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

Latest revision as of 12:30, 26 April 2008

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

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

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 $\mathrm{B}$ .

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