Math off the grid

Two Hinged Triangle Geometry Walk Through

noreply@blogger.com (Benjamin Leis) — Tue, 18 Jul 2023 05:30:00 +0000

Setup

Its time for another geometry walkthrough motivated this week by this interesting problem from James Tanton.

"Two isosceles right triangles are hinged at corners as shown. Line segment connecting midpoints of their hypotenuses is used as the hypotenuse of yet another isosceles right triangle.

Prove A, B, C lie on a straight line.
Can anything be said about where B sits on segment AC?"

Initial Impressions

My first impression was that this reminded me a bit of the 3 hinged squares problem where among the key key observations was that the area of the triangles between the hinges was always the same. This problem has right triangles which are effectively half squares and involves the midpoints (or centers of the squares if they were there) so its not quite the same at least at first but keep that thought in the back of your head.

Secondly I wondered if it could be as easy as an angle chase. Sadly after working things out that appeared to not be the case. So this would take some more serious work.

The third observation I had is midpoints tend to generate similar triangles:

And if you draw in the extra line BD that is indeed the case here. Triangle BCD is similar to triangle BCG. BD is parallel to FG and the scale factor is 2:1.

But I wasn't quite sure yet what to with all of that. Colinearity is one of those slippery properties that is less directly proven. I hadn't been able to find the angles to show that they all added up to 180 via an angle chase which would be the most direct method.

Analytic Approach

So out of ideas I started going down an analytic path.

First I set C, the hinge point, to (0,0) and oriented one triangle ABC on the X and Y axes and assigned it length p. Then I squared off the second triangle and assigned those triangles length q and r. From there I eventually derived an expression for H that showed it was indeed on the line AE, This was a bit complicated and I don't tend to find the analytic solutions to be quite as explanatory as the synthetic ones.

However, via @jimsimons@mathstodon.xyz there is a nice improvement on this approach using the complex plane.

"Put C the point where the triangles meet at the origin, and let the points A and E be the complex numbers 4a and 4e (the 4's avoids fractions later).

The remaining vertices are B at 4a(1-i) and D at 4e(1+i). Note: how natural the 90 degree rotations are via complex multiplication.

The midpoints of the hypotenuses are 2a(1-i) and 2e(1+i).
The midpoint of the line joining them is a(1-i)+e(1+i).

So B is a(1-i)+e(1+i) + i(a(1-i)-e((1+i)) = 2a+2e, which is H the midpoint of AE!"

This makes all the calculations quite simple and is quite lovely.

Synthetic Approach

Back during my own investigation I still wanted a synthetic approach. Two more things occurred to me next.

The two parallel lines BD and FG meant I could make a second right triangle oriented at exactly the same way as FGH as so:

After staring a while (pobably over an hour or two in between other activities) it sure looked like H was the midpoint of CJ and J looked like it was exactly at the same vertical coordinate as E, A bit later after experimenting with variations of hinged triangles and seeing that these seemed to be invariants I added some more lines and came up with this lemma:

Given the same original hinged triangle ABC and CDE the big right isosceles triangle off of BD, BDJ forms the orange parallelogram ACEJ.

Process wise I assumed this and went onto to the next step and verified later but for clarity I'm going to show that is the case here and it turns out to be simpler than the original problem.

Lets go backwards and form the parallelogram ACEJ first and show BDJ is.a right isosceles triangle.

* First triangles BAJ and JED are congruent via SAS so the two edges DJ and BJ are congruent as well.

* Then a quick angle chase shows BJD is indeed 90 degrees.

Note: the similarities to the other problem I mentioned with the hinged squares. We once again are adding in congruent sides from each of the original figures to create 2 new triangles that share a congruent side from each of the originals

With that in hand and remembering that H the intersection of the parallelograms diagonals bisects both diagonals we can see:

Triangle FGH is surrounded by three similar triangles all in a 1:2 ratio. Or in other words FGH is a dilation of BDJ around point C scaling by a factor of 1:2.

Since via the lemma above. BDJ was a right triangle FGH must also be one! And further by definition its end point is on the diagonal AC and in fact at its midpoint. QED

Trig identity Three Ways (at least)

noreply@blogger.com (Benjamin Leis) — Sat, 04 Feb 2023 01:43:00 +0000

This post starts with reading elsewhere about someone struggling with the following trig identity given a triangle with three angles x,y,z show:

$$ \tan{x} + \tan{y} + \tan{z} = \tan{x}\cdot\tan{y}\cdot\tan{z} $$

The Straightforward approach

The first way I ended up doing this which was colored by some similar problems was just by working left to right and simplifying the expression with the hope I ended up with the right hand side.

Before getting going though its useful to examine the relationship between x, y, and z and how that plays out with the basic trig functions

1. $ \sin{z} = \sin{(\pi - (x + y))} = \sin{(x + y)} $

2. $ \cos{z} = \cos{(\pi - (x + y))} = -\cos{(x + y)} $

First looking at the first two terms restate in terms of sine and cosine:

$$ \tan{x} + \tan{y} = \frac{\sin{x}}{\cos{x}} + \frac{\sin{y}}{\cos{y}} = \frac{\sin{x}\cdot\cos{y} + \sin{y}\cdot\cos{x}}{\cos{x}\cdot\cos{y}} $$

That top expression should look familiar: its sine addition

$$ \tan{x} + \tan{y} = \frac{\sin(x + y)}{\cos{x}\cdot\cos{y}} $$

And we can then restate in terms of z

$$ \tan{x} + \tan{y} = \frac{\sin{z}}{\cos{x}\cdot\cos{y}} $$

That feels like progress since we've converted everything into products and we have part of the right hand side already. Next we need to combine the remaining piece on the left hand side:

$$ \tan{x} + \tan{y} + \tan{z} = \frac{\sin{z}}{\cos{x}\cdot\cos{y}} + \tan{z} = \frac{\sin(z)\cdot\cos{z} + \sin{z}\cdot\cos{x}\cdot\cos{y}}{\cos{x}\cdot\cos{y}\cdot\cos{z}} $$

I'm going to factor out $ \sin{z} $

$$ \tan{x} + \tan{y} + \tan{z} = \sin{z} \cdot \frac{\cos{z} + \cos{x}\cdot\cos{y}}{\cos{x}\cdot\cos{y}\cdot\cos{z}} $$

And now we'll once again replace $ \cos{z} $ with its equivalent in terms of x and y (after all we only want x and y now in the remaining expression)

$$ \tan{x} + \tan{y} + \tan{z} = \sin{z} \cdot \frac{-\cos{(x + y)} + \cos{x}\cdot\cos{y}}{\cos{x}\cdot\cos{y}\cdot\cos{z}} $$

That's convenient for when we expand the cosine addition we get:

$$ \tan{x} + \tan{y} + \tan{z} = \sin{z} \cdot \frac{-\cos{x}\cdot\cos{y} + \sin{x}\cdot\sin{y} + \cos{x}\cdot\cos{y}}{\cos{x}\cdot\cos{y}\cdot\cos{z}} = \sin{z} \cdot \frac{\sin{x}\cdot\sin{y}}{\cos{x}\cdot\cos{y}\cdot\cos{z}} $$

And that's exactly the right hand sign expression!

But wait ...

I translated to sine and cosine reflexively when starting before because it usually makes things more approachable. But can we directly manipulate the tangents and follow the same general flow?

To start I'm going to add the a third equivalence relation

3. $ \tan{z} = \tan{(\pi - (x + y))} = -\tan{(x + y)} $

Lets substitute that in

$$ \tan{x} + \tan{y} + \tan{z} = \tan{x} + \tan{y} - \tan{(x+y)} $$

And then apply the tangent addition identity

$$ \tan{x} + \tan{y} + \tan{z} = \tan{x} + \tan{y} - \frac{\tan{x} + \tan{y}}{1 - \tan{x}\cdot\tan{y}} $$

Simplifying that into a common fraction we get a few cancellations:

$$ \tan{x} + \tan{y} + \tan{z} = \frac{ -(\tan{x} + \tan{y}) \cdot\tan{x}\cdot\tan{y} }{1 - \tan{x}\cdot\tan{y}} $$

And then we can just apply 3. again and do everything in reverse to get the desired result

$$ \tan{x} + \tan{y} + \tan{z} = \cdot\tan{x}\cdot\tan{y}\cdot \frac{ -(\tan{x} + \tan{y}) }{1 - \tan{x}\cdot\tan{y}} $$

$$ \tan{x} + \tan{y} + \tan{z} = \tan{x}\cdot\tan{y}\cdot -\tan{(x + y)} = \cdot\tan{x}\cdot\tan{y}\cdot\tan{z} $$

Mastodon: The Wild Wild Neolithic West

noreply@blogger.com (Benjamin Leis) — Tue, 08 Nov 2022 17:20:00 +0000

Its been a long while since my last social media post: https://blog.mathoffthegrid.com/2017/08/how-i-use-twitter.html and everything is all of a sudden in huge flux. With all the turmoil on Twitter I've been exploring Mastodon on the math focused server a friend runs: mathstodon.xyz. However, a new platform means starting over again

You have to rebuild your network of follows and followers. This is huge and discovering people has made my previous two attempts at using Mastodon unsatisfying. But this time is different due to the chaos at Twitter. Large enough groups of people I know have migrated that I could start with a core group of folks and participate in enough conversations to have fun while finding new people.

Why things are vital this time - The extraordinary growth of the network

Peculiarities. Mastodon isn't twitter and has a few quirks that you have to get used. No quote tweets due to a fear of harassment (which seems overly paranoid to me - since you're just a screenshot away from the same effect) and a general stance that makes discoverability harder. You have to use hashtags since full text search doesn't exist across the fediverse. And sometimes the other instance's data is an extra hop away from your server . For instance, you're surfing a profile on another server and sometime have to click to it to get full info. Crucially, since each instance only stores the posts that users on it follows - there is a deep effect on search even for hashtags. You can only see what you know about or someone else on your instance knows about. As a consequence, the larger the server grows the more useful it becomes if you're interested in finding things . There's also an overly precious stance on content warnings that doesn't fit my theory of action. But I can live with that.

Usability. Once you have enough people in your network Mastodon is quite usable despite the large discoverability issues. Can it be your only microblogging platform? That remains an open question for me. The overall network is much smaller than twitter. I don't need most of the twitterverse though just the parts I read. And on that front - the missing piece is probably government and media accounts. For now there are bridge sites like birdsite.wilde.cloud that will publish tweets to toots. But they aren't quite realtime or completely reliable. But the network is growing very rapidly (See above) so the situation could very easily shift in the upcoming months.

Trust. This is a huge general issue. I went with mathstodon because I knew the admins and could implicitly assume they would operate in good faith. But how are millions of people going to make that leap? I feel like there needs to be some type of vetting process or change in structure for many of the instances to answer this question. Perhaps there will be multitudes of small sites where everyone personally knows someone involved - perhaps companies will enter the space and you'll pay for some additional guarantees of stability/security? This also remains to be seen and I expect new developments as growth continues. But that's the key as well - the network is growing despite this issue so its not an adoption blocker yet.

Scalability. In addition to the question of trust across large networks Mastodon is untested technically at twitter-like scales. What's going to happen when instead of 50k posts per hour there are 1M? I could easily imagine a rewrite in its future off Ruby for instance. But even if you make technical fixes when everyone runs there own server you can't easily distribute them. The same goes for scaling up infrastructure when your not running in a single companies datafarm. Its a huge distributed process.

Maintainability. I don't see how longer term, all the infrastructure can continue to be supplied by volunteers. Keeping servers up and running at scale is real wor and I'm not sure if enough new people are going to step up to provide sites for a 10x increase in users. Practically, it all costs $$$ to run larger sites. Perhaps this is going to be solved with donations but that will be a long term issue. As will governance and long term succession plans. How will each site organize to spread out serveradmin tasks, make sure if someone retires there is plan going forward, make crucial moderation decisions etc.

But nevertheless, I'm having fun and experimenting. I think people are going to tackle most of the issues I raise out of necessity. So despite all the open questions I'm in for the ride for now and we'll see where things go ... You can fine me @benleis@mathstodon.xyz

Unexpected Binomial Theorem Connection

noreply@blogger.com (Benjamin Leis) — Sat, 03 Sep 2022 18:32:00 +0000

Source: @ElyemGercek

I saw this post and immediately thought: "What an odd connection to Pascal's Triangle. What's going on?"

Due to time constraints I didn't get a chance to look into this for several weeks and while its not quite as geometric as I hoped the connection is fairly natural.

First just as in the triangle you need an initial row of values from which the relation will then generate all the rest of the rows. In this case its the triangle with a single median. Stewart's Theorem is helpful here for finding the initial equation. What's novel is while I've done this before I've never rearranged in this fashion.

$ a^2 \cdot m + c^2 \cdot m = 2m \cdot (m^2 + b^2) $

$a^2 + c^2 = 2 \cdot (m^2 + b^2) $

$ a^2 - 2b^2 + c^2 = 2m^2 $

What then happens in each iteration is we split the triangle into two smaller versions which we can then find find an equation for using the previous step. Note: how there are two inputs just like in Pascal's triangle.

For example:

From the left hand triangle we get $ a^2 - 2b^2 + c^2 = 2m^2 $. Then we can do the same on the right hand side:

The right hand triangle gives us $b^2 - 2c^2 + d^2 = 2m^2 $. Since both equations are equal to $2m^2 $ we can then form the following new equality:

$a^2 - 2b^2 + c^2 = b^2 - 2c^2 + d^2 $

$ a^2 + -2b^2 - b^2 + c^2 + 2c^2 - d^2 = 0 $

$ a^2 - 3b^2 + 3c^2 - d^2 = 0$

This same pattern can be now repeated each time we divide into a left and right hand triangle and use the previous result. Each of these expressions are equal to 0 (after the first median triangle). We can then equate them and simplify. Each time we get the same shifting of the variable over by one between the two triangles and so just like in pascal's triangle we're adding the adjacent terms to generate the next row.

Properly fitting Problems

noreply@blogger.com (Benjamin Leis) — Sun, 24 Apr 2022 01:07:00 +0000

What motivated this essay was two posts I recently saw. The first started with the idea that no student should take more than five minutes on a problem. If stuck it was unlikely to be productive and they should just move on. In case, there was any confusion this expanded to the idea that investigations or inquiry based learning was ineffective. Secondly, was a note from an acquaintance about how he was about to start a class for pre-service teachers about the pedagogy of problems in the classroom which piqued my interest about which problems he would be using. Together these readings motivated me to think again about "Where do mathematical problems fit?"

To start though you have to step back and define the problem. The classic "word problems" of elementary school are generally a misnomer. Students typically (but there are exceptions) know all the operational procedures to solve them and are being asked to practice translating from natural language into mathematical statements. Instead, I'm fond of the dichotomy between an exercise versus a problem. An exercise is a mathematical task that one knows how to do or has seen demonstrated and is aimed at practicing a skill or procedure in the service of mastery. On the other hand, a problem is a mathematical task where the solution is unknown at the start and one must reason how to solve it along the way. Importantly, this distinction is relative to the practitioner. What may be an exercise for me, can be a problem for someone else or vice versa depending on what prior knowledge we each have. That also means problems exist at every grade level but because they reflect where the solver is, have almost infinitely different and evolving forms.

For me, problems are also central to learning mathematics. Skills like basic arithmetic are important in of themselves but also are only stepping stones and tools towards the service of mathematical exploration and understanding. This creates some difficult tensions though. Especially when beginning what is the proper balance between mastering skills and thinking about unsolved questions? Channeling the first author, the answer would be initially its mostly useless to try problems, the learner doesn't have enough prior knowledge and domain experience. Problems just waste time and more seriously are ineffective for knowledge acquisition. In other words, even when one solves one, the process is not well synthesized. There's usually some discussion of how experts approach problems differently but not much discussion of how or when one reaches that state (other than usually the impression its postsecondary at least)

On the other hand, there are plenty of Inquiry Based Learning advocates that are all in with problems on a daily basis. Its easy to find a class vignette with everyone standing around whiteboards working on some communal task that will take a class period. All of which seems delightful but then the sneaking suspicions enter about the amount of mathematical ground being covered or if enough practice is occurring for growth to occur. After seeing yet another task involving rearranging the digits 1-10 to fit into some equations I often end up thinking how its very easy to idle in the domain of pre-algebra and never really get anywhere.

Instead, I inhabit a middle ground. Time is limited but its important to constantly encounter problems through school in the service of developing patterns of thinking. At the same time, there needs to be a large amount of exercises along the way or you flounder without ever mastering concepts confidently enough to utilize them later. That's a very tricky space given the constraints. So I think you always have to be strategic in what you choose not only in how to to balance the hours between longer problems and practice exercises but what forms each take. To keep to the flow, the problems should riff on the concepts being worked on in an exercise or require one to practice a skill in the service of the solution so you are getting double duty out of it. They should also advance the arc of the subject being explored. That's constraining in the sense there are plenty of interesting non-curricular tasks one could do but only a subset align well with the overall goals for a class. Like many artforms though within those limits there still is tremendous opportunity for creativity and interesting experiences.

Example:

I'm going to draw from my own experience doing math circle activities where I'm very purposely fairly random about activities because we exist outside the curriculum. But I often run into things that would work really well in a classroom setting paired with a learning goal. Recently we tried out determining the number of rectangles that could be fit in one larger one.

There are 60 sub rectangles in the above 3x5

The process for exploring this was fairly messy and time consuming and involved a lot of counting strategies and looking for patterns. But the ultimate solution has deep connections to combinatorics and would work really well embedded in a unit on combinations, pascal's triangle (or both). If doing so you'd want to combine it with simpler exercises on calculating combinations and using their properties along with other simpler problems that built towards being able to tackle this one.

Creating, integrating and curating these tasks is the crux of the instructor "problem". Its easy to go a little astray and end up with a problem that falls flat and no one can solve or is too easy or fails to make the crucial connection or is boring. The chance for failure is fairly large. But practice helps and keeping one's eye on this ultimate goal and reflecting on how each attempt at problem integration into a sessions plays out aids in more effectively using them. And to return to my central philosophy, problems are too essential to Mathematics to not keep tilting at figuring out how to properly use them.

Carnival of Mathematics 203

noreply@blogger.com (Benjamin Leis) — Sat, 02 Apr 2022 19:45:00 +0000

Graphic scores by John De Cesare (1890–1972).

Welcome to the 203rd Carnival. For all the other carnivals future and past, visit The Aperiodical where you can also submit future posts. This is my fifth go around hosting a carnival and the first time I've done so during March. So hold onto your hats and prepare to be inundated with Pi Day material. For those jaded souls among you, there will be plenty of interesting math and no digit reciting contests. (If you're in the other camp I found this Pi Day Digit Song quite amusing this year)

Before we really get going via wikipedia.org here are few facts about the number 203. (We're getting higher up the ordinals and sadly the list of references is getting shorter)

203 is the seventh Bell number, giving the number of partitions of a set of size 6, 203 different triangles can be made from three rods with integer lengths of at most 12, and 203 integer squares (not necessarily of unit size) can be found in a staircase-shaped polyomino formed by stacks of unit squares of heights ranging from 1 to 12.

All things π

David Berardo - Pi Digit Distribution

I really enjoyed the set of mostly new pi videos I watched this year. We're going to start with @TedG's well done animated presentation using Gregory's theorem to approximate the value of Pi. The idea is similar to Archimedes approximation but uses area instead of perimeters The video is based on an article he wrote with a collaborator about 3-4 years ago. I never tire of another way to find pi.

https://youtu.be/-LVpjnaqtwU

Speaking of folks attempting to calculate pi, Matt Parker continued his longstanding series of pi approximations this year with a stab at using Shank's method. The project involved human computers and watching how they self organized to maximize efficiency and accuracy was thought provoking. In the end, as per tradition, the attempt was heartbreakingly derailed by calculation error but I feel that's half the fun of these videos. For another background piece on what's going on I found this writeup on Mancin's formula http://personalpages.to.infn.it/~zaninett/pdf/machin.pdf quite useful.

On a more artistic theme, Vi Hart had a new video up as well where she worked with mapping the digits of pi onto the musical scale and translated it into music. I enjoyed the composition process and hearing her thinking as she went along and the end result was more musical than I would have ever dreamt given the source material.

Somewhere in the middle of all this serious and/or beautiful material I also found the Talking Numbers blog a welcome bit of light humor:

https://talkingnumbers.tumblr.com/post/677995936069320704/pi-me-a-river

We have another blog up next from Gianluigi Filippelli on the various infinite series that lead to pi . This time it examines the historical development of expressions for $ \dfrac{\pi}{2} $ and $ \dfrac{2}{\pi} $ Not surprisingly infinite series again are at the heart of all this. (Series are fundamental would be my personal take away this year)

http://docmadhattan.fieldofscience.com/2022/03/pi-day-viete-infinite-series.html

Penultimately, I'm including one slightly older video that I also watched that ties back into the previous one on Shank's method. Here the Mathologer, Burkard Polster, goes through another well animated proof of pi's irrationality. And once again the arctangent plays a prominent role. The background material led me to a long digression down the derivation of the Arctangent Taylor Series and then trying out converting it by hand into a continued fraction. Watch this one in a pair with Matt Parker's one.

Rounding things out is this performance piece / poem from Harry Baker.

I wrote a poem about circles for #PiDay then I put on my fave circle t-shirt and filmed it walking around in a big circle. Enjoy x pic.twitter.com/jzxwEOBZKv
— Harry Baker (@harrybakerpoet) March 14, 2022

Prime Number of the Day - Danesh Forouhari

All the Rest Of Mathematics

And now for something completely different ... First up is a lovely post on Noether's Therorem by Gianluigi Filippelli: http://docmadhattan.fieldofscience.com/2022/03/noethers-theorem.html On my bucket list is to someday go back and really understand Lie groups.

Patrick Honner has a new article up on Quanta Magazine:

https://www.quantamagazine.org/what-a-math-party-game-tells-us-about-graph-theory-20220324/

Exploring the hand shaking problem and graph theory. This is one is accessible and comes with some great exercises at the end to play around with.

Daniel Piker - Cubes

On a related topic Tivadar Danka offered this fascinating thread on Twitter on the intesection between graph theory and linear algebra which I had never seen before and found really cool.

The single most undervalued fact of linear algebra: matrices are graphs, and graphs are matrices.

Encoding matrices as graphs is a cheat code, making complex behavior simple to study.

Let me show you how! pic.twitter.com/8rBIkA8ZbZ
— Tivadar Danka 🇺🇦 (@TivadarDanka) March 11, 2022

Here's another more metaphysical podcast from Tattva Deep from one of their podcasts recently:

" In this short video, we share some common and some not-so-common questions that each of them answered. All our guests seemed to possess a unique way of looking at maths and understanding what it might be about. And it changed the way we looked at maths and broadened our perception of it. We hope you derived pleasure and insights from these conversations as well. We'd love to hear what you have to say. together several interesting answers from their conversations about maths."

Link: https://www.youtube.com/watch?v=OYp-HXEFsdg

Matt Henderson - Mandelbrot Set

Cameron Son writes this contemplative piece about his numeric visualization process:

"When doing any sort of mental arithmetic or visualization of quantities, I picture numbers as shapes. In my head, operations like addition and subtraction manifest themselves as physical interaction between these shapes, and are complete with sounds and tactile feelings."

http://www.csun.io/2022/03/03/how-i-see-numbers.html

And in our final entry for the month we have another well done video this time on Artificial Intelligence and pure mathematics.

"The usefulness of machine learning (ML) in pure mathematics is still the subject of much scepticism. However, a collaboration between representation theorist Geordie Williamson (University of Sydney) and AI lab DeepMind shows that ML algorithms can do more than just analyse large data sets. Geordie used DeepMind's AI to suggest a possible line of attack on a 40-year-old conjecture about Kazhdan-Lusztig polynomials (the results were published in Nature in November 2021). In this video by the Sydney Mathematical Research Institute (SMRI), Geordie explains how AI can help
with the intuitive elements of pure mathematics."

Thanks again for reading. As usual the process of preparing this write up led me down so many interesting mathematical rabbit holes.

15-75-90 on the quarter circle

noreply@blogger.com (Benjamin Leis) — Wed, 30 Mar 2022 15:44:00 +0000

I just saw a nice visual proof of the ratios for the 15-75-90 on the internet via @ilarrosac. This one works via symmetry and the Pythagorean theorem.

Two Circumcircles Walkthrough

noreply@blogger.com (Benjamin Leis) — Fri, 04 Mar 2022 20:45:00 +0000

I really enjoyed working through this problem from Stanley Rabinowitz I saw recently on Twitter. This was one of those problems where I circled around it and after finding the first solution realized I could improve it a lot and simplify the algebra fairly drastically.

One first impression, I saw two things. First there was power of the point at D which could be used to give expressions for DQ and AD. On the other hand AP looked like it would be hardest segment to get a purchase on.

My second thought was around the midpoints and filling in the similar triangle DEF looking important as well finding all the congruent and parallel segments it created.

I started also filling in the cycle quadrilaterals as well at this point but didn't immediately see to do with all of this.

Next I fired up geogebra and modelled a few thing. One I was curious about the circumcenters and if they could help. That didn't immediately feel promising. The alignment in particular didn't give me anything. However, I did notice the centroids completely aligned as expected but I hadn't though about that yet. (Others online went from there to homothety)

At this point I went back to a combination of a few of my initial experiments with the cyclic quadrilaterals and similar triangles and started just working out the ratios.

You can see the basic algebra I worked out with the similar triangles and power of the point around the 2 intersections solved the proof. But it wasn't entirely satisfying. Yes the calculation for the two parts were equal but why? It didn't really feel like it was elucidating the structure.

At this point I could see that the 2x scaling for a and b above was key but I wondering what would happen if I started from that point. What I found below was I could drop all the ratios and just rely on the cyclic quads.

But I kept thinking about the problem some more and had my final conceptual breakthrough. Above I ended up proving GH = 2EO at which point I could just do segment addition and subtraction but what did that imply? Everything I kept looking at was 2x scaled. So what about the whole cyclic quads and their diagonals? As soon as I adjusted my mental model the final version fell out which is almost completely algebra free and focuses directly on scaling throughout.

The midpoints create an inner triangle DEF that is scaled 1:2 and rotated 180 degrees
When circumscribed we then get 2 cyclic quadrilateral DEPF and ABQC that are also scaled 1:2 and rotated 180 degrees

That means the central diagonals AQ and PD are also scaled 1:2
If AQ = 2PD then AQ - PD = PD or recasting AP + DQ = PD

Ptolemy's Theorem via a bit of Trig and similar triangles

noreply@blogger.com (Benjamin Leis) — Wed, 09 Feb 2022 01:53:00 +0000

Here's a straightforward trigonometric proof of Ptolemy's Theorem using scale factors and the Law of Cosines.

First we start with a cyclic quadrilateral and assign it segment lengths:

1. $ \triangle{CDE} $ has edges of length x, y and z.

2. $ \triangle{BEA} \sim \triangle{CDE} $ so we give a scale factor k and then edges of length kx, ky, kz.

3. $ \triangle{BEC} \sim \triangle{AED} $ and from the given values the scale factor is x:y so we add the last side w and scale it appropriately to wy and wx.

Next we compute $AD \cdot BC + CD \cdot AB = w^2xy + kz^2 $

And then we apply the Law of Cosines twice on the central point E

$ w^2 = k^2 + 1 - 2k\cos{(\pi - \theta)} = k^2 + 1 + 2k\cos{\theta}$

$ z^2 = x^2 + y^2 - 2xy\cos{\theta}$

(Note for the second one I scaled everything down.)

Finally we substitute these expression into the original one which leads to a happy cancellation and we get:

$$ AD \cdot BC + CD \cdot AB = (k^2 + 1 + 2k\cos{\theta}) \cdot xy + k \cdot (x^2 + y^2 - 2xy\cos{\theta})$$

$$ AD \cdot BC + CD \cdot AB = k^2xy + xy + kx^2 + ky^2$$

$$ AD \cdot BC + CD \cdot AB = (ky + x)\cdot(kx + y)$$

$$ AD \cdot BC + CD \cdot AB = AC \cdot BD$$

Cyclic Quad Area Problem

noreply@blogger.com (Benjamin Leis) — Sun, 06 Feb 2022 04:49:00 +0000

What follows is a walkthrough of the problem above which I had a lot of fun playing around with. Most of the solutions I saw online used trigonometry and so just to be different I'm going to approach this solely via scaled similar triangles relying on the cyclic quadrilateral. Note: because the answer involves the golden ratio I was hoping to emphasize it more in the proof but so far I don't see a geometric avenue to bring that forward more prominently. If you happen to find one please comment and tell me about it. Also I did spend a fair amount of time playing around with different which is condensed here into a straight forward narrative.

Basic Triangle

Because of the cycle quadrilateral ACBD and the fact that AC is is the diameter of the circle we have a series of similar right triangles in the figuring including $ \triangle{BCE}$ and $\triangle{ADE} $. I also above add a perpendicular segment OH from the center of the circle which creates a third similar triangle $\triangle{OBH}$

Letting r be the radius of the circle, the length of BC and AB are $ r \sqrt{2} $. With tat all established I am next going to setup the ratio of sides of our basic triangle as follows for convenience when doing everything that follows (Stickler may object isn't this trig in disguise and in a sense it is but for the most part I find this elucidates things more clearly)

This setup means for BCE where the basic triangle is scaled up by $ r\sqrt{2} $ that $ [\triangle{BCE}] = \frac{1}{2} BC \cdot BE = x \cdot r^2 $ . In addition using these scale factors on the triangle OBH since OB = r we find BH = $ \dfrac{xr}{y} $ and BD = $ \dfrac{2xr}{y} $ and finally $ [BDFG] = \dfrac{4x^2r^2}{y^2} $ If we let A = [BCE] then $ [BDFG] = 4 \cdot\dfrac{x}{y^2} A $

At this point using this setup in order to solve the problem it seems likely we're going to need to find r, x and probably y as well but we have the Pythagorean relationship between x and y.

Scaling from the Cyclic Quadrilateral part 1

Let's start with original equality now: [BCE] + [ADE] = [BDFG] . We already know formulas for two of those polygons and are just missing the third. However, BCE and ADE are similar and we actually know enough to find their scale factor. We already found $ CE = y \cdot r \sqrt{2} $. On the other side AE = AB - BE which becomes $ AE = r \sqrt{2} - x r \sqrt{2} = (1-x) r\sqrt{2} $. Combining we find the second ratio $ \dfrac{AE}{CE} = \dfrac{1-x}{y} $ In terms of area then

$$ [ADE] = \dfrac{(1-x)^2}{y^2} \cdot [BCE] $$

Again in terms of $ [BCE] = xr^2 = A $ we can recast $ [ADE] = \dfrac{(1-x)^2}{y^2} A $

We're now set to look at the given equality [BCE] + [ADE] = [BDFG] and plug our formulas in

$$ \dfrac{(1-x)^2}{y^2} \cdot A + A = \dfrac{4x}{y^2}A $$

This one simplifies down quite a bit to a quadratic

$$ (1-x)^2 + y^2 = 4x $$ $$ (1-x)^2 + (1 + x^2) = 4x $$

$$ x^2 - 3x + 1 = 0 \rightarrow x = \dfrac{3 \pm \sqrt{5}}{2}$$

You could continue with these values but I'm going to recast in terms of the golden ratio which will turn out to be convenient so $x = 2 - \phi $ or $ x = 1 + \phi $

Digression

Looking back at quadratic via the vieta formulas its apparent that the two roots multiply to 1. In other words $2 - \phi $ and $ 1 + \phi $ are reciprocals.$ 1 + \phi = \phi^2$ via the basic golden ratio identity. So as its reciprocal $2 - \phi = \dfrac{1}{\phi^2} $

In either case we already know something else interesting $ x^2 + 1 = 3x \rightarrow y^2 = 3x $ and we can remove y^2 from our previous diagram to get $ [BDFG] = \dfrac{4x}{3x} A = \dfrac{4}{3} A $

Since ADE + BCE = BDFG that means we know ADE as well must be $ \dfrac{1}{3}A $

Scaling from the Cyclic Quadrilateral part 2

I'm now going to continue using the similar triangles within the cyclic quadrilateral to establish some additional ratios. In particular I'll start with $ \triangle{BDE} \sim \triangle{CAE} $ and the side length scale factor $ \dfrac{AC}{BD} = \dfrac{y}{x} $ Since we know the area of BDE is 1, based on the scale factor $ [CAE] = \dfrac{y^2}{x^2} $. We can now use our substitution for y^2 and get $ [CAE] = \dfrac{3x}{x^2} = \dfrac{3}{x} $.

We can then also note since ABC is a right isosceles triangle: $ [\triangle{ABC}] = r^2 $ and arrive at our second basic equation from [ABC] = [CAE] + [BCE] $$ r^2 = \dfrac{3}{x} + A $$

$$ A = 3 + Ax $$ $$ A = \frac{3}{1-x} $$

This is a good opportunity to pin x down. Since both sides of the equation above must be positive, 1 - x must also be positive and therefore we need to use the smaller of the two roots $x = 2 - \phi $

We're ready to put everything together now and substitute x and y into the equation above

$$ A = \dfrac{3}{1 - (2 - \phi)} $$

$$ A = \dfrac{3}{\phi - 1}$$

We can also apply a basic fact about the golden ratio here $ \phi^2 - \phi = 1 \rightarrow \phi(\phi -1) = 1 $ in other words $\phi$ and $\phi -1 $ are reciprocals. Which lets us jump to the answer (I could have kept the original rational expression for x and done the same work but this is cleaner)

$$ A = 3 \phi $$ and the rest of areas scale from there $[ADE] = \phi $ and $ [BDFG] = 4 \phi $ Now that we're at the end, it also turns out despite expectations we never needed to r^2 after all and barely found y!

Linear Algebra Retrospective

noreply@blogger.com (Benjamin Leis) — Wed, 08 Sep 2021 00:08:00 +0000

Everything I discovered after doing the linear algebra section of PreCalc last year

I really liked both 3Blue1Brown's Essence of Linear Algebra series: Youtube link and Gilbert Strang's OCW series: OCW Link
From 3Blue1Brown - I took away the emphasis on matrices as linear transformations, the focus on basic vectors and the illustration of the geometric interpretation of the determinant.
From Strang: I picked up the idea of focusing on column vectors and deriving the determinant via properties first.
I thought it was useful to come to stress what was general vs what was not when moving from 2d to 3d problems.
Linear dependence/independence was another key concept that I stressed.
Stress on the correspondence between row transforms and how one solves general system equations.
I de-emphasized minors and in favor of diagonalizing.
Eigen vectors made more sense after all this preparation.

AP PreCalculus?

noreply@blogger.com (Benjamin Leis) — Sun, 04 Jul 2021 19:50:00 +0000

I read a really interesting post by David Bressound on joining the advisory board for AP PreCalculus: link. The gist of which was despite misgivings he thought setting up the course would do more good than harm by providing an opportunity to standardize the curriculum. And in general I think this part of his analysis is probably true. Given an AP Pre-Calculus test, a lot of high schools that already teach AP Calculus are likely to switch over and replace their own course with one aligned to the AP standard. I'm much less sure if it will prove attractive to schools that don't already do AP Calculus or AP Statistics as he asserts. Anecdotally watching trends around Seattle, it feels like dual enrollment classes at the community colleges are gaining traction for upper level high school math classes. For those out of the state this is partly because Running Start makes it particularly easy for students to access this option.

The meat of his article revolves around what should such a course focus on? First he rules out just copying college level paradigms which have not been particularly effective according to the data and are heavily focused on preparing for Calculus. There are a few points which immediately occur to me here. If you believe in multiple pathways and that Calculus is not the right capstone class for high school then reforming Pre-Calculus seems too narrow. By its very nature, precalc means to prepare students for the next year. Maybe its better to throw energy directly at discrete math, or data modelling or even reforming AP Statistics. Secondly, he discusses support classes that introduce precalculus concepts in tandem with a calculus class. While I share his feeling that the existence of college algebra and precalc classes are mostly an indictment of inadequate high school classes, I have trepidations about this model too. It consumes an entire class slot and I would like to see more data on long term outcomes. Would it make more sense to slow classes down rather than doing too separate ones at once? Why are students taking precalc/calc in this situation? Do they really need to be here or would they be better served by focusing on a different branch of mathematics?

But the main part that I've been considering is what would I do with the curriculum? To give some context, unlike middle school math which I've had the chance to watch multiple kids doing, for precalc I only really have my own son as a sample. But his experience this year drove home several shortcomings which I think are more general. To start with the in school class he took used "Advanced Mathematics: Precalculus with Discrete Mathematics and Data Analysis" and occasional pieces from the ck12 site. Both of the resources were somewhere between boring to awful. The textbook spiraled so heavily that perhaps 2/3rds of the year were review of Algebra 2. There were units on linear functions, quadratics, polynomial, exponents, basic combinatorics, sequences etc. Its not just that this material didn't extend anything from previous years. It also replicates my least favorite structure from Algebra 2: the "tour of functions". This generally dull survey of function types without any direction or purpose doesn't seem to stick well or we wouldn't see so many students and classes repeating the process. Then the ck12 pages were often slip shod in quality with an overly mechanical/procedural focus that occasionally avoided discussing the reasons behind the procedure with the dreaded "The proof of this concept is beyond the scope of this page..." All the time focused on review meant most of the new material was shoe horned into a few months at the end of the year. Inevitably time ran out before everything could be finished even within the limited scope the class had set.

Due to these shortcomings, I ended up home schooling the AoPS textbook through the whole year at the same time. What I liked about it was the focus on trigonometry, complex functions and linear algebra as well as the proof and problem based structure. As material these were fun to teach and compensated for a very thin treatment of the material in school. I particularly liked the sections on De Moivre's theorem, Euler's formula and the roots of unity. But there were several issues with the book that still don't work well. For one, the chapters can drag and I saw his interest vary from point to point. There was a moment when it was clear we really had done enough complex transformation and the extension was not serving a good purpose. Secondly, while I had a lot of fun with linear algebra (about a half of the text) I don't find it very well connected to the curriculum that follows. My concern remains that it will be forgotten before its reused multiple years later. I want a stronger connection in PreCalculus to Calculus without actually teaching derivatives, a trend I see more often in modern textbooks. As a result, I introduced several ideas from elsewhere that I thought were more narrative focused and focused on concepts that would recur in the next class.

Using polynomial division to find tangent lines
Newtons difference formula.

Ultimately that's what I'm seeking out of precalculus. Discuss trigonometry which must be covered before Calculus and topics that directly relate to the notion of change, limits and series. Tie that all together by discussing a little more why this material was investigated in the first place. And have problems that reuse previous material like exponents without actually doing a unit on exponents. Most importantly, make a course that you want to take and is completely distinct from Algebra 2.

Sine Hunt

noreply@blogger.com (Benjamin Leis) — Sat, 22 May 2021 22:43:00 +0000

Sine Hunt

This is a working post to collect problems themed around sine. It all started during the last MathJam when a few came up:

Criteria: only sine and no other trig functions should be present. Also the problem needs to be novel and nonstandard

Examine the value of $ \sin(.0000001) $ (in radians) Why are there so many repeated digits?

Examine the value of $ \sin\left(\frac{1}{5555555}\right) $ (in degrees) What's going on?

Prove $ \sin^2(a) + \sin^2(b) = \sin(a+b)$ where a and b are acute angles iff triangle abc is a right triangle.

Trig Cheat Sheet Project

noreply@blogger.com (Benjamin Leis) — Thu, 13 May 2021 19:17:00 +0000

A few months ago, I saw a thread about writing a trigonometry cheat sheet for someone studying Pre-Calculus. I decided that I wanted to do a visual version that emphasized some of the structure I think is important. What follows is the set of images I created and posted over a few weeks. Its not complete so I hope to eventually do a followup post.

Facilitating a Virtual Math Circle

noreply@blogger.com (Benjamin Leis) — Sat, 08 May 2021 20:28:00 +0000

This year I haven't written much about how my Math Circle is doing. That doesn't mean it hasn't been running the whole time. My routines were shaken by the pandemic and initially I didn't have the same strong need to document everything. For some reason I can't really explain, it was both more nerve wracking and draining, at least initially, to do a zoom session with 20 kids than even in my first year working in person with 15. But over time, I've grown more accustomed to how things work when everything is over the internet. I still miss working in person, especially the feedback and sense of immediacy that web conferences do not generate. However, I don't approach each afternoon with the same sense of nervous energy that I did last Autumn and I do have a better sense of how things will likely flow.

One of the routines I've built up is, keeping a common document across all the weeks. I give out the same link each week and I add on all the materials and agenda for the day at the end each time. That way the students can save the link and are more likely to have it each week without me needing to resend it and they can refer back as well if they want to work on something offline. During each session its my default shared backdrop and workspace. So I leave lots of white space where everyone can jointly annotate on it. Sometimes this happens in a group and other times I setup break rooms and have one member share the same document so each room can work together.

Sample screen shot of my joint "workbook"

Its also useful since I can refer back to what I was doing and reconstruct things off the notes and my memory.

That brings me to the session above. Sometime around February we started to shift and the kids had more requests for weekly topics to discuss as opposed to me setting the agenda each time. If I have ideas in mind I like to survey interest ahead of time but its gratifying to see evidence of natural curiosity. Two weeks prior to this one, a few kids had asked to talk about the imaginary number i. This is actually a fairly hard thing to do given the range of backgrounds: everyone is in 6th-8th grade and taking anything from Math7 to Geometry. Even at the tail end, that's not a lot of background to easily approach the topic. And for those still at the pre-algebra level you can't even assume any formal encounters with i yet like the quadratic formula. But they key point for me is that kids aren't living in a vacuum. They know about a lot of these topics, although usually not any specifics and they are curious about them.

I wanted to honor their request so I spent a lot of time thinking about what would be accessible and how to handle the different levels of background knowledge. Part of that preparation was reminding myself to let the kids talk as much as possible and to also continually ask Socratic style questions. What has worked best for me is just working my way in order through the participant list when doing this type of activity. It more or less functions like visual random grouping. All the kids know I'm going to do it so its normalized and it lets me hear feedback from the shyer ones. (Incidentally, there's always this moment of trepidation when you worry that someone won't answer back but in practice its fairly rare which is reassuring.) I try to also ask a lot of "neutral" questions i.e. what strategies are you trying to do X, what do you already notice about this diagram etc. that are designed to encourage discussion.

Simultaneously, I lean into the chat pane a lot. The kids independently use it more than talk (there are comments occurring almost every minute) when we're doing things as a large group. Interestingly, that reverses in the break out rooms. So I keep an eye on what people are typing and often stop, narrate the question or point and pivot off of it. This is one of the few areas of online work that feels genuinely useful. I don't know how you'd replicate this type of interaction in a live setting.

Returning to specifics, I decided to concentrate on two areas around i that I thought would be really interesting and that would be manageable with least amount of hand waving. First we delved into "ghost" parabolas, a topic I learned about via Zoe Griffith. Full details can be found: https://chalkdustmagazine.com/features/the-phantom-parabola/. What works nicely is we could have a discussion about parabolas and imaginary numbers first where I could have the kids talk about what they already knew. That's helpful for getting everyone on a more even starting point. Notably, it was generally well known that i was the square root of -1.

In the online, format I could also have the various graphs pre-inserted into my "workbook" and then I could just scroll when we wanted to refer to them. This type of graph analysis actually works fairly well over zoom. I created a table with annotation tools and after talking about what happens with the imaginary numbers I split everyone up to find particularly values. Next I called on them to fill in our communal chart and annotated again right on top of the workbook. After that was done, we could all share reactions via me doing a round-robin "What do you notice?" style question. And then everything was primed to reveal the 3-d graphs which managed to generate some genuine virtual "Wows". Timewise, everything seems to go slower in this format. About a half an hour had elapsed by the end of this section.

I was then ready to pivot to the second half where I wanted to focus on multiplying by i as a rotation in the complex plane. This again is not to hard to explore because everyone already can use the distributive law to compute values and then we can talk about the graphs produced and look for patterns. To make things easier, I literally embedded a blank graph on one page to overwrite points on. My aim was to discover the rotational dynamics so I picked various complex numbers to try out and graph. For each one, I had various kids compute the results and then plotted them on the shared page. I picked value that seemed likely to produce patterns and connected lines with various colors to try to emphasize those connections. Overall, this also worked as expected. One tangent we ended up on, that I did not expect was discussing what the square root of i would be? Due to the lack of time, we speculated as a group and I put this down for a topic to more fully explore in the next week. Its not something, I would have aimed to talk about on my own but I followed the students lead.

In sum, I've been surprised by both how mathematically rich, these type discussions can be and how the kids genuinely have enjoyed them. I've actually been more ambitious in some ways than I was in previous years.

Carnival of Mathematics 193

noreply@blogger.com (Benjamin Leis) — Sun, 02 May 2021 23:56:00 +0000

(Ancient Roman mosaic floor patterns - @HistoryJourney)

Welcome to the 193rd Carnival and my fourth! time hosting. For all the other carnivals future and past, visit The Aperiodical where you can also submit future posts. It's been a wild ride since Carnival 181, spent mostly in the confines of my home office. Thankfully, I've been lucky enough to avoid the worst suffering of the pandemic. Life goes on so as the initial shock passed I've learned a great deal about doing remote math with students, epidemiology, the nature of resilience, and the comparative advantages of Zoom vs WebEx vs Google Meetings.

Before going farther, as is traditional, here are a few arcana about the number 193 courtesy of the Wikipedia:

193 is an odd number
193 is a centered 32-gonal number
193 is a happy number
193 is a lucky number
193 is an odious number
193 is a square-free number
193 is the sum of 2 square numbers: 12² + 7² (or 144 + 49)
193 is the difference of 2 square numbers: 97² - 96² (or 9409 - 9216)
193 is a prime number
- 193 is a Cuban prime
- 193 is a full reptend prime
- 193 is a Pierpont prime^[1]
- 193 is a Pillai prime^[2]
- 193 is a Proth prime, as it is 3 × 2⁶ + 1
- 193 is a twin prime with 191
- 193 can be written as the difference between the product and the sum of the first four primes
193 is a twin number
193/71 is the most accurate approximation of Euler's Number (e) such that both the numerator and the denominator are integers less than 200 (approx. 99.99897% accurate)

And while 193 doesn't resonate particularly strongly with me, in honor of this being my fourth go about, here's one my favorite geometric facts about squares: Can you spot the golden ratio embedded in this figure?

First up for this month's submissions is a blog post from physicist Sabine Hossenfelder: "Should Stephen Hawking have won the Nobel Prize?". I've really enjoyed Hossenfelder's books and videos in the past. She has a divergent but compelling view of the direction of the field that's worth checking out. This piece is more of a tour of the highlights of Hawking's career from a physics perspective than dive into the math involved But if like me you are more aware of Hawking as a celebrity versus his actual work it serves as a good overview.

Geometry Interlude

Next, there were a few submissions from GanitCharcha. Including this lovely video on a slew of number theory around the number 9

https://youtu.be/ENfuCar4zvI

You should check out the entire video but I particularly liked this highlight at the end. Exercise for the reader. Try this out and puzzle out Why does this work?

For something a little more advanced but visually quite engrossing Anna Seigal has a post up on "Three Correlations and an algebraic classification" https://picturethismaths.wordpress.com/2021/03/18/three-correlations-and-an-algebraic-classification/

This has a linear algebra connection which provides a somewhat natural transition to this fascinating tweet from Tamas Gorbe:

I took a stab at understanding why it works in my last post and after trying it out on several friends and I can attest to it being fun. Which brings me to a video series from Timothy Gowers on YouTube. He's been working through problem like the one above in real time and its been fascinating to watch. I'll highlight the latest one:

All told, both the problems and the working process are a fascinating watch. This one I'm still working on and holding out before watching the full video.

Prove: $ \text{If n, a, b, c, d are positive integers with } n^2 < a < b < c < d < (n+1)^2 \text{ then ad cannot equal bc.} $

Sam Hartburn wrote a post this month at https://samhartburn.co.uk/sh/can-you-sing-two-notes-at-once/ which delves into the maths behind a vocal trick: singing overtones. There's some compelling trigonometry/wave physics involved as well as some mind bending audio samples. That boundary between math and art continually fascinates me.

Also on the boundary between disciplines is this recent interview in the Guardian with author and roboticist, Catherine Menon musing about the similarities between the creative writing process and proof creation.

https://www.theguardian.com/books/2021/apr/10/catherine-menon-pure-mathematics-and-writing-come-out-of-the-same-creative-space

"I find that the “feeling” of constructing a mathematical proof is identical to the feeling of constructing a sentence, paragraph or plot. This is largely down to the nature of pure maths as opposed to applied maths. Pure mathematics is very abstract: it’s a language of ideas rather than numbers. The value of a proof will often lie in how elegant or interesting it is, just like the value of writing. Obviously, proofs do have to be mathematically correct, but within those constraints there’s enormous scope for expressing ideas in different ways"

Hopefully you're own creative process is inspired by Menon and you're ready for Colin Beveridge's recent blog posts on his site: https://flyingcoloursmaths.co.uk/a-pretty-puzzle/ highlighting a problem from Matt Enlow:

Full confession, I have a small part in the introduction, but the star here is both Matt's problem and Colin's process. This one is a ton of fun and I recommend a go at it first before reading Colin's writeup.

Cellular automata are always fun. But I have to say Sophia Wood's latest one on her blog

https://fractalkitty.com/101-days-of-creative-coding-docc/day-91-of-101-docc/ is a stunner.

More please!

Next we have a cool video / math project from Jo Nakasima. a Magnetic Rubik's cube. To quote the prolific Sam Harburn:
"Ooh, Jo Nakashima, who designed the slinkies that I talked about at MathsJam a few years ago, has a tutorial for a working origami Rubik's cube, held together by magnets."

This may be a bit beyond my level of craftiness but its a reminder that I still have a bucket list item to learn how to solve a Rubik's cube.

Meanwhile over on R J Lipton's blog there is an overview of Lola Thompson's recent work:

https://rjlipton.wpcomstaging.com/2021/04/18/summing-up-the-primes/

There's a lot to digest in this post but one of the most interesting parts is the discussion the Mobius and Merten's functions

Continuing with one more formal post, Several folks on twitter including Tai-Danae Bradley were recently recommending Tom Leinsters Galois theory notes.

"I'm a big fan of Tom Leinster's expository writings and just learned he's written 100+ pages of notes on Galois Theory for undergrads—lovely! https://maths.ed.ac.uk/~tl/gt/gt.pdf" This is one of those areas I never studied I never studied in school and I keep coming back to as a gap I want to explore. I've been having fun working through this extensive writeup.

Let's close things out with a very clever kenken variant

They said any number could be in a Kenken... pic.twitter.com/1UbeHQ21Dc
— Bowen Kerins 🔗 (@bowenkerins) April 2, 2021

Are you still thirsting for more? Checkout some sister carnivals:

Pascal's Determinant

noreply@blogger.com (Benjamin Leis) — Mon, 26 Apr 2021 20:02:00 +0000

Tamas Gorbe posted the preceding tweet a few days ago and it caught my attention for two reasons. First, I've been doing a lot of linear algebra recently anyway and secondly Timothy Gowers posted a video of his problem solving process while trying to make sense of this mat

Before following along, I always find its useful to take a crack at solving a problem like this on one's own. Even if you don't fully unwind what's going on, the attempt itself makes seeing a solution more understandable. So this is the process I followed.

A few thoughts almost immediately came to mind:

1. I wasn't go to get much mileage out of using minors to calculate the determinant. They were too unaligned with Pascal's Triangle properties and didn't really seem to help much for building any kind of induction.

2. Instead I wanted to focus on determinant properties that aligned better i.e. row operations like addition of two rows.

3. Finally I wonder if you could glom onto anything from the n x n case to go inductively to n + 1 x n +1.

Determinant properties

1. $ \det \begin{bmatrix} I \end{bmatrix} = 1 $

2. $ \det \begin{bmatrix}a + c & b + d & ...\\c & d & ... \\ ... \\ \end{bmatrix} = \det \begin{bmatrix}a & b &...\\c & d & ... \\ ... \\ \end{bmatrix} $

3. $ \det \begin{bmatrix}a_1 + a_2 & b_1 + b_2 & ...\\c & d & ... \\ ... \\ \end{bmatrix} = \det \begin{bmatrix}a_1 & b_1 & ...\\c & d & ... \\ ... \\ \end{bmatrix} + \det \begin{bmatrix}a_2 & b_2 & ...\\c & d & ... \\ ... \\ \end{bmatrix} $

I then started by working out a few determinants by diagonalizing (i.e. once a matrix is in triangular form, the determinant is the product of the diagonal) My hope/strong instincts was that they all reduced to a diagonal of 1s and not some odd set of rational numbers that multiplied out to 1 or a set of 1s and -1s. Since all the entries were integers its easy enough to prove that the first case is impossible and the second can be further reduced to just 1s but I also wanted to watch what happened to the rest of the matrix.

Sure enough the 2x2, 3x3 and 4x4 cases all reduced nicely and I was satisfied that was a reasonable assumption to make

$$ \det \begin{bmatrix}1 & 1\\1 & 2\\ \end{bmatrix} = \det \begin{bmatrix}1 & 1\\0 & 1\\ \end{bmatrix} $$

$$ \det \begin{bmatrix}1 & 1 & 1\\1 & 2 & 3\\1 & 3 & 6 \\ \end{bmatrix} = \det \begin{bmatrix}1 & 1 & 1\\0 & 1 & 2\\0 & 0 & 1 \\ \end{bmatrix} $$

$$ \det \begin{bmatrix}1 & 1 & 1 & 1\\1 & 2 & 3 & 4\\1 & 3 & 6 & 10\\1 & 4 & 10 & 20 \\ \end{bmatrix} = \det \begin{bmatrix}1 & 1 & 1 & 1 \\0 & 1 & 2 & 3\\0 & 0 & 1 & 3\\0 & 0 & 0 &1\\ \end{bmatrix} $$

$$ \det \begin{bmatrix}1 & 1 & 1 & 1 & 1\\1 & 2 & 3 & 4 & 5\\1 & 3 & 6 & 10 & 15\\1 & 4 & 10 & 20 & 35 \\ 1 & 5 & 15 & 35 & 70 \\ \end{bmatrix} = \det \begin{bmatrix}1 & 1 & 1 & 1 & 1 \\0 & 1 & 2 & 3 & 4\\0 & 0 & 1 & 3 & 6\\0 & 0 & 0 &1 & 4\\ 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix} $$

(I didn't see it immediately at this point - but there is a very distinct pattern going on with the upper triangles)

At this point I decided to focus on the Pascal's Triangle property that each cell in the matrix is the sum of the cell above it and to the left

So I then decided to reverse the process by subtracting the 1st row from every other row and a few other permutations of this. What I kept tripping on was going from the top down doesn't work for reversing things.

What's better is to go bottom up: Example for each row starting in the bottom of the 5x5 the result after subtracting the row above it:

$$ \det \begin{bmatrix}1 & 1 & 1 & 1 & 1\\1 & 2 & 3 & 4 & 5\\1 & 3 & 6 & 10 & 15\\1 & 4 & 10 & 20 & 35 \\ 1 & 5 & 15 & 35 & 70 \\ \end{bmatrix} = \det \begin{bmatrix}1 & 1 & 1 & 1 & 1\\0 &1 & 2 & 3 & 4\\0 & 1 & 3 & 6 & 10\\0 & 1 & 4 & 10 & 20 \\ 0 &1 & 5 & 15 & 35\\ \end{bmatrix} $$

I actually thought a bit about minors at this point since the entire left column is almost entirely zeroes. And there also was an interesting pattern with the sub matrices along the diagonal all having determinants of 1 that I stared at a bit but I couldn't quite figure a way to use them. And then I realized the pattern I ended up utilizing. Since each entry is the sum of the entry above and to the left of it after subtracting the upper row, you're left with just the left elements in each row.

In other words, the result is a rightward shift of the entire matrix with zeros showing up on the left hand side.

So for any arbitrary size of the Pascal matrix, this process can be repeated. Each iteration you only subtract up to the topmost column that isn't diagonalized yet (1st time row n to row 2, then row n to row 3 etc.) shifting all the entries until the matrix is in upper triangular form with only 1's in the diagonal! At this point the informal proof is done.

One more interesting property via my coworker Dan: if you do a full LU decomposition of the matrix you will find:

$$ \begin{bmatrix}1 & 1 & 1 & 1 & 1\\1 & 2 & 3 & 4 & 5\\1 & 3 & 6 & 10 & 15\\1 & 4 & 10 & 20 & 35 \\ 1 & 5 & 15 & 35 & 70 \\ \end{bmatrix} = \begin{bmatrix}1 & 0 & 0 & 0 & 0 \\1 & 1 & 0& 0 & 0\\1 & 2 & 1 & 0 & 0\\1 & 3 & 3 &1 & 0\\ 1 & 4 & 6 & 4 & 1 \\ \end{bmatrix} \cdot \begin{bmatrix}1 & 1 & 1 & 1 & 1 \\0 & 1 & 2 & 3 & 4\\0 & 0 & 1 & 3 & 6\\0 & 0 & 0 &1 & 4\\ 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix} $$

Thinking about it a bit more, the determinant property is not very dependent on the state of the upper triangle. Conjecture: You could actually put arbitrary values in there repeat the generation process and end up with multiple other matrices with determinants of 1.

3x3 Determinant

noreply@blogger.com (Benjamin Leis) — Tue, 23 Mar 2021 19:55:00 +0000

There are definitely more general and less messy ways to derive the 3x3 matrix determinant (breaking it apart using the linearity property and finding the 6 determinants left that are non-zero, simplifying ...) but its interesting to do it once via diagonalizing and confirm the equivalence:

$ \det \begin{bmatrix}a & b & c\\d & e & f\\g & h & i\\\end{bmatrix} = a \cdot \det \begin{bmatrix}e & f \\ h & i\\\end{bmatrix} - b \cdot \det \begin{bmatrix}d & f \\ g & i\\\end{bmatrix} + c \cdot \det \begin{bmatrix}d & e \\ g & h\\\end{bmatrix} $

First assume the matrix is linearly independent and all entries are nonzero so we not have to do row exchanges. Those cases can all be handled but I'm going to keep to the general case here.

First subtract a scaled version of row 1 from the other two in order to setup the first column.

$ \det \begin{bmatrix}a & b & c\\d & e & f\\g & h & i\\ \end{bmatrix} = \det \begin{bmatrix}a & b & c\\0 & e - \dfrac{bd}{a} & f - \dfrac{cd}{a}\\0 & h - \dfrac{cd}{a} & i - \dfrac{cg}{a}\\ \end{bmatrix} $

Next subtract a scaled version of row 2 from the row 3 to setup the second column

$ \det \begin{bmatrix}a & b & c\\0 & e - \dfrac{bd}{a} & f - \dfrac{cd}{a}\\0 & h - \dfrac{cd}{a} & i - \dfrac{cg}{a}\\ \end{bmatrix} = \det \begin{bmatrix}a & b & c\\0 & e - \dfrac{bd}{a} & f - \dfrac{cd}{a}\\0 & 0 & i - \dfrac{cg}{a} - \dfrac{\left(f - \dfrac{cd}{a}\right)\left(h - \dfrac{bg}{a}\right)}{e - \dfrac{bd}{a}} \\ \end{bmatrix} $

Back substitute to remove the upper triangle entries and you get

$ \det \begin{bmatrix}a & b & c\\0 & e - \dfrac{bd}{a} & f - \dfrac{cd}{a}\\0 & 0 & i - \dfrac{cg}{a} - \dfrac{\left(f - \dfrac{cd}{a}\right)\left(h - \dfrac{bg}{a}\right)}{e - \dfrac{bd}{a}} \\ \end{bmatrix} = \det \begin{bmatrix}a & 0 & 0\\0 & e - \dfrac{bd}{a} & 0\\0 & 0 & i - \dfrac{cg}{a} - \dfrac{\left(f - \dfrac{cd}{a}\right)\left(h - \dfrac{bg}{a}\right)}{e - \dfrac{bd}{a}} \\ \end{bmatrix} $

The determinant of a diagonal matrix is just the product of the diagonal entries so

$ \det \begin{bmatrix}a & b & c\\d & e & f\\g & h & i\\\end{bmatrix} = a \cdot \left( e - \dfrac{bd}{a} \right) \cdot \left( i - \dfrac{cg}{a} - \dfrac{\left(f - \dfrac{cd}{a}\right)\left(h - \dfrac{bg}{a}\right)}{e - \dfrac{bd}{a}} \right ) $

This looks a bit messy but we'll distribute the a into the following term

$ \det \begin{bmatrix}a & b & c\\d & e & f\\g & h & i\\\end{bmatrix} = \left( ae - bd \right) \cdot \left( i - \dfrac{cg}{a} - \dfrac{\left(f - \dfrac{cd}{a}\right)\left(h - \dfrac{bg}{a}\right)}{e - \dfrac{bd}{a}} \right ) $

And simplify last term a bit:

$ \det \begin{bmatrix}a & b & c\\d & e & f\\g & h & i\\\end{bmatrix} = \left( ae - bd \right) \cdot \left( i - \dfrac{cg}{a} - \dfrac{(af - cd)(ah - bg)}{a \cdot (ae - bd)} \right) $

Next combine some fractions

$ \det \begin{bmatrix}a & b & c\\d & e & f\\g & h & i\\\end{bmatrix} = \left( ae - bd \right) \cdot \left( i - \dfrac{aceg - bcgd + (af - cd)(ah - bg)}{a \cdot (ae - bd)} \right) $

$\det \begin{bmatrix}a & b & c\\d & e & f\\g & h & i\\\end{bmatrix} = \left( ae - bd \right) \cdot \left( i - \dfrac{aceg - \mathbf{bcgd} + a^2fh - abfg - acdh + \bf{bcdg}}{a \cdot (ae - bd)} \right)$

Fortunately the bcdg terms cancel out which are the only ones without an a in them

$\det \begin{bmatrix}a & b & c\\d & e & f\\g & h & i\\\end{bmatrix} = \left( ae - bd \right) \cdot \left( i - \dfrac{aceg + a^2fh - abfg - acdh}{a \cdot (ae - bd)} \right)$

$\det \begin{bmatrix}a & b & c\\d & e & f\\g & h & i\\\end{bmatrix} = \left( ae - bd \right) \cdot \left( i - \dfrac{ceg + afh - bfg - cdh}{ae - bd} \right)$

And there's a nice cancellation on the denominator ae-bd to get

$\det \begin{bmatrix}a & b & c\\d & e & f\\g & h & i\\\end{bmatrix} = aei - bdi - ceg - afh + bfg +cdh$

Now we'll just rearrange a bit:

$\det \begin{bmatrix}a & b & c\\d & e & f\\g & h & i\\\end{bmatrix} = aei -afh - (bdi - bfg) + cdh - ceg$

And the desired result falls out:

$aei -afh - (bdi - bfg) + cdh - ceg = a(ei-fh) - b(di - fg) + c(dh - eg)$

$a(ei-fh) - b(di - fg) + c(dh - eg) = a \cdot \det \begin{bmatrix}e & f \\ h & i\\\end{bmatrix} - b \cdot \det \begin{bmatrix}d & f \\ g & i\\\end{bmatrix} + c \cdot \det \begin{bmatrix}d & e \\ g & h\\\end{bmatrix} $

Roots of Unity Application

noreply@blogger.com (Benjamin Leis) — Wed, 10 Feb 2021 04:50:00 +0000

Show that $ \sin\left(\dfrac{\pi}{n}\right) \cdot \sin\left(\dfrac{2\pi}{n}\right) ... \sin\left(\dfrac{(n-1)\pi}{n}\right) = \dfrac{n}{2^{n-1}} $

One of my first thoughts on seeing a problem like this is this must involves roots of unity somehow. From a geometric perspective we're asked to multiply the y coordinates of all the points on the unit circle in quadrants 1 and 2 as shown below: Each of these points is an 2nth root of unity i.e. satisfies the equation $x^{2n} = 1 $

We know a lot about how to manipulate the complex roots but less about individual parts like the sin so first we're going to take advantage of the standard derivation for sine from the full root letting $ \omega $ be any root (or point on the diagram above) and introducing its conjugate. Note: geometrically the conjugates are just the reflections across the horizontal access.

$ \omega = \cos \theta + i \sin \theta $

$ \overline{\omega} = \cos \theta - i \sin \theta $

$ \sin \theta = \dfrac{\omega - \overline{\omega}}{2i} $

So now we have a translation for the original express in terms of the full roots and their conjugates. This is already nice because it starts to introduces powers of two which we want to see in the final expression. But what's even more convenient is that in general we get magnitudes when we multiple conjugates $ \omega \cdot \overline{\omega} = | \omega | $ and on the unit circle the magnitude is by definition 1. So in fact for any point on the unit circle $ \overline{\omega} = \dfrac{1}{\omega} $

We also can use the standard notions of complex multiplication and De Moivre's theorem to translate any root $ \omega_n $ into $\omega^n $ in terms of the first root.

Applying this to our expression above we get

$ \sin \theta = \dfrac{\omega ^2 - 1}{2i \cdot \omega} $

Now we're ready to put things together and apply these transforms to the original equation:

$\sin\left(\dfrac{\pi}{n}\right) \cdot \sin\left(\dfrac{2\pi}{n}\right) ... \sin\left(\dfrac{(n-1)\pi}{n}\right) = \dfrac{\omega ^2 - 1}{2i \cdot \omega} \cdot \dfrac{\omega ^4 - 1}{2i \cdot \omega^2} ... \dfrac{\omega ^{2n-2} - 1}{2i \cdot \omega^{n-1}} = \dfrac{1}{(2i)^{n-1}} \cdot \dfrac{\omega ^2 - 1}{\omega} \cdot \dfrac{\omega ^4 - 1}{\omega^2} ... \dfrac{\omega ^{2n-2} - 1}{\omega^{n-1}} $

This is already progress since the $2^{n-1} $ has started to appear. Next lets examine the numerator:

Conceptually rather than a product involving half of the 2nth roots of unit what we now have is a product involving of all of the nth roots of unity except 1:

And in fact the denominator is almost something familiar: see Chords and the roots of unity the product of the chords. All that remains is to reverse the signs to completely get it into right form:

$ \dfrac{1}{(2i)^{n-1}} \cdot \dfrac{\omega ^2 - 1}{\omega} \cdot \dfrac{\omega ^4 - 1}{\omega^2} ... \dfrac{\omega ^{2n-2} - 1}{\omega^{n-1}} = \dfrac{-1^{n-1}}{(2i)^{n-1}} \cdot \dfrac{1 - \omega ^2 }{\omega} \cdot \dfrac{1 - \omega ^4 }{\omega^2} ... \dfrac{1 - \omega ^{2n-2}}{\omega^{n-1}} = \dfrac{-1^{n-1}}{(2i)^{n-1}} \cdot n \cdot \dfrac{1}{\omega} \cdot \dfrac{1}{\omega^2} ... \dfrac{1}{\omega^{n-1}} $

We can then coalesce all the remaining denominators to:

$ \dfrac{-1^{n-1} \cdot n}{(2i)^{n-1}} \cdot \dfrac{1}{\omega^{\frac{n(n-1)}{2}}} $

Next we turn our attention to i which is problematic since our goal is a real number but we're in luck since i also can expressed in terms of $ \omega $. i is $ \frac{\pi}{2}$ on the unit circle and $\omega $ is $ \frac{\pi}{n} $ so $ i = \omega^{\frac{n}{2}} $ That means either

$ i^{n-1} = \omega^{\frac{n(n-1)}{2}} $ or vice versa

We have both forms in our expression which we'll collapse to

$ \dfrac{-1^{n-1} \cdot n }{(2i)^{n-1}} \cdot \dfrac{1}{\omega^{\frac{n(n-1)}{2}}} = \dfrac{-1^{n-1} \cdot n }{(2i)^{n-1}} \cdot \dfrac{1}{i^{n-1}} = \dfrac{-1^{n-1} \cdot n}{2^{n-1} \cdot (i^{n-1})^2} $

$ -1 = i^2 $ so $ -1^{n-1} = (i^2)^{n-1} = (i^{n-1})^2 $ Or in other words the top and bottom terms cancel out to leave use with

$ \dfrac{-1^{n-1} \cdot n}{2^{n-1} \cdot (i^{n-1})^2} = \dfrac{n}{2^{n-1}} $

And we're done!

Justifying complex number exponential notation i.e. Euler's Formula without Calculus

noreply@blogger.com (Benjamin Leis) — Fri, 15 Jan 2021 04:28:00 +0000

Background

I'm assuming that you're already familiar with polar coordinates and cis notation for complex numbers. If not look here: Link

So we start with any complex number being represented in polar coordinate on the complex plane as: $ r ( \cos(\theta) + i \sin(\theta)) $

By directly evaluating the expressions we can start off by showing the complex numbers have the all same properties as an exponential function

$ f(0) = 1 $
$ f(x + y) = f(x) \cdot f(y) $
$ f(x - y) = \dfrac{f(x)}{f(y)} $
$ f(-x) = \dfrac{1}{f(x)} $
$ f(x)^n = f(n\cdot x) $

That's very suggestive but what exponential expression is equivalent in this case? In fact Euler's Formula says that $ r ( \cos(\theta) + i \sin(\theta)) = r \cdot e^{i \theta} $

Usually that's justified via Calculus and Taylor Series but its possible to mostly ground with simpler techniques.

Lets start by looking at how e is defined: $ \displaystyle{e = \lim_{n \to -\infty} \left(1 + \dfrac{1}{n}\right)^n} $ or conceptually the result of infinitely compounding a value. That's usually been introduced before in the context of interest or other proportional growth.

We'd also like to have an expression for $e^x $ so we raise both sides to the xth power. Then on the right hand side via the binomial theorem we get:

$ ( 1 + {x \choose 1} \cdot \frac{1}{n} + {x \choose 2} \cdot \frac{1}{n^2} ... )^n $

But as x goes to infinity all of the smaller powers of n drop off and we're only left with the first two terms giving us:

$ \displaystyle{e^x = \lim_{n \to -\infty} \left(1 + \dfrac{x}{n}\right)^n} $

So how to connect these two seemingly unrelated concepts? We're going to start with De Moivre's Formula

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

We'd like this to look similar in structure to the definition of e and the normal cis notation so first step is to scale everything down:

$ (\cos(x) + i \sin(x)) = \left((\cos(\dfrac{x}{n}) + i \sin(\dfrac{x}{n})\right)^n $

Now what happens as n heads towards infinity?

For the cosine value $\frac{x}{n}$ heads towards so 0 so in this case $ \cos(\frac{x}{n}) \approx \cos(0) = 1 $. Likewise as n goes to infinity $\frac{x}{n}$ is getting closer and closer to 0 and so is the sine value: $ \sin(\frac{x}{n}) \approx\ \sin(0) \approx \frac{x}{n} $

So combining things we get as $ \displaystyle{ \lim_{n \to -\infty}} (\cos(x) + i \sin(x)) \approx \left(1 + i \cdot \frac{x}{n} \right)^n $

We then look back and realize this is exactly the same as the definition for: $ e^{ix} $ !

Pushing the limits

noreply@blogger.com (Benjamin Leis) — Sun, 29 Nov 2020 01:25:00 +0000

I'm still playing around with the ideas from https://blog.mathoffthegrid.com/2020/11/going-one-step-beyond-herons-formula.html And this time I looked at a more complex problem. Overall the approach leaned heavily on the multiple area formulas this time (that lets Heron's formula comes in). But there was a fair amount of algebra in the middle so I'm not sure I like this method as much here. The algebraic simplification may take as much logic as going for a more conventional approach. The one upside is the conversion between various area formulas is quite interesting.

More Heron's Formula Extensions

noreply@blogger.com (Benjamin Leis) — Mon, 23 Nov 2020 01:10:00 +0000

Continuing on the theme from the last post: https://blog.mathoffthegrid.com/2020/11/going-one-step-beyond-herons-formula.html here's another problem showing the power of the full conceptual framework. Heron's formula itself even makes an appearance again at the end.

Problem

Given a $\triangle{ABC} $ show that $ \tan\left(\dfrac{a}{2}\right)\tan\left(\dfrac{b}{2}\right) + \tan\left(\dfrac{a}{2}\right)\tan\left(\dfrac{c}{2}\right) + \tan\left(\dfrac{b}{2}\right)\tan\left(\dfrac{c}{2}\right) = 1 $

Traditional Approach

There are many ways to attack this problem using trig identities I'm going to demonstrate one that is very linear in the progression of ideas and that focus on a direct simplification of the left hand side. To do that I'm going to take advantage of the triangle angles.

We know that a + b + c = 180 so $ \dfrac{a}{2} + \dfrac{b}{2} + \dfrac{c}{2} = 90 \rightarrow \dfrac{c}{2} = 90 - \left( \dfrac{a}{2} + \dfrac{b}{2} \right) $

So first isolate $ \dfrac{c}{2} $

$ \tan\left(\dfrac{a}{2}\right)\tan\left(\dfrac{b}{2}\right) + \tan\left(\dfrac{a}{2}\right)\tan\left(\dfrac{c}{2}\right) + \tan\left(\dfrac{b}{2}\right)\tan\left(\dfrac{c}{2}\right) = $

$ \tan\left(\dfrac{a}{2}\right)\tan\left(\dfrac{b}{2}\right) + \tan\left(\dfrac{c}{2}\right) \cdot \left( \tan\left(\dfrac{a}{2}\right) + \tan\left(\dfrac{b}{2}\right) \right) $

Now looking at $ \tan\left(\dfrac{c}{2}\right) $ we already know that's $ \tan\left(\ 90 - \left( \dfrac{a}{2} + \dfrac{b}{2} \right ) \right) $

But $\tan(90-x) = \dfrac{\sin(90 - x)}{\cos(90 -x)} = \dfrac{\cos x}{\sin{x}} = \cot{x} $

We can also now use tangent addition: $ \tan\left(\dfrac{a}{2} + \dfrac{b}{2}\right) = \dfrac{\tan\left(\dfrac{a}{2}\right) + \tan\left(\dfrac{b}{2}\right)} {1 - \tan\left(\dfrac{a}{2}\right) \cdot \tan\left(\dfrac{b}{2}\right)} $

So together we get $ \tan\left(\dfrac{c}{2}\right) = \dfrac{1 - \tan\left(\dfrac{a}{2}\right) \cdot \tan\left(\dfrac{b}{2}\right)} {\tan\left(\dfrac{a}{2}\right) + \tan\left(\dfrac{b}{2}\right)} $

Substituting into our equation we get:

$ \tan\left(\dfrac{a}{2}\right)\tan\left(\dfrac{b}{2}\right) + \dfrac{1 - \tan\left(\dfrac{a}{2}\right) \cdot \tan\left(\dfrac{b}{2}\right)} {\tan\left(\dfrac{a}{2}\right) + \tan\left(\dfrac{b}{2}\right)} \cdot \left( \tan\left(\dfrac{a}{2}\right) + \tan\left(\dfrac{b}{2}\right) \right) $

That has a very convenient cancellation:

$ \tan\left(\dfrac{a}{2}\right)\tan\left(\dfrac{b}{2}\right) + 1 - \tan\left(\dfrac{a}{2}\right) \cdot \tan\left(\dfrac{b}{2}\right) = 1 $

Overall this is fairly neat but it needs a few identities and doesn't give any geometric notion about what's going on.

Using Ratios Instead

This time we're going to go back to the diagram and express the tangents as a ratio. In general $ \tan\left(\dfrac{a}{2}\right) = \dfrac{r}{s-a} $ where r is the inradius and s is the semiperimeter.

So substituting in we get:

$ \tan\left(\dfrac{a}{2}\right) \tan\left(\dfrac{b}{2}\right) + \tan\left(\dfrac{a}{2}\right)\tan\left(\dfrac{c}{2}\right) + \tan\left(\dfrac{b}{2}\right)\tan\left(\dfrac{c}{2}\right)$

$ = \dfrac{r^2}{(s-a)(s-b)} + \dfrac{r^2}{(s-a)(s-c)} + \dfrac{r^2}{(s-b)(s-c)} $

$ = \dfrac{r^2 \cdot ((s - a) + (s -b ) + (s -c))}{(s-a)(s-b)(s-c)} $

$ = \dfrac{r^2 \cdot s }{(s-a)(s-b)(s-c)} = \dfrac{r^2 \cdot s^2 }{s(s-a)(s-b)(s-c)} $

But both the numerator and denominator are expressions for the area of the triangle: (semiperimeter $\cdot$ inradius and heron's formula)

So we get: $ \dfrac{[ABC]^2}{[ABC]^2} = 1 $

This way, for me at least, shows the symmetry of what's going on is very direct and the intermediate expressions are all recognizable in a geometric sense.

Going one step beyond Heron's Formula

noreply@blogger.com (Benjamin Leis) — Sun, 15 Nov 2020 22:58:00 +0000

Above I've found the incenter of the triangle at the intersection of the angle bisectors and rewritten all segments lengths in terms of the semi-perimeter and and side lengths. This is a great exercise to have someone work through. For me the process highlighted a lot of the structure of how the incircle works.

Fundamental tool: The law of cosines gives us an expression for cos θ in terms of only the side lengths

$ a^2 + b^2 - 2ab \cdot \cos \theta = c^2 $ $\rightarrow$ $ \cos \theta = \dfrac{a^2 + b^2 - c^2}{2ab} $

The half angle formulas let us break that down
A quick chase of the side lengths on the outside via the congruent triangles will find their formulas in terms of s and a,b,c

Application

I have a fairly traditional derivation of Heron's Law here: https://blog.mathoffthegrid.com/2017/10/herons-formula.html but the identities above allow this more unconventional approach.

1. Start with standard trigonometric definition for $ [ABC] = \dfrac{1}{2}bc \cdot \sin(A) $

2. Expand using the double sine formula to: $ [ABC] = bc \cdot \sin(A/2) \cdot \cos(A/2) $

3. Substitute in the values from above to get $ [ABC] = bc \cdot \sqrt{\dfrac{s(s-a)}{bc}} \cdot \sqrt{\dfrac{(s-b)(s-c)}{bc}} $

4. Which simplifies to the familiar Heron's Formula $ [ABC] = \sqrt{s \cdot (s-a)(s-b)(s-c)} $

11/3 Math in the time of Covid

noreply@blogger.com (Benjamin Leis) — Sat, 07 Nov 2020 23:22:00 +0000

It took nearly a month longer than normal but I finally have the virtual math club setup for the year and we met for the first time on Tuesday. Given the weirdness and difficulty getting communication out I wasn't sure how many students would show up but I was pleasantly surprised to reach 15 on the first day . I expect this number to rise a bit in the next few weeks and while normally I would want a firm roster, in these uncertain times I'm okay with being less rigid.

Looking over who showed up or has registered so far, I'm gratified that most kids from last year came back. That's the ultimate vote of confidence for me. I also have about a third of new sixth graders which is needed for continuity. Gender balance continues to be really good and hover around 50%. However, the club does continue to skew Asian as has been the trend for the last few years. That's fine as long as everyone works with each other and remains welcoming but I would like to penetrate a bit farther into the rest of the community. Given the difficulty doing outreach, I'm somewhat at the mercy of parents and students choosing to join I don't think I'm going to make much progress on that front until school is once again occurring in the real building.

On the bright side, it was a ton of fun too see everyone from last year who came back. The main reason I kept pushing to complete what was a very bureaucratic process this year was the thought of all the kids. Because I initially was unsure how many students would be there. I did only some light planning for this meeting. My basic agenda was:

1. Go over procedures and goals for the year.

2. Have everyone introduce themselves.

3. Do a group discussion about my favorite key questions. What does Math mean to you? and What do you do when you get stuck?

4. Do a group "Which One Doesn't Belong" exercise.

5. Close with a group puzzle from a few years back: Pythagorean Triple Puzzle

This is not super ambitious but flexible enough to work with only a few or as it turns out 15 kids.

Observations

The most common response from everyone when asked about why they rejoined was missing socializing with their friends. I'm taking this to heart and trying to provide a space where that is possible. My main strategy will be to focus on small break out rooms and having kids chat and work together on a few problems each week. Also initially only about a quarter of the kids had cameras on. I'm hoping that more will do so and I'll probably encourage that over the next few week.

Right now I'm using the district standard MS Teams for the web conferences but I'm not the meeting host. That prevents me from creating break out rooms and also restricts the tools I can use in the web conference. There's also a strange phenomena that if you share a window but open the chat window, teams reverts to sharing the whole screen (which is really annoying). Long term, I really need both the breakout rooms and preferably the ability to annotate shared documents. I may look into something like Jamboard for use down the road. What I do find effective already is using the chat window. I'm going to lean on that even more than having everyone speak as a low stakes way to drive engagement. I also like using the raise your hand button for group questions.

Being virtual does make it a lot harder to track who is talking / engaged that being in person. I think next week I'm going to set up a roster and literally mark when someone speaks to make it easier for me to notice patterns.

But despite the limitations the first day went really well and I'm excited to design more ambitious sessions in upcoming weeks.

Fantasy High School Part 4 PreCalc

noreply@blogger.com (Benjamin Leis) — Tue, 27 Oct 2020 00:19:00 +0000

I've been looking around at sample curriculum sequences due to odd seeming topics I've seen at home via the Pre-Calculus class my older son is taking, In the past I've noted that Algebra 2 is extremely non-standard when it comes to curriculum. The same goes for Pre-Calculus but I think after checking a bunch of samples, it particularly lacks coherence and integration with the classes that go before and after it.

So I thought I'd dig into a reasonably typical representative curriculum:

Semester 1
Functions Analysis - Unit 1
Linear and Quadratic Functions - Unit 2
Polynomial and Rational Functions - Unit 3
Inverse, Exponential and Logarithmic Functions - Unit 4
Conic Sections - Unit 5
Matrices - Unit 6

Semester 2
Trigonometric Functions - Unit 7
Analytic Trigonometry - Unit 8
Applications of Trigonometry - Unit 9
Vectors and Polar Coordinates - Unit 10
Sequences and Series - Unit 11
Limits - Unit 12

What's striking about Semester 1 is that is mostly duplicates and spirals over topics that were just taught in a typical Algebra 2 class. There isn't potentially a new topic introduced until Conic Sections and/or Matrices. And in many cases, Algebra 2 will have covered Conic Sections already as well.

Based on some chapter samples, these really are review topics and don't go any deeper than was done the previous year. The units on linear functions and then factoring quadratics are often particularly noteworthy in this regard.

Then we come to Matrices. While Linear Algebra is definitely interesting it doesn't in anyway connect with Calculus I. This material seems more likely to be forgotten after its introduced in a typically short and cursory unit in the middle of all the other topics. What's worse is there is no motivation for why its being taught which I expect most students to notice.

Semester 2 is more promising and includes the needed trigonometry. There's a review of Series and Sequences which may be justified by their central role in Calculus. And then comes a unit on Limits. While this is probably an extension of work with asymptotes and very closely linked to the foundations of Calculus, the problem here is that almost certainly any Calculus class is going to repeat this material all over again.

So in sum, more that 50% of the material probably doesn't really make sense. That may be an opportunity because, as mentioned before Last Post there is constant demand for more topics on Discrete Math or Statistics etc. Perhaps rationalizing this class is part of the overall solution.

Or in the cold light of morning perhaps this also indicates more problems in the previous years classes. You don't hear a lot of complaining about the spiral. Is it necessary right now because either the Alg I / Alg 2 classes don't cover what they are intended to or the material doesn't stick for enough students?