Difference between revisions of "1971 AHSME Problems/Problem 17"

(Created page with "== Problem == A circular disk is divided by <math>2n</math> equally spaced radii(<math>n>0</math>) and one secant line. The maximum number of non-overlapping areas into whi...")
 
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It seems that for <math>2n</math> radii, there are <math>3n+1</math> distinct areas. The secant line must pass through <math>n</math> radii for this to occur.
 
It seems that for <math>2n</math> radii, there are <math>3n+1</math> distinct areas. The secant line must pass through <math>n</math> radii for this to occur.
  
The answer is <math>\textbf{(E)} = 3n+1.</math>
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The answer is <math>\boxed{\textbf{(E) } 3n+1}.</math>
  
 
-edited by coolmath34
 
-edited by coolmath34
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 +
== See Also ==
 +
{{AHSME 35p box|year=1971|num-b=16|num-a=18}}
 +
{{MAA Notice}}

Revision as of 09:04, 5 August 2024

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

See Also

1971 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AHSME Problems and Solutions

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