June 23rd, 2014, 10:19 AM  #11 
Banned Camp Joined: Feb 2013 Posts: 224 Thanks: 6 
Addition, subtraction. Got it. Multiplication and Division. Got it. La Géométrie (1637)  René Descartes Square Roots. Got it. Combo. $1.00 x (1+1/7)^7 Constructable angles by straight edge and compass: 12° Degrees. Hence possible 6° degrees, and 3° degrees. I think that just about covers it. 
September 5th, 2014, 06:00 AM  #12 
Banned Camp Joined: Feb 2013 Posts: 224 Thanks: 6  The solution
The solution of the diagram comes from here, but I don't know enough math to understand it. Descartes did it in La Geometrie 1637. http://apollonius.math.nthu.edu.tw/d...ction/tri3.pdf Last edited by long_quach; September 5th, 2014 at 06:03 AM. 
February 9th, 2017, 10:58 PM  #13 
Newbie Joined: Jul 2016 From: Kenya Posts: 3 Thanks: 0  Trisection
Dear friend. Am glad to have solved the trisection of an angle, in a very simple way using only compassstraightedge construction. I can help you trisect the 60 degree angle, 48 deg, as we do for 45 deg. Please contact me through; alexkimuya23@gmail.com. Thank you.

February 10th, 2017, 02:48 AM  #14  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,659 Thanks: 652 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
February 11th, 2017, 10:51 PM  #15 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
I would like to add that it is really , really , really insulting to offer a mere hundred dollars for the solution to an impossible problem. LMAO 
February 12th, 2017, 11:59 PM  #16 
Newbie Joined: Jan 2017 From: mwanza tanzania Posts: 20 Thanks: 0 
Draw a straight horizontal line, draw three circles with equal radius, centers of these circles must lie on a drawn line. These circles must touch each other at tangent points {two}. Draw a vertical line which passes at the center of middle circle, make use of divider to obtain a distance from initial point of first circle {where it crosses a line} to a final point {where a third circle crosses a line} in other words, three times diameter of a circle is a required distance between two points of a divider. From initial point of horizontal line, put one end of a divider and draw an arc which crosses a vertical line and from the other final end do the same (aim here is draw equilateral triangle whose side length is equal to 3 times diameter of circles drawn) which gives 60 degrees. To trisect this angle, draw a first line from intersection of two arcs on a vertical line to tangent point between first and second circle, and another line starting from where arcs meet to tangent point between second and third circles; there will be three formed angles with each 20 degrees, hence trisection of 60 degrees. kadomole
Last edited by skipjack; February 13th, 2017 at 05:28 AM. 
February 13th, 2017, 04:43 AM  #17 
Senior Member Joined: Feb 2010 Posts: 633 Thanks: 104 
Using trigonometry, the two "outside" angles are approximately 19.1 degrees and the "middle" one is approximately 21.8 degrees. Let's try it one more time: Using only a collapsible compass and unmarked straightedge, the trisection of an arbitrary angle is impossible. It can't be done. It has been proven ... many times ... by many people. Give it up. You might have better luck with this problem ... try touching your left elbow with your left hand. 
February 13th, 2017, 03:46 PM  #18  
Senior Member Joined: Sep 2016 From: USA Posts: 227 Thanks: 122 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
If you want to do math....learn math.  

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