Challenge Questions

There must be a large number of ways to do this?

So I thought about some selection criteria:
Economy of construction
Resistance to constructional errors
Accessibility to historic geometric analysis

There are probably many more...

But so far I've signally failed to succeed in even two of these simultaneously.

First, a lawyer's approach to the problem.

Make L = R. (L²/R²)-∏ = 1 – ∏ = –2.14159... < 0.000000005.

Okay, that really is not a solution. All good engineers and scientists would know the absolute value of the difference needs to be less than 0.000000005.

One approach would be to construct a line of length L that is ∏ time R. One approximation to ∏ (that was used in the Resistance of Pi Challenge) is 335/113. This however would give an error of 0.00000027 which is 53 time too large. Srinivasa Ramanujan's approximation is (9² + 19²/22)^1/4 or (2143/22)^1/4 which I wouldn't know how to construct given the available tools.

Good Challenge Question. I'll keep trying but I don't expect to find a solution.

Thanks,

Jim

Checked.

Beautiful!

GA (so far)

If we then mark off on a straight line, n of these line segment lengths.

I think that needs to be 2ⁿ of these line segments. I'm afraid that makes this construction a bit impractical.

(check my math please)

Yep this is a standard way to construct a square the same area as a rectangle.

It's fairly easy to prove from first principles, but, it's also just a special case of the crossing chords thing a*b=x*y

Thanks,

Yes it should be 2ⁿ line segments. That is 16384 line segments and yes that would be impractical. But anything with a compass and straight edge that has an error of less than one is something like a half billion is impractical.

I am sure there must be methods to get a factor of ∏ with less steps but they elude me.

Thanks,

Jim

GA - but some thoughts:
. This requires a minimum angle of less than a minute; but it could be the seed for a method that requires a minimum angle of about 1.5^O, which in turn can be used to develop a method that can work at about 10^O. (This would seem good enough to me, but perhaps larger minimum angles would be possible if we also use the exterior polygon - I haven't checked).
. The last part looks to occupy a lot of surface - might we use the geometric mean of 2 and ∏/2 to develop √∏?

"compass that holds its size..." I assume that means all arcs or circles used in construct must be same radius as circle. Is this correct?

Wow, I am usually pretty good at this sort of thing. Math simple on this one, but the construct with only one measurement... Uffda.

Coming up (so far) with construct that matches math to within .15, nowhere near required 8 decimal places. Not sure my proof is correct, anyway.

If I knew there was gonna be a test, I woulda studied!

I think it means can be set to what you wish and holds its size thereafter (in contrast to Euclid's compass that collapses when you take it off the paper). The intent would presumably be to simplify the drawing process - and its description.

One way to "square the circle" is shown here. It is a quite lengthy procedure, but it appears that it the square area represents the circle area to within 0.000000001.

Srinivasa Ramanujan. construct.

Regards JD.

after squaring a circle you will expect me to trisect an angle?

Make the square Pi times Pi which equals pi x D/4.

this any help to someone not at work like I am

No help, unfortunately. That would give pi²xD²/4.

I think the question is not being construed correctly by those trying to answer. We are challenged to find a construct with "construct to have the same required-precision as Srinivasa Ramanujan". SR construct uses 355/113 for P), and any solution should presumably do the same. We are, in my opinion not required to do better. This means we must somehow generate 255/113

A brute force way of doing this would be to

draw a unit length (arbitrary) line, use this to generate binary lengths

1 Form line 355 from binary lengths 256,128,8,4,1 and line 113 from 64,32,16,1

2 Divide the lengths by Similar triangles 355/113 = ~Pi/1

3 Find √~Pi by crossing chords (Randall #11) Start with rectangle ~Pi x 1

4 a = √~Pi*r by similar triangles.

I am sure a more efficient way can be devised by other members. Perhaps the old challenge of forming Pi with resistors can somehow be turned into geometry?

355/113 predates SR by over 1400 years*, and only gives an accuracy of about 3.10^-7. The requested accuracy is 5.10^-9, which corresponds to a practical truncation of one of SR's own geometrical methods.

*What Ramanujan did in respect of the 355/113 approximation can be found here (page5). Basically, he showed a practical way to produce a Euclidean construction for this.

(355/113-∏)/∏<8.492x10exp(-8)

The challenge was quite specific: the error measure was L²/R²-∏.

This gives (335/113-∏), which is as previously stated - 2.66764E-7, or about 3.10^-7

I stand corrected.

Kudos - it is often harder to recognise and admit an error than to be right in the first place.

I will try to do this without diagrams so please excuse me if it is a bit confusing. Consider the following three points A, B & C. A is any point on the circumference of a circle. B is a point at end of straight line of length (pi.R)/2 from A which passes through the centre of the circle. C is one of the two points (doesn't matter which one) where line perpendicular to line AB which intersects AB at B crosses the circumference. It can be shown (I can do the working out if someone wants but would prefer not to spend the space here) that the length of AC is (sqrt pi).R so a square with sides equal to this would have the same volume as the circle. So now all we have to do is construct a line whose length is equal to (pi.R)/2 accurate to 6 decimal places. Luckily someone has already worked out a value for pi which is the fraction 355/113 which is accurate to 6 decimal places. Now given any arbitrary circle radius (R) it is relatively easy to construct a length which is 355/113 of this using a straight edge and compass. To start draw a very long line and starting at a point (say P1) measure out 355 lengths of R along the line using the compass and call that point P2. The line P1P2 is 355R long. Now divide this line into 113 equal segments (method for doing this with straight edge and compass is relatively simple and can be found on the net). Each of these segments has length (355R)/113. Now divide one of these segments in half and apply in the initial construction to get an approximation to (sqrt pi).R that has the required accuracy.

"B is a point at end of straight line of length (pi.R)/2 from A"

I have only a compass and an unmarked straight edge. I don't have a measuring stick.

From BobD's post:-

So now all we have to do is construct a line whose length is equal to (pi.R)/2 accurate to 6 decimal places.

So, can I use the Poincare plane?

This is more of an exercise in how not to do it rather than: how to do it, but, it does nearly work.

Root 10 is 3.162 so first note that √(10-3/(23+1/144)) is close to pi.

In fact I'm a bit confused because

√(10-3/(23+1/144))–п = 0.0000000297

Where as 355/113-п is = 0.0000002667

So although I don't meet the requirements of the challenge I think this is nearly ten times as good as Ramanujan's construction (When I say good I mean in terms of accuracy: obviously this is hideous where as Ramanujan's is beautiful).

First copy R 9 times so that we have a line of length 10R, and draw a random line to use to create a "divider" for the last 3 Rs

Divide the random line into quarters (23 is one less than 24 which is ¾ of 32)

Divide the third quarter by 8 (one 32^nd part of the random line).

Copy this length to the "origin" of the line.

Tidy everything up.

Focus on the 24^th section and draw a new random line to use as a new "divider".

Divide the new random line by eight.

Extend the line then copy one eighth to the extension (144 – 128 is 16 which is 1/8^th of 128)

Tidy up again

Focus on bottom section and divide into 128 of second random line. I wish I hadn't started this!

Tidy up again, you can't quite see the crucial section near the "origin" of the new random line.

Join the end of the random line to the "top" of the 24^th section of the original random line, and then draw a line parallel to it at the 1/144 point.

Detail:-

Truncate the original random line.

And Tidy up again

Join the end of truncated line to the point 3R from the end of the 10R line and draw a parallel from the ~1/23 point.

Truncate the original 10 R line.

And tidy up again: the line is now approximately pi²R long.

Draw the rectangle with area pi² R²

And construct the square with area pi² R². See post #11

At this point I thought it had all gone horribly wrong, but, I consulted my original post it note. The rectangle. At the bottom of the square has area ~ pi R².

Construct the square with the same area.

Phew. It's easy to see that I could have further divided one of those 1/144 line segments to get better resolution.

Clearly this would be impossible on a sheet of paper.

Ramanujan created many constructions - and the one referred to in the challenge is a truncated multiple-square-root arrangement whose original maths was also due to Ramanujan. This contrasts with the 355/113 construction which was just(?) an elegant tour-de-force to show that apparently difficult numbers could sometimes be created tidily.

I beleive that the interest in seeing how the 355/113 approximation could be achieved tidily was mainly due to its antiquity.

This is an interesting answer, nonetheless.

why should i think about things i could never do?

You would certainly have a dull life if you never thought about anything you could never do.

Of course the other reason is that some "things you could never do" are useful to develop or refresh techniques that are relevant to "things an engineer should be able to do" - but in a form that removes distractions that make it difficult to concentrate on the technique itself.

In this particular case the two methods found in this thread are 'successive approximation' and 'higher order cancellation'; both are methods that are extensively used in the solution of critical engineering problems.

I have not found a single method that equals Ramajudan for all the criteria in post #1, but I have found some that I find interesting. What is distinct about these is that they are optimised for accuracy using drawing - and none of them would be particularly helpful if you were using numerical methods.

First family (roughly described "pour encourager Les Autres" (anyone know him):
These are extensions of the proposal in jim's post #6 (as first hinted in post #12).
The second in the family (post #6 being the first) uses the fact that each time we halve the angle the accuracy of the approximation improves by roughly a factor of four; this allows us to use weighted differences of successive approximations to obtain an improved set of approximations where each halving of the angles pair gives a result that improves by a factor 16x. The idea can be repeated on this second set of approximations to obtain a third set that improves 64x for each halving of angles.
If we combine this third set with the exterior polygon we can in principle cancel the 6th-order term, and produce a set that improves at a rate of 256x for each halving of the angles. However, I haven't checked whether the initial error is small enough to mean that we get a reduction in the total number of angles to be combined.
But hopefully this is enough of a hint that someone else will get there before me - so I'll leave this for another day.

The second family goes directly to the side or to the diagonal of the square - R.sqrt(pi). It is based on numerical approximations, but these are selected to avoid the difficulties of drawing very small circles (my not-recent experience is that this is rather tricky using a practical compass). To this end, the largest compass setting is less than 2.6.R, and the smallest is greater than R/8. The number of iterations is minimised by (compared with with linear approximations) by using quite unequal sides of right-angled triangles to generate the third side.
I will give one example, that produces to the diagonal of the square (nominal length = √(2.∏).R ≈ 2.506628274631.R). In what follows, all lengths are taken as multiples of R (omitted due to laziness):
Triangle1: sqrt((1+1/6)²-((1+1/5)/8)²) = 2.161468...
Triangle2: sqrt(1²+1²) = 1.414214...
Triangle3: sqrt(1.4142136²+(2.161468/16)²) = 0.35516282746...
Triangle4: sqrt(1.5²-0.3551628²) = 1.4573567...
Triangle5: sqrt(2.5²+(1.4573567/8)²) = 2.5066282738...
This is in no way a fundamental method, nor is it as compact as the best of Ramanujan's nor as elegant as any that he published. However, it is less subject to drawing error than any method that takes the geometric mean of two lengths. Indeed, it is hard to see how the drawing errors could be further reduced, as every stage divides the errors due to the previous stages; most importantly, triangle 5 divides the constructional errors derived from triangle 4 by a factor of more than 100.

As it's nearly time for the official answer, and no-one has risen to the bait, here are some (slightly) more practical methods (than in post #6), but still based on chords

In post #6, Jim effectively used the approximation:
π/2 ~ a/2.sin(π/a)
Where in his case a was an integer (the number of sides of the regular inscribed polygon).

Taylor series expansion of pi gives this approximation as:
π/2 ~ a/2.[(π/a) - (π/a)³/3! + (π/a)⁵/5! - (π/a)⁷/7! + (π/a)⁹/9! - ...]
or (to give an idea of the error)
π/2 ~ π /2.[1 - (π/a)²/3! + (π/a)⁴/5! - (π/a)⁶/7! + (π/a)⁸/9! - ...] . . . (1)

Clearly, there is no problem with doubling tripling quadrupling (etc) the angle of the polygon. So we can obtain some improved estimates:

π/2 ~ a/2.[4.sin(π/a)/3 – sin(2.π/a)/6] eliminate the second order error term in equation (1) above.
This enables a polygon with 240-sides (1.5O) as the basis, compared with more than 32000 sides originally. Just 90 chords must be added/subtracted to produce π.r/2 with a theoretical error (absolute) of 3.07.10^-9/2

We can re-apply the principle to eliminate the fourth-order error term:
π/2 ~ a/2.{4[4.sin(π/a)/3 – sin(2.π/a)/6]/3 - [4.sin(2.π/a)/6 – sin(4.π/a)/12]/3}
This allows us to use a polygon with just 48 sides as the basis. We need to add/subtract 26 full chords plus a third of the result from three chords to produce π.r/2 with a theoretical (absolute) error of 1.76.10^-9/2

Theoretically, we could extend this to eliminate the sixth-order term, but we can see that the 8.π/a will inevitably be greater than unity, which limits any possible benefit. Smaller multiples (e.g. 1, 2, 3, 4) would doubtless be more useful in this regard – but the constructions become progressively more complex.
Alternatively, we could use the exterior polygon (side = r.tan(π/a)) to generate some of the segments.
These extensions do not appear worthwhile at the present level of precision – but they would come into their own beyond about 11 digits.

Although I describe the above equations as "Taylor series", they can also be accessed using basic algebra and similar triangles (this requires a degree of persistence, however). So – at least in principle - both the methods and an upper bound on the errors could in principle have been accessible to Euclid's contemporaries.

Author's comments:
Economical: modestly so
Resistance to constructional errors: modest
Historically accessible: probably

P.S. Other techniques needed (e.g. for the square root) are described elsewhere in this thread - e.g. by Jim and Randall

Were it possible, I'd give a GA to JD's "official challenge solution".
Not only is it technically correct, it is relatively compact.

Perhaps it's an added attraction that it still leaves the way open for other ideas, as it was not (so far as I could see) in any way unique or fundamental.

Other notes: The use of 1/cos near the end will divide preceding errors by about 3, so the construction of the diagonal of the square (= (D+AB) ≈ √(2.pi)) will be quite accurate. (Curiously, post #31 uses the same technique at the end of the process, but is based there on the 2.5xR instead of on 0.5xR - this smaller angle in #31 reduces preceding errors by a factor around 6)

an very simple method - the excat accuracy is unknown at the moment:

draw rectangle axis as coordinate system

draw a circle of radius R with center in the coordinaqte-zero-point and an outer square with the same sidelength

draw a line from the circle-center to the edge of the outer square (45 degrees)

half the line from the circle-peripheral to the edge of the outer square

draw a 60 degrees line from the circle center into the same direction as the 45 degree line

draw a line from the halfed line (circle peripheral to the edge of the outer square) thru the crosspoint of the 60 degree line with the circle peripheral

the line on the coordinatesystem to the coordinate zero point is half the length of a square with (nearly) the area equal the circle with radius R

I tried quite hard but could not be certain what these instructions meant.

"Outer square with same sidelength." Same sidelength as circle radius? Is this square centred on the origin? If so it's inside the circle. First guess it had one corner at the origin, and two of the sides coincide with coordinate axes. (This is the same as making the side of the square the same as the diameter)

But now we have a "45-degree line to the edge"... Perhaps you mean the corner?

If that's what you meant, you can use similar triangles and simple arithmetic to deduce the lengths of the side of your final square, and its area is 3.123445.R², giving an error slightly exceeding 0.018.

draw a square with the diameter D of the circle around the circle with sidelength's parallel to the axis of the coordinate systems.

I've some problems here to read the pages correctly, most times i have an error messages so i cannot read the comment complete.

That's what I guessed. The only reasons for not believing that this was what you meant were that the error was completely outside the range requested in the challenge (given that there are sqrt(3)*(1+1/42) equally quick ways to obtain much more accurate answers), and the error calculation or this construction was so straightforward.