Difference between revisions of "2013 AMC 10A Problems/Problem 25"
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Let the number of intersections be <math>x</math>. We know that <math>x\le \dbinom{8}{4} = 70</math>, as every <math>4</math> vertices on the octagon forms a quadrilateral with intersecting diagonals which is an intersection point. 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>. Note that diagonals like <math>\overline{AD}</math>, <math>\overline{CG}</math>, and <math>\overline{BE}</math> 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>. | Let the number of intersections be <math>x</math>. We know that <math>x\le \dbinom{8}{4} = 70</math>, as every <math>4</math> vertices on the octagon forms a quadrilateral with intersecting diagonals which is an intersection point. 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>. Note that diagonals like <math>\overline{AD}</math>, <math>\overline{CG}</math>, and <math>\overline{BE}</math> 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>. | ||
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+ | ==Solution 3 (Using the answer choices)== | ||
+ | We know that the amount of intersection points is at most <math>\dbinom{8}{4} = 70</math>, as in solution <math>2</math>. There's probably going to be more than <math>5</math> (to get <math>(B) 65</math>), leading us to the only reasonable answer, <math>\boxed{\textbf{(A) }49}</math>. | ||
+ | -Lcz | ||
==See Also== | ==See Also== |
Revision as of 12:14, 7 January 2020
Contents
Problem
All 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 clear diagram like the one below, it is easy to see that there are points.
Solution 2 (Working Backwards)
Let the number of intersections be . We know that , as every vertices on the octagon forms a quadrilateral with intersecting diagonals which is an intersection point. However, four diagonals intersect in the center, so we need to subtract from this count, . Note that diagonals like , , and 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 .
Solution 3 (Using the answer choices)
We know that the amount of intersection points is at most , as in solution . There's probably going to be more than (to get ), leading us to the only reasonable answer, . -Lcz
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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