Difference between revisions of "2010 AIME II Problems/Problem 12"
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| − | <math>\begin{array}{ | + | <math>\begin{array}{ccc} |
2a+14c&=&2b+16c\\ | 2a+14c&=&2b+16c\\ | ||
a+7c&=&b+8c\\ | a+7c&=&b+8c\\ | ||
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<center> | <center> | ||
| − | <math>\begin{array}{ | + | <math>\begin{array}{cccc} |
7c(\sqrt{a^2-(7c)^2})&=&8c(\sqrt{b^2-(8c)^2})&{}\\ | 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)(a-7c)})&=&8(\sqrt{b+8c)(b-8c)})&{}\\ | ||
| − | 7(\sqrt{(a-7c)})&=&8(\sqrt(b-8c)})&\text{ | + | 7(\sqrt{(a-7c)})&=&8(\sqrt(b-8c)})&\text{(Note that} a+7c=b+8c)\\ |
49a-343c&=&64b-512c&{}\\ | 49a-343c&=&64b-512c&{}\\ | ||
49a+169c&=&64b&{}\\ | 49a+169c&=&64b&{}\\ | ||
| − | 49a+169(a-b)&=&64b&\text{ | + | 49a+169(a-b)&=&64b&\text{(Note that} c=a-b)\\ |
218a&=&233b&{}\\ | 218a&=&233b&{}\\ | ||
\end{array}</math> | \end{array}</math> | ||
Revision as of 20:29, 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 
. Find the minimum possible value of their common perimeter.
Solution
Let the first triangle has side lengths 
, , 
,
and the second triangle has side lengths 
, , 
,
where 
.
Equal perimeter: 
Equal Area:
$\begin{array}{cccc} 7c(\sqrt{a^2-(7c)^2})&=&8c(\sqrt{b^2-(8c)^2})&{}\\ 7(\sqrt{(a+7c)(a-7c)})&=&8(\sqrt{b+8c)(b-8c)})&{}\\ 7(\sqrt{(a-7c)})&=&8(\sqrt(b-8c)})&\text{(Note that} a+7c=b+8c)\\ 49a-343c&=&64b-512c&{}\\ 49a+169c&=&64b&{}\\ 49a+169(a-b)&=&64b&\text{(Note that} c=a-b)\\ 218a&=&233b&{}\\ \end{array}$ (Error compiling LaTeX. Unknown error_msg)
Since and are integer, the minimum occurs when 
, 
, and 
Perimeter 
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 | ||
