Relation between sides of three triangles

Let $\Delta_1, \Delta_2$ and $\Delta_3$ be three triangles.

Triangle $\Delta_1$ has sides equal to $x,y$ and $l_1$. Triangle $\Delta_2$ has sides equal to $y,z$ and $l_2$, and triangle $\Delta_3$ has sides $x,z$ and $l_3$. My question is:

For known values of $l_1, l_2$ and $l_3$, what is the relation between $x,y$ and $z$?

I cannot use the law of cosines because I have no information about angles. So, I do not know how to deal with this problem.

Thanks in advance for any help!

• The only information you can get is what you get from the $9$ triangle inequalities (three for each triangle). Are there particular numerical values of $l_1,l_2,l_3$ that interest you? – quasi Feb 10 '17 at 11:27
• @quasi, No particular values of $l_1,l_2$ and $l_3$. – Alex Silva Feb 10 '17 at 12:47
• @Khosrotash, I have no other inofrmation about the triangles. – Alex Silva Feb 10 '17 at 12:47
• @Alex Silva: Where does the problem come from? Or what does it relate to? – quasi Feb 10 '17 at 12:49

You can recover the triangle identity for $\ell_1, \ell_2, \ell_3$, since you can glue the three triangles with the corresponding sides together and obtain a tetrahedron whose fourth side is the triangle with sides $\ell_1, \ell_2, \ell_3$.
• Associating a tetrahedron with the three triangles is a cool idea. But I have 2 questions: (1) What do you mean by "the triangle identity for $\ell_1, \ell_2, \ell_3$"?; (2) The question asks for the relation between $x,y,z$ when $\ell_1, \ell_2, \ell_3$" are known. Aren't you reversing the question? – quasi Feb 10 '17 at 12:26
• I do not understand either what you mean by "recover the triangle identity". Moreover, you cannot ensure that $l_1, l_2$ and $l_3$ form a triangle. – Alex Silva Feb 10 '17 at 12:49
• @quasi Oh yeah, I've reversed the $\ell$'s and $x,y,z$..but still, you can glue those triangles into a tetrahedron. I mean all the three relations $\ell_i + \ell_j \leq \ell_k$.. – pepa.dvorak Feb 10 '17 at 13:02
• But there will be some more information we can obtain, since if you maintain the idea of a tetrahedron, the assignment tells we have a given base of the solid and now, for given values of $x,y$ (i.e. for given another face of the solid), there will be a lower bound $b(x,y)$ of $z$ to close the tetrahedron. – pepa.dvorak Feb 10 '17 at 13:23