Napoleon-like theorem concerning squares erected on sides of midpoint polygon of octogon
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 6 hours ago
greedoid
38.2k114797
asked 2 days ago
Tanny Sieben
32118
|
show 2 more comments
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 6 hours ago
greedoid
38.2k114797
asked 2 days ago
Tanny Sieben
32118
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]
– Blue
2 days ago
[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.
– Blue
2 days ago
@Blue, aren't you describing the Petr-Douglas-Neumann Theorem? ( en.wikipedia.org/wiki/Petr-Douglas-Neumann_theorem )
– Tanny Sieben
2 days ago
Ah, yes. That seems to be it.
– Blue
2 days ago
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.
– Blue
2 days ago
|
show 2 more comments
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 6 hours ago
greedoid
38.2k114797
asked 2 days ago
Tanny Sieben
32118
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 6 hours ago
greedoid
38.2k114797
asked 2 days ago
Tanny Sieben
32118
edited 6 hours ago
greedoid
38.2k114797
asked 2 days ago
Tanny Sieben
32118
edited 6 hours ago
greedoid
38.2k114797
edited 6 hours ago
greedoid
38.2k114797
edited 6 hours ago
greedoid
38.2k114797
38.2k114797
asked 2 days ago
Tanny Sieben
32118
asked 2 days ago
Tanny Sieben
32118
asked 2 days ago
Tanny Sieben
32118
32118
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]
– Blue
2 days ago
[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.
– Blue
2 days ago
@Blue, aren't you describing the Petr-Douglas-Neumann Theorem? ( en.wikipedia.org/wiki/Petr-Douglas-Neumann_theorem )
– Tanny Sieben
2 days ago
Ah, yes. That seems to be it.
– Blue
2 days ago
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.
– Blue
2 days ago
|
show 2 more comments
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]
– Blue
2 days ago
[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.
– Blue
2 days ago
@Blue, aren't you describing the Petr-Douglas-Neumann Theorem? ( en.wikipedia.org/wiki/Petr-Douglas-Neumann_theorem )
– Tanny Sieben
2 days ago
Ah, yes. That seems to be it.
– Blue
2 days ago
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.
– Blue
2 days ago
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]
– Blue
2 days ago
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]
– Blue
2 days ago
[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.
– Blue
2 days ago
[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.
– Blue
2 days ago
@Blue, aren't you describing the Petr-Douglas-Neumann Theorem? ( en.wikipedia.org/wiki/Petr-Douglas-Neumann_theorem )
– Tanny Sieben
2 days ago
@Blue, aren't you describing the Petr-Douglas-Neumann Theorem? ( en.wikipedia.org/wiki/Petr-Douglas-Neumann_theorem )
– Tanny Sieben
2 days ago
Ah, yes. That seems to be it.
– Blue
2 days ago
Ah, yes. That seems to be it.
– Blue
2 days ago
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.
– Blue
2 days ago
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.
– Blue
2 days ago
|
show 2 more comments
1 Answer
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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 yesterday
greedoid
38.2k114797
add a comment |
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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 yesterday
greedoid
38.2k114797
add a comment |
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 yesterday
greedoid
38.2k114797
add a comment |
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 yesterday
greedoid
38.2k114797
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 yesterday
greedoid
38.2k114797
answered yesterday
greedoid
38.2k114797
answered yesterday
greedoid
38.2k114797
answered yesterday
greedoid
38.2k114797
38.2k114797
add a comment |
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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]
– Blue
2 days ago
[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.
– Blue
2 days ago
@Blue, aren't you describing the Petr-Douglas-Neumann Theorem? ( en.wikipedia.org/wiki/Petr-Douglas-Neumann_theorem )
– Tanny Sieben
2 days ago
Ah, yes. That seems to be it.
– Blue
2 days ago
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.
– Blue
2 days ago