The ball shall be awarded in two equal parts, obtained half contains two areas, the circle (represents a planar geometry) and a semi-sphere (represents spherical geometry), the circle is the boundary between the circuit and semi-sphere

View photo (below)

CIRCLE

within a given arbitrary angle BAC,straightedge (straightedge is flexible, it can draw on a sphere)

straight line the BA to extend the circle to give the point D

SPHERE

straightedge connect points B and D, you get the curve BD

straightedge and divider - a procedure divisions curve into two equal parts is the same as the process of division shall exceed the level in two equal parts, we get the point E

straightedge - connect points C and E and get the curve CE

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Proportion longer exists in plane geometry, I found that the process can be applied to the sphere

choose point G

divider EC, from point G we get a point Hdivider EC, from point H we get a point I

divider EC, from point E twe get a point Jdivider EC, from point J get the point K

divider EC, from point K get the point L

straightedge point L and point I and connect, we get curve LI

EG divider, from the point L we get a point P

EG divider, from the point P and get the point Ostraightedge join the dots E and P and proceed to the circle, we get the point Q

straightedge merge points E and O and proceed to the circle, we get the point R

CIRCLE

straightedge connect point A and point Q, we get straight line AQ

straightedge connect point A and point R, we get straight line AR

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these have carried out the production of a given trisection angle, is obtained from the rest of the (n-section, a regular n-polygon) ...

Now proclaim everywhere that I decided 2- millennium math problems