Difference between revisions of "2007 AMC 10B Problems/Problem 18"
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==Solution 1== | ==Solution 1== | ||
− | You can express the line connecting the centers of an outer circle and the inner circle in two different ways. You can add the radius of both circles to get <math>r+1.</math> You can also add the radius of two outer circles and use a <math>45-45-90</math> triangle to get <math>\frac{2r}{\sqrt{2}} = r\sqrt{2}.</math> Since both | + | You can express the line connecting the centers of an outer circle and the inner circle in two different ways. You can add the radius of both circles to get <math>r+1.</math> You can also add the radius of two outer circles and use a <math>45-45-90</math> triangle to get <math>\frac{2r}{\sqrt{2}} = r\sqrt{2}.</math> Since both expressions represent the same length, you can set them equal to each other. |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
r+1&=r\sqrt{2}\\ | r+1&=r\sqrt{2}\\ | ||
1&=r(\sqrt{2}-1)\end{align*}</cmath> | 1&=r(\sqrt{2}-1)\end{align*}</cmath> | ||
− | <cmath>r = \frac{1}{\sqrt{2}-1} = \frac{\sqrt{2}+1}{2-1} = \boxed{\mathrm{(B) \ } 1 + \sqrt{2}}</cmath> | + | <cmath>r = \frac{1}{\sqrt{2}-1} = \frac{1(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)} = \frac{\sqrt{2}+1}{2-1} = \boxed{\mathrm{(B) \ } 1 + \sqrt{2}}</cmath> |
==Solution 2== | ==Solution 2== | ||
− | You can solve this problem by setting up a simple equation with the Pythagorean Theorem. The hypotenuse would be a segment that includes the radius of two circles on opposite corners and the diameter of the middle circle. This results in <math>2r+2</math>. The two legs are each the length between two large, adjacent circles, thus <math>2r</math>. | + | You can solve this problem by setting up a simple equation with the Pythagorean Theorem. The hypotenuse would be a segment that includes the radius of two circles on opposite corners and the diameter of the middle circle. This results in a segment of length <math>2r+2</math>. The two legs are each the length between the centers of two large, adjacent circles, thus they are each equal to <math>2r</math>. |
Using the Pythagorean Theorem: | Using the Pythagorean Theorem: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
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r^2+2r+1=2r^2\\ | r^2+2r+1=2r^2\\ | ||
r^2-2r-1=0\\ | r^2-2r-1=0\\ | ||
− | r=\frac{2+\sqrt{4-(-4)}}{2}=\frac{2+\sqrt{2}}{2}=\boxed{\mathrm{(B) \ } 1 + \sqrt{2}} | + | r=\frac{2+\sqrt{4-(-4)}}{2}=\frac{2+2\sqrt{2}}{2}=\boxed{\mathrm{(B) \ } 1 + \sqrt{2}} |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> |
Latest revision as of 14:17, 4 June 2021
Problem
A circle of radius is surrounded by
circles of radius
as shown. What is
?
![[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(7pt)); dotfactor=4; real r1=1, r2=1+sqrt(2); pair A=(0,0), B=(1+sqrt(2),1+sqrt(2)), C=(-1-sqrt(2),1+sqrt(2)), D=(-1-sqrt(2),-1-sqrt(2)), E=(1+sqrt(2),-1-sqrt(2)); pair A1=(1,0), B1=(2+2sqrt(2),1+sqrt(2)), C1=(0,1+sqrt(2)), D1=(0,-1-sqrt(2)), E1=(2+2sqrt(2),-1-sqrt(2)); path circleA=Circle(A,r1); path circleB=Circle(B,r2); path circleC=Circle(C,r2); path circleD=Circle(D,r2); path circleE=Circle(E,r2); draw(circleA); draw(circleB); draw(circleC); draw(circleD); draw(circleE); draw(A--A1); draw(B--B1); draw(C--C1); draw(D--D1); draw(E--E1); label("$1$",midpoint(A--A1),N); label("$r$",midpoint(B--B1),N); label("$r$",midpoint(C--C1),N); label("$r$",midpoint(D--D1),N); label("$r$",midpoint(E--E1),N); [/asy]](http://latex.artofproblemsolving.com/0/9/1/0916dbd96f1ec23e5b075b5070e690e9970968fa.png)
Solutions
Solution 1
You can express the line connecting the centers of an outer circle and the inner circle in two different ways. You can add the radius of both circles to get You can also add the radius of two outer circles and use a
triangle to get
Since both expressions represent the same length, you can set them equal to each other.
Solution 2
You can solve this problem by setting up a simple equation with the Pythagorean Theorem. The hypotenuse would be a segment that includes the radius of two circles on opposite corners and the diameter of the middle circle. This results in a segment of length . The two legs are each the length between the centers of two large, adjacent circles, thus they are each equal to
.
Using the Pythagorean Theorem:
See Also
2007 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.