Listen Up Chucklefucks, I Just Gotta Say. I'm Not Defending Zir, But I'm Sad Zie Deactivated. Like, I

the fact that pro-monarchy arguments have degenerated, over the past few centuries, from “the king rules by divine right and is accountable to nobody but god”, to “uhm the royals generate a lot of income from tourism” will never stop being extremely funny to me

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.

Big Mood : Bigger Mood :: Bad Analogies : Badger Analogies

in school they used to call me semipermeable membrane for my habit of allowing some things to pass through me

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.

being as i am an idiot, and having been one my whole life, i just wanna say that i find it very easy to do nothing, and go nowhere. i eat chocolate late at night in the dark. i stand in the garden also. and i’m often waiting for something to happen. and i’m stupid.

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.

Les Mis Shipping Showdown: Round of 32

turnchetta art by felicitymildradeworthington (deactivated)

I have tried for years to discover something, anything, about this card with no success.