Folklore
$a$ and $b$ are the given sides of a triangle.
How to choose the third side $c$ so that the points of tangency of the incircle and the excircle with this side divide it into three equal segments?
For which $a$ and $b$ does such a side exist
(We consider the excircle that touches side $c$ and the extensions of sides $a$ and $b$.)
Consider the case $ab$ "symmetric" to the one discussed: $c=3(a-b)$, the solvability condition is $b<a<2b$.

## Answer

$c=3|a-b|$, when $b<a<2b$ or $a<b<2a$.