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Number of equations needed to define a rectangle?
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I've been wondering whether there is some sort of fundamental rule (such as n equations are needed to solve for n variables) underlying the seeming need for the definitions of different shapes.
For example, to define a right angled triangle given points in a plane, we need only the right angle - its characteristic - hence we can write an equation using Pythagoras' theorem which defines it:
$$AB^2 + AC^2 = BC^2$$
However, what about the definition of a rectangle? What fundamental properties does a rectangle possess? I tried the idea of right angles for each of the vertices, but it seems that you can also define a square in just 3 equations. For example:
$$AC=BD$$
$$AB=CD$$
$$AB^2 + AC^2 = BC^2$$
As well as:
$$AD=BC$$
$$AD^2 + BA^2 = BD^2$$
$$CD^2 + CB^2 = BD^2$$
What fundamental truth requires three properties to define a rectangle in terms of points in a plane? Is it related to solving a problem in n variables? Can this rule, if any, be extended to shapes in general in arbitrary dimensions?
algebra-precalculus geometry
asked 46 mins ago
Resquiens
20518
add a comment |Â
up vote
3
down vote
favorite
I've been wondering whether there is some sort of fundamental rule (such as n equations are needed to solve for n variables) underlying the seeming need for the definitions of different shapes.
For example, to define a right angled triangle given points in a plane, we need only the right angle - its characteristic - hence we can write an equation using Pythagoras' theorem which defines it:
$$AB^2 + AC^2 = BC^2$$
However, what about the definition of a rectangle? What fundamental properties does a rectangle possess? I tried the idea of right angles for each of the vertices, but it seems that you can also define a square in just 3 equations. For example:
$$AC=BD$$
$$AB=CD$$
$$AB^2 + AC^2 = BC^2$$
As well as:
$$AD=BC$$
$$AD^2 + BA^2 = BD^2$$
$$CD^2 + CB^2 = BD^2$$
What fundamental truth requires three properties to define a rectangle in terms of points in a plane? Is it related to solving a problem in n variables? Can this rule, if any, be extended to shapes in general in arbitrary dimensions?
algebra-precalculus geometry
asked 46 mins ago
Resquiens
20518
2
You are basically asking about a special case of "rigid graphs" (where "graph" is used in the combinatorial sense, not the $xy$-plotting sense). There is extensive literature on this subject, some (most?) of it pretty sophisticated. Web searches for "rigid graph" and "graph rigidity" should get you started.
– Blue
37 mins ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I've been wondering whether there is some sort of fundamental rule (such as n equations are needed to solve for n variables) underlying the seeming need for the definitions of different shapes.
For example, to define a right angled triangle given points in a plane, we need only the right angle - its characteristic - hence we can write an equation using Pythagoras' theorem which defines it:
$$AB^2 + AC^2 = BC^2$$
However, what about the definition of a rectangle? What fundamental properties does a rectangle possess? I tried the idea of right angles for each of the vertices, but it seems that you can also define a square in just 3 equations. For example:
$$AC=BD$$
$$AB=CD$$
$$AB^2 + AC^2 = BC^2$$
As well as:
$$AD=BC$$
$$AD^2 + BA^2 = BD^2$$
$$CD^2 + CB^2 = BD^2$$
What fundamental truth requires three properties to define a rectangle in terms of points in a plane? Is it related to solving a problem in n variables? Can this rule, if any, be extended to shapes in general in arbitrary dimensions?
algebra-precalculus geometry
asked 46 mins ago
Resquiens
20518
I've been wondering whether there is some sort of fundamental rule (such as n equations are needed to solve for n variables) underlying the seeming need for the definitions of different shapes.
For example, to define a right angled triangle given points in a plane, we need only the right angle - its characteristic - hence we can write an equation using Pythagoras' theorem which defines it:
$$AB^2 + AC^2 = BC^2$$
However, what about the definition of a rectangle? What fundamental properties does a rectangle possess? I tried the idea of right angles for each of the vertices, but it seems that you can also define a square in just 3 equations. For example:
$$AC=BD$$
$$AB=CD$$
$$AB^2 + AC^2 = BC^2$$
As well as:
$$AD=BC$$
$$AD^2 + BA^2 = BD^2$$
$$CD^2 + CB^2 = BD^2$$
What fundamental truth requires three properties to define a rectangle in terms of points in a plane? Is it related to solving a problem in n variables? Can this rule, if any, be extended to shapes in general in arbitrary dimensions?
algebra-precalculus geometry
algebra-precalculus geometry
asked 46 mins ago
Resquiens
20518
asked 46 mins ago
Resquiens
20518
edited 26 mins ago
asked 46 mins ago
Resquiens
20518
asked 46 mins ago
Resquiens
20518
asked 46 mins ago
Resquiens
20518
20518
2
You are basically asking about a special case of "rigid graphs" (where "graph" is used in the combinatorial sense, not the $xy$-plotting sense). There is extensive literature on this subject, some (most?) of it pretty sophisticated. Web searches for "rigid graph" and "graph rigidity" should get you started.
– Blue
37 mins ago
add a comment |Â
2
You are basically asking about a special case of "rigid graphs" (where "graph" is used in the combinatorial sense, not the $xy$-plotting sense). There is extensive literature on this subject, some (most?) of it pretty sophisticated. Web searches for "rigid graph" and "graph rigidity" should get you started.
– Blue
37 mins ago
2
2
You are basically asking about a special case of "rigid graphs" (where "graph" is used in the combinatorial sense, not the $xy$-plotting sense). There is extensive literature on this subject, some (most?) of it pretty sophisticated. Web searches for "rigid graph" and "graph rigidity" should get you started.
– Blue
37 mins ago
You are basically asking about a special case of "rigid graphs" (where "graph" is used in the combinatorial sense, not the $xy$-plotting sense). There is extensive literature on this subject, some (most?) of it pretty sophisticated. Web searches for "rigid graph" and "graph rigidity" should get you started.
– Blue
37 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
Think about
$$(AC-BD)^2+(AB-CD)^2+(AB^2+AC^2-BC^2)^2=0.$$
answered 44 mins ago
Carl Schildkraut
10.3k11438
I'm not sure what squaring all the terms does? I see that you've got a rearranged version of each of the statements in each of the three clauses and that since each is 0 their squares should add up to 0, but what am I missing?
– Resquiens
33 mins ago
@Resquiens It's a way to encode three equations in one. Your three equations are equivalent to this one equation.
– Cheerful Parsnip
25 mins ago
But could it not be the case that the first two terms cancelled out the third term?
– Resquiens
24 mins ago
add a comment |Â
up vote
2
down vote
While not exactly what the OP wants, I would suggest the coordinate approach. Then the "triangle equation" becomes a circle equation in the form $x^2+y^2=R^2$ which completely defines a circle centered at the point $(0,0)$.
Because of the corners, there's no such nice equation for a square. But we can make some "square-like" shapes in a similar way using the equation:
$$x^2n+y^2n=R^2n, qquad n>1$$
Here's an illustration for a few $n$, starting with a circle:
"But none of these are squares!" - anyone would say, and be right.
To define a square we can't just use a single equation. We'd have to deal with the absolute value function.
For example, here's a square with the diagonal $D$, rotated by $pi/4$:
$$|x|+|y|=D$$
We can easily rotate this one right back, and expand it so it fits with the others:
$$|x+y|+|x-y|=a$$
Where $a$ is now the length of the side.
For a rectangle (the edited question) we just need to scale one of the coordinates. For a general rectangle with the sides $a$ and $b$ we have:
$$left|fracxa+fracyb right|+left|fracxa-fracyb right|=1$$
To get rid of the absolute value we can try squaring twice:
$$left(fracxa+fracyb right)^2+left(fracxa-fracyb right)^2+2left|fracx^2a^2-fracy^2b^2 right|=1$$
$$left(left(fracxa+fracyb right)^2+left(fracxa-fracyb right)^2-1right)^2=4left(fracx^2a^2-fracy^2b^2 right)^2$$
Unfortunately, as it often happens with squaring, we get extra solutions (the dashed lines), which don't lie on the original rectangle:
answered 17 mins ago
Yuriy S
14k331110
This is quite a nice approach, and it helped me understand the absolute value in Mohammad Riazi-Kermani's answer. Thank you.
– Resquiens
13 mins ago
@Resquiens, you are welcome. See also the edit
– Yuriy S
1 min ago
add a comment |Â
up vote
1
down vote
Note that with your $$AC=BD, AB=CD,AB^2+AC^2=BC^2$$ or
$$AD=BC,
AD^2+BA^2=BD^2,CD^2+CB^2=BD^2$$
You do not get a square, you only get a rectangle.
You need to modify your equations.
It would be helpful if you use absolute value for the square as well.
answered 30 mins ago
Mohammad Riazi-Kermani
37k41956
Yes, I do see that. It seems that a square requires an additional reduction in its degree of freedom (possibly equating $BD=CA$). But why is this? In any case, I will modify the question to ask about a rectangle rather than a square.
– Resquiens
28 mins ago
Of course more restriction requires more equations. I am not sure if you really need four equations for the square, it sounds like too many. Think about it some more and see if you can modify your equations to just three and get a square.
– Mohammad Riazi-Kermani
13 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Think about
$$(AC-BD)^2+(AB-CD)^2+(AB^2+AC^2-BC^2)^2=0.$$
answered 44 mins ago
Carl Schildkraut
10.3k11438
I'm not sure what squaring all the terms does? I see that you've got a rearranged version of each of the statements in each of the three clauses and that since each is 0 their squares should add up to 0, but what am I missing?
– Resquiens
33 mins ago
@Resquiens It's a way to encode three equations in one. Your three equations are equivalent to this one equation.
– Cheerful Parsnip
25 mins ago
But could it not be the case that the first two terms cancelled out the third term?
– Resquiens
24 mins ago
add a comment |Â
up vote
3
down vote
Think about
$$(AC-BD)^2+(AB-CD)^2+(AB^2+AC^2-BC^2)^2=0.$$
answered 44 mins ago
Carl Schildkraut
10.3k11438
I'm not sure what squaring all the terms does? I see that you've got a rearranged version of each of the statements in each of the three clauses and that since each is 0 their squares should add up to 0, but what am I missing?
– Resquiens
33 mins ago
@Resquiens It's a way to encode three equations in one. Your three equations are equivalent to this one equation.
– Cheerful Parsnip
25 mins ago
But could it not be the case that the first two terms cancelled out the third term?
– Resquiens
24 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Think about
$$(AC-BD)^2+(AB-CD)^2+(AB^2+AC^2-BC^2)^2=0.$$
answered 44 mins ago
Carl Schildkraut
10.3k11438
Think about
$$(AC-BD)^2+(AB-CD)^2+(AB^2+AC^2-BC^2)^2=0.$$
answered 44 mins ago
Carl Schildkraut
10.3k11438
answered 44 mins ago
Carl Schildkraut
10.3k11438
answered 44 mins ago
Carl Schildkraut
10.3k11438
answered 44 mins ago
Carl Schildkraut
10.3k11438
10.3k11438
I'm not sure what squaring all the terms does? I see that you've got a rearranged version of each of the statements in each of the three clauses and that since each is 0 their squares should add up to 0, but what am I missing?
– Resquiens
33 mins ago
@Resquiens It's a way to encode three equations in one. Your three equations are equivalent to this one equation.
– Cheerful Parsnip
25 mins ago
But could it not be the case that the first two terms cancelled out the third term?
– Resquiens
24 mins ago
add a comment |Â
I'm not sure what squaring all the terms does? I see that you've got a rearranged version of each of the statements in each of the three clauses and that since each is 0 their squares should add up to 0, but what am I missing?
– Resquiens
33 mins ago
@Resquiens It's a way to encode three equations in one. Your three equations are equivalent to this one equation.
– Cheerful Parsnip
25 mins ago
But could it not be the case that the first two terms cancelled out the third term?
– Resquiens
24 mins ago
I'm not sure what squaring all the terms does? I see that you've got a rearranged version of each of the statements in each of the three clauses and that since each is 0 their squares should add up to 0, but what am I missing?
– Resquiens
33 mins ago
I'm not sure what squaring all the terms does? I see that you've got a rearranged version of each of the statements in each of the three clauses and that since each is 0 their squares should add up to 0, but what am I missing?
– Resquiens
33 mins ago
@Resquiens It's a way to encode three equations in one. Your three equations are equivalent to this one equation.
– Cheerful Parsnip
25 mins ago
@Resquiens It's a way to encode three equations in one. Your three equations are equivalent to this one equation.
– Cheerful Parsnip
25 mins ago
But could it not be the case that the first two terms cancelled out the third term?
– Resquiens
24 mins ago
But could it not be the case that the first two terms cancelled out the third term?
– Resquiens
24 mins ago
add a comment |Â
up vote
2
down vote
While not exactly what the OP wants, I would suggest the coordinate approach. Then the "triangle equation" becomes a circle equation in the form $x^2+y^2=R^2$ which completely defines a circle centered at the point $(0,0)$.
Because of the corners, there's no such nice equation for a square. But we can make some "square-like" shapes in a similar way using the equation:
$$x^2n+y^2n=R^2n, qquad n>1$$
Here's an illustration for a few $n$, starting with a circle:
"But none of these are squares!" - anyone would say, and be right.
To define a square we can't just use a single equation. We'd have to deal with the absolute value function.
For example, here's a square with the diagonal $D$, rotated by $pi/4$:
$$|x|+|y|=D$$
We can easily rotate this one right back, and expand it so it fits with the others:
$$|x+y|+|x-y|=a$$
Where $a$ is now the length of the side.
For a rectangle (the edited question) we just need to scale one of the coordinates. For a general rectangle with the sides $a$ and $b$ we have:
$$left|fracxa+fracyb right|+left|fracxa-fracyb right|=1$$
To get rid of the absolute value we can try squaring twice:
$$left(fracxa+fracyb right)^2+left(fracxa-fracyb right)^2+2left|fracx^2a^2-fracy^2b^2 right|=1$$
$$left(left(fracxa+fracyb right)^2+left(fracxa-fracyb right)^2-1right)^2=4left(fracx^2a^2-fracy^2b^2 right)^2$$
Unfortunately, as it often happens with squaring, we get extra solutions (the dashed lines), which don't lie on the original rectangle:
answered 17 mins ago
Yuriy S
14k331110
This is quite a nice approach, and it helped me understand the absolute value in Mohammad Riazi-Kermani's answer. Thank you.
– Resquiens
13 mins ago
@Resquiens, you are welcome. See also the edit
– Yuriy S
1 min ago
add a comment |Â
up vote
2
down vote
While not exactly what the OP wants, I would suggest the coordinate approach. Then the "triangle equation" becomes a circle equation in the form $x^2+y^2=R^2$ which completely defines a circle centered at the point $(0,0)$.
Because of the corners, there's no such nice equation for a square. But we can make some "square-like" shapes in a similar way using the equation:
$$x^2n+y^2n=R^2n, qquad n>1$$
Here's an illustration for a few $n$, starting with a circle:
"But none of these are squares!" - anyone would say, and be right.
To define a square we can't just use a single equation. We'd have to deal with the absolute value function.
For example, here's a square with the diagonal $D$, rotated by $pi/4$:
$$|x|+|y|=D$$
We can easily rotate this one right back, and expand it so it fits with the others:
$$|x+y|+|x-y|=a$$
Where $a$ is now the length of the side.
For a rectangle (the edited question) we just need to scale one of the coordinates. For a general rectangle with the sides $a$ and $b$ we have:
$$left|fracxa+fracyb right|+left|fracxa-fracyb right|=1$$
To get rid of the absolute value we can try squaring twice:
$$left(fracxa+fracyb right)^2+left(fracxa-fracyb right)^2+2left|fracx^2a^2-fracy^2b^2 right|=1$$
$$left(left(fracxa+fracyb right)^2+left(fracxa-fracyb right)^2-1right)^2=4left(fracx^2a^2-fracy^2b^2 right)^2$$
Unfortunately, as it often happens with squaring, we get extra solutions (the dashed lines), which don't lie on the original rectangle:
answered 17 mins ago
Yuriy S
14k331110
This is quite a nice approach, and it helped me understand the absolute value in Mohammad Riazi-Kermani's answer. Thank you.
– Resquiens
13 mins ago
@Resquiens, you are welcome. See also the edit
– Yuriy S
1 min ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
While not exactly what the OP wants, I would suggest the coordinate approach. Then the "triangle equation" becomes a circle equation in the form $x^2+y^2=R^2$ which completely defines a circle centered at the point $(0,0)$.
Because of the corners, there's no such nice equation for a square. But we can make some "square-like" shapes in a similar way using the equation:
$$x^2n+y^2n=R^2n, qquad n>1$$
Here's an illustration for a few $n$, starting with a circle:
"But none of these are squares!" - anyone would say, and be right.
To define a square we can't just use a single equation. We'd have to deal with the absolute value function.
For example, here's a square with the diagonal $D$, rotated by $pi/4$:
$$|x|+|y|=D$$
We can easily rotate this one right back, and expand it so it fits with the others:
$$|x+y|+|x-y|=a$$
Where $a$ is now the length of the side.
For a rectangle (the edited question) we just need to scale one of the coordinates. For a general rectangle with the sides $a$ and $b$ we have:
$$left|fracxa+fracyb right|+left|fracxa-fracyb right|=1$$
To get rid of the absolute value we can try squaring twice:
$$left(fracxa+fracyb right)^2+left(fracxa-fracyb right)^2+2left|fracx^2a^2-fracy^2b^2 right|=1$$
$$left(left(fracxa+fracyb right)^2+left(fracxa-fracyb right)^2-1right)^2=4left(fracx^2a^2-fracy^2b^2 right)^2$$
Unfortunately, as it often happens with squaring, we get extra solutions (the dashed lines), which don't lie on the original rectangle:
answered 17 mins ago
Yuriy S
14k331110
While not exactly what the OP wants, I would suggest the coordinate approach. Then the "triangle equation" becomes a circle equation in the form $x^2+y^2=R^2$ which completely defines a circle centered at the point $(0,0)$.
Because of the corners, there's no such nice equation for a square. But we can make some "square-like" shapes in a similar way using the equation:
$$x^2n+y^2n=R^2n, qquad n>1$$
Here's an illustration for a few $n$, starting with a circle:
"But none of these are squares!" - anyone would say, and be right.
To define a square we can't just use a single equation. We'd have to deal with the absolute value function.
For example, here's a square with the diagonal $D$, rotated by $pi/4$:
$$|x|+|y|=D$$
We can easily rotate this one right back, and expand it so it fits with the others:
$$|x+y|+|x-y|=a$$
Where $a$ is now the length of the side.
For a rectangle (the edited question) we just need to scale one of the coordinates. For a general rectangle with the sides $a$ and $b$ we have:
$$left|fracxa+fracyb right|+left|fracxa-fracyb right|=1$$
To get rid of the absolute value we can try squaring twice:
$$left(fracxa+fracyb right)^2+left(fracxa-fracyb right)^2+2left|fracx^2a^2-fracy^2b^2 right|=1$$
$$left(left(fracxa+fracyb right)^2+left(fracxa-fracyb right)^2-1right)^2=4left(fracx^2a^2-fracy^2b^2 right)^2$$
Unfortunately, as it often happens with squaring, we get extra solutions (the dashed lines), which don't lie on the original rectangle:
answered 17 mins ago
Yuriy S
14k331110
edited 1 min ago
answered 17 mins ago
Yuriy S
14k331110
answered 17 mins ago
Yuriy S
14k331110
answered 17 mins ago
Yuriy S
14k331110
14k331110
This is quite a nice approach, and it helped me understand the absolute value in Mohammad Riazi-Kermani's answer. Thank you.
– Resquiens
13 mins ago
@Resquiens, you are welcome. See also the edit
– Yuriy S
1 min ago
add a comment |Â
This is quite a nice approach, and it helped me understand the absolute value in Mohammad Riazi-Kermani's answer. Thank you.
– Resquiens
13 mins ago
@Resquiens, you are welcome. See also the edit
– Yuriy S
1 min ago
This is quite a nice approach, and it helped me understand the absolute value in Mohammad Riazi-Kermani's answer. Thank you.
– Resquiens
13 mins ago
This is quite a nice approach, and it helped me understand the absolute value in Mohammad Riazi-Kermani's answer. Thank you.
– Resquiens
13 mins ago
@Resquiens, you are welcome. See also the edit
– Yuriy S
1 min ago
@Resquiens, you are welcome. See also the edit
– Yuriy S
1 min ago
add a comment |Â
up vote
1
down vote
Note that with your $$AC=BD, AB=CD,AB^2+AC^2=BC^2$$ or
$$AD=BC,
AD^2+BA^2=BD^2,CD^2+CB^2=BD^2$$
You do not get a square, you only get a rectangle.
You need to modify your equations.
It would be helpful if you use absolute value for the square as well.
answered 30 mins ago
Mohammad Riazi-Kermani
37k41956
Yes, I do see that. It seems that a square requires an additional reduction in its degree of freedom (possibly equating $BD=CA$). But why is this? In any case, I will modify the question to ask about a rectangle rather than a square.
– Resquiens
28 mins ago
Of course more restriction requires more equations. I am not sure if you really need four equations for the square, it sounds like too many. Think about it some more and see if you can modify your equations to just three and get a square.
– Mohammad Riazi-Kermani
13 mins ago
add a comment |Â
up vote
1
down vote
Note that with your $$AC=BD, AB=CD,AB^2+AC^2=BC^2$$ or
$$AD=BC,
AD^2+BA^2=BD^2,CD^2+CB^2=BD^2$$
You do not get a square, you only get a rectangle.
You need to modify your equations.
It would be helpful if you use absolute value for the square as well.
answered 30 mins ago
Mohammad Riazi-Kermani
37k41956
Yes, I do see that. It seems that a square requires an additional reduction in its degree of freedom (possibly equating $BD=CA$). But why is this? In any case, I will modify the question to ask about a rectangle rather than a square.
– Resquiens
28 mins ago
Of course more restriction requires more equations. I am not sure if you really need four equations for the square, it sounds like too many. Think about it some more and see if you can modify your equations to just three and get a square.
– Mohammad Riazi-Kermani
13 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note that with your $$AC=BD, AB=CD,AB^2+AC^2=BC^2$$ or
$$AD=BC,
AD^2+BA^2=BD^2,CD^2+CB^2=BD^2$$
You do not get a square, you only get a rectangle.
You need to modify your equations.
It would be helpful if you use absolute value for the square as well.
answered 30 mins ago
Mohammad Riazi-Kermani
37k41956
Note that with your $$AC=BD, AB=CD,AB^2+AC^2=BC^2$$ or
$$AD=BC,
AD^2+BA^2=BD^2,CD^2+CB^2=BD^2$$
You do not get a square, you only get a rectangle.
You need to modify your equations.
It would be helpful if you use absolute value for the square as well.
answered 30 mins ago
Mohammad Riazi-Kermani
37k41956
answered 30 mins ago
Mohammad Riazi-Kermani
37k41956
answered 30 mins ago
Mohammad Riazi-Kermani
37k41956
answered 30 mins ago
Mohammad Riazi-Kermani
37k41956
37k41956
Yes, I do see that. It seems that a square requires an additional reduction in its degree of freedom (possibly equating $BD=CA$). But why is this? In any case, I will modify the question to ask about a rectangle rather than a square.
– Resquiens
28 mins ago
Of course more restriction requires more equations. I am not sure if you really need four equations for the square, it sounds like too many. Think about it some more and see if you can modify your equations to just three and get a square.
– Mohammad Riazi-Kermani
13 mins ago
add a comment |Â
Yes, I do see that. It seems that a square requires an additional reduction in its degree of freedom (possibly equating $BD=CA$). But why is this? In any case, I will modify the question to ask about a rectangle rather than a square.
– Resquiens
28 mins ago
Of course more restriction requires more equations. I am not sure if you really need four equations for the square, it sounds like too many. Think about it some more and see if you can modify your equations to just three and get a square.
– Mohammad Riazi-Kermani
13 mins ago
Yes, I do see that. It seems that a square requires an additional reduction in its degree of freedom (possibly equating $BD=CA$). But why is this? In any case, I will modify the question to ask about a rectangle rather than a square.
– Resquiens
28 mins ago
Yes, I do see that. It seems that a square requires an additional reduction in its degree of freedom (possibly equating $BD=CA$). But why is this? In any case, I will modify the question to ask about a rectangle rather than a square.
– Resquiens
28 mins ago
Of course more restriction requires more equations. I am not sure if you really need four equations for the square, it sounds like too many. Think about it some more and see if you can modify your equations to just three and get a square.
– Mohammad Riazi-Kermani
13 mins ago
Of course more restriction requires more equations. I am not sure if you really need four equations for the square, it sounds like too many. Think about it some more and see if you can modify your equations to just three and get a square.
– Mohammad Riazi-Kermani
13 mins ago
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2
You are basically asking about a special case of "rigid graphs" (where "graph" is used in the combinatorial sense, not the $xy$-plotting sense). There is extensive literature on this subject, some (most?) of it pretty sophisticated. Web searches for "rigid graph" and "graph rigidity" should get you started.
– Blue
37 mins ago