1971 AHSME Problems/Problem 17


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

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


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 $\textbf{(E)} = 3n+1.$

-edited by coolmath34