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November 1

Dividing a circle into thirds using two parallel lines.

For a circle of radius r and two parallel lines that split the circle into three sections each of size 1/3 pi r^2, how far are they apart? (and thus from that, it can be calculated how far is each from the ends of the perpendicular diameter) Can lines of that type be constructed with compass and straight edge from the original circle? And other than a single diameter dividing it into two sections, what can be done for divisions by n parallel lines into n+1 equal sections? Naraht (talk) 12:55, 1 November 2022 (UTC)[reply]

It's fairly easy to calculate the areas if you divide the figure into triangles and wedges, but I'd need to draw a figure to explain how I did it exactly. I get the condition (hopefully I made no mistake...)

r^{2}\arcsin {\frac {h}{r}}+2h{\sqrt {r^{2}-h^{2}}}={\frac {\pi }{3}}r^{2}

(wrong, see below)

Here, h is the distance of a line from the centre of the circle (2h is what you're looking for). The radius can be set to r=1 for convenience.

Solving numerically, I find h = 0.3627133 r. --Wrongfilter (talk) 13:23, 1 November 2022 (UTC)[reply]

OK, I agree that the center piece can be divided into two halves and then what is on one side of the diameter is two wedges and a triangle. But that doesn't seem to work as it would make the center stripe 72% of the diameter if I'm reading the numbers correctly. I would expect the center stripe to be less than one third since it is the one that goes all the way from one side to the other.Naraht (talk) 13:27, 1 November 2022 (UTC)[reply]

72% of the radius, 36% of the diameter. --Wrongfilter (talk) 13:35, 1 November 2022 (UTC)[reply]

Still seems wrong for it to be 36% of the diameter. Remember that is the longest stripe.Naraht (talk) 13:47, 1 November 2022 (UTC)[reply]

Should it be

2\left(\arcsin \left({\frac {h}{r}}\right)+{\frac {h}{r}}{\sqrt {1-\left({\frac {h}{r}}\right)^{2}}}\right)={\frac {\pi }{3}}

? catslash (talk) 14:04, 1 November 2022 (UTC)[reply]

Yes, you're right, I forgot to double the wedge... This solves to h = 0.2649249 r, which is more plausible. There's the article Circular segment, which should have everything (different notation though). --Wrongfilter (talk) 14:25, 1 November 2022 (UTC)[reply]

Using the formula for area "[i]n terms of

R

and

a

" at Circular segment § Arc length and area, I find (using your notation) h = 0.2649320846 r. The height of each of the two circular segments is then 0.7350679154 r, whereas the central strip has width 0.5298641692 r. I very much doubt that any case of dividing into three or more equal-area strips lends itself to straightedge and compass construction. --Lambiam 17:10, 1 November 2022 (UTC)[reply]

OK, so basically the ratio in the perpendicular diameter is *roughly* 3::2::3 (73.5::53::73.5)) that *feels* about right. And if arccos/arcsin are in the equation, straightedge/compass goes *right* out the window. Thanx!Naraht (talk) 17:19, 1 November 2022 (UTC)[reply]

The last statement is not necessarily true. The equation

\arcsin \left({\frac {h}{r}}\right)={\frac {\pi }{3}}

is solved by

h={\frac {r}{2}}{\sqrt {3}}.

A combination of inverse trigonometric functions and square roots tends to have only transcendental zeros, though. --Lambiam 19:34, 1 November 2022 (UTC)[reply]

7::5::7 and 43::31::43 are nicer one- and two-digit rational approximations. If we had sufficient digits for h, would it be possible to guess that it is the solution to a transcendental equation by examining its rational approximates? catslash (talk) 13:55, 2 November 2022 (UTC)[reply]

The sequence of best rational approximations (assuming Lambiam's numbers are accurate to the last digit shown) begins 1:1, 3:2, 4:3, 7:5, 18:13, 25:18, 43:31, 240:173, 1003:723, 1243:896, 2246:1619, 21457:15467, 23703:17086 —Tamfang (talk) 16:29, 2 November 2022 (UTC)[reply]

All convergents of the continued-fraction expansion are best rational approximations, but there may be best rational approximations that do not occur in the sequence of convergents. In this case, 11 : 8 is also a best rational approximation, better than 7 : 5. --Lambiam 17:12, 2 November 2022 (UTC)[reply]

I tend to forget that the method of continued fractions embodies mediant rounding, which minimizes not the absolute error but (roughly) the product of that error with sqrt(num,den). —Tamfang (talk) 18:18, 2 November 2022 (UTC)[reply]

You need to find a circular segment whose area is 1/3 of the whole circle. The section Circular segment#Arc length and area might help :) CiaPan (talk) 18:30, 2 November 2022 (UTC)[reply]

if only someone had thought of that yesterday —Tamfang (talk) 19:05, 2 November 2022 (UTC)[reply]