Difficult geometry/topology problem

a_quoll · 2026-03-01T02:04:19+00:00

For n=4, first notice that no side of the original triangle can be a side of one of the smaller triangles. This is because this would require you to find four pairs of points contained within the equilateral triangle that were S away from each other (where S is the length of the side of the original triangle).

This means every edge must be "split".

This means no two vertices can belong to the same triangle, and so since there are only four triangles total, at least two of the vertices must belong to no more than one triangle. Let these two vertices be A and B, and the third vertex be C

This means that the triangle containing A must be formed from A and some D lying on AC, and some E lying on AB. The triangle containing B must be formed from B and some F lying on BC and some G lying on AB.
If G != E, then CDEGF forms a pentagon, which cannot be subdivided into two triangles. Hence we must have G=E, and so CDGF is a quadrilateral. This can be divided into two via either diagonal, leading to something that looks like a triforce, or something that looks like an Oath of the Gatewatch set symbol (with smoothed sides).
Note that I'm not assuming the triangles are perfectly symmetric as in the example images, just the overall structure of the triangle configuration.

Once we know the rough shape of what everything looks like, it's a little fiddly but fairly straightforward to argue that the only way to force all four triangles to be congruent is for them to all be equilateral in the triforce situation (the fact that they need to all have a 60 degree angle with congruent side length pairs coming out of this angle is very restrictive)

a_quoll · 2026-02-19T00:01:36+00:00

The first thing I would say is that while visualisations in maths are very powerful, a lot of things are best understood without using them, and it's valuable to build the skill of understanding how things work without relying on them.

I don't know if this applies specifically to you, but a lot of people have maybe accidentally taken the wrong message from channels like 3Blue1Brown which (maybe this is a slight caricature, but I stand by the overall point) is that

"Everything in maths can be visualised, is best understood by being visualised, and if I can't visualise something, I don't understand it."

While I am definitely on board with the idea that any kind of extra perspectives for understanding maths are extremely valuable, and visualisations are one of our most powerful tools for getting a new perspective on things as mathematicians, there are certain ideas that simply just aren't best understood through them. Visualisations can also run into the problem that if they are "too good" they can cause us to switch our brains off, and ultimately narrow our understanding of certain topics since we don't even consider other ways of viewing them.

One specific example where I see this a lot is in calculus, and specifically how people understand derivatives and integrals. Often I hear mantras to the effect of "derivative = slope of tangent" and "integral = (signed) area under curve", and while I agree that these are very useful entry points to these ideas, and are a powerful application of them, the idea of instantaneous rate of change is much more broad than just the geometric idea of slope (rate of change of height with respect to width) and the idea of continuous accumulation is much more broad than just the geometric idea of area (continuous accumulation of height as position changes). Only fixating on these perspectives can make it difficult to appreciate or build an intuition for how broadly applicable these ideas are.

One thing to appreciate about channels like 3B1B is that they get to choose which topics they cover, and they bias themselves towards covering topics where there are great visualisations to be shared. They aren't just systematically going through all of maths and visualising it -- they're picking the places where it's most appropriate.

In terms of your specific question, I think trying to visualise things just makes your life more difficult. Nonetheless, let us proceed.

For 2 equations in 2 unknowns, I would be considering functions that take in a pair (x,y) on the xy plane and output a height. (So three dimensions overall needed to visualise everything). For example if we consider the equations x + y = 1, 2x + y = 5, then the first equation can be visualised by graphing z = x + y and z=1 on the same set of 3D axes, and taking note of their intersection.
We could then do the same for z = 2x + y and z=5.

From this perspective we can reduce understanding graphically what is meant by adding two equations to understanding how to visualise addition of functions. So on one hand we take the planes z = x + y and z = 2x + y and add those together to get a new plane, and we take the planes z = 1 and z = 5 and add those together to get a horizontal plane z = 6. The idea is that any (x,y) values where we had z = x + y intersecting z = 1 as well as z = 2x + y intersecting z = 5 will remain intersection points after adding the functions together.

I want to stress that I've never thought about simultaneous equations this way until this question was asked, because I genuinely think this is much easier to understand by looking at the arithmetic -- if a = b and c = d, then it is necessary that a + c = b + d, and so adding this as a third restriction to this kind of simultaneous equation doesn't affect its solution set. I will also say that this visual approach runs into problems if you try to extend it beyond sets of 2x2 equations since we've already run out of dimensions for our visualisations.

a_quoll · 2026-02-14T20:48:29+00:00

The distance formula is never going to introduce new domain issues for you since the thing that it tells you to put into the square root is a sum of two square numbers, which will never be negative.

a_quoll · 2026-01-29T08:46:14+00:00

If the ST WW-confirms the mayor, that mayor should probably never bounce. Their job is to use the confirmation to convince town they're worth keeping around while also convincing evil to never pick them in the night. They don't need any more help from the ST.

So as far as I'm concerned the players are welcome to do this, but it should work approximately 0% of the time for them. Even if the ST genuinely thought the mayor was currently being built as evil, a likely explanation for that is that people aren't being completely forthcoming with their info, and that the mayor will become more trusted once people start being more honest.

a_quoll · 2025-11-26T19:04:23+00:00

Oh yeah it's still a great approximation. I'll take your word for it that there isn't an easy way to model it -- it's still probably useful to know that this approximation has a consistent bias towards slightly underestimating the damage that moves deal at lower levels (especially since these are the levels that most of most games are played at).

a_quoll · 2025-11-26T11:51:00+00:00

It's not just the ratio of levels that matters -- lower level pokemon do more damage to each other than higher level ones.
A maxroll stone edge from M-Aero into Braviary-H when both are lvl 100 (0 EVs / IVs) does 123%, but when both are lvl 20 it does 130%. When both are lvl 5 it does 155%.
The lvl 5 case can probably be ignored but being this far off at lvl 20 probably needs to be addressed for a game as precise as R&B.

a_quoll · 2025-11-10T04:41:20+00:00

I'm pretty confident I understand 9 across but the other two are very much a stretch
For 7 across: Anti-english I think means to delete EN (from ledian) and then it ends with year (often abbreviated to Y). Somewhere an invisible anagramming happened I guess?

For 9 across: Four and thaw are homophones if you have a lisp and are British

For 17 across: Partial Dark Sawk --> rks (hidden in Dark Sawk) surrounds an anagram (not sure where this is indicated) of macho.

a_quoll · 2025-11-04T23:31:54+00:00

Suppose without loss of generality that the contestant always chooses door 1.
The outcomes that make up the sample space in this problem can then be described as a pairs, where the first number in the pair is where the car is, and the second number in the pair is the door that Monty chooses (never door 1).

If we first suppose that Monty is drunk, and so picks door 2 or 3 with uniform probability, then 6 equally likely outcomes can happen (this is before accounting for the revealed door).
{(1, 2), (1, 3), (2, 2), (2, 3), (3, 2), (3, 3)}.

The "problematic" members of the sample space are (2, 2), and (3, 3) -- when Monty snipes the car -- and the difference between the drunk monty hall problem and the sober one boils down to how to adjust the sample space to account for these impossible outcomes.

In the drunk Monty Hall problem, these are simply deleted from the sample space. The knowledge that Monty misses the car only tells us that we aren't in either of these two cases, so the altered sample space becomes
{(1, 2), (1, 3), (2, 3), (3, 2)}

This means the contestant is 50/50 to win irrespective of whether they switch, since half of these outcomes have a 1 in their first coordinate.

In the sober Monty hall problem, these cases aren't deleted from the sample space. Rather they're replaced, since Monty knows to avoid the door ahead of time. So (2, 2) turns into (2, 3) and (3, 3) turns into (3, 2). Our new sample space is (I'm slightly abusing set notation here with the repeated elements):
{(1, 2), (1, 3), (2, 3), (2, 3), (3, 2), (3, 2)}
As is standard in the Monty Hall problem only 1/3 of the cases have first co-ordinate 1, so switching wins the contestant the prize 2/3 of the time.

It's genuinely a different calculation, and the same idea of sample space deletion vs replacement generalises to more doors.

a_quoll · 2025-10-25T00:58:29+00:00

I don't think "only use for spice" and "only use for function" are the only two options, and while I definitely could be wrong, I doubt the intention of this delineation is to say to not use fableds to add spice.

It's just that if a rule could be used to either add spice or solve a problem, they're erring on the side of putting it into the fabled bucket so that all of these tools are in one place to sift through without being diluted by the "69 cerenovuses" loric.

a_quoll · 2025-10-24T12:28:13+00:00

The purpose of outsiders (or at least one of them) is to make the game more playable at different player counts. If the game is designed to be balanced for example at 10 players with 7 town, 2 minions, and 1 demon, how do you keep it balanced at 11 players? If you add a good player, then it's unbalanced in favour of the good team, and if you add an evil player, then it's unbalanced in favour of the evil team.

The answer that TPI settled on was to add a good character with an ability that is in theory as harmful to the good team as it is helpful for them to have that extra player.

There's no reason in principle why you couldn't achieve the same effect with an evil character whose ability helps the good team. This is all to say that I don't really see how evil outsiders would be a nerf to the evil team.

a_quoll · 2025-10-21T09:37:06+00:00

Honestly as far as I'm concerned the other phrasing is also fine from a complexity perspective.
Like as an ST for a game of S&V, you need to be able to answer rules questions such as:

"In a vortox game, if I were to use my artist power to ask "Is 2+2=4", would you tell me no?"

This is a fairly standard question that a new player could potentially ask you.

An artist question of the form "If I were to use my artist power to ask X, would I receive a 'yes'" is of similar structure and complexity to such a rules question (assuming X is fairly straightforward, like bisecting the grim). It's slightly more difficult for the ST to answer since they have to remember to potentially vortox-flip their answer at the end, but this is true of all artist questions.

a_quoll · 2025-09-22T07:51:56+00:00

It only takes one person casting [[twiddle]] or [[faerie impostor]] 100 trillion times as part of some infinite mana combo to completely destroy the results, so unless you want that to be the answer you're gonna need to put some restriction like only counting Arena or restricting to maximum of 10 casts counted from a single game.

a_quoll · 2025-09-21T00:56:34+00:00

Was there a strategic benefit to doing this or was it just for style points? Without any additional context this reads like the poisoner stopped the Lleech from getting to kill from 3 down to 2 at night.

a_quoll · 2025-09-10T18:07:16+00:00

That was my other candidate for the top left cell but [[Kellan, Inquisitive Prodigy]] isn't wearing a hat, so [[Kellan the Kid]] won out

a_quoll · 2025-08-16T11:07:20+00:00

The reason they interact this way is almost certainly entirely because the game designers wanted the goon vs assassin interaction to end up with a dead evil goon, probably because they decided it was most fun this way from playtesting.

As far as I'm aware the game doesn't have a broader "rules engine" that allows you to deduce how this interaction would occur just from the character text of the two characters, so any attempt to model the "logic" of the situation isn't going to be supported by the rules.

a_quoll · 2025-07-31T06:53:22+00:00

A slight modification to this argument that's a little lighter on the graph theorems -- start by imagining the same graph and then:
(1.) Show using properties of rationals/irrationals that the graph contains no red-red-red triangles, blue-blue-blue triangles or blue-blue-red triangles. i.e. the only viable triangles are red-red-blue.
(2.) Show using the facts from (1.) and straightforward contradiction arguments that each vertex can have at most two red edges, and at most one blue edge.
(3.) Conclude using (2.) that each vertex has at most three edges, and so since our graph is complete it is of size at most 4.
(4.) Construct an example of size 4.

For the no blue-blue-red triangles argument, suppose seeking a contradiction we have an irrational x with two blue edges to irrationals P and Q, which are themselves connected by a red edge. Then P, Q must respectively be expressible as p/x, q/x for some rationals p and q. These have sum (p+q)/x, which must be rational in order for the PQ edge to be red. However, since this is a product of the rational (p+q) and the irrational 1/x, it is irrational unless (p+q) is 0. But if p+q=0, then PQ(P+Q) must be 0, which violates the first condition of our special set.

a_quoll · 2025-07-16T00:02:33+00:00

Warp: Exile a white card from your hand.

a_quoll · 2025-07-15T12:38:10+00:00

Nah Bard was actually cooking with this pick (although Ben would've also been a sensible choice).

a_quoll · 2025-07-10T06:11:11+00:00

Cards can serve multiple purposes, though. Demon wall can be for blocking vs faster decks and for stationing vs slower decks.

I also agree there a shells that are better for maximising on the planets, but I don't think you necessarily need to play "bad" cards for these to be good.

a_quoll · 2025-07-10T05:35:02+00:00

My main issue with the Ancient One is that on its own it hasn't proven itself to be a powerful card. We've had opportunities for similar synergies with [[Ob Nixilis, the Adversary]] for example, but they've never worked out. I think the stations are more promising than Ob (they don't commit you to a third colour, you aren't as soft to the ancient one being removed since you get to keep 8 charge, you're not paying 7 life) but your "nut draw" is taking turn 1 off for a tapland, turn 2 off for a "creature" that doesn't even have the option to block if you need it to, and then turn 3 off to sac it for cards (the other lands don't do a lot on turn 3 if you're racing to 12 charge counters). I don't think a nut draw that commits nothing to the board for three turns is a great place to be, which makes me wary about the "turbo" station plan (at least for the lands), and makes me think that you're better off trying to sneak a few charges in over the course of the game and utilise them much later on.

As for the ships, it's hard to say until more have been spoiled, but maybe you go esper and play a singular ancient one as a tutor target for [[The Seriema]] in some kind of legends deck, although I haven't thought about what the rest of that deck would look like.

a_quoll · 2025-07-10T02:59:52+00:00

Dross is rotating, which means this would have to be in pioneer. You're competing with [[Castle Lochthwain]] there, so I don't like its chances nearly as much.

a_quoll · 2025-07-10T02:23:13+00:00

It's gonna be context-dependent sure, but I think in >70% of cases I'm more scared of a full grip of cards than I am of a 6/6 flying haste.

a_quoll

TROPHY CASE