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Construction of a regular pentagon
Dr Andrew French
Const. Pent. p1 of 10. A.French 2012
x
y
222 Ryx
R
O
‘Construction’ means one can only draw lines using the following equipment
• A straight edge• A compass
Step 1
Draw a circle and divide itvertically and horizontally to form the y and x axes.
Use the line bisection methodto find the x axis, given the y axis.
Const. Pent. p2 of 10. A.French 2012
O
R
x
y
222 Ryx
241
222
21
2R
RyRx
0,21 R
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.
X
Const. Pent. p3 of 10. A.French 2012
O
R
x
y
222 Ryx
24122
21 RyRx
0,21 R
ba,
R,0
R21
R
Rxy
RccR
cxy
2
20
2
21
Rxy 2
Step 3. Draw a line from the base of the larger circle and through the centre of thesmaller circle. Find where this crosses the upper edge of the smaller circle.
Const. Pent. p4 of 10. A.French 2012
O
R
x
y222 Ryx
0,21 R
ba,
R,0
Rxy 2
k
k
222 kRyx
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. p5 of 10. A.French 2012
24122
21 RyRx
K
O x
y
k
Step 5
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 aregular pentagon! Use a compassto 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.
Const. Pent. p6 of 10. A.French 2012
K
O
R
x
y222 Ryx
24122
21 RyRx
0,21 R
ba,
R,0
Rxy 2
k
222 kRyx
24122
21
222
2
RbRa
Rab
kRba
5
2 222
ak
kaa
10
55
10
20255
055
44
2
22
22
241222
412
24122
21
Ra
RRRa
RaRa
RRaRaRaRa
RRaRa
5110
55521
RkRk
is the –ve solution
Const. Pent. p7 of 10. A.French 2012
ba,
O x
y
k
R,0
R,0
ooo
oo
542
72180
2180
725
360
R2
k
o54
o90
kR
kR
o
o
36cos2
54sin2
o36
Circle theorem
Now let’s consider the regular pentagon separately. To demonstrate why the construction works we needan expression for the triangle side k, derived only fromproperties of the regular pentagon. i.e. independent of theconstruction. If everything agrees then the construction works!
Const. Pent. p8 of 10. A.French 2012
O x
y
k
R,0
R,0
R2
k
o54
o90
kR
kR
o
o
36cos2
54sin2
o36
5121 Rk
k
809.04
5154sin36cos oo
2
51
R
kis the GOLDEN RATIOFrom the construction
By properties of a regular pentagon
By equating k we find
which indeed is the case!
Note
Const. Pent. p9 of 10. A.French 2012
a
ba
a
2
51
01
11
2
a
ba
b
aTypically take a > b soGOLDEN RATIO
618.12
51
Define ‘Golden rectangle’ by ratio of sidesa/b
The Golden Ratio
Const. Pent. p10 of 10. A.French 2012
2
2
51
o54
o90
o36
The construction of the regular pentagon therefore gives us another exact right angled triangle relationship
From Pythagoras’ theorem
5210
5521164
55214
4
514
4
514
21
21
2
2
2
w
w
w
521021 w
5210
5154tan
51
521036tan
2
5154sin
521054cos
521036sin
4
5136cos
o
o
o
41o
41o
o