Drum Roll Please: The Debut of TC's Total Challenge

As you may have read in an earlier comment, I've invited one of MathNotations' most dedicated and talented contributors to go beyond commenting and share some of his creative ideas and insights by being an occasional guest blogger - he has graciously accepted.

For his inaugural offering, tc is challenging you and/or your students to solve a classic calculus problem using non-calculus methods. I have made a few minor edits, but the activity is essentially what tc sent to me.
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I give you tc's Total Challenge I:

One of my math professors in college used to say there were three ways
of tackling any problem: the right way, the wrong way and the Navy way
(correct, but extremely roundabout).

In this exercise, we will look at three ways (not necessarily the ones
named above) of doing the following problem:

Determine the rectangle of maximum area that can be inscribed
in a circle of given radius r.

Let the inscribed rectangle have sides a and b. The diagonal of the rectangle passes through the center of the circle (this can be shown, but you can assume it is true).

(1) Express r in terms of a and b.

(2) Express the area in terms of a and r.

(3) Instead of maximizing the area, we can maximize the square of the area.
(a) Express the square of the area as a quadratic in a² (you may want to substitute c for a²).
(b) By completing the square, determine the value of a for which the area is a maximum.
(c) Determine the value of b and the maximum area.
(d) What conclusion can you draw about the rectangle of maximum area?
(This is the first way, which I call the Algebra way)

(4) Divide the rectangle into 2 congruent triangles, using a diagonal. Draw a half
diagonal that intersects this diagonal.
(a) Write an inequality for the area of one of these triangles in terms of r alone. The inequality should be of the form Area ≤ _______.
(b) If you can achieve equality, then you have maximized the area of the rectangle! Find out when this occurs, and if it does, find the lengths of a and b. (The Geometry way).

(5) Method 3 - the Calculus way of course.
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Additional comments from DM:
(i) thanks, tc!
(ii) tc's geometric approach in (4) also suggests a connection to the famous AM-GM Inequality. Visit this link and see if you can make the connection. This is not obvious.
Hint: Apply the AM-GM to a² and b².

An In(de)scribable Rectangle in a Rectangle - A Geometry Challenge

Now that the summer months have arrived, I thought it was time for a geometry challenge problem to chew on. Although this is a departure from the lesson plans I have been writing, it's still an enrichment experience. I'm often asked by students and parents how one becomes better at solving 'hard' math problems. My response is: "Keep trying hard problems!" One can only improve at problem-solving by challenging one's mind. Also, learn from others - we all learn from good models. There are no shortcuts here. Some frustration is healthy and if you want more cliches, let me know!

OVERVIEW
This question should definitely challenge your geometry students. It was brought to my attention by a teacher via a student who was given this by his honors geometry teacher. I'd provide proper attribution if I knew the original source. However, it is possible to go beyond this question and generalize. There are endless problems one could generate from inscribing Figure A in Figure B. Rectangles in rectangles, other than special cases (square in a square) are not often seen by students.

In addition, strong algebra skill and a graphing calculator would be useful. Use of Geometer's Sketchpad (or traditional drawing tools) would also make sense here as a fairly accurate construction of the diagram (better than my crude attempt) would be highly instructive and students enjoy 'solving' the problem this way. Of course they need to understand that such a solution is not mathematically valid!

THE PROBLEM
In the diagram below, ABCD is a rectangle with AB = 8 and BC = 6. Rectangle PQRS is inscribed in ABCD, i.e., the vertices of PQRS lie on the sides of ABCD. If PQ = 8, what is the length of QR?

Notes:
(a) Figure not drawn to scale! Drawing this was not fun!
(b) Someone out there will argue that side PQ could coincide with side AB by my definition of inscribed. After all, the diagram is not drawn accurately! I should have added that the rectangles share only those vertices in common!
(c) Students often begin by assuming that PQ is parallel to the diagonal AC. Careless use of similar triangles could lead to an answer of 2. The only problem is that the actual answer is 2.2085 rounded! Does PQ have to be parallel to AC? In fact, is it even possible here? The instructor might begin with assuming parallelism and asking students to see where that leads and if the conclusion makes sense.
(d) What might a mathematician do to extend this numerical problem? Would they consider the issue of a unique solution here, i.e., is the given length of PQ enough to produce only one such inscribed rectangle? What is the range of possible values for PQ (assuming that PQ represents the longer dimension)? Could PQ be 10 or more? Explain. Could PQ be 6 or less? How could we generalize this result further?
(e) As always the disclaimer: My results need independent verification - I depend on my astute readers to check them and correct any careless errors. You are always my best editors!