Difference between revisions of "1950 AHSME Problems/Problem 45"
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==Problem== | ==Problem== | ||
− | The number of diagonals that can be drawn in a polygon of 100 sides is: | + | The number of diagonals that can be drawn in a polygon of <math>100</math> sides is: |
<math>\textbf{(A)}\ 4850 \qquad | <math>\textbf{(A)}\ 4850 \qquad |
Latest revision as of 20:42, 3 May 2024
Contents
[hide]Problem
The number of diagonals that can be drawn in a polygon of 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 .
Solution 2
We can choose vertices for each vertex to draw the diagonal, as we cannot connect a vertex to itself or any of its two adjacent vertices. Thus, there are diagonals, because we are overcounting by a factor of (we are counting each diagonal twice - one for each endpoint). So, our answer is .
Solution 3
The formula for the number of diagonals of a polygon with sides is . Taking , we see that the number of diagonals that may be drawn in this polygon is or .
~ cxsmi
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 44 |
Followed by Problem 46 | |
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All AHSME Problems and Solutions |
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