1950 AHSME Problems/Problem 45: Difference between revisions
Mrdavid445 (talk | contribs) Created page with "==Problem== The number of diagonals that can be drawn in a polygon of 100 sides is: <math>\textbf{(A)}\ 4850 \qquad \textbf{(B)}\ 4950\qquad \textbf{(C)}\ 9900 \qquad \textbf{(..." |
Added a solution to the problem. Lots of these problems don't even have solution sections. |
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\textbf{(D)}\ 98 \qquad | \textbf{(D)}\ 98 \qquad | ||
\textbf{(E)}\ 8800</math> | \textbf{(E)}\ 8800</math> | ||
== Solution == | |||
Each diagonal has its two endpoints as vertices of the 100-gon. Each pair of vertices determines exactly one diagonal. Therefore the answer should be <math>\binom{100}{2}=4950</math>. However this also counts the 100 sides of the polygon, so the actual answer is <math>4950-100=\boxed{\textbf{(A)}\ 4850 }</math>. | |||
== See Also == | |||
{{AHSME 50p box|year=1950|num-b=44|num-a=46}} | |||
[[Category:Introductory Combinatorics Problems]] | |||
Revision as of 10:20, 23 April 2012
Problem
The number of diagonals that can be drawn in a polygon of 100 sides is:
Solution
Each diagonal has its two endpoints as vertices of the 100-gon. Each pair of vertices determines exactly one diagonal. Therefore the answer should be 
. However this also counts the 100 sides of the polygon, so the actual answer is 
.
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 44 |
Followed by Problem 46 | |
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