This excellent solution came from Shu Cao of Oxford High School. Well done Shu!

FlexiQuads Image

A convex quadrilateral

Q

is 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

a_{1}

a_{2}

a_{3}

and

a_{4}

be vectors representing the sides (in this order) so that

a_{1} + a_{2} + a_{3} + a_{4} = 0

(the zero vector). Now let

d_{1}

and

d_{2}

be the vectors representing the diagonals of

Q

. We may choose these so that

d_{1} = a_{4} + a_{1}

and

d_{2} = a_{3} + a_{4}

.

As

d_{1} = a_{4} + a_{1}

and

d_{2} = a_{3} + a_{4}

it follows that

a_{1} + a_{2} = - d_{2}, a_{2} + a_{3} = - d_{1} .

\begin{matrix} a_{2}^{2} + a_{4}^{2} - a_{1}^{2} - a_{3}^{2} & = (a_{2}^{2} - a_{1}^{2}) + (a_{4}^{2} - a_{3}^{2}) \\ = (a_{2} - a_{1}) (a_{2} + a_{1}) + (a_{4} - a_{3}) (a_{4} + a_{3}) \\ = - d_{2} (a_{2} - a_{1}) + d_{2} (a_{4} - a_{3}) \\ = d_{2} (a_{4} - a_{3} - a_{2} + a_{1}) \\ = d_{2} ((a_{4} + a_{1}) - (a_{3} + a_{2})) \\ = d_{2} (d_{1} + d_{1}) \\ = 2 d_{2} . d_{1} \end{matrix} .

Now

a_{1} + a_{2} + a_{3} + a_{4} = 0

implies that

a_{4} = - a_{1} - a_{2} - a_{3}

\begin{matrix} a_{1} \cdot a_{3} - a_{2} . a_{4} & = a_{1} . a_{3} - a_{2} (- a_{1} - a_{2} - a_{3}) \\ = a_{1} . a_{3} + a_{2} . a_{1} + a_{2} . a_{2} + a_{2} . a_{3} \\ = a_{1} (a_{2} + a_{3}) + a_{2} (a_{2} + a_{3}) \\ = (a_{1} + a_{2}) (a_{3} + a_{2}) \\ = (- d_{1}) (- d_{2}) \\ = d_{1} . d_{2} \end{matrix} .

Hence

\begin{matrix} 2 (a_{1} . a_{3} - a_{2} . a_{4}) & = 2 d_{1} . d_{2} \\ = a_{2}^{2} + a_{4}^{2} - a_{1}^{2} - a_{3}^{2} . \end{matrix}

If the diagonals of

Q

are perpendicular in one position of

Q

, then

2 d_{1} . d_{2} = a_{2}^{2} + a_{4}^{2} - a_{1}^{2} - a_{3}^{2} = 0

. As

a_{1}, a_{2}, a_{3}, a_{4}

are constant in length

a_{2}^{2} + a_{4}^{2} - a_{1}^{2} - a_{3}^{2}

will always be zero which implies that

d_{1} . d_{2} = 0

, so they are perpendicular in all variations of

Q