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That's Danish; IKEA is Swedish.

the bluetooth chip in my beloved ikea eneby speaker (with the gay pride front cover) decided abruptly to stop connecting to devices today TT__TT i cracked the thing open of course and had a look but unplugging and replugging various cables had no effect; next course of action is probably to try to resolder the bluetooth daughter board (which i HAVE identified, thank you ifixit) except i still haven't unpacked my soldering iron post move and now it's buried in the shed in one of several boxes all of which are behind the woodworking bench and like three bikes. fortunately the 3.5mm jack still works so i THINK my strat is going to be to just fucking plug in one of those bluetooth 3.5mm adapters until i can get my tools unfucked enough to hopefully fix it properly. and of course i can't just get a replacement because ikea has discontinued the eneby because fuck you that's why

I love Golurk's looming, awkward appearance, and how it seems like it's held up by ancient magitek. All of its details contribute to an aesthetic of ruin and age. It seems like such a mournful, protective force. I particularly like the jagged crack in its chest that has been repaired with the same substance as its wristbands. Sometimes I wonder whether it repaired itself, or if another entity did -- and who created it in the first place?

Best Underdog Whose Name Starts With "G" 📂🇬🖋️

Round 2 - Match 4

Our Contestants:

This poll is part of an event that allows the early eliminees from the main tournament have more time in the spotlight!

happy holiday everyone!

On this day, July 27th in 1987, a single was released that would change the world forever.

It's Rick Astley's debut single, Never Gonna Give You Up!

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.

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.

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

this is good and correct. that scene is in fact extremely horny

My toxic trait is finding O'Brien's interrogation of Winston to be extremely horny

This is a good example of what are called "Barnum statements" -- things which are true of almost everyone, but which seem insightful when you read them because they are true of *you*. Furthermore, each result contains many statements (I counted 56 in mine), so that if any one of them matches your current preoccupations or self-concept, you'll forget the ones that didn't apply or only slightly applied.

A fun exercise when you encounter something like this is to tally up all the statements and see how many were really true. In my results, I counted 20 statements that were true about me and 25 that were false.

This doesn't mean that the results are useless, since you may not have thought about your life and emotions in these terms before, and it is still useful to consider e.g. whether you harbor resentment about someone in your life, or whether peace is truly achievable with your current approach. It just means that the usefulness of the results depends entirely on you and your self-knowledge, and the test has no special insight about you.

Quotes from my results under the cut for spoilers.

[disliked olive green] ...inhibiting limitations. Difficult circumstances limit your opportunities for experience and your freedom of action. You feel deprived because you have to do without some of the things that would make life pleasant. You expect far too much understanding for your needs from other people, and as a result, you often feel disappointed. You might ask yourself how much understanding and empathy you extend to others. You would like to be free of your...

I think it's quite clever in the way it mixes different motivations and traits. Here, the main topic is "inhibiting limitations", which makes me think of anxiety and an aversion to breaking rules. But then the next two sentences are about being constrained by non-omnipotence and unable to afford everything you want, which is a universal complaint among finite beings, but not one that seems particularly related to inhibitions. Then, the next two sentences are about wanting others to understand you but not being very understanding of others, which again is very common among humans, but which has even less to do with inhibitions.

As a result, someone whose main problem is any one of {anxiety, poverty, ingratitude} might feel like this paragraph applies to them "uncannily well". And there are 12 such paragraphs! It's no wonder that so many people upthread have felt understood.

where’s that quiz where you choose lke 4 colours u like and 4 u dont and it hands your ass on a plate

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

black locust / bur oak / can't tell but possibly musclewood (hop-hornbeam and elm also have similar leaves)

sassafras / tuliptree / hickory (does not appear to be shagbark hickory, but i can't tell the other hickories apart just from their leaf)

cottonwood / horsechestnut (or buckeye; i can't tell them apart by their leaf either) / some kind of willow

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.

listen up chucklefucks, i just gotta say. I'm not defending zir, but I'm sad zie deactivated. Like, i get that trauma lasts a long time and the good stuff is maybe easy to forget?? so maybe it's just like that. And my beloved mutual @/pompeyspuppygirl made a post about zir clout chasing behavior, which is pretty shitty behavior if it's true (and if we're canceling someone it had better be pretty severe). anyways now that zie's gone pompeyspuppygirl said it was okay to make this post (again, thanks ppg everyone go follow her --really everyone in this whole drama is worth a follow)

ANYways yeah zie was my mutual and like, reblogged a lot my smaller posts. (that isn't to discredit what my mutual pompeyspuppygirl is saying about zie clout chasing ofc). AND idk zie was always reblogging art from new and undiscovered artists and reblogging donation posts (which if you don't know is really bad if you're trying to clout chase...) (again, though, ppg is my mutual i believe her.) and like, remember on valentines day i tried to blaze zir posts and zie told me to stop because zie didn't want the posts to go viral? (but again ppg is my mutual and has a lot of proof in the Google doc I'm not trying to disprove that I'm just saying what else I know)

Idk, like i feel like a lot of people loved zir's blog a while back, bc like zie DID make some good posts?? So idk why everybody's acting like they aren't even a little bit sad.,. like ngl this feels like maybe all the reasonable people left to Twitter and all the Twitter refugees who love drama came here??? shdfhhdhdhdhdh haha but idk...look idk, i just, julie i do miss you. idk. more thoughts later sorry I'm getting worked up shshs

He's unaware of a lot of things, but he does intentionally choose to align himself with the police -- I am inclined to say that he willingly incarnates and represents the Hunt.

Would Javert from Les Misèrables become an avatar of the Hunt

Clouds for scenery (1783) PNGs

(source: desimonewayland)

do you see the people post, posting the color of the sky it is the shitpost of the people who are not afraid to die

“average person eats 3 spiders a year” factoid actualy just statistical error. average person eats 0 spiders per year. Spiders Georg, who lives in cave & eats over 10,000 each day, is an outlier adn should not have been counted

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

anti AI guys keep making AI sound so fucking cool

I agree that the phrase "being normal about [group]" can be used to mean "behaving like a typical person (which is good) with respect to [group]", which I dislike. In fact, while writing the above reply, I was thinking of another common usage of the phrase as meaning "having the correct opinions about [group]", which bothers me even more.

If "normal" is being used to mean "correct, popular among people I respect, typical, admirable, common sense", that is a bad way to use words, because it conflates concepts which are important to distinguish.

However, in this particular context, "normal" can also be read as "everyday, chill, neutral, default, forgettable", which does not strike me as a pernicious usage. If you read it this way, then "being normal about [group]" points at an important aspect of tolerating and respecting the group in question.

This concept of "capable of neutral, casual interactions" is particularly useful when assessing a potential friend (or someone you might invite to a groupchat, or someone whose party you might attend, etc.). In that circumstance, it's usually less relevant what their political beliefs are, how much they know about [group], or how much they care about the welfare of [group] -- what you want to know is whether they can treat you like any other person in the friend group. It is awkward and uncomfortable when the prospective acquaintance has very strong positive feelings about your demographic group, or when they are very concerned about interacting with you respectfully, even though those things are probably good in an abstract sense.

To inquire about this by asking "are they normal about [group]?" is suboptimal because of the ambiguity with other meanings of "being normal about", but it is a way to express something that needs to be expressed, and as such I am sympathetic to it.

Hate how people talk about “being normal” about something. That only applies to like, being weirdly obsessed with something unusual. You can tell me to please be normal about riding a train, or watching an Anne Hathaway movie. Things that I KNOW I’m weird about.

If you’re using it to describe whether someone is a bigot or not, it’s completely incoherent. Bigotry is normal to bigots. When I hear someone say “I’m normal about X group” I don’t assume that means they share my beliefs. I assume that means they’re uncritical about their own.

Is there something I’m missing here??

I don't consider "normal" a desirable or praiseworthy state, so the usage of the word to (ironically) describe unusual obsession tends to rub me the wrong way.

That said, I do think that the application of the term to a person's feelings about minorities is pointing at something real. Having strong and unusual emotions about people you interact with on the basis of their demographics is generally awkward, counterproductive, and destructive of empathy and solidarity, even if the emotions are positive.

Personally, I notice that when I have negative aliefs or inclinations related to a demographic group, they prevent me from perceiving that group as "normal" and "just people" -- I feel like I should "balance it out" with positive evaluations of the group, and end up thinking about whether I am being bigoted more than actually interacting with them as a person.

If you decide how to act towards someone based primarily on their demographics, that is the same mistake as bigots make, even if you treat members of othered minorities unusually well instead of unusually poorly. "Being normal about" a group can mean treating members of that group like normal people and interacting with them without having an unusually strong emotional reaction to their membership in a given demographic.

Hate how people talk about “being normal” about something. That only applies to like, being weirdly obsessed with something unusual. You can tell me to please be normal about riding a train, or watching an Anne Hathaway movie. Things that I KNOW I’m weird about.

If you’re using it to describe whether someone is a bigot or not, it’s completely incoherent. Bigotry is normal to bigots. When I hear someone say “I’m normal about X group” I don’t assume that means they share my beliefs. I assume that means they’re uncritical about their own.

Is there something I’m missing here??

New stickers for my pals who love (trans) men soon

It's simple.

If you give it defender, then it can't attack. But if you have an effect that says it can attack "as though it didn't have defender", then it can attack again. But then, once it has been declared as an attacker and is now an attacking creature, its ability means that it can't be declared as an attacker. But this doesn't matter, because no attacking creature can be declared as an attacker anyway, since the declare attackers step happens before any creature becomes attacking.

It's a perfectly meaningful, and perfectly useless, ability.

Akolia, Cruel Duskmaster, submission by @WillWeaverRVA