Clouds For Scenery (1783) PNGs

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!

just had a nice sudden cry thinking about the song "tempus est iocundum" and how much i wish i could meet the anonymous author of that song and many others from the remote past so i could tell them how much i like it, feeling sad that they may never know that even 800 years later someone sings their music every day in the shower and while opening at work and stuff... but i think my tears were tears of gladness too, because maybe through some mystery i will be able to speak to them someday! at times in my past it has felt really certain to me that one day all of time will be gathered up into one perfect moment and we will all get to greet each other forever

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

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

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’m sick of these SJWs telling me not to buy bottled water

i propose a new hashtag

#watergate

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.

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.

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