Sliding Squares

Instead of the case when both squares are moving, consider the case when only one is moving twice faster and the other is stationary. Since the square is moving along the diagonal, we have $\Delta x = \Delta y$ .

The average area of overlap is then:

$\int_{0}^{1}x^{2}\,dx = \left.\dfrac{x^{3}}{3}\right|_{0}^{1} = \dfrac{1}{3}$

Let $x$ be the distance between the left side of the left unit square (and by symmetry, the right side of the right unit square) and let $s$ be the length of the overlapped square formed in the center.

Then comparing horizontal segments, we have $2 = 2x + (1 - s) + s + (1 - s)$ , which simplifies to $s = 2x$ , for an overlapping area of $A = 4x^2$ for $0 \leq x \leq \frac{1}{2}$ . Since the process is symmetrical, we only need to consider this interval.

The average area is then $\frac{\int_{0}^{\frac{1}{2}} 4x^2 dx}{\frac{1}{2} - 0} = \boxed{\frac{1}{3}}$ .

Let's introduce a time variable t and a(t)=the area at time t. Set t=0 when the red is at its lowest position and t=4 when the animation has completed a cycle. a(0)=0, a(1)=1. The second second t=1 to t=2 is just the reverse of the first second as the red reaches the upper corner and the final two seconds are the same as the first two. Each of the four seconds is symmetrical in terms of overlap, so we can just consider the overlap from t=0 to t=1.

The overlap area $a(t)=t^{2}$ . Integrating this from 0 to 1 gives the solution $\boxed{\frac{1}{3}}$

Let $x$ be the lenght of the overlapped square's side. Area changes with a function $g(x)$ such that:

$g(x)= \begin{Bmatrix} x^2 \; \; \; \; \; \; \; \; \; \; if \;\; 0 \leq x < 1 \\ (x-2)^2 \; if \; \; 1 \leq x \leq 2 \end{Bmatrix}$

It's clear that $\int_0^2g(x)dx=\int_0^2(x-1)^2dx$ :

Now, what the problem is asking is to find $f(c)$ where $f(x)=(x-1)^2$ such that $\int_0^2f(x)dx=(2-0)f(c)=2f(c)$ . (If you don't understand this, you may check Mean Value Theorem for Integrals , which states that if $f(x)$ is a continuous function on $[a,b]$ there exists $c \in [a,b]$ such that $\frac{1}{b-a} \int_a^bf(x)dx=f(c)$ ). Therefore,

$f(c)=\frac{1}{2}\int_0^2(x-1)^2dx=\frac{1}{2} \cdot \frac{2}{3}=\boxed{\frac{1}{3}}$

Is it overly simplistic to see it like this: at the beginning of each movement the overlap is 0%, at the middle 100% and at the end 0%. The overlap varies smoothly between these three, which are equally spaced, so the average overlap is the average of 0%, 100% and 0%, i.e. 33. 333...% Looks logical to me but am I oversimplifying the problem?

Simply taking <x^2> from 0 to 1, gives us the required Answer=(1/3)

It suffices to find the average until the squares coincide. If the corner of each square has moved a distance t along the diagonal, the overlapped area is one half times the product of the diagonals ( which have length 2t) which is 2 times t squared. If we integrate this from 0 to the square root of 2 divided by 2 and divide by the square root of 2 divided by 2 we get one third

At the beginning of the animation, the red square is in the top right corner. At this specific time, the area of overlap between the two squares is 0. Halfway along its path, the red square and blue square completely overlap. 100% of the red square is on the blue square, so $\frac{1}{1}$ , or 1 of the square overlaps. Lastly, when the red squares reaches the bottom left, the area of overlap between the two squares is 0. Taking the average of 0, 1, and 0 across these three time interveals yeilds $\frac{1}{3}$

I cheated, just a little. First I assumed that the squares move at constant speed. Now, the integrand is obviously the square of something, and integrating it will yield a third of something else. There was just one answer that had a 3 in the denominator.