Difference between revisions of "2014 AMC 10B Problems/Problem 22"
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[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
+ | ==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> | ||
+ | |||
+ | We will start by creating an equation by the Pythagorean theorem: <cmath>\sqrt{1^2 + \left(\frac12\right)^2} = \sqrt{\frac54} = \frac{\sqrt5}{2}.</cmath> | ||
+ | |||
+ | 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}} |
Latest revision as of 23:36, 30 December 2019
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?
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.
We will start by creating an equation by the Pythagorean theorem:
Let's call as the radius of the circle that we want to find. We see that the hypotenuse of the bold right triangle is , and thus is .
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 |
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