https://www.geogebra.org/m/CukhmEVy
straightedge slip on the point B - line i
divider FIG , slides on straightedge , after slipping point F straighte line BC , point G describes lokus1
section lokus1 and line k point J , when changing the angle, the point J must be manually set to the intersection
construction of regular polygon is possible by means of the proportion of the angle
Whether you are from the previous understand how it works proportions of angles, or that you explain step by step?
load the attachment
https://www.geogebra.org/m/KrYWuNEv
see the description of the construction
slider -[TEX] alpha[/TEX] -select the angle
slider - point P - ruler with a socket, point Q must be line n
valid for the odd [tex]a={3,5,7,9,11,...}[/tex]
Proper ninth angle
- straight line [tex]A_1A_2[/tex]
- straightedge and compass ,[tex]\frac{A_1A_2}{10}[/tex] , point [tex]A_4[/tex] , [tex]a+1[/tex] , [tex]a=9. followed by .9+1=10[/tex]
- straightedge and compass , [tex]A_1A_3[/tex] normal [tex]A_1A_2 [/tex] , angle [tex]C_3C_1C_2=90^o[/tex]
- compass [tex]A_1A_4[/tex] , from point [tex]A_5[/tex]
- straightedge , straight line [tex]A_4A_5[/tex]
- straightedge and compass , bisection arc [tex]A_2A_3[/tex] , point [tex]A_6[/tex]
YOU TRY TO KEEP ... Figure down
- compass [tex]C_2F_6[/tex] , from point [tex]E_6[/tex] , point [tex]A_{12}[/tex]
- compass [tex]C_2F_6[/tex] , from point [tex]E_7 [/tex], point [tex]A_{13}[/tex]
- straightedge , semi-line [tex]C_2A_{11}[/tex]
- straightedge , semi-line [tex]C_2A_{12}[/tex]
trisection is complete, any error !!!
this is true for angles [tex]180^o<\alpha<0^o [/tex], larger angles of first division of the [tex]180^o[/tex]
are you ready for the process of construction of the regular polygon
- straightedge , straight line [tex]C_2F_1[/tex] , [tex]C_2F_1=C_2C_3[/tex]
- compass [tex]C_2E_5[/tex] , from point [tex]C_2[/tex] , point[tex]F_3[/tex]
- straightedge and compass , straight line the normal to [tex]C_2F_3[/tex]
- compass [tex]D_6D_3[/tex] , from point [tex]C_2[/tex] , point[tex]F_4[/tex]
- straightedge ,straight line extension [tex]C_2F_4[/tex]
- compass [tex]D_7D_2[/tex] , from point [tex]C_2[/tex] , point [tex]F_5[/tex]
- straightedge and compass , normal from point [tex]F_5[/tex] na duž [tex]C_2F_1[/tex] , point [tex]F_6[/tex]
Solution - in the picture below
- straightedge and compass , perpendicular [tex] b_1[/tex] straight line [tex]C_2D_5[/tex]
- straightedge and compass , perpendicular [tex]b_2[/tex] on the [tex]b_1[/tex] from point [tex]D_3[/tex] , straight line [tex]D_6D_3[/tex]
- straightedge and compass , perpendicular [tex]b_3[/tex] on the [tex]b_1[/tex] from point [tex]D_2[/tex] , straight line [tex]D_7D_2[/tex]
YOU TRY TO KEEP ... Figure down
[tex]F_1[/tex] is located on the arc [tex]C_3C_1 [/tex], [tex]C_3F_1=C_1F_1[/tex]
- straightedge and compass , perpendicular to the line [tex]a_1[/tex] straight line [tex]C_2C_7[/tex]
- compass [tex]C_3D_5[/tex] , in point [tex]C_2[/tex] , points [tex]E_1 and E_2[/tex]
- straightedge and compass , perpendicular to the line [tex]a_2[/tex] line [tex]a_1[/tex] , point [tex]E_3[/tex]
- straightedge and compass , perpendicular to the line [tex]a_3[/tex] line [tex]a_1[/tex] , point [tex]E_3[/tex]
- straighedge , straight line [tex]E_3E_4[/tex] , point [tex]E_5[/tex]
- straightedge and compass , perpendicular to the line [tex]a_4[/tex] straight line [tex]C_5C_6[/tex] , point [tex]E_6[/tex]
- straightedge and compass , perpendicular to the line [tex]a_5[/tex] straight line [tex]C_5C_6[/tex] , point [tex]E_7[/tex]
YOU TRY TO KEEP ... Figure down
that's the way, without the knowledge of what is happening in the sphere of
- given the angle [tex]C_1C_2C_3[/tex]
- straightedge and compass , straight line [tex]C_2C_3[/tex] , is divided into two equal parts, point [tex]C_4[/tex]
- straightedge and compass , straight line [tex]C_2C_4[/tex] , is divided into two equal parts, point [tex]C_5[/tex]
- compass [tex]C_2C_5[/tex] , from the point [tex]C_2[/tex], point [tex]C_6[/tex]
- straightedge and compass, angle bisection [tex]C_1C_2C_3[/tex] , point [tex]C_7[/tex]
- straightedge , straight line [tex]C_2C_7[/tex]
- compass [tex]C_2C_3[/tex] , from the point [tex]C_2[/tex] , arc [tex]C_3C_1[/tex]
- compass [tex]C_5C_6[/tex] , from the point [tex]C_3[/tex] , point [tex]D_1[/tex]
- compass [tex]C_5C_6[/tex] , from the point [tex]D_1[/tex] , point [tex]D_2[/tex]
- compass [tex]C_5C_6[/tex] , from the point [tex]D_2[/tex] , point[tex]D_3[/tex]
- straightedge , straight line [tex]C_3D_3[/tex]
- straightedge and compass, angle bisection [tex]C_3D_3[/tex] , point [tex]D_4[/tex]
- straightedge , straight line [tex]C_2D_4[/tex] , point [tex]D_5[/tex]
YOU TRY TO KEEP ... Figure down
- point [tex]A_1[/tex]
- compass , from point [tex]A_1[/tex] , circular arc [tex]A_2A_5[/tex]
- straightedge , in points [tex]A_1 , A_2[/tex] , straight line [tex]A_1A_2[/tex]
- straightedge , in points [tex]A_1 , A_5[/tex] , straight line [tex]A_1A_5[/tex]
- point [tex]A_7[/tex] , requirement [tex]A_1 A_7<\frac{A_1A_2}{3}[/tex]
- compass [tex]A_1 , A_7[/tex] , from point [tex]A_1 [/tex] , point [tex]A_8[/tex]
- straightedge , in points [tex]A_7, A_8[/tex] , straight line [tex]A_7A_8[/tex]
- bisection circular arc [tex]A_2A_5[/tex] , point [tex]B_1[/tex]
- straightedge , in points [tex]A_1, B_1[/tex] , straight line [tex]A_1B_1[/tex] , point [tex]B_2[/tex]
- compass [tex]A_1A_2[/tex] , from point [tex]A_1[/tex] , circular arc [tex]A_9B_3[/tex]
- compass [tex]A_7A_8[/tex] , from point [tex]A_9[/tex] , point [tex]A_{11}[/tex]
- compass [tex]A_7A_8[/tex] , from point [tex]A_{11}[/tex] , point [tex]A_{12}[/tex]
- compass [tex]A_7A_8[/tex] , from point [tex]A_{12}[/tex] , point [tex]A_{10}[/tex]
- straighedge , in point [tex]A_9 ,A_{10}[/tex] , straigt line [tex]A_9A_{10}[/tex]
- bisection circular arc [tex]A_9A_{10}[/tex] , ppint [tex]B_4[/tex]
- straightedge , in points [tex]A_1, B_4[/tex] , straight line [tex]A_1B_4[/tex] , point [tex]B_5[/tex]
To be continued ...
tendon [tex]A_7A_8[/tex] , [tex]A_9A_{10}[/tex] ,[tex]A_{11}A_{12}[/tex] arcs are parallel to each other
when we look at the top sphere of circular arcs
[tex]A_1A_6A_2,A_1A_6A_3,A_1A_6A_4,A_1A_6A_5[/tex]
- seem straight[tex]A_1A_2,A_1A_3,A_1A_4,A_1A_5[/tex] or [tex]A_6A_2,A_6A_3,A_6A_4,A_6A_5[/tex]
[tex]A_1[/tex] - center ball
[tex]A_1 A_2[/tex] - radius ball
[tex]A_1A_7A_8[/tex] - circular arc
[tex]A_1A_9A_{10}[/tex] - circular arc three times higher than [tex]A_1A_7A_8[/tex] [tex]A_1A_7A_8=A_1A_9A_{11}=A_1A_{11}A_{12}=A_1A_{12}A_{10}[/tex]
[tex]A_1A_2A_5[/tex] - circular arc
[tex]A_1A_2A_3=A_1A_3A_4=A_1A_4A_5[/tex] , points [tex]A_2 , A_3 , A_4 , A_5[/tex] on the best circle the ball (or sphere)
[tex]A_1A_6A_2=A_1A_6A_3=A_1A_6A_4=A_1A_6A_5[/tex] - circular arcs , are circular arcs on a spher
whether the circular arc that looks like straight?
applies this photo
bisection angle DAC is obtained by point J
along AJ
GF section circular arc and along the AJ, obtained point L
AF divider, from the point J, we get the point O
divider AF, from point A circle c1
divider GL, from the point J, the circuit d1, get the points P and Q
basic angle CAB [tex]45^o[/tex]
starting angle EAB [tex]135^o[/tex] consists of the sum of the angles CAB [tex]45^o[/tex] DAC [tex]45^o [/tex]EAD [tex]45^o[/tex]
DC straightedge the normal to the point D , gets the point F
AF divider from point A, we get the point G
divider AB from point F, divider AB from point G, we get the point H
HG divider from point H, creates a circular arc FG
difference angle IAB [tex]30^o[/tex]
section IA and longer circular arc FG is a point J
- previous post was in error -
Required accessories - pencil, compass, unmarked straightedge
basic angle - can be any angle that can be construction using compass and unmarked straightedge, angle CAB [TEX] 45 ^ o[/TEX]
starting angle - sum of 2, 3, 4, 5, ... basic angles , EAB [TEX] 135 ^ o[/TEX]
sum angles CAB [TEX] 45 ^ o[/TEX] DAC [TEX] 45 ^ o[/TEX] EAD [TEX] 45 ^ o[/TEX]
difference angle - the angle which increases or decreases the starting angle. difference starting angle and the angle of whom do not know the measure , this angle is known to see a procedure HAB [TEX] 30 ^ o[/TEX]
straightedge AB is divided into three parts AF , how we have a basis in the angles starting angle
divider AF from point A the circular arc FG
section straightedge AH the circular arc FG , point I
straightedges FG , ED
--------------------------------------
will continue - if there are errors
Required accessories - pencil, compass, unmarked straightedge
Odd proportion angles (there is a proportion of the steam angles) of the element 3 (which may be 5,7,9, ...) and the base angle [TEX] 45 ^ o[/TEX] (which can be any angle which is obtainable by means of compass and straightedge)
The angle CAB [TEX]45 ^ o[/TEX] can be obtained with a compass and unmarked ruler, he added angles (each have [TEX]45 ^ o[/TEX]) DAC and EAD, obtained angle EAB [TEX]135 ^ o[/TEX]is the starting angle
Merge points E (D, C, B) and get a longer ED (DC, CB)
Along the DC from the point C draw is normal that intersects the segment AB, the intersection is a point G
Divider AG and from point G draw a circular arc to a longer EA and H get the point, and the arc GH, join the dots G and H and get along GH
Longer GH (ED, DC, CB) are equal, the arc EB's first circular arc can be made smaller or larger with a constant radius AB, arc GH is the second circular arc can be made smaller or larger with a constant radius AG
INCREASING THE ANGLE
the starting angle EAB add angle FAE [TEX]15 ^ o[/TEX] get the angle FAB [TEX]150 ^ o[/TEX] - continued in the next post
Divider DH, referred to in Clause, cuts a circular arc FG, obtained point J
should be - Divider DH, from the point I, intersects a circular arc FG, obtained point J
I found how to determine the proportion of angles, and thus solve the trisection angles
Given the angle CAB
Divider AD (Point D is the free choice of the branch AB), from point A, creates a circular arc ED
Bisection circular arc ED obtained item H
Divider AD, from point D, obtained point L
Divider AD, from the point of L, we get the point F
Divider AF, from point A, creates a circular arc FG
Divider DH, from the point F, intersects a circular arc FG, obtained point I
Divider DH, referred to in Clause, cuts a circular arc FG, obtained point J
Divider FJ, from point J, cuts a circular arc FG, obtained point K
[tex] \angle GAK = \angle KAJ = \angle JAF = {\angle GAF \over 3} [/tex]
Next - my character
- Solution of the construction of a regular n (n> 2) of the polygon
Gibt es in diesem Forum irgend jemanden, der von diesem Beitrag profitiert? Was soll der Sinn dessen sein? Kann mich da mal bitte jemand aufklären?
if there is a mathematical procedure to solve the numbers (4, 2) and the interval [TEX]\{0,2\}\cup\{4,6\}[/TEX] for a solution which I PASSED (all 12 forms of addition) then show and prove that what I have he presented is not new in mathematics
