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I stumbled upon this question while trying to solve another geometric problem. Consider a cartesian plane with three half-lines $l_1,l_2,l_3 $ starting at $(0,0)$ at angles $\theta_1,\theta_2,\theta_3$. The angle of a half-line is defined as the angle spanned by a ccw rotation of the positive x-axis until it reaches the half-line, therefore for every angle $\theta$ of a half-line $l$ starting at $(0,0)$ we have that $\theta \in [0,2\pi)$.

Consider the question of whether there exists a line $l$ through $(0,0)$ such that the lines $l_1,l_2,l_3$ are in the same half-plane with respect to $l$. I suppose there has to be some easy criterion to determine that however I'm having some trouble finding it. I would appreciate any help.

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  • $\begingroup$ In a quick thought, $max(\theta_1, \theta_2,\theta_3, \theta_l)-min(\theta_1, \theta_2,\theta_3, \theta_l) <=\pi$ $\endgroup$
    – Ripi2
    Commented Dec 14, 2019 at 17:42
  • $\begingroup$ @Ripi2 I dont think that works. Take $\theta_1=45^\circ , \theta_2=280^\circ ,\theta_3=290^\circ, \theta_l=90^\circ.$ Here $l$ works but your inequality does not hold. $\endgroup$
    – Sr52
    Commented Dec 14, 2019 at 17:54
  • $\begingroup$ Mmmm... I think it's better to work with angles between lines, like $\phi_ij=|\theta_i-\theta_j|$, and if some $\phi>\pi$ then $\phi=\phi-\pi$ $\endgroup$
    – Ripi2
    Commented Dec 14, 2019 at 17:58
  • $\begingroup$ I think that something in that direction could work. Suppose $\theta_1<\theta_2<\theta_3$ but take $\phi_{i,j}=min(|\theta_i-\theta_j|, 2\pi-|\theta_i-\theta_j|)$. Some criterion involving $\phi_{i,j} $ might work. $\endgroup$
    – Sr52
    Commented Dec 14, 2019 at 18:11
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    $\begingroup$ For angles $\theta_1=0º, \theta_2=135º, \theta_3=225º$ there's no such "half plane", so there's no possible half line $l$ $\endgroup$
    – Ripi2
    Commented Dec 14, 2019 at 19:47

2 Answers 2

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This happens if and only if the numbers $$\sin(\theta_2-\theta_1),\, \sin(\theta_3-\theta_2),\, \sin(\theta_1-\theta_3)$$ are neither all positive nor all negative.

This can be seen as follows. Take points $(\cos(\theta_k), \sin(\theta_k))$ for $k\in\{1,2,3\}$. Then these points lie not in the same half-plane if and only if the origin lies in the interior of the convex hull of these points. This happens exactly if the coordinates of the cross-product of $(\cos(\theta_1), \cos(\theta_2), \cos(\theta_3))$ and $(\sin(\theta_1), \sin(\theta_2), \sin(\theta_3))$ are either all positive or all negative. Writing out this cross-product and simplifying with trigonometric identities leads to the criterion above.

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  • $\begingroup$ Great idea to work with $sin\theta, cos\theta$ as well as the convex hull. Thanks! $\endgroup$
    – Sr52
    Commented Dec 15, 2019 at 16:01
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If the angle $\theta_l$ defines the half line $l$, then we have $\theta_{l1} = \theta_l-\pi/2$ and $\theta_{l2} = \theta_l+\pi/2$

If some of those angles is $<0$ or $>2\pi$ then adjust it to $[0,2\pi]$ range.
Let's assume $\theta_{l1} < \theta_{l2}$

Knowing if a given half line $l_i$ with angle $\theta_i$ is in the same half plane as the line $l$ is easy by just comparing: $$\theta_{l1} <= \theta_l <= \theta_{l2}$$

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