Trisect an Angle
Ok, here is a slightly more interesting one to further the progression of puzzles we have had here. Given a pencil and a straight edge, trisect a given (acute) angle. This is a classic geometry problem that has resisted the advances of the scale toting geometers for quite some time. Of late, a new and interesting way has been proposed to solve this problem which approaches geometry in a very nice and innovative way. I will post a hint if you guys have much trouble (though going by the other problems I have posted, this seems unlikely..)
Macaulay
Macaulay
posted by Macaulay at 5/23/2006 08:15:00 PM
10 Comments:
I thought it had been proved and well-accepted taht ou can't trisect arbitrary angles with straightedge and compass.
Okay, I am making the usual assumptions, I guess. If any of the axioms for the proof of the above are to be violated, I would like to check which one that is.
So, is it (1) You are allowed to make marks on the straight edge (2) the angle is drawn on a thin, flat flexible piece of paper or (3) something else?
2 -- I was thinking of the origami solution (it is pretty interesting to prove that this works in ltself) -- check out http://www.cut-the-knot.org/pythagoras/PaperFolding/AngleTrisection.shtml for a proof...
Hey,what's up with everyone?No new puzzles for a month now!!
Somewhat more precisely, the correct claim is 'given only a pencil and a straight edge, one cannot trisect a given acute angle in finitely many steps'.
It is easy to construct an infinite sequence of operations that converges to a trisection. Can you tell me how?
Elaborating on my previous comment for completeness, here is how you can get arbitrarily close to trisecting an angle.
Say you want to trisect an angle theta.
Divide the angle theta into 4 equal sectors (say, using bisections).
Take the second of these 4 sectors. Call it theta_1.
Divide theta_1 into 4 equal sectors.
Take the second of these 4 equal sectors of theta_1.
Divide that into 4 equal parts ... And so on.
This sequence of 4-sections converges to a trisection as follows.
theta/4 + theta/16 + theta/64 + ...
= theta/4/(1-0.25)
= theta/4/(0.75)
= theta/3.
Thats a very neat idea, indeed!
To generalize that, you can approximate any (real) fraction of an angle by writing out the fraction's binary notation and doing the appropriate bisections.
Keep up the good work »
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