A Solution to Crux Mathematicorum Problem 5021

The original problem statement is intentionally not reproduced here. This article follows its notation and presents only a generalization, its proof, and a diagram. The complete statement is available in the official issue of Crux Mathematicorum.

A generalization

Proposition

Suppose that $\triangle OAB\sim\triangle OCD$ with the same orientation, and let $K$ and $L$ be their circumcenters. Assume that all the points and lines below are well-defined.

The lines $AB$ and $CD$ intersect the lines $OC$ and $OB$ at $F$ and $E$ , respectively. Let $P$ be the intersection of $AE$ and $DF$ . Then $OP$ is perpendicular to $KL$ .

Proof

Let $Q$ be the intersection of $AC$ and $BD$ .

Two directly similar triangles OAB and OCD. Lines AE and DF meet at P, while AC and BD meet at Q. The points O, Q, and P are collinear. The two circumcircles have centers K and L, and OP is perpendicular to KL. — Figure 1. The Pappus configuration and the two circumcircles whose radical axis is OP.Select the diagram to enlarge it.

Applying Pappus’s theorem to the two triples $(A,B,F)$ and $(D,C,E)$ shows that $O$ , $P$ , and $Q$ are collinear.

Furthermore, $\triangle OCA\sim\triangle ODB$ with the same orientation. Indeed, the given direct similarity implies

\frac{OC}{OD}=\frac{OA}{OB}, \qquad \measuredangle COA=\measuredangle DOB.

Hence, working with oriented angles,

\measuredangle OAQ =\measuredangle OAC =\measuredangle OBD =\measuredangle OBQ,

and

\measuredangle OCQ =\measuredangle OCA =\measuredangle ODB =\measuredangle ODQ.

Consequently, the four points $O$ , $A$ , $B$ , and $Q$ are concyclic, as are $O$ , $C$ , $D$ , and $Q$ .

Hence $OP$ is the radical axis of these two circles and is therefore perpendicular to the line through their centers, namely $KL$ .

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