1950 AHSME Problems/Problem 45

Problem

The number of diagonals that can be drawn in a polygon of 100 sides is:

$\textbf{(A)}\ 4850 \qquad \textbf{(B)}\ 4950\qquad \textbf{(C)}\ 9900 \qquad \textbf{(D)}\ 98 \qquad \textbf{(E)}\ 8800$

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 $\binom{100}{2}=4950$ . However this also counts the 100 sides of the polygon, so the actual answer is $4950-100=\boxed{\textbf{(A)}\ 4850 }$ .

Solution 2

We can choose $100 - 3 = 97$ 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 $(100)(97)/2=4850$ diagonals, because we are overcounting by a factor of $2$ (we are counting each diagonal twice - one for each endpoint). So, our answer is $\fbox{A}$ .

Solution 3

The formula for the number of diagonals of a polygon with $n$ sides is $n(n-3)/2$ . Taking $n=100$ , we see that the number of diagonals that may be drawn in this polygon is $100(97)/2$ or $\boxed{\textbf{(A)}\ 4850 }$ .

~ cxsmi

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 44	Followed by Problem 46
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1950 AHSME Problems/Problem 45

Contents

Problem

Solution

Solution 2

Solution 3

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