Brain teaser

[elioruzan]

Hi,
I stumbled upon this question on Facebook and I tried to solve it unsuccessfully.
I saw in the comments people talking about symmetry in their solutions but I didn't understand them.
To me it seems there is insufficient info because the heights and widths of A and B share constraints, but there are still 2 variables.
Any help would be greatly appreciated.

 

[lev888]

Not sure which symmetry they had in mind... I would try the following approach:
Pick an arbitrary height for A, call it h. What's the area of A? Now, what is the maximum area of B? It will be a function of h. What's the total area? Again, this is a function of h. Maximize it.

 

[blamocur]

I haven't solved it, but using Lagrange multipliers looks promising.

 

[fresh_42]

If you use polar coordinates you will only have two angles as variables and $f(\varphi_a,\varphi_b)=\sin(2\varphi_a)+\sin(2\varphi_b)-2\sin(\varphi_a )\cos(\varphi_b)$ but the solution that WA has found is a bit strange (~78,7%):

maximize (sin(2x) +sin(2y) - 2*cos(y) *sin(x) ) where 0<x< pi/2 and 0<y< pi/2 - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com

 

[mario99]

Hi,
I stumbled upon this question on Facebook and I tried to solve it unsuccessfully.
I saw in the comments people talking about symmetry in their solutions but I didn't understand them.
To me it seems there is insufficient info because the heights and widths of A and B share constraints, but there are still 2 variables.
Any help would be greatly appreciated.

You can solve the problem by symmetry or Lagrange multiplier. But I prefer polar coordinate.

Let $A_1$ be the area of the first rectangle and let $A_2$ be the area of the second rectangle.

$A_1 = x_1y_1$
$A_2 = x_2y_2$

$x = r\cos\theta$
$y = r\sin\theta$

Then

$A_1 = r^2\cos\theta_1\sin\theta_1$
$A_2 = r^2\cos\theta_2\sin\theta_2$

Or

$\displaystyle A_1 = \frac{r^2}{2}\sin2\theta_1$

$\displaystyle A_2 = \frac{r^2}{2}\sin2\theta_2$

$\displaystyle A = A_1 + A_2 = \frac{r^2}{2}\sin2\theta_1 + \frac{r^2}{2}\sin2\theta_2 = \frac{r^2}{2}(\sin2\theta_1 + \sin2\theta_2)$

Now think about it. To get the maximum area, you want $\sin2\theta_1 + \sin2\theta_2$ to be maximum.

Also you must have noticed that because of the symmetry of the two rectangles (they have the same area), $\theta_1 = \theta_2$.

$\displaystyle A = \frac{r^2}{2}(\sin2\theta + \sin2\theta) = r^2\sin2\theta = r^2$

Finding the angle $\theta$ will help you to find the length of each leg of the rectangle by substituting back in $x$ and $y$.

 

[fresh_42]

Where is your negative term? I have
$$\begin{array}{lll} A&=\sin \varphi_a\cdot \cos \varphi_a\\ B&=\cos \varphi_b \cdot (\sin \varphi_b -\sin \varphi_a)\\ 2A&=2\sin \varphi_a\cdot \cos \varphi_a=\sin(2\varphi_a)\\ 2B&=2\cos \varphi_b \cdot \sin \varphi_b- 2\cos \varphi_b\sin \varphi_a\\ &=\sin(2\varphi_b)- 2\cos \varphi_b\sin \varphi_a\\ 2A&=\sin(2\varphi_b)- 2\cos \varphi_b\sin \varphi_a+\sin(2\varphi_a) \end{array}$$and WA says it is $\max \{2A\} \approx 1,23607 \text{ at } (\varphi_a, \varphi_b) \approx (0.553574, 1.01722)$ which covers about $78.7\%$ of the area.

 

[mario99]

Where is your negative term? I have
$$\begin{array}{lll} A&=\sin \varphi_a\cdot \cos \varphi_a\\ B&=\cos \varphi_b \cdot (\sin \varphi_b -\sin \varphi_a)\\ 2A&=2\sin \varphi_a\cdot \cos \varphi_a=\sin(2\varphi_a)\\ 2B&=2\cos \varphi_b \cdot \sin \varphi_b- 2\cos \varphi_b\sin \varphi_a\\ &=\sin(2\varphi_b)- 2\cos \varphi_b\sin \varphi_a\\ 2A&=\sin(2\varphi_b)- 2\cos \varphi_b\sin \varphi_a+\sin(2\varphi_a) \end{array}$$and WA says it is $\max \{2A\} \approx 1,23607 \text{ at } (\varphi_a, \varphi_b) \approx (0.553574, 1.01722)$ which covers about $78.7\%$ of the area.

I apologize. I have just realized that my approach will let the two rectangles overlapping which is wrong!

 

[mario99]

You can solve the problem by symmetry or Lagrange multiplier. But I prefer polar coordinate.

Let $A_1$ be the area of the first rectangle and let $A_2$ be the area of the second rectangle.

$A_1 = x_1y_1$
$A_2 = x_2y_2$

$x = r\cos\theta$
$y = r\sin\theta$

Then

$A_1 = r^2\cos\theta_1\sin\theta_1$
$A_2 = r^2\cos\theta_2\sin\theta_2$

Or

$\displaystyle A_1 = \frac{r^2}{2}\sin2\theta_1$

$\displaystyle A_2 = \frac{r^2}{2}\sin2\theta_2$

$\displaystyle A = A_1 + A_2 = \frac{r^2}{2}\sin2\theta_1 + \frac{r^2}{2}\sin2\theta_2 = \frac{r^2}{2}(\sin2\theta_1 + \sin2\theta_2)$

Now think about it. To get the maximum area, you want $\sin2\theta_1 + \sin2\theta_2$ to be maximum.

Also you must have noticed that because of the symmetry of the two rectangles (they have the same area), $\theta_1 = \theta_2$.

$\displaystyle A = \frac{r^2}{2}(\sin2\theta + \sin2\theta) = r^2\sin2\theta = r^2$

Finding the angle $\theta$ will help you to find the length of each leg of the rectangle by substituting back in $x$ and $y$.

I will do my calculations again, but this time I will get rid of the overlapping area.

Let $A_1$ be the area of the first rectangle and let $A_2$ be the area of the second rectangle.

$A_1 = x_1y_1$
$A_2 = x_2y_2 - x_2y_1$



$x = r\cos\theta$
$y = r\sin\theta$

Then

$A_1 = r^2\cos\theta_1\sin\theta_1$
$A_2 = r^2\cos\theta_2\sin\theta_2 - r^2\cos\theta_2\sin\theta_1$

Or

$\displaystyle A_1 = \frac{r^2}{2}\sin2\theta_1$

$\displaystyle A_2 = \frac{r^2}{2}\sin2\theta_2 - r^2\cos\theta_2\sin\theta_1$

$\displaystyle A = A_1 + A_2 = \frac{r^2}{2}\sin2\theta_1 + \frac{r^2}{2}\sin2\theta_2 - r^2\cos\theta_2\sin\theta_1$

Now let us assume that the two rectangles are inside a unit circle in the first quadrant.

$\displaystyle A = \frac{1}{2}(\sin2\theta_1 + \sin2\theta_2) - \cos\theta_2\sin\theta_1$

It is not easy to find by hand when $A$ is a maximum, so you will have to use a CAS, W|A, or any other approximation methods.

I get $\theta_1 \approx 0.553574 \ \ \ \$ and $\ \ \ \ \theta_2 \approx 1.01722$

So, the maximum area of the two rectangles when $r = 1$ is:

$\displaystyle A = \frac{1}{2}(\sin2\theta_1 + \sin2\theta_2) - \cos\theta_2\sin\theta_1$

where $\theta_1$ and $\theta_2$ as mentioned above.

Or in general, the maximum area is:

$\displaystyle A = \frac{r^2}{2}(\sin2\theta_1 + \sin2\theta_2 - 2\cos\theta_2\sin\theta_1)$

 

[mario99]

Or in general, the maximum area is:

$\displaystyle A = \frac{r^2}{2}(\sin2\theta_1 + \sin2\theta_2 - 2\cos\theta_2\sin\theta_1)$

Just to be clear about my last sentence. I meant that this is the general area of the two rectangles (inscribed inside a circle in the first quadrant) and if you want to maximize it, choose a radius and then use a CAS, W|A, or any other approximation method to get the angles.

 

[blamocur]

I get $\theta_1 \approx 0.553574$ and $\theta_2 \approx 1.01722$

BTW, $\tan \theta_1 = \frac{\sqrt{5}-1}{2}, \tan\theta_2 = \frac{\sqrt{5}+1}{2}$ and $\theta_1 + \theta_2 = \frac{\pi}{2}$, i.e. angles of the golden ratio.

 

[mario99]

BTW, $\tan \theta_1 = \frac{\sqrt{5}-1}{2}, \tan\theta_2 = \frac{\sqrt{5}+1}{2}$ and $\theta_1 + \theta_2 = \frac{\pi}{2}$, i.e. angles of the golden ratio.

Beautiful.

I did not even think about it. Nice catch!



And here is the magic: The maximum area is:

$\large A_{\text{max}} = r^2\left(\frac{\sqrt{5} - 1}{2}\right)$

where $\large r$ is the radius of the arc.

Thanks to the golden ratio!

Did anyone arrive to this conclusion?

 

[fresh_42]

BTW, $\tan \theta_1 = \frac{\sqrt{5}-1}{2}, \tan\theta_2 = \frac{\sqrt{5}+1}{2}$ and $\theta_1 + \theta_2 = \frac{\pi}{2}$, i.e. angles of the golden ratio.

Beautiful. Is there a mathematical derivation or did you get there by the numeric solutions?

I had the following chain of associations: golden ratio --> Fibanocci numbers --> Archimedean spiral --> squaring the circle.



(Image source: https://de.wikipedia.org/wiki/Archimedische_Spirale)

I have no idea whether this works for the given problem statement, but whenever Fibonacci numbers appear, there is likely something mathematical behind them.

 

[blamocur]

Beautiful. Is there a mathematical derivation or did you get there by the numeric solutions?

I got numerical solutions first, then figured out formal derivation, but the latter is somewhat hairy -- I might try cleaning it up and posting later.

 

[fresh_42]

I couldn't follow your argument about the golden ratio, but I found out how to get the exact values if we assume that $\varphi_a+\varphi_b=\dfrac{\pi}{2}.$ This appears to be the "symmetry argument" the OP spoke about. Here it goes:

Assume $\varphi_a+\varphi_b=\dfrac{\pi}{2}$ and let $f=f(\varphi_a,\varphi_b)=A+B.$ Then
$$\begin{array}{lll} f&=\sin \varphi_a\cdot \cos \varphi_a+\cos \varphi_b \cdot (\sin \varphi_b -\sin \varphi_a)\\ &=\sin\varphi_a\cdot\cos\varphi_a+\cos((\pi/2)-\varphi_a)\cdot(\sin((\pi/2)-\varphi_a)-\sin\varphi_a)\\ &=\sin\varphi_a\cdot\cos\varphi_a+\sin\varphi_a \cdot (\cos\varphi_a-\sin\varphi_a)\\ &=\sin(2\varphi_a)-\sin^2\varphi_a=\sin(2\varphi_a)-1+\cos^2\varphi_a\\ &=\sin(2\varphi_a)-1+\dfrac{1}{2}(1+\cos(2\varphi_a))\\ 2f&=-1 + \cos(2 \varphi_a) + 2 \sin(2 \varphi_a)\\ 2f'&=-2\sin(2\varphi_a)+2\cos(2\varphi_a)=0 \Longrightarrow \tan(2\varphi_a)=2\Longrightarrow \varphi_a=\dfrac{1}{2}\tan^{-1}(2)\, , \,\varphi_b=\dfrac{\pi}{2}-\dfrac{1}{2}\tan^{-1}(2) \end{array}$$
This yields the numerical values
$$\max\{2f\} \approx 1,23607\text{ and }f \approx 0.618035\approx 78.7\% \cdot \dfrac{\pi}{4}\text{ at } (\varphi_a, \varphi_b)\approx (0,553574, 1,01722).$$
Remains to prove why the sum of the two angles to the upper right corners has to be $90°.$

And, of course, why $$\tan\left(\dfrac{1}{2}\tan^{-1}(2)\right) = \dfrac{\sqrt{5}-1}{2},$$ and
$$\tan\left(\dfrac{\pi}{2}-\dfrac{1}{2}\tan^{-1}(2)\right) = \dfrac{\sqrt{5}+1}{2}.$$

 

[The Highlander]

Did anyone arrive to this conclusion?

Yes, the Japanese blogger who claims to have "created" the problem did arrive at the same answer, here.
(Based on some unsubstantiated assumptions).

I have also attached an English version of his post (for anyone whose browser doesn't offer to translate it).

PS: I note that you have generalized your solution () but the original problem did specify that the radius was 1.

 

[fresh_42]

And, of course, why $$\tan\left(\dfrac{1}{2}\tan^{-1}(2)\right) = \dfrac{\sqrt{5}-1}{2},$$ and
$$\tan\left(\dfrac{\pi}{2}-\dfrac{1}{2}\tan^{-1}(2)\right) = \dfrac{\sqrt{5}+1}{2}.$$

At least this part was easy:

Let $c=\dfrac{1}{2}\tan^{-1}(2)$ and $\tan(c)=x.$ Then
$$\begin{array}{lll} \tan(2c)&=2=\dfrac{2\tan(c)}{1-\tan^2(c)}=\dfrac{2x}{1-x^2}\\ 0&=x^2+x-1\Longrightarrow x_{1,2}=-\dfrac{1}{2}\pm\sqrt{\dfrac{1}{4}+1}=\dfrac{-1\pm\sqrt{5}}{2}\\ \tan(c)&=\dfrac{\sqrt{5}-1}{2}\\[12pt] \tan\left(\dfrac{\pi}{2}-c\right)&=-\tan\left(c-\dfrac{\pi}{2}\right)=\cot(c)=\dfrac{1}{\tan(c)}=\dfrac{1}{x}\\ &=\dfrac{2}{\sqrt{5}-1}=\dfrac{2(\sqrt{5}+1)}{5-1}=\dfrac{\sqrt{5}+1}{2} \end{array}$$
Now for the "symmetry argument"...