bob boben

angle=180° 30.straightedge (ruler) AB 31.caliper C-CD 32.caliper D-DC ,gets the point E 33.straightedge (ruler) CE 34.caliper E-ED , gets the point F 35.straightedge (ruler) CF

third part computer program-coreldraw 13 http://www.fileserve.com/file/sZpPSyf/T ... -DOKAZ.cdr png image http://www.fileserve.com/file/EpEKHxx/p ... ection.png the picture is -circle -diameter circle AB -points on the diameter of the circle A(0°),B(180°), C (AD=DC=CB) , D , F(center of the circle ,FB-radius circle) -tendon EF I went from the assumptions that there are lines in a circle which is the first point (180 °> point> 0 °), the last point is (360 °> point> 180 °) to the intersection with the circle diameter (AB) be the point C (D). I shared a circle with corners (second part , 9 to 23). I found the string (EF, 3.75 ° -187.5 °) which passes through the point C ,which means that we have the angle trisection. fourth part 24.caliper D-DD5-(22.) 25.caliper E-DD4-(19.) 26.straightedge (ruler) D4D5 , gets the point H1 27.straightedge (ruler) AH1 28.caliper H1-H1D , gets the point H2 29.straightedge (ruler) AH2

second part 9.caliper D-DH , gets the point D1 10.straightedge (ruler) HD1 11.caliper D-DH 12.caliper D1-D1D 13.straightedge ( ruler ) HI1 , gets the point D2 14.caliper D2-D2D 15.caliper D-DD2 16.straightedge (ruler) HI2 , gets the point D3 17.caliper D3-D3D 18.caliper D-DD3 19.straightedge (ruler) HI3 , gets the point D4 20.caliper D4-D4D 21.caliper D-DD4 22.straightedge (ruler) HI4 , gets the point D5 23.this procedure with a compass and a ruler is actually a series of numbers (finite, infinite)

when you look at the other three parts , then you will make surethat it is possible, see you for 20-30h,

The solution of mathematical tasks in the ancient Greek Trisection of angles angle=0° - no solution 180°>angle>0° - general solution (consists of 4 parts) the first part 1.ruler AB 2.ruler AC 3.caliper A-AD 4.ruler DE 5.caliper D-DE 6.caliper E-DE 7.ruler FG intersects DE the point H ,DH=HE 8.caliper H-HE