Le-blanc-et-la-noire - Not A Bot Just Shy

I'm not beating myself up for no reason, I have to acausally incentivize my past self not to have fucked up

Let's say there's an online community of people who all have Whatever Syndrome. They talk about all the difficulties and frustrations and issues etc related to Whatever Syndrome. They share advice, they vent, sometimes they just chat and enjoy talking to people who can relate to them properly. Sometimes they make fun relatable observational comedy-style memes about common Whatever Syndrome experiences.

Some of the experiences they make memes about will probably overlap to some extent with the experiences of people who are not on the WS spectrum at all. Let's stipulate (made-up, obviously meaningless numbers incoming) that 10% of the memes they make appeal to a non-WS audience in this way, but 90% are highly specific to the WS niche and won't really be appreciated by outsiders.

In this scenario, the 10% of universally relatable memes will, because they are universally relatable, likely spread far beyond the core WS community. The 90% of niche ones will not (why would they?). From the perspective of someone without WS who doesn't engage with the WS community directly, this will look like 100% of all WS memes seeming to be about things that are just universal human experiences being described as WS-specific experiences for no particular reason. This person might begin to suspect that WS is just a trendy diagnosis that arbitrarily groups completely normal personality traits as a medical issue and that the whole thing is maybe kind of fake. This person is not being unreasonable given the information they have, but for reasons that are hopefully obvious the information they have is very skewed.

On the other hand! If this sort of thing distorts the public perception of what WS is about strongly enough, some people are going to latch onto the relatable memes about it, relate to them (because they're relatable), and wonder if maybe that means they themselves might have WS. This person might do a bit of googling and discover that, in addition to all the relatable stuff they relate to, there are other symptoms that they don't really identify with as much... but then, no one really seems to talk about those things very often, you mostly see people talking about [relatable stuff] when WS comes up, so the latter must be like, the main part, right? So (they think) it can't be too important if the other stuff doesn't apply to me.

[also the whole medical establishment is nightmarishly hard to access and a lot of doctors suck and make diagnoses based on random whims and prejudices, blah blah blah you know all this, the point is that the most obvious solution to "how do I confirm whether I do or don't have a specific medical thing?" is often not reliable.]

Well now, given all of the above... stuff might get confusing huh!

Okay, okay, (you might say), that's all well and good as a toy model of things that might be underlying the discourse you're alluding to, but to what extent is this dynamic actually responsible for what's actually happening? Aha! I have no clue whatsoever, sorry. I'm just the ideas guy.

One thing that's interesting about Dr. Phil is that his show is a spin-off of Oprah and one thing I've discovered, that's interesting, is that it's virtually impossible to find footage from Oprah's show. Go looking and you'll find it hard to find uncut clips of even well-liked famous moments; instead they're all intercut with commentary on the event. Interviews watched by tens of millions of people now exist only in twenty seconds of clips interspersed with talking heads telling us how to feel about it. Over 4,5000 episodes, almost all of it vaulted

Remember Oprah platforming anti-vaxxers and all kinds of alt-med scams and literary hoaxes? Uh, no you don't, prove it. She doesn't even let people see uncut footage of the Good Moments of her show (since she seems to be deeply embarrassed by literally all of it?) so of course she's burying any Bad Moments.

In conclusion the Rainbow Parties segment is one of the few full clips you'll find online intact, somehow. Enjoy.

for /-yr/ i like the song La Monture from Notre-Dame de Paris

for the fun-to-say pile: méli-mélo, micmac, assujetissement, eussent été, farfelu

i'm now looking at my list of least favorite french words to pronounce and going "too many r's" for about 40% of them and "skill issue" for most of the rest. some of these are actually very fun to pronounce i just couldn't wrap my tongue around them a year or so ago, but now i can i guess??? so that's very exciting. makes me hope that someday i'll be able to pronounce the rest of them. this is a bit pie in the sky because i really don't see myself ever getting there with procureur du roi but you never know. and luckily the french abolished the monarchy so it's not like i'll ever have to use that phrase in modern conversation.

anyway here are the words i actually love pronouncing now: décaféiné diététicien filleul pneumonie

i now feel normal/neutral about these words that used to be hard for me: automne, condamner douloureux électricité, énergie inférieur, supérieur, etc. itinéraire lourdeur salmonellose sclérose subodorer succincte

words that are definitely within the realm of my current capability but i haven't practiced them enough: bugle hiérarchisation méditerranéen phtisie

words that are still the bane of my existence but i live in hope: [yʁ] plus at least one other r or [y] sound: chirurgie, fourrure, marbrure, moirure, nourriture, ordures, peinturlurer, procureur du roi, prurit, purpurin, sculpture, serrurerie, structure, sulfureux, tournure all words beginning with ur-, hur-, or sur- other difficult sequence of r's and vowels: construire and other -truire verbs; lueur and sueur; utérus too many r's: marbre, martre, meurtre, opprobre, proroger, réfrigérateur, rétrograde, rorqual difficult sequence of vowels and/or semivowels: coopérant, extraordinaire, hémorroïdal, kyrie eleison, météorologique, micro-ordinateur, micro-organisme, mouillure, quatuor, vanillier not pronounced the way i would expect from the spelling: indemne, penta-, punk just hard for some reason: humour

Triangle Tuesday 2: The circumcenter, pedal triangles, degeneracy, and what even is a triangle anyway?

The circumcenter is almost as simple an idea as the centroid, which we looked at before. To define it, you start the same way. Take triangle ABC, find the midpoints of the sides Ma, Mb, and Mc. Then instead of drawing lines to the midpoints from the vertices, draw perpendicular lines through the midpoints. These lines all coincide at a point O, which is the center of a circle that you can draw through the vertices. The circle is called the circumcircle, and that's why the point is called the circumcenter.

I say almost as simple, but in a sense the circumcenter is simpler than the centroid, because you could easily discover it by accident in the process of simply finding the midpoints. Drawing that perpendicular line, the perpendicular bisector, is the standard way of finding the midpoint of a line segment. It's covered all the way back in Book 1, Proposition 10 of Euclid's Elements, and it's simply this:

So if you find the midpoint of all three sides of a triangle with this method, you've already identified the circumcenter. But that doesn't prove that the perpendicular bisectors always coincide, nor that their point of crossing is the center of the circumcircle. For that, let's return to Euclid (Elements, book 4, proposition 5). Euclid's proof is very straightforward, and leads nicely into something interesting, so we'll follow that, but I will state the theorem differently.

Theorem: the perpendicular bisectors of a triangle coincide and their point of intersection is the center of a circle that meets all three vertices.

Let ABC be a triangle with midpoints of the sides Ma opposite A, similarly for Mb and Mc. Draw perpendiculars to sides AC and BC from their midpoints to meet at point O. Connect three segments from O to A, B, and C.

Consider the two blue triangles. Their sies AMb and CMb are equal, since Mb is the midpoint of AC. They also have OMb in common. Their angles at Mb are right angles, and therefore equal. So they have two sides and one angle the same, making them congruent, and therefore OA = OC.

The same argument applied to the green triangles shows that OB and OC are equal. By transitivity, OA = OB and O is equidistant from the three vertices. The radii of a circle are all equal, so a circle centered at O passing through A also passes through B and C.

Finally, draw a line from O perpendicular to AB. This creates two white triangles with sides OA and OB equal, side OZ in common, and equal right angles at Z. The two triangles are then congruent and the two sides AZ and BZ are equal. So Z is the midpoint Mc, showing that the perpendicular bisectors all meet.

And the same argument works when ABC is obtuse. The circumcenter lands outside the triangle, and in this coloring the white triangles are no longer white, but all the relationships between the segments are the same.

(What Euclid didn't prove is that the perpendicular bisectors of AC and BC do in fact meet somewhere, that is, that they aren't parallel. It's not difficult, but I'm not going to prove that either, at least not yet, for reasons.)

Let's develop another idea. We located the circumcenter by drawing the perpendicular bisectors, but now consider doing this construction in reverse. That is, pick a point, and then draw perpendiculars to the three sides. The intersection of the perpendicular and the side is called the foot of that point with respect to that side. If you do that with with the circumcenter, the feet are of course the midpoints, but you can find the feet for any point.

And if we connect those three feet, we get a triangle. In this case, the medial triangle, which we have seen before. For a point in general, the triangle formed by its feet is called the pedal triangle of that point. ("Pedal" meaning "related to feet," and yes, that is why a lever operated with your foot is also called a pedal.)

So let's draw the pedal triangle for an arbitrary point, move it around, and see what happens. The point is going to sometimes be outside the triangle, but that's all right. With extended sides (dashed lines) we will still be able to draw a perpendicular to find a foot, no matter where the point is.

So there's something interesting -- the three feet become colinear and the pedal triangle flattens out into a straight line when the point is on the circumcircle. Does that always happen?

Looks like it does! So let's prove that. Below is a drawing of the flattened-out pedal triangle of a point on the circumcircle, all labeled up. I've also added a couple dashed lines to make following the proof easier. What we would like to show is that ∠JKP + ∠PKL = 180°.

We're going to extract some information from this drawing based on two facts: a) in a cyclic quadrilateral (meaning it has all vertices on the same circle), opposite angles sum to 180° and b) if two right triangles have the same hypotenuse, the triangles have the same circumcircle. I'm not going to prove either of those here because this post is long enough already, but both of these results follow straightforwardly from the inscribed angle theorem.

Theorem: For a point P on the circumcircle of a triangle ABC, the feet J, K, and L with respect to ABC are colinear.

Okay. PCBA is a cyclic quadrilateral, so

1) ∠BAP + ∠PCB = 180°.

And ∠BAP is the same as ∠LAP, so

2) ∠LAP + ∠PCB = 180°.

The two triangles AKP and ALP are right triangles with the same hypotenuse (the dashed segment AP), so all four points are on the same circle and ALKP is a cyclic quadrilateral. Therefore,

3) ∠LAP + ∠PKL = 180°,

4) ∠PKL = ∠PCB.

Quadrilateral PKCJ is also cyclic (again because of right triangles sharing the same hypotenuse), so

5) ∠JKP = ∠JCP

by the inscribed angle theorem. ∠PCB is supplemental to ∠JCP, so

6) ∠JKP = 180° - ∠PCB

and then combining 4) and 6),

7) ∠JKP + ∠PKL = ∠PCB + (180° - ∠PCB) = 180°,

which means that the pedal triangle of a point on the circumference of a circle is flattened to a line segment. Can we consider such a figure to be a triangle?

Now we can return to Euclid's omission in the existence proof of the circumcircle. Proving that the perpendicular bisectors aren't parallel is equivalent to proving that no two sides of a triangle are parallel, or that the three vertices of a triangle aren't colinear. Euclid didn't do that, but it's pretty simple, so he could have. And then he would simply have said that such an arrangement of line segments isn't a triangle. Modern geometers working with projective geometry can answer differently, and might say that this is a degenerate triangle, but we haven't gotten into that yet.

Let's do one more thing. We can extend the flattened line segment into a line, called the Simson line, after Robert Simson, who never wrote anything about it. It was actually discovered by William Wallace, but not named for him, because that's how things work in math.

The set of all Simson lines from all points on the circumcircle form an envelope in the shape of a deltoid, the Steiner deltoid, named for Jakob Steiner, who for all I can tell was its actual discoverer.

The deltoid is tangent to the sides of the triangle at three points where the Simson line coincides with the sides. I'll have more to say about this lovely deltoid later, but for now, please just enjoy this gif. It took me several hours to figure out how to make it, so if people reading this could spend a collective several hours staring at it, that would be great.

If you found this interesting, please try drawing some of this stuff for yourself! You can use a compass and straightedge, or software such as Geogebra, which I used to make all my drawings. You can try it on the web here or download apps to run on your own computer here.

sygol framed poll (handle with care), 2024 mixed media on tumblr post

Triangle Tuesday 3: The orthocenter, the Euler line, and orthocentric systems

Previously, we have looked at two different ways to mark a point in a triangle. First, we drew cevians (lines from the vertices) to the midpoints of the sides and found that they all cross at a point, which is the centroid. Then we tried drawing perpendiculars to the sides from the midpoints, and those all met at the circumcenter. And you could do this with any point on the side of a triangle -- draw a cevian to it, or a perpendicular from it, and see what happens.

This time, though, we're going to do both. That is, we're going to work with the cevians that also form perpendiculars to the sides. These are the altitudes, which run from a vertex to the nearest point on the opposite side, called the foot of the altitude. The three altitudes all meet at a point H, and that's the orthocenter. (The letter H has been used to mark the orthocenter since at least the late 19th century. I believe it's from the German Höhenschnittpunkt, "altitude intersection point.") Anyway, let's prove that the orthocenter exists.

Theorem: the three altitudes of a triangle coincide.

Here's a very simple proof that the three altitudes coincide. It relies on the existence of the circumcenter, which we already proved before. Given a triangle ABC, draw a line through A parallel to the opposite side BC. Do the same at B and C. These lines cross at D, E, and F and form the antimedial triangle (in blue).

Then the altitudes of ABC are also the perpendicular bisectors of DEF. We proved before that perpendicular bisectors all meet at a point, therefore altitudes do as well.

That was easy. Let's do it again, but in a different way. It's not quite as simple, but it includes a large bonus.

Theorem: the three altitudes of a triangle coincide at a point colinear with the circumcenter and centroid, and GH = 2 GO.

Let's take triangle ABC, and let F be the midpoint of side AB. Then mark two points that we already know, the circumcenter O and the centroid G. We'll also draw the median (green) and the perpendicular bisector (blue) that run through F, leaving the other ones out to avoid cluttering the picture.

We already know from our look at the centroid that G cuts segment FC at a third of its length, so GC = 2GF. Let's extend segment OG in the direction of G by twice its length out to a point we'll label H, so that GH = 2GO.

Now consider the two triangles GOF and GHC. By construction, their two blue sides are in the ratio 1:2, and the same for their two black sides. They also meet in vertical angles at their common vertex G. So by side-angle-side, the triangles are similar, and it follows that HC is parallel to OF, and therefore perpendicular to AB. So H lies on the altitude from H to side AB.

By analogous construction, we can show that H also lies on the other two altitudes. So not only have we proved that the altitudes coincide, but also that O, G, and H all lie on a line, and furthermore that G is located one third of the way from O to H, in any triangle.

This proof is due to Leonard Euler, and the line OGH is called the Euler line. Not only these three points but many others as well fall on this line, which we will get to later on.

Let's look at some more properties of the orthocenter and the feet of the altitudes. I'm just going to look at the case of an acute triangle for now, and show how this extends to the obtuse case later.

Theorem: two vertices of a triangle and the feet of the altitudes from those vertices are concyclic.

Proof is easy: the two right triangles AHcC and AHaC share segment AC as a hypotenuse. Therefore AC is a diameter of the common circumcircle of AHcC and AHaC (following from Thales's theorem).

(Incidentally, look at the angle formed by the blue segment and the altitude CHc. It subtends the same arc as angle CAHa, so (by the inscribed angle theorem again) they must be equal. That's not a part of this theorem, so just tuck that fact away for a moment.)

Theorem: a vertex, the two adjacent feet of the altitudes, and the orthocenter are concyclic.

Same idea, but now the right triangles are AHcH and AHbH, and AH is the diameter of the common circumcircle.

(And incidentally, look at the angle formed by the new blue line and the altitude CHc. It subtends the same arc as HbAH, which is same angle as CAHa. So those angles must be equal too. Since both angles between a blue line and the altitude CHc are equal to the same thing, they are equal to each other. Again, not a part of this theorem, just something I wanted to note.)

So those are some interesting concyclicities, but now let's look at the pedal triangle of the orthocenter, which is called the orthic triangle.

Oh, hey, it's made up blue lines, just like the ones we were talking about. And we proved that the two longest blue lines make equal angles with the altitude between them. By symmetry, we can prove the same thing about all the angles made by the blue lines. So that means

Theorem: two sides of the orthic triangle make equal angles with the altitude between them.

Another way to say this is that the altitudes are the angle bisectors of the orthic triangle. And I admit that was kind of a roundabout way to introduce the orthic triangle, but I think it makes the proof of this property easier to follow.

Two other properties of the orthic triangle immediately follow from this:

In an acute triangle, the inscribed triangle with the shortest perimeter is the orthic triangle

and

In an acute triangle, the orthic triangle forms a triangular closed path for a beam of light reflected around a triangle

which are two ways of saying the same thing.

But those two properties only hold for acute triangles. What happens to the orthic triangle in an obtuse triangle? Let's push point C downward to make triangle ABC obtuse and see what happens. To make things clear, I'm going to extend the sides of ABC and the altitudes from line segments into lines. Here's the before:

And here's the after:

The orthocenter has moved outside of triangle ABC, and two of the altitudes have their feet on extensions of the sides of ABC rather than on the segments AC and BC. The orthic triangle now extends outside ABC, and certainly isn't the inscribed triangle with the shortest perimeter any more.

But look at it another way. We now have an acute triangle ABH, and the line HHc is an altitude of both the obtuse triangle ABC and the acute triangle ABH. Meanwhile, lines AC and BC have become altitudes of ABH.

So what we have is essentially the same acute triangle with two swaps: point C trades places with H, and Ha trades places with Hb. That means that our two theorems about concyclic points morph into each other as triangle ABC switches between acute and obtuse. Here's an animation to show the process:

And this is why I didn't bother with the obtuse case above -- each theorem of concyclicity is the obtuse case of the other.

So if we can just exchange the orthocenter with one of the vertices, what does this mean for their relationship? If you are given a group of vertices and lines, how can you tell which one is the orthocenter and which one are the vertices? Well, you can't.

Theorem: Given an acute or obtuse triangle ABC and its orthocenter H, A is the orthocenter of triangle BCH, B is the orthocenter of ACH, and C is the orthocenter of ABH.

The proof comes from consulting either of the "before" and "after" figures above. Take any three lines that form a triangle, red or black. The other three lines are then the altitudes of that triangle. The three feet are where a red and black line meet perpendicularly, so they are the same for all four possible triangles, which means all four share the same orthic triangle.

(Of course, if ABC is a right triangle, then we get a degenerate case, as you can see from the gif at the moment when C and H meet.)

Such an arrangement of four points is called an orthocentric system. Of the four points, one is always located inside an acute triangle formed by the other three, and it's conventional to label the interior one H and the others ABC.

Orthocentric systems pop up all over the place in triangles, so expect to see more of them as we go along. Now, let me do one little lemma about altitudes, and then I'll show something cool about orthocentric systems.

Lemma: the segment of an altitude from the orthocenter to a side of the triangle is equal to the extension of the altitude from the side to the circumcircle.

We can show this with just a little shuffling of angle identities. Extend altitude CHc to meet the circumcircle at C'. The angles CAB and CC'B, labeled in red, subtend the same arcs, so they are equal. Triangle ABHb is a right triangle, so angle HbBA, in blue, is complementary to it. The same is true for the right triangle C'BHc, so the two angles labeled in blue are equal. Then by angle-side-angle, triangles BHcH and HHcC' are congruent, and segment HHc = HcC'.

By the same argument, we can show that triangle AHHc is congruent to AC'Hc, which leads us into the next bit.

Theorem: all the circumcircles of the triangles of an orthocentric system are the same size.

The blue triangle has the same circumcircle as triangle ABC. From the foregoing, the blue and green triangles are congruent. Therefore their circumcircles are the same size as well. The same argument works for ACH and BCH.

So here is an orthocentric system with its four circumcircles.

The four circumcenters O, Oa, Ob, and Oc form another orthocentric system, congruent to the first one.

If you found this interesting, please try drawing some of this stuff for yourself! You can use a compass and straightedge, or software such as Geogebra, which I used to make all my drawings. You can try it on the web here or download apps to run on your own computer here.

anti AI guys keep making AI sound so fucking cool

GO MY GERM 🦠🦠🦠🦠🦠🦠🦠🦠🦠

I had the idea to do a cell model cosplay beamed into my forebrain like visions from an angry god a few weeks ago and started working on this immediately. and good lord this was the most fun ive had on a project in a long time. like seriously it was so fun getting creative with these materials and thinking about fun ways to do all the organelles. while working on this i probably went "hehehehohohe" at least once an hour.

I think i'm going to name this guy Prota Z. Bacilli. If you look closely I do have every basic organelle that you'd expect from a cell model. (and thats a flu shot lol.)

also please appreciate this costume iteration because. i had to get my mail in this and freaked out my mailman.