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#37
Aug1511, 01:46 PM

Math
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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 [itex]\sqrt{3}/2[/itex] times the side of the hexagon. 


#38
Aug1511, 03:21 PM

P: 28

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. 


#39
Aug1511, 03:27 PM

Mentor
P: 18,346

Please post your stuff on pi now. 


#40
Aug1511, 03:32 PM

P: 28

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. 


#41
Aug1511, 03:39 PM

Mentor
P: 18,346

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


#42
Aug1511, 03:51 PM

P: 5,462




#43
Aug1511, 04:05 PM

P: 7

@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 noone. 


#44
Aug1511, 04:20 PM

P: 5,462

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. 


#45
Aug1511, 04:21 PM

P: 16

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.
The geometric proof is simple and can be done informally as follows: A regular hexagon is an regular ngon 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 ngon 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, sidebyside 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 


#46
Aug1511, 04:48 PM

P: 28

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



#47
Aug1511, 05:02 PM

P: 16

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. 


#48
Aug1511, 05:30 PM

P: 11

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. 


#50
Aug1511, 05:38 PM

P: 5,462

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. 


#51
Aug1511, 08:10 PM

P: 120

sure is a circular argument



#52
Aug1511, 08:22 PM

P: 16




#53
Aug1511, 09:02 PM

P: 800




#54
Aug1511, 09:14 PM

P: 120




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