MATHCOUNTS
Ring Theory
videos
AMC
USA(J)MO
USAMO
geometry

Fun stuff. Some poor sportsmanship going on though...

"It doesn't matter, so I won't answer"? I'm rooting for David anyway, as he's the only competitor in the countdown I've met.

okay did david yang just walk of the stage without shaking bobby shen's hand? what are kids learning these days..

but congratulations to bobby shen and whichever team won (i am unsure of what team this is)!

Texas won. Mixxouri came in 2nd.

David, eh, he lost by sketchy answering practices. Agreed on handshaking being important. No calling matches fail :P. Although, [i]would[/i] you consider giving the same answer as the other person did (but after time) to be sketchy? Methinks I do somewhat find it necessary to give a different answer, even if wrong. *That* I'd have cheered for.

Congrats to Mewto55555 on his 3rdness! I'm unacquainted with the other top 3's usernames.

Now that is a really hard way to lose- giving the right answer, just barely too late after buzzing in.

Texas won as a team. Top ten:
1. Texas
2. Missouri
3. Washington
4. California
5. Maryland
6. Indiana
7. Massachusetts
8. North Carolina
9. New Jersey
10. Virginia.

GO ALAN! seeing as hes the only one in cd i know...congrats to all the other pplz too
i could only see the end cuz i just got back from school...and my comp is messed up so i couldnt hear any sound.

I rooted for Bobby. David said fail waaaaayyy too many times and said "oops, I misread the question" after buzzing in at least 3 times.

does anyone have a list of all the countdowners? And what happened to Darryl Wu?

wish i could have heard pythag say fail lol...

[quote="solafidefarms"]Mewto was a pleasant surprise to watch.

And, don't hate on David _too_ much :). He [i]did[/i] come in 2nd and he's got to get credit for that, even if he failed reading a few times.[/quote]

If u misread something, its better to make an educated guess. but maybe he should get some credit for getting 2nd

and also, David probably might have won if only

a. He had not tried to buzz in early (he did get the answer faster or maybe bobby was checking his work)

b. If he wasnt so rude, the judges would have been more lenient in determining the validity of his answer.

yes athunder that was quite epic, especially the final.

[quote="pythag011"]Dude according to some people David Yang is always sarcastic.

Also, he wasn't trying to be funny.[/quote]

Riiiiiiiiiiiiiiiiiiiiiiiiiiiiiight, okeydokeys.

[quote="pythag011"]Dude according to some people David Yang is always sarcastic.

Also, he wasn't trying to be funny.[/quote]

 :rotfl:

the "some people" is yourself

[quote="pinkmuskrat"]

"Let's talk about fractions."

LOL (crowd laughs) :rotfl:[/quote]

I know, it was funny. That part was challenging. Only Bobby Shen could figure it out immediately, and the others needed a little time to solve and explain it.

[quote="pinkmuskrat"][quote="remy1140"][quote="Textangle"]Because I want to watch it again? Lol.

What was the Masters Round problem?[/quote]

I think the Masters Round problem was about base -4 numbers. It was about how to express a base ten number in a base -4 system, and how we can show if a base -4 number is positive or negative when expressed in base 10, and how to do base -4 addition/subtraction. There was also a question a judge asked: how to express fractions (eg 1/2, 1/4, 1/16, 1/3) in base -4. That's all I can remember right now, but I'm sure there were a few more topics.[/quote]

"Let's talk about fractions."

LOL (crowd laughs) :rotfl:[/quote]
yeah, that was really funny. bobby and max took about the same amount of time, while calvin and david tk a little longer. they were still all quite fast, though. They also asked questions about divisibility rules. max was really fast with them, i bet he likes nt!  :D  I was slower than everyone there though..

Darn I was hoping they would ask us to prove that sqrt 2  is normal in base -4 :( No luck with that.

Err also for the fractions thing nobody took particularly long, but some people explained it better than others.

[quote="RoFlLoLcOpT"]is there a webcast of the masters round presentations? or do you have to be there in person to see it?[/quote]

If I remember correctly, [i]someone[/i] was videotaping it.

[quote="AIME15"]Weren't you the one in the other thread saying "He doesn't deserve to win, because he was so rude."[/quote]

Hm.. I never recall saying that he didn't deserve to win. But the main thing was that he didn't even shake hands with anyone.

But he still is smart.

If he's rude, he can still be awesome.

If you could pick who won the Nats, would you pick the smartest person or the person who was the best sport?

EDIT: Sorry, I meant the person who would score the highest

Officially I would choose who was the smartest. However, we should still recognize those who respected others the most and followed the rules. It would be great if the best sport and showing respect toward the judges, moderators, and spectators became the winner. And I think this year's result was one of those. :D

You have asked two different questions.

The smartest person will not necessarily win Nats, because Mathcounts is a highly trivialized competition. So...winning Nats is almost luck. Though, those consistent with answering questions at a rapid pace do deserve to win more.

But...if winning Nats had an incredible amount of prestige and was more difficult, then I would still pick the smartest person to win. In the end, they are the one that deserves to win.

EDIT:
[quote="remy1140"] And I think this year's result was one of those. :D[/quote]

Agreed.

[quote="tennis123"]If you could pick who won the Nats, would you pick the smartest person or the person who was the best sport?[/quote]

Both. If I couldn't, then I would pick the smartest person with good sport. If nobody has good sport, I would pick the smartest.

David Yang was actually really good, he just waited too long to answer that last question:

"How many 2 x 2 x 2 cubes must be added to an 8 x 8 x 8 cube to make a 10 x 10 x 10 cube?" (Ans. 61)

[quote="connaissance"][/quote][quote="pinkmuskrat"][quote="remy1140"][quote="Textangle"]Because I want to watch it again? Lol.

What was the Masters Round problem?[/quote]

I think the Masters Round problem was about base -4 numbers. It was about how to express a base ten number in a base -4 system, and how we can show if a base -4 number is positive or negative when expressed in base 10, and how to do base -4 addition/subtraction. There was also a question a judge asked: how to express fractions (eg 1/2, 1/4, 1/16, 1/3) in base -4. That's all I can remember right now, but I'm sure there were a few more topics.[/quote]

"Let's talk about fractions."

LOL (crowd laughs) :rotfl:[/quote][quote="connaissance"]
yeah, that was really funny. bobby and max took about the same amount of time, while calvin and david tk a little longer. they were still all quite fast, though. They also asked questions about divisibility rules. max was really fast with them, i bet he likes nt!  :D  I was slower than everyone there though..[/quote]

I actually did not get everything that fast, I just explained my thought processes as I went along so it looked like I got the problem and THEN explained it =P


Divisibility
number theory
binomial coefficients
IMO
IMO 1967

Let $k,m,n$ be natural numbers such that $m+k+1$ is a prime greater than $n+1$. Let $c_s=s(s+1)$. Prove that

\[(c_{m+1}-c_k)(c_{m+2}-c_k)\ldots(c_{m+n}-c_k)\]

is divisible by the product $c_1c_2\ldots c_n$.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

We have that $c_1c_2c_3...c_n=n!(n+1)$

and we have that $c_a-c_b=a^2-b^2+a-b=(a-b)(a+b+1)$

So we have that $(c_{m+1}-c_k)(c_{m+2}-c_k)\ldots(c_{m+n}-c_k)=\frac{(m+n-k)!}{(m-n)!}\frac{(m+n+k+1)!}{(m+k+1)!}$ We have to show that:

$\frac{(c_{m+1}-c_k)(c_{m+2}-c_k)\ldots(c_{m+n}-c_k)}{n!(n+1)!}=\frac{(m+n-k)!}{(m-n)!n!}\frac{(m+n+k+1)!}{(m+k)!(n+1)!} \frac 1{m+k+1}$ is an integer

But $\frac{(m+n-k)!}{(m-n)!n!}=\binom {m+n-k}n$ is an integer and ${(m+n+k+1)!}{(m+k)!(n+1)!} \frac 1{m+k+1}=\binom {m+n+k+1}{n+1}\frac 1{m+k+1}$ is an integer because $m+k+1|m+n+k+1!$ but does not divide neither $n+1!$ nor $m+n!$ because $m+k+1$ is prime and it is greater than $n+1$ (given in the hypotesis) and $m+n$

[quote="Simo_the_Wolf"]We have that $c_1c_2c_3...c_n=n!(n+1)$[/quote]I think you forgot a $! \Rightarrow c_1c_2c_3...c_n=n!(n+1)!$


Well I know that the SAT and SAT II scores are now online.  How did you all do?

lets put it this way- if you got an 800 on the Math IIC, you were in the 90th percentile. And most everyone on this site are mathematically gifted... Oh yeah.

*grin*

800 IIC
780 Writing
780 Chem (much to my dismay!)


Suzy the cat has grown up to a size which enables her to climb stairs one or two at a time. How many different ways can Suzy climb a staircase consisting of ten steps?

This is simply $ 2^{10\minus{}1}$, or $ 512$.

Or is it $ 1024$?

Please post if I'm wrong. :wink:

I don't think that's right...
We use recursion.
Let $ P_n$ denote the number of ways to climb $ n$ steps.
Then, we have
$ P_1\equal{}1$
$ P_2\equal{}2$
and
$ P_n\equal{}P_{n\minus{}1}\plus{}P_{n\minus{}2}$
because if she gets to the $ n\minus{}1$th step, she has to take $ 1$ step to get to the $ n$th step, giving her one way,
and if she gets to the $ n\minus{}2$th step but never the $ n\minus{}1$th step, then she must take $ 2$ steps to get to the $ n$th step, giving her one way.
These are the Fibonacci numbers with the index shifted once to the right, such that $ P_n\equal{}F_{n\plus{}1}$. $ P_{10}\equal{}F_{11}\equal{}\boxed{89}$.


(1) How many points are there in a plane that are 4 cm. from a given line and also 5 cm. from a given point on that line?

(2) How many points are there that are equidistant from two given points A and B and also 2 inches from the line passing through A and B?

(3) Given <ABC, what is the number of points found in the interior of the angle, equidistant form the sides of the angle, and also equidistant from points A and B?

(4) The number of points equidistant from two given parallel lines and at the same time equidistant from two given points on oneof these line is

(A) 1         (C) 3
(B) 2         (D) 4

(5) Line AB and CD are paralllel and are 6 inches apart. Point P is on line AB. The number of points equidistant from these two lines and also 5 inches from point P is

(A)1          (C)0
(B)2          (D)4

(6) The triangle whose vertices are A(1,2), B(7,2), and C(5,6).

(PART A) From the locus points equidistant from A and B, write an equation of the locus.
(PART B) Write an equation of the line passing through A and C.
(PART C) Find the coordinates of point P if P is a point on line AC that is equidistant from points A and B.

[hide="one"]There are four positions for this point. One to the right-above, right-below, left-above, left-below. [/hide]

[hide="two"]Draw the two points and the line connecting them. There are 2 points avaliable, one above the line and one below the line. [/hide]

Hey wait a second, this is in the high school forum too!  :mad:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70323


trigonometry
inequalities unsolved
inequalities

let $ x$ $ \in \mathbb R$ , find maximum of :

$ \sin(\cos x)\plus{}\cos(\sin x)$

Clearly $ \sin (\cos x)\leq \sin 1$, since $ \minus{}1\leq \cos x \leq 1$ and $ \sin$ is strictly increasing on $ [\minus{}1,1]$.  Thus, $ \sin (\cos x)\plus{}\cos (\sin x)\leq 1\plus{}\sin 1$, and this value is achieved at $ x\equal{}0$, so it is the global maximum.


analytic geometry
algebra
polynomial
calculus
calculus computations

Consider the curve defined by the equation y^5 - y - x^2 =-1
a) Find the equation of three different lines which the tangent to the curve at x=1
b) Find the coordinates of the points at which the tangent line to the curve is vertical. Express the coordinates to three decimal places.
c) Where does this curve intersect the y-axis.

All I got so far is dy/dx = 2x/(5y^4-1).

For part (a), where does the curve intersect the line $x=1$? Look at those points. For part (b), how can you have a vertical tangent? This leads to an equation, which you want to intersect with the curve (and you'll get a 5th degree polynomial equation that can't be solved exactly). Part (c) should be obvious.

where does the curve intersect the line x=1?[/quote]

And how am I supposed to do that?


AMC
AIME
factorial
function
trigonometry
floor function
geometry

This set of problem will have exactly 1 week of solving time. After 1 week has passed and I received no solution, I'll call a draw. So please at least send some answers.

The people who are eligible to compete on this are:

The Original Pi Guy
BornForMath

I'll put rest in hidden text.

[hide="Rules"]

The following problems are in level of AIME or even in some approach, higher. I used many ideas from other competitions to write these problems and you may find some similar problems. That's okay because they may have same idea but not the same problem or answer. Each problem worth 1 point so trying harder one is a bad idea. I can't guarantee if these are in level of difficulty. Some may find #1 harder than #14. That's your opinion. Also, the answers must be in exact form (That is, include $\pi$, $!$ (factorial sign), and any other sign that you need)

Today's date is 6/28. Thus, the answer is due on 7 days from today and that is 7/5.

Use of Calculator.

You're allowed to use calculators but not in number 10. Allowing the use of calculator on that problem can make people to brute force and the real way of not using calculator is why the problem is so hard but beautiful. Other than that, I allow you to use calculator. But putting in the function is not acceptable. Moreover, please do not use EquationSolver.

Feel free to pm me for the questions.

[hide="Questions"]
1. Compute the value of $S$.

$S = \sum^{2005}_{a=1} b\left(\frac{a}{2006}\right)$

where $b(x) = \frac{16^x}{4+16^x}$.

2. Find all $x$ that satisfy the equation:

$(4x+1)(6x+1)(10x+1)(60x+1) = \frac{35}{2}$

3. In $\triangle XYZ$, $x \leq y \leq z$. If the triangle satisfies $\frac{x^3+y^3+z^3}{\sin^3 X + \sin^3 Y + \sin^3 Z} = 323$, compute the maximum value of $x$.

4. Let $a_n = 11^n + 13^n$. If $a_k$ leaves remainder of 72 when divided by 144 and $1986 \leq k \leq 1992$, find $k$.

5. The international math school has 2000 lockers and 2000 students. To celebrate their last day of the school, they decided to play the game called "reverse." The rule is simple. Each student will be given a number and they have to "reverse" every locker of their number. The game starts with all lockers closed. So:

Student 1 would open every lockers (since his number is 1)
Student 2 would close every locker with an even number (since he needs to reverse the locker.)
Student 3 would reverse (it will be closed/opened) every third locker.

After 2000 students finished the process, they had $N$ lockers open. Compute $N$.
(So for the 6th locker, it will be first opened by Student 1, closed by Student 2, opened by Student 3, and closed by Student 6)

6. $\left(\frac{66......64}{16....6}\right)^2 + \left(\frac{16....6}{66......64}\right)^2 = m$. Find $\lfloor {100m} \rfloor$.

(In both denominator and numerator, there is same amount of number of 6's)

7. Solve for $y$:

$\sqrt{\frac{y+1}{y-1}} + \frac{y-1}{y+1} - \sqrt{\frac{y-1}{y+1}} - \frac{y}{y+1} = \frac{3}{y+1}$

8. Amy received A+ on her last semester's final. To give her a bad start on the new semester, evil teacher Dr. Evil gave her 50-point assignment (harder than she can manage). Can you help Amy? The problem is this: ($e(x)$ stands for Evil Function)

$e(x) = (-4 \cos x - 10 + 3 \sin x)(3 \sin x + 4 \cos x - 10)$

Her problem asks to compute the minimum value of $e(x)$. So, find the minimum value of $e(x)$.

9. A sequence of integers $b_1,b_2,b_3,...$ is "best" sequence because for each $m$ and $n$, the following statements hold true.

[color=blue]If $m$ divides $n$ and $m < n$, then $a_m$ divides $a_n$ with $a_m < a_n$.[/color]

In this best sequence, compute the least possible vlaue of $a_{5488}$.

10. Find the remainder when $(\tan^2 1^\circ + \tan^2 3^\circ + \tan^2 5^\circ + \cdots + \tan^2 89^\circ)$ is divided by 24.

11. For $a,b,c > 0$, find the minimum value of

\[\left(a+\frac{1}{b}\right)^2 + \left(b+\frac{1}{c}\right)^2 + \left(c+\frac{1}{a}\right)^2\]

if $a+b+c = 9$.

12. Twelve kids brought their favorite CDs to the math camp. Each kid brought one CD. Those twelve CDs (congruent on size) are placed on the circle $C$ of radius such that twelve CDs cover the circle $C$. Also, they were placed so that each are tangent to each other and none of them overlaps. Let $r$ be the sum of areas of twelve CDs. Let $R$ be the area of the circle $C$. If $\frac{R}{r} = \frac{a + b \sqrt c}{d}$ where $c$ is not divisible by square of any prime, find $a+b+c+d$.

13. A group of men and women went to pizza restaurant where each pizza is made up of 13 small slices. Each man can eat up to 8 or 9 slices while woman can eat up to 3 or 4 slices. If 6 pizzas weren't enough and 7 pizzas were too much, compute the largest amount of people (sum of men and women) who came to the restaurant.

14. A six-digit number is defined as "amazing" if it is made up of only using digits 1,3, and 9 (all 3 must present to form "amazing" number) and such that there are even number of 9's. How many "amazing" numbers are there?

15. A 12 x 12 x 12 cube is constructed by 1 x 1 x 1 unit cubes. A cricket is in the center $K$ of one of the corner cubes. At noon, it jumps to the center of the other cubes which has common face with the cube it sits, as long as its jump increases the distance from $K$ to its current position (thus, a cricket can't jump around a circle since its jump may not increase the distance from $K$ to current position). Let $C$ be the number of ways for cricket to reach the unit cube at its opposite corner. Find $C$. Express your answer as $\frac{j}{k^m}$. Do not reduce your answer.[/hide][/hide]

In number 6, is that the greatest integer function?

I apologize to two contestants on Championship Round since they haven't got any replies on their answers.

As soon as computer's fixed, I'll post the winner and correct answers.

It's fixed now and I already have the solutions made (handwritten though).

I'm going to school shopping now so hopefully, I can announce the winner tonight and remove this thread.  :)


geometry
rectangle
calculus
calculus computations

You are planning to make an open-top rectangular box from a 10 by 18 cm piece of cardboard by cutting congruent squares from the corners and folding up the sides.

What are the dimensions of the box of largest volume you can make?

Draw a picture!

[hide]One way to do this is to denote the lengths of the sides of the squares by something like $ x$.

Then, for the base of box (the side facing opposite the open top), the width (or length) will be $ 10 \minus{} 2x$ and the length (or width) will be $ 18 \minus{} 2x$.  If the rectangles formed by cutting squares out of the folders are folded up to form the height of the box, the height will be $ x$.

Now find the volume in terms of $ x$ and optimize that expression. By the physical nature of the problem description, $ 2x < 10$ (well, also 2x < 18, but that doesn't matter given this condition) and $ x > 0$.[/hide]


superior algebra
superior algebra solved

Suppose $R$ is a commutative ring with $1$. Suppose $I$ is an irreducible ideal of $R$. (We call an ideal $J$ irreducible if $J=J_1\cap J_2$ then $J=J_1$ or $J=J_2$). Also suppose for $t\in R$ we define $(I: t)=\{x\in R|xt\in I\}$. Prove that if $r$ is an element of $R$ that $(I: r)\neq I$ and the set $\{(I: r),(I: r^2),\dots\}$ is finite, then there exist $n\in\mathbb N$ which  $(I: r^n)=R$

I'm not sure this wasn't posted before.

Since the set $\{(I: r^n)|n\in\mathbb N^*\}$ is finite, there exists $n$ such that $(I: r^n)=(I: r^{n+1})$. In this case, I prove that $I=(I: r)\cap(I+r^nR)$.
Obviously $I\subseteq(I: r)$ and $I\subseteq(I+r^nR)$.
Let $i+r^nx\in I+r^nR\cap(I: r)$ with $i\in I$ and $x\in R$. Then $r(i+r^nx)\in I\Rightarrow ri+r^{n+1}x\in I\Rightarrow r^{n+1}x\in I\Rightarrow x\in (I: r^{n+1})=(I: r^n)\Rightarrow r^nx\in I\Rightarrow i+r^nx\in I$. This proves the reverse inclusion $I\subseteq (I: r)\cap(I+r^nR)$.

Therefore $I=(I: r)\cap(I+r^nR)$. From the irreducibility of $I$ and from the assumption $I\neq(I: r)$, we get $I=I+r^nR\Rightarrow r^n\in I\Rightarrow (I: r^n)=R$.


There are two Ls in your name. Does Si[b]l[/b]verfa[u]l[/u]con equal Si[u]l[/u]verfa[b]l[/b]con?

That's the point. It's not simply n!. You have to do something about my two L's ;)

So the answer to is question is yes, barring the fact that one L is in bold?

Can we use calculators?  And we have to get all of the problems right, right?

I'm sort of confused on what question was asking.

This is just if you have a word "BOB"

It's 3!/2! = 6/2 = 3.

Same thing for here.

----
For this round, no, only two top scores will be headed into the higher division.

For number two, can we leave it in factorial form? It's kinda hard to compute a factorial w/o a calculator.

For number four, can we leave it in radical form? If not, I'd never finish typing my message in time.

you can use calculators.

So, don't leave it as factorial form.

[quote="Silverfalcon"]5. In the Christmas, good kids get 2 presents with one pair of socks. The bad kids get 1 present with one sock. Last year, the Santa Claus had to deliver 120 presents and some number of socks. How many pairs did Santa Claus delivered?[/quote]

So good kids get two socks and bad kids get one sock?

I haven't decided yet.

I'm kind of busy right now (really busy) that I might even stop doing math chats.

I'll try to post rest of the ten questions soon. I have two already but I'm trying to get 8 more.

I changed the rule.

7 questions are the total. Send me 7 questions by Dec. 24th.

Early Merry Christmas everyone!

Silverfalcon

[quote="LynnelleYe"]Wait...are you sure you gave enough info on #5? I think we ought to know the number of kids there are, or how many more good kids there are than bad kids, or something along those lines, too...because there are a lot of possible answers otherwise...[/quote]

I think theres enough info there.

at the risk of sounding stupid, what exactly is question seven asking for?

So does that mean the bad kids' socks count toward the pairs as well?
Alan--I think 7 is asking for the distance between the two points.

Yes. All socks count to the sock pair.

And like someone posted, all enough information is given.

For #7, this is actually simple question. Find the magnitude.

the 7th question is *damn* easy
but I don't know the rules, am I supposed to post my answer here?

[quote="paganinio"]the 7th question is *-* easy
but I don't know the rules, am I supposed to post my answer here?[/quote]
PM answers to Silverfalcon.

P.S. Silverfalcon, when are you going to be done grading them?

No.

Don't pm to me those questions.

You didn't pass the first round so it's not allowed.

For answers, I'll grade them when I get all answers. It takes less than minute to do it but I like to keep people excited. ;)

(Laugh) :P

[quote="LynnelleYe"][quote="Silverfalcon"]For answers, I'll grade them when I get all answers. It takes less than minute to do it but I like to keep people excited. ;)

(Laugh) :P[/quote]
You're evil! But how do you know when you'll get all the answers? Have you set a due date yet?[/quote]

Thank you and Merry Christmas :)

Dec.24th is the due date. But if you want to see how you got it, pm me. I'll tell you as Christmas gift.

Regards,

Well, I sent them. Good luck everyone. Happy Holidays.


linear algebra
matrix
linear algebra unsolved

Let $q\in\mathcal{C},\, q\ne1.$ Prove that for every non-singular matrix $A\in Mat_{n\times n}(\mathcal{C})$ there exists non-singular matrix $B\in Mat_{n\times n}(\mathcal{C})$ such that $AB-qBA=I.$
(V. Mazorchuk)


modular arithmetic
induction
ceiling function
inequalities
number theory

Let $A$ be a set of $N$ residues $\pmod{N^2}$. Prove that there exists a set $B$ of $N$ residues $\pmod{N^2}$ such that the set $A+B=\{a+b \vert a \in A, b \in B \}$ contains at least half of all the residues $\pmod{N^2}$.

We will construct the set $ B$ in $ n$ steps. In the $ k$-th step we will add an element to set $ B$ in such a way that the size of the set $ A \plus{} B$ will be at least $ kN \minus{} \frac {(k \minus{} 1)k}{2} \equal{} N \plus{} (N \minus{} 1) \plus{} (N \minus{} 2) \plus{} ... \plus{} (N \minus{} k \plus{} 1)$. We prove by induction that this is possible.

For $ k \equal{} 1$ it's obvious: after adding an arbitrary residue the set $ A \plus{} B$ will have $ N$ elements.

Suppose that we have already $ k \minus{} 1$ elements in the set $ B$ and the set $ A \plus{} B$ consists of $ m \geq N \plus{} (N \minus{} 1) \plus{} ... \plus{} (N \minus{} k \plus{} 2) \equal{} (k \minus{} 1)N \minus{} \frac {(k \minus{} 2)(k \minus{} 1)}{2}$ elements.
Consider the sets $ B_0, B_1, ..., B_{N^2 \minus{} 1}$ where $ B_i \equal{} \{a \plus{} i: a \in A\}$. Suppose that $ l$ is minimum number such that we can't increase the size of the set $ A \plus{} B$ by $ l$. It means that every set $ B_i$ has at least $ N \minus{} l \plus{} 1$ elements in common with $ A \plus{} B$, so:
$ |(A \plus{} B) \cap B_0| \plus{} |(A \plus{} B) \cap B_1| \plus{} ... \plus{} |(A \plus{} B) \cap B_{N^2 \minus{} 1}| \geq N^2(N \minus{} l \plus{} 1)$.
There are $ m$ elements in the set $ A \plus{} B$ so one of them appears in this intersections at least $ \big \lceil \frac {N^2(N \minus{} l \plus{} 1)}{m} \big \rceil \equal{} s$ times. Note that $ s \leq N$. Indeed, otherwise there would be a residue with at least $ N \plus{} 1$ representations in the form $ a \plus{} i$, where $ a \in A$, but this is impossible since $ |A| \equal{} N$.
So we have the inequalities:
$ \frac {N^2(N \minus{} l \plus{} 1)}{m} \leq s \leq N$ so it follows that $ l \geq N \plus{} 1 \minus{} \frac {m}{N}$.

Therefore
\[ m \plus{} l \minus{} 1 \geq m \plus{} N \minus{} \frac {m}{N} \equal{} m\frac {N \minus{} 1}{N} \plus{} N \geq ((k \minus{} 1)N \minus{} \frac {(k \minus{} 2)(k \minus{} 1)}{2})\frac {N \minus{} 1}{N} \plus{} N \equal{} (k \minus{} 1)(N \minus{} 1) \plus{} N \minus{} (1 \plus{} 2 \plus{} ... \plus{} (k \minus{} 2))\frac {N \minus{} 1}{N} \equal{} kN \minus{} (k \minus{} 1) \minus{} (1 \plus{} 2 \plus{} ... \plus{} (k \minus{} 2))\frac {N \minus{} 1}{N} \geq kN \minus{} (k \minus{} 1) \minus{} (1 \plus{} 2 \plus{} ... \plus{} (k \minus{} 2)) \equal{} kN \minus{} \frac {(k \minus{} 1)k}{2}
\]
and the inductive proof is completed, because we can increase size of $ A \plus{} B$ by $ l \minus{} 1$.

Hence, after the $ n$-th step the set $ A \plus{} B$ will consist of at least $ N^2 \minus{} \frac {N^2 \minus{} N}{2} \equal{} \frac {N^2 \plus{} N}{2} > \frac {N^2}{2}$ elements and the solution is finished.

Note: as you can see this problem has completely nothing to do with the number theory and residues. The better way to look at this problem is as follows:
Suppose that we have an $ N^2$ points regularly placed on the circle and a template, consisting of $ N$ of the given points, which we can use to paint the points of the circle (in one painting we paint $ N$ points). Prove that we can paint at least half of the given points within the $ N$ paintings.

Very very nice your solution!!!
I am deeply impressed by the construction of $ \frac{N^{2}(N\minus{}l\plus{}1)}{m}\leq s\leq N$
Many thanks :lol:


let $ ABC$ be an isoceles with vertex $ A$ .let $ P$ be the midpoint of the arc $ BC$ and the point $ M$ is on the side $ BC$ such that $ \frac{BM}{MC}\equal{}\frac12$ and let $ N$ be the midpoint of the side $ AC$ .prove that $ PM\bot BN$

 this problem has an easy solution :wink:

Call $ G: PA \cap BN$, we know that $ GM \parallel AB$ and so $ GM \perp BP$ so M is the orthocenter of $ \triangle BPN$ and $ PM \perp BN$.

Nice solution, I had a feeling it would involve showing M is the orthocenter of a triangle, but for a while I was confused why $ GM\parallel AB$. To clarify, G is the centriod which is 2/3 way from A to BC, so since 1/2:1/3 as 3/2:1, so $ GM\parallel AB$ by similarity, is there an easier way to see this?


calculus
trigonometry
function
inequalities unsolved
inequalities

Let be given triange [tex] ABC[/tex];[tex]x,y,z[/tex] are real positive numbers.
Find max of
[tex]sin^xAsin^yBsin^zC[/tex].

can I ask what is the meaning by  "Crux" ?I have seen it many times :(

I think he means the Canadian journal "Crux mathematicorum". You can learn more about it [url=http://www.journals.cms.math.ca/CRUX/]here[/url]

[quote="nttu"]Put $ F = sin^xA.sin^yB.sin^zC $
 F max when $ xcotgA=ycotgB =zcotgC $
  Dear Erdos , Could you please show me how I can register to read Crux mathematicorum ?[/quote]

cot or csc? :?

[quote="nttu"]
  Dear Erdos , Could you please show me how I can register to read Crux mathematicorum ?[/quote]
You can register in the site of the journal [url=http://www.journals.cms.math.ca/CRUX/]here[/url] , but you need to pay. Even though , there are some that you can read without registering.

I have a solution for this problems (by me)
 I'll post it soon  ;)

[quote="nttu"]I meaned $ x.cot A = y.cot B = z.cot C $ . I used calculus for this problem[/quote]
ah yes you are right. How stupid i was. I think this can be solved by lagrange multiplier.

I think we can use Cauchy

The same one:

Find  Max of

$P=(\cos A)^x(\cos B)^y (\cos C)^z$

This is my solution:

[b]Problem 1[/b] Let a,b,x,y are positive number such that x,y>1
Find the Minumum value of

               $H=\displaystyle\frac{(a^2+1)^x(b^2+1)^y}{(a+b)^2}$

[b]Solution:[/b]
   By Cauchy we have

        $ \alpha u+\beta v \ge  u^{\alpha} v^{\beta} $

        Where $\alpha,\beta >0$ and $\alpha+\beta=1$

 Now we have

$\displaystyle
  (a^2+1)^x=x^x\left[\left(a^2+\frac{1}{x+y-1}\right)\cdot\frac{1}{x}+\left(\frac{x+y-2}{(x+y-1)(x-1)}\right)\cdot\frac{x-1}{x}\right]^x \\
\ge \left(a^2+\frac{1}{x+y-1}\right)\cdot x^x\cdot\left[\frac{x+y-2}{(x+y-1)(x-1)}\right]^{x-1}$

And

$\displaystyle
  (b^2+1)^y\ge\left(b^2+\frac{1}{x+y-1}\right)\cdot y^y\cdot\left[\frac{x+y-2}{(x+y-1)(y-1)}\right]^{y-1}$

  :arrow:  :arrow:

$\displaystyle (a^2+1)^x(b^2+1)^y $

$\ge\left(a^2+\frac{1}{x+y-1}\right)\cdot\left(b^2+\frac{1}{x+y-1}\right)\frac{x^x}{(x-1)^{x-1}}\cdot\frac{y^y}{(y-1)^{y-1}}\cdot\left(\frac{x+y-2}{x+y-1}\right)^{x+y-2}$
$\ge\frac{(a+b)^2}{x+y-1}\cdot\frac{x^x}{(x-1)^{x-1}}\cdot\frac{y^y}{(y-1)^{y-1}}\cdot\left(\frac{x+y-2}{x+y-1}\right)^{x+y-2}$

  Then we have

$\displaystyle \frac{(a^2+1)^x(b^2+1)^y}{(a+b)^2}
\ge\frac{x^x}{(x-1)^{x-1}}\cdot\frac{y^y}{(y-1)^{y-1}}\cdot\frac{(x+y-2)^{x+y-2}}{(x+y-1)^{x+y-1}}$

  The equater when

     $\displaystyle a^2=\frac {y-1}{(x-1)(x+y-1)}$
     $\displaystyle b^2=\frac{x-1}{(y-1)(x+y-1)}$

[b]The Answer:[/b]

Put $=(\sin A)^x(\sin B)^y(\sin C)^z$
      $ \displaystyle a=cotg A$
      $ \displaystyle b=cotg B $

We have

    $\displaystyle (\sin A)^2=\frac{1}{cotg A^2+1}=\frac{1}{a^2+1}$
    $\displaystyle (\sin B)^2=\frac{1}{b^2+1}$

 Easy to see

    $\displaystyle (\sin C)^2=\frac{(a^2+1)(b^2+1)}{(a+b)^2}$
Then
$H=\frac{1}{F^2}=\frac {(a^2+1)^{x+z}(b^2+1)^{y+z}}{(a+b)^{2z}}
                           =\left[\frac {(a^2+1)^u(b^2+1)^v}{(a+b)^2}\right]^z$

Where
            $\displaystyle u=\frac{x+z}{z}>1$
            $\displaystyle v=\frac{y+z}{z}>1$

By proplem1 thus giving us

$\displaystyle H \ge \left[\frac {u^u}{(u-1)^{u-1}}\cdot \frac {v^v}{(v-1)^{v-1}}\cdot\frac{(u+v-2)^{u+v-2}}{(u+v-1)^{u+v-1}}\right]^z
=\frac{(x+y)^{x+y}(y+z)^{y+z}(z+x)^{z+x}}{x^x y^y z^z (x+y+z)^{x+y+z}}$

Finally:

[quote]$\large \displaystyle F \le \sqrt{ \frac{x^x y^y z^z (x+y+z)^{x+y+z}}{(x+y)^{x+y}(y+z)^{y+z}(z+x)^{z+x}}}$[/quote]

The equator when

       $\displaystyle \sin A^2=\frac{x(x+y+z)}{(x+y)(x+z)}$
       $\displaystyle \sin B^2=\frac{y(x+y+z)}{(y+z)(y+x)}$
       $\displaystyle \sin C^2=\frac{z(x+y+z)}{(z+x)(z+y)}$

For Example

When $x=1, y=2, z=3$
Then  $ \sin A\sin B^2 \sin C^3 \leq \frac{27}{100\sqrt{5}}$ (Math & Young)

Did I have any mistake?

╔═════════╗
║ Viet  Math |VA|  :D  :D  :D
╚═════════╝

[quote="N.T.TUAN"][quote="nttu"]I meaned $ x.cot A = y.cot B = z.cot C $ . I used calculus for this problem[/quote]
Can you post your solution?[/quote]

I guess he used lagrange multiplier. It's pretty straightforward using it.

[quote="Viet Math"]How about my solution?

 True  :P  or False  :(    ???

 I will find another solution,shorter  ;)[/quote]

I think it 's true, but it's rather ad-hoc to me... can you tell us how did you think of the solution?

Yes,it take me 3 hours  :D  :D  :D

 The first , I put a=cotg(a).. and Now we must find the Min of

 $\displaystyle H=\frac{(a^2+1)^x(b^2+1)^y}{(a+b)^2}$

 when x,y>1

 I try to solve by this way

 + We have $ \displaystyle (a+b)^2 \le \frac{(a^2+v)(b^2+v)}{v}$ (1)
    (And the equator when ab=v)
 + We have
    $\displaystyle (a^2+1)^x = u^x(\frac{a^2+\alpha}{u}+\frac{1-\alpha}{u-1}\frac{u-1}{u})^x
                     \ge(a^2+\alpha)^{\frac{x}{u}}(1-\alpha)^{\frac{(u-1)x}{u}}$
    From (1) we given u=x
    (Bacause $\frac{x}{u}$ must be =1)
    Solve the equator we have
           $ \displaystyle a^2+\alpha= \frac{1-\alpha}{u-1}
                               =\frac{1-\alpha}{x-1}$ (*)
    The same for b
           $ \displaystyle b^2+\frac{1-\alpha}{y-1}$

    But a,b such that the equator (1) so that

     $\displaystyle (\alpha)^2=a^2.b^2=(\frac{1-\alpha}{x-1}-\alpha)(\frac{1-\alpha}{y-1}-\alpha)$
     $ \displaystyle (\alpha)^2(x-1)(y-1)=(1-\alpha x)(1-\alpha y)$
     $ \displaystyle (\alpha)^2*(x+y-1)-(x+y)\alpha+1=0$
     $ \displaystyle (\alpha)=1$ or $ \alpha=\frac{1}{x+y-1})$

  From (*) we have $\displaystyle \alpha=\frac{1}{x+y-1}$

  And have a solution

  (I think It's nice because it don't use function and easy to know)

-------------------------------------------------------------------
I hard to choice the word to Write (My E Bad)
but I think Math is an International  language so I must try

[quote="Viet Math"]The same one (But easier)

 Find the Maximum value of

 $(\cos A)^x(\cos B)^y(\cos C)^z$

 Where $A,B,C \le frac{\pi}{2}$

 Can anybody post the solution ?  :D  :D  :D[/quote]

for me, your solution is too hard to find. So i still prefer calculus.
equality holds when $x \tan A=y \tan B=z \tan C$

To sinhochung:

 You can use Cauchy (The same way as my solution)

 Good Luck !  :D  :D  :D

Hint

 The equality hold when

$\displaystyle(\cos A)^2=\frac{x^2}{(x+y)(x+z)}$
$\displaystyle(\cos B)^2=\frac{y^2}{(y+x)(y+z)}$
$\displaystyle(\cos C)^2=\frac{z^2}{(z+x)(z+y)}$

Or

$\displaystyle\frac{\cot A}{x}=\frac{\cot B}{y}=\frac{\cot C}{z}$

Good Luck!!!  :D  :D  :D

Put $ f(m) = x.ln sin(A+m) + y ln sin(B-m) + z.ln sinC $
        $ f'(m) = x.cotg(A+m) - y.cotg(B-m) $
   f max  <=> $ f'(0) =0 $ <=> $ x.cotgA=y.cotgB $
   $ h(n)= x.ln sinA + y ln sin(B+n) + z.ln sin(C-n) $
  Similarly , h max <=> $ h'(0) =0 $ <=> $ y.cotgB = z.cotgC $


agha inam baraye chand jomlei ha va moadele tabei
dar har marhale az har kodoom ye masale midim va bad hamin tor 2 ta 2 ta mirim jelo...khoobe? ;)

agha yeki midam garm shin vali jozve marathon nist...
hamaye tavaabee$f: Z\longrightarrow Z$ ke dashte bashim:
$f(m+f(n))=f(m)+n, m,n\in Z$
hala ino bezanid too rag(jozve marathone):
first problem:...
1)hameye tavaabee $f: R\longrightarrow R$ ke baraye hame ye $x,y,z$ haghighi dashte bashim:
$f(x^2+yf(z))=xf(x)+zf(y)$
2)chand jomlei haye $P(x)$ va$Q(x)$ chanaanad ke be ezaaye har adade haghighi darim:
$P(x^2-x+1)=Q(x^2+x+1)$
saabet konid ke har do chandjomlei meghdari saabet ast.
khosh bashin... :D


inequalities
triangle inequality
probability
expected value
AMC

Is it worth taking the time to read the question twice, organize the information neatly (draw a neat diagram), do the calulations very slowly, then read it again to make sure it answers the question? Or should you do it quickly and wait till you've finished all of them and then go back and check? Working carefully can increase my accuracy by as much as 25% on a hard test, but it seems like it takes me forever to get through one problem. Otherwise, I make a ton of dumb mistakes that cost me.

Kay in my opinion, i think mistakes aren't caused by carelessness, but rather by skill and how much you are familiar with the topics
I personally tend to do careful and slow calculations on like school tests and stuff but then it takes too long -___-
But, I wouldn't say do EVERY SINGLE problem so "carefully", but you should try to zoom through the easy ones and if you have enough practice, hopefully you'll make no mistakes >.> and then make sure to be careful on like the harder ones.

Rereading the question is definitely worth it. Rushing calculations is sometimes necessary. If you get a gut feeling that something has gone wrong, check the problem over later - it'll help to get distance from it for a few minutes.

When doing the problems, just check it over once when you're done with each one (basically check as you go along). Then if you have time at the end, recheck over the problems that you aren't sure about.

If you suddenly get stuck on one of the first 15-20 problems just skip it and do it later. Like for the triangle inequality one this year on the 10A I came back to it and then realized to use the triangle inequality.

No, I did not take 10a, my school had the Florida Writes that day, I'm taking 10b.

You should definitely read each problem twice.  In the past most of my mistakes were caused by misreading the problem, but I'm being more careful about that now and it's helping a lot.  Think about it, if spending an extra ten seconds per problem can consistently save you from a mistake or two, it's worth it.

Sometimes it helps to check your work right after you finish a problem, but probably not always.  If you finish solving a problem and you're a little worried you might have made a mistake, then check it right away.  Sometimes I make a step in solving a problem and I get a gut feeling that it might not have been right, and if this happens I check my work right away.

When you're trying to decide whether to check your work or continue trying to solve problems, think about expected value (Yes, I'm serious :)).  Fixing a mistake will give you 6 points, and solving a problem will give you 4.5 points.  Therefore, unless you feel like you're significantly more likely to solve another problem than you are to find a mistake, you should check your work.


I've heard of this school anyone know much about it?

A few days ago I looked at their website, and if memory serves:
[list][*]It's quite new, only 4 classes have gone through the program.
[*]Very small, 75 students a year.
[*]All who get accepted get a 30k/year scholarship, which leaves ~15k for the student.[/list]
You should check out their website if you want to know other statistics.[/list]

Two of the former members of the Southern California ARML team have attended Olin; one was a freshman in 2002 (I think that was the college's first year of operation) and the other is a frehman now in 2005. I haven't heard back from either of them, but if I do, I'll pass it along.

A few of my former students have attended Olin, but I haven't heard anything directly from them.  One is currently there and a couple of months ago I ran into his older brother, who said that it's a great place and that his younger brother is loving it.


algebra
polynomial
number theory proposed
number theory

Let P(x) be a polynomial with integer coeficients. If P(x) is irreducible and P(x²) is reducible, prove that P(x²)=Q(x)Q(-x) when Q(x) is some polinomial with integer coeficients.

In other words: P(x²) can only be factoriced in polynomials of the form h(x²) and of the form h(x)h(-x).

I see this property in a solution, where this is supoused well-know. For my this isn't well-know. I would like to know if this is well-know and if there are some others facts like this.

Idea:  let $ r$ be a root of $ P(x)$ and consider the degree of $ \mathbb{Q}[\sqrt{r}]: \mathbb{Q}[r]$.

Please, this is just that I want. I am looking for facts about minimal polinomials, conjugates, and this things. I was studied this informaly, and I would like to know some book or material that make a complete development of this theory.

Thanks for yur reply, and please help me.

I haven't studied this formally, either, but $ \mathbb{Q}[\sqrt {r}] : \mathbb{Q}[r] \equal{} 2$ implies $ P(x^2)$ is irreducible, so $ \mathbb{Q}[\sqrt {r}] : \mathbb{Q}[r] \equal{} 1$ and $ \sqrt {r}$ has a minimal polynomial of degree $ n$; call it $ Q(x)$.  Then plainly $ \minus{} \sqrt {r}$ has a minimal polynomial $ Q( \minus{} x)$.

The above argument applies directly when $ P(x)$ is monic.  I am not sure if the statement is true when the leading coefficient is not a square.

Edit:  The statement is false for, say, $ P(x) \equal{} x \minus{} 1$ (the leading coefficients of $ Q(x), Q(\minus{}x)$ are wrong).  I believe it should read $ P(x^2) \equal{} (\minus{}1)^n Q(x) Q(\minus{}x)$ where $ n \equal{} \text{deg } P$.


There are many primes of the form n^2+1. :wink:

this problem is still unsolved !!!  :blush:
we do not know any thing about such a primes :(


geometry
circumcircle
geometric transformation
rotation
geometry proposed
Miquel point
Brocard

Given any triangle $ABC$, show how to construct $A'$ on the side $AB$, $B'$ on the side $BC$ and $C'$ on the side $CA$, such that $ABC$ and $A'B'C'$ are similar (with $\angle A = \angle A', \angle B = \angle B', \angle C = \angle C'$) and $A'B'C'$ has the least possible area.

After some angle-chasing, it is easy to verify that if you have $A',B',C'$ on $AB,BC,AC$ so that $ABC$ and $A'B'C'$ are similar, you will have $\angle BA'B'=\angle CB'C'=\angle AC'A'$ and vice versa.
Let's prove another lemma:
[b]Lemma[/b]:The circumcentre of $A'B'C'$ is the same as circumcentre of $ABC$.
[b]Proof[/b]: Let's draw a circle with centre $O$ and radius less than or equal than the circumradius of $ABC$. Let the circle intersect $AB$ at $M,A'$, $BC$ at $N,B'$, $CA$ at $K,C'$. Now let $A''B''C''$ be a triangle inscribed in the given circle so that $A''B''C''$ is similar to $ABC$, and $A''B'',B''C'',C''A''$ is parallel to $AB,BC,CA$ respectively. Now because $A'B'C'$ and $A''B''C''$ are similar and have same circumradius they are congrent, so $A'B'C'$ is a rotation of $A''B''C''$. But that would mean the angle $A'B'$ makes with $A''B''$ is same as angle $B'C'$ makes with $B''C''$ is same as angle $A'C'$ makes with $A''C''$. But because $A''B'',B''C'',C''A''$ are parallel to $AB,BC,AC$ respectively we will have $\angle BA'B'=\angle CB'C'=\angle AC'A'$, so $A'B'C'$ will be a required triangle.
It just remains to note that for given $\angle BA'B'$ there is only one possible triangle $A'B'C'$ to be formed. therefore the circumcentre of $A'B'C'$ is the circumcentre of $ABC$. QED

Now, we will just find $O$, the circmucentre, drop the perpendiculars from $O$ to $AB,BC,CA$ let them be $X,Y,Z$ respectively. We will choose the largest one of $OX,OY,OZ$ wolog $OX$ and draw the circle with centre at $O$ and raius $OX$. The circle will intersect $AB$ at $M,A'$, $BC$ at $N,B'$, $CA$ at $K,C'$. Let's prove $A'B'C'$ and $MNK$ are the required triangles.
$A'B'C'$ is obvously similar(see the lemma above) to $ABC$. It must have minimum area, so its circumradius must be minimum. ley's prove it is. If the circumradius is greater than $OX$ the new triangle $A'B'C'$ will obvously have a greater ares. If the circumradius is less than $OX$, then because the circumcentre is fixed, we will have the circumcircle not reaching $AB$(because in the case when circumradius is equal to $OX$ the circle touches $AB$, giving a single value for pt $A'$), so pt $A'$ will not exist, and we will not get the required triangle.

it's not true. Try to draw this in cabri.

In fact, the "circumcentre" is the Brocard Point. Prove it!

in fact,the so-called circumcenter is the similarity centre of ABC and A'B'C',as well as the first Brocard point of triangle ABC.
the proof is easy~


Prove that : With every possitive integer n , the number of the possitive integers [u][b]which has the sum of the digits is $ 2n $ and the digits are just 1 ,3 ,4 [/b][/u]is perfect squares

.. if n = 1
there is only a number : 11 (1+1=2=2n)
but 11 is not a perfect squares...

Well, if you denote the number of such numbers for $k$ by $a_k$, we get $a_n=a_{n-1}+a_{n-3}+a_{n-4}$ (according to whether the first digit is $1, 3$ or $4$). Shouldn't it be a pretty standard exercise from here?

Actually, it is quite boring.
This problem was used on Russian TST 1995.

Where can I find the solution of this problem ?

[quote="Myth"]:lol:  :lol:  :lol:
He asked for number of such numbers. For $n=1$ we have only one such number: 11. And, of course, 1 is a perfect square![/quote]
may be he fixed it .. i remember that he wrote :"the sum of the possitive integers " .. not "the number of the possitive integers "

To romano : ( it is a personal question .. )
[hide].. u from VietNam ??? U post many problem ... in our lesson last week ...
If not , where do u get it ? Answer me plz [/hide]

You are right , Myth !
  And the problem hasn't been solved !
  Dear Intotherainbow , this is a problem from a Vietnamese friend on Mathlinks ( his nickname is [b]vmo[/b]) . Best wishes !

I got it! Without boring calculations! Pure combinatorics! :D

[b]Lemma.[/b] $a_{n}a_{n-2}=a_{n-1}a_{n-3}+a_{n-1}a_{n-2}$.
Proof. By induction. It is true for $n=4$ and
\[a_{n-1}a_{n-3}+a_{n-1}a_{n-2}=(a_{n-2}a_{n-4}+a_{n-2}a_{n-3})+a_{n-1}a_{n-2}=a_{n-2}(a_{n-1}+a_{n-3}+a_{n-4})=a_na_{n-2}.\]

Consider a row of $2n$ balls. We need to break into a pieces of 1, 3 or 4 balls. Consider a piece, which contain middle of the row. There some possibilities:
[b]1)[/b] there is no such piece, then number of corresponding partitions is $a_n^2$;
[b]2)[/b] the middle belongs to piece of 3 balls, then number of corresponding partitions is $2a_{n-1}a_{n-3}$;
[b]3)[/b] the middle belongs to piece of 4 balls, then number of corresponding partitions is $a_{n-2}^2+2a_{n-1}a_{n-3}$.
So, finally, we have
\[a_{2n}=a_n^2+a_{n-2}^2+2(a_{n-1}a_{n-3}+a_{n-1}a_{n-2}),\]
but applying lemma we see
\[a_{2n}=(a_n+a_{n-2})^2.\]

That's all.  :D  :D  :D

All question you can address to Grobber! :lol:


geometry
3D geometry
number theory
prime numbers

Do there exist $2004$ consecutive integers with each one divisible by a perfect cube larger than $1$?

If we find such number $x$ and distinct prime numbers $p_1,p_2,...,p_{2004}$ for witch stands \[x\equiv_{p_1^3} -1\] \[x\equiv_{p_2^3} -2\] \[.\] \[.\] \[.\]
 \[x\equiv_{p_{2004}^3} -2004\]

then we have the wanted number.
Such $x$ exist and you can find it by usin $Chineas.remainder.theorem$


Five people are sitting at a round table. Let $f \geq 0$ be the number of people sitting next to at least one female and let $m \geq 0$ be the number of people sitting next to at least one male. The number of possible ordered pairs $(f,m)$ is

A. 7
B. 8
C. 9
D. 10
E. 11

I don't understand the question. :huh: Is it just saying that there is some number of females and males such that $f+m=5$, and there are at least one of each?

[quote="K81o7"]I get A.
And now the thing wants the message to be longer, so grjk;anags;lgfna;skfjvnal;skdfnlajks,.nma.,sdn.c,sand.,cnsa,dcvasndv.[/quote]

What?

I don't see why the problem is confusing. For clarification, $f+m = 5$.

If all 5 people are female, then the ordered pair is clearly (5,0).  If there is one male and four females, the ordered pair must be (5,2).  If there are two males, they are either next to each other, for an ordered pair of (5,3), or there is a female between them, so the ordered pair would be (4,3).  If there are more males than females, we can reverse all the ordered pairs, so there are 8 possible ordered pairs. The answer is B.

I get that if we have all 5 as female, or all five as male, we get $(5,0)$ or $(0,5)$.  If there is one male and 4 female, or vice versa, then we have $(5,2)$ or $(2,5)$.  If there are two males and three females (or vice versa) then we can have to two odd ones out sitting next to each other to get $(5,4)$ or $(4,5)$.  Or the odd ones out can be separate, in which case we get $(4,3)$ and $(3,4)$.  Yeah, I made a stupid mistake.  I assumed a person was sitting next to a person of some gender as long as that person was of that gender for some reason...


ratio function calculus integration geometry perimeter trigonometry

Prove that circle ratio is greater than 3.05.

Are you asking us to prove that $\pi > 3.05$?

[quote="Misha123"]You can prove it for example with the $\zeta$-function  :D. $\zeta(4)=\frac{\pi^4}{90}$ and is greater than the integral $\int_1^{\infty}\frac{1}{x^4}dx=1$.
From $\frac{\pi^4}{90}>1$ it follows $\pi>3.08...$.

Misha[/quote]

You are wrong in calculation of integral.

kunny

Can anyone solve this problem more basically?

[quote="Myth"]Consider perimetr $p$ of regular octagon inscribed in unit circle. We know that $p$ is less than $2\pi$, and $p=16\sqrt{\frac{1-1/\sqrt{2}}{2}}>3.05$.[/quote]

I think $p=8\sqrt{\frac{1-1/\sqrt{2}}{2}}$, right?

This problem was posted by 2003 Tokyo University/Science.

[color=blue]Solution 1[/color]
Consider perimeter $l$ of regular dodecagon inscribed in unit circle.
So, $l=12\times 2 \times \sin 15^\circ = 24 \sin 15^\circ$
Therefore, $\pi>\frac{l}{2}=12\sin 15^\circ$
$\sin 15^\circ = \sin(45^\circ-30^\circ)=\frac{\sqrt{2}(\sqrt{3}-1)}{4}$ with an addition theorem.
$\therefore \ \pi>12\sin 15^\circ = 3\sqrt{2}(\sqrt{3}-1) \approx 3.105144>3.05$
$(\sqrt{2}\approx 1.414, \ \sqrt{3}\approx 1.732)$

[color=blue]Solution 2[/color]
$\frac{\pi}{6}=\int_{0}^{\frac{1}{\sqrt{3}}}\frac{dx}{1+x^{2}}>\int_{0}^{\frac{1}{\sqrt{3}}}(1-x^{2})dx=\frac{8}{9\sqrt{3}}$
$\therefore \ \pi>\frac{16}{3\sqrt{3}}=\frac{16\sqrt{3}}{9}\approx3.079>3.05 \ \ (\sqrt{3}\approx 1.732)$

I saw this question When I was watching something. I'll upload it for everyone to see. It's an explanation of how pi can be proven to be greater than 3.05. This is very similar for the first method in the previous post, however, i don't understand one bit of it.


pigeonhole principle
linear algebra
matrix
combinatorics unsolved
combinatorics

In a $10$ by $10$ table are written the integers from $1$ to $100$.
From each row we choose the third largest integer.  Show that
the sum of these numbers is not less than the sum of the
numbers in some row.

someone......?
okey, someone please prove that the sum of the third largest numbers of each row is $\ge 522=8+18+28+...+98$ :help:

Hm.. Let $x_{i,j}$ be the $i^{th}$ greatest integer in row $j$. Since exchanging two rows is not a problem, then we can assume that $x_{3,1}> x_{3,2}> \cdots > x_{3,10}$
-) $x_{3,10}> x_{i,10}$ for all $i \in \{ 4,5,\cdots,10 \}$. Then, $x_{3,10}\geq 1+7 = 8$
-) For $l \leq 9$, $x_{3,l}> x_{i,j}$ for all $i \in \{ 4,5,\cdots,10 \},j \in \{ l,10 \}$ and $x_{3,l}> x_{3,k}$ for all $k \in \{l+1,\cdots,10 \}$. Then, $x_{3,l}\geq 1+(11-l) \cdot 7+(10-l) = 88-8l$.
-) $\sum_{i=1}^{10}a_{3,i}\geq 80+72+\cdots+16+8 = 440$

Actually, $522$ is too high, since this condition can be achieved...
And are u sure that $522 = 8+18+\cdots+98$ ? I think it's $530$...  :maybe:

yes cause there are an odd number of $8$s.

it's easy to find an arrangement where the sum is smaller (i think that it's the arrangement with the smallest sum but i'm not really sure.)

let $x_{i,j}$ the $i^{th}$ largest number in row $j$.
now define: $x_{1,1}=99, x_{2,1}=98, x_{1,2}=97, x_{2,2}=96, ..., x_{1,10}=81, x_{2,10}=80$. Now start in the first row to count down, so $x_{3,1}=79, x_{4,1}=78, ....., x_{3,2}=71, ...., x_{10,10}=1$

When you now sum up the 3rd largest number in each row, the sum is $79+71+...+15+7=79*10-8*\frac{9*10}{2}=430<530$

Kathrin

[b]Solution.[/b] Let $\bold{M_{10}}=(x_{i,j})$ be the entries in the $10\times 10$ matrix. Suppose $1\leq x_{i,1}<x_{i,2}<\hdots <x_{i,10}\leq 100$.  Assume to the contrary that $x_{1,8}+\hdots+x_{10,8}< x_{i,1}+\hdots+x_{i,10}$. Hence,
\[\begin{matrix}x_{1,8}+\hdots+x_{10,8}< x_{1,1}+\hdots+x_{1,10}\\ \vdots\\ x_{1,8}+\hdots+x_{10,8}< x_{10,1}+\hdots+x_{10,10}\end{matrix}\]
As a result,
\[\begin{matrix}10(x_{1,8}+\hdots+x_{10,8})<\sum_{1\leq i\leq j\leq 10}x_{i,j}=\frac{100(101)}{2}\\ \\ x_{1,8}+\hdots+x_{10,8}<505\end{matrix}\]
But the ordering that minimizes the total sum is one that satisfies $x_{A,B}=(A-1) 10^{A-1}+B$, along with our initial ordering.  This means that $x_{i,8}=(i-1)10^{i-1}+8$; that is, $\min\sum_{1\leq i\leq 10}x_{i,8}=8+18+28+\cdot+98=530$. Contradiction!$\blacksquare$

[b]Solution.[/b] Let $1\leq x_{1}<\cdots <x_{10}\leq 100$ be all third largest numbers from each row. Since $x_{10}$ is the third largest number of a given row, then the two larger numbers in its row exceed it.  Also, at most $2(10)=20$ numbers exceed it, where the $20$ numbers may come from each row.As a result, $x_{10}\geq 80$.  Using the same argument for $x_{9}$, we have that the $20$ numbers larger than the third largest numbers and all other numbers in the row containing $x_{10}$ may be $\geq x_{9}$, that is $x_{9}\geq 100-20-8=72$.  As a result,
\[S(10)=\sum_{i=1}^{10}x_{i}\geq x_{1}+(x_{1}+1)+(x_{1}+2)+\cdots+(x_{1}+7)+72+80=8x_{1}+180\]
The row with the minimal sum of values is the row containing $x_{1}$, which has a maximal sum of $8x_{1}+171<S(10)$.

And the conclusion follows.$\blacksquare$

Wow, now that's a good solution...  :thumbup:
But, even the solution is good, I don't agree if the minimal sum of values is the row containing $x_{1}$..  :(
Counter Example : $x_{1}= 15, x_{2}= 16, x_{k}= 8k$ for $k \geq 3$
Row containing $x_{1}$ : $8-9-10-11-12-13-14-15-99-100$
Row containing $x_{2}$ : $1-2-3-4-5-6-7-16-81-82$
Row containing $x_{k}$ : $8k-7,8k-6,8k-5,8k-4,8k-3,8k-2,8k-1,8k,2k+77,2k+78$

[quote="zgorkster"]Tell me if I'm wrong, but what about the arrangement like this:

$\left ( \begin{array}[b]{cccccccccc}1&2&3&4&5&6&7&8&81&82\\ 9&10&11&12&13&14&15&16&83&84\\ \cdots \\ 73&74&75&76&77&78&79&80&99&100 \end{array}\right )$.

Wouldn't the sum of the third biggest numbers in each row be $8+16+\cdots+80=\frac{10}{2}\cdot (8+80)=440$?  If im not mistaken, this problem needs a more clever argument...[/quote]

yes, thats my arrangement, but i forgot the number 100, so my highest numbe rwas 99.... but the idea was the same (and because i forgot 100, my sum was 430 not 440. of course, 440 is right!)


$a$ and $b$ are non-neqative numbers such that $a+b\geq2.$
Prove that $\sqrt{1+4a}+\sqrt{1+4b}\geq4.$

i dont know my proof is true or not but
[hide]here is my solution:$(\sqrt{1+4a}+\sqrt{1+4b})^{2}\ge 10+2\sqrt{9+16ab}\ge16 \Longleftrightarrow 9+16ab\ge9 \Longleftrightarrow 16ab\ge0$and its clear.[/hide]

[quote="parinaz.sa"]i dont know my proof is true or not but
[hide]here is my solution:$(\sqrt{1+4a}+\sqrt{1+4b})^{2}\ge 10+2\sqrt{9+16ab}\ge16\Longrightarrow 9+16ab\ge9\Longrightarrow 16ab\ge0$and its clear.[/hide][/quote]
The proof is right, but please, mind the arrows! Now you're just proving that if the statement is right, we would have that $ab\geq 0$. Anyway, reversing the signs (or making them $\Longleftrightarrow$s) would make the proof valid.

you mean that i should change signes and use$\Longleftrightarrow$instead of$\Longrightarrow$??

Yes, because you are working backwords.  However, the proof is correct.:)


function
algebra
functional equation
algebra unsolved

Let $f: \mathbb{N}\to \mathbb{N}$ and \[\forall m,n \in \mathbb{N}: f(m^{2}+n^{2}) = f(m)^{2}+f(n)^{2}.\] If $f(1)=1,$ show that \[\forall n \in \mathbb{N}, f(n)=n.\]

[quote="googleplex"]Prove first over extension to non-negative reals.
Let m = 1, n = 0.
f(1) = f(1)^2 + f(0)  ==> f(0) = 0
Now let m vary, n = 0.
f(m^2) = f(m)^2 + f(0)^2 = f(m)^2 all m.
Then f(m^2 + n^2) = f(m)^2 + f(n)^2 = f(m^2) + f(n^2) all m, n in R.
Every element of R+ is a square of some element of R+.  Let m^2 = x, n^2 = y.
Then f(x + y) = f(x) + f(y) all x, y in R+.  This is the Cauchy functional equation which has solution f(x) = x over R.
Therefore in particular f(n) = n over N.

The question can still be proved for f a general integer function such that f(1) is positive rather than specifically = 1.  (Try it) Extension to the reals yields an additional solution f(x) = 1/2 for all x.  This is connected to the Reimann zeta function critical line.[/quote]

[quote="pleurestique"]how is this proof valid? it assumes that any solution to a functional equation on the naturals can be extended to one on the reals. if this is in fact true, it's definitely not obvious, and i think it calls for justification.[/quote]

Pleurestique is right. You cannot just simply extend a function on natural numbers to one on the reals, since the original function is only defined for natural numbers. This also means that $f(0)$ is not defined, so you cannot obtain or use it. Furthermore you cannot use Cauchy Equations either because they require some condition such as the function is monotonic or continuous, which has no meaning when the function is over natural numbers.

That all makes sense.  I was mistaken to extend to the reals in the first place.  However surely if I had established f(x) + f(y) = f(x + y) for the reals it woud not be illicit to infer the same result for the natural numbers.

Also, is n = 0 definitely not a natural number.  Some authors treat it as a natural number.  I am happy to discontinue my use of it as a natural number of course if that is a strictly applied convention.

I can prove this without extending to reals if n = 0 is a natural number.
Otherwise not.

[quote="googleplex"]Also, is n = 0 definitely not a natural number.  Some authors treat it as a natural number.  I am happy to discontinue my use of it as a natural number of course if that is a strictly applied convention.[/quote]

Hmm.. I have never heard of 0 as a natural number but of course I cannot say that it has never been used as such. But I think here we use $\mathbb{N}=\mathbb{Z}^{+}$


Determine all functions $f$, defined for all real numbers and taking real numbers as values, for which $f((x-y)^2) = x^2 - 2yf(x) + (f(y))^2.$

$x=y=0$ implies that $f(0)=f(0)^2$ and so $f(0)=0$ or $f(0)=1$.

$x=y$ implies that $f(x)^2-2xf(x)+(x^2-f(0))=0$

So, $f(x)= x \pm \sqrt{f(0)}$.

If $f(0)=0$, then $f(x)=x$ for all $x \in R$. (1)
If $f(0)=1$, then $f(x)=x \pm 1$ for all $x \in R$. (2)

For solve the second case, we use the following equations :
$x=0$ and $y=a$ implies that $f(a^2)=-2af(0)+f(a)^2 = -2a+f(a)^2$
$y=0$ and $x=a$ implies that $f(a^2)= a^2 +f(0)^2=a^2+1$

So, $f(a)^2=(a+1)^2$. Hence, $f(a)=a+1$ (because $f(a)=-a-1$ is impossible since (1) and (2) )

So, we have to solutions : $f(x)=x$ and $f(x)=x+1$.


inequalities
search
inequalities unsolved

prove the following inequality
for a,b,c>=0 and k>=0
(a/b+c)^k+(b/c+a)^k+(c/a+b)^k>=min(2,3/2^k)

I'm sorry, but I can't find any of the posts.
Could anyone post a link?


Hello. i had some questions about legendre polynomials. I have solved most of them but i had just two not answered question. I tried to solve this problem by rodriguez rule but it was really hard for me. Could anyone help me or give me some hints for this question?
i tried also generating function.but i couldnt find the solution.sorry but i really got troubled with these questions.
can anyone give me suggestion or show me a way to solve these problems?
I just solved 10 of 12 proofs that my teacher gave me for homework.and just two these questions remained.

[url=http://img139.imageshack.us/my.php?image=mathkw4.jpg][img]http://img139.imageshack.us/img139/7731/mathkw4.th.jpg[/img][/url]

alligator123, please to not make the same post multiple times. If you have something to post, please post it only once.

To the moderators of this forum: I have deleted the copies of this that appeared in the College Playground so that there would be fewer copies on the site. If you think that this is more appropriate for the College Playground anyway, then move this copy.

which part of the forum is good for this problem? so i can send it there.


geometry
trapezoid
symmetry
trigonometry
number theory
relatively prime
trig identities

A hexagon inscribed in a circle has three consecutive sides each of length $3$ and three consecutive sides each of length $5$. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length $3$ and the other with three sides each of length $5$, has length equal to $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

$\text{(A)}\ 309 \qquad \text{(B)}\ 349 \qquad \text{(C)}\ 369 \qquad \text{(D)}\ 389\qquad \text{(E)}\ 409$

Hmm, if nobody wants to post the solution for this problem, I will.  :D

[hide="Solution"]In hexagon $ABCDEF$, let $AB = BC = CD = 3$ and let $DE = EF = FA = 5$. Notice that $\angle BCF$ and $\angle BEF$ subtend the same arc, so that $\angle BCF=\angle BEF$. Additionally, arc $BAF$ is one third of the circle's circumference, so that $\angle BCF=\angle BEF=60^\circ$. Similarly, $\angle CBE = \angle CFE = 60^\circ$.

Let $P$ be the intersection of $BE$ and $CF$, $Q$ that of $BE$ and $AD$, and $R$ that of $CF$ and $AD$. Triangles $EFP$ and $BCP$ are equilateral, and by symmetry, triangle $PQR$ is isosceles and thus also equilateral. Additionally, $\angle BAD$ and $\angle BED$ subtend the same arc, as do $\angle ABE$ and $\angle ADE$. Hence triangles ABQ and EDQ are similar.

Thus, $\frac{AQ}{EQ}=\frac{BQ}{DQ}=\frac{AB}{DE}=\frac 35$. It follows that $\frac{\frac 12(AD-PQ)}{5+PQ}=\frac{3-PQ}{\frac12(AD+PD)}=\frac 35$. Solving this equation, $AD=\frac{360}{49}$, and $m+n=409$.[/hide]

[hide="Solution"]Ptolemy!

Call the hexagon ABCDEF, where AB=BC=CD=3, DE=EF=FA=5. $ \angle FAB \equal{} \angle EDC \equal{} 120$, since both intercept 2/3 of the circumference=240 degrees. By Law of Cosines, BF=CE=7.

Ptolemy on BCEF gives BE=CF=8. Ptolemy on BCED gives BD=AC=39/7. Ptolemy on ABCD gives AD=360/49.[/hide]

[hide="Big trash .... actually its a trig bash"]
Let the radius of the circle be $ r$. Let $ 2\alpha$ be the measure of each of the angles subtended by segments of length $ 3$, and let $ 2\beta$ be the measure for $ 5$. Then $ 6\alpha\plus{}6\beta\equal{}360^{\circ}$ so $ \alpha\plus{}\beta\equal{}60$. Furthermore, bisecting a $ 2\alpha$ angle shows that $ 2r\sin\alpha \equal{} 3$, and likewise $ 2r\sin\beta \equal{} 5$.

Now, the length we seek is just \[ 2r\sin3\alpha \equal{} 2r\sin3\beta \equal{} 2r \sin\beta[\cos2\beta\plus{}2\cos^2\beta] \equal{} 5 (4\cos^2\beta\minus{}1)\]

Solving for $ \beta$,\[ \frac35 \equal{} \frac{\sin\alpha}{\sin\beta}\equal{}\frac{\sqrt{3}}{2}\cot\beta\minus{}\frac12\]
Thus \[ \cot^2\beta\equal{}\frac{121}{75}\equal{}\frac{1}{\sin^2\beta}\minus{}1\implies\frac{11}{14}\equal{}\cos\beta\]
The answer follows after substituting into $ 5(4\cos^2\beta\minus{}1)\equal{}\boxed{\frac{360}{49}}$ :)

[/hide]


combinatorial geometry
geometry
euclidean distance
IMO
point set
IMO 1971

Prove that for every positive integer $m$ we can find a finite set $S$ of points in the plane, such that given any point $A$ of $S$, there are exactly $m$ points in $S$ at unit distance from $A$.

i shall prove a more general statement about the unit distance graph($V=\mathbb{R}^2$, adjacency iff the Euclidean distance between the points is $1$) and this will follow as a consequence.
 if $G,H$ occur as unit distance graphs, then so does $G\times H$( here $\G\times H$ is described as $V(G\times H)=V(G)\times V(H), (v_1,w_1)\leftrightarrow (v_2,w_2) \Leftrightarrow v_1=v_2,w_1\leftrightarrow w_2$ or $v_1\leftrightarrow v_2,w_1=w_2$).
this is seen by describing the vertices by complex numbers. suppose there is an embedding of $G$ by the complex numbers $v_1,v_2\ldots,v_n$ and for $H$ by the numbers $w_1,w_2,\ldots, w_m$. we claim that for some choice of $0\leq\theta<2\pi$, $v_i+e^{i\theta}w_j$ will do the job(a suitable rotation).
 what we need is that $(v_i+e^{i\theta}w_j)\leftrightarrow (v_k+e^{i\theta}w_l)$ iff either $v_i=v_k,|w_j-w_l|=1$ or $w_j=w_l,|v_i-v_k|=1$. clearly if the condition holds then the adjacency is satisfied. suppose $i\neq k,j\neq l$ and that the corresponding complex numbers are at a distance $1$ from one another. Then this gives a quadratic in $e^{i\theta}$ and hence $\theta$ can take only finitely many values here.ruling this out for each such set of $i,j,k,l$ hence rules out finitely many values of $\theta$ and therefore a suitable choice exists.
for the given problem we need a unit distance graph which is regular of degree $m$ for any given $m$. since we can form the graph $K_2$, we can form $Q_n=Q_{n-1}\times K_2$(the unit cube) and that solves the problem.

Suppose $S$ has the form $$S_m=\left\{\sum_{\mathbf v \in U} \mathbf v \;\middle\vert\; U \subseteq V_m\right\}=\left\{\sum_{i=1}^m b_i \mathbf v_i \;\middle\vert\; b_i \in \{0,1\}\right\}\!,$$ where $V_m=\{\mathbf v_1,\mathbf v_2,\dots,\mathbf v_m\}$ is unknown set of distinct unit vectors in $\mathbb R^2$. We can build $V_m$ inductively, starting from the empty set and adding vectors to it, one by one. We just need to make sure that two thing are respected:
1. All $2^m$ vectors in $S_m$ are distinct;
2. Two vector sums are at unit distance from one another if and only if they differ in presence of exactly one summand (i.e. one and only one coefficient $b_i$ in the sum changes from $0$ to $1$).
If these two conditions are kept, then each of $2^m$ points at $S_m$ will be at unit distance from exactly $m$ points corresponding to sums at which one and only one of $m$ coefficients differs from coefficients of this point. However, respecting these conditions is not hard because $\left\|\sum_{\mathbf v \in U_1} \mathbf v - \sum_{\mathbf v \in U_2} \mathbf v\right\|=\left\|\sum_{\mathbf v \in U_1 \setminus U_2} \mathbf v \,-\! \sum_{\mathbf v \in U_2 \setminus U_1} \mathbf v\right\|$ and for each new vector being added to $V_m$ there is at most some finite set of forbidden endpoints given by sums/differences of already determined vectors but the rest of the (infinite) unit circle is permissible.

We will use induction on m.
For m=1 take two points with the distance 1
Let P be a set of k distinct points on the plane such that the given statement holds for m=k.
Let P' be a set of distinct points on the same plane, disjoint with P, so that if X,Y∈P then there exist X', Y'∈P' such that XYY'X' is a parallelogram.
and XX'=YY'=1. We call image of X is X'. Note that there are infinitely many choices for P'.
 Now there are at least k+1 points at a unit distance from X∈P: k points in P and the point X'.
There are at least k+1 points at a unit distance from X'∈P': k points in P'(image points of the k points, which is in P, which are at a unit distance from X , are at a unit distance from X') and the point X.
Now we have to avoid situations like X'Y=1 where X'∈P' and Y∈P while X' is not the image of Y(i.e.X≠Y).
Note X'X=X'Y=1. So X' lies on the perpendicular bisector of XY. So there are at most 2 possible choices for X' .This means there are at most two possible choices for P' so that X'Y=1, for particular points X and Y. If |P|=p then there are at most 2×(p[b]C[/b]2) possible choices for P' such that X'Y=1 for some points X'∈P' and Y∈P.But there are infinitely many choices for P'.
So we can avoid the above mentioned situations. Therefore we are done.


analytic geometry
Galois Theory
algebra unsolved
algebra

Sorry for the typo.But can you complete it?

No. I have no idea what to do with the 2nd factor.

[quote="Absolut"]If $a^{8}+a+1=0$ then $a^{24}+a^{3}-3a^{9}+1=0$, or equivalently $a^{26}+a^{5}-3a^{11}+a^{2}=0$, we have also that $a^{6}-a^{5}+a^{3}-a^{2}+1=0$ sum up and we need to solve $a^{26}-3a^{11}+a^{6}+a^{3}+1=0$...[/quote]

Err... why do you want to do that? You will now have 26 solutions, most of which are wrong!

I found the 8 exact complex solutions.

But they don't lie on a circle or something, so I cannot determine any relation between the coordinates  :(

Any hint?

If n=2 mod 3 (n=3k+2) then the decomposition of x^n+x+1 in irreducible factors over Q is:
x^n+x+1=(x^2+x+1)(x^(3k)-x^(3k-1)+x^(3k-3)-x^(3k-4)+......
+x^3-x^2+1).
Here k=2 and the second factor is p(x)=x^6-x^5+x^3-x^2+1.
The Galois group of p is S6 (use Maple); S6 is not solvable thus we can't solve p(x)=0 by radicals. Only the roots of x^2+x+1=0 can be done explicitly.


how did you get from the second to last step to the last step?  :maybe:
I have a feeling that this is a really stupid question  :D

Since the sum is telescopic, by observing the sum, you notice that what is left over is the last step.  Try the first values.  The pattern is obviously true.


limit
logarithms
calculus
calculus computations

Find the limit
\[ \underset{y\to \,\infty }{\mathop{\lim }}\,{{y}^{\ln \left( y \right)}}.{{\left| {{x}^{y}}\minus{}1 \right|}^{y}}\,\,\forall x\in \left( 0,1 \right)\]

Hi :) !!

that's obvious that:

$ y^{ln(y)}\left|x^y \minus{} 1\right|^y \equal{} e^{(ln(y))^2 \plus{} yln \left|e^{yln(x)} \minus{} 1 \right|}$ with $ y \succ 0$ and $ 0 \prec x \prec 1$

than the limit is (after some simplification) take to $ \plus{} \infty$ ....


If a,b,c be Positive numbers  prove that:
                    (Π a)( Π(a+2))≤(1+(2/3)*(Σ ab)).^3

This problem purposed in Korean Mo-2006.  I find 2 solution for it and post them soon !

By AM-GM:

   $ \prod \sqrt{bc}(a+2) \le (\frac 1{3}\sum \sqrt{bc}(a+2))^{3}$.


When December 1 is on a Tuesday, on what day of the week is
January 25 of the following year?

The last day of December that is on a Tuesday is $ 7\times 4 \plus{}1\equal{}29$.  That means the $ 30^\text{th}$ is a Wednesday and $ 31^\text{st}$  Thursday, meaning that January the $ 1^\text{st}$ is a Friday.  Doing almost the same thing as before, we do $ 7\times 3 \plus{}1\equal{}22$, and get that the $ 22^\text{nd}$ of January is also a Friday.  The $ 25^\text{th}$ is three days afterward, so we get $ \boxed{\text{Monday}}$


inequalities
inequalities proposed
3-variable inequality
algebra

[quote="lasha"]Let $ a,b,c$ be positive numbers, satisfying $ abc > \equal{} 1$.
Prove that $ a^{3} \plus{} b^{3} \plus{} c^{3} > \equal{} ab \plus{} bc \plus{} ca$.[/quote]

By Muirhead we have:

$ a^3\plus{}b^3\plus{}c^3\equal{}\frac{1}{2} \sum_{sym}{a^3} \ge \frac{1}{2}\sum_{sym}{a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{1}{3}}}\equal{}\sqrt[3]{abc}(ab\plus{}ac\plus{}bc) \ge ab\plus{}ac\plus{}bc$

An easier approach is the following: Chebyshev's inequality implies \[ a^3 \plus{}b^3 \plus{}c^3 \geq \frac{a\plus{}b\plus{}c}{3}(a^2 \plus{}b^2 \plus{}c^2).\] By AM-GM we have $ \frac{a\plus{}b\plus{}c}{3}\geq (abc)^{\frac{1}{3}}\geq 1,$ and clearly we have $ a^2 \plus{}b^2 \plus{}c^2 \geq ab\plus{}bc\plus{}ca$. The result follows.

by letting$a=x^3,b=y^3,c=z^3$
we get$x^9+y^9+z^9\ge x^4y^4z+x^4yz^4+xy^4z^4$
which is trivial by Muirhead

By Hölder, $9(a^{3} \plus{} b^{3} \plus{} c^{3}) \geq (a+b+c)^3$. By Maclaurin (or trivial expanding), $(a+b+c)^2 \geq 3(ab+bc+ca)$. By AM-GM, $a+b+c \geq 3\sqrt[3]{abc} \geq 3$. So  $(a+b+c)^3 \geq 9(ab+bc+ca)$, and we are done.

By AM-GM, $\frac{a^3 + b^3 + 1}{3} \ge ab$ and $a^3 + b^3 \ge 3ab - 1$. Cyclic summing gives $2(a^3 + b^3 + c^3) \ge 3(ab+bc+ac)-3$ and $a^3 + b^3 + c^3 \ge \frac{3}{2} (ab+bc+ac) - \frac{3}{2}$.

Again by AM-GM, $\frac{ab+bc+ac}{3} \ge (a^2 b^2 c^2)^\frac{1}{3} \ge 1$ and $\frac{1}{2}(ab+bc+ac) \ge \frac{3}{2}$.

Finally, combining both inequalities, we have $a^3 + b^3 + c^3 \ge \frac{1}{2}(ab+bc+ac) + (ab+bc+ac) - \frac{3}{2} \ge ab+bc+ac$ and we win.

if we prove $a^{3}+b^{3}+c^{3}\geqslant a^{2}+b^{2}+c^{2}$ then it will be done. Let's rewrite it as $\frac{a^{4}}{a}+\frac{b^{4}}{b}+\frac{c^{4}}{c}\geqslant a^{2}+b^{2}+c^{2}$, by Cauchy $(a^{2}+b^{2}+c^{2})^2\geqslant (a+b+c)(a^2+b^2+c^2)$ then we have to prove $a^2+b^2+c^2\geqslant a+b+c$ again applying Cauchy $\frac{(a+b+c)^2}{3}\geqslant a+b+c$ and get $a+b+c\geqslant 3$ and by AM-GM  $a+b+c\geqslant 3\sqrt[3]{abc}\geqslant 3$, Q.E.D.

\[\Longleftarrow \sum x^9 \ge \sum x^4y^4z \]
\[\Longleftrightarrow \sum (x^9+x^9+x^9+x^9+y^9+y^9+y^9+y^9+z^9-9x^4y^4z) \ge 0\]
Which is trivial by $ AM-GM \ inequality   $

We know $(abc)^{1/3}(ab+bc+ca)\ge1(ab+bc+ca)$ is true by the given inequality. By Muirhead $2(a^3+b^3+c^3)\ge2(a^{4/3}b^{1/3}c^{1/3}+a^{1/3}b^{4/3}c^{1/3}+a^{1/3}b^{1/3}c^{4/3})$ is true. Dividing both sides of the inequality by 2 tells us that $a^3+b^3+c^3\ge(abc)^{1/3}(ab+bc+ca)\ge1(ab+bc+ca)$ so we are done.

[quote=trkac][quote="lasha"]Let $ a,b,c$ be positive numbers, satisfying $ abc\geq 1$.  Prove that
\[ a^{3} \plus{} b^{3} \plus{} c^{3} \geq ab \plus{} bc \plus{} ca.\][/quote]

$3(a^3+b^3+c^3) \ge a^3+b^3+c^3+6abc \ge$

$\ge (a+b+c)(ab+bc+ca) \ge 3(ab+bc +ca)$[/quote]
Sorry, how to prove it?
$$a^3+b^3+c^3+6abc \ge (a+b+c)(ab+bc+ca) $$

[quote=Mathskidd][quote=trkac][quote="lasha"]Let $ a,b,c$ be positive numbers, satisfying $ abc\geq 1$.  Prove that
\[ a^{3} \plus{} b^{3} \plus{} c^{3} \geq ab \plus{} bc \plus{} ca.\][/quote]

$3(a^3+b^3+c^3) \ge a^3+b^3+c^3+6abc \ge$

$\ge (a+b+c)(ab+bc+ca) \ge 3(ab+bc +ca)$[/quote]
Sorry, how to prove it?
$a^3+b^3+c^3+6abc \ge (a+b+c)(ab+bc+ca) $[/quote]

Simplifying it we see it is equivalent to $a^3+b^3+c^3+3abc \ge \sum a^2b$. Note that if we prove $3abc\ge a^3+b^3+c^3$ then we will be done as it will prove that $a^3+b^3+c^3+3abc \ge 2(a^3+b^3+c^3) \ge \sum a^2b$. Not sure how to do this give restriction $abc\ge1$, for a second I got the inequality was not true oops

@anandlyer
Are you sure
$3abc\ge a^3+b^3+c^3$
But AM_GM
$ a^3+b^3+c^3\ge 3abc$

[quote=Mathskidd]
Sorry, how to prove it?
$$a^3+b^3+c^3+6abc \ge (a+b+c)(ab+bc+ca) $$[/quote]
It's just Schur.
We can prove it by very many ways. For example:
Since this inequality is symmetrical, we can assume $a\geq b\geq c$.
Hence, $a^3+b^3+c^3+6abc - (a+b+c)(ab+bc+ca)=\sum_{cyc}(a^3-a^2b-a^2c+abc)=$
$=\sum_{cyc}a(a-b)(a-c)\geq a(a-b)(a-c)+b(b-a)(b-c)=(a-b)^2(a+b-c)\geq0$.

[quote=lasha]Let $ a,b,c$ be positive numbers, satisfying $ abc\geq 1$.  Prove that
\[ a^{3} \plus{} b^{3} \plus{} c^{3} \geq ab \plus{} bc \plus{} ca.\][/quote]

We have
\[\sum{a^3}-\sum{bc}=\frac{1}{3}\sum{(2a+1)(a-1)^2}+\frac{3+a+b+c}{6}\sum{(b-c)^2}+(abc-1)\ge{0}\]

By AM-GM, $a+b+c \ge 3 \sqrt[3]{abc} \ge 3$, so $a^2+b^2+c^2 \ge \frac{(a+b+c)^2}{3} \ge a+b+c$.
Because $a^2+b^2+c^2 \ge ab+bc+ca$, it follows by Cauchy-Schwarz that
$a^3+b^3+c^3 \ge \frac{(a^2+b^2+c^2)^2}{a+b+c} \ge ab+bc+ca$.

[quote=lasha]Let $ a,b,c$ be positive numbers, satisfying $ abc\geq 1$.  Prove that
\[ a^{3} \plus{} b^{3} \plus{} c^{3} \geq ab \plus{} bc \plus{} ca.\][/quote]
The following inequality is also true :
Let $ a,b,c$ be positive numbers, satisfying $ abc\geq 1$.  Prove that
\[ 2( a^{3} \plus{} b^{3} \plus{} c^{3} )+3 \geq 3( ab \plus{} bc \plus{} ca ).\]

[quote=abdelkrim]
The following inequality is also true :
Let $ a,b,c$ be positive numbers, satisfying $ abc\geq 1$.  Prove that
\[ 2( a^{3} \plus{} b^{3} \plus{} c^{3} )+3 \geq 3( ab \plus{} bc \plus{} ca ).\][/quote]

$\sum_{cyc}(a^3+a^3+1)\ge \sum_{cyc}(3a^2)\ge  \sum_{cyc}(3ab)$

We can even use sophie Germain's identity

[quote=arqady][quote=sqing]Let $ a,b,c$ be positive numbers, satisfying $ abc= 1$. Prove that
\[ a^{3} + b^{3} + c^{3} \geq \frac{9}{8}(a+ b+ c)^2-\frac{57}{8}.\][/quote]
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that $f(v^2)\geq0,$ where
$$f(v^2)=9u^3-9uv^2+w^3-\frac{27}{8}u^2w+\frac{19}{8}w^3.$$
But we see that $f$ decreases, which says that it's enough to prove our inequality for the maximal value of $v^2$,
which happens for equality case of two variables.
Let $b=a$.
Hence, $c=\frac{1}{a^2}$ and we need to prove that
$$2a^3+\frac{1}{a^6}\geq\frac{9}{8}\left(2a+\frac{1}{a^2}\right)^2-\frac{57}{8}$$ or
$$(a-1)^2(16a^7-4a^6-24a^5+13a^4+14a^3+15a^2+16a+9)\geq0,$$ which is obvious.

The following inequality is also true.
Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$a^3+b^3+c^3\geq\frac{4}{3}(a+b+c)^2-9.$$[/quote]
Thanks.

My Solution:

Just note,that due to Muirhead's inequality,we have:

$a^{3}+b^{3}+c^{3}\geq\sqrt[3]{abc}(ab+bc+ca)$ because it's equivalent to $a^{3}+b^{3}+c^{3}\geq\sqrt[3]{a^{4}b^{4}c}+\sqrt[3]{b^{4}c^{4}a}+\sqrt[3]{c^{4}a^{4}b}$ and we know,that$(3,0,0)\succ(\frac{4}{3},\frac{4}{3},\frac{1}{3})$

And since $abc\geq1$,we have:

$LHS\geq\sqrt[3]{abc}(ab+bc+ca)\geq ab+bc+ca$

$Q.E.D$

Let $ a,b,c$ be positive numbers satisfying $ abc\geq 1$.  Prove that
$$a^3+ b^2+ c^2 \geq  a^2 +b+c$$
$$a^3+ b^2+ c \geq a^2 +b+1$$

[quote=sqing]Let $ a,b,c$ be positive numbers satisfying $ abc\geq 1$.  Prove that
$$a^3+ b^2+ c^2 \geq  a^2 +b+c$$[/quote]

[quote=sqing]Let $ a,b,c$ be positive numbers satisfying $ abc\geq 1$.  Prove that
$$a^3+ b^2+ c \geq a^2 +b+1$$[/quote]
$$\frac{a^2+1}{2}+b+c\geq a+b+c\geq 3$$
$$ \frac{3}{2}a^2+2b+c- \frac{3}{2}\geq a^2 +b+1$$
$$a^3+ b^2+ c\geq \frac{3}{2}a^2+2b+c- \frac{3}{2}\geq a^2 +b+1$$

[hide=Thanks.][quote=GravitoelectromagnetismTM]Homogenize by multiplying $\sqrt[3]{abc}$, then trivial by Muirhead[/quote]

[/hide]

[quote=lasha]Let $ a,b,c$ be positive numbers, satisfying $ abc\geq 1$.  Prove that
\[ a^{3} \plus{} b^{3} \plus{} c^{3} \geq ab \plus{} bc \plus{} ca.\][/quote]
$\Leftrightarrow \sum_{sym}{a^3}\ge \sum_{sym}{ab}$
As: $abc\ge 1 \Rightarrow \sum_{sym}{a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{1}{3}}}\ge\sum_{sym}{ab}$
It is enough to prove that:
$\sum_{sym}{a^3}\ge \sum_{sym}{a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{1}{3}}}$
As:
$3\ge \frac{4}{3}$
$3\ge \frac{8}{3}$
$3=3$
By Muirhead:
$\sum_{sym}{a^3}\ge \sum_{sym}{a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{1}{3}}}_\blacksquare$


Source: ACoPS
(Canada, 1970) Let $P(x)=x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}$ be a polynomial with intgral coefficients. Suppose that there exist four distinct intgers $a, b, c, d$ with $P(a)=P(b)=P(c)=P(d)=5$. Prove that there is no integer $k$ with $P(k)=8$.

[hide="my solution"]
We start this pbm. by defining a new function $Q(x)=P(x)-5$.
Then it is easy to observe that $Q(x)$ has four solutions, namely $a,b,c,d$.
$Q(x)=(x-a)(x-b)(x-c)(x-d)R(x)$.
Now lets assume that there's an integer $k$ with $P(k)=8$.
So $Q(k)=P(k)-5=3$.
consequently, $Q(k)=3=(k-a)(k-b)(k-c)(k-d)R(k)$ with $(k-a),(k-b),(k-c),(k-d),R(k)$ are integers.
However, it is impossible to have $3$ as the product of more than three integers. Contradiction.
[/hide]

i think he just wanted to see if anyone else solved it a different way.

and the solution you gave is the way that i also solved this problem.

[quote="Andreas"]Discussed here [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=511519#p511519[/url]
I don't understand why to propose a problem with the solution below. It doesn't make sense, I think.  :maybe:[/quote]
Sorry I didn't know that it was posted before...
well, ACoPS doesn't have solutions to the problems and I want to know other approaches (if they exist) as maokid7 said....


If $n,k$ are integer and $k>0$, prove

$\left\lfloor{n\over k}\right\rfloor+\left\lfloor{{n+1}\over k}\right\rfloor+\dots+\left\lfloor{{n+k-1}\over k}\right\rfloor=n$

[quote="Nerdy_girl_number1"]One of the numbers n/k,(n+1)/k,...,(n+k-1)/k must be integer.
Assume it is (n+i)/k, then L.H.S=i[n/k]+(k-i)[(n+i)/k]
Next, it is not hard to prove that L.H.S=n
(I am sorry but it is really hard for me to go on since I don't know how to type in the symbols :( )[/quote]

You have the right idea, but I think your simplified LHS is a bit off
I believe it should be:

$i\left(\frac{n+i}{k}-1\right)+(k-i)\frac{n+i}k$

and then that simplifies to $n$

Just quote someone to see how they write it.[quote="Nerdy_girl_number1"]
L.H.S=i[n/k]+(k-i)[(n+i)/k][/quote]
This would be written as i\left[\frac{n}{k}\right]+(k-i)\left[\frac{n+i}{k}\right]. Inside dollar signs, it becomes $i\left[\frac{n}{k}\right]+(k-i)\left[\frac{n+i}{k}\right]$.

The \left and \right are not necessary. It just makes it look better. For example, $i\left[\frac{n}{k}\right]$ rather than $i[\frac{n}{k}]$

http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideLay.php

[quote="Nerdy_girl_number1"]yeah,that is what I mean,but I really hate to post replies.Can anyone be a kind guy(or girl) and teach me how to type in the formulas[/quote]

All of the math formatting is done in LaTeX, which you can read about if you click the button on the panel on the left side of your screen that says "LaTeX".  In general you can type commands between $\$$ signs.  There's also another forum that covers just LaTeX, but for an example, here are some basic things you could do:

[list]$\$$ 4x^2 + 3x + 6 $\$$ becomes $4x^{2}+3x+6$[/list]
[list]$\$$ \frac{3x+7}{y+2} $\$$ becomes $\frac{3x+7}{y+2}$[/list]

Here's another idea, based on induction:

[hide]Show that it holds for $n=0$, and then show that if $n$ increases by $1$, the value of the expression also increases by $1$, and conversely, if $n$ decreases by $1$, then the expression also decreases by $1$.[/hide]


Fie $f:R_ + ^*  \to R $ strict crescatoare astfel incat :$\mathop {\lim }\limits_{x \to \infty } \frac{{f(x)}}{x} = 0
$.
Sa se arate ca exista $\mathop {\lim }\limits_{x \to \infty } \frac{{f(x + 1) - f(x)}}{x} = 0$ atunci acesta limita este zero.

Poate nu inteleg ??!!    :cool: The problem seems to be too trivial  for a final step ! :cool:

Foloseste criteriul cu $\epsilon$ si $\delta$. Dar nu inteleg unde intervine monotonia. Poate ca ai omis ceva.


AMC
AMC 10
MATHCOUNTS
geometry
trigonometry
algebra
polynomial
\/closed

[quote="worthawholebean"]I'm a freshman in precalculus and I'm wondering whether I should get volume 1 or 2 of AoPS. I got a 124 on the AMC10 last year and a 103.5 on the 12 this year. I got a 25 at National MathCounts last year.[/quote]
Vol.1 would be mostly review to you, it might have a few new concepts. I'd suggest vol. 2 if you can only get one of them.

This is a bit of a tough call - in general, we'd recommend starting with V1 if your AMC score is below 110-120.  This is particularly true if your weakness is geometry.  If geometry is a strength, but topics like trig are a weakness, then you might be fine skipping Vol. 2.  (Another option if geometry is a weakness is the Intro Geometry book.)

what if your weaknesses are polynomials and functions ?

Polynomials are covered in Volume 2.  Functions are in both volumes, but for what you're looking for, Volume 2 is likely the answer.

I'm a freshman in pre-calc too and I use Volume 2.

Don't be fooled by the tables of contents of the books - when we say 'similar triangles', for example, we are going to do much more challenging problems by the end of that chapter than you'll see in a typical textbook.  So, you may think, "I know similar triangles," but in the book, you'll see strategies to apply similar triangles in much harder problems than you're already familiar with.

My suggestion is to borrow your friend's books for a week.  Look through them, particularly the problems at the end of each chapter.  Then, invest in the ones in which you can't do more than 20-30% of the harder problems (Challenge Problems for the Intro series).

[quote="piac31"]I'm in a similar situation too. I'm a freshman in Alg 2. I scored 121.5 on the AMC 10 and only 2 on the AIME (I made a lot of stupid mistakes). I really want to make USAMO next year, so I'm planning on doing some serious practice this summer. I really need to learn some things like recursive equations and summations because I haven't ever really learned them. In geometry, I know the basics from geometry class, but I have a hard time seeing how to apply them. When I look at the table of contents for AoPS I it looks like stuff I already know.

One of my friends has a bunch of books and offered to let me borrow them, but I'm think I want to have a book for reference. Should I buy AoPS I or just borrow it for a little while and buy AoPS II or buy AoPS II and borrow intro to geo?

It's so hard to tell which book. :([/quote]

I'll be a freshman in Alg 2 also. I am thinking of just using a regular algebra 2 textbook to learn basics, then progress to AoPS Vol. 2 . And then, Intro to NT and C+P maybe.

[quote="Phelpedo"][quote="worthawholebean"]Geometry is definitely not my weakness. I took it twice because my old school sucked and I retook it at my new school.[/quote]

I might add that most school curriculum geometry is very basic and probably insufficient for harder problems that you might encounter on the AIME. After you do a bunch of problems, you'll probably encounter a type that makes you go "Darn it, not another ____ problem..." For me it's geometry and functional equations. I've had pretty much no experience with the latter, but I have gone through quite a bit of geometry and it still bugs me.[/quote]

I found most of the problem's from our schools "Problem Set C" in the Geometry book were comparable with the AoPS ones in Vol 1. Plus, Vol 1 Geometry isn't really going to help you at all for AIMEs #6-10 which is what he's worrying about the most.

[quote="archimedes1"][quote="now a ranger"] I am thinking of just using a regular algebra 2 textbook to learn basics...[/quote]
No!  "The basics" can be memorized in about 30 minutes. Unless you have an [i]exceptionally[/i] amazing textbook (like AoPS), DO NOT just go through the book on your own (I speak from experience).  I went through the whole Alg II textbook and did every "challenge" problem, and[i] still [/i]when I looked through AoPS 2 it was like a whole new world.  I felt like all that hard work at home was completely useless.  If you really want to learn, use AoPS 2.[/quote]

Seeing as AoPS is mostly Pre-Calc topics, I wouldn't be surprised. You can't really berate an Alg II book for not teaching you mostly Pre-Calc concepts.

But I would definitely borrow the Vol 1 book and look through areas which you think you'll be weakest, if it's too hard, buy it and Vol II, if it's not, look for the areas you are weakest in, focus on those, and then move on to Vol II with possible supplemental things.

You see, the main difference between your typical textbook and AoPS is the amount of [i]problems[/i] in AoPS, as opposed to [i]exercises[/i]. (For those of you who have Zeitz, you know where I'm getting this from.) Speaking from my experience with mainstream math textbooks, even the most challenging of the "problems" in the book are, without a doubt, exercises; problems that require little or no additional thinking besides utilising the concept learned a few pages previously. Sure, AoPS has these too, but the sheer amount of problems, that is, problems that require you to actually think rather than apply, at the end of the chapters is what makes AoPS different, and just that much more useful, not only for math contests, but just for learning the concepts in general as well.

If you think you have learned something thoroughly, and you learned it from a conventional textbook, I beg you to reconsider.


calculus
integration
derivative
function
calculus computations

I learned this method of integrating, but I'm not sure what it's called or if it's an "official" way of doing it..
Is there a real name for this?

For example:
$ \int x\sqrt{x^2 \plus{} 2}\ dx$

$ \equal{} \frac{1}{2}\int\sqrt{x^2 \plus{} 2}\ d(x^2\plus{}2)$

(since the derivative of $ x^2\plus{}2$ is $ 2x$, and there is a $ x$ in the integrand so a $ \frac{1}{2}$ must be multiplied in front as a constant. The $ x$ inside the integrand is removed because we are now integrating with respect to $ x^2\plus{}2$)

$ \equal{} \frac{1}{2} * \frac{2}{3}(x^2 \plus{} 2)^\frac{3}{2} \plus{} C$

$ \equal{} \frac{1}{3}(x^2 \plus{} 2)^\frac{3}{2} \plus{} C$

It looks similar to u-substitution, but I'm wondering what it's called when you integrate "with respect"? to something else other than $ x$, such as a function of $ x$ as in the problem. Thanks! :D

its called" u-substitution in your head"

It is a substitution. The only difference is that when you do a substitution on paper, you usually define another variable so that it's easier to follow visually.

The following two are equivalent:

$ \frac {1}{2}\int\sqrt {x^2 \plus{} 2}\ d(x^2 \plus{} 2)$

$ \frac {1}{2} \int \sqrt {u} \; du \qquad u \equal{} x^2 \plus{} 2$

The latter is arguably easier to read.

When the functions get more complicated, it is almost always easier to read and follow if you just create a new variable.

Such u-substitutions are results of the chain rule.  When you think you have an integrand that was produced from the chain rule, you should use such a substitution.

hello, setting $ t\equal{}\sqrt{x^2\plus{}2}$ and $ dt\equal{}\frac{x}{\sqrt{x^2\plus{}2}}dx$ you will get $ \int t^2 dt$.
Sonnhard.

You've actually stumbled upon a pretty interesting idea that provides a way to generalize the Riemann integral, namely, the [url=http://en.wikipedia.org/wiki/Riemann-Stieltjes_integral]Riemann-Stieltjes integral[/url].  As a technique to evaluate Riemann integrals, though, it's just $ u$-substitution as has already been stated.


$x_{1},x_{2},\ldots ,x_{n}$ are positive integers satysfaying condition:$\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}}=1$.
Determine the minimal value of
$x_{1}+\frac{x_{2}^{2}}{2}+\frac{x_{3}^{2}}{3}+\ldots+\frac{x_{n}^{n}}{n}$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=16636
http://www.mathlinks.ro/Forum/viewtopic.php?t=51416

  darij


[quote="rituraj007"][quote="Aravind Srinivas L"]$ a(a \plus{} d)(a \plus{} 2d)(a \plus{} 3d) \plus{} d^4 \equal{} (a^2 \plus{} 3ad \plus{} d^2)^2$
where $ a,d$ are the integral first term of the four consecutive terms  and common difference respectively.[/quote]

$ Hence Proved$
 :)  :P  :)[/quote] :spam:

[quote="rituraj007"]there is no spam in that..i was just telling how the problem solving is done on this forum..the classic way... :P  :)[/quote]
 :D  :harhar:

[quote="rituraj007"]there is no spam in that..i was just telling how the problem solving is done on this forum..the classic way... :P  :)[/quote]
how do you define classic?? :rotfl:  :?:


Anyone have the answer keys for 2007 AMC 10B?

Please... I'm in need of them...


I looked in the pdf provided and could not find how to make the not equal to sign.

\notequal ??

That doesn't work
It says it's a dangerous function, haha


I was looking through the sets I have and doing 2005 when I realized that that page was missing.   :o I'd appreciate it if someone could send me that page or a link to it.  Thanks.

idk can't you get it from michael on...oh wait you're not going to be there...just tell him to e-mail it to you

I looked on both the CD and email and the page was gone.  Frost sent me a copy though, so i'm good.


How do I determine how many positive integer solutions there are to the diophantine equation 3x+7y where x and y are also positive integers. I can figure out for example 2x+5y=1 but that involves x and y being negative.

For your equation: 3x+7y, there seems to be no R.H.S., so , how can you ask the number of positive integer solutions without specifying the R.H.S?

Sorry about that what i meant was 3x+7y=n where n is a positive integer. What are all the positive integer possibilities of n? So basically what numbers can you get from 3x+7y. Its not really a diophantine equation. And by solutions i didn't mean x and y. I meant n.

If we denote by $ a\mathbb{Z}$ the set of all multiples of $ a$(for example,$ 3\mathbb{Z} \equal{} \{..., \minus{} 3,0,3,6,...\}$) then it's well known fact that $ a\mathbb{Z} \plus{} b\mathbb{Z} \equal{} d\mathbb{Z}$, where $ d \equal{} gcd(a,b)$.This means that linear diophantine equations $ ax \plus{} by \equal{} n$ have solutions if and only if $ gcd(a,b)|n$

$ 3x + 7y = n\stackrel{(3 < 7)}{\Longleftrightarrow} x = \frac {n - 7y}{3} = - 2y + \frac {n - y}{3}$ . Since $ x$ and $ - 2y$ are integers obtain $ n - y = 3k$ , $ k\in Z$ , i.e. $ y = n - 3k$ .

Therefore, $ x = - 2(n - 3k) + k$ , i.e. $ x = - 2n + 7k$ , where $ k\in Z$ . In conclusion, $ \boxed {x = 7k - 2n}$ and $ \boxed {y = - 3k + n}$ , where $ k\in\mathbb Z$ .

[b]Another example[/b] : $ \boxed {5x - 13y = 4}$ .

$ \boxed {\ \begin{array}{ccc} 5x - 13y = 4 & \stackrel {(5 < 13)}{\Longleftrightarrow} & x = 2y + \frac {3y + 4}{5} \\
 \\
3y + 4 = 5z & \stackrel{(3 < 5)}{\Longleftrightarrow} & y = z + \frac {2z - 4}{3} \\
 \\
2z - 4 = 3t & \stackrel{(2 < 3)}{\Longleftrightarrow} & z = t + 2 + \frac t2 \\
 \\
t = 2k & \mathrm{where} & k\in\mathbb Z\end{array}\ }$ $ \begin{array}{c} \uparrow \\
 \\
\uparrow \\
 \\
\uparrow \\
 \\
\uparrow\end{array}$ $ \implies$ $ \boxed {\ \begin{array}{ccc} t = 2k & \mathrm{where} & k\in\mathbb Z \\
 \\
z = t + 2 + k & \Longleftrightarrow & z = 3k + 2 \\
 \\
y = z + t & \Longleftrightarrow & y = 5k + 2 \\
 \\
x = 2y + z & \Longleftrightarrow & x = 13k + 6\end{array}\ }$ $ \implies$ $ \boxed {\ \begin{array}{c} x = 13k + 6 \\
 \\
y = 5k + 2\end{array}\ }$ , where $ k\in\mathbb Z$ .

[u]Now you can verify[/u] : $ 5x - 13y = 5(6 + 13k) - 13(2 + 5k)\implies 5x - 13y = 4$ .

[b][u]Remark[/u].[/b] The equation $ ax + by = c$ , where $ \{a,b,c\}\subset \mathbb Z$ and $ abc\ne 0$ has at least a integer solution iff $ (a,b)$ divides $ c$

and in this case the general integer solution is $ \boxed {\ \begin{array}{c} x = bk + x_0 \\
\ y = - ak + y_0\end{array}\ }$ , where $ k\in \mathbb Z$ and $ (x_0,y_0)$ is a particular integer solution of it.

For the above equation $ 5x - 13y = 4$ obtain easily the particular solution $ (6,2)$ and the general solution is $ \boxed {\ \begin{array}{c} x = - 13k + 6 \\
\ y = - 5k + 2\end{array}\ }$ , where $ k\in\mathbb Z$ .

For $ k: = - k$ we refind the previous equivalent general sulution.


A 3 by 3 by 3 cube is formed by gluing together 27 standard cubical dice. The smallest possible sum of all the numbers showing on the surface of the 3 by 3 by 3 cube is what?

[hide="Solution I think"]
There are 8 dice with 3 sides showing, 12 dice with 2 sides showing, and 6 dice with 1 side showing. The smallest sum of the dice with 3 sides showing is 6 (3+2+1). The smallest sum with of the dice with 2 sides show is 3, (1+2). The smallest sum of the dice with 1 side showing is 1. Therefore, the minimum sum is
\[ (8\cdot3)\plus{}(12\cdot3)\plus{}(6\cdot1)\equal{}\boxed{66}
\]
[/hide]

I worked on the problem further, and I think your method of using the 6 cubes with one face showing, 12 cubes with 2 faces showing, and 8 cubes with 3 faces showing is correct.

From then you have:

The 6 cubes with 1 side showing can all be oriented to have 1 showing. The cubes with 2 faces showing can show a 1 and 2. The six cubes with 3 faces showing can have 1, 2, and 3, are showing.

6(1) + 12 (1+2) + 8(1+2+3) = 6 + 36 + 48 = 90

[quote="hunter34"]
There are 8 dice with 3 sides showing, 12 dice with 2 sides showing, and 6 dice with 1 side showing. The smallest sum of the dice with 3 sides showing is 6 (3+2+1). The smallest sum with of the dice with 2 sides show is 3, (1+2). The smallest sum of the dice with 1 side showing is 1. Therefore, the minimum sum is
\[ (8\cdot\boxed{3}) \plus{} (12\cdot3) \plus{} (6\cdot1) \equal{} \boxed{66}
\]
[/quote]

Yeah he just wrote it wrong.

Yep, 66 is correct:\[ (6\times1)\plus{}(12\times3)\plus{}(4\times6)\equal{}\fbox{66}\]

No, 90 is correct! Your last term should be 8x6 as there are 8 corner pieces, not 4!