# Pascal's theorem for spherical hexagon

I draw a cyclic spherical hexagon and I check by geogebra that Pascal's theorem is true in this case.

My question 1. Is there simple proof for this?

My question 2. Can we change the circle on sphere by the curve like "conic" in plane?

My question 3. If we use orthogonal projection to project this configuration onto a plane. Which do the plane problem we obtain?

• Dear Tran Quang Hung, is the Pascal line through six points? Jul 6 '18 at 4:04
• Dear Mr Dao, three points "collinear" on sphere, this means there is a great circle passes through them, actually we have six points with symmetry. Jul 6 '18 at 4:11
• Does it not follow by taking the 2-1 map to the real projective plane, where Pascal's theorem has its usual proof? Jul 6 '18 at 13:41
• You can do a central projection onto a plane. Then the result follows from the planar case. Jul 6 '18 at 18:01

This is really a comment but will be too long and, anyway, I'm not entitled. There is a standard approach to this problem but I haven't carried through the computations. Let me indicate the method for euclidean case first. Suppose we have points $A_1,\dots,A_6$ in the plane, with the first five fixed, say $A_1=(a_{11},a_{12})$ etc. but $A_6=(x,y)$ variable. Now compute $X,Y,Z$,the intersections of $A_2 A_6$ and $A_5A_3$, etc. Then apply the condition for these three points to be collinear. This turns out to be a quadratic in $x$ and $y$ which proves Pascal's theorem AND its converse. These computations are rather tedious but can be carried in a few minutes with, say, Mathematica. They can even be done by hand by assuming wlog that three of the original points are $(0,0),(1,0),(0,1)$.