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We are given a tetrahedron $ABCD$ with three known edges: $|AB|=a$, $|BC|=b$ and $|CD|=c$. We also know that all four lines that connect a vertex to the center of the incircle of the opposite face has a mutual point. Does this information define tetrahedron uniquely? If so, how do we find the rest of the edges?

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  • $\begingroup$ did you try anything? $\endgroup$ – Arnaldo Jan 10 '17 at 17:23
  • $\begingroup$ @ArnaldoNascimento, I don't see how to approach it. We can set $x=|AC|$, $y=|AD|$ and $z=|BD|$ as unknowns and try to derive three equations from the property but I don't see how yet. $\endgroup$ – Maxim Jan 10 '17 at 17:30
  • $\begingroup$ Barycentric coordinates! $\endgroup$ – Jack D'Aurizio Jan 10 '17 at 18:28
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The barycentric coordinates of the incenter are straightforward to find. If $A,B,C$ are three points in the plane and $a=BC, b=AC, c=AB$, the incenter $I$ fulfills $$ I = \frac{aA+bB+cC}{a+b+c} $$ but the same holds if $A,B,C,D$ are the vertices of a tetrahedron with $AB=a, BC=b, AC=c_1, BD=a_1, CD=c$. In such a case the incenter of $BCD$ lies at $$ I_{BCD} = \frac{a_1 C + b D + c B}{a_1+b+c} $$ and the incenter of $ABC$ lies at $$ I_{ABC} = \frac{a C + b A + c_1 B}{a+b+c_1}. $$ The lines $AI_{BCD}$ and $DI_{ABC}$ meet iff the four points $$ A,\qquad D,\qquad \frac{a_1 C + b D + c B}{a_1+b+c},\qquad \frac{a C + b A + c_1 B}{a+b+c_1}$$ are coplanar. By saying that through a determinant, iff $ac=a_1 c_1$.
Here it comes the real wonder:

In a tetrahedron $ABCD$, the lines joining each vertex with the incenter of the opposite face are concurrent (i.e. there is a sort of 3D-analogue of the Gergonne point) iff the product of the lenghts of opposite edges is constant.

enter image description here In our case, that gives $AD=\frac{ac}{b}$.

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