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

## 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 $\boxed{\textbf{(B)} \frac{\sqrt{5}-1}{2}}$

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 $\frac{1}{2}$ into the square, we need to do $1 \cdot \frac{1}{2}$ to get the $\frac{1}{2}$ distance on the bottom.

## See Also

 2014 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 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.

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