Chord partition of regular polygon: same fraction of area and perimeter?












4












$begingroup$


This is a variation of a question posed by
James Tanton on Twitter.



Let $P$ be a regular $n$-gon, $n ge 3$. A chord $c$ of $P$ is a segment connecting
two distinct points of the boundary of $P$, on two distinct edges (so not two points on one edge). Such a chord partitions both the perimeter and area of $P$ into two non-zero parts.
For a given chord $c$, let $a(c)$ and $p(c)$ be the smaller area fractions and
smaller perimeter fractions of the parts.
In other words, $a(c)$ is the area to the smaller side of $c$ divided by the area of $P$. And similarly for $p(c)$.




Q. For which $n$-gons do there exist chords $c$
and fractions $a(c) = p(c)$ not equal to $frac{1}{2}$?




Tanton's question asks for $n=4$ (a square), can $frac{1}{3}$ be achieved? "Simultaneously divide off one-third of the area of a square and one-third of the perimeter of the square?" And the answer is No.



         
Tanton_1/3

         

(Image from Tanton.)

The generalization asks for all those non-half fractions achievable
for regular $n$-gons.
Perhaps the answer to Q is: For none?






geometry polygons plane-geometry







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    This is a variation of a question posed by
    James Tanton on Twitter.



    Let $P$ be a regular $n$-gon, $n ge 3$. A chord $c$ of $P$ is a segment connecting
    two distinct points of the boundary of $P$, on two distinct edges (so not two points on one edge). Such a chord partitions both the perimeter and area of $P$ into two non-zero parts.
    For a given chord $c$, let $a(c)$ and $p(c)$ be the smaller area fractions and
    smaller perimeter fractions of the parts.
    In other words, $a(c)$ is the area to the smaller side of $c$ divided by the area of $P$. And similarly for $p(c)$.




    Q. For which $n$-gons do there exist chords $c$
    and fractions $a(c) = p(c)$ not equal to $frac{1}{2}$?




    Tanton's question asks for $n=4$ (a square), can $frac{1}{3}$ be achieved? "Simultaneously divide off one-third of the area of a square and one-third of the perimeter of the square?" And the answer is No.



             
    Tanton_1/3

             

    (Image from Tanton.)

    The generalization asks for all those non-half fractions achievable
    for regular $n$-gons.
    Perhaps the answer to Q is: For none?






    geometry polygons plane-geometry







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      0



      $begingroup$


      This is a variation of a question posed by
      James Tanton on Twitter.



      Let $P$ be a regular $n$-gon, $n ge 3$. A chord $c$ of $P$ is a segment connecting
      two distinct points of the boundary of $P$, on two distinct edges (so not two points on one edge). Such a chord partitions both the perimeter and area of $P$ into two non-zero parts.
      For a given chord $c$, let $a(c)$ and $p(c)$ be the smaller area fractions and
      smaller perimeter fractions of the parts.
      In other words, $a(c)$ is the area to the smaller side of $c$ divided by the area of $P$. And similarly for $p(c)$.




      Q. For which $n$-gons do there exist chords $c$
      and fractions $a(c) = p(c)$ not equal to $frac{1}{2}$?




      Tanton's question asks for $n=4$ (a square), can $frac{1}{3}$ be achieved? "Simultaneously divide off one-third of the area of a square and one-third of the perimeter of the square?" And the answer is No.



               
      Tanton_1/3

               

      (Image from Tanton.)

      The generalization asks for all those non-half fractions achievable
      for regular $n$-gons.
      Perhaps the answer to Q is: For none?






      geometry polygons plane-geometry







      share|cite|improve this question











      $endgroup$




      This is a variation of a question posed by
      James Tanton on Twitter.



      Let $P$ be a regular $n$-gon, $n ge 3$. A chord $c$ of $P$ is a segment connecting
      two distinct points of the boundary of $P$, on two distinct edges (so not two points on one edge). Such a chord partitions both the perimeter and area of $P$ into two non-zero parts.
      For a given chord $c$, let $a(c)$ and $p(c)$ be the smaller area fractions and
      smaller perimeter fractions of the parts.
      In other words, $a(c)$ is the area to the smaller side of $c$ divided by the area of $P$. And similarly for $p(c)$.




      Q. For which $n$-gons do there exist chords $c$
      and fractions $a(c) = p(c)$ not equal to $frac{1}{2}$?




      Tanton's question asks for $n=4$ (a square), can $frac{1}{3}$ be achieved? "Simultaneously divide off one-third of the area of a square and one-third of the perimeter of the square?" And the answer is No.



               
      Tanton_1/3

               

      (Image from Tanton.)

      The generalization asks for all those non-half fractions achievable
      for regular $n$-gons.
      Perhaps the answer to Q is: For none?






      geometry polygons plane-geometry




      geometry polygons plane-geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 6 at 1:02







      Joseph O'Rourke

















      asked Jan 6 at 0:03









      Joseph O'RourkeJoseph O'Rourke

      18k349109




      18k349109






















          1 Answer
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          1












          $begingroup$

          Partial answer: no good chord for the square (with any ratio different from $frac{1}{2}$).



          Proof:
          Assume that we connect two points on opposite sides. Then we obtain a trapezoid.
          Denote the bases by $x,y$. Then $a(c)= frac{x+y}{2}$ and $p(c)=frac{x+y+1}{4}$, which yields $x+y=1$, and thus $a(c)=p(c)=frac{1}{2}$.



          If the two points are on adjacent sides, then we obtain a right triangle.
          Let $x,y$ denote the legs this time. Then $a(c)=frac{xy}{2}$ and $p(c)=frac{x+y}{4}$, which yields $(x-frac{1}{2})(y-frac{1}{2})=frac{1}{4}$.
          As $0leq x,yleq 1$, we have $|x-frac{1}{2}|leq frac{1}{2}$ and $|y-frac{1}{2}|leq frac{1}{2}$, thus $(x-frac{1}{2})(y-frac{1}{2})=frac{1}{4}$ is possible only if $|x-frac{1}{2}|=|y-frac{1}{2}|=frac{1}{2}$.
          Hence, either the triangle is degenerate (a point), or $x=y=1$, in which case $a(c)=p(c)=frac{1}{2}$ again.



          The fact that we needed very different arguments in the two cases suggests that the general question might be hard to handle.






          share|cite|improve this answer









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            1 Answer
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            oldest

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            active

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            1












            $begingroup$

            Partial answer: no good chord for the square (with any ratio different from $frac{1}{2}$).



            Proof:
            Assume that we connect two points on opposite sides. Then we obtain a trapezoid.
            Denote the bases by $x,y$. Then $a(c)= frac{x+y}{2}$ and $p(c)=frac{x+y+1}{4}$, which yields $x+y=1$, and thus $a(c)=p(c)=frac{1}{2}$.



            If the two points are on adjacent sides, then we obtain a right triangle.
            Let $x,y$ denote the legs this time. Then $a(c)=frac{xy}{2}$ and $p(c)=frac{x+y}{4}$, which yields $(x-frac{1}{2})(y-frac{1}{2})=frac{1}{4}$.
            As $0leq x,yleq 1$, we have $|x-frac{1}{2}|leq frac{1}{2}$ and $|y-frac{1}{2}|leq frac{1}{2}$, thus $(x-frac{1}{2})(y-frac{1}{2})=frac{1}{4}$ is possible only if $|x-frac{1}{2}|=|y-frac{1}{2}|=frac{1}{2}$.
            Hence, either the triangle is degenerate (a point), or $x=y=1$, in which case $a(c)=p(c)=frac{1}{2}$ again.



            The fact that we needed very different arguments in the two cases suggests that the general question might be hard to handle.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Partial answer: no good chord for the square (with any ratio different from $frac{1}{2}$).



              Proof:
              Assume that we connect two points on opposite sides. Then we obtain a trapezoid.
              Denote the bases by $x,y$. Then $a(c)= frac{x+y}{2}$ and $p(c)=frac{x+y+1}{4}$, which yields $x+y=1$, and thus $a(c)=p(c)=frac{1}{2}$.



              If the two points are on adjacent sides, then we obtain a right triangle.
              Let $x,y$ denote the legs this time. Then $a(c)=frac{xy}{2}$ and $p(c)=frac{x+y}{4}$, which yields $(x-frac{1}{2})(y-frac{1}{2})=frac{1}{4}$.
              As $0leq x,yleq 1$, we have $|x-frac{1}{2}|leq frac{1}{2}$ and $|y-frac{1}{2}|leq frac{1}{2}$, thus $(x-frac{1}{2})(y-frac{1}{2})=frac{1}{4}$ is possible only if $|x-frac{1}{2}|=|y-frac{1}{2}|=frac{1}{2}$.
              Hence, either the triangle is degenerate (a point), or $x=y=1$, in which case $a(c)=p(c)=frac{1}{2}$ again.



              The fact that we needed very different arguments in the two cases suggests that the general question might be hard to handle.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Partial answer: no good chord for the square (with any ratio different from $frac{1}{2}$).



                Proof:
                Assume that we connect two points on opposite sides. Then we obtain a trapezoid.
                Denote the bases by $x,y$. Then $a(c)= frac{x+y}{2}$ and $p(c)=frac{x+y+1}{4}$, which yields $x+y=1$, and thus $a(c)=p(c)=frac{1}{2}$.



                If the two points are on adjacent sides, then we obtain a right triangle.
                Let $x,y$ denote the legs this time. Then $a(c)=frac{xy}{2}$ and $p(c)=frac{x+y}{4}$, which yields $(x-frac{1}{2})(y-frac{1}{2})=frac{1}{4}$.
                As $0leq x,yleq 1$, we have $|x-frac{1}{2}|leq frac{1}{2}$ and $|y-frac{1}{2}|leq frac{1}{2}$, thus $(x-frac{1}{2})(y-frac{1}{2})=frac{1}{4}$ is possible only if $|x-frac{1}{2}|=|y-frac{1}{2}|=frac{1}{2}$.
                Hence, either the triangle is degenerate (a point), or $x=y=1$, in which case $a(c)=p(c)=frac{1}{2}$ again.



                The fact that we needed very different arguments in the two cases suggests that the general question might be hard to handle.






                share|cite|improve this answer









                $endgroup$



                Partial answer: no good chord for the square (with any ratio different from $frac{1}{2}$).



                Proof:
                Assume that we connect two points on opposite sides. Then we obtain a trapezoid.
                Denote the bases by $x,y$. Then $a(c)= frac{x+y}{2}$ and $p(c)=frac{x+y+1}{4}$, which yields $x+y=1$, and thus $a(c)=p(c)=frac{1}{2}$.



                If the two points are on adjacent sides, then we obtain a right triangle.
                Let $x,y$ denote the legs this time. Then $a(c)=frac{xy}{2}$ and $p(c)=frac{x+y}{4}$, which yields $(x-frac{1}{2})(y-frac{1}{2})=frac{1}{4}$.
                As $0leq x,yleq 1$, we have $|x-frac{1}{2}|leq frac{1}{2}$ and $|y-frac{1}{2}|leq frac{1}{2}$, thus $(x-frac{1}{2})(y-frac{1}{2})=frac{1}{4}$ is possible only if $|x-frac{1}{2}|=|y-frac{1}{2}|=frac{1}{2}$.
                Hence, either the triangle is degenerate (a point), or $x=y=1$, in which case $a(c)=p(c)=frac{1}{2}$ again.



                The fact that we needed very different arguments in the two cases suggests that the general question might be hard to handle.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 0:28









                A. PongráczA. Pongrácz

                5,9531929




                5,9531929






























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