Geometry - Proof of Construction of regular pentagon by using compass and straightedge.











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How to prove that the polygon constructed by the method mentioned in the following link is indeed a regular pentagon.



https://www.youtube.com/watch?v=9VceAA2qBPA






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    up vote
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    down vote

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    How to prove that the polygon constructed by the method mentioned in the following link is indeed a regular pentagon.



    https://www.youtube.com/watch?v=9VceAA2qBPA






    geometry geometric-construction







    share|cite|improve this question







    New contributor




    Mak Man Tung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      up vote
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      favorite









      up vote
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      down vote

      favorite











      How to prove that the polygon constructed by the method mentioned in the following link is indeed a regular pentagon.



      https://www.youtube.com/watch?v=9VceAA2qBPA






      geometry geometric-construction







      share|cite|improve this question







      New contributor




      Mak Man Tung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      How to prove that the polygon constructed by the method mentioned in the following link is indeed a regular pentagon.



      https://www.youtube.com/watch?v=9VceAA2qBPA






      geometry geometric-construction




      geometry geometric-construction






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      Mak Man Tung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          After some setup of perpendicular segments (here, $overline{OA}$ and $overline{OC}$), the first few steps are the standard construction of the midpoint ($B$) of $overline{OA}$, so let's start there; and let's say that $|OB| = |BA| = 1$.



          enter image description here



          By the Pythagorean theorem,
          $$|BC| = sqrt{|OB|^2 + |OC|^2} = sqrt{1+2^2} = sqrt{5}$$
          and we construct $bigcirc B$ with that radius, meeting the "horizontal" axis at $D$, so that $|OD| = |BC|-|OB| = sqrt{5}-1$. Thus,
          $$|CD| = sqrt{|OD|^2 + |OC|^2} = sqrt{(sqrt{5}-1)^2+2^2} = sqrt{10-2sqrt{5}}$$



          It is "known" that the ratio of the side-length of a regular pentagon to its circumradius is
          $$sqrt{frac{5-sqrt{5}}{2}}= frac{sqrt{10-2sqrt{5}}}{2}= frac{|CD|}{|OA|}$$ so $|CD|$ is precisely the side-length needed for this regular pentagon.



          Then, $bigcirc C$ through $D$ transfers this side-length to $|CE|$, with $E$ on $bigcirc O$. Congruent circles construct additional points at the same distances about the circle, giving the perfect regular pentagon. $square$






          share|cite|improve this answer





















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            After some setup of perpendicular segments (here, $overline{OA}$ and $overline{OC}$), the first few steps are the standard construction of the midpoint ($B$) of $overline{OA}$, so let's start there; and let's say that $|OB| = |BA| = 1$.



            enter image description here



            By the Pythagorean theorem,
            $$|BC| = sqrt{|OB|^2 + |OC|^2} = sqrt{1+2^2} = sqrt{5}$$
            and we construct $bigcirc B$ with that radius, meeting the "horizontal" axis at $D$, so that $|OD| = |BC|-|OB| = sqrt{5}-1$. Thus,
            $$|CD| = sqrt{|OD|^2 + |OC|^2} = sqrt{(sqrt{5}-1)^2+2^2} = sqrt{10-2sqrt{5}}$$



            It is "known" that the ratio of the side-length of a regular pentagon to its circumradius is
            $$sqrt{frac{5-sqrt{5}}{2}}= frac{sqrt{10-2sqrt{5}}}{2}= frac{|CD|}{|OA|}$$ so $|CD|$ is precisely the side-length needed for this regular pentagon.



            Then, $bigcirc C$ through $D$ transfers this side-length to $|CE|$, with $E$ on $bigcirc O$. Congruent circles construct additional points at the same distances about the circle, giving the perfect regular pentagon. $square$






            share|cite|improve this answer

























              up vote
              1
              down vote













              After some setup of perpendicular segments (here, $overline{OA}$ and $overline{OC}$), the first few steps are the standard construction of the midpoint ($B$) of $overline{OA}$, so let's start there; and let's say that $|OB| = |BA| = 1$.



              enter image description here



              By the Pythagorean theorem,
              $$|BC| = sqrt{|OB|^2 + |OC|^2} = sqrt{1+2^2} = sqrt{5}$$
              and we construct $bigcirc B$ with that radius, meeting the "horizontal" axis at $D$, so that $|OD| = |BC|-|OB| = sqrt{5}-1$. Thus,
              $$|CD| = sqrt{|OD|^2 + |OC|^2} = sqrt{(sqrt{5}-1)^2+2^2} = sqrt{10-2sqrt{5}}$$



              It is "known" that the ratio of the side-length of a regular pentagon to its circumradius is
              $$sqrt{frac{5-sqrt{5}}{2}}= frac{sqrt{10-2sqrt{5}}}{2}= frac{|CD|}{|OA|}$$ so $|CD|$ is precisely the side-length needed for this regular pentagon.



              Then, $bigcirc C$ through $D$ transfers this side-length to $|CE|$, with $E$ on $bigcirc O$. Congruent circles construct additional points at the same distances about the circle, giving the perfect regular pentagon. $square$






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                After some setup of perpendicular segments (here, $overline{OA}$ and $overline{OC}$), the first few steps are the standard construction of the midpoint ($B$) of $overline{OA}$, so let's start there; and let's say that $|OB| = |BA| = 1$.



                enter image description here



                By the Pythagorean theorem,
                $$|BC| = sqrt{|OB|^2 + |OC|^2} = sqrt{1+2^2} = sqrt{5}$$
                and we construct $bigcirc B$ with that radius, meeting the "horizontal" axis at $D$, so that $|OD| = |BC|-|OB| = sqrt{5}-1$. Thus,
                $$|CD| = sqrt{|OD|^2 + |OC|^2} = sqrt{(sqrt{5}-1)^2+2^2} = sqrt{10-2sqrt{5}}$$



                It is "known" that the ratio of the side-length of a regular pentagon to its circumradius is
                $$sqrt{frac{5-sqrt{5}}{2}}= frac{sqrt{10-2sqrt{5}}}{2}= frac{|CD|}{|OA|}$$ so $|CD|$ is precisely the side-length needed for this regular pentagon.



                Then, $bigcirc C$ through $D$ transfers this side-length to $|CE|$, with $E$ on $bigcirc O$. Congruent circles construct additional points at the same distances about the circle, giving the perfect regular pentagon. $square$






                share|cite|improve this answer












                After some setup of perpendicular segments (here, $overline{OA}$ and $overline{OC}$), the first few steps are the standard construction of the midpoint ($B$) of $overline{OA}$, so let's start there; and let's say that $|OB| = |BA| = 1$.



                enter image description here



                By the Pythagorean theorem,
                $$|BC| = sqrt{|OB|^2 + |OC|^2} = sqrt{1+2^2} = sqrt{5}$$
                and we construct $bigcirc B$ with that radius, meeting the "horizontal" axis at $D$, so that $|OD| = |BC|-|OB| = sqrt{5}-1$. Thus,
                $$|CD| = sqrt{|OD|^2 + |OC|^2} = sqrt{(sqrt{5}-1)^2+2^2} = sqrt{10-2sqrt{5}}$$



                It is "known" that the ratio of the side-length of a regular pentagon to its circumradius is
                $$sqrt{frac{5-sqrt{5}}{2}}= frac{sqrt{10-2sqrt{5}}}{2}= frac{|CD|}{|OA|}$$ so $|CD|$ is precisely the side-length needed for this regular pentagon.



                Then, $bigcirc C$ through $D$ transfers this side-length to $|CE|$, with $E$ on $bigcirc O$. Congruent circles construct additional points at the same distances about the circle, giving the perfect regular pentagon. $square$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










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