Saw a neat tweet from Dave Radciffe earlier in the week:

I was reminded of this problem again this morning when I asked my older son what he wanted to talk about for our Family Math project today. The topic he chose was similarity and congruence. I was a little surprised since it has been a while since we’ve talked about any geometry, but Dave’s problem fit the bill.

Although this problem is a little to difficult for my younger son to understand, I thought it would be fun to talk through it with both kids. He did struggle with a few of the ideas, but I think that overall it was an interesting problem for him. For my older son, this was a nice geometry review.

We started by talking through the problem and finding one easy idea:

Next we returned to the more general case to explore the ideas of similarity and congruence that help solve this problem:

For the next part of the project we tried to apply the ideas from the last part of the talk to our problem. This piece was difficult for my younger son, but we went slowly. It was nice to see him begin to understand some of the algebraic expressions.

Finally, we wrapped up by finishing the calculations and found that the area the shaded region in the problem was $4 \pi$ no matter where the segment of height 4 is placed!

So, a fun project and a excellent bit of luck that Dave had posted this problem earlier in the week. I still don’t know what made my son think of today’s topic, but it was a good geometry review for him and a nice algebra / geometry introduction for my younger son.

I could sort of picture it and decided to try to build it with our Zometool set. The unfortunate difficulty is that I don’t have a lot of green struts.

Here’s a tetrahedron with side length equal to two long green struts:

Here are the two smaller tetrahedrons that we’ll remove for the first part of the dissection:

Finally, here’s how to chop up the remaining shape into two triangular prisms, with an extra bonus of me not being able to figure out how to put the shape back together!

Interestingly, the triangular prisms here are a shape that we’ve explored previously because they were part of a famous / infamous test question:

Octahedrons Tetrahedrons and the Pyramid Puzzle

# A fun fractal project: exploring the Gosper curve

Over the last few days I’ve been preparing a project with the boys based a great fractal geometry example I found in this wonderful book:

We finally got going with that project this morning. The starting point was watching this Vi Hart video which gives a “proof” that $\pi = 4:$

After we watched the video we sat down to talk about the strange result and what they thought was going on. They seemed to gravitate to the idea that the jagged edges were causing problems, but the fact that the zigs and zags were getting smaller and smaller – and would eventually have a height of 0 – was still a bit confusing:

After talking about the Vi Hart video I introduced the kids to the Gosper curve by showing them the figures in the book that inspired this example. We also made use of an amazing program that Dave Radcliffe shared when I asked for a little help on Twitter:

Playing around with this program really helped the boys see the first couple of shapes in the sequence that eventually leads to the Gosper curve. I definitely owe Dave a big favor!

The next part of the project was to build the first couple of shapes that lead to the Gosper curve out of our Zometool set. The initial hexagon was easy, obviously, but the shape at step 2 gave them a little difficulty. In the video below they talk about building the shapes and then explore a connection between the hexagon from step 1 and the shape from step 2. The fun part here is that the boys saw some of the important connections that lead you to the Gosper curve.

Next we built a level 3 shape. It was lucky that we had the program from Dave Radcliffe since that allowed the boys to a little more confident that they had the right shape. It is interesting to see the 6 new level 2 shapes surrounding the original level 2 shape. Too bad our living room isn’t big enough to make a level 4 shape!

One interesting comment from my younger son is that he thinks that as we increase the size we’ll get closer and closer to a shape that looks like the original hexagon.

For the final part of the project we used our 3d printer to make 7 of the (approximate) Gosper curves. Here’s the shape we used from Thingiverse (our shape is the 2nd of the three shapes, but I can’t get that one to link properly):

The Gosper Curve on Thingiverse (I printed the middle one)

The punch line for the project is the same punch line that caught my attention in the book – when you increase the linear size of the Gosper curve by 3, the area inside the curve increases by a factor of 7 rather than a by a factor of 9. Everything that the boys have learned about scaling up to this point is that area scales as the square of the linear factors, but fractals have a different property. Pretty amazing!

Also, sorry for not explaining the analogy between the two boundaries right. Felt as if I was wrong as I was explaining it, but didn’t see what I got wrong until just now.

As a fun end to the project, I showed them Dan Anderson’s modification of the Gosper Island shape – sort of a combination of the Sierpenski Triangle / Menger Sponge shapes and the Gosper Island:

So, a fun project giving the kids an introduction to fractal dimension – a concept that I never would have guessed could be made accessible to kids. Really happy to have had the luck of running into this fun idea last week.

Today I learned a really neat (and new to me) basic algebraic technique from Dave Radcliffe on twitter.

This morning James Tanton posed this interesting problem:

I started thinking about it, but work got in the way. Checking back a little later I saw an amazing solution (that fits in a tweet!) from Dave Radcliffe:

Super clever. If you want a circle, you’ll need an equation like $x^2 + y^2 = r^2$ and the conditions of the problem essentially allow you to force that equation!

For fun I plugged these three equations into Wolfram Alpha to take a peek at the solution:

(1) $y = x^2 + x - 10$,

(2) $x = y^2 + 3y- 4$, and

(3) $x^2 + (y+ 1)^2 = 15.$

The third equation comes from applying Dave’s idea to the first two equations. Sure enough, you get this picture:

The Wolfram Alpha code is here:

Wolfram Alpha code for drawing the above picture

Not every day you learn a new high school algebra technique from a tweet – thanks for posting the cool solution, Dave!

[Post publishing note]

Two great Desmos programs help give you a feel for the problem. First from Chris Lusto:

Chris Lusto’s Desmos Program

and second from Justin Lanier:

Justin Lanier’s Desmos Program

# A neat number theory problem from David Radcliffe

[sorry if this isn’t edited well, limited editing time due to lots of kid activities today]

Saw this post on twitter earlier in the week:

It made for a great follow up to a previous Family Math which also happened to arise from a problem we saw on twitter:

http://mikesmathpage.wordpress.com/2014/10/19/a-neat-number-theory-problem-for-kids-from-tracy-johnston-zager/

The first thing that we did this morning was talk through the problem to be clear about a couple of math terms – proper divisors, for example.   We also looked at a few other integers to help make sure that we really did have our arms around the problem.   One neat thing is that the kids were able to see that when we listed the factors of a number, the solution to Dave’s problem required exactly four factors (or exactly two rows of divisors the way were were listing them).

The next step was to try to understand how you could sort out which numbers had exactly two rows of divisors.  My older son noticed that numbers that were the product of two primes (to the first power) would have exactly four divisors.  We worked through a couple of examples of integers of this form and found that they did indeed have that property.   My younger son was able to explain why this was the case – thanks to Art of Problem Solving’s Introduction to Number Theory book!  The last thing we talked about in this part was whether or not numbers of this form comprised all of the solutions to Dave’s problem:

Next up we started searching for other potential solutions.  We started off down the path of looking at powers of primes – it turns out this was a lucky road to head down.  We looked at squares and fourth powers, but neither of these types of numbers seemed to solve the problem.  It did give us the idea to look at cubes of primes, though, and that showed us one more set of solutions:

The last task was to see why the two types of numbers that we’d found – products of exactly two primes, and a single prime cubed – formed a complete set of solutions to the problem.     This part of the talk has a little more theory in it, but I think the lesson here is important – how do we know that our solution is complete?   We talk about how you count the number of divisors, and then how that counting process could arrive at exactly 4 divisors.   Funny enough, the divisors of 4 play an important role in answering that question.   The fact that there are only two ways to multiply integers together to arrive at 4 (2×2 and 4×1) tells us that we have a complete solution to the problem.  Yay!

So, in a couple of weeks we went from a neat problem shared by Tracy Johnston Zager about the sum of the divisors to a neat problem shared by David Radcliffe about products of proper divisors of integers.  Of course this journey was helped by the fact that I’m going through a number theory book with my younger son (and have gone through the same book with my older son previously).  These are challenging problems for kids to think through for sure, but I think that kids will enjoy the challenge.  These problems also do a great job of building up number sense because as you work through them you are constantly thinking about factoring, multiplying, and integers that share certain types of properties.    I forget the exact phrase, but in my mind problems like these definitely belong in the “low entry point / high exit point” problems that the people who study math education seem to really like.

As always, I’m glad that people are sharing problems like these ones on Twitter!

# Dave Radcliffe’s amazing Fibonacci / GCD post

I’ve been covering the concept of greatest common divisor with my younger son this week, so Dave’s post had some great timing.  It is such a neat formula that I thought I’d use it for a little talk with my kids tonight instead of doing our usual weekend Family Math.

We started by talking through Dave’s formula and writing down the first 15 Fibonacci numbers.  I then worked through a simple example to make sure that the kids understood the language of the formula:

next I had my younger son see if he could work through the formula using the 8th and 12th Fibonacci numbers.  He’s just learned the Euclidean algorithm, so I was assuming that he’d use that approach to find the greatest common divisor.  This formula turns out to be a great way to practice the Euclidean algorithm.

I wanted my older son to work through an example, too, and picked one with slightly larger numbers – the 10th and 15th Fibonacci numbers.  I have covered a bit more number theory with him and suspected that he would remember that you could find the greatest common divisor of two numbers by factoring.  I was glad that he did since it was a nice example for my younger son to see:

Finally, to do a few more complicated examples we went to the computer.  Since Mathematica has built in functions to calculate the GCD of any two numbers as well as to calculate any Fibonacci number, we could easily work with much larger numbers.  We did a couple of examples, including one where we forced a large greatest common divisor.

I’ve given no thought at all to how to prove Dave’s formula, but I do have a little bit of driving to do this weekend so at least I’ll have something to think about!  The boys seem to enjoy seeing fun little math facts like these, so despite not having a lot of context around this amazing formula I’m happy with how this project went.  Definitely not your standard greatest common divisor exercise!

# Problem Solving

I was lucky to have my early development in math shaped by an incredible high school math teacher – Mr. Waterman.  He lived and breathed math and got out of bed everyday thinking about how to teaching problem solving to high school kids.   I ate it all up back then and the problem solving techniques that I first learned in his classes remain important pieces of my mathematical tool kit even today.   It is actually pretty fun to look back and see how ideas originating in high school math contests can come into play 25+ years later at work.  Although maybe the fact that I still use these ideas every day explains why I think problem solving is such an important part of math education.

For the last few weeks I’ve been seeing a lot of posts on twitter about problem solving.  Yesterday, for example, I ran across this wonderful post from Fawn Nguyen:

Making Problem Solving Part of the Math Curriculum

While others may not, I find it difficult to talk about problem solving in the abstract.  In talking or learning about problem solving a specific example is much more valuable to me than 1,000 words of abstract discussion.  Two of the most amazing specific examples I’ve seen lately come from the Field’s Medalist Tim Gowers and from the head of Art of Problem Solving, Richard Rusczyk.    Gowers decided to “live blog” his attempt at working through problem 1 in this year’s International Mathematics Olympiad.   It is an all too rare treat to see the thought process of one of the top mathematicians in the world in real time:

Tim Gowers walks through an IMO problem

Richard Rusczyk’s example is also a math contest problem – in this case problem #24 from the 2013 AMC 12.  The problem asks a question about the quadratic function  $f(z) = z^2 + i z + 1.$  I’m sure that as a kid a quadratic with an imaginary component would have really intimidated me, but Rusczyk’s calm and systematic approach to walking through the problem takes all of that intimidation away.  It is such a great example of problem solving – in fact this blog post has been delayed by about 40 minutes as I’ve watched the video twice because I enjoy it so much!

With these two examples in mind, I thought it would be fun to do my own example, though believe me, it isn’t within a million miles of the quality of either of the examples above.  However, what I think is incredibly important about the Gowers and Rusczyk examples is that they show the problem solving process, and I think that the more examples of that process that are out there the better.  So, neither an IMO problem nor a problem from a big US math contest, but here’s the process that I went through thinking about a fun little problem I saw posted on twitter a week ago:

I was actually up in Boston running around with the family when I saw the original tweet, but for some reason the problem stuck with me and I spent the next few days sort of daydreaming about it.

The first thing I did was think through cases that I hoped would be easy – quadratic equations (and also simplified the problem a tiny bit by assuming that the infinitely many perfect squares arose from positive integers n).  Take a polynomial like $x^2 + 2x + 5$, for example.  We can rewrite this expression as $(x + 1)^2 + 4$.   Since the only perfect square than is 4 more than another perfect square is 4, this new expression helps us see that $x^2 + 2x + 5$ will not be a perfect square for infinitely many integers $x$.

What if the linear term has an odd coefficient, though – say $x^2 + 3x + 1$.  Writing this expression as  $(x + 1)^2 + x$ solves the problem in a slightly different way than the prior case.  For $x > 1$, we see that the new expression is greater $(x + 1)^2$ but less than $(x + 2)^2$.  Since  it lies in between two consecutive perfect squares, it cannot be a perfect square and so it will not be a perfect square for infinitely many positive integers x.

Since the coefficient of the linear term will either be even or odd, that takes care of the quadratic case. The only way that a quadratic polynomial can satisfy the conditions of the problem is if it can be written as $(x + a)^2$ for some integer a.  Great, but how does this fact generalize to higher degree polynomials?

The next example I thought about was the 4th degree polynomial $x^4 + 3x^3 + 3x^2 + 2x + 1 = (x^2 + x + 1)^2 + x^3$. From the quadratic case I thought that transforming the original polymonial into a perfect square plus a polynomial with a degree of 2n – 1 would be the way to go. Unfortunately I couldn’t see what to do, and certainly couldn’t see anything that would help in the case where the remainder was a general cubic polynomial.  What’s stopping a perfect square plus a number described by a cubic polynomial from being a perfect square just by accident? Hmmmm.

As we ran around Boston the problem stayed in the back of my mind. One day I got the idea to look for a different approach to the quadratic case and that turned out to be the key idea for getting my head around the problem. Writing the polynomial $x^2 + 3x + 1$ as $(x + 3/2)^2 - 5/4$ also helps you see that this polynomial cannot be a square for infinitely many integers.  Really for the same reason as above – the value is trapped between $(x + 1)^2$ and $(x + 2)^2$. This new expression led me to stumble on a similar statement for higher degree polynomials that was a nice little surprise (to me anyway!).

It turns out that completing the square generalizes in a neat way. For the case we are looking at – a monic polynomial of degree 2n with integer coefficients – the generalization is that you can always find a monic polynomial of degree n with rational coefficients whose square matches the first n + 1 terms of the original polynomial. In math symbols, I mean that we can write:

$x^{2n} + a_{2n-1} x^{2n-1} + a_{2n-2} x^{2n-2} + . . . + a_0$

as

$( x^n + b_{n-1} x^{n-1} + . . . + b_0 )^2$ + $c_{n-1} x^{n-1} + c_{n-1} x^{n-1} + . . . + c_0$

where the a’s are integers, and the b’s and c’s are rational.  The the 4th degree case I was looking at above gives an illustrative example:

$x^4 + 3x^3 + 3x^2 + 2x + 1 = (x^2 + (3/2) x + 3/8)^2 + 7x/8 + 55/64.$

Convincing yourself that this generalization of completing the square is true isn’t all that difficult.  When you square the polynomial with the $b_{i}$‘s above, you see that $2b_{n-1}$ has to be equal to $a_{2n-1}$, so solving for $b_{n-1}$ is exactly the same exercise you do when you complete the square for a quadratic.  Once you have the value for $b_{n-1}$ you see that $a_{2n-2}$ has to equal  $(b_{n-1})^2 + 2b_{n-2}$, which give you the value of $b_{n-2}.$  Similarly, once you have the first k $b_{i}$‘s, solving for the next one just involves solving a linear equation in $b_{n - k - 1}$.  Basically, you just end up dividing by 2 a lot.

Although the rational coefficients present a small problem, we can use  ideas similar to the ones  we used in the quadratic case to show that for large values of x the above expression cannot be a perfect square.  In the above example of the 4th degree equation, assume that the expression is equal to $z^2$ when x and z are integers.  Multiply everything by 64 to arrive at the equation:

$(8x^2 + 12x + 3)^2 + 56x + 55 = (8z)^2$

For large values of x, the expression on the left hand side lies between the consecutive perfect squares $(8x^2 + 12x + 3)^2$ and $(8x^2 + 12x + 4)^2$.   For clarity, note that $(8x^2 + 12x + 4)^2 = (8x^2 + 12x + 3)^2 + 2*(8x^2 + 12x + 3) + 1$.    As long as x is sufficiently large, $2*(8x^2 + 12x + 3) + 1$ will be larger than $56x + 55$ since the $x^2$ term will dominate the $x$ term.  Thus for sufficiently large integers x, the value of $(8x^2 + 12x + 3)^2 + 56x + 55$ (which we’ve assumed to equal $(8z)^2$) lies between two perfect squares and cannot be a perfect square itself.  So neither $(8z)^2$ nor $z^2$ can be a perfect square for sufficiently large values of x which, in turn, means that the original expression cannot be a perfect square for infinitely many positive integers x.

Essentially the same argument will work for any monic polynomial of degree 2n with integer coefficients, and also essentially the same argument works for showing that the polynomials cannot take on infinitely many perfect square values for negative integer values of x.

So, the generalization of completing the square shows that the only time a monic polynomial in x of degree 2n will take on infinitely many perfect square values for integer values of x is when the polynomial itself is a perfect square.  In that case, all of the values will be perfect squares as desired!

I doubt that my solution is the best or the most elegant, but I had a lot of fun thinking through this problem.  I’m also happy to have stumbled on this generalization for completing the square which I’m surprised to have either never seen previously or (more likely) just forgotten about.

As I said at the beginning of this post, the lessons about problems solving that I learned in the three years I spent in Mr. Waterman’s classes were such an important foundation in my own mathematical development.  Even though I’m not longer in academic math these problem solving strategies play a critical role in my work just about every day.  Away from work I try to communicate those lessons to my kids when we talk about math.

I hope that more mathematicians will follow the lead of Gowers and Rusczyk and give some public examples  of their own problem solving process so that everyone – and especially students – get lots of different looks at the problem solving process.  Seeing that work will help show that mathematical thinking  isn’t  about finding answers instantly and effortlessly, but often involves lots of trial and error, false starts, and most importantly joy at making a little progress.  These are important lessons from math with applications that go far beyond whatever specific problem you happen to be working on at the time.

# If you want to get people talking about math – talk about how to divide fractions!

http://online.wsj.com/articles/marina-ratner-making-math-education-even-worse-1407283282

Dividing fractions is a subject with a bit of history for me. On the funny side, as a kid I missed the week of school when this particular subject was taught and I never seemed to be able to catch up from that missed week. My pals on my high school math team loved giving me a hard time about always having to go back and figure out how dividing fractions worked. Even now they’ll needle me about it when it comes up in one of the videos I do with the boys.

When I was going through this subject with my older son I’m sure my approach was pretty much all over the place. The primary reason is that I’d never had to explain dividing fractions in detail to anyone – much less a kid. That alone assures a lot of stumbling around. Another reason is that although we were following Art of Problem Solving’s Prealgebra book when the subject came up formally, much of the teaching I do with my kids doesn’t follow a textbook and many important subjects come up almost out of the blue as we discuss various math problems. I certainly wasn’t aware of the different (and sometimes strongly held) beliefs about teaching fraction division when I was talking about it with my son.

Teaching the same subject to my younger son was a little different. Hopefully I’d learned a little bit from going through this subject once already (ha!), but also I’d begun to follow lots of teachers and math ed folks online so I’d seen some approaches to teaching fractions that were different that what I’d done on my own. I still used the approach in Art of Problem Solving’s Prealgebra book as the starting point, but I supplemented it with a couple of other ideas. Here are those three approaches from back in January. Having watched all three of this videos again just now, I’m perfectly happy with what we did and I believe that all three approaches have merit:

(1) Art of Problem Solving – define division as the reciprocal of multiplication and understanding fraction division just boils down to understanding what the reciprocal of a fraction is. Note also that he notices that you need multiplication to be commutative for this approach to work (!):

(2) Talking about patterns. Here we look at this sequences of divisions: 8 / 8, 8 / 4, 8 / 2, and 8 / 1 to help form a guess about what the value of 8 / (1/2) might be. We also use dimes and nickels to illustrate the division:

(3) Drawing rectangles / using snap cubes to talk about division. This is the approach that appears to have motivated today’s WSJ article -“Who would draw a picture to divide 2/3 by 3/4?”

The best response that I saw to the WSJ piece was this simple tweet from David Radcliffe:

Finally, if you don’t find fraction division to be an interesting topic to think about, perhaps you’ll be more interested in this delightful problem that Radcliffe posted earlier in the week:

# Follow up to Prime Numbers and Infinity

I got a nice question from an old high school classmate asking about yesterday’s blog post on prime numbers and infinity.  He asked me how you prove that the sum of the inverse squares was pi^2 / 6 and that the sum of the inverse primes was infinity.

Unfortunately both proofs are a little too advanced for kids.  The first requires understanding of the Taylor series for sin(x)  (at least the proof that I know), and the second requires a bit of number theory.  I wasn’t trying to present the ideas as something for the kids to prove, but rather just trying to show some fun facts that will come down the road (way down the road in this case).

However, there is one series that is pretty easy to understand that also has an infinite sum -> 1 + 1/2 + 1/3 + 1/4 + 1/5 + . . . .

After the question yesterday I toyed around with the idea of doing a second video, but decided to do it another time.  Then, as with yesterday’s blog, some funny coincidences happened today.  The first being that I started a new section on fractions with my younger son and the second one being a short proof of the divergence of the above sum that Dave Radcliffe put on twitter:

I’d never seen this clever little trick before and couldn’t wait to get home to show it to the boys:

I’ve been following a bunch of math folks on Twitter for about a year now and just can’t believe how many fun  examples I’ve found to share with the boys.  This little community on twitter has been a great resource for me.