# Monthly Archives: April 2013

## Arrangements Connections for Young Students

Mathematics is not arithmetic.

The latter is a set of symbol manipulation rules that dominates most of what we teach in school.  Mathematics, on the other hand, is a science of patterns.  It is a way of logical thinking, making sense of forms and arrangements–sometimes applied and sometimes purely imagined.  It involves looking at the implications of what we know and pushing that knowledge as far as we can to see what else can be learned based solely upon connections we can make from our assumptions.

Within the last few weeks, I’ve discovered a great daily ‘blog run by @Five_Triangles “for (but not limited to) school years 6-8.”  I’d argue that those posts are great for a broader range of ages. I gave my 3rd grade daughter one of the puzzles during breakfast.  We had some great conversations then and on the way to school.  I share those below.  Another offering extends that thinking in a way that may not be immediately obvious to young people.

Here’s the part of the post I used at breakfast.  For my daughter, I saw this problem presenting two different possibilities–the obvious arithmetic problem and a mathematics extension.  The arithmetic requires very basic subtraction facts and wee bit of trial-and-error (a GREAT mathematics skill!) to tease out a solution.  Part of the mathematics here, in my opinion, involves asking a “What if?” question.

I posed this problem to my 3rd grade daughter and after randomly dropping in some numbers at first and seeing some frustration, I said to her, “I wonder what sorts of numbers subtract to give 3.”  Her frustration evaporated as she started making a list of several possibilities for such digits. She noted that there were far more possibilities for these difference than space in the problem allowed.  I encouraged her to keep trying.  We never explicitly discussed the problem’s set up with a four-digit number subtracted from a five-digit number, but I saw her try a couple different first digits before realizing that the first character of the five-digit number clearly had to be “1”.  A little more experimentation and she had an answer.

She thought the puzzle was over–after all, school has trained her to think that once she had “an” answer, she must have found “the” answer.

That’s when I prompted some mathematics.  I asked if she could find another answer.  A few other prompts and she had found 6 different solutions.  I asked her how she found them.  “Easy,” she replied.  “You just put the number pairs in different orders.”  She found through trial-and-error that the five-digit number always started “12…” and therefore the four-digit number started “9…”.  Checking her list of differences leading to 3 left no other possibilities.  Everything else was flexible, thus her six different answers.

• Can you explain why the five-digit number must start “12…”?
• Once I had the “12…” and “9…”, I knew there were at least 6 solutions  before I had found even the first one.  My daughter wasn’t ready for this thought, but can you explain why this is true?
• Can you find all 6 answers?
• Better: Can you explain why there cannot be any more?

The second part of the problem (with the same rules and a different result) is definitely tougher. You can quickly conclude that the first digit of the five-digit number must be 4 or 3, but it’s definitely more challenging to tease out the rest.  Rather than dealing with the entire problem at once, I suggest another great mathematics strategy:  Simplify the problem.  Using only the digits 1 to 9, can you find all possibilities that would result in the beginning of the problem? If this is part of an answer, the six digits not used in those three boxes must have an arrangement that subtracts to 333.  Unfortunately, none of these actually pan out.  Convince yourself why this must be true.  Students need to learn that not finding an answer is OK.  Knowing that there’s not a solution is actually a solution–you’ve learned something.

Extending the beginning of the problem to eventually shows that the five-digit number could start “412..” with the four-digit number starting “79..”.  That means the remaining four digits must have exactly two arrangements for precisely the same reasons that the first problem had six solutions.  Can you find the two arrangements that satisfy the 33333 problem?  In case you want to check, I list the answers at the end of this post.

The next week provided another puzzle using the arrangements idea. The problem doesn’t yield a straightforward solution that can be solved.  Instead, laying out all possible finishing arrangements and testing the veracity of the claims leads to a solution.  Again, there are three entries, so this problem is (mathematically) just like the 3333 subtraction problem above–both have six possible arrangements.  Helping a young person see this connection would be a great thought achievement.

Start by listing the six possible 1st, 2nd, and 3rd place arrangements of the letters A, B, and C:  A-B-C, A-C-B, B-A-C, B-C-A, C-A-B, & C-B-A.   As an example, if the boys finished A-B-C, all three boys would have told the truth, so that finish doesn’t satisfy the problem requirement of one false statement.  Comparing each arrangement to the boys’ statements eventually shows that only one of these arrangements satisfies the problem’s requirement that exactly one of the three boys made a false statement.

A good mathematical extension would be to see if there are any other questions that could be asked from the boys’ statements.  Is it possible that all three told the truth?  Is it possible that only one was truthful?  Are there any other possible outcomes?  Do any of these have unique outcomes given the boys’ statements, or do some have multiple possibilities?

CONCLUSION:  I fear that too often school and students stop at a single answer and don’t explore other possibilities.  Asking “What if” is a critical question in all of science and mathematics.  It inspires creativity, wonder, and exploration.  It doesn’t always yield results, so it also helps motivate stamina.  Convincing yourself that there are no (more) solutions is itself an intellectual accomplishment.

We need more of this.

SOLUTIONS:

• 3333 solutions: 12678-9345, 12687-9354, 12768-9435, 12786-9453, 12867-9534, & 12876-9543.
• 33333 solutions: 41268-7935 & 41286-7953.
• Competition solution:  A-C-B

## A Student’s Powerful Polar Exploration

I posted last summer on a surprising discovery of a polar function that appeared to be a horizontal translation of another polar function.  Translations happen all the time, but not really in polar coordinates.  The polar coordinate system just isn’t constructed in a way that makes translations appear in any clear way.

That’s why I was so surprised when I first saw a graph of $\displaystyle r=cos \left( \frac{\theta}{3} \right)$. It looks just like a 0.5 left translation of $r=\frac{1}{2} +cos( \theta )$ . But that’s not supposed to happen so cleanly in polar coordinates.  AND, the equation forms don’t suggest at all that a translation is happening.  So is it real or is it a graphical illusion?

I proved in my earlier post that the effect was real.  In my approach, I dealt with the different periods of the two equations and converted into parametric equations to establish the proof.  Because I was working in parametrics, I had to solve two different identities to establish the individual equalities of the parametric version of the Cartesian x- and y-coordinates.

As a challenge to my precalculus students this year, I pitched the problem to see what they could discover. What follows is a solution from about a month ago by one of my juniors, S.  I paraphrase her solution, but the basic gist is that S managed her proof while avoiding the differing periods and parametric equations I had employed, and she did so by leveraging the power of CAS.  The result was that S’s solution was briefer and far more elegant than mine, in my opinion.

S’s Proof:

Multiply both sides of $r = \frac{1}{2} + cos(\theta )$ by r and translate to Cartesian. $r^2 = \frac{1}{2} r+r\cdot cos(\theta )$ $x^2 + y^2 = \frac{1}{2} \sqrt{x^2+y^2} +x$ $\left( 2\left( x^2 + y^2 -x \right) \right) ^2= \sqrt{x^2+y^2} ^2$

At this point, S employed some CAS power. [Full disclosure: That final CAS step is actually mine, but it dovetails so nicely with S’s brilliant approach. I am always delightfully surprised when my students return using a tool (technological or mental) I have been promoting but hadn’t seen to apply in a particular situation.]

S had used her CAS to accomplish the translation in a more convenient coordinate system before moving the equation back into polar.

Clearly, $r \ne 0$, so $4r^3 - 3r = cos(\theta )$ .

In an attachment (included below), S proved an identity she had never seen, $\displaystyle cos(\theta) = 4cos^3 \left( \frac{\theta }{3} \right) - 3cos \left( \frac{\theta }{3} \right)$ , which she now applied to her CAS result. $\displaystyle 4r^3 - 3r = 4cos^3 \left( \frac{\theta }{3} \right) - 3cos \left( \frac{\theta }{3} \right)$

So, $\displaystyle r = cos \left( \frac{\theta }{3} \right)$

Therefore, $\displaystyle r = cos \left( \frac{\theta }{3} \right)$ is the image of $\displaystyle r = \frac{1}{2} + cos(\theta )$ after translating $\displaystyle \frac{1}{2}$ unit left.  QED

Simple. Beautiful.

Obviously, this could have been accomplished using lots of by-hand manipulations.  But, in my opinion, that would have been a horrible, potentially error-prone waste of time for a problem that wasn’t concerned at all about whether one knew some Algebra I arithmetic skills.  Great job, S!

S’s proof of her identity, $\displaystyle cos(\theta) = 4cos^3 \left( \frac{\theta }{3} \right) - 3cos \left( \frac{\theta }{3} \right)$ : ## Fun with Series

Two days ago, one of my students (P) wandered into my room after school to share a problem he had encountered at the 2013 Walton MathFest, but didn’t know how to crack.  We found one solution.  I’d love to hear if anyone discovers a different approach.  Here’s our answer.

PROBLEM:  What is the sum of $\displaystyle \sum_{n=1}^{\infty} \left( \frac{n^2}{2^n} \right) = \frac{1^2}{2^1} + \frac{2^2}{2^2} + \frac{3^2}{3^3} + ...$ ?

Without the $n^2$, this would be a simple geometric series, but the quadratic and exponential terms can’t be combined in any way we knew, so the solution must require rewriting.  After some thought, we remembered that perfect squares can be found by adding odd integers.  I suggested rewriting the series as where each column adds to the one of the terms in the original series.  Each row was now a geometric series which we knew how to sum.  That  meant we could rewrite the original series as We had lost the quadratic term, but we still couldn’t sum the series with both a linear and an exponential term.  At this point, P asked if we could use the same approach to rewrite the series again.  Because the numerators were all odd numbers and each could be written as a sum of 1 and some number of 2s, we got where each column now added to the one of the terms in our secondary series.  Each row was again a geometric series, allowing us to rewrite the secondary series as Ignoring the first term, this was finally a single geometric series, and we had found the sum. Does anyone have another way?

That was fun.  Thanks, P.