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09-19-2003
03:00 AM

09-19-2003
03:00 AM

Heptadecagon Construction

i Can construc a regular heptadecagon, 17 sided, but I need to PROVE mathmatically it is REGULAR. i constructed it somewhat like this http://www.geocities.com/RainForest/Vines/2977/gauss/formulae/heptadecagon.html

but can someone help with the numerical proof please, thasnks

but can someone help with the numerical proof please, thasnks

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09-21-2003
03:00 AM

09-21-2003
03:00 AM

Heptadecagon Construction

On 9/19/2003 8:35:14 PM, endproc wrote:

>i Can construc a regular

>heptadecagon, 17 sided, but I

>need to PROVE mathmatically it

>is REGULAR. i constructed it

>somewhat like this

>http://www.geocities.com/RainF

>orest/Vines/2977/gauss/formula

>e/heptadecagon.html

>

>but can someone help with the

>numerical proof please,

>thasnks

The first thing you must verify is this:

cos(2pi/17) = -1/16 + 1/16 sqrt(17) + 1/16 sqrt[34 - 2sqrt(17)] + 1/8 sqrt[17 + 3sqrt(17) - sqrt(34-2sqrt(17)) - 2sqrt(34+2sqrt(17))]

How could you possibly verify that? Well, lets consider the properties of 2pi/17. Note this:

(2pi/17) + 16(2pi/17) = 2pi

cos[(2pi/17) + 16(2pi/17)] = cos(2pi)

cos[A + 16A] = 1

Now, expand that out using the sum formulas and the double angle formulas. It will be messy. StudyWorks can make quick work of this. Express the answers in terms of just cos(A). Now, let B = cos(A). The result is a polynomial of degree 17 in terms of B. Subtract the 1 on the right to get the form:

(17-degree polynomial in B) = 0

This can be factored! For purposes of sanity, let StudyWorks do this if you can.

(B-1)(8-degree polynomial in B)^2 = 0

Since B=1 is certainly not what we want (since the cosine of 2pi/17 is not 1), we are left with this:

(8-degree polynomial in B) = 0

I don't see an easier way at this point than to just plug in that huge formula for cos(2pi/17) in place of B and verify the identity. Studyworks can do this, but it will struggle over it. You must "coax" it to simplify the expression down.

Once you get this to work, you are almost done. YOu have proved that the expression solves cos(17A) = 1. All that is left is to show that the expression is numerically equal to cos(2pi/17), not some other root. (Comparing it numerically to a few places is enough to show this.) Since you now know it is a solution, and you know it is the right solution, you have proven the identity.

The second stage is to prove that the construction actually constructs that number. Law of cosines on V3-O-V5 (which is an angle of measure 2pi/17) seems like a promising start. It will be ugly, so let StudyWorks do what it can.

Craig

>i Can construc a regular

>heptadecagon, 17 sided, but I

>need to PROVE mathmatically it

>is REGULAR. i constructed it

>somewhat like this

>http://www.geocities.com/RainF

>orest/Vines/2977/gauss/formula

>e/heptadecagon.html

>

>but can someone help with the

>numerical proof please,

>thasnks

The first thing you must verify is this:

cos(2pi/17) = -1/16 + 1/16 sqrt(17) + 1/16 sqrt[34 - 2sqrt(17)] + 1/8 sqrt[17 + 3sqrt(17) - sqrt(34-2sqrt(17)) - 2sqrt(34+2sqrt(17))]

How could you possibly verify that? Well, lets consider the properties of 2pi/17. Note this:

(2pi/17) + 16(2pi/17) = 2pi

cos[(2pi/17) + 16(2pi/17)] = cos(2pi)

cos[A + 16A] = 1

Now, expand that out using the sum formulas and the double angle formulas. It will be messy. StudyWorks can make quick work of this. Express the answers in terms of just cos(A). Now, let B = cos(A). The result is a polynomial of degree 17 in terms of B. Subtract the 1 on the right to get the form:

(17-degree polynomial in B) = 0

This can be factored! For purposes of sanity, let StudyWorks do this if you can.

(B-1)(8-degree polynomial in B)^2 = 0

Since B=1 is certainly not what we want (since the cosine of 2pi/17 is not 1), we are left with this:

(8-degree polynomial in B) = 0

I don't see an easier way at this point than to just plug in that huge formula for cos(2pi/17) in place of B and verify the identity. Studyworks can do this, but it will struggle over it. You must "coax" it to simplify the expression down.

Once you get this to work, you are almost done. YOu have proved that the expression solves cos(17A) = 1. All that is left is to show that the expression is numerically equal to cos(2pi/17), not some other root. (Comparing it numerically to a few places is enough to show this.) Since you now know it is a solution, and you know it is the right solution, you have proven the identity.

The second stage is to prove that the construction actually constructs that number. Law of cosines on V3-O-V5 (which is an angle of measure 2pi/17) seems like a promising start. It will be ugly, so let StudyWorks do what it can.

Craig