Navigation Panel: (IMAGE)(IMAGE)(IMAGE)(IMAGE) (SWITCH TO TEXT-ONLY VERSION) (IMAGE)(IMAGE) (These buttons explained below)

UNIVERSITY OF TORONTO
MATHEMATICS NETWORK

Question Corner and Discussion Area


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  (IMAGE) (this can be found by starting with the equation  (IMAGE) , using trigonometric identities to write  (IMAGE) as a polynomial in  (IMAGE) , factoring and solving the resulting polynomial equation for  (IMAGE) ).

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  (IMAGE) . 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  (IMAGE) ).

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

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  (IMAGE) . 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:

 (IMAGE)

[ Submit Your Own Question ] [ Create a Discussion Topic ]

This part of the site maintained by (No Current Maintainers)
Last updated: April 19, 1999
Original Web Site Creator / Mathematical Content Developer: Philip Spencer
Current Network Coordinator and Contact Person: Joel Chan - mathnet@math.toronto.edu


Navigation Panel: (IMAGE)(IMAGE)(IMAGE)(IMAGE) (SWITCH TO TEXT-ONLY VERSION) (IMAGE)(IMAGE)

(IMAGE) Go backward to Do Parallel Lines Meet At Infinity?
(IMAGE) Go up to Question Corner Index
(IMAGE) Go forward to The Three Classical Impossible Constructions of Geometry
 (SWITCH TO TEXT-ONLY VERSION) Switch to text-only version (no graphics)
(IMAGE) Access printed version in PostScript format (requires PostScript printer)
(IMAGE) Go to University of Toronto Mathematics Network Home Page