Generalizing Heron's Formula for Cyclic $n$-gons
$begingroup$
Consider the following extension of Heron's Formula.
For a cyclic $n$-gon $C$ with side lengths $x_1, x_2, dots, x_n$ and semi-perimeter $P = frac{1}{2} left( x_1 + x_2 + dots + x_nright)$ define:
$$ M = sqrt{P^{4-n} (P-x_1)(P-x_2) dots (P-x_n)} $$
After some experimentation in GeoGebra, it turns out that $M$ is pretty close (but not equal) to the usual area for cyclic $n$-gons. This suggests that $M$ is less than a constant multiple of area $A$. Does anyone have an idea of how to prove this?
To get a sense of what might happen, for regular $n$-gons $R_n$ it turns out that: $displaystyle lim_{n rightarrow infty} M(R_n)/A(R_n) = pi/e$.
Observe that $M$ is a homogeneous function $M(lambda C) = lambda^2 M(C)$. Area is also homogenous in the same degree $A(lambda C) = lambda^2 A(C)$. Thus, $M/A$ is scale invariant. We can pick a scale to work. Can anyone show that $M$ is bounded on cyclic polygons of area one?
geometry inequality euclidean-geometry geometric-inequalities
asked Jan 2 at 17:54
Parker Glynn-AdeyParker Glynn-Adey
634
$endgroup$
add a comment |
$begingroup$
Consider the following extension of Heron's Formula.
For a cyclic $n$-gon $C$ with side lengths $x_1, x_2, dots, x_n$ and semi-perimeter $P = frac{1}{2} left( x_1 + x_2 + dots + x_nright)$ define:
$$ M = sqrt{P^{4-n} (P-x_1)(P-x_2) dots (P-x_n)} $$
After some experimentation in GeoGebra, it turns out that $M$ is pretty close (but not equal) to the usual area for cyclic $n$-gons. This suggests that $M$ is less than a constant multiple of area $A$. Does anyone have an idea of how to prove this?
To get a sense of what might happen, for regular $n$-gons $R_n$ it turns out that: $displaystyle lim_{n rightarrow infty} M(R_n)/A(R_n) = pi/e$.
Observe that $M$ is a homogeneous function $M(lambda C) = lambda^2 M(C)$. Area is also homogenous in the same degree $A(lambda C) = lambda^2 A(C)$. Thus, $M/A$ is scale invariant. We can pick a scale to work. Can anyone show that $M$ is bounded on cyclic polygons of area one?
geometry inequality euclidean-geometry geometric-inequalities
asked Jan 2 at 17:54
Parker Glynn-AdeyParker Glynn-Adey
634
$endgroup$
$begingroup$
This paper and some of the references therein appear relevant.
$endgroup$
– saulspatz
Jan 2 at 18:18
$begingroup$
Thanks, Robbin's Conjecture and that direction of work is certainly relevant.
$endgroup$
– Parker Glynn-Adey
Jan 2 at 18:31
add a comment |
$begingroup$
Consider the following extension of Heron's Formula.
For a cyclic $n$-gon $C$ with side lengths $x_1, x_2, dots, x_n$ and semi-perimeter $P = frac{1}{2} left( x_1 + x_2 + dots + x_nright)$ define:
$$ M = sqrt{P^{4-n} (P-x_1)(P-x_2) dots (P-x_n)} $$
After some experimentation in GeoGebra, it turns out that $M$ is pretty close (but not equal) to the usual area for cyclic $n$-gons. This suggests that $M$ is less than a constant multiple of area $A$. Does anyone have an idea of how to prove this?
To get a sense of what might happen, for regular $n$-gons $R_n$ it turns out that: $displaystyle lim_{n rightarrow infty} M(R_n)/A(R_n) = pi/e$.
Observe that $M$ is a homogeneous function $M(lambda C) = lambda^2 M(C)$. Area is also homogenous in the same degree $A(lambda C) = lambda^2 A(C)$. Thus, $M/A$ is scale invariant. We can pick a scale to work. Can anyone show that $M$ is bounded on cyclic polygons of area one?
geometry inequality euclidean-geometry geometric-inequalities
asked Jan 2 at 17:54
Parker Glynn-AdeyParker Glynn-Adey
634
$endgroup$
Consider the following extension of Heron's Formula.
For a cyclic $n$-gon $C$ with side lengths $x_1, x_2, dots, x_n$ and semi-perimeter $P = frac{1}{2} left( x_1 + x_2 + dots + x_nright)$ define:
$$ M = sqrt{P^{4-n} (P-x_1)(P-x_2) dots (P-x_n)} $$
After some experimentation in GeoGebra, it turns out that $M$ is pretty close (but not equal) to the usual area for cyclic $n$-gons. This suggests that $M$ is less than a constant multiple of area $A$. Does anyone have an idea of how to prove this?
To get a sense of what might happen, for regular $n$-gons $R_n$ it turns out that: $displaystyle lim_{n rightarrow infty} M(R_n)/A(R_n) = pi/e$.
Observe that $M$ is a homogeneous function $M(lambda C) = lambda^2 M(C)$. Area is also homogenous in the same degree $A(lambda C) = lambda^2 A(C)$. Thus, $M/A$ is scale invariant. We can pick a scale to work. Can anyone show that $M$ is bounded on cyclic polygons of area one?
geometry inequality euclidean-geometry geometric-inequalities
geometry inequality euclidean-geometry geometric-inequalities
asked Jan 2 at 17:54
Parker Glynn-AdeyParker Glynn-Adey
634
asked Jan 2 at 17:54
Parker Glynn-AdeyParker Glynn-Adey
634
asked Jan 2 at 17:54
Parker Glynn-AdeyParker Glynn-Adey
634
asked Jan 2 at 17:54
Parker Glynn-AdeyParker Glynn-Adey
634
asked Jan 2 at 17:54
Parker Glynn-AdeyParker Glynn-Adey
634
634
$begingroup$
This paper and some of the references therein appear relevant.
$endgroup$
– saulspatz
Jan 2 at 18:18
$begingroup$
Thanks, Robbin's Conjecture and that direction of work is certainly relevant.
$endgroup$
– Parker Glynn-Adey
Jan 2 at 18:31
add a comment |
$begingroup$
This paper and some of the references therein appear relevant.
$endgroup$
– saulspatz
Jan 2 at 18:18
$begingroup$
Thanks, Robbin's Conjecture and that direction of work is certainly relevant.
$endgroup$
– Parker Glynn-Adey
Jan 2 at 18:31
$begingroup$
This paper and some of the references therein appear relevant.
$endgroup$
– saulspatz
Jan 2 at 18:18
$begingroup$
This paper and some of the references therein appear relevant.
$endgroup$
– saulspatz
Jan 2 at 18:18
$begingroup$
Thanks, Robbin's Conjecture and that direction of work is certainly relevant.
$endgroup$
– Parker Glynn-Adey
Jan 2 at 18:31
$begingroup$
Thanks, Robbin's Conjecture and that direction of work is certainly relevant.
$endgroup$
– Parker Glynn-Adey
Jan 2 at 18:31
add a comment |
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$begingroup$
This paper and some of the references therein appear relevant.
$endgroup$
– saulspatz
Jan 2 at 18:18
$begingroup$
Thanks, Robbin's Conjecture and that direction of work is certainly relevant.
$endgroup$
– Parker Glynn-Adey
Jan 2 at 18:31