# Mathematics marathon



## y3k9 (Dec 31, 2010)

Ok, this was quite interesting on aops, and I learned quite recently that there's a lot math people like me on the forum, I am starting the mathematics marathon.

The basic idea is that you go on post an interesting problem and solve the one posted above you. It's quite simple. However, on aops there's marathons for multiple skill levels, and I don't want multiple threads so just say for what level the problem is intended and solve only those that are of your level. In this thread abstract math is strongly encouraged. Here are rules:

1. Have an included solution to the question you answered and *bold* the answer.
2. Make use of spoilers, make sure to have a separate spoiler for both your answer and solution.
3. If your really into the REALLY abstract math title your questions "ABSTRACT"
4. Try to post multiple problems if possible.
5. Frequent this thread, and help someone if they answer your question wrong.

That's about it, I'll go first:
Level: intermediate (8th grade)
Problem 1:


Spoiler



For what value x is \( 2+x-x^2 = 0 \)



As of 31/12/10:
This thread seems to be doing good, might I ask however that you either box or bold your answer in a solution to a problem as well as number your problems. It was intention for each problem to be numbered, so please number the next problem you write.


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## aronpm (Dec 31, 2010)

Answer *TO PROBLEM 1*:


Spoiler



Divide by -1 to get \( x^2 - x - 2 = 0 \) and factorize to get \( (x - 2)(x + 1) = 0 \). Therefore either \( x - 2 = 0 \) (so \( x = 2 \)) or \( x + 1 = 0 \) (\( x = -1 \))

IDK why you asked for me to write a proof to that question since my hypothetical dog could do it in his sleep.



Problem *2*:


Spoiler



Provide a unique proof of Fermat's Last Theorem


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## Christopher Mowla (Dec 31, 2010)

aronpm said:


> Provide a unique proof of Fermat's Last Theorem


You have a very good sense of humor, my friend.


y3k9 said:


> Level: intermediate (8th grade)


Sorry, but in the math world, basic algebra is not intermediate: it is knowledge that should already be second nature for you to have any hope to solve other problems.

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


Spoiler



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


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## Rinfiyks (Dec 31, 2010)

Answer:


Spoiler



\( \sin^2(x)+\cos^2(x) = 1 \)
\( \sin^2(x) - 1 = -\cos^2(x) \)
So it becomes
\( \frac{d}{dx} \left[ \frac{1}{\sqrt{-\cos^2(x)}} \right] \)
Which... er... I thought that you'd have designed it so the minus sign isn't in the square root!
I can't do it


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## TimMc (Dec 31, 2010)

cmowla said:


> Sorry, but in the math world, basic algebra is not intermediate: it is knowledge that should already be second nature for you to have any hope to solve other problems.



They didn't say:


> I hereby declare the following problem to be of an intermediate level of difficulty in the world of mathematics.



They might very well have said:


> The average 8th grade student could solve this problem because it's likely to be a part of their curriculum.



Hooray for Maths!

Tim.


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## Sa967St (Dec 31, 2010)

Answer to cmowla's problem:


Spoiler









Still not too sure if it's right.



Problem:


Spoiler



Find the derivative of:


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## Christopher Mowla (Dec 31, 2010)

Answer to Sa967St's problem


Spoiler



\( \frac{d}{dx}\left[ \sqrt{x^{2}+9}+2\ln \left( x+\sqrt{x^{2}+9} \right) \right] \)

\( =\frac{d}{dx}\left[ \left( x^{2}+9 \right)^{1/2}+2\ln \left( x+\left( x^{2}+9 \right)^{1/2} \right) \right] \)

\( =\frac{1}{2}\left( x^{2}+9 \right)^{-1/2}\left( 2x \right)+2\frac{1+\frac{1}{2}\left( x^{2}+9 \right)^{-1/2}\left( 2x \right)}{x+\left( x^{2}+9 \right)^{1/2}} \)

\( =\frac{x}{\sqrt{x^{2}+9}}+2\frac{1+\frac{x}{\sqrt{x^{2}+9}}}{x+\sqrt{x^{2}+9}} \)

\( =\frac{x}{\sqrt{x^{2}+9}}+2\frac{\frac{\sqrt{x^{2}+9}}{\sqrt{x^{2}+9}}+\frac{x}{\sqrt{x^{2}+9}}}{x+\sqrt{x^{2}+9}} \)

\( =\frac{x}{\sqrt{x^{2}+9}}+2\frac{\sqrt{x^{2}+9}+x}{\sqrt{x^{2}+9}}\frac{1}{x+\sqrt{x^{2}+9}} \)

\( =\frac{x}{\sqrt{x^{2}+9}}+\frac{2}{\sqrt{x^{2}+9}} \)

\( =\frac{x+2}{\sqrt{x^{2}+9}} \)



Problem:


Spoiler



The same one I already posted


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## Sa967St (Dec 31, 2010)

Cmowla got mine correct. 
I edited my answer to his question, but I'm not too sure if it's correct.


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## Christopher Mowla (Dec 31, 2010)

Sa967St said:


> Still not too sure if it's right.


It's not. I noticed that, where I posted the problem, latex makes the superscript 2 so small. Try again.

This is the solution to check your answer:


Spoiler



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

Does not exist for real numbers. The function is not even defined on its own domain. (Which is a problem because we need a function to be continuous at least somewhere in order to calculate the derivative at that x-value). See its graph. Notice that the orange is the imaginary portion, and the blue is the real portion.

\( \text{Reason:} \)

\( \sin ^{2}\left( x \right)\le 1\text{ for all }x\in \left( -\infty ,\infty \right). \)

\( \text{Hence, if }\sin ^{2}\left( x \right)=1,\text{ that is, if }x=\frac{n\pi }{2}\text{ for }n\text{ being non-zero integers,} \)

\( \text{then we have }\frac{d}{dx}\left[ \frac{1}{\sqrt{1-1}} \right]=\frac{d}{dx}\left[ \frac{1}{\sqrt{0}} \right]. \)

\( \text{If sin}^{2}\left( x \right)<1,\text{ then we have a negative in the square root}\text{.} \)


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## Rinfiyks (Dec 31, 2010)

But it still has a solution in complex numbers. I thought that was what we were supposed to find?


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## keemy (Dec 31, 2010)

Answer to cmowla


Spoiler



\( (-cos^2(x))^{\frac{-1}{2}} \)
(i guess this is a complex valued function?) 
use chain rule
\( \frac{-2cos(x)sin(x)}{2(-cos^2(x))^{\frac{3}{2}}} \)
\( \frac{cos(x)sin(x)}{(cos^2(x))^{\frac{3}{2}}} \)



Problem:


Spoiler



Prove:\( \tbinom {2n}{n} = \sum_{k=0}^{n} {\tbinom {n}{k}}^2 \)


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## Whippopotamus (Dec 31, 2010)

Hi, I'm doing As Level Maths at the minute, and I have a fair interest in maths, and I would like to pose this question, its just something I noticed when I was bored in a physics lesson:



Spoiler



Why is it that:
sin cos cos sin (n)
sin cos cos sin cos cos cos sin (n)
sin cos cos sin cos cos cos sin cos cos cos cos sin (n)

Firstly, for any value of n, tend towards the same value, and secondly, why do they tend towards a more specific value as you work down the list (as the number of sin and cos functions are increased in that pattern)?



P.S Im not 100% sure of my answer...


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## Christopher Mowla (Dec 31, 2010)

Rinfiyks said:


> But it still has a solution in complex numbers. I thought that was what we were supposed to find?


You are correct. We can express the solution as a complex function if we wanted to.


Spoiler



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

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

\( =\frac{1}{i}\left( -\frac{1}{2}\left( 1-\sin ^{2}\left( x \right) \right)^{-3/2}\left( -2\sin \left( x \right)\cos \left( x \right) \right) \right) \)

\( =\frac{1}{i}\frac{\sin \left( x \right)\cos \left( x \right)}{\left( 1-\sin ^{2}\left( x \right) \right)^{3/2}} \)

\( =\frac{1}{i}\frac{\sin \left( x \right)\cos \left( x \right)}{\left( \cos ^{2}\left( x \right) \right)^{3/2}}=\frac{1}{i}\frac{\sin \left( x \right)\cos \left( x \right)}{\left| \cos ^{3}\left( x \right) \right|} \)

\( =\frac{1}{i}\frac{\sin \left( x \right)\cos \left( x \right)}{\left( \cos ^{2}\left( x \right) \right)^{3/2}}=\frac{1}{i}\frac{\sin \left( x \right)\cos \left( x \right)}{\left| \cos ^{3}\left( x \right) \right|} \)

\( =\frac{1}{i}\tan \left( x \right)\left| \frac{1}{\cos \left( x \right)} \right|\times \left( \frac{i}{i} \right)=-i\tan \left( x \right)\left| \frac{1}{\cos \left( x \right)} \right| \)

\( (\text{For }x\ne \frac{n\pi }{2},\text{ where }n\text{ is any non-zero integer)}\text{.} \)


EDIT:


keemy said:


> (i guess this is a complex valued function?)


Yes. See my spoiler.


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## macky (Dec 31, 2010)

Answer to keemy



Spoiler



Rewrite as
\( \binom{2n}{n} = \sum_{k = 0}^n \binom{n}{k}\binom{n}{n-k}. \)
Partition of set of 2n objects into two sets of n. Then this says that one chooses n objects by choosing k from the first set and n-k from the second, where k runs over 0 through n.



[edit]
Can I not center displays? No align environment?

[edit]
A problem:


Spoiler



Prove that
\( \frac{a_3^2}{a_3^2 + a_4^2 + a_5^2 - a_3a_4 - a_4a_5} \)
is maximized for real \( a_3, a_4, a_5 \) when \( a_3 = 3a_5, a_4 = 2a_5 \).

I'll reveal the source later.


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## vcuber13 (Dec 31, 2010)

Sa967St said:


> Answer to cmowla's problem:
> 
> 
> Spoiler
> ...


woulnt it be the square root of the sinx -1 cubed?


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## Sa967St (Dec 31, 2010)

vcuber13 said:


> woulnt it be the square root of the sinx -1 cubed?


Squared, but the ^2 wasn't there when I answered it:


cmowla said:


> I noticed that, where I posted the problem, latex makes the superscript 2 so small.


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## vcuber13 (Dec 31, 2010)

i mean wouldn't n^(3/2) be sqrt(n^3) not cuberoot(n^2)?


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## Sa967St (Dec 31, 2010)

vcuber13 said:


> i mean wouldn't n^(3/2) be sqrt(n^3) not cuberoot(n^2)?


Yes. I always make dumb mistakes. 
Still wasn't correct since there are no real answers as cmwola explained.


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## keemy (Dec 31, 2010)

ya the latex on SS is kind of annoying.

answer to macky


Spoiler



Prove that
k going to just let x=a2, y=a4, z=a5.
plugging in we find we want to prove max is 3/2. So set up inequality.


\( \frac {x^2}{x^2+y^2+z^2-xy-yz} \leq \frac {3}{2} \)
\( 2x^2 \leq 3x^2 +3 y^2 +3z^2 -3xy - 3yz \)
\( (x-y)(x-2y) + (y-z)(y-2z) +z^2 \geq 0 \)

so the only way this could be false would be if \( y<x<2y \) and/or \( z<y<2z \) (so one or both of first terms are negative) note that the least the first 2 terms could be (in said range) are when \( x=1.5y \), \( y=1.5z \) but plugging in gets the \( LHS=.1875z^2 \) which clearly can't be negative


I feel like I missed something...

problem uhh I think there is still one left unanswered also don't want to find one now.


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## macky (Dec 31, 2010)

That works. Simpler:


Spoiler



\( \frac{a_3^2}{a_3^2 + a_4^2 + a_5^2 - a_3a_4 - a_4a_5} \le \frac32 \)
\( \Leftrightarrow 2a_3^2 \le 3a_3^2 + 3a_4^2 + 3a_5^2 - 3a_3a_4 - 3a_4a_5 \)
\( \Leftrightarrow 0 \le \frac14(4a_3^2 + 9a_4^2 - 12a_3a_4) + \frac34(a_4^2 + 4a_5^2 - 4a_4a_5) \)
\( \Leftrightarrow 0 \le \frac14(2a_3 - 3a_4)^2 + \frac34(a_4 - 2a_5)^2 \)

And this proof easily generalizes to more variables.

This was part of Exercise 21.14 from Fulton and Harris's _A First Course in Representation Theory_. These maximization problems come up when ruling out Dynkin diagrams of simple Lie algebras (the above corresponds to a branch of length 3). For example, they can be used to show that if a node with three branches with length \( p, q, r \), then \( 1/p + 1/q + 1/r \le 1 \). This rules out all diagrams of this kind except \( D_5, E_6, E_7, E_8 \).

Also, there's no good reason why the index starts at 3.


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## cmhardw (Dec 31, 2010)

I love the idea of this thread. It appears that there is no question to answer right now, so I'll post one.

Problem Goal: Spot the error in the proof
Required Pre-requisite: Up to differential calculus. Probably easy/intermediate for most people on this forum.



Spoiler



Examine the function \( f(x) = x^2 \) on the open interval \( (0,\infty) \)

On this interval we know that the following equation holds true:
\( x^2 = x^2 \)

Now let's rewrite things a bit for fun:

\( x^2 = x + x + x + \ldots + x \)
Where there are \( x \) instances of the variable \( x \). This is by the definition of multiplication as a repeated addition.

\( \frac{d}{dx} \left( x^2 \right) = \frac{d}{dx} \left( x + x + x + \ldots + x \right) \)
Differentiate both sides.

\( 2x = 1 + 1 + 1 + \ldots + 1 \)
Where there are \( x \) instances of the number 1.

\( 2x = x \)
Repeated addition can be simplified to a multiplication

Since we are considering the open interval \( (0,\infty) \), then we know that \( x \) will never have the value 0 anywhere on this interval. This means that dividing by \( x \) will be defined.

\( \frac{2x}{x} = \frac{x}{x} \)

\( 2 = 1 \)
All math is meaningless, because there is really only one number anyway.


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## Rinfiyks (Dec 31, 2010)

Attempt at an answer


Spoiler



When you differentiate x + x + x + . . . + x, you're not taking into account the x that tells you how many xs there are. You're just assuming it's a constant amount of xs, and not a variable amount as it is. Not a very rigorous disproof 



Problem


Spoiler



Without evaluating either of them (e.g. with Maclaurin sequences or whatever those things are called), find which number is largest.
\( \sqrt{2} \)

\( \sqrt[3]{3} \)


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## einstein00 (Dec 31, 2010)

answer


Spoiler



raise each to the 6th power. \( (\sqrt{2})^6 = 8 \), and \( (\sqrt[3]{3})^6 = 9 \). Clearly, the latter is larger.



problem


Spoiler



Ann and Betty initially each have $50. Ann then receives $10 from her mother, and Betty receives $20 from her mother. Together, they only have $120. They neither gained nor lost any money in any other way. Explain.


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## cmhardw (Dec 31, 2010)

Rinfiyks said:


> Attempt at an answer
> 
> 
> Spoiler
> ...


 
Without giving away too much I can tell you that this is not the first error I made in the proof. Try to find the first error.


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## einstein00 (Dec 31, 2010)

another attempt at an answer to cmhardw:


Spoiler



If x is not an integer, how can there be x instances of x. For example, 0.123 instances of 0.123 doesn't make sense (unless you're multiplying 0.123 by 0.123, but that doesn't follow the form x+x+x+...+x, it's x*x)


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## cmhardw (Dec 31, 2010)

einstein00 said:


> another attempt at an answer to cmhardw:
> 
> 
> Spoiler
> ...


 
Yes, this is the first, and largest, error in the proof. The other errors are all consequences of this false assumption. Also one comment to your statement:


Spoiler






> If x is not an integer, how can there be x instances of x.



The identity \( x^2 = x + x + x + \ldots + x \) with x instances of x does not hold for all integers. It holds true only for the natural numbers.


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## einstein00 (Dec 31, 2010)

cmhardw said:


> Yes, this is the first, and largest, error in the proof. The other errors are all consequences of this false assumption. Also one comment to your statement:
> 
> 
> Spoiler
> ...



probably already obvious.. 


Spoiler



all integers > 0 are natural numbers.


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## Baian Liu (Dec 31, 2010)

einstein00 said:


> problem
> 
> 
> Spoiler
> ...


 


Spoiler



Betty is Ann's mother, and Betty receives $20. They each have $60.


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## Keroma12 (Dec 31, 2010)

Problem:


Spoiler



\( n^{(-2)} \int_{-n}^{n} (n-x)^{\frac{1}{2}}(x+n)^{\frac{1}{2}}\,dx \)



Edit: Thanks Chris, I wasn't sure how to write it this way.


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## Christopher Mowla (Dec 31, 2010)

Sa967St said:


> Squared, but the ^2 wasn't there when I answered it:
> 
> 
> cmowla said:
> ...


The ^2 was always there. I just was implying that since it was made so small by the latex, I could understand why you missed it. When I first edited that post, it was because I wrote the problem 
\( \frac{d}{dx}\left[ \frac{1}{\sqrt{1-\sin ^{2}\left( x \right)}} \right] \) instead of \( \frac{d}{dx}\left[ \frac{1}{\sqrt{\sin ^{2}\left( x \right)-1}} \right] \). But you obviously saw that edit before you answered.

No worries. Latex has some weird ways of displaying symbolic logic (although I can't complain about most of the output).


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## cmhardw (Dec 31, 2010)

Keroma12 said:


> Problem:
> 
> 
> Spoiler
> ...



@Keroma

Do you mean?


Spoiler



\( n^{(-2)} \int_{-n}^{n} (n-x)^{\frac{1}{2}}(x+n)^{\frac{1}{2}}\,dx \)


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## oprah62 (Dec 31, 2010)

Spoiler



Every month, a girl gets allowance. Assume last year she had no money, and kept it up to now. Then she spends 1/2 of her money on clothes, then 1/3 of the remaining money on games, and then 1/4 of the remaining money on toys. After she bought all of that, she had $7777 left. Assuming she only gets money by allowance, how much money does she earn every month?


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## qqwref (Dec 31, 2010)

cmhardw said:


> Spoiler
> 
> 
> 
> \( n^{(-2)} \int_{-n}^{n} (n-x)^{\frac{1}{2}}(x+n)^{\frac{1}{2}}\,dx \)


 


Spoiler



= \( n^{-2} \int_{-n}^{n} ((n-x)(x+n))^{\frac{1}{2}} dx \)
= \( n^{-2} \int_{-n}^{n} (n^2 - x^2)^{\frac{1}{2}} dx \)
= \( 2 n^{-2} \int_{0}^{n} (n^2 - x^2)^{\frac{1}{2}} dx \)
= \( 2 n^{-2} \) * area under quarter-circle of radius n
= \( 2 n^{-2} * \frac{\pi n^2}{4} \)
= \( \frac{\pi}{2} \)




New problem:


Spoiler



Let x be some real number from 0 to 1. Imagine we take a square ABCD, of side length 1 (and area 1), and draw a line segment from A to a point x of the way from B to C, then draw a line segment from B to a point x of the way from C to D, and so on, until there are 4 line segments drawn. Let y be the area of the smaller square that is now formed.

If x is 1/2, what is y? If y is 1/2, what is x? If x is rational, does y have to be rational, and vice versa?


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## Keroma12 (Dec 31, 2010)

qqwref said:


> Spoiler
> 
> 
> 
> ...





Spoiler



What if n<0?


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## qqwref (Dec 31, 2010)

Keroma12 said:


> Spoiler
> 
> 
> 
> What if n<0?





Spoiler



Fair enough. If n<0, then after the second line we replace n with m (defined as -n) which doesn't change the look of the problem at all, so we can continue as normal.


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## Keroma12 (Dec 31, 2010)

qqwref said:


> Spoiler
> 
> 
> 
> Fair enough. If n<0, then after the second line we replace n with m (defined as -n) which doesn't change the look of the problem at all, so we can continue as normal.


 


Spoiler



But if you you replace -n with m in the first step:
\( n^{-2} \int_{-n}^{n} (n^2 - x^2)^{\frac{1}{2}} dx \)
\( m^{-2} \int_{m}^{-m} (m^2 - x^2)^{\frac{1}{2}} dx \)
\( -m^{-2} \int_{-m}^{m} (m^2 - x^2)^{\frac{1}{2}} dx \)
And if you continue from there don't you get -\( \frac{\pi}{2} \)

If I'm wrong, where am I making my mistake?


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## Keroma12 (Dec 31, 2010)

qqwref said:


> New problem:
> 
> 
> Spoiler
> ...


 
*Answer to qqwref*


Spoiler



y = \( \frac{x^2-2x+1}{x^2+1} \)
If x = 1/2, y = 1/5. If y = 1/2, x = \( \frac{2-3^{1/2}}{2} \)
If x is rational, y must be also, but not vice versa.
Correct?


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## y3k9 (Dec 31, 2010)

Bold your answers and number your problem!


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## qqwref (Dec 31, 2010)

Keroma12 said:


> Spoiler
> 
> 
> 
> ...


Good point.



Keroma12 said:


> *Answer to qqwref*
> 
> 
> Spoiler
> ...





Spoiler



How did you derive the first equation?


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## JBCM627 (Jan 1, 2011)

Here are a couple tricky but useful little integrals:

Problem JBCM-1:


Spoiler



This is neatly related to Keroma12's integral:
\( \int_{a}^{b} \frac{1}{x} (x-a)^{1/2}(b-x)^{1/2} dx \)
Where \( a \) and \( b \) are arbitrary real constants. Hint: complete the square. Though I'd be interested if you can evaluate it using Mathematica, too 


Problem JBCM-2:


Spoiler



\( \int_{-\infty}^{\infty} \frac{e^{\imath\omega\tau}}{1+(\frac{t}{\tau})^2} dt \)
Where \( \imath = \sqrt{-1} \), and \( \omega \) and \( \tau \) are arbitrary constants. Hint: contour integral.


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## Keroma12 (Jan 1, 2011)

qqwref said:


> Spoiler
> 
> 
> 
> How did you derive the first equation?





Spoiler



Draw the diagram, then the area of the inner square will be the area of the whole square (1) minus the 4 larger triangles [(x)(1)/2] plus the 4 smaller triangles [(xcos\( \theta \))(xsin\( \theta \))/2]
So y = 1-2x+2x^2sin\( \theta \)cos\( \theta \)
And from the larger triangle, sin\( \theta \) = \( \frac{x}{\sqrt{x^2+1}} \) and cos\( \theta \) = \( \frac{1}{\sqrt{x^2+1}} \)
Plug those in and then simplify.
This is what my first instinct was to do. Is there a better way? And am I even right?


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## qqwref (Jan 1, 2011)

Keroma12 said:


> Spoiler
> 
> 
> 
> ...





Spoiler



Sure, that works fine. A possibly easier way is:
y = 1 - 4*(area of small triangle + trapezoid, a shape which is itself a right triangle)
= \( 1 - 4*(\frac{1}{2} \sin(\theta) \cos(\theta)) \) (those are the two non-hypotenuse sides of the triangle)
= \( 1 - 2*(\frac{x}{\sqrt{x^2+1}} * \frac{1}{\sqrt{x^2+1}}) \)
= \( 1 - \frac{2x}{x^2+1} \)
= \( \frac{x^2 - 2x + 1}{x^2 + 1} \)


.


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## Rinfiyks (Jan 1, 2011)

A problem with no nasty calculus or algebra:


Spoiler



It's a unit square. B and F are midpoints. Find the area of the shaded region.


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## jerry (Jan 1, 2011)

Whippopotamus said:


> Hi, I'm doing As Level Maths at the minute, and I have a fair interest in maths, and I would like to pose this question, its just something I noticed when I was bored in a physics lesson:
> 
> 
> 
> ...



I'm not sure if the following qualifies as a full/legitimate answer, but hopefully it'll give you a rough idea of what's happening. I encountered the same thing with my friend while messing with a calculator 



Spoiler



I tried this too using cos(cos(cos(...x))))...). It tended towards about 0.7931. Compounding sine also tends to 0.7931, except it does so slower (I think). I learned the reason for this behavior with a google search. It turns out that 0.7931 is the "steady state" of the function, which is when the input equals the output. 

I suggest you read about it. It's a bit interesting.
http://books.google.com/books?id=Oi...AA#v=onepage&q=cosine of cosine .7391&f=false


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## maggot (Jan 1, 2011)

Given that \( f(x)=x \) is the solution to the DE 
\( y'' - 2xy + 2y = 0 \)
determine a second linearly independent soln.

no need to break down, this is no elementary integration. ; w;


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## vcuber13 (Jan 1, 2011)

Rinfiyks said:


> A problem with no nasty calculus or algebra:
> 
> 
> Spoiler
> ...


 
it may be off by a bit because i only used like 3 decimals


Spoiler



\( 
2.75 \sqrt{74}
\)
or about
\( 
23.6564
\)


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## keemy (Jan 1, 2011)

maggot: your problem is wrong y=x isn't a solution (unless you mean 2xy' for second term) in that case the solution is really nasty because it ends up being something like


Spoiler



\( x \int x^{-2} e^{x^2} dx \)


 so pretty lame (someone can try for themselves it's pretty easy redux of order then solve the first order DE with an integrating factor or as a separable equation but then integral of that is as stated above).

and a problem (not that hard)


Spoiler



let \( a_0=c \) and \( a_1=k \) solve the recurrence \( a_{n+2} = \frac {-a_n}{2(n+2)} \) (if you want you can give separate solutions for even and odd)


----------



## Quadrescence (Jan 1, 2011)

No calculus required.

Solve for \( x \):

\( 4 = x + \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} \)

where the pattern continues _ad infinitum_.

Look at this hint if you're stuck.


Spoiler



There is an awful lot of repetition going on here. If I said \( y = 2 - y \), then it'd also be true that \( y = 2 - (2 - y) \), and furthermore, \( y = 2 - (2 - (2 - y)) \), _ad infinitum._ Try to use this idea here.


----------



## keemy (Jan 1, 2011)

quad's


Spoiler



*2*
\( 4 = x + \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} \)
\( (4-x)^2 = x + \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} \)
\( 16-8x+x^2= 4 \)
\( (x-6)(x-2)=0 \)
6 dun werk so 2.


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## Rinfiyks (Jan 1, 2011)

vcuber13 said:


> it may be off by a bit because i only used like 3 decimals
> 
> 
> Spoiler
> ...



Hmm, that's not the answer I have. I got a fraction. And it's not 1/3.



Quadrescence said:


> No calculus required.
> 
> Solve for \( x \):
> 
> ...





Spoiler



\( 4 = x + \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} \)
\( 4 = x + \sqrt{4} \)
\( x = 2 \)
I like problems like this.


 
A slightly harder problem:


Spoiler



I have 5 equally sized spheres. I put 4 of them on a flat surface. These 4 all touch the adjacent ones, like in the picture below, which is a view from above.




I then place the 5th sphere in the middle of the others. How high is it off the surface?


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## Quadrescence (Jan 1, 2011)

keemy said:


> quad's
> 
> 
> Spoiler
> ...


 
keemoop [and rinfiyks] are right, here is another way, basically equivalent to rinf's


Spoiler



\( 4 = x + \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} \)
\( 4 - x = \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} \)
\( 4 - x = \sqrt{x + (4 - x)} \)
\( 4 - x = \sqrt{4} \)
\( 4 - x = 2 \)
\( x = 2 \)


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## Quadrescence (Jan 1, 2011)

Rinfiyks said:


> A slightly harder problem:
> 
> 
> Spoiler
> ...


 
Here's my initial stab at it.


Spoiler



Suppose the radii are \( r \). Then, connecting the spheres' centers, we have a square with every side length equal to \( 2r \) (because each sphere is tangent to one another, two centers of two such spheres will be no further than \( 2r \)). Since all faces (except the base) are equilateral triangles, this is called the first Johnson solid (\( J_1 \)). Not that that is relevant, but neato trivia.

Anyway to find the height, consider the distance to the center of the view given in the problem. The distance from the center of the image and any one of the four circle's centers is \( 2r/\sqrt{2} = r\sqrt{2} \) (why? construct a right triangle). Now we have the distance from the center of the pyramid's base, to one of its base vertexes. Now we have a right triangle if we travel along the edge of the pyramid to the apex. This is a right triangle, with base \( r\sqrt{2} \) and hypotenuse \( 2r \). Since \( \mathrm{base}^2+\mathrm{height}^2=\mathrm{hypotenuse}^2 \), we have 

\( (r\sqrt{2})^2 + h^2 = (2r)^2 \)
\( 2r^2 + h^2 = 4r^2 \)
\( h^2 = 2r^2 \)
\( h = r\sqrt{2} \)

This is only the height from the plane which the centers of the four base spheres form, which is itself at a height of \( r \). So the center of the top sphere is at \( (1+\sqrt{2})r \). But if we want to know the distance the bottom of the sphere is from the ground, we just take off \( r \) from this, giving \( r\sqrt{2} \).

Not sure if this is correct, but it's my initial intuition.


----------



## Quadrescence (Jan 1, 2011)

keemy said:


> Spoiler
> 
> 
> 
> let \( a_0=c \) and \( a_1=k \) solve the recurrence \( a_{n+2} = \frac {-a_n}{2(n+2)} \) (if you want you can give separate solutions for even and odd)


 


Spoiler



Let's see here. My gut feeling is

\( a_n = 
\begin{cases}
\displaystyle\left(-\frac{1}{2}\right)^{n/2}\frac{c}{n!!} & \text{even } n\\
\displaystyle\left(-\frac{1}{2}\right)^{(n-1)/2}\frac{k}{n!!} & \text{odd } n
\end{cases} \)​
where \( n!! = n(n-2)(n-4)\cdots \) is the *double factorial* ending at \( 2 \) if \( n \) is even, \( 1 \) if odd. This can be "simplified" to

\( a_n = \left(-\frac{1}{2}\right)^{\lfloor n/2\rfloor}\frac{[(n+1)\bmod 2]c+(n\bmod 2)k}{n!!} \)​ 
where \( n\bmod m \) is modular reduction.


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## Quadrescence (Jan 1, 2011)

New problem: Define a rational function as one which is a quotient of polynomials. Let \( \Delta f(x) = f(x+1)-f(x) \), Show that no rational function \( f \) exists satisfying

\( \Delta f(x) = \frac{1}{x}. \)

If you're daring, show that

\( \Delta_k f(x) = \frac{1}{x} \)
​ has no rational solution, where \( \Delta_k f(x) = f(x+k)-f(x) \) for non-zero \( k\in\mathbb{Z} \).​


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## cyoubx (Jan 1, 2011)

vcuber13 said:


> it may be off by a bit because i only used like 3 decimals
> 
> 
> Spoiler
> ...


 
How exactly did you get an answer that was greater than 1 seeing as the region is a subset of the unit square, haha?

I got a very sketchy answer of 0.2529, but I may have done something wrong in my haste.

EDIT: I think I did it wrong...I got 0.266 the second time.

EDIT 2: *Final Answer: 0.267* which is actually *4/15* without rounding errors.


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## keemy (Jan 1, 2011)

\( \lim_{x \to 0} \frac{ \ln (1-x) -\sin {x} } {1 - \cos^2{x}} \) oMg


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## joey (Jan 1, 2011)

keemy: -1


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## qqwref (Jan 1, 2011)

Rinfiyks said:


> A problem with no nasty calculus or algebra:
> 
> 
> Spoiler
> ...





Spoiler



Let G be the intersection of BD and CF, and H and I as the other vertices of the shaded region so it is labeled region DGHI.
- Shaded region = ACDE - CGD - EID - AHC - AHE
- Area of shaded region = 1 - 2*(area of CGD) - 2*(area of AHC)
To find the area of those two triangles, let's establish a coordinate system where E = (0,0), D = (1,0), A = (0,1).
Now G is at the intersection of lines y = -2x + 2 and y = x/2 + 1/2. Thus G = (3/5, 4/5), so it's a distance of 2/5 from line CD, and the area of CGD is 1/2 (1)(2/5).
Similarly, H is at the intersection of lines y = x/2 + 1/2 and y = 2x. Thus H = (1/3, 2/3), so it's a distance of 1/3 from line AC, and the area of AHC is 1/2 (1)(1/3).

Finally we get:
- Area of shaded region = 1 - 2*(1/5) - 2*(1/6)
- Area of shaded region = 4/15.


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## vcuber13 (Jan 1, 2011)

cyoubx said:


> How exactly did you get an answer that was greater than 1 seeing as the region is a subset of the unit square, haha?
> 
> I got a very sketchy answer of 0.2529, but I may have done something wrong in my haste.
> 
> ...


 
ya um i multiplied by 10 so it is 100 times bigger than i meant, so it should be like .236564


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## LewisJ (Jan 2, 2011)

For something less proof-centered than quad's last problem:

For what values of M, N can an MxN checkerboard be tiled with dominoes? Explain.


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## Rinfiyks (Jan 2, 2011)

Quadrescence said:


> Here's my initial stab at it.
> 
> 
> Spoiler
> ...


You are correct 
You might want to try it for 3 spheres on the base instead of 4.
Answer (I think):


Spoiler



\( 
\frac{2\sqrt{6}}{3}r
\)







LewisJ said:


> For something less proof-centered than quad's last problem:
> 
> For what values of M, N can an MxN checkerboard be tiled with dominoes? Explain.


 


Spoiler



My guess:
M and N must both be greater than 1, and at least one must be greater than 2.
Each domino will occupy 1 light square and 1 dark square. So both M and N cannot be odd.



Another problem:


Spoiler



You may have heard the riddle about the man who walks 1 mile south, 1 mile east, 1 mile north, and ends back where he started. The question is, where did he start? The celebrated answer is the North Pole.
However, there is another answer in which he starts slightly north of the South Pole. How far north does he start? And how does it work?


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## Kynit (Jan 2, 2011)

Rinfiyks said:


> Another problem:
> 
> 
> Spoiler
> ...





Spoiler



He is one mile north of the point where the distance around the earth is 1 mile. I'm not sure how to figure this out exactly, though


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## keemy (Jan 2, 2011)

lewis:


Spoiler



wlog when M is even. with M even you can just place dominoes in a MX1 row N times. if both are odd then there are an odd # of tiles and since placing dominoes can only cover an even # of tiles wont work



quad (incomplete)


Spoiler



let \( f(x)= \frac {P(x)}{Q(x)} \) were P and Q are polynomials an Q not 0.

assume for sake of contradiction
\( \Delta f(x) = \frac{P(x+1)Q(x) - P(x)Q(x+1)}{Q(x+1)Q(x)} = \frac{1}{x} \)

so either 1.\( Q(x)=xR(x) \) or 2.\( Q(x+1)= xR(x) \) R being another non 0 polynomial

1. \( \Delta f(x) = \frac{xP(x+1)R(x) - (x+1)P(x)R(x+1)}{(x(x+1)R(x+1)R(x)} = \frac{1}{x} \)
so you can see that \( P(x+1)R(x) \) must have a factor of \( x+1 \) basically you can keep going down this path showing stuff needs more factors over and over so the polynomials can't be finite at least 

2. same as above basically.


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## Christopher Mowla (Jan 2, 2011)

keemy said:


> \( \lim_{x \to 0} \frac{ \ln (1-x) -\sin {x} } {1 - \cos^2{x}} \) oMg



Answer


Spoiler



\( \underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{\ln \left( 1-x \right)-\sin \left( x \right)}{1-\cos ^{2}\left( x \right)} \right]=\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{\ln \left( 1-x \right)-\sin \left( x \right)}{\sin ^{2}\left( x \right)} \right] \)

\( \text{This has the form }\left( \frac{0}{0} \right),\text{ so we can use L }\!\!'\!\!\text{ hospital }\!\!'\!\!\text{ s rule:} \)

\( \overset{L}{\mathop{=}}\,\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{\frac{d}{dx}\left[ \ln \left( 1-x \right)-\sin \left( x \right) \right]}{\frac{d}{dx}\left[ \sin ^{2}\left( x \right) \right]} \right] \)

\( =\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{\frac{1}{x-1}-\cos \left( x \right)}{2\sin \left( x \right)\cos \left( x \right)} \right]=\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{\frac{1}{x-1}-\cos \left( x \right)}{\sin \left( 2x \right)} \right] \)


_Clearly this limit *does not exist* due to the denominator._


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## LewisJ (Jan 2, 2011)

You guys got that, now for the interesting version with the restriction which I forgot previously: 
For what values of M, N can you cover an MxN checkerboard with an equal number of horizontal and vertical dominoes?


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## keemy (Jan 2, 2011)

cmowla is the new state champion!!!!

lewis


Spoiler



when N and M r both even whoops


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## einstein00 (Jan 2, 2011)

i'm just wondering... how old are you guys? I think i need to step up my math a bit...


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## Rinfiyks (Jan 2, 2011)

LewisJ said:


> You guys got that, now for the interesting version with the restriction which I forgot previously:
> For what values of M, N can you cover an MxN checkerboard with an equal number of horizontal and vertical dominoes?


 


Spoiler



Both M and N must be even, and at least one must be a multiple of 4?




Edit:


einstein00 said:


> i'm just wondering... how old are you guys? I think i need to step up my math a bit...


 
I'm 18, so I have little experience with high level maths involving calculus and limits. That's why I'm posting the style of puzzles that only requires a small understanding of maths.


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## maggot (Jan 2, 2011)

keemy said:


> maggot: your problem is wrong y=x isn't a solution (unless you mean 2xy' for second term) in that case the solution is really nasty because it ends up being something like
> 
> 
> Spoiler
> ...



thx for pointing out the mistake in the problem! it was late when i wrote the post lol. . .

i broke it down for the approx for fun. . 
\( y(x) = -1 + x^2 + \frac {x^4}{6} + . . . + \frac {x^{2n}}{n!(2n-1)} + . . . \)

and for the above limit, i got it wrong lmao! good problem! i had to actually think about differentiating the trig functions lmao.. . you guys are making me think so hard! this thread is a good test of my memory!


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## LewisJ (Jan 2, 2011)

keemy, rinfiyks:


Spoiler



They don't have to both be even; 3x8 is coverable.



einstein:


Spoiler



I'm 16, took multivar calc last semester, ODE next semester


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## cyoubx (Jan 2, 2011)

einstein00 said:


> i'm just wondering... how old are you guys? I think i need to step up my math a bit...


 
I finished linear algebra and differential equations freshman year, and multivariable calc as a sophomore in High School.


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## einstein00 (Jan 3, 2011)

:O i'm taking single variable calc this year (sophomore) and i guess i'm behind lolz

i got an 8 on the AIME last year though


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## dbax0999 (Jan 4, 2011)

A spherical oil tank with a radius of 10 feet is half full of oil that weighs 60 pounds per cubic foot. Its top is 6 feet under ground. How much work is needed to pump the oil out of the tank through a hole in its top?


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## Keroma12 (Jan 4, 2011)

einstein00 said:


> i'm just wondering... how old are you guys? I think i need to step up my math a bit...


 
17. All I've been taught from school is up to the first half of high school calculus, derivatives. But I've already done some stuff on my own, not much though. I'm reading On Numbers and Games by John Conway right now. Anybody else here read it? What do you think?


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## LewisJ (Jan 4, 2011)

dbax0999 said:


> A spherical oil tank with a radius of 10 feet is half full of oil that weighs 60 pounds per cubic foot. Its top is 6 feet under ground. How much work is needed to pump the oil out of the tank through a hole in its top?


 
Woo, I haven't done a problem like this in nearly a year. Let's see how well my memory holds.

I will assume the oil is to be pumped to ground level, so a constant 6 will be added to pumping distance. 

I will let the bottom of the tank be y=0 and integrate wrt y
Our limits of integration are thus 0 and 10 since the tank is half full
We must integrate the work required to pump out a "slice" of the tank over this region

Work = Force * displacement
The displacement at height y is 20+6-y feet
The force is 60 * volume of "slice"
Volume is dy * pi * r^2 (r being slice radius)
Radius is found with a right triangle with hypotenuse being sphere radius, one leg along y-axis, other leg being slice radius, and we get
slice radius = sqrt(100 - y^2)

Plug everything back in and we see what we must integrate:
60pi(26-y)(100-y^2)dy
from 0 to 10.

To hell with integration by hand.

And we get 890000pi foot-pounds of work required.

Edit: I totally just did someone's math homework. Meh.


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## cmhardw (Jan 4, 2011)

I always liked this problem.

There is a deck of cards that contains only 3 cards. One card is red on both sides. One card is black on both sides. One card is red on one side and black on the other. The deck is shuffled randomly and one card is produced from the deck and laid on the table. This card shows red on top. What is the probability that when you flip this card over that it will be black on the the other side?


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## LewisJ (Jan 4, 2011)

cmhardw said:


> I always liked this problem.
> 
> There is a deck of cards that contains only 3 cards. One card is red on both sides. One card is black on both sides. One card is red on one side and black on the other. The deck is shuffled randomly and one card is produced from the deck and laid on the table. This card shows red on top. What is the probability that when you flip this card over that it will be black on the the other side?


 


Spoiler



1/3; there are three cases with red on top; each side of the all-red card, plus the one side of the red/black card. Only one has black on the other side.


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## Rinfiyks (Jan 4, 2011)

LewisJ said:


> Spoiler
> 
> 
> 
> 1/3; there are three cases with red on top; each side of the all-red card, plus the one side of the red/black card. Only one has black on the other side.


 


Spoiler



Really? I thought it would be 1/2. There are 2 cards with red on them. You've either got one or the other.


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## cmhardw (Jan 4, 2011)

Rinfiyks said:


> Spoiler
> 
> 
> 
> Really? I thought it would be 1/2. There are 2 cards with red on them. You've either got one or the other.


 


Spoiler



1/3 actually is the correct answer tu LewisJ). To convince you that this is true, it's very similar to the Monty Hall problem, another _very_ interesting probability question.


----------



## Rinfiyks (Jan 4, 2011)

cmhardw said:


> Spoiler
> 
> 
> 
> 1/3 actually is the correct answer tu LewisJ). To convince you that this is true, it's very similar to the Monty Hall problem, another _very_ interesting probability question.


 


Spoiler



Ah yeah, you're given a random face, not a random card! I should've thought "what's the probability that the card has the same colour on both sides", which is effectively what it's asking! Awesome problem  more!


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## cmhardw (Jan 4, 2011)

Rinfiyks said:


> Spoiler
> 
> 
> 
> Ah yeah, you're given a random face, not a random card! I should've thought "what's the probability that the card has the same colour on both sides", which is effectively what it's asking! Awesome problem  more!


 
Yes exactly! If you like this style problem, and now that you have the basic idea, try some more.

There are 6 cards in a deck. 1 card is red on both sides, 2 cards are black on both sides, and 3 cards have red on one side and black on the other side. If this deck is shuffled randomly, and one card is produced that shows red on top, what's the probability that when you flip this card over it shows black on the other side?

And now for the fun question, what's the probability that if you shuffle this deck randomly and produce a card, that flipping this card over shows the other color?


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## Rinfiyks (Jan 5, 2011)

cmhardw said:


> Yes exactly! If you like this style problem, and now that you have the basic idea, try some more.
> 
> There are 6 cards in a deck. 1 card is red on both sides, 2 cards are black on both sides, and 3 cards have red on one side and black on the other side. If this deck is shuffled randomly, and one card is produced that shows red on top, what's the probability that when you flip this card over it shows black on the other side?
> 
> And now for the fun question, what's the probability that if you shuffle this deck randomly and produce a card, that flipping this card over shows the other color?


 


Spoiler



First question: 3/5?
Second question: 1/2. Part of me wants it not to be 1/2 so that my mind is blown again like before!


----------



## cmhardw (Jan 5, 2011)

Rinfiyks said:


> Spoiler
> 
> 
> 
> First question: 3/5?



Yep 

Second question:


Spoiler






> Second question: 1/2. Part of me wants it not to be 1/2 so that my mind is blown again like before!



Ok I can't let this one slide, you have to be confident in your answer  Is it 1/2 or is it not 1/2?


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## Rinfiyks (Jan 5, 2011)

cmhardw said:


> Yep
> 
> Second question:
> 
> ...


 


Spoiler



Haha, yes I am confident


----------



## cmhardw (Jan 5, 2011)

Rinfiyks said:


> Spoiler
> 
> 
> 
> Haha, yes I am confident


 


Spoiler



Yep, 1/2 is right here. There are two common ways to prove this. I like the method that says that only 3 of the 6 cards are multi-colored, so 1/2 chance you picked one of those cards from the deck.

The other method is to evaluate the probabilities in the following formula:
P(red on top side) * P(black on other side | red on top side) + P(black on top side) * P(red on other side | black on top side)

Both methods yield 1/2 as the answer, so I agree


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## cmhardw (Jan 5, 2011)

Another probability question that I always loved:

What's the probability that at least two people out of a group of 23 people have the same birth month and birth day, assuming that no one in the group has a birthday on Feb. 29th?


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## Rinfiyks (Jan 5, 2011)

cmhardw said:


> Spoiler
> 
> 
> 
> ...


 


Spoiler



I prefer the former too  it's the way I did it.
Reminds me of another probability problem:
You have two boxes. You know that one of the boxes contains a certain amount of money while the other one contains twice as much. But you do not know which is which.
You pick a box and assume it contains a value of money we shall call x. (You don't open it yet)
You then reason that the other box contains half as much or twice as much: x/2 or 2x.
You then work out the "average" value of the other box.
(x/2)/2 + 2x/2
(1/4)x + x
(5/4)x.
So statistically the other box contains, on average, more money. So you should switch!
You could then apply the same reasoning and switch again and again...


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## keemy (Jan 5, 2011)

Spoiler



\( 1- \frac {23! \times \binom{365}{23}}{365^{23}} \)



since we are on combo already

lets say you have a 3 by n binary matrix what is the minimum n required to ensure that some row has has either 3 1s or 3 0s


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## Rinfiyks (Jan 5, 2011)

cmhardw said:


> Another probability question that I always loved:
> 
> What's the probability that at least two people out of a group of 23 people have the same birth month and birth day, assuming that no one in the group has a birthday on Feb. 29th?


 
I've heard this one before! I don't know how to calculate it exactly but


Spoiler



It's >50%



Edit:


keemy said:


> Spoiler
> 
> 
> 
> ...


 


Spoiler



5?


----------



## keemy (Jan 5, 2011)

that's correct (though you should try to prove it) also in the extend the question further find a function \( f \) such that \( f(k)= \) minimum n to ensure there are either k 1s or k 0s in some row.


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## Rinfiyks (Jan 5, 2011)

keemy said:


> that's correct (though you should try to prove it) also in the extend the question further find a function \( f \) such that \( f(k)= \) minimum n to ensure there are either k 1s or k 0s in some row.





Spoiler



You can get up to k - 1 of both 0s and 1s in a row safely. This makes 2(k - 1). Then if you add another column, that row must have either k 0s or k 1s. So it's 2(k - 1) + 1 = 2k - 1.


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## keemy (Jan 5, 2011)

true but not a proof.


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## qqwref (Jan 5, 2011)

Here's a cool one:

Find a polynomial p(x,y) so that, when x and y are integers, p gives a unique value for each possible value of x and y. So, if you know x and y are integers, and you are given a value of p, you should be able to figure out the single possible (x,y) pair that gives that value.


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## cmhardw (Jan 5, 2011)

qqwref said:


> Here's a cool one:
> 
> Find a polynomial p(x,y) so that, when x and y are integers, p gives a unique value for each possible value of x and y. So, if you know x and y are integers, and you are given a value of p, you should be able to figure out the single possible (x,y) pair that gives that value.


 
If p is irrational for some x and some y, then must it be written in decimal form?


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## keemy (Jan 5, 2011)

A polynomial over what field? I mean I can easily choose


Spoiler



\( p(x,y)= x+iy \)


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## qqwref (Jan 5, 2011)

Pshh, alright. All the coefficients in p(x,y) must be integers.


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## cmhardw (Jan 5, 2011)

qqwref said:


> Pshh, alright. All the coefficients in p(x,y) must be integers.


 
Ok, I like this. I was thinking of:


Spoiler



\( p(x,y)=x\sqrt{2} + y\sqrt{3} \)
personally.



With integer coefficients... *scratches head*


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## Keroma12 (Jan 5, 2011)

cmhardw said:


> I always liked this problem.
> 
> There is a deck of cards that contains only 3 cards. One card is red on both sides. One card is black on both sides. One card is red on one side and black on the other. The deck is shuffled randomly and one card is produced from the deck and laid on the table. This card shows red on top. What is the probability that when you flip this card over that it will be black on the the other side?


 
I love these problems! Got any more? (Yes, I got this one right.)


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## cmhardw (Jan 5, 2011)

I'm still working on Michael's, but I have another good one I just remembered (and Michael you may like this one).

Prove that for any 5 points placed randomly on the surface of a sphere, that at least four of those points are always contained on the same hemisphere.


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## keemy (Jan 5, 2011)

ok I think I have one (and can probably make some other with the same idea) 



Spoiler



\( P(x,y)=y^2+2yx+y+x^2+3x = (x+y)(x+y+1) +2x= \) double the (x+y)th triangular number + 2x uhh I am not positive if this will work for all integers but it should be good for positive ones


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## LewisJ (Jan 5, 2011)

It is also simple to find a polynomial in \( \mathbb{R}^2 \) rather than using imaginaries:
\( p(x,y) = \pi*x + y \)
but by requiring coefficients to be integers, this easy solution is also eliminated.


----------



## qqwref (Jan 5, 2011)

LewisJ said:


> It is also simple to find a polynomial in \( \mathbb{R}^2 \) rather than using imaginaries:
> \( p(x,y) = \pi*x + y \)
> but by requiring coefficients to be integers, this easy solution is also eliminated.


Right, solutions which use pi, i, square roots, etc. are trivial and boring, and they eliminate the point of the problem.

keemy:


Spoiler



Can you think of a polynomial p(x) that uniquely converts an integer x into a positive integer?


----------



## keemy (Jan 5, 2011)

oh yeah lol so I guess just square each term I had or w/e


----------



## FMC (Jan 5, 2011)

An optimization problem,calculus not needed,level:high school.
find the minimum value of sqrt(x^2+1) + sqrt((x-y)^2+1 ) + sqrt((y-z)^2+1) + sqrt((z-10)^2+9), what values of x,y and z give that minimum


----------



## qqwref (Jan 5, 2011)

keemy said:


> Spoiler
> 
> 
> 
> oh yeah lol so I guess just square each term I had or w/e


No, it isn't quite that simple.


----------



## Rinfiyks (Jan 5, 2011)

qqwref said:


> No, it isn't quite that simple.


 


Spoiler



\( p(x) = x^2 + x/2 \)


----------



## blah (Jan 5, 2011)

cmhardw said:


> Prove that for any 5 points placed randomly on the surface of a sphere, that at least four of those points are always contained on the same hemisphere.


Putnam 

Hint:


Spoiler



Big round things and birds.


----------



## keemy (Jan 5, 2011)

blah said:


> Putnam
> 
> Hint:
> 
> ...





Spoiler



I don't think you even to do that much if you just use the fact that any 3 points are on the same hemisphere


----------



## blah (Jan 5, 2011)

keemy said:


> Spoiler
> 
> 
> 
> I don't think you even to do that much if you just use the fact that any 3 points are on the same hemisphere





Spoiler



I think that fact comes from the fact about great circles?


----------



## hawkmp4 (Jan 5, 2011)

Chris':


Spoiler



Any 3 points are on a hemisphere...call that hemisphere A...then the 4th point must either be in A or not in A. If it is in A, 4 points in the same hemisphere. Done. If the 4th point is not in A, then it is in B, C, or D, the hemispheres formed by the 4th point and one of the pairs possible from the original 3 points.

Grr...
Don't know where to go from here...



EDIT:
Michael's:


Spoiler



So we're looking for a antisymmetric polynomial relation, of degree 2 or greater. A linear function won't work since there are no two distinct numbers that don't span the integers. What about \( x^3+y^5 \)? No, there are surely numbers that are both cubes and...quintics? Hm. I'll return to this later.


----------



## blakedacuber (Jan 5, 2011)

why is pi 3.14 or 22 over 7?


----------



## kvaele (Jan 5, 2011)

22/7 is an approximation. Pi is irrational, meaning it cannot be put into a fraction.


----------



## hawkmp4 (Jan 5, 2011)

blakedacuber said:


> why is pi 3.14 or 22 over 7?


 
Pi isn't either of those things


----------



## blakedacuber (Jan 5, 2011)

hawkmp4 said:


> Pi isn't either of those things


 
in maths it is

i dont care why its a letter


----------



## blakedacuber (Jan 5, 2011)

kvaele said:


> 22/7 is an approximation. Pi is irrational, meaning it cannot be put into a fraction.


 
but why those numbers?

who said PI is 3.14 or 22/7 but more importantly(and my question) why?


----------



## nathanajah (Jan 5, 2011)

blakedacuber said:


> in maths it is
> 
> i dont care why its a letter


 
It is not *exactly* 22/7 or 3.14, both of it are just approximations.
Pi is an irrational number.

edit: ninja'd, I'll answer the second question.
Pi, counted to 50 decimal digit is 3.14159265358979323846264338327950288419716939937510 (wikipedia.
We could see that 22/7 = 3.142857142857142857.... which is close enough (at least the first 3 digits are close enough) to pi, while 3.14 is clearly the truncation of pi.

http://en.wikipedia.org/wiki/Numerical_approximations_of_π (calculating pi)


----------



## hawkmp4 (Jan 5, 2011)

blakedacuber said:


> why is pi 3.14 or 22 over 7?


 


blakedacuber said:


> but why those numbers?
> 
> who said PI is 3.14 or 22/7 but more importantly(and my question) why?


You did...


----------



## Rinfiyks (Jan 5, 2011)

blakedacuber said:


> but why those numbers?
> 
> who said PI is 3.14 or 22/7 but more importantly(and my question) why?


 
When you're using pi for whatever reason (working with circles or spheres maybe), you don't usually need it that accurate. So 3.14 is a good _approximation_. 22/7 is even better, and something you can easily do on a calculator.
But pi isn't 3.14 or 22/7. It's 3.1415926535... digits go on forever with no known pattern.


----------



## kinch2002 (Jan 5, 2011)

Someone just answer properly! It's the ratio between the circumference and the diameter of a circle


----------



## JBCM627 (Jan 5, 2011)

blah said:


> Spoiler
> 
> 
> 
> I think that fact comes from the fact about great circles?


What is this about birds?


Spoiler



2 points on a sphere and a point in the center of the sphere define a plane slicing the sphere into hemispheres. If the points coincide, there is some freedom to pick whatever plane you like. Then, at least 2 of the 3 remaining points must lie on either hemisphere (or also on the plane). So 4 points are on that hemisphere. Of course, that hemisphere must include the great circle drawn out by the plane, so the problem becomes somewhat ill defined here, since the hemisphere is not expressly open or closed.





Rinfiyks said:


> Spoiler
> 
> 
> 
> \( p(x) = x^2 + x/2 \)


That doesn't return integers.


Spoiler



\( 4(2x+1/2)^2 \)



I'll need to think of a problem more like these, I guess. Nobody likes my integrals


----------



## Rinfiyks (Jan 5, 2011)

JBCM627 said:


> I'll need to think of a problem more like these, I guess. Nobody likes my integrals



I'm sure I would if I knew anything about them!


Here's a problem:

On each corner of an equilateral triangle is an ant. Let's call each ant A, B and C.
They then start to crawl towards each other at the same speed. A crawls towards B, B crawls towards C, C crawls towards A.
How far have they crawled, in relation to the side length of the triangle, when they meet?


----------



## dPod121 (Jan 5, 2011)

They meet at the incenter of the triangle. They have crawled 1/2 the length of one of the sides of the triangle.


----------



## Kynit (Jan 6, 2011)

They don't just walk into the center of the triangle, do they?


----------



## dPod121 (Jan 6, 2011)

I think they do.


----------



## Kynit (Jan 6, 2011)

At the very start of the problem, wouldn't they at least begin walking along the triangle's side, and turn inwards?


----------



## Rinfiyks (Jan 6, 2011)

Yes, they would start off walking in the direction of the triangle's side and walk in a logarithmic spiral I think.
And no, it's not 1/2 the length of the side, try again


----------



## keemy (Jan 6, 2011)

dPod121 said:


> They do.


 
nope. and even if they did that's more than 1/2 the length of an edge it is \( \frac {\sqrt{3}}{3} \) or w/e.


----------



## blah (Jan 6, 2011)

JBCM627 said:


> What is this about birds?





Spoiler



Pigeonhole.


----------



## qqwref (Jan 6, 2011)

Still haven't seen a fully written out/proved solution for the polynomial problem  (And if you're done - can you make one for p(x,y,z), or even one with an arbitrary number of integer inputs? What's the lowest degree polynomial you can think of, for the 2- and 3-variable cases? Is it possible to make a generalized set of polynomials, one for each number of inputs, so that if we encode some finite ordered list of integers, then *without knowing which polynomial we used* the final number will tell us how long the list was as well as letting us figure out the list's contents?)


I don't think I've tried the triangle ant problem before, so here goes (not sure if this is at all the right approach):


Spoiler



Since the ants start on an equilateral triangle and do the same things, just rotated 120 degrees from each other around the center, their locations will always form an equilateral triangle. Now suppose at some point they are on a triangle of side length x and they move a length of Δ. Then by the law of cosines the new side length should be \( \sqrt{(x-\Delta)^2 + \Delta^2 - \Delta(x-\Delta)} = \sqrt{x^2 - 3 \Delta x + 3 \Delta^2} \approx \sqrt{(x - \frac{3}{2}\Delta)^2} = x - \frac{3}{2}\Delta \).

So it looks like (as Δ approaches zero) the side length of the ants' triangle decreases 3/2 as fast as the ants themselves move. Since the side length decreases from the side length of the original triangle down to 0, the ants should each have to move 2/3 of that original side length.


----------



## blah (Jan 6, 2011)

Jimmy:


Spoiler



\( \int_a^b\frac{\sqrt{x-a}\sqrt{b-x}}{x}\,dx=\left(\frac{a+b}{2}-\sqrt{ab}\right)\pi \)



Hope it's right because I cheated


----------



## JBCM627 (Jan 6, 2011)

Rinfiyks said:


> On each corner of an equilateral triangle is an ant. Let's call each ant A, B and C.
> They then start to crawl towards each other at the same speed. A crawls towards B, B crawls towards C, C crawls towards A.
> How far have they crawled, in relation to the side length of the triangle, when they meet?


I like this one 


Spoiler



I don't have an exact answer, but I guess I could work on that. Mathematica sort of sucks at solving differential equations, so it still might be possible to come up with an analytic solution (Mathematica couldn't handle what I gave it). So instead, numerically:


```
X1 = {1, 0}; X2 = {-1, 0}; X3 = {0, Sqrt[3]};
tri = ListPlot[{X1, X2, X3, X1}, AspectRatio -> 1, Axes -> False, 
   PlotRange -> {{-2, 2}, {-1, 3}}, Joined -> True];

sol = NDSolve[{
     x1'[t] == (x2[t] - x1[t])/
       Sqrt[(x2[t] - x1[t])^2 + (y2[t] - y1[t])^2],
     y1'[t] == (y2[t] - y1[t])/
       Sqrt[(x2[t] - x1[t])^2 + (y2[t] - y1[t])^2],
     x2'[t] == (x3[t] - x2[t])/
       Sqrt[(x3[t] - x2[t])^2 + (y3[t] - y2[t])^2],
     y2'[t] == (y3[t] - y2[t])/
       Sqrt[(x3[t] - x2[t])^2 + (y3[t] - y2[t])^2],
     x3'[t] == (x1[t] - x3[t])/
       Sqrt[(x1[t] - x3[t])^2 + (y1[t] - y3[t])^2],
     y3'[t] == (y1[t] - y3[t])/
       Sqrt[(x1[t] - x3[t])^2 + (y1[t] - y3[t])^2],
     x1[0] == X1[[1]], x2[0] == X2[[1]], x3[0] == X3[[1]],
     y1[0] == X1[[2]], y2[0] == X2[[2]], y3[0] == X3[[2]]
     }, {x1, x2, x3, y1, y2, y3}, {t, 0, 4/3}][[1]];

Show[
 tri,
 ParametricPlot[
  {{x1[t], y1[t]}, {x2[t], y2[t]}, {x3[t], y3[t]}} /. sol,
  {t, 0, 4/3}]
 ]

NIntegrate[Sqrt[
    ((x2[t] - x1[t])/Sqrt[(x2[t] - x1[t])^2 + (y2[t] - y1[t])^2])^2 +
     ((x2[t] - x1[t])/Sqrt[(x2[t] - x1[t])^2 + (y2[t] - y1[t])^2])^2
    ] /. sol, {t, 0, 4/3}]/2
```
A bit messy, but I'm not sure how to condense. This returns the attached image:

And gives a path length of 0.76109x, which disagrees with qq's value of 2/3.


----------



## Rinfiyks (Jan 6, 2011)

Spoiler



The correct answer is 2/3 though!


----------



## blakedacuber (Jan 6, 2011)

kinch2002 said:


> Someone just answer properly! It's the ratio between the circumference and the diameter of a circle


 
THANK YOU Daniel


----------



## JBCM627 (Jan 6, 2011)

blah said:


> Jimmy:
> 
> 
> Spoiler
> ...


Chestey wins!



Rinfiyks said:


> Spoiler
> 
> 
> 
> The correct answer is 2/3 though!





Spoiler



Ah, I agree now. Found a typo in my code. The integral should be:

```
NIntegrate[
  Sqrt[((x2[t] - x1[t])/
        Sqrt[(x2[t] - x1[t])^2 + (y2[t] - y1[t])^2])^2 + ((y2[t] - 
          y1[t])/Sqrt[(x2[t] - x1[t])^2 + (y2[t] - y1[t])^2])^2] /. 
   sol,
  {t, 0, 4/3}]/2
```
Which does give 0.666667x.


----------



## einstein00 (Jan 6, 2011)

here's a problem (that i do not know how to solve, but I know the answer to)

A) An ant is at the origin of a number line. Every second, it randomly moves one unit left or right. If this continues infinitely, what is the probability that the ant returns to the origin at some point? If the probability is 1, what is the expected number of seconds that pass before the ant first returns to the origin?

B) An ant is at the origin of a rectangular coordinate plane. Every second, it randomly moves one unit left, right, up, or down. If this continues infinitely, what is the probability that the ant returns to the origin at some point? If the probability is 1, what is the expected number of seconds that pass before the ant first returns to the origin?

C) An ant is at the origin of a three dimensional space. Every second, it randomly moves one unit in a direction parallel to one of the three axes. If this continues infinitely, what is the probability that the ant returns to the origin at some point? If the probability is 1, what is the expected number of seconds that pass before the ant first returns to the origin?


----------



## Rinfiyks (Jan 6, 2011)

einstein00 said:


> here's a problem (that i do not know how to solve, but I know the answer to)
> 
> A) An ant is at the origin of a number line. Every second, it randomly moves one unit left or right. If this continues infinitely, what is the probability that the ant returns to the origin at some point? If the probability is 1, what is the expected number of seconds that pass before the ant first returns to the origin?
> 
> ...


 
Here you go


----------



## blah (Jan 6, 2011)

kinch2002 said:


> Someone just answer properly! It's the ratio between the circumference and the diameter of a circle


How do you know that the ratio is a constant?


----------



## einstein00 (Jan 6, 2011)

blah said:


> How do you know that the ratio is a constant?


 
because all circles are similar.


----------



## blah (Jan 6, 2011)

einstein00 said:


> because all circles are similar.


How do you know that?

Also, how do you know that that's not a circular argument (pun intended)? What is the definition of similarity?


----------



## Rinfiyks (Jan 6, 2011)

blah said:


> How do you know that the ratio is a constant?


 
Because circumference and diameter are both 1-dimensional measures, so they are proportional to each other?


----------



## einstein00 (Jan 6, 2011)

Two plane figures are defined to be similar if one is able to make both figures perfectly overlap (congruent) by applying any or all of the following:
1. a rotation of the figure around a point,
2. a proportional enlargement or reduction, and
3. a linear translation.

A property of all similar figures is the constant ratio between corresponding linear parts. For example, if _a_ and _b_ are the lengths of two sides of a triangle, and _A_ and _B_ are the lengths of the corresponding two sides of a similar triangle, then _A/a = B/b_. This holds true because all of these values are linear values (measurements of length, instead of area or volume).

A circle, by definition, is the locus of points equidistant from a given point (the center). All circles are similar because one can apply an enlargement/reduction and a translation to overlap any two circles. Because all circles are similar, and because the radius, diameter, and circumference of a circle are all linear measures, any circle will have the same ratios between these linear values. Pi is defined to be the ratio of the circumference to the diameter in any circle, and since this ratio is always constant, pi is always constant.


----------



## maggot (Jan 6, 2011)

blakedacuber said:


> why is pi 3.14 or 22 over 7?



pi is actually the a proportion of the diameter of a circle to its circumference. you can look it up on wiki. 

http://en.wikipedia.org/wiki/Pi

and like many poster have already stated. pi is not 3.14 or 22/7. it is: (click spoiler) 


Spoiler



3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 8214808651 3282306647 0938446095 5058223172 5359408128 4811174502 8410270193 8521105559 6446229489 5493038196 4428810975 6659334461 2847564823 3786783165 2712019091 4564856692 3460348610 4543266482 1339360726 0249141273 7245870066 0631558817 4881520920 9628292540 9171536436 7892590360 0113305305 4882046652 1384146951 9415116094 3305727036 5759591953 0921861173 8193261179 3105118548 0744623799 6274956735 1885752724 8912279381 8301194912 9833673362 4406566430 8602139494 6395224737 1907021798 6094370277 0539217176 2931767523 8467481846 7669405132 0005681271 4526356082 7785771342 7577896091 7363717872 1468440901 2249534301 4654958537 1050792279 6892589235 4201995611 2129021960 8640344181 5981362977 4771309960 5187072113 4999999837 2978049951 0597317328 1609631859 5024459455 3469083026 4252230825 3344685035 2619311881 7101000313 7838752886 5875332083 8142061717 7669147303 5982534904 2875546873 1159562863 8823537875 9375195778 1857780532 1712268066 1300192787 6611195909 2164201989. . .
is a better approximation of Pi.


----------



## blah (Jan 7, 2011)

einstein00 said:


> Two plane figures are defined to be similar ...


I do see your argument/justification. I'm not trying to be obnoxious or anything, but imho there's too much vagueness in "rotation", "enlargement/reduction", "translation", etc. I prefer a more analytic definition, such as the area of a unit circle, or more precisely, \( \pi=2\int_{-1}^1 \sqrt{1-x^2}\,dx \) (and the trig functions could be defined from here). Or if the trig functions have been defined by series or other means, then something like \( \pi=\inf\{x\in\mathbb{R}: x>0,\sin x=0\} \) would do. I'm aware that such definitions are not very accessible to most people but I'm not trying to be a douche or anything, it's just that the ratio definition has never settled well with me when I was in high school because of it's heavy reliance on intuition. I'm just bringing this up for people to realize that there are other more "solid" definitions lying around.


----------



## qqwref (Jan 7, 2011)

How, exactly, do you plan to prove pi is a constant using such advanced techniques? (And if you ask me, intuition is fine in geometry, even though it should often be avoided in analysis.)


----------



## blah (Jan 7, 2011)

qqwref said:


> How, exactly, do you plan to prove pi is a constant using such advanced techniques? (And if you ask me, intuition is fine in geometry, even though it should often be avoided in analysis.)


I was simply defining pi as an integral, which is obviously a constant. To show that the ratio A/r^2 is a constant (which happens to be pi) is then (almost) trivial.

First, \( x=ru\Rightarrow dx=r\,du \). Then \( \frac{A}{r^2}=\frac{2}{r^2}\int_{-r}^r \sqrt{r^2-x^2}\,dx=\frac{2}{r^2}\int_{-1}^1 \sqrt{r^2-r^2u^2}\cdot r\,du=2\int_{-1}^1 \sqrt{1-u^2}\,du=\pi \).


----------



## qqwref (Jan 7, 2011)

I'm a lot more interested in the C/2r ratio than the A/r^2 ratio, but I suppose the derivation would be more or less similar.

(And sure, while you technically could define pi as the area of a unit circle as expressed by an integral, or as the sum of some infinite series, etc., it is a LOT less interesting and useful to do so - and you then have to explain why that integral or series also happens to be the circumference/diameter ratio, and half the period of a sine, and so on.)


----------



## blah (Jan 7, 2011)

Intuition is what makes math interesting. It also happens to make math less rigorous/more "questionable".


----------



## qqwref (Jan 7, 2011)

You can always make intuition rigorous if you try, but you can't always make rigor intuitive.


----------



## keemy (Jan 7, 2011)

blah said:


> I was simply defining pi as an integral, which is obviously a constant.


 
Pi is defined as the ratio of the circumference to the diameter (it can also be defined as the ratio of the area to the radius squared). If anything you were taking the definition and then using it to find and analytic expression of pi. It would be like if I defined \( \binom nk \) as \( \frac{n!}{(n-k)!k!} \) when it's definition is the number of ways to choose k elements from a set of n elements or the coefficient of of the \( x^k \) term in the binomial expansion of \( (x+1)^n \)


----------



## malcolm (Jan 7, 2011)

qqwref said:


> Here's a cool one:
> 
> Find a polynomial p(x,y) so that, when x and y are integers, p gives a unique value for each possible value of x and y. So, if you know x and y are integers, and you are given a value of p, you should be able to figure out the single possible (x,y) pair that gives that value.


 

\( P(x,y)=(2x^2+x+1)^2((2y^2+y+1)^2+1) \) works.
Note that \( Q(x)=2x^2+x+1 \) is distinct for each integer \( x \) and is strictly positive, so \( P(x,y) \) is always a positive integer.
Then if \( P(x,y)=p=a^2b \) where \( a,b >0 \) and \( b \) is squarefree, we must have \( a^2=(2x^2+x+1)^2, b=(2y^2+y+1)^2+1 \) and since \( Q(x) \) is distinct for each integer \( x \), we can work out \( x \) and \( y \) from here.


----------



## qqwref (Jan 7, 2011)

malcolm said:


> \( P(x,y)=(2x^2+x+1)^2((2y^2+y+1)^2+1) \) works.
> Note that \( Q(x)=2x^2+x+1 \) is distinct for each integer \( x \) and is strictly positive, so \( P(x,y) \) is always a positive integer.
> Then if \( P(x,y)=p=a^2b \) where \( a,b >0 \) and \( b \) is squarefree, we must have \( a^2=(2x^2+x+1)^2, b=(2y^2+y+1)^2+1 \) and since \( Q(x) \) is distinct for each integer \( x \), we can work out \( x \) and \( y \) from here.


Careful. \( (2y^2+y+1)^2+1 \) isn't necessarily squarefree - if y = -2, it gives 50 (= 2*5*5). I'm not sure if this leads to paired x and y values or not, but it means the approach you used isn't quite perfect.


----------



## blah (Jan 7, 2011)

keemy said:


> Pi is defined as the ratio of the circumference to the diameter (it can also be defined as the ratio of the area to the radius squared). If anything you were taking the definition and then using it to find and analytic expression of pi.


Are you serious? :fp There's a "correct" definition now? If two statements are equivalent, what's wrong with using one as the definition and then proving its equivalence with the other?

I personally like how integrals are rigorously defined from the ground up in calculus. Further manipulation convinces me that the ratios C/2r and A/r^2 are indeed constants. Simple as that.

On the other hand, terms like "rotation" and "translation" don't feel rigorous enough to me, so the fact that all circles are similar is questionable if I were skeptical enough. The implication that ratios of one-dimensional objects should be constant when two shapes are similar, without rigorous justification, is also questionable for an extreme skeptic.

I would like to mention that I *am* perfectly fine with the ratio definition. After all, what's intuition there for?

When someone keeps asking why, people generally label him/her as being insufferable. But the impenetrable truth is what characterizes mathematics. I don't know about you, but if someone asks me what pi is and I give them the ratio definition and they asked the very same questions that I asked, sure I'd be annoyed, but at the same time I'd also feel inadequate, like I don't know my stuff well enough. How else do you think people like Hilbert and Weierstrass laid down the foundations of modern math? Keep asking questions and finding answers, until the most annoying person in the world has no more questions to ask, that's mathematics, that's impenetrable truth.

tl;dr: Giving an answer that you're prepared to defend to the end is imo the better approach to mathematics.


----------



## y3k9 (Jan 7, 2011)

Ok, I am going to ask a question, that is not really a problem but I still like an answer and explanation:


Spoiler



How do you find a complex cube root of a number?


----------



## blah (Jan 7, 2011)

y3k9 said:


> Ok, I am going to ask a question, that is not really a problem but I still like an answer and explanation:
> 
> 
> Spoiler
> ...


What's with the spoiler? First you draw a circle. Find the cube root of its radius. Trisect the angle. Voila.


----------



## einstein00 (Jan 7, 2011)

blah said:


> Intuition is what makes math interesting. It also happens to make math less rigorous/more "questionable".


 
look here please. Not everything in the world can be proven rigorously. Try proving Angle-Angle triangle similarity to see what I mean.




blah said:


> Are you serious? :fp There's a "correct" definition now? If two statements are equivalent, what's wrong with using one as the definition and then proving its equivalence with the other?


 
This is a very serious logic flaw. Pi was first _defined_ as the ratio C/D. Your "equivalent" statement is not a _definition_. You only know that pi is the area of a unit circle because a unit circle has a radius 1, and thus its area is pi * 1^2 = pi. But here, you are using a previous definition of pi. Without the first definition of pi, the area formula would not exist, and thus your "equivalent" statement cannot stand alone as a _definition_.

Edit:

If we define A and use it to prove B, we _cannot_ use B to prove A. We can only assume B is true _because_ it is derived from A. If we wish to use B to prove A, then we must first assume that A is not necessarily true (or else it's circular logic). However, since A is not necessarily true, then B is not necessarily true, either, and the proof breaks down immediately because of a false initial assumption.

This is because logic is like a building - you start with your definitions as the foundation, and then you build upward. You can't use the second floor to "prove" the first floor because the second floor exists _only because_ the first floor exists already.

I'm not sure how else I can explain this, so I hope you understand this analogy.

Edit 2:

Unless you're stating that this is your new definition of pi, not some formula devised depending on the existing historical definition of pi. At this point, I guess you can define anything you want, as long as it all makes sense. No one can argue with a definition. I'm just sticking to the historically first definition of pi being C/D.


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## blah (Jan 7, 2011)

einstein00 said:


> look here please. Not everything in the world can be proven rigorously. Try proving Angle-Angle triangle similarity to see what I mean.


Are you making the assumption that I don't know what axioms are? If you are, then I'm offended.



einstein00 said:


> This is a very serious logic flaw. Pi was first _defined_ as the ratio C/D. Your "equivalent" statement is not a _definition_. You only know that pi is the area of a unit circle because a unit circle has a radius 1, and thus its area is pi * 1^2 = pi. But here, you are using a previous definition of pi. *Without the first definition of pi, the area formula would not exist, and thus your "equivalent" statement cannot stand alone as a definition.* <--- lolololololololololol


Now _this_ whole chunk of text is a very serious logical flaw.

In response to your additional comments that you just edited in, thank you very much for your explanation of a circular argument. I think I know pretty well what that is. But there is nothing circular about defining pi to be the area of a unit circle. I don't know you but from all your explanations and arguments, I'm going to make the assumption that you're still in high school. Is this true? If you're gonna ask why this matters, it's because in short, they lie to you a lot in high school.


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## keemy (Jan 7, 2011)

einstein00 said:


> look here please. Not everything in the world can be proven rigorously. Try proving Angle-Angle triangle similarity to see what I mean.


 
I am loling pretty hard right now.


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## einstein00 (Jan 7, 2011)

blah said:


> Are you making the assumption that I don't know what axioms are? If you are, then I'm offended.


 
Hey hey, no one's trying to offend anyone here. I certainly don't want this to be a flame war. If you already know something I've pointed out, it's wise to simply ignore it.

After thinking some more, I edited the post (a lot). I hope you understand what I'm trying to say. And let's not start a flame war, that just makes life miserable. 



blah said:


> But there is nothing circular about defining pi to be the area of a unit circle.


 
That's cleared up in my edit 2.  problem solved


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## blah (Jan 7, 2011)

I'm not trying to start anything. I'm just saying that you should be more open-minded. Real math isn't what they teach you in high school. Real math isn't even what they teach you in the calculus sequence in college. Just accept that.


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## qqwref (Jan 7, 2011)

blah said:


> On the other hand, terms like "rotation" and "translation" don't feel rigorous enough to me,


In what way are they less rigorous than integrals? In a 2-dimensional system, they are transformations on functions, relations, sets, etc. of points:
Rotation (by angle t): (x,y) -> (x cos(t) - y sin(t), x sin(t) + y cos(t))
Translation (by (a,b)): (x,y) -> (x + a, y + b)
Dilation (by factor r): (x,y) -> (rx, ry)
And I'm pretty sure you can prove invariance of straight and arc lengths by rotation/translation, and length ratios and angles by all rotation/translation/dilation, if you just define them in a mathematically meaningful way.



blah said:


> I don't know about you, but if someone asks me what pi is and I give them the ratio definition and they asked the very same questions that I asked, sure I'd be annoyed, but at the same time I'd also feel inadequate, like I don't know my stuff well enough.


But if your problem is that you don't know your stuff well enough, you'd learn a lot more (and end up with a better answer) by finding a way to rigorously fix the intuitive issues you saw, as opposed to defining pi as something totally different, then using advanced techniques to prove the invariance you wanted to start with.


einstein00: blah was defining pi to be "the area of a unit circle", or more specifically "this integral", which clearly exists, so that wasn't a problem. But it IS awkward to redefine something as the value of an integral just to solve a basic issue you had, especially since historically people were asking "I wonder what is this constant C/d ratio" a huge number of years earlier than people were asking "what is the value of this integral".



blah said:


> Real math isn't what they teach you in high school. Real math isn't even what they teach you in the calculus sequence in college. Just accept that.


Too bad real math is often boring :'(


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## blah (Jan 7, 2011)

I actually sat down and did everything you just did there a couple of hours ago. My issue is that the OP didn't.

Of course, I failed to define an integral and a bunch of other stuff as well, but that's a different story...

As for your second point, yes, that is what I would do. But for now, I don't know enough about euclidean geometry to defend anything; however, I do know enough about calculus/analysis to defend my definition of pi as an integral, hence the preference.


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## einstein00 (Jan 7, 2011)

blah said:


> I'm just saying that you should be more open-minded.


 
I actually am a very open minded person. I just didn't immediately realize that you wanted to use that formula as _your definition_ of pi. My bad. I understand and agree that no one can argue with a definition, because by definition, a definition is defined to be true. Of course, that also means you have to "reprove" everything else based off of this new definition... but that's totally besides the point.

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.


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## keemy (Jan 7, 2011)

next question


Spoiler



how many trolls does it take to define pi?


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## Ranzha (Jan 7, 2011)

keemy said:


> next question
> 
> 
> Spoiler
> ...


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


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## blah (Jan 7, 2011)

einstein00 said:


> 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].


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## aronpm (Jan 7, 2011)

@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.


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## qqwref (Jan 7, 2011)

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



blah said:


> 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.


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## einstein00 (Jan 7, 2011)

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.


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## Rinfiyks (Jan 27, 2011)

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 = ?


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## peterbone (Jan 27, 2011)

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?


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## Rinfiyks (Jan 28, 2011)

peterbone said:


> 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?


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## peterbone (Jan 28, 2011)

Rinfiyks said:


> Spoiler
> 
> 
> 
> ...



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.


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## cmhardw (Jan 28, 2011)

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


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## vcuber13 (Jan 28, 2011)

cmhardw said:


> 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


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## Ordos_Koala (Jan 28, 2011)

cmowla said:


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


 
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


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## Rune (Jan 28, 2011)

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


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## cmhardw (Jan 28, 2011)

vcuber13 said:


> Spoiler
> 
> 
> 
> ...



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


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## Ordos_Koala (Jan 28, 2011)

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.


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## Ordos_Koala (Jan 28, 2011)

cmhardw said:


> 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


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## qqwref (Jan 29, 2011)

cmhardw said:


> 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.


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## whauk (Jan 29, 2011)

Rinfiyks said:


> Spoiler
> 
> 
> 
> ...





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... )


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## cmhardw (Jan 29, 2011)

Ordos_Koala said:


> Spoiler
> 
> 
> 
> ...


 


qqwref said:


> I think this should work:
> 
> 
> Spoiler
> ...


 
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!


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## Ordos_Koala (Jan 29, 2011)

cmhardw said:


> 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.


 
sry, next time, I will  

whauk's problem:


Spoiler



ratio for 2 people is 365:365^2=1:365
so now for three peple, we have again 365 right answers (for each day, i won't count leap-years), but now there are 365^3 possiblities
So from what we see, the formula is 365:365^x
( =1:365^(x-1) in my opinion this one looks better, but it's not so obvious)



for my problem, look at my earlier posts

btw for the next time, could someone pls help me how to write superscripts?


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## whauk (Jan 29, 2011)

Ordos_Koala said:


> whauk's problem:
> 
> 
> Spoiler
> ...


 


Spoiler



your approach is totally wrong.
take a look into http://en.wikipedia.org/wiki/Birthday_problem


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## pr*****t (Jan 29, 2011)

but im unable to understand any mistake in Ordos_Koala's solution of birthday probability problem....... can u explain it to me where is he wrong?


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## MTGjumper (Jan 29, 2011)

He's answering the wrong question. He's answering "given n people, what is the probability that they all share a birthday?". Whauk's question is "given n people, what is the probability that a birthday is shared by x people?" (often x=2)


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## pr*****t (Jan 29, 2011)

MTGjumper said:


> He's answering the wrong question. He's answering "given n people, what is the probability that they all share a birthday?". Whauk's question is "given n people, what is the probability that a birthday is shared by x people?" (often x=2)


ok thanx ... i shud have read the question again


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## Ordos_Koala (Jan 29, 2011)

ok, sorry I didn't understand you properly for the first time...


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## Kynit (Jan 29, 2011)

Hm... (Whaulk's problem)



Spoiler



There are xCn ways to choose x people out of a group of n. This gives you xCn possibilities of finding a matching birthday.

Finding a matching birthday for x people:
365 / 365^x
or 
1 / 365 ^ x-1

Therefore, the chance of finding a matching birthday is xCn / 365^x-1

Does this sound right? I'm not too familiar with the choose function.


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## qqwref (Jan 30, 2011)

cmhardw said:


> 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!





Spoiler



This is less amazing than it might seem. It is precisely *because* the white centers of a normal 4x4 are indistinguishable that the position (ignoring the logo) and the placement of the logo (on which of the 4 white centers) are independent. The position itself can give you no information on the logo placement, and the logo placement itself can give you no information on the position. So solving a logoed 4x4 is not actually any different (as long as you don't care where the logo goes, of course), because the logo neither helps nor hinders a solution. This also means that any set of positions on a normal 4x4 is bijectively related to a set of four times as many positions on a logoed 4x4 - this is true for the number of solved states as well as for the number of positions in total.

Incidentally, the same exact thing is true for many other supercube-like concepts, such as a 3x3 with the orientation of one or more centers marked, or any higher-order supercube. The position and the center placements are independent, so the number of solved positions and the number of total positions will always be multiplied by the same number when you add extra markings.





Ordos_Koala said:


> 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.





Spoiler



There are clearly \( a^3 \) cubes in total. Consider the bigger cube like an a*a*a Rubik's cube, with corners, edges, and centers (as well as hidden interior pieces). The first pile contains the centers, for a total of \( 6(a-2)^2 \) pieces; the second pile contains the edges, for a total of \( 12(a-2) \) pieces; the third pile contains the corners, for a total of 8 pieces; and the last pile contains the hidden pieces, for a total of \( (a-2)^3 \) pieces.

a) The problem here is: \( 9*6(a-2)^2 = 4*(a-2)^3 \), or \( 54 = 4*(a-2) \), but there is no integer solution for a, because 54 isn't divisible by 4.
b) The problem here is: \( 6(a-2)^2 = 3*12(a-2) \), or \( 6(a-2) = 36 \), which means a=8.


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## Ordos_Koala (Jan 30, 2011)

qqwref said:


> Spoiler
> 
> 
> 
> ...



nicely done 



> Kynit
> Hm... (Whaulk's problem)
> 
> 
> ...



read posts before, this isn't what he asked (I thoght he did too  )

whauk's problem v.2 


Spoiler



it's much easier just to count chances that they DON'T have birthday on same day... so for two people it's:
365*364*363*...*(365-n+1)/365^n
it means the first one can be born whenever in year, but the second one can't be born at the same day like the one before him and we go on like this (it's how many possiblities are right for that condition) slash all possiblities including the wrong ones

for three people it's similar:
above in this fraction will be again 365 beacause the first one can be born at any day again, so can be the second, because that condition is broken only when *3* people are born at the same day. So if these two would be born at same day, the third one can't be, so now it's only 364, but it's again same for the fourth, because these three people can make only one day inappropriate... below in fraction, it is same as for 2 people sharing birthday

365*365*364*364*363*.../365^n
(above are n elements)

for any higher number of people for the same birthday there will be more 365s above, but I hope you understand what I mean. I realize there is probably nice equasion with some factorials and no dots, but I don't know much about factorials, so I'd fail somewhere

of course at the end you have to calculate the ratio for that they ARE born at the same day: 1-x=y
where x is probability that they won't (that's what we have counted earilier) and y the number we are seeking


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## Hiero (Jan 31, 2011)

The following problem is from an elementary mental math test we had here in Texas last year. Each problem should take about 5-10 seconds. I know the solution to this but have a hard time seeing how an elementary student would see the answer so quickly, but maybe there is another pattern that I am missing.

"The next number in the sequence 0, 7, 26, 63,...is___________"

Give the next number and the pattern you used. Try not to google the answer.


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## MTGjumper (Feb 1, 2011)

124. I thought it was fairly obvious (hopefully I'm not shooting myself in the foot here). How old is an elementary student?

(Out of interest, could anyone point me in the direction of a question and its solution which concerns a question where there doesn't appear to be enough information to answer it, but with additional information such as "the second brother has a ginger beard" there is a solution. There used to be a similar problem on Shelley's site.)


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## Hiero (Feb 1, 2011)

MTGjumper said:


> 124. I thought it was fairly obvious (hopefully I'm not shooting myself in the foot here). How old is an elementary student?
> 
> (Out of interest, could anyone point me in the direction of a question and its solution which concerns a question where there doesn't appear to be enough information to answer it, but with additional information such as "the second brother has a ginger beard" there is a solution. There used to be a similar problem on Shelley's site.)



It is 124. What was your reasoning? This would be for 9-11 years old.


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## Ordos_Koala (Feb 1, 2011)

if you know that murderer has a ginger beard, who killed this guy?  sry, but I don't really get your point...


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## Rinfiyks (Feb 1, 2011)

Had this optional brain teaser at the end of a question sheet for theory of programming:



Spoiler



There are three gods, Angel, Demon and Chaos. Angel will always answer truthfully, Demon will always lie, and Chaos will answer your questions randomly. The gods are standing before you, but you do not know which is which. Your task is to determine the identity of the gods, by asking them three questions. To make matters worse, the gods will answer your questions only with 'da' or 'ja', which mean true and false in their own language, but you don't know which is which.

Clarifications:
Each question can be directed only at one god, but you may ask two (or three) of your questions to the same god, if you so wish.
The gods, being gods, are all-knowing and perfect logicians.
There's no "trick", like asking them to answer paradoxes.
The three gods all speak the same language.


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## cmhardw (Feb 1, 2011)

I was watching Deal or No Deal a couple days ago, and an interesting probability situation came up. It's very similar to another problem I've asked on here before.

The lady playing had gotten down to only 2 cases, the one with $750,000 and the one with $1,000,000 dollars. Before she was to open her case, Howie gave her the option to switch cases and open the one still on stage rather than the one she picked at the beginning of the game. She had been having such good luck throughout the game that she decided not to switch cases.

Did she make the right decision? Or did it not matter one way or the other whether she swapped cases or not?


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## Keroma12 (Feb 2, 2011)

cmhardw said:


> I was watching Deal or No Deal a couple days ago, and an interesting probability situation came up. It's very similar to another problem I've asked on here before.
> 
> The lady playing had gotten down to only 2 cases, the one with $750,000 and the one with $1,000,000 dollars. Before she was to open her case, Howie gave her the option to switch cases and open the one still on stage rather than the one she picked at the beginning of the game. She had been having such good luck throughout the game that she decided not to switch cases.
> 
> Did she make the right decision? Or did it not matter one way or the other whether she swapped cases or not?


 


Spoiler



Wrong decision. When there are n cases, and you pick one, there is a 1/n chance you are right, and so a (n-1)/n chance you are wrong and it is in another case. Since it wasn't in the cases already opened, there is a (n-1)/n chance she should have swapped. I don't watch this show, but I assume n > 2


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## qqwref (Feb 2, 2011)

Spoiler



Since nobody told the lady what the contents of any given box was until after she opened it, we could have switched the $750,000 and $1,000,000 boxes before the game started, and had the exact same probability of ending up in this situation at the end. So switching is irrelevant, because each box is equally likely to be the $750,000 or the $1,000,000, given that we end up with these two boxes left.


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## cmhardw (Feb 2, 2011)

qqwref said:


> Spoiler
> 
> 
> 
> Since nobody told the lady what the contents of any given box was until after she opened it, we could have switched the $750,000 and $1,000,000 boxes before the game started, and had the exact same probability of ending up in this situation at the end. So switching is irrelevant, because each box is equally likely to be the $750,000 or the $1,000,000, given that we end up with these two boxes left.


 


Spoiler



Michael, my hat's off. I initially thought of this as a Monty Hall style problem, but I can see the flaw in that now because of the same probability of ending up with the same two cases at the end even if they were swapped.

I'm imagining a different version of Deal or No Deal where there are 26 cases, one case contains $1,000,000 and the other 25 are empty. Imagine in this game if you pick a case at the beginning, then proceed to open 24 empty cases. At this point Howie asks if you would like to swap cases or not. Since you are down to two cases, one empty and one the $1,000,000 I wonder if this works the same way.

My impression is to try to calculate it.

The probability of winning the $1,000,000 case assuming you do not switch would be 1/26 since you open the case that you initially picked.

The probability of winning the $1,000,000 assuming that you do switch is:
\( \frac{25}{26} * \frac{24}{25} * \frac{23}{24} * \ldots * \frac{2}{3} * \frac{1}{2} = \frac{1}{26} \)

Wow, so it does not matter whether you switch or not even in this variation, the probability of winning the million is still 1/26. Very neat result! I definitely got it wrong when watching the show. I thought she should have switched. Cool, you learn something new every day.


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## qqwref (Feb 2, 2011)

cmhardw said:


> Spoiler
> 
> 
> 
> I'm imagining a different version of Deal or No Deal where there are 26 cases, one case contains $1,000,000 and the other 25 are empty. Imagine in this game if you pick a case at the beginning, then proceed to open 24 empty cases. At this point Howie asks if you would like to swap cases or not. Since you are down to two cases, one empty and one the $1,000,000 I wonder if this works the same way.





Spoiler



Once you've managed to open 24 empty cases, you are indeed less likely to have chosen the million dollars than in the beginning - in 25 of 26 possibilities the million dollar case will be in the last of the 25 non-chosen ones, but in only 1 possibility will you have chosen the million dollars. Of course, you're a lot more likely to have opened 24 empty cases if you started with the million, but once you've done it the probability of the past event isn't relevant anymore.

The difference between this version and the version you saw on the real show was that the $750,000 and $1,000,000 cases function identically, so neither is biased against by the act of removing every other case. In this example, the $1,000,000 case and empty case you are left with are NOT identical, because there were more cases just like the empty one at the start.


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## LewisJ (Feb 2, 2011)

Hiero said:


> It is 124. What was your reasoning? This would be for 9-11 years old.


 
The differences between terms are 7, 19, 37; the differences between these are 12 and 18; the difference between 37 and the next increased amount is thus 24, so the next increased amount is 61, which added to 63 gets 124. This is not a 5-10 second problem for 9-11 year olds, but it is certainly solvable within a minute or two by a determined individual in that age group.


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## JonWhite (Feb 2, 2011)

LewisJ said:


> The differences between terms are 7, 19, 37; the differences between these are 12 and 18; the difference between 37 and the next increased amount is thus 24, so the next increased amount is 61, which added to 63 gets 124. This is not a 5-10 second problem for 9-11 year olds, but it is certainly solvable within a minute or two by a determined individual in that age group.


 
I believe I know exactly where this problem came from. If I am correct, then this problem must be solved mentally without any scratchwork. It's possible, but indeed very hard to do it this way without scratchwork. I'm also pretty sure there's another simpler way to solve this problem.



cmhardw said:


> I was watching Deal or No Deal a couple days ago, and an interesting probability situation came up. It's very similar to another problem I've asked on here before.
> 
> The lady playing had gotten down to only 2 cases, the one with $750,000 and the one with $1,000,000 dollars. Before she was to open her case, Howie gave her the option to switch cases and open the one still on stage rather than the one she picked at the beginning of the game. She had been having such good luck throughout the game that she decided not to switch cases.
> 
> Did she make the right decision? Or did it not matter one way or the other whether she swapped cases or not?


 
I haven't watched this show, so I have a question: Did the lady pick which cases to open, and thus was lucky (or unlucky?) to not have opened the million dollar case, or did the host open cases for her, intentionally picking cases that did not contain the million, and thus the lady was "fated" to be left with the million unopened?


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## qqwref (Feb 2, 2011)

I'd consider it a stretch to say that the sequence "12, 18, ..." is obviously followed by 24. I mean, sure, that's arguably the most obvious choice, but at that point you only just have enough information to even figure out an arithmetic series, which means it's a pretty iffy solution. At least "perfect cube minus one" gives you enough terms to verify that the idea isn't coming out of nowhere.

(Incidentally, I gotta say that I'm not fond of "follow the sequence" questions at all, just because there are many possibilities - any sufficiently difficult one which doesn't give a huge number of terms will have more than one plausible solution.)


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## LewisJ (Feb 2, 2011)

qqwref said:


> I'd consider it a stretch to say that the sequence "12, 18, ..." is obviously followed by 24. I mean, sure, that's arguably the most obvious choice, but at that point you only just have enough information to even figure out an arithmetic series, which means it's a pretty iffy solution. At least "perfect cube minus one" gives you enough terms to verify that the idea isn't coming out of nowhere.
> 
> (Incidentally, I gotta say that I'm not fond of "follow the sequence" questions at all, just because there are many possibilities - any sufficiently difficult one which doesn't give a huge number of terms will have more than one plausible solution.)


 
I feel really stupid now for not noticing the perfect cubes part. I did feel the 12, 18, therefore 24 thing was a stretch, but it fit the solution. 

Although I sort of share your sentiment about sequence problems, I have a rather reasonable one:
Find the next term in
2, 3, 10, 21, 55, 104, 221, 399, ___


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## scylla (Feb 2, 2011)

sorry I don't know how the spoiler function works



> Michael, my hat's off. I initially thought of this as a Monty Hall style problem, but I can see the flaw in that now because of the same probability of ending up with the same two cases at the end even if they were swapped.
> 
> I'm imagining a different version of Deal or No Deal where there are 26 cases, one case contains $1,000,000 and the other 25 are empty. Imagine in this game if you pick a case at the beginning, then proceed to open 24 empty cases. At this point Howie asks if you would like to swap cases or not. Since you are down to two cases, one empty and one the $1,000,000 I wonder if this works the same way.
> 
> ...



Yes it doesn't matter in this case. The problem looks very like the 3-doors problem (in which switching is better) but it's different at 1 point.

In the 3-doors problem the quizmaster opens a door without the jackpot (because he knows where the jackpot is), in this problem the person chooses the boxes to be opened without knowing where the jackpot is. That leaves just a simple propability calculation that every price has the same propability to be in every box. So the choice is indifferent. 

This problem I invented by myself in a pub when I was playing with (round) coasters:

Suppose you have three coasters with a diameter of 1. What is the biggest circle you can totally cover with those 3 coasters?


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## keemy (Feb 2, 2011)

Lewis:


Spoiler



782 (nth prime)*(nth Fibonacci)


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## Rinfiyks (Feb 2, 2011)

Rinfiyks said:


> Had this optional brain teaser at the end of a question sheet for theory of programming:
> 
> 
> 
> ...


 
Hint


Spoiler



You don't need to know what 'da' and 'ja' mean.


Bigger hint


Spoiler



You can word a question in such a way that if asked to Angel or Demon you will know if it is true or false without knowing da or ja.


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## Ordos_Koala (Feb 3, 2011)

qqwref said:


> Spoiler
> 
> 
> 
> ...


 


Spoiler



you're wrong... at first you have chance 1/26. But when you open first case, your chance incrase to 1/25 since there are now only 25 cases... at the end, it doesn't matter wheter you opened million, twenty-four or any of these empty cases, your chance to win at the end is 1/2
What you tink is common mistake...


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## cmhardw (Mar 11, 2011)

This seemed the best place to post this question without starting a new thread.

What would your next move be in this sudoku, and more importantly why?









Spoiler



If we label the columns A-I going from left to right and the rows 1-9 going from bottom to top, then here is how I decided my next move:

I placed a "3" in cell I6, then continued to solve the puzzle correctly. I don't like how I reasoned that a 3 belonged here, it feels a bit like cheating.

From earlier studying of the puzzle I found that in the upper left 3x3 grid that the 6 and 8 must be in cells B8 and B9. This will be useful information in a moment. After some time, I came to study where the 3s could go in the puzzle as a whole. All possible options as seen on the picture are:

A2, A4, A6, B2, B4, I4, I6

I noticed that if there was a 3 in cell I4 that it would force the position of all the other 3s. At this point I decided to search to see if this would lead to a contradiction. I always do this mentally, and I never write down the possible values that would result from "guessing" a number like this. I consider guessing like this cheating in a sense, so I prefer to limit myself to what my short term memory can handle, rather than writing down what happens.

If I4 is a 3, then so too will be cells A6 and B2. Now, cells A2, B2, C2 must be the 3,5,8 in the lower left 3x3 grid. With this extra information I can reason that cell A2 must be an 8 and cell C2 must be a 5. However, let's look at the entirety of row 4. If cell I4 is a 3, then I could reason that since cells B8 and B9 are the 6 and 8, then cell A4 must be an 8. This places the 8 in two different places in column A. This is a contradiction. Since the 3 can only occupy either I4 or I6 in column I, then I knew I must play it to cell I6.

I don't like what I did here, it feels like cheating. How _should_ I have decided what number to play next?


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## qqwref (Mar 11, 2011)

Spoiler



Your solution does make sense - another way to see it is that I4=3 forces the rest of column 4 (B4=4, E4=6, A4=8), and A4=8 forces A2=3, but then there's no place to put 3 in column B.

Alternate way to get a clue, using your notation:
- Notice that the only places on row 8 that can hold the 2 are G8 and I8. This means cell I9 can't be a 2, and since it can't be 1,3,4,5,6,9 either it is either 7 or 8.
- In the bottom-right area, row 3 tells us that H3 and I3 are both either 5 or 6, and this means that G1 and I1 are both either 7 or 8.
- So now we have I1 and I9 are both either 7 or 8. This eliminates the possibility of being an 8 from cell I4. But now the only place on row 4 that an 8 can be is A4, so there you go. Then you can solve A2 and go from there.

BTW, the logic you (could have) used to get that B8 and B9 were 6 and 8, if you forgot: B8 can be 6 or 8 already, but B9 can be 5, 6, or 8. However, B7 and C7 are the only cells in row 7 that can be a 5, so B9 can't be a 5.


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## cmhardw (Mar 11, 2011)

Spoiler






qqwref said:


> Alternate way to get a clue, using your notation:
> - Notice that the only places on row 8 that can hold the 2 are G8 and I8. This means cell I9 can't be a 2, and since it can't be 1,3,4,5,6,9 either it is either 7 or 8.
> - In the bottom-right area, row 3 tells us that H3 and I3 are both either 5 or 6, and this means that G1 and I1 are both either 7 or 8.
> - So now we have I1 and I9 are both either 7 or 8. This eliminates the possibility of being an 8 from cell I4. But now the only place on row 4 that an 8 can be is A4, so there you go. Then you can solve A2 and go from there.



I see this now. I had already noticed that G1 and I1 were the 7 and 8 in the lower right grid, but I never ruled out the 2 as a possibility for cell I9. I like this line of logic better, and I see how the rest follows from there to get A4 as an 8 (assuming you remember that B8 and B9 are the 6 and 8).


 
Thanks Michael!


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## qqwref (Mar 28, 2011)

Here's an interesting problem: How many playable chords are there on a piano, for a typical person, if they only use one hand? Okay, I know, that's really vague. Let's approximate, in rough order of difficulty/accuracy:

First approximation: A piano has 88 keys. How many possible chords are there if you can play any combination of keys? (Two combinations of the same keys in a different order count as the same.)

Second: A typical person can reach about an octave (two keys separated by exactly 11 keys in between), so let's use that. How many chords are there that are no more than an octave wide?

Third: A typical person has five fingers on each hand, and will only press one key with each finger. How many chords from part 2 have 5 or fewer keys pressed?


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## mr. giggums (Mar 28, 2011)

Spoiler



There are 88 ways to get one key
87+86+85+84+83+82+81+80+79+78+77+76=978 ways for 2 keys
86+2*85+3*84+4*83+5*82+6*81+7*80+8*79+9*78+10*77=4400 for 3
85+3*84+6*83+10*82+15*81+21*80+28*79+36*78=9570 for 4
84+4*83+10*82+20*81+35*80+56*79=10782 for 5
88+978+4400+9570+10782=25818
My answer is 25818 combinations


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## Rinfiyks (Mar 28, 2011)

qqwref said:


> Here's an interesting problem: How many playable chords are there on a piano, for a typical person, if they only use one hand? Okay, I know, that's really vague. Let's approximate, in rough order of difficulty/accuracy:
> 
> First approximation: A piano has 88 keys. How many possible chords are there if you can play any combination of keys? (Two combinations of the same keys in a different order count as the same.)





Spoiler



Pascal's triangle 
Take 4 keys on a piano as an example. 4th row on Pascal's triangle (if the first row is the 0th row) is 1 4 6 4 1
1 way to play a chord with 0 keys, 4 ways to play a chord with 1 keys, 6 ways to play a chord with 2 keys, 4 ways to play a chord with 3 keys, 1 way to play a chord with 4 keys.
The sum of the nth row is 2^n.
But a chord with 0 keys doesn't really count 
So it's 2^88 - 1.





> Second: A typical person can reach about an octave (two keys separated by exactly 11 keys in between), so let's use that. How many chords are there that are no more than an octave wide?





Spoiler



For the first 11 notes, 2^11 - 1 chords.
Then if you move 1 note to the right, you need to add another 2^11 - 1, but discount all the chords that don't include the rightmost note, because they have been covered by the first 11 notes. The number of chords that don't include the rightmost note is 2^10 - 1. Then we move 1 to the right again, and more discounting by the same amount, and so on, right to the end. So you subtract (88 - 5)(2^10 - 1) from the final answer of (88 - 4)(2^11 - 1).
(88 - 4)(2^11 - 1) - (88 - 5)(2^10 - 1) = 87039.
I don't know if this is right so I won't attempt the final question until I know.





> Third: A typical person has five fingers on each hand, and will only press one key with each finger. How many chords from part 2 have 5 or fewer keys pressed?





Spoiler



I don't know if I got part 2 right, but I'll assume I did 
If we're limiting it to 5 keys, then we only add up the first 5 (not including the 1 at the start) values on the 11th row of Pascal's triangle.
11+55+165+330+462 = 1023 = 2^10 - 1.
To work out how many we discount, add up the first 5 values (ex. 1) on the 10th row.
10+45+120+210+252 = 637.
So we have
(88 - 4)(2^10 - 1) - (88 - 5)(637) = 33061.
I hope.


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## Keroma12 (Apr 13, 2011)

Problems I'm stuck on:

\( \int ln(sin(x))dx \)

\( \int ln(cos(x)+1)dx \)

My attempt:


Spoiler



For the first, I said \( sinx=u \) and \( cos(x)dx=du \). This eventually leads to \( \int \frac{ln(u)du}{sqrt[1-u^2]} \), but this doesn't really help as far as I can see.

Also, how do I make a square-root symbol?


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## Sa967St (Apr 13, 2011)

Here are two questions from a math contest I did earlier today that I had trouble with.
1.





2.
A 75 year old has a 50% probability of living for another 10 years.
A 75 year old has a 20% probability of living for another 15 years.
An 80 year old has a 10% probability of living for another 10 years.
What is the probability that an 80 year old will live for another 5 years?


My fail attempt at the first one:


Spoiler



lol. I didn't get much further than this.






I had a few answers for the second one. This one was my best guess: 


Spoiler



1.5(.50/.20)= 2(x/.10)
x=0.1875 or 18.75%


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## Keroma12 (Apr 13, 2011)

Sa967St said:


> Here are two questions from a math contest I did earlier today that I had trouble with.
> 1.


 
I did the same contest.


Spoiler



From where you got to, replace \( 2^{logx} \) with \( n \) and you have a quadratic: \( n^2-16=6n \)
So \( (n-8)(n+2)=0, n=8 \) or \( n=-2 \)
Then \( 2^{logx}=8 \) (the other has no real solutions)
So \( logx=3, x=1000 \)


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## cmhardw (Apr 13, 2011)

Keroma12 said:


> Problems I'm stuck on:
> 
> \( \int ln(sin(x))dx \)
> 
> ...


 


Spoiler



I tried the same thing you did, as well as some integration by parts and I never got anywhere. I found this interesting though, perhaps this may be useful?

\( \int ln(sin(x))dx = \int \frac{1}{2}ln\left[sin^2(x)\right]dx \)
\( \int ln(sin(x))dx = \frac{1}{2} \int ln\left[1-cos^2(x)\right]dx \)
\( \int ln(sin(x))dx = \frac{1}{2} \int ln\left[(1+cos(x))(1-cos(x))\right]dx \)
\( \int ln(sin(x))dx = \frac{1}{2}\left[\int ln(1+cos(x))dx + \int ln(1-cos(x))dx\right] \)

I don't see where to go from here, unless we could figure out the right hand side integrals relatively easily. I will try to keep looking into this as well. This is interesting.



P.S.

\( \sqrt{5} \)

\( \sqrt[n]{(x+1)^{2n+1}} \)


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## Sa967St (Apr 13, 2011)

Keroma12 said:


> I did the same contest.
> 
> 
> Spoiler
> ...


Wow, I definitely went the wrong direction with this. I tried stuff like taking the log of both sides, which got me no where. I didn't even think of changing 3(2^(1+logx)) to 3(2^1)(2^logx)=6(2^logx). :fp Thanks. 
My math teacher couldn't even get it btw.


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## qqwref (Apr 13, 2011)

Sa967St said:


> 1.





Spoiler



Let \( y = \log_{10}x \). Then the equation can be simplified to \( 2^{2y} = 6 * 2^y + 16 \). Now let \( z = 2^y \) and this becomes \( z^2 = 6z + 16 \), which is a quadratic, so we get that z must be -2 or 8, and -2 doesn't make sense, so z = 8. Then y = 3 and x = 1000.


.



Sa967St said:


> 2.
> A 75 year old has a 50% probability of living for another 10 years.
> A 75 year old has a 20% probability of living for another 15 years.
> An 80 year old has a 10% probability of living for another 10 years.
> What is the probability that an 80 year old will live for another 5 years?





Spoiler



Wait a minute. 20% of 75-year-olds live to age 90, but only 10% of 80-year-olds do? That means, for every living 90-year-old, there were 10 living 80-year-olds 10 years ago, but only 5 living 75-year-olds 15 years ago. How do you get to be 80 without getting to be 75 first?


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## Keroma12 (Apr 13, 2011)

cmhardw said:


> Spoiler
> 
> 
> 
> ...


 
Thanks for the square root. Well we've only done u substitutions at school so I don't know many techniques. Actually if you could find the sum of the two, if each own its own is too difficult, that would be just as useful.


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## Sa967St (Apr 13, 2011)

qqwref said:


> Spoiler
> 
> 
> 
> Wait a minute. 20% of 75-year-olds live to age 90, but only 10% of 80-year-olds do? That means, for every living 90-year-old, there were 10 living 80-year-olds 10 years ago, but only 5 living 75-year-olds 15 years ago. How do you get to be 80 without getting to be 75 first?


Yeah I have no idea. The wording is lol.


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## cmhardw (Apr 13, 2011)

Sa967St said:


> 2.
> A 75 year old has a 50% probability of living for another 10 years.
> A 75 year old has a 20% probability of living for another 15 years.
> An 80 year old has a 10% probability of living for another 10 years.
> What is the probability that an 80 year old will live for another 5 years?



I know that what I am doing is not working, because I am getting a bogus answer, but here is what I did:


Spoiler



Bayes' Theorem approach

\( p(A|B)=\frac{p(B|A)p(A)}{p(B)} \)

A = A 75 year old lives to 85
B = A 75 year old lives to 80

p(A|B) = probability that a 75 year old lives to 85 given that they lived to 80 (what the problem is asking)
p(B|A) = probability that a 75 year old lives to 80 given that they lived to 85 = 1
p(A) = 0.5
p(B) = unknown

\( p(A|B)=\frac{p(B|A)p(A)}{p(B)} \)
\( p(A|B)=\frac{1*0.5}{p(B)} \)
\( p(A|B)=\frac{1}{2*p(B)} \)

Now examining p(B) we will examine how people live to 90.

A 75 year old has a 20% chance to live to 90.
an 80 year old has a 10% chance to live to 90.

Imagine there are N (N>0) 75 year olds in a group. 0.2N will survive to 90.

OR

some unknown portion, p(B), of them will survive to 80. Then, 10% of that group will live to 90.

So in an original group of 75 year olds, 0.2N of them live to 90 and N*p(B)*0.1 live to 90.

\( 0.2N = N*p(B)*0.1 \)
\( 2=p(B) \)

Which is clearly bogus. I'm thinking that I am either somehow not working the partition of people correctly, and am overcounting the number that survive, or otherwise I am not sure where I am going wrong.

That's my stab at it so far.


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## Keroma12 (Apr 13, 2011)

Sa967St said:


> 2.
> A 75 year old has a 50% probability of living for another 10 years.
> A 75 year old has a 20% probability of living for another 15 years.
> An 80 year old has a 10% probability of living for another 10 years.
> What is the probability that an 80 year old will live for another 5 years?


 


Spoiler



Not sure I'm right, but I did:
P(75-->90)=P(75-->85)xP(85-->90)
P(80-->90)=P(80-->85)xP(85-->90)
So P(75-->90)xP(80-->85)=P(75-->85)xP(80-->90)
So [.2]x[n]=[.5]x[.1]
So n=.25


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## yamahammer08 (Apr 13, 2011)

This thread makes me feel stupid. I always used to be good at math, well at least I thought I was, till I hit college. I just stared at Integral (t+1)/t for a good 10 minutes before I realized what to do haha


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## cmhardw (Apr 13, 2011)

Keroma12 said:


> Spoiler
> 
> 
> 
> ...


 


Spoiler



Keroma,

I essentially did the same thing, but this result means that 200% of 75 year olds live to be 80, the same error I ran into.

Using your notation:
p(75-->80)*p(80-->85)*p(85-->90)=0.2

Now we need p(85-->90) which we will get from:
p(80-->90)=p(80-->85)*p(85-->90)

Rearranging we get:
p(80-->90) / p(80-->85) = p(85-->90)

Now that we have calculated that p(80-->85)=0.25 we can substitute:
0.1/0.25 = p(85-->90)
2/5 = p(85-->90)

Now going back to:
p(75-->80)*p(80-->85)*p(85-->90)=0.2

We will solve for p(75-->80)

p(75-->80) = 0.2 / [p(80-->85)*p(85-->90)] 
p(75-->80) = 0.2 / [0.25 * 2/5]
p(75-->80) = 2

So 200% of all 75 year olds survive to 80. Or, the probability of being 75 and living to 80 is 200%. Take a look at my post for another take on the same approach, but I run into the same error.






yamahammer08 said:


> This thread makes me feel stupid. I always used to be good at math, well at least I thought I was, till I hit college. I just stared at Integral (t+1)/t for a good 10 minutes before I realized what to do haha


 
Reading posts by Michael, Lucas, and Bruce sometimes makes me feel stupid as well. It's all relative. Math is very much about practice.

"An expert is a person who has made all the mistakes that can be made in a very narrow field." -Niels Bohr

The only way to make those mistakes is to practice it, and making the mistakes is the process of learning it.


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## Rinfiyks (Apr 13, 2011)

Rubik's cube inspired problem:

Imagine looking at a Rubik's cube from above - you'd just see the U face. But if you rotated the U face less than 90 degrees, you'd get something that looks like this:




What is the *formula for the area* of the black part of the image, with the angle the U face is rotated as a variable?


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## MaeLSTRoM (Apr 13, 2011)

Things we know
At all multiples of 90, the area is 0.
the area of the black section is 4x the area of each triangle, since each one would be identical.
The path traveled by the cube is a circle.

Calculating the area


Spoiler



We know that the area starts at (0,0) [angle,area]
It then goes to (45,max)
and then to (90,0)

I would then say that the area = 
\( (sin(x).cos(x)) \) where x is the angle. 
Thiis is true when the length of the cube is 1 becuase it would be 4 lost of the above formula, one for each triangle. Then you would have to divide by 4 becuase it would have to be less than 1 (the area never exceeds the area of the cube)


Edit: Yeah i think this is wrong...


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## qqwref (Apr 13, 2011)

Interesting idea, Rinfiyks.



Spoiler



There are 4 black triangles, so we'll calculate the area of one of them and multiply by 4. Assume (for convenience) that the face is rotated clockwise by t degrees (t<90) and that the length of each side of the face/cube is 1 unit. I can't think of any really clever way to do this right now, so let's use a coordinate approach: center is (0,0), bottom right is (.5, -.5), and top right is (.5, .5).

Now, let's look at a triangle, say the bottom-right one. The triangle is defined by three lines: x=0.5, y=-0.5, and this line that goes between the top right corner rotated by t degrees, and the bottom right corner rotated by t degrees. So that third line goes through these two points:
(.5 cos(t) + .5 sin(t), -.5 sin(t) + .5 cos(t))
(.5 cos(t) - .5 sin(t), -.5 sin(t) - .5 cos(t))
The slope of this line (defined by (y2-y1)/(x2-x1)) is cot(t), and using the equation y = cot(t) x + b we can get that b = -.5 csc(t).

So our third line should be y = cot(t) x - .5 csc(t). It intersects the two other lines at the points (.5, .5 cot(t) - .5 csc(t)) and (-.5 tan(t) + .5 sec(t), -.5). This means that the height of the triangle is (.5 cot(t) - .5 csc(t) + .5) and the width is (.5 + .5 tan(t) - .5 sec(t)). The area of the triangle is (h*w)/2, or, in simplified form:
1/4 (1 - csc(t))(1 - sec(t)).

So the area of the entire black space would be 4 times that, or (1 - csc(t))(1 - sec(t)). A little icky, but it does have the appropriate property (from 0 to 90 degrees) of symmetry with the ends at 0. The maximum (at 45 degrees) evaluates to 3 - 2 sqrt(2), or about 0.172.


----------



## Carrot (Apr 13, 2011)

Sa967St said:


> 2.
> A 75 year old has a 50% probability of living for another 10 years.
> A 75 year old has a 20% probability of living for another 15 years.
> An 80 year old has a 10% probability of living for another 10 years.
> What is the probability that an 80 year old will live for another 5 years?


 
so..


Spoiler



75->85=0.5
75->90=0.2
80->90=0.1
85->90=[75->90]/[75->85]=0.2/0.5=0.4
75->80=[75->90]/[80->90]=0.2/0.1=2 wtf?? (why are people cloning themself ??)
80->85=[75->85]/[75->80]=0.5/2=0.25


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## StachuK1992 (Apr 13, 2011)

Keroma12 said:


> Problems I'm stuck on:
> 
> \( \int ln(sin(x))dx \)
> 
> \( \int ln(cos(x)+1)dx \)


 Gah. I've tried both of these, for so long now. I can't figure them out.


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## qqwref (Apr 13, 2011)

Keroma12 said:


> Problems I'm stuck on:
> 
> \( \int ln(sin(x))dx \)
> 
> \( \int ln(cos(x)+1)dx \)


Not to spoil your fun, but you should probably stop trying:


Spoiler



WolframAlpha's solutions for both of these involve the Dilogarithm function, which strongly suggests that no solution with only simple functions (ln, trig, arc-trig) exists.


----------



## MaeLSTRoM (Apr 13, 2011)

StachuK1992 said:


> Gah. I've tried both of these, for so long now. I can't figure them out.



The first is 


Spoiler



\( -tan(x)(ln(sin(x))-1) \)
This is becuase \( \int ln(x) \) \( = x(ln(x)-1) \)
substitute in \( sin(x) \) and use inverse chain rule to get:
\( (sin(x)/-cos(x))(ln(sin(x)-1)) \) which then simplifies.


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## Deluchie (Apr 13, 2011)

t=-0.5x2 + 24x + 13, for x in terms of t


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## Keroma12 (Apr 14, 2011)

qqwref said:


> Not to spoil your fun, but you should probably stop trying:
> 
> 
> Spoiler
> ...


 


Spoiler



Ok, well it says \( -\int \frac{ln(1-x)}{x} = Li2(x) \) so if we can split it up so that it contains a \( \frac{ln(1-x)}{x} \) as one term, it should be possible.


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## qqwref (Apr 14, 2011)

MaeLSTRoM said:


> The first is
> 
> 
> Spoiler
> ...


I don't think you're allowed to do this.


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## caseyd (Apr 14, 2011)

you know, I used to like math
but this, this
ugh


----------



## Kynit (Apr 16, 2011)

I had a math competition today and one of the more difficult problems was this:

Find the sum of the coefficients of all odd powers of x in the expansion of:
\( (x^2 + 2x + 3)^3 - (x^2 - 2x - 3)^3 \)


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## Rinfiyks (Apr 16, 2011)

Kynit said:


> I had a math competition today and one of the more difficult problems was this:
> 
> Find the sum of the coefficients of all odd powers of x in the expansion of:
> \( (x^2 + 2x + 3)^3 - (x^2 - 2x - 3)^3 \)


 


Spoiler



In general,
\( (a+b+c)^3 = (a^3 + b^3 + c^3) + 3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) + 6abc \)
I can't think of a clever way to go from here other than analyse each term and substitute in your head. I'll write a 0 if it's not an odd power of x. If it is, I'll write the coefficient. Starting from the left.
First bracket: 0 + 8 + 0 + 3(2 + 0 + 0 + 0 + 0 + 18) + 36
Second bracket: 0 - 8 + 0 + 3(- 2 + 0 + 0 + 0 + 0 - 18) + 36
So... (8 + 60 + 36) - (-8 - 60 + 36)
= 136


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## Kynit (Apr 16, 2011)

Rinfiyks said:


> Spoiler
> 
> 
> 
> = 136


That looks right to me! I can't remember the answer exactly but that sounds familiar. I think I see where I went wrong in this now.


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## Rinfiyks (Apr 16, 2011)

Kynit said:


> That looks right to me! I can't remember the answer exactly but that sounds familiar. I think I see where I went wrong in this now.


 
If you know how to expand (a+b)^3, you can substitute to expand (a+b+c)^3:


Spoiler



\( (a+b+c)^3 \)
Let \( b + c = d \)
\( (a+d)^3 \)
\( = a^3 + d^3 + 3(a^2d + d^2a) \)
Then substitute d out and simplify.
\( = a^3 + (b+c)^3 + 3(a^2(b+c) + (b+c)^2a \)
\( = a^3 + (b^3 + c^3 + 3(b^2c + c^2b)) + 3(a^2b+a^2c + (b^2+c^2 + 2bc)a) \)
\( = a^3 + b^3 + c^3 + 3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) + 6abc \)
Lots of places to make a mistake though (especially when typing it up, which I did, and took me a while to find it )


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## Rinfiyks (Apr 16, 2011)

I have another problem!
There are two poles in the ground. They are both vertical. The height of both poles (from the ground) is 100 metres. A 50 metre rope is attatched to the top of both poles at either end. It is allowed to hang freely. The rope's closest point to the ground is 75 metres from the ground. What is the distance between the poles?


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## qqwref (Apr 17, 2011)

Hehe.



Spoiler



In general this would be a pretty hard problem. But we see that the rope's lowest point (75m elevated) is 25m from its highest point (100m elevated), and its length is only twice that or 50m, so it must be completely doubled up. The poles are touching.




Here's one: If we took a solid regular dodecahedron, and cut it (using only planar cuts that go all the way through the dodecahedron) like a Megaminx, how many pieces would there be? How about a Gigaminx? Pyraminx Crystal? Starminx? Pentultimate?


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## Sa967St (May 10, 2011)

Sa967St said:


> 2.
> A 75 year old has a 50% probability of living for another 10 years.
> A 75 year old has a 20% probability of living for another 15 years.
> An 80 year old has a 10% probability of living for another 10 years.
> What is the probability that an 80 year old will live for another 5 years?


Bump. 

I got the numbers in the question wrong. I just did it from memory, which is why people's answers weren't making any sense. :/

The original question was:
A 75 year old person has a 50% chance of living at least another 10 years.
A 75 year old person has a 20% chance of living at least another 15 years.
An 80 year old person has a 25% chance of living at least another 10 years.
What is the probability that an 80 year old person will live at least another 5
years? (question 7a)

And the answer: 62.5,clicky


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## Rinfiyks (May 12, 2011)

The diagram shows a square with side length 1. It's divided into 4 rectangles which all have the same area. Find x.


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## MaeLSTRoM (May 12, 2011)

Got It!


Spoiler



Area of each rectangle = 0.25
height of bottom rectangle = 0.25 (1*0.25)
there fore remaining area = 0.75
Left rectangle is full height, so width is 1/3 (0.25/0.75)
Width 2 rectangles on right is 2/3
x=2/3 (same as width of rectangles on right)


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## Igora (May 12, 2011)

Interestingly I got the same thing using a different method:


Spoiler


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## Rinfiyks (May 12, 2011)

Well done  now for a more difficult one.

The diagram shows two semicircles inside a square of side length 1. The common centre of both semicircles lies on a diagonal of the square.
What is the shaded area?


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## qqwref (May 13, 2011)

I don't get it, what is stopping me from constructing a circle (cut in half) inside the square and saying that circle satisfies the problem? Are there any added constraints you didn't mention?


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## Rinfiyks (May 13, 2011)

qqwref said:


> I don't get it, what is stopping me from constructing a circle (cut in half) inside the square and saying that circle satisfies the problem? Are there any added constraints you didn't mention?


 
The large semicircle must have its *corners touching two adjacent edges* of the square, and its *arc must also touch two adjacent edges* of the square. There's only one way to acheive this (that I can see.) The smaller one fits snugly in the remaining gap. Sorry for not making it clear.


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## samehsameh (May 13, 2011)

Spoiler



Area =3Pi/2 * ( 3 - 2sqrt(2) )
or 0.8085 if u prefer numerical values


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## qqwref (May 14, 2011)

Rinfiyks said:


> The diagram shows two semicircles inside a square of side length 1. The common centre of both semicircles lies on a diagonal of the square.
> What is the shaded area?





Spoiler



Call the radii of the two circles R (bigger) and r (smaller). Also, let A be the common center, B be the tangent point between the smaller semicircle and top line, and C be the intersection point of the larger semicircle's top corner with the top edge. Let D be the tangent point between the larger semicircle and the bottom line.

Now, a circle tangent to a horizontal line has its center directly above its point of tangency. This means that B, A, and D are on the same vertical line (and thus BD=1); since AB=r and AD=R, we have R+r=1. This also means that the angle ABC is a right angle. Now, by symmetry across the top-left/bottom-right diagonal, angle BCA must be a 45 degree angle. So triangle ABC is a right isosceles triangle. But AC=R (and again AB=r); this means R = r * sqrt(2).

So we have R+r=1 and R = r * sqrt(2). We can solve to find that r = 1/(sqrt(2)+1) = sqrt(2)-1 and R = 1-r = 2-sqrt(2). The shaded area is 1/2 pi r^2 + 1/2 pi R^2 = 1/2 pi (9-6sqrt(2)) ~= 0.80852.


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## uberCuber (Jun 2, 2011)

Let \( P(a, a^2) \) be a point on the parabola \( y = x^2, a>0 \). Let \( O \) be the origin and \( (0,b) \) the y-intercept of the perpendicular bisector of line segment \( OP \). Find \( lim_{P \to O} b. \)


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## Akash Rupela (Jun 2, 2011)

uberCuber said:


> Let \( P(a, a^2) \) be a point on the parabola \( y = x^2, a>0 \). Let \( O \) be the origin and \( (0,b) \) the y-intercept of the perpendicular bisector of line segment \( OP \). Find \( lim_{P \to O} b. \)


 


Spoiler



Slope of OP= (a^2-0)/(a-0)=a
Thus slope of perpendicular bisector = -1/a
Also perpendicular bisector passes through midpoint of OP , i.e (a/2 , a^2/2)
so equation of perpendicular bisector is (y-a^2/2)/(x-a/2)=-1/a
Put x=0 in this to get y intercept b=(1+a^2)/2
as P tends to O, a tends to 0
Thus limit= 1/2


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## qqwref (Jul 11, 2011)

Bump. I've been having trouble with this one, maybe someone can solve it... It doesn't originally exist in this wording, but I rewrote it to make it feel more like a math word problem.

There are C cabinets, each with D initially empty drawers in them. You place T things (0<T<C*D) into the drawers, randomly, but such that no drawer has more than one thing in it. What is the chance that exactly one cabinet is completely empty of things? Two cabinets? N cabinets?


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## Akash Rupela (Jul 11, 2011)

Well, i dont really understand the question completely, is it like C and D are constants while T is variable which takes all values from 1 to CD-1, and due to random placing, all are equally likely . Or is T also fixed(then i dont make sense of this)?

Also is the answer 1/(CD-1) for all parts ? or maybe (CDcn)/2^(CD) ? where CDcn=CD!/(n! * (CD-n)! )
(i can see this problem in multiple ways, so trying to put all possible answers i can think of, if any is correct, i will elaborate)


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## kprox1994 (Jul 11, 2011)

A math thread? This isn't school! I have a question, if the answer to life, the universe and everything is 42; then what is the answer to death, the antiuniverse, and nothing?


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## Akash Rupela (Jul 11, 2011)

24? or maybe 1/42?


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## qqwref (Jul 11, 2011)

Akash Rupela said:


> Well, i dont really understand the question completely, is it like C and D are constants while T is variable which takes all values from 1 to CD-1, and due to random placing, all are equally likely . Or is T also fixed(then i dont make sense of this)?


C, D, and T are all constants, although you don't know what they are. All possible placements of the T things in the drawers are equally likely.



Akash Rupela said:


> Also is the answer 1/(CD-1) for all parts ? or maybe (CDcn)/2^(CD) ? where CDcn=CD!/(n! * (CD-n)! )


I don't have the answer, so what's your logic behind these guesses?


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## Akash Rupela (Jul 11, 2011)

Oh i m so sorry i misread it , i read empty cabinets as drawers
so now i have T drawers full while cd-T drawers empty (i wil use lowercase c here to avoid confusion)
in 1st case i want 1 cabinet empty, means i have to distribute T objects among (cd-d) drawers.
for number of ways in which this can happen, i solve the integer equation, x1+x2+x3......x(cd-d)=t where each of the cd-d drawers have value 0 or 1
this can happen if i give a value of 1 to any random t drawers, and 0 to cd-d-t drawers
So i select those drawers with value=1 in (cd-d)Ct ways .
But we can also select the cabinet which will be empty in c ways
Also the total number of ways of selecting t drawers from cd drawers is equal to (cd)Ct
So answer of the 1st part should be c* (cd-d)Ct / (cd)Ct

Now for the general expression, first i select which n cabinets will be empty in cCn ways( in 1st part cC1= c)
then from the remaining (cd-nd) drawers, i select any t drawers which will be full while others empty, in (cd-nd)Ct ways
and total number of ways will still remain the same
so answer of N cabinets part should be cCn (cd-nd)Ct/ (cd)Ct = (if i simplify this , it will become too dirty without equation editor)

Put n=2 to get answer of 2nd part

I hope this is right


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## qqwref (Jul 11, 2011)

Akash Rupela said:


> So i select those drawers with value=1 in (cd-d)Ct ways .
> But we can also select the cabinet which will be empty in c ways
> Also the total number of ways of selecting t drawers from cd drawers is equal to (cd)Ct
> So answer of the 1st part should be c* (cd-d)Ct / (cd)Ct


You can't just do this though, because you are not taking into account whether other cabinets will be empty or not. If exactly one cabinet is empty, you need to make sure the other ones are NOT empty.

As an example, let's have C=5, D=5, T=5. Your equation gives 5*(20 choose 5)/(25 choose 5) = 1.46, which is clearly wrong because probabilities can't be more than 1.


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## SpacePanda15 (Jul 11, 2011)

Ummm I have a question(not math) How old are you guys. I only understand some of this stuff. I'm 13. I don't want ot fell that I am very slow


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## Akash Rupela (Jul 11, 2011)

aaaah 

Spacepanda15, i m 18


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## MTGjumper (Aug 4, 2011)

This has been annoying me:

Using the Chinese Remainder Theorem, prove that there are a million consecutive integers, each of which is divisible by at least a million different prime numbers.

Some kind of hint would be nice, but feel free to include your solution


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## y235 (Aug 4, 2011)

I have a problem:
Let A be a group in the plane that its area is bigger than 1. prove that you can move it such that, so it will contain 2 integer points (integer point - point (a,b) and a,b are integers)


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## Erdos (Aug 5, 2011)

MTGjumper said:


> This has been annoying me:
> 
> Using the Chinese Remainder Theorem, prove that there are a million consecutive integers, each of which is divisible by at least a million different prime numbers.
> 
> Some kind of hint would be nice, but feel free to include your solution


Denote the set of consecutive integers as X, so |X| = 10^6., and n is the first of the consecutive integers. Now suppose n is divisible by at least a million different prime numbers. Therefore, consider the group Z_n. Using CRT, what is this isomorphic to? How does Z_n+1 relate? Then use induction for formalities, although since |X| is a finite number, you don't really need to.


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## macky (Aug 5, 2011)

MTGjumper said:


> This has been annoying me:
> 
> Using the Chinese Remainder Theorem, prove that there are a million consecutive integers, each of which is divisible by at least a million different prime numbers.
> 
> Some kind of hint would be nice, but feel free to include your solution



Hint: Use more than a million primes.

A solution:


Spoiler



It's no harder to show that there are \( N \) consecutive integers, each divisible by at least \( M \) different primes. Choose \( NM \) primes indexed as \( p_{ij}, 1 \le i \le N, 1 \le j \le M \). We will take the string \( a+1, \ldots, a+N \) so that \( a+i \) is divisible by \( p_{ij}, 1 \le j \le M \). For this it suffices to take \( a \equiv -i \pmod{p_{ij}} \) using CRT.






y235 said:


> I have a problem:
> Let A be a group in the plane that its area is bigger than 1. prove that you can move it such that, so it will contain 2 integer points (integer point - point (a,b) and a,b are integers)



Reworded: Given a region in the plane whose area [whatever that means] is greater than 1, prove that some translation contains 2 integer points.

A solution:


Spoiler



Reduce mod \Z \times \Z so that the region lies in [0, 1) \times [0, 1). Since it has area greater than 1, some two points have the same reduction, and these can be translated to integer points.


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## hic0057 (Aug 8, 2011)

Got given this on a math comp and I have no idea how to work it out.

*The first digit of a six-digit number is 1. This digit is now moved from the first digit position to the end, so it become the last digit. The new six-digit number is now 3 times larger than the original number. What are the last three digits of the original number.*

Can you please explain how to get the answer so that a fairly smart 15 years old could understand. Thanks


----------



## y235 (Aug 8, 2011)

hic0057 said:


> Got given this on a math comp and I have no idea how to work it out.
> 
> *The first digit of a six-digit number is 1. This digit is now moved from the first digit position to the end, so it become the last digit. The new six-digit number is now 3 times larger than the original number. What are the last three digits of the original number.*
> 
> Can you please explain how to get the answer so that a fairly smart 15 years old could understand. Thanks



A clue: try to think how the numbers are related to each other.



Spoiler



the new number is 3 times bigger than the original one, which means that the last digit of the original one must be 7 (because 7 is the only digit that if you multiply you get 10X+1). the second digit from the end must be 5, because 5 is the only digit that if you multiply it by 3 and add 2 (because 7*3=21) you get 7.
in the same way the third digit from the end is 8.


Sorry for my bad english, Hope you'll understand.


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## hic0057 (Aug 8, 2011)

y235 said:


> Spoiler
> 
> 
> 
> the last digit of the original one must be 7 (because 7 is the only digit that if you multiply you get *10X+1)*





Spoiler



Thanks for that, really helpful


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## chicken9290 (Aug 8, 2011)

wow i see problems like this every wednesday at math team.


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## hic0057 (Aug 9, 2011)

Got another one that my maths teacher couldn't help me with.

_*The diagram shows the net of a cube. On each face there is an integer: 1, w, 2011, x, y and z.*_

[w]
[x][y][2011][z]
[1]



_*IF each of the numbers w, x, y and z equals the average of the numbers written on the four faces of the cube adjacent to it, find the value of x.*_

Could you please explain how to work this out so that a fairly smart 15 years old could understand.


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## blah (Aug 9, 2011)

hic0057 said:


> Got another one that my maths teacher couldn't help me with.
> 
> _*The diagram shows the net of a cube. On each face there is an integer: 1, w, 2011, x, y and z.*_
> 
> ...





Spoiler



Opposite faces must have the same value. If you can't figure out why, then you haven't given this problem the thought it deserves, which really isn't much. So w = 1, x = 2011, and y = z = 2012/2 = 1006.


Your math teacher probably didn't even try >_>


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## hic0057 (Aug 9, 2011)

blah said:


> Spoiler
> 
> 
> 
> ...


 
Umm, The answer is only 3 digits.


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## blah (Aug 9, 2011)

hic0057 said:


> Umm, The answer is only 3 digits.


And what does that tell you?


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## qqwref (Aug 9, 2011)

blah said:


> And what does that tell you?


That 1 = 2011?


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## hic0057 (Aug 9, 2011)

That you are wrong.

This question was on an Australian National Mathematics Competition.

Quote from the text paper
_"Requires a whole number answer between o and 999"_


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## y235 (Aug 9, 2011)

hic0057 said:


> That you are wrong.
> 
> This question was on an Australian National Mathematics Competition.
> 
> ...


Blah's solution seemes correct to me.


EDIT:


aronpm said:


> The question is worded kinda poorly. It says that w,x,y,z are equal to the average of their adjacent sides but doesn't say 1 and 2011 are.


 forgot about that


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## hic0057 (Aug 9, 2011)

Would someone be able to explain how to get an answer between 0 and 999?


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## aronpm (Aug 9, 2011)

The question is worded kinda poorly. It says that w,x,y,z are equal to the average of their adjacent sides but doesn't say 1 and 2011 are.


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## hic0057 (Aug 9, 2011)

aronpm said:


> The question is worded kinda poorly. It says that w,x,y,z are equal to the average of their adjacent sides but doesn't say 1 and 2011 are.


 
Just wondering, did you do the test.


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## aronpm (Aug 9, 2011)

hic0057 said:


> Just wondering, did you do the test.


 
I haven't had the opportunity to do these math-test-things in years


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## qqwref (Aug 9, 2011)

I guess the official solution would be


Spoiler



to just write out a bunch of equations:
4w = x+y+2011+z
4x = w+y+1+z
4y = w+x+2012
4z = w+x+2012

From the last two we get y = z, and we can insert those equations into the first two to get:
4w = x+2011+(w+x+2012)/2
4x = w+1+(w+x+2012)/2

Simplifying a bit:
7w = 3x+6034
7x = 3w+2014

And solving for x:
7x = 3(3x+6034)/7+2014
49x = 9x+18102+14098
40x = 32200
x = 805


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## hic0057 (Aug 9, 2011)

Thanks qqwref for that


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## hic0057 (Aug 9, 2011)

Another one...

*A 40x40 white square is divided into 1x1 squares by lines parallel to its sides. Some of these 1x1 squares are coloured red so that each of the 1x1 squares, regardless of whether it is coloured red or not, shares a side with at most one red square (not counting itself.) What is the largest possible number of red squares?*

The answer is between 000 and 999.

Also can you please explain it so that a 15 years old could understand. Thanks


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## Erdos (Aug 11, 2011)

This is a simple case of using induction to find the formula for an nxn white square. In other words, try the coloring with a 1x1 square, then 2x2, 3x3,... You should find a pretty simple pattern.


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## JonWhite (Aug 11, 2011)

hic0057 said:


> Another one...
> 
> *A 40x40 white square is divided into 1x1 squares by lines parallel to its sides. Some of these 1x1 squares are coloured red so that each of the 1x1 squares, regardless of whether it is coloured red or not, shares a side with at most one red square (not counting itself.) What is the largest possible number of red squares?*
> 
> ...


 
do you know what year AIME this problem was taken from

Also, try coloring one red square and see what other squares MUST be colored white.


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## Tim Major (Aug 11, 2011)

The 40x40 grid was on the yr 10/11 test I did. I got 400 though not in a mathematical way, I just drew a much smaller grid and found the percentage.
I might have had the adjacent one too, but I left about 15 blank near the start to do the 10 point ones, and then had my test taking off me for talking (albeit it was a totally unrelated topic)
Did you get the one about climbing the mountain, 2 people 12km apart? I was stupid, I drew 20~ dashes and moved 2 torn pieces along them. I got the right answer though


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## asportking (Aug 12, 2011)

Ok, I'm not sure if this is where I should ask it, but does anyone know a good way to understand normal form of a linear equation? I've been taking pre-calculus online, and I just learnt that there are a few things I need to know to pass the pre-calculus test that aren't on the online course, like normal form of a linear equation. I looked through a pre-calculus textbook, but I just can't understand what it's talking about. If anyone knows a video or a website that explains it well (or maybe could explain it themselves? ), it would be much appreciated.


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## cmhardw (Aug 12, 2011)

asportking said:


> Ok, I'm not sure if this is where I should ask it, but does anyone know a good way to understand normal form of a linear equation? I've been taking pre-calculus online, and I just learnt that there are a few things I need to know to pass the pre-calculus test that aren't on the online course, like normal form of a linear equation. I looked through a pre-calculus textbook, but I just can't understand what it's talking about. If anyone knows a video or a website that explains it well (or maybe could explain it themselves? ), it would be much appreciated.


 
I have to be honest that I have never studied normal form of a linear equation before today. You learn something new every day I suppose.

Ok, so here is what I have gathered from the Wikipedia article as well as looking into it myself.

I don't want to give too much away, as the self discovery of math is beautiful for one, and also you will remember this better if you derive most of this yourself.

Here is what I did to study normal form:


Spoiler



Start with the equation:
\( Ax+By=C \)

We seek to find the point on this line that also lies on the normal of the line. Wikipedia defines the normal as the shortest distance between the given line and the origin. The shortest distance between a given point and a given line is the line through that point which is perpendicular to the given line.

The slope of the equation:
\( Ax+By=C \)

is \( \frac{-A}{B} \)

Therefore the normal will have slope: \( \frac{B}{A} \)

The normal must go through the origin, so the equation on which the normal lies is:
\( y=\frac{B}{A}x \)

We now seek to find the point on the given line, which also lies on the normal. This point is the solution to the following system of equations.

\( Ax+By=C \)
\( y=\frac{B}{A}x \)

I strongly suggest you solve this system yourself, but the solution comes out to be:
\( \left(\frac{AC}{A^2+B^2} , \frac{BC}{A^2+B^2}\right) \)

The length of the normal, \( p \), is the distance from the above point to the origin. Using the distance formula you should get:
\( p=\sqrt{\left(\frac{AC}{A^2+B^2}-0\right)^2+\left(\frac{BC}{A^2+B^2}-0\right)^2} \)

Again, I strongly suggest you work this out on your own, but this simplifies to:
\( p=\left|\frac{C}{\sqrt{A^2+B^2}}\right| \)

To find the angle \( \theta \) between the normal and the x-axis, we simply need to realize that the normal line has slope \( \frac{B}{A} \) and so:
\( \theta=\arctan\left(\frac{B}{A}\right) \)

Now, Normal Form is listed on Wikipedia as:
\( y\sin\theta+x\cos\theta-p=0 \)

Substitute in what we know:
\( y\sin\left(\arctan\left(\frac{B}{A}\right)\right)+x\cos\left(\arctan\left(\frac{B}{A}\right)\right)-\left|\frac{C}{A^2+B^2}\right|=0 \)
and simplify to get:

\( y\frac{B}{\sqrt{A^2+B^2}}+x\frac{A}{\sqrt{A^2+B^2}}-\left|\frac{C}{\sqrt{A^2+B^2}}\right|=0 \)

Notice that this looks very similar to the standard form equation, just divided by \( \sqrt{A^2+B^2} \). You will also have to pay attention to the sign of the C term as well, since p is defined as the distance (so positive) of the normal, and C might be negative.

I'll leave you to connect the final dots, but basically given a standard form equation:
\( Ax+By=C \) you should at the very least be able to find \( sin\theta \), \( cos\theta \), and \( p \)


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## asportking (Aug 12, 2011)

cmhardw said:


> I have to be honest that I have never studied normal form of a linear equation before today. You learn something new every day I suppose.
> 
> Ok, so here is what I have gathered from the Wikipedia article as well as looking into it myself.
> 
> ...


 
Thanks, that really helped! I've been really stressed out lately, since I've got so much new stuff to learn now, no where to learn it from, and only a week to do it. It's awfully scary thinking that I might have just wasted the whole summer studying this only to have to take it all over again at school. I got a hold of last years lesson plan for pre-calc, so it helps knowing exactly what I need to learn.
Although I wonder, what exactly is the point of knowing normal form? I imagine it will become clear once I get into more advanced stuff, but I can't help but questioning the usefulness of this...


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## qqwref (Aug 12, 2011)

asportking said:


> Although I wonder, what exactly is the point of knowing normal form? I imagine it will become clear once I get into more advanced stuff, but I can't help but questioning the usefulness of this...


It isn't really useful, and I'm not sure why you're learning it. It probably has a few very specific applications, but in general you'll never see it. There is sort of an unofficial rule in actual math where you use whatever requires the least work for the problem/application at hand (hence the whole idea of 'simplest form') so you will see the ax+by+c=0 or ax+by=c forms almost 100% of the time.


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## asportking (Aug 12, 2011)

qqwref said:


> It isn't really useful, and I'm not sure why you're learning it. It probably has a few very specific applications, but in general you'll never see it. There is sort of an unofficial rule in actual math where you use whatever requires the least work for the problem/application at hand (hence the whole idea of 'simplest form') so you will see the ax+by+c=0 or ax+by=c forms almost 100% of the time.


 Oh thank god. That formula did NOT look fun to use all the time.


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## Erdos (Aug 12, 2011)

I've never seen this before, but it seems like a trivial result of Gram-Schmidt, i.e. producing an orthogonal vector given another vector. Yeah don't think you'll ever need to know that for high school or even college. However, the concept is pretty important in all linear algebra: producing linearly independent equations.


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## hic0057 (Aug 18, 2011)

JonWhite said:


> do you know what year AIME this problem was taken from
> 
> Also, try coloring one red square and see what other squares MUST be colored white.



This question was in the 2011 AIME test that was taken about 2 weeks ago in the Intermediate (year 9 and 10) and Senior (year 11 and 12) division.



Tim Major said:


> The 40x40 grid was on the *yr 10/11* test I did. I got 400 though not in a mathematical way, I just drew a much smaller grid and found the percentage.
> I might have had the adjacent one too, but I left about 15 blank near the start to do the 10 point ones, and then had my test taking off me for talking (albeit it was a totally unrelated topic)
> Did you get the one about climbing the mountain, 2 people 12km apart? I was stupid, I drew 20~ dashes and moved 2 torn pieces along them. I got the right answer though


 
There wasn't a year 10/11 AIME test but a Year 9/10 test and a 11/12 test.

I found the mountain one fairly easy.

*Two tourist are walking 12km apart along a flat track at a constant speed of 4km/h. When each tourist reaches the slope of a mountain, she begins to climb a constant speed of 3km/h.

What is the distance, in km, between the two tourist during the climb.*

They way I worked it out was finding how long the second person would take to get to the bottom of the mountain after the person ahead. It was 3hrs than I multiply that to the person ahead speed per hour to get the distance of 9km.


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## vcuber13 (Aug 18, 2011)

Find an equation to solve the area of any regular polygon.
what i got:


Spoiler



\( \frac{b^2 n \cdot cot(\frac{180}{n})}{2} \)

where b is the length of the base, and n is the number of sides.

doesnt work, time to retry


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## Erdos (Aug 18, 2011)

Area of n-gon.


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## vcuber13 (Aug 19, 2011)

vcuber13 said:


> Spoiler
> 
> 
> 
> ...


 
ive been thinking why this wasnt working, i figured it out:


Spoiler



the way i worked it out i got the 180/n and then tanned it, but because its angles and 180 degree its pi radians and so it should be:
\( \frac{b^2 n \cdot cot(\frac{\pi}{n})}{4} \)
oh, and i also missed a *.5 so it should be over 4 not over 2


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## Keroma12 (Mar 8, 2012)

Anybody know how to prove det(AB) = det(A)*det(B) using only the permutation definition of the determinant function? Completely forgotten how to type in math here, otherwise I would type out the definition. http://en.wikipedia.org/wiki/Leibniz_formula_for_determinants

(I am aware that this is a terrible way to prove this and that there are much much nicer ways.)


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## jblake17 (Nov 24, 2013)

Did anyone do the CIMC 2013?


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