Trisection of angles-similarly

[bob boben]

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

 

[Deleted member 4993]

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

View attachment 1349

1.ruler AB
2.ruler AC
3.caliper A-AD
4.ruler DE

View attachment 1350

5.caliper D-DE
6.caliper E-DE
7.ruler FG intersects DE the point H ,DH=HE

View attachment 1351

8.caliper H-HE

If think you have trisected a general angle - with straight-edge and collapsible compass - send your proof to mathematics department of Cambridge University (UK). I think they have a prize (1 million pounds) for the person who can do so. The prize remained without valid claim since the days of Hilbert (I think).

 

[mmm4444bot]

I'm not sure why you posted this information. Are you waiting for somebody to supply part 4? :?

 

[bob boben]

If think you have trisected a general angle - with straight-edge and collapsible compass - send your proof to mathematics department of Cambridge University (UK). I think they have a prize (1 million pounds) for the person who can do so. The prize remained without valid claim since the days of Hilbert (I think).

can you send me link (proof to mathematics department of Cambridge University (UK)), I have notfound

 

[bob boben]

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)
After this follows an experiment I discovered that it canbe trisection (part 3, part 4) This is just an introduction,

 

[bob boben]

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

:idea::idea::idea::idea::idea:

 

[bob boben]

<font color="#0040ff">third part </font><br>
computer program-coreldraw 13<br>
<a href="http://www.fileserve.com/file/sZpPSyf/TRISEKIJA_UGLA-DOKAZ.cdr" target="_blank">http://www.fileserve.com/file/sZpPSyf/T ... -DOKAZ.cdr</a><br>
png image<br>
<a href="http://www.fileserve.com/file/EpEKHxx/png.slika.trisection.png" target="_blank">http://www.fileserve.com/file/EpEKHxx/p ... ection.png</a><br>
the picture is<br>
-circle<br>
-diameter circle AB<br>
-points on the diameter of the circle A(0°),B(180°), C (AD=DC=CB) , D , F(center of the circle ,FB-radius circle)<br>
-tendon EF<br>
<br>
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).<br>
I shared a circle with corners (second part , 9 to 23).<br>
I found the string (EF, 3.75 ° -187.5 °) which passes through the point C ,which means that we have the angle trisection.<br>
<font color="#0080ff">fourth part</font><br>
<img src="http://www.freemathhelp.com/forum/attachment.php?attachmentid=1355&stc=1" attachmentid="1355" alt="" id="vbattach_1355" class="previewthumb"><br>
24.caliper D-DD5-(22.)<br>
25.caliper E-DD4-(19.)<br>
26.straightedge (ruler) D4D5 , gets the point H1<br>
27.straightedge (ruler) AH1<br>
28.caliper H1-H1D , gets the point H2<br>
29.straightedge (ruler) AH2<br>
<br>
<img src="images/smilies/icon_idea.gif" alt="" title="" smilieid="70" class="inlineimg" border="0"><img src="images/smilies/icon_idea.gif" alt="" title="" smilieid="70" class="inlineimg" border="0"><img src="images/smilies/icon_idea.gif" alt="" title="" smilieid="70" class="inlineimg" border="0"><img src="images/smilies/icon_idea.gif" alt="" title="" smilieid="70" class="inlineimg" border="0"><img src="images/smilies/icon_idea.gif" alt="" title="" smilieid="70" class="inlineimg" border="0">

 

[bob boben]

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