Constructing the external golden section points: Phi

This time we find a point outside of our line segment AB so that the new point defines a line which is Phi (1·618..) times as long as the original one.

Here's how to find the new line Phi times as long as the original:

1. First repeat the steps 1 and 2 above so that we have found the mid-point of AB and also have a line at right angles at point B.
2. Now place the compass point on B and open them out to touch A so that you can mark a point T on the vertical line which is as long as the original line.
3. Placing the compass point on the mid-point M of AB, open them out to the new point T on the vertical line and draw an arc on the original line extended past point B to a new point G.
4. The line AG is now Phi times as long as the original line AB.

Why does this work?
If you followed the reasoning for why the first construction (for phi) worked, you should find it quite easy to prove that AG is Phi times the length of AB, that is, that AG = (√5/2 + 1/2) times AB.

Hint:
Let AB have length 1 again and so AM=MB=1/2. Since BT is now also 1, how long is MT? This is the same length as MG so you can now find out how long AG is since AG=AM+MG.

Hofstetter's 3 Constructions of Gold Points

Using only circles

Kurt Hofstetter has found a beautifully simple construction for a line and its golden section point using only compasses to draw 4 circles:

1. On any straight line S, pick two points X and Y.
   With each as centre draw a circle (green) through the other point labelling their points of intersection G (top) and B (bottom) and the points where they meet line S as P and Q;
2. With centre X, draw a circle through Q (blue);
3. With centre Y, draw a circle through P (blue); labelling the top point of intersection of the blue circles A;

G is a gold point of AB (G is at the golden section of AB).
A Simple Construction of the Golden Section, Kurt Hofstetter in Forum Geometricorum Vol 2 (2002) pages 65-66 which has the proof too.

Another even simpler method!

Kurt Hofstetter has discovered a very simple constructions of the gold point on a line AB just using circles and one line to find the gold point(s) on a given line segment AB:

1. With centre A, draw a circle through B;
2. With centre B, draw a circle through A;
3. Extend BA to meet the first circle (centre A) at C;
4. Label the lower point on both circles as D;
5. With centre C draw another circle through B;
6. Where the larger circle meets the circle centred on B, label the upper point E;

Line DE crosses line AB at G.
G is a gold point of AB
The other gold point on AB can be constructed similarly.

Uli Eggermann of Germany points out that we also have A as the gold point of CG!
It is quite simple to prove too using the properties of Phi and phi that phi=1/Phi and 1+phi=Phi:

• If AB is 1 then so is CA: they are both radii of the same circle
• Above we constructed G, the gold point of AB (=1), so that AG is phi
• CG = CA + AG = 1 + phi = Phi
• Since CA is 1 and CG is Phi then CA/CG = 1/Phi = phi
• so A is the gold point of CG

Kurt Hofstetter's proof was published as Another 5-step Division of a Segment in the Golden Section in Forum Geometricorum Vol 4 (2004) pages 21-22.

Lemoine's Construction of the Golden Section

In this construction to find the gold point of a given line segment AB rediscovered by K Hofstetter but going back to 1902, we again need to construct only one line:

1. With A as centre, draw the first circle through B;
2. With B as centre, draw a second circle through A using labels C (top) and D (bottom) for the two points of intersection with the first circle;
3. Through the top point, C, draw a third circle through A and B and use label E for its other intersection with the first circle;
4. Draw the line CD and where it meets the third circle, label this point F;
5. With centre E draw a circle though F; where it meets the line AB label this point G;

G is a gold point of AB.
Also, if BA is extended to meet the larger circle with centre E at G', then
A is a gold point of G'B.

A 5-step Division of a Segment in the Golden Section K Hofstetter in Forum Geometricorum Vol 3 (2003) 205-206.
The earliest reference to this construction appears to be in Géométrographie ou Art des Constructions Géométriques by E. Lemoine, published by C. Naud, Paris, 1902, page 51.

Phi and Pentagons

There is an intimate connection between the golden section and the regular 5-sided shape called a pentagon and its variation - the pentagram - that we explore first.

Pentagons and Pentagrams

The pentagram is a symmetrical 5-pointed star that fits inside a pentagon. Starting from a pentagon, by joining each vertex to the next-but-one you can draw a pentagram without taking your pen off the paper.

The pentagram has 5 triangles on the edges of another pentagon at its centre. Let's focus on one of the triangles and the central pentagon as shown here.

All the orange angles at the vertices of the pentagon are equal. They are called the external angles of the polygon. What size are they? This practical demonstration will give us the answer:

• Take a pen and lay it along the top edge pointing right
• Turn it about the top right vertex through the orange angle so that it points down to the lower right
• Move the pen down that side of the pentagon to the next vertex and turn it through the next orange angle
• Repeat moving it along the sides and turning through the rest of the orange angles until it lies back on the top edge
• The pen is now back in its starting position, pointing to the right so it has turned through one compete turn.
• It has also turned through each of the 5 orange angles
• So the sum of the 5 orange angles is one turn or 360°
• Each orange angle is therefore 360/5=72°

The green angle is the same size as the orange angle so that the two "base" angles of the blue triangle are both 72°.
Since the angles in a triangle sum to 180° the yellow angle is 36° so that 72° + 72° + 36° = 180°.

The basic geometrical facts we have used here are:

So the pentagram triangle has angles of 36 °, 72° and 72°. Now let's find out how long its sides are.

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