On Sun, 29 Oct 2006 07:02:21 GMT, "CW" wrote:

Well said.

Tom Veatch wrote in message

Interior lines from the center of a hexagon to each vertex form 6
Isosceles triangles, each with an included angle of 60 degrees. The
length of each of those interior lines is equal to the radius of the
circle concentric with the hexagon and passing through the vertices.

Length of the base of an Isosceles triangle is 2 X length of side X
sine of 1/2 the included angle.

L = 2 R sin (1/2 A)

whe
L= length of the base
R = length of the sides of the triangle = radius of the circle
A = angle included by the sides

But since A is 60 degrees, 1/2 A = 30 degrees and sin(30) = 0.5000

So, L = 2R X (.5) = R

From which the length of the side of a hexagon is identical to the
radius of the hexagon's circumscribing circle. (and those 6 included
angles are not only isosceles, but are, in fact, equilateral)


I can't argue with the trig- I can immediately see that it is correct.
Here's my only gripe with proving Euclid's method via trig- When you
prove it using trig, you are starting with the hexagon, and
circumscribing a circle around it. It's accurate and mathematically
perfect. And it works just fine. I can draft a regular hexagon, find
the center and circumscribe a circle about it very simply, and it
works great- with no fudging about compass placement error. But I
still can't get Euclid's technique to work, and I haven't seen anybody
do it with full sucess, even when it was being taught.

I am missing something here, though, and just realised it. The arc
length is s = r alpha Pi / 180, right?

So if you do the problem via trig, you come up with a cord length of
five on a circle with a radius of five.

Which is what the compass is drawing, right?

So if I'm looking at the arc length by working backwards from the
formula for the circumference of a circle, I'll come up with:

c = 3.1415 x 10 = 31.415

31.415 / 6 = 5.2358 - the arc length

Now for the missing link:

s = [5 x 60 x 3.1415] / 180
s = 5.2358

Finally.

You have no idea how long that damn thing has been bugging me. At
least I learned something this evening that will put that problem to
rest. Must have missed that day in class somewhere along the line.
Still not using Euclid for that one, though- I can't get it to come
out.

Anyhow, the method I described for drafting a hexgon from a segement
works too, and you don't need to figure out what size circle you need
in the first place if you know how wide you want it to be. And the
jig works, too- I've used it several times for the bottoms of
segemented turnings.

Thanks for the correction, CW and Tom. I just have an aversion to
taking things on faith if I can't see all the pieces fitting together
nicely.