Difference between revisions of "2013 AMC 10A Problems/Problem 25"
m (→Solution 2 (elimination): lines) |
Csasomachar (talk | contribs) (→Solution 1 (Drawing)) |
||
Line 7: | Line 7: | ||
[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
− | ==Solution 1 (Drawing)== | + | ==Solution 1 (Drawing: The REALLY Dumb YET "it still works" Way)== |
If you draw a good diagram like the one below, it is easy to see that there are <math>\boxed{\textbf{(A) }49}</math> points. | If you draw a good diagram like the one below, it is easy to see that there are <math>\boxed{\textbf{(A) }49}</math> points. |
Revision as of 15:13, 3 July 2016
Contents
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: The REALLY Dumb YET "it still works" Way)
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
points forms a quadrilateral with intersecting diagonals. 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
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.