How to calculate the center of a regular polygon?

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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.







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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.







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    migrated from mathoverflow.net Dec 25 '14 at 17:27


    This question came from our site for professional mathematicians.
















      up vote
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      down vote

      favorite









      up vote
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      down vote

      favorite











      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.







      share|cite|improve this question












      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.









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      asked Dec 25 '14 at 15:12







      Vahagn











      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.






















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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
          $$((a1+b1)/2+(a2-b2)/2T,(a2+b2)/2+(b1-a1)/2T)\
          ((a1+b1)/2+(b2-a2)/2T,(a2+b2)/2+(a1-b1)/2T)$$






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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
            $$((a1+b1)/2+(a2-b2)/2T,(a2+b2)/2+(b1-a1)/2T)\
            ((a1+b1)/2+(b2-a2)/2T,(a2+b2)/2+(a1-b1)/2T)$$






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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
              $$((a1+b1)/2+(a2-b2)/2T,(a2+b2)/2+(b1-a1)/2T)\
              ((a1+b1)/2+(b2-a2)/2T,(a2+b2)/2+(a1-b1)/2T)$$






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                up vote
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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
                $$((a1+b1)/2+(a2-b2)/2T,(a2+b2)/2+(b1-a1)/2T)\
                ((a1+b1)/2+(b2-a2)/2T,(a2+b2)/2+(a1-b1)/2T)$$






                share|cite|improve this answer












                Let $T=tan(180^circ/n)$. The midpoint of $AB$ is $((a1+b1)/2,(a2+b2)/2)$, the centres would be
                $$((a1+b1)/2+(a2-b2)/2T,(a2+b2)/2+(b1-a1)/2T)\
                ((a1+b1)/2+(b2-a2)/2T,(a2+b2)/2+(a1-b1)/2T)$$







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                share|cite|improve this answer










                answered Dec 25 '14 at 17:40









                Empy2

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