Permutations & Combinations
Regions formed by circles
Grade 11
Question:
<p>Let \(f(n)\) be the number of regions in which <i>n</i> coplanar circles can divide the plane. If it is known that each pair of circles intersect in two different points and no three of them have common point of intersection, then</p>
<p>\(f(20) = 382\)</p>
<p>\(f(n)\) is always an even number</p>
<p>\(f^{-1}(92) = 10\)</p>
<p>\(f(n)\) can be odd</p>
Step-by-Step Solution
Key Concept: Each new circle intersects all previous circles at exactly 2 points each, creating 2n new regions. The recurrence relation is f(n) = f(n-1) + 2n, which solves to f(n) = n² - n + 2.
<p><strong>Step 1:</strong> Start with base cases.</p><p>• f(0) = 1 (no circles, whole plane is 1 region)</p><p>• f(1) = 2 (one circle divides plane into 2 regions)</p><p><strong>Step 2:</strong> Establish the recurrence relation.</p><p>When the nth circle is added, it intersects each of the (n-1) existing circles at exactly 2 points (given condition). This creates 2(n-1) intersection points on the nth circle.</p><p>These 2(n-1) points divide the nth circle into 2(n-1) arcs. Each arc splits an existing region into two, adding exactly 2(n-1) new regions.</p><p>Therefore: f(n) = f(n-1) + 2(n-1)</p><p><strong>Step 3:</strong> Solve the recurrence relation.</p><p>f(n) = f(n-1) + 2(n-1)</p><p>f(n) = f(n-2) + 2(n-2) + 2(n-1)</p><p>f(n) = f(0) + 2[1 + 2 + 3 + ... + (n-1)]</p><p>f(n) = 1 + 2 · [(n-1)n/2]</p><p>f(n) = 1 + n(n-1)</p><p><strong>f(n) = n² - n + 2</strong></p><p>∴ Answer: AC</p>
Correct Answer: AC