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From: ljdickey@water.UUCP (Lee Dickey)
Newsgroups: net.math
Subject: Re: A geometrical construction poser
Message-ID: <209@water.UUCP>
Date: Tue, 11-Feb-86 16:24:56 EST
Article-I.D.: water.209
Posted: Tue Feb 11 16:24:56 1986
Date-Received: Wed, 12-Feb-86 20:22:22 EST
References: <964@decwrl.DEC.COM>
Organization: U of Waterloo, Ontario
Lines: 69

The question is this:
Does this proceedure (paraphrased) give the vertices of a regular polygon?

> Procedure:
> 
> Let  O:(0,0) be the center of the circle, radius 1.
> Then A:(1,0) is on the circle and
>      B:(0,1) is on the circle.
> 
> 1.  Bisect OA to locate C.
> 2.  From C, with CB as radius, strike an arc to locate D (on x-axis, x <0).
> 3.  From B, with BD as radius, strike an arc to locate E (on circle).
> 4.  From E, still using BD as radius, locate F (on circle).
> 5.  Repeat step 4, moving around circumference, to locate G and H.
> 6.  BEFGHB is the required regular inscribed pentagon.

------------------------------------------------------------------------
Answer:
	Yes, the construction is correct, and provable.
I think you will find that this proof is not too difficult.
------------------------------------------------------------------------
Proof:
	By simple calculations, the square of the distance from B
to D is found to be 

	dist(B,D) = sqrt ( ( 5 - sqrt(5) )/ 2) .

Now, is that really the length of the side of a regular pentagon?

I propose to show it is by finding the coordinates of the regular
polygon whose vertices are the roots of the complex equation z^5=1.

We will find the complex number x+iy which is the root of z^5 - 1 for
which both x and y positive.  Then  (x+iy)^5 = 1.  This equation, when
expanded, can be separated into two parts, the real part and the
imaginary part.  The imaginary part has a factor of y.  Since y>0,
that factor can be removed.  Call the two resulting equations (1),
for the real part, and (2), for the imaginary part.  There is a third
equation (3), which says that (x,y) has to be on the circle:  x^2 + y^2 = 1. 

	Now, both (1) and (2) have only even powers of y, so (3) can be
used to advantage.  From (1) and (3), you can derive this:

	(1)':	16x^5 - 20x^3 + 5x - 1 = 0.

But the value x=1 is a root of this equation (corresponding to y=0),
but (1,0) is not the point we want (we want both x and y positive). 
Divide (1)' by the polynomial x-1 to get:

	(1)":	16x^4 + 16x^3 - 4x^2 - 4x + 1 = 0.

Similarly, working with (2) and (3) gives:

	(2)':	16x^4 - 12x^2 + 1 = 0.

Now, use (1)" and (2)' to get:

	4x^2 + 2x - 1 = 0.

Since x>0, this gives exactly the x-coordinate of the point (x,y),

	x = ( sqrt(5) - 1 ) / 4

Knowing this, the square of the distance from (x,y) to (1,0) is 
found to be  (x-1)^2 + y^2 = 2 - 2x = ( 5 - sqrt (5) ) / 2.

	This is exactly what we wanted to know.  Since the length
of the side BD agrees with the distance from (x,y) to (1,0), the
figure given by the construction is a regular pentagon.