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