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Difference between revisions of "1950 AHSME Problems/Problem 45"

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==Problem==
 
==Problem==
  
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The number of diagonals that can be drawn in a polygon of 100 sides is:
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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
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== Solution 2 ==
 
== Solution 2 ==
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For each vertex we can choose <math>100 - 3 = 97</math> vertices to draw the diagonal, as we cannot connect a vertex to itself or either adjacent vertices. Thus, the answer is <math>(100)(97)/2=4850</math>, as we are overcounting by a factor of 2 (we are counting each diagonal twice - one for each endpoint).
+
We can choose <math>100 - 3 = 97</math> 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 <math>(100)(97)/2=4850</math> diagonals, because we are overcounting by a factor of <math>2</math> (we are counting each diagonal twice - one for each endpoint). So, our answer is <math>\fbox{A}</math>.
 +
 
 +
== Solution 3==
 +
The formula for the number of diagonals of a polygon with <math>n</math> sides is <math>n(n-3)/2</math>. Taking <math>n=100</math>, we see that the number of diagonals that may be drawn in this polygon is <math>100(97)/2</math> or <math>\boxed{\textbf{(A)}\ 4850 }</math>.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
  
 
== See Also ==
 
== See Also ==

Latest revision as of 20:42, 3 May 2024

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

See Also

1950 AHSC (Problems • Answer Key • Resources)
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

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