# 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 $\boxed{\textbf{(A)}\ 4850 }$.