12

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: $2a+14c=2b+16c \rightarrow a+7c=b+8c \rightarrow c=a-b$

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

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 $c=15$

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

@@ Line 37: / Line 37: @@
 </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>
+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>
-Perimeter <math>=2a+14c=2(223)+14(15)=\boxed{676}</math>
+Perimeter <math>=2a+14c=2(233)+14(15)=\boxed{676}</math>
 == See also ==
 {{AIME box|year=2010|num-b=11|num-a=13|n=II}}

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

Revision as of 20:31, 3 April 2010

Problem 12

Solution

See also