## Saturday, June 23, 2007

### 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!

Totally_clueless said...

Hi Dave,

Thanks for the problem; it turned out to be quite a challenge in the general case. In the course of my meanderings, I modified the problem as follows: With your figure, let AP=x and AS=y. Let AB=a, and BC=b. Find an equation involving x, y, a and b such that PQRS is an inscribed rectangle.

This turns out to be a hyperbola, which also gives you values of x that are impermissible. All values of y between 0 and 6 are possible.

The answer to your question involves the intersection of the hyperbola with a circle to find x and y such that the length of the longer side is a given quantity. Then, you can find the length of the smaller side, for which I was able to verify your answer.

Heath said...

Wow-wee, such a simple problem definition, and such a fiendishly difficult solution. After working on this for some time, I have an answer, but could not find an exact solution.

Initially I used 3 geometric identies:

Line segments:
AP + PB = AB = 8
BQ + QC = BC = 6

Pythagoras:
PB^2 + BQ^2 = PQ^2 = 8^2
QC^2 + AP^2 = QR^2

And finally I gave in and included similar triangles:
PQ/QR = PB/QC = BQ/AP

This was looking promising, but I found nothing. Instead, I turned to Cartesian geometry and plotted the lines of the inner PQRS rectangle.

I placed point D at the origin, AD along the y-axis and DC along the x-axis. PQ then became "y = mx + b" and PS became "y = (-1/m)x + 6 - (b-6)/m^2". By doing similar for SR (simply the gradient of PQ and the y-intercept of PS), I arrived at two possible equations for QR. One described QR coinciding with SR at the x-axis and one described QR coinciding with PQ along BC.

Simply equation these two possibilities for QR then, gave me a relationship between m and b which produces rectangles PQRS inside ABCD. The relationship is as follows:

b = (8m^5 + 8m^3 + 6m^2 + 6)/(1-m^4)

Not very easy to work with at all! Nonetheless, plotting the result and animating the parameter m produced a very interesting picture of all possible rectangles PQRS.

After observing that a range of rectangles were possible, and indeed two of them have PQ = 8 (one is the trivial PQRS == ABCD), I then constrained the distance formula for PQ to equal 8. This is where the limits of algebraic accuracy broken down. Plotting the distance constraint I could clearly see the where the two solutions lay (as well as two more that didn't make physical sense), but was unable to produce an exact solution. At this point I resorted to numerical methods.

I found that a gradient, m = -0.59768393, gives an intercept, b = 6.6772057 and a PQ length of 8.000000026.

Plugging this into a distance formula for QR gives this final value:

||QR|| = 2.2085313

Of course, without an exact solution, this is slightly unfulfilling. I wonder if anyone has any comments on the method. For reference, the equation for ||PQ|| which lead to the breakdown was as follows:

((8-(6-b)/m)^2 + (8*m+b-6)^2)^(1/2)

Regards,
Heath

Dave Marain said...

Excellent work, Heath and totally clueless!
Sorry I haven't replied for awhile on this.

My algebraic approach was similar to TC's, although I like the analytical method used by Heath. Whenever I struggle with a geometry problem, I tend to go to a coordinate set-up. This almost always produces algebraic relationships that I might have missed synthetically (using Euclidean methods).

My labeling was similar to TC, but not quite the same. Here goes:

Let PB = x, therefore AP = 8-x.
Let BQ = y, therefore QC = 6-y.
From this point, the key to the problem (geometrically) is to recognize how the angles match up. For example, angle BPQ is complementary to angle BQP. Similarly, angle BQP is comp to angle RQC because of the inscribed right angle! Continuing in this way, around the diagram, we obtain FOUR similar right triangles at the corners.

From here, we can set up ratios of sides of similar triangles. Since Triangle PBQ is similar to triangle QCR, we obtain:
y/x = (8-x)/(6-y). This leads to our first xy-relationship:
(*) x^2 - y^2 = 8x - 6y

Since PQ = 8, we obtain
(**) x^2 + y^2 = 64
from Pythagoras.

This system of equations is what TC meant by the intersection of a circle (**) and a hyperbola (*).

Algebraically, there's probably a neater solution but I used brute force:
Solving for y from (**) and restricting y to be positive, we have
(***) y = √(64-x^2).
Substituting into (*), leads to some 4th degree equation (circles and hyperbolas can intersect in at most 4 points!). However, I chose to go to Solver on my TI-84 directly from the radical equation to obtain the value x = 6.8670 rounded.
Subbing back into (***) produced
y = 4.1043. Using the Pythag Thm on triangle QCR produced the stated value for the width of the rectangle, 2.2085 rounded.

I really feel there's more going on in this problem. There's still the issue of the common error students might make in assuming that PQ is parallel to diagonal AC (not drawn). The first time I looked at this question, I began making that assumption and I had to prove to myself why it was wrong! Making that false assumption, leads to the width = 2, which is close! The logic error is worth discussing in class.
Heath, by an exact solution I assume you meant an exact radical expression for the width, yes? I'm sure this can be done since every 4th degree equation is solvable, but I had no desire to revisit my Theory of Equations course!

Heath said...

Wonderful! Yes, by exact I meant rational, and I'm afraid I couldn't rest until I at least had a closed algebraic form for the solution, which was really only a few careful steps beyond what you've presented.

Using linear combinations of your equations (*), (**), I arrived at equations for x and y in terms of only a single instance of each other. From there I picked the equation for x and substituted the equation for y in, eliminating all unknowns from the equation, except x. Through lots of careful elimination I able to get rid of all the square roots, and sure enough, eventually ended up with a quartic in x. Here it is:

x^4 - 8x^3 - 39x^2 + 256x + 448 = 0

Really quite pretty, considering the process. The delightful thing about this one is that plotting it equal to y gives a lovely quartic curve with two local minimums and four roots. Two are negative, one is the trivial x=8 solution, and the other is the x=6.8670 you discovered.

Sure, applying Ferrari's algebraic solution to quartics might give me my rational solution, but I'm pretty content right here, thank you ;)

By the way, I think the key for me to see PQ /| AC is that the slope of PQ is dependent on the length of BC. Given there's nothing special about BC being 6 while AB is 8, it would only be coincidence if they were indeed parallel.

Anonymous said...

Hi,
Can someone tell me how you came up with a 4th degree equation? I am coming up with the following equation:
2x^2 - 8x + 6*((64-x^2)^0.5) - 64 = 0

Once I get this equation, I have to solve it in Excel using a equation and just trying out numbers until it equals 0. I did come up with x = 6.86695. I also get the width of the rectangle as 2.208531......
I am just wondering how anyone came up with a 4th degree equation.
Any insight?

Kelsey Miner said...

I solved a very similar problem in the general case. I assumed the large rectangle dimensions are known and the width (RQ) of the smaller rectangle is known.

This simulates trying to find the length to cut a board (2x4, etc) in a door frame to brace it.

Let Angle SRD = α
Let Angle QRC = ß
Let CD = H
Let RQ = T
Let RC = y
Let CQ = x

Tan(α) = (W-x)/(H-y) (eq 1)

T^2 = x^2 + y^2
x = (T^2-Y^2)^1/2

Substituting x back into eq 1

tan(α) = (W-(T^2-y^2)^1/2)/(H-y) (eq2)

Getting α in terms of y.
α = 90-ß
cos(ß) = y/T
ß = cos^-1(y/T)
α = 90-cos^-1(y/T)

Substituting α back in eq2

tan(90-cos^-1(y/T)) = (W-(T^2-y^2)^1/2)/(H-y) (eq3)

Then graph or use numerical approximation to solve for y.

Then find L (length of the board)

L = ((W-x)^2+(H-y)^2)^1/2

L= ((W-(T^2-y^2)^1/2)^2 + (H-y)^2)^1/2

Enter the value for y solved previously, and you have L.

Dave Marain said...

Nice, Kelsey! I need to verify it but it looks very good.

Isn't it awesome that the problem lives on over 5 yrs after orig publ.

Are you an educator or is math just a passion?
Thanks again for sharing your generalization!
Dave