Why Pi is wrong


by crocque
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crocque
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#19
Aug15-11, 07:22 AM
P: 28
Quote Quote by disregardthat View Post
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
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#20
Aug15-11, 07:23 AM
P: 28
Where's a professor when you need one?
daveyp225
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#21
Aug15-11, 07:31 AM
P: 88
Quote Quote by crocque View Post
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
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#22
Aug15-11, 07:31 AM
P: 28
Quote Quote by cmb View Post
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
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#23
Aug15-11, 07:49 AM
P: 28
Quote Quote by daveyp225 View Post
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
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#24
Aug15-11, 07:55 AM
P: 28
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
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#25
Aug15-11, 08:15 AM
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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
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#26
Aug15-11, 09:41 AM
P: 28
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
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#27
Aug15-11, 09:41 AM
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Quote Quote by crocque View Post
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 [itex]\sqrt{x^2- x^2/4}= x\sqrt{3}/2[/itex].

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

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
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#28
Aug15-11, 09:55 AM
P: 28
Quote Quote by HallsofIvy View Post
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 [itex]\sqrt{x^2- x^2/4}= x\sqrt{3}/2[/itex].

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

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
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#29
Aug15-11, 09:57 AM
P: 28
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
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#30
Aug15-11, 09:58 AM
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Quote Quote by crocque View Post
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
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#31
Aug15-11, 09:59 AM
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Quote Quote by crocque View Post
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
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#32
Aug15-11, 10:07 AM
P: 88
When a hexagon is inscribed inside the circle, it is true. Not the other way around.
BloodyFrozen
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#33
Aug15-11, 11:57 AM
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I couldn't find Chad's Theorem anywhere. Perhaps a link?

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

(not in there either)
aeson25
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#34
Aug15-11, 12:22 PM
P: 7
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.....
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cmb
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#35
Aug15-11, 12:40 PM
P: 623
Quote Quote by crocque View Post
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
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#36
Aug15-11, 01:32 PM
P: 327
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 is a good one and it took a pretty powerful proof to show it was impossible.


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