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