# An Euclidean Construction with Inversion

Here is a problem from The Amewrican Mathematical Monthly (11572, v 120, February 2013, p 176-7). The problem was proposed by Sam Sakmar, University of South Florida, Tampa, FL

Given a circle $C(O)$ and two points $A$ and $B$ outside $C(O)$, give an Euclidean construction to find a point $P$ on $C(O)$ such that if $Q$ and $S$ are the second intersections with $C(O)$ of $AP$ and $BP$ respectively, then $QS$ is perpendicular to $AB$. (Special configurations, including the case that $A$, $B$, and the center of $C(O)$ are collinear, are excluded.)

The applet below serves a dynamic illustration:

### Proof

Solution is by Robert A. Russel, New York, NY.

In general, the problem has two solutions. These are the two points of intersection of the given circle $C(O)$ with the circle through $A$ and $B$ perpendicular to $C(O)$.

How to construct such a circle? Inversion in circle $C(O)$ preserves points on $C(O)$, lines through its center $O$, and the circles orthogonal to $C(O)$. If $C(F)$ is the circle through $A$ and $B$ orthogonal to $C(O)$, then the imgages $A'$ and $B'$ of $A$ and $B$ lie on $C(F)$. It follows that $C(F)$ can be constructed as the circumcircle of, say, $\Delta ABA'$.

Let $P$ be one of the intersections of the two circles. (The proof refers to the following diagram.)

In the diagram there are several auxiliary points: $G$ is the intersectio of $AB$ and $OP$; $J$ the intersection of $AB$ and $SQ$; $I$ and $H$ are points $FP$ on two sides from $P$. On the other hand, the center $F$ of the constructed circle is not shown. Since the two circles are orthogonal, $OP\perp FP$ and $OP$ is tangent to $C(F)$ whereas $FP$ is tangent to $C(O)$. We thus can use the properties of inscribed angles and angles formed by a tangent and a chord.

Observe that

$\angle BPG=\angle BAP =\angle JAQ$

and

$\angle BPH=\angle IPS =\angle PQS=\angle AQJ,$

Now $\angle BPH+\angle BPG=\angle GPH = 90^{\circ}$. It follows that also $\angle JAQ+\angle AQJ= 90^{\circ}$. Therefore, in $\Delta AJQ$, $\angle AJQ=90^{\circ}$, as required.