\(A_1\) - center ball \(A_1 A_2\) - radius ball \(A_1A_7A_8\) - circular arc \(A_1A_9A_{10}\) - circular arc three times higher than \(A_1A_7A_8\) \(A_1A_7A_8=A_1A_9A_{11}=A_1A_{11}A_{12}=A_1A_{12}A_{10}\) \(A_1A_2A_5\) - circular arc \(A_1A_2A_3=A_1A_3A_4=A_1A_4A_5\) , points \(A_2 , A_3 , A_4 , A_5\) on the best circle the ball (or sphere) \(A_1A_6A_2=A_1A_6A_3=A_1A_6A_4=A_1A_6A_5\) - circular arcs , are circular arcs on a spher Please Register or Log in to view the hidden image! whether the circular arc that looks like straigh ?

I recognize that English is not your first language, and that is not a problem, but I think we need a little more description of the actual question you are asking. Can you rephrase? Also, your subject line is non-descriptive.

blame the google translator I found the process of proportion angles However, I will gradually become familiar with the procedures of when we look at the top sphere of circular arcs \(A_1A_6A_2,A_1A_6A_3,A_1A_6A_4,A_1A_6A_5\) - seem straight line\(A_1A_2,A_1A_3,A_1A_4,A_1A_5\) or \(A_6A_2,A_6A_3,A_6A_4,A_6A_5\) Please Register or Log in to view the hidden image!

trisection angle and n-regular polygon using a compass and straightedge tendon \(A_7A_8\) , \(A_9A_{10}\) ,\(A_{11}A_{12}\) arcs are parallel to each other Please Register or Log in to view the hidden image!

Please post your method for trisecting an angle using compass and straightedge. A list of the steps that you use would be good. Thankyou.

- point \(A_1\) - compass , from point \(A_1\) , circular arc \(A_2A_5\) - straightedge , in points \(A_1 , A_2\) , straight line \(A_1A_2\) - straightedge , in points \(A_1 , A_5\) , straight line \(A_1A_5\) - point \(A_7\) , requirement \(A_1 A_7<\frac{A_1A_2}{3}\) - compass \(A_1 , A_7\) , from point \(A_1 \) , point \(A_8\) - straightedge , in points \(A_7, A_8\) , straight line \(A_7A_8\) - bisection circular arc \(A_2A_5\) , piont \(B_1\) - straightedge , in points \(A_1, B_1\) , straight line \(A_1B_1\) , point \(B_2\) - compass \(A_1A_2\) , from point \(A_1\) , circular arc \(A_9B_3\) - compass \(A_7A_8\) , from point \(A_9\) , point \(A_{11}\) - compass \(A_7A_8\) , from point \(A_{11}\) , point \(A_{12}\) - compass \(A_7A_8\) , from point \(A_{12}\) , point \(A_{10}\) - straighedge , in point \(A_9 ,A_{10}\) , straigt line \(A_9A_{10}\) - bisection circular arc \(A_9A_{10}\) , piont \(B_4\) - straightedge , in points \(A_1, B_4\) , straight line \(A_1B_4\) , point \(B_5\) Please Register or Log in to view the hidden image! To be continued ...

Probably for the same reason, I don't understand this, either. How are points A7,A8 and A9 used to determine the location of A11?

- view http://www.sciforums.com/attachments/a-3d-sfera-png.1261/ , what is on the ball switch to plane Please Register or Log in to view the hidden image! - \(A_1\) the center of the circle, whose parts are circular arcs \(A_7A_8\) ,\(A_9A_{10}\), \(A_2A_5\) - \(A_9B_3\) , We know that the circular arc three times higher than \(A_7A_8\) ,since we do not know what the arc is listed point \(B_3\) - point \(A_2\) has nothing to do with the circular arc \(A_9B_3\) - \(A_9A_{10}\) - circular arc three times higher than \(A_7A_8\) ,\(A_7A_8=A_9A_{11}=A_{11}A_{12}=A_{12}A_{10}\) I hope that you understand

Is it your assertion that the points A9, A11, A12 and A10 are equally spaced and therefore that angles such as A9A1A11 and A11A1A12 are equal? It is not clear to me how you split the arc A9B3 into three equal parts using the compass and straightedge. Are you saying that A1A7 has length (1/3) A1A2? If so, tell me how you did the 1/3 division using compass and straightedge.

ball arc \( A_7A_8=A_9A_{11}=A_{11}A_{12}=A_{12}A_{10}\) \( A_9A_{11}+A_{11}A_{12}+A_{12}A_{10}=A_9A_{10}\) \( A_2A_3=A_3A_4=A_4A_5\) \( A_2A_3+A_3A_4+A_4A_5=A_2A_5\) \( A_6A_2=A_6A_3=A_6A_4=A_6A_5\) angle \( A_7A_1A_8=A_9A_1A_{11}=A_{11}A_1A_{12}=A_{12}A_1A_{10}\) \( A_9A_1A_{11}+A_{11}A_1A_{12}+A_{12}A_1A_{10}=A_9A_1A_{10}\) \( A_2A_1A_3=A_3A_1A_4=A_4A_1A_5\) \( A_2A_1A_3+A_3A_1A_4+A_4A_1A_5=A_2A_1A_5\) \( A_6A_1A_2=A_6A_1A_3=A_6A_1A_4=A_6A_1A_5\) arc \( A_9B_3\) is not divided into three parts , It is applied to tendons \( A_7A_8\) three times , to give a tendon \( A_9A_{10}\) proportions staight line is known in primary school , I gave \( A_1A_7<\frac{A_1A_2}{3}\) , points \( A_{11}A_{12}\) should be in angle \( A_2A_1A_5\) in plane

that's the way, without the knowledge of what is happening in the sphere of Please Register or Log in to view the hidden image! - given the angle \(C_1C_2C_3\) - straightedge and compass , straight line \(C_2C_3\) , is divided into two equal parts, point \(C_4\) - straightedge and compass , straight line \(C_2C_4\) , is divided into two equal parts, point \(C_5\) - compass \(C_2C_5\) , from the point \(C_2\), point \(C_6\) - straightedge and compass, angle bisection \(C_1C_2C_3\) , point \(C_7\) - straightedge , straight line \(C_2C_7\) - compass \(C_2C_3\) , from the point \(C_2\) , arc \(C_3C_1\) - compass \(C_5C_6\) , from the point \(C_3\) , point \(D_1\) - compass \(C_5C_6\) , from the point \(D_1\) , point \(D_2\) - compass \(C_5C_6\) , from the point \(D_2\) , point\(D_3\) - straightedge , straight line \(C_3D_3\) - straightedge and compass, angle bisection \(C_3D_3\) , point \(D_4\) - straightedge , straight line \(C_2D_4\) , point \(D_5\) YOU TRY TO KEEP ... Figure down Please Register or Log in to view the hidden image!

- straightedge and compass , perpendicular to the line \(a_1\) straight line \(C_2C_7\) - compass \(C_3D_5\) , in point \(C_2\) , points \(E_1 and E_2\) - straightedge and compass , perpendicular to the line \(a_2\) line \(a_1\) , point \(E_3\) - straightedge and compass , perpendicular to the line \(a_3\) line \(a_1\) , point \(E_3\) - straighedge , straight line \(E_3E_4\) , point \(E_5\) - straightedge and compass , perpendicular to the line \(a_4\) straight line \(C_5C_6\) , point \(E_6\) - straightedge and compass , perpendicular to the line \(a_5\) straight line \(C_5C_6\) , point \(E_7\) YOU TRY TO KEEP ... Figure down Please Register or Log in to view the hidden image!

- straightedge and compass , perpendicular \( b_1\) straight line \(C_2D_5\) - straightedge and compass , perpendicular \(b_2\) on the \(b_1\) from point \(D_3\) , straight line \(D_6D_3\) - straightedge and compass , perpendicular \(b_3\) on the \(b_1\) from point \(D_2\) , straight line \(D_7D_2\) YOU TRY TO KEEP ... Figure down \(F_1\) is located on the arc \(C_3C_1 \), \(C_3F_1=C_1F_1\) Please Register or Log in to view the hidden image!

- straightedge , straight line \(C_2F_1\) , \(C_2F_1=C_2C_3\) - compass \(C_2E_5\) , from point \(C_2\) , point\(F_3\) - straightedge and compass , straight line the normal to \(C_2F_3\) - compass \(D_6D_3\) , from point \(C_2\) , point\(F_4\) - straightedge ,straight line extension \(C_2F_4\) - compass \(D_7D_2\) , from point \(C_2\) , point \(F_5\) - straightedge and compass , normal from point \(F_5\) na duž \(C_2F_1\) , point \(F_6\) Solution - in the picture below Please Register or Log in to view the hidden image!

- compass \(C_2F_6\) , from point \(E_6\) , point \(A_{12}\) - compass \(C_2F_6\) , from point \(E_7 \), point \(A_{13}\) - straightedge , semi-line \(C_2A_{11}\) - straightedge , semi-line \(C_2A_{12}\) trisection is complete, any error !!! this is true for angles \(180^o<\alpha<0^o \), larger angles of first division of the \(180^o\) are you ready for the process of construction of the regular polygon

valid for the odd \(a={3,5,7,9,11,...}\) Proper ninth angle Please Register or Log in to view the hidden image! - straight line \(A_1A_2\) - straightedge and compass ,\(\frac{A_1A_2}{10}\) , point \(A_4\) , \(a+1\) , \(a=9. followed by .9+1=10\) - straightedge and compass , \(A_1A_3\) normal \(A_1A_2 \) , angle \(C_3C_1C_2=90^o\) - compass \(A_1A_4\) , from point \(A_5\) - straightedge , straight line \(A_4A_5\) - straightedge and compass , bisection arc \(A_2A_3\) , point \(A_6\) YOU TRY TO KEEP ... Figure down Please Register or Log in to view the hidden image!

This is making no sense to me. I don't know if it is just a language problem or what. How can you divide lines into 2 equal parts with just a straight edge and a compass? Seems to me that you are using a ruler somewhere in this analysis.

the proportion of straight line (to be in Serbia to enter primary school in mathematics) Please Register or Log in to view the hidden image! - straightedge , straight line \(AB\) - straightedge , straight line \(AC\) , angle \(CAB\) - point \(D\) - compass \(AD\) , from point \(D\) , point \(F\) - compass \(AD\) , from point \(F\) , point \(H\) - compass \(AD\) , from point \(H\) , point \(J\) - compass \(AD\) , from point \(A\) , point \(E\) - compass \(AD\) , from point \(E\) , point \(G\) - compass \(AD\) , from point \(G\) , point \(I\) - compass \(AD\) , from point \(I\) , point \(K\) - straightedge , straight line \(ED\) - straightedge , straight line \(GF\) - straightedge , straight line \(IH\) - straightedge , straight line \(KJ\) - compass \(ED\), from point \(F\) , point \(L\) - compass \(ED\), from point \(H\) , point \(M\) - compass \(ED\), from point \(M\) , point \(N\) - compass \(ED\), from point \(J\) , point \(O\) - compass \(ED\), from point \(O\) , point \(P\) - compass \(ED\), from point \(P\) , point \(Q\) To be continued ...