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 1

Note that, before adding the secant line, we have $2n$ regions, and, starting from the edge of the disk, every time the secant line intersects a radius or the other end of the disk, a new region is created. Thus, we desire to maximize the number of radii that the secant line passes through.

The angle between any two adjacent radii must be $\tfrac{360^{\circ}}{2n}=\tfrac{180^{\circ}}n$ . A given secant cannot pass through more than $180^{\circ}$ of the radii, and passing through exactly $180^{\circ}$ would imply that the secant must go through the center of the circle, and so it would not produce any new regions. Thus, to maximize the number of regions, the secant line must pass through $n$ radii (spanning $\tfrac{180}n(n-1)$ degrees) and thereby create $n+1$ new regions (one for each radius and one for exiting the disk). Thus, we have a maximum of $2n+n+1=\boxed{\textbf{(E) }3n+1}$ regions.

Solution 2

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)
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1971 AHSME Problems/Problem 17

Contents

Problem

Solution 1

Solution 2

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