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mfgoode 2007-01-26 16:50

The Tangent and the golden mean

Intro: It is amazing that by a simple construction of 4 points on a line, elegant ratios of the of the GM can be obtained.

Problem: If P is a point on the chord AB (extended) of a circle such that the tangent at P which touches the circle at T is equal to AB. Draw C on AB such that PT =PC.

Prove: 1) AP/AB = phi

2) CA/CB =Phi

3 ) AP/AC = Phi^2

4) AP/BC = phi^3

5) How will you peform the construction in question ? For the problem take it that the construction is done.

Mally :coffee:

mfgoode 2007-01-31 16:26

A start.

Well since no one has tackled even the first proof I will give the full derivation of 1) AP/AB = phi

Please follow the figure as I have described with the same lettering.

PT^2 = AP.BP (theorem)

i.e. AB^2 = AP ( AP -AB )

Whence AP^2 - AP.AB - AB^2 = 0

or (AP/AB)^2 - (AP/AB) -1 = 0

Thus AP/ AB = (1 + sq.rt.5 ) /2 = phi QED

Now, if C is a point in PA such that PC = PT, find CA/CB

HINT : It is also equal to phi. Please try to prove it.

Mally :coffee:

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