# construct inverse point with respect to the circle by the use of the compass alone

If the given point P lies inside a circle C ，with center O，the circle of radius OP about P intersects C in two points. How to construct point P' inverse to point P with respect to the circle C by the use of the compass alone?

This question is raised from page 144，What is mathematics—2nd Edition

The following content is stated there:

“If the given point P lies inside C the same construction and proof hold, provided that the circle of radius OP about P intersects C in two points.”

I couldn’t understand why does the same construction and proof hold, could you explain it specifically to me?

• I can do it with compass and straight edge, but you want the construction without straight edge, right? – robjohn May 17 '12 at 15:04
• @robjohn You could always just compose your construction with the standard techniques for converting straightedge-and-compass constructions into compass-only? – Steven Stadnicki May 17 '12 at 18:07

## 2 Answers

According to the Mohr-Mascheroni Theorem and my construction below, this can be done. However, as with most "compass-only" constructions, the process would probably be extremely convoluted.

Since the proof I have uses some ideas from , I will post it even though it uses both straight-edge and compass.

The main idea of the construction is that the inverse of a point with respect to a line is just the reflection of the point across the line. First, add $Q$ at $\overrightarrow{OP}\cap C$ (below, on the left), and then consider the inverse with respect to the gray circle centered at $Q$ (below, on the right).

$\hspace{5mm}$ Since $C$ passes vertically through $Q$ on the left, it becomes a vertical line on the right. Since the line that contains $P$ and $Q$ passes horizontally through $Q$ on the left, it becomes a line passing horizontally through $P$ and the image of $\infty$ on the right.

The inverse of $P$ with respect to $C$ on the right is $I$, the reflection of $P$ across $C$. Notice that the blue circle on the right passes through $P$ and intersects both $C$ and the line containing $P$ and the image of $\infty$ at right angles. Since inversion is conformal, the blue circle on the left passes through $P$ and intersects both $C$ and the line containing $P$ and $Q$ at right angles.

Consider the green line on the right. It passes through $P$ where it crosses the line containing $P$ and the image of $\infty$ at $45^\circ$. It intersects $C$ at $R$, where $C$ intersects the blue circle. This means that below, it becomes a green circle, passing through $P$ and $Q$ and crossing the line containing $P$ and $Q$ at $45^\circ$.

$\hspace{3cm}$ Since the green circle passes through $P$ and $Q$ and crosses the line containing $P$ and $Q$ at $45^\circ$, the center of the green circle, $E$, forms the $45{-}45{-}90$ $\triangle PQE$. That is, $E$ lies at the intersection of the red circle whose diameter is $\overline{PQ}$ and the perpendicular bisector of $\overline{PQ}$.

Summary:

Given $O$, $P$, and $C$, extend $\overrightarrow{OP}$ to where it intersects $C$ at $Q$. Draw the perpendicular bisector of $\overline{PQ}$ with $D$ at the midpoint of $\overline{PQ}$. Draw the red circle centered at $D$ and passing through $P$.

Place $E$ at either intersection of the perpendicular bisector of $\overline{PQ}$ and the red circle. Draw the green circle centered at $E$ and passing through $P$. Place $R$ at other intersection of $C$ and the green circle.

Place $F$ at the intersection of the perpendicular bisector of $\overline{PR}$ and the line containing $P$ and $Q$. Draw the blue circle centered at $F$ and passing through $P$. $I$ is at the other intersection of the blue circle and the line containing $P$ and $Q$.

• That's a very nice explanation (+1), but you don't exploit the given hypothesis that the circle with center $P$ through $O$ intersects the circle at which you reflect, which makes the construction of the reflection $P'$ of $P$ much easier (and with compass only), see my answer below. – t.b. May 19 '12 at 14:06
• Yeah, I didn't read close enough to see that we could assume that. :-) – robjohn May 19 '12 at 14:45

The construction in the special case you ask about is very simple:

Draw the circle around $P$ through $O$ and by the given hypothesis it intersects the circle $c$ in two points $X$ and $Y$. Draw circles with centers $X$ and $Y$ through $O$. Call $P'$ the second point of intersection of those circles. Then $P'$ is the inverse of $P$ with respect to the circle $c$: The reason why this works is as follows: The point $P'$ lies on the line through $OP$ which is fixed under inversion with respect to $c$. Note that $OP$ is the perpendicular bisector of the segment $XY$.

Since $P$ is the circumcenter of the triangle $OXY$, it is the intersection of the perpendicular bisector $\color{red}{x'}$ of $OX$ and the perpendicular bisector $OP'$ of $XY$. After inversion with respect to $c$ those perpendicular bisectors are the circle $\color{red}{x}$ and the line $OP$, and they intersect at $P'$. Now finish off by symmetry.

Alternatively, as robjohn pointed out, you can use similar triangles, not using inversion at all, apart from the definition: write $$\frac{|OX|/2}{|OP|} = \frac{|OP'|/2}{|OX|}$$ in order to see that $P$ and $P'$ are inverse to each other with respect to $c$.

• If the length of $\overline{OP}$ is less than half the radius of $c$, then the dotted circle won't intersect $c$. – robjohn May 19 '12 at 14:33
• Yes, of course. Thanks for the feedback! However, it is assumed in the OP that the dotted circle intersects $c$ in two points. – t.b. May 19 '12 at 14:35
• I know it is possible, but is there an easy way to divide a segment in half with only a compass? – robjohn May 20 '12 at 3:46
• I don't know if that's "easy", but here's one method: 1. Find the point $C$ on the ray from $A$ through $B$ such that $|AC| = 2|AB|$ using my previous comment. 2. Intersect the circle with center $C$ through $A$ with the circle with center $A$ through $B$ to find $D_1,D_2$. 3. The midpoint of $AB$ is the second point of intersection of the two circles with center $D_i$ through $A$. – t.b. May 20 '12 at 9:28
• Here's a picture of what I have in mind. – t.b. May 20 '12 at 12:38