# The space of triangles that fit inside a given triangle, parametrized by edge lengths

Given a triangle T with sides a, b, and c, describe its "fitting set," the set of all points (x,y,z) in 3-dimensions for which a triangle with sides x, y, z exists that fits in T.

Such a set lies in the positive octant, is star-shaped with respect to the origin, and probably is a polyhedron; but it seems difficult to describe.

• Welcome to MO! I tried to make your title a bit more informative, hope that's all right. – j.c. Oct 17 '17 at 17:58
• This problem can be found in the book "100 problems" by Hugo Steinhaus, but without solution. – user82588 Oct 17 '17 at 18:22
• This space is not convex. Consider a triangle with sides $(1,1,2-\epsilon_1)$ for $0 < \epsilon_1 <1$. For small $\epsilon_2$, a triangle with sides $(1,1,\epsilon_2)$ will fit inside, but an equilateral triangle with sides $(1,1,1)$ will not. Another way to see failure of convexity is that if triangles of sides $(3/2,3/2,\epsilon), (3/2,\epsilon, 3/2), (\epsilon,3/2,3/2)$ fit, this does not imply an equilateral $(1,1,1)$ fits. – Douglas Zare Oct 17 '17 at 18:29
• It seems the case of two right angled triangles has been considered in the literature, see J.H. Jepsen, V. Vulpe "Fitting One Right Triangle in Another" maa.org/sites/default/files/jepsen6-200737759.pdf – user82588 Oct 17 '17 at 18:51
• @j.c. The isosceles triangle with legs of length $1$ and apex angle $30^\circ$ doesn't fit into the equilateral $(1,1,1)$. – MTyson Oct 17 '17 at 20:45

## 1 Answer

Here's the abstract of K.A. Post, "Triangle in a triangle: On a problem of Steinhaus", Geom Dedicata (1993) 45: 115; this paper was cited in the one given in the comment by Nemo.

A necessary and sufficient condition on the sides $p, q, r$ of a triangle $PQR$ and the sides $a, b, c$ of a triangle $ABC$ in order that $ABC$ contains a congruent copy of $PQR$ is the following: At least one of the 18 inequalities obtained by cyclic permutation of $\{a, b, c\}$ and arbitrary permutation of $\{p, q, r\}$ in the formula: \begin{array}{l} Max\{ F(q^2 + r^2 - p^2 ), F'(b^2 + c^2 - a^2 )\} \\ + Max\{ F(p^2 + r^2 - q^2 ), F'(a^2 + c^2 - b^2 )\} \le 2Fcr \\ \end{array} is satisfied. In this formula $F$ and $F'$ denote the surface areas of the triangles, i.e. \begin{array}{l} F = {\textstyle{1 \over 4}}(2a^2 b^2 + 2b^2 c^2 + 2c^2 a^2 - a^4 - b^4 - c^4 )^{1/2} \\ F' = {\textstyle{1 \over 4}}(2p^2 q^2 + 2q^2 r^2 + 2r^2 p^2 - p^4 - q^4 - r^4 )^{1/2} . \\ \end{array}

Here's a picture for the case $a=b=c=1$, which I plotted in this SageMath Jupyter notebook. Note that in this case we need only consider 3 of the above inequalities (plus the triangle inequalities): 