Burrowed In The Furrows 'tween The Eyebrows Of The Cliff Face Is A Woman, Dark And Narrow, With Her Posture

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

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

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.

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!

en.wikipedia.org

Christmas truce - Wikipedia

happy 110th to the christmas truce!

:3

And looking at all these kinsmen so arrayed, Arjuna, the son of Kunti,

Was overcome by deep compassion; and in despair he said: Krishna, when I see these my own people eager to fight, on the brink,

My limbs grow heavy, and my mouth is parched, my body trembles and my hair bristles,

My bow, Gandiva, falls from my hand, my skin’s on fire, I can no longer stand—my mind is reeling,

I see evil omens, Krishna: nothing good can come from slaughtering one’s own family in battle—I foresee it!

I have no desire for victory, Krishna, or kingship, or pleasures. What should we do with kingship, Govinda? What are pleasures to us? What is life?

The men for whose sake we desire kingship, enjoyment, and pleasures are precisely those drawn up for this battle, having abandoned their lives and riches.

Teachers, fathers, sons, as well as grandfathers, maternal uncles, fathers-in-law, grandsons, brothers-in-law, and kinsmen—

I have no desire to kill them, Madhusudana, though they are killers themselves—no, not for the lordship of the three worlds, let alone the earth!

Where is the joy for us, Janardana, in destroying Dhritarashtra's people? Having killed these murderers, evil would attach itself to us.

It follows, therefore, that we are not required to kill the sons of Dhritarashtra—they are our own kinsmen, and having killed our own people, how could we be happy, Madhava?

And even if, because their minds are overwhelmed by greed, they cannot see the evil incurred by destroying one’s own family, and the degradation involved in the betrayal of a friend,

How can we be so ignorant as not to recoil from this wrong?

~the bhagavad gita 1:27-39, johnson trans

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.

Fuck you, City of Ur!

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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??