bob boben Posted September 4, 2011 Report Share Posted September 4, 2011 The solution of mathematical tasks in the ancient GreekTrisection of anglesangle=0° - no solution180°>angle>0° - general solution (consists of 4 parts) the first part 1.ruler AB2.ruler AC3.caliper A-AD4.ruler DE 5.caliper D-DE6.caliper E-DE7.ruler FG intersects DE the point H ,DH=HE 8.caliper H-HE Quote Link to comment Share on other sites More sharing options...
Turtle Posted September 4, 2011 Report Share Posted September 4, 2011 well, i think you should have written "straightedge", not ruler for one thing. for another, you have in the second page constructed a vesica piscis and since it is a 120º arc it trisects the angle. this is a special case however and the trisection problem specifies an arbitrary angle and using only a compass & straightedge. it is proven impossible. :( angle trisectionAngle trisection is a classic problem of compass and straightedge constructions of ancient Greek mathematics. It concerns construction of an angle equal to one-third of a given arbitrary angle, using only two tools: an un-marked straightedge, and a compass. With such tools, the task of angle trisection is generally impossible, as shown by Pierre Wantzel (1837). Wantzel's proof relies on ideas from the field of Galois theory—in particular, trisection of an angle corresponds to the solution of a certain cubic equation, which is not possible using the given tools. Note that the fact that there is no way to trisect an angle in general with just a compass and a straightedge does not mean that it is impossible to trisect all angles: for example, it is relatively straightforward to trisect a right angle (that is, to construct an angle of measure 30 degrees). ... Quote Link to comment Share on other sites More sharing options...
CraigD Posted September 4, 2011 Report Share Posted September 4, 2011 Hi Bob – welcome to hypography. :) You seem to have omitted the finishing steps of your construction. :( Notwithstanding the problem of construction by straightedge only being able to generate surd-valued (rational numbers, their irrational square roots, and square roots of them) points, while the value of points on lines of the trisection of some angles may require irrational cube roots, the construction in post #1 doesn’t draw 2 lines through point A, evenly or unevenly trisecting angle CAD. To trisect an angle, you’ve gotta ultimately draw these 2 lines. Do you have some more steps you failed to include in your first post? Quote Link to comment Share on other sites More sharing options...
bob boben Posted September 5, 2011 Author Report Share Posted September 5, 2011 second part9.caliper D-DH , gets the point D110.straightedge (ruler) HD111.caliper D-DH12.caliper D1-D1D13.straightedge ( ruler ) HI1 , gets the point D214.caliper D2-D2D15.caliper D-DD216.straightedge (ruler) HI2 , gets the point D317.caliper D3-D3D18.caliper D-DD319.straightedge (ruler) HI3 , gets the point D420.caliper D4-D4D21.caliper D-DD422.straightedge (ruler) HI4 , gets the point D523.this procedure with a compass and a ruler is actually a series of numbers (finite, infinite) Quote Link to comment Share on other sites More sharing options...
bob boben Posted September 7, 2011 Author Report Share Posted September 7, 2011 angle=180°30.straightedge (ruler) AB31.caliper C-CD32.caliper D-DC ,gets the point E33.straightedge (ruler) CE34.caliper E-ED , gets the point F35.straightedge (ruler) CF Quote Link to comment Share on other sites More sharing options...
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