Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 21"

 
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==Problem==
 
==Problem==
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A convex polygon has <math>n</math> sides and <math>740</math> diagonals. Then <math>n</math> equals
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<math>\mathrm{(A)}\ 30\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 50\qquad\mathrm{(D)}\ 60\qquad\mathrm{(E)}\ \text{None of these}</math>
  
 
==Solution==
 
==Solution==
{{solution}}
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The number of diagonals in a polygon is <math>\frac{n(n-3)}{2}</math>. In this case, <math>\frac{n(n-3)}{2}=740</math>, so <math>n(n-3)=1480</math>.
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By solving the [[quadratic equation]], we find <math>n = 40</math>, so the answer is <math>\mathrm{B}</math>.
  
 
==See also==
 
==See also==
{{CYMO box|year=2006|l=Lyceum|num-b=21|num-a=22}}
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{{CYMO box|year=2006|l=Lyceum|num-b=20|num-a=22}}
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[[Category:Introductory Combinatorics Problems]]

Latest revision as of 12:30, 26 April 2008

Problem

A convex polygon has $n$ sides and $740$ diagonals. Then $n$ equals

$\mathrm{(A)}\ 30\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 50\qquad\mathrm{(D)}\ 60\qquad\mathrm{(E)}\ \text{None of these}$

Solution

The number of diagonals in a polygon is $\frac{n(n-3)}{2}$. In this case, $\frac{n(n-3)}{2}=740$, so $n(n-3)=1480$.

By solving the quadratic equation, we find $n = 40$, so the answer is $\mathrm{B}$.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 20
Followed by
Problem 22
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