Difference between revisions of "2001 AIME II Problems/Problem 6"
Fuzzy growl (talk | contribs) m (→Solution) |
|||
Line 32: | Line 32: | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 19:35, 4 July 2013
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
.
![[asy] size(150); pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("4 4") + blue + linewidth(0.7); pair C=(1,1), D=(1,-1), B=(-1,1), A=(-1,-1), E= (1, -0.2), F=(1, 0.2), G=(1.4, 0.2), H=(1.4, -0.2); D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle); D(MP("E",E,SW)--MP("F",F,NW)--MP("G",G,NE)--MP("H",H,SE)--cycle); D(CP(D(MP("O",(0,0))), A)); D((0,0) -- (2^.5, 0), d); D((0,0) -- G -- (G.x,0), d); [/asy]](http://latex.artofproblemsolving.com/a/1/4/a14d609f4250e7310e06648a5d575443252da2bb.png)
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
, 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 • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.