compass and straightedge construction of inverse point

Let $c$ be a circle in the Euclidean plane with center $O$ and let $P\neq O$. One can construct the inverse point $P^{\prime}$ of $P$ using compass and straightedge.

If $P\in c$, then $P=P^{\prime}$. Thus, it will be assumed that $P\notin c$.

The construction of $P^{\prime}$ depends on whether $P$ is in the interior of $c$ or not. The case that $P$ is in the interior of $c$ will be dealt with first.

1. 1.

Draw the ray $\overrightarrow{OP}$.

2. 2.

Determine $Q\in\overrightarrow{OP}$ such that $Q\neq O$ and $\overline{OP}\cong\overline{PQ}$.

3. 3.

Construct the perpendicular bisector of $\overline{OQ}$ in order to find one point $T$ where it intersects $c$.

4. 4.

Draw the ray $\overrightarrow{OT}$.

5. 5.

Determine $U\in\overrightarrow{OP}$ such that $U\neq O$ and $\overline{OT}\cong\overline{TU}$.

6. 6.

Construct the perpendicular bisector of $\overline{OU}$ in order to find the point where it intersects $\overrightarrow{OP}$. This is $P^{\prime}$.

Now the case in which $P$ is not in the interior of $c$ will be dealt with.

1. 1.

Connect $O$ and $P$ with a line segment.

2. 2.

Construct the perpendicular bisector of $\overline{OP}$ in order to determine the midpoint $M$ of $\overline{OP}$.

3. 3.

Draw an arc of the circle with center $M$ and radius $\overline{OM}$ in order to find one point $T$ where it intersects $C$. By Thales’ theorem, the angle $\angle OTP$ is a right angle; however, it does not need to be drawn.

4. 4.

Drop the perpendicular from $T$ to $\overline{OP}$. The point of intersection is $P^{\prime}$.

A justification for these constructions is supplied in the entry inversion of plane.

If you are interested in seeing the rules for compass and straightedge constructions, click on the provided.

Title compass and straightedge construction of inverse point CompassAndStraightedgeConstructionOfInversePoint 2013-03-22 17:13:17 2013-03-22 17:13:17 Wkbj79 (1863) Wkbj79 (1863) 12 Wkbj79 (1863) Algorithm msc 51K99 msc 53A30 msc 51M15