We cannot find solutions to all problems in the universe.

And if we supposedly did find all solutions to all problems, life would have no purpose because there are no problems to solve, a vegetable life. Nothing can be made or created because new ideas would be created, potentially damaging ones. Feelings, including jealously, would also be aroused.

Such a utopia would only exist if no one had feelings, a nonexistential life.

And while people might say that they have the solution to all the problems in the world and implement it, creativity and freedom would then be severely controlled, such as depicted in novels [u]Fahrenheit 451[/u] and [u]1984[/u]. Corruption would still occur.

We can't find solutions to all problems.
Sometimes when you fix one problem, you cause another.
So, we can usually make things better, but can't fix everything.

We couldn't even find Bin Laden -- how do you think we could end ignorance, abuse of power, crime, environmental destruction, poverty/malnutrition/hunger, disease, crime, hate, terrorism, and war (to name a few)?

Here are a few more related questions:

1. Do you think a problem-free state is stable?
2. In particular, do you think that the individually-held notion of self-interest will ever be compatible with a problem-free world within any social context?
3. If so, under what kind of social contexts/social conditioning would this be possible?

The authors of the books mentioned by [b]lightning[/b] attempted to answer that final question.

[quote="nobelium"]Can we find solution to all the conflicts(war, etc) and problems prevalent in this world? Or is it too impractical?
Please express your views on this.[/quote]

Probably not. As long as there is hegemony, and Weapons Of Mass Destruction (WMD)-- peace can't be attained. Also, peace in this world is contingent upon some countries or superpower(s) foreign policies.

So, As of now, your idea or notion of utopia is pretty far-fetched.

[quote="TZF"]We couldn't even find Bin Laden -- how do you think we could end ignorance, abuse of power, crime, environmental destruction, poverty/malnutrition/hunger, disease, crime, hate, terrorism, and war (to name a few)? [/quote]

I really find it pretty unbelievable that the U.S. (or rather the CIA) couldn't find Bin Laden even after 8 years of relentless pursuit. It is well known that he resides in Pakistan or rather operates from Pakistan.

http://www.dawn.com/wps/wcm/connect/dawn-content-library/dawn/news/pakistan/03-osama-bin-laden-operating-from-pakistan-ss-07

So, I think there are some vested interests which are protecting Bin Laden.

[quote="Gen8"]As long as there is hegemony, and Weapons Of Mass Destruction (WMD)-- peace can't be attained.[/quote]  This seems like overly strong conditions to reach the conclusion; I would have said, "As long as there are preceived inequities, ambitious, charismatic people, and tools more deadly than a dinner knife, someone somewhere will be doing something unpeaceful."

[quote="JBL"] This seems like overly strong conditions to reach the conclusion; I would have said, "As long as there are preceived inequities, ambitious, charismatic people, and tools more deadly than a dinner knife, someone somewhere will be doing something unpeaceful." [/quote]

Dear [b] JBL [/b] :roll: , these aren't overly strong conditions. Elimination of WMD-- which implicitly means "global nuclear disarmament"-- is one of the most important preconditions for peace in this world. I reiterate again, as long as there are WMD--peace can't be attained.

Another precondition for world peace is a radical change in some countries' foreign policies. Some country or countries advocate nuclear disarmament whilst at the same time they possess thousands of nuclear weapons. Another important precondition is a radical change in people's mindsets-- which as an offshoot could result in the end of "terrorism". Again, a radical change in people's mindsets is contingent upon a radical change in some countries' (or country's) foreign policies.

I also assert that hegemony should be curbed because it results in escalation of conflict thereby destabilizing world peace. But, I will  acquiesce with some of the points (witty?) [b] JBL [/b] had mentioned.

You misunderstood his point -- he's saying that even if WMD's and such weighty issues never existed, there would still not be a "problem free" world.

This is because the problems of humanity are much more deeply rooted and fundamental -- namely, basic human nature and general patterns in equilibrium social structure (which result from basic human nature).

[quote="nobelium"]Can we find solution to all the conflicts(war, etc) and problems prevalent in this world? Or is it too impractical?
Please express your views on this.[/quote]

Not only is it impractical, it's meaningless. But you're right, we could achieve greater peace if these things were under better control.

I really hope this isn't spam but I'm a little scared because this is the first time I am posting in this forum.

Honestly there is no possible way whatsoever to get rid of all potential problems in the world. There is always someone who is unsatisfied. (I recommend reading the book [u]The Giver[/u] by Lois Lowry for an example of a utopia that still has faults.)

Also, why should we get rid of all the problems in this world? In my opinion the only people that live on Earth that don't have problems are the people who are dead, buried in a cemetery, and in heaven. If you really don't have any problems, I really think you should get down on you knees, pray, and beg God for some problems because you're going to leaving soon.=]

i agree, i see no reason to be in a perfect world. though it seems nice to us right now, imagine being same as everyone else. This concept and its flaws are explained by writer Kurt Vonnegut in "Harrison Bergeron", a great story. basically, everyone has been equalized, except Diana Moon Glampers, who does the equalizing, and the methods of equalization are reall cruel. those with great physical strength must carry bags of birdshot, pretty people must wear ugly masks, singers must purposely sound awful, reporters must be monotonous, and the intelligent people wear headgear that sends sirens into their ears every 10 minutes to disrupt their thoughts. being in a perfect world would be boring, awful, and impossible. there will always be problems in this world, because imperfect people cannot create a perfect, problem-free society.

The world can never be problem free. If there were no problems, that in itself is a problem, because to have total peace would mean that feelings of people would have to be removed. As lightning said, Fahrenheit 451 and 1984 show the problematic "problem-free" societies of the future, where feelings and thoughts are suppressed. A real-world example would be the Soviet Union. The founders of the USSR set out to make a problem-free state, where everyone was equal, and it turned out to be a disaster. People revolted because they weren't allowed their full freedom. Suppression of freedom is necessary for a problem-free world, but then is has been proved no one can bear the thought of that.

HJ :lol:

[quote="inoit"]Honestly there is no possible way whatsoever to get rid of all potential problems in the world. There is always someone who is unsatisfied. (I recommend reading the book [u]The Giver[/u] by Lois Lowry for an example of a utopia that still has faults.)[/quote]

Love that book! Gives great insight on trying to "fix" problems.

I think the answer should be NO. Different people have different points of view. Thus contradiction between always exists.

Just like geometry. Given different systems of axioms, we can obtain different kind of geometry. They are not consistent with each other, but no one can say either of them is wrong.

[quote="infinite101"]Obama won the peace award...And he is still sending troops to Afghanistan.[/quote]

He won the award before he did [i]anything[/i].


function
Princeton
college
algebra
functional equation
algebra unsolved

Find all functions $ f:R \longrightarrow R$ satisfying the condition:
$ [f(x)+f(y)].[f(u)+f(v)] = f(xu-yv)+f(xv+yu)$ for all $ x,y,u,v \in R$

i think it is a hard problem too
But i think you shoulds write "my notebook"==> "my teacher"
Solution will come soon

I think this problem is not too hard . I have a solution :
       $ f(x) = x^2  or f(x) = 0  or f(x) = 1/2 $

Wow, my friends (master and NTTu), I 'm sure that we both meet this one before (from our teacher), but I post it here so that we can receive many solutions,OK?

I'll try to be short, cause I'm pretty tired and the longer I write, the more I am bound to make mistakes. Set all 0, and we get $2f^2(0)=f(0)$. If $f(0)=\dsp \frac 12$ then by setting $u=v=0$ we get $f(x)+f(y)=1$ for all $x,y$. This ibviously leads to $f=\dsp \frac 12$

If $f(0)=0$ the either $f=0$ or $f \neq 0$ (extraordinary logical argument, don't you think) In the latter case, make $y,v=0$ and we will get $f(xu)=f(x)f(u)$ for all $x,u$. Thus, $f$ is multiplicative, and as we assumed it non-zero, we will have $f(1)=1$. By switching the two pairs of variables $(x,y)$ and $(u,v)$ we will get $f(xu-yv)=f(yv-xu)$ for all $x,y,u,v$ and this implies that $f$ is even. Now, let $x$ and $y$ be non-zero and let $u=\dsp \frac 1x, v=\dsp \frac 1y$. Keeping multiplicativity in mind and denoting $a=\dsp \frac xy$, we get

\[f(a)+f(\dsp \frac 1a)+2=f(a+\dsp \frac 1a)\]

Multipliy this by $f(b)$ and we get

\[f(ab)+f(\dsp \frac ba)+2f(b)=f(ab+\frac ba)\]

For any positive $x,y$ we will find $a$ and $b$ s.t. $ab=x^2$, $\dsp \frac ba=y^2$, so the above relation becomes

\[f(x^2)+f(y^2)+2f(xy)=f(x^2+y^2) \Rightarrow f^2(x)+f^2(y)+2f(x)f(y)=f(x^2+y^2) \Rightarrow (f(x)+f(y))^2=f(x^2+y^2) \eqno (1)\]

This last equality and the even-ness of $f$ implies that $f$ is always non-negative. Thus, let $f(x)=g^2(x)$ where $g(x) \geq 0$ only for positive $x$. $g$ is obviously multiplicative and relation (1) implies  that $g^2(x)+g^2(y)=g(x^2+y^2) \Rightarrow g(a)+g(b)=g(a+b)$ if we set $a=x^2$ and $b=y^2$. $g$ is thus aditive too, and since it is multiplicative, it will be non-zero and therefore positive for $x \neq 0$. Thus $g$ is a positive aditive function, and thus it is increasing, and we know that this implies that it is linear. The fact that it is multiplicative implies that it is $x$. So $f(x)=x^2$.

As ussualy, Andrei forgot to check that the function does respect the functional equation, so, if he remembers IMO2002, he lost 1 point and gold medal? :)
Anyway, this is ironic. Andrei, are you at Princeton now? How is there?


algebra
polynomial
abstract algebra
calculus
integration
number theory
Ring Theory

[i]Let $ K$ be a number field. Assume that $ \mathcal{O}_K$ has only finitely many irreducibles $ p_1,\cdots,p_n$. Then $ 1 \plus{} p_1\cdots p_n$ must be divisible by some irreducible $ q$, and this can't be any of $ p_1,\cdots,p_n$. Contradiction. Of course the argument breaks down unless we can find at least one irreducible element in $ \mathcal{O}_K$. But since not every element of $ \mathcal{O}_K$ is a unit this is easy: let $ x$ be any non-unit and let $ p$ be some irreducible factor of $ x$. [/i]

(a) Is the result about an infinity of irreducibles correct? Why or why not?
(b) Is the proof given above correct? Why or why not?

[quote="Arne"](a) Is the result about an infinity of irreducibles correct? Why or why not?[/quote]
Yes with a tiny amendment - [color=red]if I haven't overlooked the following (see below)[/color].  :whistling:  Cover the case that $ u \equal{} 1 \plus{} p_1\cdots p_n$ is a unit by taking some sufficiently high power of $ u$ (use that the group of units is finite by assumption). The other case that the element is a non-unit can be treated as proposed.

[color=red]Correction: The given proof seems to be flawed if $ u$ is a unit. Then we have $ u^k\equal{}1$ for some $ k>0$. This implies $ x \cdot f(x) \equal{}0$ with $ x\equal{}p_1\cdots p_n$ for a non-trivial polynomial $ f(X)\in Z[X]$. Now it is not obvious that $ f(x)\neq 0$ to have $ x\equal{}0$ in order to get a contradiction. It seems to me that this cannot be deduced in the given setting. Unfortunately I do not have a counterexample at hand. [/color]

[quote](b) Is the proof given above correct? Why or why not?[/quote]
True since $ \mathcal O_K$ is a noetherian integral domain.

[color=red]Added sketch of proof: Let R be a noetherian integral domain. Assume that there is a non-unit $ a \in R$ which is not divided by an irreducible element. Then construct inductively a sequence of elements $ (a_n)$ in $ R$ with $ a_0\equal{}a$ and $ a_{n\plus{}1}$ is a divisor of $ a_n$ where in the representation $ a_n\equal{}a_{n\plus{}1}b_{n\plus{}1}$ both $ a_{n\plus{}1}$ and $ b_{n\plus{}1}$ are non-units. Then the family of ideals  is increasing, hence stationary. This yields to a contradiction.[/color]

Do you have any other doubts? :roll:

Well I'm confused. One of the books on algebraic number theory I'm using gives this as an exercise.

The question is: find the mistake in the proof.

Here a proof which is not flawed as the Euclidean type proof given in the exercise - just for completeness:
If there were only finitely many irreducibles in $ \mathcal O_K$, there would be finitely many prime ideals in $ \mathcal O_K$ such that their union would contain all irreducibles. Since every non-unit can be decomposed into irreducibles in a Noetherian ring, every prime integer (a real integer here) would be contained in the union. But this would imply that there would be only finitely many prime integers. AND this is absurd due to the Euclidean proof.


inequalities
linear algebra
linear algebra unsolved

$A,B \in M_n(R)$ such that $0<rank(A)<rank(B)$ and $AB=B^2; BA^2=A^3$

Prove that $rank(A^2+B) \leq n-1$

------------------------------------------

 How difficult do you find this one ?
thx

By Sylvester's inequality we have that :
$rank(A^2 +B) + rank(B) \leq n + rank(BA^2 +B^2) = n + rank(A^3 + AB) = n + rank(A \cdot (A^2 + B) ) \leq n + rank(A) \leq n + rank(B) - 1 $ .
The rest follows.

B(A^2+B) = BA^2+B^2 = A^3+AB = A(A^2+B), whence (B-A)(A^2+B) = 0. But if rank(A^2+B) = n, then B = A, which is absurd.


floor function
number theory
least common multiple

How many of the first 100 integers are divisible by all of the numbers 2, 3, 4, 5?

I assume you meant positive integers. ;)

[hide]$\lfloor \frac{100}{\text {lcm } (2,3,4,5)} \rfloor = \lfloor \frac{100}{60} \rfloor = 1$, only $60$.[/hide]

[hide=answer]First, multiply 3,4, and 5 to determine the smallest integer that satisfies the condition.  We need not multiply by 2 because it's already included in the 4.

The smallest integer that meets the conditions is 60. The next smallest would be 120, but it can't be larger than 100.

There is 1 such integer.


How do you make the following in latex?

1. sigma

2. logarithms

3. dollar signs

Any help is greatly appreciated.  :)


Prove that $\binom{pm}{m}$ is always a multiple of $p$.

Show work.

[quote="mathgeniuse^ln(x)"]Could you explain it better?

I can't comprehend what you are trying to say.[/quote]

I wish I knew how to use latex.
Anyways,

(pm)!/(m!(pm-m)!)
=(pm)x(pm-1)x...x(m+1)/(pm-m)!

In this step, I know that pm>m

So, all terms in pm! get cancelled up to m+1

That leaves us with:

=(pm)x(pm-1)x...x(m+1)/(pm-m)!

We know that since m>0 and pm>m, p>1.

So, we know  that pm-m = m(p-1)[u]>[/u]m+1

So, All terms cancel up to pm-m

=(pm)x(pm-1)x...x(m+1)/(pm-m)!
=(pm)x(pm-1)x...x(pm-m+1)

Hence, since all of the terms are being multiplied, if any one of the terms contains a factor of p, p is a factor of (pm  m)

[quote="neelnal"](pm)!/(m!(pm-m)!)
=(pm)x(pm-1)x...x(m+1)/(pm-m)!

We know pm>m, hence,

pm-m[u]>[/u]m+1

It will delete all terms up to some term pm-m+1

=(pm)x(pm-1)x...x(m+1)/(pm-m)!
=(pm)x(pm-1)x...x(pm-m+1)

pm is our first term, and since all terms are being multiplied, p must be a factor.[/quote]
If I understand you clearly, I'm not thinks you are right. I think what you have tried to say is that some of the term will be the same in the numerator and the deminstrator (after the first reduction) but you still have terms that aren't the same.
anyway, here is my work
[hide]$\binom{pm}{m}=\frac{(pm)!}{m!(pm-m)!}=\frac{(m+1)\cdots(pm)}{1\cdot2\cdots(pm-m)}$
we can say that $p>1$ and we want to show that there are more $p$ in the numerator than in the denominator-in order it will be a multiple of $p$. so, we want to show that $\left\lfloor{\frac{pm-m}{p}}\right\rfloor<\left\lfloor{\frac{pm}{p}}\right\rfloor-\left\lfloor{\frac{m}{p}}\right\rfloor$ which is obviously true if $p\nmid m$ if $p|m$ so we check $p^{2}$ and so on until $p^{n}\nmid m$[/hide]


Theres a new movie coming out on October 7th this year called Wallce and Gromit: The Curse of the Were-Rabbit. You might be familiar with this popular animated short. They take photographs for 1/24 of a second of the film, and repeat this process numerous times. If they did this process about 100,000 times, about how long is the movie?

[hide="ClIcK t0 rEaVeAl sTufF"]
if you divide 100,000 by 24 you will get the answer in seconds. So the film is $4166.\bar{6}$ seconds
So if you divide that by 60 you will get the film in minutes.  SO the film is $69.\bar{4}$ minutes.
That can be converted to [b]1 hour and 9 minutes and about 27 seconds[/b]
that looks wrong though.[/hide]

thats actually correct. The article in the magazine said rougly 100,000 times. So its probably more, but thats the answer i was looking for ^.^

its about 1 hour 9 minutes and 27 seconds.  It looks weird though and that is not an exact answer.  Were you actually trying to figure out how long a movie you were going to watch was? :rotfl:


let $ [{{E_{n},n\geq 1}}]$ be a set if increasing events.

         then prove that  $ \lim_{n\rightarrow infinity}P(E_{n})$=$ P(\lim_{n\rightarrow infinity}E_{n})$

where $ P(E_{n}$) denotes probability of the event $ E_{n}$.

[quote="JBL"]The command \to saves some writing and the command \infty saves some ugliness ;)[/quote]

oh! thank u sir...

Let $ E: \equal{}\lim_{n\to \infty}E_n\equal{}\bigcup_{n}E_n$

Then $ E\equal{}E_1\cup(E_2\minus{}E_1)\cup(E_3\minus{}E_2)\cup.\, .\, .$

 and $ P(E)\equal{}P(E_1)\plus{}P(E_2)\minus{}P(E_1)\plus{}P(E_3)\minus{}P(E_2)\plus{}.\, .\, .\equal{}\lim_{n\to \infty}P(E_n)$


function
topology
real analysis
real analysis unsolved

The following is a technical issue I would like to use in another proof. Consider the Borel sets $B(\mathbb{R},d)$ on the real line, where $d$ is the standard metric. Let $d'$ be the restriction of $d$ to $(a,b)\times (a,b)$. Is it true that $B((a,b), d')$ is equal to the sets of  $B(\mathbb{R},d)$ contained in $(a,b)$. This seems likely, but I don't see how to show it.

[quote="mlok"][quote="Kalle"]Let $d'$ be the restriction of $d$ to $(a,b)\times (a,b)$.[/quote]
Is this a typo, or did you really want to consider a product space?

The answer is yes, because the map $A\mapsto A\cap (a,b)$ [from the power set of $\mathbb R$ to the power set of $(a,b)$] preserves all of the relevant structures: it sends open sets to open, unions to unions, complements to complements, etc.[/quote]
It is not a typo. I was thinking of $d$ as a function from $\mathbb{R}\times \mathbb{R}$, because the "input" in the metric is a pair of real numbers. I am only trying to express that I want to consider $(a,b)$ as a subspace in the usual way.

Anyway, the justification you give is more or less why I believed this result to be true. However, my definition of the borel sets in a metric space $(X,d)$ is the smallest $\sigma$-algebra containing all open sets of $X$. I suppose that it would be possible to express this as "finite compositions of countable unions and set differences" or something like that. However, this seems to be very hard to carry out without doing a considerable amount of hand-waving. Isn't there a more succinct argument?

The intersection map $i$ sends $\sigma$-algebras to $\sigma$-algebras. So, $i(B(\mathbb R))$ is a $\sigma$-algebra that contains all open subsets of $(a,b)$. Conversely, the union of $B((a,b))$ with $B(\mathbb R\setminus (a,b))$ generates a $\sigma$-algebra $B'$ that contains all open subsets of $\mathbb R$ and satisfies $i(B')=B((a,b))$.

This works in more generality: for open subsets in a topological space.


AMC
AIME
geometry
geometric transformation
reflection
power of a point

Um, pardon me, but I thought I found really a much easier way to solve it.

Aren't QR and lets say ZR just secants through circle A?
The power of a point thereom for secants states:
The product of the external length times the total length of the secant is equal to the product of the external length of another secant times its total length.

So:
PR = QR = x
x(2x) = 2x :^2:

2x :^2: = (8 + 6 + 2 + 4 + 6) (4 + 6)
2x :^2: = 260
x :^2: = 130

Were you responding to something?  What are you talking about?

I'm sorry, I accidently posted this in response to the problem a bit further, accidental misclick.

I'll repost it here:
The distance AB is 12. The circle center A radius 8 and the circle center B radius 6 meet at P (and another point). A line through P meets the circles again at Q and R (with Q on the larger circle), so that QP = PR. Find QP :^2:   .

How did you get that huge sum? the 8+6+4+2+6 thing....

I think that Xbiz is assuming that R is also collinear with A and B,
and if that is the case

- that sum would be the total length of the secant.
- the solution is coincidental

might be wrong though

Well I am sure you can just choose R wherever you want it to be... So I suppose you could define R to be collinear.  Nice solution!

R is colinear with A and B ... but I'm not entirely sure how you would prove it.  I made some comment about that in the thread that this was supposed to be part of.  It may be a result of Zabelman's reflections, or it may not.

Ragingg -- you can't choose R whereever you want because it has to be such that RP = PQ, which doesn't happen for nearly every choice of R.

Xbizkitl -- how did you decide that BAR was a line?


function
number theory proposed
number theory

[quote="Max D.R."]Find all pairs of positive integers $ (m,n)$ with $ m\neq n$ such that $ m^{\phi( n)} \equal{} n^{\phi( m)}$.[/quote]

Clearly, $ m$ and $ n$ have the same set of prime divisors.

Then, since $ \varphi(m)\equal{}m\prod_{p|m}(1\minus{}\frac 1p)$ and $ \varphi(n)\equal{}n\prod_{p|n}(1\minus{}\frac 1p)$, the equation becomes $ m^n\equal{}n^m$

This equation is very classical and has general solution in $ \mathbb R$ :  $ (u,u)$ and $ (u^{\frac{1}{u\minus{}1}},u^{\frac{u}{u\minus{}1}})$

The set of non trivial solutions is reduced in $ \mathbb N$ to $ u\equal{}2$ and the couple $ (2,4)$


I have only found one solution for this problem so far, could you please help me find out more solutions?

When a certain positive integer N is divided by a positive integer d, the remaider is 7. If 2N+3 is divided by d, the remainder is 1. Find all possible values of d.

The only solution I got was D=16 and N=23, is there more solutions?

(1) N:d -> remainder=7
(2) (2N+3):d -> remainder=1

From (1) we get N=ad+7 where a is a positive integer and d>7 (because 7 is the remainder)

Let's now find 2N+3

2N+3=

2(ad+7) +3 =

2ad +14 +3 =

2ad +17

If we try to divide this number by the number d, the only remainder will come from 17, because the mumber d divides the first term 2ad

Hence, the remainder of the division (2ad+17):d must be equal to the remainder of the division 17:d

So, 17 = rd+1 => 16=rd

So d divides the number 16.

The dividers of 16 are: 1,2,4,8,16

But d>7, so the only possible values for d are 8 and 16

e.g. d=8 and N=15

Notice that, once you choose the number d, there are infinite values for N, because if (1) holds then:

2N+3 = 2ad+17 = 2ad+16+1 = 2ad+rd+1 = (2a+r)d+1

(e.g. for d=8, N may be 15,23,31,39,47,...)

Exactly the same question was solved earlier today on this board :lol: . [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42051]Here.[/url] Since both solutions are the same, we can be pretty sure the answer is correct :)


AMC USA(J)MO USAMO geometry trigonometry algorithm AIME

There are a LOT of 28s being thrown around, but I have a feeling a lot of them will be lower than that.  Still, I think (these estimates may be a touch high)

Black cutoff: Probably around 28
Blue cutoff: Probably around 22 or so
Red cutoff: I have no idea.

[quote="physisnu"]So....people.

We finished the 37th annual USAMO!!!  :P

I didn't take the test last year, but was able to get 4 questions in a mock test...

Got 3 for this year.

Some people say that 2008 was harder than 2007....

What do you guys think about the cut-off line?

(I know it's not that meningful a thing to do, but predicting is fun  :lol: )

Mine:

Black - 24~26 (But, alas, there are SO many people who claim to have gotten 4 questions...that's scary.)

Blue - 20~21 (I somewhat hoped for Blue, but...yeah, it's gonna be hard when you solved only 3 problems.)

Red - Somewhere around 9 (Geez, I really hope I get in to Red...)[/quote]

Why are you allowed to go to MOP? (Your location says Seoul).

In my opinion, it is much easier than last few years. I think black will >=29, blue will>=22. (Only my opinion) And my prediction of my score is around 25-27. so... :D

Yeah it kinda worries me that I can name 30 people who think they got > 28.

there's an 8th grader i know who thinks he got a 30
and a 7th grader who claims a 23

I think I got 5 (12345)... would that be good enough for black MOP (I'm a senior)? It's definitely much better than I've done the last three years (3, 2, 2, respectively), so that led me to think scores would be higher this year than previous years. I definitely don't think this year was any harder than last year. It's hard to top a test on which Brian Lawrence got a 32.

wahhh what
this thread is making me really insecure now
i guess im gonna be either barely in blue or miss by 1 or 2 points

[quote="jb05"]I think blue will be about 3.5 problems and black 4.5 problems. There is no way a 21 can make blue, I think... 1, 4, 5 were all easy... But then, a lot of people who I trust disagree, so we will see.

In response to the last post, people taking the USAMO for the first time don't estimate their scores too well ;). I don't know that many people who legitimately claimed 4 problems, so I think it will be okay. But I definitely think there are more than 12 who got 4 problems.[/quote]

This post WILDLY exaggerates the number of high scorers.

Bleh whatever the case, I have talked to many many people who claim to have gotten 1,4,5, though perhaps I mostly talk to people who would make mop anyway... I dunno. I definitely think extra points above 21 will be necessary, if only one or two.

Yeah I think too many people solved 4 problems so cutoff will be like 29 or 30. :(

While everyone exaggerates the scores, I don't think Red will be a 9 this year. Maybe 13 or 12...

[quote="Math Geek"][quote="physisnu"]Red - Somewhere around 9 (Geez, I really hope I get in to Red...)[/quote]

Wait, wuh?  I got numbers 3, 4, and 5 kinda, with like 3 or 4 points on each.  Would that qualify me for red?  That would be awesome.  I probably got like at most one point for my proofs for 1, 2, and 6, but I think my score could be around a nine.  Not sure though...[/quote]
You're too young for red; it's only for freshmen. :wink:

[quote="Ignite168"][quote="nealth"]Does anyone know how Evan O'Dorney did? He never replied my email.[/quote]

Are you one of his friends?[/quote]

nealth = Delong

MOPpers tend to want to know how each other did.

[quote="Ignite168"]How was meeting Evan O'Dorney?
Was it as great as imagined?[/quote]

That depends how great you'd imagine it to be.  I personally didn't think it was as great as going to DC and meeting, say, Alex Zhai.

Gradually, I realized that this year is really going to surpass year 2002(black will reach 35 and blue will reach 28). There are already a bunch of people who solved more than 5 problems, and these people are always tend to estimate their score correctly.

By the way, how did you do. nealth? if you doesn't mind telling :D

I'm doubtful of that...
I would be very surprised if black passed 30 or if blue passed 21 (or HM pass 23).

I have not heard of many people who have solved "more than 5" problems. Heck, I have only heard of 12-14 who have solved $ \geq 4$ problems.

In addition, if we go by the 5 returning red mop statistic, it seems like the cutoff would have to be around 21. I think I have the highest (of the red moppers with a possible exception for Vlad) score estimate, and that's 25.

So apparently we're thinking Vlad got about 21, which means that for there to be 5 returning red moppers, the cutoff'll have to be around 21.

[quote="CatalystOfNostalgia"]If I remember correctly, MOP phone calls (it was an email last year I believe) went out and winners/HM were released 11 days after the USAMO (on a Sunday), and we got our scores 3 days later, on a Wednesday. I'm quite sure about the second one because I remember finding out the night I arrived at MathCounts.

So by that pattern, winners/HMs and MOPpers will be announced on 5/11, this Sunday, and scores will be available 5/14, this Wednesday. Obviously, the pattern may not follow exactly the same way as last year. I hear in past years, scores were available before MOP phone calls were made.

(edit) all right, apparently in the USAMO manual it says both will happen on 5/13...that makes the above completely pointless.[/quote]
Now u're making me very nervous~~ :D

[quote="Sean Gu"][quote="Phelpedo"]Why don't we just wait like the 3 days until the results come out.[/quote]
we only have to wait 3 days?.. I thought it was 5.13 :D[/quote]

Hence the "like" in his sentence.

[quote="davidyko"][color=green]Please stop pyramid quoting. -mod[/color]

Hence the "like" in his sentence.[/quote]
oh...no I'm just figuring since I'll be out whatsoever,,,,it's better to know earlier~~ :lol:

This topic is getting very spammy very quickly, guys...

[quote="matt276eagles"]Also, you'd be surprised how immature some very, very smart people are.[/quote]

lol

:(

oh, and i'm totally not making blue mop. i guess i'm gonna join the giant mob of people who got #1 and 4 right and pity points on the other four questions.


A rubber ball is dropped from a height of 27 feet onto a concrete floor. Each time it hits the floor, it bounces back up to a height 2/3 of the height from which it fell.

   1. Calculate the height of each of the first four bounces.

   2. Find a formula that gives the height of the nth bounce.

   3. Use your formula to find the height of the 12th bounce to the nearest inch.

Extra: Suppose a different ball is dropped from 40 feet, and on its tenth bounce it reaches a height of 1 foot. What percent of the height from which it fell does that ball bounce back up each time it hits the floor?

Extension: Tell if you agree or disagree with the following statement and explain why. Support your opinion with any relevant mathematical ideas.

1. $18, 12, 8, \frac{16}{3}$
2. $f(n)=(\frac{2}{3})^n\cdot 27$
3. $\approx 2.49$
Extra. $40\cdot x^10=1\Longrightarrow x^10=1/40\Longrightarrow x=\sqrt[10]{1/40}$

[color=darkblue][size=75]<Edited to correct formatting problems by moderator rcv>[/size][/color]

thanks for answering

i cant believe only one person aswered

how did you get those answers? and are they right?

Yes, I think they're correct.  This was from [b]May[/b] of [b]2005[/b]:D .

[quote="robinhe"]This was from [b]May[/b] of [b]2005[/b]:D .[/quote]

In other words, don't bump topics this old. ;)


I was wondering if any body could suggest a good book for elementry topology?

Everybody these days seems to recommend starting with the Munkres textbook. If you can't get that from a library, there are some inexpensive Dover reprints that are older and thus not as modern in approach, but still good.

I second the vote for Munkres.  Also, Basic Topology by Armstrong is good.

I looked at one of the dover books and didnt really like it but ill try the Munkres book. Thank you.

There are some other books you might look at.  [i]One is A Combinatorial Introduction to Topology[/i] by Michael Henle.  I wouldn't use that alone, but in conjunction with other books.

[i]Topology Now![/i]  by Phil Straffin and Robert Messer just came out.  I haven't seen it yet, but the table of contents combined with the excellent pedagogical reputation of the authors suggests it'll be great.  The book's website is at
http://www.albion.edu/math/ram/TopologyNow!/

There's another new topology text which either just came out or is about to come out, but I can't remember the title/author.  (If someone else posts about it, I can confirm/deny, though...)

For less basic topology, the standard used to be Brown Munrkes but is now Hatcher's [i]Algebraic Topology[/i].  Red Munkres (now published with a green cover) is still the best point-set topology text, as tokenadult says.


I'm from the 715 area code of the upper half of Wisconsin.

So what I really want to know is if anyone else is too, or at least from where people are planning to fly.

I'm coming from Long Island NY, but I'll probably have to fly out of NYC.

I hail from Westchester, NY (an overall large county, but north of NYC).

Ideally, i'll be going to PROMYS, but I can't lose out going to Mathcamp either.  Now I wait for the PROMYS decision...

I have a question on the flight plans. In fact it is about the AA company.

If I call the company to order my ticket, they only request to charge me the luggage fee and booking fee that time, right? They won't charge that 326 from me, but from Mathcamp (after I mail the Mathcamp my check). So I still need to mail the 326 to North Carolina?

I want to make this sure so that I won't spend twice 326 on the ticket.  :P

Hi,

[quote="Go Beyond"]I have a question on the flight plans. In fact it is about the AA company.[/quote]

Please send your travel questions to travel@mathcamp.org.

I will have my plane connection in Chicago flight #AA1611 at 8.35am
Anybody can go with me?

My connection to seattle will be from Atlanta, on Delta flight #823, on the off chance that anyone uses the same flight.

For me, yet to be confirmed...

My friend has reserved his ticket already. I think he's taking Asiana OZ752/OZ272 from Singapore, transit in Seoul.


In this game, someone posts up a word like "good", "jump", or "book".  Then, the next user posts up as many synonyms as they can, earning the number of synonyms they post up as points.

I'll start first:

[b]wise[/b]

clever, smart, intelligent, knowing, knowledgeable, cunning,

New word, anyone?  OK, I will..........lets do [b]capable[/b]

able, competent. good, qualified, proficient, skillfull

NW

[b]message[/b]

[b]reply, news, result, outcome on paper

How about "ignorant"?[/b]

From now on, everyone please post up scores.

[b]careless, unmindful, inexperienced, oblivious[/b]

[hide="Scores"]

inoit 6
tennis123 6
professordad 4
maybach 4

[/hide]

Since professordad didn't post a new word, I will:
[b]liberal(one of my favorite words and conservative is one of my least favorite:))[/b]

I bet you're conservative. :P

/me looks up liberal just to make sure me knows what it means.

/me does not look in a thesaurus.

[b]open-minded, undedicated, hypocritical[/b]

I don't know about the last 2, so you don't have to put 3 points if you don't want to.

[b]justice[/b]

:D
[hide="Scores"]

inoit 6
tennis123 6
professordad 4
maybach 4
1=2 ?

[/hide]

code law rule truth
[hide="scores"]
inoit 6
tennis123 6
professordad 4
maybach 4
SuperNerd123 4
1=2 3

[/hide]
Nw = Echolocation

[b]locating, finding[/b]   sigh, only two.

NW:allocation

appropriation, quota, ration, share
Nw = Antidysenteric
[hide="scores"]
inoit 6
tennis123 6
professordad 4
maybach 4
SuperNerd123 8
1=2 3
AIME15USAMO 2[/hide]


[quote="ffdbzathf"]Two positive point charges +q are on the y axis at y = +a and y = -a.  A bead of mass m carrying a negtive charge -q slides without friction along a thread that runs along the x axis.  (a) Show that for small displacements of x << a, the bead experiences a restoring force that is proprtional to x and therefore undergoes simple harmonic motion.  (b) Find the period of the motion.[/quote]

The collective E-field of the two point charges on the y-axis is given by:

$\vec{E}(x)= k \{ q \frac{\vec{x}-a\hat{y}}{|\vec{x}-a\hat{y}|^3}+q \frac{\vec{x}+a\hat{y}}{|\vec{x}+a\hat{y}|^3}\}$

When $ a >> x$ this reduces to $\vec{E}(x)= 2 k q \frac{\vec{x}}{a^3}$

Multiplying this expression by $-q$ to obtain the coulomb force on the negative charge one obtains: $\vec{F}(x)= -2 k q^2 \frac{\vec{x}}{a^3}$

This force is along the x-direction, due to the vector nature of $\vec{x}$ itself, restorative because of the minus sign, and $\propto \vec{x}$; therefore, this coulomb force is an example of Hooke's law.

The angular frequency for this sort of arrangement is given by: $\omega^2 =\frac{K}{m}$

$K= \frac{2 k q^2}{a^3}$

$\therefore \omega = \frac{2\pi}{T} = \sqrt{\frac{K}{m}}=\sqrt{\frac{2 k q^2}{m a^3}}$

$T = 2 \pi \sqrt{\frac{m a^3}{2 k q^2}}$

Let $A(+q),B(-q)$ and $P(-q)$ at distance of $OP=x$, $\angle {APO}=\angle {BPO}=\theta$.For $a\gg x$, we can consider that $PA=PB\approx a$,

yielding $\cos \theta \approx \frac{x}{a}$. Since we have $\overrightarrow {E}=\overrightarrow {E_{A}}+\overrightarrow {E_{B}}$, we obtain $|\overrightarrow {F_{P}}|\approx 2|\overrightarrow {E_{A}}|\cos \theta\approx 2\cdot k\cdot\frac{q^2}{a^2}\cdot \frac{x}{a}=\frac{2kq^2}{a^3}x$, where $k$ is Coulomb's constant.Therefore the equation of motion for the bead at $P$ is $m\frac{d^2 x}{dt^2}=-Kx$, where $K=\frac{2kq^2}{a^3}$,yielding the desired period is that one which is derived by Dr.No.


My wii isn't connected to the internet, but I wish it was.

[quote="7h3.D3m0n.117"]Isn't online Wii free?[/quote]

I don't know.

The wii isn't really that bad. It is really fun. I just got Wii Fit.

You can connect your wii to your internet. That can be done in setting. You can search out the wireless signal from your router, or you can attach the LAN to your wii with an adapter. The internet channel, which is an opera browser, used to be free, but you now have to pay 500 wii pts for it ( $ \$$ 5 ).

What are wii points. I have never heard of these before.

I'm assuming they're similar to the Microsoft Points made by...Microsoft. Basically, you can buy them either online using a credit card, or at your local store on prepaid cards. You can use them to buy things such as downloadable games, or in this case, membership.

Hmm, that is interesting. I will have to take a look at that.

My wii is connected to the internet, but my dad is the only one who uses it...  :mad:

I'm getting a wii game for christmas but i'm not sure which game to get. I'll usually play by myself so should i choose mario party 8, super smash brothers brawl, or Ncaa football09??

SSBB. At least you can train for MLG in the long run...

I only have SSMB and Mario Party 8 out of the three jjx1 listed, but I haven't gotten enough time to play Mario Party 8 yet, but I have seen my brother play it. I think you should get SSMB.

I have Mario Kart wii. Whenever my connection fails during an online race, I automatically lose. Kinda like a forefit. It messes up my rating.

[quote="Wickedestjr"]I just got mine connected to the internet. What kinds of things can you do without any wii points?[/quote]

Wi-fi games (Call of duty, Super smash bros. brawl, Mario Kart, ect.ect) and you can download some of the other free channels such as voting channel, and the mii channel also has online capabilities. Other than that their really isn't alot to do with Wii online.

i have a wii, and it's really good. but i'm a casual gamer, and the whole wii system and almost all its games are casual. so the system isn't for people who like hardcore games, like Halo and GTA and stuff like that. but if you don't have enough games, it gets really boring after a while.  :juggle:


Can we find $a, b, c, d \in N$ s.t $a<b<c<d$

and $a^2, b^2, c^2, d^2$ makes an arithmetic sequence?

No, this is a famous result due to Fermat, with a quite long proof, that there is no 4 squares in arithmetic progression.

Pierre.

Isn't it a famous result in the Mathlinks contest as well? :)


In some of my country 's math forums, there are several problems that  no body solves. Here are some of them:
Problem1: Have:
$a^{2}+b^{2}\geq c^{2}, b^{2}+c^{2}\geqq^{2}, c^{2}+a^{2}\geq b^{2}$.
Prove that:
$(a+b+c)(a^{2}+b^{2}+c^{2})(a^{3}+b^{3}+c^{3}) \geq 4(a^{6}+b^{6}+c^{6})$

Problem2: a,b,c > 0. Prove that: $\sum \frac{1}{7a^{2}+2bc}\geq \frac{3}{(a+b+c)^{2}}$

I hope that every one can solve these problem with pleasure.

[quote="rem"]Here is the solution for problem $1$:
We have $a^{3}b^{3}+a^{3}c^{2}b=(a^{3}b)(b^{2}+c^{2})\ge a^{5}b$
Therefore $\sum_{sym}a^{3}b^{3}+\sum_{sym}a^{3}c^{2}b\ge \sum_{sym}a^{5}b$
Also, by Muirhead, $\sum_{sym}a^{5}b\ge \sum_{sym}a^{4}b^{2}$
Therefore:
$(a+b+c)(a^{2}+b^{2}+c^{2})(a^{3}+b^{3}+c^{3})=$
$=a^{6}+b^{6}+c^{6}+\sum_{sym}a^{4}b^{2}+\sum_{sym}a^{5}b+\sum_{sym}a^{3}b^{3}+\sum_{sym}a^{3}c^{2}b$
$\ge a^{6}+b^{6}+c^{6}+2\sum_{sym}a^{5}b+\sum_{sym}a^{4}b^{2}$
$\ge a^{6}+b^{6}+c^{6}+3\sum_{sym}a^{4}b^{2}$
$=a^{6}+b^{6}+c^{6}+3\sum_{cyc}a^{4}(b^{2}+c^{2})$
$\ge 4(a^{6}+b^{6}+c^{6})$ QED[/quote]

So nice a solution. I am not know much about Muirhead inequality, so my solution was not as beutiful as you. Thank 's alot.
And how about the second problem, I think it's nice one. Or it so easy for you to prove? I am waiting for a nice solution.

[quote="rem"]Thank you.
Unfortunately the second problem has been already posted on Aops and here is the link:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=frac+geq+2bc+7a+sum&t=33538[/url]
The solution by Barra is amazing:
[quote="barra"]The above inequality is equivalent to
$\sum \frac{a^{2}+b^{2}+28ab}{(7a^{2}+2bc)(7b^{2}+2ac)}(b-a)^{2}\\+\sum \frac{42(a^{2}+ac+c^{2})}{(7a^{2}+2bc)(7b^{2}+2ca)(7c^{2}+2ab)}(a-c)^{2}(b-a)^{2}\geq 0$[/quote]
This is just beautiful.[/quote]

Unfortunately, his solution is not correct (the very last post at the end of your link)

* $\sum \frac{1}{7a^{2}+2bc}\geq \frac{3}{(a+b+c)^{2}}$

Assume that: $a \geq b \geq c$
We have the inequalaty with new form:
$\sum (a-b)^{2}\frac{7a^{2}+7b^{2}+28ab-6ac-6bc}{ (a^{2}+2bc)(b^{2}+2ac)}\geq 0$
 Have: $a \geq b \geq c$ so that :
$7a^{2}+7b^{2}+28ab-6ac-6bc \geq 0$
$7a^{2}+7c^{2}+28ac-6bc-6ab \geq 0$
So we need to prove:$(a-c)^{2}\frac{7a^{2}+7c^{2}+28ab-6ac-6bc}{ (a^{2}+2bc)(c^{2}+2ab) }+$$(b-c)^{2}\frac{7b^{2}+7c^{2}+28bc-6ab-6ac}{ (b^{2}+2ac)(c^{2}+2ba) }$
But we also have$(a-c)/(b-c) \geq a/b$ with $a \geq b \geq c$
So we just only need to prove : $49a^{4}b^{2}+14a^{5}c+98a^{2}b^{2}c^{2}+14a^{3}c^{3}+154a^{3}b^{2}c+56a^{4}c^{2}-84a^{3}b^{3}-12a^{4}bc+154a^{2}b^{3}c-12a^{3}bc^{2}+49a^{2}b^{4}+14b^{5}c+14b^{3}c^{3}+56b^{4}c^{2}-12b^{4}ca-12ab^{3}c^{2}\geq 0$
The last inequality is of course right.
So we can prove the problem.

@: When I try to prove the last inequality, I try Mple software to factor it, but it 's not suceed. But we can easily see that it's so easy to prove, just only so "ugly". I post this here to find out another solution. And thanks you rem.

[quote="timlaiminh"][quote="Lovasz"][quote="timlaiminh"]In some of my country 's math forums, there are several problems that  no body solves. Here are some of them:
Problem1: Have:
$a^{2}+b^{2}\geq c^{2}, b^{2}+c^{2}\geqq^{2}, c^{2}+a^{2}\geq b^{2}$.
Prove that:
$(a+b+c)(a^{2}+b^{2}+c^{2})(a^{3}+b^{3}+c^{3}) \geq 4(a^{6}+b^{6}+c^{6})$

[/quote]
Applying Mink's inequality, \[(a+b+c)(a^{2}+b^{2}+c^{2})(a^{3}+b^{3}+c^{3}) \ge (a^{2}+b^{2}+c^{2})^{3}.\] It remains to show that: \[(x+y+z)^{3}\ge 4(x^{3}+y^{3}+z^{3}),\] where $x,y,z$ are sides of a triangle. Done.[/quote]

Can you explain more about the last inequalty?
If we are expand it, what will happen? I think that it might be wrong.
Can you explain it more clearly?[/quote]
put $x=r+s, y=s+t, z=t+r$. expand it. we get \[2\sum r^{3}+6\sum_{\textrm{sym}}rs^{2}+12rst \geq 2\sum r^{3}+3\sum_{\textrm{sym}}rs^{2}\] which is equivalent to \[3(\sum_{\textrm{sym}}rs^{2}+4rst) \geq 0\] Also, you can call the mink's inequality to "Holder's inequality".
ah.... This inequality is not sharp. I think we can put some restrictions for it.

But I don't like Holder's inequality as its display.


In LaTeX, is there a symbol for the long division sign? Also, can you underline equations? Thanks.


trigonometry
geometry
geometric transformation
reflection
symmetry
algebra
polynomial

For what values of $ x$ does the sequence $ f_{n}\left(x\right)\equal{}\sin 7^{n}\pi x$ converge and what is its limit?

for which real numbers $ r$ does $ \{7^nr\}$ converge? ($ \{\cdot\}$ denotes fractional part.) where's our resident expert on fractional parts of powers? :)

It converges if and only if it's eventually constant; all equilibria are unstable.

jmerry, is that a response to my question?

It's an answer to both versions, which are slightly different. The $ \sin$ has reflection symmetry as well.

There is a polynomial $ p$ such that $ \sin(7x)\equal{}p(\sin x)$. We're iterating $ p$, and all the usual tricks for dealing with iterations apply.


Putnam
combinatorics unsolved
combinatorics

Can you draw circles on the plane so that every line intersects at least one of them but no more than 100 of them?

I believe so. Construct circles along the graph of $ y \equal{} x^3$.

why do you need $ x\equal{}\minus{}y^3$?

Sorry; I was thinking of a Putnam problem where one solution required constructing circles near both axes, but clearly that's not necessary here.

but wait... something is weird here. it's too easy to be a M.S. problem... it might be a bad translation.
Another solution would be to draw unit circles along the x and y axis. (whoops! not that simple  :lol: )

How about if we restrict that the circles must not intersect?

It's correct according to [url=http://www.math.u-szeged.hu/~mmaroti/schweitzer/]this site[/url], but I agree.  Wouldn't the $ x$ and $ y$-axes themselves intersect the circles you describe?

And the non-intersecting condition doesn't seem to add anything to the problem.  One can just place a circle of some fixed radius $ r$ at the origin and then place circles of radius $ r$ tangent to your previous circles with centers on $ y \equal{} x^3$.

I meant circles with no points in common..

The centers of the circles, instead of following a curve, should zigzag above and below the curve.  For example, one can alternate placing circles on $ y \equal{} x^3 \minus{} 1$ and $ y \equal{} x^3 \plus{} 1$ of the appropriate radius.

I might be missing something but i don't see how to do that without making the circles tangent...

If I'm understanding you correctly, the only issue with adding the constraint that the circles can't be tangent is that there exist lines that pass between two consecutive circles.  But in the zigzag construction the circles don't need to be tangent to guarantee that lines can't pass between them.

In the simplest case, two tangent unit circles can be replaced with three unit circles less than $ \sqrt{3} \minus{} 1$ apart pairwise; then the common external tangents of any two of them passes through the third.  The zigzag construction should be done so that this always occurs.

I still feel like the argument is not full.
When you add more and more circles you might not be able to keep the distance between them that small while on the two mentioned curves. (unless you increase their radius, in which case how do we know that there is no line intersecting infinitely many of them)

This problem recently appeared on MathOverflow as well (though my reaction was that Olympiad problems are 1) for AoPS, 2) for spending a month or more trying to solve them before posting). Since it would be nice to have a solution somewhere for those who already did 2) without any effect, I'm including it here. Note that you may learn quite a few things trying to solve the problem by yourself, so don't read the spoiler unless you decided to give up completely.
[hide="Solution"]
[hide="Answer"]
It is impossible. We will argue by contradiction. Assume that such a configuration exists. I'll be generous and count touching as an intersection in the "should intersect at least one circle" requirement but not in the "should intersect not more than $100$ circles" requirement.
[/hide]

[hide="Idea 1: almost circle  free angle"]
Let $N$ be the maximal number of circles that can be intersected by one line and let $\ell$ be the line intersecting exactly $N$ circles $C_1,\dots,C_N$. Without loss of generality, $\ell$ is the $y$-axis. Then every line passing through the origin and making sufficiently small angle with $\ell$ intersects all these circles too and, thereby, no others. Hence, all other circles are entirely contained either in the angle $A=\{(x,y):|y|\le Bx\}$ with some large $B>0$ or in the symmetric angle $-A$.
[/hide]

[hide="Idea 2: CAT scan"]
Note that if a circle is contained in the angle $A$ and intersects the line $x=r$, then it is also contained in the strip $Q^{-1}r<x<Qr$ with some large $Q=Q(B)>0$. Now choose $R$ so large that no line at distance $R$ or more from the origin can intersect any of the circles $C_1,\dots,C_N$ and consider the lines $x=r$ with $R<r<2R$. Each such line has to intersect a circle contained in $A$, so the sum of diameters of circles contained in the intersection $I(R)$ of $A$ and the strip $Q^{-1}R<x<2QR$ is at least $R$.
[/hide]

[hide="Idea 3: Radar scan"]
Now take a circle $C$ of diameter $d$ contained in $I(R)$ and look at the lines $y=bx$ with $-B<b<B$. The interval $J(C)$ of $b$ for which the line $y=bx$ intersects $C$ is at least $cd/R$ with some $c=\frac 1{2Q}>0$. Since the sum of diameters of the circles contained in $I(R)$ is at least $R$, we conclude that the average number of circles contained in $I(R)$ a line $y=bx$ intersects is at least $c'=\frac c{2B}$.
[/hide]

[hide="End of proof"]
Now just choose sufficiently large $R_1,\dots,R_K$ so that $I(R_j)$ are disjoint. Then the average number of circles contained in $\cup_j I(R_j)$ a line $y=bx$ intersects is at least $Kc'$. If $K$ is so large that $Kc'>100$, we are in trouble.
[/hide]
[/hide]

On the graph of x^3, eventually the graph starts getting straight up, then eventually there will be 100 circles on 1 line.


induction
number theory proposed
number theory

Determine all positive integers a for which there exist infinitley many squarefree positive integers n such that n divides a^n-1.

Bomb

A very very beautiful problem, bomb! I think the answer is all numbers except 2 and 3. For 2 actually there are no numbers $n$ except 1 and for 3 it is not difficult to prove that if $n|3^n-1$ then $n$ is a multiple of 4. So, suppose $a>3$ and take $p_1$ an odd prime factor of $a-1$. Next, take $p_2$ an odd prime factor of $ a^{p_1}-1$ different from $p_1$, next $p_3$ an odd prime factor of  $a^{p_1p_2}-1$  and so on. The idea is that we will always have the chance to choose these primes and this is a consequence of a more general property:
   If $p| x-1$ is an odd prime, then there is an odd prime dividing $ x^p-1$, different from $p$.
  The property follows by choosing any prime factor of $\frac{x^p-1}{p(x-1)}$ (exercise!).
   There is stil some problem, namely when $a-1$ is a power of 2. But then $a^2-1$ is not a power of 2 and apply the same reasoning with this new value.

Harazi, nice proof!!! Your exercise is quite easy, expansion by binomial thorem does it,

I proved the existence ogf such a prime by induction on coprime factors... while my other post on the primes of form nk+1 provides a stronger than needed identity..

Bomb


In a convex polygon, exactly five of the interior angles are obtuse. The largest possible number of sides for this polygon is ?

Okay suppose that we have a $k+5$ sided polygon. There are $5$ angles less than $180$, there are $k$ angles less than $90$. Since it is convex, we have that all the angles are less than 180.

Sum of the angles = $180*(k+5-2)<180*5+k*90\implies k<4$. Hence the max of $k$ is $3$, and the max number of sides is $3+5=\boxed{8}$ sides.

[hide]Let n-gon $ Q$ have $ n$ sides, and $ n$ chillin' little angles of measure $ \theta$. $ \frac{2\pi}{n}= \theta$.  Of the three angles constructing a theoretically isosceles triangle, $ \theta$ is unique.  $ \theta$'s two sisters (who are identical) sum to be obtuse.

$ \theta = \frac{2\pi}{n}< \frac{\pi}{4}\Rightarrow n = 8$[/hide]
Ta da?  :huh:


pigeonhole principle
geometry
rectangle
probability

Kevin is saying "Susie sells seashells by the seashore" in which that each word is rearranged in some way such that two consecutive words don't have the same starting letter. How many ways can this sentence be arranged?

 :D

0, by the Pigeonhole Principle (pair up the 6 words into 3 pairs).

hm doesnt

ussie ellss seashells by the seashore

work?

starting letters are U, E, S, B, T, S.

OH! I thought he said the words are rearranged... :(

I foresee nasty casework... :|

That will take a lot of casework to solve.

oh. could this one be solved with less casework? mark wanted to go to mcdonalds, and his house is on the bottom left corner of a 5x6 grid (each square in the grid has a side length that's a block). if mcdonalds is on the right upper corner and he wants to walk a path such that no more than two consecutive Norths are made and no more than three consecutive Easts are made (he cannot go backwards: down/left, and what i mean by norths and easts is like some path: NNNEEEEE is a path cuz it gets to mcdonalds.), what is the largest area of the polygon that his path forms with the left side of the 5x6 rectangle and the top side (this is kinda confusing but o well). oh yeah, he can only walk N or E, not like half N, half E, then one third N,....

second problem: on the way back, mark walks some other way. what is the largest possible area of the polygon that can be formed by the two paths? (he can only walk no more than two consecutive S's and no more than three consecutive Ws, and no Ns or Es.)

 :D

if 8 kids are in a row and if two consecutive kids are wearing the same colored shirt, they would be in a "funny" arrangement. if each kid has 6 shirts: red, blue, green, yellow, orange, and black, then what is the probability that they will be arranged in a "funny" arrangement if they each randomly choose a shirt the night before?

 :D


projective geometry
geometry proposed
geometry

hello.
in this picture :
BD is diameter of ABCD and XM=YM . O is center of the circle.
prove:LQ and PS and DM are crossover.

$ M$ can move on $ BD$ (don't need $ MX \equal{} MY$)
Let $ PQ\cap AC \equal{} \{I\}, LI\cap (O) \equal{} \{S'\}$
We will show that $ S'\equiv S$:
This result is well-known:" $ BD$ is the polar of $ I$". Let $ PL\cap QS' \equal{} \{N\}$ then $ N$ lies on the polar of $ I$ so $ N,B,D$ are collinear.
Since $ P \equal{} AD\cap NL, Q \equal{} DC\cap NS', I \equal{} LS'\cap AC$ are collinear then applying [b]Desargues[/b]'s theorem for triangle $ ADC$ and $ LNS'$ we have $ AL, ND, CS'$ are concurrent at $ M$.
So $ C,S',M$ are collinear or $ S'\equiv S$.
Thus $ L,S,I$ are collinear
Let $ J \equal{} QL\cap PS$ then $ J$ lies on the polar of $ I$ or $ J,B,D$ are collinear.
$ \rightarrow$ QED


algebra
polynomial
Rational Root Theorem

[quote="i_like_pie"][quote="inom"]Solve the equatıon
$x^{3}+x^{2}-x-1=0$[/quote]
[hide]$x=1$[/hide][/quote]

that's incorrect :wink:

[hide]moving the x to the other side of the equation, we can factor it as $x(x^{2}+x-1)=1$.  this is equal true when $x=x^{2}+x-1=\boxed{\pm1}$[/hide]

Oh, I forgot about odd powers. :oops:
Excuse: I haven't eaten breakfast yet. :rotfl:

[quote="inom"]Solve the equatıon
$x^{3}+x^{2}-x-1=0$[/quote]

[hide="classic"]
Move the one over to the other side and get $\pm 1$ (It is a classic factorign sum)[/hide]

[hide]Factor this way:

$x^{3}+x^{2}-x-1=x^{2}(x+1)-(x+1)=(x^{2}-1)(x+1)=0$ (do you see why?)

Then it goes from there.[/hide]

[quote="nutz_for2.718281828"][hide="solution"]By inspection, x=1[/hide][/quote]
[hide="Incorrect"]$\pm1$[/hide]

Just by looking at it, I realize that 2 answers are

[hide]1,-1[/hide]

After doing a bit of work, I don't think that there are any more.

The lesson to learn is that if you get an answer with "inspection" divide the equation by the solution...

[quote="srulikbd"]what about $x=0$?[/quote]
$-1\ne0$
$x=0$ is a solution iff the constant term is $0$.

[quote="neelnal"]x^3 + x^2 - x- 1= 0

So, let us find all possible rational roots.

ax^3 + bx^2 + cx + d= 0

a: 1
d: -1,1

So, rational roots are: -1/1, 1/1 or -1, 1

Plug these into the equation and you get x= +/-1.[/quote]
This is the Rational Root Theorem (aka RRT). Is it in high school math?

[quote="bpms"]This is the Rational Root Theorem (aka RRT). Is it in high school math?[/quote]
I learned it in Pre-Calc. :)


search
combinatorics unsolved
combinatorics

[quote="nktp"]Let 4 pots: 18 litres pot with full of milk, 7,7,4 litres pot with empty of milk. Please find the minimum number steps of measure milk satisfy in each 7 litres pot we have 6 litres of milk.[/quote]

At most 13 steps :

00 : 18 0 0 0
01 : 11 0 7 0
02 : 11 0 3 4
03 : 15 0 3 0
04 : 15 0 0 3
05 : 08 0 7 3
06 : 08 0 6 4
07 : 12 0 6 0
08 : 05 7 6 0
09 : 05 3 6 4
10 : 09 3 6 0
11 : 09 0 6 3
12 : 02 7 6 3
13 : 02 6 6 4

I also find 2 solution for this problem and this is 13 steps, too. But I can't show that is best number. Who can help me?

One way is to do a breadth-first search starting from the initial state to compute all possible outcomes reachable in at most 12 moves.

For example:
[code]
0:  18 0 0 0
1:  11 0 7 0
    14 0 0 4
2:  04 7 7 0
    07 0 7 4
    11 0 3 4
    14 0 4 0
3:  00 7 7 4
    04 3 7 4
    07 4 7 0
    10 0 4 4
    11 4 3 0
    15 0 3 0
...
[/code]
It will probably be quite a lot of work, but I doubt there is an easier way, at least for the general case, unless there is some special property intrinsic to this particular instance.


find all answer of$x^2+y^2+z^2=3xyz$.
$(x,y,z\in \mathbb{N})$

This is a special case ($n=3$) of Markov' equation, which can be solved by descent.
This appeared several times on ML.

Pierre.


Let $ AD$ be the altitude of an acute-angled triangle $ ABC$ and let $ H$ be any interior point on $ AD$. Lines $ BH$ and $ CH$, intersect $ AC$ and $ AB$ at $ E$ and $ F$, respectively. Prove that $ \angle EDH\equal{}\angle FDH$.

Dear Mathlinkers,
see for example
Honsberger R., Mathematical Gems, Two-Year College Mathematics Journal, vol. 14, 2, p. 154-155.
Sincerely
Jean-Louis


algebra
polynomial
superior algebra
superior algebra unsolved

Suppose $ \alpha$ is a root of the irreducible polynomial $ f(x)=x^{3}-3x^{2}-4$.  Find the minimal polynomial for $ 1/\alpha$.  Also, find the $ N(\alpha^{3})$ and $ Tr(\alpha^{3})$ (norm and trace of $ \mathbb{Q}(\alpha)$ over $ \mathbb{Q}$, respectively).

[quote="baz"]Suppose $ \alpha$ is a root of the irreducible polynomial $ f(x)=x^{3}-3x^{2}-4$.  Find the minimal polynomial for $ 1/\alpha$.[/quote]

That means $ f\left( \frac{1}{x}\right) =\frac{1}{x^{3}}-\frac{3}{x^{2}}-4$

Now $ \frac{1}{\alpha}$ is a zero of $ f(1/x)$. Thus, it seems that $ 1-3x-4x^{2}$ is a polynomial having $ 1/\alpha$ as a zero.

EDIT: Fixed mistake mentioned by -oo-

@sylow: yes, $ 1-3x-4x^{2}$ is the minimal polynomial of $ 1/\alpha$, but of course, $ 1/\alpha^{3}-3/\alpha^{2}-4=0$ is wrong.
@baz: what have you tried so far, and what do you know about norm and trace? An approach would be using $ N(xy)=N(x) N(y)$ and calculating the minimal polynomial of $ \alpha^{3}$.

Yes, I was able to see that the minimal polynomial for $ 1/\alpha$ is $ x^{3}+(3/4)x-(1/4)$ (where my definition of minimal polynomial includes monic with minimal degree).  Also, I know $ N(xy)=N(x) N(y)$ gives us $ N(\alpha^{3})=N(\alpha)^{3}=64$ since the field polynomial for $ 1/\alpha$ equals $ f(x)$ in this case and the norm equals the negative of the constant term in the field polynomial.  But, how does one find $ Tr(\alpha^{3})$ without too much work?  (I did not see if the other parts of the problem mentioned above provide any help toward the question.)

For those interested, I found  :idea: that $ Tr(\alpha^{3})\equal{}Tr(3\alpha^{2}\plus{}4)\equal{}3Tr(\alpha^{2})\plus{}4Tr(1)$.  To find the $ Tr(\alpha^{2})$, we find the minimal polynomial for $ \alpha^{2}$.  Notice that $ f(x)(x^{2}(x\plus{}3)\plus{}4)\equal{}(x^{2}(x\minus{}3)\minus{}4)(x^{2}(x\plus{}3)\plus{}4)\equal{}x^{4}(x^{2}\minus{}9)\plus{}4x^{2}(\minus{}9)\minus{}16$ kills $ \alpha$ so that in turn
$ x^{2}(x\minus{}9)\plus{}4x(\minus{}9)\minus{}16\equal{}x^{3}\minus{}9x^{2}\minus{}36x\minus{}16$ kills $ \alpha^{2}$.  Moreover, it is easy to see that this is its minimal polynomial and also its field polynomial (well, given some knowledge/theorems on the subject).  So, the $ Tr(\alpha^{2})\equal{}9$ yields that $ Tr(\alpha^{3})\equal{}3Tr(\alpha^{2})\plus{}4Tr(1)\equal{}3\cdot 9\plus{}4\cdot 3\equal{}39$.  :D happy that this is done.

Thanks to the moderators and people here

[quote="baz"]For those interested, I found  :idea: that $ Tr(\alpha^{3}) \equal{} Tr(3\alpha^{2}\plus{}4) \equal{} 3Tr(\alpha^{2})\plus{}4Tr(1)$.  To find the $ Tr(\alpha^{2})$, we find the minimal polynomial for $ \alpha^{2}$.  Notice that $ f(x)(x^{2}(x\plus{}3)\plus{}4) \equal{} (x^{2}(x\minus{}3)\minus{}4)(x^{2}(x\plus{}3)\plus{}4) \equal{} x^{4}(x^{2}\minus{}9)\plus{}4x^{2}(\minus{}9)\minus{}16$ kills $ \alpha$ so that in turn
$ x^{2}(x\minus{}9)\plus{}4x(\minus{}9)\minus{}16 \equal{} x^{3}\minus{}9x^{2}\minus{}36x\minus{}16$ kills $ \alpha^{2}$.  Moreover, it is easy to see that this is its minimal polynomial and also its field polynomial (well, given some knowledge/theorems on the subject).  So, the $ Tr(\alpha^{2}) \equal{} 9$ yields that $ Tr(\alpha^{3}) \equal{} 3Tr(\alpha^{2})\plus{}4Tr(1) \equal{} 3\cdot 9\plus{}4\cdot 3 \equal{} 39$.  :D happy that this is done.
[/quote]

$ \rm{SIMPLER}\quad Tr(a^{2}) \equal{} a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} (a\plus{}b\plus{}c)^{2}\minus{}2(ab\plus{}bc\plus{}ca) \equal{} 3^{2}\minus{}2(0)$


[quote]Please enter the student's current Art of Problem Solving Username.
This is the same name that the student uses in the forums and/or has used in previous classes.

Enter the student's username:

If the student does not have a username, click here to create a new account. [/quote]

familiar?
what do i do after i enter my username. when do we have to pay the money?

You enter your username there, then click Enroll on that page.  Follow the directions on the pages that follow.


calculus
derivative
function
algebra
binomial theorem
calculus computations

Find a formula for $ f^{(n)}(x)$, where $ f(x)\equal{}\sqrt{x}$.

I have

$ f^{(n)}(x)\equal{}\frac{(\minus{}1)^{n\plus{}1}(2n\minus{}2)!}{2^{2n\minus{}1}(n\minus{}1)!x^{n\minus{}\frac{1}{2}}}$.

Is there a better way to write this?

More find the ${ f^{(n)}(x)}$ where $ f(x)=x^a$

Does it look SOMETHING like the binomial theorem? I have a solution, but I'm not sure if it's right.

Mine looks VERY similar to the binomial theorem.

The binomial theorem, you say? Let's consider Newton's version of that.

For notation, we can make the following definition:

If $ \alpha\in\mathbb{R}$ and $ n$ a nonnegative integer, then let

$ \binom{\alpha}{n}\equal{}\frac{\alpha(\alpha\minus{}1)(\alpha\minus{}2)\cdots
(\alpha\minus{}n\plus{}1)}{n!}.$

We have this power series centered at zero:

$ (1\plus{}x)^{\alpha}\equal{}\sum_{n\equal{}0}^{\infty}\binom{\alpha}{n}x^n$

Shift this and stretch it by letting $ x\equal{}\frac{y\minus{}b}{b}$ so that $ 1\plus{}x\equal{}\frac{y}{b}.$

$ \left(\frac{y}{b}\right)^{\alpha}\equal{}
 \sum_{n\equal{}0}^{\infty}\binom{\alpha}{n}\left(\frac{y\minus{}b}{b}\right)^n$

$ y^{\alpha}\equal{}\sum_{n\equal{}0}^{\infty}\binom{\alpha}{n}b^{\alpha\minus{}n}(y\minus{}b)^n$

But by Taylor's theorem, we should have $ f(y)\equal{}\sum_{n\equal{}0}^{\infty}\frac{f^{[n]}(b)}{n!}(y\minus{}b)^n$

Hence, if $ f(y)\equal{}y^{\alpha}$, then

$ f^{[n]}(b)\equal{}n!\binom{\alpha}{n}b^{\alpha\minus{}n}$

$ \equal{}\alpha(\alpha\minus{}1)(\alpha\minus{}2)\cdots(\alpha\minus{}n\plus{}1)b^{\alpha\minus{}n}$

(Which, of course, we could have gotten directly.)

You can also write $ \binom{\alpha}{n}$ in terms of Gamma functions - I leave that as an exercise.

Nifty!

So they are sort of related!

:-)

KM, your posts are always informative and helpful! Thank you for being such a great resource.

[quote="Kent Merryfield"]
Hence, if $ f(y) \equal{} y^{\alpha}$, then

$ f^{[n]}(b) \equal{} n!\binom{\alpha}{n}b^{\alpha \minus{} n}$

$ \equal{} \alpha(\alpha \minus{} 1)(\alpha \minus{} 2)\cdots(\alpha \minus{} n \plus{} 1)b^{\alpha \minus{} n}$

(Which, of course, we could have gotten directly.)

You can also write $ \binom{\alpha}{n}$ in terms of Gamma functions - I leave that as an exercise.[/quote]

This is one approach to defining fractional derivatives! As above, we have that

\[ \frac{d^\mu}{dy^\mu}\left( y^\lambda\right) \equal{} \frac{\Gamma (\lambda \plus{}1)}{\Gamma (\lambda \minus{}\mu \plus{}1)}y^{\lambda\minus{}\mu}\]

For suitable $ \lambda , \mu \in\mathbb{C}$ (e.g. $ \lambda \neq \minus{}1$), so that we have defined fractional derivatives (derivatives of non-integer order--even complex orders) of a power function. I suppose (though I haven't tried it) that one might begin a treatment of fractional calculus based on this via power series.


Is there a nicer solution than SOS (sum of squares)?

[quote="Russian_Man"]a,b,c >0 prove $ \sum \frac{ab}{a\plus{}b\plus{}2c} \le \frac{a\plus{}b\plus{}c}{4}$[/quote]
 $ a\plus{}b\plus{}2c\equal{}x$
  $ a\plus{}2b\plus{}c\equal{}y$
  $ 2a\plus{}b\plus{}c\equal{}z$
then we must prove $ \sum \frac{(\frac{3(x\plus{}y\plus{}z)}{4}\minus{}x\minus{}z)(\frac{3(x\plus{}y\plus{}z)}{4}\minus{}x\minus{}y)}{x}\le \frac{x\plus{}y\plus{}z}{16}$ or
$ \sum \frac{(3y\minus{}x\minus{}z)(3z\minus{}x\minus{}y)}{x}\le x\plus{}y\plus{}z$or
$ \sum \frac{10yz\minus{}3z^2\minus{}3y^2}{x}\le 4(x\plus{}y\plus{}z)$ or
$ 5\sum_{sym} x^2y^2\le 2\sum_{sym} x^2yz\plus{}3\sum_{sym}x^3y$ which is true

Let the left side be K, by Cauchy, we need to show
$ K^2 \le (ab \plus{} bc \plus{} ca)(\sum \frac {ab}{(a \plus{} b \plus{} 2c)^2}) \le \frac {(a \plus{} b \plus{} c)^2}{16}$
By AM-GM, $ (a \plus{} b \plus{} 2c)^2 \ge 4(a \plus{} c)(b \plus{} c)$, so we have to show
$ (ab \plus{} bc \plus{} ca)(\sum \frac {ab}{(a \plus{} c)(b \plus{} c)}) \le \frac {(a \plus{} b \plus{} c)^2}{4}$
$ 4(ab \plus{} bc \plus{} ca)(a^2b \plus{} ab^2 \plus{} b^2c \plus{} bc^2 \plus{} c^2a \plus{} ca^2) \le (a \plus{} b \plus{} c)^2(a \plus{} b)(b \plus{} c)(c \plus{} a)$
Which I believe is true (Can someone check it?).
EDIT: Whoops, it is false.

[quote="dima ukraine"]$ 5\sum_{sym} x^2y^2\le 2\sum_{sym} x^2yz \plus{} 3\sum_{sym}x^3y$[/quote]
What about x=y=1, z=0? I know this is impossible for the definitions of x,y,z, but how do you prove it?

[quote="Russian_Man"]$ a,b,c >0$ prove $ ab/(a\plus{}b\plus{}2c)\plus{}bc/(b\plus{}c\plus{}2a)\plus{}ca/(c\plus{}a\plus{}2b) <\equal{}(a\plus{}b\plus{}c)/4$[/quote]
Yes, we can kill it by SOS.
$ \sum_{cyc}\frac{ab}{a\plus{}b\plus{}2c}\leq\frac{a\plus{}b\plus{}c}{4}\Leftrightarrow\sum_{cyc}\left(\frac{c}{4}\minus{}\frac{ab}{a\plus{}b\plus{}2c}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{2c^2\plus{}ac\plus{}bc\minus{}4ab}{a\plus{}b\plus{}2c}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{(c\minus{}a)(2b\plus{}c)}{a\plus{}b\plus{}2c}\minus{}\frac{(b\minus{}c)(2a\plus{}c)}{a\plus{}b\plus{}2c}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a\minus{}b)\left(\frac{2c\plus{}a}{b\plus{}c\plus{}2a}\minus{}\frac{2c\plus{}b}{a\plus{}c\plus{}2b}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a\minus{}b)^2(a\plus{}b\minus{}c)(a\plus{}b\plus{}2c)\geq0.$
Let $ a\geq b\geq c.$ Hence, $ S_b\geq0,$ $ S_c\geq0$ and $ b^2S_a\plus{}a^2S_b\geq0.$
Hence, $ \sum_{cyc}(a\minus{}b)^2S_c\geq(a\minus{}c)^2S_b\plus{}(b\minus{}c)^2S_a\geq\frac{a^2}{b^2}(b\minus{}c)^2S_b\plus{}(b\minus{}c)^2S_a\equal{}$
$ \equal{}\frac{(b\minus{}c)^2}{b^2}(a^2S_b\plus{}b^2S_a)\geq0.$ Done!

[b]Problem:[/b]

Given $ a,b,c \geq 0$, Prove that:
\[ \frac {ab}{a + b + 2c} + \frac {bc}{b + c + 2a} + \frac {ca}{c + a + 2b} \leq \frac {a + b + c}{4}
\]
Solution (This is Can_Hang's solution, that i saw in other topic):
By Cauchy Schwarz:
\[ \sum_{cyc}\left(\frac {ab}{a + b + 2c}\right) \le \frac {1}{4}\cdot{\sum_{cyc}\left(\frac {ab}{a + c} + \frac {ab}{b + c}\right) = \frac {1}{4}\cdot{\sum_{cyc}\left(\frac {ab}{a + c} + \frac {bc}{a + c}\right) = \frac {1}{4}\cdot\sum_{cyc}\left(\frac {b(a + c)}{a + c}\right) = \frac {1}{4}\cdot\sum_{cyc}b = \frac {a + b + c}{4}
}}\]

[quote="dgreenb801"]
$ 4(ab \plus{} bc \plus{} ca)(a^2b \plus{} ab^2 \plus{} b^2c \plus{} bc^2 \plus{} c^2a \plus{} ca^2) \le (a \plus{} b \plus{} c)^2(a \plus{} b)(b \plus{} c)(c \plus{} a)$
Which I believe is true (Can someone check it?).
[/quote]
Your statement is valid and we can prove it as follow:
This inequality is equivalent to
\[ 4(ab\plus{}bc\plus{}ca)[(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\minus{}3abc] \le (a\plus{}b\plus{}c)^2[(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca) \minus{}abc]\]
\[ abc [12(ab\plus{}bc\plus{}ca) \minus{}(a\plus{}b\plus{}c)^2] \ge (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)[4(ab\plus{}bc\plus{}ca)\minus{}(a\plus{}b\plus{}c)^2]\]
If $ 4(ab\plus{}bc\plus{}ca) \le (a\plus{}b\plus{}c)^2$, then
\[ (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)[4(ab\plus{}bc\plus{}ca) \minus{}(a\plus{}b\plus{}c)^2] \le 9abc[4(ab\plus{}bc\plus{}ca)\minus{}(a\plus{}b\plus{}c)^2]\]
And we just need to prove that
\[ 9[4(ab\plus{}bc\plus{}ca)\minus{}(a\plus{}b\plus{}c)^2] \le 12(ab\plus{}bc\plus{}ca)\minus{}(a\plus{}b\plus{}c)^2\]
\[ (a\plus{}b\plus{}c)^2 \ge 3(ab\plus{}bc\plus{}ca)\]
which is true.
If $ 4(ab\plus{}bc\plus{}ca) \ge (a\plus{}b\plus{}c)^2$, applying Schur's Inequality, we have
\[ abc \ge \frac{[(a\plus{}b\plus{}c)^2\minus{}(ab\plus{}bc\plus{}ca)][4(ab\plus{}bc\plus{}ca)\minus{}(a\plus{}b\plus{}c)^2]}{6(a\plus{}b\plus{}c)}\]
We can deduce our inequality to
\[ [(a\plus{}b\plus{}c)^2\minus{}(ab\plus{}bc\plus{}ca)][12(ab\plus{}bc\plus{}ca)\minus{}(a\plus{}b\plus{}c)^2] \ge 6(a\plus{}b\plus{}c)^2(ab\plus{}bc\plus{}ca)\]
Setting $ p\equal{}a\plus{}b\plus{}c,q\equal{}ab\plus{}bc\plus{}ca$, then we may write the above inequality as
\[ (p^2\minus{}q)(12q\minus{}p^2) \ge 6p^2q\]
\[ \minus{}p^4\plus{}7p^2q \minus{}12q^2 \ge 0\]
\[ (p^2\minus{}3q)(4q\minus{}p^2) \ge 0\]
which is true since $ 4q \ge p^2 \ge 3q$.

shaaam, your solution is very nice and simple.


Find all integers solutions of $a^{3}+2b^{3}=4c^{3}$

[b]Solution.[/b] If $a_{1}^{3}+2b_{1}^{3}=4c_{1}^{3}$, then after checking parity, one obtains that $\left(\frac{a_{1}}{2},\frac{b_{1}}{2},\frac{c_{1}}{2}\right)$ is also a solution.  Similarly, one obtains that $\left(\frac{a_{1}}{2^{k}},\frac{b_{1}}{2^{k}},\frac{c_{1}}{2^{k}}\right)$ is a solution for any integer $k$.  This implies that the only integral solution to this equation is $(a,b,c)=(0,0,0)$. $\Box$

The proof is nicer with contradiction:

Assume for sake of contradiction that there is a solution set where $|a|+|b|+|c|>0$. By WOP there is a set where that value is minimal.

Now go through the argument.

Contradiction; you a solution where that value is smaller.

Hence $|a|+|b|+|c|=0$ is the only other option; i.e. $a=b=c=0$.

This manner of arguing is much cleaner than the whole: "a/2^k" stuff.


factorial
calculus
calculus computations

Can anyone find a closed form for $ \sum_{k\equal{}1}^n k!$?

Hum, we can find $ \sum_{k\equal{}1}^n k\cdot k!$.

hello, it is
$ \sum_{k=1}^n k\cdot k!=(n+1)!-1$.
Sonnhard.

[quote="igiul"]Can anyone find a closed form for $ \sum_{k \equal{} 1}^n k!$?[/quote]

I think it's hard, and there are lot of other similar sums for example

$ \sum \sqrt{k}$

But it might be possible to compute them.


It's a transcendental equation: but we can get something out of it. It's trivial to show that it has two roots. The bigger one is obviously $ 2$, and the other one can be found only numerically (correct me if I'm wrong, but you're dealing with Lambert W function here, so I don't expect anything elegant/without using Newton's method or sth like that.)

Mathematica's result was:

$ 0.15495346619034528$

and the one I got using second degree McLaurin approximation was

$ \frac {4 \minus{} \ln 2 \minus{} \sqrt {16 \minus{} 8\ln 2 \minus{} (\ln 2)^2}}{2(\ln{2})^2}\approx 0.15467723825798213$

The error is not too big  ($ 0.17\%$);)

For integers:

[hide="solution"]Taking the expression log 4, we get $ x\equal{}\log_{4}8\plus{}\log_{4}x$, or $ x\minus{}\log_{4}x\equal{}3/2$

Since this is increasing when $ x>1$, there should be either 1 or 0 values above 1 such that this expression is true. Let's try some small values of $ x$:
$ 1\minus{}\log_{4}1\equal{}1$
$ 2\minus{}\log_{4}2\equal{}3/2$
BINGO!

Now if $ x<1$, then we can let $ \frac{1}{x}\equal{}y>1$. Thus we have $ x\minus{}\log_{4}x\equal{}\frac{1}{y}\plus{}\log_{4}y$ for some positive $ y>1$. Since there is no rational $ y$ such that the expression equals two, I'm just going to state a bound: $ y\in (8,16)$.[/hide]


calculus
derivative
function
conics
parabola
calculus computations

Find the values of any relative extrema.
f(x) = x^2 + 2x - 3

[quote="Kent Merryfield"](This can be done with no calculus at all, but ...)

Show us what you've done so far, please.[/quote]

f'(x) = 2x + 2
2x + 2 = 0
x = - 1
if this is correct what is ( or what are) answer

Your extreme point is simply $ (\minus{}1, f(\minus{}1))$ but you must do either the first or second derivative test to determine whether you have a max, min or straddle point.

[quote="Kent Merryfield"]OK, why did you take the derivative and why did you set it equal to zero? What is the purpose of those actions?[/quo

Tryed to find derivative....that isn't  correct :(

[quote="Prime factorization"]Your extreme point is simply $ ( \minus{} 1, f( \minus{} 1))$ but you must do either the first or second derivative test to determine whether you have a max, min or straddle point.[/quote]
How can i do that?

There were no computational errors in your post in which you found $ f'(x)$ and solved for when it was zero.

As I told a class of mine the other day after I handed a test back: being able to calculate does matter and will always matter. But being able to calculate will never be enough; it will never be anywhere close to enough. You have to know [i]what[/i] to compute, and you have to properly interpret the result of any such computation.

That's what's lacking here. You did do the correct computation, but showed no sign of knowing why you chose to do that. And then, you did not know how to interpret the result. Any computation whose result you can't interpret might as well not have been done.

Some basic questions:

What is the geometric meaning of the derivative of a function?

What, exactly, is the relationship between zeros of the derivative and extreme values? There a one-way implication there: if (one thing) then (another thing), and the converse of that isn't true. You have to know that if-then statement.

And that function is a pretty simple formula. Do you know what its graph looks like, generally speaking?

Well, your second derivative is 2, which is >0, thus you have a min point. Just to check, in the derivative you have the points (-2,-2) and (0,2) from plugging in. Since the first derivative goes from -ve to +ve relative to the critical point, the function itself goes from decreasing to increasing. Draw that on a graph, and you'll get a parabola opening upwards, which obviously has a min point as stated earlier.

Edit: Just to add to what Jmerry said, also keep in mind that you have a critical point if the derivative leads to an indeterminate form. As an example |x| (which we know has a min) has derivative x for x > 0 and -x for x<0, but the derivative at x=0 doesn't exist.


[quote="ProblemSets"]i did an estimate on my calculator and was getting that x should be about 1.5ish...someplace around there. i typed your given answers and it did not satisfy the equation. am i crazy?[/quote]
It's about 1.58, so you might have made a mistake there.  Anyways, if $x=2^\frac{2}{3}$, then $x^\frac{3}{2}=2^{(\frac{3}{2})(\frac{2}{3})}=2^{1}=2$.  Thus, $\frac{1}{x}=\frac{1}{2}$, so $x-\frac{1}{x}=2-\frac{1}{2}=\frac{3}{2}$, which is what we are given, so the two answers work.

[quote="ProblemSets"]i did an estimate on my calculator and was getting that x should be about 1.5ish...someplace around there. i typed your given answers and it did not satisfy the equation. am i crazy?[/quote]

yes.

jk...but when i checked on my computer calculator (dont have my ti-89 with me right now) i got 1.84-.54 which isnt [i]that[/i] close to 1.5...

i figured what i did wrong. i read 2^(3/2) as 2 and 2/3

my bad. sorry about that. my eyes are getting bad :)


Can anyone help me out on how to solve this? It's driving me crazy.  :rotfl:

Two circles are internally tangent. The smaller circle is also tangent to two
perpendicular radii of the larger circle. What is the ratio of the area of the small circle to
the area of the large circle? Express your answer as a decimal rounded to the nearest
hundredth.

[geogebra]63e178a92816325337e39522fe0cf04f5bee8a4f[/geogebra]
Sorry, I'm not too good with Geogebra.
EDIT: Thanks for pointing that out.  It's past 10:30 here, so at least I have an excuse (kind of)
Ok, I think this is right though.
To find the radius of the smaller circle, take the radius of the larger circle and divide by two.  Now you have the length of the diagonal of the square.  Dividing by the square root of two, we have the radius of the smaller circle.  Thus the answer is $ (\frac{1}{2\sqrt{2}})^2=\frac{1/8}$, or $ 0.13$ to the nearest hundredth.

Unfortunately your answer could not be correct. It's more complicated than that. Clearly the smaller circle is less than one quarter of the larger one.

But how to calculate???

Let the radius of the larger circle be $ x$ and the radius of the smaller circle $ r$.  From 45-45-90 triangle relationships, and looking at the radii, we can conclude that $ r(1\plus{}\sqrt2)\equal{}x$.  We just square this ratio, then take the reciprocal to find the desired area ration, we find the answer is $ \boxed{0.17}$


geometry
rectangle
combinatorics open
combinatorics

let us consider a square containing of square identical of format 4*4 how much rectangles can one find in the large square one

Do you mean how many rectangles can be found in a 4x4 grid?
The answer is $\frac{25 * 16}{4}= 100$


Prove or disprove this statement:
There exist integer solution (x,y) except (4,6) such that
$ x^2\plus{}x^3\equal{}2^y\plus{}16$.

Here is my start.

$ x^2(x\plus{}1)\equal{}2^y\plus{}16$

We easily find that $ y\ge6$

1) x divisible by 4
2) x+1 divisible by 16

First case has one solution which is included into statement.

But how to solve second case??  :maybe:

Thanks in advance.

Idea for second case: We know that x+1 isn't divisible by 32, since the RHS isn't divisible by 32. Dividing both sides by 16 gives us

$ x^2(oddinteger) \equal{} 2^{y \minus{} 4} \plus{} 1$

Maybe we could prove that no squares other than 1 divide the RHS...
EDIT: That method fails because sometimes $ 2^{y\minus{}4}\equiv 8\bmod{9}$ for some $ y$.

[hide]
Checking $ y\equal{}0$ gives us no solution, thus $ x$ must be even, so substitute $ x \equal{} 2k$ to get:

$ 4k^2 \plus{} 8k^3 \equal{} 2^y \plus{} 16$

$ k^2 \plus{} 2k^3 \equal{} 2^{y \minus{} 2} \plus{} 4$

$ k^2(2k \plus{} 1) \equal{} 2^{y \minus{} 2} \plus{} 4 \implies k \equal{} 2m$ (checking $ y \equal{} 2$ gives us no solution, thus $ y > 2$)

$ 4m^2(4m \plus{} 1) \equal{} 2^{y \minus{} 2} \plus{} 4$

$ m^2(4m \plus{} 1) \equal{} 2^{y \minus{} 4} \plus{} 1$

quickly checking $ y \equal{} 4$ gives no integer solution, thus RHS is odd $ \implies m \equal{} 2n \plus{} 1$

$ (4n^2 \plus{} 4n \plus{} 1)(8n \plus{} 5) \equal{} 2^{y \minus{} 4} \plus{} 1$

checking LHS  we see that it's remainder is $ 1 \pmod 4 \implies y \ge 6$

$ y \equal{} 6$ gives us solution so let's check $ y > 6$

$ 2^{y \minus{} 4} \plus{} 1 \equiv 1 \pmod 8$

Knowing that square of an odd number always gives remainder $ 1 \pmod 8 \implies (4n^2 \plus{} 4n \plus{} 1)(8n \plus{} 5) \equiv 5 \pmod 8$ which is contradiction. Thus the only solution is $ (4, 6)$.[/hide]

Flame, you solved only first case of my problem (look at post #2, x is not neccessary even) :wink:

We would just do same as in my previous post, shouldn't be too hard to finish it. :)

I had to use divisibility of 32, then 64, then 128, then 256... but it still wasn't enough, so I don't think it's as easy as in the first case :)

Is it really to hard for pre-olympiad level?


Let M be a metric space with the discrete metric.
1) show that every subset of M is clopen
2) show that every function defined on M is continuous.

I don't know where to start with this problem. I know the definition of closed and open sets, continuous function, but I just don't see how to approach this. Hope someone can give me some help.

1) If $ \varepsilon<1$ and $ x\epsilon M$ then the $ \varepsilon$-Ball $ B_\varepsilon(x)$ under the discrete metric is just the set $ \{x\}$. The fact that all subsets of M are open follows trivially from the definition of open sets under the metric topology.

2) If $ f: M\rightarrow Y$, then for any subset $ A\subset Y$, and thus any open subset, $ f^{\minus{}1}(A)$ is a subset of M, and therefore by 1), is open. So f satisfies the definition of continuity.

Thanks for your help.
1)The singleton set {x} is closed, but I don't see how all subsets of M are open is trivial from the definition of open sets. I'm slow at this. If epsilon is greater than 1, then M_epsilon(x) is just M. We have a theorem saying that M_epsilon(x) is open for any epsilon greater than 0 and any x in M.

2) I see that the pre-image of A is a subset of M, then by part 1) it is open. But why does this satisfy the definition of continuity. We have a theorem stating that if the pre-image of any open set V in N is open, then f is continuous. Did you apply that theorem here?

1) let $ X \subset M$ and let $ x \epsilon X$, then $ B_{\varepsilon}(x)\subset X$, so long as $ {0 < \varepsilon < 1}$.

2) In topology, that property is taken as the definition of continuity, not as a theorem. I was working under that definition.


modular arithmetic
number theory unsolved
number theory

Solve the equation in natural numbers
$ 2^{x\minus{}1}\plus{}1\equal{}y^{z\plus{}1}$

We have $ y^{z \plus{} 1} \minus{} 2^{x \minus{} 1} \equal{} 1$ so by Mihailescu's theorem, the solutions where $ z\geq 1$ and $ x\geq 3$ are $ x \equal{} 4,y \equal{} 3,z \equal{} 1$.

Assume now that $ z \equal{} 1$.

Then $ y^2 \minus{} 2^{x \minus{} 1} \equal{} 1$. But Ko Chao's result says that if $ x\geq 4$ and $ y^2 \minus{} 2^{x \minus{} 1} \equal{} 1$ has a solution then $ 2\geq 2^{x \minus{} 2} \minus{} 2$. But this means that $ x \equal{} 4$.

So let us check the cases $ x\in \{1,2,3\}$.

If $ x \equal{} 1$ then $ y^2 \equal{} 2$ having no solutions in integers.
If $ x \equal{} 2$ then $ y^2 \equal{} 3$ having no solutions in integers.
If $ x \equal{} 3$ then $ y^2 \equal{} 5$ having no solutions in integers.

So $ (x,y,z) \equal{} (4,3,1)$ is the only positive solution.

It is equivalent to $ 2^{x-1}=(y-1)(1+y+\dots+y^z)$. Hence $ y=2^a+1$ for some $ a\in\mathbb N$. Then $ 2^{x-1-a}=1+y+\dots+y^z\equiv z+1\pmod 2$. Thus $ z$ is odd. Therefore
\begin{align*}2^{x-1-a}&=1+y+\dots+y^z=(1+y)(1+y^2+\dots+y^{z-1})\\
&=2(2^{a-1}+1)(1+y^2+\dots+y^{z-1}).\end{align*}
Hence $ 2^{a-1}+1$ must be a power of $ 2$, which is possible only when $ a=1$. Then $ y=3$ and we get
\[ 2^{x-4}=1+3^2+\dots+3^{z-1}.\]
If $ x=4$ we get $ z=1$. Otherwise assume that $ x>4$ and then we get $ 0\equiv \frac{z+1}{2}\pmod 2$. Thus $ z\equiv -1\pmod 4$. Then
\[ 2^{x-4}=(1+3^2)(1+3^4+\dots+3^{z-3}),\]
which is impossible as $ 5\not |\,2^{x-4}$ but $ 5$ divides $ 1+3^2=10$. Hence the only solution is $ (x,y,z)=(4,3,1)$.


Prove that there doen't exist function which satisfying all following conditions:
$ f(f(1))\equal{}5$, $ f(f(2))\equal{}6$, $ f(f(3))\equal{}4$, $ f(f(4))\equal{}3$, $ f(f(n))\equal{}n\plus{}2$ for every $ n\in N, n\ge 5$

Let $ a\equal{}f(3),b\equal{}f(4)$. It is easy to chek, that $ a\not \equal{}3,4, \ b\not \equal{}3,4, \ a\not \equal{}b$  and $ f(f(a))\equal{}b,f(f(b))\equal{}a$, therefore $ a<5,b<5$.
It give $ a\equal{}1,b\equal{}2$ or $ a\equal{}2,b\equal{}1$ - contradition.


inequalities
geometry
inequalities proposed

H  is the  orthocenter  of  a  triangle  ABC. Prove that
AH*BH*CH <=(8*ha*hb*hc)/27

HAPPY  NEW  YEAR!

It's easy to see (area computation) that HA/ha+HB/hb+HC/hc=2. We just apply AM-GM to this: 2/3>=[HA*HB*HC/(ha*hb*hc)] 1/3 iff 8/27 >= HA*HB*HC/(ha*hb*hc) iff HA*HB*HC<=(8*ha*hb*hc)/27.

It's obvious, for the same reasons, that if AD, BE, CF are cevians intersecting at X (D on BC, etc.), then XA*XB*XC<=(8*AD*BE*CF)/27, which is more general.


trigonometry
algebra
polynomial
quadratics
calculus
Kettering MO
geometry

Today was the 5th Kettering Olympiad - and here are the problems, which are very good intermediate problems.

1. Find all real $x$ so that $(1+x^2)(1+x^4)=4x^3$

2. Mark and John play a game. They have $100$ pebbles on a table. They take turns taking at least one at at most eight pebbles away. The person to claim the last pebble wins. Mark goes first. Can you find a way for Mark to always win? What about John?

3. Prove that

$\sin x + \sin 3x + \sin 5x + ... + \sin 11 x = (1-\cos 12 x)/(2 \sin x)$

4. Mark has $7$ pieces of paper. He takes some of them and splits each into $7$ pieces of paper. He repeats this process some number of times. He then tells John he has $2000$ pieces of paper. John tells him he is wrong. Why is John right?

5. In a triangle $ABC$, the altitude, angle bisector, and median split angle $A$ into four equal angles. Find the angles of $ABC.$

6. There are $100$ cities. There exist airlines connecting pairs of cities.

a) Find the minimal number of airlines such that with at most $k$ plane changes, one can go from any city to any other city.

b) Given that there are $4852$ airlines, show that, given any schematic, one can go from any city to any other city.

stop being so n00b fred, we all know it was Nick and John </rant>

nah, j/k, it looks right to me besides that

hopefully i didnt screw up too much

[quote="amirhtlusa"]1.
[hide]$x=1$[/hide][/quote]How can you prove that is the only solution?

On 6.b., can there be several airlines connecting a pair of cities?
[hide="6b"]Since there are $100$ cities, and airlines connect $2$ of them, if there are $\binom{100}{2}=\frac{100\times99}{2}=4950$ airlines, then all cities will be connected.  So since there are $4952$ airlines, all cities must be connected, and one can travel from any city to another.[/hide]

[quote="tetrahedr0n"]amirhtlusa, solutions are required.[/quote]
isn't that obvious??? ;)
hint
[hide] rewrite the whole thing like this:
$(1+x^2)((1+x^2)^2-2x^2)=4x^3$
$(1+x^2)^3-(2x^2(1+x^2))=4x^3$
$(1+x^2)^3-2x^2(1+x)^2=0$
you do the rest... :lol: [/hide]

[quote="tetrahedr0n"]Well, obvious or not, soutions are always required.

But can you explain the last step of your hint? Something just seems to misterously dissappear.[/quote]
bring the $4x^3$ to the other side and then factor out a $2x^2$, you will end up with a $....2x^2(1+x^2+2x)$.
 :)

[quote="tetrahedr0n"]On 6b) at most one airline connects each pair of cities. But check your work, eryaman.[/quote]Woopsies.  I misread the question, thought it said 4952 instead of 4852.

[quote="tetrahedr0n"]Ok, (your 3rd line confused me, it just had the $4x^3$ gone completely.) But write your complete solution amirhtlusa. Thats why I posted the problem - as a challenge for you guys to solve.[/quote]
 ;)
here is the third line:
[hide]$(1+x^2)^3-(2x^2(1+x^2))=4x^3$
$(1+x^2)^3-(2x^2(1+x^2))-4x^3=0$
$(1+x^2)^3-2x^2(1+x^2+2x)=0$
$(1+x^2)^3-2x^2(1+x)^2=0$[/hide]

I have a solution for problem 2.
[hide="Problem 2"]For Mark to win, at the end of the game, he must leave John with the last pebble.  For there to be 1 pebble left for John, there must be 10 left when John draws pebbles.  If John draws one, then Mark will draw 8, so there will be one left for John.  If John draws 2, then mark will draw 7, so there will be one left for John, and so on until, if John draws 8, then Mark will draw 1, leaving the last pebble for John.  So for there to be 10 left when John draws pebbles, John also must draw the 19, 28, 37, etc. pebbles.  But this goes all the way up to 100, the total number, so there is no strategy for Mark to win.

For John to win, he must make Mark draw when there are 10 pebbles left.  Going up to the pattern adding 9 each time again, Mark must draw when there are 91 pebbles.  So if Mark draws 1 pebble, John draws 8, if Mark draws 2 pebbles, John draws 7, and so on until if Mark draws 8 pebbles, John draws 1.  John must keep the numbers that Mark draws at 82, 73, 64, 55, 46, 37, 28, 19, 10, then 1 using the above method. [/hide]

[hide="Problem 4"]
He starts with 7 sheets of paper. Each time he rips on into 7 pieces, the total number of pieces increases by 6, so he can never change the number of pieces he has in mod 6. Initially, he has 7 sheets, congruent to 1 mod 6, so he'll always have a number congruent to 1 mod 6. Because 2000 is congruent to 2 mod 6, he can never have 2000 sheets of paper.
[/hide]

eryaman, for Mark to win, he has to take the last pebble, not vice-versa.

amirhtlusa, I'm with you so far. What's next?

[quote="tetrahedr0n"]eryaman, for Mark to win, he has to take the last pebble, not vice-versa. [/quote]*needs new glasses* wow, second problem in a row I've misread.  Just take my solution, and replace the word "Mark" with "John, and "John" with "Mark".

[hide="Problem 1"]
$(1+x^2)(1+x^4)=4x^3$
Since the left hand side must be positive, $x^3>0$.
$1+x^2+x^4+x^6=4x^3$
$\frac{1}{x^3}+\frac{1}{x}+x+x^3=4$
$\left(\frac{1}{x^3}+x^3\right)+\left(\frac{1}{x}+x\right)=4$
Let $y=\frac{1}{x}+x$.
Then $y^3=\frac{1}{x^3}+\frac{3}{x}+3x+x^3$.
$(y^3-3y)+(y)=4$
$y^3-2y-4=0$

I don't know the method for depressed cubics, so must resort to guess and check. (Good thing $y=2$ gives a root)
$(y-2)(y^2+2y+2)=0$
$y-2$ then $y=2$
Or $(y^2+2y+2)=0$ then $y=\frac{-2\pm\sqrt{2^2-4(1)(2)}}{2(1)}$
Since the latter produces imaginary roots, the only possibility is $y=2$.

$2=\frac{1}{x}+x$.
$x^2-2x+1=(x-1)^2=0$

So $x=1$ is the only solution.
[/hide]

I had essentially the same solution, except that I first I divided the given polynomial by $(x-1)^2$ then used the symmetric trick getting a quadratic in $x+1/x.$

Anyone use derivatives in 1? Find (1,4) is an inersection point; then show that 4x^3 increases/decreases more/less rapidly than (1+x^2)(1+x^4) on certain ntervals, proving that they never intersect again? I did that, and it took a long, long tme.
I got 1, 2, 4, and 6a.

On 6a, I made a slight mistake in thinking that 1 change of flight meant only 1 flight, not 2. So I wrote for k=1, a= 4950 and for 1<k<99, a = 99. How much will I get marked off?
I just didn't understand what 6b asked for.
On 5, I used some trig, and found that cos(<A) = cos(<3A) for 0<A<180
So, being the mistake maker I am, I thoght there was no solution. But, after the competition, I saw that I had overlooked an obvious solution ; A= 90. How dumb could I be! Then, the rest would esily follow (B= 22.5, C= 67.5). But, in all fairness to the difficulty of the problem; I used lots of trig (Law of Sines espcially) ; until I was left with cos a = cos 3a. I can't believe I was that dumb!
-Question to people: on 3, how did you recognize you should use cos(a+b) - cos (a-b)? Practice? If I could get that insight, the problem was trivial. Darn!

- Hey, tetrahedr0n, I think I saw you!

Predictions anyone? What scores did you need last year to be in the top 7?

What were the rest of your scores?

For no. 1, just separate 4x^3 into 2x^3 + 2x^3. Then you have (1-x^3)^2 and x^2(1-x)^2, and go from there.
No. 5, i got 2tan(a)=tan(3a) + tan(a) for one fourth of the big angle A, and ended up with tan(a)=sqrt(2) +1 and sqrt(2)-1. But i used +1 instead of -1. hopefully i'll get most of the credit.


Let $A_{1}, B_{1}, C_{1}$ 	be the midpoints of the segments $BC, AC, AB$ of the triangle $ABC$
respectively. Let $H_{1}, H_{2}, H_{3}$ be the intersection points of the altitudes of the triangles
$AB_{1}C_{1}, BA_{1}C_{1}, CA_{1}B_{1}.$ Prove that the lines $A_{1}H_{1}, B_{1}H_{2}, C_{1}H_{3}$ are concurrent.

Nice!

I proved that $AOA_{1}H_{1}$, $BOB_{1}H_{2}$ and $COC_{1}H_{3}$ are parallelograms.
(only see that $AH_{1}= AH/2 = OA_{1}$, and the analogous)

with this, the problem is easy!  :)