I have an album on my facebook page with three jpeg graphic displays of my method for finding the 2/3rds slope of three different given angles using a method that can be completed with a straight-edge and compass. I could add complication by showing how to find the 1/3rds slopes of the three different angles, but the finding of any third gives the same size of the other thirds, so I kept the method very simple. http://www.facebook.com/album.php?aid=25451&id=100000573024201&l=e2ae1c95e4 Very Respectfully, Ray Donald Pratt
Finding a trisection of certain angles is easy (trisecting 180 degrees is easy) but it is impossible to come up with a method which will trisect any angle in finitely many steps using a straight edge and compass. It's along the same logic as squaring a circle.
Reconstructed from pictures, see if you agree that this is Pratt's alleged method. Given an angle ABC, construct a circle at B and extend rays BA and BC to intesections with the circle at D and E, respectively. Analysis: B = (0,0), E =(1,0), D = (cos θ, sin θ) Construct M, the bisector of DE, extend ray M B to intersection with circle at F. Analysis: M = ( (1/2) ( 1 + cos θ ), (1/2) sin θ ) = ( (cos θ/2)^2 , (cos θ/2)(sin θ/2) ) , F = ( -(cos θ/2) , -(sin θ/2) ) Construct N, the bisector of BD. Construct G on circle such that BG is parallel to the ray FN. Analysis: N = ( (1/2) cos θ, (1/2) sin θ) G = (N - F) / length(N -F) = ( (1/2) cos θ + (cos θ/2) , (1/2) sin θ + (sin θ/2) ) / sqrt( 5/4 + (cos θ/2) ) Claim: 3 times Angle GBE - 2 times Angle DBE = 0 Analysis: 3 arctan(((1/2) sin θ + (sin θ/2))/((1/2) cos θ + (cos θ/2))) - 2 θ = -θ^3/216-θ^5/10368 - .... So while the construction gives a close approximation ( ~ 0.5%), it does not satisfy the claim unless θ is trivially 0. θ -> expected -> error 45° -> 30° -> -0.0434° 90° -> 60° -> -0.3612° 135° -> 90° -> -1.3063° The last two are where the trisectors are explicitly and independently constructable and where careful construction will easily reveal that the claimed 2/3s angle is not comeasurable with the trisector. So in addition to the general demonstration that no general method for trisection exists referred to above, here is are two independent verifications in algebraic geometry and in geometric construction suitable for 13-year-olds.
From visual inspection, this method has large errors for large angles and small errors for small angles (and zero error at zero angle). It's actually approximating 1/6 of the angle by approximating 1/3 of the half angle, thus, the remainder of the half angle is about 1/3 (that is, 2/3 of 1/2 is 1/3). Any trisection method only needs to consider angles less than 90° because 90° can be trisected perfectly, and the remaining angle (beyond 90°) can be trisected with the invented method.
Gentlemen, thank you. I went back into my geometry program and put in segments in the trisection triangle bases and checked their equality, and I would find two equal and one disequal, etc. I don't trust the trigonometric arguments because of the numerical inexactness in trigonometric calculations, but if your math is actually dealing with exact logic, then I do not know enough trigonometry yet. I am going to go back into the geometry program and artificially create trisected angles and then double check the relations around several of them without moving the constructions with the computer. I am surprised that there were any errors because that is how I initially constructed my movable test angles which always stayed trisected no matter how I changed the angle -- or so I thought.
No, he didn't. He might of thought he did or he might have known a way to get a very close approximation but he didn't know an exact way. And how do we know this? Because using some pretty nifty mathematics which I don't pretend to understand you can prove that if you have only the operations which can be done using a straight edge and a compass then you cannot construct a finite length algorithm which leads to the construction of the square with the same area of a given circle. A more understandable is the example of Galois proving that there is not version of the quadratic formula for polynomials of order 5 or higher. The formula is only allowed to involve addition, multiplication, division, subtraction, n'th roots etc. For the quadratic case of \(ax^{2}+bx+c=0\) you have \(x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\), the use of the square root is sufficient, in that if someone gives you a value Y you can tell them the value y such that \(Y = y^{2}\). For a 4th order polynomial you need to be able to give the y such that \(y^{4} = Y\). For a polynomial of the form \(ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f=0\) you need to be able to give someone the value y such that \(y+y^{5}=Y\) and this cannot be done using finitely many steps using only roots etc for a general setup. Its a bit like computer logic. In computer logic you need to build up every algorithm from the logic gates of NOT, AND, OR etc. There's a great many things you can do with the appropriate combinations (as computers demonstrate) but there are some things you simply cannot do using that logic construction. The limits of what Turing machines can do is extremely interesting, if not a little abstract for my tastes.
If by "square the circle" you mean "exhibit a rectilinear shape which has the same area as a given circle" (which is what the Greeks meant when they said "square the circle") then Archimedes knew how to square the circle. He even had a proof, using the law of the excluded middle, that the rectangle he constructed had the same area as the circle (though he never published his method - it was only very recently that we discovered a text containing that.) Agreed. It's great that you know all this stuff (or at least that you can talk about it in a way that suggest you know it) but I don't see the relevance.
Citation required. The use of the spiral \(r \propto \theta\) does allow you to trisect angles or square the circle, but since the spiral cannot be constructed with straightedge and compass, the proof that something cannot be done with straightedge and compass remains unmolested. http://en.wikipedia.org/wiki/On_Spirals#Squaring_the_circle http://en.wikipedia.org/wiki/Archimedean_spiral http://mathworld.wolfram.com/ArchimedesSpiral.html http://mathworld.wolfram.com/GeometricProblemsofAntiquity.html
The link below is to a jpeg graphic of an artificially trisected angle where the relations are correct for any angle except the last step of finding the false 2/3rds of angle slope. The error is so slight that it can't be seen without extreme magnification (the line doesn't actually pass through the center of the tiny circle that I used). http://www.facebook.com/photo.php?pid=332855&l=c4a9403e94&id=100000573024201 Very Respectfully, Ray Donald Pratt
Hmm. Apparently I don't have the required post count to post a link. However, if you enter the text is.gd into your address bar, followed by a forward slash and then the text e7ZRe you will find yourself at Archimedes' proof. Link added by moderator: is.gd/e7ZRe.
Quick bump to ask if anyone has looked at this since I posted it. Do you agree that Archimedes knew how to square the circle?
He did not know how to construct, using only straightedge and compass, a right triangle with legs in proportion \(1:2 \pi\), so he did not know how to square the circle using only straightedge and compass, which was the challenge defined in post #2.
No, he didn't. He might have provided an algorithm whose limit, if you applied it infinitely many times, squared the circle but the challenge is to do it in finitely many steps. There's ways to work out pi involving inscribed regular polygons which the Greeks used to find pi to a few decimal places but that too would need infinitely many steps to get exactly pi. I was giving an example of how particular kinds of mathematics can be used to disprove the existence of such a method. In functional analysis there are things like \(f(x) = x^{2}\), \(f(x) = e^{x}\), \(f(x) = \sin x\) etc which are known as 'elementary functions', the sort of basic alphabet of calculus. Its possible to prove that some functions cannot be written in closed form using this 'alphabet', such as \(\Phi(x) = \int_{-\infty}^{x}e^{-y^{2}}dy\). It cannot be written in terms of polynomials, trig functions, roots, logs and exponentials. Squaring the circle and trisecting the angle are both geometric things but the principle is the same. If your 'alphabet' only involves a straight edge and a compass there are some things you simply cannot do and squaring the circle and trisecting a general angle are two. This isn't a matter of "No one knows a way" its a matter of "We know there is no way". Using a compass and a straight edge you can turn a line of length L into a line of length \(\sqrt{2}L\) because you can draw a right angled triangle whose angles are 45, 45, 90 and the hypotenuse is \(\sqrt{2}\) the length of each of the other two edges. Other roots can be done without too much hassle. There are constructable numbers, the numbers you can get using the 'alphabet' of only a compass and a straight edge. Doubling the cube amounts to making a length of size L into a length of size \(\sqrt[3]{2}L\) which cannot be done. So thanks to this powerful mathematics I don't need to look at your link, it is certain not to justify your claim as your claimed result of Archimedes is known to be impossible.
I didn't say anything about compass and ruler construction - for the Greeks, who weren't even comfortable with the idea that curvilinear shapes might have rectilinear areas, Archimedes' achievement of exhibiting a triangle with the same area as a circle was impressive enough to be considered revolutionary.
Apparently you haven't actually read the 'if' part of my 'if-then' statement. I doubt it. The Greeks didn't know what 'decimal places' were. Archimedes was able to show that pi is between 3 10/71 and 3 1/7, which I presume is what you're referring to? Sadly you don't seem to understand what I'm claiming - which is very simple and uncontroversial. Simply that Archimedes was able to exhibit a triangle with the same area as a given circle, and prove (via the method of exhaustion, aka law of the excluded middle) that the triangle had the area he said it did.
Alpha, you should have looked at the link! Roubini did not claim that Archimedes's result was based on straightedge and compass: Measurement of a circle, Proposition 1: The area of any circle is equal to a right-angled triangle in which one of the sidea bout the right angle is equal to the radius, and the other to the circumference, of the circle. If the circumference of a circle may be rolled out onto a line (something that may be done in practice for a given circle, but not with a straightedge and compass), then this is an achievable result, right?
