>No. Without loss of generality we assume that all circles are pairwise
>different. (If two circles are identical, we remove one of them.)
>Since there exists a point that lies inside all circles, two of them
>intersect in two points. There are N circles and therefore N*(N-1)/2
>unordered pairs of circles. This may lead to up to N*(N-1) intersections.
>

But most of these intersections are not interesting, because they are
obscured by other circles. On the _perimeter_ of the big buch of circles,
there are a maximum og N+1 points to check.


Try to draw circles on a paper so that all of them overlap at some point,
and see what happens. How many points around the edge of the figure do you
need to count?

>> It sould not be difficult to locate those points, checking the distance
to P
>> and chose the one(s) closest.
>
>The closest intersection point may not be the correct solution. It may
>lie inside another (third) circle.
>

No, because all the intersection points in question are outside all the
circles. (See above.)

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