2001 AIME II Problems/Problem 6
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Problem
Square 
is inscribed in a circle. Square 
has vertices 
and 
on 
and vertices 
and 
on the circle. The ratio of the area of square to the area of square can be expressed as 
where 
and 
are relatively prime positive integers and 
. Find 
.
Solution
Let 
be the center of the circle, and 
be the side length of , 
be the side length of . By the Pythagorean Theorem, the radius of 
.
Now consider right triangle 
, where 
is the midpoint of 
. Then, by the Pythagorean Theorem,
Thus 
(since lengths are positive, we discard the other root). The ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, so ![\frac{[EFGH]}{[ABCD]} = \left(\frac 15\right)^2 = \frac{1}{25}](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/7/7/2/772ce048cabfd5caec117955ca220e79749c94c5.gif)
, and the answer is 
.
Another way to proceed from 
is to note that 
is the quantity we need; thus, we divide by 
to get
This is a quadratic in , and solving it gives 
. The negative solution is extraneous, and so the ratio of the areas is 
and the answer is 
.
See also
| 2001 AIME II (Problems • Resources) | ||
| Preceded by Problem 5 | Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
