There are a number of constructions given on the internet for a regular pentagon, some fairly efficient, some quite complex. I have not seen one that allows easy construction of a regular pentagon with the length of the side specified. Here is a fairly simple approach.

Construct segment AB on line *l*. (This may be a
segment of given length.)
Draw a circle C_{1} with center A and radius AB.

Draw a circle C_{2} with center B and radius AB.

Draw the perpendicular bisector of AB through the intersections of
these two circles. Identify the midpoint M of segment AB.
On the bisector of AB mark off segment MP congruent to AB.

On ray AB mark off segment MQ congruent to AP.

Draw circle C_{3} with center A and radius AQ.

Draw circle C_{4} with center B and radius AQ.

The intersection of C_{2} and C_{3} is point C.

The intersection of C_{1} and C_{4} is point D.

The intersection of C_{3} and C_{4} is point E.

ABCDE is the required pentagon.

Note that it is not necessary to draw segment AP, although I did include it in the figure. We only need to set the compass to length AP to mark off MQ.

It is possible to do this with less "junk" inside the pentagon.
Rather than drawing circles C_{1} and C_{2}, just make
the usual small arcs to define the perpendicular bisector. Draw MP on
one side of segment AB only. Then construct point Q as above. Now on
the opposite side of AB from P, make small arcs from A (length AQ) and
B (length AB) to find C. Make small arcs from B (length AQ) and A
(length AB) to find E. Then finish with small arcs from A (length AQ)
and E (length AB) to find D. The interior of the pentagon is nearly
empty, except for a few arcs.

You may be wondering why this construction works. The ratio of the
diagonal of a pentagon to a side is the golden ratio. The first part
of this construction gives a segment AQ whose ratio to AB is the golden
ratio. Then circle C_{3} contains the endpoints of both
diagonals which have one endpoint at A (that's diagonals AC and AD),
and circle C_{4}
contains the endpoints of both diagonals which have one endpoint at B
(that's BE and BD). You should be able to work out the details from
there.

A second construction of a regular pentagon (but which does not allow the length of the sides to be specified in advance) is given here.

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Date last modified: 8/12/2007

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