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.