### 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.

### 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?

### Which Twin Is Older?

A simplified account of special relativity and the twins paradox.

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