So today, I’m going to finally delve into it. I’ve seen a lot of variants on this in terms of orientation, the triangle areas and whether the outer rectangle is a square. Spoiler it doesn’t matter. The areas by themselves uniquely constrain the area of the remaining triangle tot he same value regardless of the rectangles proportions as long as they fit.

There are several approaches you can take but the simplest and most direct one since you’re given areas is to try to work in terms of areas and not drop down to segment lengths. (That will work but can be much more complicated)

So like above the main conceptual technique is to draw into the central dotted axes. You’ll see that divides the box into 2 reflected triangles of area 5 in the lower right. The other two outer triangles are made of half of 2 of the boxes where they both share the common one in the upper left.

We can formalize this as so

What’s nifty is because the various rectangles share the same side lengths they are proportional. I.e [upper left] : [lowerleft] = [upper right] : [lower right]. Or you could equivalently say [upper left] : [upper right] = [lower left] : [lower right]

From there you arrive at a quadratic equation. $x^2 = 24x + 48 = 0$. When solved you find $ x = 12 - 4 \sqrt{6}$ The other conjugate solution is discarded because it would lead to negative areas in some of the original rectangles. With this in hand: its easy to calculate the total area of the square $12 + 4\sqrt{6}$ and then the inner triangle itself which is $4\sqrt{6}$

Since none of this assumed anything more than an outer rectangle - we also see the solution is unique independent of that outer rectangle’s proportions.

The following problem came up on reddit:

Show that the lines tangent to the parabola at the ends of a focal chord intersect at right angles on the directrix.

This has a nice proof based on the groundwork from post

Setup

Everything that follows will build on the fact the tangents to a point on a parabola lie on the perpendicular bisector of the segment from the focus to the perpendicular with the directrix shown in the previous post.

So to start let’s graph out the basic layout.

• We have a chord AB that goes through the focus at F.
• I’ve added in the segments DF and CF from the focus to the perpendiculars and the perpendicular bisectors of both segments which are the respective tangents to A and B.
• Also from the definition of a parabola, AF = AD and BF = BC.

Initial angle chase

Since the perpendicular BC and AD are parallel, the red angles at A and B are congruent. Let’s set that angle to x.

• Then FBC = 180 - x and since FBC is isosceles, BFC = BCF = 90 - x/2.
• Likewise FAD is isosceles and FDA = ADF = x/2.

That implies BFC and AFD are complementary and the remaining angle at that point CFD is a right angle.

Completion

• Point O is where the perpendicular bisector OA intersects the directrix. By definition its equidistant from F and D so FO = DO.
• We can now angle chase the triangle CFO. Let OFD = ODF = y, then CFO = 90 - y, and FOC = 2y so the remaining angle in the triangle FCO must be 90 - y as well.
• CFO is therefore also isosceles and CO = FO = DO. Another way of thinking about this is FCD defines a circle centered at O because its a right triangles and CO, FO, and DO are all radii.
• We now go the other way around because CO = OF it must lie on the perpendicular bisector of CF i.e. O is the intersection of both bisectors.
• Finally because of all the right angles CF is parallel to OA, and BO is parallel to FD. Since three of the four angles on our parallelogram are right angles the fourth one at angle AOB is also a right angle. QED

Another year draws to a close

Today was the last day of Husky Math Academy for the year and like all such closures its bittersweet for me. By the end of the year the weekly preparation starts to wear a bit thin. (I really should keep my workings somewhere so I can reuse them in future years rather than deriving from scratch again.) But also typically within about two weeks I start to miss the weekly interactions with the students. This year, the main section I was leading (6 girls and 5 boys) had really good flow most of the time. With a minimum of guidance, everyone in the room, worked really hard and often surprised me with interesting analyses. On the flip side, they were quiet enough that I often found myself cold calling on kids or surveying the room one by one more often as the months went by to encourage more talking. By contrast, the second section was more typical and you would enter the room and it would be quite noisy at first and take a few moments to really settle in.

What did the kids remember?

For the last day I decided to go around and ask everyone what they enjoyed the most during the Spring and two main themes emerged.

1. Examining parabolas from the perspective of the Directrix and Focus
2. Deriving the Cubic Equation via Cardano’s Method.

There were a lot of comments how everyone enjoyed when I went off the book and looked for extensions of the material.

Curriculum Retrospect

As mentioned above one of the challenges of this format was we were covering Algebra and Geometry but everyone had generally done it once in school. In fact a few of the kids were already officially taking PreCalculus. To a large extent, I often felt like I was fighting the curriculum, IDEA Math. It was either not novel enough, had a very quirky notion of sequencing or didn’t quite take topics as far as I wanted. For example, the geometry section was all over the place. One section would focus on right triangle trigonometry and then it would immediately jump to quadrilaterals and then it would switch to one of the circle centers etc. While I didn’t rearrange things very aggressively this year, if I repeated I definitely would. Instead I often focused on telling a narrative about the material. For instance, when we discussed a topic that had been split across multiple weeks I always referred back to the sequence we were working on and what we had looked at before and how this was building on it. Occasionally, some topics were included with little explanation like partial fraction decomposition. For those ones, I would say frankly this may not seem that useful right now but its being introduced for use later on calculus where it will be one of the tools we use to break down problems. That’s not particularly satisfying so if I do this again, I will probably defer that subject until the more natural moment during the teaching of integration where it makes much more sense why you’d need to do it.

Cardano’s Method

One of my main goals for the Spring was to get through this topic which isn’t often covered in High School. I ended up splitting over about 4 sessions.

1. Looking at symmetry of the cubic equation
2. Discussing the suppressed form: why its convenient - how to derive it.
3. Looking for the u+v substitution to crack the problem and working through that.
4. Roots of unity and the casus irreducibilis.

Overall things went well for the first time doing this but for the future I think I would focus on including a few more problems along the way especially for part 4. And I still have to think about encouraging more experimentation along the way. Every one is so used to mostly having all the techniques laid out and you want to breakout of that mind set.

UWMO

Also last week I judged at the UW Math Olympiad which was a first for me. Overall, it was a great experience. Me and my partner judge listened to about a dozen kids over three hours presenting proofs to the various problems. You have to then probe the arguments until either they reach a satisfactory conclusion or they go back to the room to think things over before trying again. I had been worried about thinking on my feet given a hard to follow proof and as a result had spent a few hours reviewing all the problems, trying them out, looking at proofs and thinking about what kids might try to do. That helped a lot on the day of the event and I felt mostly like I was able to keep up with the kids and give the proper feedback. Its a really good experience for the students and I will likely do this again in the future.

Summer

I have nothing particular, mathwise planned for the Summer. As I said above, I know I will be eager to jump back in by the Fall but I’m hoping to find interesting topics to blog here now that I have things up and running again on Jekyll+github

Puzzle Box that I gave out to everyone. These went over surprisingly well.

Our test parabola $ 2x^2 + 3x + 2$ and the point P at $(-\frac{1}{2},1)$

Algebra

With only the tools from Algebra I, we can already attack this problem.

1. First let’s assume there is some tangent line $y = mx + b$ that goes through b. In this case its more convenient to use point slope notation. $ m = \frac{y - 1}{x + \frac{1}{2}}$

2. We want this to intersect the parabola so we set the two expressions equal to each other: $mx + \frac{m}{2} + 1 = 2x^2 + 3x + 2$

3. That simplifies to: $0 = 2x^2 + (3-m)x + (1 - \frac{m}{2})$

4. Now we want a single solution to this equation since the tangent intersects only once and to do that we need the discriminant to be 0. That works out to $ (3-m)^2 - 4 \cdot 2 \cdot (1 - \frac{m}{2}) = 0$

5. Conveniently (why?) that quadratic simplifies to $ (m-1)^2 = 0$. We can now plug the slope of 1 back into our equation in step 1. This gives us $1 = \frac{y - 1}{x + \frac{1}{2}}$ or reorganizing a bit $y=x + \frac{3}{2} $

Geometry via the directrix-focus formulation.

In my last post post I delved into how the perpendicular bisector of the segment between the focus and a perpendicular down to the directrix is also the tangent line.

For this equation, the focus is at $(-\frac{3}{4}, 1)$ and the directrix is $y= \frac{3}{4}$ The slope of this segment is $ \frac{1 - \frac{3}{4}}{-\frac{3}{4} + \frac{1}{2}} = -1$

The perpendicular to this segment is its inverse reciprocal and has a slope of 1. So we can again utilize point slope form to get $1 =\frac{y -1}{x + \frac{1}{2}}$ This simplies again to $ y = x + \frac{3}{2}$

Polynomial division

Its not as widely known but you can find the tangent line for any polynomial at $x=x_0$ by dividing by $(x-x_0)^2$ See: post with discussion about this

So in this case since we’re look at $x_0 = -.5$.

Or in other words the tangent line is $ y = x + \frac{3}{2}$

Calculus

Finally for comparison via the derivative:

$f’(x) = 4x + 3$ and $f’(.5) = 1$ That’s consistent with all of above and if plugged into point slope from will give us the equation $ y = x + \frac{3}{2}$ for the fourth time

We’ve been going the reflective property for a parabola in class and I wanted to make my own diagram to illustrate a few nice points (and all without calculus).

First without loss of generality I’m going to use the simplest quadratic equation $y=x^2$ but remember “There is only true parabola”

That said to set the context, for this parabola the directrix is at $y=-\frac{1}{4}$ and the focus F is at $(0,\frac{1}{4})$

1. By definition every point P on the parabola is equidistant from the focus F and a perpendicular line down to the directrix $\overline{\rm PN}$

2. That means by definition P is also on the perpendicular bisector of FN (every point on the perpendicular bisector is also equidistant from the two endpoints).

3. What’s super nice its easy to prove that no other point on the parabola also lies on the $\perp$ bisector.

   

   P’ is equidistant from F and N’ while P’N is the hypotenuse of a triangle with P’N’ and therefore P’ is not equidistant from P and N, not on the $\perp$ bisector.

4. That means the perpendicular bisector by definition is also the tangent to the parabola at point P since no other point beside P on the parabola lies on it.

5. Via the congruent triangles one can see the tangent line intersects the horizontal segment through vertex to the $\overline{\rm PN}$ exactly at its midpoint.

6. All three grey angles are congruent (two are from the perpendicular bisector and the last is from the transversal.

7. $\therefore$ that means a vertical wave or ray heading down will bounce off the parabola and hit the focus or put another way every vertical ray will converge at the focus.

In the course of the evening, we noodled on this for a while and initially made no progress. So I decided to model it in geogebra to confirm whether there was a unique solution. What I found visually was that there appeared to be one when the smaller box was tilted 45 degrees. And that was in fact not hard to confirm because the triangle DFA inscribes the diameter in that case, along with the triangle FGA. You can then solve with the Pythagorean theorem and verify that they give the same result in this configuration.

\(\text{the diameter: } AF^2 = AD^2 + DF^2 = 16 + (4 + 2\sqrt{2})^2 = 40 + 16 \sqrt{2}\) \(\text{and also the diameter: } AF^2 = FG^2 + AG^2 = 4 + (2 + 4 \sqrt{2})^2 = 40 + 16 \sqrt{2}\)

But this isn’t entirely satisfying because it basically presumes that answer and confirms it works and it doesn’t settle whether this is the only answer.

Trying to prove the configuration

One of the central things the above process used was the inscribed right angles embedded in the circle at D and G. This seemed like an important part of the solution. I tried out a bunch of variants including adding lines through the center of the circle to middle of the squares and to the outer corners. Nothing seemed to prove that interior angle was 45 like I hoped. So I turned to a different approach: trying to prove uniqueness of the solution and settling for a known solution for existence.

The Spirograph Approach

Imagine taking the smaller box and rotating it fully around the circle. Because the box is rigid and a fixed length you end up tracing a smaller circle like below

Its clear the smaller square can only intersect when either one of the corners overlaps point C and only two configuration work for that case. Every other position will place the ends of the square at other locations on the smaller circle. This provides a uniqueness proof of sorts* Although it actually suggests there are 2 configurations depending on which corner of the two corners of the small square is used and the 45 degree rotation is only one of them.

A difficult construction that works

Later on I realized that the inner circle was interesting because it created another inscribed angle at C.

This creates the triangle HCN and if we let r be the radius of the inner circle: $(2r)^2 = 4^2 + CN^2$. And CN is discoverable using the radius of the outer circle which I’ll call R.

From triangle AOQ we can get $R^2 = AQ^2 + 2^2$ and $AQ = 4 - \frac{1}{2}CN$

All I needed was one more relationship between r and R to solve. So I introduced the addtional triangle OPF from the first figure. That has a Pythagorean relationship $R^2 = 1^2 + OP^2$ and we can OP in terms of r as well by introducing yet another triangle.

Triangle OQE gives us $r^2 = OQ^2 + 1^2$ and $OP = OQ + 2$. So we eventually end up with a more complicated than desired system but it is solvable:

*I’ll leave the algebra as an exercise for the reader.

Breakthrough

That felt decent but more complicated than should be possible so eventually I went back to a simpler variant of on my original experiments: looking at the intersecting chords formed by the two squares.

Let consider extending CD to F’ and CG to A’. Note we don’t know if A’ = A or F’ = F yet. This gives us two intersecting chords A’G and DF’.

These must satisfy the relationship: $DC \cdot CF’ = CG \cdot CA’$ which simplifies to $2CF’ = CA’$

Let w = CF’ for the rest of this proof. We now can go back to our original inscribed angles. We don’t know if they overlap each other this time but each inscribed angle D and G defines a triangle on a diameter of the circle and those diameters are the same length. So:

That’s a lot easier than the previous systems and simplifies to $3w^2=24$ or $w=2\sqrt{2}$ Note that’s exactly the diagonal of our smaller square as expected.

We don’t have to really prove this is the case because w is enough to calculate the radius by just plugging into the Pythagorean formula for either inscribing triangle like I did initially.

But for fun let’s look at all points that are $4\sqrt2$ from C and and since that forms a circle there must be a second intersection.

Clearly that second intersection isn’t collinear with either of the segments CD or CG and the diagonal on AC really is.

But if we extend to see what it’s connecting to - voila the second case I alluded to at the top pops out. We can also have the smaller triangle touching and slightly overlapping the larger one.

It’s been a long 3 years. In the course of Covid and life changes, I took a hiatus from my the old blog site and in the interim it broke down. But the itch to post math related posts has been growing along with the desire to migrate off of blogger and onto to a more open source platform.

Math Updates

I’ve switched my focus for the last 2 years and am now leading classes at HMA. This has been fun because in this format, I have a block of kids who are following a curriculum with homework and unlike in a math circle, its appropriate to have a sequence that builds on itself. The kids are a mixture of 7th to 10th graders currently and room sizes are around 15 kids with me and an undergraduate or graduate IA. Currently I’m leading a Algebra block and assisting on a Counting Theory one. I’m thinking a lot about curriculum sequences as I’ve become acquainted with IDEA math, our textbook. You can see my official bio here: Bio

As usual there is a bit of the normal tension to whether I think of myself as a teacher. Despite being a volunteer position 3 hours a week, the format is not much different than any other math classroom. I prefer the terms “leading a room” or sometimes “instructor” to avoid misconstruing the work.

Blog Migration

As I mentioned above, this time around I wanted more control. Blogspot was on life support under google and had basically no tech support and the configuration was what it was. I also wanted to write things in markdown rather than a proprietary markup language so that I wasn’t as tightly bound to a platform.

In my first attempt, I took a look at hugo but wasn’t able to easily get it up and running. The second time around, I decided to try out jekyll instead given this post . With some minor work to get ruby installed on my Mac that required some digging around, the basic instructions worked fairly well and my test posts were working. Deployment on github also was smooth.

I then discovered that jekyll themes are not seamlessly interchangeable. My sample was setup on the basic, minimal theme and I wanted to try out chirpy which seemed promising based on web research. That was most easily done by just starting from scratch with a starter template on github. This also required changing to github actions for deployment but once that was understood the process again was smooth enough. What I did a bit differently next was leverage claude to munge the config files directly rather than the starter pack. After twiddling things around I again was up but the UI wasn’t exactly what I wanted. It took some work to get tags going, turn off categories, make the summaries include a picture etc. During this process, I became a bit more familiar with jekyll and realized chirpy wasn’t as flexible as I would have liked and I was cloning off various configuration elements to customize. I also couldn’t get it to present the view I wanted: a list of full or mostly full blog posts rather than summaries on the home page.

So I researched again and arrived at the minimal mistakes theme. The docs for this one were more detailed than chirpy and it had more accessible tuning knobs built in. I used claude again to migrate the main configuration. Then began the slog of exporting all the blog entries from google and converting them over. Among other things I had to:

• Remove mathjax and css headers
• Fix up the tags and get them into the front matter
• Fix up the images and properly organize them under assets.
• Change all mathjax inline separators to $ which was more natural for jekyll’s built in mathjax support.
• Fiddle with image scaling.
• Tinker with the amount of paragraphs to include in the archive summaries. I’m still not completely happy with where this stands.

Things left to do

I still have to manually confirm the edits on a bunch of the older posts and fix up anything that looks broken. Also I have learned that it was a mistake to embed twitter links like I did for a period of a time. Many of those have broken over time. Moral of the story: keep a local copy of images instead if you want more permanence. Its possible I can find these on the wayback machine and I’ll have to test and try for that.

I also want to improve my local editing setup. I currently use emacs, in markdown mode. I need to set up a template for the front matter to save some time and also see if I can find a way for it to inline display images. That’s currently broken by the format that uses a template {site.baseurl} rather than a full path.

It’s 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 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 it’s 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 do 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 (probably 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.

Let’s 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 parallelogram’s 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

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

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:

That top expression should look familiar: its sine addition

And we can then restate in terms of z

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:

I’m going to factor out $\sin{z}$

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)

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

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

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

Lets substitute that in

And then apply the tangent addition identity

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

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

• 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