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crocque
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I'm not sure it's safe to post real theorems here.
Is it?
Is it?
crocque said: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.
crocque said: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 spatial 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.
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.daveyp225 said: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 said:It simply is true. I proved it in the 10th grade. Check it out, disprove it, if you wish.
crocque said:It simply is true. I proved it in the 10th grade. Check it out, disprove it, if you wish.
I asked you nicely if you do not believe me that disprove it. Please do not tell me what I already know.disregardthat said:No, it's not. You probably proved it when the circle is circumscribed by the hexagon, not inscribed.
crocque said: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.
You're getting close. Just think a bit more outside the circle.cmb said: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?
daveyp225 said: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?
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].crocque said:It simply is true. I proved it in the 10th grade. Check it out, disprove it, if you wish.
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, you know the point from the center of a circle to the perimeter? what are you saying? Ty for your time.HallsofIvy said: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 said: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.
crocque said: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.
crocque said:You're getting close. Just think a bit more outside the circle.