Trisection of angles-similarly
Posted: 04 September 2011 06:22 AM   [ Ignore ]
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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

c8ay0p1o0iufughx3o81.png

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

djwbdd6r48czpuqorsxf.png

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

49vx9rqw92tr0mynuc2.png

8.caliper H-HE

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Posted: 04 September 2011 05:57 PM   [ Ignore ]   [ # 1 ]
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OK, nice job of bisecting the angle, but a bit more steps that usual.  smile

I’m waiting for the trisection.

Occam

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Posted: 04 September 2011 06:19 PM   [ Ignore ]   [ # 2 ]
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I remember as a kid I made an attempt at trisecting an angle for a science fair project when I was in, oh, maybe sixth grade. I didn’t know anything about geometry, but details like that never stopped me in the past (or future). Anyway, I came up with a clever scheme that kinda-sorta worked, but was totally wacko.

No, I didn’t win any prizes.

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Posted: 04 September 2011 06:29 PM   [ Ignore ]   [ # 3 ]
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As I recall one must not use a ruler because measurements are never infinitely precise.  Instead, one must use a straight-edge. 

When I very young and first heard about it, I started out trying to a 90 degree angle and did it quite easily.  I was pissed when I showed it to my teacher and she said condescendingly, “Very good.  You’ve found the only angle that can be trisected.” 

Occam

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Posted: 05 September 2011 02:23 AM   [ Ignore ]   [ # 4 ]
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Occam. - 04 September 2011 05:57 PM

OK, nice job of bisecting the angle, but a bit more steps that usual.  smile

I’m waiting for the trisection.

Occam

It’s a trisection of angle A. He just didn’t continue two of the lines through A after hitting it.

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Posted: 05 September 2011 05:08 AM   [ Ignore ]   [ # 5 ]
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For those who don’t know, angle trisection with compass (or caliper) and straightedge has been proven impossible for all but a very small subset of cases. For those not familiar with the issue, check out (e.g.) the Wiki page, in particular:

“Because it is defined in simple terms, but complex to prove unsolvable, the problem of angle trisection is a frequent subject of pseudomathematical attempts at solution by naive enthusiasts.”

It can be done with a marked straightedge (i.e. a ruler), as mentioned in the Wiki.

[ Edited: 05 September 2011 05:10 AM by dougsmith ]
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Posted: 05 September 2011 07:15 AM   [ Ignore ]   [ # 6 ]
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Aha, that’s what you guys are talking about. Neat.

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Posted: 05 September 2011 09:13 AM   [ Ignore ]   [ # 7 ]
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second part
k805u2rf67l5c5gaj7qx.png
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
wagamsh3t8pjrngk8ej.png
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
r4kpwyxdb3w4vz0b3x2i.png
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 can be trisection (part 3, part 4) This is just an introduction,

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Posted: 06 September 2011 04:18 AM   [ Ignore ]   [ # 8 ]
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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
uo41ab2zkvwmzujbwc3b.png
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

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Posted: 06 September 2011 08:53 AM   [ Ignore ]   [ # 9 ]
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Points D4 and D5 are determined independently of Angle A; therefore point H1 is also independent of Angle A. In other words, as Angle A varies, point H1 does not move, meaning that it cannot trisect all values of Angle A.

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Posted: 06 September 2011 09:26 AM   [ Ignore ]   [ # 10 ]
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Chris Crawford - 06 September 2011 08:53 AM

Points D4 and D5 are determined independently of Angle A; therefore point H1 is also independent of Angle A. In other words, as Angle A varies, point H1 does not move, meaning that it cannot trisect all values of Angle A.

only 180 °> angle A> 0 °
point H1 is dependent on angle A (180 °> angle A> 0 ° ) but is determined through a long DE (diameter of circle) ,
and points (D4 , D5) are dependent on the angle obtained by the procedure (second part , 9 to 23)
angle A=0° - no solution
angle A=180° -following solution
360°>angle A>180° -following solution
angle A=360° -following solution

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Posted: 07 September 2011 03:50 AM   [ Ignore ]   [ # 11 ]
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angle=180°
djswrqryz59x9b3ev8n4.png
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

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Posted: 08 September 2011 12:09 PM   [ Ignore ]   [ # 12 ]
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The trick about geometry is knowing to do it accurately in your brain even when your drawing tools aren’t accurate, the ancient Greeks with their math drawn in the sand amazingly did that trick, and did it correctly, and did it thousands of years before computers!  The trick here is that if bisect means dividing by exactly 2, then trisect means dividing by exactly 3.

The last diagram is not exactly trisecting, it is showing the old mistake of approximately trisecting the angle.  Looking at the diagram of step 35. you started with a 180° angle ACB, a semi-circle is drawn from one end of line ACB to the other (approximately).  The caliper is set to the length CD in step 31., which is the radius of the semi-circle.  In the next few step the caliper is walked from point D to E to F around the circumference, only two steps which makes the diagram look like one more will trisect ACB, but that step is missing from the diagram and if they had drawn the third step we’d see that it would fall short of the semi-circular circumference.  There is a factor that relates circumference to radius, can anyone guess what fator?  π = 3.14159, and notice that is aprroximately three, but a bit more.  The circumference of the semi-circle is a multiple of π, and when the caliper only walks three steps (of r), it falls short by π - 3 = 0.14159, and so the angle is not trisected exactlydowner

  C = 2*π*r   for the full circle.
  C = π*r   for the semi-circle, 3*r isn’t enough to make it around the semi-circle.

You have nice diagrams bob boben smile, but they don’t add up with what you’re saying about trisecting.

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Posted: 08 September 2011 05:05 PM   [ Ignore ]   [ # 13 ]
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I fail to see how Bob’s 180deg angle solution is incorrect. CD has same length as CE, which has same length as DE, which makes CDE an equilateral triangle, meaning it has internal angles of 60deg. 60deg * 3 = 180deg.

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Posted: 08 September 2011 06:38 PM   [ Ignore ]   [ # 14 ]
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trisection

Yes domokato I agree, you are right.  I was mistaken because as the calipers walk from one point on the semi-circle to the next, the distance between them is a strait line, that is the radius, so these strait lines form the perimeter of a semi-hexagon (red, half-way round, 180 degrees) made of equilateral triangles.  The calipers don’t measure the circumference of the semi-circle (green).  The difference between the perimeter and the circumference is π - 3 = 0.14159 (times the radius).  Diagram 35. is a trisection, bob boben.

[ Edited: 08 September 2011 06:49 PM by jump_in_the_pit ]
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Posted: 29 October 2011 11:30 PM   [ Ignore ]   [ # 15 ]
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dougsmith - 05 September 2011 05:08 AM

For those who don’t know, angle trisection with compass (or caliper) and straightedge has been proven impossible for all but a very small subset of cases. For those not familiar with the issue, check out (e.g.) the Wiki page, in particular:

“Because it is defined in simple terms, but complex to prove unsolvable, the problem of angle trisection is a frequent subject of pseudomathematical attempts at solution by naive enthusiasts.”

It can be done with a marked straightedge (i.e. a ruler), as mentioned in the Wiki.


Hi - I am new here.

Is this of interest: http://users.tpg.com.au/musodata/trisection/trisecting_any_angle.htm ?

Jayeskay.

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