Category Archives: problem-solving

Inscribed Triangle Challenge

@MathCeyhun posed an interesting geometry problem yesterday.

Even more interesting is that, as @MathCeyhun noted in a subsequent tweet, none of the posted solutions is correct.  There have been a few posted answers, but no solutions, so I thought I’d give it a try.

OBSERVATIONS

  • The perpendicular bisector of each chord was given, and perpendicular bisectors of chords always lie on radii of the circle.
  • If r is the radius of the circle, then the lengths of the extensions of the perpendicular bisectors are r-1, r-2, and r-3.
  • Nothing given guarantees anything special about the triangle, so I assumed it was scalene.  I called the side lengths 2x, 2y, and 2z to simplify the bisection labels.
  • Adding the bisector extensions, the radii to the vertices, and segment names and labels gave me this.

Inscribed_Triangle2

[Irrelevant to this problem, but I just realized by looking at this image that every triangle can be dissected into three isosceles triangles with congruent sides and a shared vertex point at the triangle’s circumcenter.  Pretty.]

SETTING UP MY SOLUTION 

Each isosceles triangle is bisected by the perpendicular bisector of its base from which I extracted three relationships from the Pythagorean theorem.

x^2+(r-1)^2=r^2 \longrightarrow x^2=(2r-1) \cdot 1

y^2+(r-2)^2=r^2 \longrightarrow y^2=(2r-2) \cdot 2

z^2+(r-3)^2=r^2 \longrightarrow z^2=(2r-3) \cdot 3

[The relationship between the half-sides, the extension of the radius beyond the triangle, and the radius is another pretty pattern.]

That gives 3 equations in 4 variables.  I needed one more to solve….

 

The area of \Delta ABC can be expressed two ways:  as the sum of the areas of the isosceles triangles, and using Heron’s formula.  From the areas of the isosceles triangles,

Area( \Delta ABC) = \frac{1}{2}(2x)(r-1) + \frac{1}{2}(2y)(r-2) + \frac{1}{2}(2z)(r-3)
Area( \Delta ABC) = x \cdot (r-1) + y \cdot (r-2) + z \cdot (r-3)

The sides of \Delta ABC are 2x, 2y, and 2z, so its semiperimeter is x+y+z and Heron’s formula gives its area as

Area( \Delta ABC) = \sqrt{(x+y+z)(-x+y+z)(x-y+z)(x+y-z)}.

The area of a given triangle is unique, so the two different area expressions are equivalent, giving a fourth equation.

SOLVING A SYSTEM & ANSWERING THE QUESTION

With four equations in four variables, I had a system of equations.  The algebra was messy, so I invoked my CAS to crunch it for me.

Inscribed_Triangle3

The question asked for the area of the triangle, so I just substituted my values back into the area formulas.

Inscribed_Triangle4

And 17.186… is clearly not one of the choices in the original problem.

 

A PLEA…

Recognizing the perpendicular bisectors, seeing all the right triangles, and connecting the multiple ways to describe the area of a triangle made this solution reasonably easy to find with the help of my computer algebra system (CAS), but I know the background algebra is, at best, cumbersome.  I hope there’s a more elegant solution, but I don’t see it.  Can anyone offer a suggestion?

Either way … this is definitely becoming a challenge problem for my Quantitative Reasoning class this coming week!

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Implicit Derivative Question

Yesterday, this question posted within the AP Calculus Community about implicit derivatives.

implicit

Below, I argue why the derivatives MUST be the same, show how four different variations can all be shown to give the same derivative, and provide a final conclusion.

INITIAL INTUITION

The Desmos graph of the given relation, is y^2 = \frac{x-1}{x+1} , is shown below.  Logically, it seems that even when the terms of the relation are algebraically rearranged, the graph should be invariant.   The other two forms mentioned in the Community post are on lines 2 and 3.  Lines 4, 5, and 6 show three other variations.  Here is the link to my Desmos graph allowing you to change between the forms to visually confirm the graphical invariance intuition.

implicit2.jpg

If calculus “works”, it also shouldn’t matter how one calculates a derivative.   While the forms of the derivative certainly could LOOK different,  because any point on the invariant graph has the same tangent line no matter what the form of its equation, and the derivative of a relation at a point is the slope of that invariant tangent line, then the derivative also MUST be invariant.

CALCULATING “DIFFERENT” DERIVATIVES

To show the derivatives are fundamentally all the same (as suspected by the initial post), I calculate the derivatives of the equations on lines 1 and 3 given in the initial post as well as my variations on lines 4 and 6.

LINE 1:

Using the Chain Rule on the left and the Quotient Rule on the right gives

implicit3

LINE 3:

This version is more complicated, requiring the Product Rule in addition to the earlier Chain and Quotient Rules.  In the penultimate line, I used the original equation to substitute for y^2 to transform the derivative into the same form as line 1.

implicit4.jpg

LINE 4:

This time, differentiation requires only the Chain and Product Rules.

implicit5

After the usual substitution for y^2, I multiplied both sides by (x+1) to clear the denominator and solved for y', returning the same result.

implicit6

LINE 6:

This time, the relation is solved for x, resulting in a much more complicated Quotient+Chain Rule calculation, but substituting for y^2 and changing the form leads once again to the same answer.

implicit7

Hopefully this is convincing evidence that all derivative forms can be shown to be equivalent.   If you’re still learning implicit differentiation, I encourage you to show the derivatives from the lines 2 and 5 variations are also equivalent.

CONCLUSION

So which approach is “best”?  In my opinion, it all depends on your personal comfort with algebraic manipulations.  Some prefer to just take a derivative from the given form of y^2 = \frac{x-1}{x+1}.  I avoid the more complicated quotient rule whenever I can, so the variation from line 4 would have been my approach.

The cool part is that it doesn’t matter what approach you use, so long as your algebraic manipulations are sound.  You don’t have to accept the form in which a problem is given; CHANGE IT to a form that works for you!

Pythagorean Investigation

 

Here’s a challenge @jamestanton tweeted yesterday:

Visually, Tanton is asking if there is an integer right triangle (like the standard version shown on the left below) for which the integer triangle on the right exists.

Pythag

The algebraic equivalent to this question is, for some a^2+b^2=c^2, does there exist a Natural number d so that b^2+c^2=d^2?

I invoked Euclid’s formula in my investigation to show that there is no value of d to make this possible.  I’d love to hear of any other proof variations.

INVOKING EUCLID’S FORMULA

For any coprime natural numbers m & n where m>n and m-n is odd, then every primitive Pythagorean triple can be generated by \left\{ m^2 - n^2, 2mn, m^2 + n^2 \right\}.

For any Natural number kevery Pythagorean triple can be generated by \left\{ k \cdot \left( m^2 - n^2 \right), k \cdot \left( 2mn \right), k \cdot \left( m^2 + n^2 \right) \right\}.

The generator term k \cdot \left( m^2 + n^2 \right) must be the original hypotenuse (side c), but either k \cdot \left( m^2 - n^2 \right) or k \cdot \left( 2mn \right) can be side b.  So, if Tanton’s scenario is true, I needed to check two possible cases.  Does there exist a Natural number d such that

d^2 = \left( k \cdot \left( m^2 - n^2 \right) \right)^2 + \left( k \cdot \left( m^2 + n^2 \right) \right)^2 = 2k^2 \left( m^4 + n^4 \right)

or

d^2 = \left( k \cdot \left( 2mn \right) \right)^2 + \left( k \cdot \left( m^2 + n^2 \right) \right)^2 = k^2 \left( m^4 +6m^2n^2 + n^4 \right)

is true?

EVALUATING THE POSSIBILITIES

For the first equation, there is a single factor of 2 on the right, and there is no way to extract an odd number of factors of 2 from \left( m^4 + n^4 \right) or k^2, so 2k^2 \left( m^4 + n^4 \right) can’t represent a perfect square.

For the second equation, there is no way to factor \left( m^4 +6m^2n^2 + n^4 \right) over Integers, so k^2 \left( m^4 +6m^2n^2 + n^4 \right) can’t be a perfect square either.

Since neither equation can create a perfect square, there is no Natural value of d that makes {b, c, d} a Pythagorean triple.  Tanton’s challenge is impossible.

Does anyone have a different approach?

FiveThirtyEight Riddler Express Solution

I’d not peeked at FiveThirtyEight’s The Riddler in a while when I saw Neema Salimi ‘s post about the June 22, 2018 Riddler Express   Neema argued his solution was accessible to Algebra 1 students–not always possible for FiveThirtyEight’s great logic puzzles–so I knew it was past time to return.

After my exploration, I’ve concluded this is DEFINITELY worth posing to any middle school learners (or others) in search of an interesting problem variation.

Following is my solution, three retrospective insights about the problem, a comparison to Neema’s solution, and a proposed alternative numeric approach I think many Algebra 1 students might actually attempt.

THE PROBLEM:

Here is a screenshot of the original problem posted on FiveThirtyEight (2nd item on this page).

If you’re familiar with rate problems from Algebra 1, this should strike you immediately as a “complexification” of D=R*T type problems.  (“Complexification” is what a former student from a couple years ago said happened to otherwise simple problems when I made them “more interesting.”)

MY SOLUTION:

My first thought simplified my later algebra and made for a much more dramatic ending!!  Since Michelle caught up with her boarding pass with 10 meters left on the walkway, I recognized those extra 10 meters as irrelevant, so I changed the initial problem to an equivalent question–from an initial 100 m to 90 m–having Michelle catch up with her boarding pass just as the belt was about to roll back under the floor!

Let W = the speed of the walkway in m/s.  Because Michelle’s boarding pass then traveled a distance of 90 m in W m/s, her boarding pass traveled a total \displaystyle \frac{90}{W} seconds.

If M = Michelle’s walking speed, then her total distance traveled is the initial 90 meters PLUS the distance she traveled in the additional 90 seconds after dropping her boarding pass.  Her speed at this time was (M-W)  m/s (subtracting W because she was moving against the walkway), so the additional distance she traveled was D = (M-W) \cdot 90, making her her total distance D_{Michelle} = 90 + 90(M-W).

Then Michelle realized she had dropped her boarding pass and turned to run back at (2M+W) m/s (adding to show moving with the walkway this time), and she had \displaystyle \frac{90}{W} - 90 seconds to catch it before it disappeared beneath the belt.  The subtraction is the time difference between losing the pass and realizing she lost it.  Substituting into D = R*T gives

\displaystyle 90 + 90(M-W)=(2M+W)* \left( \frac{90}{W} - 90 \right)

A little expansion and algebra cleanup …

\displaystyle 90 + 90M - 90W = 180 \frac{M}{W} - 180M + 90 - 90W

\displaystyle 90M = 180 \frac{M}{W} - 180M

\displaystyle 270M = 180 \frac{M}{W}

And multiplying by \displaystyle \frac{W}{270M} solves the problem:

\displaystyle W = \frac{2}{3}

INSIGHTS:

Insight #1:  Solving a problem is always nice, but I was thinking all along that I pulled off my solution because I’m very comfortable with algebraic notation.  This is certainly NOT true of most Algebra 1 students.

Insight #2:  This made me wonder about the viability of a numeric solution to the problem–an approach many first-year algebra students attempt when frutstrated.

Insight #3:  In the very last solution step, Michelle’s rate, M, completely dropped out of the problem.  That means the solution to this problem is independent of Michelle’s walking speed.

Wondering if other terms might be superfluous, too, I generalized my initial algebraic solution further with A = the initial distance before Michelle dropped her boarding pass and B = the additional time Michelle walked before realizing she had dropped the pass.

\displaystyle A + B(M-W)=(2M+W)* \left( \frac{A}{W} - B \right)

And solving for gives \displaystyle W = \frac{2A}{3B}.

So, the solution does depend on the initial distance traveled and the time before Michelle turns around, and it was simplified in the initial statement with A=B=90.  That all made sense after a few quick thought experiments.  With one more variation you can show that the scale factor between her walking and jogging speed is relevant, but not her walking speed.  But now it was clear that in all cases, Michelle’s walking speed is irrelevant!

COMPARING TO NEEMA:

My initial conclusion matched Neema’s solution, but I really liked my separate discovery that the answer was completely independent of Michelle’s walking speed.  In my opinion, those cool insights are not at all intuitive.

AN ALTERNATIVE NUMERIC APPROACH:

While this approach is just a series of special cases of the generic approach, I suspect many Algebra 1 students would likely get frustrated quickly by the multiple variable and attempt

Ignoring everything above, but using the same variables for simplicity, perhaps the easiest start is to assume the walkway moves at W=1 m/s and Michelle’s walking speed is M=2 m/s.  That means her outward speed against the walkway is (2-1) = 1 m/s.  She drops the pass at 90 meters after 90 seconds.  So the pass will be back at the start after 90 seconds, the additional time that Michelle walks before realizing her loss.

I could imagine many students I’ve taught working from this through some sort of intelligent numeric guess-and-check, adjusting the values of M and W until landing at \displaystyle W=\frac{2}{3}.  The fractional value of W would slow them down, but many would get it this way.

CONCLUSION:

I’m definitely pitching this question to my students in just a few weeks.  (Where did summer go?)  I’m deeply curious about how they’ll approach their solutions.  I”m convinced many will attempt–at least initially–playing with more comprehensible numbers.  Such approaches often give young learners the insights they need to handle algebra’s generalizations.

Painting and Probability

 

Here’s a cool probability problem the start of which is accessible to middle and high school students with careful reasoning.  It was posted by Alexander Bogomolny on Cut the Knot a week ago. I offer what I think is a cool extension of the problem following the initial credits.

The next day Mike Lawler tweeted another phenomenal video solution worked out with his boys during Family Math.

Mike’s videos show his boys thinking through simpler discrete cases of the problem and their eventual case-by-case solution to the nxnxn question.  The last video on the page gives a lovely alternative solution completely bypassing the case-by-case analyses.  This is also “Solution 1” on Alexander’s page.

EXTENDING THE PROBLEM:

When I first encountered the problem, I wanted to think about it before reading any solutions.  As with Mike’s boys, I was surprised early on when the probability for a 2x2x2 cube was \displaystyle \frac{1}{2} and the probability for a 3x3x3 cube was \displaystyle \frac{1}{3}.  That was far too pretty to be a coincidence.  My solution exactly mirrored the nxnxn case-by-case analysis in Mike’s videos:  the probability of rolling a painted red face up from a randomly selected smaller cube is \displaystyle \frac{1}{n}.

Surely something this simple and clean could be generalized.  Since a cube can be considered a “3-dimensional square”, I re-phrased Alexander’s question into two dimensions.  The trickier part was thinking what it would mean to “roll” a 2-dimensional shape.

The outside of an nxn square is painted red and is chopped into n^2 unit squares.  The latter are thoroughly mixed up and put into a bag.  One small square is withdrawn at random from the bag and spun on a flat surface.  What is the probability that the spinner stops with a red side facing you?

Shown below is a 4×4 square, but in all sizes of 2-dimensional squares, there are three possible types in the bag:  those with 2, 1, or 0 sides painted.

Square1

I solved my first variation case by case.  In any nxn square,

  • There are 4 corner squares with 2 sides painted.  The probability of picking one of those squares and then spinning a red side is \displaystyle \frac{4}{n^2} \cdot \frac{2}{4} = \frac{2}{n^2}.
  • There are 4(n-2) edge squares not in a corner with 1 side painted.  The probability of picking one of those squares and then spinning a red side is \displaystyle \frac{4(n-2)}{n^2} \cdot \frac{1}{4} = \frac{n-2}{n^2}.
  • All other squares have 0 sides painted, so the probability of picking one of those squares and then spinning a red side is 0.
  • Adding the probabilities for the separate cases gives the total probability:  \displaystyle \frac{2}{n^2}+\frac{n-2}{n^2}=\frac{1}{n}

After reading Mike’s and Alexander’s posts, I saw a much easier approach.

  • Paint all 4 edges of an nxn square, and divide each painted edge into n painted unit segments.  This creates 4 \cdot n total painted small segments.
  • Decompose the nxn original square into n^2 unit squares.  Each unit square has 4 edges giving 4 \cdot n^2 total edges.
  • Because every edge of every unit square is equally likely to be spun, the total probability of randomly selecting a smaller square and spinning a red side is \displaystyle \frac{4n}{4n^2}=\frac{1}{n}.

The dimensions of the “square” don’t seem to matter!

WARNING: 

Oliver Wendell Holmes noted, “A mind that is stretched by a new experience can never go back to its old dimensions.”  The math after this point has the potential to stretch…

EXTENDING THE PROBLEM MORE:

I now wondered whether this was a general property.

In the 2-dimensional square, 1-dimensional edges were painted and the square was spun to find the probability of a red edge facing.  With the originally posed cube, 2-dimensional faces were painted and the cube was tossed to find the probability of an upward-facing red face.  These cases suggest that when a cube of some dimension with edge length n is painted, is decomposed into unit cubes of the original dimension, and is spun/tossed to show a cube of one smaller dimension, then the probability of getting a painted smaller-dimensional cube of is always \displaystyle \frac{1}{n}, independent of the dimensions occupied by the cube.

Going beyond the experiences of typical middle or high school students, I calculated this probability for a 4-dimensional hypercube (a tesseract).

  • The exterior of a tesseract is 8 cubes.  Ignore the philosophical difficulty of what it means to “paint” (perhaps fill?) an entire cube.  After all, we’re already beyond the experience of our 3-dimensions.
  • Paint/fill all 8 cubes on the surface of the tesseract, and divide each painted cube into n^3 painted unit cubes.  This creates 8 \cdot n^3 total painted unit cubes.
  • Decompose the original tesseract into n^4 unit tesseracts.  Each unit tesseract has 8 cubes giving 8 \cdot n^4 total unit cubes.
  • Because every unit cube on every unit tesseract is equally likely to be “rolled”, the total probability of randomly selecting a smaller tesseract and rolling a red cube is \displaystyle \frac{8n^3}{8n^4}=\frac{1}{n}.

The probability is independent of dimension! 

More formally,

The exterior of a d-dimensional hypercube with edge length n is painted red and is chopped into n^d unit d-dimensional hypercubes.  The latter are put into a bag of sufficient dimension to hold them and thoroughly mixed up.  A unit d-dimensional hypercube is withdrawn at random from the bag and tossed.  The probability that the unit d-dimensional hypercube lands with a red (d-1)-dimensional hypercube showing is \displaystyle \frac{1}{n} .

PROOF:

  • The exterior of a d-dimensional hypercube is comprised of 2d (d-1)-dimensional hypercubes of dimension (d-1).  Paint/fill all 2d surface hypercubes and divide each painted (d-1)-dimensional hypercube into n^{d-1} painted unit hypercubes.  This creates 2d \cdot n^{d-1} total painted unit hypercubes.
  • Decompose the original tesseract into n^d unit d-dimensional hypercubes.  Each unit d-dimensional hypercube has 2d surface (d-1)-dimensional hypercubes giving 2d \cdot n^d total surface unit d-dimensional hypercubes.
  • Because every unit (d-1)-dimensional hypercube on the surface of every unit d-dimensional hypercube is equally likely to be “rolled”, the total probability of randomly selecting a unit d-dimensional hypercube and rolling a (d-1)-dimensional red-painted hypercube is \displaystyle \frac{2d \cdot n^{d-1}}{2d \cdot n^d}=\frac{1}{n}.

 

I hope you can now return to something close to your old dimensions.

Inscribed Right Angle Proof Without Words

Earlier this past week, I assigned the following problem to my 8th grade Geometry class for homework.  They had not explored the relationships between circles and inscribed angles, so I added dashed auxiliary segment AD as a hint.

What follows first is the algebraic solution I expected most to find and then an elegant transformational explanation one of my students produced.

PROOF 1:

Given circle A with diameter BC and point D on the circle.  Prove triangle BCD is a right triangle.

RightAngle1

After some initial explorations on GeoGebra sliding point D around to discover that its angle measure was always 90^{\circ} independent of the location of D, most successful solutions recognized congruent radii AB, AC, and AD, creating isosceles triangles CAD and BAD.  That gave congruent base angles x in triangle CAD, and y in BAD.

RightAngle2

The interior angle sum of a triangle gave (x)+(x+y)+(y)=180^{\circ}, or m \angle CDB = x+y = 90^{\circ}, confirming that BCD was a right triangle.

PROOF 2:

Then, one student surprised us.  She marked the isosceles base angles as above before rotating \Delta BCD 180^{\circ} about point A.

RightAngle3

Because the diameter rotated onto itself, the image and pre-image combined to form an quadrilateral with all angles congruent.  Because every equiangular quadrilateral is a rectangle, M had confirmed BCD was a right triangle.

CONCLUSION:

I don’t recall seeing M’s proof before, but I found it a delightfully elegant application of quadrilateral properties.  In my opinion, her rotation is a beautiful proof without words solution.

Encourage freedom, flexibility of thought, and creativity, and be prepared to be surprised by your students’ discoveries!

Binomial Expansion Variation

Several years ago, I posed on this ‘blog a problem I learned from Natalie Jackucyn:

For some integers A, B, and n, one term of the expansion of (Ax+By)^n is 27869184x^5y^3.  What are the values of A, B, and n?

In this post, I reflect for a moment on what I’ve learned from the problem and outline a solution approach before sharing a clever alternative solution one of my students this year leveraged through her CAS-enabled investigation.

WHAT I LEARNED BEFORE THIS YEAR

Mostly, I’ve loved this problem for its “reversal” of traditional binomial expansion problems that typically give A, B, and n values and ask for either complete expansions or specific terms of the polynomial.  Both of these traditional tasks are easily managed via today’s technology.  In Natalie’s variation, neither the answer nor how you would proceed are immediately obvious.

The first great part of the problem is that it doesn’t seem to give enough information.  Second, it requires solvers to understand deeply the process of polynomial expansion.  Third, unlike traditional formulations, Natalie’s version doesn’t allow students to avoid deep thinking by using technology.

In the comments to my original post, Christopher Olah and a former student, Bryan Spellman, solved the problem via factoring and an Excel document, respectively.  Given my algebraic tendencies, I hadn’t considered Bryan’s Excel “search” approach, but one could relatively easily program Excel to provide an exhaustive search.  I now think of Bryan’s approach as a coding approach to a reasonably efficient search of the sample space of possible solutions.  Most of my students’ solutions over the years essentially approach the problem the same way, but less efficiently, by using one-case-at-a-time expansions via CAS commands until they stumble upon good values for A, B, and n.  Understandably, students taking this approach typically become the most frustrated.

Christopher’s approach paralleled my own.  The x and y exponents from the expanded term show that n=5+3=8.  Expanding a generic (Ax+By)^8 then gives a bit more information.  From my TI-Nspire CAS,

binomial1

so there are 56 ways an x^5y^3 term appears in this expansion before combining like terms (explained here, if needed).  Dividing the original coefficient by 56 gives a^5b^3=497,664, the coefficient of x^5y^3.

binomial2

The values of a and b are integers, so factoring 497,664 shows these coefficients are both co-multiples of 2 and 3, but which ones?  In essence, this defines a system of equations.  The 3 has an exponent of 5, so it can easily be attributed to a, but the 11 is not a multiple of either 5 or 3, so it must be a combination.  Quick experimentation with the exponents leads to 11=5*1+3*2, so 2^1 goes to a and 2^2 goes to b.  This results in a=3*2=6 and b=2^2=4.

WHAT A STUDENT TAUGHT ME THIS YEAR

After my student, NB, arrived at a^5b^3=497,664 , she focused on roots–not factors–for her solution.  The exponents of a and b suggested using either a cubed or a fifth root.

binomial3
binomial3

The fifth root would extract only the value of a if b had only singleton factors–essentially isolating the a and b values–while the cubed root would extract a combination of a and b factors, leaving only excess a factors inside the radical.  Her investigation was simplified by the exact answers from her Nspire CAS software.

binomial4

From the fifth root output, the irrational term had exponent 1/5, not the expected 3/5, so b must have had at least one prime factor with non-singular multiplicity.  But the cubed root played out perfectly.   The exponent–2/3–matched expectation, giving a=6, and the coefficient, 24, was the product of a and b, making b=4.  Clever.

EXTENSIONS & CONCLUSION

Admittedly, NB’s solution would have been complicated if the parameter was composed of something other than singleton prime factors, but it did present a fresh, alternative approach to what was becoming a comfortable problem for me.  I’m curious about exploring other arrangements of the parameters of (Ax+By)^n to see how NB’s root-based reasoning could be extended and how it would compare to the factor solutions I used before.  I wonder which would be “easier” … whatever “easier” means.

As a ‘blog topic for another day, I’ve learned much by sharing this particular problem with several teachers over the years.  In particular, the initial “not enough information” feel of the problem statement actually indicates the presence of some variations that lead to multiple solutions.  If you think about it, NB’s root variation of the solution suggests some direct paths to such possible formulations.  As intriguing as the possibilities here are, I’ve never assigned such a variation of the problem to my students.

As I finish this post, I’m questioning why I haven’t yet taken advantage of these possibilities.  That will change. Until then, perhaps you can find some interesting or alternative approaches to the underlying systems of equations in this problem.  Can you create a variation that has multiple solutions?  Under what conditions would such a variation exist?  How many distinct solutions could a problem like this have?