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Astonishingly, no mathematician ever could give a "Mr. Foobar invented this" whenever I came up with this construction, although it is very elementary.

Given are 3 circles C1,C2,C3 (avoid degenerate configurations for now). Let L be the geometric locus of the centers of all circles C which intersect C1, C2 and C3 under the same angle @ (which may be non-real - doesn't hurt!)

Clearly the radical center (@=90°) and the all-outer/inner Apollonius center (@=0/180°) lie on L, and some analytic geometry immediately shows L is a straight line.

Bonus Track (only if you have too much time): Calculate @ for the Gergonne point when L is the Soddy line of C1, C2, C3. A most surprising result awaits. (Purely geometric proof, anyone?)

Edit: (Added from comments)

Here's an image:

enter image description here

The dotted circle is for @=120° (of course everything is drawn only approximate!)

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Mathematicians are not responsible for memorizing the entire history of mathematics. I don't understand what you mean by $L$. –  Qiaochu Yuan Aug 16 '11 at 15:36
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Welcome to the fray, Hauke! It's an interesting idea, one new to me. –  hardmath Aug 16 '11 at 15:47
    
Here's an image: imgur.com/ttjN4 The dotted circle is for @=120° (of course everything is drawn only approximate!) –  Hauke Reddmann Aug 16 '11 at 16:08
    
L is not a line but 4 lines, right? (E.g. when C1-C3 become straight lines L is 4 points: centers of in/ex-circles.) –  Grigory M Aug 19 '11 at 14:20
    
@Grigory: If we specify a certain direction for directed angles, as I assumed in my solution, then there is one line for each of the 4 configurations. (Fix the direction of $C$ and $C_1$, then there are $2 \times 2$ configurations for $C_2$ and $C_3$.) –  user21820 Dec 31 '11 at 4:49
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1 Answer

I have no idea what the Soddy line is, but I think I have solved the first part:

(I only consider the cases where $C_1$, $C_2$, $C_3$ are pair-wise distinct)
Let the centre of $C$ be $O$ and the centres of $C_1$, $C_2$, $C_3$ be $O_1$, $O_2$, $O_3$
If $C_1$, $C_2$, $C_3$ have the same size,

    L is not a straight line but a point because the angle at the intersection of $C$ and $C_1$ is monotonic in $\overline{OO_1}$
    Thus $\overline{OO_1} = \overline{OO_2} = \overline{OO_3}$ and $O$ is unique

If $C_1$, $C_2$, $C_3$ do not all have the same size,

    I think no two Apollonius circles can be concentric
    Thus there is a point at which inversion maps two corresponding ones to concentric circles
    In that case $C_1$, $C_2$, $C_3$ map to circles of the same size, so we are back to the earlier case!
    $C$ must then map to a circle with centre $P$ uniquely defined by the images of $C_1$, $C_2$, $C_3$
    Thus the centre of $C$ must lie on the line uniquely defined by the inversion centre and $P$

(QED)

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I forgot to mention that inversion preserves angles at intersections. –  user21820 Dec 30 '11 at 15:56
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