I Have Tried For Years To Discover Something, Anything, About This Card With No Success.

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

horsethoughtbarn 5 name

if horses werent called horses what do you think they should be called

I'm glad that people are still having fun on tumblr even after we found out about the frightening ghoul that reblogs posts but doesn't say anything

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

i propose a new hashtag

#watergate

academia.edu

Straightening the Queer: Gender and Sexuality in Truman Capote’s Novella Breakfast at Tiffany’s and Blake Edwards’ 1961 Film Adaptation

Straightening the Queer: Gender and Sexuality in Truman Capote’s Novella Breakfast at Tiffany’s and Blake Edwards’ 1961 Film Adaptation

Unnamed Narrator

from Breakfast at Tiffany's by Truman Capote (1958)

Otome Isekai Roundup

With the new year, I feel the need to make some kind of year-in-review list. So even though I've largely stopped reading comics (the desire to read ebbs and flows every six month), here are the otoisekai that stuck out to me the most in 2024

Crimson Lady/Resetting Lady

What can I even say about the GOAT? As we crawl towards a conclusion, the despair only grows. Death is the only answer, death must be avoided at all costs.

The Villainess Who is No Better than an Extra Cross-dresses to Be a Love Interest

I never talked about this one, namely because I'm not sure how to describe it without focusing on its wokeness. But it is woke. It's bizarrely, strangely woke, as well as genuinely enjoyable, but I keep getting fixated on all the genre-unusual progressiveness, like

BELIEVABLE female crossdressing in a shojo manga

her older brother is fat, a good person, and nobody ever comments on his weight (this point is the most shocking to me honestly)

Older brother is loudly supportive of his gnc sister and male cousin

Unclear if MC is gay, transmasc, or just doing this out of survival, but it says something that becoming a love interest was her *first* response

Boy the MC bewitched in girl form as a child is thrown off by her handsome appearance as an adult, yet awkwardly asks if she wants help breaking her engagement with a man

I forgot, originally he was the stock "sexist love interest who dials it back for his one true love" type, but this one shatters the mold not by making her the one girl he isn't mean to, but instead having him step back from his feelings and step UP as a genuine, actual, fr ally and continuing their friendship. Insane upgrade.

And it all feels pretty organic! They'll hint at complicated feelings from the people around her without grandstanding or molding medieval-y types into 21st century values

Like I can't emphasize enough how weird it is to see a child drawn like this and they're not evil

Behind Her Highness's Smile

I never talked about this one for the opposite reasons of the above. Um. Extremely horny ethicsplay* thriller about a mentally-challenged princess forced to marry a duke (who, in turn, was forced to marry a mentally-challenged princess). What if you were the abused sexy sensual prisoner princess forcibly married to a smoldering tall dark-haired duke AND you had brain damage.

Let me say that I enjoy this story sincerely and that her issues are not played lightly, but it is absolutely going for eroticism. Like oh nooo, you're not mentally competent enough to consent 😉😉 the duke could do whatever he wants and you're too dumb and doll-like to do anything about it😉😉😉 also your maids roofied you😉did I mention your sick bastard brother-king made you like this? And that he takes immense pleasure in that fact?🤫

But I know what you're thinking. Because I was thinking it. For chapters and chapters. "Surely this is a ruse. They didn't actually write a mentally handicapped female lead, not when it's already so horny. This is just to add to the fucked up atmosphere that feeds into everything because there's no fucking way anyone would keep to this premise. Wow she's really committing to the bit lol, not giving an inch. If I were gullible, I would believe they're actually going to fuck. Wait. Was that guy supposed to represent the writer talking to the reader? Hey?"

Did they? Read to find out :)

*I've always hated describing stories as problematic (positive). Ethicsplay, like the story is fucked up. They know it's fucked up, YOU know it's fucked up, and that's why you're here reading it.

A Splendid Revenge Story of a Super-Dreadnought Cheat Villainess

I don't have a lot to say about this one other ​than it's a refreshing revenge-centric OI. It's not treading new ground, but it executes the genre's tropes well. The villains are exaggerated caricatures of hubris, brazen leeches who've forgotten whose blood they've been surviving on, each with their own distinctive brand of arrogance introduced at a measured pace to keep the true hero's OPness from getting boring, all with a unique stylistic flair.

Princess revenge stories are frequently derailed by dull romance or the desire to reinvent capitalism, so the fact that Super-Dreadnought commits itself to smiting Lunaria's enemies without straying from the path makes it a high recommend.

Turning the Mad Dog into a Genteel Lord

I realized I don't have enough screenshots to prove my controversial opinion (that this is less puppy dog bf fantasy and more crypto-age reggressor/caretaker right up until they knew it would mess with the overall light and goofy tone), and I can't say those kinds of things without proof. so. I'll save that for another day.

Anyway. Plot: Priestess Diarin, who is so hot I have more screenshots of her than any of the men, has to tame an abused beast-like 6'8 shredded ex-child soldier into a noble. She's the only one who can change him, she's the only one he can be vulnerable with. Middling plot, heavy slapstick/reaction face-based humor, but everyone is sexy and there are mild to moderate sadist-on-the-art-team impies, so. Recommend.

On an unrelated note can tappytoons please stop picking up manwha with good art? Their translation choices make me unreasonably mad.

If they draw shoes like this, please let Lezhin handle it? I can't take the comma stutters and obvious tone neutering.

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.

do you guys know about the internet roadtrip? right now somewhere between 500 and 900 people are collectively 'driving' a car on google street view trying to make it to canada. it's fun i recommend it

The Hanged Man by Victor Hugo