Const. Pent. p1 of 10. A.French 2012
Construction of a regular
pentagon
Dr Andrew French
Const. Pent. p2 of 10. A.French 2012
y
‘Construction’ means one can only
draw lines using the following
equipment
x2  y 2  R2
O
R
• A straight edge
• A compass
x
Step 1
Draw a circle and divide it
vertically and horizontally to
form the y and x axes.
Use the line bisection method
to find the x axis, given the y axis.
Const. Pent. p3 of 10. A.French 2012
y
Step 2
Construct a circle with half the
radius of the larger circle. Find the
centre of this circle by bisecting
the OX line as indicated.
x2  y 2  R2
O
 12 R,0
X
x
R
x  12 R2  y 2   R 
2
2
 14 R 2
Const. Pent. p4 of 10. A.French 2012
y
y  2x  R
x2  y 2  R2
R
a, b 
 12 R,0
O
R
1
2
x
R
y  2x  c
0  2 12 R   c  c   R
 y  2x  R
0, R 
x  12 R2  y2  14 R2
Step 3. Draw a line from the base of the larger circle and through the centre of the
smaller circle. Find where this crosses the upper edge of the smaller circle.
y  2x  R
Const. Pent. p5 of 10. A.French 2012
y
x2  y 2  R2
k
a, b 
 12 R,0
O
R
K
k
x
x 2   y  R  k 2
2
0, R 
x  12 R2  y2  14 R2
Step 4. Draw an arc of a
large circle with radius from
the base of the first circle to
the small circle crossing point
found in step 3. Find position
K where this crosses the
original circle.
Const. Pent. p6 of 10. A.French 2012
Step 5
y
K
O
x
k
Draw a straight line from the
intersection of the y axis and
the original circle and the
crossing point K found in
step 4. This is an edge of a
regular pentagon! Use a
compass
to step round the original
circle to find the other three
points.
To minimize the effects of
drawing errors, work out the
lower vertices from the left
and right large circular arc
intersections rather than
step round using a compass
from one side only.
y  2x  R
Const. Pent. p7 of 10. A.French 2012
a, b 
x2  y 2  R2
a, b 
 12 R,0
O
R
a  12 R 2  b 2  14 R 2
a 2  2a   k 2
2
x
k
a 2  b  R   k 2
b  2a  R
2
y
k a 5
a  12 R 2  2a  R 2  14 R 2
a 2  aR  14 R 2  4a 2  4aR  R 2  14 R 2
5a 2  5aR  R 2  0
0, R 
x  12 R2  y2  14 R2
x 2   y  R  k 2
2
5 R  25R 2  20R 2
a 
10
5 5 

 a  R

 10 
is the –ve
solution

5 5 5
  k  12 R 1  5
k  R

 10 

Const. Pent. p8 of 10. A.French 2012
y
360o

 72o
5
180    2
0, R 

180o  72o
 
 54o
2

O 
x
Circle theorem
o
54
k
90o
0, R 
2R
k
Now let’s consider the regular pentagon separately.
To demonstrate why the construction works we need
an expression for the triangle side k, derived only from
properties of the regular pentagon. i.e. independent of the
construction. If everything agrees then the construction works!
36o
2 R sin 54o  k
2 R cos36o  k
Const. Pent. p9 of 10. A.French 2012
y
0, R 
By properties of a
regular pentagon
54o
90o
2R
O
k
x
36o
k
2 R sin 54o  k
2 R cos36o  k
0, R 
By equating k we find
cos36o  sin 54o 
k

k  12 R 1  5

From the construction
1 5
 0.809
4
which indeed is the case!
Note
k 1 5

is the GOLDEN RATIO
R
2
Const. Pent. p10 of 10. A.French 2012
b
a
a
a
Define
‘Golden rectangle’
by ratio of sides
a/b
a ab
1
 
 1
b
a

  2   1  0
1 5
 
2
The
Golden
Ratio
Typically take a > b so
GOLDEN RATIO
1 5

 1.618
2
The construction of the regular pentagon therefore gives
us another exact right angled triangle relationship
54o
w
10  2 5
1
2
o
90
cos36o 
1 5
4
sin 36o 
1
4
10  2 5
cos54o 
1
4
10  2 5
sin 54o 
2
36o
1 5
2
10  2 5
t an36 
1 5
o
t an54 
o
1  5 
4
w
 w  4
1  5 
2
From
Pythagoras’
theorem
4
2
2
w
1
2
1 5
2
4
10  2 5
 4
1 2 5  5

4
1
2
1 5
10  2 5
16  1  2 5  5