You may also like

problem icon

A Knight's Journey

This article looks at knight's moves on a chess board and introduces you to the idea of vectors and vector addition.

problem icon

8 Methods for Three by One

This problem in geometry has been solved in no less than EIGHT ways by a pair of students. How would you solve it? How many of their solutions can you follow? How are they the same or different? Which do you like best?

problem icon

Which Twin Is Older?

A simplified account of special relativity and the twins paradox.

Flexi Quads

Age 16 to 18 Challenge Level:

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 $Q$.