Difference between revisions of "2014 AMC 10B Problems/Problem 22"
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==Problem== | ==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== | + | <math>\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</math> |
| + | |||
| + | <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> | ||
| + | [[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>. | ||
| + | |||
| + | -- LORD_ERTY09 | ||
==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 08:32, 5 October 2024
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?
![[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]](http://latex.artofproblemsolving.com/a/a/3/aa3449d7a1f5549a72c1e5be22ee89bd4499ee51.png)
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]](http://latex.artofproblemsolving.com/f/5/f/f5f12073b2d89af4810a457ff1ec79f235f2098e.png)
We will start by creating an equation by the Pythagorean theorem: ![\[\sqrt{1^2 + \left(\frac12\right)^2} = \sqrt{\frac54} = \frac{\sqrt5}{2}.\]](http://latex.artofproblemsolving.com/1/e/1/1e1ac7515df8c8dee73da68c2981b421de64d860.png)
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 
.
-- LORD_ERTY09
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. 
