AN EVEN SIMPLER PROOF OF PASCAL’S HEXAGON THEOREM

René DESCARTES^{1} and Shalosh B. EKHAD^{2}

We give an even shorter, simpler, and more elegant proof than [vY] of: Pascal’s Theorem: 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. Proof:Type maple (CR), and then^{3} the following:

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);

Reference

[vY] J. van Yzeren, A simple proof of Pascal’s theorem, Amer. Math. Monthly 100(1993) 930-931.

1 Queen Christine’s Court, Stockholm, Sweden. Current address: Panthéon, Paris, France. 2 Department of Mathematics, Temple University, Philadelphia, PA 19122, USA. ekhad@math.temple.edu .

3 Inserting 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.