Geometry
Posted: Mon Jan 02, 2017 6:57 pm
Maximum how many intersecting points can be found by joining all the diagonals of a 2013-gon?
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Re: Geometry
Posted: Mon Jan 02, 2017 8:26 pm
Under the assumption that no two diagonals are parallel and no three diagonals are concurrent, the answer is clearly $3\dbinom{2013}{4}$ as choosig $4$ points yield $3$ intersections of diagonals.
Re: Geometry
Posted: Sat Jan 21, 2017 6:24 pm
please give a small description how it happened...
Re: Geometry
Posted: Sun Jan 22, 2017 4:34 pm
Take any $4$ points, you will get three intersection points. And how many ways can you take 4 points? Yes, in $ \binom {2013}{4}$ ways.
Re: Geometry
Posted: Sun Jan 22, 2017 6:47 pm
Why will any 4 points give 3 intersecting points of diagonals? Any group of 4 points will generate 2 unique diagonals, and the 2 diagonals will intersect at 1 point.
Re: Geometry
Posted: Mon Jan 30, 2017 6:37 pm
Let $ABCD$ be the four points. $AB$ and $CD$ intersect at point $1$.
$AC$ and $BD$ intersect at point $2$.
$AD$ and $BC$ intersect at point $3$.
A total of $3$ points.
$AC$ and $BD$ intersect at point $2$.
$AD$ and $BC$ intersect at point $3$.
A total of $3$ points.
Re: Geometry
Posted: Mon Jan 30, 2017 9:30 pm
The way you are counting
1. views certain side(s) as diagonal(s).
For example, if A and B are adjacent vertices, then AB is a side, not a diagonal. You might argue that no pair of the four points (A, B, C and D) is adjacent, but when you generalize the counting for 2013 sides by choosing any group of 4 vertices, some pairs of points, when connected, will definitely form sides, not diagonals.
2. assumes that most of the intersecting points are formed by extension of diagonals outside the polygon, not by direct intersection inside the polygon.
For example, if the points A, B, C and D are in the order (clockwise or anti-clockwise) that they are written, then the only valid intersection point is formed by diagonal AC and diagonal BD. The other two intersections are not valid, because remember a diagonal by definition is a line-segment, not a line.
Hope it clarifies all: one valid intersection point for each group of 4 vertices.
1. views certain side(s) as diagonal(s).
For example, if A and B are adjacent vertices, then AB is a side, not a diagonal. You might argue that no pair of the four points (A, B, C and D) is adjacent, but when you generalize the counting for 2013 sides by choosing any group of 4 vertices, some pairs of points, when connected, will definitely form sides, not diagonals.
2. assumes that most of the intersecting points are formed by extension of diagonals outside the polygon, not by direct intersection inside the polygon.
For example, if the points A, B, C and D are in the order (clockwise or anti-clockwise) that they are written, then the only valid intersection point is formed by diagonal AC and diagonal BD. The other two intersections are not valid, because remember a diagonal by definition is a line-segment, not a line.
Hope it clarifies all: one valid intersection point for each group of 4 vertices.
Re: Geometry
Posted: Mon Jan 30, 2017 9:37 pm
It depends on your consideration of a diagonal. I computed the total number of intersection points.
Re: Geometry
Posted: Mon Jan 30, 2017 9:48 pm
No, it doesn't depend on anyone's personal consideration of diagonals. Sides are not diagonals, and diagonals are not lines. These two facts don't change with our opinions. 
Re: Geometry
Posted: Mon Jan 30, 2017 9:55 pm
Oh, right. I don't know much English terms. Sorry about that. We are actually used to consider sides as diagonals too.