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Three circles intersect pairwise but let's assume there is no point shared by all three of them. There are three pairs of circles. Two circles in any pair share a chord. The problem is to prove that the three chords meet at a point.

This page gives a solution:

Consider each circle as the equator of a sphere. Let the plane of the circles be horizontal. It cuts the spheres in two (of course symmetric) halves so that a circle is the projection of the corresponding sphere onto the horizontal plane. Any two spheres intersect at a circle. The chords at hand are the projections of three such (vertical) circles. However, the three spheres share two points which are symmetric with respect to the plane. Those points belong to all three pairwise intersections of the spheres - vertical circles. Therefore, projections of the three vertical circles - our chords - share a point which is the projection of the common points of the spheres.

The part(in bold type) I don’t understand is why it is certain that the three vertical circles(pairwise intersection of the spheres) intersect at a common vertical line. Isn’t it possible that these three vertical circles only intersect pairwise but don’t have common intersection by all three?

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  • $\begingroup$ The sphere argument doesn't seem to make a lot of sense in general. After all, its conclusion is that the three chords intersect at interior points, which isn't true for three arbitrary pair-wise-intersecting circles. I believe there's an implicit condition that there's a region common to all three circle interiors. (That is, the circles make a proper three-set Venn diagram.) Curiously, the problem statement excludes the case of the circles having a common point, even though the statement is obviously true in that case: the chords have that common point in common. Go figure. $\endgroup$ – Blue Jan 13 '18 at 23:00
  • $\begingroup$ The Cut-the-Knot page mentions in a Note that the proposed proof doesn't hold when the centers of the circles are collinear. Yet the author, having been made aware of the flaw, didn't adjust the problem description to rule-out this case. This, together with your observation, suggests that the author is being uncharacteristically sloppy with this puzzle. $\endgroup$ – Blue Jan 13 '18 at 23:13
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Since the sphere argument seems a bit involute, I give you a very simple proof based on radical axis. Recall that the radical axis of two circle is the locus of points which have the same power respect to both circles.

Let $\Gamma_1,\Gamma_2,\Gamma_3$ be the three given circles, $r_{ij}$ the radical axis of $\Gamma_i$ and $\Gamma_j$ for each $i<j$. Let $P\in r_{12}\cap r_{23}$ and let $p_i$ denote the power of $P$ respect to $\Gamma_i$. Then $P\in r_{12}$ implies $p_1=p_2$ while $P\in r_{23}$ implies $p_2=p_3$, thus by transitivity, $p_1=p_3$, hence $P\in r_{13}$. This proves $P\in r_{12}\cap r_{23}\cap r_{13}$.

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  • $\begingroup$ what does it mean by $P\cap r_{12}\in r_{23}$, shouldn't it be $P\in r_{12}\cap r_{23}$? and what is power of $P$ respect to a circle? $\endgroup$ – athos Mar 28 '18 at 12:12
  • $\begingroup$ @athos Thanks for your comment. I fixed then mistake and added a link to the definition of the power of a point respect to a circle. $\endgroup$ – Fabio Lucchini Mar 28 '18 at 12:43
  • $\begingroup$ thx now i got it. $\endgroup$ – athos Apr 17 '18 at 4:50

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