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# Practice Paper 3 Question 10

$$N$$ circles in a plane all intersect each other such that every circle intersects every other circle at exactly 2 points. Find in terms of $$N$$ the minimum and maximum number of disjoint closed regions that can be formed. Hint: You may wish to check your answers (visually) for greater $$N,$$ e.g. $$N=4.$$

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## Warm-up Questions

1. Given that $$a_1=1$$ and $$a_n=2a_{n-1}$$ find $$a_{20}.$$

## Hints

• Hint 1
• Hint 2
Consider the number of different points at which circles intersect.
• Hint 3
Try to achieve the maximum and the minimum number of intersection points.
• Hint 4
How many existing regions can a newly added circle pass through?
• Hint 5
Note that a newly added circle does not (necessarily) pass through every previous region. In fact, after the $$3^{rd}$$ circle that's impossible.
• Hint 6
How does the number of new intersection points relate to the number of new regions?

## Solution

Denote with $$N_k$$ the number of disjoint regions after $$k$$ circles have been added. The base case is $$N_1=1.$$

The minimum occurs when the 2 points are the same for all pairs of circles, and each new circle is bigger than the last one added. Adding a new circle slices one internal region in two and adds a new region outside of the previous union. Thus we have $$N_{k+1} = N_k+2.$$ Hence $$N_{k}=2k-1.$$

The maximum occurs when each new added circle intersects each existing circle at two different points for every circle (maximum number of intersection points). Note that a newly added circle does not necessarily pass through every previous region; in fact, after the 3rd circle that's impossible. In this case, adding a new circle will add as many new regions as twice the number of existing circles. In other words, $$N_{k+1}=N_k+2k.$$ We have $$N_{k+1}=1+2\sum_{i=1}^{k}i=1+k(k+1)$$ or $$N_k=1+k(k-1).$$

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