Can four different sized yet proportional rectangles fit into a square?

Question

Can anyone find a way to prove that four quadrilaterals of different sizes, but equally proportional dimensions can fit exactly into a square.

So $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{g}{e}$

and, of course, b = a+c = d+e = a+f+g etc.

You can set any dimension to "1", if that helps you.

From the image I made it sure seems that it can be done, but I haven't been able to verify it mathematically yet. Can you? What lengths would those sides be?

As a bonus, can you find any other shape so that four of them, all different sizes but proportional can fill a square? Something along the lines of this crude example:

Answer 1

The solution shown here follows the hint given by @CalvinLin, has a different numerical value to the other answer.

Suppose the configuration of the rectangles is as shown:

As the rectangles are similar to each other, each has a ratio of $1:x$.

We let $AD = 1$ and $DG = x$ for the moment.

Since the rectangles are distinct, we must have $AB = x^2$.

Here comes the fun part. There are two possible values of $GI$, since both $\dfrac {GI}{GC} = x$ or $\dfrac {GC}{GI} = x$ are valid.

Since $GC = 1+x^2$, $GI = \dfrac 1x + x$ or $x + x^3$.

Similarly, there are two values of $BK$ for each value of $GI$.

If $GI = \dfrac 1x + x$, $DI = \dfrac 1x + 2x$, and we could have $BK = 1 + 2x^2$ or $\dfrac 1{x^2} + 2$.

If $GI = x + x^3$, $DI = 2x + x^3$, and we could have $BK = 2x^2 + x^4$ or $2 + x^2$.

Since a square is formed, we have $DI = DK$. This results in four equations, two cubic and two quartic:

\begin{align} \frac 1x + 2x &= 2 + 3x^2\\ \frac 1x + 2x &= \frac 1{x^2} + 3 + x^2\\ 2x + x^3 &= 1 + 3x^2 + x^4\\ 2x + x^3 &= 3 + 2x^2 \end{align}

(Un)surprisingly, the pairs of equations with the same degree are related by a substitution $x \mapsto \frac 1x$.

The quartic equation $2x + x^3 = 1 + 3x^2 + x^4$ can be written as: $$x^4 - x^3 + \left(\frac14x^2 + \frac74 x^2 + x^2\right) - 2x + 1 = 0$$ $$(x-1)^2 + x^2(x-\frac 12)^2 + \frac74x^2 = 0$$ which is a sum of squares that cannot be zero at the same time, so this quartic equation has no solutions.

The unique real solution to the cubic equation $2x + x^3 = 3 + 2x^2$ is:

$$\frac 13 \left(2 - 2 \sqrt[3]{\frac2{61 + 3 \sqrt{417}}} + \sqrt[3]{\frac12 (61 + 3 \sqrt{417})}\right) \approx 1.8105357$$

Correspondingly we have, approximately: \begin{align} AD &= 1 &\text{or} \quad0.1046454\\ DG &= 1.8105357 &\text{or} \quad0.1894643\\ AB &= 3.2780396 &\text{or} \quad0.3430319\\ GC &= 4.2780396 &\text{or} \quad0.4476773\\ GI &= 7.7455434 &\text{or} \quad0.8105357\\ BK &= 5.2780396 &\text{or} \quad0.5523227\\ KL &= 9.5560791 &\text{or} \quad1 \end{align}

The second column is the side lengths of each rectangle if the side length $KL$ of the square is set to $1$, and we can check that: $$\frac {DG}{GA} = \frac {AB}{DG} = \frac {GI}{GC} = \frac {KL}{BK} \approx 1.8105357$$

and the diagram above is constructed using this data.

Answer 2

Consider, for the moment, only the three rectangles with side lengths $a,b$ and $c,d$ and the rectangle formed by removing these two from your diagram. I claim that, in this configuration, all of the labels can be replaced with functions of $a$, as follows

Now, I claim that the ratio of the side lengths of the third rectangle is given by $$f(a)=\frac{1-\frac{1-a}{a}}{1-a}$$

Suppose then, that $f(a)=na$, then we could partition the remainder into 2 rectangles of the appropriate side length. This gives a cubic equation for each $n$, and general solutions are known for this case, but ugly. Setting $n=2$ in wolfram alpha gives