Why Pi is wrong

[crocque]

I'm not sure it's safe to post real theorems here.

Is it?

[crocque]

I'm new and not sure how to even ask. I feel like I'm on to something, but... there's that uncertainty.

[Pengwuino]

Dear god I hope your theorem is "pi = 4" or "pi is rational".

[crocque]

Certainly not, but this is why I'm hesitant to post. I'm in a beginning stage. I don't have the answer yet. I just see it.

[disregardthat]

What are you asking exactly? Post your theorems if you have any!

[Bourbaki1123]

Is it wrong because Tau is right?

I think you're going to need to explain what you even mean. I mean, Pi exists, there are algebraic methods for computing it out to n decimal places, so it's well defined. Do you mean that you don't think that Pi is the right constant to use in the equation Pi*radius^2 = area of the geometric object called a circle?

[daveyp225]

In what way is Pi wrong? There must be some specific context or definition of Pi that you are referring to?

[crocque]

I came up with this "Inside a regular inscribed hexagon, the radius of the circle is equal to the sides of the hexagon".

It was my beginning. I believe Pi is not needed and that it is not accurate. The area of a circle, or even better a sphere, can be done with no numbers, just variables. It's a work in progress. I feel I need a partner with good object and spacial orientation to understand where I am headed, less everyone will think I'm crazy, lol.

If I could just find one person that understands this part. I might convice you of the rest. I saw the theorem, but it took others besides me to prove it for me, with much persistence on my part. Now that I'm working on a new one, I feel it's ground breaking and it scares me.

I'm not asking anyone to care, but if you think there's any value in studying what I'm saying, by all means, talk to me.

[wukunlin]

that sounds very interesting

[cmb]

I came up with this "Inside a regular inscribed hexagon, the radius of the circle is equal to the sides of the hexagon".

It was my beginning. I believe Pi is not needed and that it is not accurate.

If you are talking about a regular hexagon in which a circle is inscribed that just grazes the sides of the hexagon (at their centres) then the radius of the circle is [SQRT(3)]/2 (~0.866) times the length of the sides of the hexagon.

If you are talking about a regular hexagon in which a circle is inscribed that passes through the apexes of the hexagon, then the radius of the circle is clearly the same length as the sides of the hexagon, because the line from the centre to an apex is one side of two of the equilateral triangles that form a set of 6 nested equilateral triangles forming the hexagon, and it is also a radius of the circle, so radius and all sides are identical.

This used to be the classic way of drawing a hexagon, with ruler and compasses, before there was an excess amount of computer power to make your brain go soft.

What's your point?

[disregardthat]

It is simply not true that a circle inscribed in a regular hexagon has a radius equal to the sides of the hexagon.

[crocque]

It simply is true. I proved it in the 10th grade. Check it out, disprove it, if you wish.

[daveyp225]

I came up with this "Inside a regular inscribed hexagon, the radius of the circle is equal to the sides of the hexagon".

It was my beginning. I believe Pi is not needed and that it is not accurate. The area of a circle, or even better a sphere, can be done with no numbers, just variables. It's a work in progress. I feel I need a partner with good object and spacial orientation to understand where I am headed, less everyone will think I'm crazy, lol.

If I could just find one person that understands this part. I might convice you of the rest. I saw the theorem, but it took others besides me to prove it for me, with much persistence on my part. Now that I'm working on a new one, I feel it's ground breaking and it scares me.

I'm not asking anyone to care, but if you think there's any value in studying what I'm saying, by all means, talk to me.

I'm just curious as to what you mean.

Do you believe Pi is not the correct number to describe the area of a circle of radius 1? That the ratio of the circumference to the diameter is not constant? What does the hexagon's perimeter imply about the area of the circle? Have you found anything "wrong" with current proofs?

[crocque]

See this is why I get scared, if someone does realize what I'm sayng they will run with it. Then again, it's my theorem. No one wanted to believe me then or now, still true.

[crocque]

I'm just curious as to what you mean.

Do you believe Pi is not the correct number to describe the area of a circle of radius 1? That the ratio of the circumference to the diameter is not constant? What does the hexagon's perimeter imply about the area of the circle? Have you found anything "wrong" with current proofs? No Pi isn't the right number, in my mind. I'm looking outside the cricle. No circle is perfect. That's why gemotery is the only way. Yeah we can use algebra to form curvatures, doen't make it correct. Think 3 dimensional.

I don't want to say everything because I feel only a certain type of person can see this.

[Jobrag]

It simply is true. I proved it in the 10th grade. Check it out, disprove it, if you wish.

Please post your proof

[disregardthat]

It simply is true. I proved it in the 10th grade. Check it out, disprove it, if you wish.

No, it's not. You probably proved it when the circle is circumscribed by the hexagon, not inscribed.

[crocque]

I didn't come to these forums to show off, more to meet people that are thinkers like myself. I don't have the background to discuss this with education on my side, it's something I see.

I really need to get into college again, but I'm 40, still I love learning.

[crocque]

No, it's not. You probably proved it when the circle is circumscribed by the hexagon, not inscribed. I asked you nicely if you do not belive me that disprove it. Please do not tell me what I already know.

I promise you they are equal. No need to get defensive.

[crocque]

Where's a professor when you need one?

[daveyp225]

No Pi isn't the right number, in my mind. I'm looking outside the cricle. No circle is perfect. That's why gemotery is the only way. Yeah we can use algebra to form curvatures, doen't make it correct. Think 3 dimensional.

I don't want to say everything because I feel only a certain type of person can see this.

What do you mean by "no circle is perfect"? Pi has many different definitions, so it looks like you've narrowed your disagreement down to: C/D is not a constant number. If you don't accept the calculus proofs, why not? I understand you don't want to reveal your "theorem" but you seem to be unintentionally trolling the forum. If I made a thread saying that I can clearly "see" that gravity actually did not exist but didn't want to communicate my idea, what would you think?

[crocque]

If you are talking about a regular hexagon in which a circle is inscribed that just grazes the sides of the hexagon (at their centres) then the radius of the circle is [SQRT(3)]/2 (~0.866) times the length of the sides of the hexagon.

If you are talking about a regular hexagon in which a circle is inscribed that passes through the apexes of the hexagon, then the radius of the circle is clearly the same length as the sides of the hexagon, because the line from the centre to an apex is one side of two of the equilateral triangles that form a set of 6 nested equilateral triangles forming the hexagon, and it is also a radius of the circle, so radius and all sides are identical.

This used to be the classic way of drawing a hexagon, with ruler and compasses, before there was an excess amount of computer power to make your brain go soft.

What's your point? You're getting close. Just think a bit more outside the circle.

[crocque]

What do you mean by "no circle is perfect"? Pi has many different definitions, so it looks like you've narrowed your disagreement down to: C/D is not a constant number. If you don't accept the calculus proofs, why not? I understand you don't want to reveal your "theorem" but you seem to be unintentionally trolling the forum. If I made a thread saying that I can clearly "see" that gravity actually did not exist but didn't want to communicate my idea, what would you think?

troll, really?, I just joined the forums. I joined to ask about this math problem that has been in my head forever. Thought maybe I should actually ask some knowledgable people. Forgive me, I'm just saying I think this can be done geometrically. This is not my first time debating this in life. I'm not trying to argue with anyone. I'm looking for help.

Really I wish this was a trool and I was crazy, it's not a troll but I might be crazy. I usually talk on the xbox forums, I came here to get real ideas.

But ok, see, if you have the triangular base of a sphere and know the radius. There is a way to use that radius and find out what the missing area is whether it is a circle or a sphere. I am working on it. Need more specifics? I've thought about it for 20 years.

It's not Pi. no 22/7

[crocque]

Circle seems to be different, not like a square, but it is. It follows the same rules. Just another geometrical object. This is what I am trying to achieve.

[micromass]

Crocque, could you please clearly state what you're thinking about??

And I would love to see the proof with that hexagon, do you mind giving it??

[crocque]

The proof to my theorem is simple. 1 line down, 1 line across, another line down to form one side of a hexagon. One radius up, one sideways, another one down. They both equal 3.

I'm not going to write out a proof, you end up with 6 equilateral triangles. I just saw it, I didn't need a proof, my teacher's helped me prove it. To be honest, I don't even remember the proof, but, like I said, prove me wrong. I think we did something with the angles and proved it. Was 20 years ago.

Chad's Theorem "Inside a regular inscribed hexagon, the radius of the circle is equal to the sides of the hexagon."

[HallsofIvy]

It simply is true. I proved it in the 10th grade. Check it out, disprove it, if you wish.
No, it's not. The radius of the inscribed circle is the distance from the center of the hexagon to the center of one side. The length of one side of the regular hexagon is equal to the distance from the center of the hexagon to a vertex of the hexagon. If we take "x" as the length of a side of the hexagon, then drawing the line segment from the center of the hexagon to the center of a side gives a right triangle with one leg of length x/2 and hypotenuse of length x. The length of the other leg is \sqrt{x^2- x^2/4}= x\sqrt{3}/2.

That is, the radius of a circle inscribed in a hexagon of side length x is
\frac{x\sqrt{3}}{2}

NOT equal to the side length. It is the radius on the circle circumscribed about a hexagon that is equal to a side of the hexagon.

[crocque]

No, it's not. The radius of the inscribed circle is the distance from the center of the hexagon to the center of one side. The length of one side of the regular hexagon is equal to the distance from the center of the hexagon to a vertex of the hexagon. If we take "x" as the length of a side of the hexagon, then drawing the line segment from the center of the hexagon to the center of a side gives a right triangle with one leg of length x/2 and hypotenuse of length x. The length of the other leg is \sqrt{x^2- x^2/4}= x\sqrt{3}/2.

That is, the radius of a circle inscribed in a hexagon of side length x is
\frac{x\sqrt{3}}{2}

NOT equal to the side length. It is the radius on the circle circumscribed about a hexagon that is equal to a side of the hexagon. Draw it out measure it, whatever you need to do. You're saying the same things I've heard before. It's is not almost equal, nearly equal, it is equal. Debate it all you want. you're overlooking the fact that when the hexagon is inscribed, the radius is right there, already equal, just as a hexagon is. The radius of the circle is just that, just because it happens to be one side of a triangle, it's still the radius, ya know the point from the center of a circle to the perimeter? what are you saying? Ty for your time.

Will anyone take me seriously please? If not disprove me. But you can't, there's the problem. I know I'm right. I just want help going further. This theorem is old news.

[crocque]

I'm trying to remember but I think I proved it by angles. If this angle and that angle are such a degree, you have an equitateral triangle. Hard to remember. This talk of circumscribed is nonsense, sides would be larger than the radius.

[micromass]

Draw it out measure it, whatever you need to do. You're saying the same things I've heard before. It's is not almost equal, nearly equal, it is equal. Debate it all you want. you're overlooking the fact that when the hexagon is inscribed, the radius is right there, already equal, just as a hexagon is. Ty for your time.

Will anyone take me seriously please? If not disprove me. But you can't, there's the problem. I know I'm right. I just want help going further. This theorem is old news.

You claim it's true, so you prove it. That's how it works. The two things are not equal.

And no, as long as you don't prove this, nobody will take you serious.

Perhaps download http://www.geogebra.org/cms/ and draw it out. If it comes out to be equal, then you're likely right. But it won't come out equal.

[micromass]

I'm trying to remember but I think I proved it by angles. If this angle and that angle are such a degree, you have an equitateral triangle. Hard to remember.

You say you've been working on this for 20 years already?? Hard to believe...

[daveyp225]

When a hexagon is inscribed inside the circle, it is true. Not the other way around.

[BloodyFrozen]

I couldn't find Chad's Theorem anywhere. Perhaps a link?

http://en.wikipedia.org/wiki/List_of_theorems

(not in there either)

[aeson25]

Visual aid to the discussion. Radius = 1. Therefore side of triangle = 1. Sum of hexagon sides = 6x1=6. Circumference of circle = 2*pi*r = 6.2832.....

[cmb]

You're getting close. Just think a bit more outside the circle.

If you tell us in what way I am 'close', then maybe we can guess what the hecky you're talking about!

As I said, and others repeated even with a diagram, yes a hexgaon that inscribes a circle has sides of the same length as the circle's radius. This is known. You've no right to try to ensnare folks in your wierd guessing game. You've got our attention [more than you deserve] so now dish up or clam up.

[Bourbaki1123]

No one is going to steal your theorem, because it is wrong. You're trying to claim that you've made a discovery in geometry that no one else has realized during 2000+ years of mathematical thought, using the same simple methods that have been available for those 2000+ years. You're claiming that you've not only got more geometric insight and raw ability than Euclid, Archemedes, Gauss etc. but that despite the simplicity of your proof, no one during the last couple of millennia has been able to think of the same construction and devise the proof.

Post the proof. You've got a wrong proof, and you might learn a thing or two by posting it up here and having people pick it apart. That is the issue here.

Also, you might want to look at the history of trying to prove impossible, but not immediately obviously so, methods using geometry and algebra. Here, trying to square the circle (http://en.wikipedia.org/wiki/Squaring_the_circle) is a good one and it took a pretty powerful proof to show it was impossible.

[HallsofIvy]

troll, really?, I just joined the forums. I joined to ask about this math problem that has been in my head forever. Thought maybe I should actually ask some knowledgable people. Forgive me, I'm just saying I think this can be done geometrically. This is not my first time debating this in life. I'm not trying to argue with anyone. I'm looking for help.

Really I wish this was a trool and I was crazy, it's not a troll but I might be crazy. I usually talk on the xbox forums, I came here to get real ideas.

But ok, see, if you have the triangular base of a sphere and know the radius. There is a way to use that radius and find out what the missing area is whether it is a circle or a sphere. I am working on it. Need more specifics? I've thought about it for 20 years.
I don't know what you mean by "the triangular base of a sphere". This is getting very strange.

Do you understand the difference between "inscribed" and "circumscribed". A circle circumscribed about a hexagon has radius equal to the side of the hexagon. A circle inscribed in a hexagon, which is what you said, has radius equal to \sqrt{3}/2 times the side of the hexagon.


It's not Pi. no 22/7
What? pi is not equal to 22/7. Where did that come from?

[crocque]

Maybe I am confused all. I mean that the hexagon is inside the circle. I thought that was an inscribred hexagon? Sorry for any confusion. I haven't been in school in 20 years. I am asking for help. I do not remember my own proof. This has been in my head for this long though. I however, know I proved it.

All my teacher's thought I was wrong too. I suppose since I was in advanced classes, they gave me a shot and helped me. There was a banner down the math hall with Chad's Theorem on it for over half the year. (Independence High school, Charlotte, NC)

I didn't say I was in some math archive, lol. Help me out guys.


I'll talk about Pi once we can agree that this theorem is sound. Which it is.

[micromass]

Maybe I am confused all. I mean that the hexagon is inside the circle. I thought that was an inscribred hexagon? Sorry for any confusion. I haven't been in school in 20 years. I am asking for help. I do not remember my own proof. This has been in my head for this long though. I however, know I proved it.

All my teacher's thought I was wrong too. I suppose since I was in advanced classes, they gave me a shot and helped me. There was a banner down the math hall with Chad's Theorem on it for over half the year. (Independence High school, Charlotte, NC)

I didn't say I was in some math archive, lol. Help me out guys.


I'll talk about Pi once we can agree that this theorem is sound. Which it is.

Fine, if you mean that the haxagon is inside the circle, then your theorem is correct. The radius of the circle indeed equals the side of the triangle. I think we can all agree on that.
Please post your stuff on pi now.

[crocque]

I really didn't expect this much backlash on my original theorem. If that fails, I'm nowhere. When you break it down you have 6 equilateral triangles. Now you can make that a sphere and you have pyramids around the center. All that is left are the curves, but I have the radius already. I'm not going into that until someone can take this theorem as truth or help me reprove it so people will believe it.

If it takes another college class, so be it. I'm retired anyway. Math is just fun.

[micromass]

I really didn't expect this much backlash on my original theorem. If that fails, I'm nowhere. When you break it down you have 6 equilateral triangles. Now you can make that a sphere and you have pyramids around the center. All that is left are the curves, but I have the radius already. I'm not going into that until someone can take this theorem as truth or help me reprove it so people will believe it.

If it takes another college class, so be it. I'm retired anyway. Math is just fun.

Ok, now it stops being fun. Please post everything at once. I don't appreciate you playing with us like that.

I'm starting to think "troll" here...

[Studiot]

When you break it down you have 6 equilateral triangles.

How do you know the triangles are equilateral? Can you prove it?

[aeson25]

@Studiot
crocque already admitted he made an error on terminology. We're talking about a hexagon circumscribed by a circle. If you don't believe it, see my attached picture for visual confirmation of the 6 triangles, see wiki for verbal confirmation.

@crocque
Are you taking this into 3D? You need to improve on your descriptions. If I'm understanding you correctly, you are growing pyramids out of the triangles. Is this on both sides so you end up with 12 x 3 sided pyramids? (Egyptian pyramids have 4 sides btw) If you're looking at my posted picture and consider the direction coming out of the screen to be 'z' then you'll be left with z=0 ->1 completely unfilled.

Please post pictures if you can so people can understand where you're coming from...Even if you're only trying to take people one step at a time. Being cryptic helps no-one.

[Studiot]

crocque already admitted he made an error on terminology. We're talking about a hexagon circumscribed by a circle. If you don't believe it, see my attached picture for visual confirmation of the 6 triangles, see wiki for verbal confirmation.


This is not about terminology.

If someone is going to resile existings proofs and offer alternatives they need to be careful they don't (inadvertantly) incorporate the result they have already rejected into their working.

I note you mentioned the ancient Egyptians. The inadvertantly used pi in their measurements because they performed linear measurement with a wheel.
This was a famous puzzle for many years until this was understood since they did not know about pi.

[{???}]

I can't believe that on Physics Forums, of all places, two and a half pages worth of posters managed to miss this on the first page.
"Inside a regular inscribed hexagon, the radius of the circle is equal to the sides of the hexagon".
Picking at the use of "inside" would have made sense, but he correctly stated that the hexagon was inscribed. Barking up the wrong tree, if you ask me.
The geometric proof is simple and can be done informally as follows:

A regular hexagon is an regular n-gon with even n. That means that:
1. You can connect every pair of opposite vertices using a diagonal, and all these diagonals will intersect at the same point.
2. You can inscribe it within a circle with all vertices of the n-gon touching the circle.
3. The internal angle between two sides of a regular polygon is bisected by the diagonal.

With these preliminaries out of the way, draw the three diagonals between the three pairs of opposite vertices.
It isn't hard to see that the internal angles are 120 degrees, and the bisecting diagonal turns this into two angles of 60 degrees, side-by-side around the bisector.

From 1 we have that the hexagon has been split into six triangles. From 3 we have that, for each of these triangles, two angles are 60 degrees.

Thus it holds necessarily that these triangles are equilateral. Next we use the lemma that the radius of the circle is equal to the distance from the center of the inscribed polygon to any vertex thereof. My proof is as follows:

We note that, from 2, we know that the vertices of the hexagon all lie on the circle, and from our knowledge that the triangles are equal, we see that each of the six vertices is equidistant from the hexagon's center (described by the intersection of the diagonals).

Now we look at this from the perspective of the circle in which the hexagon is inscribed. Every point on the circle is equidistant from one particular point within the circle; this unique point is the center and the unique distance the radius.

As a finite subset of the infinite set of points on the circumference of the circle, there is a set of six points which form the vertices of the hexagon. These too must be equidistant from the center of the circle.

However, both their distance from the hexagon's center and the hexagon's center itself are known, and the distance and point must be unique. Thus we can conclude that the length of one side of the equilateral triangle is equal to the radius and the center coincides with the center of the circle.

Given this it is trivial to prove that, since two sides of any of the equilateral triangles are radii of the circle, and the third (due to our construction) is one of the sides of the hexagon, the radius of the circle is equal to any side of the hexagon. QED

[crocque]

Exactly why I'm going no further without someone to protect my interests.

[{???}]

Exactly why I'm going no further without someone to protect my interests.

I'm not quite sure I follow...so far I believe your theory is correct through above proof (or probably reproof). I also believe that this is precisely what you requested so that others can begin to accept it. As far as I know my informal proof used only undisputed axioms for components, so others who read it should not encounter difficulty.

That being said, you should elaborate on your theory about removal of pi from the area/volume/circumference/surface area calculations. I'm intrigued by your hypothesis and I'm finding it difficult to stave off the urge to test it.

[PSz]

It looks like not so fresh theorem of qantized circles, which has been proven to be usless in mathematics, as it does not offer approximations better than Pi does. Your quite trite notice about looking outside a circle suggests that.

If it's not (though I really think it is), the real question is how long you are going to lead others by the nose and keep this guessing game run. Either you have something or you don't. The one and only way to find it out is to share your theorem in the greatest details possible.

[wukunlin]

you people are going around in circles

[Studiot]

I can't believe that on Physics Forums, of all places, two and a half pages worth of posters managed to miss this on the first page.


I didn't miss it.

Your 'proof' assumes a great deal of underlying geometry.
The whole purpose of my question was to try to probe the supporting background crocque was proposing to use.

It would have been useful to have seen his explanation of equilaterial triangles.

[Dr. Seafood]

sure is a circular argument

[{???}]

sure is a circular argument
In a roundabout manner of speaking.

[SteveL27]

Exactly why I'm going no further without someone to protect my interests.

You tell 'em. They're just out to steal your idea. Don't tell 'em a thing. Euclid knew pi was a fake and look what happened to him! LOL. This is a funny thread.

[Dr. Seafood]

In a roundabout manner of speaking.

Agreed. I think he's being a bit irrational about this whole thing.

[micromass]

Agreed. I think he's being a bit irrational about this whole thing.

Well, it's a complex topic...

[mishrashubham]

Well, it's a complex topic...

one that requires imaginary polygons...

[Bohrok]

Well, it's a complex topic...

It's very transcendental; much higher than anyone else is capable of coming to terms with.

[D H]

No one is going to steal your theorem, because it is wrong. You're trying to claim that you've made a discovery in geometry that no one else has realized during 2000+ years of mathematical thought, using the same simple methods that have been available for those 2000+ years. You're claiming that you've not only got more geometric insight and raw ability than Euclid, Archemedes, Gauss etc. but that despite the simplicity of your proof, no one during the last couple of millennia has been able to think of the same construction and devise the proof.
Exactly.

This thread has gone on for far too long. It is time to put it to rest. Thread locked.

crocque: Please read our rules about overly speculative posts. The rules are available in every window, right at the center of the menu bar that is right below the Physics Forums logo at the top of the page.