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See here I am a little confused. I never really understood why one couldn't do this: construct an equilateral triangle atop the angle to be trisected bisect an
by phoobahr 15y ago
See here I am a little confused. I never really understood why one couldn't do this:
construct an equilateral triangle atop the angle to be trisected
bisect any side, construct a new triangle with said 30deg angle giving a line segment 1/sqrt(3) of the original eq tri
bisect the new 60deg angle, constructing a new 30/60/90 tri
voila, 1/[sqrt(3)sqrt(3)] == 1/3. You have trisected a line with a compass and unmarked straightedge
- sp332 15y agoIf I'm starting with a 40 degree angle, how do I construct an equilateral triangle on it?
- phoobahr 15y agoGiven angle(PQR) begin as you would when bisecting the angle: http://www.mathopenref.com/constbisectangle.html http://www.mathopenref.com/constbisectangle.html. Your arc intersections with |PQ| and |RQ| (call them p & q) give you an isosceles triangle pQr. Construct an equilateral triangle on |pr| similar to: http://www.mathopenref.com/constequilateral.html http://www.mathopenref.com/constequilateral.html