Page **4** of **4**

Posted:

**Fri Jul 26, 2019 1:42 pm**
by **geometry**

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.

Posted:

**Fri Jul 26, 2019 2:08 pm**
by **Tazaro**

In post 72, Kagami wrote:If we assume the super-being is indeed a super-being with super prediction powers, then we accept that she can predict what our philosophical response is to the dilemma when it's presented and so our ultimate decision

What if one doesn't have a prepared strategy at the time the dilemma gets presented and the box content has already been dealt with by her -- like the box dilemma's being something fun for her to do that she just springs up on a random person in a flower field somewhere, and at the end of the day the mystery box will be shown to have either 0 or 1,000,000 so just take both after simply figuring out that the best thing is to add 1000 to whatever the irrevocable amount in the mystery box is?

Though, I do acknowledge that if a person if prepared to meet her and her dilemma with a committed strategy already in play, then her prediction could Subsequently take that into account and synchronize with the "ultimate decision" that takes place AFTER boxes' content are offered for selection.

Moreover, it's interesting that the evidence for her predictive power is technically inductively grounded (like merely justifying the proposition that "sun rises every day" by saying it has always done so/merely making the assessment that she is a good predictor because she has gotten previous things right), and that power may actually be thwarted this one time (among possible myriad times) by her actually doing an interaction with the objects/experimental person/events that are attempted to be predicted.

Posted:

**Fri Jul 26, 2019 2:16 pm**
by **Kagami**

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.

Posted:

**Fri Jul 26, 2019 2:33 pm**
by **Kagami**

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.

Posted:

**Fri Jul 26, 2019 2:36 pm**
by **Felissan**

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.

Posted:

**Fri Jul 26, 2019 3:21 pm**
by **Who**

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.

Posted:

**Sun Jul 28, 2019 11:46 am**
by **StrangerCoug**

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?

Posted:

**Sun Jul 28, 2019 2:28 pm**
by **chennisden**

Doh

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

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

If these are equal n+2=3

Posted:

**Sun Jul 28, 2019 2:29 pm**
by **chennisden**

Ugh people these days just think about alg and NT

and geo gets ignored

Posted:

**Sun Jul 28, 2019 2:31 pm**
by **Kagami**

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

Posted:

**Sun Jul 28, 2019 3:10 pm**
by **chennisden**

Hmm.

Posted:

**Sun Jul 28, 2019 3:17 pm**
by **chennisden**

(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

Posted:

**Sun Jul 28, 2019 3:54 pm**
by **Mitillos**

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

Posted:

**Sun Jul 28, 2019 6:21 pm**
by **StrangerCoug**

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.

Posted:

**Sun Jul 28, 2019 7:15 pm**
by **Mitillos**

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

Posted:

**Mon Jul 29, 2019 1:53 am**
by **BTD6_maker**

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.

Posted:

**Mon Jul 29, 2019 11:54 am**
by **Kagami**

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.

Posted:

**Sat Aug 03, 2019 11:20 pm**
by **chennisden**

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

Posted:

**Sat Aug 03, 2019 11:20 pm**
by **chennisden**

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?

Posted:

**Sat Aug 03, 2019 11:22 pm**
by **chennisden**

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

Posted:

**Sun Aug 04, 2019 5:13 pm**
by **Kagami**

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.