# Inscribing hexagon in circle

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• posted on May 20, 2009, 1:07 am
Regarding that business up above about dividing a circle into 3 equal parts using a square, blah blah blah ... maybe I don't remember my geometry so well. Checked back in that table I mentioned in "Proven Shop Tips" for dividing a circle into n equal parts, and sure enough, the number to multiply the diameter of a circle by to divide into 6 equal parts is ... exactly 0.5.
So anyone got the proof handy that a hexagon with sides of length s can be inscribed in a circle whose radius equals s? I have my old algebra and calculus books, but no geometry.
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• posted on May 20, 2009, 1:47 am
On Tue, 19 May 2009 18:07:58 -0700, David Nebenzahl
...

It's easy enough to show using Trigonometry, but that doesn't constitute a geometric proof, which as I recall has to be done with only compass and straight edge. But it's simple enough to demonstrate with a compass and straight edge. I don't know whether demonstration by construction constitutes a formal geometric proof, or not.
Set your compass to a convenient radius and draw a circle. Without changing the compass setting strike an arc from any point on the circle that intersects the circle. From that intersection, strike another intersecting arc. Continue around the circle and, if done carefully enough, the 6th arc will pass through the original point. Since all arcs have the same radius, all the chords connecting the intersections are the same length and equal to the radius of the arc which is also the radius of the circle. Connect each point of intersection with its neighbors using a straight line. By definition, 6 sides, all of the same length, constitute a regular hexagon.
Tom Veatch Wichita, KS USA
An armed society is a polite society. Manners are good when one may have to back up his acts with his life. Robert A. Heinlein
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• posted on May 20, 2009, 1:54 am

Given your demonstration, I believe you can safely append:
QED
Regards,
Tom Watson http://home.comcast.net/~tjwatson1 /
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• posted on May 20, 2009, 2:19 am
On 5/19/2009 6:47 PM Tom Veatch spake thus:

That's a demonstration, not a proof. But you knew that.
I'm interested in the proof. It can't be all that complicated.
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• posted on May 20, 2009, 2:29 am
wrote:

It is well-known that one cannot trisect an angle with a straight-edge and compass, so I don't think you'll get a proof with that approach. On the other hand, using division you can divide 360 degrees, or 2*Pi radians by 6 to get the angle for each slice of the pie The rest has already been discussed (side-side-side).
Bill
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• posted on May 20, 2009, 11:45 am
wrote:

Nobody's attempting to trisect an angle in that approach; in fact, it's essentially the same method as Euclid's proof, a link to which was already posted up-thread.
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• posted on May 20, 2009, 2:02 am
"David Nebenzahl" wrote:

Remember the first three (3) plane geometry proofs?
1) Side-Angle-Side 2) Side-Side-Side 3) Angle-Side-Angle
Side-Side-Side works for me.
Lew
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• posted on May 20, 2009, 2:20 am
On 5/19/2009 7:02 PM Lew Hodgett spake thus:

Expand, please. Don't know exactly how this proof works.
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• posted on May 20, 2009, 2:34 am
"David Nebenzahl" wrote:

Plug " Side-Side-Side" into Google, should keep you out of trouble for a couple of hours, especially the congruent triangle proofs.
Lew
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• posted on May 20, 2009, 4:29 am
"David Nebenzahl" wrote:

http://mathworld.wolfram.com/Hexagon.html
Lew
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• posted on May 20, 2009, 5:01 am

The illustration with the A, B, C, D and E circles gives the OP his 3 pie pieces, if he just puts the triangle inside the hexagon.
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-MIKE-

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• posted on May 20, 2009, 2:18 am

An equilateral triangle has equal sides and three 60-degree angles. Arrange six of them with sides s in an array with one common vertex, and you will have the inscribed hexagon.
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• posted on May 20, 2009, 2:42 am

Or, put another way:
Consider a regular hexagon. Draw line segments from the center of the hexagon to each of the six vertices. These six equal angles at the center must add up to 360, so each is 60. Since the triangles are isosceles, and their angles add to 180, they are also equilateral. So the side of the hexagon is equal to the length of the line form the center to a vertex on the hexagon, which is the radius of the circle.
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• posted on May 20, 2009, 3:33 am
On 5/19/2009 7:42 PM alexy spake thus:

That sounds better. (Don't know if it constitutes a rigorous proof or not, but it satisfies my "itching".)
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• posted on May 20, 2009, 11:42 am

OK so far...

.. but you just went astray there.
That's not sufficient to prove that the triangles are equilateral, since the angles add to 180 in *all* triangles.
This may be what you meant to say:
Since the angle at the vertex of each triangle is 60 degrees, the sum of the angles at the base is 180 - 60 = 120 degrees. Since each triangle is isosceles, the angles at the base are equal, and (since they add to 120) therefore also 60 degrees. The triangles are therefore equiangular, and therefore equilateral.

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• posted on May 20, 2009, 2:44 pm
snipped-for-privacy@milmac.com (Doug Miller) wrote:

one 60 degree angle are necessary and sufficient conditions for an equilateral triangle (in Euclidean geometry).

That is what I said. You just have to read between the lines <g>.

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• posted on May 20, 2009, 3:30 am
On 5/19/2009 7:18 PM alexy spake thus:

I don't see any proof in there, only an assertion.
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• posted on May 20, 2009, 3:41 am

Look deeper. This is a proof of the "from which it can be clearly seen..." type that occasionally drove me batty!
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• posted on May 20, 2009, 8:13 pm
On Tue, 19 May 2009 18:07:58 -0700, David Nebenzahl

Well, I havn't seen the proof but dividing a circle into thirds is very easy with a compass. Set your radius, and keep it there. Draw the circle, set the compass point on the circle and draw an arc inside the circle. Repeat 5 times using an intersect as another pivot point. You will get a perfect 6-petal flower.