<s>[INST] Two rectangles: The $1$st has twice the perimeter of the $2$nd and the $2$nd has twice the area of the $1$st. How can this be solved using just algebra, where the first rectangle has sides $a$ and $b$, and the second rectangle has sides $c$ and $d$?
These are the two equations that follow:
$$a + b = 2(c + d)$$
and
$$2ab = cd$$
 [/INST] Given $a,b,c,d>0$, the condition is $$\begin{cases}c+d=\frac{a+b}2\\ cd=2ab\end{cases}$$
Let's adopt the convention $c\le d$, so that the system is satisfied if and only if $$\begin{cases}c=\frac{a+b}4-\frac12\sqrt{\left(\frac{a+b}2\right)^2-8ab}\\ d=\frac{a+b}4+\frac12\sqrt{\left(\frac{a+b}2\right)^2-8ab}\end{cases}\\ \begin{cases}c=\frac{a+b-\sqrt{a^2-30ab+b^2}}4\\ d=\frac{a+b+\sqrt{a^2-30ab+b^2}}4\end{cases}$$
Notice that, under the hypothesis, both those numbers are strictly positive, provided that the square root exists, i.e. provided that $a^2-30ab+b^2\ge 0$. Under the condition $a,b>0$ this is the case if and only if $\frac ab\le 15-4\sqrt{14}$ or $\frac ab\ge15+4\sqrt{14}$.
Therefore the set of solutions $(c,d)$ with $c\le d$ is parametrised by the region of plane $\begin{cases} \frac ab\ge 15+4\sqrt{14}\\ b>0 \end{cases}\lor\begin{cases} \frac ab\le 15-4\sqrt{14}\\ b>0 \end{cases}$ via the function $$(a,b)\mapsto \left(\frac{a+b-\sqrt{a^2-30ab+b^2}}4,\frac{a+b+\sqrt{a^2-30ab+b^2}}4\right)$$
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