Difference between revisions of "2007 AMC 10A Problems/Problem 15"
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== Solution 2 == | == Solution 2 == | ||
| − | We draw the long diagonal of the square. This diagonal yields <math>2\sqrt{2}+1+1+2+2=2\sqrt{2}+6</math> | + | We draw the long diagonal of the square. This diagonal yields <math>2\sqrt{2}+1+1+2+2=2\sqrt{2}+6</math>. We know that the side length in terms o the diagonal <math>d</math> is <math>\frac{d}{\sqrt(2)</math>, so our side length is <math>\frac{\sqrt{2}+6}{\sqrt{2}}</math> |
==See Also== | ==See Also== | ||
Revision as of 10:58, 24 December 2019
Three circles are inscribed in a rectangle of width w and height h as shown. Two of the circles are congruent and are each tangent to two adjacent sides of the rectangle and to each other. The other circle is larger and is tangent to three sides of the rectangle and to the two smaller circles. What the ratio of h to w? Express your answer as a decimal to the nearest hundredth.
Contents
Problem
Four circles of radius 
are each tangent to two sides of a square and externally tangent to a circle of radius 
, as shown. What is the area of the square?
![[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); real h=3*sqrt(2)/2; pair O0=(0,0), O1=(h,h), O2=(-h,h), O3=(-h,-h), O4=(h,-h); pair X=O0+2*dir(30), Y=O2+dir(45); draw((-h-1,-h-1)--(-h-1,h+1)--(h+1,h+1)--(h+1,-h-1)--cycle); draw(Circle(O0,2)); draw(Circle(O1,1)); draw(Circle(O2,1)); draw(Circle(O3,1)); draw(Circle(O4,1)); draw(O0--X); draw(O2--Y); label("$2$",midpoint(O0--X),NW); label("$1$",midpoint(O2--Y),SE); [/asy]](http://latex.artofproblemsolving.com/2/3/7/23759fb5d0c074107da579ab5ce358f0aa30b5a0.png)
Solution
Draw a square connecting the centers of the four small circles of radius . This square has a diagonal of length 
, as it includes the diameter of the big circle of radius and two radii of the small circles of radius . Therefore, the side length of this square is ![\[\frac{6}{\sqrt{2}} = 3\sqrt{2}.\]](http://latex.artofproblemsolving.com/9/a/e/9ae6a28635f7bc2f1d620576f68cea4d8a917152.png)
The radius of the large square has a side length units larger than the one found by connecting the midpoints, so its side length is ![\[2 + 3\sqrt{2}.\]](http://latex.artofproblemsolving.com/2/d/8/2d8d2199d03dc28d51bf35318bfa8e414d50d207.png)
The area of this square is 
Solution 2
We draw the long diagonal of the square. This diagonal yields 
. We know that the side length in terms o the diagonal 
is $\frac{d}{\sqrt(2)$ (Error compiling LaTeX. Unknown error_msg), so our side length is 
See Also
| 2007 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
