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December 29th, 2016, 08:37 AM   #1
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From: leskovac

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misli

1a.jpg
- given the angle $\displaystyle C_1C_2C_3$
- straightedge and compass , straight line $\displaystyle C_2C_3$ , is divided into two equal parts, point $\displaystyle C_4$
- straightedge and compass , straight line $\displaystyle C_2C_4$ , is divided into two equal parts, point $\displaystyle C_5$
- compass $\displaystyle C_2C_5$ , from the point $\displaystyle C_2$, point $\displaystyle C_6$
- straightedge and compass, angle bisection $\displaystyle C_1C_2C_3$ , point $\displaystyle C_7$
- straightedge , straight line $\displaystyle C_2C_7$

- compass $\displaystyle C_2C_3$ , from the point $\displaystyle C_2$ , arc $\displaystyle C_3C_1$
- compass $\displaystyle C_5C_6$ , from the point $\displaystyle C_3$ , point $\displaystyle D_1$
- compass $\displaystyle C_5C_6$ , from the point $\displaystyle D_1$ , point $\displaystyle D_2$
- compass $\displaystyle C_5C_6$ , from the point $\displaystyle D_2$ , point$\displaystyle D_3$
- straightedge , straight line $\displaystyle C_3D_3$
- straightedge and compass, angle bisection $\displaystyle C_3D_3$ , point $\displaystyle D_4$
- straightedge , straight line $\displaystyle C_2D_4$ , point $\displaystyle D_5$

YOU TRY TO KEEP ... Figure down
1b.jpg
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December 29th, 2016, 03:58 PM   #2
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What is the question?
Thanks from 123qwerty
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December 30th, 2016, 02:47 AM   #3
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Quote:
Originally Posted by mathman View Post
What is the question?
- what I told you he presented represents a method for trisection angle (
the process is similar to a regular n-polygon )

- whether you can proceed in the picture http://mymathforum.com/attachments/m...8-misli-1b.jpg ...
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January 1st, 2017, 03:45 AM   #4
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- straightedge and compass , perpendicular to the line $\displaystyle a_1$ straight line $\displaystyle C_2C_7$
- compass $\displaystyle C_3D_5$ , in point $\displaystyle C_2$ , points $\displaystyle E_1 and E_2$
- straightedge and compass , perpendicular to the line $\displaystyle a_2$ line $\displaystyle a_1$ , point $\displaystyle E_3$
- straightedge and compass , perpendicular to the line $\displaystyle a_3$ line $\displaystyle a_1$ , point $\displaystyle E_3$
- straighedge , straight line $\displaystyle E_3E_4$ , point $\displaystyle E_5$
- straightedge and compass , perpendicular to the line $\displaystyle a_4$ straight line $\displaystyle C_5C_6$ , point $\displaystyle E_6$
- straightedge and compass , perpendicular to the line $\displaystyle a_5$ straight line $\displaystyle C_5C_6$ , point $\displaystyle E_7$

YOU TRY TO KEEP ... Figure down
1cc.jpg
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January 1st, 2017, 04:06 PM   #5
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How do you know that the angle has been trisected? I would be surprised to say the least.
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January 1st, 2017, 11:43 PM   #6
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proportioned angles, which is possible on a sphere
arc $\displaystyle A_7A_8=A_9A_{11}=A_{11}A_{12}=A_{12}A_{10}=\frac{A _9A_{10}}{3}$ , picture below
http://drive.google.com/file/d/0B54I...U4Xy1hakk/view

when you look at the sphere above, looks like the angle divided into three equal parts ,
http://drive.google.com/file/d/0B54I...FBY0E3ZVk/view

I found that this procedure can be performed in a plane
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January 2nd, 2017, 06:54 AM   #7
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First, it has been known for literally millennia how to trisect an angle. Second, it has been proved that it is impossible to trisect an angle in a plane using ONLY a collapsing compass and unmarked straight edge. Third, a "proof" that relies on "looks like" is not a proof.
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January 2nd, 2017, 08:23 AM   #8
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- straightedge and compass , perpendicular $\displaystyle b_1$ straight line $\displaystyle C_2D_5$
- straightedge and compass , perpendicular $\displaystyle b_2$ on the $\displaystyle b_1$ from point $\displaystyle D_3$ , straight line $\displaystyle D_6D_3$
- straightedge and compass , perpendicular $\displaystyle b_3$ on the $\displaystyle b_1$ from point $\displaystyle D_2$ , straight line $\displaystyle D_7D_2$

YOU TRY TO KEEP ... Figure down
$\displaystyle F_1$ is located on the arc $\displaystyle C_3C_1 $, $\displaystyle C_3F_1=C_1F_1$
1d.jpg
Quote:
Originally Posted by JeffM1 View Post
Second, it has been proved that it is impossible to trisect an angle in a plane using ONLY a collapsing compass and unmarked straight edge
there is one picture and two text, and check whether my action is correct or has errors
Quote:
Originally Posted by JeffM1 View Post
Third, a "proof" that relies on "looks like" is not a proof.
I should not have to use the name "looks like"
January 2nd, 2017 12:43 AM - Thread
$\displaystyle A_6A_2$ the angle arc $\displaystyle 90^o$
$\displaystyle A_6A_2$ the straight line $\displaystyle sin90^o$
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January 2nd, 2017, 09:33 AM   #9
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There has not been any question for millennia that a right angle can be trisected. The question is not about some finite set of special angles, but about any arbitrary angle whatsoever.

You have not taken an arbitrary angle and then demonstrated that your construction trisects that angle. You simply describe a construction (although it only seems to work if you can keep the figure supine) but never explain why the resulting angle is one third the original angle.
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January 4th, 2017, 01:48 AM   #10
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- straightedge , straight line $\displaystyle C_2F_1$ , $\displaystyle C_2F_1=C_2C_3$
- compass $\displaystyle C_2E_5$ , from point $\displaystyle C_2$ , point$\displaystyle F_3$
- straightedge and compass , straight line the normal to $\displaystyle C_2F_3$
- compass $\displaystyle D_6D_3$ , from point $\displaystyle C_2$ , point$\displaystyle F_4$
- straightedge ,straight line extension $\displaystyle C_2F_4$
- compass $\displaystyle D_7D_2$ , from point $\displaystyle C_2$ , point $\displaystyle F_5$
- straightedge and compass , normal from point $\displaystyle F_5$ na duž $\displaystyle C_2F_1$ , point $\displaystyle F_6$

Solution - in the picture below
1e.jpg
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