Dot Products Around the Octagon

There are a total of $8\choose2$ $= 28$ pairs of unit vectors: 8 pairs make an angle of $45^\circ$ , 8 pairs make an angle of $90^\circ$ , 8 pairs make an angle of $135^\circ$ , and 4 pairs make an angle of $180^\circ$ . Using the formula $\vec{u}\cdot\vec{v} = \|u\|\|v\|\cos(\theta)$ and since each vector is has magnitude 1, the contribution by each type is:

$\begin{aligned} 8\cos(45^\circ) &= 4\sqrt{2} \\ 8\cos(90^\circ) &= 0 \\ 8\cos(135^\circ) &= -4\sqrt{2} \\ 4\cos(180^\circ) &= -4 \end{aligned}$

The sum of all these values is $\boxed{-4}$ .

Let the vectors be represented by $v_i$ .

We have $\sum v_i = 0$ (since pairs of them cancel out).
We also have $\sum v_i ^2 = 8$ since each of them are unit vectors.
Since

$( \sum v_i ) ^2 = \sum v_i^2 + 2 \sum_{i < j} v_i v_j.$

Hence, we conclude that $\sum v_i v_j = \frac{0-8}{2} = -4$ .

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Note: All that we needed was $\sum v_i = 0$ . We didn't need to know the exact shape of the octagon.

Let the vectors be $v_1, \dots, v_8$ . For any given vector $v_i$ we have $\sum_j v_i\cdot v_j = v_i\cdot\left(\sum_j v_j\right) = v_i\cdot 0 = 0.$ But we must sum pairs of distinct vectors, so we subtract the product of $v_i$ with itself: $\sum_{j\neq i} v_i\cdot v_j = \left(\sum_j v_i\cdot v_j\right) - v_i\cdot v_i = 0 - v_i^2 = -1.$ Adding this for all eight vectors we get $\sum_i \sum_{j\neq i} v_i\cdot v_j = 8 (-1) = -8.$ However, now we have counted each pair $i\neq j$ twice, so divide by two. The answer is $\frac{-8}2 = -4$ .

The dot product of vectors is distributive over addition. As such, for a given pair $(v_i, v_j)$ , the two products $v_i . v_j$ and $v_i . -v_j$ cancel out, except that we have an unpaired term where $v_i = -v_j$ (since only distinct pairs are counted).

So it's just $v_1 . v_5 + v_2 . v_6 + v_3 . v_7 + v_4 . v_8$ . Each of these dot products is equal to $1 \times 1 \cos \pi = -1$ , giving a total of $\boxed{-4}$ .