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

m (Solution: LaTeXed numbers)
(Problem)
 
(4 intermediate revisions by 2 users not shown)
Line 16: Line 16:
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
  
==Solution==
 
  
 +
==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 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.
  
Line 39: Line 39:
 
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>
 
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>
 
  
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.
+
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 21:29, 27 October 2022

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]

We will start by creating 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png