Copyright © University of Cambridge. All rights reserved.

## 'Flexi Quads' printed from http://nrich.maths.org/

Consider a convex quadrilateral $Q$ made from four rigid rods with
flexible joints at the vertices so that the shape of $Q$ can be
changed while keeping the lengths of the sides constant. Let ${\bf
a}_1$, ${\bf a}_2$, ${\bf a}_3$ and ${\bf a}_4$ be vectors
representing the sides (in this order) of an arbitrary
quadrilateral $Q$, so that ${\bf a}_1+{\bf a}_2+{\bf a}_3+{\bf a}_4
= {\bf 0}$ (the zero vector). Now let ${\bf d}_1$ and ${\bf d}_2$
be the vectors representing the diagonals of $Q$. We may choose
these so that ${\bf d}_1={\bf a}_4+{\bf a}_1$ and ${\bf d}_2={\bf
a}_3+{\bf a}_4$. Prove that

$${\bf a}_2^2+{\bf a}_4^2-{\bf a}_1^2-{\bf a}_3^2= 2({\bf
a}_1{\cdot}{\bf a}_3-{\bf a}_2{\cdot}{\bf a}_4).\quad (1)$$

and that the scalar product of the diagonals is constant and given
by:

$$2{\bf d}_1{\cdot}{\bf d}_2 = {\bf a}_2^2+{\bf a}_4^2-{\bf
a}_1^2-{\bf a}_3^2.\quad (2)$$

Use these results to show that, as the shape of the quadrilateral
is changed, if the diagonals of $Q$ are perpendicular in one
position of $Q$, then they are perpendicular in all variations of
$Q$.