modular arithmetic
number theory unsolved
number theory

Why "wrong problem"¿

Zsigmondy's back again by the way :D
But I've the feeling to have posted a solution (without Zsigmondy, just standard stuff).

[quote="e.lopes"]wrong problem!

$p^{2}| 2^{p-1}-1$ for many values of p![/quote]
It is new problem. How many primes (I know only 2 primes) satisfyed $p^{2}|2^{p-1}-1$?

Sory for my stupidity, but I don't see how we can use Zsigmondy's theorem here. Could you please post the solution.

[quote="lajanugen"]sandu2508: Using Zsigmondy's theorem and the fact that the RHS cannot have two distinct prime factors, we are done

Suppose $ n$ is atleast $ 2$.
$ p$ is odd. $ n$ must be odd since if it were even, LHS would be $ 1 (mod4)$ which is impossible for $ m > 1$.
But then $ \frac {p^n \plus{} 1}{p \plus{} 1}$ is odd and we have a contradiction

A similar problem is this, from a JBMO : Let m,n>1 be integers. Solve in positive integers the equation $ x^n \plus{} y^n \equal{} 2^m$
and a solution very similar to the above is possible[/quote]

Thank you, but I still didn't understand what we do when $ n$ is prime

Here is what i understood: If $ m$ is composite $ m \equal{} dk$, By Zsigmondy;s theorem we get $ p$ is the primitive prime of $ 2^m \minus{} 1$ so $ p\nmid 2^d \minus{} 1$ then $ p^n \equal{} (2^d \minus{} 1)\cdot X$, which is impossible.  But I don't know what to do when $ n$ is prime.

oops, im sorry I didn't see that comin, as you say, Zsigmondy only forces m to be a prime

I haven't found an elementary proof yet but of course one can use Mihalescu's theorem(the only two consecutive powers are 8 and 9) and the problem is done :)

I fear I am missing something, but here is what I believe to be an elementary solution. Suppose $ p^n \equal{} 2^m \minus{} 1$ for integers $ n$ and $ m$ and a prime $ p$. Then $ n$ must be odd (unless $ m < 2$, in which case there are no solutions), and as $ p^n \plus{} 1 \equal{} 2^m$ and because $ n$ is odd, the left hand side is divisible by $ p \plus{} 1$. This must mean that either $ p \plus{} 1 \equal{} p^n \minus{} 1$ and we are done, or $ p \equal{} 2^k \minus{} 1$ for some $ k < n$.

Then we have $ (2^k \minus{} 1)^n \equiv \minus{} 1 \pmod{2^m}$, and $ (2^k \minus{} 1)^{2n} \equiv 1 \pmod{2^m}$. Letting $ d$ be the order of $ 2^k \minus{} 1 \pmod{2^m}$, we see that we must have $ d|\gcd(2n, \phi(2^m))$, and as $ \phi(2^m) \equal{} 2^{m \minus{} 1}$ we must have $ d|\gcd(2n, 2^{m \minus{} 1}) \equal{} 2\gcd(n,2^{m \minus{} 1}) \equal{} 2$ (as $ n$ is odd). Then $ d \equal{} 2$ or $ d \equal{} 1$, which means that either $ 2^k \minus{} 1 \equiv 1 \pmod {2^m}$ or $ (2^k \minus{} 1)^2 \equiv 2^{2k} \minus{} 2^{k \plus{} 1} \plus{} 1 \equiv 1 \pmod {2^m}$.

The first case yields $ 2(2^{k \minus{} 1} \minus{} 1) \equiv 0 \pmod {2^m}$ and $ m \leq 1$ and we have no solutions, and the second case yield $ 2^{k \plus{} 1}(2^{k \minus{} 1} \minus{} 1) \equiv 0 \pmod{2^m}$ and $ k \plus{} 1 \geq m$, but as $ k < m$ we must then have $ k \equal{} m \minus{} 1$.

Then $ p^n \equal{} (2^k \minus{} 1)^n \equal{} (2^{m \minus{} 1} \minus{} 1)^n \equal{} 2^m \minus{} 1$, or $ 2^m \minus{} 1 \equiv 0 \pmod {2^{m \minus{} 1} \minus{} 1}$, but as $ 2^m \minus{} 1 \equal{} 2(2^{m \minus{} 1} \minus{} 1) \plus{} 1 \equiv 1 \pmod {2^{m \minus{} 1} \minus{} 1}$ we must have $ 0 \equiv 1 \pmod {2^{m \minus{} 1} \minus{} 1}$, which is impossible unless $ m \equal{} 2$, in which case $ p \equal{} 3$ and $ n \equal{} 1$, and we are done - the only possible case is that $ n \equal{} 1$. I do not believe the hypothesis that $ p$ is prime has been used here, so there might be some simpler solution.

[quote="math12061992"]Can anyone please provide a link for the hugely mentioned ZSIGMONDY's THEOREM.[hide]I don't know it  :blush: [/hide][/quote]

Here it is: [url=http://en.wikipedia.org/wiki/Zsigmondy's_theorem]link[/url]
A proof for the case $ b\equal{}1$ can be found [url=http://www.ams.org/proc/1997-125-07/S0002-9939-97-03981-6/S0002-9939-97-03981-6.pdf]here[/url] and [url=http://www.math.dartmouth.edu/~thompson/Zsigmondy's%20Theorem.pdf]here[/url]

[quote="KarlErik"]... integers $ n$ and $ m$ and a prime $ p$. Then $ n$ must be odd (unless $ m < 2$, in which case there are no solutions), and as $ p^n \plus{} 1 \equal{} 2^m$ and because $ n$ is odd ...
[/quote]

Why must n be odd?

Because if $ n$ is even, the left side of the equation must be $ 1 \pmod 4$. However, the left hand side is $ 0 \pmod 4$ if $ m\equal{}0$, $ 1 \pmod 4$ if $ m\equal{}1$ and $ \minus{}1 \pmod 4$ if $ m \geq 2$. As the cases $ m\equal{}0$ and $ m\equal{}1$ are trivial, the only possibility is that $ n$ is odd.


geometry
3D geometry
tetrahedron
geometry solved

hi,
here's a funny little prob:

4 triangles of same area are the faces of a tetrahedron. Prove that these 4 triangles are isometric ..

actually they are congruent triangles if I'm not mistaking ...

Let ABCD be an our tetrahedron. Let M and N be the midpoints of AD and BC. Since the dist(A,BC)=dist(D,BC) then MN is perpendicular to AD and so to BC.

Now if we consider the sircumscribed parallelepiped of ABDC we found that  one of its edges is perpendicular to one of the faces. In this way we can prove, that all its edges are perpendicular to corresponding faces. So the parallelepiped is rectangular, so the faces of tetrahedron are congruent.


induction
superior algebra
superior algebra unsolved

Show there is a unique monomial order on $ \mathbb{C}[x]$

Two cases: $ x>1$ and $ x<1$.  In the former case, use induction to show $ x^{n\plus{}1}>x^n$.  In the latter case, do the same thing but show that this fails to be a well-ordering.


probability
geometry
3D geometry
function
conditional probability

Yes, I've heard this question many times before, however, assuming biology, I would have to say that it is $1/2$ because the chances are the same.

I may be wrong though because a question like this about flipping heads or tails in a coin was also the same way. Someone said it was 1/3 due to some principle. However, I believe that it is $1/2$

[quote="bjones23"]hello,

this is my first time on this board.   I have a question of probability.

Suppose we know that parents have 2 kids and we know that 1 is a boy.  What is the chance that the other is also a boy?

some ppl say the chance is 1/3 given that the combinations are BB, BG, GB and GG and since we want BB and we know GG is out.

Others say it is 1/2 since all we care about is the gender of the second child.

This was a hot topic on another forum and lots of controversy.  They are not math experts so I came here to get the right answer.

thanks!![/quote]

However, aren't $BG$ and $GB$ essentially the same because order does not matter in this question? Because then there would only be 2 choices left, thus being $1/2$.

It doesn't matter. The probability is 1/2 regardless of the other children.

but scientifically, i've heard somwhere that the probability of a boy being born is greater than that of a girl being born. i'm not sure though.

in bio class last year, we had this on a test (or something along the lines of it):

there was a couple who gave birth two boys who each both gave birth to one boy and a girl who, in turn, all gave birth to boys. take one of the greatgrandsons of the original couple. what is the probability that they will give birth to a girl?

and indeed the answer was 1/2. the key here is that all offspring are independent of each other, because the genotypes of the parents will always be the same: XX and XY.  a simple punnet square shows that each gender is equally likely.

does this answer your question?

[quote="furious"]since one child's gender is known, it is just a $\frac{1}{2}$ probability since the first child's gender has no effect on the second's.[/quote]

but even if gender does have an effect, there will still be only 2 possibilities, $BG$ and $BB$, since the other possibilities are knocked out because we know the other child is a boy.

thanks, I think it's 1/2 ... although others on the other forum insist  :roll:  it's 1/3 since they feel there are 3 possibilities of BG, GB and BB.   But all we want to know is the chance of the other child is a boy ... equals 1/2 IMO.

choice really is -- (1) both kids are boys  (2) its a split pair

[quote="Bojan Basic"]No, it is [u]not[/u] 1/2, it is 1/3.

Consider the following problem:

We know that parents have 2 kids and we know that [color=red]older[/color] is a boy. What is the chance that the other is also a boy?

Do you see the differnce between this question and yours? In this case the answer is 1/2 because there are only two possibilities - BB and BG, but in your case order of the kids [u]does[/u] matter (because we don't know whether the "known boy" is older or younger, so there are 3 possibilities (namely BB, BG and GB) and hence the answer is 1/3.[/quote]

Yes, exactly. I remember we had a huge discussion on it this year in class, and you worded it perfectly (I knew it was $\frac{1}{3}$, but I couldn't explain it lol).

[quote="Bojan Basic"]No, it is [u]not[/u] 1/2, it is 1/3.

Consider the following problem:

We know that parents have 2 kids and we know that [color=red]older[/color] is a boy. What is the chance that the other is also a boy?

Do you see the differnce between this question and yours? In this case the answer is 1/2 because there are only two possibilities - BB and BG, but in your case order of the kids [u]does[/u] matter (because we don't know whether the "known boy" is older or younger, so there are 3 possibilities (namely BB, BG and GB) and hence the answer is 1/3.[/quote]

Hmm...I don't seem to understand that...what's the difference between BG and GB? Age doesn't necessarily matter, does it?

Hey, Bojan Basic, how can age be related to this question? That is impossible? :o  or am I wrong  :(  please tell me how?

[quote="dakyru"]If you're going to treat it that way, there are not 3 possibilities, but 4.  $B_1B_2$, $B_2B_1$, $BG$, and $GB$.  Because you do not know whether the known boy came first or second either.  In that case, the probability still is $\frac{1}{2}$.[/quote]

Thanx darkyru...see, it is $1/2$ because age does not necessarily matter in this type of question.

its 1/2 because the 2 children have to be split separately

[quote="236factorial"]I think the problem is right now is believing whether BG and GB are the same or not. It's the door to deciding the answer.[/quote]

I think BG and GB are the same because age does not matter in this question, thus order doesn't. :bruce:

it is 1/2 as reproduction is a memoryless function

I think the answer is $\frac{1}{2}$ since it's mainly the gender that counts.

Interesting... :mad:

I think it is ALWAYS 1/2. It is only the gender of the new child that matters.

[quote="Singular"]"Suppose we know that parents have 2 kids and we know that 1 is a boy. What is the chance that the other is also a boy? "

if parents have two kids, then 25% chance of each of the following:  $\{ BB, BG, GB, GG \}$

Now you know that there is one boy, so obviously 1/3.[/quote]

if you can say GB and BG, then i can say BB twice , because you do NOT know what order the children were born in

thus the given child (the boy) could have been born before or AFTER his brother...

thus you have {B1-B2, B2-B1, GB, BG}
where B1 is the first given boy

you are NOT assuming independence, because after a child is born, its not like we cross out one of those possible choices in your set

imagine, 1/3 boy, 1/3 girl, and 1/3 something else  :lol:

[quote="Priapus"]My in-depth (and possibly completely wrong) analysis of the problem:
[quote]hello,

this is my first time on this board. I have a question of probability.

Suppose we know that parents have 2 kids and we know that 1 is a boy. What is the chance that the other is also a boy?

some ppl say the chance is 1/3 given that the combinations are BB, BG, GB and GG and since we want BB and we know GG is out.

Others say it is 1/2 since all we care about is the gender of the second child.

This was a hot topic on another forum and lots of controversy. They are not math experts so I came here to get the right answer.

thanks!![/quote]

[quote]Suppose we know that parents have 2 kids[/quote]

Ok, we know that there are 2 kids, and the parents don't matter unless one (or more) is sterile or their both the same sex or...etc. (but I won't factor those possibilities in).

[quote]...and we know that 1 is a boy.[/quote]

Ok, so one is a boy. First of all, we are assuming the two siblings are not identical twins (otherwise they would both be boys) or fraternal twins (in which case the chances of a girl would be slightly higher because one sperm with a Y chromosome has already fertilized an egg).

If you interpret the above to mean at least one is a boy, then either the condition that at least one is a boy is satisfied or it isn't (no duh). If the one we know is a boy satisfies the condition that the problem-writer wants, then the chances are 1/2 because there is no direct link between the two siblings' sexes. If it doesn't satisfy the condition, then the other child must be a boy to satisfy the condition (and the probability is one).

If you interpret the above to mean that you know one child is a boy, then the sex of the other 1/2 because they are independent events. The BB BG GB argument doesn't work because it the age doesn't matter. The age of the first child of whom we know the sex (although he may not necessarily be the older one) doesn't affect the sex of the second (though not necessarily younger) child.

Punnett square for second child:
...M.O.T.H.E.R
F........X......X
A.X....XX....XX
T...................
H..................
E.Y.....XY....XY
R

(hope you can figure out the chart, dots are there for spacing)

Two XX (girl) and two XY (boy)

so probability is 1/2[/quote]
Your little diagram there shows only for one child, and the probability that it will be a boy is 1/2.  You must consider that the other is a boy, and not leave it totally alone

[quote="236Factorial"]
I think the problem is right now is believing whether BG and GB are the same or not. It's the door to deciding the answer.[/quote]
Of course there not  Look at Pascal 1 + 2 + 1 = 4.  1/4 for both boys, 2/4 for 1 of each, 1/4 for both girls.    The probability of having both boys divided by the probability of having at least one boy.   Again: (1/4)/(3/4)=1/3.

How can some of you think that the part of knowing that one is a boy does not matter, of course it does, that will obviously lower the prob of the other being a boy

[quote="nat mc"]How can some of you think that the part of knowing that one is a boy does not matter, of course it does, that will obviously lower the prob of the other being a boy[/quote]
No it doesn't... it's not like the chromosomes will say: "oh, they have a boy already... now we should make the probability of a girl higher"  :lol:

I believe we are getting different answers because we are interpreting the problem differently.

Those folks (like nat mc and me) who are getting an answer of 1/3 are interpreting the problem as follows: "Given that at least one child is a boy, what is the probability that both children are boys?".

The folks who are getting an answer of 1/2 are interpreting the problem as follows: "Given that this particular child is a boy, what is the probability that both children are boys?".

When discussing the problem further, please specify which interpretation you are using.

That clears it up. Maybe the problem should be more clear to begin with. I just go with the interpretation of the 1/2 answer. But, the other one isn't wrong.

You have fallen into a well-known trap.  Let $A$ be the event "both children are boys".  Let $X$ be the event "the older child is a boy".  Let $Y$ be the event "the younger child is a boy".  You are correct that both $\Pr(A \mid X)$ and $\Pr(A \mid Y)$ are equal to $\frac{1}{2}$.  From there, however, it does not follow that $\Pr(A \mid X \textrm{ or } Y)$ is equal to $\frac{1}{2}$.

[quote="aidan"]But if at least one child is a boy, wouldn't it still be $ \frac {1}{2}$? The child concerned can either be the elder or younger child. If he is elder, the younger child can be a boy or girl. If he is younger, the older child can either be a boy or girl as well. That leaves the probability as $ \frac {1}{2}$ doesn't it?[/quote]
Why would age matter?  It is the probability of both being a boy over the probability of at least one being a boy or $\frac13$


It is obviosly:
$ 1\ge (a_1\plus{}...\plus{}a_n)^2\equal{}\sum_{j\equal{}1}^n a_j (a_j\plus{}2\sum_{i<jj}a_i)\ge \sum_j (2j\minus{}1)a_j^2,$


find all functions from N to N (N doesnt contain zero)
satsfying
f(m 2 +f(n))=f(m) 2 +n
the only thing we need is f(1) and i am not able to find that
sorry if the question is being reposed
best regards


[color=purple] I was known the proff but any one can give the similar or the stronger ? [/color]

[quote="evarist"][color=purple]Let ABC is a triangle.Prove [/color]
 $\frac{R}{r}\geq 2+\frac{9[(\sqrt{a}-\sqrt{b})^{2}+(\sqrt{b}-\sqrt{c})^{2}+(\sqrt{c}-\sqrt{a})^{2}]}{4(\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}}$[/quote]
See here my result (posted: Mon Aug 28, 2006 12:43 am ):
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107858
It's stronger. :wink:

Can you post the proff for the last inequality in the above link arqady? thanks


I was wondering how to take a sketch from GSP and post in on the forum.  I tried savive the sketch as a .GIF, but that didn't work, am I forgetting something?

Thanks.

i guess you have to save the picture as .JPEG

This question has been asked and answered before.  GSP will only save its files in whatever weird format it uses, but you can select the diagram, copy it into Paint (or any other image-editor) and then save it there in whatever format you prefer.


Can anyone help with these two questions?

1) The line y=-3x+k is a tangent to the circle $ {x^2} \plus{} {y^2} \equal{} 10$. Find the possibilities of y.

2) The straight line y=2x+k meets the circle $ {x^2} \plus{} {y^2} \minus{}2x \plus{} 4y \equal{} 0$ at two points. Find the set of possible values of k.

Plug y in the equation, then put the condition that the discriminant of the quadratic equation is 0 and solve for k.


Alex has 10 different kinds of lunch meat and 9 different kinds of cheese. If he wants to make a sandwich with one kind of meat and two kinds of cheese, how many different sandwiches could he make?

He can choose $ \binom{10}{1}\equal{}10$ types of lunch meat and pick $ \binom{9}{2}\equal{}36$ types of cheese. You must use combination and not permutation because the order in which he chooses the meat/cheese does not matter.

$ \binom{10}{1}\binom{9}{2}\equal{}\boxed{360}$

Nobody is going to understand the reasoning behind his explanation.

The logic is obvious if people understand combinations. Your post assumes that people know about combinations (as does mine), so the explanations are essentially trivial to those who already understand combinations, and to those who don't understand combinations, neither of our solutions would make sense.

Shouldn't the problem specify two [i]different[/i] kinds of cheese? You could have two of the same cheeses in a sandwich.

Could yall hide your answers?

@above I don't think so


I was wondering how one can space text on a multiple-paged document so that the lines dont squish together, kinda like double spacing in Word, etc.

Use the setspace package. Details are at [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=linespace]TeX FAQ[/url]. This FAQ is a mine of useful information and should be near the top of every LaTeX users' links.


inequalities
vector
function
integration
calculus
triangle inequality
real analysis

Big Lp is a normed vectors space of functions with finite value under the Lp norm. $ \parallel{}f\parallel{}_p \equal{} \left( \int_E |f|^p \right)^{1/p}$ Little lp is the same idea, but with sequences and instead of an integral the norm uses a sum. $ \parallel{}<a_n>\parallel{}_p \equal{} \sum_{n\equal{}1}^{\infty} |a_n|^p$ a sequence in in the space if it is finite under this norm. I want to establish the minkowski inequality: $ \parallel{}<a_n\plus{}b_n>\parallel{}_p \leq \parallel{}<a_n>\parallel{}_p \plus{} \parallel{}<b_n>\parallel{}_p$.

I tried to write a proof by looking at the partial sums of the sequences in the inequality. Since $ |a_i \plus{}b_i| \leq |a_i| \plus{} |b_i|$ for each i, we can say the same about the partial sums of each of these. I think that the minkowski inequality should follow from there-- but, I can't quite make the connection and I don't have much confidence in what I've done so far.

 :(

You can translate Minkowski's inequality in $ L_p$ to $ l_p$ by considering a counting measure.

Here's a proof, using Holder's inequality (I will not consider the case when $ p \equal{} 1$ or when things are trivially zero):

It should state $ \| (a_n)_{n\in\mathbb{N}} \|_p \equal{} \left(\sum\limits_{n \equal{} 1}^\infty |a_n|^p\right)^{\frac1p}$ (I will suppress the $ {}_{n\in\mathbb{N}}$ from now on), and Minkowski's inequality then reads

$ \|(a_n \plus{} b_n)_{n\in\mathbb{N}}\|_p \leq \| (a_n)_{n\in\mathbb{N}} \|_p \plus{} \| (a_n)_{n\in\mathbb{N}} \|_p$.

Now, we know that $ \|(a_n \plus{} b_n)\|_p$ is finite by the estimate $ |a_n \plus{} b_n|^p \leq 2^{p \minus{} 1} (|a_n|^p \plus{} |b_n|^p)$.

We can thus write:

$ \|(a_n \plus{} b_n)\|_p^p \equal{} \sum\limits_{n \equal{} 1}^\infty |a_n \plus{} b_n|^p$

$ \equal{} \sum\limits_{n \equal{} 1}^\infty |a_n \plus{} b_n| \, \cdot \, |a_n \plus{} b_n|^{p \minus{} 1}$         (applying the triangle inequality gives)

$ \leq \sum\limits_{n \equal{} 1}^\infty (|a_n| \plus{} |b_n|) \, \cdot \, |a_n \plus{} b_n|^{p \minus{} 1}$.

Set $ q \equal{} \frac {p}{p \minus{} 1}$, giving $ \frac1p \plus{} \frac1q \equal{} 1$. We use Holder's inequality on the product of the sequences $ a_n \plus{} b_n$ and $ (a_n \plus{} b_n)^{p \minus{} 1}$ giving

$ \leq \left[\left(\sum\limits_{n \equal{} 1}^\infty (|a_n|^p\right)^\frac1p \plus{} \left(\sum\limits_{n \equal{} 1}^\infty (|b_n|^p\right)^\frac1p\right]\cdot\left(\sum\limits_{n \equal{} 1}^\infty (|a_n \plus{} b_n|^{(p \minus{} 1)q}\right)^\frac1q$

Now, $ (p \minus{} 1)q \equal{} p$ and $ \frac1q \equal{} 1 \minus{} \frac1p$, which means this last line is equal to

$ \equal{} \left[ \| (a_n) \|_p \plus{} \| (b_n)\|_p\right] \cdot \frac {\| (a_n \plus{} b_n)\|^p_p}{\| (a_n \plus{} b_n)\|_p}$

We thus have upon multiplying both sides by $ \frac {\| (a_n \plus{} b_n) \|_p}{\| (a_n \plus{} b_n) \|_p^p}$ that

$ \|(a_n \plus{} b_n)\|_p \leq \| (a_n) \|_p \plus{} \| (b_n) \|_p$.

(Note: this proof is by no means original.)

[quote="WWW"]The norm in $ l^p$ is precisely  $ (\int_E |f|^p d\mu)^{1/p}$, where $ E$ is the natural numbers, $ \mu$ assigns measure $ 1$ to each $ n \in E$, and $ f(n) \equal{} a_n$ for all $ n$. There's nothing extra you need to prove.[/quote]

That's a bright way of looking at it! I like it!

 :D

The proof is literally exactly the same as the one used for $ L^{p}$; just replace integrals with sigmas and $ f$s and $ g$s with $ x_{n}$s and $ y_{n}$s. :wink:


My wife and I were invited to a party recently, a party attended by four other couples. Some of the 10 knew some of the others and some did not, and some were polite and some were not. As a result, a certain amount of handshaking took place in an unpredictable way, subject only to two obvious conditions: No one shook his or her own hand and no husband shook his wife’s hand. When it was all over, I became curious, and I went around the party asking each person: "How many hands did you shake?…And you? …And you?" What answers could I have received? Conceivably, some people could have said None, and others could have given me any number between 1 and 8 inclusive. That’s right isn’t it? Since self-handshakes and spouse-handshakes were ruled out, eight is the maximum number of hands that any one of the party of 10 could have shaken.

I asked nine people (everybody, including my own wife), and each answer could have been any one of the nine numbers 0 to 8 inclusive. I was interested to note, and I hereby report, that the nine different people gave me nine different answers: someone said None, someone said One, and so on and, finally, someone said Eight. Next morning, I told the story to my colleagues, and I challenged them, on the basis of the information just given, to tell me how many hands my wife shook.

 :D

isnt it like the sum of the number of hands shaken by the wife and the husband = 8.. So like:

Couple # 1: 0, 8
Couple # 2: 1, 7
Couple # 3: 2, 6
Couple # 4: 3, 5
Couple # 5: 4, 4

You said they all gave u different answers, therefore u and ur wife are Couple # 5..so your wife shook 4 hands >???..


[quote="dduclam"]If $ a,b,c$ be sidelengths of a triangle. Prove that
\[ \frac {1}{a^{2} \plus{} bc} \plus{} \frac {1}{b^{2} \plus{} ca} \plus{} \frac {1}{c^{2} \plus{} ab}\ge\frac {2(a \plus{} b \plus{} c)^2}{3(a^2 \plus{} b^2 \plus{} c^2)}\left(\frac1{(a \plus{} b)^2} \plus{} \frac1{(b \plus{} c)^2} \plus{} \frac1{(c \plus{} a)^2}\right)
\]
[hide]Using this result, we can prove my nother Inequality here http://www.mathlinks.ro/viewtopic.php?t=272947[/hide][/quote]
It is equivalent to
$ \sum \left[ 4a\plus{}3b\plus{}3c\minus{}\frac{3(b^3\plus{}c^3)}{a^2\plus{}bc}\minus{}\frac{2a^2}{b\plus{}c}\right] (a\minus{}b)(a\minus{}c) \ge 0.$
Now, Vornicu Schur solves it very well. :)

[quote="can_hang2007"][quote="dduclam"]If $ a,b,c$ be sidelengths of a triangle. Prove that
\[ \frac {1}{a^{2} \plus{} bc} \plus{} \frac {1}{b^{2} \plus{} ca} \plus{} \frac {1}{c^{2} \plus{} ab}\ge\frac {2(a \plus{} b \plus{} c)^2}{3(a^2 \plus{} b^2 \plus{} c^2)}\left(\frac1{(a \plus{} b)^2} \plus{} \frac1{(b \plus{} c)^2} \plus{} \frac1{(c \plus{} a)^2}\right)
\]
[hide]Using this result, we can prove my nother Inequality here http://www.mathlinks.ro/viewtopic.php?t=272947[/hide][/quote]
It is equivalent to
$ \sum \left[ 4a \plus{} 3b \plus{} 3c \minus{} \frac {3(b^3 \plus{} c^3)}{a^2 \plus{} bc} \minus{} \frac {2a^2}{b \plus{} c}\right] (a \minus{} b)(a \minus{} c) \ge 0.$
Now, Vornicu Schur solves it very well. :)[/quote]
I see this time I usually get mistake in calculating, so I am not so confident with my proof. And thus, it is better to post my full proof here. Thank you very much if someone can help me to check my proof. Here is the proof of this problem

I checked your nice proof, my dear friend.  :)

It is OK. :)

P.S. Have you proved, by this inequality, the other inequality

[quote="dduclam"]Let $ a,b,c$ be sidelengths of a triangle. Prove that
\[ \sum_{cyc}\frac1{a^2 \plus{} bc}\ge\frac3{4}(a^2 \plus{} b^2 \plus{} c^2)\left(\frac1{a^3b \plus{} b^3c \plus{} c^3a} \plus{} \frac1{ab^3 \plus{} bc^3 \plus{} ca^3}\right)
\]
[/quote]

Thank you very much.

Dear dduclam,

can you please explain how to prove your nice inequality posted at

http://www.mathlinks.ro/viewtopic.php?t=272947

using the other your nice inequality

\[ \frac {1}{a^{2} \plus{} bc} \plus{} \frac {1}{b^{2} \plus{} ca} \plus{} \frac {1}{c^{2} \plus{} ab}\ge\frac {2(a \plus{} b \plus{} c)^2}{3(a^2 \plus{} b^2 \plus{} c^2)}\left(\frac1{(a \plus{} b)^2} \plus{} \frac1{(b \plus{} c)^2} \plus{} \frac1{(c \plus{} a)^2}\right)
\]

I tried by full expansion but it is ugly. :(

Thank you very much. :)

Dear manlio,

I am extremely sorry for detected some errors when using this Inequality to prove my other at http://www.mathlinks.ro/viewtopic.php?t=272947 .  However I still think that the following Inequality is corect for all $ a,b,c$ be sidelengths of a triangle:
\[ \frac1{(a\plus{}b)^2}\plus{}\frac1{(b\plus{}c)^2}\plus{}\frac1{(c\plus{}a)^2}\ge\frac{9(a^2\plus{}b^2\plus{}c^2)^2}{8(a\plus{}b\plus{}c)^2}\left(\frac1{a^3b\plus{}b^3c\plus{}c^3a}\plus{}\frac1{ab^3\plus{}bc^3\plus{}ca^3}\right)
\]
Do you think so ?

[hide]This time, I do not have much time to online and solve Inequality, sorry.  :oops: [/hide]

Dear dduclam,

thank you very much for your answer. :)

I will try to prove it. :wink:


calculus
integration
logarithms
trigonometry
calculus computations

I think I may have succeeded in solving this problem, but my rough work is quite huge and cumbersome. I will just recheck my calculations and hopefully post them soon.

[color=darkblue]Suppose $x>0\ ,\ y>0$ and denote $I(x,y)\equiv\int_{0}^{\frac{\pi}{2}}\ln\left(\sqrt{x^{2}+y^{2}\sin^{2}t}+y\sin t\right)\ dt\ .$ Then the our integral is $I=I(x,y)-I(y,x)\ .$ Thus,
$\boxed{\ \begin{array}{ccc}u(t)=\ln\left(\sqrt{x^{2}+y^{2}\sin^{2}t}+y\sin t\right) & \Longrightarrow & u'(t)=y\cdot\frac{\cos t}{\sqrt{x^{2}+y^{2}\sin^{2}t}}\\ v'(t)=1 & \Longrightarrow & v(t)=t\end{array}\ }$ $\Longrightarrow$ $I(x,y)=\left[t\cdot\ln\left(\sqrt{x^{2}+y^{2}\sin^{2}t}+y\sin t\right)\right]_{0}^{\frac{\pi}{2}}-y\cdot K(x,y)$, where $\boxed{\ K(x,y)=\int_{0}^{\frac{\pi}{2}}\frac{t\cos t}{\sqrt{x^{2}+y^{2}\sin^{2}t}}\ dt\ }\ .$ Therefore, $I(x,y)=\frac{\pi}{2}\cdot\ln\left(\sqrt{x^{2}+y^{2}}+y\right)-y\cdot K(x,y)\ .$ [b]Ready, sorry, I am blocked !!

I"ll begin again.[/b] Observe that $I(x,y)=\frac{\pi}{2}\cdot\ln\frac{y}{x}+F\left(\frac{x}{y}\right)-F\left(\frac{y}{x}\right)$, where $F(k)=\int_{0}^{\frac{\pi}{2}}\ln\left(\sqrt{k^{2}+\sin^{2}t}+\sin t\right)\ dt=$ $\frac{\pi}{2}\cdot \ln\left(1+\sqrt{1+k^{2}}\right)-L(k)$, where $L(k)=\int_{0}^{\frac{\pi}{2}}\frac{t\cos t}{\sqrt{k^{2}+\sin^{2}t}}\ dt\ .$ [b]Ready, sorry, I am blocked again !! I renounce and wait the yours solutions for this difficult (for me) exercise.[/b][/color]


The integer $n$ is the smallest positive multiple of 15 such that every digit of $n$ is either 8 or 0.  Compute $\frac{n}{15}$.

[hide="Solution"]Let's consider divisibility by $3$ and $5$ seperately. For $n$ to be divisible by $5$, its units digit must be $0$ or $5$. However, we are restricted to $0$ or $8$ as our units digits, so we must use $0$.

For $n$ to be divisible by $3$, the sum of its digits must be divisible by $4$. We need at least $3$ eights to do this, so the smallest possible $n$ is $n=8880$. Our answer is therefore $\frac{n}{15}=592$. [/hide]

[hide="Answer"]The number must end in $0$, since it is a multiple of $5$, and the other digits must sum to a multiple of $3$, so we have $8880\Rightarrow \frac{8880}{15}=\boxed{592}$.[/hide]

[hide]I plugged and chugged $880, 8800, 8880$, and arrived at $8880$ and $\frac{8880}{15} = \boxed{592}$[/hide]

has to be a multiple of 3 and a multiple of 5 to be divisible by 15. So it has to end in 80, and and be divisble by 3. Zero doesnt affect the sum of the digits so the least possible sum is 8x3 =24. The number that satisifies both conditions is 8880. 8880/15 = 592

Obviously the last digit will be 0 and since it is a multiple of 15, it is also a multiple of 3 and hence the first 3 digits can be 888. So, n = 8880. Hence, n/15 = 592

If it is divisible by 15, then it must be be a multiple of both 3 and 5.

If it divisible by 5, then the last digit must be 0 or 5. But we are restricted to 8 and 0. So the last digit is 0 of this number.

If it is divisible by 3, then the sum of the digits must be divisible by 3. The Lcm of 8 and 3 is 24. Hence, the number of 8's at minimum will be 3.

So, our minimum number will be $\boxed{8880}$.

Dividing 15 yields:
$$\boxed{592}$$

Oops! I forgot to divide by 15. Check it now!

I'm international so Idk rules of aime :)

[hide]the number has to be a multiple of 15, and it only consists of 8and 0. It ends with 0. I just id some guess and check. The you get n=8880. THe rest is easy.[/hide]

15 = 3*5 so if it is multiple of 3 it is 3 8s and multiple of 5 it has to end in 0
so 8880/15=592

[quote=AwesomeLife_Math][quote=bookstuffthanks]If it is divisible by 15, then it must be be a multiple of both 3 and 5.

If it divisible by 5, then the last digit must be 0 or 5. But we are restricted to 8 and 0. So the last digit is 0 of this number.

If it is divisible by 3, then the sum of the digits must be divisible by 3. The Lcm of 8 and 3 is 24. Hence, the number of 8's at minimum will be 3.

So, our minimum number will be $\boxed{8880}$[/quote]

Forgot to divide by $15$ (although you can't put in $8880$ on an AIME)[/quote]

This wouldn't be aime leveled

[quote=OlympusHero]Dude, how in the world is this AIME?

[hide=Solution]We must end in a $0$ to be a multiple of $5$, since we can't use the digit $5$. Therefore, we just try to minimize the amount of $8$'s that we need in order to get a multiple of $3$ (putting in a $0$ affects nothing). Since $3|8 \cdot 3$, the smallest possible number is $8880 \implies \boxed{592}$.[/hide][/quote]

Exactly

Case 1: $n$ has $1$ digit.
No possibility.

Case 2: $n$ has two digits.
No possibility.

Case 3: $n$ has three digits.
No possibility.

Case 4: $n$ has four digits.
Then $n=8880$. $\frac{8880}{15}=\boxed{592}$.

since $15|n$
so $3|n$ and $5|n$
using $5|n$ ,the first digit only may 5 or 0, and in this problem we only can use 0 and 8, so first digit is 0
using $3|n$, the necessity and sufficiency condition is 3 is factor of n's digit sum
if n is two digit number, second digit number must is 8 $3|n$ means $3|8$, its false
let n is k digit number, $3|n$ means $3|8x$.
x is number in set$ {1,2,...,k-1}$, Only one x satisfies we can conclude there have a n satisfy $15|n$ in k digit
let solve $3|8(k-1)$, when $k=4, 3|24$
so only n is four digit number.
using $3|n, n=8880$

$\color{blue} \boxed{\textbf{SOLUTION}}$

$3|n \implies n$ has atleast $3, 8's$
$5|n \implies n's$ last digit is $0$

Smallest $n=8880 \implies \frac{n}{15}=592$

[quote=OlympusHero]Dude, how in the world is this AIME?

[hide=Solution]We must end in a $0$ to be a multiple of $5$, since we can't use the digit $5$. Therefore, we just try to minimize the amount of $8$'s that we need in order to get a multiple of $3$ (putting in a $0$ affects nothing). Since $3|8 \cdot 3$, the smallest possible number is $8880 \implies \boxed{592}$.[/hide][/quote]

1984 problem 2.

Old year, early problem, makes sense its easy.

[hide]We know that the number has to be divisible by 3 and 5 so 8880 works and is the smallest. 8880/15 = 592.[/hide]

Also rn it would be like a number 6 on amc 10

[hide=solution]
The units digit is $0$, now we need the sum to be a multiple of $3$, adding $0$ won't help, so we have that $\frac{8880}{15}=\boxed{592}.$

[/hide]


In the diagram, there are more than three triangles. If each triangle has the same probability of being selected, what is the probability that a selected triangle has all or part of its interior shaded? Express your answer as a common fraction.

[asy]draw((0,0)--(1,0)--(0,1)--(0,0)--cycle,linewidth(1));
draw((0,0)--(.5,0)--(.5,.5)--(0,0)--cycle,linewidth(1));

label("A",(0,1),NW);
label("B",(.5,.5),NE);
label("C",(1,0),SE);
label("D",(.5,0),S);
label("E",(0,0),SW);

filldraw((.5,0)--(1,0)--(.5,.5)--(.5,0)--cycle,gray,black);[/asy]

There are 5 different triangles that can be selected. Of these, only 3 have the shaded triangle. So our fraction is $ \boxed{\frac{3}{5}}$.


This might seem kind of stupid but its my first time using LaTeX

After, I compile and then click output, nothing happens. The tutorial says I have to change something to LaTeX => PDF but I don't know where that is. I know I installed everything.

Maybe it's just me.

There is a little dropdown window on the top (below the toolbar, above the body, like where all of the symbols are in Microsoft Word).  Anyway, it shouldn't matter if it's a PDF or DVI, it should render anyway.  Maybe there's something else.

[i]I have moved your post to the correct area[/i]

Are you using TeXnicCenter? Have you told it where to find MiKTeX?
1. If the answers are yes and no then
Go to Build, Define Output Profiles and press the Wizard button at the bottom left. Then navigate to where MiKTeX binary files are installed probably something like C:\Program Files\MiKTeX 2.6\miktex\bin and complete the Wizard.
When you have completed it, change the LaTeX=>DVI in the toolbar to LaTeX=>PDF then compile.

2, if the answers are different you will need to give much more detail.


function symmetry counting derangement algebra polynomial probability

Let $S_n$ be the set of all permutations on $n$ elements. For $s \in S_n$, let $Fix(s)$ be the number of elements fixed by the permutation $s$.
Now define for positive integers $n,k$:
\[{ a_{n,k} : =\frac{1}{n!}\sum_{s \in S_n} Fix(s)^k}  \]
a) all $a_{n,k}$ are integers
b) for given $k$ and $m,n \geq k$, we have that $a_{m,k} = a_{n,k}$
c) find a way to express $a_{n,n}$, e.g as recurrence or as a combinatorial problem.

Okay, I remember someone asked me this question via PM a few months ago.  As I recall, $a_{n,k}$ happens to be the number of partitions of $1..k$ into at most $n$ nonempty subsets.  In particular, for $n \ge k$, the answer is the $k$th [url=http://mathworld.wolfram.com/BellNumber.html]Bell Number[/url].

I'll try to post a proof later.

[quote="Ravi B"]  As I recall, $a_{n,k}$ happens to be the number of partitions of $1..k$ into at most $n$ nonempty subsets.  [/quote]

After reading your posting, I confirmed this by a (not difficult) calculation with generating functions.  It would be interesting to see a more direct argument.

My prove (also giving the full statement with the partitions) uses group theory...
But as fleeting_guest, I'm interested in a direct combinatorial approach.

That's interesting.  I tend not to think of sophisticated ideas such as group theory or generating functions unless I'm stuck.  I would be interested in hearing more about your proofs.  Anyway, here's the argument I had in mind.

First, I find it convenient to restate the problem in probabilistic terms.  Given a random permutation $\pi$ of $1..n$, what is the expected value of $\mathrm{Fix}(\pi)^k$?

I will use the "falling powers" notation
\[ x^{\underline{j}} = x(x-1) \cdot \cdots \cdot (x - j + 1) \, .  \]
Using a probabilistic technique I used in an earlier thread called [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23684]Permutations and fixed points[/url], we can show that the expected value of $\mathrm{Fix}(\pi)^{\underline{j}}$ is $1$ if $n \ge j$, and $0$ otherwise.

Now we can translate from falling powers to ordinary powers by using the [url=http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html]well-known formula[/url]
\[ x^k = \sum_{j = 0}^\infty S(k, j) x^{\underline{j}} \, .  \]
Here $S(k, j)$, the Stirling number (of the second kind),  is the number of ways of partitioning $1..k$ into exactly $j$ nonempty sets.  Let's plug in $x = \mathrm{Fix(\pi)}$ and take expected value of both sides.  We will then get the promised formula.

Since nobody seems to have found the following combinatorial argument, I will post it:

Consider the set of $k$-tuples $x=(x_1,...,x_k)$ with $x_i\in N=\{1,...,n\}$. Denote by $a_x$ the number of permutations $\sigma$ with $\sigma(x_i)=x_i$ for all $i=1,...,k$. Then it is immediate to see that

$a_{n,k} = \frac 1{n!} \sum_{\sigma \in S_n} {\text Fix}(\sigma)^k = \frac 1{n!} \sum_{x\in N^k} a_x$.

Now consider any partition of $\left\{1,2,...,k\right\}$ into $l$ nonempty subsets $P=\{N_1,...,N_l\}$ with $l\leq n$. Consider the set $X(P)$ of those $x$ with $x_i=x_j$ if $i,j\in N_m$ for some $m$, else $x_i\neq x_j$. For each such $x$, we have $a_x = (n-l)!$. On the other hand, $|X(P)| = \frac{n!}{(n-l)!}$. It follows that

$\sum_{x\in X(P)} a_x = \sum_{x\in X(P)} (n-l)! = n!$.

Since every $x\in N^k$ lies in $X(P)$ for one and only one $P$ with $l\leq n$, the conclusion follows.

Peter

Here goes mine (very similar the Peter's when I look back):
Let the group $S_n$ of the permutations of $N: =\{1,2,...,n\}$ act on $N^k$ by $\sigma(x_1,x_2,...,x_k) : = (\sigma (x_1),\sigma (x_2),...,\sigma (x_k))$ for all $\sigma \in S_n$.
If $g$ had $f$ fixpoints under the action on $N$, it will have exactly $f^k$ fixpoints under the action on $N^k$. By the nice theorem in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60832 , $a_{n,k}$ equals now the number of orbits generated by this action, so is indeed an integer.

To get now the interpretation as the partitions of sets, see that (since $S_n$ acts $n$-times transitively on $N$) two elements $(x_i),(y_i) \in N^k$ are in the same orbit iff $x_i=x_j \iff y_i=y_j$, so the number of orbits equals the number of partitions of a set with $k$ elements into at most $N$ subsets (the bijection is given by: $k_i,k_j \in \{ 1,2,...,k\}$ are in the same subset iff $x_i=x_j$).

This problem came up in my graduate algebra course this semester.  My prof tackled it with generating functions, but I had a way which I thought was pretty nice for small values, but which doesn't generalize well.

If $k=1$, then
\[ \sum_{\sigma\in S_n} \text{fix}(\sigma)=n!  \]
Proof: Clearly, there are $(n-1)!$ permutations which fix $1$.  By symmetry, there are $(n-1)!$ permutations which fix the $i$th place, for $i=1,2,\ldots,n$.  Define $\text{fix}_i(\sigma)=1$ if $\sigma$ fixes the $i$th place, and $\text{fix}_i(\sigma)=0$ otherwise.  Then
\[ \text{fix}(\sigma)=\sum^n_{i=1} \text{fix}_i(\sigma)  \]
and so
\[ \sum_{\sigma\in S_n}\sum^n_{i=1} \text{fix}_i(\sigma)=\sum^n_{i=1}\sum_{\sigma\in S_n}\text{fix}_i(\sigma)=\sum^n_{i=1}(n-1)!=n(n-1)!=n!  \]
as desired.

For $k=2$, we get
\begin{eqnarray*} \sum_{\sigma\in S_n} \text{fix}(\sigma)^2&=&\sum_{\sigma\in S_n}\left(\sum^n_{i=1} \text{fix}_i(\sigma)\right)^2 \\ &=& \sum_{\sigma\in S_n} \left(\sum^n_{i=1} \text{fix}_i(\sigma)^2+\sum_{i\not=j} \text{fix}_i(\sigma)\text{fix}_j(\sigma)\right) \\ &=& n!+\sum_{i\not=j}\sum_{\sigma\in S_n} \text{fix}_i(\sigma)\text{fix}_j(\sigma) \\ &=& n!+n(n-1)\cdot (n-2)! \\ &=&2(n!) \end{eqnarray*}
if $n\geq 2$.

[quote="Ravi B"]I would be interested in hearing more about your proofs.[/quote]From the definition (writing a permutation as a derangement on the non-fixed points and the identity on the fixed points), $a_{n,k}$ is the $X^n$ coefficient in $(\Sigma {D_m}X^m/m!)(\Sigma f^k X^f/ f!)$ and the rest is equivalent to the other solutions posted here.

[quote]First, I find it convenient to restate the problem in probabilistic terms.  Given a random permutation $\pi$ of $1..n$, what is the expected value of $\mathrm{Fix}(\pi)^k$? [/quote]Expected value is how I viewed it as well, but that seems to motivate only the problem, not the solution.

[quote] $x^{\underline{j}} = x(x-1) \cdot \cdots \cdot (x - j + 1)$. We can show that the expected value of $\mathrm{Fix}(\pi)^{\underline{j}}$ is $1$ if $n \ge j$, and $0$ otherwise. [/quote]That is the key step.  How does one see in advance that the $F^{\underline{j}}$ is the correct basis to use for polynomials in $F$ (i.e. that the expectation of these elements has a very simple form)?

Well, if we try to expand the expected value of $\mathrm{Fix}(\pi)^k$, we get a sum of probabilities of the form $\Pr(\pi \textrm{ fixes } i_1, \ldots, i_k)$.  Unfortunately, this probability depends on how many of the $i_1$, $\ldots$, $i_k$ are distinct.  That makes the sum an awkward sum of cases.  It's possible to compute this sum -- Peter shows how -- but it requires some care.

The advantage of the falling power $\mathrm{Fix}(\pi)^{\underline{k}}$ is that we get the same sum of probabilities, except that $i_1$, $\ldots$, $i_k$ are guaranteed to be [i]distinct[/i].  That means each probability term is equal, which makes the sum a lot easier to compute, at least for me.

So I was [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=257646]led to this question[/url] by the following proof in the $ k \equal{} 2$ case:

Let $ S_n$ act on $ \mathbb{C}^n$ by permutation matrices.  The character of this representation is just $ \text{Fix}(\pi)$.   This representation decomposes into a direct sum $ V \oplus V^{\perp}$ where $ V \equal{} \{ \left< 1, 1, 1, ... 1 \right> \}$ and it is not too hard to show that $ V^{\perp}$ is irreducible, so $ \text{Fix}$ is a sum of two irreducible characters and by the orthogonality relations it follows that $ \left< \text{Fix}, \text{Fix} \right> \equal{} 2$.

Does this proof generalize?  I rather like the combinatorial proof, but generalizing the above might say something interesting about $ L^k$ norms.

Also, I am curious as to what the answer is when $ k$ is a non-integer; in particular, is it still independent of $ n$?  In what sense is this a meaningful generalization of the Bell numbers?

[quote="t0rajir0u"]Also, I am curious as to what the answer is when $ k$ is a non-integer; in particular, is it still independent of $ n$?[/quote]  No.  However, we can take the limit and get some value out that would presumably be the Bell analogue you're suggesting.  I have no intuitive sense of its meaningfulness.  The limit seems to converge pretty quickly -- for $ k \equal{} 1/2$ it seems to be 0.77319... while for $ k \equal{} 3/2$ it's 1.372732....

By the way, t0r, I think billzhao proposed a similar method here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=129712[/url].

Alright:  let $ \rho_n$ be the representation of $ S_n$ on $ \mathbb{C}^n$ by permutation  matrices and define $ \rho \equal{} \rho_n^{\otimes k}$.  The character of this representation is $ \chi(\pi) \equal{} \text{Fix}(\pi)^k$, so the desired result is equivalent to the claim that the trivial representation occurs as a summand $ B_{k, n}$ times in $ \rho$, or equivalently that there are $ B_{k,n}$ invariant subspaces of dimension $ 1$.  But I think (although I don't understand $ \rho$ enough to be sure) these are precisely the averages over the orbits in ZetaX's solution, so this is not essentially different from the combinatorial solution.


a game in which a coin is flipped n times and stopped when either heads or tails is flipped 10 times. there's a 50 dollar prize. if the coin is flipped heads 10 times first, A wins the prize. and if the coin is flipped tails 10 times first, B wins the prize. However this time the game stops at heads 8 times, and tails 7 times. how should A and B split the 50 dollar prize.

I assume they should split the prize based on the probability of each player winning if the coin was flipped until a winner was found.

After 4 more flips, somebody must win.  So we have _ _ _ _, where we must fill each _ with A or B.  There are $2^{4}=16$ ways to fill in the blanks.  For player B to win, we must fill at least 3 of the blanks with B.  There are 5 ways to do this.  Therefore, player B should get 5/16 of the $\$50$, and player A should get 11/16 of the $\$50$.  Player A should get $\$34.37$ and Player B should get $\$15.62$.  Use the other cent to buy something expensive. :P

is there a general solution to this? say if they stopped at 4 heads and 3 tails, how would u do it then?


ratio
geometry
circumcircle
geometry solved

Given a triangle $ABC$.
Let $M$ and $N$ be the points where the angle bisectors of the angles $ABC$ and $BCA$ intersect the sides $CA$ and $AB$, respectively.
Let $D$ be the point where the ray $MN$ intersects the circumcircle of triangle $ABC$.

Prove that $\frac{1}{BD}=\frac{1}{AD}+\frac{1}{CD}$.

Hint:
A_1, B_1, C_1 the orthogonal projection of D onto BC, CA, AB
express
DB_1 in terms of DA, DC, R the circumradius of ABC
DA_1 in terms  of DB, DC
DC_1 in terms of DA, DB

Draw the perpendiculars [tex]DX=x,DY=y,DZ=z[/tex] from [tex]D[/tex] to [tex]CB,BA,AC[/tex].
Then:[tex]x+y=z[/tex]...(1)
(Some time ago I have posted this problem in:
http://www.personal.us.es/rbarroso/trianguloscabri/pro150.htm )
Sorry for the spanish.
Also by similar pairwise triangles:
[tex]BYD,CZD: \frac{y}{BD}=\frac{z}{CD}[/tex] ..(2)
[tex]BXD,AZD: \frac{x}{BD}=\frac{z}{AD}[/tex] ..(3)
From (1),(2) and (3):
[tex]\frac{1}{BD}=\frac{1}{AD}+\frac{1}{CD}[/tex]       QED.

I can see that a solution has been posted, but I took the time to write this, so..:)

At first I struggled really hard to find a non-computational proof, but I failed. However, the computational proof I found is far from nasty.

Let $T=AD\cap BC$, and let $a,b,c=BC,CA,AB$ respectively. We use three things: $(1)$: triangles $TBD,TAC$ are similar, so $\frac {TD}{BD}=\frac{TC}b$; $(2)$: triangles $TDC,TBA$ are similar, so $\frac{TD}{CD}=\frac{TB}c$; $(3)$ from the transversal theorem we get $\frac {BN}{NA}TC=\frac{CM}{MA}TB+\frac{TD}{AD}BC\iff \frac ab TC=\frac ac TB+\frac{TD}{AD}a$. we deivde everything by $a$ and we get $\frac {TC}b=\frac{TB}c+\frac{TD}{AD}$.

From $(1),(3)$ we get $\frac {TB}c=TD(\frac 1{BD}-\frac 1{AD})\ (*)$, and from $(*),(2)$ we get $\frac{TD}{CD}=TD(\frac 1{BD}-\frac 1{AD})$, and by dividing this last equality with $TD$ we get exactly what we want.

I think it has been posted before, btw, because I remember struggling to find an inverseive proof some time ago. This idea was inspired by the fact that after an inversion of pole $D$ we would have to show that $DB'=DA'+DC'$, where, in general, $X'$ is the image of $X$ under the inversion, but this seems much harder and I could make no progress whatsoever.


analytic geometry
ratio
geometric sequence
real analysis
real analysis unsolved

Let number sequence $ (u_n)$ so that: $ \begin {cases} u_1\equal{}1 \\ u_{n\plus{}1}\equal{}\frac{1}{2}u_n\minus{}1 \end {cases}$

Find formula of $ u_n$ by $ n$?

Note that $ u_{n\plus{}1} \plus{} 2 \equal{} \frac{1}{2} (u_n \plus{} 2)$.  What can you do from here?

$ (u_n,\ u_{n \plus{} 1})\in y \equal{} \frac {1}{2}x \plus{} 1$, cosnider the $ x$ coordinate $ \alpha$ of the intersection point of the two lines $ y \equal{} \frac {1}{2}x \minus{} 1$ and $ y \equal{} x$, we have $ \alpha \equal{} \frac {1}{2}\alpha \minus{} 1$ and $ \frac {u_{n \plus{} 1} \minus{} \alpha}{u_n \minus{} \alpha} \equal{} \frac {1}{2}$
that is to say, $ \alpha \equal{} \minus{} 2$ leads to the recurrsive formula $ u_{n \plus{} 1} \plus{} 2 \equal{} \frac {1}{2}(u_n \plus{} 2)$. Thus $ \{u_n \plus{} 2\}$ is a geometric sequence with common ratio $ \frac {1}{2}$, yielding $ u_n \plus{} 2 \equal{} \left(\frac {1}{2}\right)^{n \minus{} 1}(u_1 \plus{} 2) \equal{} 3\left(\frac {1}{2}\right)^{n \minus{} 1}$, yielding $ u_n \equal{} 3\left(\frac {1}{2}\right)^{n \minus{} 1} \minus{} 2\ (n\geq 1)$.

Thank [u][b]kunny[/b][/u] and [u][b]t0rajir0u[/b][/u] very much.

why cant we define

$ y_n\equal{}u_n\plus{}2$

then $ y_{n\plus{}1}\equal{}\frac{y_n}{2}$

$ u_1\equal{}1 \to y_1\equal{}3$

$ y_n \equal{} \frac{1}{2} y_{n\minus{}1} \equal{} ... \equal{} \frac{1}{2^k} y_{n\minus{}k}\equal{}\frac{1}{2^{n\minus{}1}}y_1\equal{}\frac{6}{2^n}$

so $ u_n \equal{} y_n \minus{}2 \equal{} 2[\frac{3}{2^n}\minus{}1]$

i think kunny has a typo (first time for everything)

[quote="kenn4000"]why cant we define

$ y_n \equal{} u_n \plus{} 2$

then $ y_{n \plus{} 1} \equal{} \frac {y_n}{2}$

$ u_1 \equal{} 1 \to y_1 \equal{} 3$

$ y_n \equal{} \frac {1}{2} y_{n \minus{} 1} \equal{} ... \equal{} \frac {1}{2^k} y_{n \minus{} k} \equal{} \frac {1}{2^{n \minus{} 1}}y_1 \equal{} \frac {6}{2^n}$

so $ u_n \equal{} y_n \minus{} 2 \equal{} 2[\frac {3}{2^n} \minus{} 1]$

i think kunny has a typo (first time for everything)[/quote]

Thank kenn4000!!


limit
Sequence
algebra
recurrence relation
IMO Shortlist

Define a sequence $\langle f(n)\rangle^{\infty}_{n=1}$ of positive integers by $f(1) = 1$ and \[f(n) =  \begin{cases} f(n-1) - n & \text{ if } f(n-1) > n;\\ f(n-1) + n & \text{ if } f(n-1) \leq n, \end{cases}\]
for $n \geq 2.$  Let $S = \{n \in \mathbb{N} \;\mid\; f(n) = 1993\}.$

[b](i)[/b] Prove that $S$ is an infinite set.
[b](ii)[/b] Find the least positive integer in $S.$
[b](iii)[/b] If all the elements of $S$ are written in ascending order as \[ n_1 < n_2 < n_3 < \ldots ,  \] show that \[ \lim_{i\rightarrow\infty} \frac{n_{i+1}}{n_i} = 3.  \]

i think it is very easy.I have $f(3)=6$ and if $f(n)=2n$ then $f(3n-1)=2(3n-1)$

Let $m_1=3,m_{k+1}=3m_k-1$, then first mean when $f(n_1)=a>0$ is $3m_l-1-2a$, were l is first suth that $m_l>a$. In this case (a=1993) $n_1=12417$. All terms satisfyed recurent equation $n_{k+1}=3n_k+4a-1$. Therefore $n_k=3^{k-1}n_1+(3^{k-1}-1)\frac{4a-1}{2}=\frac{5*3^{k+7}-7971}{2}$.

Umm.... I have a doubt I am trying to prove the first part.
Suppose $f(n-1)\leq n$. Then, $f(n)=f(n-1)+n$, which will be greater than $n$. Thus, $f(n+1)=f(n)-n=f(n-1)$. Hence, the sequence is periodic with period 2. But, this is not happening here. What is the mistake?

$f(n)$ is OEIS [url=https://oeis.org/A046901]A046901[/url]

It is a concatenation $S_0,S_1,S_2,S_3,\cdots$ where $S_n$ is the sequence $b_0,b_1,\cdots,b_{k-1}$ for $k=5\cdot 3^n$ where $b_0=1$, $b_{2j}=k+j$, and $b_{2j+1}=\frac{k+1}{2}-j$

In each block $S_n$ with $n\geq 7$, we have $b_{5\cdot 3^n-3984}=1993$. So for $n\geq 7$ we have $f\left(\frac{3(5\cdot 3^n-2657)}{2}\right)=1993$.

[quote=InkretBear]It is greater than $n$ but not necessarily greater than $n+1$.[/quote]

Oops! Right. Thanks Inkretbear :)


prove that if n divides  2^(n)+1 then n=3 or 9 divides n

[quote="behzad"][quote="Inspired By Nature"]Which is the question:
-prove that if $n$ divides $2^{n}^{+}^{1}$ then $n=3$ or $n$ divides $9$
OR
--prove that if $n$ divides $2^{n}+1$ then $n=3$ or $n$ divides $9$[/quote]
Really isn't it obvious? :o[/quote]
I have to make sure what the author means behzad. Your interpretation might be different than mahyar_math's. :|

كاملا واضح است كه منظورم دوميه

--prove that if n divides 2^{n}+1 then n=3 or 9 divides n

[quote="mahyar_math"]كاملا واضح است كه منظورم دوميه

--prove that if n divides 2^{n}+1 then n=3 or n divides 9[/quote]
oh, I see.
well if $n=3$, then $2^{n}+1$=9. This is obviously divisible by 9 or 3.
This isn't a proper proof, but you gave the anwser when you said "or n divides 9". :)

[quote="mahyar_math"]manzorm in bood ke 9 divides n
excuse me :oops:[/quote]
No problem


calculus
integration
trigonometry
function
calculus computations

Calculate: $ \int \frac{x^7\plus{}1}{cos^2x} dx$

First step: it equals $ \int x^7\sec^2x\,dx\plus{}\int\sec^2x\,dx.$

The second integral is well known, so you may focus your efforts on the first integral.

I thing the problem is interesting. It seems easy but it is not so easy. May you continue the solution? (This integral is form Bulgarian National Students Olympiad 2000)

[quote="Kent Merryfield"]
First step: it equals $ \int x^7\sec^2x\,dx \plus{} \int\sec^2x\,dx.$
[/quote]

Use:
$ \int uv \equal{} u\int v \plus{} u'\int \int v \plus{} u''\int \int \int v \plus{} u''' \int \int \int \int v \plus{} \ldots$

Choose the $ u$ wisely, here $ u$ being $ x^7$ so that the terms terminate ultimately.

Rajiv, this observation is very interesting. Could you (or someone else) post a detailed solution?

In the first place, that formula is not true (there should be minus signs in the even ordered terms of the right member). Second, it does not help here, since for $ v \equal{} \sec^2 x$,

$ \int v \equal{} \int \sec^2x\,dx \equal{} \tan x$

$ \int \int v \equal{} \int \tan x \,dx \equal{} \minus{} \ln|\cos x|$

and now you cannot proceed to compute $ \int \int \int v$.

[quote="Carcul"]
In the first place, that formula is not true (there should be [b]minus signs[/b] in the even ordered terms of the right member.

[/quote]

 :oops: Yeh right!!! Slip of the finger!!! :D

[hide="The proof:"]
$ \boxed{\int uv \equal{} u\int v \minus{} u'\int \int v \plus{} u''\int \int \int v \minus{} u''' \int \int \int \int v \plus{} \ldots}$
We know,

$ \int uv \equal{} u\int \minus{} \int\int v u'
\\
\equal{}u\int v \minus{} (u'\int\int v \minus{} \int\int v u'')$
And so on...
[/hide]

But as pointed out by Carcul, this won't help in this case.

I'm interested if the integral I posted is solvable in elementary functions.

Doubt it. Mathematica spits out a bunch of polylog stuff:
The $ x^7 \sec x$ part works out to
$ \minus{}\frac{141 i \pi ^8}{2048}\minus{}\frac{15}{128} i \pi ^7 x\plus{}\frac{35}{128} i \pi ^6 x^2\minus{}\frac{7}{32} i \pi ^5 x^3\plus{}\frac{i x^8}{8}\minus{}\frac{7}{64} \pi ^7 \text{Log}\left[1\minus{}i e^{\minus{}i x}\right]\plus{}\frac{35}{64} \pi ^6 x \text{Log}\left[1\minus{}i e^{\minus{}i x}\right]\minus{}\frac{21}{32} \pi ^5 x^2 \text{Log}\left[1\minus{}i e^{\minus{}i x}\right]\minus{}\frac{1}{128} \pi ^7 \text{Log}\left[1\plus{}i e^{\minus{}i x}\right]\plus{}x^7 \text{Log}\left[1\plus{}i e^{\minus{}i x}\right]\plus{}\frac{15}{128} \pi ^7 \text{Log}\left[1\plus{}i e^{i x}\right]\minus{}\frac{35}{64} \pi ^6 x \text{Log}\left[1\plus{}i e^{i x}\right]\plus{}\frac{21}{32} \pi ^5 x^2 \text{Log}\left[1\plus{}i e^{i x}\right]\minus{}x^7 \text{Log}\left[1\plus{}i e^{i x}\right]\plus{}\frac{1}{128} \pi ^7 \text{Log}\left[\text{Cot}\left[\frac{1}{4} (\pi \minus{}2 x)\right]\right]\plus{}7 i x^6 \text{PolyLog}\left[2,\minus{}i e^{\minus{}i x}\right]\plus{}\frac{7}{64} i \pi ^5 (5 \pi \minus{}12 x) \text{PolyLog}\left[2,i e^{\minus{}i x}\right]\plus{}\frac{35}{64} i \pi ^6 \text{PolyLog}\left[2,\minus{}i e^{i x}\right]\minus{}\frac{21}{16} i \pi ^5 x \text{PolyLog}\left[2,\minus{}i e^{i x}\right]\plus{}7 i x^6 \text{PolyLog}\left[2,\minus{}i e^{i x}\right]\plus{}42 x^5 \text{PolyLog}\left[3,\minus{}i e^{\minus{}i x}\right]\minus{}\frac{21}{16} \pi ^5 \text{PolyLog}\left[3,i e^{\minus{}i x}\right]\plus{}\frac{21}{16} \pi ^5 \text{PolyLog}\left[3,\minus{}i e^{i x}\right]\minus{}42 x^5 \text{PolyLog}\left[3,\minus{}i e^{i x}\right]\minus{}210 i x^4 \text{PolyLog}\left[4,\minus{}i e^{\minus{}i x}\right]\minus{}210 i x^4 \text{PolyLog}\left[4,\minus{}i e^{i x}\right]\minus{}840 x^3 \text{PolyLog}\left[5,\minus{}i e^{\minus{}i x}\right]\plus{}840 x^3 \text{PolyLog}\left[5,\minus{}i e^{i x}\right]\plus{}2520 i x^2 \text{PolyLog}\left[6,\minus{}i e^{\minus{}i x}\right]\plus{}2520 i x^2 \text{PolyLog}\left[6,\minus{}i e^{i x}\right]\plus{}5040 x \text{PolyLog}\left[7,\minus{}i e^{\minus{}i x}\right]\minus{}5040 x \text{PolyLog}\left[7,\minus{}i e^{i x}\right]\minus{}5040 i \text{PolyLog}\left[8,\minus{}i e^{\minus{}i x}\right]\minus{}5040 i \text{PolyLog}\left[8,\minus{}i e^{i x}\right]$

[quote="borislav_mirchev"]
This integral is form Bulgarian National Students Olympiad 2000

[/quote]

Do you have the answer???

No, unfortunately I haven't the brochure. I found the theme in a Bulgarian Math Magazine. I have few ideas:
1. Integral is not easy at all.
2. There are some bounds not mentioned.
3. There is some other typo or gap in the idea for solving.

I'd go for number 3, as it's 99% certain that no elementary antiderivative of the original function exists, and I don't see some nice boundaries for it (notice that the cosine in the denominator becomes $ 0$ every now and then).

[quote="borislav_mirchev"]
No, unfortunately I haven't the brochure.

[/quote]
Are you sure it is $ x^7$? :D

I wrote the integral as it was in the magazine. What if it isn't x^7?

$ \cos^2 x \equal{} \frac {1 \plus{} \cos 2x}{2}$

$ \frac {x^7}{\cos^2 x} \equal{} \frac {2x^7}{1 \plus{} \cos 2x} \equal{} 2x^7\sum_{k \equal{} 0}^{\infty}{( \minus{} 1)^k\cos^{k}2x}$

$ \int{\frac {x^7}{\cos^2 x}\text{d}x} \equal{} 2\sum_{k \equal{} 0}^{\infty}{( \minus{} 1)^{k} \int{x^7\cos^{k}2x\;\text{d}x}}$ Each integral we can evaluate :D

Another way: $ \int{x^7\text{d}(\tan x)} \equal{} x^7 \tan x \minus{} 7\int{x^6\tan x \;\text{d}x}$
Last integral we can't evaluate because we even can't evaluate well-known integral $ \int{x\tan x \;\text{d}x} \equal{} \sum_{n \equal{} 1}^{\infty}{\frac {2^{2n}(2^{2n} \minus{} 1)\text{B}_{n}x^{2n \plus{} 1}}{(2n \plus{} 1)!}}$, so maybe there was error in magazine.

[quote="borislav_mirchev"]
I wrote the integral as it was in the magazine. What if it isn't $ x^7$?

[/quote]

I'm just pointing out a possibility. It [b]might[/b] be $ x^2$. And $ 2$ in a small font at times appears to be $ 7$, but thats only if the magazine font is very small, maybe 12 or 14.


function
number theory
least common multiple
modular arithmetic
algebra
binomial theorem

Find the last three digits of $ 17^{256}$

[hide]$ \lambda (1000) \equal{} 100$, so $ 17^{256} \equiv 17^{56}$

This is equal to $ (7\plus{}10)^{56} \equiv \binom{56}{0} 7^{56} \plus{} \binom{56}{1} 7^{55} 10 \plus{} \binom{56}{2} 7^{54} 10^2$ by the Binomial Theorem, since $ 10^n \equiv 0$ for $ n > 2$.

Now it's just some minimal brute force... (especially the third term, since you already have two zeros in there!)[/hide]

oops... how did I get $ lcm(2,100) \equal{} 200$?  Anyway nicely done Zuton.  Your solution gives the same thing.  And the third term of your expansion drops out since the binomial coefficient $ \binom{56}{2}$ is a multiple of $ 10$.


[hide]I started playing around with Euler's identity a bit and got this:
\[\begin{tabular}{rcll} e^{i\pi} + 1 & = & 0 \\
 \\
e^{i \pi} & = & - 1 & \qquad \qquad \qquad \text{subtracting 1} \\
 \\
i \pi & = & \ln - 1 & \qquad \qquad \qquad \text{taking natural log} \\
 \\
i^2 \pi^2 & = & (\ln - 1)^2 & \qquad \qquad \qquad \text{squaring} \\
 \\
- \pi^2 & = & 2 \ln - 1 & \qquad \qquad \qquad \text{simplifying using }i^2 = - 1\text{ and log identity} \\
 \\
\pi^2 & = & - 2 \ln - 1 & \qquad \qquad \qquad \text{multiplying by - 1} \\
 \\
e^{(\pi^2)} & = & e^{ - 2 \ln - 1} & \qquad \qquad \qquad \text{raising }e\text{ to both sides} \\
 \\
e^{(\pi^2)} & = & (e^{\ln - 1})^{ - 2} & \qquad \qquad \qquad \text{simplifying using exponent identity} \\
 \\
e^{(\pi^2)} & = & ( - 1)^{ - 2} & \qquad \qquad \qquad \text{simplifying using log identity} \\
 \\
e^{(\pi^2)} & = & 1 & \qquad \qquad \qquad \text{evaluating} \\
 \\
\pi^2 & = & \ln 1 & \qquad \qquad \qquad \text{taking natural log} \\
 \\
\pi^2 & = & 0 & \qquad \qquad \qquad \text{evaluating} \\
 \\
\pi & = & 0 & \qquad \qquad \qquad \text{taking square root} \end{eqnarray*}\]
Obviously, there must be something wrong. I couldn't seem to find it though, other than one thing: $ \ln - 1$ isn't defined, and I have that in my third step. How does that cause problems? If it doesn't, where is it wrong?

If this needs to be moved up to Intermediate topics, someone please do so.
[/hide]

EDIT: Wow. $ (\ln - 1)^2 \not= 2 \ln - 1$

In your sixth step you did:

$ (a^b)^c$ when it should be $ a^{(b^c)}$.

I'm not understanding you. You're talking about the step "raising $ e$ to both sides", or from $ \pi^2 \equal{} \minus{} 2 \ln \minus{} 1$ to $ e^{(\pi^2)} \equal{} e^{ \minus{} 2 \ln \minus{} 2}$. I don't see anything wrong there.

EDIT: Oohh, are you talking about the step after that, where I changed $ e^{ \minus{} 2 \ln \minus{} 1}$ to $ (e^{\ln \minus{}1})^{\minus{}2}$? That uses the exponent law that $ (a^b)^c\equal{}a^{bc}$.

$ 2\text{ln} \minus{} 1$ multiplied by $ \minus{} 1$ is not $ \minus{} 2\text{ln} \minus{} 1$.

Oh hmm apparently I did not realize the whole thing was in the logarithm. Ignore.

Thanks to [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=319829]FantasyLover[/url], who caught it.

[quote="xpmath"]$ 2\text{ln} \minus{} 1$ multiplied by $ \minus{} 1$ is not $ \minus{} 2\text{ln} \minus{} 1$.[/quote]

How is it not?

Going back even more, $ (\ln(\minus{}1))^2\neq2\ln(\minus{}1).$ I think you should enclose the $ \minus{}1$ in parentheses to let people know that the entire quantity is the argument of the logarithm.

You have to be careful when dealing with complex stuffs. E.g.:

$ e^{i\pi} \equal{} \minus{} 1$

$ i\pi \equal{} \ln{ (\minus{} 1)} \equal{} \ln{ (\minus{} 1^3)} \equal{} 3\ln{ (\minus{} 1)} \equal{} 3i\pi \implies 2i\pi \equal{} 0$

ZOMG?!?!

im not sure about this, but i thought that you couldnt have natural logs when you're dealing with e^xi, because where we have re^xi on teh imaginary plane, $ re^xi\equal{}re^{2\pi k\plus{}xi}$ where k is an integer


Let triangle ABC be right with AB = 3, BC = 4, and AC = 5.  An altitude is drawn from side AC, an angle bisector is drawn from A to side BC, and a median is drawn from angle C to side AB.  What is the area of the resulting triangle?

im having some trouble drawing medium from angle C to side BC

i guess he means the median from $\angle C$ to $AB$.

This problem is the same as

http://www.mathlinks.ro/Forum/viewtopic.php?t=27303 ?


linear algebra
matrix
absolute value
superior algebra
superior algebra unsolved

1. Prove that $ \text{Gl}(n,\mathbb{R})$ is not compact.
2. Prove that $ \text{O}(n)$ is compact.

1) GL(n,R) is not bounded
2) O(n) is bounded and closed, which can be seen from the continuous (defining) equation $ O^{t}O\equal{}1_{n}$

Okay. The compactness of groups is a new concept for me, so is it right to say that in 1) it is not bounded because one can find a matrix with arbitrary large determinant?

In two, how one can see the group is closed and bounded. I guess that $ O^{t}O\equal{}1$ implies that the absolute value of the elements of $ O$ is $ \leq 1$ but how to prove it.

as for 1) I had more in mind $ \mathbb{R}\subset GL(n,\mathbb{R})$ via the identification of diagonal matrices.

as for 2) evaluate the diagonal coefficients in $ O^{t}O\equal{}1_{n}\Longrightarrow\forall i\equal{}1,\ldots n:\,1\equal{}\sum_{j\equal{}1}^{n}O_{ji}^{2}\Longrightarrow\forall i,j\equal{}1,\ldots,n : |O_{ij}|\leq 1$ and note, that it is important to work over the real numbers (for instance $ O(n,\mathbb{C})$ is not compact). This proves the boundedness and you can identify $ O(n,\mathbb{R})$  as a closed subset, because the defining equations are continuous functions.


You are a robber and are robbing a house. You go into a bedroom, and when you close the door, it locks by itself! There is a hole for inserting dollar bills, and a computer screen for showing how much money you have inserted and how much money it needs to unlock(200 dollars). There is also a clock showing "20:00" minutes and it is counting down. You must find your way out before that time.

Basically, the people playing will PM me what they want to do in-game, and I will tell them the results. For example:

woser456: Pace up and down.
me: You pace up and down, but you don't get any brilliant ideas. You have just wasted a minute.

I will send prods if people don't respond for a while(say, 2 days). This is essentially a race through riddles to the finish.

Signups here! Once I have 8 people I will begin.

[quote="SimonM"]Open a variety of drawers, cupboards and wardrooms in a frantic search for cash

Q: Is the door the only way out (Windows, thin walls etc)
Q: Does the counting down clock have any relevance to [b]the game[/b] (I just lost)
Q: Will there be any external events which could change the situation which we simply have to wait for
Q: Could a sufficiently skilled engineer find a way to get out of the room by altering the lock[/quote]

Basically you do that, but you need skills to do that.

Q1: Yes.
Q2: Yes. Once the clock gets to zero, the alarms around the house activate(not a common house!)
Q3: No.
Q4: This lock was MADE by a skilled engineer, and he made a failsafe that if someone tried to do that, then the alarm would go off automatically.

This sounds fun and interesting. I'll sign up.

You'll display time, right, in bold? [b]20:00 remaining.[/b] Something like that?

Also:

< Do you have any weapons, such as a gun or knife?
< Are there any animals?
< Are there any materials or supplies in the room (i.e. Pencil, etc)
< Is the clock built into a wall?

You have no weapons, but I will try to incorporate items into the game.

There are no animals.

the clock is in the wall above the door.

And about how large is the clock? What is the estimated area/circumference of it?

It's a clock, and you can't do anything to it. It is there to tell you how much time you have left.

P.S. I have also decided to set traps around the room to subtract various amounts of time ranging from one minute to 5 minutes.

You're in a bedroom, so there has to be jewels or some cash in there, right? (Would be most helpful if there were...) How about clothes?

Only cash. There are clothes. Also, there is a boy sleeping in the bed, with his hand on a knife attached to the side of the bed. He seems to be pretty intense when it comes to burglers...

would taking the boy's knife and possibly killing him be of anyuse.

maybe he could be used as some sort of decoy to set off some sort of pressure sensor?(thinking dumb ideas aloud)

/in for this game!

>Can this skilled engineer help you? Will he?
>Is the clock wired to the lock? Can the wire be cut? Can
the wall be cut?

/me would also like to join

(I forgot to mention it earlier)

ugh.

The boy seems so intense that he is probably not a sound sleeper.
There is only you, and the skilled engineer is sleeping; you are robbing him; he doesn't like being robbed. (NO, REALLY?!)
The clock is wired to the alarm system all around the house, and the wires cannot be cut without drilling into the wall which would wake up the boy.

signups:
pythag011
D3m0n Shad0w
BOGTRO
SimonM

Also, if you do something stupid like pace around the room, I will take off a full minute, and you might need all the time to get out of there.

Can you kill the boy then cut the wires in time?

I'm playing... that's why I ddin't PM you.

I need a confirmation from only I Hate Watermelons. Once I get it, I'll start.

If I don't get a confirmation by 6 PM, I will start.

WTH I thought I posted here! I'll post it again.

You are a burgler in a house, but for some reason you are locked in a room with an elaborate lock system(insert 200 dollars here), so you need to bust it. You see a closet, a bookshelf, a bureau, and a computer desk.

PM your actions in!

Will you post their decisions on the thread so nonplayers can read?

that would fill up the topic, but I'll do it at the end.

[quote="mathblitz"]its too bad no one else can join. i was going too but too late. for us nonplayers could u tel us what the room looks like?or better u could just pm those that ask.(AKA ME!!!!) :D[/quote]

But you could then offer advice to your fellow players who are actually involved.

Is this thread still alive?

if yes, I have a few questions:

1) What is there exactly in the room besides the computer, door, and clock?
2) Do I post here or PM you?

I think this thread died. I was playing, but I suddenly stopped getting replies from 1=2.

i will sign up
is this in the real world? is there magic?
is there a gun? a lock?
can you push a button for money to fall out?
Ah! *Hint Hint
You use the computer to make a robot to break the door and produce real money if that's possible.

I wish someone can make another thread like this. It seems interesting.

Can 1=2, if he/she is still active, please post the solution for the benefit of the nonplayers?

This is almost 2 years old. Don't revive extremely old topics.


are you asking whether or not the opposite of 0 is 0, or stating it? your post isn't really phrased like a question ;).
i'm pretty sure the opposite of 0 is indeed 0.

[quote="joyofpi"]Well I was just asking if it was 0; I wasn't sure for a moment and yeah... I'm an idiot... o_o[/quote]"Opposite of 0" sounded like a vague phrase when I first read it but after a quick thought I'm sure you were asking about the 'sign' of the number. I'm compelled to say that 0 has no opposite but it depends on what definition you like more.

Also, you are not an idiot. All mathematicians always ask dumb questions and never afraid to do that. If I ever ask a 'smart' question, then... isn't it idiotic of me that I ask a 'smart' question where I should of known the answer in the first place if it was 'smart'? :)

[quote="PowerOfPi"]Opposite can also mean inverse, like the reciprocal of something. The reciprocal of 0 is undefined. If you mean the negative value of 0, it would be 0 again.[/quote]

Isn't opposite defined as the additive inverse?

[quote="jjfun1"]I don't think that there's such thing as the "opposite of zero" exactly... lol[/quote]

Isn't the opposite of $ x$ defined as the additive inverse of $ x$, that is, a number $ c$ such that $ x \plus{} c \equal{} 0$, resulting in $ c \equal{} \minus{}x$.

In this case, $ c \equal{} \minus{}1 \cdot 0 \equal{} 0$.

Well, in quadratics, there's a double root if the discriminent is 0, so it becomes something $ \pm{0}$.

So 0 is the opposite of 0. :maybe:

[quote="Funkymunk"]
Isn't the opposite of $ x$ defined as the additive inverse of $ x$, that is, a number $ c$ such that $ x \plus{} c \equal{} 0$, resulting in $ c \equal{} \minus{} x$.
[/quote]

Yes, that's the definition of opposite.

[quote="joyofpi"]The opposite of 0 would be 0 wouldn't it considering the fact that it's not negative nor positive in a sense...[/quote]
Yes, 0 is the opposite of 0, but normally in math I don't think that this is ever used or needed.


algebra
difference of squares
special factorizations

this is the best i can get to.

(r + x)^(3/2) * (r - x)^(3/2)

Whenever the exponent is a fraction $\frac nm$ then you raise the number to the $n\text {th}$ power and then take the $m \text {th}$ root.

In your case you raise $r^2 - x^2$ to the third power and take the square root.

$\sqrt {(r^2 - x^2)(r^2-x^2)(r^2-x^2)} = (r^2-x^2)\sqrt {r^2-x^2}$

You can go farther by factoring $r^2-x^2$ using difference of squares $(r+x)(r-x) = r^2 - x^2$.

Hope this helps.

try to understand the difficulty of this forum...

and

special thanks to tarquin for showing the work....

and if it doesn't make any sense to you, try plugging in simple numbers


p+1=2x^2
p^2+1=2y^2
where x,y are integers and p is a prime.

thank you... :)

Clearly $ p$ is odd. Subtracting gives $ p(p\minus{}1)\equal{}2(y\plus{}x)(y\minus{}x)$, so $ p\cdot \frac{p\minus{}1}{2}\equal{}(y\plus{}x)(y\minus{}x)$. As $ p$ is prime, $ p$ must divide either $ y\plus{}x$ or $ y\minus{}x$. However, it's easy to see that $ 0<x<y<p$, so we must have $ p\equal{}y\plus{}x$. This gives
\[ y\plus{}x\equal{}p,y\minus{}x\equal{}\frac{p\minus{}1}{2}\Rightarrow x\equal{}\frac{p\plus{}1}{4}\]
\[ \Rightarrow p\plus{}1\equal{}2\left( \frac{p\plus{}1}{4}\right)^2\Rightarrow 8\equal{}p\plus{}1\Rightarrow p\equal{}\boxed{7}\]


This is actually just (one form of) the hockey stick identity: it says $ \sum_{i \equal{} 0}^n \binom{ \minus{} x \minus{} 1 \plus{} i}{i} \equal{} \binom{ \minus{} x \plus{} n}{n}$.


ABCD is a convex quadrilateral. RA is the circumradius of the three vertices other than A. RB, RC and RD are defined similarly. Show that RA + RC > RB + RD iff angle A + angle C > angle B + angle D.

a solution at :
http://www.kalva.demon.co.uk/short/soln/sh9617.html


factorial
number theory
prime factorization

NOTE: None of these require the use of a calculator.

1. Billy wants to buy a gift. He sees hundreds of items, all a different positive integer cost. Unfortunately, he only has 13 dollar bills and 11 dollar bills. What is the price of the most expensive item he cannot buy without recieving change?

2. Evaluate:

$\displaystyle\sum_{n=1}^{20} n^2-1$

For clarification purposes, the -1 is part of the sigma, not independent.

3. Right triangle(s) exist with with side lengths 9, 41, and x. Find the sum of all possible x.

4. How many 0's are at the end of $42!$?

5. The sum of the first 3n positive integers is 150 more than the sum of the first n positive integers. Find the sum of the first 4n positive integers.

6. Find the 11th term of $(x-y^2)^{25}$ (Do not simplify any factorials)

7. How many positive factors are there of $2006^5$?

8. A regular polygon has 35 diagonals. Find the measurement of one of its angles.

Only read the answers below when you're done.

[hide="Answers"]1. 119
2. 2850
3. 40+$\sqrt{1762}$
4. 9
5. 300
6. $\frac{25!x^{15}y^{20}} {10!15!}$
7. 216
8. 144 degrees[/hide]

If only I could post all these as separate posts *sigh*. They would add a lot towards my post count :D .
[hide="1"]By the Chicken McNugget Theorom, the answer is 13*11-13-11=119.[/hide]
[hide="2?"]Who knows how many times this has been asked, but what does that sign mean?[/hide]
[hide="3"]x can either be a leg with length 40 or the hypotenuse with length sqrt1762 (by the Pythagorean theorom). Answer: 40+sqrt1762.[/hide]
[hide="4"]5 goes into 42 8 times. 25 goes into 1 time. 125 is bigger than 42, so there are 8+1=9 zeros at the end of 42!. Someone posted a formula for this under the 'Formulas' topic.[/hide]
[hide="5"]Write the equation n(n+1)/2+150=3n(3n+1)/2. Solve and you find n=6. The sum of the first 4n (24) numbers is 300.[/hide]
[hide="7"]The prime factorization of 2006 is 2*17*59, so the prime factorization of $2006^5$ is $2^5*17^5*59^5$. Add one to all the exponents, and multiply to get 216.[/hide]
[hide="8"]The number of diagonals for an n gon is n(n-3)/2,so making that equal 35 we get n=10. The measure of each interior angle for an n gon is 180(n-2)/n, so substitute in 10 for n and get 144 degrees.[/hide]

[quote="Zenith Xan"]The thing in #2 is called Sigma.

Refer to [url=http://en.wikipedia.org/wiki/Summation]this article.[/url][/quote]
That's neat. :) [hide="answer for 2"]It's easier to find the sum of the first 20 squares, then subtract 20:2870-20=2850.[/hide]


A solid of density $d_{1}$ is sinking into a liquid of density $d_{2}$. The refractive index of the liquid is $n$. What is the apparent acceleration of image formed in the air (given that the acceleration due to gravity is $g$)?

[hide="Answer"] $\frac{(d_{1}-d_{2})g}{nd_{1}}$[/hide]

The force accelerating the body is $(d_{1}-d_{2})Vg$, where $V$ is the volume of the body, so the acceleration of the body is $(d_{1}-d_{2})g/d_{1}$.

Let's consider the block when it is at a depth $h$. We'll find the position of the image of the block by tracing two rays emanating from the body and finding their intersection. Let the first ray be the one perpendicular to the surface -- this ray doesn't get refracted. Let the second ray make a small angle $\phi$ with the line perpendicular to the surface (ie. $\phi$ is the angle of incidence). Upon refraction, the new angle, $\theta$, is given by Snell's law
\[n\sin\phi = \sin\theta.\]
The intersection point of these rays is at a distance $h'$ below the surface, somewhere along the line of the first ray. Trivial geometry gives $h' = h \tan\phi/\tan\theta$, which after using the small-angle formula and Snell's law gives
\[h' = \frac{h}{n}.\]
Differentiating twice we get
\[\ddot h' = \frac{\ddot h}{n}= \frac{d_{1}-d_{2}}{nd_{1}}g.\]
Note that this is the acceleration of the image directly above the body.


Think about replacing that 3 pi/ 2 with an x. Does that expression look familiar now?

because cos 3pi/2 = 0
cos(3pi/2 + h ) - cos(3pi/2) = cos 3pi/2 . cos h - sin 3pi/2 . sin h - cos 3pi/2
                                           = sin h
when h ~0, sin h ~ 0
so
 the limit sinh / h = 1
the answer is A

other way to think about, think it's as derivatives of $\cos(x)$ at $3\pi/2$. which equal to $ -\sin(3\pi/2)=1$


Find all non-negative integer solutions of the equation
\[ 2^x\plus{}2009\equal{}3^y5^z.\]

$ x > 0$ or $ y > 0$  so $ 2^x \equiv 1 \bmod{3}$ or $ 2^x \equiv 1 \bmod{5}$, so $ x$ is even.
We can also check that $ n \ge 3$ so $ 3^y5^z \equiv 1 \bmod{8}$, therefore $ y$ and $ z$ are both even.

Then $ 2009 \equal{} 3^y5^z \minus{} 2^x \equal{} a^2 \minus{} b^2 \equal{} (a \minus{} b)(a \plus{} b)$

The factorisations of 2009 are $ 2009 \equal{} 1.2009 \equal{} 7.287 \equal{} 41.49$

Checking each case, the only factorisation that gives a solution is $ a \plus{} b \equal{} 49, a \minus{} b \equal{} 41 \implies a \equal{} 45 \equal{} 3^2.5^1, b \equal{} 4 \equal{} 2^2$

So the only solution is $ (x,y,z) \equal{} (4,4,2)$

x should be A MULTIPLE OF 4
BECAUSE
RIGHT HAND SIDE MUST BE A MULTIPLE OF 5
NOW ON LEFT HAND SIDE WE HAVE A ODD NUMBER + AN EVEN NUMBER

WHICH CAN BE A MULTIPLE OF 5 ONLI WHEN THE FINAL NUMBER ON LEFT HAND SIDE HAS A LAST DIGIT OF 0 OR 5
YA IT SHOULD BE ODD
BUT IT SHOULD ALSO BE A MULTIPLE OF 5
SO THE ONLI WAY TO MAKE THE FINAL LHS NUMBER TO END IN 0 OR 5 IS FOR 2 raised to the power
TO HAVE THE LAST DIGIT AS 6
SO WE HAVE
1
2
4
16-----
32
34
128
256------
AND SO ON
16=2^4
256=2^8
AND SO ON

NOW U TRY X=4
THAT GIVES U
2009+16
=2025
=3^4*5^2
SO CASE 1:
X=4,Y=4,Z=2

so the onli way is that the number on LHS must end with a 5


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linear algebra
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number theory

I was looking over the topics for the Math IIC, and it said that I need to know what is an interquartile range and what is a standard deviation. Can anyone tell me what these are in simple terms? Are there any calculations that you would need to be able to do with these? Thanks.

By the way, I am surprised that the above is on the exam, as well as number theory and matrices. I thought it would a lot easier...

Also, I've already tried using google to search, but I can't find a simple explanation and the applications that would apply to the SATII. If anyone knows of one please let me know ASAP since the SATII is tomorrow.

Regarding standard deviation, they occasionally ask a qualitative question about it.  They don't seem to ask quantitative questions about it.  All you need to know is that standard deviation is a measure of how "spread out" a list of numbers is (i.e., how far away they are from the mean).  The closer together the numbers are, the smaller the standard deviation is, and vice versa.  For example, you might be given 5 lists of numbers, and asked which list has the smallest (or largest) standard deviation.

Regarding percentiles and quartiles, they again seem to ask mostly qualitative questions.  You should know that being at the 75th percentile means roughly speaking that you did better than 75% of the sample.

Regarding matrices, you don't need to know much.  If you know how to multiply 2 by 2 matrices, or find the determinant of a 2 by 2 matrix, or know what you get when you multiply a 3 by 4 matrix with a 4 by 5 matrix, then you should be okay.

I would expect topics like that .  For standard deviation if they ask you to evaluate one it's not that hard although the formula is intimidating.

First you find the average of the set of numbers.  Then you subtract this from every number in the set, square each of these results and add them.  Divide this by the the # of #'s in the set and take the sqrt.  It looks a whole lot better in a formula but is harder to understand.  Also, variance is equal to the standard deviation squared.


function
ceiling function
real analysis
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limit
real analysis unsolved

Let $ \{f_n\}$ be a sequence of continuous functions on $ \mathbb{R}$. Is it possible for $ \{f_n(x)\}$ to be unbounded if and only if $ x\in\mathbb{Q}$?

Massively edited: I was thinking about this last night, and this morning tried to recreate my question without looking at my notes. Sorry, this belongs in Real Analysis.

Things I've tried (but I seem to find fault with them all).

Suppose there does exist such a sequence. Then choose $ M>0$ and $ x\in\mathbb{R}\setminus\mathbb{Q}$. Now we can find a sequence of rationals $ x_k$ such that $ x_k\to x$.

Since $ \{f_n(x_k)\}$ is unbounded for each $ x_k$ we can find some $ N\in\mathbb{N}$ such that $ f_n(x_k)>M$ for each $ n>N$. (If we can't, then one of the sequences is bounded contradicting the assumption). So for any fixed $ n>N$ we have that $ f_n(x_k)>M$ and by continuity we get that $ f_n(x)\geq M$. But since $ M$ was arbitrary we have $ \{f_n(x)\}$ unbounded contradicting irrationals being bounded.

Potential problem: If we explicitly try to construct $ N$ say by letting $ N_k$ by the corresponding integer such that $ f_n(x_k)>M$ for all $ n>N_k$ we can do something like $ N\equal{}\lceil\sup\{N_1, N_2, \ldots\}\rceil\plus{}1$. But just because each $ N_k$ is a finite integer doesn't mean $ N$ has to be finite.

I also tried to construct it by taking $ F_n(x)$ to be some continuous function whose limit is the dirac delta function at $ 0$. Then let $ f_n$ be the countably superimposed $ F_n$ shifted so that the center on every rational. This would give that each $ f_n$ is continuous and is creates an unbounded sequence only at the rationals. The problem is that I don't think this is a legitimate construction.

Lastly, neither of these use a single thing from the chapter in which this problem comes from (Rudin, Banach Spaces). So my guess would be I'm supposed to assume you can do it and construct a set of linear operators that violate Banach-Steinhaus, or suppose you can do it and construct from it a countable collection of nowhere dense sets whose union is $ \mathbb{R}$ to violate Baire.

It has been a week and I still haven't gotten any further (and this is still in the wrong section  :wink: sorry, my fault).

Just wondering if anyone could comment on either of my attempted solutions (if those problems are fixable). Or if anyone had any ideas...not even solutions, but maybe just an idea that you think might work and haven't tried that I can start fiddling with.

[hide="oops this is wrong"]Consider functions $ g_n$, where
$ g_n(x) \equal{} 0$ if x is irrational, and if $ x \equal{} \frac {p}{q}$ (simplified fraction), then $ g_n(x) \equal{} \frac {1}{q}$.
This function has the property that it is not continuous on $ \mathbb{Q}$ but is continuous on $ \mathbb{R} \minus{} \mathbb{Q}$.

Now, let $ f_n$ be a continuous function on $ \mathbb{R}$, where $ | f_n \minus{} ng_n| < \frac {1}{n^2}$ (the norm is sup norm).

If $ x \in \mathbb{R} \minus{} \mathbb{Q}$, $ |f_n(x)| < \frac {1}{n^2}$, so the set $ \{f_n(x)\}$ has to be bounded.
If $ x \in \mathbb{Q}$, say $ x \equal{} \frac {p}{q}$, simplified, then $ |f_n(x) \minus{} \frac {n}{q}| < \frac {1}{n^2}$. Now $ q$ is fixed and n going to infinity, $ \{f_n(x)\}$ is unbounded.[/hide]

Soarer's proof has an obvious flaw: you cannot approximate discontinuous functions by continuous ones in the uniform norm. Anyway, look at the following two threads:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22177[/url]
where this particular problem is solved, and
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=202444875&t=25745[/url]
where a slightly different problem is discussed ;)

I don't want to reply to such an old post, but if people read the second link, the error with grobber's approach is that $ f_n$ are supposed to be continuous, but the given $ f_n$ are clearly not since for any fixed $ n$ we have $ \lim_{x\to q_1}f_n(x)\equal{}0$, but $ f_n(q_1)\equal{}n$.

No, there is no error in grobber's solution. You just didn't read it attentively or misunderstood what he said for some other reason. Note that he defines $ f_n$ explicitly just at the points and outside small but nondegenerate intervals around those points. He doesn't say what the function is inside those intervals; that the reader is assumed to guess himself and your guess is wrong; it is not $ 0$ there ;)

By the way, there is nothing wrong with reviving all topics if you discover a problem with the solution or want to post some other meaningful comment. On the other hand, there is nothing wrong with the way you did it either :).

In that case, I do have a concern with when he lets $ n$ tend to infinity. He has $ \varepsilon_n > 0$ decreasing to $ 0$ quickly. Then we have $ t \in (q_k \minus{} \varepsilon_n, q_k \plus{} \varepsilon)$ for all $ n$. It doesn't seem like we have enough to conclude that $ t$ is in the limit, though.

Don't they have to be closed intervals for the limit to be guaranteed nonempty? We have $ \cap (0, 1/n)\equal{}\emptyset$.

Although, this seems to work for my delta function approach in post #2...I think.

Edited: for dignity?

[quote="HilbertThm90"]
Don't they have to be closed intervals for the limit to be guaranteed nonempty?
[/quote]
It is, indeed, true that the nested intervals must be closed to [b]guarantee[/b] that the intersection is nonempty without any additional assumptions but it is not true that if they are open, then the intersection must be empty.
[quote]
We have $ 0\in ( \minus{} 1/n, 1/n)$ for all $ n$, but the limit is $ \cap ( \minus{} 1/n, 1/n) \equal{} \emptyset$.
[/quote]
:what?: :o I'm speechless...


the polynomial $p(x)=ax^n-ax^{n-1}+{c_2}x^{n-2}+...+{c_{n-2}}x^{2}-n^2 {bx}+b$
has n real roots and positive prove that all of these roots are like

let $r_1, ..., r_n$ be the roots to the equation.
$\sum{r_i}=1$
$\sum{r_1r_2...r_{n-1}}=(-1)^n\frac{n^2b}{a}$
$r_1r_2...r_n=(-1)^n\frac{b}{a}$

and so $\sum \frac{1}{r_i}=n^2$.

by cauchy, we have $\sum{\frac{1}{r_i}}\cdot\sum{r_i}\ge n^2$ and so we have the equality case of cauchy, which since they're all positive implies that all the roots are equal.


Year by year, there are more and more country bringing team mascot. Before there is only Canadians bringing the moose, later few other countries start bringing mascots too, like last year, there are Russians with the tiger and Norwayians with the polar bear and some one with a cow(unknown country)...
    What mascots you guys plan to bring this year?

some one from an European country...
the same guy brought the cow to Ljubljana too

LOL  :lol: that would be a cool mascot ^^

Mexicans kidnapped the polar bear in Vietnam!!!!!

[quote="Ilthigore"]We (United Kingdom) brought Newton the newt last year, but he was abducted at the closing ceremony, so we are in search of some other potential famous UK mathematician who is an animal.

Hamilton the Camel...ton? Maybe not...[/quote]

Ramsey the Ram?

[quote="Naphthalin"]and i think the german team will bring the MathemaTigerin to spain - if she survives until july ;)

Naphthalin[/quote]

This was the one that took the medal of Peter at Vietnam at the closing ceremony ? :lol:

[quote="Valentin Vornicu"]Who had the cow?[/quote]

The cow was from Austria and we had an additional mascot as well - "Waldi", the rider of the cow. However, "Waldi" finally made the IMO qualification this year so that we probably have to change our mascot since many of us oppose having someone as mascot and at the same time as participant as well (like the USA team did one or two years ago if I'm not mistaken^^).

I think that the ''crazy'' team from Sweden had also a maskot at Vietnam .
And that maskot was fight with the German team's mascot at the aeroport of Bankong . :D
Who was the winner ?  :P


a nice probability puzzle (from mathoverflow, where a beautiful solution has been given):
Throw $ N$ points (uniformly) at random in $ [0;1]$ and call $ d$ the minimum distance between any pair of these points. What is the probability that $ D$ is greater than $ d$ ?

The distances between these $ 0,$ these points in order, and $ 1$ are uniformly distributed on the simplex $ x_0\plus{}x_1\plus{}\cdots\plus{}x_n\equal{}1,$ $ 0\le x_0,x_1,\cdots,x_n$. The probability we seek is the probability $ p(D)$ that at least one of $ x_1,x_2,\cdots,x_{n\minus{}1}$ is $ < D$. That's a bit tricky, so we look at the complementary probability $ 1\minus{}p(D)$ that all of $ x_1,x_2,\cdots,x_{n\minus{}1}$ are $ \ge D$.
What does that look like? If we set $ y_i\equal{}x_i\minus{}D$, we have $ 0\le y_i$. We also have $ x_0\plus{}y_1\plus{}y_2\plus{}\cdots\plus{}y_{n\minus{}1}\plus{}x_n\equal{}1\minus{}(n\minus{}1)D$, and the possible region is the simplex $ 0\le x_0,x_n,y_1,y_2,\cdots, y_{n\minus{}1}$ and $ x_0\plus{}y_1\plus{}y_2\plus{}\cdots\plus{}y_{n\minus{}1}\plus{}x_n\equal{}1\minus{}(n\minus{}1)D$. This simplex is similar to the original with scale factor $ 1\minus{}(n\minus{}1)D$, so its $ n$-volume is $ (1\minus{}(n\minus{}1)D)^n$ times the volume of the original. By the uniform distribution, $ 1\minus{}p(D)\equal{}(1\minus{}(n\minus{}1)D)^n$ and $ p(D)\equal{}1\minus{}(1\minus{}(n\minus{}1)D)^n$.


Compute $ \sin^4{\theta}\plus{}\cos^4{\theta}$ if

$ a) \sin{2\theta}\equal{}\frac{1}{4}$

$ b) \cos{2\theta}\equal{}\frac{1}{4}$

From (a),
\[ \sin \theta\cdot \cos \theta \equal{} \frac {1}{8}
\]
and from (b),
\[ \cos^2 \theta \minus{} \sin^2\theta \equal{} \frac {1}{4}
\]
So
\[ \sin^4\theta \plus{} \cos^4 \theta \equal{} (\cos^2\theta \minus{} \sin^2\theta)^2 \plus{} 2\sin^2\theta\cos^2\theta \equal{} \frac {1}{16} \plus{} 2\frac {1}{64} \equal{} \boxed{\frac {3}{32}}
\]

a) and b) are separate problems.  (The two conditions clearly contradict each other.)

Indeed, they cannot be true at the same time: $ \cos^{2}\theta \minus{} \sin^{2}\theta \equal{} \frac {1}{4} \Longrightarrow \sin\theta\cos\theta \equal{} \frac {\sqrt {15}}{8}$.

(I'm assuming the reader is familiar with the trigonometric identities used — this is not very commendable, I know, but it's 2 a.m. If you don't understand how I arrived at a certain conclusion, feel free to tell me so.)

[hide="When (a) is true"]$ (\sin^{2} \theta \plus{} \cos^{2} \theta)^{2} \equal{} \sin^{4} \theta \plus{} 2 \sin^{2}\theta \cos^{2} \theta \plus{} \cos^{4} \theta \equal{} 1 \Longleftrightarrow \\
\sin^{4} \theta \plus{} \cos^{4} \theta \equal{} 1 \minus{} 2 \sin^{2} \theta \cos^{2} \theta \equal{} 1 \minus{} (2 \sin \theta \cos \theta)(\sin \theta \cos \theta)$.

We know that $ \sin 2\theta \equal{} 2\sin\theta\cos\theta \equal{} \frac {1}{4}$ and, consequently, that $ \sin\theta\cos\theta \equal{} \frac {\sin 2\theta}{2} \equal{} \frac {1}{8}$; thus,

$ \sin^{4} \theta \plus{} \cos^{4} \theta \equal{} 1 \minus{} \frac {1}{32} \equal{} \boxed{\frac {31}{32}}$[/hide]

[hide="When (b) is true"]$ \cos 2\theta \equal{} \cos^{2} \theta \minus{} \sin^{2} \theta \equal{} 1 \minus{} 2\sin^{2}\theta \equal{} \frac {1}{4} \Longleftrightarrow \\
2\sin^{2}\theta \equal{} \frac {3}{4} \Longrightarrow \sin^{2}\theta \equal{} \frac {3}{8}$.

Now, $ \sin^{2} \theta \plus{} \cos^{2} \theta \equal{} \frac {3}{8} \plus{} \cos^{2} \theta \equal{} 1$, therefore $ \cos^{2}\theta \equal{} 1 \minus{} \frac {3}{8} \equal{} \frac {5}{8}$. Now that we have found the values of $ \sin^{2}\theta$ and $ \cos^{2}\theta$, we can solve the problem, since $ \sin^{4}\theta \plus{} \cos^{4}\theta \equal{} (\sin^{2}\theta)^{2} \plus{} (\cos^{2}\theta)^{2} \equal{} \left(\frac {3}{8}\right)^{2} \plus{} \left(\frac {5}{8}\right)^{2} \equal{} \frac {9}{64} \plus{} \frac {25}{64} \equal{} \frac {34}{64} \equal{} \boxed{\frac {17}{32}}$[/hide]

Any corrections appreciated. :P

[quote="metafor"]Indeed, they cannot be true at the same time: $ \cos^{2}\theta \minus{} \sin^{2}\theta \equal{} \frac {1}{4} \Longrightarrow \sin\theta\cos\theta \equal{} \frac {\sqrt {15}}{8}$.[/quote]

More simply, $ \cos^2 2 \theta \plus{} \sin^2 2 \theta \equal{} \frac{1}{8} < 1$.

At least you know trig identities, [b]jli[/b] — when they show those suckers to my classmates next year, they're going to think the teacher converted to Satanism.

I am oblivious to the obvious, [b]t0rajir0u[/b] — it could have taken me [i]days[/i] to realize that. :P

Its very useful to learn how to derive identitiys from the most common and fundamental ones.

[quote="Illos"]Compute $ \sin^4{\theta} \plus{} \cos^4{\theta}$ if

$ a) \sin{2\theta} \equal{} \frac {1}{4}$

$ b) \cos{2\theta} \equal{} \frac {1}{4}$[/quote]
a)      sinX^4 + cosX^4= 1-2sinX^2*cosX^2
                   = 1- 2 {(sin2X)^2/4}
                   = 1-2 < 1/64>
                   =1-1/32
                     =31/32

and when u have cos2X for part b , find sin2x using sin^2 + cos^2 =1
and solve it the same way as above.
Simple.

This problem is essentially the same as the HMMT 2008 Algebra #1.


$ a,b,c,d$ are positive integers where $ a > b > c > d$ such that $ a \plus{} b \plus{} c \plus{} d \equal{} 2004$ and $ a^{2} \minus{} b^{2} \plus{} c^{2} \minus{} d^{2} \equal{} 2004$.Find the minimum possible value of $ a$.

[hide]We have $ a \plus{} b \plus{} c \plus{} d \equal{} (a \minus{} b)(a \plus{} b) \plus{} (c \minus{} d)(c \plus{} d) \Longleftrightarrow (a \minus{} b \minus{} 1)(a \plus{} b) \plus{} (c \minus{} d \minus{} 1)(c \plus{} d) \equal{} 0$
$ a > b \implies a\ge b \plus{} 1\implies a \minus{} b \minus{} 1\ge 0$
Similary $ c \minus{} d \minus{} 1\ge 0$
$ \therefore a \equal{} b \plus{} 1, c \equal{} d \plus{} 1$
$ \implies 2a \plus{} 2d \equal{} 2004$
$ a\ge b \plus{} 1 \ge c \plus{} 2\ge d \plus{} 3 \implies 4a\ge 2a \plus{} 2d \plus{} 6 \equal{} 2010$
$ \implies \boxed{a\ge 503}$[/hide]


[hide="Hint"]The left hand side needs to cancel out somehow. (positives must have their negative inverses somewhere)[/hide]

[hide="Solution"]Define $f(y)=|x_{y}-y|$.  So we have $f(1)+f(2)+\cdots+f(2003)=0$.  We know $f(y)$ is non negative for all $y$ because it is an absolute value.  Now we prove that $f(1)=0, f(2)=0,...,f(2003)=0$.

Proof by contradiction:
suppose $f(s)>0$. Solving for $f(s)$ in the equation gives us $-(f(1)+f(2)+\cdots+f(2003))=f(s)$.  Because every term in $f(1)+f(2)+\cdots+f(2003)$ is nonnegative, $-(f(1)+f(2)+\cdots+f(2003))$ is nonpositive implying that $f(s)\le0$  Contradiction!

Therefore we have $|x_{y}-y|=0$ for $y=1,2,3...,2003$ meaning that $x_{y}=y$ for $y=1,2,3,...,2003$.  Now we can find the value of $2^{x_{1}}-2^{x_{2}}-2^{x_{3}}-\cdots-2^{x_{2}002}+2^{x_{2}003}$
$=2^{1}-2^{2}-2^{3}-\cdots-2^{2002}+2^{2003}$
$=2^{1}-(-4(1-2^{2003}))+2^{2003}$
$=2+4(1-2^{2003})+2^{2003}$
$=6-3(2^{2003})$

I think this is right
[/hide]


If $ f(x)\equal{}\sqrt[3]{x^2\plus{}x\plus{}{1\over 3}}$, then $ a\equal{}f(b), b\equal{}f(c), c\equal{}f(a)$, giving $ a\equal{}f(f(f(a)))$. Since $ f$ is strictly increasing (easy to show), that gives $ a\equal{}f(a)\implies a\equal{}b\equal{}c$. Then the first equation gives $ 4a^3\equal{}(a\plus{}1)^3\implies a\equal{}b\equal{}c\equal{}{1\over\sqrt[3]{4}\minus{}1}$

WLOG, assume $ a$<=$ b$<=$ c$

hence $ 3a^3$ <= $ 3c^3$
$ 3b^2\plus{}3b\plus{}1$ <= $ 3a^2\plus{}3a\plus{}1$
therefore $ a\equal{}b$

repeat to show that $ b\equal{}c$ and the result follows.

I understand your argument correctly, it relies on the premise that $ 3a^2 \plus{} 3a \plus{} 1 \geq 3b^2 \plus{} 3b \plus{} 1 \implies a \geq b$. You then use your initial assumption $ a \leq b$ to conclude that $ a \equal{} b$, and the symmetric.

However, note that $ 3x^2 \plus{} 3x \plus{} 1$ is a parabola and therefore not strictly increasing.

For example, if $ a \equal{} \minus{} 5$ and $ b \equal{} 0$, then $ 3a^2 \plus{} 3a \plus{} 1 \equal{} 61$ and $ 3b^2 \plus{} 3b \plus{} 1 \equal{} 1$.
Both $ a^3 \leq b^3$ and $ 3a^2 \plus{} 3a \plus{} 1 \geq 3b^2 \plus{} 3b \plus{} 1$ are satisifed, but $ a \not \equal{} b$.

Unfortunately, you will need to directly involve all three equations somehow (for example, how Farenhajt did) to cycle around and say that all numbers are equal.

Yes, what I need to add to my solution is this:

Since $ 3x^2 \plus{} 3x \plus{} 1$ is always positive for real $ x$, we have that $ a,b,c$ are positive. Hence it's enough to observe $ f(x)$ for $ x > 0$. Since $ g(x) \equal{} 3x^2 \plus{} 3x \plus{} 1$ reaches its minimum at $ x \equal{} \minus{} {1\over 2}$, it is strictly increasing IN THE DOMAIN we're interested in.

Zuton, with that appendix (that we observe $ 3x^2 \plus{} 3x \plus{} 1$ only for positive $ x$), amitie's solution stands as well, and is actually quite elegant.

That it is  :) I'm just being a pain  :P