Talk:Area of a circle

addition proposal

WP:NOTFORUM The following discussion has been closed. Please do not modify it.

@CiaPan: Thank you for thinking along. I've added some extra explanation to support my theorem. Please note, that I'm not calculating $\pi$, what I'm doing, is comparing the circle directly to a square, instead of other polygons. Gmac4247 (talk) 20:42, 19 January 2021 (UTC)

@Gmac4247: You seem to miss the fact the proportionality coefficient between the area of a circle and the square of its radius is the same as between the circumference of the circle and its diameter (which is known at least since Archimedes). And that coefficient IS denoted by the Greek letter π. --CiaPan (talk) 20:55, 19 January 2021 (UTC)

@CiaPan: Yeah. Yet, they're still proportional. Gmac4247 (talk) 09:53, 20 January 2021 (UTC)

@Gmac4247: Please indent your entries properly. This helps much in following the talk when it forks into sub-threads.
Compare the content and the visible results in these three edits: Special:Diff/1001583645, Special:Diff/1001583999 and Special:Diff/1001584067. Please find more description at Help:Talk pages § Indentation. --CiaPan (talk) 10:05, 20 January 2021 (UTC)

@CiaPan:

okGmac4247 (talk) 11:32, 20 January 2021 (UTC)

@Gmac4247: In your "proof", you do not explain how c follows from a and b. And in fact, c is false: the sum of the areas of the four quarter circles (centered at the vertices and passing through a point halfway between the square's center and a side) is strictly less than the area of the square.

If you want to contribute to mathematical research, the proper way to do it would be to do what all people do when they are starting out: First discuss your work with your friends and colleagues to make sure that your argument is written so clearly that no one could misunderstood it. If you are not able to make the people around you understand it, then you should not proceed any further before rewriting. It may involve several rounds of back and forth before your argument is polished enough to be understandable. Once that is done, you should submit it to a mathematics journal. Posting your own research onto a Wikipedia article is something that you should never do, even if the research is correct - that should be left to others. Best wishes, Ebony Jackson (talk) 16:12, 20 January 2021 (UTC)

@Ebony Jackson:

Thank you. I've extended the proof on its talk page.Gmac4247 (talk) 20:58, 23 April 2021 (UTC)

 — Preceding unsigned comment added by Gmac4247 (talk • contribs) 08:00, 10 May 2021 (UTC) Gmac4247 (talk) 09:35, 15 May 2023 (UTC)

Disproval of the mathematical constant pi

WP:NOTFORUM The following discussion has been closed. Please do not modify it.

The idea of the mathematical constant $\pi$ is based on the assumption, that the circumference of a circle can be calculated from the difference of the perimeters of an inscribed and a circumscribed polygon. To test this theory, I start with squares.

300px

r1=radius of the small circle; r3=radius of the larger circle; a=side of the small polygon; A1/P1=area/perimeter of the small circle; A2/P2=area/perimeter of the small polygon; A3/P3=area/perimeter of the larger circle; A4/P4=area/perimeter of the larger polygon; c=coefficient of the area/perimeter of the circle $r1={\tfrac {a}{2}};r3={\tfrac {\sqrt {2}}{2}}*a;P1=2c*{\tfrac {a}{2}}=c*a;P2=4*a;P3=2c*{\tfrac {\sqrt {2}}{2}}*a=c*{\sqrt {2}}*a;P4=4*{\sqrt {2}}*a$

I continue with hexagons.

300px

$r1={\tfrac {\sqrt {3}}{2}}*a;r3=a;P1=2c*{\tfrac {\sqrt {3}}{2}}*a=c*{\sqrt {3}}*a;P2=6*a;P3=2c*a;P4=6*{\tfrac {2}{\sqrt {3}}}*a={\tfrac {12}{\sqrt {3}}}*a$

The number of the polygons' sides can be increased to infinite.

Problem #1: The mathematical constant $\pi$ is based on calculating with 71 side polygons. Such determination is a rough guess between 4 and infinite.

Problem #2: Despite of the difference decreases between the polygons' perimeter, as the number of their sides increases, the actual value of their perimeter can only be calculated with endless fractions. (See ${\sqrt {3}}$ above for instance.) That means decrease of accuracy.

Take the areas of the squares and the inscribed circles instead:

300px

$A1=c*{\tfrac {a^{2}}{4}};A2=a^{2};A3=c*{\tfrac {a^{2}}{2}};A4=2*a^{2};$

This proportion enables to exactly determine the area of the circle between the squares and vice versa: the square between the inscribed and the circumscribed circles.

Gmac4247 (talk) 09:27, 15 May 2023 (UTC)

As you have been repeatedly told, this is original research, which is forbidden by Wikipedia's core policies. You need to get your research published. After that, it needs to become notable. (Most research, including my own, is not.) After that, it can be incorporated into Wikipedia. Until then, your research should not be discussed on Wikipedia talk pages, because talk pages are for discussing impending edits to articles. Mgnbar (talk) 12:45, 16 May 2021 (UTC)

Your definition of c and value of P4 is wrong, but no serious math journal would publish your work even if you fix it. PrimeHunter (talk) 13:00, 16 May 2021 (UTC)

Five points.

Point 1. Sentence #1 is false. Twice.

The idea of the mathematical constant \pi is based on the assumption, that the circumference of a circle can be calculated from the difference of the perimeters of an inscribed and a circumscribed polygon.

There is NO such assumption, and the coefficient named pi is NOT calculated from a difference of any perimeters.

The properties of two sequences of perimeters allow us to infer the coefficient exists and to estimate (bound from both sides) its actual value. Additionally, the more sides of polygons, the better estimation (shorter interval in which the sought value is).

Point 2. Sentence #5 is clearly invented.

The mathematical constant \pi is based on calculating with 71 side polygons.

No grounds for this claim.

The number 71 appears in the well-known result of Archimedes, who calculated the bounding $3{\frac {10}{\color {darkred}71}}<\pi <3{\frac {1}{7}}$, but it is not a number of sides!

Point 3. Sentences #7 & 8 reveal an obvious misunderstanding.

the actual value of their perimeter can only be calculated with endless fractions. (See {\sqrt {3}} above for instance.) That means decrease of accuracy.

One doesn't need a square root to get an 'endless fraction', a simple ordinary quotient is enough: 1/3 = 0.3333333...

And this has nothing to do with accuracy, because we can calculate values of roots as accurate (as many exact digits) as we want.

Point 4. There is a mistake in the first drawing: a side of the big square is $a{\sqrt {2}}$, not $2a$.
As a result, the fourth equation in the second set should read:

$A1=c\cdot {\tfrac {a^{2}}{4}};A2=a^{2};$

$A3=c\cdot {\tfrac {a^{2}}{2}};\color {darkred}A4=2\cdot a^{2};$

and all you can infer from this system is:

$A3:A1=A4:A2=2$

$A1:A2=A3:A4={\tfrac {c}{4}}$

and

$A2<A3<A4$

hence

$a^{2}<c\cdot {\frac {a^{2}}{2}}<2a^{2}$

and finally

$2<c<4$

Honestly, not an impressive result. Especially today, in days of super-computers, compared to a twenty-two–centuries–old result by Archimedes (who died circa 212 BC).

Point 5. Please, please, PLEASE: cease bragging about your mind-blowing results, and start learning from old masters instead.
--CiaPan (talk) 21:20, 16 May 2021 (UTC)

{re|Mgnbar|CiaPan}

Thank you, the large square is corrected.

I don't know, if it was a 71, or a 96 side polygon, but that's not the point.

Gmac4247 (talk) 09:27, 17 May 2021 (UTC)

You have been asked not to edit earlier posts in talk page threads. Again, please just post a new, corrected version at the bottom of the thread. If this process seems cumbersome, because you are revising a large block of text, then please instead post that text in your personal userspace and edit it there. Mgnbar (talk) 12:03, 17 May 2021 (UTC)

I've asked two supercomputers about this. Here are their responses:

Screenshot of discussing the area of a circle with a supercomputer (Bing)

Screenshot of discussing the area of a circle with a supercomputer (Bard) Gmac4247 (talk) 13:25, 10 November 2023 (UTC)

A wonderful example of why LLMs cannot be relied on for facts. MrOllie (talk) 14:00, 10 November 2023 (UTC)

Unfortunately, I can't ask Mr Archimedes about it. I didn't find any notes about if he considered and dismissed this approach. Gmac4247 (talk) 16:03, 11 November 2023 (UTC)

Most probably Archimedes did not considered this approach, because he apparently understood, what he was trying to do. And if he tried it and then he left no notes about it, then it would mean the approach turned out counterproductive. --CiaPan (talk) 17:37, 13 November 2023 (UTC)

You've asked two digital superidiots. Please stop doing this. They may easily produce as long laudatory enunciations as you wish, based just on a statistical correlation of words you used in your question, but they have ZERO idea about abstract notions you use and their mathematical definitions. --CiaPan (talk) 17:34, 13 November 2023 (UTC)

Can this ref be considered valid?

Please see this addition by User:Ebony Jackson on 7 January 2014:

Special:Diff/589555823.

Is the page linked there (published at shreevatsa.wordpress.com) a reliable source? I'm afraid it counts as a primary source and as such it should be considered WP:OR.
--CiaPan (talk) 17:05, 18 January 2023 (UTC)

Hi CiaPan, you can remove the reference if you want. But I would leave the fact (that there is no better approximation with denominator <16604) there. The fact is routine to verify for anyone with computer literacy, so I would think it would fall under WP:CALC. Ebony Jackson (talk) 23:23, 22 January 2023 (UTC)

First paragraphs of onion proof and triangle proof

The first paragraph of onion proof refers to shell integration, but the derivation of shell integration in turn depends on the formula of area of ring which apparently relies on the formula of area of circle. Without referring to shell integration, it is actually complicated to justify "one can approximate this ring by a rectangle". Anyway, it's still a good introduction paragraph to onion proof.

Similar problem lies in the first paragraph of triangle proof that it is complicated to justify "unwrapping the concentric circles to straight strips". But in this case it is even worse because the next paragraph (dividing up a circle into triangles) follows a completely different idea. It would be better to let the reader be aware of this difference, like adding another picture of "dividing up a circle into triangles". Shenyqwilliam (talk) 05:17, 8 June 2024 (UTC)

I think the disk / circle unambiguation can go

Since a circle has no area, as the author points out, then anybody talking about the area of a "circle" must necessarily be talking about a disk. And we all know that. There's no confusion even possible about the area of a circle and of a disk, since circles don't have one. I've never heard of this being a problem. I'm a fan of "yes, but technically" as anyone, but it doesn't need saying in such a foundational article. It's more likely to create confusion than solve it. 2A00:23C5:2E2:1A01:A18D:F142:FCC0:F2E9 (talk) 22:09, 2 May 2025 (UTC)

In other words, the note is pedantic? I tend to agree. But I have learned over the years that Wikipedia math articles should err on the side of pedantry, when it serves the bigger purpose of correctness. Your "we all know that" applies only to a very narrow meaning of "we". So I support keeping the note. It's in its own little section, easily skipped over by the uninterested reader. Regards, Mgnbar (talk) 22:41, 2 May 2025 (UTC)

Even if you take the pedantry out, there are enough pedants reading Wikipedia that it will inevitably come back unless you carefully guard it. It's a losing battle, and the best bet is to include at least enough pedantic disclaimer to satisfy most of the would-be drive-by fixers. With that said though, "area enclosed by a circle" is pretty unambiguous and really doesn't need the disclaimer. –jacobolus (t) 03:18, 3 May 2025 (UTC)

Yes, the "area enclosed" in the first sentence is a great wording. But that's not the title of the article. And no, I am not trying to reopen that argument. :) Mgnbar (talk) 16:30, 3 May 2025 (UTC)