Mafiascum.net

here's a geometry question (quite easy):

Consider circle W and points A,B outside circle W. Let the tangents from A to W intersect W at P,Q and let the midpoint of PQ be A'. Define B' similarly. Prove that AA'B'B is cyclic.

there is a sense in which we can recognize the superior utility of ending up choosing the one box, but I have an explanation for why this doesn't actually come from the "math" value of your DECISION. The nature of this problem suggests a different origin of this supposed utility, and this addition to the context negates the validity of appealing to "expected utility" as a basis for determining your best option. Please read what I have written.
If the predicting entity ends up having made the right prediction, then given the awesome predicting power of the genie that suggests you had no real chance to oppose his prediction... we can conclude that you had virtually no real choice--nor egress via decision-making--to cause Mr. Right / Mr. G'Nie(Nius) to be in error; that is to say, for all the talk about wielding the versatile instrument of human will, you could not escape the fate that corresponds to the inner truth contained in the erstwhile unopened mystery box. That would mean the whole situation was simply an (approximately) inevitable gift to you and not a game of efficacious decision-making (fleshing this out with more description, we can say that a weird conceptual object is in the thick of things with the salient trait that it "impersonates" expected utility; it appears to be the case that, whatever its formal name could be, it is NOT maximized by your opting to exclusively choose the one box--so it is not the the so-called expected utility--but is established by the predictor's judgment w.r.t. whether to fill the mystery box or not, in essence signifying your fate in advance )... THUS all of us should disregard the expected utility argument and avail ourselves of the remaining, second perspective in service to our reasoning out how both our guesswork in this decision puzzle and our genie's magically powered guesswork can coexist/interrelate in a meaningful framing of this puzzle. An interesting thing to ask here is this question: is it possible to upgrade FROM sleepwalking through this conundrum, our guessing game conundrum, and being merely satisfied with whatever happens at the end TO the weal of realizing an unambiguously optimal single solution (either the option to one-box or to two-box) in the event that one presents itself as potentially of efficacious use to us in the course of this two boxes game?
Insofar as there is a bona fide chance to actually "play the game" in terms of the puzzle's context, which would be the chance that the magical entity will be wrong, there consequently is a rather simple but crucial detail to notice. The auspicious element of fallibility in this equation of guesses and magic makes possible a situation in which acting according to the dominance perspective wins out over the former perspective as the "right" strategy. Just in case there is an actual game in which you can genuinely make and carry out a decision, the right one is to take both boxes. A successful framing of this puzzle seems finally to be at hand when all the preceding text (which I forged) is understood as holding key insight into the mysterious paradox and is trusted as a proper source of its exact means of resolution. This is thanks to the perfect fit of our forged key into this infamous treasure chest of paradox, our money-offering thought experiment that has entertained us.

Regardless of whether you have prepared a strategy, the super-being can divine the strategy you will arrive at once presented with the box, and she will arrange the contents knowing that strategy. It presupposes that she is not a super predicting super-being to say that she can't do so.

I'm treating the evidence of her being a super being in the article as designed for entertainment purposes, and that this should be approached philosophically, accepting her superness as a given.

If this were a practical exercise, I assume her to be a fraud and any prior predictive prowess she's shown is likely driven by survivorship bias (we wouldn't be in this situation if she had been documented being wrong in the past). The correct choice would be to take both if I had to and assume I'm getting 1k pounds, or, if allowed, to sell the chance to choose to a more gullible individual.

The concept of something being able to predict anything in the future with perfect accuracy has so many paradoxes inherent to it that it feels impossible to give a correct answer to this question.

In this specific case it doesn't have paradoxes because anything the super being could accomplish by predicting they could also accomplish by cheating (Have the box set up so that it automatically burns the money if you take both or some other similar thing). If we accept the super being's superness as fact, then the problem is the same as facing a cheating being, so obviously don't take the second box.

Other than the fact that it may be possible to use her to break quantum mechanics.

Here's an interesting one in addition to #76 above for those who like proofs: Prove that there are only finitely many triangular numbers that are also tetrahedral numbers. How many are there, and what are they?

Doh

Tetrahedral: (n * (n + 1) * (n + 2)) / 6

Triangular: (n * (n + 1)) / 2

If these are equal n+2=3

Ugh people these days just think about alg and NT

and geo gets ignored

The index of the tetrahedral and triangular number don't have to be the same.

Hmm.

(n * (n + 1) * (n + 2)) / 6=(m * (m + 1)) / 2

n(n+1)(n+2)=3m(m+1)

Hmm.

There are only so many ways to get 3m(m+1) to be three consecutive numbers. Uh

Do the factors on the two sides have to be the same, or can you have different factors giving you the same product?

It only matters that both sides of the equation come out to the same number. You can use one variable on one side and another on the other.

Yeah, that was my point, in that 3, m, and m+1 potentially need not be consecutive. :Þ

Part of the difficulty in this problem is that there is no easy way to make tetrahedral numbers into nth powers. I have a proof that, for any n >= 3, there are only finitely many triangular numbers that are nth powers, but I haven't yet got it to work for tetrahedral numbers. This is an interesting problem in itself, so you can try this problem (as well as my two previous combinatorics problems).

For geometry's geometry problem, let O be the centre of W and let it have radius R. Let one of the tangents from A touch W at C, and let theta be the angle AOC. Then OA' = OC cos theta and OC = OA cos theta so OA' OA = OC^2 = R^2. Likewise, OB' OB = R^2. Also, O, A, A' are collinear, and so are O, B, B'. Since OA' OA = OB' OB, we have AA'B'B is cyclic.

The triangular number = tetrahedral number proof is not something anyone should be banging their head over.

I don't think (please correct me if I'm wrong) anyone's presented an elegant proof of it, and the inelegant proof is arduous with steps that are far from straightforward.

In post 91, Kagami wrote:The triangular number = tetrahedral number proof is not something anyone should be banging their head over.

I don't think (please correct me if I'm wrong) anyone's presented an elegant proof of it, and the inelegant proof is arduous with steps that are far from straightforward.

i agree and now i feel worser about not finding it

here's a nice problem:

A secret spy organization needs to spread some secret knowledge to all of its members. In the beginning, only 1 member is informed. Every informed spy will call an uninformed spy such that every informed spy is calling a different uninformed spy. After being called, an uninformed spy becomes informed. The call takes 1 minute, but since the spies are running low on time, they call the next spy directly afterward. However, to avoid being caught, after the third call an informed spy makes, the spy stops calling. How many minutes will it take for every spy to be informed, provided that the organization has 600 spies?

In post 90, BTD6_maker wrote:Part of the difficulty in this problem is that there is no easy way to make tetrahedral numbers into nth powers. I have a proof that, for any n >= 3, there are only finitely many triangular numbers that are nth powers, but I haven't yet got it to work for tetrahedral numbers. This is an interesting problem in itself, so you can try this problem (as well as my two previous combinatorics problems).

For geometry's geometry problem, let O be the centre of W and let it have radius R. Let one of the tangents from A touch W at C, and let theta be the angle AOC. Then OA' = OC cos theta and OC = OA cos theta so OA' OA = OC^2 = R^2. Likewise, OB' OB = R^2. Also, O, A, A' are collinear, and so are O, B, B'. Since OA' OA = OB' OB, we have AA'B'B is cyclic.

yup! A' is called the "inversion" of A about W. defined such that OA' * OA = r^2

In post 93, chennisden wrote:here's a nice problem:

A secret spy organization needs to spread some secret knowledge to all of its members. In the beginning, only 1 member is informed. Every informed spy will call an uninformed spy such that every informed spy is calling a different uninformed spy. After being called, an uninformed spy becomes informed. The call takes 1 minute, but since the spies are running low on time, they call the next spy directly afterward. However, to avoid being caught, after the third call an informed spy makes, the spy stops calling. How many minutes will it take for every spy to be informed, provided that the organization has 600 spies?

S0 = 1
S1 = 2
S2 = 4
S3 = 8
and
Si = 2*Si-1 - Si-4 for i > 3

So you get a goofy fibonacci-esque recursive sequence. For something as low as 600, you can just do it manually or programatically to get the sequence:

1, 2, 4, 8, 15, 28, 52, 96, 177, 326, 600, 1104, 2031, 3736, 6872, 12640, 23249, 42762, 78652, 144664, 266079, 489396, 900140, 1655616

And you see that you've informed exactly 600 spies in 10 minutes. To solve it for an arbitrary N, there should be a closed form solution of the form sum(c_i a_i ^ N) with at most four terms, which wouldn't be terribly difficult to sort out.

I've got a fun little maths problem you might want to get your head around... and chances are you probably already know about the puzzle. Welp!

A long time ago lived a very rich trader in the desert who had three sons. The rich trader had a whopping seventeen camels. The trader decided that after his death, his oldest son should inherit half of his camels, the middle one would get a third and the youngest son received a ninth of the seventeen camels.

Your task is to calculate how you're going to divide the camels among the three sons, and tell me how you got there. It'd be very helpful if you didn't have to cut up some of the camels..

1/2 + 1/3 + 1/9 = 17/18

In post 97, Mitillos wrote:1/2 + 1/3 + 1/9 = 17/18

Yes.
But it's not the answer I want to hear. How will you divide the camels among the sons?

Lend the dead man a camel.