Difference between revisions of "2014 AMC 10B Problems/Problem 22"

(Solution: added diagram and solution)
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draw(scale((sqrt(5)-1)/4)*unitcircle);
 
draw(scale((sqrt(5)-1)/4)*unitcircle);
 
</asy>
 
</asy>
 +
[[Category: Introductory Geometry Problems]]
  
 
==Solution==
 
==Solution==

Revision as of 10:53, 13 August 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 (ProblemsAnswer KeyResources)
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

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