Saturday, September 8, 2018

Golden Ratios in Regular Polygons

Here's the first 3: for the triangle, square and pentagon.

Equilateral Triangle 

The triangle ABC is similar DEC so \( \frac{AC}{BC} = \frac{a}{b} = \frac{CD}{CE} = \frac{a+b}{a} = \phi \)


golden ratio by Tran Quang Hung, construction #1

In square ABCD M,N,P,Q are midpoints of the sides, as shown below; F the center of the square. Then circle (AP) on AP as a diameter, cuts NF (point X) in Golden Ratio. Chord AX cuts MF in Golden Ratio:


Image result for pentagon and golden ratio

Wednesday, August 29, 2018

Seattle MathJam or Pints and Polynomials

Last night I launched a small project I've been thinking about for a while.   I'm now officially the host for the Seattle MathJam.         

"MathsJam is a monthly opportunity for like-minded self-confessed maths enthusiasts to get together in a pub and share stuff they like. Puzzles, games, problems, or just anything they think is cool or interesting. We don't have organised talks, planned activities or even strict timings - just turn up and join in."

The events were started by a group of Brits whom I admire, Colin Wright, Katie Steckles, Matt Parker et al. and have spread fairly far.  There's actually a group down in Tacoma but traffic being what it is, I've never been able to get down there. So as they say "If the mountain will not come to Muhammad, then Muhammad must go to the mountain"  After thinking about hosting a Seattle version for several months I finally jumped in.

I'm still figuring out ways to draw people to the events but with the places I tried, I already have a small mailing list and 4 people showed up at the inaugural night:

To give a flavor of what a MathJam is like (or at least our version) here's some of what happened. First, each site takes turns providing a starter set of puzzles and problems for all the groups. Last night's set was from Bristol, UK.  Because its my default instinct, I also brought a few things I had seen recently and thought would be interesting.

We started with the make 21 puzzle which took  us quite a while to crack.  Hint: you need to use fractions.

I then suggested we work this puzzle I found from @mpershan:

I actually had done this one a few weeks ago but conveniently forgotten my solution and it was still pleasant to rework out with a group.  I will definitely use this one again with kids.

We then had conversation about the egg timer puzzle. I'm still not sure which answer is correct but after considering some of the factors (suspended sand vs the force of the particles hitting the bottom vs a closed system for mass) it seems like a great science project.

Next for a change of pace, we discussed Colin Wright's Polynomial puzzle:

"Think of a polynomial with positive integer coefficients then pick an integer greater than any of coefficients and give me its value. With only that, I will tell you the original polynomial:  Example:  p(5) = 413."

I had thought about this algorithmically  previously. If you use a greedy strategy, you will always generate a unique polynomial. i.e. 5^4 is > 413 so the largest exponent is 5^3 and we can 3*5^3 = 375 then check 5^2 and then 5 and finally 1 to get  p(x) 3x^3 + x^2 + 2x + 3

Last night, one of the attendees immediately said isn't this just about converting the number into the base N?   And after thinking about it a bit I suddenly realized that was exactly what the greedy algorithm was doing  (413 in base 5 is 3123)   From there we had a lovely conversation riffing about exotic base systems like base 3/2 or Phinary:

And that was the fun of the night, we socialized a bit, talked about math in our lives, did some puzzles and went off on mathematical tangents.   I'm  expectantly looking forward to the second gathering next month. My goal is to attract a few more folks.

Tuesday, August 28, 2018

Parabola Coordinates

I put this together partly because I've been thinking about: Vieta Formula Brainstorming but mostly because I haven't seen it elsewhere.  The symmetry is more obvious when everything is on one diagram as well as the interesting correspondences once everything is in terms of roots. 

Interesting notes from the Vieta formulas  b = - (p+q) and c = pq.  Also the vertex form is particularly elegant looking because its position neutral.

Standard Form

Root Form

Vertex Form

Monday, August 20, 2018

Cyclotomic Polynomials

Most of the treatments of this topic are fairly grounded in Abstract Algebra and for this post I wanted to record my hopefully simpler conceptual framework.


Factor:  \( x^6 - 1 \)  I don't remember when I first saw this question but what was interesting about it was there were two paths.

If you broke it apart as a cube:  \( (x^2)^3 - 1  \) you'd get \(x^6 - 1 = (x^2 - 1)(x^4 + x^2 + 1) = (x+1)(x-1)(x^4 + x^2 + 1) \)


 If you broke it apart as a square you'd get:  \(x^6 - 1 = (x^3)^2 - 1 = (x^3 - 1)(x^3 + 1) = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1) \)

The second form was better but it wasn't clear to me at the time how you'd pick which way to go or the relationship between the two forms of factorization.  Although its more obvious now that since the expressions are equal  \((x^2 - x + 1)(x^2 + x + 1) =  (x^4 + x^2 + 1) \) 

There is some cool geometry underlying this factorization and all related forms \(x^n - 1\) that ultimately leads to some deeper understanding of what's going on here.  First rewritten \(x^n = 1\)  its clear we're looking for the n roots of one or as its usually termed the roots of unity.  (There must be n of them due to the fundamental theory of algebra)   We can then graph the solutions on the complex plane and soon some patterns emerge

$x = 1$
Solution: \( (1,0) \)

$x^2 = 1$
Solutions: \( (1,0), (-1,0) \)

$x^3 = 1$
Solutions: \( (1,0), (-\frac{1}{2}, \frac{\sqrt{3}}{2}i), (-\frac{1}{2}, \frac{-\sqrt{3}}{2}i) \)

$x^4 = 1$
Solutions: \( (1,0), (0,i), (-1,0), (0,-i) \)

What's noticeable?

  • 1 is always a root.
  • There's a lot of symmetry going on: sometime in one and sometimes across 2 axes.
  • Also although its not quite apparent yet, the roots are evenly spaced.

De Moivre's Theorem:

The reason for the symmetry and spacing comes directly from De Moivre's Theorem

$(\cos(x) + i \sin(x))^n = \cos(nx) + i \sin(nx) $

Consider sets of the form \( 2 \pi \) divided into n equal segments i.e. : \( 0, 1 \cdot \frac{2 \pi}{n},  2 \cdot \frac{2 \pi}{n}, 3 \cdot \frac{2 \pi}{n} \) ... \( (n-2) \cdot \frac{2 \pi}{n} , (n-1) \cdot  \frac{2 \pi}{n}  \)

When you apply De Moivre's theorem to each of these element raised to the nth power you get:

\( 1, \cos(2 \pi )  + i sin (2 \pi),   \cos(4 \pi )  + i sin (4 \pi) ... \cos((2n-1) \cdot 2 \pi )  + i sin ((2n - 1) \cdot2 \pi \)

and since the trigonometric are cyclic with period \( 2 \pi \) these are all equivalent and equal to 1.

In other words the n evenly spaced points that around the circle that split it up also form the set of primitive roots of degree n.

Example: Without directly solving as we did earlier we can now see the 5th roots:

Note: these are symmetrical reflected across the x-axis.

The roots are all related

The second interesting fallout of De Moivre's Theorem is that the roots are all powers of each other. The simplest case is to look at them all in terms of the first root \( \omega = \frac{ 2 \pi} {n} \)

So for example: \( \omega ^2 = 2 \cos{2 \cdot  \frac{ 2 \pi} {n}} + i \sin 2 \cos{2 \cdot  \frac{ 2 \pi} {n}} \) which is exactly the second root. This simple relation continues for all powers of n up to n-1.
(Likewise, you can raise the other roots to various powers and keep moving around the circle i..e this is a closed group)  Its often convenient to then think about the set of roots as \( \{ 1, \omega, \omega ^2 ...  \omega^{n-2}, \omega^{n-1} \} \)

Addition Results

Intuitively we  can see from the graph above that in cases like n = 2 or n = 4 that the roots appear evenly balanced and will add up to 0. Given this observation above we can show this is generally the case even in the odd powers where its clear from the graph that the points balance each other vertically but its not as easy to see the horizontal case.

Adding the roots we get a simple geometric series  \( 1 + \omega^2 .... + \omega^{n-1} = \sum_{i=0}^{n-1} \omega^i \).

Applying the sum of a geometric formula produces:
$ \sum_{i=0}^{n-1} \omega^i  =   \frac{ \omega^n  - 1}  {\omega - 1} $

But \( \omega \) is an nth root of 1 so by definition \( \omega^n = 1 \) and so:

$ \sum_{i=0}^{n-1} \omega^i  =   \frac{ \omega^n  - 1}  {\omega - 1} =  \frac{1 - 1}  {\omega - 1} = 0$

What's nice about this result is that we can even further break it apart on the complex plane and say that the cosines of the roots sum to 0 and the sines do the same. This allows the calculation of some interesting relations like on the last graph of the fifth roots:

$ \cos{0} + \cos{ 1 \cdot \frac{ 2 \pi}{5}} +  \cos{ 2 \cdot \frac{ 2 \pi}{5}} +   \cos{ 3 \cdot \frac{ 2 \pi}{5}} +  \cos{ 4 \cdot \frac{ 2 \pi}{5}} = 0 $

$ 1 + \cos{ \frac{ 2 \pi}{5}} +  \cos{ 2 \cdot \frac{ 2 \pi}{5}} +   \cos{ 3 \cdot \frac{ 2 \pi}{5}} +  \cos{ 4 \cdot \frac{ 2 \pi}{5}} = 0 $

And since from the symmetry \( \cos(n \cdot \frac{2 \pi}{5} ) = \cos( (5-n) \cdot \frac{ 2 \pi}{5}) \)  we can consolidate a little further:

$ 1 +  2 \cdot \cos{ \frac{ 2 \pi}{5}} +  2\cdot  \cos{ 2 \cdot \frac{ 2 \pi}{5}}  = 0 $
$ \cos{ \frac{ 2 \pi}{5}} +  \cos{ \frac{ 4 \pi}{5}}  = - \frac {1}{2} $

Multiplication Results

Since all the roots are powers of each other, its  clear that multiplying any of them together will just result in another one of the roots. (you may cycle around the circle a few times) This begs the parallel question to the previous section: what happens if you multiple all of them together?

Example: multiply all the 4th roots of unity starting at (1,0) and ending at (-1, 0)

We can also apply the same line of reasoning to the multiplication of all the roots and see what falls out:

so \( \prod_{i=0}^{n-1}\omega^i = 1 \cdot \omega  \cdot \omega^2 ..... \omega^{n-2} \cdot \omega^{n-1} \)

Consolidating the powers of  \( \omega\) we get  \( \prod_{i=0}^{n-1}\omega^i  = \omega^{1 + 2 ... + (n -1)} \)   The exponent in this case is a triangle number and using the formula for such sums \( \sum_{i=1}^{n} i   = \frac{n(n +1)}{2} \)  We get  \( \prod_{i=0}^{n-1}\omega^i  = \omega^{\frac{(n-1) \cdot n}{2}}  =  ( \omega^n ) ^ {\frac{n - 1}{2}}\)

But again \( \omega \) is an nth root of 1 so \( \omega^n  = 1 \) which means the whole product
simplifies to \( 1^ {\frac{n - 1}{2}} = \sqrt{1} = \pm 1 \)  Further its positive when n is odd and negative when n is even.

We can also generally break the roots apart like with addition to see what the product of the cos or sin of them are.   The general way to do this is just to expand the original equation \(x^n  - 1 = 0\) by substituting in  \( \omega = \cos(x)  + i \sin(x) \) and then breaking out the real and imaginary parts, substituting \( \cos^2 = 1 - \sin^2 \) or vice versa and solving.  Caution: be careful to solve both parts since extra solutions may appear in either half that are not common to both.  Note: also regular trig multiplication to addition rules can usually also be applied to reduce these type problems back into the addition cases.

This is a nice parallel to addition. When you add all the roots you get 0, when you multiply them all you get \( \pm 1 \).

Cyclotomic Polynomials

Returning to our family of functions: \( x^n - 1 = 0 \)  We can now look at the factoring problem more closely.

First we'll reuse the sum of a geometric series for n > 2 formula

$ \sum_{i=0}^{n-1} x^i  =   \frac{ x^n  - 1}  {x - 1} $

So we can always factor as following:

$ x^n  - 1 =  (x- 1)(1 + x + x^2 ...  + x^{n-1}) $

This reaffirms that 1 is always a root of any of these functions.  But when can we further factor? The graphs of the unit circle are again helpful in pointing out the solution:

Here we're looking at  roots of \(x^6 - 1 \) the original problem at the top.  Because the roots are evenly spaced they are at  (1,0) and then 1/6, 2/6, 3/6, 4/6 and 5/6 of the way around the circle. But some of these fractions reduce. For example,  2/6 is also 1/3 or one of the roots of \( x^3 - 1 \) and 3/6 is also 1/2 one of the roots of \( x^2 - 1 \)  In other words, factoring the polynomial is tied to factoring the degree of the polynomial. In \(x^n - 1\) if  degree n is not prime and can be factored \( p_1^{d_1} \cdot p_2^{d_2} \cdot ... p_k^{d_k} \)  we can find all the primitive roots for its factors in the \(x^n - 1\) itself.     If we then multiply \((x - \omega_n) \) together for all the roots from a \(p_k\) we'll get back the original polynomial \(x^{p_k} - 1\)

This gives us an algorithm to factorize the polynomials of form \(x^n - 1\).

  • First factorize the degree of the polynomial
  • Since 1 is a common of root of all the polynomials involved divide all of them by x -1.
  • Then divide the remainder \(x^n - 1\) by \(x^k - 1\) for all factors k. to accumulate sub polynomials.
  • Combine all the polynomials found including x -1 back together to get the full factorization.

The irreducible polynomials that result in the end have a special name, cyclotomic polynomials.  As the model above suggests if the degree of the exponent n  for all n > 1 is  prime then \(\frac{x^n - 1}{x -1}\)  will be cyclotomic.

Example:  \( x^{21} - 1 \)

1. 21 factors to \( 3 \cdot 7 \) 
2. So we have 3 subpolynomials:  x - 1, \(x^3 - 1 , x^7 - 1\)
3. Dividing x - 1 out of the latter two we get: \(x^2 + x + 1,  x^6 +x^5 + x^4 +x^3 + x^2 + x  + 1\)
4. Dividing each of these in turn from  \(x^{21} -1\) We are left with: \( x^{12} - x^{11} + x^9 - x^8 + x^6 - x^4 +  x^3 - x + 1 \)   

All together \( x^{21} - 1 = (x-1)(x^2 + x + 1)(x^6 +x^5 + x^4 +x^3 + x^2 + x  + 1)(x^{12} - x^{11} + x^9 - x^8 + x^6 - x^4 +  x^3 - x + 1) \)   

and in formal notation we call the final residual polynomial \(\Phi_{21}\)

First Few Cyclotomic Polynomials

{\displaystyle {\begin{aligned}\Phi _{1}(x)&=x-1\\\Phi _{2}(x)&=x+1\\\Phi _{3}(x)&=x^{2}+x+1\\\Phi _{4}(x)&=x^{2}+1\\\Phi _{5}(x)&=x^{4}+x^{3}+x^{2}+x+1\\\Phi _{6}(x)&=x^{2}-x+1\\\Phi _{7}(x)&=x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\\\Phi _{8}(x)&=x^{4}+1\\\Phi _{9}(x)&=x^{6}+x^{3}+1\\\Phi _{10}(x)&=x^{4}-x^{3}+x^{2}-x+1\\\Phi _{11}(x)&=x^{10}+x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\\\Phi _{12}(x)&=x^{4}-x^{2}+1\\\Phi _{13}(x)&=x^{12}+x^{11}+x^{10}+x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\\\Phi _{14}(x)&=x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1\\\Phi _{15}(x)&=x^{8}-x^{7}+x^{5}-x^{4}+x^{3}-x+1\\\Phi _{16}(x)&=x^{8}+1\\\Phi _{17}(x)&=x^{16}+x^{15}+x^{14}+x^{13}+x^{12}+x^{11}+x^{10}+x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\\\Phi _{18}(x)&=x^{6}-x^{3}+1\\\Phi _{19}(x)&=x^{18}+x^{17}+x^{16}+x^{15}+x^{14}+x^{13}+x^{12}+x^{11}+x^{10}+x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\\\Phi _{20}(x)&=x^{8}-x^{6}+x^{4}-x^{2}+1\\\Phi _{21}(x)&=x^{12}-x^{11}+x^{9}-x^{8}+x^{6}-x^{4}+x^{3}-x+1\\\Phi _{22}(x)&=x^{10}-x^{9}+x^{8}-x^{7}+x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1\\\Phi _{23}(x)&=x^{22}+x^{21}+x^{20}+x^{19}+x^{18}+x^{17}+x^{16}+x^{15}+x^{14}+x^{13}+x^{12}+x^{11}\\&\qquad \quad +x^{10}+x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\\\Phi _{24}(x)&=x^{8}-x^{4}+1\\\Phi _{25}(x)&=x^{20}+x^{15}+x^{10}+x^{5}+1\\\Phi _{26}(x)&=x^{12}-x^{11}+x^{10}-x^{9}+x^{8}-x^{7}+x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1\\\Phi _{27}(x)&=x^{18}+x^{9}+1\\\Phi _{28}(x)&=x^{12}-x^{10}+x^{8}-x^{6}+x^{4}-x^{2}+1\\\Phi _{29}(x)&=x^{28}+x^{27}+x^{26}+x^{25}+x^{24}+x^{23}+x^{22}+x^{21}+x^{20}+x^{19}+x^{18}+x^{17}+x^{16}+x^{15}+x^{14}\\&\qquad \quad +x^{13}+x^{12}+x^{11}+x^{10}+x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\\\Phi _{30}(x)&=x^{8}+x^{7}-x^{5}-x^{4}-x^{3}+x+1.\end{aligned}}}

TODO: flesh out

To complete the model its only necessary to motivate why the cyclotomic polynomials cannot be reduced further.

The hand wavy  argument goes like this:  We've already removed every symmetric pair via factoring out the other cyclotomic polynomials and we're now left with the totient set.  If a set of these produced a smaller polynomial with integer coefficients it would have to have conjugate pairs to cancel out the complex parts of the roots. But those conjugate pairs would imply a symmetry that would imply another smaller factor existed that we didn't already find which is a contradiction since we fully factored already.

I don't fully have a good model here yet so here's a pointer to a standard approach:

Tuesday, August 14, 2018

Getting Back into Gear

Never put off till to-morrow what you can do day after to-morrow just as well.—B. F   "Mark Twain falsely attributing a quote to Benjamin Franklin"
Its that time of year again when I start to think about logistics and planning.  Unlike last year, this time feels like a known quantity.   As a result of experience there are a few things I want to changes.

  • Utilize the facebook page and PTSA newsletter.
  • Get a blurb in the school newspaper if it publishes early enough.
  • Get one of last year's kids to make a morning announcement on the PA.
  • Reach out to the Alg2 teacher.  (This is the biggest stretch since I don't know him.)

My expectation is I'll keep most of the kids who were there last year and didn't graduate and I'll find a new crop of sixth graders.  But because of the odd demographics I'm not sure about 8th graders. So that's my motivation for trying to sync up with Alg2 teacher and see if he would hand out a flyer for me or nudge some students.   Male / Female balance was decent last year but could be better. But I'd really have to engage all the teachers to actively do something about that. I'm going to wait until the first responses come back to decide whether I need to be active here.

New ideas:
  • I'm going to put up a self study spreadsheet for AMC8 per the request of a few kids. I'll track and see if anyone is participating.  What can I do to optimize usage?
  • Procuring permission to photograph.  I want to create photos for my talk later this fall. So for the first time, I'm going to ask permission to take them.
  • Being explicit about pre-Algebra vs. Algebra when advertising. My idea is to say we may use algebra/geometry from time to time but its ok to join as long as you're curious and willing to ask questions.

Gathering Ideas:
I spend the summer just recording whatever looks interesting for use later on.  It looks very reactive when you jot it down at first but six months later I really like looking back.

graph war -   Can this be adapted back to paper?

fractals (The kids asked for more on this theme at the end of last year)
logs (This is just me)
graph coloring quanta article.
Q8 group experiment:

PCMI problem sets

String Art / Parabolas

NRICH: Haga's Theorem

Tantons video on geom series:
Cool egyptian fraction / infinite geom sequence metaphor

Balanced Ternary: Robjlow chalkdust issue 6 article  (

Several units organized around math history. Maybe the Euclid Algorithm with direct pieces of the Elements would work.

Collatz conjecture:  Harris's visualization is really cool.    Another one: abd there is a numberphile video

Revisit cutting mobius strips plus gluing mobius strips to get a klein bottle

Could we do parts of this?

Choose an integer bigger than all the coefficients, and evaluate. so, for example, p(9)=26260. Then tweet x and p(x). I'll then guess your polynomial.
via Colin Wright.
Greedy Algorithm works and its unique because of the second constraint.

Proof that sqrt(2)^sqrt(2) = 2

graceful labelling of graphs:

Several different art/math project ideas "Curves of Pursuit"

Rubiks cube: is this the year to do it? Its about $8 a cube and I bet a bunch of kids already own one.

A golden ratio/pentagram art idea (although we covered this partly last year)

Euclidean Algorithm: Need some problems around it - also a candidate for a talk on Euclid

Potato Paradox:
You have 100kg of potatoes, which are 99% water by weight. You let them dehydrate until they're 98% water. How much do they weigh now?"

Math4Love's Hindu in School columns:

I'm also in preliminary discussions with Richard Rusczyk at Art of Problem Solving. If all the stars align he may give a talk at the school when he's out here this fall.

Wednesday, August 8, 2018

Vieta Formula Brainstorming

I've on and off thought a bit about Vieta's Formulas over the last few years.  The AoPS Algebra textbook introduces them and has a few beginner problems. They initially seemed both obvious and a bit contrived in their application. Then I learned about Vieta Jumping as a technique in the IMO and bit by bit various problems showed up where they were very useful indeed.  But a chance encounter with a paper by Ben Blum Smith @ linking them to Galois Field Theory has me extremely interested in them now and I'm going through the brainstorming process here to see if I could build a day out of them. This is definitely still a work in progress.


  • This could only be in the Spring after any students in Algebra have quadratics under their belt.
  • How would I handle any pre-algebra students?
  • How much complexity can we explore in 1 hour?

Math History:
What more can I find?

Quadratic Exposition:
[This is condensed from a twitter discussion with Ben]

Graphically speaking the quadratic formula can be interpreted as there is a point equidistant
between the two roots (and incidentally aligned with the vertex).   That distance is
\( \frac{b^2 - 4ac}{2a} \)

We derive that by completing the square but using the Vieta formulas we get another interpretation.

For this discussion lets only consider monic quadratics where a  = 1 and let p and q be the 2 roots.

  • c  = pq
  • b = -(p + q)
So \(  \frac{b^2 - 4c}{2}  =  \frac{(-p -q)^2 - 4pq}{2} = \frac{(p-q)^2}{2}  \)

This means the discriminant is precisely the average of the square of the distance between the 2 roots!

Likewise \( \frac{-b}{2} = \frac{x + y}{2} \) or the average the 2 roots.

[Probably I would have the kids state the quadratic formula and then try to label the graph first and then ask "yes but why?"]

Unsolved tangent: The vertex in terms of  p and q is  \(- (\frac{p-q}{2})^2 \).  How does that fit in geometrically speaking?

Entry Problems:
  • What is \( \frac{1}{x} + \frac{1}{y} \)  if you only know \(xy = k_1 \) and \(x + y = k_2 \)
  • let a and b  denote the roots of \( 18x^2 + 3x - 28 = 0 \) Find the value of \( (a - 1)(b - 1) \)
  • Some basic i.e. the root sum is x and the root's product is y. What is the polynomial?
  • Problem from
  • If m an n are nonzero roots of x^2 + mx + n = 0 What does m+n  equal?

  • Basic Formula expansions  - quadratic, cubic and maybe symmetric in x,y and low degree
  • Reducing various sample polynomials to elementary symmetric polynomials. 

I think I would probably lead with 1 group problem - then a bit of math history. Can I find some media to enhance this and then break back into the whiteboard problems?

Then Stop to do some group work on the idea all symmetric polynomials can be reduced and then try out some expansions. (Maybe tie this further ala Ben's paper or maybe introduce the numberphile video)

Problem: That already is probably more than 1 hour of material.

Friday, August 3, 2018

Trig Problem Walk Through

John Rowe posted the fun puzzle pictured above on twitter.  The goal is to find x.

One reasonable brute force approach is to start with lower part of the diagram and apply trigonometry to find various angles and sides and work your way up until you arrive at x.

This looks a bit more complex than it really is but its a lot of steady work.  So I immediately thought it might be possible to apply synthetic techniques and save some work.

The actual process I went through

1. Angle Chase
2. Look at all similar triangles (and extend to make one more at the top).  This made finding all the of the 22.5-67.5-90 triangles a bit easier.

Note split between the top and bottom. You need to start angle chasing near the bottom middle and the work counter clockwise but once its done the rest of the problem only needs the top part of the diagram to be solved. The lower 6x6 isosceles triangle is superfluous.

False start: I thought I had found more congruent triangles than there really were and after testing I realized something was bogus.  So I went back and realized there was only one angle bisector indicated on the diagram while I had assumed there were 3 for some reason.

3. I then noticed the similar triangles at the top. From them I could find the ratio for base of the desired triangle and apply trig. I did this like below but found an easier way later on from online.

4. Discover all the hidden 45-45-90 triangle and draw them in.

(Total Time: 1 day a burst before work and then a refinement when going to bed. As usual a bit of incubation often helps)

Final Method

First always start by angle chasing.  After doing so I noticed the large number of 22.5-67.5-90 triangles as highlighted above.

The second thing to notice (although this is not the order I did it in is all the explicit and implicit 45-45-90 triangles. In particular x lies on OE which is at a 45 angle.

What this means is that all the interesting information is actually in the upper left hand corner:

Here I've blown up and squared off the top right corner. In addition I've added replaced one diagonal line with some horizontal and vertical ones to highlight another small right isosceles triangle.

What this shows is:

  • The total length on the side is 10 so IJ = HI - HJ = 10 - 6  = 4.
  • The two lower triangles AEJ and AJI are congruent by SAS and therefore EJ is 4 as well.  You could stop right here and use trigonometry to find x.   \( \frac{4}{x} = \tan{\frac{\pi}{8}} \)
  • We can use basic special triangle properties to find AE by just working our way around the square from EF to FG to AC to AE.


Final Note: My original mistake actually lend itself to a related but interesting problem. If the top triangles are congruent then the 67.5 angle is actually trisected. The key difference is there is no perpendicular angle at E and the two lower angles are congruent instead.

Thursday, August 2, 2018

Fantasy High School Part 2

[See: for part 1]

Recently I listened to a podcast  interview with Conrad Wolfram: Link.  I've been aware of some of his thinking especially his popular: Ted Talk. But in the past it seemed to hazy and unfleshed out to get my mind around and decide if I agree.  In the intervening 8 years, he's moved forward and there's a website: and the beginnings of some concrete curricular ideas.

So I'm going to parse and explore some of the questions I thought while listening again especially in the context of "Is Wolfram's vision the right path forward for High School?"

To paraphrase his basic arguments:

Math is all about 4 steps:

1. Define the problem
2. Can we turn it into a symbolic representation?
3. Take the question to an answer
4. What does it mean? Is that right?

He points out that Step 3 is where technology can be most useful, yet we spend 80% of our math time in schools on that step by teaching hand calculating. Instead, he says, we should be focused on steps 1, 2, and 4 which are what people really need to be good at today.

What this means in practice is using all the tools of Mathematica to explore more complex problems that don't have neat Diophantine solutions or simple characterizing equations. "Why not use a cubic?"

An example problem: 

"2. "Should I insure my laptop?"
Students begin by playing lotteries and calculating the costs and probabilities of winning. They learn how insurance works and then determine their personal utility curves for their laptop and compare the losses and gains made from insuring or playing a lottery. They role–play a competitive insurance market, with student insurance agents determining policy premiums to offer to their student clients. The teacher is able to see the big picture and determine an overall winning insurance agent."
I think there are some interesting ideas brought up here.
  • Are we overemphasizing algebra 1 techniques like factoring?
  • Is Mathematica the right tool to use instead vs. Desmos or Python, Sage or ...?
  • Is coding more fundamental? Do we need to emphasize this first?
  • The sample material presented is pretty small and still has a very strong statistics bent. It remains to be seen whether this approach is comprehensive and links together.
For a concrete example: Let's think about number talks. To channel Wolfram focusing on the various ways we use the distributive law etc. to solve a problem like 23 x 5 mentally is the wrong use of time. In real life its going to be 23.049 x 4.67 and we won't mentally solve it exactly at all. Instead, what's useful is knowing how to estimate this is approximately 20 x 5 (or any other good estimation) and realizing how to quickly calculate on a phone/calculator/desmos etc and confirm the answer is reasonable.

I'm going to focus on the thorniest question for me.  "Can we get here from there?"  How much conceptual understanding and practice with simpler problems is needed to effectively use a sophisticated tool like Mathematica? I'd argue we don't really have an examples yet of this since we all still are educated in the old path. I'd really like to see experimental data on real kids to see how this all comes together. [And in Wolfram's defense there seem to be pilot projects ongoing]   Likewise, this is fundamentally intertwined with which topics should be explored during High School. All these tools can be applied in multiple domains.  Which ones are fundamental and should be done universally? 

At this point I see this more viably as an approach to enrich the curriculum. In other words:
  • teach enough coding to use tools like Mathematica
  • tackle more complicated applied problems using tools.
  • Thread these explorations and visualizations through the regular curriculum.
  • This has particular relevance to Calculus which I'm still avoiding but in a nutshell what is the value of learning to integrate if this can be done by tools?

Tuesday, July 24, 2018

Visual Proof: The mediant is in the middle

Usually I've seen this done algebraically. See:   But from a vector addition standpoint (or a parallelogram) you can see visually that the slope of the diagonal which is the mediant falls between the original slopes.

Sunday, July 22, 2018

Forward to sender

This week I had been thinking of writing about the Big Internet Math Off but instead was invited to write about it on itself. So go to and give it a read.

Tuesday, July 17, 2018

Fantasy High School

@carloliwitter was tweeting recently about re-imagining High School mathematics and asking for ideas. The topic has caught my fancy even though its a bit off message for the blog.  High School is ripe territory for change while at the same time rife with difficulties for anyone attempting to do so. I'll start with a model that I wish I could have had myself:  proof school.  The high schoolers there do the following:

2 math blocks every afternoon covering a single subject over 6 weeks including:
  • Problem Solving + Combinatorics 
  • Algebra + Number Systems
  • Geometry + Topology 
  • Analysis + Statistics
  • Number Theory + Computer Science
The course descriptions sound interesting and they deep dive into number theory, abstract algebra, more advanced Geometry topics etc.

Note: Calculus is not covered.

Ok so that's where my heart's at but I don't think that within a public school setting you could achieve the same thing and the more general challenge is providing a interesting and relevant path for everyone in the system.

So here's some more ideas about changes with the caveat that these are only my impressions and desires and don't reflect a systematic view of the state of the world.

  • Reinvigorate Geometry. There's enough material to split into 2 years on its own. De-emphasize all the taxonomy, volumes and surface areas. Instead elevate a proof and computer aided exploration approach. I'd particularly like to see more time to get to cyclic quadrilaterals, powers of a point, Ceva's theorem etc.  That only seems possible if you stretch the course over more than 1 year.   Then its easy to branch out to circle inversions, tessellations etc.
  • Explode Algebra II.  The narrative structure isn't there and worse it often ends up being a significant  rehash of Algebra I.   What's core here and what can we add in to make this connect more clearly to future topics? I don't think touring absolute value, exponential functions,  polynomials etc. is working very well.  What's worse is that pre-calc often bleeds into this material.
  • Integrate programming more fully as first class component. Maybe the capstone of H.S. should be based around numerical computing.
  • Stop doing stats as an add-on over several years. Make this a first class subject and perhaps a joint responsibility of the Science and Math departments. 
  • Create more "labs" i.e. exploration opportunities.
  • Have an applied pathway like Canada. Let's figure out for real what we really need and no more. So Mathematics is not a barrier to those who aren't going into STEM.
  • More overall arc to the pathway. Students should have more of an idea of what they are heading towards and what the overall field looks like as well as some of the history of how it developed.
On the flip side what are the systemic issues making this hard to do:

  • Choice models. Any model that envisions multiple pathways runs into several immediate concerns. First is that finding teachers to teach all these new subjects well is hard and anytime you have to schedule more classes it makes the logistical process more difficult. Its always going to be easier to schedule math where everyone does the same thing each year so all the teachers and period slots are more easily interchangeable.  Secondly, these models work best at large schools where there is a sufficient number of students to actually support them.  Small schools especially in low population areas may never be able to pull this off. Finally, status issues among classes easily subvert the aim of offering them. If one pathway is perceived to be more useful to getting into college then any such system will devolve into a high/low pathway with all the more ambitious students making the same choices. See: Calculus.
  • Integration with College. Any such system has to not impede admission chances and that creates tremendous inertia as well as concern among students and parents.
  • Tensions over tracking.   This feeds into my previous point about choice. More radical anti-tracking ideas mean you really cannot provide anything but a single pathway because one of them is likely to be come the de facto "high" pathway if left to develop naturally. 
  • The realities of the wide difference in relative performance by high school. Kids come into ninth grade at vastly different places within the curriculum and their own personal journey. The system needs to move everyone forward but finding a way to do this is really hard. Most of the solutions so far involve slowing down significantly and basically hanging out in the pre-algebra and then algebra space for much longer than we currently do.  We really need a way to remove math as a gateway to everything post 12th grade. Because systemically we know that isn't working.

[I deliberately didn't really tackle Calculus's place in the curriculum here. That deserves a followup.]

Monday, July 9, 2018

Book Review: Introduction to Number Theory and An Illustrated Theory of Numbers

Over the last 5 months,  I worked through both Introduction To Number Theory and  parts of An Illustrated Theory of Numbers with my son at home so I thought while the experience is still fresh I would write down my impressions.

We started with the Art of Problem solving textbook.  This is considerably shorter than some of their other volumes and its generally expected to take only half a year working through it. (Scale everything by how long it took you to do the Algebra book)  In fact, this was the first AOPS textbook I ever purchased.  A few years back, my neighbor's son was working through a problem set for the online version of this book and needed some help on the problems. I spent a weekend going through the problem set and at the end thought to myself, "Wow these are really interesting problems."  The one below was my favorite.

There are unique integers \(a_2, a_3,   ... a_7 \) such that
$$\frac{5}{7} = \frac{a_2}{2!} + \frac{a_3}{3!} + \frac{a_4}{4!} + \frac{a_5}{5!} + \frac{a_6}{6!} + \frac{a_7}{7!}$$
where \( 0 <= a_i < i\) for i = 2,3...,7.    Find  \(a_2, a_3 ... a_7 \)

So I ended up purchasing book then and reading and trying out some of the exercises over the winter in the ski lodge while my kids were taking lessons.  Five years later, I've actually had experience with a few of the other books and I now wanted something to do with my son over the back half of the year.  

The book itself informally divides into 3 major sections. The beginning sections on integers, divisibility and factorization which culminate in the Euclidean Algorithm for finding the greatest common divisor. There is a middle section  which includes different base number systems and a review of some applications with decimals and fractions. Followed by the last section which deals with modular arithmetic and linear congruences.

Overall the pacing and material is a bit more uneven than some of the other textbooks in the series. I found portions of the earlier and middle chapters repeated topics from the pre-algebra book and some of the problem sets at the end felt a bit repetitive. So we tended to skim some of these chapters.  In particular, if I were revising the book I would replace the decimal/fraction chapters and perhaps even some of the focus on different number bases with a dive into the Chinese Remainder Theorem and a discussion of Diophantine equations and how they relate to the linear congruences.

However, the payoff really was in the last 4 chapters starting with the introduction of modular arithmetic. These are all  well done and have a nice ratio of practice to theory. If one were limited in time, I would focus on this section. I particularly like the work around exponent towers as a motivating problem.

Because there were a few weeks left in the school year when we finished the AoPS book and after seeing a favorable review by Mike Lawler I also purchased Weissman's book.  This is actually a very nice companion to the previous book.  What's most striking is the extremely strong conceptual/visual framework.

We ended up doing the introductory chapter and skipping ahead to modular arithmetic parts. My hope was to make it to the topograph sections but we ran out of time.  The very strong narrative strands really make this book. There are interesting conceptual pieces and data visualizations almost every few pages along with illustrations to go with them.  As  Weissman notes "Most of our proofs are given with visual explanations; geometric and dynamical proofs are preferred"  which makes the treatment fairly different from other texts.

For example, the early chapters have some lovely figurate number illustrations just with the number 100:

and then a section on using Hasse Diagrams to visualize factorization (rather than the more normal trees).

Where this really works well though is in the later chapters.   I love conceiving of the "modular world" and having it bound horizontally via factorization and vertically via powers of prime.

The final chapters which use John Conways topograph construct rather than more familiar approaches to discuss quadratic residues are particularly fun.

Overall, the only weakness I felt in the book was a need for more problems at the end of the chapter. From time to time I also found myself missing interleaved problems ala AOPS as well.  For older more experienced students I might just use the Wiessman book alone but otherwise it made for a very nice complement to the AoPS text.

Wednesday, June 27, 2018

Sometimes representation does matter

This is a small observation based on a post from @samjshah on the topic of the trig double angle formulas:

  • \( \sin{2\theta} = 2 \sin{\theta} \cdot cos{\theta} \)
  • \( \cos{2\theta} = \cos^2{\theta} - \sin^2{\theta} \)

Sam used the construction below to derive the formulas which is a bit different than how I usually think of doing it.


I immediately thought to myself, I can assign x to the base of one triangle and all the relationships must fallout from either the Pythagorean Theorem or similar triangles.  This is certainly true and it looked like this:

So for example using the direct definition of sin and cosine:
$$\sin{2\theta} = \frac{\sqrt{1-x^2}}{1} = \sqrt{1 - x^2} $$

which you can compare to:

$$ 2 \cdot \sin{\theta} \cdot \cos{\theta} = 2 \cdot \frac{1+x}{\sqrt{2} \cdot \sqrt{1+x}} \cdot \frac{\sqrt{1-x^2}}{\sqrt{2} \cdot \sqrt{1+x}} =  \frac{2 \cdot (1 + x) \sqrt{1 -x^2}}{2 \cdot ( 1 + x )} = \sqrt{1 - x^2} $$

That was the expected result but didn't really seem that exciting to me.  But then I looked at  how Sam had done it. Rather than assigning x to the missing piece he directly used the trig value and did this:

This time from the similar triangles BDE and ACF you get the following relationship:

$$\frac{BD}{BE} = \frac{AC}{CF}$$  or
$$\frac{2\sin{\theta}}{\sin{2\theta}} = \frac{1}{\cos{\theta}}$$ which after cross multiplication directly gives
$$\sin{2\theta} = 2 \sin{\theta} \cdot \cos{\theta} $$

So with this representation something much more insightful for me falls out.  Trig while offering a lot of power often has the ability to obscure geometric relationships. But as seen here sometimes with careful usage it can be quite interesting. And of course the importance of the problem setup towards simple vs complex solutions to the same problem is much more universal.