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
$$2ab = cd$$
 A: Let $a,b,c,d$ be as described. Note that this implies
$$a,b,c,d>0$$
else we wouldn't have a rectangle
Now
$$ab=A_1=\frac{1}{2}A_2=\frac{1}{2}cd$$
$$2a+2b=P_1=2P_2=4c+4d$$
This gives us the set of equations
$$2ab=cd$$
$$a+b=2(c+d)$$
Now, this has an infinite number of solutions as we have four unknowns and two equations. Solving for $a$ and $c$ gives us
$$a=\frac{d (b-2 d)}{4 b-d}$$
$$c=\frac{2 b (b-2 d)}{4 b-d}$$
Since these are positive, we know
$$b>2d$$
$$d< 4b$$
which implies $b>2d$. However, we can also write this as 
$$b<2d$$
$$d>4b$$
which implies $d>4b$. Thus, one infinite family of solutions might be $b=3d$ which gives us
$$a=\frac{d}{11}$$
$$b=3d$$
$$c=\frac{6d}{11}$$
$$d=d$$
Then for any $d>0$ we get
$$A_1=ab=\frac{6d^2}{22}$$
$$A_2=cd=\frac{6d^2}{11}$$
$$P_1=2a+2b=\frac{68 d}{11}$$
$$P_2=2c+2d=\frac{34 d}{11}$$
which satisfy the original problem. For example, $a=1, b=33, c=6,$ and $d=11$ would give areas of $33$ and $66$ and perimeters of $68$ and $34$, respectively.
A: 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 $$(a,b)\mapsto \left(\frac{a+b-\sqrt{a^2-30ab+b^2}}4,\frac{a+b+\sqrt{a^2-30ab+b^2}}4\right)$$
