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Expert: - 7/10/2009


Question
I need to explain using abstract algebra how to geometrically construct a square with an area equal to that of a given circle and use a geometric construction to trisect an arbitrary angle. I was told you could use abstract algebra to solve these proofs. I was thinking that you need to use the theorem of field extension. I am lost on how to do it. I don't need to show the proof but explain it. I think each problem is solved different.

Answer
The is a way to find an angle that is 1/3 of an angle less than 90 degrees.  http://planetmath.org/encyclopedia/TrisectionOfAngle.html
Note that it finds an angle that is 1/3 as much, but that the angle is not in the angle itself.

If the angle is between 90 and 180 degrees, take off the 90 degrees and trisect what is now less than 90 and add on 30 degrees.

If the angle is between 180 and 270 degrees, take off 180 degrees and trisect what is left.  Then add a 60 degree angle to it.

If the angle is between 270 and 180 degrees, subtract off 180 degrees and deal with what is left as an angle between 90 and 180 degrees.

To find out how much to add on, construct a equilateral triangle.
Each of the angles is 60 degrees.  Bisect one of the angles to get 30 degrees.

Thus, any angle, no how great it is, could be trisected if this original problem could be done.

That may seem like it can be done, but it can't.
Places to look are http://mathforum.org/mathgrepform.html
or http://daisy.uwaterloo.ca/~alopez-o/math-faq/

In higher geometrical mathematics it is learned to be an impossibility.

Also, people have tried to equate the square and the circle,
but it can't be done either.  That might have something to do with pi being a nonrepeating number.  From what I've seen the first 60 places of pi past the decimal are
3.14159 26535 89793 23846 26433 83279
 50288 41971 69399 37510 58209 74944 ...

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