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?
Best Answer
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:
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:
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: