How fat is a triangle? The slimness factor of a geometric shape in 2 dimensions is the ratio between the side-length of its smallest containing square and its largest contained square. This is an important factor in computational geometry. So the slimness of a square is 1, the slimness of a circle is $\sqrt 2$, but what is the slimness factor of a given triangle?
After trying unsuccessfully to solve this geometrically, I decided to try a purely analytical solution. Here it is:
The triangle
The fatness obviously does not depend on scale. Hence two parameters are sufficient to define the triangle. I choose as parameters an angle $\theta$ adjacent to the longest side and the ratio $D$ between the side next to that angle and the longest side.  We have to find the fatness as a function of $\theta$ and $D$.
Normalize the triangle so that its longest side lies on the x axis, between $(0,0)$ and $(1,0)$. Define $\theta$ as the angle at the origin. Rotate the triangle such that $\theta$ is between $0^\circ$ and $90^\circ$. For brevity, let $C=\cos \theta$ and $S=\sin \theta$. $D$ is the length of the side starting at the origin, so that the 3rd vertex of the triangle is at $(DC,DS)$, where $0<D<1$, $0\leq C<1$ and $0\leq S<1$.
The slimness factor should be calculated as a function of the parameters $D$, $C$ and $S$. 
The containing square
Consider the unit square $[0,1]\times[0,1]$. We want to transform it so that it contains the triangle. A general transformation of a point $(x,y)$ is: $(rcx+rsy+h, rsx-rcy+v)$ where: $r$ is the dilation factor, $s$ and $c$ are sine and cosine of the rotation angle (we can assume they are both in $[0,1]$ because of the rotational symmetry of the square), $h$ is horizontal translation and $v$ is vertical translation. So a general transformation of the unit square has the following corners:


*

*$(h,v)$

*$(rc+h,rs+v)$

*$(rs+h,-rc+v)$

*$(rc+rs+h, rs-rc+v)$


and the following sides:


*

*$s(x-h)=c(y-v)$

*$s(x-h-rs)=c(y-v+rc)$ which is the same as $s(x-h)=c(y-v)+r$

*$c(x-h)=-s(y-v)$

*$c(x-h-rc)=-s(y-v-rs)$ which is the same as $c(x-h)=-s(y-v)+r$


In order to contain the triangle, each vertex $(x,y)$ of the triangle must be in the 4 half-planes defined by the 4 sides of the square:


*

*$c(y-v)\leq s(x-h)\leq c(y-v)+r$

*$-s(y-v)\leq c(x-h)\leq -s(y-v)+r$


Now, substitute each of the 3 vertices of the triangle and get, for $(0,0)$, $(1,0)$ and $(DC,DS)$ respectively:


*

*$-cv\leq -sh\leq -cv+r$

*$sv\leq -ch\leq sv+r$

*$-cv\leq s-sh\leq -cv+r$

*$sv\leq c-ch\leq sv+r$

*$c(DS-v)\leq s(DC-h)\leq c(DS-v)+r$

*$-s(DS-v)\leq c(DC-h)\leq -s(DS-v)+r$


There are a total of 12 inequalities here. Our goal is to find the smallest $r$ such that there are $c,s,h,v$ satisfying all these 12 inequalities.
Removing some redundant inequalities and ordering, we get:


*

*$r\geq s+cv-sh$

*$cv-sh\geq 0$

*$r\geq c-ch-sv$

*$-ch-sv\geq 0$

*$r\geq s(DC-h)-c(DS-v)=sDC-cDS+cv-sh$

*$r\geq c(DC-h)+s(DS-v)=cDC+sDS-ch-sv$


To make $r$ as small as possible, we should select $h$ and $v$ such that inequalities 2 and 4 become zero. Then we remain with the following inequalities which $r$ must satisfy:


*

*$r\geq s$

*$r\geq c$

*$r\geq cDC+sDS$


Here I am stuck: how can I solve this optimization problem?
The contained square
Again consider the unit square $[0,1]\times[0,1]$. Now we want to transform it so that it is contained in the triangle. The 4 corners of the square must satisfy the 3 inequalities dictated by the sides of the triangle, which are:


*

*$y\geq 0$

*$xDS\geq yDC$

*$(1-x)DS\geq (1-DC)y$


Substituting the 4 corners of the square gives the following 12 inequalities which r must satisfy, some of which are redundant:


*

*$-rc+v\geq 0$

*$v\geq 0$ (redundant)

*$rs+v\geq 0$ (redundant)

*$rs-rc+v\geq 0$ (redundant)

*$hDS\geq vDC$

*$(rc+h)DS\geq (rs+v)DC$

*$(rs+h)DS\geq (-rc+v)DC$ (redundant)

*$(rc+rs+h)DS\geq (rs-rc+v)DC$ (redundant)

*$(1-rc-h)DS\geq (1-DC)(rs+v)$

*$(1-h)DS\geq (1-DC)v$ (redundant)

*$(1-rc-rs-h)DS\geq (1-DC)(rs-rc+v)$

*$(1-rs-h)DS\geq (1-DC)(-rc+v)$ (redundant)


These imply the following 5 inequalities (note that this time the direction of the inequalities is inversed because we are looking to maximize $r$ subject to the inequalities):


*

*$cr \leq v$

*$0\leq hDS-vDC$

*$(sDC-cDS)r  \leq hDS-vDC$

*$(s-sDC+cDS)r\leq DS-v-hDS+vDC$

*$(s-c-sDC+cDC+sDS+cDS)r\leq DS-v-hDS+vDC$


Here, again, I am stuck...
The last step is just to divide the two $r$'s, but, how do I find each $r$?
Note
I am looking for a formula that gives fatness as a function of $\theta$ and $D$. However, if you think there is another pair of parameters by which it is more convenient to represent the fatness (e.g. the two smaller angles), then this is also welcome.
 A: I do not think using squares is a good idea. If you use instead the ratio $r/R$ of the inradius to the circumradius, you get a very clean formula:
$$
4\sin(A/2)\sin(B/2)\sin(C/2),
$$
where $A, B, C$ are the angles of the triangle. See for instance here. 
A: The containing square
If we imagine rotating the triangle by an angle $\theta$ and measuring the width (minimum $x$ coordinate to maximum $x$ coordinate, as could be measured with a vernier caliper),
we get a width function $w(\theta)$ which can be expressed in terms of the side lengths $s_1,s_2,s_3$ and side headings $\phi_1, \phi_2, \phi_3$ as
$$w(\theta)=\max\left(
s_1\left|\cos(\phi_1-\theta)\right|,
s_2\left|\cos(\phi_2-\theta)\right|,
s_3\left|\cos(\phi_3-\theta)\right|
\right).$$
(Note that the maximum of those three values, representing the side that extends from the minimum $x$ coordinate to the maximum $x$ coordinate, always equals the sum of the other two.)
A square whose sides have headings $\theta$ and $\theta+\pi/2$ can fit around the triangle iff its sides have length at least
$q(\theta)=\max\left(w(\theta),w(\theta+\pi/2)\right)$. 
The smallest containing square is $\min_\theta q(\theta)$.
Since $q$ is the maximum of six rectified sinusoids, its minimum must be at one of the values of $\theta$ where (at least) two maximal sinusoids meet. 
There are at most six such values in $[0,\pi/2]$ ($q$ is periodic with period $\pi/2$) so they can be enumerated to find the minimum.
Geometric configurations of the minimal square: For triangles having at most one angle below $\pi/4$, there are two possible configurations: (1) One triangle vertex is in a corner of the square, with the other two vertices on the two far sides, and/or (2) two vertices of the triangle are on one side of the square, with the third vertex  on the opposite side. 
Triangles with two angles below $\pi/4$ use a third configuration: (3) The square has the triangle's long side as its diagonal. 
For example, a triangle with $(s_1,s_2,s_3)=(6,9,9)$ yields configuration (1), while $(8,9,9)$ yields configuration (2), and $(14,9,9)$ yields configuration (3).
The contained square
Since every triangle is convex, the square is inside the triangle iff its four corners are in the triangle.
If a side of the triangle is untouched by the square's corners, a homothety (centered at the vertex where the two other sides meet) can enlarge the square while keeping it in the triangle.  Therefore, the maximal square's vertices touch all three sides of the triangle.
If two of the triangle sides are touched only at their common vertex (by square corner $A$), then the third triangle side can only be touched by corner $C$ (the corner diagonally opposite $A$). 
In this case, the square is not maximal, as it can be rotated and enlarged while staying inside the triangle. 
Therefore, at least three of the square's corners lie on the triangle.
If two corners of the square are on the same side (say $b$) of the triangle, then the square might be maximal. 
It is easy to find the largest square that sits on side $b$: Either $b$ is one of the short sides of an obtuse triangle, in which case the largest square has a vertex at the obtuse angle, or else the altitude $h_b$ to side $b$ lies inside the triangle, in which case the largest square has its other two vertices on the other two sides of the triangle, and has a side length of half the geometric mean of the lengths of $h_b$ and $b$, namely $s=\frac{h_b b}{h_b + b}$. 
By finding the largest square for each of the three sides of the triangle, we can easily find the largest of these three squares, which as we will soon see is indeed the maximal possible.
The only remaining case is that three corners of the square lie in the interiors of the three sides of the triangle, with the fourth corner strictly inside the triangle. 
In this case, the square can again be rotated and enlarged. 
This can be seen by considering the paths of the other corners of the square as two corners slide back and forth on two sides of the triangle:
The paths are elliptical. 
This means that the square can be moved so that two corners stay on the triangle, and (in at least one direction) the third corner will move into the interior of the triangle, so (like the first case above) the square is not maximal.
A: Only one circle can be superscribed on a given triangle (i.e. only one circle can pass through the corner points of a triangle) and only one circle can  be inscribed in any given triangle . If R1 and R2 are  the radii of the the two circles , superscribed and inscribed , respectively ,then the slimness factor will equal R2/R1.................................When you inscribe (or superscribe ) a square in a triangle ,there can be many squares that can be inscribed (or superscribed)into or contained by(or contain) a triangle .So the concept of inscribed and superscribed squares , cannot be used to define the slimness of a triangle -- as this not yield an unique value .
