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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?

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    $\begingroup$ 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. $\endgroup$ – Blue Oct 29 '18 at 18:34
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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:

enter image description here

"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 $\pi/4$:

$$|x|+|y|=D$$

enter image description here

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.

enter image description here

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:

$$\left|\frac{x}{a}+\frac{y}{b} \right|+\left|\frac{x}{a}-\frac{y}{b} \right|=1$$


To get rid of the absolute value we can try squaring twice:

$$\left(\frac{x}{a}+\frac{y}{b} \right)^2+\left(\frac{x}{a}-\frac{y}{b} \right)^2+2\left|\frac{x^2}{a^2}-\frac{y^2}{b^2} \right|=1$$

$$\left(\left(\frac{x}{a}+\frac{y}{b} \right)^2+\left(\frac{x}{a}-\frac{y}{b} \right)^2-1\right)^2=4\left(\frac{x^2}{a^2}-\frac{y^2}{b^2} \right)^2$$

Unfortunately, as it often happens with squaring, we get extra solutions (the dashed lines), which don't lie on the original rectangle:

enter image description here

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  • $\begingroup$ This is quite a nice approach, and it helped me understand the absolute value in Mohammad Riazi-Kermani's answer. Thank you. $\endgroup$ – Resquiens Oct 29 '18 at 18:58
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    $\begingroup$ @Resquiens, you are welcome. See also the edit $\endgroup$ – Yuriy S Oct 29 '18 at 19:10
  • $\begingroup$ You could also use a modified version of the uniform norm (en.wikipedia.org/wiki/Uniform_norm) : max(|x-x0|/a, |y-y0|/b) = 1 with a the width and b the height. $\endgroup$ – Faibbus Oct 30 '18 at 13:12
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Think about

$$(AC-BD)^2+(AB-CD)^2+(AB^2+AC^2-BC^2)^2=0.$$

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  • $\begingroup$ 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? $\endgroup$ – Resquiens Oct 29 '18 at 18:38
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    $\begingroup$ @Resquiens It's a way to encode three equations in one. Your three equations are equivalent to this one equation. $\endgroup$ – Cheerful Parsnip Oct 29 '18 at 18:46
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    $\begingroup$ @Resquiens how could they cancel? Once you square a term, it can't be negative, so... $\endgroup$ – mathmandan Oct 29 '18 at 19:29
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    $\begingroup$ @Resquiens For all real $a$, $a^2\geq 0$, with equality if and only if $a=0$. Thus, $a^2+b^2=0$ if and only if both $a$ and $b$ are $0$... this method can be generalized to any such problem. $\endgroup$ – Carl Schildkraut Oct 29 '18 at 20:15
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    $\begingroup$ The answer would be improved by including information from the comments. $\endgroup$ – Tommi Brander Oct 30 '18 at 9:11
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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.

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  • $\begingroup$ 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. $\endgroup$ – Resquiens Oct 29 '18 at 18:44
  • $\begingroup$ 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. $\endgroup$ – Mohammad Riazi-Kermani Oct 29 '18 at 18:59
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The solution of the equation $1-x^2=0$ is two vertical lines, $x=\pm1$. Similarly, $1-y^2=0$ gives the horizontal lines $y=\pm1$. We can combine these to get a set of four lines: $$(1-x^2)(1-y^2)=0$$ To get a rectangle from this we need only restrict the domain to $|x| \leq 1, |y| \leq 1$. We can do this with square roots: $$\sqrt{1-x^2}\sqrt{1-y^2}=0$$

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    $\begingroup$ @YuriyS No, it covers the edges of the square too. Suppose $x = 1$, for example. Then the equation $\sqrt{1-x^2}\sqrt{1-y^2}=0$ simplifies to $0=0$, which is satisfied by any $y$ in the domain, i.e., $-1 \le y \le 1$. That's the right edge; the other three edges are similar. An equation that corresponds only to the corners would be, for example, $$(1-x^2)^2 + (1-y^2)^2 = 0$$ because the LHS is $0$ only when $1-x^2$ and $1-y^2$ are both equal to $0$, i.e., when $x=\pm1$ and $y=\pm1$. $\endgroup$ – Théophile Nov 5 '18 at 15:47
  • $\begingroup$ You are right of course, +1 $\endgroup$ – Yuriy S Nov 5 '18 at 15:48
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Various congruence relations area applicable. In general if you know any 3 pieces of information about a triangle, two angles and a side, two sides and an angle, 3 sides.

In Euclidean Geometry-

SAS: Given two sides of a triangle and the angle they make, you can determine the other angles and the third side. Essentially by the cosine rule.

AAS: Given two angles and a side opposite one of those angles, you can determine the third angle and the remaining side.

Hypotenuse leg: Given hypotenuse and a leg of a right triangle, one can determine the angles and the other leg.

ASS: Ambiguous case. Given two sides and and non-included angle, only two triangles up to congruency can be formed.

These have implications regarding other polygons as well.

FYI: AAA only guarantees similarity and not congruence in Euclidean Geometry.

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The number of equations needed to define a rectangle must equal the number of unknowns needed to define a rectangle. Three points can uniquely define a rectangle, and each point consists of two coordinates, so there are six unknowns, and therefore six equations.

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    $\begingroup$ By the same reasoning: two points uniquely determine a line, and each point consists of two co-ordinates, therefore we need four equations for a line? $\endgroup$ – Théophile Oct 29 '18 at 19:38
  • $\begingroup$ After thinking about it some more any rectangle located on the 2D plane can be uniquely defined by the width, height, rotation, and its center location, which would be 5 unknowns requiring 5 equations. $\endgroup$ – Thomas Oct 31 '18 at 11:27
  • $\begingroup$ You certainly don't need five or six equations; you can do it in one. $\endgroup$ – Théophile Oct 31 '18 at 17:17

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