There is a carnival that comes to town at Apple Blossom time in
Winchester, Virginia. There is one carnival concession in which, for
the price of a ticket, you drop three identical disks several
feet onto a painted circle. If they cover the target circle so that no part
of it is exposed, you win a prize.
The operator usually lays the disks
down first to prove that they can
cover the circle completely. The
challenge is difficult because the
circle is the maximum size that the
disks can cover.
Imagine that you're setting up one
of these concessions. If your disks
are four inches in diameter, what
should be the diameter of the target
circle?
Solution to the Problem:
The maximum diameter of the target circle is about 4.61 inches.
Points A, B and C are on the circumference of the target circle
(shown here on top of the disks). A, B and C are also the ends
of three lines across the disks. The longest possible line across a disk is
its diameter. So lines AB, BC, and CA must equal the diameters of the
three disks (4") if the target circle is to be the maximum size that can be covered
by the disks.
The radius of the target circle is the distance from A, B, or C to the center of
the circle at D. AE and BF extend from A and B to the centers of the
opposite sides of triangle ABC, and their intersection at D is the center of the
target circle. AEC is a right triangle with hypotenuse AC of 4" (the
circumference of a disk) and side EC of 2". So, by the Pythagorean Theorem,
AE is the square root of 12 (42 - 22).
