Difference between revisions of "2014 AMC 10B Problems/Problem 22"
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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> | 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> | ||
| − | Why does this tangent work? It's because the semicircle has two tangents of the same "height", the other to the left (imagine a line of that "height). But since they are symmetrical about a line 1/ | + | Why does this tangent work? It's because the semicircle has two tangents of the same "height", the other to the left (imagine a line of that "height). But since they are symmetrical about a line <math>\frac{1}{2}</math> into the square, we need to do <math>1 \cdot \frac{1}{2}</math> to get the <math>\frac{1}{2}</math> distance on the bottom. |
==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 12:38, 28 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?
![[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 
Why does this tangent work? It's because the semicircle has two tangents of the same "height", the other to the left (imagine a line of that "height). But since they are symmetrical about a line 
into the square, we need to do 
to get the distance on the bottom.
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. 
