Put point O and draw circle with radius 1 around it. Pick some point d on this circle and draw the diameter through Od. Add the perpendicular diameter and call one of its endpoints E. Now |Od| = |EO| = 1. Label the midpoint of Od with P and draw the line EP. Now |OP | = |P d| = \frac{1}{2}. Draw a circle around P through E and label its intersection with the diameter through Od with a. Now |aP | = |EP | = \frac{1}{2} \sqrt{5}. Moreover, we have constructed the golden number Φ := \frac{1}{2}(\sqrt{5} + 1) as |ad| = Φ. Moreover, note that |aO| = |ad| − 1 = \frac{1}{2}(\sqrt{5} − 1) = \frac{1}{Φ}. Draw a circle round d through a, and name one of its intersection points with the original circle b, and draw the diameter of this original circle through b, as well as the line bd. Now the triangle Obd has cathedes of length 1 and hypothenuse of length Φ. This is the basic element of the pentagramm, which we will construct in the next step.

O d

E

P a

b

First we remove the line EP and the point P. We add an circle around b through d, label its intersection with the diameter through b with e, and add the lines baed. Draw two circles through O around b and d, and label their intersection that is different from O with c. Draw the pentagon abcdea and its inscribed pentagramme acebda.

O d

E

a

b

e

c

What are the coordinates of the points? We have O = (0, 0), E = (0, 1), d = (1, 0), and a = (\frac{1}{Φ}, 0) ≈ (0.618033988, 0). The first coordinate of e is \frac{Φ}{2} − \frac{1}{Φ} = \frac{1}{8Φ}((2Φ)^{2} − 8) = \frac{1}{8Φ}((\sqrt{5} + 1)^{2} − 8) = \frac{1}{4Φ}(\sqrt{5} − 1) = \frac{1}{2Φ^{2}} . By the Pythagorean triangle, the second coordi-

nate of e is \sqrt{\frac{1}{Φ^{2}} − \frac{1}{4Φ^{4}}} = \frac{1}{2Φ^{2}} \sqrt{(2Φ)^{2} − 1} = \frac{1}{2Φ^{2}} \sqrt{(\sqrt{5} + 1)^{2} − 1} = \frac{1}{2Φ^{2}} \sqrt{2(\sqrt{5} + 1) + 3} =

\frac{1}{2Φ^{2}} \sqrt{4Φ + 3}. Thus, we have e = \frac{1}{2Φ^{2}} (1, \sqrt{4Φ + 3}) ≈ (0.190983005, 0.587785252). The first coordinate of b is −\frac{1}{2Φ}. Thus, by the Pythagorean triangle, the second coordi-

nate of b is −\sqrt{1 − \frac{1}{4Φ^{2}}} = −\frac{1}{2Φ} \sqrt{(2Φ)^{2} − 1} = −\frac{1}{2Φ} \sqrt{4Φ + 3} . Thus, we have b =

−\frac{1}{2Φ}(1, \sqrt{4Φ + 3}) ≈ (−0.309016994, −0.951056516), and c = −\frac{1}{2Φ}(2Φ − 1, \sqrt{4Φ + 3}) ≈ (0.690983005, −0.951056516).

Now we want to construct a larger pentagon inscribed into the original circle.

O d

E

a
b

e

c

O H G  P A B  C X  X′ D  E F

Φ := \frac{1}{2}(\sqrt{5} + 1). \frac{1}{Φ} = \frac{1}{2}(\sqrt{5} − 1). AO = OX = OX′ = 1. OP = P X′ = \frac{1}{2}. AP = GP = \frac{1}{2} \sqrt{5}. GX′ = GP + P X′ = Φ. GO = GP − OP = \frac{1}{Φ}. AG = \frac{1}{Φ} \sqrt{Φ^{2} + 1}.

O H G  P A B  C X  X′ D  E F

5

B  C  A  D  E

6