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Constructing a Pentagon

Asked by Ting Ting Wu, student, State College Area High School on January 27, 1998:
I want to know how to construct a pentagon. I have done it before, but I have forgotten how. I remember it is similar to constructing a hexagon, but a bit more difficult. Thanks.
There are several ways to do it. Unfortunately we are very short-staffed right now and cannot spare the resources to hunt down the easiest and most elegant construction. However, the following method will work:

Constructing a pentagon is equivalent to dividing a circle (a full 360 degrees) up into five equal parts (angle 72 degrees each). The cosine of 72 degrees is (sqrt(5)-1)/4 (this can be found by starting with the equation cos(5t) = cos(360 degrees) = 1, using trigonometric identities to write cos(5t) as a polynomial in cos(t), factoring and solving the resulting polynomial equation for cos(t)).

Therefore, this angle of 72 degrees can be constructed by building a right-angled triangle whose hypotenuse is 4 and whose adjacent side is of length sqrt(5)-1. This latter length can be constructed by taking hypotenuse of a right triangle whose other sides have lengths 1 and 2, and subtracting length 1 from it.

The following procedure uses this idea to construct a pentagon:

Start with a circle C, with centre point O. Let P be a point on C. Draw the perpendicular bisector L to segment OP (bisecting it at point Q). Construct the midpoint R of OQ. (RQ is going to be our unit length).

With centre Q and radius RQ, draw an arc intersecting L at point S. Draw segment OS. (This is the hypotenuse of a right triangle OQS whose other sides have length 1 and 2, so OS has length sqrt(5)).

With centre S and radius RQ (= QS), draw an arc intersecting OS at point T. (Now OT has length sqrt(5)-1).

Construct the line passing through point T at right angles to OT. Let it intersect the circle C at point U.

Now the triangle OTU has hypotenuse of length OU = radius of C = 4, and side length OT = sqrt(5)-1. Therefore, angle UOT is 72 degrees. Extend segment OT past S until it meets the circle C at point V; you have now constructed two vertices (U and V) of the pentagon.

To construct the remaining vertices: with centre V and radius UV draw an arc intersecting C at point W. With centre W and the same radius, draw an arc intersecting C at point X. Finally, with centre X and the same radius, draw an arc intersecting C at point Y. UVWXY will be a pentagon.

There are probably much more efficient ways to do it, but the above procedure will certainly work, for the reasons described. The procedure is illustrated below:

                              |L        * C
                              |         *
                              |         *
                          T   |         *
                         /    S          
                        /              *
                       /              V
                      /             *
                     /            *
            *       /           *
              **** U ********* 

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