## Constructions for the Golden Ratio

Let's start by showing how to construct the golden section points on any line: first a line phi (0·618..) times as long as the original and then a line Phi (1·618..) times as long. John Turner has nicknamed the two points that divide a line at the golden ratio (0.618 of the way from either end) as gold points .

## Constructing the internal golden section points: phi

If we have a line with end-points A and B, how can we find the point which divides it at the golden section point?

We can do this using **compasses** for drawing circles and a **set-square** for drawing lines at right-angles to other lines, and we **don't need a ruler at all** for measuring lengths!

(In fact we can do it with just the compasses, but how to do it without the set-square is left as an exercise for you.)

We want to find a point G between A and B so that AG:AB = phi (0·61803...) by which we mean that G is phi of the way along the line. This will also mean that the smaller segment GB is 0·61803.. times the size of the longer segment AG too.

AG | = | GB | = phi = 0·618033.. = | √5 – 1 |

AB | AG | 2 |

**Here's how to construct point G using set-square and compasses only:**

- First we find the mid point of AB. To do this
**without a ruler**, put your compasses on one end, open them out to be somewhere near the other end of the line and draw a semicircle over the line AB. Repeat this at the other end of the line**without altering the compass size**. The two points where the semicircles cross can then be joined and this new line will cross AB at its mid point. - Now we are going to draw a line half the length of AB at point B, but at right-angles to the original line. This is where you use the set-square (but you CAN do this just using your compasses too - how?). So first draw a line at right angles to AB at end B.
- Put your compasses on B, open them to the mid-point of AB and draw an arc to find the point on your new line which is half as long as AB. Now you have a new line at B at right angles to the AB and BC is half as long as the original line AB.
- Join the point just found to the other end of the original line (A) to make a triangle. Putting the compass point at the top point of the triangle and opening it out to point B (so it has a radius along the right-angle line) mark out a point on the diagonal which will also be half as long as the original line.
- Finally, put the compass point at point A, open it out to the new point just found on the diagonal and mark a point the same distance along the original line. This point is now divides the original line AB into two parts, where the longer part AG is phi (0·61803..) times as long as the original line AB.

**Why does this work?**

It works because, if we call the top point of the triangle T, then BT is half as long as AB. So suppose we say AB has length 1. Then BT will have length 1/2. We can find the length of the other side of the triangle, the diagonal AT, by using Pythagoras' Theorem:ATi.e.^{2}= AB^{2}+ BT^{2}AT^{2}= 1^{2}+ (1/2)^{2}ATNow, taking the square-root of each side gives:^{2}= 1 + 1/4 = 5/4AT = (√5)/2Point V was drawn so that TV is the same length as TB = AB/2 = 1/2.

So AV is just AT - TV = (√5)/2 – 1/2 = phi.

The final construction is to mark a point G which is same distance (AV) along the original line (AB) which we do using the compasses.So AG is phi times as long as AB!

1·61803 39887 49894 84820 45868 34365 63811 77203 09179 80576 ..More..