MATHCOUNTS Exercises – 16

Solution First, connect the center of the circle with the center of the semicircle. Then drop a perpendicular from the center of the circle to side AB, as shown.

Let \(r\) be the radius of the circle. So, segment \(IJ=2+r\), since the side of square \(ABCD\) is 4. Also, segment \(IK=4-r\) and \(KJ=2-r\). Since \(IK \perp KJ\), we can set up a Pythagorean:
$$(4-r)^2+(2-r)^2=(2+r)^2$$
This simplifies to $$16-8r+r^2+4-4r+r^2=4+4r+r^2$$ Simplifying further, we get a quadratic equation:
$$r^2-16r+16=0$$
Applying the quadratic formula, we get \(r=8-4\sqrt{3}\).