Pat Ballew ran with my earlier WCYDWTRP post over on his blog, so I thought I would return the favor.
Pat's take was an old theorem about the vertices of a triangle lying on a rectangular hyperbola. It turns out the orthocenter of these three points will also lie on that hyperbola, as the dynamic applet below illustrates.
Note: Double clicking on any of the applets that follow will open it in a separate GeoGebra window
All of these following applets were made with the FREE Dynamic Mathematics Software GeoGebra.
An interesting property of three points and their orthocenter is this: Select any one of these four points. The selected point is the orthocenter of the remaining three. These four points are sometimes called an orthocentric quartet. Drag the slider in the applet below to see what I mean.
Anyway, I began to wonder, what would happen if the vertices of a triangle were restricted to some other conic? An ellipse, perhaps? Drag the red points in the applet below around. Right-click (ctrl-click) in the green orthocenter to turn the trace of the point on.
What about restricting the vertices of the triangle to a parabola?
Subscribe to:
Post Comments (Atom)
Steve,
ReplyDelete(this may be a little long for a comment, if you want to break out as another blog, and interject comments, straight on boy...
A partial answer to the last situation... Consider first, a special (but different) case. What happens if we have two points (call them A and B) of a triangle fixed, and move the third vertex, C, of course, along a line parallel to the base AB. It is not too difficult to prove that the locus of points of the orthocenter of ABC is a parabola that passes through A and B. Take the trivial case where C moves on the x-axis and A and B are the points (-r,s) and (r,s). The perpendicular to C will have the same x coordinate as C (let's call the coordinates of C (p,0) since AB is perpendicular to the x-axis...
The line from A to C has a slope of s/(p+r) so the slope of the perpendicular thorough B will have a slope of (p+r)/s...
so the equation of this altitude through B perpendicular to AC is y-s = (p+r)(x-r)
and since we know the x-coordinate where it intersects the altitude through C must be p, we know that the y-coordinate if the orthocenter must be the solution to y-s= (p+r/s) (p-r) or
at y=(p^2-r^2)/s + s.. but that is the equation of a parabola that goes through (r,s) and (-r,s)... so the locus is a parabola..
But wait, the question was what happens if the three points are all on a parabola... but as you pointed out in the blog.. the three points and the orthocenter form a quartet of in which any three can be the vertices, and the fourth the vertex... SOOOOOO, if the three points are on a parabola and we hold two of them to be on a segment perpendicular to the axis of symmetry, then the third will move along a straight line parallel to the segment AB...
And I THINK (but have not proved to myself) that in any other case, the locus will be a hyperbola, I came to this thought by considering the case when Point C moves on a line that is not parallel to AB.. at the point where the line it moves on and AB intersect, ABC all lie on a straight line, and the orthocenter diverges to infinity... still have a little work to do to see if that is right, but for now... papers to grade... ahh, the life of a HS teacher...
Thanks for a great diversion.