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Problem 865

AMC 12 late, AIME early

Geometry Difficulty 4.6 Prove it Brazilian Mathematical Olympiad · Brazil

The vertex $C$ of the triangle $ABC$ is allowed to vary along a line parallel to $AB$ . Find the locus of the orthocenter.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Take axes so that $A = (-a, 0)$ , $B = (a, 0)$ and $C = (k, b)$ . Then the orthocenter lies on the line $x = k$ . The line $AC$ has gradient $\frac{b}{k+a}$ , so the perpendicular has gradient $-\frac{k+a}{b}$ . Hence the altitude from $B$ has equation $y + \frac{(x-a)(k+a)}{b} = 0$ . So the intersection is $x = k, y = -\frac{(k-a)(k+a)}{b}$ . So the locus is all or part of the parabola $by = a^2 - x^2$ . But we can get an orthocenter with any x-coordinate (by taking $C$ to have the same x-coordinate), so we can get all points on the parabola.

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