Review question

# If $ABCDEFGH$ is a regular octagon, what do these vectors add to? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5992

## Solution

$ABCDEFGH$ is a regular octagon. Express in terms of a single vector the sum of the vectors

1. $\mathbf{AC},\mathbf{AH},\mathbf{HG}$,

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 $\mathbf{AC} + \mathbf{AH} + \mathbf{HG} = \mathbf{AE}$.

1. $\mathbf{AB},\mathbf{AC},\mathbf{AF},\mathbf{AH},\mathbf{HG}$.

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 $\big \vert \mathbf{AF} \big \vert = 2 + 2\times \frac{2}{\sqrt{2}} = 2 + 2\sqrt{2}.$