A Property of Pascal's Hexagon Pascal May Have Overlooked
On January 3, 2014, Dao Thanh Oai posted the following theorem at CutTheKnot facebook page. He dubbed it Viet Nam Theorem as well he might; for, I believe the theorem that deals with Pascal's hexagon has been likely overlooked by Pascal himself and the generations of geometers ever since.
Theorem
Given hexagon $ABCDEF,$ let $AB$ meet $DE$ at $G,$ $BC$ meet $EF$ at $H,$ $AF$ meets $CD$ at $I:$
Then, as well known, the six vertices of the hexagon lie on a conic iff the points $G,$ $H,$ $I$ are collinear. In addition, the two conditions are equivalent to
$\displaystyle\frac{AG}{GB}\cdot\frac{BH}{HC}\cdot\frac{CI}{ID}\cdot\frac{DG}{GE}\cdot\frac{EH}{HF}\cdot\frac{FI}{IA}=1.$
Proof
The assertion of the theorem has been discovered experimentally with the GeoGebra. Its veracity is very likely to be implied by the general property of what Howard Eves termed h-expressions [Survey, p. 247]. The details can be found on an earlier page at this site. The expression being a projective invariant, one needs to prove the assertion for a simple configuration that couls be obtained from the above by a projective mapping. This would be similar to the proof found by Hubert Shutrick for a different claim (which arose from the original confused formulation - luckily for every one):
For a hexagon $ABCDEF$ where $AE\cap BD=P,$ $BF\cap CE=Q$ and $CA\cap DF=R,$ the vertices lie on a conic iff
$\displaystyle\frac{PB}{AP}\frac{BQ}{QC}\frac{CR}{RD}\frac{PE}{DP}\frac{EQ}{QF}\frac{RA}{FR}=1.$
It is interesting to note that these points $P,$ $Q,$ $R$ do not arise from Pascal's theorem in any permutation of the vertices, so these points do not even have to be collinear.
References
- H. Eves, A Survey of Geometry, Allyn and Bacon, Inc. 1972
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