equal quadrants..

[collbee]

A large number of members will probably already know this - but just in case.........

The radius of a circle, when used on the circumference of the said circle, will always dissect the circle into six(6) equal quadrants.

This can be very useful when marking out a six spoked flywheel.

Collbee.....

[Bluechip]

Six equal sectors surely ???

Unless you're making one and a half flywheels ...

Dave

[mklotz]

It's true, of course, (with quadrant replaced by sextant) but you're only allowed to make use of the fact if you can prove that it's true using only geometry.   

[Jim Nic]

Are we sure??  (At the risk of exposing how long ago I went to skule ...) The circumference of a circle is 2 x Pi x R and since Pi  used to be 3.142 Collbee's original statement was not exactly correct all those years ago.  Perhaps it's different now that we're metric.   
Jim

[tel]

You're not allowing for the thickness of the pencil line mate - that makes it right. 

[mklotz]

Well, you were right about the exposure.

The statement is exactly correct and its truth can be proved without the use of any value for pi.

 BTW, 3.142 is not the exact value of pi.  The expansion of pi goes on forever which means that there is an infinite number of digits after the decimal point.

To 14 digits, the value is 3.1415926535898, a value any IBM 709 programmer memorized ages ago.

[tvoght]

Set a pair of dividers to the radius of the circle. Walk the dividers around the circumference. You'll be setting out the vertices of a hexagon inscribed on the circle. The hexagon is composed of 6 equilateral triangles, each side of each triangle being equal to the radius of the circle.

--Tim

[mklotz]

Set a pair of dividers to the radius of the circle. Walk the dividers around the circumference. You'll be setting out the vertices of a hexagon inscribed on the circle. The hexagon is composed of 6 equilateral triangles, each side of each triangle being equal to the radius of the circle.

All true but how do you prove that the triangles are equilateral?

[tvoght]

I can't give a proper proof. Geometry class goes back too far.

My reasoning would be that all the angles meeting at the center must add up to 360 degrees. There are six of them, so those angles are 60 degrees. I already know the sides adjoining those angles are equal. Here it "becomes obvious to me" that the triangles are equilateral.

Sorry, best I can do.

--Tim

By the way, my original response was just to bring the concept of using dividers into the picture. It's clearer that way.

[Allen Smithee]

Set a pair of dividers to the radius of the circle. Walk the dividers around the circumference. You'll be setting out the vertices of a hexagon inscribed on the circle. The hexagon is composed of 6 equilateral triangles, each side of each triangle being equal to the radius of the circle.

All true but how do you prove that the triangles are equilateral?

By simply observing that the three sides are of the same length - two of them being radii and the third being a radius length transferred to a chord joining two points on the circumference. QED.

AS

[tvoght]

Thank you Allen. Your explanation had just occurred to me.

--Tim

[mklotz]

Even if one were not aware of the construction that makes the chord equal to the radius one could reason as follows...

The two sides are equal since they are radii of the circle.  Thus the triangle is isosceles.  The base angles of an isosceles triangle are equal.  [Technically, this requires a further proof but it's such a staple of elementary geometry that we'll let it as an exercise.]

The sum of the two equal angles must equal 180 - 60 = 120 because the central angle is 60 and the internal angles of a triangle sum to 180.  Thus each base angle is 60.

All the angles of the triangle are equal.  The sides of an equiangular triangle are equal, therefore the chord must equal the radius of the circle.

That's a geometry proof.  For a trigonometry proof, simply apply the law of cosines to find the chord...

c^2 = r^2 + r^2 -2*r^2*cos(60) = r^2  => c = r

[derekwarner]

Guys......this is also they way they built the Pyramid's all those years ago.... ....but  .....Derek

No trig tables...
No slide rules...
No calculators...
No computers...


"That's a geometry proof.  For a trigonometry proof, simply apply the law of cosines to find the chord"...

c^2 = r^2 + r^2 -2*r^2*cos(60) = r^2  => c = r

[mklotz]

Derek,

I have no idea what your post is trying to say, if anything.

First off, the discussion we have had here, so far, has not involved the use of anything you mention - trig tables, slide rules, calculators, or computers.  It's been purely a geometry exercise done mentally much as the ancient Greeks scratching drawings in the sand would have done.

Secondly, don't kid yourself about the mathematical sophistication of the Egyptians.  Despite lacking our compact notation and theoretical understanding they managed to do some very clever and complex math.  If you don't believe me, take the time to read about Egyptian fractions...

http://en.wikipedia.org/wiki/Egyptian_fraction

So, please, what was the point of your post?

[IanR]

The original post puzzled me, as Pi x 2r is something I remember from school, and it doesn't matter how many decimal places you approximate Pi to, it's still 3 and a bit.
Tim's post made it clear. You're not dividing the circumference into arcs, but measuring out chords.