Solution
Ten Circles
Show the problem again
Ten circles are drawn in the plane so that every two of them intersect in exactly two points and no three pass through a common point. Into how many regions (bounded or unbounded) do they divide the plane? Give an integer.
Worked solution
The answer is 92. Add the circles one at a time. The $k$-th circle meets each of the previous $k - 1$ circles in 2 points, giving $2(k - 1)$ points on it that cut it into $2(k - 1)$ arcs (for $k \ge 2$). Each arc splits a region, so it adds $2(k - 1)$ regions; the first circle adds 1. Total: $1 + 1 + \sum_{k=2}^{10} 2(k-1) = 2 + 2 \cdot 45 = 92$. In general $n$ such circles give $n^2 - n + 2$ regions. Euler's formula agrees: $V = 90$, $E = 180$, so $F = E - V + 2 = 92$.
Source: Classic exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.