1950 AHSME Problems/Problem 45

Revision as of 16:02, 24 December 2022 by Xiaoxiaoxiong (talk | contribs) (Solution 2: grammar clarity issue fixed)

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 $\fbox{A}$.

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 44
Followed by
Problem 46
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png