WildTrig29: Trisecting angles and Hadley's theorem

[benthurston27]

I noticed a 1,2,1 for Q1,Q2,Q3 respectively satisfies Hadley's theorem but doesn't s3 have to be larger than s2?

[benthurston27]

@benthurston27 I see, a 1,2,1 triangle is actually 3 colinear points so the spread's would all be zero which does satisfy the 3 equations.

[benthurston27]

@benthurston27 Ok I've got it now a 1,2,1 quadrances triangle is not 3 colinear points, it is a 1,square root of 2,1 in distances or a right triangle. which does make Hadley's theorem say the spreads would be 1/2, 1, 1/2 which is fine. Also the 3rd spread polynomial can be less than the second because sometimes the spread of an angle greater than 90 degrees is less than 1. In this case it would be trisecting the spread of .5 into 3 equal spreads of .5 which is fine.

[basalduat]

Dear Sir,

For a simple example of trisection visit

youtube.com/watch?v=D2D20RIGqm­Y

[basalduat]

Dear Sir,

Look into Thomas Hull's book, Project Origami.

On page 47, he describes a very simple way to trisect any angle <= 90 degrees.

Moreover he explains why you can trisect with "folding paper geometry."

Tim Basaldua

[linelites]

Is Hadley's theorem the same as the Lost Theorem, described in my video 'Trisecting This" on youtube user LineLItes