Difference between revisions of "2013 AMC 10A Problems/Problem 25"
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− | 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>. | + | 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>\boxed{\textbf{(B) }65}</math>), leading us to the only reasonable answer, <math>\boxed{\textbf{(A) }49}</math>. |
-Lcz | -Lcz | ||
Revision as of 18:43, 3 April 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 (Answer choices and reasoning)
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
Solution 4 (Drawing but easier)
Like solution one, we may draw. Except note that the octagon has eight regions, and each region has an equal number of points, so drawing only one of the eight regions and the intersection points suffices. One of the eight regions contains points (not including the octagon center). However each adjacent region share one side in common and that side contains intersection points, so in actuality there are points per region. We multiply this by to get and add the one center point to get .
~skyscraper
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=3ovuMRYs--8
~dolphin7
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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