The current mathematics is limited and sinful.
I'll show you the six examples, who I am, present a review of mathematics
1. trisection angle
image.1. OA ruler - along
image.2. caliper O-OA- circular arc
image.3. caliper A-AO-I walk a circular arc to get the point A1
image.4. OA1 ruler - along
image.5. caliper A1-A1A - get round
image.6. caliper A-AA1 - get the second round, we get the point B at the intersection of two circles
image.7. OB ruler - along, I walk AA1 circular arc, we get the point A2
this process with a ruler and compass (angle bisection) is a set of numbers listed in the picture (infinite , image 8 )
We have been the angle of the picture that we do not know how many degrees.
We can solve the reconstruction of the sum (image 9) and the above series of numbers
image.10. given angle BOA
image.11. Caliper O-OA-draw circular arc
---a1 (a member of the sum) for n = 1-AA1 = 60 °
image.12.caliper A-AA1-I walk a circular arc C we get the point
---a2 (a member of the sum) for n = a-AA1 = 60 °, the requested item is not in the circular arc AB, as in the following procedure will be left
image.13. caliper C-AA1, I walk a circular arc, we get the point C1
---a2, for n = 2 - AA2 = 30 °, the requested item is not in the circular arc AB, as in the following procedure will be left
image.14. caliper C-AA2, I walk a circular arc, we get the point C2
---a2, for n = 3 - AA3 = 15 °
image.15. caliper C-AA3, I walk a circular arc, we get the point C3
---a3, for n = 3 - AA3 = 15 °, the requested item is not in the circular arc AB, as in the following procedure will be left
image.16.. Caliper C3-AA3, I walk a circular arc, we get the point C4
---a3, n = 4-AA4 = 7.5 °, the required point C5 and point B are in the same place, which means that the sum of the reconstruction was completed and that the angle is 82.5 °
image.17. Caliper C3-AA4, I walk a circular arc, we get the point C5
I introduced the concept subtractionak ( subtraction+ak=subtractionak , image 18), we trisection subtractionak reconstruction performed using the conditions (image 19)
image.20. initial conditions, the angle of 82.5 °
a1 = 82.5° ÷ 3 = 27.5 ° is the third arc
a2 , for n = 1 - AA1 - 60 ° - did not match the conditions 60 °> 27.5 °
------for n = 2 - AA2 - 30 ° - does not match the conditions of 30 °> 27.5 °
------for n = 3 - AA3 -15 ° - 15 ° correspond to the conditions <27.5 °, 27.5 ° -15 ° = 12.5 °
imagine.21. caliper A-AA3, cut an arc AB, we get the point C
a3 , for n = 4-AA4 - 7.5 ° - 7.5 ° corresponding to the conditions <12.5 °, 12.5 ° -7.5 ° = 5 °
imagine.22. caliper C-AA4, cut an arc AB, we get the point C1
a4 , for n = 5 - AA5 - 3.75 °, corresponding to conditions of 3.75 ° <5 °, 5 ° -3.75 ° ˝ = 1.25 °
imagine.23.caliper C1-AA5 , cut a circular arc, you get the point C2
a5, for n = 6 - AA6 -1.875 ° - does not match the conditions of 1875 °> 1.25 °
-----for n = 7 - AA7 - 0.9375 ° - comply with the terms 0.9375 ° <1.25 °, 1.25 ° -0.9375 ° = 0.3125 °
imagine.24.caliper C2-AA7, cut an arc AB, we get the point C3
Now we have a result with an error below 1 ° (3x0.3125° = 0.9375°), if you want more precise results continue reconstruction
imagine.25. ruler 0C3 - along
imagine.26. caliper C3-C3A , cut a circular arc AB, we get the point C4
imagine.27. along the ruler-0C4
