Date: Nov 8, 1997 4:36 AM
Author: Jon Roberts
Subject: Re: construction of triangle of given perimeter, given point and angle



Joshua Zucker wrote:

> This is an interesting question!
> Of course you can always solve it by analytic geometry, and then see
> that all the quantities there are constructible, and construct them.
> But that's pretty ugly! And I don't yet see any better way to do it.
>
> I have another construction problem which is similarly irritating to
> me (where I can prove with analytic geometry that it's constructible,
> but I can't find a nice construction, though the person who gave me
> the problem implied there ought to be a nice one):
>
> Given triangle ABC, find points X and Y on AB and AC such that
> BX = XY = YC.
>
> --Joshua Zucker


I am in a similar quandary with regards to the vertex of a parabola. In
a "recent" posting:
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

y = a(x-r)(x-s) and then equating with y = ax^2 + bx + c but my question
is about constructing the vertex geometrically - can it be done with the
classical instruments or are more sopisticated tools needed. The vertex
lies on the perpendicular bisector of the segment joining the
x-intercepts, and this axis can be used to produce a fourth point on the
parabola by reflecting the y-intercept, but what does one do to locate
the one and only one point on the axis which then gives a parabola. Any
other point gives an ellipse or an hyperbola - I'm stuck at this point.

hoping you can help

Jon Roberts.