Trisecting an AngleDate: 4/16/96 at 18:30:1 From: David Owens Subject: help me please! Dear Dr. Math, My teacher gave me an assignment asking me to trisect a 90 degree angle using only a compass and a straightedge. He said that we could use outside sources. I have looked everywhere and I can't find an answer. I can bisect it easily but I can't trisect it perfectly. Would you please send me instructions on how I would do this? Thanks! Aaron Isaacson aarona@creighton.knightnet.edu Date: 4/22/96 at 20:35:18 From: Doctor Joshua Subject: Re: help me please! Hi Aaron, Okay, I think I've figured this out, but it might be hard to explain with limited computer drawings, so bear with me, and get paper and a pencil: First draw your 90 degree angle, making sure that the lengths of the two segments extending from the center are the same length (this can be done with the compass). Call the 90 angle point P, end of the North-South segment point Q, and the East-West segment point R: Q | <-- ray of length 1 | | |________ <-- ray of length 1 P R Then draw three arcs whose radius is the same length as the segments you've drawn, and whose centers are located at point P (call this arc A), point Q (call this arc B), and point R (call this arc C). Arcs B and C should go through point P (I'm not going to try drawing this). Call point M the intersection of Arcs A and B, and call point N the intersection of Arcs A and C. Now with a straight edge, draw four straight lines: from M to P, M to Q, N to P, and N to R. Because the lengths of MQ=PQ=PM, and NP=PR=RN, you have two equilateral triangles (NPR and MPQ). Because every angle in an equilateral triangle is 60 degrees, angle MPQ is 60 degrees, so angle MPR must be 30 (=90-60). Similarly, angle NPQ is 30, so the last remaining angle, NPM, must also be 30 (=90-30-30), and our right angle is trisected. Good luck! -Doctor Joshua, The Math Forum |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/