\(ABCDEFGH\) is a regular octagon. Express in terms of a single vector the sum of the vectors

  1. \(\mathbf{AC},\mathbf{AH},\mathbf{HG}\),
the octagon plus five vectors showing an inscribed square

Vector addition tells us that \(\mathbf{AH} + \mathbf{HG} = \mathbf{AG}\).

From the symmetry of the octagon, \(ACEG\) must be a square, so the vector \(\mathbf{AG}\) is the same as \(\mathbf{CE}\). The vector sum of \(\mathbf{AC}\) and \(\mathbf{CE}\) is \(\mathbf{AE}\).

Thus we have \(\mathbf{AC} + \mathbf{AH} + \mathbf{HG} = \mathbf{AE}\).

  1. \(\mathbf{AB},\mathbf{AC},\mathbf{AF},\mathbf{AH},\mathbf{HG}\).
the octagon plus three vectors showing an inscribed rectangle

Using the first part, we can say immediately that \(\mathbf{AC} + \mathbf{AH} + \mathbf{HG} = \mathbf{AE}\) so we are left to sum \(\mathbf{AB}\), \(\mathbf{AE}\) and \(\mathbf{AF}\).

But the diagram above shows us that \(\mathbf{AB} + \mathbf{AF} = \mathbf{AE}\), since \(\mathbf{AE}\) is the diagonal of the rectangle \(ABEF\).

Thus \(\mathbf{AB} + \mathbf{AC} + \mathbf{A} + \mathbf{AH} + \mathbf{HG} = 2\mathbf{AE}\).

Given that \(\big \vert \mathbf{AB} \big \vert = 2,\) calculate \(\big \vert \mathbf{AF} \big \vert\).

the length AF divided into three parts

From the symmetry of the octagon, we can see that \(AF\) is parallel to \(GH\). The small triangles on \(AH\) and \(FG\) are isoscelese and right-angled, and so \[\big \vert \mathbf{AF} \big \vert = 2 + 2\times \frac{2}{\sqrt{2}} = 2 + 2\sqrt{2}.\]