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# 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. Draw the ray $\overrightarrow{OP}$.

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

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

4. Draw the ray $\overrightarrow{OT}$.

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

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

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

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. 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 link provided.

## Mathematics Subject Classification

51K99*no label found*53A30

*no label found*51M15

*no label found*

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