# Is there any polygon exists such that sum of length of square of two adjacent sides is equal to another side/diagonal?

In Right angle triangle we have $a^2 + b^2 = c^2$
where $a^2 = (x_1-x_2)^2 + (y_1-y_2)^2 ,$
$b^2 = (x_3-x_2)^2 + (y_3-y_2)^2$
and $c^2 = (x_1-x_3)^2 + (y_1-y_3)^2$

And in Square we have

$a^2 + b^2 = c^2$
$d^2 + e^2 = c^2$
$a^2 + d^2 = f^2$
$b^2 + e^2 = f^2$
and $a=b=d=e , c=f$
where $a^2 = (x_1-x_2)^2 + (y_1-y_2)^2 ,$
$b^2 = (x_3-x_2)^2 + (y_3-y_2)^2$ ,
$c^2 = (x_1-x_3)^2 + (y_1-y_3)^2$ ,
$d^2= (x_1-x_4)^2+(y_1-y_4)^2$ ,
$e^2= (x_3-x_4)^2+(y_3-y_4)^2$ ,
and $f^2= (x_2-x_4)^2+(y_2-y_4)^2$

Is there any other polygon exists which posses this property (sum of length of square of two adjacent sides is equal to another side/diagonal ) ?
It's a sufficient condition to say if this property is satisfied by a polygon then it's a right angle triangle or square ?

• Could you give an example of what the conditions would look like for pentagon and a hexagon (odd number of sides and even number of sides)? Do you want this to apply for every pair of sides, or every pair of adjacent sides - and do you really mean sides or diagonals (note also that a hexagon has two types of diagonal)? – Mark Bennet Oct 5 '14 at 20:22

Two adjacant sides and the corresponding diagonal fulfill this equality iff they form a right triangle. Some simple examples of such polygons are polyominoes, or any polygon with only horizontal and vertical edges. If you allow the "$c$" to be an unrelated diagonal or side, many many other polygons are possible.
Perhaps you're looking for something like the hexagon $$(3,4,12,84,3612,3613)$$ where $$3^2+4^2+12^2+84^2+3612^2=3613^2$$. Polygons like this are found by matching side $$C$$ of one triple with side $$A$$ of another. There are infinite combinations with the number of sides ranging from $$3$$-to-$$\infty$$. I can show you formulas for finding them if you like.