Diametrically Opposite Points: What Is This About?
A Mathematical Droodle

Explanation

Copyright © 1996-2008 Alexander Bogomolny

Let P lie on the circumcircle of triangle ABC. Let Q be the point on the circumcircle diametrically opposite P. Then the Simson lines of P and Q are perpendicular. As P changes, the points of intersection of the two simsons traces the 9-point circle of ABC.

If you change one of the vertices A, B, or C, the intersection point still traverses a circle.

Hubert Shutrick sent me the following proof. Let P_a, P_b, P_c and Q_a, Q_b, Q_c be the feet of the perpendiculars from P and Q onto BC, AC, AB.

In the diagram, angles CP_aP and CP_bP are right which makes quadrilateral CPP_aP_b cyclic. It follows that the inscribed angles CP_aP_b and PP_aP_b are equal. A similar argument applies to show that quadrilateral CQQ_bQ_a is also cyclic and that angles QCQ_b and QQ_aQ_b are equal.

Now, angles CPP_b and QCQ_b have perpendicular sides and are therefore equal. It follows that

CPaPb = QQaQb

But one pair of their side, viz., CP_a and QQ_A are orthogonal which implies that the so is the second pair:

PaPb QaQb.

This proves that the simsons of P and Q are indeed orthogonal.

Let A'B'C' be the medial triangle of ΔABC. Since the perpendicular to BC at A' passes through the circumcenter of ABC, A' is also the midpoint of P_aQa and the right angle gives that A'M = A'P_a so there is an angle θ at M also.  By the same token there is the same angle φ at M and P_c:

MPcC' = MC'Pc.

However,

A'B'C' = ABC = θ + φ = A'MC',

so that M is on the circumcircle of ΔA'B'C', the nine point circle of ABC. Conversely, given a point M such that A'B'C' =  A'MC, we have candidates for P_a and Q_a whose distances from A' are A'M. From them we construct two diagonally opposite pairs P, Q and P', Q' on the circumcircle of ABC, where, in the diagram, the other candidate for P' for P would be above the line BC.  The Simson lines of each pair intersect at right angles and pass through P_a and Q_a and so their points of intersection are the points of intersection of the circle with diameter P_aQ_a with the nine point circle. One of these must be the given M. We have ignored the interchange of P with Q that gives the same point M and the fact that the nine point circle is described twice as P goes round the circumcircle. Q(uite) E(asily) D(one)

References

1. F. G.-M., Exercise de Géométrie, Éditions Jacques Gabay, 1991, p. 330

Copyright © 1996-2008 Alexander Bogomolny