Pascal theorem

Given six points ABCDEF on any conic section,
U intersection point of AB and DE, V intersection of CD and AF, and W intersection of EF and BC
U,V,W in line

This is true for any conic section, and also to a pair of straight lines (Pappus theorem).
We need just to prove it for a circle and then transform the property by central projection.

Apply the Menelaüs theorem to triangle GHI with secants : DVC, AUB and EWF.
DVC : VH/VI · CI/CG · DG/DH = 1
AUB : AH/AI . BI/BG . UG/UH = 1
EWF : FH/FI . WI/WG . EG/EH = 1

Multiplying these relations and rearranging the terms :
VH/VI . WI/WG . UG/UH . (CI.BI/AI.FI) . (AH.FH/DH.EH) . (DG.EG/CG.BG) = 1
But CI.BI = IB.IC = IA.IF = AI.FI = power of I.
And similarily for other terms in parenthesis.
Remains VH/VI . WI/WG . UG/UH = 1 which says U,V,W are in line.

The six points being given on the conic section, the "numbering" of these points is arbitrary and there are 60 different ways of choosing the hexagon built by the six vertices. Stared as in previous case, or convex as this one.

Brianchon theorem

It is a dual version of Pascal theorem.
Let an hexagon IJKLMN circumscribed to a conic section
(that is all sides IJ etc. tangent to the conic section)
IL JM KN intersect in one point

Let ABCDEF the contact points.
From Pascal theorem, U=AB^DE, V=CD^AF and W=EF^BC are in line.
AB is the polar of I and DE the polar of L.
That is U is conjuggate of I and L. Hence the polar of U is IL.
Similarily, the polar of V is KN and the polar of W is JM.
Hence pole of UV is intersection point G of KN and IL
As it is also the pole of the same line UW, it is intersection point of IL and JM. QED.

 

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