\(ABCDEFGH\) is a regular octagon. Express in terms of a single vector the sum of the vectors
- \(\mathbf{AC},\mathbf{AH},\mathbf{HG}\),
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 \(\mathbf{AC} + \mathbf{AH} + \mathbf{HG} = \mathbf{AE}\).
- \(\mathbf{AB},\mathbf{AC},\mathbf{AF},\mathbf{AH},\mathbf{HG}\).
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 \(\big \vert \mathbf{AF} \big \vert\).
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 \[\big \vert \mathbf{AF} \big \vert = 2 + 2\times \frac{2}{\sqrt{2}} = 2 + 2\sqrt{2}.\]
