### 14th Putnam 1954

Problem B4

Let F be a point, and L and D lines, in the plane. Show how to construct the point of intersection (if any) between L and the parabola with focus F and directrix D.

Solution

Fairly hard.

If L and D are parallel and a distance d apart, then take the circle center F radius d. If it meets L, then the point(s) of intersection are the required points. If it does not, then there are none.

So assume L and D meet at O. If F lies on D, then the (degenerate) parabola is the line through F perpendicular to D. If it meets L, then the point of intersection is the required point. So assume F does not lie on D.

If L and D are perpendicular, then there is just one point. Take any point P (except O) on L and let the circle center P radius PO cut the line OF again at Q. Let the line through F parallel to PQ cut L at G. Then G is the required point (because an expansion center O takes the circle center P radius PO into a circle center G radius GO = GF).

So assume L and D are neither perpendicular nor parallel and meet at O. Take D' through O making the same angle with L as D. D and D' divide the plane into 4 sectors. If F lies in one of the two sectors containing L, then the required points exist. There are two required points unless F lies on D', when there is just one. [Any circle with center on L, touching D, must lie entirely within these two sectors, and hence F must also if the points exist.]

Take any point P on L (except O) and draw a circle center P touching D. Let the circle meet the line OF at Q. Let the line through F parallel to PQ cut L at G. Then G is the required point. If F does not lie on D', then there will be two possibilities for Q and hence two possibilities for G.