If $ABCDEFGH$ is a regular octagon, what do these vectors add to?

If $ABCDEFGH$ is a regular octagon, what do these vectors add to?

Question

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

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

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

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

Suggestion

We can start by choosing one convenient pair to add. Remember that as long as we keep the sense correct we can add them in whatever order we like.

What can we say about the lines $AF$ and $GH$?

Solution

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

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 $\mathbf{AB} + \mathbf{AC} + \mathbf{A} + \mathbf{AH} + \mathbf{HG} = 2\mathbf{AE}$.

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