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Square with area of circle: CR4 Challenge (09/29/09)

Posted September 27, 2009 5:01 PM

This week's Challenge Question:

Given a circle of any given radius "R", and an adjustable compass that holds its size and not calibrated in any way, plus a straight edge also not calibrated or marked in any way, and the ruler and compass to be used separately. Show how to construct a square of lengths "L" that has an area that approximates the area of the circle of diameter "D". Your construct is to differ from that of Srinivasa Ramanujan construct. But your construct to have the same required precision as Srinivasa Ramanujan, ( L²/R²)-∏ < 0.000000005.)

Thanks to jdretired for providing us with this puzzle.

And the Answer is...

Scribe circle C1 with centre O, scribe horizontal line WE and vertical line NS both passing through circle C1 centre O, scribe line VZ at 45 deg and line XY at 45 deg both passing through the centre of circle C1 at O, construct square VXYZ. Scribe lines S1,W1 and W1,N1 and N1,E1, divide O,W1,V,N1 into 4 equal parts by joining up the intersections, and include the diagonal bottom right and top left. With compass set to radius V,T scribe circle C2, extend line Z,V to intersect C2 at M.

With compass set to C1 radius, at centre S scribe an arc intersecting circle C1 at R1, scribe a line from O to R1 intersecting line V,Y at P. Scribe a line from P parallel to W,E to intersect circle C1 at Q. Scribe a line from Q to M, intersecting lines W1,N1 at A, line Y,X at B and the bottom right diagonal at C, with compass set to radius C,A at centre Z scribe circle C3, extend line V,Z to intersect circle C3 at R2. With compass set to radius O,R2 scribe an arc from R2 to intersect an extension of E,W at F. and an extend of line N,S at G. The distance from F to G equals L being the length of a side of a square that would approximate the area of the circle C1.

Maths:

In the maths example D = 10, therefore L using ∏ = 8.862269255.

Calculate length Y,Z. = √ ((10²)/2) = 7.071067812

Calculate length H,S2 = ((Y,Z)/8)x5) = 4.419417382

Calculate height S2,O = ((Y,Z)/2) x tan 30 = 2.041241452

Calculate S2,Q = √(R² -(S2,O)²) = 4.564354646

Length H,Q = H,S2 + S2,Q = 8.983772028.

Height H,M = H,S2 + S2,O = 6.460658835.

Find angle Z,M,Q = atan(H,Q/H,M) - 45 = 9.278287358 deg

Length A,B = (R/2)/cosM = 2.533141376

Note angle M = 9.278287358 deg is also a constant for this construct.

Find length L = √((D+AB)²)/2) = 8.862269257

Therefore the length L has an error of 0.000000002

Accuracy ratio (L²/R² = 3.141592655) - (∏ = 3.141592653) = 0.000000002

Therefore ratio error to pi = 0.000000002

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#1

Re: Square with area of circle: CR4 Challenge (09/29/09)

09/28/2009 11:24 AM

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.

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#2

Re: Square with area of circle: CR4 Challenge (09/29/09)

09/28/2009 12:46 PM

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

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#6
In reply to #2

Re: Square with area of circle: CR4 Challenge (09/29/09)

09/28/2009 6:41 PM

Consider the quarter circle in the following figure. It has a radius of R. If a line is drawn connecting the two ends of the quarter circle arc (X1), this is a crude approximation of the arc length (21/2R versus ∏/2*R or approximately 1.4142 R versus 1.5708 R). If we bisect this angle, we can construct the line X2, which has a length of approximately 0.7654 R versus 0.7854 for the arc it subtends. If we keep bisecting the angles and comparing the linear approximations to the subtended arcs, we get closer and closer, percentage wise. After enough continued steps, the length of the line segment is close enough to the arc length to be good enough. If the angle is defined as Θ = ∏/2n, then for n = 14 our error is less than that required of the challenge. If we then mark off on a straight line, n of these line segment lengths, we get our approximation to ∏ R.

Now we need to get a line of length ~∏1/2 R. Draw a rectangle of width ~∏ R and height R. Construct a circle with radius ~∏ R/2 centered at the middle of the bottom side. Also construct a circle with radius R centered at the bottom left hand corner. Where the last circle intersects the bottom side, construct a perpendicular line to the top side and extend to the first circle. By magic (check my math please), the length of the line from the center of the small circle to the intersection on the large circle is ~∏1/2 R.

Construct a square with a side of this length and the area is ~ ∏ R2.

Thanks,

Jim

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#10
In reply to #6

Re: Square with area of circle: CR4 Challenge (09/29/09)

09/29/2009 5:24 AM

Checked.

Beautiful!

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#11
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Re: Square with area of circle: CR4 Challenge (09/29/09)

09/29/2009 5:35 AM

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

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#15
In reply to #11

Re: Square with area of circle: CR4 Challenge (09/29/09)

09/29/2009 10:37 AM

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

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#12
In reply to #6

Re: Square with area of circle: CR4 Challenge (09/29/09)

09/29/2009 6:14 AM

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 1.5O, which in turn can be used to develop a method that can work at about 10O. (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 √∏?

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#33
In reply to #2

Re: Square with area of circle: CR4 Challenge (09/29/09)

10/05/2009 4:18 PM

There does seem to be a small problem with the dimensions if we follow Ramanujan's equation too literally, and this wouldn't be an acceptable solution anyway. I don't know how Ramanujan would have performed this construction (anyone have a reference?), but perhaps we could aim initially to generate it as 3.(R.(1+(192/92/22))1/4).

The following brute-force method should then be OK:

Starting from the innermost bracket:

To generate R.192/92/22
r.19/18=r.(1+1/18) is constructible and of reasonable size.
We can square this by drawing a tangent of that length to the original circle of radius R, drawing a circle radius one centred on the end of the tangent, and joining the end of the tangent to to the intersection of the circles. The distance between the second intersection of the circles and the end of the tangent is the length we require.

Doubling this length and then dividing by 11 should be straightforward.
Extend by this length by the length of a single radius R, and we have R.(1+(192/92/22)).

Take the square root of R.(R.(1+(192/92/22)) using your established method.
Repeat the procedure on the result..

You now have a line of length approximately pi.R/3. Multiply by 1.5 to generate (theoretically) R.(p - 1.00715E-9)/2
Now use your "square-rooting" technique to take the geometric mean of 2.R and your approximation to R.pi/2 - the side of your square.

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#34
In reply to #33

Re: Square with area of circle: CR4 Challenge (09/29/09)

10/06/2009 3:08 AM

In case anyone else is trying to follow this

........"To generate R.192/92/22"..........

Should read ........"To generate R.192/92/22"..........

(Left me baffled for ages, but that doesn't take much)

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#35
In reply to #33

Re: Square with area of circle: CR4 Challenge (09/29/09)

10/06/2009 4:15 AM

Again for any one else trying to follow, this is the construction to square a line:-

The grid is 1 unit; the original circle has radius 2; the tangent at the bottom of the original circle is 2.111111... units long (= 2*(1+1/18), and, the highlighted line is the required square of that tangent.

For me (at least) this is a great thing to take away from this thread:- Thanks Fyz.

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#36
In reply to #35

Re: Square with area of circle: CR4 Challenge (09/29/09)

10/06/2009 5:37 AM

Thanks, Randall.
Though actually my intention was r.(19/18)2x2 rather than r.(19/9)2/2. Combined with using r.pi/2 as the intermediary instead of r.pi, this avoids the requirement for using any dimension that is greater than 2.R (and we can even use the original circle to generate the square if we wish).

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#3

Re: Square with area of circle: CR4 Challenge (09/29/09)

09/28/2009 5:41 PM

"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!

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#7
In reply to #3

Re: Square with area of circle: CR4 Challenge (09/29/09)

09/28/2009 6:57 PM

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.

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#4

Re: Square with area of circle: CR4 Challenge (09/29/09)

09/28/2009 6:31 PM

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.

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#5

Re: Square with area of circle: CR4 Challenge (09/29/09)

09/28/2009 6:40 PM

Srinivasa Ramanujan. construct.

Regards JD.

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#8

Re: Square with area of circle: CR4 Challenge (09/29/09)

09/28/2009 11:40 PM

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

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#9
In reply to #8

Re: Square with area of circle: CR4 Challenge (09/29/09)

09/29/2009 4:28 AM
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#13
In reply to #8

Re: Square with area of circle: CR4 Challenge (09/29/09)

09/29/2009 9:14 AM

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

this any help to someone not at work like I am

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#14
In reply to #13

Re: Square with area of circle: CR4 Challenge (09/29/09)

09/29/2009 9:54 AM

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

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#20
In reply to #8

Re: Square with area of circle: CR4 Challenge (09/29/09)

10/01/2009 6:18 PM

That's easy within the limits of this challenge.

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#16

Re: Square with area of circle: CR4 Challenge (09/29/09)

09/30/2009 1:56 PM

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?

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#24
In reply to #16

Re: Square with area of circle: CR4 Challenge (09/29/09)

10/02/2009 6:07 AM

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.

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#25
In reply to #24

Re: Square with area of circle: CR4 Challenge (09/29/09)

10/02/2009 7:14 AM

(355/113-∏)/∏<8.492x10exp(-8)

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#26
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Re: Square with area of circle: CR4 Challenge (09/29/09)

10/02/2009 7:37 AM

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

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#27
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Re: Square with area of circle: CR4 Challenge (09/29/09)

10/02/2009 12:28 PM

I stand corrected.

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#28
In reply to #27

Re: Square with area of circle: CR4 Challenge (09/29/09)

10/02/2009 12:41 PM

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

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#17

Re: Square with area of circle: CR4 Challenge (09/29/09)

09/30/2009 11:55 PM

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.

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#18
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Re: Square with area of circle: CR4 Challenge (09/29/09)

10/01/2009 8:05 AM

"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.

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#19
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Re: Square with area of circle: CR4 Challenge (09/29/09)

10/01/2009 10:05 AM

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.

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#21

Re: Square with area of circle: CR4 Challenge (09/29/09)

10/01/2009 6:20 PM

So, can I use the Poincare plane?

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#22
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Re: Square with area of circle: CR4 Challenge (09/29/09)

10/01/2009 7:28 PM

I had to look that up. At my age there's no damned way am I going to mess up my head with the black art of the tensors.

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#23
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Re: Square with area of circle: CR4 Challenge (09/29/09)

10/01/2009 7:36 PM

I agree. I have enough trouble trying to keep Euclidean space straight in my alleged mind. I was just "poking him in the eye with a sharp stick" as the old folks say. I think he appreciates good clean fun (I hope so).

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#29

Re: Square with area of circle: CR4 Challenge (09/29/09)

10/02/2009 1:28 PM

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 32nd part of the random line).

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

Tidy everything up.

Focus on the 24th 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 1/8th 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 24th 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.

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#30
In reply to #29

Re: Square with area of circle: CR4 Challenge (09/29/09)

10/02/2009 5:52 PM

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.

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#46
In reply to #29

Re: Square with area of circle: CR4 Challenge (09/29/09)

11/05/2009 12:31 AM

why should i think about things i could never do?

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#47
In reply to #46

Re: Square with area of circle: CR4 Challenge (09/29/09)

11/05/2009 4:16 AM

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.

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#50
In reply to #47

Re: Square with area of circle: CR4 Challenge (09/29/09)

12/08/2009 2:18 PM

oh - i've problems enough to solve in my every day work that i could do in reality, if i had nothing to do all the days i could think about anything i don't have to do (or just anything else).

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#31

Re: Square with area of circle: CR4 Challenge (09/29/09)

10/02/2009 6:16 PM

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.

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#32

Re: Square with area of circle: CR4 Challenge (09/29/09)

10/05/2009 11:53 AM

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

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#37

Re: Square with area of circle: CR4 Challenge (09/29/09)

10/06/2009 1:12 PM

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)

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#38
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Re: Square with area of circle: CR4 Challenge (09/29/09)

10/07/2009 4:44 AM

Perhaps JD could enter a reply with the single word "BOW" and at least two of us would award it a good answer.

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#39
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Re: Square with area of circle: CR4 Challenge (09/29/09)

10/07/2009 4:56 AM

Thank you gentlemen, from two very, very worthy people.

Regards JD.

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Re: Square with area of circle: CR4 Challenge (09/29/09)

10/07/2009 5:03 AM

Leaving the off-topic checked rather defeats the objective of Randall's request...
(But I've given it a GA anyway).

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#41

Re: Square with area of circle: CR4 Challenge (09/29/09)

10/07/2009 5:40 AM

I'd like to thank those that have contributed to this challenge, particularly the remarks of Fyz ad Randall (BOW), and those that have put in some time working on the problem but did not not come up with a answer, frustrating isn't it?

Regards JD.

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#42

Re: Square with area of circle: CR4 Challenge (09/29/09)

11/01/2009 2:07 AM

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

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#43
In reply to #42

Re: Square with area of circle: CR4 Challenge (09/29/09)

11/01/2009 4:40 PM

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.

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#44
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Re: Square with area of circle: CR4 Challenge (09/29/09)

11/01/2009 11:58 PM

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.

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#45
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Re: Square with area of circle: CR4 Challenge (09/29/09)

11/02/2009 9:36 AM

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.

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#48

Re: Square with area of circle: CR4 Challenge (09/29/09)

12/06/2009 11:34 AM

Here is another problem:

gibven a rectangular coordinate system, draw a circle with the radius 100, draw an outer square to this circle (with a sidelength of 100, droaw a (diagonal) line from one edge of the outer square trough the coordinate zero point (the center f the radius) to the opposite site of the circle circumference, divide the line from the edge of the outer square to the opposite site of the circle into eight equal parts; one part of this 8part line from the edge of the outer square (where the diagonal line begins) on the diagonal line is an edge of an square with the the circumference as the circle.

for an circle with the radius 100 there is a difference for a half diagonal of the square of +.17 (or +.12 for a half squareside)

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#49
In reply to #48

Re: Square with area of circle: CR4 Challenge (09/29/09)

12/06/2009 11:47 AM

Presumably the square has a side-length of 200, not 100 as written?

If I draw the diagonal line, the length from the corner of the square to the further edge of the circle is 100*(1+√2). After this point, I cannot follow your directions

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#51
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Re: Square with area of circle: CR4 Challenge (09/29/09)

12/08/2009 2:29 PM

the circle has a radius of 100 - that is a diameter of 200, the square has the same corcumference 2*pi*r=3.1415927*200=628.31854, the square has a sidelength of a quarter of the circles circumference, 157.079635 or half of that is nearly 78.504, the suare results from the descrition has a sidelength of nearly 157.32 (the difference of 0.17 in half the diagonal is 0.12 in half the sidelength or 0.24 in the sidelength).

The difference to the circumference is less than +1 percent; this is lower than most tolerances in microelectronics - and that works fine!

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#52
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Re: Square with area of circle: CR4 Challenge (09/29/09)

12/08/2009 3:27 PM

What I wrote is that I've no idea how to draw what you are proposing.
Instead of clarifying what your construction is you've written a heap of claims for the quality.

Although what you wrote is opaque (at least to me), I imagine that what you are doing is one of the many ways of writing ∏ ≈1+1.5.√2 + 0.020272. Simple, and effective if that is all that is required.

Naturally, non-critical digital electronics is perfectly happy with Voltage tolerances of 10%. "High quality" audio DACs start at 14 bits encoding, but differences can (supposedly) be heard up to 20-bits. That's about 1-ppm. And other areas of electronics (such as navigation timing) work at the ppb level (timing). Horses for courses.

The "course requirement" for this challenge was about 1.5-ppb. By that definition your 0.6% "achievement" is way off topic.

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#53
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Re: Square with area of circle: CR4 Challenge (09/29/09)

12/09/2009 12:25 AM

divide that line of length 2.414 into eight equal parts and take the first part from the edge of the outer square as edge of the circumference equal square to the circle.

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#54
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Re: Square with area of circle: CR4 Challenge (09/29/09)

12/13/2009 5:22 PM

Please explain more carefully.

As written, I can only interpret this to mean using 1/8th of the length of the line as the side of the square - which gives a perimeter of about 1.2071 units.

I would assume that what you actually intend will be equivalent to adding 1/2 of the radius to 3/4 of the diagonal of the square (4(1/2+3/4√2) ≈6.2426.

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#55
In reply to #54

Re: Square with area of circle: CR4 Challenge (09/29/09)

12/13/2009 11:59 PM

The line from one edge of the outer square to the opposite site fo the circle diameter divided into eight equal parts and one part of this line from the edge of the outer square results in an edge of a square with nearly the same sidelength like the circumferences quarter.

this linelength divided into 8 equal parts is (1.414+1)/8 =2,414/8 times the radius of the circle; 7 eights of this is 7/8*(1.414+1) times r, half the diagonal of the resulting square is (7/8*(1.414+1) - 1)r and half the sidelength is 0.707*(7/8*(1.414+1)-1)r!

the exact value of that sidelength is a=2*pi*r/4.

adebar

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#56
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Re: Square with area of circle: CR4 Challenge (09/29/09)

12/14/2009 6:39 AM

OK, your side length is 7/4-1/4/√2, which gives an error just over 0.15%. That is a lot of construction to get a length error that is nearly 4x greater than the schoolboy's (1+4/7) [0.04%].

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#57
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Re: Square with area of circle: CR4 Challenge (09/29/09)

01/15/2010 11:10 PM

Geogebra as research aid

Introduction: During the late 20th century and the early 21st century India became a power house in software development. We believe this happened because India embarked on an educational program producing a large number of B.E.s and M.C.A.s. Both these programs require mathematical skill. Both had emphasis on programming, developing a machine-executable version of a specification. Program development requires deductive reasoning. Now the field of software development is moving away from programming to developing specifications. Specifications involve generalization from examples. In other words the future software development requires skill in inductive reasoning.

One educational tool recently introduced is GeoGebra. This allows interaction using pictorial inputs and algebraic expressions. We give references to sources of GeoGebra system and also to articles explaining its use. Since the need of future is inductive reasoning, we give an instant of using GeoGebra in research. A challenging problem in geometry is to develop a procedure to convert a given circle to a square of equal area. The mathematical prodigy Ramanujam solved this. We have solved the inverse problem of finding a circle for a given square. It is further demonstrated that for functions with inverse, it is possible to use the value of the function and its inverse to get greater accuracy.

We hope that GeoGebra would find extensive use among our school and college students. We expect that our computer experts would develop the GeoGebra system and its extensions in multiple languages in conformance with the educational system in India.

Literature: Aykan Kursat Erbas [1] in a dissertation towards Ph.D. states:

The value of algebra goes beyond academic study. Algebraic literacy is also a medium for social justice and job and later opportunities.

Like Aykan, John Cyril Ridd [2] in his Ph.D. dissertation considers the contribution of algebra for 'problem solving.'

Here the consequence of algebraic literacy is that not only are all questions involving cost price, selling price and percentage profit or loss all possible, but they can all be solved by the same approach: form an equation, substitute the information and solve. Because that general approach is applicable to an enormous variety of situations, the implications of algebraic literacy are so large as to become a fundamental change in a student's approach to mathematics, in particular to 'problem solving.'

Notwithstanding the importance of algebra, mass literacy in algebra is lacking. Marya D [3] gives the following explanation.

When industrialization was taking place, universal literacy became important for progress. What "literacy" meant for math, back then, was basic arithmetic. Now it looks like the civilization needs, among other things, mass algebraic literacy to progress and to solve its problems. Governments have been working toward that goal, by their usual means (influencing school curricula). Despite this mass algebra instruction, we don't see mass algebraic literacy.

I think one of the reasons why we don't see mass algebraic literacy is that algebra isn't viewed as something you MESS with. Kids doodle or compose texts for their myspace, but they don't MAKE (construct, create, build) any algebra entities, ever.

A system to 'mess with algebra' is GeoGebra. A brief description from the 'about' page [4] of the GeoGebra web is given below.

GeoGebra is a dynamic mathematics software for schools that joins geometry, algebra and calculus.

On the one hand, GeoGebra is an interactive geometry system. You can do constructions with points, vectors, segments, lines, conic sections as well as functions and change them dynamically afterwards.

On the other hand, equations and coordinates can be entered directly. Thus, GeoGebra has the ability to deal with variables for numbers, vectors and points, finds derivatives and integrals of functions and offers commands like Root or Extremum.

These two views are characteristic of GeoGebra: an expression in the algebra window corresponds to an object in the geometry window and vice versa.

Research is facilitated by a system that helps answer 'what if?' questions. GeoGebra is designed to mess with algebra. Thus using GeoGebra helps the students learn and practice inductive reasoning.

Research using GeoGebra: Algebra has infinite precision. However any computer system by necessity has limited precision. GeoGebra offers two views of the same object: a geometric view and an algebraic view. Because of limitations of the display device the geometric view is limited in accuracy while the algebraic equations suffer limited precision due to the limited precision of the numbers that are given as constants in the algebraic expressions. We urge the reader to experiment with the GeoGebra system to feel the limitations. In this paper we give the steps to create the objects using GeoGebra.

Circle à Square: Ramanujam [5] gave an intuitive procedure to draw a square of equal area to a given circle. In view of the difficulty in getting the above paper, we quote the procedure below (from page 367 of the article):

Let AB … be a diameter of a circle whose center is O.

Bisect AO at M and trisect OB at T.

Draw TP perpendicular to AB and meeting the circumference at P.

Draw a chord BQ equal to PT and join AQ.

Draw OS and TR parallel to BQ and meeting AQ at S and R respectively.

Draw a chord AD equal to AS and a tangent AC = RS.

Join BC, BD, and CD and draw EX, parallel to CD, meeting BC at X.

Then the square on BX is very nearly equal to the area of the circle, the error being less than a tenth of an inch when the diameter is 40 miles long.

We draw equivalent squares for circles with areas of 1, 1000000, and 1000000000000 respectively. A circle of radius 0.56418958354775628694807945156077 has an area of 1. The 'Construction Protocol' for this example follows.

No. Name Definition Algebra
1 Point A A = (0, 0)
2 Circle c Circle with center A and Radius 0.56 c : x

2 + y2 = 0.32

We gave the radius to be 0.56418958354775628694807945156077 and the system reports just two digits to the right of the decimal point. It is possible that the system stores more digits but reports just 2 digits in the definition.
3 Number areac Area[c] areac = 1
4 Text Textc "Area=" + areac Textc = "Area = 1"
5 Point B Intersection point of c, xAxis B = (-0.56, 0)
5 Point C Intersection point of c, xAxis C = (0.56, 0)
Geogebra gives all results of a single geometric operation with the same row number.
6 Point D D = (0, 0.3)
7 Point E E = (0, 0.4)
8 Point F F = (0, 0.6)
9 Segment a Segment[F,C] a = 0.82
We need to select two points to define the segment. We selected point F first and thus the definition lists F followed by C. The length of the segment is 0.82. This information is of no use for us.
10 Line b Line through E parallel to a b: 0.6x + 0.56y = 0.23
11 Point G Intersection point of b, xAxis G = (0.38, 0)
Triangle FAC is similar to triangle EAG and thus the point G trisects CA.
12 Segment d Segment[F,B] d = 0.82
13 Line e Line through D parallel to d e: 0.6x –0.56y = -0.17
14 Point H Intersection point of e, xAxis H = (-0.28, 0)
Triangles FAB and DAH are similar and H bisects AB. Our points of interest are G and H. Thus hide points D, E, F and segments a, and d and line b, and e.
15 Line f Line through G perpendicular to xAxis f: x = 0.38
16 Point I Intersection point of c, f I = (0.38, -0.42)
16 Point J Intersection point of c, f J = (0.38, 0.42)
Point J is of interest to us. Hide line f and point I.
17 Segment g Segment[J,G] g = 0.42
We have connected the segment between points J and G to show the figure given by Ramanujam in his 1914 paper as far as possible.
18 Line h Line through J parallel to xAxis h: y = 0.42
19 Line i. Line through C parallel to yAxis i: x = 0.56
20 Point K Intersection point of h, i. K = (0.56, 0.42)
21 Circle k Circle with center C through K k: (x – 0.56)

2 + y2 = 0.18

22 Point L Intersection point of c, k L = (0.41, 0.39)
22 Point M Intersection point of c, k M = (0.41, -0.39)
23 Segment j Segment[L,C] j = 0.42
Segment j is of interest to us. Hide lines I, and h; points K, and M; and circle k.
24 Segment l Segment[B,L] l = 1.05
25 Line m Line through A parallel to j m: 0.39x + 0.16y = 0
26 Line n Line through G parallel to j n: 0.23x + 0.09y = 0.09
27 Point N Intersection point of m, l N = (-0.08, 0.2)
28 Point O Intersection point of m, l O = (0.25, 0.33)
29 Segment p Segment[N,A] p = 0.21
30 Segment q Segment[O,G] q = 0.35
Hide lines m, and n.
31 Circle r Circle with center B through N r: (x + 0.56)

2 + y2 = 0.27

32 Point P Intersection of r, c P = (-0.32, 0.46)
32 Point Q Intersection of r, c Q = (-0.32, -0.46)
33 Segment s Segment[B,Q] s = 0.52
Hide circle r, and point p.
34 Line t Line through N parallel to yAxis t: x = -0.08
35 Line a

1

Line through B parallel to yAxis a

1: x = -0.56

36 Circle c

1

Circle with center N through O c

1: (x + 0.08)2 + (y – 0.2)2 = 0.12

37 Point R Intersection point of c

1, t

R = (-0.08, 0.54)
37 Point S Intersection point of c

1, t

S = (-0.08, -0.15)
38 Line b

1

Line through S parallel to l b

1: -0.39x + 0.97y = -0.12

39 Point T Intersection point of a

1, b1

T = (-0.56, -0.35)
40 Segment d

1

Segment[B,T] d

1 = 0.35

Hide lines t, a

1, and b1; circle c1; and points R and S.

41 Segment e

1

Segment[T,Q] e

1 = 0.27

42 Segment f

1

Segment[C,Q] f

1 = 1

43 Segment g

1

Segment[C,T] g

1 = 1.18

We are unable to proceed now. Ramanujam has not given any instruction for fixing the point E on the chord BD. A review of the instructions reveals that there is no action taken on point M which bisects the chord AO. We presume that performing some action on point M gets point E. We first try to see whether E is on the line perpendicular to the diagonal AB on the chord BD.

No. Name Definition Algebra
44 Line h

1

Line through H perpendicular to xAxis h

1: x = -0.28

45 Point U Intersection point of h

1, f1

U = (-0.28, 0.44)
46 Line i

1

Line through U parallel to e

1

i

1: 0.06x + 0.13y = -0.07

47 Point V Intersection point of g

1, i1

V = (-0.51, -0.33)
48 Quadrilateral poly1 Polygon[V,C,4] Poly1 = 1.27
48 Point W Polygon[V,C,4] W = (0.23, 1.08)
48 Point Z Polygon[V,C,4] Z = (-0.85, 0.74)
48 Segment v Segment[V,C] of Quadrilateral poly1 v = 1.13
48 Segment k

1

Segment[C,W] of Quadrilateral poly1 k

1 = 1.13

48 Segment w Segment[W,Z] of Quadrilateral poly1 w = 1.13
48 Segment z Segment[Z,V] of Quadrilateral poly1 z = 1.13
49 Text Textpoly1 "Area =" + poly1 Textpoly1 = "Area = 1.27"

Since the area of the resultant square is 1.27, our assumption that point E lies on the perpendicular to the diagonal through M is wrong. We undo all the steps from 44 till 49 and try the case when E lies on a line perpendicular to the chord BD through point M.

No. Name Definition Algebra
44 Line h

1

Line through H perpendicular to f

1

h

1: 0.53x + 0.28y = -0.15

45 Point U Intersection point of h

1, f1

U = (-0.1,-0.35)
46 Line i

1

Line through U parallel to e

1

i

1: 0.07x + 0.15y = -0.06

47 Point V Intersection point of g

1, i1

V = (-0.28, -0.26)
48 Quadrilateral poly1 Polygon[V,C,4] Poly1 = 0.78
48 Point W Polygon[V,C,4] W = (0.3, 0.85)
48 Point Z Polygon[V,C,4] Z = (-0.54, 0.58)
48 Segment v Segment[V,C] of Quadrilateral poly1 v = 0.89
48 Segment k

1

Segment[C,W] of Quadrilateral poly1 k

1 = 0.89

48 Segment w Segment[W,Z] of Quadrilateral poly1 w = 0.89
48 Segment z Segment[Z,V] of Quadrilateral poly1 z = 0.89
49 Text Textpoly1 "Area =" + poly1 Textpoly1 = "Area = 0.78"

We have not succeeded as the area is just 0.78. It is possible that BE equals BM. We try this option now after undoing steps from 44 on. This yields a square with unit area. Thus we complete the procedure given by Ramanujam as below.

Let AB … be a diameter of a circle whose center is O.

Bisect AO at M and trisect OB at T.

Draw TP perpendicular to AB and meeting the circumference at P.

Draw a chord BQ equal to PT and join AQ.

Draw OS and TR parallel to BQ and meeting AQ at S and R respectively.

Draw a chord AD equal to AS and a tangent AC = RS.

Join BC, BD, and CD.

Mark E on BD such that BE equals BM and draw EX, parallel to CD, meeting BC at X.

Then the square on BX is very nearly equal to the area of the circle, the error being less than a tenth of an inch when the diameter is 40 miles long.

The remaining steps of the 'Construction Protocol' follow.

No. Name Definition Algebra
44 Circle h

1

Circle with center C through H h

1: (x – 0.56)2 + y2 = 0.72

45 Point U Intersection point of h

1, f1

U = (-0.19,-0.39)
46 Line i

1

Line through U parallel to e

1

i

1: 0.11x + 0.24y = -0.12

47 Point V Intersection point of g

1, i1

V = (-0.39, -0.3)
48 Quadrilateral poly1 Polygon[V,C,4] Poly1 = 1
48 Point W Polygon[V,C,4] W = (0.27, 0.96)
48 Point Z Polygon[V,C,4] Z = (-0.69, 0.66)
48 Segment v Segment[V,C] of Quadrilateral poly1 v = 1
48 Segment k

1

Segment[C,W] of Quadrilateral poly1 k

1 = 1

48 Segment w Segment[W,Z] of Quadrilateral poly1 w = 1
48 Segment z Segment[Z,V] of Quadrilateral poly1 z = 1
49 Text Textpoly1 "Area =" + poly1 Textpoly1 = "Area = 1"

Starting with a circle of unit area, we get a square of unit area. Still Ramanujam indicates that his method is not perfect. The precision used by GeoGebra for this example is not high enough to indicate the mismatch. We try with a circle of area 1000000. We need not rework the problem from scratch. In the GeoGebra file for the unit area circle, we redefine step 2. Instead of giving the radius as 0.56418958354775628694807945156077, we give the radius as 564.18958354775628694807945156077. We need to 'zoom out' to get the display right. The resultant square has an area of 1000000.08 giving 8-digit accuracy for Ramanujam's method. Similarly for a circle of area 1000000000000 Ramanujam's method gives a square of area 1000000084913.68 with 8-digit accuracy. We conclude that Ramanujam's method gives an accuracy of 8-digits and this accuracy is not limited by the implementation of GeoGebra.

Square à Circle: We have developed the inverse method of getting a circle of equal area for a given square. Initially we describe the intuitive method. We work out examples for squares of areas 1, 1000000, and 1000000000000 respectively.

The steps to get circle from a given square are:

ABCD is the given square.

E, F, G, and H are midpoints of AB, AD, BC, and AC respectively.

Draw quarter circle e with center A passing through E, and F.

Draw quarter circle f with center B passing through G, and E.

Draw semi circle g with AB as diameter.

Point I is intersection of quarter circle e and semi circle g.

Point J is intersection of quarter circle f and semi circle g.

Draw circle h with center I and passing through J.

Draw circle k with center J and passing through I.

Point K is intersection of circles h, and k.

Draw segment i joining points D, and H.

Draw segment j joining points C, and H.

Draw circle p with center K and passing through H.

Point M is intersection of circle h with segment i.

Draw ray l from point M through point K.

Mark the intersections of circle p with ray l as point N (near point M) and point O (away from point M).

Draw ray m from point J through point O.

Mark the intersections of circle p with side CD of square as point Q (near M) and point P (away from M).

Point R is the intersection of circle k and segment j.

Draw segment n between points Q and R.

Point S is the intersection of segment n, and ray m.

Draw circle q with center H passing through S.

Circle q is the result.

We give the 'Construction Protocol' for drawing a circle of area 1 starting with a square of area 1 below.

No. Name Definition Algebra
1 Point A A = (-0.5, -0.5)
2 Point B B = (0.5, -0.5)
3 Quadrilateral poly1 Polygon[A, B, 4] poly1 = 1
3 Point C Polygon[A, B, 4] C = (0.5, 0.5)
3 Point D Polygon[A, B, 4] D = (-0.5, 0.5)
3 Segment a Segment{A, B] of Quadrilateral poly1 a = 1
3 Segment b Segment[B, C] of Quadrilateral poly1 b = 1
3 Segment c Segment[C, D] of Quadrilateral poly1 c = 1
3 Segment d Segment[D, A] of Quadrilateral poly1 d = 1
4 Point E midpoint of A, B E = (0, -0.5)
5 Point F midpoint of A, D F = (-0.5, 0)
6 Point G midpoint of B, C G = (0.5, 0)
7 Point H midpoint of A, C H = (0, 0)
8 Arc e CircularArc[A, E, F] e = 0.79
9 Arc f CircularArc[B, G, E] f = 0.79
10 Arc g Semicircle through A and B g = 1.57
11 Point I Intersection point of g, e I = (-0.25, -0.07)
12 Point J Intersection point of f, g J = (0.25, -0.07)
13 Circle h Circle with center I through J h: (x + 0.25)

2 + (y + 0.07)2 = 0.25

14 Circle k Circle with center J through I K: (x – 0.25)

2 + (y + 0.07)2 = 0.25

15 Point K Intersection of h, k K = (0, 0.37)
15 Point L Intersection of h, k L = (0, -0.5)
16 Segment i. Segment[D, H] i = 0.71
17 Segment j Segment[C, H] j = 0.71
18 Circle p Circle with center K through H P: x

2 + (y – 0.37)2 = 0.13

19 Point M Intersection point of h, i. M = (-0.41, 0.41)
20 Ray l Ray through M, K l: 0.04x + 0.41y = 0.15
No. Name Definition Algebra
21 Point N Intersection point of p, l N = (-0.36, 0.4)
21 Point O Intersection point of p, l O = (0.36, 0.33)
22 Ray m Ray through J, O M: -0.4x + 0.11y = -0.11
23 Point P Intersection point of p, c P = (0.34, 0.5)
23 Point Q Intersection point of p, c Q = (-0.34, 0.5)
24 Point R Intersection point of k, j R = (0.41, 0.41)
25 Segment n Segment[Q, R] n = 0.75
26 Point S Intersection point of n, m S = (0.39, 0.41)
27 Circle q Circle with center H through S Q: x

2 + y2 = 0.32

28 Text Textpoly1 "Area = " + poly1 Textpoly1 = "Area = 1"
29 Number areaq Area[q] areaq = 1
30 Text Textq "Area = " + areaq Textq = "Area = 1"

Redefining the points A and B as (-500, -500) and (500, -500) gives the area of the square as 1000000, while that of the resultant circle is just 999924.59 giving at best 5-digit accuracy. Changing the points A and B to (-500000, -500000) and (500000, -500000) gives a square of area 1000000000000 and the equivalent circle of area 999924591399.52 again with 5-digit accuracy only.

Increasing the accuracy – concept: We propose the following hypothesis. Let functions f and g be inverse of each other. That is, y = f(x); and x = g(y). Given the finite precision of computers, we would rarely find x = g( f(x) ). Let y1 = f (x1). Calculate x2 = g (y1). Now define a new function g( f(x) ) = ( x + g( f(x) ) )/2. Mathematically this is always true. This is because simplifying the definition, we get g( f(x) ) / 2 = x / 2. However due to numerical errors, the value g ( f(x1) ) would be different from the value ( x1 + g( f(x1) ) )/2.

For example consider the drawing of the square given a circle (Ramanujam's method) as function f, and the drawing of a circle given a square (our method) as g, then we define a new method of finding a circle for a square as follows.

1. Find a square given a circle using Ramanujam's method.

2. For the resultant square, find a circle using our method.

3. The new function defines a circle with average of the radii of the initial circle and the final circle as the more accurate result for the square.

4. Use similar triangles and find the radius of the resultant circle for the given square.

Ramanujam's method gives a square of area 1000000084913.68 for a circle of area 1000000000000. Our method gives a circle of area 999924591399.52 for a square of area 1000000000000. The circles of both methods are less in area than the corresponding squares. In this situation, finding the average improves the accuracy of our method while the result is worse than the accuracy of Ramanujam's method. This is because the errors are non-compensating. To get compensating errors, one method should give a circle of area larger than the square. Since our method has less accuracy, we fine tune the intuitive method algorithmically to get a circle that is larger in area than the given square.

Modified square à circle: The resultant circle in our method has point S, (point of intersection of segment QR with ray JO) on the circle with the center at point H. If we rotate the segment QR with center at R in a clockwise direction, and consider the new point of intersection with ray JO, the area of the resultant circle would be larger. We continue with our method below.

No. Name Definition Algebra
31 Point T Intersection point of q, c T = (0.26, 0.5)
31 Point U Intersection point of q, c U = (-0.26, 0.5)
32 Segment r Segment[U,R] r = 0.68
33 Point V Intersection point of r, m V = (0.39, 0.41)
34 Circle s Circle with center H through V S: x

2 + y2 = 0.32

35 Number areas Area[s] Areas = 1
36 Text Texts "Area = " + areas Texts = "Area = 1"

We redefine the points A and B to be (-500000, -500000), and (500000, -500000) and find that the area of the circle s is 1000849918452.57. The area of circle s is greater than the area of the square. Thus, we need to identify a point between the points Q, and U such that the average of the circles has an area as close to that of the square as possible.

Increasing accuracy – demonstration: We start with a circle; find the square using Ramanujam's method; find a circle for this square using our method; select a point between the two points corresponding to points Q, and U and get a circle with area larger than the square; we find the average of the radii of the circles and draw the result circle. We compare the accuracy of the transformations. The 'Construction Protocol' is given below.

No. Name Definition Algebra
1 Point A A = (0, 0)
2 Circle c Circle with center A and radius 40000000 C: x

2 + y2 = 1600000000000000

3 Point B Intersection point of c, xAxis B = (-40000000, 0)
3 Point C Intersection point of c, xAxis C = (40000000, 0)
4 Point D Midpoint of B, A D = (-20000000, 0)
5 Point E E = (0, 30000000)
6 Point F F = (0, 20000000)
7 Segment a Segment[E, C] a = 50000000
8 Line b Line through F parallel to a B: 30000000x + 40000000y = 800000000000000
9 Point G Intersection point of b, xAxis G = (26666666.67, 0)
Hide points E, F; segment a; and line b
10 Line d Line through G perpendicular to xAxis d: x = 26666666.67
11 Point H Intersection point of c, d H = (26666666.67, -29814239.7)
11 Point I Intersection point of c, d I = (26666666.67, 29814239.7)
12 Line e Line through C parallel to yAxis e: x = 40000000
13 Line f Line through I parallel to xAxis f: y = 29814239.7
14 Point J Intersection point of f, e J = (40000000, 29814239.7)
15 Circle g Circle with center C through J g: (x – 40000000)

2 + y2 = 888888888888888.8

16 Point K Intersection point of c, g K = (28888888.89, 27666443.55)
16 Point L Intersection point of c, g L = (28888888.89, -27666443.55)
17 Segment h Segment[K, C] h = 29814239.7
Hide lines d, e, and f; circle g; and points H, I, J, and L.
18 Segment i. Segment[B, K] i = 74236858.17
19 Line j Line through G parallel to h J: 829993306532582.1x + 333333333333333.4y = 22133154840868860000000
20 Line k Line through A parallel to h k: 27666443.55x + 11111111.11y = 0.0
21 Point M Intersection point of i, j M = (17407407.41, 23055369.63)
22 Point N Intersection point of i, k N = (-5555555.56, 13833221.78)
Hide lines j, and k.
23 Segment l Segment[M, G] l = 24845199.75
24 Segment m Segment[N, A] m = 14907119.85
25 Circle p Circle with center B through N p: (x + 40000000)

2 + y2 = 1377777777777778.2

26 Point O Intersection point of p, c O = (-22777777.78, 32881192.79)
26 Point P Intersection point of p, c P = (-22777777.78, -32881192.79)
27 Segment n Segment[B, P] n = 37118429.09
Hide circle p, and point O.
28 Line q Line through N parallel to yAxis q: x = -5555555.56
29 Line r Line through B parallel to yAxis b: x = -40000000
30 Circle s Circle with center N through M s: (x + 5555555.56)

2 + (y – 13833221.78)2 = 612345679012345.9

31 Point Q Intersection point of s, q Q = (-5555555.56, 38578841.17)
31 Point R Intersection point of s, q R = (-5555555.56, -10912397.61)
32 Line t Line through R parallel to I t: -27666443.55x + 68888888.89y = 598040482625502.5
33 Point S Intersection point of r, t S = (-40000000, -24745619.39)
34 Segment a

1

Segment[B, S] a

1 = 24745619.39

Hide lines q, r, and t; circle s; and points Q, and R.
No. Name Definition Algebra
35 Segment b

1

Segment[S, P] b

1 = 19047112.45

36 Segment c

1

Segment[C, P] c

1 = 70867638.75

37 Segment d

1

Segment[C, S] d

1 = 83739749.7

38 Circle e

1

Circle with center C through D e

1: (x – 40000000)2 + y2 = 3600000000000000

39 Point T Intersection point of e

1, c1

T = (-13150729.06, -27838821,81)
40 Line f

1

Line through T parallel to b

1

F

1: 8135573.4x + 17222222.22y = -586435097287090

41 Point U Intersection point of d

1, f1

U = (-27731902.52, -20950848.51)
42 Point V Midpoint of U, C V = (6134048.74, -10475424.25)
43 Segment g

1

Segment[V, A] g

1 = 12139236.68

44 Line h

1

Line through U parallel to g

1

h

1: -11282245916307449000000x – 6606495799819703000000y = 451289836652298000000000000000

45 Line j

1

Line through A parallel to d

1

j

1: 24745619.39x – 80000000y = 0

46 Point W Intersection point of h

1, j1

W = (-33865951.26, -10475424.25)
47 Circle k

1

Circle with center A through W k

1: x2 + y2 = 1256637168141593.2

48 Point Z Intersection point of k

1, xAxis

Z = (-35449078.52, 0)
48 Point A

1

Intersection point of k

1, xAxis

A

1 = (35449078.52, 0)

49 Point B

1

Intersection point of k

1, yAxis

B

1 = (0, 35449078.52)

49 Point C

1

Intersection point of k

1, yAxis

C

1 = (0, -35449078.52)

50 Line i

1

Line through C

1 parallel to xAxis

i

1: y = -35449078.52

51 Line l

1

Line through Z parallel to yAxis l

1: x = -35449078.52

52 Line m

1

Line through A

1 parallel to yAxis

M

1: x = 35449078.52

53 Point D

1

Intersection point of l

1, i1

D

1 = (-35449078.52, -35449078.52)

54 Point E

1

Intersection point of m

1, i1

E

1 = (35449078.52, -35449078.52)

55 Quadrilateral poly1 Polygon[D

1, E1, 4]

poly1 = 5026548672566372
55 Point F

1

Polygon[D

1, E1, 4]

F

1 = (35449078.52, 35449078.52)

55 Point G

1

Polygon[D

1, E1, 4]

G

1 = (-35449078.52, 35449078.52)

55 Segment n

1

Segment[D

1, E1] of quadrilateral poly1

n

1 = 70898157.05

55 Segment p

1

Segment[E

1, F1] of quadrilateral poly1

p

1 = 70898157.05

55 Segment q

1

Segment[F

1, G1] of quadrilateral poly1

q

1 = 70898157.05

55 Segment r

1

Segment[G

1, D1] of quadrilateral poly1

r

1 = 70898157.05

56 Number areac Area[c] areac = 5026548245743669
57 Text Textc "Area =" + areac Textc = "Area = 5026548245743669"
58 Text Textpoly1 "Area =" + poly1 Textpoly1 = "Area = 5026548672566372"
Hide all except the points A, B, C, D

1, E1, F1, and G1; circle c; and square poly1.

59 Point H

1

Midpoint of D

1, E1

H

1 = (0, -35449078.52)

60 Point I

1

Intersection point of r

1, xAis

I

1 = (-35449078.52, 0)

61 Point J

1

Intersection point of p

1, xAxis

J

1 = (35449078.52, 0)

62 Arc s

1

CircularArc[D

1, H1, I1]

s

1 = 55683282.33

63 Arc t

1

CircularArc[E

1, J1, H1]

t

1 = 55683282.33

64 Arc c

2

Semicircle through D

1 and E1

c

2 = 111366564.66

65 Point K

1

Intersection point of s

1, c2

K

1 = (-17724539.26, -4749275.98)

66 Point L

1

Intersection point of t

1, c2

L

1 = (17724539.26, -4749275.98)

67 Circle d

2

Circle with center K

1 through L1

d

2: (x + 17724539.26)2 + (y + 4749275.98)2 = 1256637168141593.2

68 Circle e

2

Circle with center L

1 through K1

e

2: (x - 17724539.26)2 + (y + 4749275.98)2 = 1256637168141593.2

69 Point M

1

Intersection point of d

2, e2

M

1 = (0, 25950526.56)

69 Point N

1

Intersection point of d

2, e2

N

1 = (0, -35449078.52)

70 Circle f

2

Circle with center M

1 through A

f

2: x2 + (y – 25950526.56)2 = 673429828764945.6

71 Segment a

2

Segment[G

1, A]

a

2 = 50132567.62

72 Segment b

2

Segment[F

1, A]

b

2 = 50132567.62

73 Point O

1

Intersection point of d

2, a2

O

1 = (-28894114.88, 28894114.88)

No Name Definition Algebra
74 Ray g

2

Ray through O

1, M1

g

2: 2943588.32x + 28894114.88y = 749817495633202.2

75 Point P

1

Intersection point of f

2, g2

P

1 = (-25816901.78, 28580623.76)

75 Point Q

1

Intersection point of f

2, g2

Q

1 = (25816901.78, 23320429.36)

76 Ray h

2

Ray through L

1, Q1

h

2: -28069705.35x + 8092362.52y = -535955457404115

77 Point R

1

Intersection point of f

2 , q1

R

1 = (24149686.11, 35449078.52)

77 Point S

1

Intersection point of f

2 , q1

S

1 = (-24149686.11, 35449078.52)

78 Point T

1

Intersection point of e

2 , b2

T

1 = (28894114.88, 28894114.88)

79 Point U

1

Intersection point of c , q

1

U

1 = (18530052.13, 35449078.52)

79 Point V

1

Intersection point of c , q

1

V

1 = (-18530052.13, 35449078.52)

80 Line i

2

Line through S

1 parallel to yAxis

i

2: x = -24149686.11

81 Circle k

2

Circle with center S

1 and Radius 500000

k

2: (x + 24149686.11)2 + (y – 35449078.52)2 = 250000000000

82 Circle p

2

Circle with center S

1 and Radius 5547706.49

p

2: (x + 24149686.11)2 + (y – 35449078.52)2 = 30777047281435.46

The actual value of the radius of Circle p

2 was 5547706.4884 . Since GeoGebra rounds off the values to two decimal places it was reported as 5547706.49 in the Construction Protocol. See the table below for the sequence of the radii of circle p2 in the table that follows.

83 Point W

1

Intersection point of k

2 , i2

W

1 = (-24149686.11, 3594078.52)

83 Point Z

1

Intersection point of k

2 , i2

Z

1 = (-24149686.11, 3494078.52)

84 Point A

2

Intersection point of p

2 , i2

A

2 = (-24149686.11, 40996785.01)

84 Point B

2

Intersection point of p

2 , i2

B

2 = (-24149686.11, 29901372.03)

85 Segment j

2

Segment[B

2 , V1]

j

2 = 7896665.97

86 Line l

2

Line through Z

1 parallel to j2

l

2: -5547706.49x + 5619633.99y = 330376399779621.75

87 Point C

2

Intersection point of l

2 , q1

C

2 = (-23643203.48, 35449078.52)

88 Segment m

2

Segment[C

2 , T1]

m

2 = 52944663.27

89 Point D

2

Intersection point of m

2 , h2

D

2 = (27474820.74, 29071197.03)

90 Circle q

2

Circle with center A through D

2

q

2: x2 + y2 = 1600000271723777.2

91 Point E

2

Intersection point of q

2 , xAxis

E

2 = (-40000003.4, 0)

91 Point F

2

Intersection point of q

2 , xAxis

F

2 = (40000003.4, 0)

92 Point G

2

Midpoint of B, E

2

G

2 = (-40000001.7, 0)

93 Circle r

2

Circle with center A through G

2

r

2: x2 + y2 = 1600000135861886

94 Number arear

2

Area[r

2]

arear

2 = 5026548672566372

95 Text Textr

2

"Area = " + arear

2

Textr

2 = "Area = 5026548672566372"

96 Text text1 "Area = " + poly1 Text1 = "Area = 5026548672566372"

The radius of circle p2 used in the above Construction Protocol was found after 24 iterations. The first value was 5500000 giving the area of the resultant circle as 5026550333806841, which was greater than that of the square. We tried 5550000 and found that the resulting circle had an area of 5026548593427671 less than that of the square. Since GeoGebra allows us to redefine the inputs given by us, we need to just make one change using the Construction Protocol and GeoGebra recalculates all the dependent values and we get the final answer. We have manually followed an approximate binary search. The search process is documented below. We see that the last row gives the area of the circle to be 5026548672566372 giving a 16-digit accuracy.

Area of square is 5026548672566372

No. Radius of circle p

2

Area of final circle
1 5500000

5026550333806841

2 5550000

5026548593427671

The correct circle has radius between 5500000 and 5550000.
3 5549000

5026548627925024

The desired radius is between 5549000 and 5500000.
4 5548500

5026548645178410

The answer is between 5548500 and 5500000.
5 5548000

5026548662434930

Try between 5548000 and 5500000.
6 5547500

5026548679694593

The answer is larger than 5547500 but less than 5548000.
7 5547700

5026548672790349

The desired radius is between 5547700 and 5548000.
8 5547750

5026548671064369

Try between 5547700 and 5547750.
9 5547710

5026548672445152

Check between 5547700 and 5547710.
10 5547705

5026548672617750

The answer is larger than 5547705 but smaller than 5547710.
11 5547708

5026548672514192

The desired radius is between 5547705 and 5547708.
12 5547707

5026548672548711

Try a value greater than 5547705 but less than 5547707.
13 5547706

5026548672583231

Check between 5547706 and 5547707. We need to add fractions now.
14 5547706.5

5026548672565970

The desired radius is between 5547706 and 5547706.5.
15 5547706.3

5026548672572873

Check between 5547706.3 and 5547706.5.
16 5547706.4

5026548672569422

Try a value greater than 5547706.4 but less than 5547706.5.
17 5547706.45

5026548672567698

The answer is larger than 5547706.45 but smaller than 5547706.5.
18 5547706.48

5026548672566660

The desired radius is between 5547706.48 and 5547706.5.
19 5547706.49

5026548672566316

Check between 5547706.48 and 5547706.49.
20 5547706.485

5026548672566488

Try a value greater than 5547706.485 but less than 5547706.49. The GeoGebra does not display more than two decimals to the right of the decimal point. We need to remember the digits not displayed and continue. We record the value to be tried in this table first and then redefine in the GeoGebra system.
21 5547706.488

5026548672566385

The answer is larger than 547706.488 but smaller than 5547706.49.
22 5547706.489

5026548672566350

The desired radius is between 5547706.488 and 5547706.489.
23 5547706.4885

5026548672566368

Check between 5547706.488 and 5547706.4885.
24 5547706.4883

5026548672566376

Try a value greater than 5547706.4883 but less than 5547706.4885.
25 5547706.4884

5026548672566372

Before demonstrating the derivation of a circle of area equal to a given square, we calculate the errors due to Ramanujam's method and our method. The starting circle with a radius of 40000000 had an area of 5026548245743669. Ramanujam's method gave a square of area 5026548672566372 which is larger by 426822703. Starting with a square of area 5026548672566372 our modified method gave a circle of area 5026549099389091. The square is smaller than this circle by 426822719. The resultant circle was drawn using the average of the radii of the two circles giving the area equal to that of the square. In terms of accuracy, Ramanujam's method differs on the eighth digit, as also our method. However the average of the radii of the two circles gives 16 digit accuracy. We have demonstrated our hypothesis. It is hoped that other researchers and programmers would exploit the hypothesis in future.

Accurate square à circle: Let us find a circle with area equal to a square of side 60000000. The area of the square is 3600000000000000. Draw a right angled triangle AI1I2 where I2 is the point of intersection of the result circle and a line perpendicular to the x-Axis through the point I1. In the above triangle the side AI1 is equal to half the side of the square and the side AI2 is equal to the radius of a circle of equal area. We define point J2 at (-30000000, 0) making the side AJ2 equal to half the side of the given square. A perpendicular to x-Axis through J2 intersects the side AI2 at K2. The triangle AJ2K2 is similar to triangle AI1I2. Thus the side AK2 is equal to the radius of a circle with area of 3600000000000000. A circle with center A passing through point K2 is circle t2. The area of this circle is 3599999999999998.5 giving 16-digit accuracy the error being just 1,5 out of 3600000000000000. The steps from Construction Protocol are given below.

No. Name Definition Algebra
97 Point H

2

Intersection point of r

2 , r1

H

2 = (-35449078.52, 18530055.79)

97 Point I

2

Intersection point of r

2 , r1

I

2 = (-35449078.52, -18530055.79)

98 Segment n

2

Segment[I

2 , A]

n

2 = 40000001.7

99 Point J

2

J

2 = (-30000000, 0)

100 Line s

2

Line through J

2 perpendicular to xAxis

s

2: x = -30000000

101 Point K

2

Intersection point of s

2 , n2

K

2 = (-30000000, -15681696.03)

102 Circle t

2

Circle with center A through K

2

T

2: x2 + y2 = 1145915590261646

103 Number areat

2

Area[t

2]

Areat

2 = 3599999999999998.5

104 Text Textt

2

"Area = " + areat

2

Textt

2 = "Area = 3599999999999998.5"

We have the option to set the number of decimal places to be displayed in Construction Protocol. We worked with 2 decimal places. The current version allows up to 5 decimal places to be displayed. Thus selecting 5 decimals to be displayed would invalidate suggestion 4 that follows. We demonstrate that there are problems requiring more decimal places than any fixed limit given in GeoGebra. Instead of starting with the initial circle with radius 40000000 let us start with a circle with radius 4000000. The independent variables and their new values are given below.

No. Parameter Value Row in Construction Protocol
1 Point E (0, 3000000) 5
2 Point F (0, 2000000) 6
3 Point J

2

(-3000000, 0) 99
4 Radius of Circle k

2

50000 81
5 Radius of Circle p

2

554770.64884 82

If we set the initial radius to 400000, we get the following values.

No. Parameter Value Row in Construction Protocol
1 Point E (0, 300000) 5
2 Point F (0, 200000) 6
3 Point J

2

(-300000, 0) 99
4 Radius of Circle k

2

5000 81
5 Radius of Circle p

2

55477.064884 82

GeoGebra reports the radius of circle p2 as 55477.06488 because we have set the number of decimals to 5. In this problem itself, decreasing the radius of circle c to smaller and smaller values shall require more decimals to be reported. Thus it is required that GeoGebra reports the user input without any modification in the Construction Protocol.

Suggested improvements to GeoGebra: We have demonstrated that GeoGebra permits messing with algebra. The following suggestions are made based on our experience in writing this paper.

1. We spent lot more time in copying the 'Construction Protocol' than in using GeoGebra to solve the problems. This is because the 'Construction Protocol' could not be copied; it could be printed alone. The 'Construction Protocol' should be made exportable to a document as a table.

2. To avoid keying in the 'Construction Protocol' given as a table in a document, it should be possible to import a 'Construction Protocol.'

3. As a picture is worth a thousand words, to follow the steps involved in developing a solution, the 'Construction Protocol' of which was imported, we need to have the facility to execute a 'Construction Protocol' one step at a time.

4. We performed an approximate binary search on the parameter 'radius of circle p2.' It should be possible to specify a range for a number of desired input parameters and request the GeoGebra system to optimize some objective function.

In the next section we consider some techniques to modify the GeoGebra system.

Modifying GeoGebra:

GeoGebra could be downloaded from [6] freely. If there is to freedom change the GeoGebra system by any expert, there would be chaos. There is a forum to discuss possible changes to GeoGebra. The user forum is accessed through [7]. The forum has two sections, one dedicated to use issues, and the other dedicated to technical issues. Konrad Remus [8] discusses the possibility of using recursive functions in GeoGebra. From the discussion it is clear that the changes to the GeoGebra system are under the control of the system administrator.

We have suggested the possibility of exporting and importing the Construction Protocol to and from documents. GeoGebra community provides an online facility to upload GeoGebra programs, web pages and the like. The steps to be followed in uploading are described in [9]. Some of the uploaded materials are found in [10]. To facilitate the training in inductive reasoning, Markus Hohenwarter and Judith Preiner [11] have suggested methods to create dynamic web pages. Notwithstanding these online resources, we believe the modification of GeoGebra system to export and import the Construction Protocol is justified as the permanency of a journal article is orders of magnitude more than an online resource. An online resource becomes obsolete when the supporting technology becomes obsolete.

Conclusion: Computer code development is more or less automated. Thus future computer applications would be oriented towards developing the specifications of applications. Specifications development needs generalization requiring inductive reasoning. GeoGebra permits messing with algebra and geometry and helps the teachers in developing lesson plans to train the students in inductive reasoning. We have demonstrated that GeoGebra is a research aid too.

Dedication:

To be included after refereeing is completed but before publication.

References:

1. http://jwilson.coe.uga.edu/Pers/erbas_ayhan_k_200408_phd.pdf p19 of 366.

2. http://eprints.jcu.edu.au/1176/02/02whole.pdf p 81 of 222.

3. http://www.squarecirclez.com/blog/21st-century-computer-algebra-literacies/937

4. http://www.geogebra.org/cms/index.php?option=com_content&task=blogcategory&id=67&Itemid=63

5. Ramanujam, Approximate geometrical constructions for p, Quarterly Journal of Mathematics, XLV (1914) 350 – 374.

6. http://www.geogebra.org/cms/index.php?option=com_content&task=blogcategory&id=71&Itemid=55

7. http://www.geogebra.org/forum/

8. http://www.geogebra.org/forum/viewtopic.php?f=13&t=4913

9. http://www.geogebra.org/en/wiki/index.php/Upload_Materials

10. http://www.geogebra.org/en/upload/

11. Marcus Hohenwarter and Judith Preiner, Creating Mathlets with Open Source Tools, The Journal of Online Mathematics and Its Applications, 7 (July 2007) Article ID 1574. http://www.maa.org/joma/Volume7/Hohenwarter2/index.html

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