Difference between revisions of "2014 AMC 10B Problems/Problem 22"
Line 15: | Line 15: | ||
</asy> | </asy> | ||
[[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> | ||
+ | |||
+ | 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> | ||
Revision as of 23:30, 15 October 2015
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.
Let's make 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.