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\quad\quad AN EVEN SIMPLER PROOF OF PASCAL'S HEXAGON THEOREM
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\it
\qquad\qquad\qquad\qquad\qquad Ren\'e DESCARTES\footnote{$^1$}
{\eightrm Queen Christine's Court, Stockholm, Sweden. Current address:
Panth\'eon, Paris, France.} 
and Shalosh B. EKHAD\footnote{$^2$}
{\eightrm 
Department of Mathematics, Temple University,
Philadelphia, PA 19122, USA. 
{\eighttt ekhad@math.temple.edu .} 
}
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\rm
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We give an even shorter, simpler, and more elegant proof than [vY] of:
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{\bf Pascal's Theorem:} {\it If the vertices of a hexagon lie on a circle
and the three pairs of opposite sides intersect, then the three points of
intersection are collinear.}
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{\bf Proof:}Type
{\tt maple} (CR), and then\footnote{$^3$}{\eightrm Inserting
{\eighttt cos:=proc(t):(t+1/t)/2:end: sin:=proc(t):(t-1/t)/2/I:end: }
at the beginning
will speed up the computation twenty-fold.} the following:
{\eighttt
\obeylines
with(geometry): for i from 0 to 5 do point(A.i,cos (t.i),sin(t.i)) od:
for i from 0 to 2 do line(L.i,[A.i,A.(i+1)]):line(M.i,[A.(i+3) ,A.(i+4 mod 6)]): 
point( P.i,coordinates(inter(L.i,M.i))):od:triangle(T,[P0,P1,P2]):evalb(simplify(area(T))=0);
}
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{\bf Reference}
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[vY] J. van Yzeren, {\it A simple proof of Pascal's theorem},
Amer. Math. Monthly {\bf 100}(1993) 930-931.
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