Bill

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Thanks everyone
Bill

1. A diagram (page 3) shows a square of side 1 nested inside a square of side cuberoot(2), in the manner found here.

2. It is noted that in the right angled triangle so formed, the non-right angles are approximately 18 degrees and 72 degrees. (The discrepancy is about 1/70th of a degree).

3. It is observed (page 6) that we may construct a right angled triangle with angles 90, 18 and 72 degrees with ruler and compass alone.

4. Some calculation (page 7) shows that this construction gives an approximation to the cube root of two with a relative error of only about 0.000103

So the aim of approximating the cube root of two very well with ruler and compasses is achieved. However, the classical delian problem asks for a construction not of an approximation, however good, but rather of the exact side length of the larger cube, and this unfortunately has not been achieved. Indeed, this classical delian problem has now been proved to have no solution.

Thankyou

sfwc
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Bill

I was checking on the angles and how close they are. I worked it out that the discrepancy is not 1/70 but closer to 1/371.

Let me know what you think. I get angel discrepancy of 0.00269977 of a degree.

You can check this calculation by pasting the string "2-2Y1.5R=i@+2Y3R=/2=iS-72=" into the windows calculator.

To check this result was correct, I worked out (sin(72 + e) + cos(72 + e))^3 for e = 1/73 and 1/74:
e = 1/73: 1.9999948662234549429307173... < 2
e = 1/74: 2.0000047493598750317211042... > 2
So the correct value for e lies between 1/74 and 1/73, as stated above.

Thankyou

sfwc
<><

Thanks Bill

You stated:
x + y = s
x^2 + y^2 = 1
=> x^2 + (s-x)^2 = 1
=> 2x^2 - 2sx + s^2 - 1 = 0
=> x = 1/2(s +_ sqrt(2 - s^2)

You used x + y = s and x^2 + y^2 = 1 where the hypotenuses "s" and 1 will arrive at the desired lengths of x and y. So, there can be any number of workable x and y to arrive at "s" but since you plugged in 1 for the second equation it can only have one x and y. Did I get that right?

>You used x + y = s and x^2 + y^2 = 1 where the hypotenuses
>"s" and 1 will arrive at the desired lengths of x and y.
1 is certainly a hypotenuse; it is the side length of the inner square and so the hypotenuse of each of the 4 surrounding triangles. So we have x^2 + y^2 = 1 by pythagoras' theorem. However, s is not the hypotenuse of any of the triangles in the figure. Rather, it is the side length of the outer square. We have s = x + y since each side of the outer square is divided into two parts by a vertex of the inner square; these two parts have lengths x and y.

>So,
>there can be any number of workable x and y to arrive at "s"
>but since you plugged in 1 for the second equation it can
>only have one x and y.
There are of course many solutions to x + y = s. But if we also impose the condition x^2 + y^2 = 1 there are only two possible solutions. In one of them cos x is about 18 degrees and cos y about 72 degrees. In the other, cos x is about 72 degrees and cos y is about 18 degrees. This symmetry results from the fact that I did not specify which of the sides of the triangle was x and which was y.

Thankyou

sfwc
<><

The solution is faulty because it's not exact. Your construction, while ingenious, only leads to a good approximation that might have satisfied the Greek oracles and gods, but will not satisfy a mathematician.

Bill Branch
Flagstaff, AZ
bill.branch@gmail.com

I do like it because the error is finer than any printing or writing utility can represent. Better yet, it does not require a ruler-just a straight edge and compass.

Thanks for looking at it!

Bill

If and when it'll come to that I'll look up this thread. This will not happen in the next few days, though.