AIME 1999 Problem 14
Given three points A, B and C,
find the point P such that
angle(PAB) = angle(PBC) = angle(PCA).
Express tan(PAB) in terms of A, B and C.
Sat May 8 10:52:00 EDT 1999
- I came across this problem while leisurely recreating solutions for
the AIME 1999 that Henry
brought home from school.
Of the 15 problems, this one, number 14, is the one that intrigued me the most.
- The solution is not one of the well know centers of the triangle.
I still don't know what special name it has, but this will pop up some day.
Note added in 2001: that day was some day last year when John Sharp told me these were the Brocard points of the triangle ; see his article.
- What I find intriguing about it is that the construction is asymmetric in
an unusual way. There are two such points, as the definition could have read angle(PCB) = angle(PBA) = angle(PAC), traversing the points in the other
direction. Sets of points in the triangle usually come in ones and threes,
for obvious symmetry reasons. But twos? Intriguing ...
- The other point P, called O here, is the isogonal conjugate of P.
I hadn't heard of such a thing before yesterday. If you take the 3 lines from the vertices A, B and C to P, then reflect them across the angle bissectors, you get a new triple of lines. They concur on the isogonal conjugate.
- And the best is saved for last: every point in the plane has an isogonal
conjugate wrt. the triangle. Every line is isogonal conjugate to a certain
A line that crosses the circumcircle is isogonal to a
hyperbola, a tangent to a parabola, a line that is wide of
the circumcircle is conjugate to a circle.
Thus the conjugate of the circumcircle is the line at infinity.
He's written a book documenting 400 different centers of a triangle ... and keeps it up to date!
Eric Weinstein's article on Wolfram's MathWorld site:
D. E. Joyce's The Geometry Applet.
This is a wondeful applet. If I could take only 1 applet to a desert island,
I'd take this one.
Make sure you also visit his Euclid's Elements.