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Re: construction of triangle of given perimeter, given point and angle
Posted:
Nov 8, 1997 4:36 AM



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(xr)(xs) 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 xintercepts, and this axis can be used to produce a fourth point on the parabola by reflecting the yintercept, 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.



