Mixing
Chord partition of regular polygon: same fraction of area and perimeter?
$begingroup$
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.
(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
asked Jan 6 at 0:03
Joseph O'RourkeJoseph O'Rourke
17.9k348107
$endgroup$
add a comment |
$begingroup$
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.
(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
asked Jan 6 at 0:03
Joseph O'RourkeJoseph O'Rourke
17.9k348107
$endgroup$
add a comment |
$begingroup$
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.
(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
asked Jan 6 at 0:03
Joseph O'RourkeJoseph O'Rourke
17.9k348107
$endgroup$
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.
(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
asked Jan 6 at 0:03
Joseph O'RourkeJoseph O'Rourke
17.9k348107
asked Jan 6 at 0:03
Joseph O'RourkeJoseph O'Rourke
17.9k348107
edited Jan 6 at 1:02
Joseph O'Rourke
asked Jan 6 at 0:03
Joseph O'RourkeJoseph O'Rourke
17.9k348107
asked Jan 6 at 0:03
Joseph O'RourkeJoseph O'Rourke
17.9k348107
asked Jan 6 at 0:03
Joseph O'RourkeJoseph O'Rourke
17.9k348107
17.9k348107
add a comment |
add a comment |
1 Answer
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$begingroup$
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.
answered Jan 6 at 0:28
A. PongráczA. Pongrácz
5,9731929
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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.
answered Jan 6 at 0:28
A. PongráczA. Pongrácz
5,9731929
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 6 at 0:28
A. PongráczA. Pongrácz
5,9731929
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 6 at 0:28
A. PongráczA. Pongrácz
5,9731929
$endgroup$
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.
answered Jan 6 at 0:28
A. PongráczA. Pongrácz
5,9731929
answered Jan 6 at 0:28
A. PongráczA. Pongrácz
5,9731929
answered Jan 6 at 0:28
A. PongráczA. Pongrácz
5,9731929
answered Jan 6 at 0:28
A. PongráczA. Pongrácz
5,9731929
5,9731929
add a comment |
add a comment |
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