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I'm thinking about Descartes' theorem "Wikipedia".

I understood how to find the radius with algebra. Now I'm trying to use ruler and compass to find the midpoint of the 4th circle.

I thought about using the midpoints of $k_1-k_3$ to get a triangle and the centroid as midpoint, but it's not the right approach. Any ideas, what point I have to find to get the midpoint?

And further more: Is it possible using ruler and compass to get the midpoint of a $4$th circle touching the first three, if $k_1-k_3$ are not touching?

Any toughts about this are welcome.

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A convenient transformation, when you are dealing with circles, is inversion.

Inversion takes circles and straight lines to circles and straight lines. Moreover, if a circle passes through the center of inversion then it turns into a line. Here we have circles tangent to each other. So, if we use as center the tangency point of two of the circles, they both will turn into lines. Even more, since inversion preserves angles, these two lines will be parallel.

Take the tangency point, $P$, of two of the given circles, $c_1$ and $c_2$, and apply inversion with center $P$ and radius, say $1$.

(Actually it might be better to use a radius, if possible, such that the circles get intersected at two points. That way you know that those two points are not going to move during inversion.)

The two circles $c_1$ and $c_2$ are going to become parallel lines, $L_1$ and $L_2$, so the center of the circle you are looking for lies in the middle line, $L_3$. Notice that $L_1$ and $L_2$ are easy to construct since they pass through the intersection points of a circle with center $P$ and radius $1$ and each circle $c_1$ and $c_2$. In the worst case you can always invert two or three points of the circle, as needed, and the inverse line or circle will pass through those inverted points.

The other circle $c_3$ becomes a circle $C_3$ in between the two lines because tangency (i.e. angles) is preserved. So, the tangency points of the inversion $C_4$, of the circle to be constructed to that circle $C_3$ are the intersection of the middle line $L_3$ with that third circle, $C_3$.

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  • $\begingroup$ Can this be construed as a ruler-and-compass construction (per Question)? $\endgroup$ – hardmath Jul 9 '13 at 13:32
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    $\begingroup$ Of course! That is why instead of linking to the Wikipedia article about inversion I linked to a page that shows how to invert using ruler and compass. (Also because the Wikipedia article seem so poorly written). $\endgroup$ – OR. Jul 9 '13 at 13:34
  • $\begingroup$ Excellent answer, thanks a lot. It's quite what I was looking for $\endgroup$ – ulead86 Jul 14 '13 at 11:11

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