Question

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? \t\t

$\dfrac{1+\sqrt2}4$

$\dfrac{\sqrt5-1}2$

$\dfrac{\sqrt3+1}4$

$\dfrac{2\sqrt3}5$

$\dfrac{\sqrt5}3$

Solution:

The distance from the center of the square to the center of the semicircles can be found as a hypotenuse of a right triangle.

One of the legs is from the center of the square to the center of one of the sides which is of distance $1.$

The other leg is from the center of the side to the center of one of the semicircles which is of distance $\dfrac 12.$ This also shows that the radius of the semicircles is $\dfrac 12.$

Therefore, the distance from the center of the square to the center of the semicircle is $\sqrt{1^2 + \dfrac 12 ^2} = \dfrac {\sqrt 5}2.$ Then , we subtract $\dfrac 12$ for the radius of the semicircle. This makes the radius of the circle $\dfrac{\sqrt 5 -1}2 .$

Thus, the correct answer is B .

2014 AMC 10B Problem 22

Solution:

Problem 22 in Other Years