Let $ f(x)$ be a polynomial of degree $ n\in\mathbb{N}$ whose all coefficients are real and the leading coefficient is equal to $ 1$.
If all $ f(q_1),f(q_2),\cdots,f(q_n)$ are rationals for $ n$ distinct rationals $ q_1,q_2,\cdots,q_n$,then show that all the coefficients of $ f(x)$ are rationals.

See [url=http://en.wikipedia.org/wiki/Lagrange_polynomial]Lagrange interpolation[/url], although to take advantage of the monic condition it will be easier to use [url=http://en.wikipedia.org/wiki/Newton_polynomial]Newton interpolation[/url].


algebra
polynomial
Vieta
inequalities
inequalities proposed

Let $ a,b,c$ be positive real numbers. Prove that
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \plus{}\frac{28(ab\plus{}bc\plus{}ca)}{(a\plus{}b\plus{}c)^2} \ge 12.$

There is a proof using only AM-GM Inequality. Can someone figure it out?

Remark. Equality holds if and only if
$ \left\{\begin{array}{l} (a\minus{}b)(b\minus{}c)(a\minus{}c)>0\\(a \plus{} b \plus{} c)^3 \equal{} 56abc \\
a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(ab \plus{} bc \plus{} ca)\end{array}\right.$

[quote="can_hang2007"]Remark. Equality holds if and only if
$ \left\{\begin{array}{l} (a \plus{} b \plus{} c)^3 \equal{} 56abc \\
a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(ab \plus{} bc \plus{} ca)\end{array}\right.$[/quote]
$ \Longleftrightarrow a: b: c\equal{}\left(\sqrt{7}\minus{}tan\frac{4\pi}7\right): \left(\sqrt{7}\minus{}tan\frac{\pi}7\right): \left(\sqrt{7}\minus{}tan\frac{2\pi}7\right)$
This is because by Vieta, $ a,b,c$ are in the equality case proportional to the roots of $ x^3\minus{}14 x^2\plus{}49x\minus{}49$.
They can be found in a similar way than in [url=http://www.mathlinks.ro/viewtopic.php?p=253494&search_id=399239273#253494]this thread[/url] (from post #42 onwards) :)
  :!: In your "if and only if" statement you must add a condition like $ a>b>c$, as the inequality is only cyclic.  :wink:

[quote="spanferkel"]In your "if and only if" statement you must add a condition like $ a > b > c$, as the inequality is only cyclic.  :wink:[/quote]
Thank you, I've edited it.  :blush: Do you have a solution for this problem?

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers. Prove that
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \frac {28(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2} \ge 12.$[/quote]
Very nice inequality. :lol:
It is equivalent to

$ \sum a(b^2\minus{}ab\minus{}3bc\plus{}2ca)^2 \ge 0$.

[quote="Vasc"][quote="can_hang2007"]Let $ a,b,c$ be positive real numbers. Prove that
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \frac {28(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2} \ge 12.$[/quote]
Very nice inequality. :lol:
It is equivalent to

$ \sum a(b^2 \minus{} ab \minus{} 3bc \plus{} 2ca)^2 \ge 0$.[/quote]
Great solution, my solution is exactly the same.  :lol:


integration
trigonometry
real analysis
real analysis unsolved

Let $U=\{x\in\mathbb{R}^{2}: 0<|x|<1\}$, and consider the equation (1)
$-\Delta u=0$, $x\in U$
$u(0)=0$, $u_{|x|=1}=1$.

First, if $u$ is a solution show that
$v(r)=\frac{1}{2\pi}\int_{0}^{2\pi}u(r,\theta)d\theta$ is also a solution.

Second, show that the equation (1) has no solution.

This should be quite easy, but I'm getting a bit confused with the polar coordinates.

[quote="Xtaao"]Let $U=\{x\in\mathbb{R}^{2}: 0<|x|<1\}$, and consider the equation (1)
$-\Delta u=0$, $x\in U$
$u(0)=0$, $u_{|x|=1}=1$.

First, [skip]

Second, show that the equation (1) has no solution.
[/quote]

First, $\Delta=\frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial}{\partial r})+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}$.

Second, by Poisson formula
$u(r,\varphi)=\frac{1}{2\pi}\int \frac{(a^{2}-r^{2})f(\alpha)d\alpha}{a^{2}-2ar\cos(\varphi-\alpha)+r^{2}}$.

Here $f(\alpha)=1$, $a=1$. We receive $u(0)=1\ne 0$.


analytic geometry
advanced fields
advanced fields theorems

As the text develops the procedure to convert General PDE form to canonical form, the author arrives at:

(omitting a number of steps)

$ Au_{\xi} + Bu_{\eta} + cu + d = 0$

$ A = a\xi_x + b \xi_y$  and $ B = a\eta_x + b\eta_y$

He then sets $ B = a\eta_x + b\eta_y = 0\;\;\; (a)$

And then he states that "eq. (a) has infinite many solutions. We can find one of them by assiging initial data on a non-characteristic curve and solving the resulting initial value problem."
He does so and arrives at:

$ \begin{align} & \eta(x_0,y) = y \\
& \xi(x,y) = x\end{align}\;$

Oddly enough, I kinda understood the discussion...  What I do not understand is do we always solve eq (a) for every problem and "design" a coordinate transformation for each problem or does the author's development (eq's 1 and 2) already address each situation?

This has nothing to do with your question, but it does have to do with an error you've made several times in different threads: the word is "canonical," not "canical."

Made as spelling corrections indicated. Thank you for pointing it out...


Let $p$ be a positive real number given. Find the minimun vale of $x^3+y^3$, knowing that $x$ and $y$ are positive real numbers such that $xy(x+y)=p$.

I think the proof is immediate.

Assume w.l.o.g. $x\le y$.  We have by Chebyshev's Inequality that

$x^3+y^3=x\cdot x^2+y\cdot y^2\ge y(x^2)+x(y^2)=xy(x+y)=p$.

There is equality when $x=y=\displaystyle (\frac{p}{2})^{\frac{1}{3}}$.

[quote="José"]Let $p$ be a positive real number given. Find the minimun vale of $x^{3}+y^{3}$, knowing that $x$ and $y$ are positive real numbers such that $xy(x+y)=p$.[/quote]
 It's easy to prove that: Let x,y >0, then we have:
   x^3+y^3>=xy(x+y),
Prof: x^3+y^3=(x+y)(x^2+y^2-xy)
From: x^2+y^2>=2xy
then (x^2+y^2-xy)>=xy. From that it's proved.[quote][/quote]

This can be done using Schur's ineq, which states that:
For any non-negative reals $x,y,z$ and any positive integer $t$ it holds that:
$$x^t(x-y)(x-z)+y^t(y-x)(y-z)+z^t(z-x)(z-y) \ge 0$$
Where equality holds when $x=y=z$ or when any $2$ of them are equal and the last one is $0$.
[b][color=#f00]Proof of the main problem:[/color][/b] i claim that the minimun is $p$ and its reached when $x=y=\sqrt[3]{\frac{p}{2}}$
Now its enough to show that $x^3+y^3 \ge xy(x+y)$ which is equivant to $x(x)(x-y)+y(y)(y-x) \ge 0$ but this holds by Schur ineq when $z=0$ and $t=1$ hence the we are done  :blush:


How many 9’s are in the decimal expansion of 999999899992?

What do you mean by decimal expansion? This is an integer...

I have a feeling it might be $ 99999989999^2$.

Try this:

How many digit 9 are there in the decimal representation of $ 999999999899999999999989^2$?

[hide="answer"]
There are $ 29$ digit 9 in the decimal representation of $ 999999999899999999999989^2$.
[/hide]

What about this: How many digit 9 are there in the decimal representation of $ 9999999998999999999999899909^2$?

I'll leave that for you to ponder on.

I'm too lazy to actually mathematically do it, but Yacas tells me the the decimal expansion is:
[hide]99999999980000000000997998180000200182000000010018208281, which has 13 zeroes, if I counted right.
EDIT: I mean 13 nines![/hide]

[quote="Brut3Forc3"]I'm too lazy to actually mathematically do it, but Yacas tells me the the decimal expansion is:
[hide]99999999980000000000997998180000200182000000010018208281, which has 13 zeroes, if I counted right.[/hide][/quote]

It says 9, which is actually 13, and there are way more than 13 0s.

Oh, I meant 13 9's. Yeah, there's 13 of them...


If you are applying Early Action, they say the deadline is November 1st -- Does that mean the last possible moment to turn it in is 11:59PM as of TONIGHT, or 11:59 PM as of November 1st?

The preposition kills me.

November 1st is the [i]postmark[/i] deadline, so it's tomorrow.


The computers here don't allow us to play math games....

Maybe the teachers or directors might let us.. but we have to install flash but it doesn't work.

Weird. At '08 people there played FTW almost every day. And a few played runescape. And other games.

The Loomis computer lab doesn't have flash. You can't install it either.

If you want to play FTW go to the library after dinner.

I tried to play primeshooter in the computer lab in worner, and i played for a while, but I got really annoyed and gave up because the connection was so slow.

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

  mathpath
one ^ brownie point!

We aren't allowed to play games... :(And I don't want to sneak and play games)'

Maybe, I could ask the counselors if we could do something educational and interactive on the weekends.)

[quote="mathluver2.718"]I tried to play primeshooter in the computer lab in worner, and i played for a while[/quote]
You did what now? :)

They don't want you playing games at MathPath. Other staff have even said no to FTW because that's something you can do at home. The MathPath staff would like you to experience things you can not do at home (e.g., work and socialize with other top math students throughout the nation).

ok, I won't play primeshooter (or any other game) while i'm here.

so we are no longer allowed to play FTW?

the rules were you could play anything math related...

also, is that why you muted everyone at mathpath, you are abusing you powers by blaming me for swearing when i didnt.

[quote="k4mc"]so we are no longer allowed to play FTW?[/quote]
That's up to the staff at MathPath, not AoPS. Please discuss with them.

[quote="k4mc"]also, is that why you muted everyone at mathpath, you are abusing you powers by blaming me for swearing when i didnt.[/quote]
No one is abusing powers. There is a student at MathPath who was acting inappropriately and was muted. Since mutes are based on IP address, it would mute everyone at the campus. I have removed the IP based mute but it may be re-instituted if we have further problems with MathPath students.

thanks for answering me, i dint know why we were all muted


geometry
geometric transformation
reflection

Some thoughts...

Most healthy people have a source of pride that gives them self-confidence and a feeling of accomplishment.

Here are some examples of sources of pride:
1. Physically Attractive
2. Athletic
3. Adaptable
4. Funny
5. Impactful
6. Needed by Friends/Family/etc.
7. Intelligent (Emotional/Intellectual)

There are many more, but I will focus on the last one to illustrate my point.

In my opinion, intelligence means IQ, however this conclusion in and of itself is a reflection of my level. I don't refer to the score one gets on IQ tests, but on what those tests attempt to measure -- the ability to learn, adapt, and create.
IQ has three important aspects for me:

Depth (the level of complexity one is able to comprehend and build upon)
Accuracy (the reliablility of one's observations and conclusions), and
Speed (the rate at which one adapts, learns, or creates).

However, each person for whom intelligence is a major source of pride has a different idea of what intelligence means, and each person has a different idea about how the 'most intelligent' person would act. What's the more sophisticated form of showing intelligence? the most mature way of giving the impression that you are smart?

Let's say Person A has a photographic memory and knows a lot of information. Person A knows all about history, geography, mathematical, spelling, music, art, language, etc. This person finds it easy to impress others with the amount of information that he knows. Say Person A raises his hand often in class and makes it well known that he knows so much. From the perspective of others, one impression Person A gives is talkative.

Let's say Person B is someone who defines intelligence not by how much knowledge one knows but by how much one can figure out. This person takes great pride in being creative and often talks and shares what she has figured out. She argues and debates, and tries to prove others wrong. She looks upon Person A with a grin and feels proud that she has found a more sophisticated way of giving off the impression of high intelligence. Person B also gives off the impression of being talkative.

Let's say Person C is not talkative. He looks upon Persons A and B and feels that both of them are immature in their expressions of intelligence. A truly intelligent person, in Person C's opinion is one who understands all but does not need to show it. After all, the wisest person in the world knows what he knows and knows what he doesn't know. Person C feels that he has a good grasp of what he knows and what he doesn't know and feels that there is no need for expressing his own opinion. Person C feels that everything is subjective, and there is no point in arguing since arguing will not help anyone gain anything. Person C only talks when he is clarifying something he does not understand. Other than for that reason, Person C feels that it is unnecessary to talk. Person C feels that by remaining silent, the most intelligent people will understand and appreciate his complexity, which is shown through his silence.

Now, we have Person D, who shares all of Person C's views except that Person D differs on one significant point. Person D feels that although everything is subjective, it is important to judge and measure the opinions of all others, and that through conversation only can we ever hope to coming close to truth. Though Person D acknowledges that there is no such thing as absolute truth, he feels that one can still strive to get as close as possible to the whole truth. So Person D talks and asks questions, and acts just like Persons A and B (to the majority of "spectators").

You get my point. Each subsequent level of impression understands all the reasons behind all the actions of the previous group. However, the levels never end. Person W looks down upon Person V just as Person X looks down upon Person W.

I don't think there exists a highest level in general, but I think there is a highest level for a specific enough situation -- which might or might not exist.

I used to think that the highest level would be someone who doesn't consider the effects of his or her actions and acts naturally, however now I don't think so.

If one acted naturally, then a person who naturally acts selfish would continue acting selfish. As long as there is a relative hierarchy of good and bad with respect to actions, the goal to act naturally without considering the effects of one's actions will never be the highest -- it would be average in terms of complexity.

Comments?

Another way I think of intelligence is this:
Somebody is average. They act completely average. Good for them.
The smarter people are smarter and are able to get a lot of attention for bing smart. They are able to gain influence because of this but also may recieve a lot of unwanted attention. (people calling them nerd!!!)
The smartest people are those who are smart, realize they are smart, (thereby making themselves smarter) and do thier best to conceal it so that they do not stand out and attract unwanted attention. They are still able to use thier intelligence to thier advantage, but since they have considerably more of it than others they are still able to 'keep thier cover', and have an influence.
This is arguable though because some people would want the attneiton.


Asymptote
function
geometry
trapezoid
analytic geometry
graphing lines
slope

This is not exactly what I am really interested in but I believe that a solution to this greatly simplified problem can shed some light on the real question. So, imagine two nice curves $P$ and $Q$ like on the picture with common endpoints on some line and assume that for every point $X$ on that line the lengths of the intervals $AB$ and $CD$ are equal. Does it follow that $P$ and $Q$ are symmetric to each other about this line?
[asy]import cse5; size(300); pair X=MP("X",(70,0)); void MA(pair X, real a, real c, pen p, string s) { fill(CR(X,3,a,c)--X--cycle,p); MP(s,X+3*dir((a+c)/2),8,dir((a+c)/2)); } MA(X,0,45,blue,"45"); MA(X,-45,0,green,"45"); D((0,0)--(100,0),black); path P=D((0,0)..MP("P",(30,15),N)..(70,10)..(100,0)), Q=D((0,0)..MP("Q",(30,-10))..(70,-10)..(100,0),green), L=L((40,30),(100,-30)), M=L((40,-30),(100,30)); pathpen=magenta; D(MP("A",D(IP(L,P)),8,N)--MP("B",D(IP(L,Q)),8)); D(MP("C",D(IP(M,P)),8,N)--MP("D",D(IP(M,Q)),8)); D(X);[/asy]

Another way to state the problem: consider all strictly increasing continuous functions $f\colon [0,1]\to [0,1]$ such that $f(0)=0$, $f(1)=1$, and $f(x)\le x$ for $x\in [0,1]$. Is the map $f\mapsto f+f^{-1}$ one-to-one?    :maybe:

This reformulation at least excludes the counterexample of a symmetric trapezoid with base angle $<45^\circ$ where the unnamed segment in the image is chosen to be the shorter of the two parallel sides. owk

I think that by "two nice curves P and Q like on the picture" fedja meant curves with slope less than $1$ in absolute value. In the functional reformulation this assumption is built-in.

Yes, assume that the slopes are less than $1$, perhaps even some convexity if you need it :). Generally speaking, I am much more interested in learning [b]how[/b] to prove something like that than in [b]what[/b] exactly can be proven.

If you require your functions to be analytic, something even weaker than mlok's statement can easily be checked by considering power series coefficients inductively. owk

Yes, that's a very nice idea :). Thanks owk! At the very least, it shows that there are no "easy" counterexamples. Unfortunately, the endpoint analyticity assumption is too strong to be useful for my purposes :(. Let's see if we can get away assuming something less restrictive.

OK, it looks like I found a really nice solution unless my brain just melted down by the end of the day and I cannot spot my own mistakes anymore.

The conditions are that for some $h\ge 0$ we have $g=f^{-1}+h$ and $f=g^{-1}+h$ (I assume that $f,g$ are "top lines"). We need to show that $f=g$.
Now, for any two increasing functions $u$ and $v$ mapping $[0,1]$ onto itself, define the distance $D(u,v)$ as the integral
of $\frac 1{x+y+1}$ over the area bounded by the graphs of $u$ and $v$. We have two simple claims:

1) $D(u,v)=D(u^{-1},v^{-1})$. This is just because the reflection about the diagonal $x=y$ doesn't change the density.

2) $D(u+h,v+h)\le D(u,v)$ for any $h\ge 0$ with equality only when $h=0$ (the density is decreasing in $y$).

Thus, if $h$ is not identically $0$, the mapping $u\mapsto u^{-1}+h$ is a weak contraction, so a cycle of length $2$ is impossible and $g=f$.

So, am I a hopeless idiot because I cannot find an error in my faulty argument, or because I couldn't solve a trivial problem for 6 years?  :stretcher:

Looks correct. To double check, I'll rephrase the proof in the original setting: consider the symmetric difference between the set within the curves and its reflection about the horizontal axis (taken to be the $x$-axis). Let $T$ and $B$ be the top and bottom halves of the symmetric difference. By symmetry, $\iint_T x = \iint_B x$. On the other hand, every line of slope $1$ crosses $T$ and $B$ along sets of equal length, and its intersection with $T$ has greater values of $x$. By Fubini, $\iint_T x>\iint_T x$, unless $T$ and $B$ are empty.


Well, triviality is a tricky thing. To experts, they may think that $e^{\pi i}=-1$ is 'trivial', but to some people, this may not be 'trivial', and in fact, most people in the real world find that equation extremely 'non-trivial'. 'Triviality' depends on the type of audience. Another example: my friend knows how to fold with grass leaves to make a grasshopper figure and he finds that 'trivial' because he knows how to do it; but to me, I find that 'non-trivial'. Hopefully you understand what I mean by this. And no, there is nothing we can do to stop triviality from life because that's what life is.

Usually when someone uses the word "trivial" in a proof, it means that they are either too lazy, don't know how, or don't have enough space to prove or fully explain it. It does, of course, depend on your audenience, but the over-use of the word "trivial" is one of the biggest problems in mathematical writing.

[quote="mna851"]What does it mean when in proofs, people refer to parts of their proof as being trivial, or non-trivial?[/quote]

An example can be found in if and only if proofs.  For example, we might state Fermat's Last Theorem as: The equation $a^n+b^n=c^n$, $a,b,c,n\in\mathbb{N}$ has solutions if and only if $n=1$ or $n=2$.  We might say that $\Longleftarrow$ is trivial; for $n=1$, $a=1,b=1,c=2$ works, and for $n=2$, $a=3,b=4,c=5$ works.  However, the opposite direction is highly NON trivial, because it took professional mathematicians the better part of four centuries to figure it out.

I admit to (intentional) misuse of the word trivial in certain situations; if I am particuarly pleased with a proof of mine, I will preface it with 'This is trivial.'


calculus
integration
function
complex analysis
real analysis
real analysis unsolved

I can't solve this analytically and would like to (I believe it converges):

$ \int_{0}^{\infty}Log[1+\frac{1}{e^{x}\sqrt{x}}]dx$

I use "Log" even though I mean natural log because I think this problem might require the use of complex variables but I can't work through that route either.  Any suggestions?

Let I be your integral. Then, integrating by parts, we get $ I =-\frac{1}{2}\left(1-\int^{\infty}_{0}\frac{1+2x}{1+\sqrt{x}e^{x}}\,dx\right)$.
This integral converges because $ \sum_{n=0}^{\infty}\frac{1+2n}{1+\sqrt{n}e^{n}}$ converges (just compare with $ \frac{\sqrt{n}}{e^{n}}$, knowing that this series converges as
$ \frac{\sqrt{n}}{e^{n}}< \frac{n}{e^{n}}< \frac{n}{n^{3}}= \frac{1}{n^{2}}$).
To compute the integral perhaps complex variables are appropriate.

Very nice Carcul!  That's a good start.   Perhaps I can formulate an approach with the complex function:

$ f[z]=\frac{1+2z}{1+e^{z}\sqrt{z}}\; ; \; Log[z]=ln(r)+i\Theta \; ; 0<\Theta<2 \pi$

However, nothing in math is more difficult for me to understand (that I'm capable of understanding) than working real integrals through complex functions involving branch cuts . . . might take a few weeks.

Hey Carcul, looks like we both made a mistake:

$ \int udv=uv-\int vdu$

Then:

$ \int_{0}^{\infty}Log[1+\frac{1}{e^{x}\sqrt{x}}]dx=\left[xLog[1+\frac{1}{e^{x}\sqrt{x}}\right]_{0}^{\infty}+1/2\int_{0}^{\infty}\frac{1+2x}{1+e^{x}\sqrt{x}}dx =1/2\int_{0}^{\infty}\frac{1+2x}{1+e^{x}\sqrt{x}}dx$

I've since worked out a solution to the problem using residue integration which I  feel would not be appropriate to post here.  I'll post my results in the Complex Analysis forum once I make sure it's ok (actually I'm pretty sure it's ok . . . just want to make sure so I don't make any waves over there).  You guys can look at it there if you wish.

Yes, you are right, both limits are zero.


I hope you can make it!

Pro: With $ a,b,c \geq 0$ and $ a\plus{}b\plus{}c\equal{}3$. Find $ k$ good best:

$ \sqrt \frac {a}{k \minus{} a^2b} \plus{} \sqrt \frac {b}{k \minus{} b^2c} \plus{} \sqrt \frac {c}{k \minus{} c^2a} \leq \frac {3}{\sqrt {k \minus{} 1}}$

It's very nice!
 :P

[quote="vipCD_A1"]I hope you can make it!

Pro: With $ a,b,c \geq 0$. Find $ k$ good best:

$ \sqrt \frac {a}{k \minus{} a^2b} \plus{} \sqrt \frac {b}{k \minus{} b^2c} \plus{} \sqrt \frac {c}{k \minus{} c^2a} \leq \frac {3}{\sqrt {k \minus{} 1}}$

It's very nice!
 :P[/quote]
Your inequality is wrong for all $ k>1$ because there are non-negative $ a$ and $ b$ for them $ k\minus{}a^2b<0.$ :wink:

[quote="vipCD_A1"][quote="arqady"][quote="vipCD_A1"]I hope you can make it!

Pro: With $ a,b,c \geq 0$. Find $ k$ good best:

$ \sqrt \frac {a}{k \minus{} a^2b} \plus{} \sqrt \frac {b}{k \minus{} b^2c} \plus{} \sqrt \frac {c}{k \minus{} c^2a} \leq \frac {3}{\sqrt {k \minus{} 1}}$

It's very nice!
 :P[/quote]
Your inequality is wrong for all $ k > 1$ because there are non-negative $ a$ and $ b$ for them $ k \minus{} a^2b < 0.$ :wink:[/quote]
I'm sorry but you see, [b]find k good best[/b][/quote]

help me, jisen anh vasc

[quote="akai"][quote="vipCD_A1"]I hope you can make it!

Pro: With $ a,b,c \geq 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Find $ k$ good best:

$ \sqrt \frac {a}{k \minus{} a^2b} \plus{} \sqrt \frac {b}{k \minus{} b^2c} \plus{} \sqrt \frac {c}{k \minus{} c^2a} \leq \frac {3}{\sqrt {k \minus{} 1}}$

It's very nice!
 :P[/quote]

I don't like your problem because it's not nice,i think.
Have you got the solution for this problem,vipCD_a1? If yet,you must post it in box Inequalities unsolved  :mad:[/quote]
oh, I'm sorry, but it's nice, you don't see
I'm sorrow for you! :rotfl:

[quote="vipCD_A1"]I hope you can make it!

Pro: With $ a,b,c \geq 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Find $ k$ good best:

$ \sqrt \frac {a}{k \minus{} a^2b} \plus{} \sqrt \frac {b}{k \minus{} b^2c} \plus{} \sqrt \frac {c}{k \minus{} c^2a} \leq \frac {3}{\sqrt {k \minus{} 1}}$

It's very nice!
 :P[/quote]

Help me, jichen!

Don't spam vip CD, I think you should take care of your problem thoroughly before posting it in  the forum


trigonometry
calculus
integration
calculus computations

Find the integral [b]without Weierstrass substitution[/b]

$ \int \frac {1 \plus{} \sin x}{(1 \minus{} \sin x)^2}\,dx$

subsitute $ tan(\frac {x}{2}) \equal{} t$ so we have $ sin x \equal{} \frac {2t}{1 \plus{} t^2}$ and $ dx \equal{} \frac {2dt}{1 \plus{} t^2}$ .
so $ \int \frac {1 \plus{} sin x}{(1 \minus{} sin x)^2}\ dx \equal{} 2\int \frac {1 \plus{} t^{2} \plus{} 2t}{(1 \plus{} t^{2} \minus{} 2t)^2}\ dt$ ...

[quote="sepehr ahmadzadeh"]subsitute $ tan(\frac {x}{2}) \equal{} t$ so we have $ sin x \equal{} \frac {2t}{1 \plus{} t^2}$ and $ dx \equal{} \frac {2dt}{1 \plus{} t^2}$ .
so $ \int \frac {1 \plus{} sin x}{(1 \minus{} sin x)^2}\ dx \equal{} 2\int \frac {1 \plus{} t^{2} \plus{} 2t}{1 \plus{} t^{2} \minus{} 2t}\ dt$ ...[/quote]

Not, look again

[quote]Find the integral without Weierstrass substitution[/quote]

$ \int \frac{1\plus{}\sin x}{(1\minus{}\sin x)^{2}}dx$

$ \equal{}\int \frac{\left( 1\plus{}\sin x\right) ^{3}}{\cos ^{4}x}dx$

$ \equal{}\int \frac{4\sin ^{2}x\plus{}\cos ^{2}x\plus{}4\sin x\minus{}\sin x\cos ^{2}x}{\cos ^{4}x}dx\vspace{1pt}$

$ \equal{}\int \left( 4\tan ^{2}x\sec ^{2}x\plus{}\sec ^{2}x\plus{}\frac{4\sin x}{\cos ^{4}x}\minus{}\frac{\sin x}{\cos ^{2}x}\right) dx\vspace{1pt}$

$ \equal{}\allowbreak \frac{4}{3}\tan x\sec ^{2}x\plus{}\tan x\plus{}\frac{4}{3}\sec ^{3}x\minus{}\sec x\plus{}C$

[quote="DeMath"][quote="kunny"]I have just solved the problem.

[hide="Answer"]$ \minus{} \frac {1}{\tan \left(\frac {\pi}{4} \minus{} \frac {x}{2}\right)} \plus{} C.$[/hide]

[hide="hint"]$ x \equal{} \frac {\pi}{2} \minus{} t$[/hide][/quote]

[b]Incorrect[/b][/quote]

No. Correct is my answer.

Let $ x \equal{} \frac {\pi}{2} \minus{} t$, we have

$ \int \frac {1 \plus{} \sin x}{(1 \minus{} \sin x)^2}dx \equal{} \minus{} \int \frac {1 \plus{} \cos t}{(1 \minus{} \cos t)^2}dt \equal{} \minus{} \int \frac {2\cos ^ 2 \frac {t}{2}}{\left(2\sin ^ 2 \frac {t}{2}\right)^2}\ dt \equal{} \int \frac {1}{\tan ^ 2 \frac {t}{2}}\left(\tan \frac {t}{2}\right)'dt \equal{} \minus{} \frac {1}{\tan \frac {t}{2}} \plus{} C \equal{} \minus{} \frac {1}{\tan \left(\frac {\pi}{4} \minus{} \frac {x}{2}\right)} \plus{} C$.

....$ \equal{}\minus{}\int \frac{2\cos ^{2}\frac{t}{2}}{\left( 2\sin ^{2}\frac{t}{2}\right) ^{2}}\ dt\equal{}\int \frac{1}{\tan ^{2}\frac{t}{2}}\left( \tan \frac{t}{2}\right) \prime dt$ ???????????? :wink:

Regrettably your answer is incorrect.

You are right. That should be

$ \int \frac {1 \plus{} \sin x}{(1 \minus{} \sin x)^2}dx \equal{} \minus{} \int \frac {1 \plus{} \cos t}{(1 \minus{} \cos t)^2}dt \equal{} \minus{} \int \frac {2\cos ^ 2 \frac {t}{2}}{\left(2\sin ^ 2 \frac {t}{2}\right)^2}\ dt \equal{} \int \frac {1}{\tan ^ 2 \frac {t}{2}}\left(\frac {1}{\tan \frac {t}{2}}\right)'dt\equal{}\frac 13\frac{1}{\tan ^ 3 \frac{t}{2}}\plus{}C$.


[quote="Hikaru79"]Prove that \[\left(a+\frac1a\right)^{2}+\left(b+\frac1b\right)^{2}\geq\frac{25}{2}\] for all real positive $a,b$ such that $a+b=1$.
When does the equality hold?

I hope this is okay for intermediate. It may be Pre-Olympiad, but I tried to err on the side of a challenge ;)[/quote]
[hide]
Expand:

$a^{2}+\frac{1}{a^{2}}+b^{2}+\frac{1}{b^{2}}\ge \frac{17}{2}$

$(a+b)^{2}-2ab+\frac{(a+b)^{2}-2ab}{(ab)^{2}}\ge\frac{17}{2}$

$-2ab+\frac{1-2ab}{(ab)^{2}}\ge \frac{15}{2}$

Since we are trying to maximize $ab$ because $(ab)^{2}$ increases faster than $1-2ab$ by AM-GM $\frac{1}{4}\ge ab$. Substituting, we get

$-\frac{1}{2}+\frac{1-\frac{1}{2}}{\frac{1}{16}}\ge \frac{15}{2}$ which simplifies to $0\ge 0$ and we know if we decrease $ab$ then the value of $-2ab+\frac{1-2ab}{(ab)^{2}}$ would increase which would make the inequality true since when $ab$ is maximized we get equality

Not much of a proof...but...[/hide]

Well if you're really feeling like doing some algebra...
[hide]
Substitute $b = 1-a$ and multiply through by $2a^{2}(1-a)^{2}$ and the inequality we need to prove is

$4a^{6}-12a^{5}-3a^{4}+26a^{3}-11a^{2}-4a+2 \ge 0$ for $0 \le a \le 1$.

We know equality holds at $a = \frac{1}{2}$, so we factor out as many $2a-1$'s as we can to get

$(2a-1)^{2}(a^{4}-2a^{3}-3a^{2}+4a+2)$.

So we just want to show that the last term is positive for $0 \le a \le 1$.

Notice that $a^{4}-2a^{3}-3a^{2}+4a+2 = \left(\left(a-\frac{1}{2}\right)^{2}-\frac{9}{4}\right)^{2}-2$ (by experimenting). But then

$\left|\left(a-\frac{1}{2}\right)^{2}-\frac{9}{4}\right| \ge 2$

since $0 \le a \le 1$. Consequently,

$\left(\left(a-\frac{1}{2}\right)^{2}-\frac{9}{4}\right)^{2}-2 \ge 2^{2}-2 = 2 > 0$,

as desired.[/hide]

The function $f(x) = \left(x+\frac{1}{x}\right)^{2}$ is convex since $f''(x) = 2+\frac{6}{x^{4}}$. Therefore, by Jensen's inequality: \[\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}= f(a)+f(b) \geq 2f\left(\frac{a+b}{2}\right) = 2f\left(\frac{1}{2}\right) = \frac{25}{2}.\]


function
LaTeX
real analysis
real analysis unsolved

let $f:\mathbb{C} \mapsto \mathbb{C}$ be a $C^{\infty}$ function such that $|f(z)|=1$ for every $z \in \mathbb{C}$.
Prove that there exists a $C^{\infty}$ function $v:\mathbb{C} \mapsto \mathbb{R}$ such that $f(z)=e^{v(z)}$

What is C? If it is the circle, then I think I had it in khole. Is it so hard in this case? I have great problems if f is supposed to be only continuous, but many times differentiable/. (another classical problem from ENS). And I imagine that v must be required to be infinitely differentiable in this case, not?

Sorry, Alekk, what I said is non-sense. What I had in kholle was much easier that this. Moubi, can you please post a solution.?

Can someone write the exact statement of the problem?

sorry  :D  $\mathbb C$ is the set of complex number ! but i don't know latex very well.
(I think it's harder that relevement theorem.)

[i]Edited by Myth[/i]

[quote="alekk"]i don't know latex very well.[/quote]
See again :)

In the original problem, what are the requirements on $v$?  If it is just required to be a function, then the problem is trivial.

[quote="alekk"]what is the english name of the following theorem:
let $f: \mathbb{R} \mapsto \mathbb{C}$ be a continuous function such that $|f(x)|=1$. Then there exists a continuous function
$\theta: \mathbb{R} \mapsto \mathbb{R}$ such that $f=e^\theta$[/quote]
I think you mean $f(z)=e^{i\theta(z)}$.

yes, and I've just edited my original post .. sorry  :blush:

harazi, I think the problem you mentioned in the case where $f$ is only continuous (the problem you had in an oral examination) can be solved if you construct a sequence of functions $C^1$ that converges uniformly towards $f$ .


Does anyone here know where to buy the 2004 AMC test? (They only sell 2000 to 2003 in the site)

I believe their website is simply out-of-date.

If you download the PDF version of the order form off the AMC website, you are able to order the 2004 contests.


how do you evaluate

arcsin($ \frac{1}{5}$) + arcsin($ \frac{1}{7}$)

urgent help been working for this for an hour

Let $ x \equal{} \arcsin\left(\frac15\right) \plus{} \arcsin\left(\frac17\right)$.

Take the $ \cos$ of both sides (or $ \sin$, but I found it easier with $ \cos$) to get

$ \cos x \equal{} \cos\left(\arcsin\left(\frac15\right) \plus{} \arcsin\left(\frac17\right)\right)$.

We know that $ \cos(a \plus{} b) \equal{} \cos a\cos b \minus{} \sin a\sin b$ and $ \cos(\arcsin x) \equal{} \sqrt {1 \minus{} x^2}$, giving us

$ \cos x \equal{} \left(\sqrt {1 \minus{} \frac1{25}}\right)\left(\sqrt {1 \minus{} \frac1{49}}\right) \minus{} \frac15\cdot\frac17$

$ \equal{} \sqrt {\frac {24\cdot48}{25\cdot49}} \minus{} \frac1{35} \equal{} \frac {24\sqrt2 \minus{} 1}{35}$, so

$ x \equal{} \boxed{\arccos\left(\frac {24\sqrt2 \minus{} 1}{35}\right)\equal{}\arcsin\left(\frac{4\sqrt3\plus{}2\sqrt6}{35}\right)}$, whatever that is...

[hide="Hint"]Think about the unity circle, a point on it defines a right angled triangle with sides $ 1$, $ \cos(x)$ and $ \sin(x)$. Then take $ \sin$ or $ \cos$ of $ \arcsin(\frac{1}{5})\plus{}\arcsin(\frac{1}{7})$ and play around for a while.
[/hide]

I was like 30 secondes late jeje

you can use
$ \arcsin x \equal{} \minus{}i \ln \left(ix \plus{} \sqrt{1 \minus{} x^2}\right)$

convert the stuff into imaginary form then simplify


number theory
Diophantine equation
number theory open

Find the min value of n that n>1 1^2+2^+3^...n^2/n( n is common) is a perfect square

Clearly, the problem is equivalent to finding the minimal $ n\in\mathbb{N}$ such that $ (n\plus{}1)(2n\plus{}1)\equal{}6x^2$. Since the factors on the LHS are coprime and $ 2n\plus{}1$ is odd, we must have one of the two situations for a factorization $ x\equal{}x_1x_2$ with positive and coprime factors $ x_1,x_2$:

1) The case $ 2n\plus{}1\equal{}x_1^2,n\plus{}1\equal{}6x_2^2$ implies $ 1\equal{}12x_2^2\minus{}x_1^2\Longrightarrow 1\equiv \minus{}x_1^2\bmod{4}$, which is impossible.

2) The case $ 2n\plus{}1\equal{}3x_1^2,n\plus{}1\equal{}2x_2^2$ implies (*) $ \;1\equal{}(2x_2)^2\minus{}3x_1^2$, i.e. we have to find a minimal $ x_2$ such that the diophantine equation (*) is fulfilled.

The positive solutions of Pell's equation $ 1\equal{}a^2\minus{}3b^2\equal{}(a\plus{}\sqrt{3}b)(a\minus{}\sqrt{3}b)$ are given by $ a_m\plus{}b_m\sqrt{3}\equal{}(2\plus{}\sqrt{3})^m$. This yields the recursions $ a_{m\plus{}1}\equal{}2a_m\plus{}3b_m,b_{m\plus{}1}\equal{}2b_m\plus{}a_m$. From this we conclude that $ a_{m\plus{}1}\equiv b_m\bmod{2}$ and $ b_{m\plus{}1}\equiv a_m\bmod{2}$ and $ a_1\not\equiv b_1\bmod{2}$ hold and that $ a_m,b_m$ are monotone increasing series. Since $ n>1$ and $ 2|a_m$ is required, we have $ 2x_2\equal{}a_{3},x_1\equal{}b_{3}$ i.e. $ x_2\equal{}13,x_1\equal{}15$ and $ n\equal{}337$ is the minimal value.


Is there a quotient map from $ \mathbb{T}^2$ onto $ A$, where $ A$ is the annulus in 2-space? What about from $ A$ onto $ \mathbb{T}^2$?

Here is what I did.  $ \mathbb{T}^2$ is homeomorphic to the quotient of $ A$ when you identify all its boundaries, call this $ A^*$. There is a quotient map $ p: A\rightarrow A^*$ and $ f$ a homeomorphism between $ A^*$ and $ \mathbb{T}^2$. Then $ f \circ p$ is a quotient map from $ A$ to $ \mathbb{T}^2$. Is this right?

The other way around I was thinking of a function that would collapse the torus into a plane, but I don't exactly know how to say this mathematically. Any hints?

The first part is correct.

But collapsing a torus into a plane is not a quotient map. There is a problem at the points that get mapped into the boundary of $ A$.

The obstruction comes from the fact that the fundamental group of $ \mathbb T^2$ is $ \mathbb Z^2$, while the fundamental group of $ A$ is $ \mathbb Z$. There is no injective homomorphism from $ \mathbb Z^2$ into $ \mathbb Z$.

i don't agree with you, mlok.

consider a doughnut on the table, and just press it against it. you (roughly) obtain the projection of $ T^2$ onto an annulus. more formally, consider the "standard", "horizontal" embedding of $ T^2$ in $ \mathbb{R}^3$, and project to the $ xy$-plane. the image is an annulus, and restricting the projection gives a quotient map.

maybe you meant to say that there's no covering $ T^2\to A$? well.. there's no covering $ A\to T^2$ either, if $ A$ is compact with boundary (and it seems to be the case).

You're probably right, since I don't really know what is meant by the quotient map here.

Well in this case since both $ \mathbb{T}^2$ and $ A$ are compact and Hausdorff a quotient map from one to the other is just a surjective continous map.

So I guess $ p: \mathbb{T}^2\rightarrow A$, $ (x,y,z)\rightarrow (x,y)$ is a quotient map. Here I view the torus as the surface of a doughnut of radius 2 with the revolving circle of radius 1.


(i) By letting $ c \equal{} cos\theta$, show that the equation $ \cos4\theta \equal{} \cos3\theta$ can be expressed in the form $ 8c^4 \minus{} 4c^3 \minus{} 8c^2 \plus{} 3c \plus{} 1 \equal{} 0$.

(ii) Show that $ \theta \equal{} \frac {2n\pi}{7}$, where n is an integer, satisfies the equation $ \cos4\theta \equal{} \cos3\theta$

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

(i) $ cos4\theta$ = $ cos(2\theta \plus{} 2\theta)$ expand, simplify
$ ///ly$ for $ cos3\theta$ and bring to one side and the above expression will be obtained

(ii) The [b]solution[/b] has sub for n=1,2,3,4 since leading degree 4 has max. 4 roots.
[i]Is there a [b]general[/b] way to solve this question please?[/i]

thanks

A fourth degree polynomial can be solved using the quartic formula...

This isn't the type of problem that will generally show up on a contest or anything.

[hide="ii"] $ \cos 4 \theta \minus{} \cos 3 \theta \equal{} 2 \sin \frac{7 \theta}{2} \sin \frac{\theta}{2}$ by product-to-sum formula.  The conclusion follows. [/hide]

I see how you got it now.

But how does the conclusion follow from this?

$ \cos 4 \theta \minus{} \cos 3 \theta \equal{} 0 \Leftrightarrow \sin \frac{7 \theta}{2} \equal{} 0$ or $ \sin \frac{\theta}{2} \equal{} 0$.  The first gives $ \frac{7 \theta}{2} \equal{} \pi n\Leftrightarrow \theta \equal{} \frac{2 \pi n}{7}$ and the second gives $ \frac{\theta}{2} \equal{} \pi n\Leftrightarrow \theta \equal{} 2 \pi n$, a subset of the first; this is the complete solution set.


Subiectele concursurilor desfasurate in ultimii doi ani la gimnaziu

le puteti gasi la  http://mateclub.8k.ro/olimpiade_2007_2008.php

[color=darkred]Propun moderatorilor ca acest topic sa apara in amonte.[/color]

Va anunt ca s-a schimbat putin adresa, intrucat am modificat ceva la site. Totusi adresa ramane aceeasi: http://www.mateclub.8k.ro
Numai bine, Claudiu Mindrila!


On his last test, Kyle scored a 95, raising his average by one point to 89.  What must he score on his next test to raise his average to 90?

think in drag up. if all test are =
95 dragged up 88 by 1 must have been 6  to drag up 90 with 7 89's you need a 97

hi jimli, can you explain your solution a little more.  i dont understand but it seems much better than my solution which i used (algebra) that got the same answer but perhaps wasn't as insightful


Putnam AMC AIME calculus vector USA(J)MO USAMO

im not sure if this is the right place for this (if not will a mod please move it) but here goes.

i am wondering if there are any books/resoruces that could help me prepare for the putnam exam.
i am currently a highschool senior and just took up more competative math my sophmore/junior year. i am currently woking on problems that are about AIME leve.

any help would be greatly appreciated.

I just used old tests and my high school friend's olympiad material. There are many books that might be helpful though. Depends on what you want to practice really.

My favorite such book is Larson's Problem Solving Through Problems - it has a lot of nice problems and examples.  If you want more recent problems, then you might want to try Art and Craft of Problem Solving, by Zeitz, although I do not like this book nearly as much.

And of course this very forum! [url=http://www.mathlinks.ro/Forum/index.php?f=218]College Playground[/url]. There are thousands of problems that you can use in preparation :)

I posted this before, but no one answered this.......

What are books that have some contest problem solving approaches in:
1. Calculus
2. Vector Calculus
3. Linear Algbebra
4. Differential Equations
5. Real Analysis

On a side note, do skills needed to qualify for USAMO transfer over to the Putnam fairly well?

blahblahblah mentioned Larson and Zeitz; there's also a book by Engel.

[quote="1234567890"]On a side note, do skills needed to qualify for USAMO transfer over to the Putnam fairly well?[/quote]
Clearly, yes. A significant number of the Americans whose names appear high on the Putnam list did well in the AMC contests when they were in high school. Of course, in effect the competitve pool is larger because of students from other countries attending U.S. and Canadian colleges. In a number of cases, these are people who did well in the olympiad-level contests in their home countries.

You said "... skills needed to qualify for USAMO ...." The AMC and AIME are short-answer tests; the Putnam requires proofs. You do need to develop the art of explaining yourself. But with that caveat, the skills do transfer.


summer program
MathPath
AwesomeMath
AMC
AIME
USA(J)MO
USAMO

So. How do you guys normally spend your summers?

I usually go to CTY summer program at Lancaster, but my parents say this is going to be my last year. :(

What do you guys do (as of summer programs, traveling, etc)?

I went to CTY at Lancaster last year!!!   :D

Next year I want to go to AwesomMath though.  I don't know what course I'd take if I were in CTY again.  What course did you take last year?  I took Individually Paced Math Sequence.

Last year I went to a chess camp...

I'm applying to MathPath this year.

[quote="alexhhmun"]What course did you take last year?  I took Individually Paced Math Sequence.[/quote]

which session? i went session 2.

i took Fundamentals of Computer Science..

This year i'm going session 1 Mathematical Logic!!! :)

Going to AwesomeMath this year, MOP next year.

Wow, you've got some guts to say you're that sure of making MOP in 7th grade...
CTY and summer school (failed AIME, can't make MOP :() yuck.
If I don't make MOP next year, AwesomeMath.

[quote="alexhhmun"]Next year I want to go to AwesomMath though.  I don't know what course I'd take if I were in CTY again.  What course did you take last year?  I took Individually Paced Math Sequence.[/quote]

I'm going there next year too!!!

I am gonna go to AwesomeMath these next two years, and if I make USAMO the year after next year, MOP.

I went to CTY once, last year. The most advanced math they offered was Game Theory, which was basic probability(I wanted to ask permission to sleep in class), matrices, and a mix of the two.

I don't do any math-type stuff at all. Mostly video games and travel to my relatives in SOUTH Florida. Like Miami. Really hot down there. Seriously.

This summer I'm going to French immersion camp in Minnesota for four weeks. Usually I would go to multiple camps, but almost everything has overlapping dates with French camp, so that's the only camp I'm going to this year.
I'm going to start a project on Tipover...I might spend more time playing the game than working on the project, though.
I'll probably edit somebody else's novel or script...

Cool!
I am going to Yunasa, for sure, the beach, sailing camp, and I'm staying home to teach myself all I can the rest of the time!
(besides building my test tank for my model sailboats [hopefully before my 14th birthday]and hydroponics experiments)

I went to a computer camp at Emory last summer... it sucked big time.
I was planning on going to the ross program or some other camp this summer, but I changed my mind.
Now I'm staying home, doing tons of math, and finishing up a bio. course. fun  :P

im going to take some college courses and study my next year 8th grade curriculum  :P ......

[quote="7h3.D3m0n.117"]1.) Xbox

2.) Computer

3.) Math?

4.) Any other school work I need to complete for next year.

But this year, I'm going to Taiwan for my Sweet 16th celebration for like half the summer. And I've got to take a CTY Chem course so I can take AP chem next year.[/quote]
Hmmm... you know if your chem skills are pro enough, you can get the teacher to override the prerequisites...that's my plan to get into chem AP at least, and also to skip to bc calc. If that succeeds, then w00t physics, calc, and chem AP as sophmore!
I normally go to China for most of the summer actually, but this year plane tickets are too expensive :(
However that means I can really focus on improving my math/piano/physics/chem skills so I can own next year :D

[quote="serialk11r"]Hmmm... you know if your chem skills are pro enough, you can get the teacher to override the prerequisites...[/quote]

Actually, my dad has lot of chemistry background, and even the teachers say I need to take CTY, no matter how skilled we may be. Just a requirement, I guess.

Is there a list of academic camps/programs like these? Because if there is I really need to see it. There's no way I'm wasting my summer again.

John Hopkins CTY
EPGY
Those are the 2 big ones I think. CTY is a lot bigger than EPGY. Both are sorta noob though...

Hmmm...my mom had chemistry degree heh but that's off topic. At my school they're pretty chill with the overriding prerequisites thing ;) I got into AP physics because of my pro application I think. The weird thing is I got chosen over moogra, who has a better math grade record than me (I don't do homework :P too boring), and he's going into 11th grade, so he should get priority. Teachers who are nice and who think you're good can override the prerequisites and move you up I think...I almost got into Bio AP except I didn't officially learn chem so I couldn't :(

Is Johns Hopkins CTY the same as Duke TIP except for people in the Northeast? Because I live in Alabama and don't think I'm eligible for that...

[quote="serialk11r"]Well CTY is held at multiple sites in the country...and there probably isn't one near you that you can go to.[/quote]

It took 20 or 30 minutes to drive from my house to the CTY site.

PWND!

That's one of the only reasons I took it...

I am about 15 minutes from the Saratoga one and 25 from the other one - Sienna or Skidmore, I think.

Ha!

bet you can narrow down my location to 2 spots now. =D

[quote="1=2"][quote="serialk11r"]Well CTY is held at multiple sites in the country...and there probably isn't one near you that you can go to.[/quote]
It took 20 or 30 minutes to drive from my house to the CTY site.
[/quote]

me, too. but i'm in another state..

Oh. and it doesn't matter if you live in Alabama, you can still apply. You just have a heck of a long drive to the site. I know a couple people from California who come to LAN just because we're beastly, and I know one girl from Kentucky and another from Tennessee or something like that.

um nevermind. i am applying to the ross program. w00t!


prove that for natural numbers $ a,b$ ,$ (a^6)^2 \minus{} (b^6)^2$ is divisible by $ 91$ if $ a,b$ are prime to $ 91$

Fermat's little theorem. Go to High School Basic.

$ (a^6)^2 \equal{} 1^2 \equal{} 1(mod 7)$ and
$ (a^6)^2 \equal{} a^{12} \equal{} 1(mod 13)$ for all $ a$ relatively prime to $ 7 \times 13 \equal{} 91$ gives the result.


probability
combinatorics proposed
combinatorics

let $ a\in \mathbb{N^*}$ and $ n\in \mathbb{N^*}$
Some box contains $ a$ white balls and $ a \plus{} 1$ black ball.

we take $ n$ balls from the box ($ 1 \le n \le 2a \plus{} 1$) by puting back each one we take(avec remise)

How many possibilities will give us an even number of white balls.

[hide="Answer?"]n+a-1 isn't it?[/hide]

In an event of picking the ball from the bag we have $ a$ chances of getting a white ball and $ a\plus{}1$ chances of getting a black ball.

Probability of picking the white ball = $ \frac{a}{2a\plus{}1}$

Probability of picking the black ball = $ \frac{a\plus{}1}{2a\plus{}1}$

Since $ n$ is odd, $ n\equal{}2k\plus{}1$
Probability of picking even number of white balls=$ \frac{1}{(2a\plus{}1)^n}\sum_{r\equal{}0}^{r\equal{}k} {n\choose 2r} a^{2r} b^{n\minus{}2r}$

n is not necessary odd, we should discuss

and what we want is the number of possibilities, not the probability.

no one can tell me if my answer is true ??

some one help please, I need it.


What is 150% of 0.84, to the nearest hundredth?


I will try.
A  triangle ABC with AB=BC=CA=1 is covered with 6 circles with radius $r$. Prove that $r\geq \frac{\sqrt{3}}{10}$.
http://www.zaba.ru/cgi-bin/tasks.cgi?tour=national.bmo.1999.3

SORRY，I can't understand your problem :(
can you explain more?

I edited it.I hope now it`s more clear. Thank you.

Please continue discussion at http://www.mathlinks.ro/Forum/viewtopic.php?t=48067 .

  darij


geometry
incenter
geometric transformation
homothety
projective geometry
geometry proposed

[quote="Erken"]Let $ \Gamma_1,\Gamma_2$ and $ \Gamma_3$ be three non-intersecting circles,which are tangent to the circle $ \Gamma$ at points $ A_1,B_1,C_1$,respectively.Suppose that common tangent lines to $ (\Gamma_2,\Gamma_3)$,$ (\Gamma_1,\Gamma_3)$,$ (\Gamma_2,\Gamma_1)$ intersect in points $ A,B,C$.
Prove that lines $ AA_1,BB_1,CC_1$ are concurrent.[/quote]
[hide="Solution"]
Let the incenter of $ \triangle ABC$ be $ I$ and the incircle be $ \omega$.  Now, $ A_1$ is the center of homothety between $ \Gamma$ and $ \Gamma_1$ and $ A$ is the center of homothety between $ \Gamma_1$ and $ \omega$.  Then, by Monge's Theorem, $ AA_1$ passes through the center of homothety between $ \omega$ and $ \Gamma$.  Similarly do $ BB_1$ and $ CC_1$, and hence they concur. [/hide]

[quote="The QuattoMaster 6000"][quote="plane geometry"]Thank The QuattoMaster 6000 for a nice solution
but here I have a question: Monge's theorem states that for any three circles in a plane, none of which is inside one of the others, the three intersection points of the three pairs of external tangent lines are collinear.(http://en.wikipedia.org/wiki/Monge%27s_theorem)
in our problem the the three circles you mentioned don't satisfy "none of which is inside one of the others" ,could you tell me about the details?
Thank you in advance! :blush:[/quote]
Ah yes, sorry.  Monge's Theorem might not be the correct name of the theorem I am stating.  I am using some sort of extended verison of Monge's Theorem.  I don't have a diagram, but I will try to explain.  Consider three circles with centers $ A$, $ B$, and $ C$ and with respective radii $ x$, $ y$, and $ z$ (the circles can be internal or external.)  Let the center of homothety between $ (A)$ and $ (B)$ be $ Z$, that between $ (A)$ and $ (C)$ be $ Y$, and that between $ (B)$ and $ (C)$ be $ X$.  Then, $ \frac {BX}{XC} \equal{} \frac {y}{z}$.  Similarly, $ \frac {AY}{CY} \equal{} \frac {x}{z}$ and $ \frac {AZ}{BZ} \equal{} \frac {x}{y}$.  Now, if you use the directed segments correctly, $ X$, $ Y$, and $ Z$ should line up by Menelaus's Theorem (even when $ (A)$ might be inside $ (B)$, for example.)

Sorry if some of my statements might be confusing... I am really bad at explaining directed segments without a diagram.

EDIT: I just remembered that the Geogebra Applet has been released  :oops:

Here is a diagram, but it's quite inaccurate (this is my first time trying to use Geogebra.)  I hope it at least helps you get the idea: [geogebra]6efb0d213b97e721d81eb95ffffc242da77cfc84[/geogebra][/quote]
thank you dear The QuattoMaster 6000! :lol:
I guess I understand
ps:why don't you use Sketchpad，it's a strong tool for geo fans :P

[quote="The QuattoMaster 6000"][quote="plane geometry"]
ps:why don't you use Sketchpad，it's a strong tool for geo fans :P[/quote]
I don't have Sketchpad...  :(

Is it a free download?[/quote]
oh,it's a pity
I don't know whether it is free in your place but I have a chinese version,if you know chinese I can send it to you :)

For the archive: http://www.mathlinks.ro/Forum/viewtopic.php?t=5364

  darij


*sigh* People's definitions are too broad.

Hmm I counted 8 (including real life and AoPS - the ones I actually talk to every other day).

ehmm... one two three... oh I forgot it, ernn... ehmm... 0?

erm... do random people u chat with on AoPS count???

People you haven't met before do not count (although if you have 100+ pms back and forth with them that might be an exception)

Ehm, people I actually hang out with, one. People I've met and am on good terms with, quite a few.

i voted 1-2 math friends but that's only 1-2 ppls i know in real life if it counted all my aops friends i'd have a lot more :D


is he guilty?

hope this hasn't already been discussed

Although I should say he is guilty, I think that what Michael Jackson was telling about how people wanted his money was true. So, I vote that he was somewhat innocent.

i thought he would be guilt for at least one charge.

The prosecution knew they were screwed by their own witnesses. That's what made the jury vote not guilty on each charge.

yeah...the mom just wanted a lot of money

[quote="GoBraves"]Ofcourse that jackass is guilty...he probably bribed the jury[/quote]

:-) Nice thought. I don't know about bribing the jury...but the it is certainly difficult to win a case when the prosecution attorneys suck that much.

michael jackson was asking for a trial with all of the suspicious activity he was doing. having little boys stay at your house is not accepted in today's society...he should know better

i think people should just leave him alone after this

i think that jackson has commited similar "activities" in the past. however, i do not think he committed those "actions" to the boy he was on trial against. the mom wanted easy money so she lied. and even if he did do it, a jury would not convict jackon in LA. there would be riots and whatnot.

[quote="stupidkid"]I just dont' see how a man can have the urges to "molest" a boy.[/quote]

Michael Jackson never really was the norm...

[quote="eryaman"]I don't get how any of us can honestly vote "yes" or "no".  None of us were on the jury, none of us (as far as I know) personally know Jackson, the mom, or the boys.  I voted "not sure".[/quote]

Took the words out of my mouth.

[quote="G-UNIT"]needles to say, the mother who tried to sue michael jackson also sued Kobe bryant and Jay Leno on similar accounts.  However, she was unsuccessful in all cases.[/quote]

where'd you get that from?

i don't even kknow what you guy are talking about

yes, i think there is sufficient evidence

[quote="G-UNIT"]it was on the news[/quote]

you should be careful what you read/see.

[size=200]He ( or it) is and will always be guilty. : P i mean, seriously, whats with him and little boys? [/size]

wow...the same person accused Bryant and jay Leno...

 :dry:  :ewpu:  :huuh:

[quote="jhollenbeck"]whos michal Jackson?[/quote]
You joking right?

I didn't know who he was for the longest time. I came downstairs one day (this was 2 years ago) and I saw this person on tv with a pale face, pointy nose, chopped black hair, and weird red lips. I said "Man, who's that woman?!" Uhhh,  :oops: ! Well, now I know who micheal jackson is, ok... I thought he was a basketball player? NO that was micheal jordon. Whoever he is... Me has no knowledge of that stuff, nor does I.

I don't think he's guilty, I mean the person who accused him has done that before.

I feel bad for the guy. does he really like what he's done to himself?


SMO2007 Open, #2

Determine the number of those 0-1 binary sequences of ten 0's and ten 1's which do not contain three 0's together.

Comment:
After the competition yesterday, i showed my solution to my friend, and he showed me his.

Mine:
We fix the ten 1's first. Then we have eleven spaces in between these ten ones. Since at most two zeroes can be together, all zeroes in the sequence are either double or single zeroes.

If there are $k$ double-zeroes, we can choose the spot for the double zeroes in ${{11}\choose{k}}$ ways.
After this, there are $11-k$ spots left for the single zeroes to be placed, and there are $10-2k$ zeroes left, so we can choose the places for the single zeroes in ${{11-k}\choose{10-2k}}$ ways.

Since there can be any number of double zeroes from 0 to 5, the answer is \[\sum_{k=0}^{5}{{11}\choose{k}}{{11-k}\choose{10-2k}}= 24068\].

My friend's:

He fixes the sequence as follows:
$00100100100100$, or 5 groups of double zeroes separated by four 1's. He now has six 1's left to place.

Now he condenses every occurence of $01$ into a $2$, giving rise to $0202020200$ (since placing a new 1 on either side of a fixed one is the same thing and he can do this condensation), and now he can place his six ones into 11 possible spots, so he claims the answer ${{11+6}\choose{6}}$.

Both solutions seem correct, i can't find a mistake in either. However my answer is about three times his. Where is the mistake?

The error is in his: such a sequence can begin (for example) 01001... in which case it cannot be formed in the way he suggests.

Didn't notice that.. excellent observation  :)  With these confounded counting problems, it's easy to use an erroneous method..


parameterization
trigonometry
calculus
calculus computations

I need help finding the correct parametric representation for the surfaces:

1. The part of the elliptic paraboloid $x + y^2 + 2z^2 = 4$ that lies in front of the plane $x = 0$.

Lying in front of $x=0$ implies that we are looking for nonnegative x values only. So what I have is $y = y, z = z, x = 4 - y^2 - 2z^2$, where $y^2 + 2z^2 \le 4$. Would this be correct?

2. The part of the cylinder $x^2 + z^2 = 1$ that lies between the planes $y = -1$, $y = 3$.

My parametrization is: $x = \cos \theta$, $z = \sin \theta$, $y = y$, where $0 \le \theta \le 2\pi$ and $-1 \le y \le 3$

Could someone verify what I've done or point out any inconsistencies? Thanks.

Yes, both your representations are correct as far as I can see. My only remark is that one cannot really talk about [b]the[/b] correct representations because they are very many and one can choose different parametrizations for different purposes. So, your answers are perfectly right but there are plenty of other correct answers as well.


[quote="chien than"]Let $ x;y;z$ be positive real numbers.Prove that:
$ (x\plus{}y\plus{}xy)(y\plus{}z\plus{}yz)(z\plus{}x\plus{}xz)\geq xyz(x\plus{}2)(y\plus{}2)(z\plus{}2)$[/quote]
It's equivalent to the following inequality:
$ \sum_{cyc}(z^{2}\plus{}2z)(x\minus{}y)^{2}\geq0.$

[quote="chien than"]Let $ x;y;z$ be positive real numbers.Prove that:
$ (x\plus{}y\plus{}xy)(y\plus{}z\plus{}yz)(z\plus{}x\plus{}xz)\geq xyz(x\plus{}2)(y\plus{}2)(z\plus{}2)$[/quote]
[hide]Randomly substitute x=a-1, y=b-1, z=c-1
$ (ab\minus{}1)(ac\minus{}1)(bc\minus{}1)\geq (a^{2}\minus{}1)(b^{2}\minus{}1)(c^{2}\minus{}1)$
for which we can use rearrangement I think.
[/hide]

[quote="diophantient"][quote="chien than"]Let $ x;y;z$ be positive real numbers.Prove that:
$ (x \plus{} y \plus{} xy)(y \plus{} z \plus{} yz)(z \plus{} x \plus{} xz)\geq xyz(x \plus{} 2)(y \plus{} 2)(z \plus{} 2)$[/quote]
[hide]Randomly substitute x=a-1, y=b-1, z=c-1
$ (ab \minus{} 1)(ac \minus{} 1)(bc \minus{} 1)\geq (a^{2} \minus{} 1)(b^{2} \minus{} 1)(c^{2} \minus{} 1)$
for which we can use rearrangement I think.
[/hide][/quote]

err... you think? How would that work?

$ (ab \minus{} 1)(ac \minus{} 1)(bc \minus{} 1)\geq (a^{2} \minus{} 1)(b^{2} \minus{} 1)(c^{2} \minus{} 1)$

$ \Leftrightarrow a^2 \plus{} b^2 \plus{} c^2 \plus{} \sum a^2bc\ge ab \plus{} bc \plus{} ca \plus{} \sum a^2b^2$

$ \Leftrightarrow \sum (1 \minus{} a^2 )(b \minus{} c)^2\ge 0$

It is wrong because $ x,y,z > 0$ then $ a,b,c > 1$

[hide="bah"]All sums are SYMMETRIC.

Expand...
$ x^2y^2z^2 \plus{} \sum x^2y^2z \plus{} \dfrac{1}{2}\sum x^2y^2 \plus{} \dfrac{3}{2}\sum x^2yz \plus{} \sum x^2y \plus{} 2xyz$
$ \ge x^2y^2z^2 \plus{} \sum x^2y^2z \plus{} 2\sum x^2yz \plus{} 8xyz$.

Canceling, we get

$ \dfrac{1}{2}\sum x^2y^2 \plus{} \sum x^2y\ge\dfrac{1}{2}\sum x^2yz \plus{} \sum xyz$,

which follows from Muirhead.[/hide]

[quote="quangpbc"]$ (ab \minus{} 1)(ac \minus{} 1)(bc \minus{} 1)\geq (a^{2} \minus{} 1)(b^{2} \minus{} 1)(c^{2} \minus{} 1)$

$ \Leftrightarrow a^2 \plus{} b^2 \plus{} c^2 \plus{} \sum a^2bc\ge ab \plus{} bc \plus{} ca \plus{} \sum a^2b^2$
[/quote]

You switched the sign.


Please excuse my English. You can search Google to translate this. Thanks, I appreciate greatly.

According to Neils Henrik Abels Math Contest website, Norwegian Round 1 and Final Round had been going on since 1980. Is there any way I can access those problems (they do not have to be in Norwegian).

Thanks.  :)


Find all polynomials $P\in \mathbb C[X]$ such that \[P(X^{2})=P(X)^{2}+2P(X)\]

If $deg(P)=0$, then $P(x)\equiv 0$ or $P(x)\equiv-1$.
If $deg(P)>0$, then $P(z)=a\prod_{j}(z-z_{j})\Longrightarrow a=1$.
If $z$ is root P(x), then $z^{2}$ is root too. Therefore all roots are $exp(\frac{2\pi i k_{j}}{m})$.
If $z^{2}$ root, but $z=exp(\frac{\pi ik_{j}}{m})$ is not root we have $P(z)=-2$.
It give solutions $P(x)=x^{m}-1$.

Assume $P(x)\not \equiv-1$ and set $Q(X): =P(X)+1$.
Then
(*) $Q(x^{2})=Q(x)^{2}$   and $Q(x)\not\equiv 0$.
Take $n\in\mathbb{N}_{0}$ maximal with $x^{n}|Q(x)$ and write $Q(x)=x^{n}R(x)$ with $R(x)\in\mathbb{C}[x],\ R(0)\not =0$.
Then (*) gives
(**) $R(x^{2})=R(x)^{2}$.
If $R(x)\equiv c\not =0$ constant, then (**) gives $c=c^{2}$ and $c=1$.
This gives $P(x)=x^{n}-1$ with $n\in\mathbb{N}_{0}$. ($\equiv 0$ for $n=0$)

So assume $\deg R\geq 1$, and then take $m\in\mathbb{N}_{0}$ maximal with $R(x)=S(x^{2^{m}})$.  ($2^{m}\leq \deg R$)
Then (**) gives $S(x^{2^{m+1}})=S(x^{2^{m}})^{2}$ and $S(x^{2})-S(x)^{2}$ has an infinite number of roots.
So
(***) $S(x^{2})=S(x)^{2}$
and $S(-x)=S((-x)^{2})=S(x^{2})=S(x)^{2}$,
which gives $S(-x)=S(x)$ or $S(-x)=-S(x)$.
But $S(0)=R(0)\not =0$, so $S(-x)\not =-S(x)$.
And so $S$ is even and $S(x)=T(x^{2})$ for some $T\in\mathbb{C}[x]$, which contradicts the choice of $m$.

[i]A good solution , Olorin , congratulation :lol: [/i]

  [i]But i think , the problem should be in $ \mathbb{R}[x]$[/i]


Find the maximum and minimum value of $(x-2y+1)^2+(x+y-1)^2$ in the range of $0\leq x\leq 1,0\leq y\leq 1$.

$(x-2y+1)^2+(x+y-1)^2$

$=(x^2+4y^2+1-4xy+2x-4y)+(x^2+y^2+1+2xy-2x-2y)$

$=2x^2-2xy+5y^2-6y+2$

$=2\left(x^2-xy+\frac{1}{4}y^2\right)+\frac{9}{2}y^2-6y+2$

$=2\left(x-\frac{1}{2}y\right)^2+\frac{1}{2}(9y^2-12y+4)$

$=2\left(x-\frac{1}{2}y\right)^2+\frac{1}{2}(3y-2)^2$

The expression becomes minimum when $\left(x-\frac{1}{2}y\right)^2=0$ and $(3y-2)^2=0$.

Since $3y-2=0$, we obtain $y=\frac{2}{3}$ and $x=\frac{1}{3}$.
So the minimum is $\boxed{0}$ with $(x,y)=\left(\frac{1}{3},\frac{2}{3}\right)$.

For maximum, consider the maximum value of $(3y-2)^2$ first.
$0\le y\le 1\Longleftrightarrow -2\le 3y-2\le 1$, so $(3y-2)^2$ becomes the greatest value of $4$ with $3y-2=-2\Longleftrightarrow y=0$.

For the maximum value of $\left(x-\frac{1}{2}y\right)^2$, we want $x$ to be the largest and $y$ to be the smallest. When $y=0$ (to make $(3y-2)^2$ maximum), $x=1$ makes the expression the largest. That is, $\left(x-\frac{1}{2}y\right)^2=(1-0)^2=1$

Therefore, the maximum value of the entire expression is $2(1)+\frac{1}{2}(4)=\boxed{4}$ with $(x,y)=(1,0)$.

Actually, there's a shorter way to find the minimum.

Set $x-2y+1=0$ and $x+y-1=0$ and solve the system.
That gives us $(x,y)=\left(\frac{1}{3},\frac{2}{3}\right)$ and that satisfies $0\le x\le 1$ and $0\le x\le 1$. So $0$ is the minimum.

[quote="kunny"]What can you imagine  for sum of  the square :)[/quote]
I haven't put it in practice, but I think it is like a circle's formula. Also, we can try trig substitution.


There must be something wrong... ;)  Just choose $x_i = 0$ for all $i$....

Pierre.


Please answer if you're like me, who didn't mark answers on the test packet. :(

And if you're like me, and didn't even get to keep your test packet.  :mad:

Only Mr. Dunbar (AMCDirector) can answer that, but it should be in the next 2-3 weeks... I think.

I heard that it takes quite a while...?
It did for the AMC 8... :maybe:

Never, ever quickly enough. For any test. Ever.

[quote="pianoforte"]C'mon guys, they have to first get all the tests, then scan the scores of thousands of them, check any abnormals ones, and then work on statistics. Be patient! It will come when it comes.[/quote]
No, I want it now!  :(  :D

Hm...after the test the only thing they took from me was the bubble sheet...then they gave me the solutions booklet.

Please read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=189123]my announcement[/url].


let a,b be distinct real roots of (x-1)(x^2+1)(x^3-1)+30x^3=0,find a^2+b^2

$ (x\minus{}1)(x^2\plus{}1)(x^3\minus{}1)\plus{}30x^3\equal{}(x^2\plus{}3x\plus{}1)(x^4\minus{}4x^3\plus{}12x^2\minus{}4x\plus{}1)$
$ (x\minus{}1)(x^2\plus{}1)(x^3\minus{}1)\plus{}30x^3\equal{}(x^2\plus{}3x\plus{}1)((x\minus{}1)^4\plus{}6x^2)$
then $ a$ and $ b$ are the roots of $ x^2\plus{}3x\plus{}1$ with $ ab\equal{}1$ and $ a\plus{}b\equal{}\minus{}3$
then $ a^2\plus{}b^2\equal{}(a\plus{}b)^2\minus{}2 ab\equal{}7$

but can I ask how do you think of factorizing like that

Maybe this explanation helps:

Expand the expression $ P(x)\equal{}(x \minus{} 1)(x^2 \plus{} 1)(x^3 \minus{} 1) \plus{} 30x^3 \equal{} x^6\minus{}x^5\plus{}x^4\plus{}28x^3\plus{}x^2\minus{}x\plus{}1$

Check for rational roots, only possibilities are $ \plus{}1$ or $ \minus{}1$ with $ P(1)\equal{}30$ and $ P(\minus{}1)\equal{}\minus{}22$ then no rational factorization like $ P(x)\equal{}(x\minus{}a)Q(x)$

Then lets try to factor like this $ P(x)\equal{}(x^2\plus{}cx\plus{}d)(x^4\plus{}ex^3\plus{}fx^2\plus{}gx\plus{}h)$
equaling all the coefficients and solving the system for $ P(x)\equal{}(x^2\plus{}3x\plus{}1)(x^4\minus{}4x^3\plus{}12x^2\minus{}4x\plus{}1)$


If you could describe yourself in one word, what would it be?

I picked Social.

Uh, haha, I put beautiful.

Smart: Somewhat.
Truthful: No.
Responsible: Nope.
Likeable: Not sure.
Evil: Sometimes.
Curious: Definitely.
Beautiful: I don't know; I just put that.
Athletic: I have karate three times a week (1-2 hours), tennis on the weekends (two hours), and run about two miles every day. Am I athletic?
Social: Not really (especially since I started homeschooling).
Flirty: Haha, I can be a bit coquettish at times.

[quote="hunter34"]Where is WTF?[/quote]

There wasnt enough room  :wink:

Smart: Smartness is there somewhere. My brain hasn't developed.
Truthful: The lies I tell... :noo:
Responsible: Yeah, a bit.
Likeable: I'm likeable, but not insanely popular that everyone wants to know me.
Evil: A lot of times.
Curious: Sometimes.
Beautiful: Eh, I guess so.
Athletic: One of the most athletic girls in the school. I can play every sport at once and not be tired.
Social: Yeah I like to talk to my friends rather than smother myself in academics.
Flirty: A little..

There really is no one word to describe me...I'm a mix of everything.

Smart: maybe...everyone else seems to think so...
Truthful: Haha...usually to friends, at least
Responsible: Not very
Likeable: Well, people think I'm funny and nice...so I guess...
Evil: Haha...not really, at least not on purpose
Curious: Yeah, but it's not really me...
Beautiful: ...WTF, no idea...
Athletic: Sort of...like sub-6 mile and okay at swimming...
Social: Yeah...but not popular or anything
Flirty: Um...not exactly

mildly intelligent and extremely anti-social.

WELL I AM NOT AT ALL LIKEABLE BUT THAT WORKS AS AN ADVANTAGE TO ME, WHEN IT COMES TO SPRINGING SURPRISES.

I can see why... STOP SHOUTING

Why am I the only one who picked Evil???

Smart: Which one?
Truthful: 2/3 of the time
Responsible: 1/2 of the time
Likeable: For some people
Evil: All the time
Curious: A question here or there
Beautiful: What a beautiful boy!
Athletic: Now really
Social: homeschooled. Of course I have friends! There's a whole homeschool group I go to every Friday in the MathCounts season.
Flirty: And just HOW can a boy flirt?!

[quote="1=2"]
Flirty: And just HOW can a boy flirt?![/quote]

Uhhh.... guys can flirt too, guy flirting is different than girl flirting.  :wink: **cough cough**.... and besides its not all guys on these forums.

@1=2: u r NOT the only one who picked evil.

smart: compared to SOME people, possibly...
truthful: about 1/1000000000 of the time...
responsible: about 1/10000000000 of the time...
likeable: never been applied to me.
evil: that's my job, what do u expect?
curious: only about swords
beautiful: never been applied.
athletic: what's THAT mean???
social: never heard of it.
flirty: ...

i'm not exactly the ideal student/friend/daughter... hehe  :lol:

I put smart

Smart    : hm... I'm smart :)  (everybody who knows me say that , so it's not only my mind)

Truthful   : not always

Responsible  : ...

Likeable    ...

Evil    : sometimes

Curious    :  yes :)

Beautiful    : yes I think (not only I ;)

Atheletic   : may be

Social    : don't know

Flirty : ....

uhmm...

Smart: moreless...
Truthful: no.
Responsible: WHAT???
Likeable: no.
Evil: no (?)
Curious: moreless...
Beautiful: NO!
Athletic: no.
Social: not so...
Flirty: no.

Conclussion: No choice.

Smart-umm ftw makes me look retarded :D  ppls at skool seem to THINK (notice the think) i'm smart
Truthful- if i feel like it.
Responsible-umm kinda not really
Likeable- well my friends like me...but i'm not popular or anything
Evil- muahahahhahahah! no.
Curious-depends... yeah i can b curious
beautiful-nope at least i don't think so
athletic-umm no way
social-wait what?
flirty- no.

I apply to a lot of em.

Smart: Kind of.
Truthful: Sometimes.
Responsible: Sometimes
Likeable: A good amount.
Evil: Not really.
Curious: Definitely.
Beautiful: I'm not sure, I think I am.
Athletic: I think I am. I play soccer, b-ball, track and field, and other random sports at home and away. I can also run a 6:20 mile.
Social: Pretty much, have upwards of 15 friends.
Flirty: Not really.

No conclusion.

Smart: yeah, let's go with that.
Truthful: you serious?
Responsible: I procrastinate. I'm lazy.
Likeable: yep
Evil: naturally
Curious: somtimes. most of the time I couldn't care less
Beautiful: [url=http://www.artofproblemsolving.com/Forum/album.php?mode=pic&id=14637]of course[/url]
Athletic: meh not really. not if you count the occasional tennis and more common Xbox.
Social: yep
Flirty: haha sometimes  :P

Smart: (Woo for IB) The tests say yes, but I have my moments of stupidity (Woo for blonde moments!)
Truthful: Yup, I have no guilty conscience what-so-ever
Responsible: Very, see above ^, if I don't get something done (even homework) I feel horrible (which stinks)
Likeable: Depends on who you are, but mostly yes, I only have a few people who hate my guts (That I know of  :maybe: )
Evil: Only to my little brother  :wink:
Curious: About some things (like math and science related) but other than that, I don't care
Beautiful: Uh.... Pass? I really have no idea
Athletic: Remotely, my friends poke fun at my runner's legs, saying that I beat them all in athletics  :roll:
Social: Not really, I'm really shy in person
Flirty: Not exactly

I went with smart


trigonometry
inequalities
algebra unsolved
algebra

Let $ n$ be a positive integer. Find all $ x_1,x_2,\ldots,x_n$ that satisfy the relation:
\[ \sqrt{x_1\minus{}1}\plus{}2\cdot \sqrt{x_2\minus{}4}\plus{}3\cdot \sqrt{x_3\minus{}9}\plus{}\cdots\plus{}n\cdot\sqrt{x_n\minus{}n^2}\equal{}\frac{1}{2}(x_1\plus{}x_2\plus{}x_3\plus{}\cdots\plus{}x_n).\]

$ x_{n} \equal{} 2 \cdot n^2$ :)

My solution:
Put $ x_{i} \equal{} i^2 \cdot \sec^2 {y_{i}}$
Then it changes to:
$ 1^2 \cdot \tan {y_{1}} \plus{} 2^2 \cdot \tan {y_{2}} \plus{} \cdots \plus{} n^2 \cdot \tan {y_{n}} \equal{} \frac {1}{2} \cdot (1^2 \cdot \sec^2 {y_{1}} \plus{} 2^2 \cdot \sec^2 {y_{2}} \plus{} \cdots \plus{} n^2 \cdot \sec^2 {y_{n}})$
$ \implies 1^2 \cdot \tan {y_{1}} \plus{} 2^2 \cdot \tan {y_{2}} \plus{} \cdots \plus{} n^2 \cdot \tan {y_{n}}$
$ \equal{} \frac {1}{2} \cdot (1^2 \cdot (1 \plus{} \tan^2 {y_{1}} \plus{} 2^2 \cdot (1 \plus{} \tan^2 {y_{2}} \plus{} \cdots \plus{} n^2 \cdot (1 \plus{} \tan^2 {y_{n}})$

Now By AM GM inequality for each term we get the result :)

$ \sum_{k\equal{}1}^nk\sqrt{x_k^2\minus{}k^2}\equal{}\frac{1}{2}\sum_{k\equal{}1}^nx_k\iff \sum_{k\equal{}1}^n(\sqrt{x_k\minus{}k^2}\minus{}k)^2\equal{}0\iff x_k\equal{}2k^2\ \forall k\in [n]$

By AM-GM inequality ,
$ \frac{(k^2) \plus{} (x_k \minus{} k^2 )}{2} >\equal{} k  \sqrt {(x_k \minus{} k^2)}$

$ \equal{}>    \frac{x_k }{ 2} >\equal{} k \sqrt  {(x_k \minus{} k^2)}$

Sum this for all values of  k we get the given expression , with the equality sign.
This means that equality holds for each equation.

Equality holds for AM-GM inequality when , the 2 numbers are equal

So $ k^2 \equal{} x_k \minus{} k^2$

$ x_k \equal{}2 k^2$


How to prove Det(AB)=Det(A)*Det(B) ?
I need a detailed proof!

thank you!

See here:

http://www.mathlinks.ro/Forum/viewtopic.php?t=56815

what are the matrices for each of the row operation?

Illustrating with $3\times 3$ matrices:

Left-multiplying $A$ by $\begin{bmatrix}1&0&0\\0&1&0\\3&0&1\end{bmatrix}$ adds $3$ times the first row of $A$ to the third row of $A.$

Left-multiplying $A$ by $\begin{bmatrix}1&0&0\\0&-4&0\\0&0&1\end{bmatrix}$ multiplies the second row of $A$ by $-4.$

Left-multiplying $A$ by $\begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix}$ interchanges the first and second rows of $A.$


Let f(n,k ) be the number  of ways of distributing k candies to n children so that each child receives at most 2 candies
 Determine value of :
$ f(2007,1 ) \plus{} f(2007,4) \plus{} f(2007,7) \plus{} ... \plus{}f(2007,4012)$

Let $ F(x) \equal{} \sum_{k\equal{}0}^{4014} f(2007, k) x^k \equal{} (1 \plus{} x \plus{} x^2)^{2007}$.  (This is because each of the $ 2007$ children can receive $ 0, 1, 2$ pieces of candy and there are $ k$ total.)  We want to find the sum of every third coefficient, starting from the $ x$ coefficient.  Now, the constant term is evidently $ 1$; moreover, we know how to find the sum of every third coefficient starting from the $ 0$ coefficient, so consider

$ G(x) \equal{} \frac{F(x) \minus{} 1}{x}$.

The sum we are looking for is

$ \frac{G(1) \plus{} G(\omega) \plus{} G(\omega^2)}{3} \equal{} \frac{3^{2007} \minus{} 1 \minus{} \omega \minus{} \omega^2}{3} \equal{} 3^{2006}$.


calculus integration logarithms function Euler induction limit

what is 1/1+1/2+1/3+.................+1/n.Is there any formula for it ?
thank alot.

it can be proved that this series is divergent that is it doesn'tconerge to a no. as n becomes large. Although this is not rigorous, it indicates that there is no closed formula for this sum

That indicates nothing. For example form $n$ is a closed form formula for sum partial sums 1+1+1+...+1 and the series $\sum_{n=1}^\infty 1$ diverges. However, at least I don't know closed form other than $\zeta (3)$ for the sum $\sum_{n=1}^\infty n^{-3}$ but it is possible to show that this series is convergent.

What is the sharpest bound,I mean what's the smallest k that you know and
K>=1/1+1/2+1/3+...+1/n.

I dont think theres a formula to express such thing but I once saw something like this

$\frac 11+\frac 12+\frac 13+\frac 14+ ...
= \frac 11+\frac 12+(\frac 13 +\frac 14) +(\frac 15+\frac 16+\frac 17+\frac 18)+....
>1+\frac 12+\frac 12+\frac 12+\frac 12+...$

Well, integrals and logarithms should give you something better ;)

Maybe something like
[tex]ln(x)<\sum_{x=1}^n \frac 1 x<ln(x)+1[/tex] (I think this can be improved). The exact formula is
[tex]\sum_{x=1}^n \frac 1 x=\Psi(x+1)+\gamma[/tex], where  [tex]\Psi[/tex] is a digamma function (the definition is on http://mathworld.wolfram.com/DigammaFunction.html  ) and  [tex]\gamma[/tex] is Euler's constant (0.5772156649...)

We have this asymptotical (check this if you don't know the Big O notation : http://mathworld.wolfram.com/AsymptoticNotation.html) result [tex]\sum_{j = 1}^n \frac{1}{j} = \ln n + \gamma + \frac{1}{2 n} - \frac{1}{12 n^2} + \frac{1}{120 n^4} + O ( \frac{1}{n^6} )[/tex], it can be made as precise as we want using the Euler-Maclaurin Summation formula ( http://en.wikipedia.org/wiki/Euler-Maclaurin \_ formula ), and [tex]\gamma = 0,577215664901532860 \ldots[/tex] ( http://mathworld.wolfram.com/Euler-MascheroniConstant.html ) is called the gamma constant (or Euler's constant).

Well, the sequence $a_{n}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$ is divergent and its limit is $\infty$.
 This can be proved in two other ways like it follows:

  S1. It is easy to prove by induction that $a_{n} >ln(n+1)$, for any $n\geq 1$
  S2. Let's consider the sequences $b_{n}=ln n$ and $c_{n}=\frac{1+\frac{1}{2}+...+\frac{1}{n}}{ln n}$.

  By Stolz-Cesaro lema applyed to $c_{n}$ iit follows that $ c_{n}%Error. "arrow" is a bad command.
1$ and since $b_{n}%Error. "arrow" is a bad command.
\infty$ the conclusion follows. ;)

[quote="spix"]It can also be proved using the fact that $a_n=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$ is not a Cauchy sequence.[/quote]
What¿ The asymptotical behavior¿

$a_n$ converges if $\forall \epsilon > 0, \exists n_0 \in \mathcal l N$ such that $\forall m,n \in \mathcal l N$such that $m>n>n_0$ then $|a_m-a_n|<\epsilon$. If we take $\epsilon=\frac{1}{2}$, $m=2n+1$ we should write $|\frac{1}{2n+1} +\frac{1}{2n}+...+\frac{1}{n+1}|<\frac{1}{2}$ which is false so $a_n$ diverges and is increasing so $\lim_{n\to\infty} a_n=+\infty$.


how can i solve such a thing

for example how could i solve (3i + 5)^(1/3)

thanks
-nate

When you say, "Solve," do you mean evaluate?  You should try converting it to trigonometric form, or, even better, to the form $r\cdot e^{i \theta}$.

To add to JBL's clue -- do you know what De Moivre's theorem says?  One you know how to write complex numbers in the form JBL suggested, De Moivre's theorem is an extremely powerful tool for working with complex numbers.


[hide="major hint for 8"]
move all the log terms to the left side. you get $ \log^2 x\plus{}2\log x\minus{}3\equal{}$ now make the substitution $ y\equal{}\log x$. oh.. we seem to have gotten a factorable quadratic.  :wink: [/hide]

$ \ln x \equal{} \log_e x$.  It's that simple.

[hide="Hint for 17"] What you want to know is the base-change formula, which states that

$ \log_a b \equal{} \frac {\log_c b}{\log_c a}$

for [b]any[/b] $ c$.  In particular,

$ \frac {\ln b}{\ln a} \equal{} \log_a b \equal{} \frac {\log 60}{\log 72}$. [/hide]

[quote="Casa De Mathematica"]I've been through normal log material in algebra 2 class in school, but for some reason I don't know how to do these problems:

$17)$ Given that a=log2 and b=log3, find ln60 / ln72.

$8)$ Given the equation $log^{2} x + log x = 3 -$ $log$$x$, find the product of the solutions.

A lot of explanation would be awesome! I don't know much about ln, except that it has to do with e.[/quote]

$8)$ we have $log^{2} x + 2log x -3=0, so (log x -1)(log x +3), so log x=1 or log x=-3, and x=10 or 1/1000

[quote="mihail911"][hide="major hint for 8"]
move all the log terms to the left side. you get $ \log^2 x \plus{} 2\log x \minus{} 3 \equal{}$ now make the substitution $ y \equal{} \log x$. oh.. we seem to have gotten a factorable quadratic.  :wink: [/hide][/quote]

The quadratic doesn't need to be factorable.  Given the roots $ a, b$, we are asked to calculate $ ab \equal{} 10^{\log a \plus{} \log b}$, and we can calculate $ \log a \plus{} \log b$ by Vieta's.

[quote="gauss1181"][quote="Casa De Mathematica"]I've been through normal log material in algebra 2 class in school, but for some reason I don't know how to do these problems:

$ 17)$ Given that a=log2 and b=log3, find ln60 / ln72.

$ 8)$ Given the equation $ log^{2} x \plus{} log x \equal{} 3 \minus{}$ $ log$$ x$, find the product of the solutions.

A lot of explanation would be awesome! I don't know much about ln, except that it has to do with e.[/quote]

$ 8)$ we have $ log^{2} x \plus{} 2log x \minus{} 3 \equal{} 0, so (log x \minus{} 1)(log x \plus{} 3), so log x \equal{} 1 or log x \equal{} \minus{} 3, and x \equal{} 10 or 1/1000$[/quote]

ln is the natural log. We just studied that a couple of days ago....basically, a natural log ( notation : ln) has the natural base e = 2.718.....anyway..so if you had ln e 78 , the answer would be 78, because using the natural log property, the base e and exponent e would cancel out to be 1, and 1 * 78 = 78.

[hide="I Think This Is The Solution For 17"] Because of the change of base formula previously mentioned,
$ \log x \equal{} \ln x$

Now we can rewrite $ \frac {\ln 60}{\ln 72}$ as $ \frac {\log 60}{\log 72}$

Breaking down the two logs, you can write the numerator as
$ \log 2 \plus{} \log 3 \plus{} \log10$ which simplifies into $ a \plus{} b \plus{} 1$
and the denominator as $ \log 2^3 \plus{} \log 3^2$ which simplifies into $ 3a \plus{} 2b$

Thus we get the answer $ \frac {a \plus{} b \plus{} 1}{3a \plus{} 2b}$[/hide]

MathleteWannabe wrote:
[quote]Because of the change of base formula previously mentioned,
$ \log x \equal{} \ln x$[/quote]
That's true only for $ x \equal{} 1$.

His solution still applies:
$ \dfrac {\ln60}{\ln72}\equal{}\log_{60}72\equal{}\dfrac {\log60}{\log72}$

[quote="Brut3Forc3"]His solution still applies:
$ \dfrac {\ln60}{\ln72} \equal{} \log_{60}72 \equal{} \dfrac {\log60}{\log72}$[/quote]

You mean $ \dfrac {\ln60}{\ln72} \equal{} \log_{72}60 \equal{} \dfrac {\log60}{\log72}$

yes. I was wrong in saying logx = lnx, but lnx/lny = logx/logy. My mistake. It doesn't change my solution


quadratics
geometry
3D geometry
algebra
quadratic formula
factorization
difference of cubes

1. If the reciprocal of x+1 is x-1, determine x.

2. When six gallons of gasoline are put into a car's tank, the indicator goes from 1/4 of a tank to 5/8. What is the total capacity of the gasoline tank.

3. Given 2x = 8y+1 and 9y = 3x-9, find the value of x + y.

4. Find all real values of x that satisfy (x2 -5x + 5)x2-9x+2 = 1

5. Find two fractions that are evenly spaced between these two: 3/a and 7/(2a).

Challenge:
Find x2 if  :^3::sqrt:(x+9) -  :^3::sqrt:(x-9) = 3.

(If these problems look easy to you, leave them for someone else.)

1. [hide]1/(x+1)=x-1

x :^2: -1=1

x :^2: =2

x= :rt2: [/hide]

Gotta go sleep before i do the others

yif man12 wrote:1. [hide]1/(x+1)=x-1
x :^2: -1=1
x :^2: =2
x= :rt2: [/hide]

[hide]That's good so far yif, but how many numbers when squared can give you 2?[/hide]

bah. my bad, typical mistake. [hide]:pm: :rt2: [/hide]

2.[hide]6=5/8x-1/4x

6=3/8x

1/8x=2

x=16 gallons[/hide]

3.[hide]x=3y+3

substitute in to get

2y=3y+3-9

y=6

x=21[/hide]

I guess i'll leave the rest to others

4

[hide]

there are three ways to get 1.

1. If the exponent is 0 and the base is nonzero

2. If the base is 1

3. If the base is -1 and the exponent is even

for 1. setting the exponent equal to zero

x :^2: - 9x + 2 = 0.

quadratic formula gives the answers as

( 9 :pm: :sqrt: 73 ) / 2

checking in the base satisfies that the base is not equal to 0, so answers are valid.

for 2. setting the base equal to 1

x :^2: -5x+5 = 1

x :^2: -5x+4 = 0

(x-4)(x-1) = 0. so solutions are 4, 1

finally, for 3.

x :^2: -5x+5 = -1

x :^2: -5x+6 = 0

(x-3)(x-2) = 0. solutions: 3,2

checking in exponent yields even exponents.

thus, final answer is:

( 9 :pm: :sqrt: 73 ) / 2, 1, 2, 3, 4

[/hide]

Good problems, I love algebra.

5. Find two fractions that are evenly spaced between these two: 3/a and 7/(2a).

[hide]

(3/a)(2/2) = 6/(2a)

6/ 2a and 7/2a  multiply by (2/2)

6 / 2a, 6.3333333/ 2a, 6.66666666/ 2a, 7/ 2a

6 / 2a , (6+(1/3))/(2a), (6+(2/3))/(2a), 7/2a

3/ a, 19/ 6a, 20/ 6a, 7/ 2a

[/hide]

??probably wrong.

Challenge:

Find x2 if (x+9)^(1/3) - (x-9)^(1/3)= 3.

I got [hide] 7- 3rad(5) [/hide] but that looks way off...

For the second one, why not show what you did?

Well its a ton of manipulations.

[hide]

First I moved  -(x-9)^(1/3) to the other side.  Then I cubed both sides.  I simplified it out, and then made the substitution: let y = (x-9)^(1/3).  This equation has quadratic form.  Then I used the quadratic formula.  Then I cubed the y solution and added 9 to get x.  Then to get x^2 I squared my x-solution.

[/hide]

There's a trick, I just dont see it.

JBL -- is #5 right?

hint: [hide] get rid of the 1/3 powers. whats the easiest way to do taht? you dont need to take anything over to the rhs. and look for something familiar, once you rid yourself of those 1/3 powers[/hide]

I def. can't see it.  I have been thinking about it for a while.  I thought about the difference of cubes pattern but this is cube roots.  Can anyone give me another hint?

Well, I'm not entirely sure what s0mp was getting at, but this might be it:  [hide]define a and b by (x - 9) = a3, (x + 9) = b3.[/hide]  That will make things look much nicer.  Also, [hide]When you've got an equation for b, use it to simplify the expression for x2.[/hide]

Good.  The other suggestion I made was this:

x = b^3 + 9

b^2 + 3b + 1 = 0
b^3 + 3b^2 + b = 0
So x = 9 - 3b^2 - b, by subtracting 0.
3b^2 + 9b + 3 = 0
So x = 12 + 8b, by adding 0.
Then, when we solve for b and plug in to get x^2, we don't have to square the cube of something ugly, we can square something nice.

I need help with number 1.

[hide]I spent 5 minutes on this problem, became frustrated, and tried messing around and playing with numbers and I tried out x+1/1=1/x-1, which gave me x= :rt2:

But I tried this with regular numbers and it doesn't really work out.  Can anyone help me?[/hide]

Nightflarer, it seems like you got the answer. It's [hide]+/-sqrt(2)[/hide].

Night -

Quote:If the reciprocal of x+1 is x-1, determine x.

HINT:

[hide]Okay, so remember the definition of a recripocal:

(number)(its recripocal)=1

Then it's just solving a basic equation.[/hide]

[/hide]


Since no one seems to have done this, I figured I would start a poll.

So how well did everyone do (in comparison to their expectations)? I got a 12, beats the 7 I was hoping for. :lol:

Figures I got the exact lower bound for honorable mention. 19. :blush:

So wait, how many points were given for getting to the point of having to prove that $x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1$ is composite on number 5?

I got an 11. I dont know where that came from 7 for #1 how many points for a handwavey number 4 and number 5 up to the factored part?

[quote="azjps"]Just noticed that the 42 option got cut off.  Sorry for you all who (obviously) scored a 42. :P

Looking through the scores, is the highest score indeed a 32? I can't find any higher scores on the list.[/quote]
Yes, the highest score is indeed a 32 :o

11.  :huh:

Kind of unexpected, as I only solved #1 correctly.

[quote="Hamster1800"]18! Yay! Double the red MOP cutoff (yes I'm a freshman)

On another note, I'm depressed, since I missed HM by 1 point :(.[/quote]
But don't you get blue MOP anyways?

22.  :) Too bad I missed Winner by 1 pt. Hopefully I will get there next yr...

14. missed blue by 3 (soph).  but hey, that's what next year is for.  wahoo.  the joy is dripping over my mandible.

although kinda unexpected, seeing as my proof for #2 sucked, i didn't really get anywhere on 3,5,6, i know i got #1, and i thought i handwaved on #4.  beats last year's 7, however.

15, exactly as I predicted :), although 2 below MOP cutoff...

I solved 1 and 4, and i think they gave 1 point for getting up to the proving-the-polynomial-composite on problem 5

15 - just enough points to be able to check the box right under "honorable mention." Cool. I'm a senior and first-time USAMO taker who literally did no studying whatsoever, so I'm happy. Good luck to everyone next year, and good luck this summer to those who made MOP. Btw, do we ever find out how our points are distributed (I got about what I expected, but would be curious to see what recieved how many points).

[quote="matt276eagles"]Turns out I got exactly 17.  :)[/quote]

Me too.  :)

As a 16er, I feel it would be a sensitive move to separate 15-16 and 17-18 in the poll. :P

Oh wow.  I got a 23.  And the top scores are, according to my tally,

32, 28, 28, 26, 25, 25, 25, 24, 24, 23, 23, 23 22, 22, 22, 22, 22, 21, 20, 20, 19, 19, 19

That's some really close competition.  The partials really DID come through...

Yes, you are right Krishanu Sankar. The race is very close.

Blah- lowest score on this thread (8.)

Also, 2 less than last year.

Blah.

[quote="MysticTerminator"]attention david bay rush: the race is very close[/quote]

ok dude don't turn into alex...
anyway what does this have to do with me?  zuming promised he'd just pick me for IMO team anyway

I got 14.  Which is pretty cool because I was predicting 10-14.

I got 10 as a first timer in 10th grade. I predicted 9.

Graaahh... 22

one point away...

How likely do you think it is that they'll lower the black MOP cotoff?  :P

I'd really like if they did by a point (Zuming might not think it's a bad idea either) but considering the TST is like actually in DC and they'd have to fly everyone up and stuff, that doesn't seem terribly likely

so ... try convincing one of the winners to go to RSI

or physics

or something

I got a 10... Too bad I'm a sophomore and not a freshman =(

yuk...11.

Well, I solved 1, hand-waved 4, set up 5 but couldn't get the factorization, and proved that some points in problem 6 were collinear using homothety.

So probably, 7, 0, 0, 2, 1, 1.

Well, at least I increased my score by a factor of 11 compared with 2 years ago :D  (the last time I took USAMO).

btw, what did bubala get?  I know him from AMP last summer.

I find it interesting that already 6 of the 12 winners have posted on this thread.

[quote="diophantient"]I got a 9 in 8th grade...
Hopefully I can make red mop next year :)[/quote]ditto :D

[quote="B-tone_Party"]Blah- lowest score on this thread (8.)[/quote]

I "beat" you... with a 4! (which also happens to be higher than my expectations (0) so I'm really not all that 'blah' to be "winning" the lowest score race)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $S_{999}$ be a permutation group on 999 objects. Let $G$ be a subgroup of $S_{999}$ of order $1111$. Show that there is $i\in\{1,2,\cdots,999\}$ so that $\sigma(i)=i$ for every $\sigma\in G$.

the subgroup is cyclic. Let the generator be a. Its order is 1111 = 11  x 101. If there is no element being fixed, then when written as product of disjoint cycles, the cycle length has to be either 11 or 101. However, 101a + 11b = 999 does not have positive integer solution. Contradiction.

[quote="barra"]$1111$ is correct. It is possible to have a subgroup of that order since $1111\mid 999!$[/quote]
How do you know that the group is not abelian? The converse theorem of Lagrange doesnt hold.

Anyway lets assume there is such a subgroup.
Let G act on the set X={1,2,...,999}
Then by the class equation,
$X\equiv X_{G}(\mod 1111)$
(I know if works for primes, I'm not sure if it works here. I'll try to show that)
Thus,
$999\equiv X_{G}(\mod 1111)$
Thus,
$X_{G}\not = 0$
Thus, there is a fixed element.

[quote="Soarer"] the subgroup is cyclic. Let the generator be a.[/quote]
How do you know it's cyclic?

sylow_theory: yout proof doesn't work and in fact there is such a subgroup.

blue_se7en: the number of $ 11$-sylows and $ 101$-sylows is $ 1$ each. Thus the subgroup is a product of these sylows. As groups of prime order are cyclic and their group orders are coprime, the whole subgroup is cyclic.


Hey, can someone tell me whats the pick's theorem? I forgot the equation.

[url=http://mathworld.wolfram.com/PicksTheorem.html]http://mathworld.wolfram.com/PicksTheorem.html[/url]

In general, if you type in the name of any mathematical equation, concept, etc. followed by "mathworld" into google, you'll get mathworld's explanation.

I guess the link says all but just for reference:

$A = I + \frac{\ell}{2}-1$

Where $A$ is the area, $I$ is the number of lattice points inside the figure, and $\ell$ is the number of points on the sides of the polygon. This, and the shoelace theorem can make ugly coordinate geomoetry pretty quick.


an area of a semicircle with diameter of 2 inches is $ 1*1*\pi*\frac{1}{2}\equal{}\frac{\pi}{2}$

there's total 24 semicircles b/c the pattern is 12 inches long.

So the area of the pattern is $ 24*\frac{\pi}{2}\equal{}12\pi$

If we split the figure in half horizontally and shift the top half 1" to the right or left and wrap around the extra, we have 12/2=6 circles with diameter 2. The area of one of these circles is $ \pi$, so the area of the figure is $ 6\pi$.

[quote="FantasyLover"]an area of a semicircle with diameter of 2 inches is $ 1*1*\pi*\frac {1}{2} \equal{} \frac {\pi}{2}$

there's total 24 semicircles b/c the pattern is 12 inches long.

So the area of the pattern is $ 24*\frac {\pi}{2} \equal{} 12\pi$[/quote]

Yeah, dragon is right.  You have 12 semicircles, I believe.


Let $P(z)= z^{n}+c_{1}z^{n-1}+c_{2}z^{n-2}+\cdots+c_{n}$ be a polynomial in the complex variable $z$, with real coefficients $c_{k}$. Suppose that $|P(i)| < 1$. Prove that there exist real numbers $a$ and $b$ such that $P(a+bi) = 0$ and $(a^{2}+b^{2}+1)^{2}< 4 b^{2}+1$.

[hide="Proof"]
If $r_{1},r_{2},r_{3},\dots,r_{n}$ are the real roots of polynomial $P(z)\in\mathbb{C}$, then, $P(i)=(i-r_{1})(i-r_{2})(i-r_{3})\dots(i-r_{n})$. Since $P(i)<1$, $|i-r_{i}\parallel i-r_{2}\parallel i-r_{3}|\dots|i-r_{n}|<1$. The coefficients of $P(z)$ are real, and its non-real roots can be found, where pairs consists of two conjugates. However, if $r$ is in the reals, then $|i-r|=\sqrt{1+r^{2}}$. This is greater than 1. Therefore, there are 2 reals, namesay $a$ and $b$, such that $P(a+ib)=0$. Also, $|i-(a+bi)\parallel i-(a-bi)|=\sqrt{(a^{2}+b^{2}+1)^{2}-4b^{2}}<1$. Therefore, $(a^{2}+b^{2}+1)^{2}=4b^{2}+1$. $\blacksquare$
[/hide]


For how many different digits $ n$ is the two-digit number $ ``6n\text{''}$
divisible by $ n$?

We want $ 6n$ to be divisible by $ n$.  It works for $ n\equal{}1,2,3,4,5,6$ by testing.  There are $ \boxed{6}$ numbers.

Another way, although somewhat slower for this problem, is to write $ \overline{6n}$ as $ 60\plus{}n$, then we want $ \frac{60\plus{}n}{n}\equal{}\frac{60}{n}\plus{}1$ to be an integer. Therefore, $ n$ is a factor of 60, and the only digits that work are $ 1,2,3,4,5,6$, so $ \boxed{6}$.


I have discovered what I think is an error in Art of Problem Solving Volume 2: and Beyond.  I was wondering if there was an errata somewhere, or if this has been addressed before.

Please email it to us at books@artofproblemsolving.com.

yes, great, but can you answer the question: is there an errata page on this website for the Art of Problem Solving, Vol 2?   I see you have such pages for other books but I have been unable to locate the errata for this book.  thanks...sing5

To my knowledge, I don't think there is an errata page for Volume 1 either.....  :(

I've already asked this ([url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=291905]see this[/url]), but I didn't get a response......


[quote="A+MATH"]Please but the ^2 sign in the starting emoticons it's used SOOOOOO much.[/quote]

You can also use the HTML functions for superscript and subscript.

(with spaces)
In order to write x2 you can type
x  2
but without spaces.

Likewise, in order to write x2 you can type
x  2 .

This gives you the advantage of being able to type out a larger array of expressions.

just wondering, which starting emoticons did you take out (if you did take any out), or did you just add some more to make a row?

A few of the Greek letters were replaced (they're still available as emoticons, but are not among the set you see listed while you're posting).


Can be true Schur for more variables? I mean not only neccessary for three variables?

What do you think?

I think suranyi's inequality is one of generalazition of schur,Anyway,I believe Vasc has post another generalazations,But I can't find it now :(

Well, there is :
http://www.mathlinks.ro/Forum/viewtopic.php?t=40682
and
http://www.mathlinks.ro/Forum/viewtopic.php?p=165582#p165582 (the proof is here : http://www.mathlinks.ro/Forum/viewtopic.php?t=14906)

:)

I'm afraid you don't understand me. My question is, if the following inequality is true or not.

The Schur's statement is, that my suggested inequality is true for n=3.

My suggested inequality is: let $x_{i}$ is positive real for every $i=1,2,...,n$ and let $t$ is arbitrary real. Then:

$\sum_{i=1}^{n}\left(x_{i}^{t}\prod_{1 \leq j \leq n, j \not= i}(x_{i}-x_{j}) \right) \geq 0$

Is that true for every natural n? That's my question.

I don't think it is correct.
See http://www.mathlinks.ro/Forum/topic-1166.html at some posts at the last I think.


Let $ X \equal{} \sqcup^n_{i\equal{}1} \: X_i \equal{} \sqcup^n_{j\equal{}1} \: Y_j$ be two disjoint decompositions with all sets $ X_i$'s and
$ Y_j$'s having the same size. Prove that there exist distinct elements $ x_1, x_2, ..., x_n$ which are in different sets in both the decompositions.

[quote="srikanth"]What is decomposition of a set ?[/quote]

I think he means "partition".

(And the problem follows from Hall's marriage theorem, applied to the boys $ X_1$, $ X_2$, ..., $ X_n$ and the girls $ Y_1$, $ Y_2$, ..., $ Y_n$ such that $ X_i$ likes $ Y_j$ if $ X_i\cap Y_j\neq\emptyset$.)

  darij

i still don't get it. can you give me a complete detailed solution?


[quote="mathwizarddude"]Find the number of real solutions to
$ 2\cos(x^2 \plus{} x)/6 \equal{} 2^x \plus{} 1/2^x$[/quote]
I graphed it and i get none O_O.

[quote="mathwizarddude"]Find the number of real solutions to
$ 2\cos(x^2 \plus{} x)/6 \equal{} 2^x \plus{} 1/2^x$[/quote]

For $ x$ real, $ R.H.S. \ge 2$ by AM-GM
but $ L.H.S. \le \frac{2}{6}<2$, so there are no real solution for the equation.