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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 $$2ab = cd$$

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Baba.S
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  • What have you tried? Can you write down the equations involving $a,b,c,d$ which follow from the scenario you describe? – KReiser Dec 29 '19 at 23:13
  • I have edited the question showing the equations that follow. I honestly don't know what to do next. I assume that if I substitute one into the other, then the two constraints regarding perimeter and area will be joined together so that when I solve for any other variable it will account for those two constraints – Baba.S Dec 29 '19 at 23:18
  • What do you mean by "solved"? What problem is to be "solved"? you give a statement of fact: "Two rectangles: the 1st has twice the perimeter of the 2nd and the 2nd has twice the area of the 1st" but there is no question asked and no problem to be solve! What is it that you want to do? – user247327 Dec 29 '19 at 23:28
  • I asked my question in the main body rather than the title because otherwise it's too long. By solve I mean how can we come up with a systematic way of drawing these two rectangles that fit the brief using algebra. – Baba.S Dec 29 '19 at 23:42

2 Answers2

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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.

Soham Konar
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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 $$(a,b)\mapsto \left(\frac{a+b-\sqrt{a^2-30ab+b^2}}4,\frac{a+b+\sqrt{a^2-30ab+b^2}}4\right)$$