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

### Equilateral Triangle

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

### Square

In square

Running a middle school math club.

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

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

In square

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. https://mathsjam.com/what-is-mathsjam/

"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: https://en.wikipedia.org/wiki/Golden_ratio_base

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.

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

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.

Motivation:

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 \(x^6 - 1 = (x^2 - 1)(x^4 + x^2 + 1) = (x+1)(x-1)(x^4 + x^2 + 1) \)

versus

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 \((x^2 - x + 1)(x^2 + x + 1) = (x^4 + x^2 + 1) \)

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

versus

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

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?

####
De Moivre's Theorem:

$x = 1$

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.**

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

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.

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

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

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 \( \{ 1, \omega, \omega ^2 ... \omega^{n-2}, \omega^{n-1} \} \)

Intuitively we can see from the graph above that in cases like n = 2 or n = 4 that the roots

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} $

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.

This is a nice parallel to addition.

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 \(x^n - 1\).

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

**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:

https://brilliant.org/wiki/cyclotomic-polynomials/#proof-of-irreducibility

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:

https://brilliant.org/wiki/cyclotomic-polynomials/#proof-of-irreducibility

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.Advertising:

- 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 - graphwar.com 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. https://t.co/sHm2Xz8ZJN

Q8 group experiment: http://mymathclub.blogspot.com/2018/05/tangles-and-symmetry.html

PCMI problem sets projects.ias.edu/pcmi/hstp/problemsets.html

String Art / Parabolas

NRICH: Haga's Theorem https://plus.maths.org/content/folding-numbers

Tantons video on geom series: https://www.youtube.com/watch?v=Q39pDPoL0no

Cool egyptian fraction / infinite geom sequence metaphor

Balanced Ternary: Robjlow chalkdust issue 6 article (http://chalkdustmagazine.com/read/issue-06/)

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

https://www.maa.org/press/periodicals/convergence/primary-historical-sources-in-the-classroom-discrete-mathematics-and-computer-science#Projects

Collatz conjecture: Harris's visualization is really cool. Another one: https://www.youtube.com/watch?v=GJDz4kQqTV4&app=desktop 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?

http://chalkdustmagazine.com/blog/many-quadratics-factorise/

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:

fractals (The kids asked for more on this theme at the end of last year)

logs (This is just me)

graph coloring quanta article. https://t.co/sHm2Xz8ZJN

Q8 group experiment: http://mymathclub.blogspot.com/2018/05/tangles-and-symmetry.html

PCMI problem sets projects.ias.edu/pcmi/hstp/problemsets.html

NRICH: Haga's Theorem https://plus.maths.org/content/folding-numbers

Tantons video on geom series: https://www.youtube.com/watch?v=Q39pDPoL0no

Cool egyptian fraction / infinite geom sequence metaphor

Balanced Ternary: Robjlow chalkdust issue 6 article (http://chalkdustmagazine.com/read/issue-06/)

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

https://www.maa.org/press/periodicals/convergence/primary-historical-sources-in-the-classroom-discrete-mathematics-and-computer-science#Projects

Collatz conjecture: Harris's visualization is really cool. Another one: https://www.youtube.com/watch?v=GJDz4kQqTV4&app=desktop 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?

http://chalkdustmagazine.com/blog/many-quadratics-factorise/

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:

https://t.co/jxO2w7Zgbm

https://t.co/Mtvg5LEQCt

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

https://www.artfulmaths.com/mathematical-art-lessons.html

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)

http://mathhombre.blogspot.com/2018/08/golden-triangles.html

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:

https://mathforlove.com/2018/08/a-mathematician-at-play-11-unfair-numbers/

https://t.co/Mtvg5LEQCt

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

https://www.artfulmaths.com/mathematical-art-lessons.html

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)

http://mathhombre.blogspot.com/2018/08/golden-triangles.html

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:

https://mathforlove.com/2018/08/a-mathematician-at-play-11-unfair-numbers/

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.

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 @ https://arxiv.org/pdf/1301.7116.pdf 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.**

Constraints:

Math History:

https://en.wikipedia.org/wiki/Fran%C3%A7ois_Vi%C3%A8te

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.

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?

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.

Constraints:

- 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:

https://en.wikipedia.org/wiki/Fran%C3%A7ois_Vi%C3%A8te

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 http://mymathclub.blogspot.com/2018/02/in-praise-of-rational-roots-theorem.html#VietaEx
- If m an n are nonzero roots of x^2 + mx + n = 0 What does m+n equal?

Explorations:

- 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.

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.

Solution to the #trigonometry problem I put out a few hours ago. #MTBoS #iTeachMath https://t.co/ZyLzgkoRj7 pic.twitter.com/w6WYyCZJEf— John Rowe 👨🏽🚀 (@MrJohnRowe) August 2, 2018

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)

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.

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