Mathematics marathon

[keemy]

next question

Spoiler

how many trolls does it take to define pi?

 

[Ranzha]

next question

Spoiler

how many trolls does it take to define pi?

One.
Pi = 1/2(tau), where tau = 2(pi).

 

[blah]

And yes, I know the maps that qqwref posted. I just (perhaps wrongly) felt it was unnecessary to post it because it's one of the first things learned in middle school math.

But using those maps would be moving from (I don't know what you call it - traditional?) geometry to cartesian analytic geometry, where an area is described by a(n) [fill in the blank].

 

[aronpm]

@keemy:
The answer is \( \pi \)
3 to define the 3, .1 to define the .1, .04 to define the .04, .001 to define the .001, and so on.

 

[qqwref]

How many cranks does it take to prove you can't turn a lever?

But using those maps would be moving from (I don't know what you call it - traditional?) geometry to cartesian analytic geometry, where an area is described by a(n) [fill in the blank].

Greek-style geometry is intuitive and non-rigorous by nature. (Or rather, not rigorous in the modern sense, but still founded on a bunch of axioms and proven theorems/constructions/measures.) So in some sense they're different styles of writing about the exact same thing.

 

[einstein00]

I originally misunderstood, but now completely and happily accept, blah's use of integrals to explain a concept generally called geometry. Now I think blah needs to accept the "traditional" geometrical methods.

 

[Rinfiyks]

I've missed this thread.

Spoiler

7363 = 1
3777 = 0
6636 = 3
3303 = 1
9655 = 2
1121 = 0
1129 = 1
1001 = 2
1028 = 3

9801 = ?

 

[peterbone]

9801 = 4 (number of holes in the digits)

A glass rod is dropped and breaks randomly along its length in 2 places producing 3 pieces. What is the probability that the pieces can be placed end-to-end to create a triangle?

 

[Rinfiyks]

9801 = 4 (number of holes in the digits)

A glass rod is dropped and breaks randomly along its length in 2 places producing 3 pieces. What is the probability that the pieces can be placed end-to-end to create a triangle?

Spoiler

If one piece is more than half the length of the glass rod you cannot make a triangle.
So the breaks have to be one on each half.
So the probability is 1/2?
Edit: Oh that's wrong I forgot about the middle bit of glass. I think that should be 1/2 as well.
So 1/4?

Problem:

Spoiler

You have 6 ants which walk at 1 metre/minute and a 1 metre stick.
When placed on the stick, the ants will walk in the direction they are facing until they reach another ant, in which case both ants will change direction. If they reach the end of the stick, they fall off.
How should you place the ants so that time taken for the last ant(s) to fall of the stick is as long as possible?

 

[peterbone]

Spoiler

If one piece is more than half the length of the glass rod you cannot make a triangle.
So the breaks have to be one on each half.
So the probability is 1/2?
Edit: Oh that's wrong I forgot about the middle bit of glass. I think that should be 1/2 as well.
So 1/4?

Correct, it's a 1/4 - although I'm not sure your derivation was very rigorous. I'll let someone else answer your problem. I was tempted to write a quick simulation program to search for the answer but I should really be working.
Edit: Lucky I didn't write that simulation as I realise now that the answer is very simple since the result of a collision is the same as if the ants just walk through each other without colliding.

 

[cmhardw]

I got asked a very interesting question in an e-mail recently, and I found the answer quite interesting. Here is the question.

Warm-up question) A 4x4x4 cube has a picture on exactly one center piece. Write an expression for how many possible combinations this cube has (old Rubik's 4x4x4's with logo anyone? ).

And if you get that one correct, now let's have some fun with this idea

Fun question #1) A 4x4x4 cube has distinct pictures on two distinct centers that were originally of the same color. How many possible combinations does this cube have?

Fun question #2) A 4x4x4 cube has indistinct pictures on two distinct centers that were originally of the same color. How many possible combinations does this cube have?

Have fun

 

[vcuber13]

I got asked a very interesting question in an e-mail recently, and I found the answer quite interesting. Here is the question.

Warm-up question) A 4x4x4 cube has a picture on exactly one center piece. Write an expression for how many possible combinations this cube has (old Rubik's 4x4x4's with logo anyone?

Spoiler

7401196841564901869874093974498574336000000000 * 4
= 29604787366259607479496375897994297344000000000

 

[Ordos_Koala]

Not that I really expect anyone to solve this problem, but here is an "easy" problem: (not abstract, just ingenuity):

Spoiler

Find
\( \frac{d}{dx}\left[ \frac{1}{\sqrt{\sin ^{2}\left( x \right)-1}} \right] \)

I guess it can't be solved... sin x can't be higher than 1, so square sin x can't be too. So you either have negative number (root in R doesn't exist) or 1-1 which is 0 and you can't have 1 over 0

 

[Rune]

Wolframalpha gives a clear answer. (No real part though).

 

[cmhardw]

Spoiler

7401196841564901869874093974498574336000000000 * 4
= 29604787366259607479496375897994297344000000000

Yes, this is correct! If you're interested, now try the other for fun versions!

 

[Ordos_Koala]

I looked at some old tasks from math olympics and I found this, when I solved it I imagined Rubik's cubes, so I thoght I'll translete it here (if you won't understand something, please ask, I'll probably say something wrong)

You have a cube, with side "a", which is expressed in whole centimeters bigger than 2. We painted whole cube with yellow color and then we have sliced it into cubes with side of 1 centimeter. We have then sorted these little cubes into four piles: first pile has only cubes with one yellow side, in the second one were cubes with two painted sides and in the third were cubes with three painted sides. In last pile were cubes with no yellow paint at all. Now tell me how big was the original cube, when you know, that one of these claims is true:
a) Cubes in the first and fourth pile are in ratio 4:9.
b) In the first pile are three-times more cubes than in the second pile.

I solved it quickly, it isn't hard at all, but I want you to prove your answer, not just guess.

 

[Ordos_Koala]

Fun question #1) A 4x4x4 cube has distinct pictures on two distinct centers that were originally of the same color. How many possible combinations does this cube have?

Fun question #2) A 4x4x4 cube has indistinct pictures on two distinct centers that were originally of the same color. How many possible combinations does this cube have?

Have fun

1)7401196841564901869874093974498574336000000000*4*3 =88 814 362 098 778 822 438 489 127 693 982 892 032 000 000 000
2)7401196841564901869874093974498574336000000000*3*2*1=44 407 181 049 389 411 219 244 563 846 991 446 016 000 000 000
I hope i wrote the first number right in the calculator

 

[qqwref]

Warm-up question) A 4x4x4 cube has a picture on exactly one center piece.
Fun question #1) A 4x4x4 cube has distinct pictures on two distinct centers that were originally of the same color.
Fun question #2) A 4x4x4 cube has indistinct pictures on two distinct centers that were originally of the same color.

I think this should work:

Spoiler

Let x be the number of combinations of the regular 4x4. Note that you can fix one corner, so we don't need to worry about how the centers change when the puzzle is rotated (each possible placement of the 24 centers is one position). Now, normally there is 1 way to place the four white centers (let's say the pictures are on white, WLOG).

Warm-up) There are now 4 ways to place the four white centers, so there are 4x combinations.
Fun1) There are now 4*3 ways to place the four white centers, so there are 12x combinations.
Fun2) There are now (4 choose 2) ways to place the four white centers, so there are 6x combinations.

 

[whauk]

Spoiler

Problem:

Spoiler

You have 6 ants which walk at 1 metre/minute and a 1 metre stick.
When placed on the stick, the ants will walk in the direction they are facing until they reach another ant, in which case both ants will change direction. If they reach the end of the stick, they fall off.
How should you place the ants so that time taken for the last ant(s) to fall of the stick is as long as possible?

Spoiler

real solution (1 minute)

Spoiler

it makes no difference as long as you put one ant on one end and let it face the other end. (very funny sentence for reading out loudly (end, and, ant...))
when 2 ants meet each other they change direction and walk towards the other direction... but if you think as "they just walk on" (which makes absolutely no difference) the solution given at the top is the only logic consequence.

cheater solution (infinite time)

Spoiler

place all ants at the same spot on the stick and they meet each other all the time and change direction all the time and therefore never move. haha... ok this doesnt count

edit:
new problem:

Spoiler

everyone knows how to calculate the probability that 2 people out of n people are born on the same day (w/o year, only day + month).
now find a formula for the probability that x people out of n people are born on the same day. (i never figured out the solution... )

 

[cmhardw]

Spoiler

1)7401196841564901869874093974498574336000000000*4*3 =88 814 362 098 778 822 438 489 127 693 982 892 032 000 000 000
2)7401196841564901869874093974498574336000000000*3*2*1=44 407 181 049 389 411 219 244 563 846 991 446 016 000 000 000
I hope i wrote the first number right in the calculator

I think this should work:

Spoiler

Let x be the number of combinations of the regular 4x4. Note that you can fix one corner, so we don't need to worry about how the centers change when the puzzle is rotated (each possible placement of the 24 centers is one position). Now, normally there is 1 way to place the four white centers (let's say the pictures are on white, WLOG).

Warm-up) There are now 4 ways to place the four white centers, so there are 4x combinations.
Fun1) There are now 4*3 ways to place the four white centers, so there are 12x combinations.
Fun2) There are now (4 choose 2) ways to place the four white centers, so there are 6x combinations.

Yes I got the same answers Orodos you did calculate the first number correctly, but in the future please try to put your answers in spoiler tags such as not to give away information for others still working on the problem.

I'll make my next comment in the spoiler tag so as not to give away hints to others still wondering how to get these numbers:

Spoiler

I find it interesting that the Rubik's 4x4x4's actually have 4 times as many combinations as a logo-less 4x4x4, yet the Rubik's 4x4x4's also have 4 solutions instead of just 1. This means that the ratio of solved states to scrambled states is the same for a 4x4x4 with a logo and a 4x4x4 without a logo. Of course this does not present any issue from the WCA standpoint, as the regulations clearly state that one logo is allowed. I still find it interesting that someone solving a 4x4x4 with a logo and someone solving a 4x4x4 without a logo are technically not solving the exact same kind of puzzle. For some reason I had never really thought about it this way until someone asked me this question in an e-mail. Cool stuff!