Napoleon-like theorem concerning squares erected on sides of midpoint polygon of octogon
$begingroup$
Given an arbitrary octagon, construct it's midpoint polygon(the midpoint formed by the midpoints of the sides). Erect squares on the sides of the midpoint polygon, all inwards or all outwards. Consider the four segments, each connecting the centroids of two squares corresponding to opposite sides of the midpoint polygon.
The midpoint of these segments form a square.
I distinctly remember seeing this theorem in some geometry article, but I have been unable to find that article. I believe it was attributed to Van Aubel, however I am not too sure. I want to find the source of this theorem because I have found a rather powerful generalisation and I want to revisit that article.
Does anyone find this theorem familiar?
geometry reference-request complex-numbers euclidean-geometry geometric-transformation
edited Jan 6 at 7:32
Maria Mazur
49.5k1361124
asked Jan 3 at 19:18
Tanny SiebenTanny Sieben
42228
$endgroup$
|
show 3 more comments
$begingroup$
Given an arbitrary octagon, construct it's midpoint polygon(the midpoint formed by the midpoints of the sides). Erect squares on the sides of the midpoint polygon, all inwards or all outwards. Consider the four segments, each connecting the centroids of two squares corresponding to opposite sides of the midpoint polygon.
The midpoint of these segments form a square.
I distinctly remember seeing this theorem in some geometry article, but I have been unable to find that article. I believe it was attributed to Van Aubel, however I am not too sure. I want to find the source of this theorem because I have found a rather powerful generalisation and I want to revisit that article.
Does anyone find this theorem familiar?
geometry reference-request complex-numbers euclidean-geometry geometric-transformation
edited Jan 6 at 7:32
Maria Mazur
49.5k1361124
asked Jan 3 at 19:18
Tanny SiebenTanny Sieben
42228
$endgroup$
$begingroup$
Branko Grünbaum (RIP 2018) presented results of this type in a graduate course at the University of Washington many years ago. I'm not sure if they were published outside of lecture notes. There's a wonderful general result that goes something like this: Given an arbitrary $n$-gon, consider the various regular $(n/k)$-gons, defined as having vertices $(cos 2pi i k/n,sin2pi i k/n)$ for $i=0, 1, ldots, n-1$. (These include "starry" and "multiply-traced" figures, as well as the "dot" ($k=0$).) [continued]
$endgroup$
– Blue
Jan 3 at 19:37
$begingroup$
[continuation] Erecting an $(n/k)$-gon on the sides of the $n$-gon and joining centroids gives a new $n$-gon; erecting some other $(n/k)$-gon on the new sides gives another $n$-gon; etc. Iterating for all-but-one of $(n/k)$-gons, the final polygon is the remaining $(n/k)$-gon. For an $8$-gon, the $(n/k)$-gons are the dot ($k=0$), the regular octagon ($k=1$), the doubly-traced square ($k=2$), "starry octagon" ($k=3$), quad-traced segment ($k=4$, whose derived polygon is a midpoint polygon), and "reverse-direction" versions. Similar to what you have, except for the "opposite sides" thing.
$endgroup$
– Blue
Jan 3 at 19:54
$begingroup$
@Blue, aren't you describing the Petr-Douglas-Neumann Theorem? ( en.wikipedia.org/wiki/Petr-Douglas-Neumann_theorem )
$endgroup$
– Tanny Sieben
Jan 3 at 20:16
$begingroup$
Ah, yes. That seems to be it.
$endgroup$
– Blue
Jan 3 at 20:22
$begingroup$
Interestingly, the "opposite sides" thing seems attributable to a common alternative interpretation of $(n/k)$-gon. Rather than starting at $(1,0)$ of a regular "unit" $n$-gon and forming a single (possibly multiply-traced) path by jumping every $k$ vertices, possibly excluding some vertices, we connect every vertex of the $n$ gon to the vertex/vertices $k$ jumps ahead/behind. So, an $(8/4)$-gon would consist of four individual segments (perhaps best thought of as double-traced segments) instead one octo-traced segment. But your figure arguably also uses octo-traced segments. Intriguing.
$endgroup$
– Blue
Jan 3 at 20:37
|
show 3 more comments
$begingroup$
Given an arbitrary octagon, construct it's midpoint polygon(the midpoint formed by the midpoints of the sides). Erect squares on the sides of the midpoint polygon, all inwards or all outwards. Consider the four segments, each connecting the centroids of two squares corresponding to opposite sides of the midpoint polygon.
The midpoint of these segments form a square.
I distinctly remember seeing this theorem in some geometry article, but I have been unable to find that article. I believe it was attributed to Van Aubel, however I am not too sure. I want to find the source of this theorem because I have found a rather powerful generalisation and I want to revisit that article.
Does anyone find this theorem familiar?
geometry reference-request complex-numbers euclidean-geometry geometric-transformation
edited Jan 6 at 7:32
Maria Mazur
49.5k1361124
asked Jan 3 at 19:18
Tanny SiebenTanny Sieben
42228
$endgroup$
Given an arbitrary octagon, construct it's midpoint polygon(the midpoint formed by the midpoints of the sides). Erect squares on the sides of the midpoint polygon, all inwards or all outwards. Consider the four segments, each connecting the centroids of two squares corresponding to opposite sides of the midpoint polygon.
The midpoint of these segments form a square.
I distinctly remember seeing this theorem in some geometry article, but I have been unable to find that article. I believe it was attributed to Van Aubel, however I am not too sure. I want to find the source of this theorem because I have found a rather powerful generalisation and I want to revisit that article.
Does anyone find this theorem familiar?
geometry reference-request complex-numbers euclidean-geometry geometric-transformation
geometry reference-request complex-numbers euclidean-geometry geometric-transformation
edited Jan 6 at 7:32
Maria Mazur
49.5k1361124
asked Jan 3 at 19:18
Tanny SiebenTanny Sieben
42228
edited Jan 6 at 7:32
Maria Mazur
49.5k1361124
asked Jan 3 at 19:18
Tanny SiebenTanny Sieben
42228
edited Jan 6 at 7:32
Maria Mazur
49.5k1361124
edited Jan 6 at 7:32
Maria Mazur
49.5k1361124
edited Jan 6 at 7:32
Maria Mazur
49.5k1361124
49.5k1361124
asked Jan 3 at 19:18
Tanny SiebenTanny Sieben
42228
asked Jan 3 at 19:18
Tanny SiebenTanny Sieben
42228
asked Jan 3 at 19:18
Tanny SiebenTanny Sieben
42228
42228
$begingroup$
Branko Grünbaum (RIP 2018) presented results of this type in a graduate course at the University of Washington many years ago. I'm not sure if they were published outside of lecture notes. There's a wonderful general result that goes something like this: Given an arbitrary $n$-gon, consider the various regular $(n/k)$-gons, defined as having vertices $(cos 2pi i k/n,sin2pi i k/n)$ for $i=0, 1, ldots, n-1$. (These include "starry" and "multiply-traced" figures, as well as the "dot" ($k=0$).) [continued]
$endgroup$
– Blue
Jan 3 at 19:37
$begingroup$
[continuation] Erecting an $(n/k)$-gon on the sides of the $n$-gon and joining centroids gives a new $n$-gon; erecting some other $(n/k)$-gon on the new sides gives another $n$-gon; etc. Iterating for all-but-one of $(n/k)$-gons, the final polygon is the remaining $(n/k)$-gon. For an $8$-gon, the $(n/k)$-gons are the dot ($k=0$), the regular octagon ($k=1$), the doubly-traced square ($k=2$), "starry octagon" ($k=3$), quad-traced segment ($k=4$, whose derived polygon is a midpoint polygon), and "reverse-direction" versions. Similar to what you have, except for the "opposite sides" thing.
$endgroup$
– Blue
Jan 3 at 19:54
$begingroup$
@Blue, aren't you describing the Petr-Douglas-Neumann Theorem? ( en.wikipedia.org/wiki/Petr-Douglas-Neumann_theorem )
$endgroup$
– Tanny Sieben
Jan 3 at 20:16
$begingroup$
Ah, yes. That seems to be it.
$endgroup$
– Blue
Jan 3 at 20:22
$begingroup$
Interestingly, the "opposite sides" thing seems attributable to a common alternative interpretation of $(n/k)$-gon. Rather than starting at $(1,0)$ of a regular "unit" $n$-gon and forming a single (possibly multiply-traced) path by jumping every $k$ vertices, possibly excluding some vertices, we connect every vertex of the $n$ gon to the vertex/vertices $k$ jumps ahead/behind. So, an $(8/4)$-gon would consist of four individual segments (perhaps best thought of as double-traced segments) instead one octo-traced segment. But your figure arguably also uses octo-traced segments. Intriguing.
$endgroup$
– Blue
Jan 3 at 20:37
|
show 3 more comments
$begingroup$
Branko Grünbaum (RIP 2018) presented results of this type in a graduate course at the University of Washington many years ago. I'm not sure if they were published outside of lecture notes. There's a wonderful general result that goes something like this: Given an arbitrary $n$-gon, consider the various regular $(n/k)$-gons, defined as having vertices $(cos 2pi i k/n,sin2pi i k/n)$ for $i=0, 1, ldots, n-1$. (These include "starry" and "multiply-traced" figures, as well as the "dot" ($k=0$).) [continued]
$endgroup$
– Blue
Jan 3 at 19:37
$begingroup$
[continuation] Erecting an $(n/k)$-gon on the sides of the $n$-gon and joining centroids gives a new $n$-gon; erecting some other $(n/k)$-gon on the new sides gives another $n$-gon; etc. Iterating for all-but-one of $(n/k)$-gons, the final polygon is the remaining $(n/k)$-gon. For an $8$-gon, the $(n/k)$-gons are the dot ($k=0$), the regular octagon ($k=1$), the doubly-traced square ($k=2$), "starry octagon" ($k=3$), quad-traced segment ($k=4$, whose derived polygon is a midpoint polygon), and "reverse-direction" versions. Similar to what you have, except for the "opposite sides" thing.
$endgroup$
– Blue
Jan 3 at 19:54
$begingroup$
@Blue, aren't you describing the Petr-Douglas-Neumann Theorem? ( en.wikipedia.org/wiki/Petr-Douglas-Neumann_theorem )
$endgroup$
– Tanny Sieben
Jan 3 at 20:16
$begingroup$
Ah, yes. That seems to be it.
$endgroup$
– Blue
Jan 3 at 20:22
$begingroup$
Interestingly, the "opposite sides" thing seems attributable to a common alternative interpretation of $(n/k)$-gon. Rather than starting at $(1,0)$ of a regular "unit" $n$-gon and forming a single (possibly multiply-traced) path by jumping every $k$ vertices, possibly excluding some vertices, we connect every vertex of the $n$ gon to the vertex/vertices $k$ jumps ahead/behind. So, an $(8/4)$-gon would consist of four individual segments (perhaps best thought of as double-traced segments) instead one octo-traced segment. But your figure arguably also uses octo-traced segments. Intriguing.
$endgroup$
– Blue
Jan 3 at 20:37
$begingroup$
Branko Grünbaum (RIP 2018) presented results of this type in a graduate course at the University of Washington many years ago. I'm not sure if they were published outside of lecture notes. There's a wonderful general result that goes something like this: Given an arbitrary $n$-gon, consider the various regular $(n/k)$-gons, defined as having vertices $(cos 2pi i k/n,sin2pi i k/n)$ for $i=0, 1, ldots, n-1$. (These include "starry" and "multiply-traced" figures, as well as the "dot" ($k=0$).) [continued]
$endgroup$
– Blue
Jan 3 at 19:37
$begingroup$
Branko Grünbaum (RIP 2018) presented results of this type in a graduate course at the University of Washington many years ago. I'm not sure if they were published outside of lecture notes. There's a wonderful general result that goes something like this: Given an arbitrary $n$-gon, consider the various regular $(n/k)$-gons, defined as having vertices $(cos 2pi i k/n,sin2pi i k/n)$ for $i=0, 1, ldots, n-1$. (These include "starry" and "multiply-traced" figures, as well as the "dot" ($k=0$).) [continued]
$endgroup$
– Blue
Jan 3 at 19:37
$begingroup$
[continuation] Erecting an $(n/k)$-gon on the sides of the $n$-gon and joining centroids gives a new $n$-gon; erecting some other $(n/k)$-gon on the new sides gives another $n$-gon; etc. Iterating for all-but-one of $(n/k)$-gons, the final polygon is the remaining $(n/k)$-gon. For an $8$-gon, the $(n/k)$-gons are the dot ($k=0$), the regular octagon ($k=1$), the doubly-traced square ($k=2$), "starry octagon" ($k=3$), quad-traced segment ($k=4$, whose derived polygon is a midpoint polygon), and "reverse-direction" versions. Similar to what you have, except for the "opposite sides" thing.
$endgroup$
– Blue
Jan 3 at 19:54
$begingroup$
[continuation] Erecting an $(n/k)$-gon on the sides of the $n$-gon and joining centroids gives a new $n$-gon; erecting some other $(n/k)$-gon on the new sides gives another $n$-gon; etc. Iterating for all-but-one of $(n/k)$-gons, the final polygon is the remaining $(n/k)$-gon. For an $8$-gon, the $(n/k)$-gons are the dot ($k=0$), the regular octagon ($k=1$), the doubly-traced square ($k=2$), "starry octagon" ($k=3$), quad-traced segment ($k=4$, whose derived polygon is a midpoint polygon), and "reverse-direction" versions. Similar to what you have, except for the "opposite sides" thing.
$endgroup$
– Blue
Jan 3 at 19:54
$begingroup$
@Blue, aren't you describing the Petr-Douglas-Neumann Theorem? ( en.wikipedia.org/wiki/Petr-Douglas-Neumann_theorem )
$endgroup$
– Tanny Sieben
Jan 3 at 20:16
$begingroup$
@Blue, aren't you describing the Petr-Douglas-Neumann Theorem? ( en.wikipedia.org/wiki/Petr-Douglas-Neumann_theorem )
$endgroup$
– Tanny Sieben
Jan 3 at 20:16
$begingroup$
Ah, yes. That seems to be it.
$endgroup$
– Blue
Jan 3 at 20:22
$begingroup$
Ah, yes. That seems to be it.
$endgroup$
– Blue
Jan 3 at 20:22
$begingroup$
Interestingly, the "opposite sides" thing seems attributable to a common alternative interpretation of $(n/k)$-gon. Rather than starting at $(1,0)$ of a regular "unit" $n$-gon and forming a single (possibly multiply-traced) path by jumping every $k$ vertices, possibly excluding some vertices, we connect every vertex of the $n$ gon to the vertex/vertices $k$ jumps ahead/behind. So, an $(8/4)$-gon would consist of four individual segments (perhaps best thought of as double-traced segments) instead one octo-traced segment. But your figure arguably also uses octo-traced segments. Intriguing.
$endgroup$
– Blue
Jan 3 at 20:37
$begingroup$
Interestingly, the "opposite sides" thing seems attributable to a common alternative interpretation of $(n/k)$-gon. Rather than starting at $(1,0)$ of a regular "unit" $n$-gon and forming a single (possibly multiply-traced) path by jumping every $k$ vertices, possibly excluding some vertices, we connect every vertex of the $n$ gon to the vertex/vertices $k$ jumps ahead/behind. So, an $(8/4)$-gon would consist of four individual segments (perhaps best thought of as double-traced segments) instead one octo-traced segment. But your figure arguably also uses octo-traced segments. Intriguing.
$endgroup$
– Blue
Jan 3 at 20:37
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
I have a lenghty complex-vector-computational solution. The main idea is in a following lemma.
Lemma: If $ABCD$ is a square with center $S$, then $$ S = {1over 2}(B-A)i+{1over 2}(A+B)$$
where letters represent complex numbers of coresponding points (with the same name).
So we get 8 centers and then calculate all $4$ midpoints of those segments. Say we have octagon $ABCDEFGH$, then those midpoints (if we suppose that $A+B+C+D+E+F+G+H =0$) are:
$$ M_1 = {1over 4}Big[(D+H-B-F)i+(A+E-C-G)Big]=-M_3$$
$$ M_2 = {1over 4}Big[(A+E-C-G)i+(B+F-D-H)Big] =-M_4$$
Now it is easy to see that $M_2 = iM_1$ and we are done.
answered Jan 5 at 13:13
Maria MazurMaria Mazur
49.5k1361124
$endgroup$
1
$begingroup$
[+1] The approach using complex numbers' geometry is very adapted to such issues.
$endgroup$
– Jean Marie
Feb 27 at 22:44
$begingroup$
Thank you @JeanMarie, By the way, do you like this one:math.stackexchange.com/questions/3126169/…
$endgroup$
– Maria Mazur
Feb 28 at 15:42
$begingroup$
I didn't accept the answer because I am not looking for a proof (i knew how to prove it beforehand); I was looking for references to publications containing this theorem or similar results.
$endgroup$
– Tanny Sieben
Mar 4 at 18:44
$begingroup$
No problem, can you post your solution to this problem anyway?
$endgroup$
– Maria Mazur
Mar 4 at 18:49
add a comment |
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$begingroup$
I have a lenghty complex-vector-computational solution. The main idea is in a following lemma.
Lemma: If $ABCD$ is a square with center $S$, then $$ S = {1over 2}(B-A)i+{1over 2}(A+B)$$
where letters represent complex numbers of coresponding points (with the same name).
So we get 8 centers and then calculate all $4$ midpoints of those segments. Say we have octagon $ABCDEFGH$, then those midpoints (if we suppose that $A+B+C+D+E+F+G+H =0$) are:
$$ M_1 = {1over 4}Big[(D+H-B-F)i+(A+E-C-G)Big]=-M_3$$
$$ M_2 = {1over 4}Big[(A+E-C-G)i+(B+F-D-H)Big] =-M_4$$
Now it is easy to see that $M_2 = iM_1$ and we are done.
answered Jan 5 at 13:13
Maria MazurMaria Mazur
49.5k1361124
$endgroup$
1
$begingroup$
[+1] The approach using complex numbers' geometry is very adapted to such issues.
$endgroup$
– Jean Marie
Feb 27 at 22:44
$begingroup$
Thank you @JeanMarie, By the way, do you like this one:math.stackexchange.com/questions/3126169/…
$endgroup$
– Maria Mazur
Feb 28 at 15:42
$begingroup$
I didn't accept the answer because I am not looking for a proof (i knew how to prove it beforehand); I was looking for references to publications containing this theorem or similar results.
$endgroup$
– Tanny Sieben
Mar 4 at 18:44
$begingroup$
No problem, can you post your solution to this problem anyway?
$endgroup$
– Maria Mazur
Mar 4 at 18:49
add a comment |
$begingroup$
I have a lenghty complex-vector-computational solution. The main idea is in a following lemma.
Lemma: If $ABCD$ is a square with center $S$, then $$ S = {1over 2}(B-A)i+{1over 2}(A+B)$$
where letters represent complex numbers of coresponding points (with the same name).
So we get 8 centers and then calculate all $4$ midpoints of those segments. Say we have octagon $ABCDEFGH$, then those midpoints (if we suppose that $A+B+C+D+E+F+G+H =0$) are:
$$ M_1 = {1over 4}Big[(D+H-B-F)i+(A+E-C-G)Big]=-M_3$$
$$ M_2 = {1over 4}Big[(A+E-C-G)i+(B+F-D-H)Big] =-M_4$$
Now it is easy to see that $M_2 = iM_1$ and we are done.
answered Jan 5 at 13:13
Maria MazurMaria Mazur
49.5k1361124
$endgroup$
1
$begingroup$
[+1] The approach using complex numbers' geometry is very adapted to such issues.
$endgroup$
– Jean Marie
Feb 27 at 22:44
$begingroup$
Thank you @JeanMarie, By the way, do you like this one:math.stackexchange.com/questions/3126169/…
$endgroup$
– Maria Mazur
Feb 28 at 15:42
$begingroup$
I didn't accept the answer because I am not looking for a proof (i knew how to prove it beforehand); I was looking for references to publications containing this theorem or similar results.
$endgroup$
– Tanny Sieben
Mar 4 at 18:44
$begingroup$
No problem, can you post your solution to this problem anyway?
$endgroup$
– Maria Mazur
Mar 4 at 18:49
add a comment |
$begingroup$
I have a lenghty complex-vector-computational solution. The main idea is in a following lemma.
Lemma: If $ABCD$ is a square with center $S$, then $$ S = {1over 2}(B-A)i+{1over 2}(A+B)$$
where letters represent complex numbers of coresponding points (with the same name).
So we get 8 centers and then calculate all $4$ midpoints of those segments. Say we have octagon $ABCDEFGH$, then those midpoints (if we suppose that $A+B+C+D+E+F+G+H =0$) are:
$$ M_1 = {1over 4}Big[(D+H-B-F)i+(A+E-C-G)Big]=-M_3$$
$$ M_2 = {1over 4}Big[(A+E-C-G)i+(B+F-D-H)Big] =-M_4$$
Now it is easy to see that $M_2 = iM_1$ and we are done.
answered Jan 5 at 13:13
Maria MazurMaria Mazur
49.5k1361124
$endgroup$
I have a lenghty complex-vector-computational solution. The main idea is in a following lemma.
Lemma: If $ABCD$ is a square with center $S$, then $$ S = {1over 2}(B-A)i+{1over 2}(A+B)$$
where letters represent complex numbers of coresponding points (with the same name).
So we get 8 centers and then calculate all $4$ midpoints of those segments. Say we have octagon $ABCDEFGH$, then those midpoints (if we suppose that $A+B+C+D+E+F+G+H =0$) are:
$$ M_1 = {1over 4}Big[(D+H-B-F)i+(A+E-C-G)Big]=-M_3$$
$$ M_2 = {1over 4}Big[(A+E-C-G)i+(B+F-D-H)Big] =-M_4$$
Now it is easy to see that $M_2 = iM_1$ and we are done.
answered Jan 5 at 13:13
Maria MazurMaria Mazur
49.5k1361124
answered Jan 5 at 13:13
Maria MazurMaria Mazur
49.5k1361124
answered Jan 5 at 13:13
Maria MazurMaria Mazur
49.5k1361124
answered Jan 5 at 13:13
Maria MazurMaria Mazur
49.5k1361124
49.5k1361124
1
$begingroup$
[+1] The approach using complex numbers' geometry is very adapted to such issues.
$endgroup$
– Jean Marie
Feb 27 at 22:44
$begingroup$
Thank you @JeanMarie, By the way, do you like this one:math.stackexchange.com/questions/3126169/…
$endgroup$
– Maria Mazur
Feb 28 at 15:42
$begingroup$
I didn't accept the answer because I am not looking for a proof (i knew how to prove it beforehand); I was looking for references to publications containing this theorem or similar results.
$endgroup$
– Tanny Sieben
Mar 4 at 18:44
$begingroup$
No problem, can you post your solution to this problem anyway?
$endgroup$
– Maria Mazur
Mar 4 at 18:49
add a comment |
1
$begingroup$
[+1] The approach using complex numbers' geometry is very adapted to such issues.
$endgroup$
– Jean Marie
Feb 27 at 22:44
$begingroup$
Thank you @JeanMarie, By the way, do you like this one:math.stackexchange.com/questions/3126169/…
$endgroup$
– Maria Mazur
Feb 28 at 15:42
$begingroup$
I didn't accept the answer because I am not looking for a proof (i knew how to prove it beforehand); I was looking for references to publications containing this theorem or similar results.
$endgroup$
– Tanny Sieben
Mar 4 at 18:44
$begingroup$
No problem, can you post your solution to this problem anyway?
$endgroup$
– Maria Mazur
Mar 4 at 18:49
1
1
$begingroup$
[+1] The approach using complex numbers' geometry is very adapted to such issues.
$endgroup$
– Jean Marie
Feb 27 at 22:44
$begingroup$
[+1] The approach using complex numbers' geometry is very adapted to such issues.
$endgroup$
– Jean Marie
Feb 27 at 22:44
$begingroup$
Thank you @JeanMarie, By the way, do you like this one:math.stackexchange.com/questions/3126169/…
$endgroup$
– Maria Mazur
Feb 28 at 15:42
$begingroup$
Thank you @JeanMarie, By the way, do you like this one:math.stackexchange.com/questions/3126169/…
$endgroup$
– Maria Mazur
Feb 28 at 15:42
$begingroup$
I didn't accept the answer because I am not looking for a proof (i knew how to prove it beforehand); I was looking for references to publications containing this theorem or similar results.
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– Tanny Sieben
Mar 4 at 18:44
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I didn't accept the answer because I am not looking for a proof (i knew how to prove it beforehand); I was looking for references to publications containing this theorem or similar results.
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– Tanny Sieben
Mar 4 at 18:44
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No problem, can you post your solution to this problem anyway?
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– Maria Mazur
Mar 4 at 18:49
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No problem, can you post your solution to this problem anyway?
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– Maria Mazur
Mar 4 at 18:49
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Branko Grünbaum (RIP 2018) presented results of this type in a graduate course at the University of Washington many years ago. I'm not sure if they were published outside of lecture notes. There's a wonderful general result that goes something like this: Given an arbitrary $n$-gon, consider the various regular $(n/k)$-gons, defined as having vertices $(cos 2pi i k/n,sin2pi i k/n)$ for $i=0, 1, ldots, n-1$. (These include "starry" and "multiply-traced" figures, as well as the "dot" ($k=0$).) [continued]
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– Blue
Jan 3 at 19:37
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[continuation] Erecting an $(n/k)$-gon on the sides of the $n$-gon and joining centroids gives a new $n$-gon; erecting some other $(n/k)$-gon on the new sides gives another $n$-gon; etc. Iterating for all-but-one of $(n/k)$-gons, the final polygon is the remaining $(n/k)$-gon. For an $8$-gon, the $(n/k)$-gons are the dot ($k=0$), the regular octagon ($k=1$), the doubly-traced square ($k=2$), "starry octagon" ($k=3$), quad-traced segment ($k=4$, whose derived polygon is a midpoint polygon), and "reverse-direction" versions. Similar to what you have, except for the "opposite sides" thing.
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– Blue
Jan 3 at 19:54
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@Blue, aren't you describing the Petr-Douglas-Neumann Theorem? ( en.wikipedia.org/wiki/Petr-Douglas-Neumann_theorem )
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– Tanny Sieben
Jan 3 at 20:16
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Ah, yes. That seems to be it.
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– Blue
Jan 3 at 20:22
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Interestingly, the "opposite sides" thing seems attributable to a common alternative interpretation of $(n/k)$-gon. Rather than starting at $(1,0)$ of a regular "unit" $n$-gon and forming a single (possibly multiply-traced) path by jumping every $k$ vertices, possibly excluding some vertices, we connect every vertex of the $n$ gon to the vertex/vertices $k$ jumps ahead/behind. So, an $(8/4)$-gon would consist of four individual segments (perhaps best thought of as double-traced segments) instead one octo-traced segment. But your figure arguably also uses octo-traced segments. Intriguing.
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– Blue
Jan 3 at 20:37