## Math and Logic Puzzles: Redux

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geometry
Goon

Joined: July 09, 2019
Location: outted chennisden alt
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.
I signed up to play MAFIA, not so you could roll me town again.

Tazaro
Selfie

Joined: July 04, 2010
Location: I can go now without writing more
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.
Last edited by Tazaro on Fri Jul 26, 2019 2:27 pm, edited 5 times in total.
Maybe Subservience to Protocol isn't tantamount to Solution to Problem ...

Kagami

Joined: November 05, 2013
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.

Kagami

Joined: November 05, 2013
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.

Felissan
Goon

Joined: April 03, 2015
Location: France
Pronoun: She
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.
"Dammit Felissan, making someone lose the game is NOT NICE" - DeathRowKitty 2016
I believe in you, Miss Zirconium! <3

Who
Yes?

Joined: March 22, 2013
Location: Third Base
Pronoun: He
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.
Who said that?
Chamber. It's all a conspiracy.
Or is it?
5

StrangerCoug
Does not Compute

Joined: May 06, 2008
Location: El Paso, Texas
Pronoun: He
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?
STRANGERCOUG: Stranger Than You!
Current avatar by sushy of FurAffinity.

chennisden
Macho Pichu

Joined: February 11, 2019
Location: sheltered in place
Pronoun: He
Doh

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

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

If these are equal n+2=3
"I believe in our ability to somehow lose."

chennisden
Macho Pichu

Joined: February 11, 2019
Location: sheltered in place
Pronoun: He
Ugh people these days just think about alg and NT

and geo gets ignored
"I believe in our ability to somehow lose."

Kagami

Joined: November 05, 2013
The index of the tetrahedral and triangular number don't have to be the same.

chennisden
Macho Pichu

Joined: February 11, 2019
Location: sheltered in place
Pronoun: He
Hmm.
"I believe in our ability to somehow lose."

chennisden
Macho Pichu

Joined: February 11, 2019
Location: sheltered in place
Pronoun: He
(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
"I believe in our ability to somehow lose."

Mitillos
Mafia Scum

Joined: August 23, 2012
Pronoun: He
Do the factors on the two sides have to be the same, or can you have different factors giving you the same product?
You don't have ambiguity; you have options.

StrangerCoug
Does not Compute

Joined: May 06, 2008
Location: El Paso, Texas
Pronoun: He
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.
STRANGERCOUG: Stranger Than You!
Current avatar by sushy of FurAffinity.

Mitillos
Mafia Scum

Joined: August 23, 2012
Pronoun: He
Yeah, that was my point, in that 3, m, and m+1 potentially need not be consecutive. :Þ
You don't have ambiguity; you have options.

BTD6_maker
Mafia Scum

Joined: April 08, 2016
Pronoun: They
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.
"one of these days i'll read you correctly" - Transcend, Micro 714

Kagami

Joined: November 05, 2013
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.

chennisden
Macho Pichu

Joined: February 11, 2019
Location: sheltered in place
Pronoun: He
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
"I believe in our ability to somehow lose."

chennisden
Macho Pichu

Joined: February 11, 2019
Location: sheltered in place
Pronoun: He
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?
"I believe in our ability to somehow lose."

chennisden
Macho Pichu

Joined: February 11, 2019
Location: sheltered in place
Pronoun: He
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
"I believe in our ability to somehow lose."

Kagami

Joined: November 05, 2013
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.

vizIIsto
Ninja

Joined: January 23, 2019
Pronoun: It
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..

Mitillos
Mafia Scum

Joined: August 23, 2012
Pronoun: He
1/2 + 1/3 + 1/9 = 17/18
You don't have ambiguity; you have options.

vizIIsto
Ninja

Joined: January 23, 2019
Pronoun: It
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?

Mitillos
Mafia Scum

Joined: August 23, 2012
Pronoun: He
Lend the dead man a camel.
You don't have ambiguity; you have options.

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