Have fun Counting
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?
Details and assumptions
A regular octagon has ( 2 8 ) − 8 = 2 8 − 8 = 2 0 diagonals.
The answer is 49.
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4 solutions
lol made a bad bad mistake and put 65 on the test
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I got it by drawing it out, as did multiple people from my school lol.
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yeah two people did from my school ._. i missed that two '3' diagonals and one '4' diagonal met in the same place so i overcounted
geogebra ftw :)
I remember this question from practicing AMCs.
This is basically #25 from the 2013 AMC10A..
Let the number of intersections be x. We know that x ≤ ( 4 8 ) = 7 0 , as every 4 points forms a quadrilateral with intersecting diagonals. However, four diagonals intersect in the center, so we need to subtract ( 2 4 ) − 1 = 5 from this count. Note that diagonals like AD, CG, and BE all intersect at the same point. There are 8 of this type with three diagonals intersecting at the same point, so we need to subtract 2 of the ( 2 3 ) (one is kept as the actual intersection). In the end, we obtain 7 0 − 5 − 1 6 = (A) 4 9
Solution taken from: https://www.artofproblemsolving.com/Wiki/index.php/2013 AMC 10A Problems/Problem 25
There is one request:
Can it be generalized? If so, how? Please explain.
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Yes it can! See this brilliant paper by Poonen and Rubinstein! http://arxiv.org/pdf/math/9508209.pdf
I'm not sure. I don't think it can (otherwise it would probably be some kind of somewhat-obscure formula) but maybe it could. For a non-regular polygon with no diagonal intersections, it would be ( 4 x ) , and you could possibly derive a formula for a regular polygon from there.
" Note that diagonals like AD, CG, and BE all intersect at the same point."
Shouldn't it be "AD, BE, CF"?
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No, CG is right. AD and EB are mirror images of each other w.r.t. CG.
I think the hard part is to see that the answer is no less than 49.
Same approach...XD
Solutions courtesy AoPS.
This problem comes straight from 2013 AMC 10A (Problem 25). You can view two possible solutions here: http://www.artofproblemsolving.com/Wiki/index.php/2013 AMC 10A Problems/Problem 25
See 2013 AMC 10A #25 .