Difference between revisions of "2012 AMC 12A Problems/Problem 12"
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Revision as of 15:28, 12 February 2012
Problem
A square region is externally tangent to the circle with equation
at the point
on the side
. Vertices
and
are on the circle with equation
. What is the side length of this square?
Solution
![[asy] unitsize(15mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real a=1; real b=2; pair O=(0,0); pair A=(-(sqrt(19)-2)/5,1); pair B=((sqrt(19)-2)/5,1); pair C=((sqrt(19)-2)/5,1+2(sqrt(19)-2)/5); pair D=(-(sqrt(19)-2)/5,1+2(sqrt(19)-2)/5); pair E=(-(sqrt(19)-2)/5,0); path inner=Circle(O,a); path outer=Circle(O,b); draw(outer); draw(inner); draw(A--B--C--D--cycle); draw(O--D--E--cycle); label("$A$",D,NW); label("$E$",E,SW); label("$O$",O,SE); label("$s+1$",(D--E),W); label("$\frac{s}{2}$",(E--O),S); pair[] ps={A,B,C,D,E,O}; dot(ps); [/asy]](http://latex.artofproblemsolving.com/1/7/b/17b54bd31c2906599eee882f7b6902279c53365e.png)
The circles have radii of and
. Draw the triangle shown in the figure above and write expressions in terms of
(length of the side of the square) for the sides of the triangle. Because
is the radius of the larger circle, which is equal to
, we can write the Pythagorean Theorem.
Use the quadratic formula.
See Also
2012 AMC 12A (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |