2007-01-26, 16:50 | #1 |
Bronze Medalist
Jan 2004
Mumbai,India
100000000100_{2} Posts |
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 |
2007-01-31, 16:26 | #2 |
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts |
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 Last fiddled with by mfgoode on 2007-01-31 at 16:28 |
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