Given:

• Three 2 component vector $\vec{x}$, $\vec{y}$, and $\vec{z}$ such that $\vec{x} + \vec{y} = \vec{z}$ and $\|\vec{x}\| = \|\vec{y}\|$
• $\theta$ such that the angle between $\vec{x}$ and $\vec{y}$ is $\theta$
• Three more 2 component vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ such that $\vec{a} + \vec{b} + \vec{c} = \vec{z}$ and $\|\vec{a}\| = \|\vec{b}\| = \|\vec{c}\|$
• $\phi$ such that the angle between $\vec{a}$ and $\vec{b}$ is $\phi$ and the angle between $\vec{b}$ and $\vec{c}$ is $\phi$

What is the ratio of $\theta$ to $\phi$?

The English translation of what I'm asking is: Given that a vertex of an equiangular, equilateral polygon falls on the origin let the point two vertexes away be $p$. Now take an equiangular, equilateral polygon with twice as many sides which also has a vertex on the origin. The point three vertexes away is $p$. What is the ratio of the angle between the sides of the first and second polygon?

EDIT: David Quinn made the comment that this was difficult to understand. I've added a picture to help with visualization:

• According to the first part, it seems to me that there is something missing, some kind of relation between $\vec{x}$, $\vec{y}$, $\vec{z}$ and $\vec{a}$, $\vec{b}$, $\vec{c}$. If not, the ratio between $\theta$ and $\phi$ could be any. But, according to the second part, it seems that $\|\vec{x}\| = \|\vec{y}\| = \|\vec{z}\|$. And, in that case, $\theta=2\phi$. Could you clarify it, please? – AugSB May 27 '15 at 11:52
• @AugSB I've posted an answer. I expected the ratio to be simple $2\theta = \phi$ but it looks like it's more involved than that. If the answer doesn't make sense let me know how I can clarify. – Jonathan Mee May 27 '15 at 14:47

First lets prove that $\vec{a} = s\vec{x}$ and $\vec{c} = s\vec{y}$ where $s$ is some scalar:

1. Let $n$ be the number of sides of the polygon which has $\vec{x}$ and $\vec{y}$ as edges
2. Since we know that the $2n$-sided polygon has vertexes at both endpoints of $\vec{z}$ and we know that it is equiangular: $\vec{z}$ is parallel to $\vec{b}$
3. Since we know that the $n$-sided polygon has vertexes at both endpoints of $\vec{z}$ and we know it is equiangular: The angle between $\vec{x}$ and $\vec{z}$ is the same as the angle between $\vec{y}$ and $\vec{z}$
4. By the Parallel Postulate and points 2 and 3: The angle between $\vec{x}$ and $\vec{b}$ is the same as the angle between $\vec{y}$ and $\vec{b}$
5. By the Triangle Postulate: $\theta + 2 *$ the acute angle between $\vec{x}$ and $\vec{b} = \pi$
6. The obtuse angle between $\vec{x}$ and $\vec{b}$ and the acute angle between $\vec{x}$ and $\vec{b} = \pi$
7. By solving the equation in point 5 the acute angle $= \frac{\pi - \theta}{2}$
8. By substituting this into the equation in point 6 the obtuse angle $=\frac{\pi + \theta}{2}$
9. By the sum of interior polygon angles: $\pi(n - 2) = n\theta$
10. Solving point 9 for $n = \frac{2\pi}{\pi - \theta}$
11. Also by the sum of interior polygon angles: $\pi(2n - 2) = 2n\phi$
12. Substituting from point 10 into point 11: $\pi(2\frac{2\pi}{\pi - \theta} - 2) = 2\phi\frac{2\pi}{\pi - \theta}$
13. This simplifies to $\frac{\pi + \theta}{2} = \phi$
14. So the obtuse angle formed by $\vec{x}$ and $\vec{b}$ is in fact $\phi$
15. By point 14 and the fact that $\vec{a}$ and $\vec{x}$ touch the same endpoint of $\vec{z}$ and $\vec{c}$ and $\vec{y}$ touch the other endpoint of $\vec{z}$ then for some arbitrary $s$ it must be true that: $\vec{a} = s\vec{x}$ and $\vec{c} = s\vec{y}$

Now going back to point 13 we see: $\frac{\pi + \theta}{2} = \phi$

So the ratio of $\theta$ to $\phi$ is: $\pi + \theta = 2\phi$