Difference between revisions of "1950 AHSME Problems/Problem 45"
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== Solution 2 == | == Solution 2 == | ||
| − | 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). | + | 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 <math>2</math> (we are counting each diagonal twice - one for each endpoint). Thus our answer is <math>\fbox{A}</math>. |
== See Also == | == See Also == | ||
Revision as of 00:08, 29 June 2018
Contents
Problem
The number of diagonals that can be drawn in a polygon of 100 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
For each vertex we can choose 
vertices to draw the diagonal, as we cannot connect a vertex to itself or either adjacent vertices. Thus, the answer is 
, as we are overcounting by a factor of 
(we are counting each diagonal twice - one for each endpoint). Thus our answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
