2012 AMC 12A Problems/Problem 12
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.
![\[s = \frac{-8+\sqrt{8^2-4(5)(-12)}}{10} = \frac{-8+\sqrt{304}}{10} = \frac{-8+4\sqrt{19}}{10} = \boxed{\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}}\]](http://latex.artofproblemsolving.com/7/a/d/7ad6fcfea1fb8fd86a81479e331c734c0c97f65e.png)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
