> 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.