My Math Forum Angle Trisection by Geometry, $100 USD Prize  User Name Remember Me? Password  Geometry Geometry Math Forum  June 23rd, 2014, 09:19 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, 05:00 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 05:03 AM.
 February 9th, 2017, 10:58 PM #13 Newbie   Joined: Jul 2016 From: Kenya Posts: 1 Thanks: 0 Trisection Dear friend. Am glad to have solved the trisection of an angle, in a very simple way using only compass-straightedge 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: 1,555
Thanks: 597

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by Alexk Dear friend. Am glad to have solved the trisection of an angle, in a very simple way using only compass-straightedge 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.

-Dan

 February 11th, 2017, 10:51 PM #15 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,179 Thanks: 180 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 Thanks from topsquark
 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: 600 Thanks: 87 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. Thanks from topsquark
February 13th, 2017, 03:46 PM   #18
Senior Member

Joined: Sep 2016
From: USA

Posts: 114
Thanks: 44

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
 Originally Posted by kadomole simon kadomole 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
I skimmed your proof and it didn't seem to mention a fundamental mistake in Galois theory. Unless you can specify a mistake in the theorem which guarantees an arbitrary angle can't be trisected, no construction you provide will be worth the time to point out its (necessary) flaw.

If you want to do math....learn math.

 Tags \$100, angle, geometry, prize, trisection, usd

,
,

,

,

,

,

,

,

# the prize of angle trisection 2017

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post rockas Algebra 1 June 17th, 2013 02:01 PM mathbalarka Abstract Algebra 8 April 25th, 2012 04:36 AM cux Geometry 1 June 4th, 2011 05:59 PM AlexanderBrevig Algebra 2 May 23rd, 2011 08:29 PM MathIsntFun Geometry 4 May 13th, 2010 02:23 PM

 Contact - Home - Forums - Top