Example of a Problem Made Easier with Skew Coordinates Skew or oblique coordinate systems are coordinate systems where the angle between the axes is not 90 degrees. The second answer to this question has formulas to convert between these systems with an arbitrary angle, as well as a helpful diagram to illustrate the situation.  
Are there any examples of problems which would be hard (or at least harder) to solve in orthogonal coordinate systems, or at least the Cartesian coordinate system, but is reduced to an easy/easier problem when taken in skew coordinates? More generally, are there any applications of skew coordinate systems?
Since choosing a certain value of the skew angle would transform suitable parallelograms into rectangles, that could be a certain simplification to start from. The less trivial, the better.
 A: Several examples that use skew coordinates, questions & answers all very similar:

How prove this inequality $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$

How prove this inequality $\sum_{cyc}\frac{x+y}{\sqrt{x^2+xy+y^2+yz}}\ge 2+\sqrt{\frac{xy+yz+xz}{x^2+y^2+z^2}}$

How prove this inequality $\frac{2}{(a+b)(4-ab)}+\frac{2}{(b+c)(4-bc)}+\frac{2}{(a+c)(4-ac)}\ge 1$

How prove this inequality $\sum_{cyc}\frac{a^2}{b(a^2-ab+b^2)}\ge\frac{9}{a+b+c}$

This is the reference that describes the transformation:


Converting triangles to isosceles, equilateral or right???

The main result is repeated here for convenience:
$$
\left[ \begin{array}{c}   x - x_1 \\ y - y_1  \end{array} \right]  =
\left[ \begin{array}{cc} (x_2 - x_1)  &  (x_3 - x_1) \\
                         (y_2 - y_1)  &  (y_3 - y_1)
\end{array} \right]
\left[ \begin{array}{c}  \xi \\ \eta  \end{array} \right]
$$
Note that the triangle is half a parallelogram. And instead of
transforming a parallelogram into a rectangle, an arbitrary triangle is
transformed into a rectangular isosceles triangle. Therefore the above is in fact equivalent with any skew linear transformation (+ translation) :
$$
\left[ \begin{array}{c}   x \\ y \end{array} \right]  =
\left[ \begin{array}{cc}  \alpha  &  \beta \\
                          \gamma  &  \delta
\end{array} \right]
\left[ \begin{array}{c} \xi \\ \eta  \end{array} \right]
+ \left[ \begin{array}{c}  p \\ q \end{array} \right]
$$
If we define the coordinates of our transformed triangle as:$\,(x_1,y_1) = (p,q)\,$ , $\,x_2 = \alpha + p\,$ , $\,x_3 = \beta + p\,$ ,
$\,y_2 = \gamma + q\,$ , $\,y_3 = \delta + q\,$.


Note.   Via Finite Element interpolations, other parent polytopes
are associated with non-linear transformations. For example the standard
quadrilateral with vertices$(1)=(-1,-1)\, ,\, (2)=(+1,-1)\, ,\, (3)=(-1,+1)\, ,\, (4)=(+1,+1)\,$ has bilinear interpolation:
$$
   f = \frac{1}{4}(1-\xi)(1-\eta)f_1
     + \frac{1}{4}(1+\xi)(1-\eta)f_2
     + \frac{1}{4}(1-\xi)(1+\eta)f_3
     + \frac{1}{4}(1+\xi)(1+\eta)f_4 \\ 
     = \sum_{k=1}^4 N_k(\xi,\eta)\,f_k \qquad \mbox{with} \qquad N_k(\xi,\eta) = \frac{1}{4}(1\pm\xi)(1\pm\eta)
$$
The accompanying (isoparametric) transformation is found by replacing $f$ with
$x$ and $y$ :
$$
x = \sum_{k=1}^4 N_k(\xi,\eta)\,x_k \qquad ;
    \qquad y = \sum_{k=1}^4 N_k(\xi,\eta)\,y_k
$$
The bilinear transformation becomes linear again if the quadrilateral is a parallelogram ,
because then $\,x_1+x_4=x_2+x_3\,$ and $\,y_1+y_4=y_2+y_3\,$ (: diagonals property) , as substituted in:
$$
f = (f_1+f_2+f_3+f_4)/4 + \xi\,(-f_1+f_2-f_3+f_4)/4 + \eta
\, (-f_1-f_2+f_3+f_4)/4 \\
  + \xi\eta\,(f_1-f_2-f_3+f_4)
$$
and hence the last term becomes zero.
