The problem:  When some positive two-digit integers are squared, the last two digits of the square are teh same as the original two-digit integer. That is, sometimes (ab) square equals _ab or _ _ab. What is the sum of all such positive two-digit integers?

[b]101[/b]

The unit digits b that will return the same digit when squared are: 0, 1, 5, 6.

Two digit numbers ending in 0 or 1 will not meet the qualification of the digit a being the same, so I tried digits 5 and 6.

Out of those 25 and 76 are the only numbers that will fit both qualifications of the problem.

The sum is 25+76= 101

So you tried every possible number ending with 5 and 6? (Thats what i did) or did you use some rule?

I only need to test one number with a units digit of five, which made things much easier for the rest of the problem.

One math trick is when squaring numbers with a units digit of five:
[list]Begin with the multiplication of 65*65 or something similar
Next add one to the tens digit of the number being squared: 6+1=7
Multiply the result with the original number: 6*7= 42
Place this number before 25 to get 4225
This process should work each time[/list]

As you can see with this technique squaring such numbers will always end in the digits 25. Thus (25)^2 is all I need to test for units digits with 5.

For 6, all I needed to test was 46, 56, 66, 76, 86, 96.


4. In the plane are given two lines $a$ and $b$ ,and two points $A$,$B$.$X$ moves in the line $a$, and point $Y$ in the line $b$, such that $AX\|{BY}$.Find the locus of the intersection points between $AY$ and $XB$. $\clubsuit$

Denote by $S,T$ intersections of the line $AB$ with the two lines. It is easy to prove that the transformation $X\to Y$ is projective. Let $M,M'$ and $N,N'$ two pairs of transformed fixed points, and $X,X'$ two mobile tranformed pints ($X\to X'$). Hence the armonic ratio $(M,N,S,X)=(M',N',T,X')$, and $A(M,N,S,X)$ and $B(M',N',T,X')$ have a common ray: $ST$ the intersection of two omologic rays lies on a line. This line can be determined by choosing two particular cases. For example $X=\infty$ and $X'=\infty$.


geometry
perimeter
trigonometry
function
trig identities
Law of Sines
algebra unsolved

:blush:

[color=blue] [size=117]Find the biggest value of the rate $ \frac{p}{R}$ where $ p$ is the half of the perimeter and $ R$ is the radius of the outcircle of a triangle[/size].[/color]

Let $ A,B,C$ be the angles of the triangle. From the law of sines  and from the concavity of sine function , we can write :
 $ \frac pR \equal{} sin(A)\plus{}sin(B)\plus{}sin(C) \leq 3sin(\frac{A\plus{}B\plus{}C}3)\equal{} 3sin(\frac{\pi}3)\equal{}\frac 32$
The maximum is riched for the equilateral triangle.
 :cool:

[size=117][color=red]Thank you Diogene!
Very good![/color]
lapsys $ \le{3\frac{\sqrt{3}}{2}}$[/size]

[quote="fredi"]:blush:

[color=blue] [size=117]Find the biggest value of the rate $ \frac {p}{R}$ where $ p$ is the half of the perimeter and $ R$ is the radius of the outcircle of a triangle[/size].[/color][/quote]
$ IH^2 \equal{} 4R^2 \plus{} 4Rr \plus{} 3r^2 \minus{} p^2$ and $ OI^2 \equal{} R^2 \minus{} 2Rr$, and it follows that $ p \le \sqrt{4R^2\plus{}4Rr\plus{}3r^2}, r \le \frac{R}{2}$.
So: $ \frac{p}{R} \le \frac{\sqrt{4R^2\plus{}4Rr\plus{}3r^2}}{R} \le \frac{\sqrt{4R^2\plus{}2R^2\plus{}\frac{3}{4}R^2}}{R} \equal{} \frac{\sqrt{3} \cdot 3}{2}$

(Btw. $ \sin \left ( \frac{\pi}{3} \right ) \equal{} \frac{\sqrt{3}}{2}$)


[quote="zhaobin"]for $a,b,c>0,abc=1$,prove:
\[ \left(a+\frac 1 b-1\right)\left(b-\frac 1 c+1\right)+\left(b+\frac 1 c-1\right)\left(c-\frac 1 a+1\right)+\left(c+\frac 1 a-1\right)\left(a-\frac 1 b+1\right) \ge 3  \][/quote]

Isn't this exactly Vasc's inequality, in the particular case abc = 1 ? See

http://www.mathlinks.ro/Forum/viewtopic.php?t=53585
and http://www.mathlinks.ro/Forum/viewtopic.php?t=14567 (rather indirectly - see post #7 and the subsequent ones of this topic)

[b]AH! sorry[/b]

  Darij

No,I think they are different.Notice"$-$"


In a convex quadrilateral $ ABCD$, let $ E$ be the intersection point of the diagonals, and let $ F_1,F_2,$ and $ F$ be the areas of $ ABE,CDE,$ and $ ABCD,$ respectively. Prove that:

$ \sqrt {F_1}\plus{}\sqrt {F_2} \le \sqrt {F}.$


Here's a nice problem:

A clock with two hands (one for the minutes and another for the hours) marks mid-day (12h). How much time will be needed for the two hands to meet again?

$\frac{12}{11}\ \text{hour}\approx 1 \ \text{hour}\ 5 \ \text{minutes}\ 27 \ \text{seconds}$

[quote="hunter34"]How did you get that?[/quote]
additional angle minute hand moves = total angle minute hand moves - one revolution = angle hour hand moves

The purpose of the problem was to use an infinite series for the computation of the total time, that's way I posted this of Calculus forum. Unless, of course, series are taught at the middle school level in some countries.

Isn't this forum Classroom Math, not calculus? Or did someone move it?

We know that the hour hand moves $ \frac{1}{2}$ degrees every minute, and the minute hand moves $ 6$ degrees every minute.  So the minute hand move 5.5 degrees more every minute.  The hour hand and minute hand should line up once every hour (i think, maybe give or take a few, maybe like 23 times a day), but anyways, we look at the 1:00 hour.  at 1, the hour hand has already gone 30 degrees.  so 30+.5x=6x, 5.5x=30, x=60/11 (minutes).  So it happens at $ 1: 05: \frac{300}{11}$

EDIT: didnt see the date on this, but it was newest in classroom math. my bad.

speeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeds

minute = 360 per hour
hour = 30 per hour

after one hour

minute = 0
hour = 30

minute has 30 to go, at 360 - 30 = 330 speed

so it takes one hour + $ \frac{1}{11}$ hour. Which is what nakawaga said

what? i didn't get this wouldn't the answer be only 1 hour?

 :help:

dont u slove this with chicken mcnuggets formula?

[quote="Temperal"][quote="ZhangPeijin"]Do you even know what the Chicken McNuggets formula is?
And no...
That formula is when you want to find out the largest possible value that you cannot make by combining two given values.[/quote]

Uh... and you go around telling other people not to revive old threads.[/quote]
Wait this is an old thread?
Oh...
 :oops:
I didn't realize it was old because someone posted a reply in it today...

Okay, let's end this topic.

You just use the formula on AoPS book 1(Chapter 4) or just google "formula for the clock problem"

The formula is when we set $ x$ as the number of minutes needed to pass in order for the hands to coincide is this:

$ x\equal{} \text{head start}\plus{}\frac x{12}$
[hide="Reason behind this"]The reason behind this is pretty simple.
We want to find the amount of minutes needed for the clocks to coincide.
Remeber that the hour-hand moves as the minute-hand does.
Head start comes from the position of the hour-hand. If it's 4 o' clock, the minute hands has to spend 20 minutes in order to catch up. But the hour-hand is a bit ahead. (Do you get it?)
So it's basically what the hour-hand is pointing at converted(represented) to a minute.
Then we find how far the hour-hand went while the minute-hand tried to catch up. The hour hand will move $ \frac x{60}$ amount of way between 4 to 5.
Since there is 5 minutes between 4 and 5, the hour-hand travels $ \frac x{60}$ amount of way in 5 minutes.
There we have all the factors to figure out the $ x$

So the formula is $ x\equal{}\text{head start}\plus{}5 (\frac x{60})$ (<-- In AoPS book.)

But I figured that the $ 5$ is going to be always constant, I simplified to this:

$ x\equal{}\text{head start}\plus{} \frac x{12}$
------------------------------------------------------------------------[/hide]

But in this case, the hands meet at 12, which means that they are going to never coincide in between 12 through 1.

So we start from when it's 1 o' clock.

We have $ x\equal{} 5 \plus{} \frac x{12}$
You simply multiply both sides by $ 12$ to get $ x\equal{}\frac{60}{11}$

Since we have one hour gap between, we add one hour to it.

So the final answer is $ \boxed{1 \text{ hour and }\frac{60}{11}\text{ minutes}}$

oh my bad wrong formula, sorry about that and yes i know what it is


$a,b,c$ are integers such that

$\frac{a}{b}=\frac{a^2+c^2}{b^2+c^2}$, $a \neq b$

Prove that $a^2+b^2+c^2$ is never a prime.

looks known.However not so easy!
We easily obtain $c^2=ab$ so $a^2+b^2+c^2=a^2+b^2+ab=(a-b)^2+3ab$
but i dont know how to continue.........

Let's continue:
$a^2+b^2+c^2 = (a+b)^2 - c^2 = (a+b+c)(a+b-c)$ where $a+b-c=1$ cannot occure, so it's not prime.

This is a qn of Bulgarian MO 1999 i think...

Similar question.
Find all +ve integers a,b,c s.t. (a+b)^2=(a+c)(b+c) and ab+bc+ac is prime.
Essentially same conxpts.

Bomb


$ n$ is a positive integer number. Consider $ (2n \plus{} 2)$ real numbers $ a,b,a_1,a_2,...,a_n,b_1,b_2,...,b_n$
such that $ a_i \neq 0, i \equal{} 1,2...,n$ and the function $ F(x) \equal{} \sum\limits_{i \equal{} 1}^n \sqrt {a_ix \plus{} b_i} \minus{} (ax \plus{} b)$ satisfies:
There exists distinct real numbers $ m,n$ such that $ F(m) \equal{} F(p) \equal{} 0$.
Prove that $ m$ and $ p$ are the only real solutions of $ F(x)$

To Mods: Please lock this topic, this is a problem from M&Y, issue 10, and we can't discuss it now

Why does $ a_i>0$ for all $ i$?
If there is the condition "$ a_i>0$ for all $ i$", can  you solve it?

[quote="TTsphn"]$ F'(x) \equal{} \sum_{i \equal{} 1}^n\frac {a_i}{2\sqrt {ax_i \plus{} b_i}} \minus{} a$
So I think must $ a_i > 0$[/quote]
But $ F''(x)\equal{}\minus{}\frac 12 \sum_i \frac{a_i^2}{(a_ix\plus{}b_i)^{\minus{}3/2}}<0$.
Therefore it is true.


inequalities
geometry
incenter
inequalities proposed

Let  $I$ be the incenter of the triangle $\Delta{ABC}$, and let $S$ be the area of the triangle. Prove that:

$(AI+BI+CI)^2 \ge \frac{1}{2}(a^2+b^2+c^2) + 2\sqrt{3}S$.

This is equivalent to Tiks [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=85848]post[/url], I gave an algebraic proof there. But the geometric proof is much more interesting. Hope you can find it!

Sorry, guys. Tiks mentioned that it's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=85355]Armenian TST 2006[/url].


Let a,b,c be positive numbers such that: a+b+c=1. Prove that: a/(1+b^{2})+b/(1+c^{2})+c/(1+a^{2}) \ge 9/10_

Hien,

 please see my solution in :

         http://www.mathlinks.ro/Forum/viewtopic.php?t=81259

 [b]Read[/b] [color=red]post 13.[/color]

   There the condition is $a+b+c = 3$ but the solution is the same. I'm sure you will understand the few greek words in the solution. Everything there is relations.If you have some problem , write.

 [u]Babis[/u]

  [b]Comments to translate some greek !.[/b]

   << [color=red]Είναι [/color] >>  means  <<[color=blue] It is , we have [/color]>>

 << [color=red]Προσθέτουμε ... [/color]>> means << [color=blue]We add ... to take [/color]>>

[quote="hien"]I have just solved the another problem nearly your problem:
[b]Problem[/b]: let $a,b,c$ are positive number such that $a+b+c=1$. Show that: $\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\leq\frac{9}{10}$
Your Problem really is difficult, I now can't sovle. Let me try once.[/quote]

  Hien, you should try IPT of Myth. It is the best way for your problem.
  About the original problem of spring_sotana, I have no idea. Two variables are in 1 term. It seems to be more challenged.

:roll: The conjecture of spring_sotana's is not true.

For counter-example, $\,a=\frac{16}{28}, \, b=\frac{1}{28}, \, c=\frac{11}{28}.$

Thank you. Now we all know why this problem seemed so tough.
The even tougher question arises (hello Vasc ;) ;) ):
Find the least $k>0$ such that for all $a,b,c\ge0,a+b+c=1$ we have
$\boxed{ \frac{a}{kb^2+1}+\frac{b}{kc^2+1}+\frac{c}{ka^2+1}\geq\frac9{k+9}}$

[quote="spanferkel"][quote="stergiu"]   There the condition is $a+b+c = 3$ but the solution is the same. [/quote]
Your solution is very nice, but it is not the same problem. The one you solve is rather equivalent to:
$\boxed{a+b+c=1\Rightarrow \frac{a}{9b^2+1}+\frac{b}{9c^2+1}+\frac{c}{9a^2+1}\geq\frac12}$.
The ineq $\frac{a}{b^2+1} \ge a- \frac{ab}{2}$ (nice idea!!) is only useful where $b=1$ is part of the equality case. But for the problem here, it should be sharp for $b=\frac13$. So your method cannot apply here :([/quote]

  Yes, you are right! I did not see that [color=red]the inequality is not homogenous [/color]! Otherwise the condition could be changed.Now , after the last post(Guest), I think we have nothing to prove. Thanks a lot.

  [u]Babis[/u]

[quote="spanferkel"]
Find the least $k>0$ such that for all $a,b,c\ge0,a+b+c=1$ we have
$\boxed{ \frac{a}{kb^2+1}+\frac{b}{kc^2+1}+\frac{c}{ka^2+1}\geq\frac9{k+9}}$[/quote]
[hide]I conjecture $k=6\sqrt{3}-9\approx 1.3923$[/hide]

[quote="indybar"][quote="barra"]\[ \frac{a}{a^2+1}\leq \frac{72}{100}a+\frac{6}{100}\Leftrightarrow (a-\frac{1}{3})^2(a+\frac{9}{12})\geq 0  \]
[/quote]

Can somebody show me how to come up with nice things like this?  :)[/quote]
We prove that $\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\leq\frac{9}{10}$,  where $a>0,$ $b>0,$ $c>0$ and $a+b+c=1.$
Hence,
$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\leq\frac{9}{10}\Leftrightarrow\sum_{cyc}\left(\frac{a}{a^2+1}-\frac{3}{10}\right)\leq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\left(\frac{a}{a^2+1}-\frac{3}{10}+\lambda\left(a-\frac{1}{3}\right)\right)\leq0$ $\forall\lambda\in\mathbb R.$
Let $f(x)=\frac{x}{x^2+1}-\frac{3}{10}+\lambda\left(x-\frac{1}{3}\right).$
If $f'\left(\frac{1}{3}\right)=0$ then $\lambda=-\frac{72}{100}.$  :)

[quote="Vasc"]I found this conjecture a year ago and I included it in my book. I hope somebody will prove it. :)
For $k=3$ there is a nice solution.[/quote]Let me guess your nice solution:
According to Stergiu's idea, start with weighted AM-GM: $3b^2+1=3\cdot b^2+9\cdot\frac19\ge12\sqrt[12]{\frac{b^6}{9^9} }=4\sqrt{\frac{b}3}$
so $\sum_{cyc} \frac{a}{3b^2+1}=\sum_{cyc} a\big(1-\frac{3b^2}{3b^2+1}\big)\ge 1-\frac{3\sqrt{3}}4\sum_{cyc} ab\sqrt{b}$.

So it remains to show that $\sum_{cyc} ab\sqrt{b}\le\frac1{{3\sqrt{3}}}$ if $a+b+c=1$. Or equivalently $\sum_{cyc} x^3y^2\le 3$ if $x^2+y^2+z^2=3$. Now it seems to me that last one has been discussed somewhere in the forum some months ago... :?
Vasc, is that your solution? And how to complete it?


[hide]Using the hint mathgeniuse posted (and was too lazy to use), we get the three equations:

$4a+2b+c+16=1$
$9a+3b+c+54=14$
$a+b+c+2=0$

Solve to get $a=-6$, $b=5$, and $c=-1$

Therefore, the equation becomes $2x^3-6x^2+5x-1=(x-1)(2x^2-4x+1)$

$x=\boxed{1, \frac{2\pm\sqrt{2}}{2}}$[/hide]

[quote="mathgeniuse^ln(x)"][quote="coffeym"]The expression $2x^3+ax^2+bx+c$ has a factor $(x-1)$.
It has a remainder $1$ when divided by $(x-2)$
It has a remainder $14$ when divided by $(x-3)$

Using this information, solve the equation $2x^3+ax^2+bx+c=0$, leaving answers in their exact form with rationalised denominators.[/quote]

[hide]Hint:Use the remainder theorem and the factor Theorem.  The Remainder Theorem says that the remainder of x-a is f(a).  Also the Factor Theorem says that if x-b is a factor then f(b)=0.[/hide][/quote]

Has anyone checked the title of this thread lately?

[quote="mathgeniuse^ln(x)"]Yes.  This is related to the remainder theorem and the factor theorem isn't it.  Actually, no it isn't since we haven't used the factor theorem.[/quote]

Isn't the factor theorem just like the remainder theorem when the remainder is 0?

Yes, but Factor Theorem, says that if f(a)=0, then x-a is a factor of f(x).  I think.

Yes, the factor theorem is used here, as detailed above.

I suppose we have used the factor theorem here bcos we know that those things given are factors,not the remainder theorem.

@splashD-When using remainder theorem,we find the remainder but if we know what the factors are,then oyu can't call it the remainder theorem if you use the fact that f(whatever)=0


If we use a fact from say a book, can we just quote it? Could that still get full credit?

[quote="Centurian"]I have a related question:
If the problem doesn't say to prove our solution, do we just state it? Does including a program suffice?[/quote]
I'll cite one of the rules on the USAMTS website:

[quote]Except where otherwise noted in problems, participants may use calculators and computers to solve problems. In this case, the participant must explain the program that was used to solve the problem, and must prove that the method used in the program does solve the problem. [/quote]
So, if you want to use a program to solve a problem, you must explain why your program actually solves the problem, and doesn't just spit out a random number.
I'll also cite the "problems" section of the USAMTS website:

[quote]Remember to fully justify all your answers - don't just send numerical answers. [/quote]
So, you generally won't get full credit for simply stating your answer, even if the problem doesn't explicitly say "with proof".

[quote="jen7"]What if we use a fact from a book (not a program) and cite it?  Would that get full credit?[/quote]
That's a bit of a sticky subject; I've seen answers that go both ways.  If it's a "well-known" result (i.e. you can find it given in more than three sources), then you can simply say "well-known" and be done with it.  For example, on round 2, problem 2 of this year's USAMTS, some people cited a result from number theory about the continued fraction expansion of [tex]\sqrt{2}[/tex], and just put "well-known result" in their proofs. For more obscure results, I'm really not sure

[quote="jen7"]What if we use a fact from a book (not a program) and cite it?  Would that get full credit?[/quote]

As zanttrang notes, this highly depends on the fact and the grader.  If it's in many books and doesn't completely solve the problem, then you can just cite it.  If it is in one obscure book and completely blows the problem away, that probably won't get much credit without full proof.  Basically, if it's anything obscure, you're far better off proving it.  And if it is essentially equivalent to the problem at hand, you're far better off proving it.


geometry
function
trigonometry
calculus
integration
calculus computations

Well I thought of another problem today. This one should be a lot easier than some of the others I've posted

Find the area enclosed between the lines $ y\equal{}\frac{1}{9\plus{}x^2}$ and $ y\equal{}\frac{1}{3\plus{}3x^2}$

As far as method goes....

Both functions are easy to integrate via trig substitution, meaning all that remains is to find their intersection points so you know what interval(s) over which to integrate. Then you just take the difference between the higher and the lower function's integral to get your net area.

Am I overlooking something?

No, you're not overlooking anything. I could put this problem on a regular calculus test, and it wouldn't be all that hard.

I guess it really is as easy as I thought it would be then.


Find all functions $f: \mathbb{R}\to \mathbb{R}$,which satisfy the condition \[f(x)-f([x])\cdot f(\{x\})=x,\]  \[\forall x\in \mathbb{R}\]

Since  $f(0)-f(0)^{2}=0$,   $f(0)=0$  or  $f(0)=1$.  On the other hand, taking
$x=n\in\mathbb Z$, since  $[n]=n$ and  $\{n \}=0$  we get  $f(n)[1-f(0)]=n\;\;\forall n\in \mathbb Z$,  which is impossible if  $f(0)=1$,  because in that case the expression on the left is always $0$. This means  $f(0)=0$  and  $f(n)=n\;\;\forall n \in \mathbb Z$. If we pick  $x\in[0,1)$, from  $[x]=0$  we obtain   $f(x)=f(x)-f(0)f(x)=x\;\;\forall x \in [0,1)$.
Since  $[x]\in \mathbb Z\;\wedge \;\{ x\}\in [0,1)\;\;\forall x \in \mathbb R$:

$f(x)=f([x])f(\{x \})+x= [x]\{x \}+x \;\;\forall x \in \mathbb R$.


inequalities
trigonometry
inequalities unsolved

The real numbers x1,x2,...,xn all satisfies 0<xi< \pi /2 and tan(x1)*tan(x2)*...tan(xn)=2^(n/2).Find the greatest value of F=cos x1+cos x2+...+cos xn.
It's all.(Sorry about my bad English)

How can we use the fact that cos(x1)*cos(x2)*...cos(xn) <= (1/3)^(n/2) to find the greatest value of
F=cos x1+cos x2+...+cos xn?


Don't stress out too much about what you get as a score.  If you focus and do your best to make a clear and rigorous proof with the right answer, you'll probably get a 5.

As for the strictness of the grading, USAMTS graders are certainly more lenient than, say, USAMO graders.  Last year I got 5's for every submission, and some of them I had slight holes in my logic or I forgot some trivial case.  Small mistakes that are extremely easily fixed usually go unpunished.

As for the detail needed to get a 5, while I wouldn't say that just putting down the right answer and explaining the outline of the method used is enough, you don't have to explain how you did things like multiplying two numbers.  Usually, just go with your own judgement if you're deciding whether something needs to be explained further.

[quote="EricMathPath09"]This is my first time doing USAMTS. That means that I do not know how strictly they grade solutions. Can anybody tell me how strict they are? Use examples of previous results if you can. Thanks.[/quote]

Not as strict as Olympiad.

Don't worry too much about the score. If you can't solve a problem at all, just write a partial solution or something.

By the way, there is 0-5 pts per problem, 5 problems per round, 4 rounds

The grading is generally loose. The graders realize that most of these solutions are written by high school students writing a proof for the first time in their lives.

In general, if you have the right answer and some explanation, expect at least a 3 (assuming your explanation is a real explanation)

I raise a question:

What is the average score? and do many middle school students participate?

I am MS :P

You can find the answers to both of your questions on the USAMTS website. Click "Leaderboard", then choose the year, then click "Stats".
Last year, for example, there were a total of 78 12th graders, 130 11th graders, 111 in 10th grade, 64 in 9th grade, and 55 middle school.

I have another question: How come in the previous year's statistics, the highest age listed in the statistics and rankings is only 12 years old? There are no people above 12 years old.

I don't see any part about age 12.  The highest ranking is grade 12, maybe that's what you're misunderstanding :maybe:

Oh sorry. I thought that the number after the name was the age, not the grade.  :blush:

[quote="modularmarc101"]Do they also take your grade into consideration? This is also my first year I'm participating and I'm in 9th.[/quote]
No.

Unless you are above 12th grade.

[quote="Math Champion"]What do you mean by "Take it into consideration"?[/quote]

He meant grading it differently because you are younger/older.

They shouldn't. Not intentionally anyway. They do want your grade for award purposes though

[quote="AIME15"]The graders don't know your grade when they are grading your paper :wink:[/quote]

They could know. Because they know your ID and when you make an account @ USAMTS you have to put your grade.

[quote="HappyLand"]It's not too hard to get a 5. But normally they give 4 if you just get it correctly with no major errors. Or even a 3. 5 requires a very clear explanation.[/quote]

A 4 with no major errors, you mean a very minor error in your proof, right?

[quote="fwolth"][quote="HappyLand"]It's not too hard to get a 5. But normally they give 4 if you just get it correctly with no major errors. Or even a 3. 5 requires a very clear explanation.[/quote]

And from where do you draw this deep knowledge? As far as I know, that's complete nonsense; I got a 5 for some of the most obfuscating, poorly-written arguments last year (and a 4 for one that was almost complete nonsense), while getting a 3 for a very small calculation error.[/quote]

Yeah, last year I got a 5 for a completely hand-waving proof with the right answer. I didn't even know really how to prove it!  I also got a 5 for a proof that was circular but was so murky that it was hard to notice.  Getting a 5 mainly comes down to getting the right answer.  If you mess up the answer at all though you get points off.

Not always, if you put only the answer, then you only get 1 point if it is correct.

Well yeah, I assumed it was understood that you have to prove all your answers :roll:


calculus
integration
function
absolute value
calculus computations

My professor really hasn't explained too much about Calculus yet. I need help doing problems like this.

1. Find the distance traveled in 15 seconds by an object moving with a velocity of V(t)= 20 + 7cos(t) feet per second.

I assumed you want to find the secant line and then put the t=15 into the final product..... Thanks!

Distance travelled and displacement are two very different things. Your question suggests it is looking for the former.

In general:

If you take speed to the the absolute value of the velocity, then distance travelled is the integral of the speed, over the time interval specified.

The displacement on the other hand is just the distance between the endpoints of your position function, which is the integral of the velocity over the time interval.

However, in your case velocity is never negative in that time interval, so displacement = distance travelled, and you can just take the integral directly.

So:

displacement = $ \int_0^{15}{20\plus{}7\cos(t)dt}$


the remainder obtained when 7^84 is divided by 342 is
also tell me how to put superscript while posting  a topic :?:

You cannot put superscript on the topic line.

7^84 =343^28
let 343 be x
hence, we have x^28which is to be divided by x-1
applying remainder theorem we get remainder=1.......


If $ \sqrt[3]{x\plus{}5}\minus{} \sqrt[3]{x\minus{}5}\equal{}1$, Find $ x^2$

$ a \plus{} b \plus{} c \equal{} 0\Longrightarrow a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc$,

[hide="Answer"]$ x^2\equal{}52$[/hide]

hello, we find $ x\equal{}2\sqrt{13}$ and so we have $ x^2\equal{}52$.
Sonnhard.


geometry
circumcircle
perpendicular bisector
geometry proposed

[quote="thaithuan_GC"]Let a triangle ABC. The bisector of angle A cuts the circumcircle (O) at R. AR cuts the perpendicular bisector of segments AB,AC at P,Q. Denote S,T be the midpoints of segments AB,AC. Prove the areas of triangles RQT, RPS are equal to each other.[/quote]
$ \bullet$ Let $ B',\ C'$ be, the orthogonal projections of $ R,$ on $ AB,\ AC$ respectively and it is enough to show that $ BB' \equal{} CC'$ $ ,(1)$ from the congruent right triangles $ \bigtriangleup B'BR,\ \bigtriangleup CC'R$ $ ($ because they have $ RB' \equal{} RC'$ and $ \angle RBB' \equal{} \angle RCC'$ $ ).$

From $ PS\parallel RB'$ $ \Longrightarrow$ $ (RPS) \equal{} (B'PS)$ $ ,(2)$

From $ QT\parallel RC'$ $ \Longrightarrow$ $ (RQT) \equal{} (C'QT)$ $ ,(3)$

From $ (2),$ $ (3)$ it is enough to prove that $ (B'PS) \equal{} (C'QT)$ $ ,(4)$

$ \bullet$ But, $ (B'PS) \equal{} \frac {1}{2}(B'S).(PS)$ $ ,(5)$  and  $ (C'QT) \equal{} \frac {1}{2}(C'T).(QT)$ $ ,(6)$

From $ (5),$ $ (6)$ it is enough to prove that $ \frac {PS}{QT} \equal{} \frac {2(AC') \minus{} AC}{2(AB') \minus{} AB}$ $ ,(7)$

From the similar right triangles $ \bigtriangleup SAP,\ \bigtriangleup TAQ,$ we have $ \frac {PS}{QT} \equal{} \frac {AS}{AT} \equal{} \frac {AB}{AC}$ $ ,(8)$

So, from $ (7),$ $ (8),$ it is enough to prove that $ \frac {2(AB') \minus{} AC}{2(AB') \minus{} AB} \equal{} \frac {AB}{AC}$ $ ,(9)$ $ ($ because of $ AB' \equal{} AC'$ $ ).$

From $ (9)$ $ \Longrightarrow$ $ 2(AB').(AC) \minus{} (AC)^{2} \equal{} 2(AB').(AB) \minus{} (AB)^{2}$ $ \Longrightarrow$ $ 2(AB') \equal{} AB \plus{} AC$ $ ,(10)$

$ \bullet$ But, the relation $ (10)$ is true, because of $ 2(AB') \equal{} AB' \plus{} AC' \equal{} AB \plus{} BB' \plus{} AC' \equal{} AB \plus{} AC' \plus{} C'C \equal{} AB \plus{} AC$

Hence, we conclude that $ (RPS) \equal{} (RQT)$ and the proof is completed.

Kostas Vittas.


arithmetic sequence
number theory unsolved
number theory

n is the natural number which comes to give. Each other different 2n+1 anti with all buy the series where all protest product becomes the perfect square valence from in arithmetic sequence of the natural numbers which is coming to accomplish

Even after hard tries, I got no idea what this is supposed to be...
Either your translator or your buzzword bot sucks  :wink:


induction
modular arithmetic
quadratics
algorithm
continued fraction

Let $ \mathbb{Z}[\sqrt{2}]$ denote the collection of real numbers of the form $ r\equal{}a\plus{}b\sqrt{2}, a,b \in \mathbb{Z}$. For such an r, define $ N(r)\equal{}a^2\minus{}2b^2$; and show that $ N(rs)\equal{}N(r)N(s)$. Hence show that there are infinitely many pairs of integers a,b with $ a^2\minus{}2b^2\equal{}1$, and infinitely many pairs such that $ a^2\minus{}2b^2\equal{}\minus{}1$. How about pairs such that $ a^2\minus{}3b^2 \equal{} 1 or \minus{}1$?

Define $ \overline{a\plus{}b\sqrt{2}}\equal{}a\minus{}b\sqrt{2}$ and $ N(r) \equal{} r\overline{r}$ so the question reduces to $ \overline{xy}\equal{}\overline{x}\overline{y}$ which is trivial to prove (so was the first, but w/e). Also, $ N(z)\equal{}1\Rightarrow N(z^n) \equal{} 1)$ and $ N(z)\equal{}\minus{}1\Rightarrow N\left(z^{2n\plus{}1}\right)\equal{}\minus{}1.$ $ a^2\minus{}3b^2\equal{}\minus{}1$ is impossible due to mods. Did I forget anything? :)

thanks! I had the first part of it, but don't really understand how to do the second part...

If we have that $ N(z)\equal{}1$ then we can create an infinitude of solutions using the powers of $ z$. ok. but how do we find such a $ z$ to begin with? how do we know one exists?

i hope you understand :) thank you

[quote="amitie"]thanks! I had the first part of it, but don't really understand how to do the second part...

If we have that $ N(z) = 1$ then we can create an infinitude of solutions using the powers of $ z$. ok. but how do we find such a $ z$ to begin with? how do we know one exists?

i hope you understand :) thank you[/quote]

\begin{align*}
3^2-2\cdot 2^2&=1\\
1^2-2\cdot 1^2&=-1
\end{align*}

I think that generally $ a^2-xb^2=1$ has infinitely many integer solutions $ (a,b)$ but please don't ask for a proof... quoting arthur engel from my mind, only cases where it's easy to find the "first" solution occured on IMOs yet. :)
The "-1" equation can be killed by mods most of the time; if that doesn't work I think there are always infinitely many solutions. No idea of a proof again :D

The equation $ x^2\minus{}dy^2\equal{}1$ always has a nontrivial solution for squarefree $ d>1$, but the equation $ x^2\minus{}dy^2\equal{}\minus{}1$ can be more unpredictable.

For example, $ 5^2\minus{}34\equal{}\minus{}3^2$, so no mods except powers of 3 could possibly kill the equation $ x^2\minus{}34y^2\equal{}\minus{}1$. It turns out that these don't work either: it can be shown by induction that for all $ n$, there exists $ y_n$ such that $ 34y_n^2\equiv 1\pmod{3^n}$, and then $ x_n\equal{}0$ would satisfy the congruence.

However, there is no solution to $ x^2\minus{}34y^2\equal{}\minus{}1$, and this is not too difficult to show with infinite descent. (Find a nontrivial solution to $ a^2\minus{}34b^2\equal{}1$, and then note that $ (34by\minus{}ax)^2\minus{}34(ay\minus{}bx)^2\equal{}x^2\minus{}34y^2$.)

[quote="amitie"]If we have that $ N(z) \equal{} 1$ then we can create an infinitude of solutions using the powers of $ z$. ok. but how do we find such a $ z$ to begin with? [/quote]

To find $ z \equal{} x \plus{} y \sqrt {d}$ such that $ x^2 \minus{} dy^2 \equal{} 1$, let $ \frac {p_k}{q_k}$ be the [url=http://mathworld.wolfram.com/Convergent.html]convergents[/url] of the continued fraction representation of $ \sqrt {d}$.  Then

$ | p_k^2 \minus{} d q_k^2 | \le 2 \sqrt {d}$

In fact, the sequence $ p_k^2 \minus{} d q_k^2$ is periodic with the same period as the continued fraction itself, and within each period exactly one value of $ k$ exists such that $ p_k^2 \minus{} d q_k^2 \equal{} 1$.  The continued fraction of a quadratic surd is fairly easy to calculate (there is an algorithm for it which is tedious to describe, but straightforward to use), and only a finite number of cases need to be checked before the first solution is reached.


In circle $ J$, $ HO$ and $ HN$ are tangent to the circle at $ O$ and $ N$. Find the number of degrees in the sum of $ m\angle J$ and $ m\angle H$.
[asy]size(150);
import graph;
pair J = (0,0), H = (6,0), O, N;
path circ = Circle(J,3);
pair M = midpoint(J--H);
path secCirc = Circle(M,3);
pair[] tangentPoints = intersectionpoints(circ,secCirc);
O = tangentPoints[0]; N = tangentPoints[1];
draw(J--N--H--O--cycle);
draw(circ);
label("$H$",H,E);
label("$J$",J,W);
label("$N$",N,S);
label("$O$",O,W);[/asy]

By a theorem, you can say that $ m \angle J$ is double $ m \angle H$.  I think...

I'm not sure if this diagram is to scale, but it looks like $ \angle OJN$ is $ \frac{1}{3}$ of the circle, making $ m \angle J \equal{} 360 \div 3 \equal{} 120^\circ$. And if what ernie said is true, then $ m \angle H \equal{} 240^\circ$. But that can't be right...

No, what I mean is this:

$ m \angle J \equal{} 2m \angle H$

Oh! So in which case $ m \angle H \equal{} 120 \div 2 \equal{} 60^\circ$. $ 60 \plus{} 120 \equal{} \boxed{180}$.

Is $ ON =$ to $ OH$?

What's the relationship between $ JO$ and $ ON$?

Edit: if $ ON = OH$ then triangle $ HON$ is equilateral, and thus \angle $ H = 60^\circ.$

That would make \angle $ J$ $ 120^\circle.$

[color=blue]Fixed your $ \text{\LaTeX}$. -Isabella2296[/color]

Relationships are unnecessary. Since OH is tangent to circle J, and JO is a radius, angle JOH is a right angle. Similarly JNH is also a right angle. Since a quadrilateral has 360 degrees in total, answer = [b]180 degrees[/b].

Specialization is absolutely unnecessary.


Just wondering how all you guys did...

I got 37 on Galois, even though I made a mistake somewhere in 4c.

I'm not sure what point I got, but my name is on the gold medal list.

I think I got around 37-39 because I got everything up to 4b, and I didn't fully explain 4c.


logarithms
complex analysis
complex analysis unsolved

I have two questions which i can solve it until now,could anyone help me to solve it :)
1.If $ lim_{}n(\frac {u_n}{u_{n \plus{} 1}} \minus{} 1) \equal{} 1$ is the serie $ \sum\limits_{i \equal{} 1}^{ \plus{} \infty}u_n$ disconverge ?(if not please give me some example )
2.Given two series $ S_{1}$ and $ S_{2}$ and we define " product serie" of two serie $ S_{1}$ and $ S_{2}$ as follow
If $ S_1 \equal{} \sum\limits_{i \equal{} 1}^{n}u_n$ and $ S_2 \equal{} \sum\limits_{i \equal{} 1}^{n}v_n$ then $ S \equal{} \sum\limits_{i \equal{} 1}^{n}w_n$ with $ w_n \equal{} u_1.v_n \plus{} u_2.v_{n \minus{} 1}..u_n.v_1$
My second question is if two series $ S_{1}$ and $ S_{2}$ are converge(not absolute converge),what we can say about the 'prosuct serie" ?( i mean is S  disconverge or converge ?)
Sorry for my bad English :(
Thanks everyone :)

Note on English: the form you're looking for is "convergent" or "divergent". "Converge" is a verb, and we tweak the ending for the adjective form. Also, it's "series".

#1 is on the borderline between convergence and divergence. As such, it could go either way: $ \sum \frac1{n\ln n}$ diverges and $ \sum \frac1{n\ln^2 n}$ converges.

An example for #2: $ u_n\equal{}v_n\equal{}\frac{(\minus{}1)^n}{\sqrt{n}}$.

For 2, basically if the series aren't absolutely convergent then anything can happen. It's really not a good type of convergence. But if they're both absolutely convergent then the product series converges by Fubini's theorem

[quote="jmerry"]
An example for #2: $ u_n \equal{} v_n \equal{} \frac {( \minus{} 1)^n}{\sqrt {n}}$.[/quote]
So do you mean S is always disconvergent series :(( i also knew this example )


geometry
limit
induction
absolute value
algebra proposed
algebra

Find all $n\in\mathbb{N}$ s.t. there exist sequence $a_n$ in integer number and satisfying these conditions:

1)$a_1=1$
2)$a_2=a$
3)$\forall n>1 , a_n=(2n+1)a_{n-1}-(n^2-1)a_{n-2}.$
4)For every $i<j , a_i\mid a_j.$

This is number theory ,isn't it?
maybe it's Combinatoric or Geometry anyway it isn't Algebra.

There's something wrong in the statement: you can't ask for $n$ to be found, because $n$ is the index of $a_n$ in the sequence, so $n$ can take any value in $\mathbb{N}\backslash \{0\}$.
Maybe we should find $a$?

Presumably there has been an accidental doubling-up of variables and the condition (3) is meant to be
\[ \forall m>1 , a_m=(2n+1)a_{m-1}-(n^2-1)a_{m-2}.  \]
If so, we can solve the problem by noting that this recurrence has characteristic equation $t^2-(2n+1)t+(n^2-1)=0$, which has roots $\frac{2n+1\pm\sqrt{4n+5}}{2}$.  As $n$ is an integer, $4n+5$ is never 0, so the two roots are always distinct.  So the sequence has general term
\[ a_{m}=A\alpha^m+B\beta^m,  \]
where $\alpha$ and $\beta$ are the two roots.

If $n$ is such that the two roots are complex, the sequence is not increasing, so the division property does not hold.

If $n$ is such that the two roots are real, one has greater absolute value (since $2n+1$ is always nonzero).  WLOG let it be $\alpha$; then \[ \lim_{m\to\infty}\frac{a_m}{a_{m-1}}=\alpha.  \]
As $\frac{a_m}{a_{m-1}}$ is always an integer, this means that $\alpha$ is an integer as well.  So the discriminant $4n+5$ is an odd square $(2k+1)^2$, and we find that $n$ is of the form $k^2+k-1$.  Given $n$ of this form, there exists a sequence satisfying the division property: that produced by taking $a=k^2+2k$, with general term $a_m=(k^2+2k)^{m-1}$.

We can solve this recursive formula by using basic solution not by induction not by genarating function.

$a_n=\left\{1+\frac{1}{12}(a-2)(n-1)(n+4)\right\}n!\ (n=1,2,\cdots)$

kunny


linear algebra
matrix
modular arithmetic
combinatorics unsolved
combinatorics

Here are the problems:

[b]1)A "m X n matrix"  has to be filled up with "+1 or -1" in such a way that the product of the entries in each row and in each coloumn is -1.
In how many ways can this be done? :maybe:

2)A "m X n" chessboard has to be covered with "1 X 2" dominoes. In how many ways can this be done, if at least one of m and n is even?[/b] :oops:

Both double-posting and writing in unusual fonts are obnoxious.  Please don't do either in the future.

i wanna reply JBL and polarb by here that, your approaches to the human senses are very low.

[quote="polarb1"]I don't think the Olimpiad section is the place to post this "school" problems.
Answer: $ 2^{(m \minus{} 1)(n \minus{} 1)}$ when $ m\equiv n \pmod{2}$, $ 0$ else.
Hint: Consider the submatrix $ (n \minus{} 1)$x$ (m \minus{} 1)$ and the product of the whole matrix for solution.[/quote]

The following 1*2 matrix is a valid sol. for unequal paraties of m,n. :     +1  -1
May be there are more such cases. Why don't you try for a proof.

It was the second problem that i was especially interested in.

Take a look at the product of the numbers in the second column of your $ 1 \times 2$ matrix...

First, sorry if I hurted someones feelings; it wasn't my aim.
The full solution to the first problem is:
   Let us consider the product of a valid matrix. This product is the product of products of all rows, and the product of each row is $ \minus{}1$ so the product is $ (\minus{}1)^{m}$, the product of the matrix is also the product of all products of colums which is $ (\minus{}1)^{n}$. Hence, $ (\minus{}1)^{n}\equal{}(\minus{}1)^{m}$ what implies $ n \equiv m \pmod{2}$.
   From each matrix $ (m\minus{}1)$x$ (n\minus{}1)$, where $ n \equiv m \pmod{2}$ we can make only one valid matrix by putting one element on the front of each row a one element on the front of each column an finally putting the element $ (1,1)$. There is only one such putting because each of the elements, which are added is the last element put in a row/column, so to fulfill the conditions of the matrix it must has a certain value (we are able to put validly the last element because $ n \equiv m \pmod{2}$). So the number of valid matrixes $ n$x$ m$ is the number of all matrixes $ n\minus{}1$x$ m\minus{}1$, which is $ 2^{(n\minus{}1)(m\minus{}1)}$.


I noticed AoPS has movies about simon's favorite factoring trick.
http://www.artofproblemsolving.com/Movies/simon2.php
http://www.artofproblemsolving.com/Movies/simon1.php

What other movies are there?

Currently, we only have the movies that rrusczyk has mentioned in [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=1]his blog[/url], and we have no short-term plans to make more.

They were an experiment.  While we don't have short-term plans for more, we will likely have some in the longer term (6-12 months).


Let $\alpha$ and $\beta$ be positive integers such that

                    $\frac{43}{197}< \frac{\alpha}{\beta}< \frac{17}{77}.$

Find the minimum possible value of $\beta.$


For the linear transformation $\left(\begin{array}{c}x' \\ y'\end{array}\right)=\left(\begin{array}{cc}1 & 3\\2 & k \end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right),$ find the images of the following figures.

(1) The line $l: x+2y=0$
(2) The line $m: x+3y-3=0$
(3) The whole plane

All of these answers depend on the value of $k.$

(1) This line is the span of $\begin{pmatrix}-2\\1\end{pmatrix}.$ It gets mapped to the span of $\begin{pmatrix}1\\k-4\end{pmatrix}.$ This is always a line through the origin. Every direction except vertical is possible for some value of $k.$

(2) We can write this line parametrically as

$\begin{pmatrix}0\\1\end{pmatrix}+t \begin{pmatrix}-3\\1\end{pmatrix}.$

This gets transformed to

$\begin{pmatrix}3\\k\end{pmatrix}+t \begin{pmatrix}0\\k-6\end{pmatrix}.$

If $k\ne6,$ this becomes a line, not through the origin. If $k=6,$ the image is the single point $\begin{pmatrix}3\\6\end{pmatrix}$

(3) If $k\ne 6,$ the matrix is invertible and the image of the plane is the whole plane.

If $k=6,$ then the matrix has rank 1 and the image of the plane is a line, the span of $\begin{pmatrix}1\\2\end{pmatrix}.$

Ah, yes, I should have said that. It's there in my calculation, but not elegantly stated.


It's seem too hard for intermediate level. ?  :(

One way to prove it:  Assume that the xi are ordered from low to high.  For the first half of the i's, bound |xi - a| by a - xi.  For the second half, bound |xi - a| by xi - a.  When you add up the bounds, the a's will cancel out, leaving what you want.

Thanks for Ravi. B

I"m not undersatnd too much what you say.

Can you write down your proofs. Please....

Try thinking about it in this way. Lets look at the left hand side of that equation, and start with a less than all the xs. Now consider what happens as we increase a. If we increase a by a little bit, then we add a little bit for each x thats less than a, and subtract a little bit for each x thats bigger than a. So as you increase a, the left hand side gets smaller and smaller while there are still less x values less than a than there are bigger than a - and it will be a minimum when there are the same on each side (ie, the median).

Quick example to show what I mean:
1 2 3 4 5
Start with a = 0.8. When we increase a to 0.9, the sum decreases by 5*0.1. When we increase a to 1, it decreases by another 5*0.1. Now when we increase by 0.1, the sum decreases by 4*0.1 - 0.1 = 0.3. Etc. Once we get to a=3, the sum will start increasing, and thats the minimum.

If there were an even number of values, it doesn't have to be the median: anything between the 2 middle value will work by the above reasoning.

If you wanted to write up a proper proof, you would probably need to write this more formally, and thats where Ravi comes in. (its exactly the same idea).

You can find more details of a proof here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16630[/url].

:D  I got it. !!!

Thank you very much for your kind. TripleM and Ravi B.

Good luck.  :P


I know there are infinitely many primes, but how do you prove it?

If there were finitely many let's say they were $ p_1,p_2,...,p_k$, then consider the number $ p_1p_2...p_k\plus{}1$.

Then none of the primes $ p_1,p_2...,p_k$ could divide this number, so this number would be a new prime, contradicting that the only primes were $ p_1,p_2,...,p_k$.

oh yes, thnx. Just occurred to me to do an indirect proof for this one.

[quote="rofler"]If there were finitely many let's say they were $ p_1,p_2,...,p_k$, then consider the number $ p_1p_2...p_k \plus{} 1$.

Then none of the primes $ p_1,p_2...,p_k$ could divide this number, so this number would be a new prime, contradicting that the only primes were $ p_1,p_2,...,p_k$.[/quote]The reasoning is almost complete in that $ p_1p_2...p_k \plus{} 1$ may not be actually prime (in fact, it can be composite). To complete it:

Either $ p_1p_2...p_k \plus{} 1$ is a prime (follow rofler's argument), or that it is a composite of a product of primes not in the list. Show how these primes are not in the list. (Simple)

If it is composite, there must be another prime not in $ p_1\minus{}p_k$ thus contradicting our assumption.

just finishing up the proof :D

This is a useless post because this is already in Aops 2 and the person who gave the problem also knew that.

Thank you for accusing me. No, I only have the geometry and algebra books for the AoPS series, and what's the AoPS 2 book (is that the title)?

He's talking about [i]the Art of Problem Solving Volume 2: and Beyond[/i].

@varunrocks: Funny that you should mention useless posts... Anyway, the proof of the infinitude of primes belongs to Euclid, and is not unique to AoPS 2 (it's a very old problem, anyway).

@bengal: There is such a thing as [url=http://www.google.com/search?q=infinitude%20of%20primes]Google[/url]. Or you can search the forums, as this has been posted many times before.


Show that the equation
\[ x^3\plus{}2y^3\plus{}4z^3\ \equal{}\ 6xyz\]
has no rational solutions except $ (x,y,z)\equal{}(0,0,0).$

The norm from $ \mathbb Q[\sqrt [3]{2}]$ to $ \mathbb Q$ is given by $ N(a \plus{} b \sqrt [3]{2} \plus{} c \sqrt [3]{4}) \equal{} a^3 \plus{} 2b^3 \plus{} 4c^3 \minus{} 6abc$. The result and lot's of other stuff follows.
Or you can just use simple $ 2$-adic descent by looking $ \mod 2$  :wink:

Thanks. :)

I was wondering if the problem could be solved using just elementary algebra – if not, doesn’t matter.

[quote="ValentineA"] if the problem could be solved using just elementary algebra – if not, doesn’t matter.[/quote]
Let $ (x,y,z)=1$. Obviosly  $ 2|x\to 2|y\to 2z$- contadiction.

[quote="ValentineA"]I was wondering if the problem could be solved using just elementary algebra – if not, doesn’t matter.[/quote]
That is the $ 2$-adic descent I (and Rust) hinted to. You can show, by a look $ \mod 2$, that $ x$ is even (multiply with a large integer first if you don't want to work with rationals). Then rearange and get that $ y$ is even as well. Continue ad infinity, so all that number are divisible by arbitrary high powers of $ 2$, that is: they are $ 0$.


close to the border of the betatron the magnetic field changes like this :$r^{-n}$ (where $r$ is the distance from the center of the field.)
for What $n$ will the horizontal plane movement of the electron be stable?

What do you mean when you're saying stable movement? :maybe:

I mean that the electron is stable on its path. It doesnt "want" to move around.

[hide="Hint"]write the small perturbations of the path of the electron[/hide]

Here is how we get the answer:
the equations of motion:
$m(\ddot r-r \dot \varphi^{2})=-er\dot \varphi B(r)$
$\frac{1}{r}\frac{d}{dt}m(r^{2}\dot \varphi)=e\dot r B(r)$
Let the radius of the wanted path be $r_{0}$, and let the magnetic field on this path be $B_{0}=B(r_{0})$.
the angular velocity of the electron here:
$\dot \varphi_{0}=\omega_{0}=\frac{eB_{0}}{m}$
lets look at a small perturbation of this path:
$r=r_{0}+x$ and $\omega = \omega_{0}+y$
the equations of motion for this perturbation are
$\ddot x=r_{0}\omega_{0}\bigg (y-\frac{B_{01}}{B_{0}}\omega_{0}x \bigg )$
$\dot y=-\dot x \frac{\omega_{0}}{r_{0}}$
Where $B_{01}=\frac{dB}{dr}|_{r=r_{0}}$
So the $x$ perturbation satisfies the following differential equation:
$\ddot x =-\omega_{0}^{2}\bigg (1+\frac{B_{01}}{B_{0}}r_{0}\bigg ) x$
So the movement is stable if:
$B_{0}\gneqq-B_{01}r_{0}$
assuming that $B(r)=B_{1}r^{-n}$ close to the path, we get the following inequality for the condition of stability:
$n \lneqq 1$


conics
ellipse
number theory unsolved
number theory

Show that $ 2x^4\plus{}2x^2y^2\plus{}y^4\equal{}z^2$ has no integer solutions with $ x\neq 0$.

[hide="Hint"]Let $ A\equal{}\dfrac{x}{z}$, $ B\equal{}\dfrac{y}{z}$. So, Ellipse:$ 2A^2\plus{}2AB\plus{}B^2\equal{}1$.[/hide]

I found a very simple divisibility argument.
[hide="Solution"]$ 2x^4\plus{}2x^2y^2\plus{}y^4 \equal{} z^2$ iff $ 2x^2(x^2\plus{}y^2)\equal{}z^2\minus{}y^4$.  Since $ x \neq 0$, neither side is $ 0$.  Hence $ 2$ divides the left side an odd number of times and the right side an even number of times.  This is a contradiction, so no such solution exists.[/hide]

That's wrong. If $ x\equal{}y\equal{}1$ then $ 2$ divides left side in an even power.


function
abstract algebra
linear algebra
linear algebra unsolved

As title says.

f is bijective and f:A->A, with |A|=n. Prove that f^(n!) is the identity map (where f^k denotes k compositions of f).

Ok, so I realise that there are n! cycles of length n, so that would motivate the answer, but it doesn't prove anything... Also, I'm technically not supposed to know what a cycle or group is, as we haven't covered it in lectures - only maps and sets.

Any hints where to start a proof would be appreciated.

Fix an element $ a \in A$.  What do you know about the least positive integer $ k$ such that $ f^k(a) \equal{} a$?

Or, almost the same hint, in a global (rather than local) approach: any $ f^k$ is also a bijective function between $ A$ and $ A$, but these are finitely many (how many?) ...

To get at what mavroponema's doing but without referring to groups or Lagrange's theorem:

The sequence $ \text{id},f,f^2,f^3,f^4,\dots$ has to repeat itself at some point. Could there be a "tail" leading into the cycle, or does the whole thing have to be contained in one cycle?

Side question: I see no linear algebra here. Where do you think we should move this topic?


$ {x}^3 \plus{}a{x}^2 \minus{} x \minus{}2 \equal{}0$

How many solutions does the equation have?

3. By the fundamental theorem of algebra.

similar question: For which $ a$ are the solutions all real?

We can look at the threshold case when there is a double-root.

$ (x-p)^2 (x-q) = x^3+ax^2-x-2$

$ \implies \left \{ \begin{array}{c} p^2+2pq = -1 & -p^2 q = -2 \end{array} \right \;$

We want the value of $ a=-p-q$

It turns out this is disgusting, and that the only possible solution in $ (p,q)$ yields the value $ 1.70556 \ldots$ (it's gross in exact closed form).

With a little calculus, we see that the requirement is $ a \geq 1.70556 \ldots$

Alternately, we can arrive at the same value by setting the derivative equal to zero, plugging in the negative root of the resulting equation into the original equation, and solving for $ a$. From this second method, it is immediately obvious from the final equation we solve for $ a$ that we will have $ 3$ roots iff. $ a$ is at least the root of the equation.


EveryOne, please post the pictures you have taken at the IMO 2005 either in the [url=http://www.mathlinks.ro/Forum/album_cat.php?cat_id=2]general IMO 2005 category[/url] or, if you have pictures from the nice hurricane slumber party on Sunday night, please post them in the [url=http://www.mathlinks.ro/Forum/album_cat.php?cat_id=5]dedicated album category[/url].

One of the things I remembered from the "slumber party" is that after the lights were turned off after midnight, a lot of people used camera flashes as their light sources (This annoyed me greatly, and I gave up trying to sleep after receiving constant flashing for about 15 minutes). This would also mean that a LOT of photos were taken that night.  ;)

[quote="Elyot"]Yufei, you need to post all your pics!  Come on... I know you took like 2938746782634 of them![/quote]Yeah guys! Post all the pictures :P

Okay, I've posted another round of pictures from the Slumber Party. I am still waiting for all of you to do the same :)

Oh my god, I look like such a dork in some of those.

I uploaded the photos I took that night.

Btw, is there a limit to how many photos we can upload here?

The party seems nice. Why didn't we have anything like that in our hotel? :(

There are several pictures labled mathlinkers but without mentioning their name......So please show their name.

I have a little problem. I tried to upload a pic in the photo album, and I couldn't do it because it was in a vertical position, so its height was 1600 and its width 1200. I could upload it after I rotated it (it's the picture of the Polish team in the IMO 2005 section), but it should be just normal that we can view it in the normal position. Can you fix this?

Thanks.

IMO Photos of the Belgian team are being collected at [url=http://imo.stijnshome.be/]http://imo.stijnshome.be/[/url]. I will post a lot of these pictures in the Mathlinks album too.

I have uploaded my pictures [url=http://igork91.free.fr/photos/.dirsys/list.php?0=IMO&1=IMO%202005&2=Igor]here[/url]. There will probably be other photos from the french IMO team members soon.

Igor, I love that server you have for putting up photos!

Come on! What happened to all the thousands of camera flashes that put me out of sleep?! You have to post them! Otherwise I would be very angry that the photographers gave me insomnia without giving me any compensation. :furious:

Here is an interesting game: try to find yourself in the overview photos. You may have to digitally adjust the exposure a bit in order to see anything far away. Amazingly, I was able to find myself in one of the photos with the pillow over my head (to block out the millions of camera flashes of course).

U r supposewd to be super happy u got a gold medal, that is spectacular, I am refering to Yufei!

Lots of official photos (some the same as in the report, but higher quality) can now be found [url=http://erdos.fciencias.unam.mx/omm/modules.php?op=modload&name=gallery&file=index&set_albumName=album20&include=view_album.php]here[/url].

[quote="billzhao"]I've uploaded my photos [url=http://ca.pg.photos.yahoo.com/ph/z_yufei/album?.dir=7340&.src=ph&store=&prodid=&.done=http%3a//ca.pg.photos.yahoo.com/ph/z_yufei/my_photos]here[/url].[/quote]

It is really shame that only a few people here upload their photos for the sake of experience sharing! It would be nice to have different subsections, named after the users intending to upload lots of pictures. Thanks to those who provided their photos.  :)

I've uploaded some of my pics, will finish at some point but it's a bit of a job, what with all the scanning and editing involved.

The pictures of the U.S. team have been posted here: http://www.unl.edu/amc/e-exams/e9-imo/e9-1-imoarchive/2005-ia/2005imopictures.html


Does somebody know the exact site on which the personal information about every IMO2005 contestant can be found?(those things we filled in the forms they gave us)

[quote="Rushil"]I don't believe they will release that info on the net!!![/quote]
We were told they would.

i hope they do... two of mexico's contestants didn't fill theirs and the rest of us were "kind" enough to fill it for them.  if it goes on the net it's gonna be a good laugh  ;)


search
function
LaTeX
algebra
polynomial
superior algebra
superior algebra unsolved

Prove that $p(x)=(x^{2}+x)^{2^{n}}+1$ is irreducible in $\mathbb{Z}[x]$

[quote="ZetaX"]Search function ;)[/quote]how would you search for that if I may ask? In fact I searched, but latex-search doesn't work anymore since about half a year, and "irreducible polynomial" brings us nowhere...

After searching for "irreducible" and "prove" in the Algebra forum, one can find http://www.mathlinks.ro/Forum/viewtopic.php?t=70003 , leading to http://www.mathlinks.ro/Forum/viewtopic.php?t=49799 .
Don't know if I had a more direct way in mind when posting it...

But yes, the lack of LaTeX-search is annoying...


can you help me find example of a function

$f: [0,1]\to R$

such that for every x , $f(x) >0$

for every n>1

$inf f(( \frac{1}{n}, 1]) >0$

f is not continuous in $0^{+}$

i mean $lim_{x\to 0^{+}}f(x) \neq f(0)$

 and $inf f([0,1])\leq 0$

Intertwine a function that goes to zero at zero and one that doesn't.


Show that $2005^{2005}$ is the sum of two squares, but not the sum of two cubes.

[quote="ZetaX"]$2005^{1002^2}$[/quote]

This was meant to be $\left(2005^{1002}\right)^2$, I believe?


I found a site that has past USSR problems. Is this site contains the same problems as the USSR book? Thanks.

Which site? Which book? If you mean http://members.tripod.com/~PertselV/RusMath.html and the USSR Olympiad Problem Book, then no...

Yeah, that's the site I meant. So, they're totally different? (Like 100%?)

Well umm the site is just a whole bunch of Olympiads and the book is a whole bunch of selected problems so how am I supposed to know?

(Seriously though, why does it matter?  There are plenty of resources out there.)

The book you are referring to is more structured.  It it broken up into sections like Introductory Problems, Evaluating Sums and Products, and Some Distinctive Inequalities among 9 other chapters.  It seems to be much different from tests.  For example, after they ask one question they will sometimes have another related question or generalization right after that.


Guys I know that there is a topic wher the Mathlinkers have discussed the following problem,but I can't find it,I need your halp :? .
Problem.
Prove that if $2^na+b$ is perfect square for all not negative integer n($a,b\in{Z})$,then $a=0$.
Thank you.

[quote="ZetaX"]http://www.mathlinks.ro/Forum/viewtopic.php?t=23626[/quote]
Thank you so much ZetaX ;) .


Each time I try to open this thread: http://www.mathlinks.ro/Forum/viewtopic.php?t=20072 (Geometry, Unsolved Problems, "Equal"), the server seems to hang; at least, the thread is not displayed and I also get difficulties with viewing the other threads. Probably there's some endless PHP loop caused by some LaTeX garbage? I already had such a problem with a thread of mine where I believe I posted some idiot pseudo-LaTeX commands, but it was quickly repaired. Any help regarding the new hanging thread?

  Thanks,
  Darij

Indeed it took 15 minutes to load that thread but finally it did show. And indeed there was a trivial LaTeX error, repeated 4 times. Please remember that LaTeX commands like "\angleBKC" are nonsense. It should be "\angle BKC". It's faaaar not the first time I see this error, but it's one of the first times it causes such troubles.

  dg


I wonder if you can really solve this problem:

There is a repeating sequence as follows:
1) A line segment AB with length x is drawn on a plane (the geometric type of plane). 2) At AB's midpoint C, an endpoint of another segment is situated. This new segment is half as long as the original, and perpendicular to AB.  3) The new line's (CD's) midpoint is found and this we name E.  Now another segment with one endpoint as E is perpendicular to CD, and half as long as CD.  This segment is EF. 4) We repeat.  5) We repeat that again.  6)...

Imagine we repeat this infinite times.  The lines converge at a point I will call P.  (Note that the process described above, when it continues, at some point it will name one of the endpoints of a new segment "P".  Say the process skips P so that it goes like "N, O, Q, R, ...".  When the process reaches Z, it goes to A'.)  Find the distances PA and PB in terms of x.

Also, I would like to question...

Can you, given any point in the sequence mentioned, find the distance from P to the aforementioned point?

What if the sequence is administered at an arbitrary angle $\theta$?

Even worse, what if it is done in arcs which are perpendicular at the point of intersection, the arcs have lengths x(pi), x(pi)/2,... and they all take up an angle of $\theta$ out of a circle where $0<\theta <\pi$?

:huh: Why is this in basics?  It should probably be moved at least to the intermediate forum.

It isn't very difficult at all - it's just a couple infinite geometric series:

From point A, the location of P moves $\frac{x}{2}$ right, $\frac{x}{8}$ left, $\frac{x}{32}$ right, etc. The geometric scalar is $-\frac14$, meaning point P is $\frac{2x}{5}$ right of A and $\frac{3x}{5}$ left of B.

Similarly, P is $\frac{x}{5}$ above A (and B as well).

PA = $\frac{x^{2}}{5}$.

PB = $\frac{2x^{2}}{5}$.


The product of the digits of 3214 is 24. How many distinct
four-digit positive integers with unit digit not equal to 1 are such that the product of their
digits equals 18?

Case 1: 1129 and permutations
Case 2: 1136 and permutations
Case 3: 1233 and permutations

All cases have $ \frac{4!}{2!}\equal{}12$ combinations, so there are $ 36$ possible combinations. However, as I was typing this, I noticed that the unit digit can't be equal to 1. :P

Case 1 and Case 2 both have 6 numbers ending with 1. Case 3 has 3 numbers ending with 1. (Put a 1 in the units digit, then see how many possible combinations you can have with the remaining digits.) Therefore the final final answer is $ 36\minus{}6\minus{}6\minus{}3\equal{}21$.


how come when i type out my math problems it doesn't change font?
also how do i type the radical sign and exponents?

Read the complete guide of [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]LaTeX[/url].

Also for this forum, enclose your text in dollar signs. For example

[code]$\frac{2}{3}$ or $\sqrt[3]{29}$[/code]

will give

$ \frac{2}{3}$ or $ \sqrt[3]{29}$

The best place on the forum to learn about LaTeX for the purposes of using it on the forum (rather than to create your own documents) is http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690

[quote="7h3.D3m0n.117"]Read the complete guide of [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]LaTeX[/url].
[/quote]

OUTDATED.
http://www.artofproblemsolving.com/Wiki/index.php/LaTeX


Funtion $ F(x)$: $ R \to R$ such that: $ f(x\plus{}2xy)\equal{}f(x)\plus{}2f(xy)$

We know $ f(1996)\equal{}1996$, prove that $ f(1997)\equal{}1997$.

we take $ f(1)\equal{}a$
$ f(n\plus{}2)\equal{}f(n)\plus{}2a$ gives $ f(1996)\equal{}1996a$ and $ f(1997)\equal{}1996a\plus{}a$ :wink:

$ y\equal{}0\to f(0)\equal{}0$.
$ y\equal{}\frac{1}{2x}\to f(x\plus{}1)\equal{}f(x)\plus{}b, \ b\equal{}2f(\frac 12) \ \forall x\not \equal{}0$.
$ y\equal{}\frac 1x \to f(x\plus{}2)\equal{}f(x)\plus{}2f(1)\to f(1)\equal{}b$,
therefore $ f(n)\equal{}nb$ and from $ f(1996)\equal{}1996$ we get $ b\equal{}1$.


There was a Feb regional last saturday...
I'm wondering who got what place, and which school got what place- so I'll have a heads up before the results come out online.
Of course, over here, Fairview won.... *brushes shoulder*
It was a stupid test if you ask me!

I went, and got a bad score, and messed up the test. I thought the test was screwed up. Like one of the questions on our goemetry test was like which of the following angles can be constructed? The answer choices were $22\frac{1}{2}$,$82\frac{1}{2}$,$120$ and $60$. The whole competition was screwed up.

[quote="mathgeniuse^ln(x)"][quote="Elemennop"]It's obvious everything except 88.5 can be constructed..[/quote]

It says 82.5, and that can be constructed.

All angles can be constructed.[/quote]

mathgeniuse is correct, and elemenop, i put 82.5, but then realized i was wrong.

eh who's asianman

that test was incredibly queer

and the trophies were fake

edit
DUDE ITS LIKE DANIEL

who finally beat payal took you long enough

[quote="Elemennop"][quote="mathgeniuse^ln(x)"][quote="Elemennop"]It's obvious everything except 88.5 can be constructed..[/quote]

It says 82.5, and that can be constructed.

All angles can be constructed.[/quote]

oh

really

i thought it was only multiples of 3 or something

I dunno

can you half any angle? I suppose that means you can construct any angle.[/quote]
You can half any angle (how else would you construct incenter) but for integers you can only construct multiples of 3.
82.5*2=165=55*3 so first you'd need to construct a 165 degree angle then bisect it.


Try to generalize to $n$ variables this ineq

For $a,b,c,d$ positive real numbers prove that

${\displaystyle \frac{(a^2+b^2)(b^2+c^2)(c^2+a^2)(a^2+d^2)(d^2+b^2)(d^2+c^2)}{64}} \geq(\frac{ab+ac+ad+bc+bd+cd}{6})^6$

See also posts #7-#8 in http://www.mathlinks.ro/Forum/viewtopic.php?t=28730 .

  darij


Given that $ 0.05n \equal{} 1.25$, evaluate $ n \times 10^{\minus{}2}$. Express your answer as a decimal to the nearest hundredth.

divide by .05 to get rid of it and get 25

plug into the equation and get 0.25

[hide=" this is your answer"]0.25[/hide]

[hide="read beyond this"][size=200][color=#00BFFF][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH IT IS GREATLY APPRECAITED !![/u][/i][/b][/color][/size][/hide]

[size=50][hide="tHIs iS gEttINg SmaLLEr"]i have 99.9999999% confidece that im right !![/hide][/size]


function
limit
calculus
derivative
real analysis
real analysis unsolved

Let f be a function of a real variable such that one of its derivates is non-negative, say $D^+$ where

$D^+ f(x) = \overline{\lim_{h \to 0^+}} \frac{f(x+h)-f(x)}{h}$

then show that f is non-decreasing.

any ideas?

Well, you should assume something extra (say, continuity of $f$). Otherwise $f(x)=\left\{\begin{aligned} 1,&\quad x<0 \\ 0,&\quad x\ge 0\end{aligned}\right.$ is an obvious counterexample...

Then it is a classical theorem. Assume first that the upper right derivative is strictly positive at each point. If $a<b$ and $f(a)>f(b)$, consider the point $x=\sup\{t\in [a,b]\,: \, f(t)\ge f(a)\}$. Continuity implies that $f(x)=f(a)$. We also have $f(t)<f(a)$ everywhere on $(x,b]$, from where it follows immediately that the upper right derivative at $x$ must be non-positive, so we got a contradiction. Now, assume that the upper right derivative is only non-negative and consider the function $F_\delta(x)=f(x)+\delta x$ with small positive $\delta$. The above proof yields that $F_\delta$ is non-decreasing for every $\delta>0$ and it remains to pass to the limit as $\delta\to 0+$.


Find all real numbers $\alpha$ for which the equality \[\lfloor \sqrt{n}+\sqrt{n+\alpha}\rfloor =\lfloor \sqrt{4n+1}\rfloor\] holds for all positive integers $n$.

Notice that if $ \alpha <0$ then it is not true for $ n\equal{}1$
$ [\sqrt {1} \plus{} \sqrt {1 \plus{} \alpha}] < 2\equal{}[\sqrt {5}]$
Same for $ 3 \le \alpha$
$ [\sqrt {1} \plus{} \sqrt {1 \plus{} \alpha}]\equal{}[1 \plus{} \sqrt {1 \plus{} \alpha}] >2\equal{}[\sqrt {5}]$

$ 1 \le \alpha$
$ 1/2 \le \alpha /2$
$ n^2 < n^2 \plus{} n \alpha$
$ n <  \sqrt {n^2\plus{}n \alpha}$
$ n\plus{}1/2 \le \alpha /2\plus{} \sqrt {n^2\plus{}n \alpha}$
$ 2n\plus{}1 \le \alpha \plus{}2 \sqrt {n^2\plus{}n \alpha}$
$ 4n\plus{}1 \le 2n\plus{}\alpha\plus{} 2 \sqrt {n^2\plus{}n \alpha}$
$ \sqrt {4n \plus{} 1} \le \sqrt {n} \plus{} \sqrt {n \plus{} \alpha}$

$ 1\le \alpha < 3$
$ \alpha /2< 3/2$
$ n^2\plus{}n \alpha < n^2 \plus{}3 n < n^2$
$ \sqrt {n^2\plus{}n \alpha} < n$
$ \alpha /2\plus{}\sqrt {n^2\plus{}n \alpha}< 3/2\plus{}n$
$ \alpha /2\plus{}\sqrt {n^2\plus{}n \alpha}< n\plus{}1\plus{}\sqrt {4n \plus{} 1}$
$ \alpha \plus{}2 \sqrt {n^2\plus{}n \alpha}< 2n\plus{}2\plus{}2 \sqrt {4n \plus{} 1}$
$ 2n\plus{}\alpha\plus{} 2 \sqrt {n^2\plus{}n \alpha}< 4n\plus{}1\plus{}2 \sqrt {4n \plus{} 1}\plus{}1$
$ (\sqrt {n} \plus{} \sqrt {n \plus{} \alpha})^2 < (\sqrt {4n \plus{} 1}\plus{} 1)^2$
$ \sqrt {n} \plus{} \sqrt {n \plus{} \alpha}\minus{} \sqrt {4n \plus{} 1} < 1$
then $ 0 \le \sqrt {n} \plus{} \sqrt {n \plus{} \alpha}\minus{} \sqrt {4n \plus{} 1} < 1$ for $ 1\le \alpha <3$

Idem for $ 0\le \alpha \le 1$ but $ 0 \le  \sqrt {4n \plus{} 1}\minus{}\sqrt {n} \plus{} \sqrt {n \plus{} \alpha}< 1$

[quote="Peter"]Find all real numbers $ \alpha$ for which the equality
\[ \lfloor \sqrt {n} \plus{} \sqrt {n \plus{} \alpha}\rfloor \equal{} \lfloor \sqrt {4n \plus{} 1}\rfloor
\]
holds for all positive integers $ n$.[/quote]
Because $ (\sqrt x )'' < 0$ we get $ \sqrt x \plus{} \sqrt {x \plus{} \alpha } < 2\sqrt {x \plus{} \alpha/2} \equal{} \sqrt {4x \plus{} 2\alpha}$.
If $ \alpha>2$ for $ n\equal{}k^2\minus{}1$ we get
  $ \sqrt {k^2 \minus{} 1} \plus{} \sqrt {k^2 \minus{} 1 \plus{} \alpha } < 2k$.
It is equavalent to $ (\alpha \minus{}2)k^2 < \alpha ^2$ and false for big $ k$.
Therefore solution is $ 9 \minus{} 6\sqrt 2 \le \alpha \le  2$.
If it must be true for $ n \equal{} 0$ solution is $ 1\le \alpha \le 2.$

Rust: Do you have a counterexample for $ \alpha >2$ ?
Could you please find the mistake in my previous message?

If $ \alpha > 2$, then for $ n \equal{} k^2 \minus{} 1, \ k > \frac {\alpha }{\sqrt {\alpha \minus{}2}}$ give contradition.

I agree with Rust that the equality holds for all $n$ if, and only if, $9 - 6\sqrt{2} \le \alpha \le 2$. And note that $9 - 6\sqrt{2} \approx 0.51$.

[b]Proof.[/b] For $n = 1$ we can readily check that the equality holds for all $\alpha$ with $0 \le \alpha < 3$, so in particular for $\alpha \in [9 - 6\sqrt{2}, 2]$. We may therefore assume $n \ge 2$. To prove our claim, we must show both the necessity and the sufficiency of both the lower and upper bound. \\ To prove the necessity of the lower bound, we look at $n=2$. It follows that $\alpha$ must be such that $\sqrt{2+\alpha} \ge 3 - \sqrt{2}$. Squaring both sides and subtracting $2$ gives $\alpha \ge 9 - 6\sqrt{2}$. \\ To prove the necessity of the upper bound, assume by contradiction $\alpha = 2 + \epsilon$ with $\epsilon > 0$ and choose $n > \frac{1}{\epsilon}$ such that $n+1$ is a square, say $n+1 = k^2$. Then we have the following inequalities:

\begin{align*}
4n(n + 2 + \epsilon) &> 4n^2 + 8n + 4 \\
2\sqrt{n}\sqrt{n+2+\epsilon} &> 2n + 2 > 2n + 2 - \epsilon \\
2n+2+\epsilon + 2\sqrt{n}\sqrt{n+2+\epsilon} &> 4n + 4 \\
(\sqrt{n} + \sqrt{n+2+\epsilon})^2 &> 4n + 4 \\
\sqrt{n} + \sqrt{n+2+\epsilon} &> 2\sqrt{n+1} = 2k
\end{align*}

On the other hand, $\sqrt{4n+1} < \sqrt{4n+4} = 2k$, contradicting the equality. \\

To prove that these bounds are also sufficient, first we claim that $\sqrt{n} + \sqrt{n+\alpha} \ge \sqrt{4n+1}$ holds for all $n \ge 2$, whenever $\alpha \ge 9 - 6\sqrt{2}$. It suffices to prove this for $\alpha = 9 - 6\sqrt{2}$, so let us assume that for the moment. By the quadratic formula we see $8(2\alpha-1) - (\alpha-1)^2 = -\alpha^2 + 18\alpha - 9 = 0$. For $n \ge 2$ we therefore have the following inequalities:

\begin{align*}
4n(2\alpha-1) - (\alpha-1)^2 &\ge 0 \\
8\alpha n &\ge 4n + 1 - 2\alpha + \alpha^2 \\
4n^2 + 4\alpha n &\ge 4n^2 + 4n + 1 - 4\alpha n - 2\alpha + \alpha^2 \\
4n(n + \alpha) &\ge (2n+1)^2 - 2(2n+1)\alpha + \alpha^2 = ((2n+1) - \alpha)^2 \\
2\sqrt{n}\sqrt{n+\alpha} &\ge 2n + 1 - \alpha \\
2n + \alpha + 2\sqrt{n}\sqrt{n+\alpha} &\ge 4n + 1 \\
(\sqrt{n} + \sqrt{n+\alpha})^2 &\ge 4n + 1 \\
\sqrt{n} + \sqrt{n+\alpha} &\ge \sqrt{4n+1}
\end{align*}

With the above inequality in mind, we can finish the proof by showing that $\lfloor \sqrt{n} + \sqrt{n + 2} \rfloor \le \lfloor \sqrt{4n+1} \rfloor$ holds for all $n$. Assume $k^2 \le n < (k+1)^2$. We then have to distinguish between $n < k^2 + k$ or $n \ge k^2+k$. \\ If $k^2 \le n \le k^2 + k - 1$, then $\lfloor \sqrt{4n+1} \rfloor = 2k$ and $\lfloor \sqrt{n} + \sqrt{n + 2} \rfloor \le \lfloor 2\sqrt{n+1} \rfloor = 2k$, since $\sqrt{n+1} < k + \frac{1}{2}$. \\ On the other hand, if $k^2 + k \le n \le k^2 + 2k$, then $\lfloor \sqrt{4n+1} \rfloor = 2k+1$. And $\lfloor \sqrt{n} + \sqrt{n + 2} \rfloor \le 2k+1$, since $\sqrt{n} + \sqrt{n + 2} < 2\sqrt{n+1} \le 2k+2$.


a circle is inscribed in a triangle with sides length of 4cm, 13cm and 15cm, find the area of the circle.

[hide]
By Heron's, the area of the triangle is 24.

$ rs \equal{} 24 \implies r \equal{} \frac {3}{2}\implies \text{Area} \equal{} \boxed{\frac {9\pi}{4}}$
[/hide]


Solve the following trigonometric equation:

$cos12x=5sen3x+9(tanx)^{2}+(cotx)^{2}$

How many solutions does it have in $[0;2\pi]$ :?:

This has a nice solution. Just think why we have $9$, $\tan x$ and $\cot x$ on the right hand side of the equation. Think AM-GM!

Here's the complete solution anyway if you want to have a look at it.

[hide]
We first rewrite our equation: $\cos 12x-5\sin 3x = 9{\tan}^{2}x+{\cot}^{2}x .$

If $\tan x = 0$ then $\cot x$ is undefined, and if $\cot x = 0$ then $\tan x$ is undefined. So, we assume $\tan x \neq 0$ and $\cot x \neq 0$. Applying $AM-GM$ to the $RHS$, we get

$\frac{9{\tan}^{2}x+{\cot}^{2}x}{2}\geq \sqrt{9{\tan}^{2}x \cdot{\cot}^{2}x}$
$\Rightarrow RHS \geq 6$, with equality occurring iff $9{\tan}^{2}x ={\cot}^{2}x .$

Now, note that $|LHS| \leq 6$. So, the only solution possible is when the following is satisfied: $LHS=RHS=6 .$
So, ${\cos 12x-5\sin 3x}= 9{\tan}^{2}x+{\cot}^{2}x = 6$, in which case we must have $9{\tan}^{2}x ={\cot}^{2}x .$

Solving $9{\tan}^{2}x ={\cot}^{2}x$, we obtain $\tan x = \frac1{\sqrt 3},-\frac1{\sqrt 3}$. This yields the following possible values for $x$: $\frac{\pi}{6}, \pi+\frac{\pi}{6}, \pi-\frac{\pi}{6}, 2\pi-\frac{\pi}{6}.$

Out of the possible values for $x$ above, only $x=\pi+\frac{\pi}{6}$ and $x = 2\pi-\frac{\pi}{6}$ satisfy the equation ${\cos 12x-5\sin 3x}= 6 .$

Answer: $\pi+\frac{\pi}{6}, 2\pi-\frac{\pi}{6}.$
[/hide]


calculus
integration
number theory unsolved
number theory

I'm stuck on the following :huh:
Let p>3 be prime, let m = <2p/3> (where <x> is the integral part of x)
Prove that C(p,1)+C(p,2)+...+C(p,m) is divisible by p^2 (where C(n,k) - binomial coefficient "choose k from n").

thanks! i am not used to taking fractions mod p, but that seems to be amazingly reasonable sometimes :)


vector
combinatorics unsolved
combinatorics

An n x n Latin square has the numbers from $1$ to $n^2$ arranged in its cells (one per cell) so that the sum of every row and column is the same. For every pair of cells in a Latin square the centers of the cells are joined by an arrow pointing to the cell with the larger number. Show that the sum of these vectors is zero.

Posted some weeks ago (at least two times):
http://www.mathlinks.ro/Forum/viewtopic.php?t=79208


there are n urns.
each urn has a white balls and b black balls.
we extract at random a ball from the first urn and put it into the second one.
we extract at random a ball from the second urn and put it into the third one.
and so on.
at the last step we extract a ball from the last urn.
what is the probability of the last ball extracted to be white?

P.S.
I've worked on a solution but maybe there are other solutions nicer than this one.

[hide="hint"] consider the probability for the second urn [/hide]
[hide="answer"] just the same, $ \frac {a}{a \plus{} b}$ [/hide]
[hide="solution"] we find the probability of moving white ball from the second urn to the third urn:
$ p_{2} \equal{} \frac {a}{a \plus{} b}\frac {a \plus{} 1}{a \plus{} b \plus{} 1} \plus{} \frac {b}{a \plus{} b}\frac {a}{a \plus{} b \plus{} 1} \equal{} \frac {a}{a \plus{} b}$
the probability has not changed, so $ p_{100} \equal{} \frac {a}{a \plus{} b}$ [/hide]


logarithms
abstract algebra
ratio
LaTeX
limit
inequalities

okay it seems that I have to re-test this so that people will again start to post :)

Prove that [tex]\frac {x^2}{y^2}+\frac{y^2}{z^2}+\frac {z^2}{x^2} \geq \frac yx + \frac zy + \frac xz[/tex].
[/tex]

I am always happy to post mathematics but will just restrict myself to just this one:
[tex]\left(\mathbb{R}^{+}-\left\{1\right\},\circ\right)[/tex] where [tex]\circ[/tex] is defined by [tex]x\circ y=x^{\ln y}[/tex] is an abelian group.

I'm quite exciting about having a chance of learning Tex, so I'll just go wild for a while :D

[tex] \sqrt[1200]{ax^2+bx+c} \\\aleph \\\frac{1}{b+\frac{c}{d}}}[/tex]

whow!
the golden ratio:
[tex] \varphi = \frac 1 {1 + \frac 1 {1 + \frac 1 { 1 + \frac 1 \ddots }}} \\
\varphi^2 = \varphi + 1 \\
\varphi = \frac {\sqrt 5 + 1} 2 \\
1^2 + \frac 1 2 ^2 = \varphi^2 \\
\varphi : 1 = 1 : \varphi - 1[/tex]

[tex] a^2+b^2+c^2 \geq ab+bc+ca [/tex] for any [tex] a,b,c>0 [/tex]

That's strange:
 [tex] a^2+b^2> \sqrt [n] {a}+ \sqrt[n] {b} [/tex]

[quote="harazi"]How can I write idiot in latex? It may be like:
  [tex]\prod_k\frac{1}{a_k}[/tex].
 Or maybe [tex]\sqrt[2004] {ax^{2003}+bx^{2002}+1} [/tex]. And the question is: when do we have [tex]\lim_n\sqrt[n]{n!}-\frac{n} {e}}=\pi? [/tex][/quote]
or you can write  [tex]\prod_{k=1}\frac{1}{a_k}[/tex] and [tex]\lim_{n\to\infty} \sqrt[n]{n!}-\frac{n} {e}}=\pi? [/tex] ;)

And another opera of Harazi, the greatest latex writer of all times!
   Let us write one of Vasile Cartoaje's inequality:

    Prove that for any [tex] a_1,...,a_n>0 [/tex] such that [tex] a_1\cdot a_2\cdot...\cdot a_n=1 [/tex] and for any natural [tex] k>1[/tex] we have the inequality:
   [tex]\sum_i\frac{1}{\sqrt[k]{1+(n^k-1)a_i}}\geq 1 [/tex]. Yes! I'm too good!
    Some limits and relations:
    [tex]\lim_n\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}=\frac{1}{e}[/tex]
    [tex]\dim_k U+\dim_k\frac{V}{U}=\dim_k V[/tex]

Ok, let's solve Vali's problem in Latex this time. Be sure the problem will be harder this way:
  So, we must prove that [tex]\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}\geq\frac{y}{x}+\frac{z}{y}+\frac{x}{z} [/tex]. But we have the inequality [tex] a^2+b^2+c^2\geq a\cdot b+b\cdot c+c\cdot a[/tex] and if we put in this inequality [tex] a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}[/tex] we find exactly the inequality.
   Now, how can we compute [tex]\sum_{k=0,n}\k\cdot\frac{n!}{k!\cdot(n-k)!}[/tex]? Is it [tex] 2^n [/tex]? No, of course not! Ok, what have we said this evening? That the generalization of Turkevici inequality is hard to prove. But what is this generalization? Is it equivalent to:
  Prove that for any [tex] x_1,...,x_n\in {R}[/tex] such that [tex]\prod_{i=1,n}x_i=1 [/tex] we have the inequality [tex]\sum_{1\leq i<j\leq n} (x_i-x_j)^2\geq\sum_{i=1,n} x^2_i -n[/tex]

one usually uses [tex]\sum^n_{k=1} x_k [/tex] rather than [tex]\sum_{k=1,n} x_k [/tex] but I guess that's your own choice ;) :D:D

So, I suppose that writing [tex]\sum_{1\leq i_1\leq i_2\ ... \leq i_k\leq n}\frac{n!}{i_1!\cdot i_2!\cdot ... \cdot i_k!} [/tex] is not better than [tex]\sum^\frac{n}{2}_{k=1}[\frac{n}{k^2}]\cdot\phi(n^2+k^2) [/tex].

[tex]\{n \in \mbox{N*}\}[/tex]

[tex]\left\{\begin{array}{cc}\frac25, &\frac2{23}\\\sqrt{a^2+b^78} &a^b\end{array}\right][/tex]


Hi! What do you think about the following pictures?
Do you like the shape and colors, can you easily remember it? Any comments will be welcomed!

it reminds me of runes for some reason though it isn't...

i think they're both nice. enough symmetry to remember it to.

It looks like one of the runes on the title of the fourth Led Zeppelin album

Thanks for your answers. Actually, I designed it but I do not if there are any similar symbols. I am trying to make some improvements right now... adding blue... any ideas?

i know, superimpose a neon inverted cross.


I've noticed that when looking through the memberlist, it's very hard to get to some of the middle pages. For example, say you want to find the people in Connecticut. You would sort by location in ascending, and then you would have to go page by page until you get to page 16. You could of course easily get there by changing the start value in the address line to 750, but it would be nice if there was a more obvious way to jump directly to a certain page.

Are you referring to the fact that at the bottom, the numbered links only show you 1, 2, 3, ... 30, 31, 32?

[quote="Yasha"]You could of course easily get there by changing the start value in the address line to 750, but it would be nice if there was a more obvious way to jump directly to a certain page.[/quote]

Actually, your way is a nice, easy hack that I just tried out, and I liked it. I've been doing that on Amazon.com for years, but I never tried hacking URLs to jump through this site's now rather large member list. That's a good tip.


function
inequalities
calculus
calculus computations

Function $ y(x)$ is a solution of the differential equation
\[ y' \equal{} \frac {1}{\pi}\arctan(xy \plus{} 1) \plus{} \frac {1 \plus{} 2xy}{2 \plus{} 2x^2 \plus{} 2y^2}
\]
with condition $ y(0) \equal{} 0$. Prove that for every $ x > 0$ we have: $ y(x) < x$.

Maybe this is a hint for this problem.
[hide]
Write $ y(x) - x = h(x)$ then $ y(x) = h(x) + x$, rewrite the question in the form :
$ h'(x) + 1 = \frac {1}{\pi}\arctan(x(x + h(x)) + 1) + \frac {1 + 2x(x + h(x))}{2 + 2x^2 + 2(x + h(x))^2}$
with condition $ h(0) = y(0) - 0 = 0$. Prove that for every $ x > 0$ we have: $ h(x) < 0$.
Now we will prove by contra position, supposed that we have $ h(a) > 0$ for some $ a > 0$,then there exist $ b > 0$ such that
$ h'(b) + 1$ is not equal to $ \frac {1}{\pi}\arctan(b(b + h(b)) + 1) + \frac {1 + 2b(b + h(b))}{2 + 2b^2 + 2(b + h(b))^2}$
First, note that the function $ h(x)$ is continuous because $ y$ is differentiable and $ y(x) + x = h(x)$ is also differentiable and the conclusion follows.
Because $ h(x)$ is continuous and there exist $ a > 0$ such that $ h(a) > 0$ then we can claim that there is an interval $ [K,L]$, where $ K > 0$ such that $ h(x)$ is monotonic increasing in $ [K,L]$.
Proof : If $ h(x)$ is monotonc nonincreasing everywhere in $ (0,\infty)$, then $ h'(x) \geq 0$, then $ h(x) < 0$ for all $ x$, contradiction.

Now, because $ h(x)$ is continuous and there exist $ [K,L]$ such that $ K > 0$ then $ h'(m) > 0$ for all $ m \in [K,L]$
$ h'(m) + 1 > 1$.
In other side for all $ b > 0$ we have that :
$ \frac {1}{\pi}\arctan(b(b + h(b)) + 1) < \frac {1}{2}$ (because $ \arctan(b(b + h(b) + 1)) < \frac {\pi}{2}$ for all $ b > 0$ and $ \frac {1 + 2b(b + h(b))}{2 + 2b^2 + 2(b + h(b))^2} \leq \frac {1}{2}$ (because  $ 0 \leq b^2 + (b + h(b))^2 - 2(b(b + h(b)))$ then $ 1 + 2(b(b + h(b))) \leq b^2 + (b + h(b))^2 + 1$ then $ \frac {1 + 2(b(b + h(b)))}{b^2 + (b + h(b))^2 + 1} \leq 1$ and the conclusion follows.)
Adding both inequality we obtain that :
$ \frac {1}{\pi}\arctan(b(b + h(b)) + 1) + \frac {1 + 2b(b + h(b))}{2 + 2b^2 + 2(b + h(b))^2} < 1$
The left hand side in the differential equation is $ > 1$ and the right hand side in the differential equation is $ < 1$ for $ x = b > 0$, then the inequality cannot be obtained in $ x = b > 0$.
(Q.E.D)
[/hide]
Sorry if I am wrong.

The first term is strictly less than $ \frac {1}{2}$, so it suffices to show that $ \frac {1 \plus{} 2xy}{1 \plus{} x^2 \plus{} y^2} \leq 1$ for all $ (x,y)\in (0,\infty)\times \mathbb{R}$, which follows directly from $ (x \minus{} y)^2 \geq 0$.

Edit: Typos.

[quote="Qruni"]
Because $ h(x)$ is continuous and there exist $ a > 0$ such that $ h(a) > 0$ then we can claim that there is an interval $ [K,L]$, where $ K > 0$ such that $ h(x)$ is monotonic increasing in $ [K,L]$.
Proof : If $ h(x)$ is monotonc nonincreasing everywhere in $ (0,\infty)$, then $ h'(x) \geq 0$, then $ h(x) < 0$ for all $ x$, contradiction.
[/quote]

Hi, i think that's not true. For example, function $ h(x)$ can start from values $ > 0$ and non increese. This means that interval where $ h'(x) > 0$ does not exist. And if $ h(x)\leq 0$ there may not be such points where $ h(x) < 0$ :( If they exist then we can say that $ h(x) < 0$ only for all $ x > x_0$ where $ x_0$ is a solution of equation $ h(x) = 0$.

[b]JoeBlow[/b], thats the original solution :)


Solve the following equation in Z_17 (means mod 17) for both equations:
$[5][x]+[4][y] = [5]$
$[6][x]+[9][y] = [1]$

[hide="Looky here"] $Z_{17}$ is a field so you can deal with this system in usial way.
For instance it's possible by multiplication make nonzero coefficients of your choise became [1]:
$[5]\cdot[7]=[35]=[1]$
$[6]\cdot[3]=[18]=[1]$[/hide]


Suppose that $X \sim G(p)$ where $ 0\leq p \leq 1 $ and $G(p)$ is refer to the geometric distribution whish is given by
\[ Pr\{X=k\}= p(1-p)^{k-1}I_{1,2,\cdots}(k) \]
The $E_k(X)$ can be given as
\[ E_k(X)= \displaystyle\sum_{k=1}^{\infty} k p(1-p)^{k-1}= \frac{1}{p}\]
Now, how would you get the variance of $X$?

Since $\displaystyle\sum_{k=1}^\infty p(1-p)^{k}=1 $ then we can write $\displaystyle\sum_{k=1}^\infty (1-p)^{k}=\frac{1}{p} $, and by taking the first derivative we will get $\displaystyle\sum_{k=1}^\infty k (1-p)^{k}=\frac{1}{p^{2}} $. By multiplying both sides by $1-p$ and taking derivative we will get
$\displaystyle\sum_{k=1}^\infty k^{2} (1-p)^{k-1}=\frac{2-p}{p^{2}} =E(X^{2})$. Therefore $Var(X)= \frac{2-p}{p^2} - \frac{1}{p^2}=\frac{1-p}{p^2}$

thank you  :)


function
group theory
abstract algebra
algebra unsolved
algebra

Let A={1,2,3,...,m+n}, where m and n are positive integers and f:A-->A defined by:
f(i)=i+1  for  i=1,2,3,...,m-1,m+1,...,m+n-1
f(m)=1 and f(m+n)=m+1.

a) Prove that if m and n are odd then there exists a function g:A-->A such that g[g(a)]=f(a) for all a in A.
b) Prove that if m is even then m=n If there exists a function g:A-->A such that g[g(a)]=f(a) for all a in A.

If $m,n$ are odd, then the permutation is the product of the two odd cycles $(1,2,\ldots,m)$ and $(m+1,m+2,\ldots,m+n)$. An odd cycle is a square (the square of another odd cycle, actually; the subgroup of $S_m$ generated by this cycle has odd order, so $x\mapsto x^2$ is surjective), so the permutation, being the product of two squares of permutations which commute, is, itself, a square.

If $m$ is even, on the other hand, then I don't think there can be such a $g$, so this looks really confusing to me. :?


geometry
power of a point
radical axis
geometry unsolved

Let $w_b,\ w_c$ be the circles which pass through the point $A$ and are tangent to the line $BC$ in the points $B,\ C$ respectively. I note the orthocenter $H$, the midpoint $M$ of the side $BC$ and $\{A,L\}\in w_b\cap w_c$. Prove that $L\in AM,\ HL\perp AM$.
See http://www.mathlinks.ro/Forum/viewtopic.php?t=52562

the first part, about proving the collinearity, it's very simple...
let $M'$ be the intersection of $AL$ with $BC$, then, since $AL$ is the radical axis we have by power of a point that $M'B^2=M'C^2 \Rightarrow M'B=M'C$, so M' is the midpoint of BC, and then M and M' are the same point.


Find K max, G min such that $\forall \{a_i\}_{i=1}^n;a_i>0$ we have:
 $ K<\sum_{i=1}^n \displaystyle\frac{a_i}{a_i+a_{i+1}} <G $ :P

If I'm not wrong, they are 1 and n-1. I do not have a paper, but I think some brute force majorations are enough.

Yes,K=1 and G=n-1.I solved this problem yesterday!

May I ask what "LAV collection" means. Could you please describe what it is ? Thanks.  :)

Dear orl,
LAV collection = Le Anh Vinh collection.Le Anh Vinh had been award a silver medal in the 42 IMO . These problems were taken out from Le Anh Vinh's notebook. He had selected and recorded them. All of them are problems I haven't solved. :( So I send these to you, please solve them with me!:P .
To Malio:Three days ago, I was idiot and I haven't completely solved this problem(Ib_8).Malio, I'm sorry that I haven't sent solution for you.I'll send it to you when i complete this. ;)


[quote="SpaceWeather.com"]Space Weather News for Dec. 17. 2007
http://spaceweather.com

CLOSE ENCOUNTER WITH MARS: Have you noticed an intense red star rising in the east in recent evenings?  That's no star, that's Mars.  The red planet is having a close encounter with Earth this week.  At closest approach on Tuesday evening, Dec. 18th, the two worlds will lie only 55 million miles apart.  Mars won't be this nearby again until the year 2016.  To the naked eye, Mars now outshines every star in the night sky (it is slightly brighter than Sirius) and it looks great through a backyard telescope.  Visit http://spaceweather.com for photos, observing tips and sky maps.

NOTE: While, technically speaking, Mars is at its best and brightest only on Dec. 18th, the red planet will remain unusually close and beautiful for weeks to come.

GREAT GEMINIDS:  The Geminid meteor shower peaked on Dec. 14th and 15th; as predicted, it was a great show.  Some observers counted more than a hundred meteors per hour including fireballs that cast shadows like a full moon and gave off  ghostly-sounding radio echoes.  It's all chronicled in our Geminid Gallery:
http://spaceweather.com/meteors/gallery_14dec07_page4.htm[/quote]
 :) .

Time to find the mars canals! :lol:

Great I can see it polar ice caps on mars.

[quote="hunter34"]Great I can see it polar ice caps on mars.[/quote]
What power of telescope (or binoculars) are you using?

Personally, I have never seen the ice caps even with the largest telescope I have looked through.
Though only requirement is the ability to magnify objects, so a larger focal length would be necessary. :)

I may see it as a dot but I am using binoculars.

Mars is now directly above me. Seems to be a good time to carry my telescope outside  :)

Another meteor shower expected.  :)
[quote="SpaceWeather.com"]
Space Weather News for Dec. 20, 2007
http://spaceweather.com

URSID METEORS:  Earth is heading for a stream of comet dust that could produce a pleasing outburst of "shooting stars" this weekend.  Forecasters say dozens of meteors per hour could emerge from a spot in the sky near the North Star (Polaris) when Earth encounters the dust on Saturday evening, Dec. 22nd. These meteors are called "Ursids" after Ursa Minor, the constellation where the North Star is located.  If forecasting models are correct, the shower's peak will occur between 2100 and 2200 UT (4-5 pm EST) with meteors visible as much as four hours before and after that time.

The source of the dust is Comet 8P/Tuttle, which is traveling through the inner solar system this month and next. The comet itself can be seen through binoculars not far from the radiant of the shower.  This gives sky watchers a rare opportunity to see a comet and its meteors in the same observing session.

Ursid meteors, which appear in small numbers annually, have a reputation for faintness and delicacy.  Dark skies are usually required to see them; bright moonlight on Dec. 22nd will only exacerbate the problem of visibility. However, say forecasters, during an outburst of Ursids there may be a fair number of bright meteors.  No one knows what will happen--all the more reason to look!
[/quote]

[quote="AstroPhys"]What kind of telescope do you have?  :)[/quote]
A rather small one, 5'' only.
At least much to small to see the ice caps, I still cannot imagine to see them with binoculars only.


I don't really understand the aopsWiki article on it.
For an example of what I don't understand, look at the first question on
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=169518[/url], and my subsequent posts.

How is it used? How do you recognize when it would be useful?

It's basically the opposite of adding two or more fractions to get a single fraction.  Instead, you're breaking up the fraction into those smaller fractions.  One thing it's useful for is summing series.  Here's a link:
[url]http://www.purplemath.com/modules/partfrac.htm[/url]

I think Wikipedia explains it pretty well: http://en.wikipedia.org/wiki/Partial_fraction

thanks for the help!

However, in the link that I posted, I don't understand why it is
[quote="Bicameral"]${ \frac {x^3 + 9x^2 + 11x + 5}{x(x + 1)^3} = \frac {A}x} + \frac {B}{(x + 1)} + \frac {C}{(x + 1)^2} + \frac {D}{(x + 1)^3}$.
[/quote]
, rather than
${ \frac {x^3 + 9x^2 + 11x + 5}{x(x + 1)^3} = \frac {A}x} + \frac {B}{(x + 1)} + \frac {C}{(x + 1)} + \frac {D}{(x + 1)}$.

Because that's how the problem was stated? :huh:

The idea behind partial fraction decomposition is that we take all of the factors in the denominator that are [b]distinct.[/b]  $ (x \plus{} 1), (x \plus{} 1), (x \plus{} 1)$ are the same term, so your interpretation is the same thing as

$ \frac{A}{x} \plus{} \frac{(B\plus{}C\plus{}D)}{x \plus{} 1} \equal{} \frac{(A \plus{} (B\plus{}C\plus{}D))x \plus{} A}{x(x \plus{} 1)}$

which is a simpler fraction; it does not cover fractions with denominators containing an $ (x \plus{} 1)^3$.  Another way to think of the decomposition is that because $ (x \plus{} 1)^3$ is cubic, we need a decomposition like

$ \frac{A}{x} \plus{} \frac{Bx^2 \plus{} Cx \plus{} D}{(x \plus{} 1)^3}$

Both decompositions are valid, but the point is that extra terms are necessary.

[quote="t0rajir0u"]The idea behind partial fraction decomposition is that we take all of the factors in the denominator that are [b]distinct.[/b]  $ (x + 1), (x + 1), (x + 1)$ are the same term, so your interpretation is the same thing as

$ \frac {A}{x} + \frac {(B + C + D)}{x + 1} = \frac {(A + (B + C + D))x + A}{x(x + 1)}$

which is a simpler fraction; it does not cover fractions with denominators containing an $ (x + 1)^3$.  Another way to think of the decomposition is that because $ (x + 1)^3$ is cubic, we need a decomposition like

$ \frac {A}{x} + \frac {Bx^2 + Cx + D}{(x + 1)^3}$

Both decompositions are valid, but the point is that extra terms are necessary.[/quote]

Thanks for the reply, but I'm confused why it turns out to be
${ \frac {x^3 + 9x^2 + 11x + 5}{x(x + 1)^3} = \frac {A}x} + \frac {B}{(x + 1)} + \frac {C}{(x + 1)^2} + \frac {D}{(x + 1)^3}$, since $ (x)(x+1)(x+1)^2(x+1)^3 \neq x(x+1)^3$ :maybe:

When we add those fractions on the right, the denominator we end up with [b]isn't the product of the individual denominators.[/b]  Rather, it's the [b]least common multiple.[/b]  So we actually end up with

$ \frac{A}{x} \plus{} \frac{B}{x \plus{} 1} \plus{} \frac{C}{(x \plus{} 1)^2} \plus{} \frac{D}{(x \plus{} 1)^3} \equal{} \frac{A(x \plus{} 1)^3 \plus{} B(x \plus{} 1)^2 x \plus{} C(x \plus{} 1)x \plus{} Dx}{x(x \plus{} 1)^3}$.

Oh, that clarifies things (everything's so much more obvious in hindsight)... I just now realized purplemath was more than one page, which would have clarified things earlier, but still...

Thanks a lot for the help!


inequalities
search
inequalities unsolved

search "APMO 1998", you can get some links about this problem.

for example:
http://www.mathlinks.ro/viewtopic.php?t=162829


Determine all triplets $(a,b,c)$ of positive integers such that
           $a+b+c+ab+ac+bc+abc = 1$

\[\iff (a+1)(b+1)(c+1)=2 \]
kind of too easy..and certainly not ineq..

[quote="ali666"]$a+b+c+ab+ac+bc+abc\geq1+1+1+1+1+1+1=7$[/quote]
Sorry ,I can't understand it .

[quote="ashwinkrishnan"][u][b]sorry there was a small error in the problem. the problem actually reads-[/b][/u] Find all triplets $(a,b,c)$ of positive integers such that
              $a+b+c+ab+bc+ac = 1+abc$[/quote]

equivalent to (a-1)(b-1)(c-1)+2-2a-2b-2c=0
let a-1=x, b-1=y, c-1=z
then xyz=x+y+z+1 >x,y,z are not 0
therefore, 1=1/xy+1/yz+1/zx+1/xyz
there are only finite integer solutions.


limit
logarithms
calculus
calculus computations

if I have no mistake in my calculation then answer is $ \exp(\frac {1}{3\ln 3})$

The product is clearly less than 1, while
\begin{eqnarray*} 3\ln 3 > 0 & \Longrightarrow & \frac {1}{3 \ln 3} > 0 \\
& \Longrightarrow & \exp\left(\frac {1}{3 \ln 3}\right) > 1. \end{eqnarray*}

I can express this as an infinite sum using Euler's Pentagonal Number Formula, but I can't evaluate the sum, either.  Neither can the [url=http://www.research.att.com/~njas/sequences/A100220]OEIS[/url].

[Edited with some supplementary comments.]

hello, can you show us your calculations please?
Sonnhard.

as for bounds, since $ \ln (1\minus{}x)\plus{}x\leq 0$ for $ 0\leq x<1$ we get:

$ P_n\equal{}\prod_{i\equal{}1}^n(1\minus{}3^{\minus{}i})\equal{}\exp\left[\sum_{i\equal{}1}^n\ln (1\minus{}3^{\minus{}i})\right]\leq \exp\left[\minus{}\sum_{i\equal{}1}^n3^{\minus{}i}\right]\equal{}\exp\left[\frac{3^{\minus{}n}\minus{}1}{2}\right]$, i.e. $ \lim_{n\rightarrow\infty}P_n\leq\exp[\minus{}0.5]$

Generally, the theory around products $ \prod_{i\equal{}1}^{\infty}(1\minus{}q^i)$ for $ |q|<1$ is quite involving, I am not so sure, whether $ q\equal{}\frac{1}{3}$ might be accessible as a special case.

P.S for the "exact value" see for instance http://mathworld.wolfram.com/InfiniteProduct.html as a starting link

[quote="Carcul"]JBL, what is the "Euler's Pentagonal Number Formula"?[/quote]

$ \prod_{i \equal{} 1}^\infty (1 \minus{} x^i) \equal{} \sum_{n \equal{} \minus{} \infty}^\infty ( \minus{} 1)^n x^{n(3n \minus{} 1)/2}.$

It has a beautiful combinatorial interpretation: from the left side, the coefficient of $ x^n$ is the number of partitions of $ n$ into an even number of distinct parts minus the number of partitions of $ n$ into an odd number of distinct parts.  The right side tells you that this difference is always 0, 1 or -1, and that 1 and -1 occur only when $ n$ is a "generalized pentagonal number," $ n \equal{} \frac {m(3m \minus{} 1)}{2}$ for some integer $ m$.  There's a simple proof using a clever involution at Wikipedia -- I think the article is called "pentagonal number theorem" or some such.

It's also a special case of the Jacobi Triple Product.


Hey, does anyone know how the Chinese IMO Team is selected? O.O

There were several topics on this subject. Search for China TST and such.

hmm...everything I've found (or otherwise seen on this forum) has to do with what happens after the top 30 are selected and sent to camp. I was wondering about before that...

Before that, there is a national contest held to select students to go for CMO. (roughly)The best 30 contestants in CMO can enter the math camp held in March-April. Besides, usually the best 2 contestants in CWMO and CGMO can also enter the training camp.

是说数学联赛是中国奥林匹克的预赛么?
还有,希望杯是什么性质的比赛?

[quote]是说数学联赛是中国奥林匹克的预赛么?
还有,希望杯是什么性质的比赛? [/quote]
这是一种面向广大学生的竞赛（in other words,是一种比较。。。simple的比赛）
好像和IMO选拔无关
有证书
 高中数学联赛是中国奥数选拔赛。前几名会被邀请到camp，然后再进行残酷的选拔！ :huh:


can anyone please provide me an ineq with this strange condition - $ a\plus{}2b\plus{}3c\equal{}1$ .
please provide the link if u think typing the question is time consuming

hello someone please help the condition is $ a\plus{}2b\plus{}3c\equal{}1$

see my own  :)

If $ a,b,c$ are positive real numbers such that $ a \plus{} 2b \plus{} 3c \equal{} 1$ .Prove that

$ \sqrt {b} \cdot \sqrt [4]{c} \plus{} \sqrt {c}\cdot \sqrt [4]{\frac {a}{6}} \plus{}  \sqrt {\frac {a}{6}} \cdot \sqrt [4]{b} < 1$


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