Recall that the exterior angle of the regular pentagon is 360/5 =
72 degrees, which makes the interior angle equal 108 degrees. Draw the diameter of the
cicle. (In problems like this the center of the circle is usually defined. If not, we can
draw 2 arbitrary cords, and perpendicularly disect them--the point of intersection of the
two bisectors is the center of the circle.) Now, construct line OC, perpendicular to
the diameter AB (see picture at right). The midpoint of OC is D. Mark point D' on AB the
same distance away from A as D. Then, mark point E such that D'E = OD. Find point E' on
the circle the same distance away from A as E. Then, BE' is one of the sides of the
Proof: Assume that circle O is a unit circle. Then, OA = 1, and OD = 1/2,
which makes AD = sqr(5)/2. From the construction, AE' = (sqr(5) + 1)/2, and AB = 2. Then,
sin(ABE') = [sqr(5) + 1] / 4, and angle ABE' = 54 degrees, which is half the internal
angle of the pentagon.
Comment: German mathematician Gauss found all regular polygons which can be
constructed with straightedge and compass. These polygons must have the number of sides
which is either a prime Fermat's number, or a product of several prime Fermat's numbers.
Fermat's numbers have the form Fn = 2(2^n) + 1, and in Fermat's time
were considered to be prime regardless of the value of n. For instance, F0 = 3,
F1 = 5, F2 = 17, F3 = 257, F4 = 65,537.
However, Leonard Euler showed that already F5 = 4,294,967,297 = 641*6,700,417
is a composite number. Actually, so far no new prime Fermat's numbers have been found, but
neither has it been proven that such numbers do not exist.
In other words, it is possible to construct regular polygons with 3, 5, 15, 17, 51, etc.
sides, or any other number obtained by doubling these numbers.