1971 AHSME Problems/Problem 17

Problem

A circular disk is divided by $2n$ equally spaced radii( $n>0$ ) and one secant line. The maximum number of non-overlapping areas into which the disk can be divided is

$\textbf{(A) }2n+1\qquad \textbf{(B) }2n+2\qquad \textbf{(C) }3n-1\qquad \textbf{(D) }3n\qquad \textbf{(E) }3n+1$

Solution

We can draw the cases for small values of $n.$ $n = 0 \rightarrow \text{areas} = 1$ $n = 1 \rightarrow \text{areas} = 4$ $n = 2 \rightarrow \text{areas} = 7$ $n = 3 \rightarrow \text{areas} = 10$ It seems that for $2n$ radii, there are $3n+1$ distinct areas. The secant line must pass through $n$ radii for this to occur.

The answer is $\boxed{\textbf{(E) } 3n+1}.$

-edited by coolmath34

1971 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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1971 AHSME Problems/Problem 17

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