High School Area of Overlapping Squares

[bob012345]

TL;DR

Two squares overlap in a given configuration. Find the area of overlap.

Here is a little puzzle from the book 100 Geometric Games by
Pierre Berloquin.

The side of a small square is one meter long and the side of
a larger square one and a half meters long. One vertex of the
large square is at the center of the small square. The side of
the large square cuts two sides of the small square into one-
third parts and two-thirds parts.

What is the area where the squares overlap?

 

[kuruman]

A figure sure helps.

Spoiler

Let $a=0.5~$m. The area of the overlap is $0.25~\text{m}^2.$

 

[anuttarasammyak]

TL;DR: Two squares overlap in a given configuration. Find the area of overlap.

The side of
the large square cuts two sides of the small square into one-
third parts and two-thirds parts.

It seems not necessary. For any angle configuration we get the same result.

 

[bob012345]

It seems not necessary. For any angle configuration we get the same result.

The next level is if we let the length $s$ of the larger square vary, what range of values of $s$ will your statement not be true?

 

[anuttarasammyak]

$$s=\sqrt{2}/2$$is the minimum value satisfying it though not larger any more. And also $$s=\frac{1}{2\sqrt{2}} $$is the maximum.

 

[bob012345]

$$s=\sqrt{2}/2$$is the minimum value satisfying it though not larger any more.

Agreed. Now the larger square is the 1m square. The overlap depends on the relative orientation until $s$ shrinks to $\frac{1}{2\sqrt{2}}$ then it just becomes $s^2$.

Now it would be interesting to see for $s$ in that range what the overlap vs. angle function is.

 

[Gavran]

This is my interpretation of the problem statement in the original post.

Spoiler

 

[bob012345]

I worked out the overlap of the two squares as a function of angle. Given $a$ is the half edge length of the original square and $s$ is the full edge length of the second square with one corner at the center of the first ;

The area function goes as;

$$
A_{\text{overlap}} =
\begin{cases}
\mathbf{\text{For } s \ge \sqrt{2}a:} \\[10pt]
\quad \begin{cases}
a^2 &
\end{cases}
\\[20pt]
\mathbf{\text{For } a \le s \le \sqrt{2}a:} \text{ where } \theta_{\text{crit}} = \sin^{-1}\left(\frac{s}{a\sqrt{2}}\right) - \frac{\pi}{4}: \\[10pt]
\quad \begin{cases}
s^2 - \frac{\left(a - \sqrt{2}s \sin\left(\theta + \frac{\pi}{4}\right)\right)^2}{\sin(2\theta)} & \text{if } \theta_{\text{crit}} \le \theta \le \frac{\pi}{2} - \theta_{\text{crit}} \\[10pt]
a^2 & \text{otherwise}
\end{cases}
\\[20pt]
\mathbf{\text{For } a/\sqrt{2} \le s \le a:} \text{ where } \theta_{\text{crit}} = \sin^{-1}\left(\frac{a}{s\sqrt{2}}\right) - \frac{\pi}{4}: \\[10pt]
\quad \begin{cases}
s^2 - \frac{\left(a - \sqrt{2}s \sin\left(\theta + \frac{\pi}{4}\right)\right)^2}{\sin(2\theta)} & \text{if } \theta_{\text{crit}} \le \theta \le \frac{\pi}{2} - \theta_{\text{crit}} \\[10pt]
s^2 & \text{otherwise}
\end{cases}
\\[20pt]
\mathbf{\text{For } s \le a/\sqrt{2}:} \\[10pt]
\quad \begin{cases}
s^2 &
\end{cases}
\end{cases}
$$
Here is my Desmos page. This is set for $a=5$ and a range of $s$ values. Note the case when $s=a$ where the critical angle is 0 meaning the function covers the entire range of angles.

 

[anuttarasammyak]

The answers for the first case and the 4th case are symmetric for exchange of s and a. May we expect the same for the 2nd and the 3rd cases?

 

[bob012345]

The answers for the first case and the 4th case are symmetric for exchange of s and a. May we expect the same for the 2nd and the 3rd cases?

Interesting observation! It appears so if both values are within the range of the functions. Looking at $(a,s)$ being (4,5) vs, (5,4), the area of overlap is the same over the angles but they are different situations. Here are screen shots;

 

[anuttarasammyak]

The formula for the 2nd and the 3rd case is written as
$$\frac{a^2+s^2-as(\sin\theta + \cos\theta)}{2 \sin\theta \cos\theta}$$
which is obviously symmetric between s and a, and also between sin and cos, i.e. symmetric wrt y=x axis.. Thus we observe that the 3rd case solution comes from the 2nd case solution just by exchanging s and a.