Consider the following figure:

If the radius of the blue quarter-circle is 2, which is also the diameter of the larger half-circle (colored pink), then what is the radius of the smaller half-circle (colored yellow)?

Source: unknown

r=2/3.

Label the radius of the smaller (yellow) half circle r.
Draw a line joining the centers of the yellow and pink half circles.
Now there is a right triangle with legs 1 and 2-r, and hypotenuse r+1.
So (r+1)² = (2-r)²+1
r² + 2r + 1 = 4 - 4r + r² + 1
6r = 4
r = 2/3

More about this puzzle

An alternative solution to this puzzle uses Descartes' Theorem.

Imagine what would happen if you revealed the rest of the three circles, and suppose an additional pink circle were added as well.  The result might look like this:

Now you have four mutually tangent circles.  The radii of the blue and pink circles are given as 2 and 1, respectively, the only unknown circle being the yellow one.

Descartes discovered the formula that links the radii of four circles that touch each other externally.  He gave it in terms of the curvatures.  For a circle, the curvature, ε, is the reciprocal of its radius, R, and Descartes' formula appears as 

2(ε₁² + ε₂² + ε₃² + ε₄²) = (ε₁ + ε₂ + ε₃ + ε₄)²,

But in this case, the largest circle is internally tangent (that is, the inside of the largest circle is tangent) to the other circles.  Surprisingly, Descartes' formula still works, if you consider this circle to have a negative curvature.  Here, we will let ε₁ be the curvature of the internally tangent blue circle, ε₂ and ε₃ be that of the pink circles, and ε₄ be the that of the yellow circle.  So ε₁=-1/2, ε₂=ε₃=1, and our unknown is ε₄.  So Descartes' formula for our diagram is:

2(1/4+1+1+ε₄²) = (-1/2+1+1+ε₄)², or
9/2 + 2ε₄² = 9/4+3 ε₄+ ε₄²,
ε₄² - 3ε₄ + 9/4 = 0,
(ε₄-3/2)(ε₄-3/2) = 0,

So ε₄=3/2, and the radius, R₄, of the yellow circle is the reciprocal of ε₄, or 2/3.  Note that there are two solutions, both giving the same radius.  These correspond to the two possible places the yellow circle can go -- one above the pink circles, as shown, and the other below the pink circles.  Note that if the pink circles were smaller (ε₂ and ε₃ were larger) the two solutions would be different, because two different-sized yellow circles could be drawn.  Also, if the pink circles were any bigger, there would be no solutions at all, because they wouldn't both fit inside the blue circle.