Difference between revisions of "2013 AMC 10A Problems/Problem 25"
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==Solution 2 (elimination)== | ==Solution 2 (elimination)== | ||
− | Let the number of intersections be <math>x</math>. We know that <math>x\le \dbinom{8}{4} = 70</math>, as every 4 points forms a quadrilateral with intersecting diagonals. However, four diagonals intersect in the center, so we need to subtract <math>\dbinom{4}{2} -1 = 5</math> from this count. <math>70-5 = 65</math>. You might be tempted to choose 65 at this point, but note that diagonals like AD, CG, and BE all intersect at the same point. There are <math>8</math> of this type with three diagonals intersecting at the same point | + | Let the number of intersections be <math>x</math>. We know that <math>x\le \dbinom{8}{4} = 70</math>, as every 4 points forms a quadrilateral with intersecting diagonals. However, four diagonals intersect in the center, so we need to subtract <math>\dbinom{4}{2} -1 = 5</math> from this count. <math>70-5 = 65</math>. You might be tempted to choose 65 at this point, but note that diagonals like AD, CG, and BE all intersect at the same point. There are <math>8</math> of this type with three diagonals intersecting at the same point, so we need to subtract <math>2</math> of the <math>\dbinom{3}{2}</math> (one is kept as the actual intersection). In the end, we obtain <math>65 - 16 = \boxed{\textbf{(A) }49}</math> |
==See Also== | ==See Also== |
Revision as of 21:31, 28 October 2013
Problem
All 20 diagonals are drawn in a regular octagon. At how many distinct points in the interior of the octagon (not on the boundary) do two or more diagonals intersect?
Solution 1 (drawing)
If you draw a good diagram like the one below, it is easy to see that there are , points.
Solution 2 (elimination)
Let the number of intersections be . We know that , as every 4 points forms a quadrilateral with intersecting diagonals. However, four diagonals intersect in the center, so we need to subtract from this count. . You might be tempted to choose 65 at this point, but note that diagonals like AD, CG, and BE all intersect at the same point. There are of this type with three diagonals intersecting at the same point, so we need to subtract of the (one is kept as the actual intersection). In the end, we obtain
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 | ||
All AMC 10 Problems and Solutions |
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