```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. Ina "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 ofaparabola from its axes' intercepts. I can do this algebraically bycompleting the square ony = a(x-r)(x-s) and then equating with y = ax^2 + bx + c but my questionis about constructing the vertex geometrically - can it be done with theclassical instruments or are more sopisticated tools needed. The vertexlies on the perpendicular bisector of the segment joining thex-intercepts, and this axis can be used to produce a fourth point on theparabola by reflecting the y-intercept, but what does one  do to locatethe one and only one point on the axis which then gives a parabola. Anyother point gives an ellipse or an hyperbola - I'm stuck at this point.hoping you can helpJon Roberts.
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