Mixing
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
greedoid
38.9k114797
asked Jan 3 at 19:18
Tanny SiebenTanny Sieben
33818
$endgroup$
|
show 2 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
greedoid
38.9k114797
asked Jan 3 at 19:18
Tanny SiebenTanny Sieben
33818
$endgroup$
$begingroup$
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 2 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
greedoid
38.9k114797
asked Jan 3 at 19:18
Tanny SiebenTanny Sieben
33818
$endgroup$
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
greedoid
38.9k114797
asked Jan 3 at 19:18
Tanny SiebenTanny Sieben
33818
edited Jan 6 at 7:32
greedoid
38.9k114797
asked Jan 3 at 19:18
Tanny SiebenTanny Sieben
33818
edited Jan 6 at 7:32
greedoid
38.9k114797
edited Jan 6 at 7:32
greedoid
38.9k114797
edited Jan 6 at 7:32
greedoid
38.9k114797
38.9k114797
asked Jan 3 at 19:18
Tanny SiebenTanny Sieben
33818
asked Jan 3 at 19:18
Tanny SiebenTanny Sieben
33818
asked Jan 3 at 19:18
Tanny SiebenTanny Sieben
33818
33818
$begingroup$
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 2 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 2 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
greedoidgreedoid
38.9k114797
$endgroup$
$begingroup$
Is this not good enough solution to your problem?
$endgroup$
– greedoid
Jan 8 at 18:45
add a comment |
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1 Answer
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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
greedoidgreedoid
38.9k114797
$endgroup$
$begingroup$
Is this not good enough solution to your problem?
$endgroup$
– greedoid
Jan 8 at 18:45
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
greedoidgreedoid
38.9k114797
$endgroup$
$begingroup$
Is this not good enough solution to your problem?
$endgroup$
– greedoid
Jan 8 at 18:45
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
greedoidgreedoid
38.9k114797
$endgroup$
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
greedoidgreedoid
38.9k114797
answered Jan 5 at 13:13
greedoidgreedoid
38.9k114797
answered Jan 5 at 13:13
greedoidgreedoid
38.9k114797
answered Jan 5 at 13:13
greedoidgreedoid
38.9k114797
38.9k114797
$begingroup$
Is this not good enough solution to your problem?
$endgroup$
– greedoid
Jan 8 at 18:45
add a comment |
$begingroup$
Is this not good enough solution to your problem?
$endgroup$
– greedoid
Jan 8 at 18:45
$begingroup$
Is this not good enough solution to your problem?
$endgroup$
– greedoid
Jan 8 at 18:45
$begingroup$
Is this not good enough solution to your problem?
$endgroup$
– greedoid
Jan 8 at 18:45
add a comment |
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$begingroup$
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.
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– Blue
Jan 3 at 20:37