In the course of the evening, we noodled on this for a while and initially made no progress. So I decided to model it in geogebra to confirm whether there was a unique solution. What I found visually was that there appeared to be one when the smaller box was tilted 45 degrees. And that was in fact not hard to confirm because the triangle DFA inscribes the diameter in that case, along with the triangle FGA. You can then solve with the Pythagorean theorem and verify that they give the same result in this configuration.
\(\text{the diameter: } AF^2 = AD^2 + DF^2 = 16 + (4 + 2\sqrt{2})^2 = 40 + 16 \sqrt{2}\) \(\text{and also the diameter: } AF^2 = FG^2 + AG^2 = 4 + (2 + 4 \sqrt{2})^2 = 40 + 16 \sqrt{2}\)
But this isn’t entirely satisfying because it basically presumes that answer and confirms it works and it doesn’t settle whether this is the only answer.
Trying to prove the configuration
One of the central things the above process used was the inscribed right angles embedded in the circle at D and G. This seemed like an important part of the solution. I tried out a bunch of variants including adding lines through the center of the circle to middle of the squares and to the outer corners. Nothing seemed to prove that interior angle was 45 like I hoped. So I turned to a different approach: trying to prove uniqueness of the solution and settling for a known solution for existence.
The Sprirograph Approach
Imagine taking the smaller box and rotating it fully around the circle. Because the box is rigid and a fixed length you end up tracing a smaller circle like below
Its clear the smaller square can only intersect when either one of the corners overlaps point C and only two configuration work for that case. Every other position will place the ends of the square at other locations on the smaller circle. This provides a uniqueness proof of sorts* Although it actually suggests there are 2 configurations depending on which corner of the two corners of the small square is used and the 45 degree rotation is only one of them.
A difficult construction that works
Later on I realized that the inner circle was interesting because it created another inscribed angle at C.
This creates the triangle HCN and if we let r be the radius of the inner circle: $(2r)^2 = 4^2 + CN^2$. And CN is discoverable using the radius of the outer circle which I’ll call R.
From triangle AOQ we can get $R^2 = AQ^2 + 2^2$ and $AQ = 4 - \frac{1}{2}CN$
All I needed was one more relationship between r and R to solve. So I introduced the addtional triangle OPF from the first figure. That has a Pythagorean relationship $R^2 = 1^2 + OP^2$ and we can OP in terms of r as well by introducing yet another triangle.
Triangle OQE gives us $r^2 = OQ^2 + 1^2$ and $OP = OQ + 2$. So we eventually end up with a more complicated than desired system but it is solvable:
\[4r^2 = 16 + (R^2 - 4) - 8\sqrt{R^2 - 4} + 16\] \[R^2 = 1 + (r^2 - 1) + 4 \sqrt{r^2 -1} + 4\]*I’ll leave the algebra as an exercise for the reader.
Breakthrough
That felt decent but more complicated than should be possible so eventually I went back to a simpler variant of on my original experiments: looking at the intersecting chords formed by the two squares.
Let consider extending CD to F’ and CG to A’. Note we don’t know if A’ = A or F’ = F yet. This gives us two intersecting chords A’G and DF’.
These must satisfy the relationship: $DC \cdot CF’ = CG \cdot CA’$ which simplifies to $2CF’ = CA’$
Let w = CF’ for the rest of this proof. We now can go back to our original inscribed angles. We don’t know if they overlap each other this time but each inscribed angle D and G defines a triangle on a diameter of the circle and those diameters are the same length. So:
\[4^2 + (4 + w)^2 = 2^2 + (2 + 2w)^2\]That’s a lot easier than the previous systems and simplifies to $3w^2=24$ or $w=2\sqrt{2}$ Note that’s exactly the diagonal of our smaller square as expected.
We don’t have to really prove this is the case because w is enough to calculate the radius by just plugging into the Pythagorean formula for either inscribing triangle like I did initially.
But for fun let’s look at all points that are $4\sqrt2$ from C and and since that forms a circle there must be a second intersection.
Clearly that second intersection isn’t collinear with either of the segments CD or CG and the diagonal on AC really is.
But if we extend to see what it’s connecting to - voila the second case I alluded to at the top pops out. We can also have the smaller triangle touching and slightly overlapping the larger one.
Why things are vital this time - The extraordinary growth of the network
On the other hand, there are plenty of Inquiry Based Learning advocates that are all in with problems on a daily basis. Its easy to find a class vignette with everyone standing around whiteboards working on some communal task that will take a class period. All of which seems delightful but then the sneaking suspicions enter about the amount of mathematical ground being covered or if enough practice is occurring for growth to occur. After seeing yet another task involving rearranging the digits 1-10 to fit into some equations I often end up thinking how its very easy to idle in the domain of pre-algebra and never really get anywhere.
Instead, I inhabit a middle ground. Time is limited but its important to constantly encounter problems through school in the service of developing patterns of thinking. At the same time, there needs to be a large amount of exercises along the way or you flounder without ever mastering concepts confidently enough to utilize them later. That’s a very tricky space given the constraints. So I think you always have to be strategic in what you choose not only in how to to balance the hours between longer problems and practice exercises but what forms each take. To keep to the flow, the problems should riff on the concepts being worked on in an exercise or require one to practice a skill in the service of the solution so you are getting double duty out of it. They should also advance the arc of the subject being explored. That’s constraining in the sense there are plenty of interesting non-curricular tasks one could do but only a subset align well with the overall goals for a class. Like many artforms though within those limits there still is tremendous opportunity for creativity and interesting experiences.
Example:
I’m going to draw from my own experience doing math circle activities where I’m very purposely fairly random about activities because we exist outside the curriculum. But I often run into things that would work really well in a classroom setting paired with a learning goal. Recently we tried out determining the number of rectangles that could be fit in one larger one.
There are 60 sub rectangles in the above 3x5
The process for exploring this was fairly messy and time consuming and involved a lot of counting strategies and looking for patterns. But the ultimate solution has deep connections to combinatorics and would work really well embedded in a unit on combinations, pascal’s triangle (or both). If doing so you’d want to combine it with simpler exercises on calculating combinations and using their properties along with other simpler problems that built towards being able to tackle this one.
Creating, integrating and curating these tasks is the crux of the instructor “problem”. Its easy to go a little astray and end up with a problem that falls flat and no one can solve or is too easy or fails to make the crucial connection or is boring. The chance for failure is fairly large. But practice helps and keeping one’s eye on this ultimate goal and reflecting on how each attempt at problem integration into a sessions plays out aids in more effectively using them. And to return to my central philosophy, problems are too essential to Mathematics to not keep tilting at figuring out how to properly use them. 
Graphic scores by John De Cesare (1890–1972).
