Number of diagonals of a concave polygons formula?
Does this formula also apply to concave polygons?
geometry
asked Dec 4 '18 at 20:40
caasdadscaasdads
957
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Does this formula also apply to concave polygons?
geometry
asked Dec 4 '18 at 20:40
caasdadscaasdads
957
It depends on what you mean by "distinct diagonal." The expression $n(n-3)/2$ counts the number of ways to select two non-adjacent vertices of any polygon. But in a concave polygon, the lines connecting such pairs may overlap.
– angryavian
Dec 4 '18 at 20:46
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Does this formula also apply to concave polygons?
geometry
asked Dec 4 '18 at 20:40
caasdadscaasdads
957
Does this formula also apply to concave polygons?
geometry
geometry
asked Dec 4 '18 at 20:40
caasdadscaasdads
957
asked Dec 4 '18 at 20:40
caasdadscaasdads
957
asked Dec 4 '18 at 20:40
caasdadscaasdads
957
asked Dec 4 '18 at 20:40
caasdadscaasdads
957
asked Dec 4 '18 at 20:40
caasdadscaasdads
957
957
It depends on what you mean by "distinct diagonal." The expression $n(n-3)/2$ counts the number of ways to select two non-adjacent vertices of any polygon. But in a concave polygon, the lines connecting such pairs may overlap.
– angryavian
Dec 4 '18 at 20:46
add a comment |
It depends on what you mean by "distinct diagonal." The expression $n(n-3)/2$ counts the number of ways to select two non-adjacent vertices of any polygon. But in a concave polygon, the lines connecting such pairs may overlap.
– angryavian
Dec 4 '18 at 20:46
It depends on what you mean by "distinct diagonal." The expression $n(n-3)/2$ counts the number of ways to select two non-adjacent vertices of any polygon. But in a concave polygon, the lines connecting such pairs may overlap.
– angryavian
Dec 4 '18 at 20:46
It depends on what you mean by "distinct diagonal." The expression $n(n-3)/2$ counts the number of ways to select two non-adjacent vertices of any polygon. But in a concave polygon, the lines connecting such pairs may overlap.
– angryavian
Dec 4 '18 at 20:46
add a comment |
2 Answers
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Yes, no, and maybe, depending on what you mean by diagonal and distinct. For a convex polygon of $n$ vertices, you can draw a diagonal by choosing one vertex ($n$ ways), choosing a second vertex that is not the first nor connected to the first ($n-3$ ways), and dividing by $2$ because you can start at either end of the diagonal. This process works fine for a concave polygon, too, so yes. Some of the diagonals are outside the polygon, so if you require a diagonal to lie within the polygon, no. You can create a concave polygon so that more than two noncontiguous vertices are on a line. If you do, some of the diagonals will overlap, which may violate your definition of distinct, so maybe.
answered Dec 4 '18 at 20:46
Ross MillikanRoss Millikan
292k23197371
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If you count all the diagonals, including the ones that lie outside the polygon, yes.
EXCEPT that, if you have enough points and enough ingenuity, you can arrange the points of the polygon so that some diagonal somewhere will lie directly on top of some other diagonal. In that case, do you have two diagonals, with one "hidden", or do you have one diagonal that does "double duty" in connecting, not two, but three or more points? That's a question of definition and I'm not sure what the standard answer is.
answered Dec 4 '18 at 20:46
bob.sacamentobob.sacamento
2,4261819
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2 Answers
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Yes, no, and maybe, depending on what you mean by diagonal and distinct. For a convex polygon of $n$ vertices, you can draw a diagonal by choosing one vertex ($n$ ways), choosing a second vertex that is not the first nor connected to the first ($n-3$ ways), and dividing by $2$ because you can start at either end of the diagonal. This process works fine for a concave polygon, too, so yes. Some of the diagonals are outside the polygon, so if you require a diagonal to lie within the polygon, no. You can create a concave polygon so that more than two noncontiguous vertices are on a line. If you do, some of the diagonals will overlap, which may violate your definition of distinct, so maybe.
answered Dec 4 '18 at 20:46
Ross MillikanRoss Millikan
292k23197371
add a comment |
Yes, no, and maybe, depending on what you mean by diagonal and distinct. For a convex polygon of $n$ vertices, you can draw a diagonal by choosing one vertex ($n$ ways), choosing a second vertex that is not the first nor connected to the first ($n-3$ ways), and dividing by $2$ because you can start at either end of the diagonal. This process works fine for a concave polygon, too, so yes. Some of the diagonals are outside the polygon, so if you require a diagonal to lie within the polygon, no. You can create a concave polygon so that more than two noncontiguous vertices are on a line. If you do, some of the diagonals will overlap, which may violate your definition of distinct, so maybe.
answered Dec 4 '18 at 20:46
Ross MillikanRoss Millikan
292k23197371
add a comment |
Yes, no, and maybe, depending on what you mean by diagonal and distinct. For a convex polygon of $n$ vertices, you can draw a diagonal by choosing one vertex ($n$ ways), choosing a second vertex that is not the first nor connected to the first ($n-3$ ways), and dividing by $2$ because you can start at either end of the diagonal. This process works fine for a concave polygon, too, so yes. Some of the diagonals are outside the polygon, so if you require a diagonal to lie within the polygon, no. You can create a concave polygon so that more than two noncontiguous vertices are on a line. If you do, some of the diagonals will overlap, which may violate your definition of distinct, so maybe.
answered Dec 4 '18 at 20:46
Ross MillikanRoss Millikan
292k23197371
Yes, no, and maybe, depending on what you mean by diagonal and distinct. For a convex polygon of $n$ vertices, you can draw a diagonal by choosing one vertex ($n$ ways), choosing a second vertex that is not the first nor connected to the first ($n-3$ ways), and dividing by $2$ because you can start at either end of the diagonal. This process works fine for a concave polygon, too, so yes. Some of the diagonals are outside the polygon, so if you require a diagonal to lie within the polygon, no. You can create a concave polygon so that more than two noncontiguous vertices are on a line. If you do, some of the diagonals will overlap, which may violate your definition of distinct, so maybe.
answered Dec 4 '18 at 20:46
Ross MillikanRoss Millikan
292k23197371
edited Dec 4 '18 at 20:53
answered Dec 4 '18 at 20:46
Ross MillikanRoss Millikan
292k23197371
answered Dec 4 '18 at 20:46
Ross MillikanRoss Millikan
292k23197371
answered Dec 4 '18 at 20:46
Ross MillikanRoss Millikan
292k23197371
292k23197371
add a comment |
add a comment |
If you count all the diagonals, including the ones that lie outside the polygon, yes.
EXCEPT that, if you have enough points and enough ingenuity, you can arrange the points of the polygon so that some diagonal somewhere will lie directly on top of some other diagonal. In that case, do you have two diagonals, with one "hidden", or do you have one diagonal that does "double duty" in connecting, not two, but three or more points? That's a question of definition and I'm not sure what the standard answer is.
answered Dec 4 '18 at 20:46
bob.sacamentobob.sacamento
2,4261819
add a comment |
If you count all the diagonals, including the ones that lie outside the polygon, yes.
EXCEPT that, if you have enough points and enough ingenuity, you can arrange the points of the polygon so that some diagonal somewhere will lie directly on top of some other diagonal. In that case, do you have two diagonals, with one "hidden", or do you have one diagonal that does "double duty" in connecting, not two, but three or more points? That's a question of definition and I'm not sure what the standard answer is.
answered Dec 4 '18 at 20:46
bob.sacamentobob.sacamento
2,4261819
add a comment |
If you count all the diagonals, including the ones that lie outside the polygon, yes.
EXCEPT that, if you have enough points and enough ingenuity, you can arrange the points of the polygon so that some diagonal somewhere will lie directly on top of some other diagonal. In that case, do you have two diagonals, with one "hidden", or do you have one diagonal that does "double duty" in connecting, not two, but three or more points? That's a question of definition and I'm not sure what the standard answer is.
answered Dec 4 '18 at 20:46
bob.sacamentobob.sacamento
2,4261819
If you count all the diagonals, including the ones that lie outside the polygon, yes.
EXCEPT that, if you have enough points and enough ingenuity, you can arrange the points of the polygon so that some diagonal somewhere will lie directly on top of some other diagonal. In that case, do you have two diagonals, with one "hidden", or do you have one diagonal that does "double duty" in connecting, not two, but three or more points? That's a question of definition and I'm not sure what the standard answer is.
answered Dec 4 '18 at 20:46
bob.sacamentobob.sacamento
2,4261819
answered Dec 4 '18 at 20:46
bob.sacamentobob.sacamento
2,4261819
answered Dec 4 '18 at 20:46
bob.sacamentobob.sacamento
2,4261819
answered Dec 4 '18 at 20:46
bob.sacamentobob.sacamento
2,4261819
2,4261819
add a comment |
add a comment |
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It depends on what you mean by "distinct diagonal." The expression $n(n-3)/2$ counts the number of ways to select two non-adjacent vertices of any polygon. But in a concave polygon, the lines connecting such pairs may overlap.
– angryavian
Dec 4 '18 at 20:46