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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?

MarianD
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Philip
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    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

3 Answers3

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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).

$\hspace{5mm}$enter image description here

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 $45^\circ$.

$\hspace{3cm}$construction

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 $Q$.

robjohn
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  • 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
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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 $c$:

Direct construction

The reason why this works is as follows:

Reason

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 $c$.

t.b.
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  • 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
  • Ah, I hadn't noticed that. I had thought that was part of the OP's attempt at an answer, but I see you are correct. I answered the question in the title, but as restricted in the question, your answer is much simpler. (+1) – robjohn May 19 '12 at 14:44
  • It just occurred to me that we can easily reduce the general problem to the case I presented: if $r/2^{k+1} \lt |OP| \leq r/2^{k}$, then find the point $P_{k}$ on the ray $|OP|$ with $|OP_k| = 2^{k} |OP|$ so that $r/2 \lt |OP_k| \leq r$, invert $P_k$ with the above construction to find $P_{k}^\prime$ and then let $P'$ be the point on the ray $OP$ with $|OP'| = 2^{k} |OP_{k}^\prime|$. Then $$|OP|\cdot|OP'| = (|OP_k|/2^{k})\cdot(2^k|OP_{k}^\prime|) = r^2.$$ To double the distance along a ray, use the construction of a regular hexagon with vertex $O$ and center $P$; repeat $k$ times. – t.b. May 20 '12 at 02:12
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    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 03:46
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    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 09:28
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    Here's a picture of what I have in mind. – t.b. May 20 '12 at 12:38
  • Easy enough, and using similar ideas to your previous construction. Thanks! – robjohn May 20 '12 at 13:53
  • I compiled the last few comments here in this CW answer of mine I hope this is okay with you. @robjohn: maybe you are interested. – commenter Nov 02 '12 at 04:59
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Generally to invert a point with respect to a circle, construction steps depend on the position of the point with respect to the circle. If the point lies outside the circle then we follow few steps which can't be followed for the point which lies inside forest.

There exists another construction procedure named Dutta's Construction which resolves this inconvenience, i.e. we can follow same steps to invert a point whether it lies outside or inside the forest, a case distinction as in usual treatments is not needed.

Source Links: https://forumgeom.fau.edu/FG2014volume14/FG201422.pdf https://en.wikipedia.org/wiki/Inversive_geometry#Dutta's_construction

user33528
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