How to average two rectangles that would produce a rectangle with the average area of the two shapes? I know that this question doesn't have a definite answer, but I'm wondering how someone would go about averaging two shapes to produce a shape that has the mean value of a calculated property of the two shapes.
Say you have a rectangle with sides 4 x 2 and another rectangle with sides 2 x 3. If you averaged the sides together you would get a rectangle that did not have an area that was equal to the mean of the two rectangles areas. But if you did want to find a shape like that how would you go about finding it in a way that was the most logical?
 A: Well, if you have $a\times b$ and $c\times d$ and you want an area of $mn= \frac {ab + cd}2$.
If we say $c = k*a$ for some scaling factor and $b= j*d$ for some scaling factor we get $ab = adj$ and $cd = adk$ and $mn =  \frac {ab + cd}2 = ad\frac {j + k}2$.  If we let $m = a\sqrt{\frac {\frac bd + \frac ca}2}$ and $n = d\sqrt{\frac {\frac bd + \frac ca}2}$ that would be an acceptable answer.
In you example of $4\times 2$ and $2 \times 3$ we'd get:
$4\sqrt{\frac {\frac 23 + \frac 12}2} \times 3\sqrt{\frac {\frac 23 + \frac 12}2} = $
$4\sqrt{\frac 7{12}}\times 3\sqrt {\frac 7{12}}$
We end up with the sides being between the two extremes, the area being the average of the area. And... well  $\frac ba = \frac bm \frac ma$ and $\frac dc = \frac d{n}\frac {n}c$.
A: I'm assuming that given two rectangles $A$ and $B$, with $A$ having width $A_w$ and height $A_h$, and $B$ having width $B_w$ and height $B_h$, you want to find a new rectangle, $C$ with width $C_w$ and height $C_h$, such that
$$
\frac{A_w + B_w}{2}=C_w,\\
\frac{A_h + B_h}{2}=C_h,\\
\frac{A_wA_h + B_wB_h}{2}=C_wC_h.
$$
If we fill in the first two definitions for $C_w$ and $C_h$ into the third line, we get
\begin{align*}
\frac{A_wA_h + B_wB_h}{2}&=\frac{A_w + B_w}{2}\cdot\frac{A_h + B_h}{2}=\frac{A_wA_h + A_wB_h+B_wA_h+B_wB_h}{4}\\
&=\frac{A_wA_h+B_wB_h}{2}+\frac{A_wB_h+B_wA_h}{4}-\frac{A_wA_h+B_wB_h}{4}.
\end{align*}
Therefore
\begin{align*}
\frac{A_wB_h+B_wA_h}{4}-\frac{A_wA_h+B_wB_h}{4}&=0\implies\\
A_wB_h+B_wA_h-A_wA_h-B_wB_h&=0 \implies\\
(A_w-B_w)(B_h-A_h)&=0.
\end{align*}
From this last line we see that our starting rectangles must have the same width or height (or both).
In your example a 4x2 and 2x3 rectangle did not meet your criteria, but a 2x4 and a 2x3 rectangle do. If you want to go about finding more of these types of rectangles systemetically, I would consider rectangles with width always less than ore equal to height. So no 3x2 rectangles, but instead 2x3 rectangles, etc.
