How to calculate the center of a regular polygon?
What is the formula for the center of an n-edge regular polygon that has the given segment as its edge?
So, given a segment AB, with endpoints A=(a1,a2) and B=(b1,b2), I need to find out the two points X=(x1,x2) and Y=(y1,y2), such that the n-edge regular polygon with center at X, and the one with center at Y have AB as their edge.
metric-geometry polygons
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migrated from mathoverflow.net Dec 25 '14 at 17:27
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What is the formula for the center of an n-edge regular polygon that has the given segment as its edge?
So, given a segment AB, with endpoints A=(a1,a2) and B=(b1,b2), I need to find out the two points X=(x1,x2) and Y=(y1,y2), such that the n-edge regular polygon with center at X, and the one with center at Y have AB as their edge.
metric-geometry polygons
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
migrated from mathoverflow.net Dec 25 '14 at 17:27
This question came from our site for professional mathematicians.
add a comment |
What is the formula for the center of an n-edge regular polygon that has the given segment as its edge?
So, given a segment AB, with endpoints A=(a1,a2) and B=(b1,b2), I need to find out the two points X=(x1,x2) and Y=(y1,y2), such that the n-edge regular polygon with center at X, and the one with center at Y have AB as their edge.
metric-geometry polygons
What is the formula for the center of an n-edge regular polygon that has the given segment as its edge?
So, given a segment AB, with endpoints A=(a1,a2) and B=(b1,b2), I need to find out the two points X=(x1,x2) and Y=(y1,y2), such that the n-edge regular polygon with center at X, and the one with center at Y have AB as their edge.
metric-geometry polygons
metric-geometry polygons
asked Dec 25 '14 at 15:12
Vahagn
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
migrated from mathoverflow.net Dec 25 '14 at 17:27
This question came from our site for professional mathematicians.
migrated from mathoverflow.net Dec 25 '14 at 17:27
This question came from our site for professional mathematicians.
add a comment |
add a comment |
1 Answer
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Let $T=tan(180^{circ}/n)$. The midpoint of $AB$ is $((a1+b1)/2,(a2+b2)/2)$, the centres would be
$$((a1+b1)/2+(a2-b2)/2T,(a2+b2)/2+(b1-a1)/2T)\
((a1+b1)/2+(b2-a2)/2T,(a2+b2)/2+(a1-b1)/2T)$$
answered Dec 25 '14 at 17:40
Empy2
33.4k12261
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
Let $T=tan(180^{circ}/n)$. The midpoint of $AB$ is $((a1+b1)/2,(a2+b2)/2)$, the centres would be
$$((a1+b1)/2+(a2-b2)/2T,(a2+b2)/2+(b1-a1)/2T)\
((a1+b1)/2+(b2-a2)/2T,(a2+b2)/2+(a1-b1)/2T)$$
answered Dec 25 '14 at 17:40
Empy2
33.4k12261
add a comment |
Let $T=tan(180^{circ}/n)$. The midpoint of $AB$ is $((a1+b1)/2,(a2+b2)/2)$, the centres would be
$$((a1+b1)/2+(a2-b2)/2T,(a2+b2)/2+(b1-a1)/2T)\
((a1+b1)/2+(b2-a2)/2T,(a2+b2)/2+(a1-b1)/2T)$$
answered Dec 25 '14 at 17:40
Empy2
33.4k12261
add a comment |
Let $T=tan(180^{circ}/n)$. The midpoint of $AB$ is $((a1+b1)/2,(a2+b2)/2)$, the centres would be
$$((a1+b1)/2+(a2-b2)/2T,(a2+b2)/2+(b1-a1)/2T)\
((a1+b1)/2+(b2-a2)/2T,(a2+b2)/2+(a1-b1)/2T)$$
answered Dec 25 '14 at 17:40
Empy2
33.4k12261
Let $T=tan(180^{circ}/n)$. The midpoint of $AB$ is $((a1+b1)/2,(a2+b2)/2)$, the centres would be
$$((a1+b1)/2+(a2-b2)/2T,(a2+b2)/2+(b1-a1)/2T)\
((a1+b1)/2+(b2-a2)/2T,(a2+b2)/2+(a1-b1)/2T)$$
answered Dec 25 '14 at 17:40
Empy2
33.4k12261
answered Dec 25 '14 at 17:40
Empy2
33.4k12261
answered Dec 25 '14 at 17:40
Empy2
33.4k12261
answered Dec 25 '14 at 17:40
Empy2
33.4k12261
33.4k12261
add a comment |
add a comment |
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