3.4 Phi and Right-angled Triangles - the Kepler Triangle
The German astronomer and mathematician Johannes Kepler (1571-1630) had a great interest in both Pythagorean triangles and the golden ratio which was known then mainly by the term used by Euclid "the division of a line into extreme and mean ratio":
Geometry has two great treasures; one is the Theorem of Pythagoras; the other, the division of a line into extreme and mean ratio.
The first we may compare to a measure of gold; the second we may name a precious jewel.
Mysterium Cosmographicum 1596
Here we have both together in a single unique triangle.
Pythagorean triangles are right-angled triangles that have sides which are whole numbers in size. Since the golden section, Phi, is not a pure fraction (it is irrational), we will not be able to find a Pythagorean triangle with two sides in golden section ratio.
However, there is as a right-angled triangle that does have sides in the golden ratio. It arises if we ask the question:
Is there a right-angled triangle with sides in geometric progression, that is
the ratio of two of its sides is also the ratio of two different sides in the same triangle?
If there is such a triangle, let its shortest side be of length a and let's use r as the common ratio in the geometric progression so the sides of the triangle will be a, ar, ar2. Since it is right-angled, we can use Pythagoras' Theorem to get:
(ar2)2 = (ar)2 + (a)2
a2r4 = a2r2 + a2
We can divide through by a2 :
r4 = r2 + 1
and if we use R to stand for r2 we get a quadratic equation:
R2 = R + 1 or
R2 – R – 1 = 0
which we can solve to find that
| R = |
| 1 ± √5 |
 |
| 2 |
|
= Phi or –phi |
Since R is r2 we cannot have R as a negative number, so
R = r2 = Phi so
r = √Phi
The sides of the triangle are therefore
a, a √Phi, a Phi
and any right-angled triangle with sides in Geometric Progression has two pairs of sides in the same ratio √Phi and one pair of sides in the Golden Ratio!
We will meet this triangle and some interesting properties of its angles later on this page in the section on Trigonometry and Phi.
3.4.1 How to construct Kepler's Triangle
It is very easy to construct Kepler's Triangle if we start from a golden rectnagle as described above....
Make the golden rectangle...
and draw a circle centred on one corner and having the longer side of the rectangle as radius...
The point where the circle crosses the other longer side marks one vertex of Kepler's triangle, the centre of the last circle is another and the right angle of the rectangle is the third.
- A trigonometric intersection D Quadling, Math. Gaz. (2005), Note 89.70
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/phi2DGeomTrig.html 
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