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Suppose we have a 5 tuple of positive real numbers $(l_1,l_2,m_1,m_2,m_3)$, with $m_i \in (0,\pi)$ for all $i$. Now fix a point $v_1$ in the hyperbolic plane. Then consider a geodesic of length $l_1$ starting at $v_1$. suppose that ends at $v_2$. At $v_2$ draw another geodesic of length $l_2$ which makes an angle $m_1$ at $v_2$ with first line. Suppose the second geodesic ends at the point $v_3$. Then draw another geodesic of length $l_1$ at $v_3$ making an angle $m_2$ with the second geodesic. Suppose end point of the third geodesic is $v_4$. Now draw another geodesic of length $l_2$ making an angle $m_3$ at $v_4$ with the third geodesic. Let the end point of the last geodesic is $v_5$.

Then can we put some condition on the tuple to get $v_5=v_1$? If yes what that condition should be?

Thanks in advance.

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1 Answer 1

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Your broken line closes iff the triangles $(v_1,v_2,v_3)$ and $(v_3,v_4,v_5)$ are equal, and $[v_1,v_3]$ is their common side. So there are two conditions: one is that $m_1=m_3$, by the hyperbolic law of cosines.

To state the second one, Apply the hyperbolic law of cosines to find the length $x$ of the side $[v_1,v_3]$, then the hyperbolic law of sines to find the angles of the two triangles at $v_3$, and write the condition that the sum of these angles at $v_3$ is equal to your $m_2$.

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  • $\begingroup$ Did you mean the triangles $(v_1,v_2,v_3$ and $(v_2,v_3,v_5)$? $\endgroup$
    – KAK
    Oct 5, 2022 at 2:45
  • $\begingroup$ @KAK: I corrected. $\endgroup$ Oct 5, 2022 at 14:41
  • $\begingroup$ what will happen if we want to construct an octagon instead of a square? Means we have a tuple $(l_1,l_2,l_3,l_4,m_1,m_2,\cdots,m_7)$ and we form a polygonal line with these parameters then what condition should we impose to get a closed polygon? $\endgroup$
    – KAK
    Oct 13, 2022 at 10:36

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