Flexi quads

A quadrilateral changes shape with the edge lengths constant. Show the scalar product of the diagonals is constant. If the diagonals are perpendicular in one position are they always perpendicular?

Age

16 to 18

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Being curious Being resourceful Being resilient Being collaborative

Problem

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 $Q$.

Student Solutions

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

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 ${\bf a}_1$, ${\bf a}_2$, ${\bf a}_3$ and ${\bf a}_4$ be vectors representing the sides (in this order) 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 ${\bf d}_2={\bf a}_3+{\bf a}_4$.

As ${\bf d}_1={\bf a}_4 + {\bf a}_1$ and ${\bf d}_2 = {\bf a}_3 + {\bf a}_4$ it follows that ${\bf a}_1 + {\bf a}_2 = -{\bf d}_2,\ {\bf a}_2 + {\bf a}_3 = -{\bf d}_1.$

$$\eqalign{ {\bf a}_2^2+{\bf a}_4^2-{\bf a}_1^2-{\bf a}_3^2 &=({\bf a}_2^2-{\bf a}_1^2)+({\bf a}_4^2-{\bf a}_3^2) \cr &=({\bf a}_2-{\bf a}_1)({\bf a}_2+{\bf a}_1)+({\bf a}_4-{\bf a}_3)({\bf a}_4+{\bf a}_3)\cr &=-{\bf d}_2({\bf a}_2-{\bf a}_1)+{\bf d}_2({\bf a}_4-{\bf a}_3)\cr &={\bf d}_2({\bf a}_4-{\bf a}_3-{\bf a}_2+{\bf a}_1)\cr &={\bf d}_2(({\bf a}_4+{\bf a}_1)-({\bf a}_3+{\bf a}_2))\cr &={\bf d}_2({\bf d}_1+{\bf d}_1)\cr &=2{\bf d}_2 {\bf .} {\bf d}_1 }.$$

Now ${\bf a}_1+{\bf a}_2+{\bf a}_3+{\bf a}_4=0$ implies that ${\bf a}_4=-{\bf a}_1-{\bf a}_2-{\bf a}_3$.

$$\eqalign{ {\bf a}_1 \cdot {\bf a}_3-{\bf a}_2 {\bf .} {\bf a}_4 &= {\bf a}_1 {\bf .} {\bf a}_3-{\bf a}_2(-{\bf a}_1-{\bf a}_2-{\bf a}_3) \cr &={\bf a}_1 {\bf .} {\bf a}_3+{\bf a}_2 {\bf .} {\bf a}_1+{\bf a}_2{\bf .} {\bf a}_2+{\bf a}_2{\bf .} {\bf a}_3 \cr &={\bf a}_1({\bf a}_2+{\bf a}_3)+{\bf a}_2({\bf a}_2+{\bf a}_3) \cr &=({\bf a}_1+{\bf a}_2)({\bf a}_3+{\bf a}_2) \cr &=(-{\bf d}_1)(-{\bf d}_2) \cr &={\bf d}_1{\bf .} {\bf d}_2}.$$

Hence

$$\eqalign{ 2({\bf a}_1 {\bf .} {\bf a}_3-{\bf a}_2 {\bf .} {\bf a}_4) &=2{\bf d}_1 {\bf .} {\bf d}_2 \cr &={\bf a}_2^2+{\bf a}_4^2-{\bf a}_1^2-{\bf a}_3^2 .}$$

If the diagonals of $Q$ are perpendicular in one position of $Q$, then $2{\bf d}_1 {\bf .} {\bf d}_2={\bf a}_2^2+{\bf a}_4^2-{\bf a}_1^2-{\bf a}_3^2 =0$. As ${\bf a}_1,{\bf a}_2,{\bf a}_3,{\bf a}_4$ are constant in length ${\bf a}_2^2+{\bf a}_4^2-{\bf a}_1^2-{\bf a}_3^2$ will always be zero which implies that ${\bf d}_1 {\bf .} {\bf d}_2=0$, so they are perpendicular in all variations of $Q$.

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