Maths Quiz: Solve the Surd Puzzle

[jcsd]

Okay Maths quiz! normal rules apply (first to answer question correctly gets to ask the next question, etc.):

1) √(5 + √(24)) = ? in exact surd form

 

[Hurkyl]

That font doesn't work for me.

 

[jcsd]

Font corrected :)

 

[Hurkyl]

You looking for something like

√2 + √3

?

 

[jcsd]

Correct!

 

[Hurkyl]

Hrm, my favorite math puzzle is a picture puzzle, but that'd be too hard to describe in words.


Ok, here's one:

A carpeting company has taken an order to carpet an annulus-shaped room (the region between two cocentric circles)... however their drunken surveyor only measured the length of the longest straight line that can be drawn in the room! The manager complained to the resident mathematician about the problem, because this measurement didn't seem to be enough to find the area of the floor, but the mathematician said "But there's a formula for that..." to which the manager responded "Oh, there is? I know what it is then" and walked away.

What is the formula?

 

[jcsd]

πl²/2 ?

 

[Hurkyl]

not quite!

 

[jcsd]

3πl²/4?

 

[Hurkyl]

nope!

 

[Integral]

Here is how I see it.
R= Radius of Large circle
r= radius of small concentric circle
D = measurement.

The measured distance must be tangent to the smaller circle so we have
;
The Radius of the larger circle is the Hypotenuse of a right triangle with sides r and D/2 so

(D/2)² = (R² -r²)= D²/4

The Area we need is:
A = [pi] R²-[pi]r²
A = [pi](R²-r²)
so
A = [pi]D²/4

 

[ahrkron]

Since (according to the local mathematician) D is enough information to obtain the area, we have to conclude that the result is independent of R and r, which means that we can choose whichever values we like. Using r=0, we have that R = D/2, and the area is [pi]R², or [pi](D/2)², and should be the result for all cases.

 

[marcus]

Originally posted by jcsd
Okay Maths quiz! normal rules apply (first to answer question correctly gets to ask the next question, etc.):

If Maths Quiz is played according to similar rules to Astronomy Q/
A then Hurkyl should speak up and say it is Integral's turn because he was the "first to answer question correctly"

And if anybody forgets to do what theyre supposed to do, according to the Nicool rules after "2 or 3 days" someone can
do what's needed to keep the game going. Like, just go ahead and ask another question.

Because JCSD started the gamethread he can decide on how to adapt Nicool's rules as necessary (which are in Nicool's post at the start of the Astronomy gamethread)

Those were both neat quiz questions I hope Hurkyl shows up presently, and awards Ahrkron an honor point for elegance

 

[jcsd]

It's annoying because I had the right answer but I worked it out using a very roundabout method (using chords and tangents) and I made a mistake :(

 

[Integral]

Those were both neat quiz questions I hope Hurkyl shows up presently, and awards Ahrkron an honor point for elegance

I agree with this. Mine is a brute force solutoion. Hurkyl arrived at the same conclusion with simple logic. Much nicer solution. Hurkyl, if you have a problem on the par with the ones posted so far, please post it. I would have trouble comming up with a good one.

 

[Hurkyl]

Yep, Integral got the right answer, and Ahrkron got the clever solution!

I don't have any other good ones off hand, though...

 

[chroot]

As a Math Doctor on Drexel's "Ask Dr. Math" website, I see a whole lot of clever math puzzles come through the door. I will try to post some soon!

- Warren

 

[marcus]

referree call

Originally posted by Integral
I agree with this. Mine is a brute force solutoion. Hurkyl arrived at the same conclusion with simple logic. Much nicer solution. Hurkyl, if you have a problem on the par with the ones posted so far, please post it. I would have trouble coming up with a good one.

Hurkyl asked, Integral answered (seconded by Ahrkron)
The turn goes to Integral, who passes back to Hurkyl
and Hurkyl declines the turn.

Therefore it is Ahrkron's turn. Ask a good one, Ahrkron!
If you don't happen to come up with something subtle then
ask something
straightforward. This could be a good game.

 

[ahrkron]

OK, here it is:

You probably know the following riddle:


John leaves home one morning and does the following:
- Stands in front of his house, facing South
- walks straight ahead for 1 mile,
- turns left and walks 1 mile, sees a bear
- turns left and walks 1 mile,
at which point he finds himself back home.

Riddle's question: What is the color of the bear?
[/color]

The solution (the color of the bear) is unique. My question is: if we take the bear out of the question, how many different ways can such a walk be achieved? explain!

I will give some hints if necessary.

 

[Integral]

Couple of things I would like to see.

A derivation of some sort, or at least an indication of the logic used should be a part of the solution... There may be some who are wondering how to get the solution to that first problem (think quadratic !)

Also I see quality problems as important as who posts them. The winner gets first shot, but if he does not have a good proplem I would rather it be passed to someone who has one.

 

[Oxymoron]

White.

He was at the north pole therefore the only bear he saw would have been a polar bear.

 

[Oxymoron]

Oh, I didn't see the real question!

If I understand the question correctly then there would be an infinite number of places on Earth where someone could move 1 mile forward, 1 mile east then 1 mile backward and be back to where he started.

Take a point on Earth that is very close to the north-pole. (say a couple of meters, It only has to be close enough). Understand that this parallel of latitude at our chosen point circles the north pole and is 1/n miles long.

Situate John's house at a point which is exactly 1 mile south of this parallel.

Now, John walks forward (north) to our chosen line of latitude (call point P). Turns and walks east 1 mile, walking n times around that latitude and ending up back at P. Now he simply walks one mile backward to where he started.

Simply, take any point on Earth (Nth hemisphere) walk north x miles and stop at the right place before you reach the north pole. Turn east and walk that x miles again (if you stopped at the right spot your x miles should end up back at that spot) ending where you first turned east. Walk south x miles again.

Therefore there are an infinite number of ways this can be acheived. NOTE: This is possible because we are on a polar surface!

 

[marcus]

Originally posted by ahrkron
My question is: if we take the bear out of the question, how many different ways can such a walk be achieved? explain!

Oxymoron! please add to your solution that the house can be near the south pole. This will keep nitpickers quiet (of course at PF everybody is nice and there are no nitpickers)

The house is facing south. According to the original wording he must set out walking south. Then after a mile he will turn left and (presumably walk one mile East going around and around the south pole) then he will turn left again and walk north to his house.

To remind you of the precise wording I will "take the bear out of the question" and recopy the puzzle Ahrkron gave:



John leaves home one morning and does the following:
- Stands in front of his house, facing South
- walks straight ahead for 1 mile,
- turns left and walks 1 mile,
- turns left and walks 1 mile,
at which point he finds himself back home.
...how many different ways can such a walk be achieved?
[/color]

As I understand it, Oxymoron, you are an "Ozzie" that is an Australian, so probably you really mean south pole when you say north pole. We understand that Ozzies are upsidedown in some respects. And if I understand you correctly you are saying that the answer is MANY different ways

For every positive integer N there is a circle around the south pole which has circumference 1/N mile, and so one can walk for one mile to the east along this circle going N times around and be back at the same place.

This is a brilliant idea. Please edit your answer so that it will be letter-perfect in the eyes of Ahrkron and then we will get to hear what problem you can come up with!

 

[HallsofIvy]

Actually, I doubt that there are any polar bears (or any other) land animals within a mile of either the north or south poles.

 

[ahrkron]

Good job guys!

Yes, there is an infinite number of solutions. I just want to point out one family[\i] of solutions that has not been explicitly mentioned.

So far the following initial points have been described:
- The north pole,
- Points that are at such a distance from the south pole that, upon walking one mile due South and changing your heading to East, one mile walking makes you circle around the south pole an integer number of times.

The third family would be:
- Points such that the first 1-mile segment touches and passes the south pole, and deposits you in one of the (1/N mile) circles around the south pole.

After crossing the South pole, you are not waking due South anymore, but that is allowed by the wording of the problem, since it only states the heading at the beginning of the walk.

(Btw, the original wording should have said "... walks one mile, keeping his (N/S/E/W) heading[/color]").

Regarding HallsofIvy's comment, polar bears have been reported as far north as the North Pole, but they only live on the northern polar region, which is why I "took the bear out" of the original riddle.

So,

Oxymoron got the gist of the solution, and gets to post the next problem. Marcus gets an honor point for accuracy.

 

[Hurkyl]

I hope you don't mind me posting out of turn the question I wanted to post originally!


This is a geometric problem. I've attached a gif to give a visual example of the construction. Please pardon my horrible skills at drawing with a mouse!



First, draw two intersecting circles. (they needn't be the same size)

Choose one of the two intersection points. Call it 'A'. For each circle, draw the diameter that has A as one of its endpoints.

Draw the line connecting the opposite endpoints of these diameters.

This line will intersect the circles in two new points. Call them B and C. Draw the lines from A to each of these points.

Now, we have a triangle ABC. Notice that angle ABC is inscribed in a semicircle, so it must be a right angle. The same is true about angle ACB.

Thus, we have constructed a triangle with two right angles!


What went wrong??

 

[KLscilevothma]

This line will intersect the circles in two new points.

This sentence is not correct. This line will intersect the circles at one point, which is the other intersection of the circles besides point A.

I think it is the reason why you did't use tools to draw the circles, but drew them by free-hand instead.

 

[marcus]

referee call

we now have two winners
Oxymoron and Kam
and its is each one's turn to ask

All that means is that the game has branched into two
which is fine. We can handle two Q/A trains

Please Oxy and Kam, do not be alarmed, simply ask questions---we will have an Oxy track and a Kam track (until and unless one peters out)

Oh! excuse my poor timing. first Hurkyl should declare Kam's answer correct so that is is Kam's turn. Then I should say that it is both person's turn.

(BTW hurkyl congrats on solving that brainteaser puzzle
about the points on the yardstick. I looked at it and did not
see how to find a set of 8 points that would work, and I still
do not see how you contructed your set that does work!)

 

[Hurkyl]

Yes, Kam is right.

I considered drawing it with tools, and I could have drawn the incorrect picture by missing the true center of the circles with the diameters... but I'm used to drawing it by hand so I kept the tradition alive.


Oh, and on the brain teaser, I'm ashamed to admit that it's the result of half a second of brute force from a C++ program. I think I was getting the right idea when I was attacking it by hand, but I realized there were only 30 million or so possibilities and figured "why not?"

 

[marcus]

Originally posted by Hurkyl
Yes, Kam is right...

...half a second of brute force from a C++ program. I think I was getting the right idea when I was attacking it by hand, but I realized there were only 30 million or so possibilities and figured "why not?"

I say hurrah and bravo to that C++ approach since it is already very good to organize a brute force attack in a hurry as done and we are not all Ramanujan so "why not?" indeed.

I wish that both Kam and Oxy supply problems soon
clever is nice but even straightforward problems (what
did JCSD begin the thread with?) can be fun to have posed.
so nice if clever but anything will do to keep the game rolling.
It was a good idea jcsd had to start it.

Chroot goes away weekends but soon he will be back and
probably answer someone's question and then he will ask
something that stumps everbody, but we will have two QA lines going and all is well.

 

[Oxymoron]

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?

 

[KLscilevothma]

Is it ok if I post my question at the same time ?

 

[Oxymoron]

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?

 

[Oxymoron]

Go ahead by all means!

 

[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!

 

[KLscilevothma]

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 ?

 

[marcus]

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!

 

[Hurkyl]

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...

 

[KLscilevothma]

Geometry

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

 

[Oxymoron]

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

 

[Oxymoron]

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!

 

[KLscilevothma]

What should we do if nobody would like to answer our questions ?

 

[Oxymoron]

I guess we win! and get to ask another question.

Take a helium balloon (the kind that floats in air ) 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?

 

[bogdan]

Is it a convertible ? (the car)
If it's not then I guess that the baloon 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...

 

[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 ?

 

[Oxymoron]

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!

(c) is incorrect.

If anyone else would like to have a try please remember to put in your reasoning like bogdan did.

 

[bogdan]

Are you sure that c is incorrect ?
Remember:the baloon 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 baloon moves backward...
Soo...what's the problem with my answer ?

 

[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...

 

[KLscilevothma]

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.

P can be any point on AB.

Hint: Think of the special case when P=A.

 

[bogdan]

Okay...I'll think about it...