Some Special Cases

We will apply this concept to Pascal's and Brianchon's theorems, using in the first case ABCDEF for the points of the inscribed hexagon, and abcdef for the lines of the circumscribed hexagon. For this we will need to remember that lines lie on conics and can determine conics just as points do.

In Pascal's theorem, what     In Brianchon's theorem, what
happens to a line AB as B     happens to a point ab as b
moves around the conic        moves around the conic towards
towards A?  Gradually,        the point a?  Gradually, the point
line approaches the line      approaches the point which is
which is tangent at A.        the point of tangency on a.
When B and A coincide, the    When b and a coincide, the
line through A and B has      point on a and b has
only one point in common      only one line in common with
with the conic; the line      with the conic; the point is
is thus the tangent at A.     thus the point of tangency on
In this special case,         a.  In this special case,
Pascal's theorem says:        Brianchon's theorem says:

Say five points ACDEF lie     Say five lines acdef lie
on a conic.  Then T(A).CE,    on a conic.  Then (t(a))(c.e),
AC.EF, and CD.FA lie on a     (a.c)(e.f), and (c.d)(f.a) lie
line.                         on a point.

We have used T(A) to denote   We have used t(a) to denote
the tangent at A.  Note that  the point of tangency on a.
because B=A, we have used     Note that because b=a, we have
T(A) instead of AB, and have  used t(a) instead of ab, and
used AC instead of BC.        have used ac instead of bc.

This gives us a construction    This gives us a construction
with straightedge alone for     with straightedge alone for
the tangent to a conic at a     the point of tangency to a
point.  Say we have a conic     conic on a line.  Say we have
ACDEF, and want to find the     a conic acdef, and want to
tangent at A.  We first draw    find the point of tangency on
the line through AC.EF and      a.  First we draw the point on
CD.FA.  Then we intersect       (a.c)(e.f) and (c.d)(f.a).
this line with DE.  We draw     We join this point to (d.e).
a line through this             We draw a point on this line
intersection and point A, and   and line a, and by the above
by the above case of Pascal's   case of Brianchon's theorem,
theorem, this line will be      this point will be the point
the tangent to the conic at A.  of tangency to the conic on a.

If we take not only A=B but    If we taken not only a=b but
also D=E then the line AB      also d=er then the line ab be-
becomes the tangent at A,      comes the point of tangency on
and the line DE becomes the    a, and the line de becomes the
tangent at D.  So the three    point of tangency on d.  So the
points AB.DE, BC.EF, CD.FA     lines (a.b)(d.e), (b.c)(e.f),
become instead T(A).T(D),      (c.d)(f.a) become instead t(a),
AC.DF, CD.FA.  We could        .t(d), (a.c)(d.f), (c.d)(f.a).
reformulate this case:  say    We could reformulate this case:
we have a quadrilateral        say we have a quadrilateral acdf
ACDF inscribed in a conic;     circumscribed to the conic; let
let U=AC.DF and V=CD.FA.       u=(a.c)(d.f) and v=(c.d)(f.a).
Then the tangents to the       Then the points of tangency to
conic at A and C intersect     the conic on a and c have a join
on the line UV.                which lies on the point u.v.

Query: What happens in the case A=B, C=D? What about the case A=B, C=D, E=F?

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Created: Nov 30 1995 --- Last modified: Thu Nov 30 15:07:55 1995