Category Archives: problem-solving

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  One small square is withdrawn at random from the bag and spun on a flat surface.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.

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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?

Envelope Curves

My precalculus class recently returned to graphs of sinusoidal functions with an eye toward understanding them dynamically via envelope curves:  Functions that bound the extreme values of the curves. What follows are a series of curves we’ve explored over the past few weeks.  Near the end is a really cool Desmos link showing an infinite progression of periodic envelopes to a single curve–totally worth the read all by itself.

GETTING STARTED

As a simple example, my students earlier had seen the graph of f(x)=5+2sin(x) as y=sin(x) vertically stretched by a magnitude of 2 and then translated upward 5 units.  In their return, I encouraged them to envision the function behavior dynamically instead of statically.  I wanted them to see the curve (and the types of phenomena it could represent) as representing dynamic motion rather than a rigid transformation of a static curve.  In that sense, the graph of f oscillated 2 units (the coefficient of sine in f‘s equation) above and below the line y=5 (the addend in the equation for f).  The curves y=5+2=7 and y=5-2=3 define the “Envelope Curves” for y=f(x).

When you graph y=f(x) and its two envelope curves, you can picture the sinusoid “bouncing” between its envelopes.  We called these ceiling and floor functions for f.  Ceilings happen whenever the sinusoid term reaches its maximum value (+1), and floors when the sinusoidal term is at its minimum (-1).

Envelope1

Those envelope functions would be just more busy work if it stopped there, though.  The great insights were that anything you added to a sinusoid could act as a midline with the coefficient, AND anything multiplied by the sinusoid is its amplitude–the distance the curve moves above and below its midline.  The fun comes when you start to allow variable expressions for the midline and/or amplitudes.

VARIABLE MIDLINES AND ENVELOPES

For a first example, consider y= \frac{x}{2} + sin(x).  By the reasoning above, y= \frac{x}{2} is the midline.  The amplitude, 1, is the coefficient of sine, so the envelope curves are y= \frac{x}{2}+1 (ceiling) and y= \frac{x}{2}-1 (floor).

Envelope2

That got their attention!  Notice how easy it is to visualize the sine curve oscillating between its envelope curves.

For a variable amplitude, consider y=2+1.2^{-x}*sin(x).  The midline is y=2, with an “amplitude” of 1.2^{-x}.  That made a ceiling of y=2+1.2^{-x} and a floor of y=2-1.2^{-x}, basically exponential decay curves converging on an end behavior asymptote defined by the midline.

Envelope3

SINUSOIDAL MIDLINES AND ENVELOPES

Now for even more fun.  Convinced that both midlines and amplitudes could be variably defined, I asked what would happen if the midline was another sinusoid?  For y=cos(x)+sin(x), we could think of y=cos(x) as the midline, and with the coefficient of sine being 1, the envelopes are y=cos(x)+1 and y=cos(x)-1.

Envelope5

Since cosine is a sinusoid, you could get the same curve by considering y=sin(x) as the midline with envelopes y=sin(x)+1 and y=sin(x)-1.  Only the envelope curves are different!

Envelope6

The curve y=cos(x)+sin(x) raised two interesting questions:

  1. Was the addition of two sinusoids always another sinusoid?
  2. What transformations of sinusoidal curves could be defined by more than one pair of envelope curves?

For the first question, they theorized that if two sinusoids had the same period, their sum was another sinusoid of the same period, but with a different amplitude and a horizontal shift.  Mathematically, that means

A*cos(\theta ) + B*sin(\theta ) = C*cos(\theta -D)

where A & B are the original sinusoids’ amplitudes, C is the new sinusoid’s amplitude, and D is the horizontal shift.  Use the cosine difference identity to derive

A^2 + B^2 = C^2  and \displaystyle tan(D) = \frac{B}{A}.

For y = cos(x) + sin(x), this means

\displaystyle y = cos(x) + sin(x) = \sqrt{2}*cos \left( x-\frac{\pi}{4} \right),

and the new coefficient means y= \pm \sqrt{2} is a third pair of envelopes for the curve.

Envelope7

Very cool.  We explored several more sums and differences with identical periods.

WHAT HAPPENS WHEN THE PERIODS DIFFER?

Try a graph of g(x)=cos(x)+cos(3x).

Envelope8

Using the earlier concept that any function added to a sinusoid could be considered the midline of the sinusoid, we can picture the graph of g as the graph of y=cos(3x) oscillating around an oscillating midline, y=cos(x):

Envelope9

IF you can’t see the oscillations yet, the coefficient of the cos(3x) term is 1, making the envelope curves y=cos(x) \pm 1.  The next graph clear shows y=cos(3x) bouncing off its ceiling and floor as defined by its envelope curves.

Envelope10

Alternatively, the base sinusoid could have been y=cos(x) with envelope curves y=cos(3x) \pm 1.

Envelope11

Similar to the last section when we added two sinusoids with the same period, the sum of two sinusoids with different periods (but the same amplitude) can be rewritten using an identity.

cos(A) + cos(B) = 2*cos \left( \frac{A+B}{2} \right) * cos \left( \frac{A-B}{2} \right)

This can be proved in the present form, but is lots easier to prove from an equivalent form:

cos(x+y) + cos(x-y) = 2*cos(x) * cos(y).

For the current function, this means y = cos(x) + cos(3x) = 2*cos(x)*cos(2x).

Now that the sum has been rewritten as a product, we can now use the coefficient as the amplitude, defining two other pairs of envelope curves.  If y=cos(2x) is the sinusoid, then y= \pm 2cos(x) are envelopes of the original curve, and if y=cos(x) is the sinusoid, then y= \pm 2cos(2x) are envelopes.

Envelope12

Envelope13

In general, I think it’s easier to see the envelope effect with the larger period function.  A particularly nice application connection of adding sinusoids with identical amplitudes and different periods are the beats musicians hear from the constructive and destructive sound wave interference from two instruments close to, but not quite in tune.  The points where the envelopes cross on the x-axis are the quiet points in the beats.

A STUDENT WANTED MORE

In class last Friday, my students were reviewing envelope curves in advance of our final exam when one made the next logical leap and asked what would happen if both the coefficients and periods were different.  When I mentioned that the exam wouldn’t go that far, she uttered a teacher’s dream proclamation:  She didn’t care.  She wanted to learn anyway.  Making up some coefficients on the spot, we decided to explore f(x)=2sin(x)+5cos(2x).

Assuming for now that the cos(2x) term is the primary sinusoid, the envelope curves are y=2sin(x) \pm 5.

Envelope14

That was certainly cool, but at this point, we were no longer satisfied with just one answer.  If we assumed sin(x) was the primary sinusoid, the envelopes are y=5cos(2x) \pm 2.

Envelope15

Personally, I found the first set of envelopes more satisfying, but it was nice that we could so easily identify another.

With the different periods, even though the  coefficients are different, we decided to split the original function in a way that allowed us to use the cos(A)+cos(B) identity introduced earlier.  Rewriting,

f(x)=2sin(x)+5cos(2x) = 2cos \left( x - \frac{ \pi }{2} \right) + 2cos(2x) + 3cos(2x) .

After factoring out the common coefficient 2, the first two terms now fit the cos(A) + cos(B) identity with A = x - \frac{ \pi }{2} and B=2x, allowing the equation to be rewritten as

 f(x)= 2 \left( 2*cos \left( \frac{x - \frac{ \pi }{2} + 2x }{2} \right) * cos \left( \frac{x - \frac{ \pi }{2} - 2x }{2} \right) \right) + 3cos(2x)

\displaystyle = 4*  cos \left( \frac{3}{2} x - \frac{ \pi }{4} \right) * cos \left( - \frac{1}{2} x - \frac{ \pi }{4} \right) + 3cos(2x).

With the expression now containing three sinusoidal expressions, there are three more pairs of envelope curves!

Arguably, the simplest approach from this form assumes cos(2x) from the $latex $3cos(2x)$ term as the sinusoid, giving y=2sin(x)+2cos(2x) \pm 3 (the pre-identity form three equations earlier in this post) as envelopes.

Envelope16

We didn’t go there, but recognizing that new envelopes can be found simply by rewriting sums creates an infinite number of additional envelopes.  Defining these different sums with a slider lets you see an infinite spectrum of envelopes.  The image below shows one.  Here is the Desmos Calculator page that lets you play with these envelopes directly.

Envelope17

If the cos \left( \frac{3}{3} x - \frac{ \pi}{4} \right) term was the sinusoid, the envelopes would be y=3cos(2x) \pm 4cos \left( - \frac{1}{2} x - \frac{ \pi }{4} \right).  If you look closely, you will notice that this is a different type of envelope pair with the ceiling and floor curves crossing and trading places at x= \frac{\pi}{2} and every 2\pi units before and after.  The third form creates another curious type of crossing envelopes.

Envelope18

CONCLUSION:

In all, it was fun to explore with my students the many possibilities for bounding sinusoidal curves.  It was refreshing to have one student excited by just playing with the curves to see what else we could find for no other reason than just to enjoy the beauty of these periodic curves.  As I reflected on the overall process, I was even more delighted to discover the infinite spectrum of envelopes modeled above on Desmos.

I hope you’ve found something cool here for yourself.

From a Square to Ratios to a System of Equations

Here’s another ratio problem from @Five_Triangles, this time involving triangle areas bounded by a square.

Don’t read further until you’ve tried this for yourself. It’s a fun problem that, at least from my experience, doesn’t end up where or how I thought it would.

INITIAL THOUGHTS

I see two big challenges here.

First, the missing location of point P is especially interesting, but is also likely to be quite vexing for many students. This led me to the first twist I found in the problem: the introduction of multiple variables and a coordinate system. Without some problem-solving experience, I don’t see that as an intuitive step for most middle school students. Please don’t interpret this as a knock on this problem, I’m simply agreeing with @Five_Triangle’s assessment that this problem is likely to be challenging for middle school students.

The second challenge I found emerged from the introduction the coordinate system: an underlying 2×2 system of equations. There are multiple ways to tackle a solution to a linear system, but this strikes me as yet another high hurdle for younger students.

Finally, I’m a bit surprised by my current brain block on multiple approaches for this problem. I suspect I’m blinded here by my algebraic bias in problem solving; surely there are approaches that don’t require this. I’d love to hear any other possibilities.

POINT P VARIES

Because I was given properties of point P and not its location, the easiest approach I could see was to position the square on the xy-plane with point B at the origin, \overline{AB} along the y-axis, and \overline{BC} along the x-axis. That gave my point P coordinates (x,y) for some unknown values of x & y.

square2

The helpful part of this orientation is that the x & y coordinates of P are automatically the altitudes of \Delta ABP and \Delta BCP, respectively. The altitudes of the other two triangles are determined through subtraction.

AREA RATIOS BECOME A LINEAR SYSTEM

From here, I used the given ratios to establish one equation in terms of x & y.

\displaystyle \frac{\Delta ABP}{\Delta DAP} = \frac{\frac{1}{2}*12*x}{\frac{1}{2}*12*(12-y)} = \frac{3}{4}

Of course, since all four triangles have the same base lengths, the given area ratios are arithmetically equivalent to corresponding height ratios.  I used that to write a second equation.

\displaystyle \frac{\Delta BCP}{\Delta CDP} = \frac{y}{12-x} = \frac{1}{3}

Simplifying terms and clearing denominators leads to 4x=36-3y and 3y=12-x, respectively.

A VERY INTERESTING insight at this point is that there is an infinite number of locations within the square at which each ratio is true.  Specifically, the \Delta ABP : \Delta DAP = 3:4 ratio is true everywhere along the line 4x=36-3y.  This problem constrains us to only the points within the square with vertices (0,0), (12,0), (12,12), and (0,12), but setting that aside, anywhere along the line 4x=36-3y would satisfy the first constraint.  The same is true for the second line and constraint.

I think it would be very interesting for students to construct this on dynamic geometry software (e.g., GeoGebra or the TI-Nspire) and see the ratio remain constant everywhere along either line even though the triangle areas vary throughout.

Together, these lines form a 2×2 system of linear equations with the solution to both ratios being the intersection point of the two lines.  There are lots of ways to do this; I wonder how a typical 6th grader would tackle them. Assuming they have the algebraic expertise, I’d have work them by hand and confirm with a CAS.

square3

The question asks for the area of \Delta ABP = \frac{1}{2}*12*x = 6*8 = 48.

PROBLEM VARIATIONS

Just two extensions this time.  Other suggestions are welcome.

  1. What’s the ratio of the area of \Delta BCP : \Delta DAP at the point P that satisfies both ratios??
    It’s not 1:4 as an errant student might think from an errant application of the transitive property to the given ratios.  Can you show that it’s actually 1:8?
  2. If a random point is chosen within the square, is that point more likely to satisfy the area ratio of \Delta ABP : \Delta DAP or the ratio of \Delta BCP : \Delta CDP?
    The first ratio is satisfied by the line 4x=36-3y which intersects the square on the segment between (9,0) and (0,12).  At the latter point, both triangles are degenerate with area 0.  The second ratio’s line intersects the square between (12,0) and (0,4).  As the first segment is longer (how would a middle schooler prove that?), it is more likely that a randomly chosen point would satisfy the \Delta ABP : \Delta DAP  ratio.  This would be a challenging probability problem, methinks.

FURTHER EXTENSIONS?

What other possibilities do you see either for a solution to the original problem or an extension?

Party Ratios

I find LOTS of great middle school problems from @Five_Triangles on Twitter.  Their post two days ago was no exception.

The problem requires a little stamina, but can be approached many ways–two excellent criteria for worthy student explorations.  That it has some solid extensions makes it even better.  Following are a few different solution approaches some colleagues and I created.

INITIAL THOUGHTS, VISUAL ORGANIZATION, & A SOLUTION

The most challenging part of this problem is data organization.  My first thoughts were for a 2-circle Venn Diagram–one for gender and one for age.  And these types of Venn Diagrams are often more easily understood, in my experience, in 2×2 Table form with extra spaces for totals.  Here’s what I set up initially.

ratio2

The ratio of Women:Girls was 11:4, so the 24 girls meant each “unit” in this ratio accounted for 24/4=6 people.  That gave 11*6=66 women and 66+24=90 females.

ratio3

At this point, my experience working with algebraic problems tempted me to overthink the situation.  I was tempted to let B represent the unknown number of boys and set up some equations to solve.  Knowing that most 6th graders would not think about variables, I held back that instinct in an attempt to discover what a less-experienced mind might try. I present my initial algebra solution below.

The 5:3 Male:Female ratio told me that each “gender unit” represented 90/3=30 people.  That meant there were 5*30=150 men and 240 total people at the party.

ratio4

Then, the 4:1 Adult:Children ratio showed how to age-divide every group of 5 partygoers.  With 240/5=48 such groups, there were 48 children and 4*48=192 adults.  Subtracting the already known 66 women gave the requested answer:  192-66=126 men.

ratio5

While this Venn Diagram/Table approach made sense to me, I was concerned that it was a moderately sophisticated and not quite intuitive problem-solving technique for younger middle school students.

WHAT WOULD A MIDDLE SCHOOLER THINK?

A middle school teaching colleague, Becky, offered a different solution I could see students creating.

Completely independently, she solved the problem in exactly the same order I did using ratio tables to manage the scaling at each step instead of my “unit ratios”.  I liked her visual representation of the 4:1 Adults:Children ratio to find the number of adults, which gave the requested number of men.  I suspect many more students would implicitly or explicitly use some chunking strategies like the visual representation to work the ratios.

Ratio1

 

WHY HAVE JUST ONE SOLUTION?

Math problems involving ratios can usually be opened up to allow multiple, or even an infinite number of solutions.  This leads to some interesting problem extensions if you eliminate the “24 girls” restriction.  Here are a few examples and sample solutions.

What is the least number of partygoers?

For this problem, notice from the table above that all of the values have a common factor of 6.  Dividing the total partygoers by this reveals that 240/6=40 is the least number.  Any multiple of this number is also a legitimate solution.

Interestingly, the 11:4 Women:Girls ratio becomes explicitly obvious when you scale the table down to its least common value.

ratio6My former student and now colleague, Teddy, arrived at this value another way.  Paraphrasing, he noted that the 5:3 Male:Female ratio meant any valid total had to be a multiple of 5+3=8.  Likewise, the 4:1 Adult:Child ratio requires totals to be multiples of 4+1=5.  And the LCM of 8 & 5 is 40, the same value found in the preceding paragraph.

What do all total partygoer numbers have in common?

As explained above, any multiple of 40 is a legitimate number of partygoers.

If the venue could support no more than 500 attendees, what is the maximum number of women attending?

12*40=480 is the greatest multiple of 40 below 500.  Because 480 is double the initial problem’s total, 66*2=132 is the maximum number of women.

Note that this can be rephrased to accommodate any other gender/age/total target.

Under the given conditions, will the number of boys and girls at the party ever be identical?

As with all ratio problems, larger values are always multiples of the least common solution.  That means the number of boys and girls will always be identical or always be different.  From above, you can deduce that the numbers of boys and girls at the party under the given conditions will both be multiples of 4.

What variations can you and/or your students create?

RESOLVING THE INITIAL ALGEBRA

Now to the solution variation I was initially inclined to produce.  After initially determining 66 women from the given 24 girls, let B be the unknown number of boys.  That gives B+24 children.  It was given that adults are 4 times as numerous as children making the number of adults 4(B+24)=4B+96.  Subtracting the known 66 women leaves 4B+30 men.  Compiling all of this gives

ratio7

The 5:3 Male:Female ratio means \displaystyle \frac{5}{3} = \frac{5B+30}{90} \longrightarrow B=24, the same result as earlier.

 

ALGEBRA OVERKILL

Winding through all of that algebra ultimately isn’t that computationally difficult, but it certainly is more than typical 6th graders could handle.

But the problem could be generalized even further, as Teddy shared with me.  If the entire table were written in variables with W=number of women, M=men, G=girls, and B=boys, the given ratios in the problem would lead to a reasonably straightforward 4×4 system of equations.  If you understand enough to write all of those equations, I’m certain you could solve them, so I’d feel confident allowing a CAS to do that for me.  My TI-Nspire gives this.

ratio8

And that certainly isn’t work you’d expect from any 6th grader.

CONCLUSION

Given that the 11:4 Women:Girls ratio was the only “internal” ratio, it was apparent in retrospect that all solutions except the 4×4 system approach had to find the female values first.  There are still several ways to resolve the problem, but I found it interesting that while there was no “direct route”, every reasonable solution started with the same steps.

Thanks to colleagues Teddy S & Becky M for sharing their solution proposals.

Unanticipated Proof Before Algebra

I was talking with one of our 5th graders, S,  last week about the difference between showing a few examples of numerical computations and developing a way to know something was true no matter what numbers were chosen.  I hadn’t started our conversation thinking about introducing proof.  Once we turned in that direction, I anticipated scaffolding him in a completely different direction, but S went his own way and reinforced for me the importance of listening and giving students the encouragement and room to build their own reasoning.

SETUP:  S had been telling me that he “knew” the product of an even number with any other number would always be even, while the product of any two odds was always odd.  He demonstrated this by showing lots of particular products, but I asked him if he was sure that it was still true if I were to pick some numbers he hadn’t used yet.  He was.

Then I asked him how many numbers were possible to use.  He promptly replied “infinite” at which point he finally started to see the difficulty with demonstrating that every product worked.  “We don’t have enough time” to do all that, he said.  Finally, I had maneuvered him to perhaps his first ever realization for the need for proof.

ANTICIPATION:  But S knew nothing of formal algebra.  From my experiences with younger students sans algebra, I thought I would eventually need to help him translate his numerical problem into a geometric one.  But this story is about S’s reasoning, not mine.

INSIGHT:  I asked S how he would handle any numbers I asked him to multiply to prove his claims, even if I gave him some ridiculously large ones.  “It’s really not as hard as that,” S told me.  He quickly scribbled

s1

on his paper and covered up all but the one’s digit.  “You see,” he said, “all that matters is the units.  You can make the number as big as you want and I just need to look at the last digit.”  Without using this language, S was venturing into an even-odd proof via modular arithmetic.

With some more thought, he reasoned that he would focus on just the units digit through repeated multiples and see what happened.

FIFTH GRADE PROOF:  S’s math class is currently working through a multiplication unit in our 5th grade Bridges curriculum, so he was already in the mindset of multiples.  Since he said only the units digit mattered, he decided he could start with any even number and look at all of its multiples.  That is, he could keep adding the number to itself and see what happened.  As shown below, he first chose 32 and found the next four multiples, 64, 96, 128, and 160.  After that, S said the very next number in the list would end in a 2 and the loop would start all over again.

s2

He stopped talking for several seconds, and then he smiled.  “I don’t have to look at every multiple of 32.  Any multiple will end up somewhere in my cycle and I’ve already shown that every number in this cycle is even.  Every multiple of 32 must be even!”  It was a pretty powerful moment.  Since he only needed to see the last digit, and any number ending in 2 would just add 2s to the units, this cycle now represented every number ending in 2 in the universe.  The last line above was S’s use of 1002 to show that the same cycling happened for another “2 number.”

DIFFERENT KINDS OF CYCLES:  So could he use this for all multiples of even numbers?  His next try was an “8 number.”

s3

After five multiples of 18, he achieved the same cycling.  Even cooler, he noticed that the cycle for “8 numbers” was the 2 number” cycle backwards.

Also note that after S completed his 2s and 8s lists, he used only single digit seed numbers as the bigger starting numbers only complicated his examples.  He was on a roll now.

s4

I asked him how the “4 number” cycle was related.  He noticed that the 4s used every other number in the “2 number” cycle.  It was like skip counting, he said.  Another lightbulb went off.

“And that’s because 4 is twice 2, so I just take every 2nd multiple in the first cycle!”  He quickly scratched out a “6 number” example.

s5

This, too, cycled, but more importantly, because 6 is thrice 2, he said that was why this list used every 3rd number in the “2 number” cycle.  In that way, every even number multiple list was the same as the “2 number” list, you just skip-counted by different steps on your way through the list.

When I asked how he could get all the numbers in such a short list when he was counting by 3s, S said it wasn’t a problem at all.  Since it cycled, whenever you got to the end of a list, just go back to the beginning and keep counting.  We didn’t touch it last week, but he had opened the door to modular arithmetic.

I won’t show them here, but his “0 number” list always ended in 0s.  “This one isn’t very interesting,” he said.  I smiled.

ODDS:  It took a little more thought to start his odd number proof, because every other multiple was even.  After he recognized these as even numbers, S decided to list every other multiple as shown with his “1 number” and “3 number” lists.

s7

As with the evens, the odd number lists could all be seen as skip-counted versions of each other.  Also, the 1s and 9s were written backwards from each other, and so were the 3s and 7s.  “5 number” lists were declared to be as boring as “0 numbers”.  Not only did the odds ultimately end up cycling essentially the same as the evens, but they had the same sort of underlying relationships.

CONCLUSION:  At this point, S declared that since he had shown every possible case for evens and odds, then he had shown that any multiple of an even number was always even, and any odd multiple of an odd number was odd.  And he knew this because no matter how far down the list he went, eventually any multiple had to end up someplace in his cycles.  At that point I reminded S of his earlier claim that there was an infinite number of even and odd numbers.  When he realized that he had just shown a case-by-case reason for more numbers than he could ever demonstrate by hand, he sat back in his chair, exclaiming, “Whoa!  That’s cool!”

It’s not a formal mathematical proof, and when S learns some algebra, he’ll be able to accomplish his cases far more efficiently, but this was an unexpectedly nice and perfectly legitimate numerical proof of even and odd multiples for an elementary student.