#inversion

[sinnerbloom]

29 April, 2024

[noco3n-com]

Autumn of art.

[diana-andraste]

Femme-fleur, Pierre Boucher, 1935

[derseprinceoftbd]

[redfoxv]

Girlthing upside-down

Maybe the next time I'm on the table I'll have some modifications ;p

[asmrrpaddict]

Sam x Lovely headcanons

STRICTLY PLATONIC/CLOSE AS SIBLINGS

⚕️Ever since Sam healed them after Adam’s attack, he saw how distraught Vincent was and vowed then and there he would protect Lovely no matter what.

🍿Vincent and Lovely and Sam and Darlin’ have movie nights as unstated double dates.

📞If Vincent tells Lovely updates on Quinn, they call Sam to let him vent out things he doesn’t want to put on Darlin’s shoulders.

🍼Sam calls Lovely to babysit Vincent after asking him to babysit Fredrick and Bright Eyes.

🫂They hug every time they meet up.

🩸Sam held Lovely’s hand while Vincent turned them.

💬In the beginning of their relationship, Lovely would text Sam to ask Vincent’s favorite things so they could surprise him.

😠Lovely was pretty upset with Sam when he ambushed them with Porter, but after they got to hit him, they were better.

🧛🧛🧛During the Summit, when Porter started antagonizing Vincent into a fight, Lovely said Sam’s name quietly, knowing he would hear them, and he rushed over to break it up.

👩‍⚖️At the “trial” after Alexander’s murder, Sam stayed close to Lovely ready to jump in if he needed to. Especially when Christopher was explaining how it could have been one of the Solaires who did it and things were getting tense.

I really wish we could see if they have any sort of relationship.

[juliforu]

Don’t Imagine David and Asher going to sleep next to their mates

Don’t Imagine Angel and Babe smiling down at them as they sleep

Don’t imagine Angel and Babe staying up almost all night scrolling through the internet, trying to find out if humans can learn magic.

Their search for information failing, and ultimately falling into researching non-magical ways to ensure they can help their mates, to make sure they never feel as useless as they did behind that other side of the wall.

Don’t Imagine them adding medical equipment into their carts, reading articles about how to stop bleeding from a wound, how to treat scars, how to help.

Cause while they are un empowered, they’re not going to let their mates almost slip out of their hands and not even attempt to do anything about it again

[irenetherogue]

Going up upside down... That's it, just a gif of me going up haha 😂 It was my first time being fully inverted in the winch and OMG it was AMAZING. (Look at my face -- is my radiant joy showing??) I am so unspeakably happy that I've discovered a way to dance again. It's not a little bit ironic that I had to kind of invent it, and that as incredible and frankly dangerous as it is, I can dance in literally no other way. Fuck chronic illness, and fuck gravity. Today, we dance in the air.

[crimson-and-clover-1717]

Mundus Inversus (or something about feet)

In 203, Ed gives himself up to the dark blue waters of purgatory. His feet are bare, a symbol often associated with death and mourning in art and literature, and his soles are facing upwards. Ed’s literally turning up his toes; dying.

The scene cuts immediately to Stede’s slippered feet, continuing, but altering the motif. The slippers are blue, similar to Ed’s surroundings; but as well as being clothed, indicating a living state, Stede’s feet are contrary to Ed’s - soles down, signifying life. Importantly, Stede’s feet are also in contact with water.

What strikes me is the clarity of the reflection of Stede’s feet, a soles up, but clothed, living version of Ed’s. And if we think about this reflection, it might stretch all the way down, meaning there is an upside down ‘living’ Stede in a sort of Mundus Inversus or mirror world.

Stede’s feet are also getting wet, and if we believe in the truths of folklore, certain ‘monsters’ transform in such conditions.

Our inverted Stede emerges out of the darkness into Ed’s purgatory, transformed. Tail replacing feet, pointing upwards, mimicking Ed’s death plunge. Stede enters this mirror world and meets Ed where he is.

There is also a kind of evolutionary inversion to Stede’s appearance, a fetal shape in a womb-like state. And of course, This Woman’s Work is playing, indicating a chance for rebirth.

As they mirror one another, neither Stede nor Ed are inverted any longer; they are soles (tail) down, ready to ascend out of this upside-down mirror world into a state of renewal.

And as they ascend, we glimpse them in the post-credit scene, returned each by their own agency, heads above water, back to the land of the living.

They have moved from soles mirroring to souls mirroring. Choosing life, together.

(some ideas inspired by @follows-the-bees)

[pennycutenice]

Penile Inversion Blast!!!!!!

[noco3n-com]

Gaze at the beautiful morning sun from the cosmos picture frame.

[ther3alsweetheart]

Me re-listening to Milo's audio where sweetheart helps him shift

(I've listened to it so many times and I still cry)

[knottylilbrat]

In the lab with friends new and old 🩷 feeling super strong and bad fuckin ass

[mumblelard]

today, i wandered through the greening damp, ate mandarins, swept a breezy porch, and read about biosignature definition strategy, survival, and forgiveness. it was a pretty good day

[nais-doodles]

Post inversion Milo head canon. I think it would also be on his back too 

[squareallworthy]

Triangle Tuesday 13: Inversion, fiddling around with real numbers, and a far-out conclusion

This week I'd like to talk about a geometrical method called inversion. But inversion can be kind of hard to get, so first, would you like to see a geometrical method for finding the reciprocal of a real number?

We can represent a real number as a point on a line n, in this case P. Then we need to know where the points representing 0 and 1 are. Here i have marked them as Z and U (for unit, because 1 is the multiplicative unit).

To find the reciprocal of P, draw a circle through U, centered at Z. If P is outside the circle, draw a tangent to it at D, and then drop a perpendicular to n at P'. Then, interpreting n as the number line, P' is the reciprocal of P. Or, to be strict, the length of segment ZP' is the reciprocal of the length of the segment ZP. In this case ZP = 3 and ZP' = 1/3.

If P is inside the circle, simply follow the same diagram backward. Draw a perpendicular from P' to intersect the circle at D, then draw a tangent to intersect n at P.

If P is on the circle, then P = U = P', since U represents 1 and 1 is its own reciprocal.

This works because of similar triangles. Going back to the first case, since DP is a tangent, it is perpendicular to the radius ZD. Then the two triangles ZDP' and ZPD are both right and share a vertex at Z, making them similar. Therefore ZP/ZD = ZD/ZP'. But ZD is the radius of a unit circle, so ZP = 1 / ZP', and there's your reciprocal.

Here we see a this method used to find the reciprocals of 2, 3, 4, and 5. Seen as an arithmetical process, it's a way of finding the multiplicative inverse of real numbers. But seen as a purely geometrical process, we are taking a point and using a circle to find another point. This process is called inversion, or inversion in a circle. The point Z is called the center of inversion, and the circle z is the circle of inversion. We can formally define it like this:

Let P be a point and let z be a circle centered at Z with radius u. Then the inverse of P relative to the circle z is the point P' on ray ZP such that ZP * ZP' = u^2. Equivalently, we have ZP/u = u/ZP'.

(We use the ray ZP rather than the line ZP to ensure that P' is on the same side of Z as P. Alternately, we could use signed distances, such that measurement in one direction on the line ZP is positive and the other negative.)

With that definition, we can extend our idea of finding inverses to the entire plane, since we are not limited by a number line that goes only in one direction. We can also use a circle of any size centered anywhere, since unlike with reciprocation we don't have to treat 0 and 1 as special cases. But before leaving the subject of reciprocation behind, let me list some of the obvious properties of reciprocal numbers:

the obvious properties of reciprocation of real numbers have their counterparts inversion of points. For instance:

Reciprocation is symmetric: if x is the reciprocal of y, then y is the reciprocal of x.

Consequently, the reciprocal of the reciprocal of a number is the number itself: 1 / (1/x) = x.

1 is its own reciprocal, and it's the only positive number with this property. -

If a number is greater than 1, its reciprocal is less than 1, and vice versa.

Numbers close to 1 have reciprocals that are close to 1. Numbers that are very close to 0 have reciprocals that are very large, and vice versa.

I led off this post talking about reciprocal numbers because all of these properties have counterparts in properties of inverses:

If P is the inverse of Q, then Q is the inverse of P.

The inverse of the inverse of P is P.

Points on the circle of inversion are their own inverses, and they are the only points with this property.

If a point is outside the circle of inversion, then its inverse is inside the circle, and vice versa.

Points close to the circle have inverses that are close to the circle. Points very close to the center of inversion have inverses that are very far from the center, and vice versa.

With these in mind, we know what kind of general behavior to expect when we are inverting points, which makes it simpler to follow what's going on when we start inverting more complicated figures.

On difference between reciprocation and inversion is that the reciprocal of 0 is not defined. We could do the same for Z, but it will be simpler to talk about inverses of geometrical figures if we stipulate a single point at infinity, and say that it is the inverse of the center of inversion. Also, since all lines extend to infinity, we will say that all lines meet at the point at infinity.

Since that gives us an inverse for every point on a line, that lets us present our first result on geometrical figures in inversive geometry:

The inverse of a line passing through the center of inversion is the line itself.

So far, not very exciting, so let's move on to different configurations of points. What if we have a vertical arrangement of points instead of horizontal? Our construction requires each point to get its own line, but I'll use rays instead of lines to keep the picture from getting too cluttered.

As expected, as the blue points get farther from Z, their inverses (in red) get closer to Z. But also, they appear to be forming a circle. Can we prove that?

Here we have points P and Q with their inverses P' and Q', P being the closest point on line PQ to Z. By the definition of inversion,

ZP * ZP' = u^2

ZQ * ZQ' = u^2

therefore

ZP * ZP' = ZQ * ZQ'

or equivalently,

ZP / ZQ = ZQ' / ZP'.

Triangles ZPQ and ZQ'P' have a common angle at Z and the ratios of the two adjacent sides equal, so they are similar. By construction, triangle ZPQ has a right angle at P, and so triangle ZQ'P' has a right angle at Q'.

Thus, Q' lies on a circle with ZP' as diameter, as do the inverses of all points on line PQ, and this gives us our second result on inverses of figures.

The inverse of a line not passing through the center of inversion is a circle passing through the center of inversion.

Corollary: since parallel lines meet only at infinity, the inverses of parallel lines are circles that meet only at Z, meaning they must be tangent at Z. Here are two parallel lines and their inverses. Notice that since the dashed line is farther from Z than the solid line, the dashed circle that is its inverse is smaller.

What about circles that don't pass through Z? Here we have a circle in blue with diameter PQ colinear with Z, and R an arbitrary point on the circle. Their inverses are P', Q', and R'. The circle of inversion z has radius u.

By definition of inversion, ZP * ZP' = u^2 and ZR * ZR' = u^2, so ZP * ZP' = ZR * ZR'. Rewrite this as ZP/ZR' = ZR/ZP' and we have two triangles PZR and R'ZP' with a common vertex at Z and adjacent sides in proportion, which makes them similar. In the same way we can establish that QZR is similar to R'ZQ'.

The yellow angles P'R'R and QPR are exterior to corresponding vertices of the first pair of similar triangles, so they are equal. And the green angles PQR and ZR'Q' are corresponding angles from our second pair of similar triangles, and therefore also equal.

Angle PRQ subtends a diameter of the blue circle, which makes it a right angle. Therefore the other two angles in triangle PQR sum to 90 degrees. And so the other green and yellow angles likewise sum to 90 degrees, and angle P'R'Q' is a right angle as well. This shows that R' lies on a circle with Q'P' as diameter.

The same holds for any point on the blue circle, so this circle inverts to another circle, and that is our third result.

The inverse of a circle not passing through the center of inversion is another circle not passing through the center of inversion.

Corollary: a circle that crosses the circle of inversion inverts to another circle that crosses the circle of inversion at the same points, since all points on z invert to themselves.

Corollary: if P is the inverse of P' and PP' is colinear with Z, then a circle with PP' as diameter inverts to itself.

So those are the three important things to keep in mind about lines and circles in inversive geometry. To recap:

Lines through Z invert to lines through Z.

Circles not through Z invert to circles not through Z.

Circles through Z invert to lines not through Z.

Now that we have all that down, what can we do with this method? Let's try it out on a new triangle center.

Let ABC be a triangle with circumcenter O and tangential triangle A'B'C'. Then the three circles defined by AA'O, BB'O, and CC'O coincide at O and at one other place: the far-out point, numbered X(23) in the Encyclopedia of Triangle Centers.

Proof: let Ma, Mb, and Mc be the midpoints of sides a, b, and c. Consider an inversion in the circumcircle of ABC. By the construction of inverses, A' is the inverse of Ma, and A is its own inverse, since it lies on the circumcircle. The circle AA'O passes through the center of inversion, so its inverse is a line passing through A and Ma, which is one of the medians of triangle ABC.

Cycling through the other vertices, we find that the other two circles are inverses of the other two medians BMb and CMc. The three medians coincide at G, the centroid.

Since G lies on all three medians, its inverse must lie on the three inverses of the medians, and this is the far-out point. Note also that inverses are colinear with the center of inversion, so X23 lies on line OG, which is the Euler line.

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.

An index of all posts in this series is available here.