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2014 AMC 10B Problems/Problem 22: Difference between revisions

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==Solution==
==Solution==
We connect the centers of the circle and one of the semicircles, then draw the perpendicular from the center of the middle circle to that side, as shown.
<asy>
scale(200);
draw(scale(.5)*((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle));
path p = arc((.25,-.5),.25,0,180)--arc((-.25,-.5),.25,0,180);
draw(p);
p=rotate(90)*p; draw(p);
p=rotate(90)*p; draw(p);
p=rotate(90)*p; draw(p);
draw(scale((sqrt(5)-1)/4)*unitcircle);
pair OO=(0,0);
pair XX=(-.25,-.5);
pair YY=(0,-.5);
draw(YY--OO--XX--cycle,black+1bp);
label("$\frac12$",.5*(XX+YY),S);
label("$1$",.5*YY,E);
</asy>
Let's make an equation by the pythagorean theorem.
<math>\sqrt{1^2 + \left(\frac12\right)^2} = \sqrt{\frac54} = \frac{\sqrt5}{2}</math>
Let's call <math>r</math> as the radius of the circle that we want to find. We see that the hypotenuse of the bold right triangle is <math>\dfrac{1}{2}+r</math>, and thus <math>r</math> is <math>\boxed{\textbf{(B)} \frac{\sqrt{5}-1}{2}}</math>


==See Also==
==See Also==
{{AMC10 box|year=2014|ab=B|num-b=21|num-a=23}}
{{AMC10 box|year=2014|ab=B|num-b=21|num-a=23}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 19:24, 20 February 2014

Problem

Eight semicircles line the inside of a square with side length 2 as shown. What is the radius of the circle tangent to all of these semicircles?

$\text{(A) } \dfrac{1+\sqrt2}4 \quad \text{(B) } \dfrac{\sqrt5-1}2 \quad \text{(C) } \dfrac{\sqrt3+1}4 \quad \text{(D) } \dfrac{2\sqrt3}5 \quad \text{(E) } \dfrac{\sqrt5}3$

[asy] scale(200); draw(scale(.5)*((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle)); path p = arc((.25,-.5),.25,0,180)--arc((-.25,-.5),.25,0,180); draw(p); p=rotate(90)*p; draw(p); p=rotate(90)*p; draw(p); p=rotate(90)*p; draw(p); draw(scale((sqrt(5)-1)/4)*unitcircle); [/asy]

Solution

We connect the centers of the circle and one of the semicircles, then draw the perpendicular from the center of the middle circle to that side, as shown.

[asy] scale(200); draw(scale(.5)*((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle)); path p = arc((.25,-.5),.25,0,180)--arc((-.25,-.5),.25,0,180); draw(p); p=rotate(90)*p; draw(p); p=rotate(90)*p; draw(p); p=rotate(90)*p; draw(p); draw(scale((sqrt(5)-1)/4)*unitcircle); pair OO=(0,0); pair XX=(-.25,-.5); pair YY=(0,-.5); draw(YY--OO--XX--cycle,black+1bp); label("$\frac12$",.5*(XX+YY),S); label("$1$",.5*YY,E); [/asy]

Let's make an equation by the pythagorean theorem.

$\sqrt{1^2 + \left(\frac12\right)^2} = \sqrt{\frac54} = \frac{\sqrt5}{2}$

Let's call $r$ as the radius of the circle that we want to find. We see that the hypotenuse of the bold right triangle is $\dfrac{1}{2}+r$, and thus $r$ is $\boxed{\textbf{(B)} \frac{\sqrt{5}-1}{2}}$

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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

These problems are copyrighted © by the Mathematical Association of America.

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