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
*conic*. 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.

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