It is a very well known fact that a regular pentagon can be constructed with a ruler and a
compass (unlike a regular polygon with 7 sides). Such a construction is provided by a free to
download program "Ruler and Compass".
A construction below is based on the older version of program "Ruler and Compass". To see better pictures and description, download
installation program of
Step 1.
Choose a horizontal line AB. Draw a perpendicular line to AB passing through A.
Draw a circle C with a center at A and radius AB.
In this picture point A is denoted by P and point B by Q. This picture was drawn by a program
"Ruler and Compass" mentioned above. Differences in names are because program itself chooses
names. Points A and B are defined by this program by lines
A=point(250,250)
B=point(250,350)
Horizontal straight line L is defined by
L=line(A,B)
and line perpendicular (vertical) LL by
LL=Perp(L,A)
Procedure Perp(...) is provided by a library unit of
a program "Ruler and Compass". It draws perpendicular lines
using ruler and compass.
Take an intersection point of a circle C and a vertical straight line LL
E=IntersCircLin(C,LL,1)
In the picture it is denoted by computer by P2.
Take a center of the segment EA (on the picture - center of the segment with ends P and P2)
F=SegmentCenter(E,A)
Procedure SegmentCenter(...) is provided by a library unit of the program "Ruler and Compass".
Point F is denoted by E2 in the picture.
Step 2.
Draw a circle C1 with a center at F (point E2 of the picture) passing through the point B
(point Q of the picture).
Let G be the point of intersection of C1 with the vertical line LL
(it is point I2 of the picture). Corresponding lines of a program are
C1 = circle(F , B )
G=IntersCircLin(C1,LL,1)
Step 3.
Draw a circle C2 with the center at B (point Q of the picture) passing through the point G
(point I2 of the picture).
Let A1 (point M2 of the picture) be an intersection point of C with C2 . Corresponding
lines of the program are
C2 = circle(B , G )
A1=Inters2Circ(C,C2,1)
Points B and A1 (points Q and M2 of the picture) are 2 vertices of a regular pentagon inscribed
into a circle C.
Step 4.
Having 2 vertices of a regular pentagon it is easy to find other vertices A2, A3, A4,
because all of them belong to the circle C in equal distances.
Paint a received pentagon.
A procedure RegPoint(...) is used. This procedure finds a point on a circle (which is the first
variable of procedure), in the same distance to a point (second variable of procedure) as the
distance of 2 points (second and third variable of procedure).
Procedure is provided by a program, it is in a library. It uses ruler and compass only.
Step 5.
Remove all unnecessary lines.
This ends the construction.
Step 6.
We'll prove that our construction really constructs a regular pentagon.
For simplicity assume that a radius of a circle C is 1 (this is our measure unit). Hence,
we assume that length of PQ is 1. We are supposed to prove that the distance from Q to M2
is 2*sin 36, where arguments of trigonometric functions are measured in degrees. Distance
from P to E2 is equal 1/2. From Pytagorean Theorem we see that the distance from E2 to Q
is
Since the distance from E2 to Q is the same as the distance from E2 to I2, we see that the
distance from P to I2 is
From the Pythagorean Theorem the distance from I2 to Q is
Remark. In the construction we drew several lines and circles and it is hard to see
what they have in common with a regular pentagon. In fact they have nothing in common.
We just constructed a segment of a given length and we showed that calculations were correct.
Steps of a proof of ordinary construction are for example "Both triangles are similar, hence
angle A is equal to B". Literka thinks that there are no constructions of a regular pentagon,
such that proof would not require, for example, calculations with square roots.
