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Re: construction of triangle of given perimeter, given point and angle
Posted:
Nov 8, 1997 5:22 PM
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On Sat, 8 Nov 1997, Jon Roberts wrote:
>
>
> I am in a similar quandary with regards to the vertex of a parabola.
...................
> John Conway wrote:
>
> > I'm afraid that 5 is still the number of points required to determine
> > an ellipse, because ellipses are distinguished from hyperbolae merely
> > by an inequality. (So, for instance, 3 is still the right number
> > of points to determine a circle of radius larger than 1.) A parabola,
> > on the other hand, is determined by any 4 of its points. John Conway
> >
>
> This information above has lead me to attempt to construct the vertex of
> a
> parabola from its axes' intercepts. I can do this algebraically by
> completing the square on
Let me just think about this more generally. Let's try to
find the vertex from any four points. Then projectively, we're
asking something about a conic through 4 given points and touching
a given line (the line at infinity). We must find a projective
definition of the vertex V; namely it's the point whose tangent
is orthogonal to the line joining it to the contact-point of
the conic with the line at infinity. [Projectively, two lines are
orthogonal if they meet the line at infinity in points that are
divided harmonically by the circular points at infinity.]
H'm - that seems quite hard. Another thing that worries me
is that I thought the parabola through 4 points was unique,
which would imply that it could be found by rational operations;
but if the 4 points don't form a convex quadrilateral it must
be imaginary, which implies to the contrary that some equation-solving
is needed. Oh - of course it is - let the 4 points be the intersection
of two parabolae!
It seems the equation has degree 2. This is because we're
really trying to find which of the pencil of conics through 4 points
touch a given line. But a coaxal system of circles is a generic
pencil of conics, and for that the answer is that just 2 of them
touch a given line.
So at least something is rational - namely the line joining the two
vertices. It seems to me that the first task is to construct that,
and then one could construct the vertices by intersecting it with
some circle.
But obviously I'll have to think with pencil-and-paper or
chalk-and-blackboard.
John Conway
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