0
$\begingroup$

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:enter image description here

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – AugSB May 27 '15 at 11:52
  • $\begingroup$ @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. $\endgroup$ – Jonathan Mee May 27 '15 at 14:47
0
$\begingroup$

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: $\pi + \theta = 2\phi$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.