Maths Quiz: Solve the Surd Puzzle

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Discussion Overview

The discussion revolves around a math quiz featuring various puzzles and problems, including surd simplification and geometric reasoning related to movement on Earth. Participants engage with both straightforward and complex mathematical concepts, exploring different approaches to problem-solving.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • Participants are tasked with solving a surd problem: √(5 + √(24)) in exact surd form.
  • One participant suggests a possible solution of √2 + √3, which is confirmed as correct by another participant.
  • A new puzzle is introduced involving the area of an annulus-shaped room, with participants attempting to derive a formula based on a single measurement (D).
  • One participant presents a derivation involving the relationship between the radii of the circles and the measurement D, concluding that the area can be expressed as A = [pi](D/2)².
  • Another participant notes that the area result is independent of the specific radii chosen, suggesting flexibility in the values of R and r.
  • A riddle about a walk that returns to the starting point prompts discussion about the number of possible starting locations on Earth, with one participant explaining that there are infinite solutions near the North Pole.
  • Another participant emphasizes the importance of providing a derivation or logic for solutions, indicating a preference for quality problems in the quiz format.
  • Participants express varying degrees of confidence in their solutions, with some acknowledging mistakes or roundabout methods.

Areas of Agreement / Disagreement

Participants generally engage in a collaborative quiz format, with some agreement on solutions, but multiple competing views and methods remain, particularly regarding the annulus area problem and the riddle about the walk.

Contextual Notes

Some participants express the need for clarity in problem statements and derivations, indicating that assumptions about the geometry of the Earth and the nature of the puzzles may not be universally understood.

Who May Find This Useful

Readers interested in mathematical puzzles, problem-solving strategies, and collaborative discussions in mathematics may find this thread engaging.

  • #31
Sorry about the delay I have had some trouble with my internet connection lately. Anyway on to the question...(it is a 2 part question)


Qa) Two players (A and B) each have a certain number of pennies (x and y) respectively (these can be the same amount at this stage!). They begin to play a game wherein each player tosses a coin and calls it. The winner takes the penny. The probability of winning a single round is 50%. If this game was played indefinately, what would be the probability that someone wins?

Qb) What would happen if x > y?
 
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  • #32
Is it ok if I post my question at the same time ?
 
  • #33
I believe that some clarification is necessary on part a. It is saying that two players A and B each have a bank of unbiased pennies which is arbitrarily large. Player A goes first and flips one of his coins. He then calls it. If A wins he gets to keep his coin otherwise B keeps it. Then B has his turn and so on...

What is the probability that either player acquires the other players' entire holdings?

The second part is simply asking; does anything happen to this probability if we say that player A has many more pennies to start with compared to player B?
 
  • #34
Go ahead by all means!
 
  • #35
Eep, I don't know if I want to work on this one; I don't have any good ideas for the next question!
 
  • #36
Originally posted by Hurkyl
Eep, I don't know if I want to work on this one; I don't have any good ideas for the next question!

Do you mind if you answer oxy's question and I post the other question ?
 
  • #37
Guys! we are doing double threading, it is BOTH Oxy and Kam turn to ask question
and anyone can answer one or the other or both
with no preferred order

Kam please go ahead and pose a Q: we are eagerly looking forward to it.

Oxy's question is already before us and we may suppose that this
is how they play the stock market in Australia, and I would guess tho it remains to be proven, that the large investor
has a high probability of wiping out the small investor. How hard can it be? Somebody take a whack at it!
 
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  • #38
There is a finite state space and from any intermediate state, the endstate is reachable in n turns with some fixed nonzero probability for some values of n, so by the law of large numbers, the end state is reached with probability 1.

IOW you're guaranteed the game ends, no matter what x and y are.


Now, an interesting question is what is the probability the player with x pennies beats the player with y pennies... :smile:
 
  • #39
Geometry

In the figure, construct a line through p cutting triangle ABC into 2 parts with equal areas.
 

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  • #40
There is a finite state space and from any intermediate state, the endstate is reachable in n turns with some fixed nonzero probability for some values of n, so by the law of large numbers, the end state is reached with probability 1.

Correct thus far. However...

Now, an interesting question is what is the probability the player with x pennies beats the player with y pennies...

...is the real problem!

This is an example of a one-dimensional random walk (actually quite an easy one!).

If anyone needs some help just ask!

Cheers
 
  • #41
Suppose you have $20 and need $50 for a bus ride home. Your only chance to get more money is by gambling in a casino. There is only one possible bet you can make at the casino--you must bet $10, and then a fair coin is flipped. If "heads" results, you win $10 (in other words, you get your original $10 back plus an additional $10), and if it's "tails", you lose the $10 bet. The coin has exactly the same 50% probability of coming up heads or tails. What is the chance that you will get the $50 you need?

This is very similar to my original question, and I have changed it here for simplicity. In this case the casino is player A and the person is player B. Note that casino (Player A) has MUCH more coins than player B. I will now prove the question...

This problem is known as the ``Gambler's Ruin Problem''--a gambler keeps betting until he goes broke, or until he reaches a certain goal, and there are no other possibilities. Here is a nice matrix-based approach.

At any stage in the process, there are six possible states, depending on your ``fortune''. You have either $0 (and you've lost), or you have $10, $20, $30, $40 (and you're still playing), or you have $50 (and you've won). You know that when you begin playing, you are in a certain state (having $20 in the case of the latest problem). After you've played a while, lots of different things could have happened, so depending on how long you've been going, you have various probabilities of being in the various states. Call the state with $0 ``state 0'', and so on, up to ``state 5'' that represents having $50.

At any particular time, let's let P0 represent the probability of having $0, P1 the probability of having $10, and so on, up to P5 of having won with $50.

We can give an entire probabilistic description of your state as a column vector like this:

P0
P1
P2
P3
P4
P5

Now look at the individual situations. If you're in state 0, or in state 5, you will remain there for sure, because that is where the game ends. If you're in any other state, there's a 50% chance of moving up a state and a 50% chance of moving down a state. Look at the following matrix multiplication:

1 0.5 0 0 0 0 x P0 = P0 + 0.5P1
0 0 0.5 0 0 0 x P1 = 0.5P2
0 0.5 0 0.5 0 0 x P2 = 0.5P1 + 0.5P3
0 0 0.5 0 0.5 0 x P3 = 0.5P2 + 0.5P4
0 0 0 0.5 0 0 x P4 = 0.5P3
0 0 0 0 0.5 1 x P5 = 0.5P4 + P5

Clearly, multiplication by the matrix above represents the change in probabilities of being in the various states, given an initial probabalistic distribution. The chance of being in state 0 after a coin flip is the chance you were there before, plus half of the chance that you were in state 1. Check that the others make sense as well.
So if the matrix on the left in equation (3) is called P, each time you multiply the vector corresponding to your initial situation by P, you'll find the probabilities of being in the various states. So after 1024 coin-flips, your state will be represented by P1024.

Here is a probability matrix of probabilities of various states after 1024 flips:

1 0.8 0.6 0.4 0.2 0
0 0.0 0 0.0 0 0
0 0 0.0 0 0.0 0
0 0.0 0 0.0 0 0
0 0 0.0 0 0.0 0
0 0.2 0.4 0.6 0.8 1

I write 0.0 because in actual fact that coordinate is 1.549x10^-93 which is an extremely small number (EXTREMELY SMALL!)

So for practicality I will write it like this

1 0.8 0.6 0.4 0.2 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0.2 0.4 0.6 0.8 1

So what does this probability matrix tell us? Well it tells us that we have a 100% chance of winning! Look at the 1 in the bottom right hand corner. This coordinate represents where the game finishes by winning. It says 1 which represents 100% chance. But there is also a 100% of losing?!

Can anyone tell me why this is?

Also know think of why Casino's always end up on top!
 
  • #42
What should we do if nobody would like to answer our questions ?
 
  • #43
I guess we win! and get to ask another question.

Take a helium balloon (the kind that floats in air :smile: ) and attach the string to floor of you car. You should have a floating balloon tied to a string attached to the floor in your car. Got it? Good let's continue...

Now start to accelerate, which way will the balloon go?
a) BACKWARD
b) FORWARD
c) WILL NOT MOVE
d) UPWARD
e) WHAT BALLOON?

(It is not exactly a maths question, more of a physics question but let's see who can get it first!).

PS. KD I did honestly have a go at your question but was quite unsure of how to answer it?
 
  • #44
Is it a convertible ? (the car)
If it's not then I guess that the balloon should not move...
...because the air moves with the car...and if the acceleratin is not too great than the air should not compress...I guess...:smile:
 
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  • #45
For that triangle problem...KL_KAM...
I suppose that 2*AP=PB...
http://www.angelfire.com/pro/fbi/images/trio.bmp
Sorry for the delay...
Because the link above may not work...
Here is the solution:
Take on AC a segment CE=2*AC on the opposite side to A...then let F be the intersection between PE and BC...now S(ACFP)=S(PBE)...good enough ?
 
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  • #46
bogdan, the type of car doesn't really matter but for simplicity let's just say it is a normal car with all its windows up. Thanks for having a go anyway! :smile:

(c) is incorrect.

If anyone else would like to have a try please remember to put in your reasoning like bogdan did.
 
  • #47
Are you sure that c is incorrect ?
Remember:the balloon is attached to the floor...not to a weight that floats in the air and is attached to the floor by a string...soooo...why should c be incorrect ?
If it was attached to a weight than it should move forward...
And why do you think that the type of the car doesn't matter ?
If it is a convertible (opened) than the balloon moves backward...
Soo...what's the problem with my answer ?
 
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  • #48
Aaaa...and for that probability problem...the first on the 3rd page...with x and y pennies...
P(x)=x/(x+y)...P(x)->the probability that the player with x pennies wins...
P(y)=y/(x+y)...P(y)->...
Proof ? Somesort of recursivity...too long to be written here...
 
  • #49
Originally posted by bogdan
For that triangle problem...KL_KAM...
I suppose that 2*AP=PB...
http://www.angelfire.com/pro/fbi/images/trio.bmp
Sorry for the delay...
Because the link above may not work...
Here is the solution:
Take on AC a segment CE=2*AC on the opposite side to A...then let F be the intersection between PE and BC...now S(ACFP)=S(PBE)...good enough ?

I didn't say 2*AP=AC. :smile:

P can be any point on AB.

Hint: Think of the special case when P=A.
 
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  • #50
Okay...I'll think about it...
 
  • #51
Originally posted by bogdan
Aaaa...and for that probability problem...the first on the 3rd page...with x and y pennies...
P(x)=x/(x+y)...P(x)->the probability that the player with x pennies wins...
P(y)=y/(x+y)...P(y)->...
Proof ? Somesort of recursivity...too long to be written here...
I have infact found a method for proving the probability question using matrices and vectors, no recursivity. So I will be able to prove whether your answer is correct or not :wink:.
 
  • #52
b) FORWARD
 
  • #53
FINALLY! Brum is correct. Could you give me a reason why you chose B?
 
  • #54
Darn, my guess was E. :wink:

Oxymoron, care to explain the solution to your probability matrix problem since no one seems to be answering?
 
  • #55
Solution to the geometry problem

construct a line through p cutting triangle ABC into 2 parts with equal areas

Since I'll be busy, so I think I better post the solution now.

1. Construct a line AD such that CD = BD. Therefore [ADC] = [ADB]
([ADC] means area of triangle ADC)

2. Draw a line to join PD

3. Construct a line AF such that AF // PD

4. Join PF

Now, [APG] = [GFD] (note that AP does not necessary equal to FD)

and [APCF] = [PBF]

The line PF is all I want.

http://www.angelfire.com/alt2/antiwork/sol.bmp
 
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