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Geometry Difficulty 7.1 National olympiad, round 2 Prove it Greece

In the plane there are given nn different circles with the same center. In the interior of the circle with the smaller radius we consider two different points AA and BB. Next we consider kk different lines passing from point AA and mm different lines passing from point BB. All lines passing from AA intersect all lines passing from BB (there is no line passing from AA and BB) and their points of intersection are not on the given circles. Determine the maximal and the minimal number of regions bounded by the lines and the circles, lying in the interior of the circles and for the regions lying into the smaller circle at least one part of their border is an arch of the circle.

Solution

First we note that the number of points of intersection of the lines is kmk \cdot m (figure 1).

Figure 1
Figure 1

Let now all the points of intersection lie at the exterior of the circle with the greatest radius. Then the kk different lines passing through AA create 2k2k regions inside the circle with the smaller radius, while the mm different lines passing through BB create 2m2m regions inside the circle with the smaller radius. Since we have counted the shaded region (figure 2) two times, the total number of regions inside the smaller circle is 2k+2m12 \cdot k + 2 \cdot m - 1.

We have totally n1n-1 circular rings. In each circular ring is created the same number of regions 2k+2m12 \cdot k + 2 \cdot m - 1, but the shaded region is divided in two regions (figures 2α and 2β). Therefore inside any circular ring we have 2k+2m2 \cdot k + 2 \cdot m regions.

Figure 2
figure 2β

Figure 3
figure 2α

Finally in this case the total number of regions is: 2n(k+m)12n(k+m)-1.

If one or more points of intersection lie inside the circle with the smaller radius, then we have 2(k+m)2(k+m) regions inside the smaller circle (figures 3α, 3β και 3γ). In the case of figure 3α (all the points of intersection lie inside the circle with the smaller radius) we have 2(n1)(k+m)2(n-1)(k+m) regions into the n1n-1 circular rings. In the case one point of intersection lie into a circular ring the number of region increases by one. Therefore, finally we have totally
2(k+m)+2(n1)(k+m)=2n(k+m) regions. 2(k+m)+2(n-1)(k+m)=2n(k+m) \text{ regions.}

Figure 4
figure 3α

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