We write the digits of $1995$ in the following way:

$199511999955111999999555......$

1. Determine how many digits we have to write such that the sum of the written digits is $2880$.
2.Which digit is in position number $1995$?

[hide]I'm not sure.I think the solution is
a)480
b)1[/hide]

Math is my life :!:

I got a different answer for 1...
[hide="1"]

divide in "blocks" the written digits: the first is 1995, the second 11999955... clearly at the $kth$ block the sum of its digits is $24k$. So the sum of all digits written from the first to the $kth$ block is $24 \left( \frac{k(k+1)}{2} \right)$. Let it be equal to 2880, and obtain $k=15$. So we have to write 15 blocks of digits. In the first block we write 4 digits, in the second 8, in the third 12, in the $kth$ block $4k$ digits. So the number of digits written from first to $kth$ block is $4 \left( \frac{k(k+1)}{2} \right)$. Substitute $k=15$ to obtain 480 digits[/hide]
[hide="2"]
the greatest integer $\leq 1995$ that can be written in the form $4 \left( \frac{k(k+1)}{2} \right)$ is 1984 with $k=31$. So the 1984th digit is a 5. Note that in the first block the first digit is 1, in the second the first 2 digits are 1, in the $kth$ block the first $k$ digits are 1. Since
$1995-1984=15<31$, the 1995th digit is 1[/hide]

[hide]1

Let the equation $2x^2+2x$ represent the amount of digits per iteration.

$\displaystyle\sum_{n=1}^{y}24n=2880$

Factor: $24(\frac{(y)(y+1)}{2})=2880$

$y=15$

Plug back- $2(15)^2+2(15)=\boxed{480}$.

[/hide]

[hide]2
Let $2x^2+2x=1995$
$x=\frac{\sqrt{3991}-1}{2}\approx 31.087$

Last digit is $\boxed{1}$ since $.087<.25$
[/hide]

just some more confirmation. I also got [hide]$480$[/hide] and [hide]$1$[/hide]. Great problem and one of the first olympiad/pre-olympiad level problems i've been able to solve.


least common multiple
number theory unsolved
number theory

The least common multiple of positive integers $a,b,c,d$ is equal to $a+b+c+d$. Prove that $abcd$ is divisible by at least one of $3$ or $5$.

Please ask moderators for moving it next time and don't repost it.


calculus
integration
function
calculus computations

For real constant numbers $ a,\ b,\ c,\ d,$ consider the function $ f(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ such that $ f( \minus{} 1) \equal{} 0,\ f(1) \equal{} 0,\ f(x)\geq 1 \minus{} |x|$ for $ |x|\leq 1.$

Find $ f(x)$ for which $ \int_{ \minus{} 1}^1 \{f'(x) \minus{} x\}^2\ dx$　is minimized.

$ f(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ such that $ f( \minus{} 1) \equal{} 0,\ f(1) \equal{} 0$=>$ f(x)\equal{}ax^3 \plus{} bx^2 \minus{}ax \minus{}b$
And then $ \int_{ \minus{} 1}^1 \{f'(x) \minus{} x\}^2\ dx\equal{}\frac{8a^2}{5}\plus{}\frac{2}{3}(2b\minus{}1)^2$
Now take $ 0<\equal{}x<\equal{}1$ we have $ ax^3 \plus{} bx^2 \minus{}ax \minus{}b>\equal{}1\minus{}x$ and $ \minus{}ax^3 \plus{} bx^2 \plus{}ax \minus{}b>\equal{}1\minus{}x$
Then $ b(x^2\minus{}1)>\equal{}1\minus{}x$ and $ ax^3\minus{}ax>\equal{}0$
=>  $ \minus{}b(1\plus{}x)>\equal{}1$ and $ a<\equal{}0$ for all $ x$ in $ [0,1]$
=>$ b<\equal{}\minus{}1$ and $ a<\equal{}0$
But $ \int_{ \minus{} 1}^1 \{f'(x) \minus{} x\}^2\ dx$ is an  increasing function for $ a$ and $ b$
=>$ a\equal{}0$ and $ b\equal{}\minus{}1$
=>$ f(x)\equal{}\minus{}x^2\plus{}1$
=>$ \int_{ \minus{} 1}^1 \{f'(x) \minus{} x\}^2\ dx\equal{}6$


Let $ a, b, c$ be nonzero integers such that $ M \equal{} a/b \plus{} b/c \plus{} c/a$ and $ N \equal{} a/c \plus{} b/a \plus{} c/b$ are both integers. Find $ M$ and $ N$.

posted before  :roll:
[hide] http://www.mathlinks.ro/Forum/viewtopic.php?t=105855 [/hide]


Here is  mock AIME problem

$\text{Define}$
$f(x)$
$g(x)$
$h(x)$
$f(n)-f(n+1)=f(n+2)$
$g(x)$ $\text{is the sum of the digits of x.}$
$h(x)=x+2*2007$
$\text{Compute: }$
$g(h(f(2007)))$

[quote="gotztahbeazn"]maybe im just being stupid.... but don't we need like an initial case for $f(x)$? like $f(1) = 100$ or something because those functions aren't crossing out or anything[/quote]

Actually, you'll need 2 initial cases.

if you are making a mock aime, put all the problems in one test


If $ m_a,m_b,m_c$ are the medians and $ r_a,r_b,r_c$ are the radius of the escribed circles of triangle ABC, prove that:
$ m_a\cdot m_b\cdot m_c\ge r_a\cdot r_b\cdot r_c$

We have $ m_a^2\equal{}\frac{2(b^2\plus{}c^2)\minus{}a^2}{4}\ge\frac{(b\plus{}c)^2\minus{}a^2}{4}\equal{}p(p\minus{}a)$
Similarly, $ m_b^2\ge p(p\minus{}b)$
$ m_c^2\ge p(p\minus{}c)$
so that
$ m_a\cdot m_b\cdot m_c \ge p\cdot S\equal{}\frac{S^2}{r}\equal{}r_a\cdot r_b\cdot r_c$ (Q.E.D)
where p is the subperimeter, S is the area, r is the radius of inscribed circle of triangle ABC


find the nth term of thesequence 1,5,19,71,271,1055,4159,16501.... and pls give a detailed solution.

sorry, sinajackson. Can you tell me why you change the sequences into base 2 ? thanks

well I didn't change anything to base 2, I just guessed that it can be like this, actually I explained the relation between the terms of this sequence, the problem gave us only 8 terms of the sequence and wanted the $n$'th, because all of the 8 terms can be written in a regular form like :
$( 1\underbrace{00 \cdots 0}_{k}\underbrace{11 \cdots 1}_{k})_{2}= 4^{k}+2^{k}-1$ ( as Lord Wing said  :D ) so we can be sure that the $n$'th term is $4^{n}+2^{n}-1$ .

the method is good but is there any similar method to find the sequence of terms obtained from any random sequence a^n+b^n+...+c

If you know that the sequence is of that type: yes.


linear algebra
matrix
induction
superior algebra
superior algebra solved

Let U=(1 1);V=(1 0);W=(-1 0)
          (0 1)     (1 1)      (0  1)
(U,V,W are matrices of type 2,2)
Prove that for every matrix A of type 2,2 with integer elements and
detA is 1 or -1 there exists A_1,A_2,...,A_n matrices form the set
(U,V,W,U1,V1,W1) where U1,V1,W1 are the inverses of U,V,W
Such that A=A_1*A_2*...A_n.

(In other terms the matrices U,V,W generates the GL2(Z) group!).

Prove by induction that for all [b]integers[/b] k
U k =
1 k
0 1
V k=
1 0
k 1
W k=
(-1) k 0
0 1

Now look for a resemblance between the SUBJECT: QUESTIONS ABOUT DETERMINANTS and this problem.


trigonometry
AMC
AIME
quadratics
LaTeX
Diophantine equation
special factorizations

Hmm, I did this slightly less elegantly.
[hide]
We complete the square to obtain
$ (x\plus{}42)^2\plus{}244 \equal{} y^2$.

Therefore $ y^2\minus{}(x\plus{}42)^2\equal{}244$. We use the fact that the difference between two consecutive squares is the sum of their square roots to obtain values for $ y$ and $ x$. A quick check reveals that $ (60\plus{}61)\plus{}(61\plus{}62)\equal{}244$, so $ y\equal{}62$ and $ x\plus{}42\equal{}60$, or $ x\equal{}18$. We are given that the solution is unique, so we have

$ x\plus{}y\equal{}\fbox{80}$[/hide]

I did it ridiculously less elegantly.
I brute forced.
:|

Lol, me too.  I just listed out a bunch of squares and luckily eventually, plugged in 18 for x and got it.

For the sake of the archives, you should probably move your solution into a separate post.

Yea, I used difference of squares too, only I substituted $ z\equal{}x\plus{}42$ for ease.

Yeah, I basically went halfway through distracted's solution, then I was like, "meh, I probably can't solve this question using algebra anyways."
So then I started brute forcing.
Took me a good 15 minutes.

[hide="my solution"]
$ x^2 \plus{} 84x \plus{} 2008 \equal{} y^2$

$ 2008 \equal{} y^2 \minus{} x^2 \minus{} 84x \equal{} (y \minus{} x \minus{} 42)(y \plus{} x \plus{} 42) \plus{} 42^2$

$ 2008 \equal{} y^2 \minus{} x^2 \minus{} 84x \equal{} (y \minus{} x \minus{} 42)(y \plus{} x \plus{} 42) \plus{} 42^2$

$ 244 \equal{} (y \minus{} x \minus{} 42)(y \plus{} x \plus{} 42)$

Since $ 244$ is even, one of the factors is even. A quick check shows that if one of them is even, then both must be even, we should check $ 122*2$

$ y \minus{} x \minus{} 42 \equal{} 2$ and $ y \plus{} x \plus{} 42 \equal{} 122$

$ y \minus{} x \equal{} 44$

$ y \plus{} x \equal{} 80$

$ y \equal{} 62$, $ x \equal{} 18$

$ x \plus{} y \equal{} \boxed{080}$
[/hide]

[quote="distracted523"]Hmm, I did this slightly less elegantly.
[hide]
We complete the square to obtain
$ (x \plus{} 42)^2 \plus{} 244 \equal{} y^2$.

Therefore $ y^2 \minus{} (x \plus{} 42)^2 \equal{} 244$. We use the fact that the difference between two consecutive squares is the sum of their square roots to obtain values for $ y$ and $ x$. A quick check reveals that $ (60 \plus{} 61) \plus{} (61 \plus{} 62) \equal{} 244$, so $ y \equal{} 62$ and $ x \plus{} 42 \equal{} 60$, or $ x \equal{} 18$. We are given that the solution is unique, so we have

$ x \plus{} y \equal{} \fbox{80}$[/hide][/quote]
If you're going to post a solution, please post comments in a separate post so I can put the solution into the resources section.

[hide="For sake of resources, (from my post on wiki)"]
[[Completing the square]], $ y^2 \equal{} x^2 \plus{} 84x \plus{} 2008 \equal{} (x\plus{}42)^2 \plus{} 244$. Thus $ 244 \equal{} y^2 \minus{} (x\plus{}42)^2 \equal{} (y \minus{} x \minus{} 42)(y \plus{} x \plus{} 42)$ by difference of squares.

Since $ 244$ is even, one of the factors is even. A [[parity]] check shows that if one of them is even, then both must be even. Sine $ 244 \equal{} 2^2 \cdot 61$, the factors must be $ 2$ and $ 122$. Since $ x,y > 0$, we have $ y \minus{} x \minus{} 42 \equal{} 2$ and $ y \plus{} x \plus{} 42 \equal{} 122$; the latter equation implies that $ x \plus{} y \equal{} \boxed{080}$.

Indeed, by solving, we find $ (x,y) \equal{} (18,62)$ is the unique solution.
[/hide]

[quote="alkjash"]Yea, I used difference of squares too, only I substituted $ z \equal{} x \plus{} 42$ for ease.[/quote]

Me too, but I kept getting $ (y\minus{}z)(y\plus{}z)\equal{}234$ and wondering why $ y\minus{}z$ and $ y\plus{}z$ had different parities. Wasted way too much time on that problem...

[quote="mathnerd314"][quote="alkjash"]Yea, I used difference of squares too, only I substituted $ z \equal{} x \plus{} 42$ for ease.[/quote]

Me too, but I kept getting $ (y \minus{} z)(y \plus{} z) \equal{} 234$ and wondering why $ y \minus{} z$ and $ y \plus{} z$ had different parities. Wasted way too much time on that problem...[/quote]
I subtracted wrong the first time too...i got =$ 254$ and then i had to try to find what's wrong because it was impossible to get it with only 1 power of 2 if both factors had to be even since diff of squares was even.

I took the AIME unofficially and I couldn't get this problem... I spent a long time on it, playing with perfect squares modulo 3, 4, and even 5... lol I got up this morning (the morning after) and I suddenly realized distracted's solution and I solved the problem sucessfully. So I got it overnight. :)

my solution (not very "elegant")

[hide]moving $ y$ to the left, we get $ x^2 + 84x + 2008 - y^2 = 0$
Using the quadratic formula, where $ c = 2008 - y^2$, we get:
$ x = \frac { - b±\sqrt {b^2 - 4ac}}{2a}$ please ignore the Å, this is the best I could get with latex
$ x = \frac { - 84±\sqrt {7056 - 4(2008 - y^2)}}{2a}$
which simplifies to
$ x = \frac { - 84±\sqrt { - 976 + 4y^2}}{2}$

If we let $ z^2$ equal the number inside the square root. Note that since $ \sqrt {z^2}$ must be even to make $ x$ an integer, $ z$ itself must be even. So we have the equation:
$ z^2 = - 976 + 4y^2$
$ z^2 - 4y^2 = - 976$
$ z^2 - (2y)^2 = - 976$
$ (z + 2y)(z - 2y) = - 976$
$ 976$ factors into $ (2^4)(61)$.
there are many ways to split -976 into a pair of integer factors, but only one pair of these will result into $ x$ and $ y$ into positive integers. This pair is $ - 4$ and $ 244$.
If $ z + 2y$ is $ 244$and $ z - 2y$ is $ - 4$, when adding the two equations,
$ 2z = 240$ ==> $ z = 120$
so $ \sqrt {z^2}$ = $ \sqrt {120^2}$ = $ 120$.

Going back to the quadratic formula, replacing$ \sqrt {b^2 - 4ac}$ with $ 120$,
$ x = \frac { - 84±120}{2}$
Clearly the ± must be +, so
$ x = \frac { - 84 + 120}{2}$ so $ x = 18$.

$ 18^2 + 84(18) + 2008 = y^2$
$ 3844 = y^2$ ==> $ y = 62$ ==> $ 18 + 62 = [80]$
[/hide]
A lot of work, but since I don't know trig and calculus I couldn't do several of the problems, so I had a lot of time :D

The plus-or-minus symbol is just [b]\pm[/b]; for example:
\[ \frac{\minus{}b \pm \sqrt{b^2\minus{}4ac}}{2a}\]

[hide="New post"]
We complete the square to obtain
$ (x \plus{} 42)^2 \plus{} 244 \equal{} y^2$.

Therefore $ y^2 \minus{} (x \plus{} 42)^2 \equal{} 244$. We use the fact that the difference between two consecutive squares is the sum of their square roots to obtain values for $ y$ and $ x$. A quick check reveals that $ (60 \plus{} 61) \plus{} (61 \plus{} 62) \equal{} 244$, so $ y \equal{} 62$ and $ x \plus{} 42 \equal{} 60$, or $ x \equal{} 18$. We are given that the solution is unique, so we have

$ x \plus{} y \equal{} \fbox{80}$[/hide]

[hide="Quick bounding"]
$(x+42)^2 < x^2+84x+2008 < (x+45)^2$ is pretty obvious.

So $x^2+84x+2008=(x+43)^2$ or $(x+44)^2$.

The former clearly doesn't work (just consider modulo 2), so we have $x^2+84x+2008=(x+44)^2=x^2+88x+1936 \implies 4x=72 \implies x=18$ and then $x+y=2x+44=\boxed{080}$[/hide]

We can start by completing the square, $(x+42)^2 + 244 = y^2$.

We can rewrite the equation like this $y^2 - (x+42)^2 = 244$ which equals $(y-x-42)(y+x+42) = 244$.

Now we can say that $y - x - 42 = 2$ and $y + x + 42 = 122$.

So we get that $y = 62$ and $x = 18$ so our answer is $62 + 18 = 080$

I just wrote up a quick solution cause I was practicing typing solutions in latex.

[quote=aop2014]I understand, but it is generally not helpful to fill up the forums with tons of ancient problem threads, try adding solutions to the wiki instead or just writing them for yourself.[/quote]

This is actually appropriate behavior for such forums. Users are encouraged to use old problem threads to write solutions to such problems.

[quote=aop2014]Ok,
1) I've been told off before for reviving ancient threads in any way and generally avoid it
2) It really does clutter it up sometimes, not as much anymore
3) I'll keep this in mind, thanks for the advice.[/quote]

no its actually what you're meant to do....you did nothing wrong....naman12 is correct(as always)

[hide=Solution]
We complete the square to get $x^2+84x+2008=(x+42)^2+244$.

We substitute to get $y^2-(x+42)^2=244$

We use difference of squares to get $(y+x+42)(y-x-42)=244$.

We need $2$ numbers of the same parity that multiply to $244$.

We are strategically lazy so we will pick the most obvious pair $122$ and $2$.

We are strategically lazy so we will not check other pairs for more solutions.

We know that $x+y+42=122$ so $x+y=\boxed{080}$.
[/hide]

[quote=chickendude]There exist unique positive integers $ x$ and $ y$ that satisfy the equation $ x^2 \plus{} 84x \plus{} 2008 \equal{} y^2$. Find $ x \plus{} y$.[/quote]

[hide=Solution]We complete the square. We get that it would be $x^2+84+1764+244=y^2.$ We then get that $(x+42)^2=y^2-244.$ Therefore, we get that two squares are $244$ apart from each other, so we can let $x+42=z$. We get that $z^2-y^2=244$ where we see that $z>42$. We get that the prime factorization of $244$ is $61 \times 2^2$. Therefore, we see that $(z+y)(z-y)=61 \times 2^2.$ We get that we have multiple possibilities. We see that $z+y$ cannot equal $61$, since then $z<42$. We try $z+y=122$. We see that $z-y=2$. Therefore, $z=62$ and $y=60$ This works. We see that $z+y$ cannot equal 244 for parity. Therefore, we get that $z=62$ or $x+42=62$, so $x=20$. We get that the answer is $20+60=\boxed{080}$.

[quote=OlympusHero]oops I am not completing the problems in the packet in the correct order :/ well they start off easy so am getting teensy bit bored.

[hide=Solution]Rewrite as $(x+42)^2+244=y^2$, or $y^2-(x+42)^2=244 \implies (y+x+42)(y-x-42)=244$. The sum of $y+x+42$ and $y-x-42$ is $2y$, and the product is even, so each of $y+x+42$ and $y-x-42$ is even. This means $y+x+42=122, y-x-42=2$, so $2y=124 \implies y = 62$ and $x=18$. Adding gives $\boxed{80}$.[/hide][/quote]

Dude just do all the past aime 6-15s lol so you don't get bored

thankfully, I can get these done quickly, so it'll only take like a day or less to get there plus I don't really want to miss any problems I can learn from, who knows maybe one will be confusing. but yeah the later ones will be more helpful.

@above,thank you so much for being more generous!!

[hide=Solution]So $\sqrt{x^2+84x+2008}$ must be a positive integer. If $y\le{x+42}$, then $y^2\le{x^2+84x+1764}<x^2+84x+2008$. If $y\ge{x+45}$, then $y^2\ge{x^2+90x+2025}>x^2+84x+2008$ Therefore, $y=x+43$ or $x+44$. Note that $x$ and $y$ are of the same parity so $y=x+44$.

The equation now is $x^2+84x+2008=x^2+88x+1936 \implies 4x=72 \implies x=18$. So $x+y=2x+44=\boxed{080}$.  [/hide]

[hide=Difference of Squares]$x^2+84x+1764+244=y^2$
$y^2-(x+42)^2=244$
$(y-x-42)(y+x+42)=244$
We have to factorize 244 into two divisors of same parity (here, even) and the only way to do that is $244=2*122$.
This gives $y=\frac{124}{2}=62, x=18$
So $x+y=62+18=80$.
So the answer is $080$.[/hide]
Hope it helps! :)

My solution is also just similar to one of the quadratic solution i have factored my equation into x^2+84x+2008=k^2
As (k+2y)(k-2y)=-976
I thought that there would be many factors so I checked for gcd of the two products and got it as 4y but if you see that if k is odd then the gcd of the two products is one and if it's even then theor gcd is 2
So I have made two cases
Case 1 when gcd is even
(K+2y)(k-2y)=-976 can only be factored in two ways
As 61. -2^4
976.      -1
But if you see both ways do not work for y

Second case when gcd is two
Then it has three cases
8.61.  -2
4.61.  -4
2.61.   -8
Only value that's it's satisfying is the middle one getting y as 62 then putting it back in eq 1 and getting x=18

2017 AIME || P6. Basically, the same thing, but with different numbers.

[b]Solution:[/b] This just screams complete the square. So, doing that, we get $(x+42)^2 + 244 = y^2$. (Let $z=x+42$ for ease of solution writing.) Basically, you add $244$ to one square to get another square. After trial and error, it is easy to see that $y$ is $z+2$ and so then clearly $2z + 1 + 2z + 3 = 4z + 4 = 244 \Rightarrow z=60 \Rightarrow x=18, y=62$. So our answer is $\boxed{080}$.


Hello Everyone!

Join the Butterscotch Tourney! Sign-ups are open until July 31st, so everyone will have a chance to finish up all those other tourneys(this tourney starts in August). There will be 16 people in this countdown tourney, and I will make an effort to pair people up with opponents of similar levels.

[b]Please state your highest rating, your lowest rating, and your current rating at this moment.[/b]

I wanna join! highest:1357 lowest:1136 current:1274 ya!

ill join.

Highest - 1541
Current - 1404
Lowest - 1200

Can i join? im sorry im not that good but im getting better!!!

Sure. But please post your highest rating, your lowest rating, and your current rating.

Highest: 1200
Lowest: 1155
Today: 0, but only cuz i didnt play any games yet.

[quote="(^_^)"]i'll join

highest: um...13something...
lowest: um...around...1050?

too lazy to check.[/quote]

Please get back to me on that.

[color=blue]* ernie is a lazy bum.[/color]

isabella do you know what day the tourney starts?

I wanna join.
Highest--1355
Current--1312
Lowest--1136

i mean like August 12, or august 10
or watever

[quote="isabella2296"]No one's playing anymore.  :([/quote]

Make a forceful deadline, or cancel it.  I don't see any other ways.

TOURNEY MANIA! look at my old tournament. before tourney mania, first round took 3 days for all but 2 games, while eveyrone else played 2nd and 3rd round games!

izzy are you gonna cancel this tournery too?

...i beat RG for nothing?

...Darn.

[quote="isabella2296"]I'm not going to cancel this Tournament.

It's going to rot though. Everyone's dropping out like flies.  :([/quote]

Once again, tourney mania!

I had to cancel my tourneys, maybe canceling yours would be the best idea.

Maybe. But I really am disappointed...but who was I to think I could finish a tourney?  :rotfl:

it's ok izzy
most tournies don't get finished
this is just another one of them :D

[quote="isabella2296"]It's never going to happen, I can see that now.[/quote]
Aw. Sorry for you Izzy!

awww izzy...
i would play if i had someone to play! sorry if this makes you feel rushed...

ugh not again...

i beat RG for really nothing...

I am very sorry but I have no choice. No one is playing their games except for you two.


vector
linear algebra
linear algebra unsolved

Let V be a finite dimensional vector space and let $A,B : V \to V$ be linear maps. Prove that $AB-BA \neq I_{V}$ unless dim V = 0.

We have $\text{trace}(AB)=\text{trace}(BA)$, and that is all :)

That's true as long as the characteristic of the field is not 2. There are counterexamples over the field with two elements.


inequalities
number theory proposed
number theory

Let $\sqrt{23}>\frac{m}{n}$ where $ m,n$ are positive integers.
i) Prove that $ \sqrt{23}>\frac{m}{n}\plus{}\frac{3}{mn}.$
ii) Prove that $ \sqrt{23}<\frac{m}{n}\plus{}\frac{4}{mn}$ occurs infinitely often, and give at least three such examples.

[i]Dan Schwarz[/i]

[quote="Maxim Bogdan"]Let $ \sqrt {23} > \frac {m}{n}$, with $ m,n$ positive integers.
i) Prove that $ \sqrt {23} > \frac {m}{n} \plus{} \frac {3}{mn}.$
ii) Prove that $ \sqrt {23} < \frac {m}{n} \plus{} \frac {4}{mn}$ occurs infinitely often, and do exhibit(at least) three such examples.

                                                                                [b]Dan Schwartz[/b][/quote]
a) Use quadric congruence , it is easy to check that :
$ (\frac{\minus{}k}{23})\equal{}\minus{}1,\forall 1\leq k \leq 6$
Thus the equation $ 23n^2\minus{}m^2\equal{}k$ has no solution for each $ 1\leq k\leq 6$ . Therefore $ 23n^2\minus{}m^2\geq 7$ ,then inequalities (a) follows from this .
b ) Use the Pell's equation : $ 23n^2\minus{}m^2\equal{}7$ , it has solution $ (x,y)\equal{}(1,4)$ , therefore it has infinite solution . Pairs root $ (n_k,m_k)$ of this equation satisfy (b)

Sorry for my mistake .But i think the equation $ 23n^2\minus{}m^2\equal{}5$ has no integer solution . Proof of this statement use some results of Pell's equation .
[i]
A lemma  :
We known that the equation $ x^2\minus{}dy^2\equal{}1$ always has solution in set of natural number  (we only consider case $ d$ is not a perfect square . Call $ (a,b)$ is the smallest root of this equation .
Consider the equation $ x^2\minus{}dy^2\equal{}n$ (2) .Let $ (x_0,y_0)$ is the smallest root of (2) . Then we have :
$ y_0^2\leq \max\{nb^2,\frac{\minus{}na^2}{d}\}$
 [/i]
Now consider our problem . The equation $ x^2\minus{}23y^2\equal{}1$ has smallest positive solution is $ (x,y)\equal{}(24,5)$
Apply lemma , if $ (m_0,n_0)$ is the smallest solution of equation $ m^2\minus{}23n^2\equal{}\minus{}5$ then we must have :
\[ n_0^2\leq \max\{\minus{}125,126\}
\]
Thus $ n_0\leq 10$ .Check $ n\equal{}1,..,10$  ,it is easy to see that  equation (2)  has no natural solution .
My proof continues as above .

Sorry Maxim Bogdan,but can you tell me what is "Stars of Mathematics Bucharest 2008"?.If it is a contest could you please post all of its problem?  :blush:

Problems posted.

http://www.mathlinks.ro/viewtopic.php?t=1373872

there is no need for that lemma to prove that $ \minus{}5$ doesn't work, just use modulo $ 4$  :D


$I \subset R$ an interval, $|I|$ its length.

$f(x),P_{1}(x) \in L^{2}[I]$, $e_{\alpha}(x)$ is a basis of $L^{2}[0,1]$. $P_{2}(x)=P_{1}(x)+\sum_{\alpha \in A}\frac{<f-P_{1},e_{\alpha}>}{|I|}e_{\alpha}(x)$

Do we have $<P_{2}-P_{1}, f-P_{2}> =0$?

(I am not sure the given conditions are enough...)

Notationally, I would include that division by $|I|$ as part of the definition of inner product. That is, I'm going to write $\langle f,g\rangle$ for what you wrote as $\frac{\langle f,g\rangle}{|I|}.$

The condition we need on $\{e_{\alpha}\}$ is that it is an orthonormal set. That is, we need that $\langle e_{\alpha},e_{\beta}\rangle$ equals zero for $\alpha\ne\beta$ and equals $1$ for $\alpha=\beta.$ The only thing I'm going to take from the concrete identification of the setting as $L^{2}(I)$ is that an orthonormal set must be at most countable. (Hence, I'm assuming we're working in a separable inner product space.) We do not need to assume that $\{e_{\alpha}\}$ is a [i]maximal[/i] orthonormal set - that is, we don't need to say it's a basis. But we do need orthogonality. As specifically stated, we need the set to be orthonormal, but if it were orthogonal but not normalized, we could  restate the theorem to compensate. I'm also going to assume that you meant a real inner product space, as I don't feel like taking the necessary care to deal with complex inner products.

Then what you want is just a calculation. Plug it all in - no tricks.

$P_{2}-P_{1}=\sum_{\alpha\in A}\langle f-P_{1},e_{\alpha}\rangle\,e_{\alpha}$

$f-P_{2}=(f-P_{1})-\sum_{\alpha\in A}\langle f-P_{1},e_{\alpha}\rangle\,e_{\alpha}$

$\langle P_{2}-P_{1},f-P_{2}\rangle= \sum_{\alpha}(\langle f-P_{1},e_\alpha\rangle)^{2}-\sum_{\alpha}\sum_{\beta}\langle f-P_{1},e_{\alpha}\rangle\langle f-P_{1},e_{\beta}\rangle\langle e_\alpha,e_{\beta}\rangle$

$=\sum_{\alpha}(\langle f-P_{1},e_\alpha\rangle)^{2}-\sum_{\alpha}(\langle f-P_{1},e_\alpha\rangle)^{2}$ (Here we used the orthonormality.)

$=0.$


Let $ a,b,c \geq 0$ such that $ a^3\plus{}b^2\plus{}c\equal{}1$.
Find Max of $ P\equal{} \frac{a^2\plus{}1}{b^2\plus{}1}\plus{} \frac{b^2\plus{}1}{c^2\plus{}1}\plus{} \frac{c^2\plus{}1}{a^2\plus{}1}$

[quote="nguoivn"]It's only an old problem:  :wink:
Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \geq 1$. Prove that:
$ \frac {a^{2} \plus{} 1}{b^{2} \plus{} 1} \plus{} \frac {b^{2} \plus{} 1}{c^{2} \plus{} 1} \plus{} \frac {c^{2} \plus{} 1}{a^{2} \plus{} 1} \leq\ \frac {7}{2}$
And how did you solved this ineq ?[/quote]

But it is $ a^3\plus{}b^2\plus{}c\equal{}1$.

[quote="nguoivn"]From $ a^3 \plus{} b^2 \plus{} c \equal{} 1$, easily to see that $ a \plus{} b \plus{} c \geq\ 1$, my friend   :)
(because $ a^3 \leq a$ and $ b^2 \leq\ b$)[/quote]

When is the equality?

Equality occurs when (a, b, c) = (0, 0, 1).
@Stephen: Sorry but I think you should try to think before ask  :)

[quote="nguoivn"]It's only an old problem:  :wink:
Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \geq 1$. Prove that:
$ \frac {a^{2} \plus{} 1}{b^{2} \plus{} 1} \plus{} \frac {b^{2} \plus{} 1}{c^{2} \plus{} 1} \plus{} \frac {c^{2} \plus{} 1}{a^{2} \plus{} 1} \leq\ \frac {7}{2}$
And how did you solved this ineq ?[/quote]
Yes,
But I can prove this ineq with $ a,b,c \in[0;1]$ :lol:

TQuat: cho $ a,b,c \in[0;1]$
Prove that $ \sum_{cyc} \frac{a^k\plus{}1}{b^k\plus{}1} \leq 7/2  (k \in R)$


Let $ p$ be a prime number. Solve in $ \mathbb{N}_0\times\mathbb{N}_0$ the equation $ x^3\plus{}y^3\minus{}3xy\equal{}p\minus{}1$.

$ x^3 + y^3 - 3xy + 1 = (x + y + 1)(x^2 + y^2 + 1 - x - y - xy) = p$
$ x = y = 1$ doesn't give any solution so $ x + y > 0$. hence $ x^2 + y^2 + 1 - x - y - xy = 1$
if $ x$ and $ y$ are bigger than 1; ${ \frac {x^2 + y^2}{2} > = x + y}$ and $ \frac {x^2 + y^2}{2} > = xy$ so $ x^2 + y^2 > = xy + x + y$ and in the equality case $ x = y = 2$ gives $ p = 5$
if $ x$ or $ y$ is $ 1$ say $ x = 1$. $ y^2 - 2y = 0$ so $ y = 2$ or $ y = 0$. checking yields only $ y = 0$ is a solution.
so the solutions are $ (2,2,5), (0,1,2), (1,0,2)$

this reminds me of the problem:
for which $ n$ does the equation $ n\equal{}x^3\plus{}y^3\plus{}z^3\minus{}3xyz$ have a solution. :)

And what is the solution of  problem that Albanian Eagle wrote ?  Is it well-known ?  :maybe:


Euler
number theory unsolved
number theory

let $ x$ be an integer and $ p$ is prime number , prove that $ p^2$ divides $ x^{p^2}\minus{}x^p$.

a) $ p|x^p\minus{}x$ by Fermat.
b) $ p^s | x\minus{}y$ implies $ p^{s\plus{}1} | x^p\minus{}y^p$.
(Both well known and posted before)

It is quite easy :
${ p^2|x^p(x^{p(p-1)}-1}$
If $ p|x$ then $ p^2|x^p$
If $ \gcd(x,p)=1$ then $ p^2|x^{p(p-1)}-1$ (From Euler theory).


calculus
integration
trigonometry
limit
function
complex numbers
complex analysis

1. the integral from 0 to infinity of (sin x)^2/(x^2+1) dx
2. the integral from 0 to 1 of (sin x)^-3/2 dx
3. the integral from 0 to pi of (tan x)^2 dx

Is the question to evaluate these or merely whether they converge?

While evaluating the first is possible, it's not a calculus-level exercise.

the question is to determine whether this converges or not and if it converges give the bound or the value. how would you calculate the first one??

2 and 3 diverge by limit comparison to $x^{-3/2}$ and $(x-\pi/2)^{-2}$, respectively.

1 can be easily estimated from above: (see the preceding entry). It's not so easy to estimate from below, unless you are willing to accept $0$ as a lower bound.

I still don't get how I can do number 2 with comparison to x^-3/2

Exact values for #1:
[hide="Advanced material"]$\frac{\sin^{2}x}{x^{2}+1}=\frac12\cdot\frac1{x^{2}+1}-\frac12\frac{\cos 2x}{x^{2}+1}$. The integral of the first term is $\frac{\pi}{4}$. For the second term, note that $\cos 2x$ is the real part of $e^{2ix}$. We apply complex analysis and look at the integral of $\frac{e^{2iz}}{z^{2}+1}$ around a contour that covers the real axis from $-R$ to $R$, then completes the loop with a semicircle in the upper half-plane. This contour integral can be evaluated by the residue theorem; the function we are integrating is analytic except for a simple pole at $i$ with residue $\frac1{2i}\cdot e^{2i\cdot i}=\frac1{2ie^{2}}$. The integral is then $2\pi i\cdot\frac1{2ie^{2}}=\frac{\pi}{e^{2}}$. The semicircle's contribution goes to zero, and $\int_{-\infty}^{\infty}\frac{e^{2ix}}{x^{2}+1}\,dx=\frac{\pi}{e^{2}}$. Cutting it in half (we only want the integral on the positive side) and taking the real part, $\int_{0}^{\infty}\frac{\cos 2x}{x^{2}+1}\,dx=\frac{\pi}{2e^{2}}$. Combining the pieces, $\int_{0}^{\infty}\frac{\sin^{2}x}{x^{2}+1}=\frac{\pi}{4}-\frac{\pi}{4e^{2}}=\frac{\pi}{4}(1-e^{-2})$.[/hide]

[quote="EquinoX"][quote="mlok"]$\lim_{x\to 0}\frac{\sin x}{x}=1$, hence $\lim_{x\to 0}\frac{(\sin x)^{-3/2}}{x^{-3/2}}=1$[/quote]

can you explain this a bit further because I don't know what you mean here[/quote]
Let $f(x)$ be defined on $I-\{c\}$ where $I$ is an open interval containing $c$ and $g(x)$ be defined on$J-\{L\}$ where $J$ is an open interval containing $L$.

If $\lim_{x\to c}f(x) = L$ (where $L$ is finite) and $g(x)$ is continous at $L$ then $\lim_{x\to c}g(f(x))=g(L)$.

What does that mean?

The function $g(x)=x^{-3/2}$ is continous at $L=1$ where $L=\lim_{x\to 0}\frac{\sin x}{x}= 1$ so it must mean by above theorem:

$\lim_{x\to 1}g(f(x)) = \lim_{x\to 1}= \frac{(\sin x )^{-3/2}}{x^{-3/2}}=g(1)=1$

mlok means that $(\sin x)^{-3/2}$ is the "same size" as $x^{-3/2}$ as $x\to0.$ Because that limit is $1,$ we can find a positive $a<1$ (something like $a=\frac12$ would be suitable) and a $b$ such that for $0<x<b,\ (\sin x)^{-3/2}\ge ax^{-3/2}.$

Then $\int_{0}^{1}(\sin x)^{-3/2}\,dx\ge a\int_{0}^{b}x^{-3/2}\,dx+\int_{b}^{1}(\sin x)^{-3/2}\,dx.$

The second integral there, the integral on $[b,1],$ is just the integral of a continuous function on a bounded interval - it's finite, hence harmless. By direct calculation (and it's an example that you should be familiar enough with that you don't even need to see the calculation),

$\int_{0}^{b}x^{-3/2}\,dx=\infty.$

Those $\frac23$ in Kent's post should all be $-\frac32$. [color=green][Well, I was thinking 3/2 all along.][/color]

is there any way to solve this using the integral test or bound from above??

[quote="mlok"]EquinoX: Yes, you can also use the "non-limit" comparison test for #2: $(\sin x)^{-3/2}\ge x^{-3/2}$ for $x>0$, because $\sin x\le x$ for all such $x$.

[quote="kunny"]For 3, the value is $-\pi$.[/quote]
You mean $\int_{0}^{\pi}\tan^{2}x\,dx=-\pi$? :what?:[/quote]

Right!

The source of mlok's skepticism: how can a positive function have a negative integral?

[hide="The actual value of the integral."]$\int_{0}^{\pi}\tan^{2}x\,dx=+\infty.$

[hide="More."]Break it into $\int_{0}^{\frac{\pi}2}\tan^{2}x\,dx+\int_{\frac{\pi}2}^{\pi}\tan^{2}x\,dx$[/hide] [/hide]

For number 3): Let $tan^{2}(x) = 1-sec^{2}(x) = \frac{1-cos2x}{1+cos2x}$.
Then $\int_{0}^{\pi}tan^{2}(x)dx = \int_{0}^{\pi}\frac{1-cos2x}{1+cos2x}dx = \frac{1}{2}\int_{0}^{2\pi}\frac{1-cos\theta}{1+cos\theta}d\theta$ $(1)$

Now let $cos\theta = \frac{z^{2}+1}{2z}$ and $d\theta = \frac{dz}{iz}$
Then $(1)$ becomes $\frac{1}{-2i}\int_{C}\frac{(z-1)^{2}}{z(z+1)^{2}}dz$ $(2)$
Observe that if $|z| = 1$, then $z = 0$ is the only zero of $f(z) = \frac{(z-1)^{2}}{z(z+1)^{2}}$ which lies in the interior of $|z| = 1$ and it is also a simple pole. Thus by Residue Theorem, $(2)$ becomes $\frac{1}{-2i}*(2\pi i) =-\pi$

Edited: It may be that the other duplicated zero namely z = -1 which lies on the contour |z|= 1 makes it useless to apply Residue Thm. But perhaps taking a different contour one may avoid this problem. Does anyone know of a contour that works. Sorry, this is becoming a complex analysis problem in a different forum, but all math is related;-

One of the hypotheses of the Residue Theorem is that the curve on which you are integrating should not itself pass through any singularities of the function. Note that you are integrating on the unit circle $|z|=1$ and that a pole of order 2 lies on that curve at $z=-1.$ Your answer does not even make sense as a principal value.

A positive function should not have a negative integral.

[quote="Fundamental Theorem of Calculus"]If $f$ is continuous on the closed interval $[a, b]$ and $F$ is an antiderivative of $f$ then $\int_{a}^{b}f(x)dx = F(b)-F(a)$.[/quote]

$\sec^{2}$ is not continuous on $[0, \pi]$.

The moral of the story is, "Know what the theorems you're using actually say."

(The alternative moral of the story, which Kent has already covered, is, "If you integrate a positive function and get a negative answer, you've done something wrong.")

so then how would I solve no.? with what method?


find the value of the infinite product $\frac79\times\frac{26}{28}\times\frac{63}{65}\times\cdots\times\frac{k^3-1}{k^3+1}\times\cdots$.

Or for the pros: $\prod^\infty_{k=2}\frac{k^3-1}{k^3+1}$

;)

After factoring the top and bottom, we get:

(k-1)(2k+1+irad3)(2k+1-irad3) / (k+1)(2k-1+irad3)(2k-1-irad3), which simplifies to (k-1)/(k+1)*((2k+1)^2+3)/((2k-1)^2+3). The product of this from 2 to n as n tends to infinity clearly telescopes (because it is made of two telescoping products, the (k-1)/(k+1) and the ((2k+1)^2+3)/((2k-1)^2+3) part. Applying telescoping, the product from 2 to n of (k-1)/(k+1) is 2/(n*(n+1)), and the product from 2 to n of ((2k+1)^2+3)/((2k-1)^2+3) is ((2n+1)^2+3)/12. Multiplying the two together and taking the limit as n tends to infinity, we get 2/(n*(n+1))*((2n+1)^2+3)/12 =

((2n+1)^2+3)/(6*n*(n+1)) = (4n^2+4n+4)/(6n^2+6n) --> [b]2/3[/b].

I hope this is what you were looking for  :cool: .

eum... the solution is correct so I guess you'll be right... but can you latex it next time cuz it's really messy to read ;)

Performing the public service of putting JGeneson's argument into TeX, with a few edits and rearrangements:

$\frac{k^3-1}{k^3+1} = \frac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)} = \frac{k-1}{k+1}\cdot\frac{(k+\frac12)^2+\frac34}{(k-\frac12)^2+\frac34}$.

This splits the product $\prod_{k=2}^n\frac{k^3-1}{k^3+1}$ into two products, each of which telescopes.

$\prod_{k=2}^n\frac{k-1}{k+1} = \frac13\cdot\frac24\cdot\frac35\cdots\frac{n-2}n\cdot\frac{n-1}{n+1} = \frac2{n(n+1)}$

$\prod_{k=2}^n\frac{(k+\frac12)^2+\frac34}{(k-\frac12)^2+\frac34} = \frac{(n+\frac12)^2+\frac34}{(2-\frac12)^2+\frac34} = \frac{n^2+n+1}3$

So, $\prod_{k=2}^n\frac{k^3-1}{k^3+1} = \frac23\cdot\frac{n^2+n+1}{n(n+1)} \to \frac23$ as $n\to\infty$.

JGeneson went ahead and factored each cubic into its three complex factors, then recombined the conjugate pairs into (real) irreducible quadratics. In some other problem the complex factorization will be important, but if the destination is that real quadratic, the technique of completing the square suffices.

Sorry about not TeXing it. Thanks for making it nicer Kent Merryfield. I will learn TeX so my solutions don't look so nasty  :) .

[quote="JGeneson"]I will learn TeX so my solutions don't look so nasty[/quote]
One thing you might try, which works some of the time (I won't promise all of the time) is to find something posted here that looks like what you want to do, copy it (along with some text on either side so you don't miss anything), then paste it into a text editor so you can see the codes. With some trial and error, you can figure out what you need to change to do what you want. Use the "Preview" button freely and often with your own work. Always use "Preview" just before you hit "Submit."


Virginia Tech Regional Mathematics Competition

http://www.math.vt.edu/people/plinnell/Vtregional/exams.html

We can see the problems in PDF [url=http://www.math.vt.edu/people/plinnell/Vtregional/exams.pdf]here[/url],
and the [url=http://www.math.vt.edu/people/plinnell/Vtregional/solutions.pdf]Answers[/url].


Prove the inequality

$ \sqrt {2 \plus{} \sqrt [3]{3 \plus{} ... \plus{} \sqrt [{2008}]{2008}}} < 2$


Umm neither of these are even close. Hershey and Cadbury are both just cheap generic chocolate

Lol. Well, I like cheap chocolate. Can't have the rich stuff everyday.

There is nothing better than Godiva chocolate (that I know of)!!!!!

For cheaper stuff, I like linderballs a lot. 35 cents, and they are delicious!

I like Ghirardelli chocolate, especially the 60% cocoa kind.

[quote="nutz_for2.718281828"]dark. oh wait, what BRAND! ok. Godiva.[/quote]

That stuff is amazing, but it's a bit expensive.  I like that "chocolates of the world" stuff, it looks rich and is pretty good.

Valrhona and Callebaut are regarded to be the best, and I've tried Valrhona, it's reallyyyy good.  :D

[quote="fredbel6"]Belgian and Swiss chocolate.  Problem is that I've only had Belgian chocolate all my life... so I can't stand any other chocolate.[/quote]

what's wrong with belgian chocolate?! :first:  and, if you have noticed, on almost all swiss and belgian chocolates is written something in french!? :no:

[quote="mdk"]European white chocolate.[/quote]

nice.  :thumbup:

I only like Godiva. That's the only real chocolate in my area.

yeah. godiva is my favorite too. I don't like cheap chocolate like the ones I get at halloween.

Ghirardelli is amazing!

as a matter of fact, i just ate a Ghirardelli caramel square (amazing!) 20 minutes ago

...just... amazing

look at their site to see how good it is
-jorian

I like Godiva and others, but Ghirardelli is my favorite.

I have to agree with fredbel6, swiss chocolate, especially tolberone, is the best. Hershey's is good, i am not saying it isn't, but swiss is much better.  :lol:    [/hide]

tolbelrone IS amazing, but thats more of a candy bar
-jorian :P


Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?

$ \textbf{(A)}\ \left(\frac{1}{12}\right)^{12}\qquad
\textbf{(B)}\ \left(\frac{1}{6}\right)^{12}\qquad
\textbf{(C)}\ 2\left(\frac{1}{6}\right)^{11}\qquad
\textbf{(D)}\ \frac{5}{2}\left(\frac{1}{6}\right)^{11}\qquad
\textbf{(E)}\ \left(\frac{1}{6}\right)^{10}$

[hide="Click for solution"]
Exactly one die must have a prime face on top, and the other eleven must have $ 1$'s.  The prime die can be any one of the twelve, and the prime can be $ 2$, $ 3$, or $ 5$.  Thus the probability of a prime face on any one die is $ \frac{1}{2}$ and the probability of a prime product is:

\[ 12 \left (\frac{1}{2} \right ) \left (\frac{1}{6} \right )^{11}\equal{}\left (\frac{1}{6} \right )^{10}
\]

or $ \boxed{\textbf{(E)}}$.
[/hide]

[hide="What's Wrong with This Solution?"]
2 3 5 are prime. There are 3 cases - roll a 2 and 11 1s, roll a 3 and 11 1s, roll a 5 and 11 1s. Probability of each is $\left( \frac{1}{6} \right)^{12}$ and multiply by $3$ to get C...[/hide]

[hide="What's Wrong with That Solution"]
You forgot to choose which die rolls the prime number. This makes the answer $12$ times as large, which gives $\textbf{(E)}.$
[/hide]

[hide="Another reason"]If jou multiply $ \left(\frac{1}{6}\right)^{12} $ by $3$, you get $\frac{1}{2} \left (\frac{1}{6} \right) ^{11} $ not $2 \left(\frac{1}{6}\right)^{11}$.[/hide]


How many strictly positive integers have all their digits distinct?

[hide]
One digit numbers: $ 9$
Two digit numbers $ 9*9$
Three digits: $ 9*9*8$
Four digits: $ 9*9*8*7$
...
Ten digits: $ 9*9*8*7*6*5*4*3*2*1$

Sum them up and you have

$ 9\left[1+9(1+8(1+7(1+.... 2(1+1))))))))\right]$

I don't know the formula for sums of permutations >_< so I did it out by hand and got 8877690
[/hide]

There are 9 choices for the first digit (0 isn't allowed) and 9 for the second digit (0 is allowed).

the answer is right, but shouldn't there be 9*8 since all the numbers are different?

Just describing two digit numbers, the first digit must come from the set $ \{1,2,3,4,5,6,7,8,9\}$ while the second digit can come from the set $ \{0,1,2,3,4,5,6,7,8,9\}$ except for the number that was used in the first digit. This accounts for $ 9\cdot9$ possibilities.


Is there a combinatoric identity that can quickly help you calculate:

$ \binom{n}{r}\plus{}\binom{n\plus{}1}{r}\plus{}\binom{n\plus{}2}{r}...\plus{}\binom{n\plus{}x}{r}$?

it is equal to $ \binom{n\plus{}x\plus{}1}{r\plus{}1}\minus{}\binom{n}{r\plus{}1}$

Well, you can use the Hockey Stick Identity twice: \begin{align*}
\binom{n}{r}+\binom{n+1}{r}+\cdots +\binom{n+x}{r} &=
\left( \binom{r}{r}+\binom{r+1}{r}+\cdots +\binom{n+x}{r}\right) - \left( \binom{r}{r}+\binom{r+1}{r}+\cdots + \binom{n-1}{r} \right) \\
&= \binom{n+x+1}{r+1}-\binom{n}{r+1}
\end{align*}
The Hockey Stick Identity, if you don't know it, states that \[ \binom{r}{r}+\binom{r+1}{r}+\cdots + \binom{n}{r} = \binom{n+1}{r+1}\] You can try to prove this if you wish.
[hide="hint for proof"]
Induct on $ n$, with base case $ n=r$.
[/hide]
It appears I was beaten to the answer with no explanation.

But if you use $ n\equal{}r$, you will have $ \binom r r\equal{}\binom{r\plus{}1}{r\plus{}1}$. The hockey stick identity states:

\[ \binom n r\plus{}\binom{n\plus{}1}{r}\equal{}\binom{n\plus{}1}{r\plus{}1}\]

I believe you're confusing Hockey Stick with Pascal's Identity.


It seems like S={1, 2, 3,... 25} should be a counterexample since for any two a and b with a>b in S, a-b is also in S.

Gack - I'll fix it.  I meant that sum & difference are not among remaining 23 numbers.  Oops.

Do you mean the sum+difference?
I think Allison's counterexample still stands with the problem.

a = 24, b = 12 is an example of a pair that would work.

I see. b-a could be a. I didn't notice that.

Here is a proof I came up with:

[hide]

We will use contradiction.  Assume that for any a and b, a > b, a + b and/or a - b are/is equal to one of the other 23 numbers in the set.

Call our set {a_1, a_2, a_3, ... , a_25}, and WLOG let a_1 < a_2 < a_3 < ... < a_24 < a_25.  From our assumption, we have:

a_25 - a_1 = a_(m_1)

a_25 - a_2 = a_(m_2)

a_25 - a_3 = a_(m_3)

...

a_25 - a_24 = a_(m_24)

since a_25 is the largest element of the set (and thus a_25 + a_k could clearly not be in the set, so a_25 - a_k must be in the set).

We now note that:

a_(m_1) > a_(m_2) > a_(m_3) > ... > a_(m_24)      (Equation *)

since a_1 < a_2 < ... < a_24.  But this implies that:

m_1 = 24

m_2 = 23

...

m_24 = 1

since those are the only such values that would satisfy (*).  Hence, we have:

a_25 = a_1 + a_24             (Equation 1)

a_25 = a_2 + a_23             (Equation 2)

a_25 = a_3 + a_22             (Equation 3)

...                                         ...

a_25 = a_23 + a_2             (Equation 23)

a_25 = a_24 + a_1             (Equation 24)

and adding all of these equations together and rearranging yields:

12 * a_25 = a_1 + a_2 + a_3 + ... + a_24 + a_25     (Equation **)

Now, subtract equations (1) through (11) from (**) to get:

a_25 = a_12

But we stated at the beginning of the proof that a_25 > a_12, and we thus have the desired contradiction.  QED

[/hide]

[color=cyan]Everything is right up to equation **, but there's a typo in that equation (no a_25 on the left) and then the subtraction you did doesn't really work out to what you said it would.  You get a_25 = a_12 + a_13, which you already knew.  This looks really promising, though.[/color]

I think I found a way to finish the solution - here is a new, corrected version of my proof:

[hide]

We will use contradiction. Assume that for any a and b, a > b, a + b and/or a - b are/is equal to one of the other 23 numbers in the set.

Call our set {a_1, a_2, a_3, ... , a_25}, and WLOG let a_1 < a_2 < a_3 < ... < a_24 < a_25. From our assumption, we have:

a_25 - a_1 = a_(m_1)

a_25 - a_2 = a_(m_2)

a_25 - a_3 = a_(m_3)

...

a_25 - a_24 = a_(m_24)

since a_25 is the largest element of the set (and thus a_25 + a_k could clearly not be in the set, so a_25 - a_k must be in the set).

We now note that:

a_(m_1) > a_(m_2) > a_(m_3) > ... > a_(m_24) (Equation *)

since a_1 < a_2 < ... < a_24. But this implies that:

m_1 = 24

m_2 = 23

...

m_24 = 1

since those are the only such values that would satisfy (*). Hence, we have:

a_25 = a_1 + a_24 (Equation 1a)

a_25 = a_2 + a_23 (Equation 2a)

a_25 = a_3 + a_22 (Equation 3a)

... ...

a_25 = a_23 + a_2 (Equation 23a)

a_25 = a_24 + a_1 (Equation 24a)

By similar reasoning we have:

a_24 - a_2 = a_(n_1) (Equation 1b)

a_24 - a_3 = a_(n_2) (Equation 2b)

a_24 - a_4 = a_(n_3) (Equation 3b)

...

a_24 - a_23 = a_(n_22) (Equation 22b)

since a_24 + a_k cannot be another member of the set when k >= 2.  We note that:

23 >= a_(n_1) > a_(n_2) > ... > a_(n_21) > a_(n_22) >= 1

and combining this with Equations 1b and 22b, we see that n_1 = 23 and n_22 = 2.  Plugging n_1 = 23 into Equation 1b, we get:

a_24 = a_2 + a_23 = a_25 (the second equality is just Equation 2a)

But a_24 < a_25, and we thus have the desired contradiction.  QED

[/hide]

[color=cyan]Wait . . . why can't a_24 = a_1 + a_23?  I think, though, that if I'm right and you missed this case, the following happens:  you just keep forcing it smaller and smaller until you get that the set must be a_1, 2a_1, 3a_1, . . . 25a_1.  In other words, your argument, even if it doesn't force a contradiction as quickly as you want, will force us into the structure of Alison's example, which we know fails for 2 and 1.

Also, a request: if you post more, could you not do it in spoiler?  Thanks very much.[/color]

Ah, JBL, you keep catching my mistakes!  I wish I were a bit more attentive to details...  Yes, I agree with your argument and completion of proof - once it reduces to the case of a_1, 2*a_1, 3*a_1, ... , 25*a_1, we're done, since we can simply choose 24*a_1 and 12*a_1 as a and b.  This completion also gives a nice structure to the proof: we just keep reducing the equations until we get the above case, which then works out very nicely.

I won't bother to post yet another version of the proof...once your suggestion is added, it's pretty much done.  This was a really awesome problem - it was amazing how well everything worked out.

[color=cyan]Yeah, that was sweet.  I like the way you approached it -- a very clever attack.[/color]


Let ABC is a triange,(I) is a circle which is inscribed ABC.(I) meet AB at H, (I) meet AC at K.  M is the midpoint of BC. AM intersects HK at P.
Prove that BPC is an obtuse triangle. :)

[b]

['[color=darkblue][size=117][b]Nice problem ! Thank you.[/b][/size][/color]

[quote="TRAN THAI HUNG"] [color=darkred]Let $ ABC$ be a triangle with the incircle $ w \equal{} C(I,r)$ which touches $ \triangle ABC$ in $ Y\in CA$ , $ Z\in AB$ .

Denote the midpoint $ M$ of $ [BC]$ and $ P\in YZ\cap AM$ . Prove that $ m(\angle BPC)\ > \ 90^{\circ}$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ X\in w\cap BC$ , the orthocenter $ H$ of $ \triangle ABC$ , $ L\in AH\cap MI$ and $ D\in AH\cap BC$ .

I"ll use two well-known properties (we can show easily them !) : $ P\in IX$ and  $ AL \equal{} r$ . Therefore,

$ \frac {PX}{IX} \equal{} \frac {AD}{LD}\ \Longleftrightarrow\ PX \equal{} \frac {rh_a}{h_a \minus{} r} \equal{}$ $ \frac {r\cdot ah_a}{ah_a \minus{} ar} \equal{}$ $ \frac {r\cdot 2pr}{2pr \minus{} ar}$ $ \implies$ $ \boxed {\ PX \equal{} \frac {2S}{b \plus{} c}\ }$ (nice !). Otherwise

([u]without the second mentioned properties[/u]), $ \frac {PX}{AD} \equal{} \frac {MX}{MD} \equal{}$ $ \frac {\frac {|b \minus{} c|}{2}}{\frac {\left|b^2 \minus{} c^2\right|}{2a}} \equal{} \frac {a}{b \plus{} c}$ $ \implies$ $ PX \equal{} \frac {2S}{b \plus{} c}$ .

Thus, $ \boxed {\ m(\angle BPC)\ > \ 90^{\circ}\ \Longleftrightarrow\ IX^2\ < \ XB\cdot XC\ }\ \Longrightarrow\ \frac {4S^2}{(b \plus{} c)^2} < (p \minus{} b)(p \minus{} c)$ $ \Longleftrightarrow$

$ 4p(p \minus{} a) < (b \plus{} c)^2\ \Longleftrightarrow\ (b \plus{} c)^2 \minus{} a^2 < (b \plus{} c)^2\ \Longleftrightarrow\ 0 < a^2$ , what is truly.

[b][u]Remark[/u].[/b] $ \frac {PZ}{PY}\equal{}\frac {AZ}{AY}\cdot \frac {MB}{MC}\cdot \frac {AC}{AB}\implies\frac {PZ}{b}\equal{}\frac {PY}{c}\equal{}\frac {YZ}{b\plus{}c}$ . From another well-known relation

$ [ZBC]\cdot PY\plus{}[YBC]\cdot PZ\equal{}[PBC]\cdot YZ\implies [ZBC]\cdot c\plus{}[YBC]\cdot b\equal{}[PBC]\cdot (b\plus{}c)$ . Therefore,

 $ a(p\minus{}b)\sin B\cdot c\plus{}a(p\minus{}c)\sin C\cdot b\equal{}a\cdot PX\cdot (b\plus{}c)$ $ \implies$ $ PX\equal{}\frac {2S}{b\plus{}c}$ because $ h_a\equal{}b\sin C\equal{}c\sin B$ .[/color]

Thanks Virgil Nicula! :lol:
Your proof is right and very nice.My proof is entirely different(Do you want to see it?) but I still like your proof vey much!
Can you prove the same with P on YZ such that MP is perpendicular to ZY?  :) (without using the pro. before as a lemma)
P\s:thanks for your Enlish correction :)


[quote="pbornsztein"]At the upper left of the main index (and everywhere else) the link is 'IMO-Intl Math Olmypiad'.
I think 'Olympiad' should be better... :P

Pierre.[/quote]Thanks Pierre! :)

Indeed it would be better, at least for the search engines :D


calculus
integration
function
real analysis
real analysis unsolved

What's an example of a positive continuous function $ f(x,y)$ defined on the open unit square in $ \bf R^{2}$, which is integrable on $ (0,1)^{2}$ (w.r.t. the Lebesgue measure), but such that $ \int_{(0,1)}f_{a}d \lambda$ is infinite, for some $ a \in (0,1)$.

for example if $ f(x,y)=0$ for $ x \neq \frac{1}{2}$ and $ f(\frac{1}{2}, y) = \frac{1}{y}$  :)  but maybe you assums something stronger on the function $ f$ ?

Currently the integral of $ f$ over the open unit square is $ 0$, what if we desire it to be strictly positive?

Let $ f(x,y)=\frac1{\sqrt{(x-1/2)^{2}+y^{2}}}.$

This is continuous on the open square, and it is integrable. (Shift the singularity to the origin and then use polar coordinates.) But if we fix $ x=\frac12,$ then the integral on that vertical segment would be

$ \int_{0}^{1}\frac1y\,dy=\infty.$


trigonometry
function
calculus
integration
real analysis
real analysis solved

Prove that the set of those complex x on the unit circle such that $\sqrt[n]{|x^n-1|}$ goes to 1 is dense on the circle. Yesterday, Myth gave me a very nice number x that does not appear in the previous set.

Could you please post the counterexample? I think I didn't see it in the original thread :?

We can prove more, namely that the measure of those $x\in[0,2\pi)$ for which $e^{ix}$ lies in the complement of this set is zero.

Fix $m\in\mathbb N$, and put $t=1-\frac 1m$ (we may even assume $m$ is large enough for our purposes, and not be concerned with small $m$'s). If we show that the set of those $x\in I=[0,2\pi)$ which lie in infinitely many sets $A_n=\{x\in I||\sin(nx)|<t^n\}$ has measure zero, then we're done, because the total set whose negligibility we want to prove is the union of such sets as $n$ varies over the naturals.

Since the set $\{y\in [0,2n\pi)||\sin y|<t^n\}$ has measure around $nt^n$ (by this I mean that there is a universal positive constant $C$, the same for all $n$, s.t. the measure of the set mentioned above is smaller than $Cnt^n$), it means that $A_n$ has measure at most $Ct^n$. From here we get that the series $\sum_{n\ge 1}m(A_n)\ (*)$ converges.

Let's show that, in this situation, the set of those $x$ which lie in infinitely many $A_n$ has null measure. Let $\chi_n$ be the characteristic function of $A_n$, and let $f=\sum\chi_n$ (it can also take the value $\infty$; all the functions involved are clearly measurable). Since $(*)$ converges, the integral of $f$ over $[0,2\pi)$ is finite, and $x$ belongs to infinitely many $A_n$ iff $f(a)=\infty$. From here the conclusion follows.


Let's take Euler's Identity \[ e^{i\pi}+1=0  \] and let's play with it a little.
\[ \\e^{i\pi}=-1\\\\ e^{2i\pi}=1\\\\ 2i\pi=\ln{1}\\\\ i=\frac{\ln{1}}{2\pi}\\\\ i=0  \]
Obviously somewhere along the line I did something illegal. But all I did was add one and multiply by -1. What on earth is illegal about that? Is there some wierd reasons why the operations I performed here cannot be performed on $i$?

[quote="Hikaru79"]$2i\pi=\ln{1}$[/quote]

This line requires interpreting.  What it really means is $i (2\pi + 2 \pi k) = \ln 1$ for some $k$.  Remember that $e^{i \theta} = e^{i (\theta + 2 \pi k) }$ for any integer $k$.

Edit:  More generally, your mistake is in thinking that $\ln z$ returns only one number.  In fact this is not the case when $z$ is complex, the same as if you had tried to show $1 = -1$ by taking square roots.  :)

[quote="t0rajir0u"]Edit:  More generally, your mistake is in thinking that $\ln z$ returns only one number.  In fact this is not the case when $z$ is complex, the same as if you had tried to show $1 = -1$ by taking square roots.  :)[/quote]

Well, okay, before we even take $\ln$, can't we say that if $e^{2i\pi}=1$ then the exponent must be equal to $0$, and since $2\pi$ obviously isn't $0$, $i$ must be?

Or is the assumption that if $a^x=1\rightarrow x=0$ incorrect?

[quote="t0rajir0u"][quote="Hikaru79"]

Or is the assumption that if $a^x=1\rightarrow x=0$ incorrect?[/quote]

This assumption is indeed incorrect in complex numbers (for the same reason that your previous assumption about $\ln z$ was incorrect).  You can only say that $x = 2i \pi k$ for some integer $k$.[/quote]

I see; that clarifies it :) Thank you.

[quote="frengo"]
how can you explain this?[/quote]
$\sqrt{1}=+-1$ ;)

Why must you assume $\sqrt{n^2}$ is always $=+n$?

[quote="frengo"][quote="t0rajir0u"]
Edit:  More generally, your mistake is in thinking that $\ln z$ returns only one number.  In fact this is not the case when $z$ is complex, the same as if you had tried to show $1 = -1$ by taking square roots.  :)[/quote]

here is the case:

$i=i$
$\sqrt{-1}=\sqrt{-1}$

$\sqrt{\frac{1}{-1}}=\sqrt{\frac{-1}{1}}$

$\frac{\sqrt{1}}{\sqrt{-1}}=\frac{\sqrt{-1}}{\sqrt{1}}$

$\sqrt{1}\cdot\sqrt{1}=\sqrt{-1}\cdot\sqrt{-1}$
$1=-1$

but in this case we don't use $ln$ of any complex number!
how can you explain this?[/quote]

This is an entirely separate false proof, and your mistake here is in taking positive square root on one side and negative square root on the other.  :)

Edit:  I realize now my original comment may have been confusing.  When I mentioned $1 = -1$ was "the same," I meant the mistake was in assuming $\sqrt{x}$ only returned one value the same way that assuming $\ln x$ leads to the above result - but I didn't mean to imply that $\ln x$ had anything to do with the $1 = -1$ proof.  Sorry.

[quote="frengo"]
here is the case:

$i=i$
$\sqrt{-1}=\sqrt{-1}$
$\sqrt{\frac{1}{-1}}=\sqrt{\frac{-1}{1}}$
$\frac{\sqrt{1}}{\sqrt{-1}}=\frac{\sqrt{-1}}{\sqrt{1}}$
$\sqrt{1}\cdot\sqrt{1}=\sqrt{-1}\cdot\sqrt{-1}$
$1=-1$

but in this case we don't use $ln$ of any complex number!
how can you explain this?[/quote]

The identities $\sqrt{a}\sqrt{b}=\sqrt{ab}$ and $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$ only hold true for nonnegative $a$ and $b$.  The fourth line is incorrect.

Actually, Hikaru79, your assumption that 2*pi is not equal to 0 is incorrect.  Euler's Formula deals with radians; and in radians, 2pi=0.

[quote="DanK"]How did we get from $i=\frac{\ln1}{2\pi}$ to $i=0$ in the first place?

Since we already have from above the previous step that $\ln1=2i\pi$, that gives $\frac{2i\pi}{2\pi}$, so from that you get $i=i$...[/quote]I think he's saying that $\ln1=0$.

$\ln 1 = 2\pi i k \, \forall k \in \mathbb{N}$ :)

Properly speaking, if one were to actually restrict $\ln z$ to its principal values so that it would be a well defined function, the conclusion we would reach is that $\ln e^{2\pi i} = \ln e^{0} = 0$, from which we would get $0 = 0$.  :)

$\ln -1 = z$
$-1 = e^{z} = e^{ix}, x = \pi + 2\pi k, k \in \mathbb{N}$
$\ln -1 + \ln -1 = \pi + (\pi - 2\pi) = 0$

Same issue as before.  (Perhaps I should've mentioned that restricting $\ln z$ to a principal value makes some assumptions about the behavior of the function false.  :( )


geometry
circumcircle
angle bisector
geometry proposed

This is a nice problem:

Let $ABC$ be a triangle and let $II_a$ meet $BC$ at $X$, and the circumcircle of $ABC$ at $M$. Let $N$ be the midpoint of the arc $AM$ (any of them) and let $NI$ and $NI_a$ intersect the circumcircle of $ABC$ again in $S$, and $T$. Prove $S,T,X$ are colinear!

This is equivalent to the problem discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=6528 (note that the point M, being the point of intersection of the line $II_a$, i. e. of the angle bisector of the angle CAB, with the circumcircle of triangle ABC, must be the midpoint of the arc BC on this circumcircle not containing the point A).

  darij

In fact, they are the same problem (same source :D )


Show  that
[(a-b)^2]/(8a)<(a+b)/2 - sqrt(ab) < [(a-b)^2]/8b
for reals  a>b>0
25 th  swedish mo , 1985
thanks

we have 4a>( \sqrt a+ \sqrt b) 2 >4b.
Since (a+b)/2- \sqrt ab=( \sqrt a- \sqrt b) 2 /2, we have
(a-b) 2 /8a<(a-b) 2 /2( \sqrt a+ \sqrt b) 2 =(a+b)/2- \sqrt ab<(a-b) 2 /8b.
Q. E. D.


geometry
rectangle
geometric transformation
rotation
function
calculus
integration

A ball of radius 14 has a round hole of radius 8 drilled through its center. Find the volume of the resulting solid.

Looking at a cross section of the ball, we can see that the hole of radius 8 (forming a cylindrical core when drilled) will show up as a rectangle inscribed in a circle.

 [geogebra]d021f408c229e4a3b93c3ff044de1ff6c69e77c7[/geogebra]

The tops of the circle will be cut out with the cylinder, so we're just looking at the figure formed by the rotation of the left piece across the line x = 0.

First, we'll parametrize the circle. Sticking to cartesian coords, we'll want to find the upper-right intersection point.
Well, we know $ x^2 \plus{} y^2 \equal{} 14^2$ and $ x \equal{} 8$, so $ y \equal{} \sqrt{132}$.
The negative intersection on the right is $ x \equal{} 8, y \equal{} \minus{}\sqrt{132}$.

Now, we're going to want to be adding up washers here, because we're dealing with an object with a fat hole in the center.
So, we know that a washer has area $ A \equal{} \pi \cdot (R^2 \minus{} r^2)$ where R is the outer radius, and r is the inner radius.

In our case, r is constant (8) and R is a function of y.
Our integral is then
$ \pi \int_{\minus{}\sqrt{132}}^{\sqrt{132}} (\sqrt{196 \minus{} y^2} \minus{} 8^2) dy$
Which shouldn't be too tough to work out. I get
[hide="ans"]$ 4 \pi  \left(4 \sqrt{33}\plus{}49 \arcsin\left[\frac{\sqrt{33}}{7}\right]\right)$[/hide]


superior algebra
superior algebra unsolved

How many extensions to $ \mathbb{Q}(\sqrt[n]{2})$ does the archimedian valuation on $ \mathbb{Q}$ have?

[quote="ZetaX"]Only one as $ x^{n}-2$ is irreducible in $ \mathbb{Q}_{2}$ by Eisenstein's criterion.[/quote]

Why $ \mathbb Q_{2}$? For $ n=2$, there are two absolute values as should be well-known: $ \lvert a\pm b\sqrt2\rvert$. owk

Sorry I thought of the $ 2$-adic one. Well it's clear that the same applies if we take $ \mathbb{R}$ instead: we get $ 1+\frac{n}2$ if $ n$ is even and $ \frac{n+1}2$ if $ n$ is odd by the same argument (number of extending valuations = number of irreducible factors in the completion).


limit
real analysis
real analysis unsolved

Hello ladies and gents,

Hope this is the right place in the forum to post this problem... :)

Consider the following PDE:
\[ \frac {\partial u}{\partial t} \equal{} x(1 \minus{} x)(w \plus{} \frac {1}{2}\sigma^2(1 \minus{} 2x))\frac {\partial u}{\partial x} \plus{} \frac {1}{2}\sigma^2 x^2 (1 \minus{} x)^2 \frac {\partial^2 u}{\partial x^2}
\]
with $ t\geq 0$, $ x\in [0,1]$ and boundary conditions:
1. $ u(x,0) \equal{} x$
2. $ u(0,t) \equal{} 0$
3. $ u(1,t) \equal{} 1$
($ w$ and $ \sigma$ are both constants. Do whatever you want with them, just don't put either equal to 0)

The question is: does this PDE have a (unique?) solution?

I'm a bit rusty with my PDE's so any help would be very welcome :)

Cheers,

Durandal 1707

If you denote $ v(x)\equal{}u(a(x)$ where $ a$ is the solution of $ a'\equal{}a(1\minus{}a)$, the equation reduces to $ u_t\equal{}u_{xx}\plus{}gu_x$. Then you can use the standard tricks for solving the heat equation on the real line. The answer to your question is "yes, the solution does exist". Let me know if you wish to see more details :).

First off, thanks a lot, you've been great help :)

So, if we carry out your approach, we end up with the PDE:
\[ \frac {\partial v}{\partial t} \equal{} w \frac {\partial v}{\partial y} \plus{} \frac {1}{2}\sigma^2 \frac {\partial^2 v}{\partial y^2}
\]
with $ t\geq 0$, $ y\in \mathbb{R}$ and the boundary conditions:
1. $ v(y,0) \equal{} \frac {e^y}{1 \plus{} e^y}$
2. $ \lim_{y\to \minus{} \infty} v(y,t) \equal{} 0$ for all $ t$
3. $ \lim_{y\to \plus{} \infty} v(y,,t) \equal{} 1$ for all $ t$

For sure, this is a much nicer PDE but I'm still a bit unsure because of the inhomogeneous boundary conditions. For example, in the heat equation over a bounded interval you can easily get rid of them but what happens now that there is a drift term as well? I mean, there isn't even a steady-state solution satisfying 2, 3 :(

BTW, I guess that it has a (unique?) solution because of some general theorem but which one would that be? Some version of Cauchy-Kowalevski?

Sorry, I didn't wish to besiege you with questions like that, it's just that I don't really know this stuff  :oops:

Cheers and thanks again!  :)

Durandal 1707

Bump.

So, do you guys think it's solvable after all? :?:

Cheerio,

Durandal 1707


ratio
search
probability
expected value
probability and stats

You toss a fair coin and after each toss you can either stop and collect your winnings (equal to the ratio of heads to total flips) or toss again.

What is the expected value of this game?

It was discussed but not conclusively resolved [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1662613617&t=67621]here[/url].  Intuitively, it seems clear that we should stop the first time we have more heads than tails.  This gives an expected value of $ \sum_{n \equal{} 0}^\infty \frac {1}{2} \cdot \frac {C_n}{2^{2n}} \cdot \frac {n \plus{} 1}{2n \plus{} 1}$, and Mathematica tells me this evaluates to $ \frac {\pi}{4} \approx 0.785398\ldots$.

No! If you have just $ n\plus{}1$ heads after $ 2n\plus{}1$ tosses with large $ n$, you should still proceed. Intuitively, it is likely that you should stop when the excess of heads over tails is comparable with the square root of the number of tosses but I didn't attempt a formal computation for such a strategy. The simple recursion
\[ A(n,k)\equal{}\max\left\{\frac kn,\frac{A(n\plus{}1,k)\plus{}A(n\plus{}1,k\plus{}1)}2\right\}\]
and the obvious estimate $ A(n,k)\ge \max\{\tfrac kn,\tfrac12\}$ (stop immediately or continue until the law of large numbers brings you to $ \frac 12$) show that you can reach a value about $ 0.793$ (here $ A(n,k)$ is the expected outcome of the best strategy under the condition that you already made $ n$ tosses and got $ k$ heads).

I have some doubts as to whether the clean analytic solution of this problem exists at all but if it does, I would like to see it a lot.

[quote="fedja"]Intuitively, it is likely that you should stop when the excess of heads over tails is comparable with the square root of the number of tosses[/quote]
Okay, now that you've said this, I agree, but I think your intuition (at least in such a case as this) is much more sophisticated than mine ;)

Hmm... Isn't it sort of obvious? I mean at least the fact that you shouldn't stop if your excess of heads over tails is above $ 0$ but below $ \delta\sqrt n$ with very small $ \delta$. Indeed, the probability that within next $ n$ steps your excess will never be above $ 10\delta\sqrt n$ is close to $ 0$ and you never need to agree to a negative excess, so your payoff expectation if you continue for the next $ n$ steps and stop at the excess $ 10\delta\sqrt n$ if it occurs is $ 1/2$ plus $ (\approx 1)\cdot 10\delta\sqrt n/2n>\delta\sqrt n/n$. The converse statement that you do need to stop if your excess is $ C\sqrt n$ with big $ C$ is far less obvios and requires some series to converge to small values (so I haven't really checked it accurately) but it seems quite plausible too. After all, the normal distribution decays so fast ;).


modular arithmetic
combinatorics proposed
combinatorics

I found this problem here:

[url]http://www.cs.utexas.edu/users/EWD/ewd09xx/EWD904.PDF[/url]

The entire archive consisting of Dijkstra's documents can be found here:

[url]http://www.cs.utexas.edu/users/EWD/indexEWDnums.html[/url]

Given $16$ lamps in a circle and a button, whenever we press the button the following happens: if the lamps $i,i+1$ (the indices are taken $\pmod{16}$) were both on or both off, then the lamp $i$ will be on after the push of the button. If, on the other hand, $i,i+1$ had different states, then the lamp $i$ will be off after the push of the button. Show that after $16$ operations all the lamps will be on.

Nothing can be easier than this...
Let $a_i=0$ is the lamp $i$ is off and $a_i=1$ is the lamp $i$ is on. Then the described operation is $a_i\to 1+a_i+a_{i+1}$ modulo 2.
Therefore we have
$a_i\to 1+a_i+a_{i+1}$ after 1 operation
$a_i\to 1+a_i+a_{i+2}$ after 2 operations
$a_i\to 1+a_i+a_{i+4}$ after 4 operations
...
$a_i\to 1+a_i+a_{i+16}=1$ after 16 operations.


Let $\left\{x_{n}\right\}$ be a non-null Cauchy sequence.  Show that there exists $N\in\mathbb{N}$ and $c\in\mathbb{Q}^{+}$ such that $n>N\implies\left|x_{n}\right|>c$.

This question is very similar to common epsilon-delta  definition of convergence  of any sequence.

[quote="anik_andrew"]$x_{n}=\frac{1}{n}$?? Sorry, if i give a wrong interpretation of Cauchy sequence. (we call it "fundamental") sequence is Cauchy iff $\forall \varepsilon >0 \exists N \forall n,m >N: |x_{n}-x_{m}|<\varepsilon$ Right?[/quote]

That's right.  Anyway here's my idea: the seqence converges to some real number $r$, then because $\mathbb{Q}$ is dense in $\mathbb{R}$ we can choose a $c\in\mathbb{Q}^{+}$ such that $|x_{n}|<c$ for only finitely many $n$. (Not possitive this is right).

but what about $x_{n}=\frac{1}{n}$. the statement is not true for this sequence...  :?:

It's the definition: "null" here essentially means that the sequence converges to zero.

aaa.
I understand. Sorry for my english  :blush:  :blush:


real analysis
function
number theory
relatively prime

For how many ordered pairs of positive integers $ (m,n)$, $ m \cdot n$ divides $ 2008 \cdot 2009 \cdot 2010$ ?

$\textbf{(A)}\  2\cdot3^7\cdot 5 \qquad\textbf{(B)}\  2^5\cdot3\cdot 5 \qquad\textbf{(C)}\  2^5\cdot3^7\cdot 5 \qquad\textbf{(D)}\  2^3\cdot3^5\cdot 5^2 \qquad\textbf{(E)}\ \text{None}$

[hide="Remark"] This is a weighted sum over all divisors of some integer - better known as a [url=http://en.wikipedia.org/wiki/Dirichlet_convolution]Dirichlet convolution[/url]. [/hide]

Factoring gives:

$ 2^4\cdot3\cdot5\cdot7^2\cdot41\cdot67\cdot251$

So b) would be correct.

EDIT> Oh wait... Crap. Ignore this post.

Nope, that's the number of ordered pairs such that $ mn$ is [b]equal[/b] to $ 2008 \cdot 2009 \cdot 2010$.

[hide="Another remark"] Let $ a(n)$ be the answer if $ 2008 \cdot 2009 \cdot 2010$ is replaced by $ n$.  Show that $ a(n)$ is multiplicative. [/hide]

From what I can see, $ a(n)$ should the sum of the number of ways to factor each factor of $ n$ for each said factor; I don't see how that could simplify...

An integer $ u$ has some number of prime factors $ \{ a_1, a_2, a_3 ... a_v \}$ with exponents $ \{ b_1, b_2, b_3 ... b_v \}$, where v is the number of distinct prime factors of $ u$.

Define $ f(u)$ as the number of factors of $ u$, including $ 1$ and $ u$.

$ f(u) \equal{} \prod^{v}_{m \equal{} 1}(b_m \plus{} 1)$

which should imply that, say $ \{ c_1, c_2, c_3 ... c_w \}$ is the list of all $ w$ factors of n, $ c_1 \equal{} 1$, $ c_w \equal{} n$,

$ a(n) \equal{} \sum^{w}_{j \equal{} 1}f(c_w)$

I am SO doing this wrong... Would anyone be kind enough to give some pointers?

>.<

[quote="yanglunj"]$ a(n)$ should the sum of the number of ways to factor each factor of $ n$ for each said factor; I don't see how that could simplify...[/quote]
[hide="Another hint"] $ a(n)$ is the number of ordered triplets $ i, j, k$ such that $ ijk \equal{} n$.  (Compare with the definition of the divisor function.)  Why is this function multiplicative? [/hide]

Well, if they are multiplicative, then we see that $ a(mn) \equal{} a(m)a(n)$ because for each of the $ a(m)$ triplet $ \{i_1, j_1, k_1\}$ from $ a(m)$, you match it up with another $ a(n)$ triplets of ordered pairs $ \{i_2, j_2, k_2\}$ so that the resultant triplets ($ a(m)a(n)$ in total) $ \{i_1i_2, j_1j_2, k_1k_2\}$ will have product equal to $ mn$.

But I'm worried about mirror pairs of triplets like $ (\{1, 1, 2\}, \{1, 2, 1\})$ and $ (\{1, 2, 1\}, \{1, 1, 2\})$; their product is the same $ (\{1, 2, 2\})$, but is accounted for twice when you multiply $ a(m)$ and $ a(n)$. Does it magically cancel or something?

EDIT>
$ a(2) \equal{} 3$
$ a(2^2) \equal{} 6 \ne a^2(2)$

Sorry for the confusion; we call an arithmetic function, like the divisor function, multiplicative if $ a(mn) \equal{} a(m) a(n)$ for all [i]relatively prime[/i] $ m, n$.  The point is that evaluating $ a(n)$ can be done by evaluating it at prime powers, just as in the divisor case.

Oh, okay, I didn't remember that part (Sorry! Mind not working)

So for $ n \equal{} p^q, p \mbox{ prime}, q \in \mathbb{W}$, $ a(n) \equal{} \frac {(q \plus{} 1)(q \plus{} 2)}{2}$.

$ a(2008\cdot 2009 \cdot 2010) \equal{} 15 \cdot 3 \cdot 3 \cdot 6 \cdot 3 \cdot 3 \cdot 3 \equal{} 2 \cdot 3^7 \cdot 5 \rightarrow (a)$

Oh wow, that makes sense. Thank you very much! :)

I will use combination with repetition.

$ mnp \equal{} 2^{4}\cdot3\cdot5\cdot7^{2}\cdot41\cdot67\cdot251$

In how many ways 4 oranges, 2 apples, 1 banana, 1 cherry, 1 mellon, 1 tomato, 1 potato can be distributed among 3 children?

$ {{4\plus{}3\minus{}1}\choose{3\minus{}1}}{{2\plus{}3\minus{}1}\choose{3\minus{}1}}{{1\plus{}3\minus{}1}\choose{3\minus{}1}}^5 \equal{} {{6}\choose{2}}{{4}\choose{2}}{{3}\choose{2}}^5 \equal{} 15\cdot 6 \cdot 3^5 \equal{} 2\cdot 3^7 \cdot 5$


number theory
greatest common divisor
number theory proposed

obvious, we have $gcd(n, q) = gcd(n+1, q) = 1$.
by the fermat little theorem, we have $q| (n+1)^{q-1} - n^{q-1}$.
so, $q| gcd ((n+1)^p - n^p, (n+1)^{q-1} - n^{q-1}) = (n+1)^{gcd(p,q-1)} - n^{gcd(p,q-1)}$.
if $q-1$ isn´t divisible by $p, q|1$, contradiction.
so, $p|q-1$.

i prove that all prime divisors of $(n+1)^p - n^p$ are in the form $pk+1$, so, all divisors are in this form.  :)

[quote="Arne"]Let $n$ be a positive integer, let $p$ be prime and let $q$ be a divisor of $(n + 1)^p - n^p$. Show that $p$ divides $q - 1$.[/quote]
Obviously $q \nmid n$. So $(n+1)^{p} - n^p \equiv 0 \bmod q$ only if $(1 + n^{-1})^p \equiv 1 \bmod q$. Hence $p = \mbox{ord}_q(1 + n^{-1})$, and subsequently $p \mid (q-1)$, if $q \in \mathfrak{P}$. Moreover, if $q \not \in \mathfrak{P}$, however it has to be a product of primes $\equiv 1 \bmod p$, which implies $q \equiv 1 \bmod p$. The claim follows!

Simple problem, isn't it, but only very few teams solved it (Poland, St Petersburg and Germany, I believe).

It's very similar to the one hitlEULER gave. I must say that the solution by e.lopes is a bit more elegant :)


Show that the only solutions to $ y^{2}\plus{}4\equal{}x^{3}$ are $ \left(x,y\right)\equal{}\left(2,2\right)$ or $ \left(x,y\right)\equal{}\left(5,11\right)$.

It appears that this can be done similarly to post number 6 in
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=163342

More precisely, work in $ \mathbb{Z}[i]$ (the Gaussian integers) which are the ring of integers in $ \mathbb{Q}[i]$.

NOTE:  this problem is more difficult than the problem mentioned above, but the procedure is the same.
If you get stuck, you can go to
http://www.math.sc.edu/~filaseta
then go to his link on the left hand side for Algebraic Number Theory (MATH 784)
then go to Theorem 63 on page 49.
(sorry I cannot get a direct link since his page is down for the day)

Nice problem...have fun...


the sum of 1/2^n - 1/3^n when n goes from 0 to infinity

There is no common ratio
Nothing cancels out teloscopically

I'm having trouble finding an nth term expression for the sequence of its partial sums (which I would take the limit of to infinity to find the sum)

the $ k$-th partial sum is $ \sum_{n\equal{}1}^{k}\left( \frac{1}{2^n}\minus{}\frac{1}{3^n}\right) \equal{} \left( \sum_{n\equal{}1}^k \frac{1}{2^n}\right) \minus{} \left( \sum_{n\equal{}1}^k \frac{1}{3^n}\right)$, which is simply a difference of two geometric series.

oh yea didn't even realize that...heh that's kind of obvious
---thanks


modular arithmetic
number theory unsolved
number theory

Prove that there exist infinitely many pairs $ (a, b)$ of natural numbers not equal to $ 1$ such that $ b^b \plus{}a$ is divisible by $ a^a \plus{}b$.

I suppose that $ a \neq b$. Take an arbitrary even number $ a$ and $ b\equal{}a^{a\plus{}2} \minus{} a^{a} \plus{} a$. Then $ a^a \plus{} b \equal{} a^{a\plus{}2} \plus{} a$ and $ b^b \plus{} a \equiv (\minus{}a^a)^{b} \plus{} a \equiv a^{ab} \plus{} a \pmod{a^a\plus{}b}$ since $ b$ is even as $ a$ is even. So we have to prove that $ a^{a\plus{}1} \plus{} 1 | a^{ab\minus{}1} \plus{} 1$. Since $ a\plus{}1$ is odd it sufficies to show that $ a\plus{}1|ab\minus{}1$. But this is clear as $ ab \minus{} 1 \equal{} a^{a\plus{}3} \minus{} a^{a\plus{}1} \plus{} a^2 \minus{} 1 \equal{} (a\minus{}1)(a\plus{}1)(a^{a\plus{}1} \plus{} 1)$.

$ (a,b)\equal{}(a,a^{2a\minus{}2}\minus{}a^{a}\plus{}1)$ also work for $ a>1$ being of the form $ 4k\plus{}1$, where $ k$ is an integer. After easy calculations we have to prove that $ a^{2a\minus{}2}\plus{}1|a^{a(a^{2a\minus{}2}\minus{}a^{a}\plus{}1)\minus{}1}\minus{}1$, which will be proved if we show that $ 2(2a\minus{}2)|a(a^{2a\minus{}2}\minus{}a^{a}\plus{}1)\minus{}1$, but it's equivalent to $ 4(a\minus{}1)|(a\minus{}1)(1\plus{}a^{a\minus{}3}\plus{}a^{a\minus{}2}\plus{}...\plus{}a\plus{}1)$, or $ 4|1\plus{}a^{a\minus{}3}\plus{}a^{a\minus{}2}\plus{}...\plus{}a\plus{}1$, but this holds as $ a\equal{}4k\plus{}1$,from which the desired conclusion follows.


We have by AM-GM:
$1.5y^{2}+3x^{2}\ge 2xy\sqrt{4.5}\ge 4xy$
$1.5y^{2}+3z^{2}\ge 2zy\sqrt{4.5}\ge 4zy$
Adding these, we get the desired result.


calculus
derivative
trigonometry
integration
calculus computations

Can anyone here help me get the anti-derivative of 16 / ((4 + x^2)^2)?

I tried using a trig-substitution with x = 2 tan theta and dx = 2 sec^2 theta d theta, but I keep on getting the wrong anti-derivative when comparing the derivative values of 16 / ((4 + x^2)^2) with my answer - they do not match. Any help would be great, thanks.

The answer is $\tan ^{-1} \dfrac{x}{2}+\dfrac{4x}{x^2+4}+Const.$, right?

Hi, it should be 2x / (x^2 + 4) instead of 4x / (x^2 + 4). Thus,

tan^-1 (x/2) + 2x / (x^2 + 4).

Though, I have absolutely no idea on how to come about this answer. Any help would be great, thanks.

Yes, the answer is $\tan ^ {-1} \dfrac{x}{2}+\dfrac{2x}{x^2+4}+C$.

Here is solution.

Let $x=2\tan \theta$, we have $dx=\dfrac{2}{\cos ^ 2 \theta}$, thus

$\int \dfrac{16}{(4+x^2)^2}\ dx=\int \dfrac{16}{\dfrac{16}{\cos ^ 4\theta }}\cdot \dfrac{2}{\cos ^ 2 \theta}\ d\theta$

$=2\int \cos ^ 2 \theta\ d\theta$

$=2\int \dfrac{1+\cos 2\theta}{2}\ d\theta$

$=\theta +\dfrac{1}{2}\sin 2\theta +C$

$=\theta +\dfrac{1}{2}\cdot \dfrac{2\tan \theta}{1+\tan ^ 2 \theta}+C$

$=\tan ^ {-1} \dfrac{x}{2}+\dfrac{\dfrac{x}{2}}{1+\left(\dfrac{x}{2}\right)^2}+C$

$=\tan ^ {-1} \dfrac{x}{2}+\dfrac{2x}{x^2+4}+C$

Ohhh, that's how you should do it. Thanks a lot. Nice trick with the substitution.

Or you can try integration by part.
\begin{eqnarray*} \int \dfrac{dx}{(x^2+a^2)^2} &=& \dfrac{1}{a^2}\left(\int \dfrac{x^2+a^2}{(x^2+a^2)^2}dx - \int \dfrac{x^2}{(x^2+a^2)^2}dx\right) \\ &=& \frac 1{a^3}\arctan \frac xa + \dfrac{1}{2a^2}\int xd\left(\dfrac{1}{x^2+a^2}\right)\\ &=& \frac 1{a^3}\arctan \frac xa + \dfrac{1}{2a^2}\dfrac{x}{x^2+a^2} - \dfrac{1}{2a^2}\int \dfrac{dx}{x^2+a^2}\\ &=& \frac 1{2a^3}\arctan \frac xa + \dfrac{x}{2a^2(x^2+a^2)}+C \end{eqnarray*}


find all solution for this equation in $Z$IF $k\in N ,K>1$:
$y^{k}=x^{2}+x$

The only solutions are: $\ (x,y,k)=\{(0,0,k),(1,2,1),(-1,0,k),(-2,2,1)\}$

$\ y^{k}=x(x+1)$ because they are consecutive numbers, then one is odd and the other is even. Say $\ x$ is even and $\ x+1$ is odd.

So we have $\ y^{n}$ is even. Let $\ y=2m$

$\rightarrow$ $\ 2^{k}m^{k}=x(x+1)$ because $\ x+1$ is odd then $\ x=2^{k}m^{r}$ and $\ x+1=m^{k-r}$ with for some $\ r\in N$.

$\rightarrow$ $\ m^{k-r}-1=2^{k}m^{r}$

$\rightarrow$ $\ m^{k-2r}-\frac{1}{m^{r}}=2^{k}$ this equation have no solutions for $\ k>2$.

If $\ k>2r$ we have a whole number minus a fraction equals $\ 2^{k}$ (impossible)

If $\ k<2r$ we have $\ m^{k-r}<2^{k}m^{r}+1$ also impossible.

If $\ k=2r$ we have $\ 1-\frac{1}{m^{r}}=2^{k}$ no way josé!

[quote="tanlsth"]You give a fail solution[/quote]

Can you be more specific?
In my prove I assume $\ x,y$ positive, but the same reasoning can be apply it for the other cases.

[quote="parinaz.sa"][quote="El fisico"]The only solutions are: $\ (x,y,k)=\{(0,0,k),(1,2,1),(-1,0,k),(-2,2,1)\}$
but your solutions are wrong!!!!![/quote][/quote]

Are you kidding me??

no i'm not kiding because i said$k>1,k\neq1$and $(-2,2,1),(1,2,1)$are wrong

AAAAAh Sorry :oops: . But my reasoning still good. So the only solutions are $\ (0,0,k)$ and $\ (-1,0,k)$.

yeah the only solutions are $(0,0,k),(-1,0,k)$and here is my solution:
[hide]$y^{k}=x^{2}+x$and we know $gcd(x,x+1)=1$then both $x,x+1$are perfect power of $k$so we have only two pairs of nubers which are perfect power because $k>1$so our solution are $(0,0,k),(-1,0,k)$[/hide]


\[ \lim_{x\rightarrow1\minus{}0}\sum_{n\equal{}1}^{\infty}\frac{x^{n\minus{}1}}{nln(n\plus{}1)} \equal{} \sum_{n\equal{}1}^{\infty}\lim_{x\rightarrow1\minus{}0}\frac{x^{n\minus{}1}}{nln(n\plus{}1)}\]

[color=indigo]Is the above equation right? in the other words, can we take the limitation into the summation? [/color]

Yes, we can, by the monotone convergence theorem.


Evo, izvadio sam sa stranica matematičkih društava i sl. neke rasporede nacionalnih natjecanja za ovu godinu (pri tome mislim na "službena" natjecanja: na primjer, neke škole u velikim gradovima u Hrvatskoj su održale školsko natjecanje prije općinskog, ali ono nije bilo u isto vrijeme u svim školama niti su bili isti zadaci). Isto, uklopio sam samo natjecanja za srednju školu i to za "najjaču" kategoriju (tradicionalno se zove A kategorija :)).

[b]Bosna i Hercegovina[/b]
Kantonalna takmičenja
Federalno takmičenje (FBiH) [url=http://www.pmf.unsa.ba/matematika/Udruzenje%20matematicara/bilten%20federalno%202009.pdf]bilten[/url]
Republičko takmičenje (RS)
Državno takmičenje [url=http://www.pmf.unsa.ba/matematika/Udruzenje%20matematicara/I%20dan.pdf]zadaci[/url]

[b]Crna Gora[/b]
državno, zadaci i rješenja - http://www.iccg.cg.yu/

[b]Hrvatska[/b]
Školsko/općinsko/gradsko: 29.01. - [url=http://www.hmd-istra.org/2009/01/skolskogradsko-natjecanje-iz-matematike.html]zadaci i rješenja[/url]
Županijsko: 23.02. [url=http://www.hmd-istra.org/2009/02/zupanijsko-natjecanje-2009-os-porec.html]zadaci i rješenja[/url]
Državno (s izborom ekipe za IMO): 29.03.-01.04., Pula [url=http://www.matematika.hr/domaca]zadaci i rješenja (bez MO)[/url]

[b]Makedonija[/b]
regionalno 14.02.,
državno 14.03.,
Makedonska matematička olimpijada 11.04.

[b]Slovenija[/b]
Školsko ([i]Klokan[/i]): 19.03.
Izborno: 01.04.
Državno: 18.04.
Izbor ekipe za IMO: u više navrata, na više načina tijekom cijele godine - [url=http://www.dmfa.si/mat_SS-A/PripraveMMO.html]dosadašnji materijali[/url]

[b]Srbija[/b]
Općinsko: 31.01. - [url=http://www.dms.org.yu/index.php?action=competitions_mathematics_high_school]zadaci i rješenja[/url]
Okružno: 28.02. - [url=http://www.dms.org.yu/index.php?action=competitions_mathematics_high_school]zadaci i rješenja[/url]
Državno: 28.03. - [url=http://www.dms.org.yu/index.php?action=competitions_mathematics_high_school]zadaci i rješenja[/url]
Srpska matematička olimpijada: 13-14.04. - [url=http://www.dms.org.yu/index.php?action=competitions_mathematics_high_school]zadaci[/url]

Anyway, gornji podaci su nepotpuni i vjerojatno imaju nekih grešaka, pa moderatora (ako ima mogućnost editirati ovaj post) i ostale korisnike (u ime kojih će gorespomenuti moderator editiravati ovaj post  :D) molim da ih popune.

P.S. hsiljak, može ovo bit sticky? Tek tako da se osjećam važnim... :P

Piše da ne mogu editirati gornji post... pretpostavljam da je prošlo previše vremena. Uglavnom, evo zadataka i rješenja od današnjeg županijskog natjecanja u Hrvatskoj: [url=http://www.hmd-istra.org/2009/02/zupanijsko-natjecanje-2009-os-porec.html]zadaci i rješenja[/url].

Ubačeno u prvi post, tema postavljena kao sticky. Izvinjavam se na kašnjenju, ali sam nekim čudom uspio previdjeti ovaj topic u zadnjih mjesec dana  :(  Bad, bad moderator.

Jučer je bilo kantonalno takmičenje u ZDK ako se ne varam-možda i u ostalim kantonima. Kad budem imao neki news update, pišem  :)

Gotovo je državno u Hrvatskoj.

IMO ekipu čine (abecednim redom):

Ivo Božić (3.r, MEMO bronca prošle godine)
Borna Cicvarić (3.r, MEMO pohvala prošle godine)
Nina Kamčev (4.r, IMO pohvala prošle godine)
Adrian Satja Kurdija (3.r, IMO bronca prošle godine)
Matko Ljulj (2.r, MEMO bronca prošle godine)
Goran Žužić (4.r, IMO bronca prošle godine)

Svi, kao što vidite, imaju prošlogodišnje međunarodno iskustvo. Vidimo se :)

Zadaci s ovogodisnjeg Drzavnog natjecanja u Hrvatskoj (a i od proslih godina): http://www.matematika.hr/domaca.

BiH BMO tim:
Beširević Admir
Bevrnja Alija
Radović Jelena
Saković Zlatan
Uljarević Vlado
Vukojević Ivan
Džaba Tanoviću što se zove Stefan :lol:

[quote="hsiljak"]Istina, Melkior je ovo postavio u BMO thread, ali neka se nađe i ovdje:

Ekipa Makedonije:

Bodan Arsovski
Filip Talimdžioski
Predrag Gruevski
Bojan Joveski
Stefan Stojčevski
Stefan Lozanovski

Ekipa Srbije:

Teodor Von Burg
Luka Milićević
Dušan Milijančević
Mihajlo Cekić
Vukašin Stojisavljević
Stefan Stojanović.

Konstatacija broj jedan: u ekipi Srbije nema maturanata!
Konstatacija broj dva (kao što i Melkior primjeti): važno je zvati se Stefan :)[/quote]

Srbija salje drugu ekipu na BMO (ajde sad da ne kucam, ima na http://www.dms.org.yu).
Ni u drugoj ekipi nema maturanata, to je odredjeno pre takmicenja (za drugu ekipu samo) i ako se medju prvih 12 ne plasira nijeda domacin (iz Kragujevca), izbacuju 12-tog i ubacuju jednog Kragujevcana. Prilicno nefer, ali eto...

([url]http://www.dms.org.yu/bmo/index.php?action=participants[/url])

Srbija 2

Aleksandar Vasiljković
Igor Spasojević
Filip Živanović
Radomir Đoković
Stevan Gajović (Nije Stefan, nego Stevan, ali ajde... dovoljno blizu.  :lol:)
Vukan Levajac

Crna Gora

Nikola Milinković
Aleksandra Gogić
Rastko Pajković
Tanja Ivošević
Radovan Krtolica
Nevena Crnogorac

(BMO ekipe, naravno.  :ninja:)

Dio vezan za bh takmičenja editovan, ekipa slijedi:

Ademović Adnan
Beširević Admir
Radović Jelena
Uljarević Vlado
Darda Ratko
Ferizbegović Mina

Šta reći, 3 mathlinkera  :lol:

Moram priznati da ne znam tko je treći...  :blush:

Anyway, Harune, ako i samo ako ti se već da editiravati prvi post, onda bi možda bilo korisno unijeti ove linkove i informacije na pripadajuća mjesta:

SMO (zadaci) - http://www.dms.org.yu/index.php?action=competitions_mathematics_high_school
Crna Gora - državno, zadaci i rješenja - http://www.iccg.cg.yu/
Hrvatska, državno (zadaci i rješenja, bez zadataka s Male olimpijade) - http://www.matematika.hr/domaca
Makedonija (bez linkova, samo kalendar nažalost) -  regionalno 14.02., državno 14.03., Makedonska matematička olimpijada 11.04.

Editirano. Eh, pošto si vjerovatno prepoznao kolegicu i kolegu pod kodnim imenima britanskog rock banda iz Cambridgea i košarkaškog kluba iz Bileće, respektivno, vjerujem da se nisi sjetio osobe pod kodnim imenom RaleD.  :lol:


geometry
geometric transformation
rotation
geometry theorems

Is there any map that possibly alters lengths and angles but preserves the area of a given figure?

No tranlation or rotation, please. :D


Determine if there exist four polynomials such that the sum of any three of them has a real root while the sum of any two of them does not.

[hide="Answer"]No. There are no four such polynomials.
----------------------------------------------------------------------------------------------------[/hide]
[hide="Hint"]Use the two lemmas

$ \boxed{1}$ A polynomial has no real roots if and only if it has the same sign everywhere.

$ \boxed{2}$ For polynomials $ P,Q$, if $ P$ is always positive (or negative), $ Q$ is always negative (or positive), $ P + Q$ is always positive (or negative), then there exists $ n$ such that the modulus of the coefficient of $ x^n$ in $ P$ is greater than the modulus of the coefficient of $ x^n$ in $ Q$ and there is no $ m > n$ for which the modulus of the coefficient of $ x^m$ in $ Q$ is greater than the modulus of the coefficient of $ x^m$ in $ P$
----------------------------------------------------------------------------------------------------[/hide]
[hide="Proof"]In the following context a polynomial is said to be positive (or negative) if it is positive (or negative) everywhere.

Let the notation $ dc(P) > dc(Q)$ mean that
there exists $ n$ such that the modulus of the coefficient of $ x^n$ in $ P$ is greater than the modulus of the coefficient of $ x^n$ in $ Q$ and there is no $ m > n$ for which the modulus of the coefficient of $ x^m$ in $ Q$ is greater than the modulus of the coefficient of $ x^m$ in $ P$

Assume the contrary. That there exists such 4 polynomials.

Let the polynomials be $ P,Q,R,S$.

There are total $ 6$ 2-polynomial-sums: $ P + Q,P + R,P + S,Q + R,Q + S,R + S$
Each is either positive or negative.

Without loss of generality we need to consider the following cases

$ \boxed{Case\ I}$ All 2-polynomial-sums are positive
[hide="Proof of contradiction"]$ Q + R + S = \frac {(Q + R) + (R + S) + (S + Q)}{2}$ is positive, contradicting the existence of real root.
--------------------------------------------------[/hide]

$ \boxed{Case\ II}$ All 2-polynomial-sums but $ P + Q$ are positive
[hide="Proof of contradiction"]Same as Case I
--------------------------------------------------[/hide]

$ \boxed{Case\ III}$ All 2-polynomial-sums but $ P + Q,P + R$ are positive
[hide="Proof of contradiction"]Same as Case I
--------------------------------------------------[/hide]

$ \boxed{Case\ IV}$ All 2-polynomial-sums but $ P + Q,R + S$ are positive
[hide="Proof of contradiction"]$ R = \frac {(P + R) + (Q + R) - (P + Q)}{2}$ positive.
Similarly $ S$ positive
[hide="because"]$ S = \frac {(P + S) + (Q + S) - (P + Q)}{2}$ positive.
------------------------------[/hide]
$ R$ and $ S$ positive Contradicts $ R + S$ negative
--------------------------------------------------[/hide]

$ \boxed{Case\ V}$ All 2-polynomial-sums but $ P + Q,P + R,P + S$ are positive
[hide="Proof of contradiction"]Same as case I
--------------------------------------------------[/hide]

$ \boxed{Case\ VI}$ All 2-polynomial-sums but $ P + Q,P + R,Q + S$ are positive
[hide="Proof of contradiction"]$ P = \frac {(P + Q) + (P + R) - (Q + R)}{2}$ positive
Similarly $ Q$ positive, $ R,S$ negative.
[hide="because"]$ Q = \frac {(P + Q) + (Q + S) - (P + S)}{2}$ positive
$ R = \frac {(R + S) + (R + Q) - (Q + S)}{2}$ negative
$ S = \frac {(R + S) + (P + S) - (P + R)}{2}$ negative
------------------------------[/hide]
$ P$ positive,$ R$ negative,$ P + R$ positive $ \implies$
$ dc(P) > dc(R)\cdots (i)$
$ R$ negative,$ Q$ positive,$ R + Q$ negative $ \implies$
$ dc(R) > dc(Q)\cdots (ii)$
$ Q$ positive,$ S$ negative,$ Q + S$ positive $ \implies$
$ dc(Q) > dc(S)\cdots (iii)$
$ S$ negative,$ P$ positive,$ S + P$ negative $ \implies$
$ dc(S) > dc(P)\cdots (iv)$

$ (i) \& (ii) \& (iii) \& (iv) \implies contradiction$ (see definition)
--------------------------------------------------[/hide]
Hence our assumption that such $ P,Q,R,S$ exists is contradictory with itself (in all cases). Thus $ P,Q,R,S$ doesn't exist.

$ QED$
----------------------------------------------------------------------------------------------------[/hide]

I edited a few things in the above post.


A set of five different positive integers has mean 8 and median 8.  What is the greatest possible value one of the integers could have?

it should be _, _, 8, _, _.

it's about solving for greatest possible value, so let's make first two numbers 1 and 2.

1, 2, 8, _, _.

and fourth number should be 9.

1, 2, 8, 9, _.

the sum of 5 integers are 8*5=40.

so 40-1-2-8-9=20

Answer ; 20

When I solved it, I got a different answer.

The problem did not state anything about the mode, so I made the first two integers both $ 1$.

Then, I made the fourth number $ 8$.

Since the total is $ 40$, we subtract $ 40 \minus{} 18 \equal{} 22$

So wouldn't the answer be $ 22$?

Oh. I must need glasses.....

So that's why I got that question wrong!  :blush:


linear algebra
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linear algebra unsolved

I have  a few questions on $ SO(n)$

1. why $ SO(n)$ is connected?

2. $ GL(R^{n})$ is a Lie group, it is well known that its Lie algebra $ gl(R^{n})$ can be identified as
$ M_{n}(R)$ (i.e. the set of all n x n matrix). As a sub-lie group, $ SO(n)$ 's lie algebra is a subalgebra of $ gl(R^{n})$ , but how to show that $ so(n)$ is the set of all skew-symmetric n x n matrix?

#1
A constructive approach: choose a matrix $ A\in SO_{n}$. The eigenvalues come in complex-conjugate pairs on the unit circle except for some ones and an even number of $ \minus{}1$s, so we can find an orthonormal basis in which this matrix has the block-diagonal form $ \begin{bmatrix}S(\theta_{1})&0&\cdots&0&0\\0&S(\theta_{2})&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&S(\theta_{k})&0\\ 0&0&\cdots&0&I\end{bmatrix}$, where $ S(\theta)\equal{}\begin{bmatrix}\cos(\theta)&\minus{}\sin(\theta)\\ \sin(\theta)&\cos(\theta)\end{bmatrix}$. Some of the $ \theta$s may be $ \pi$, of course.

We can obviously change each $ \theta_{i}$ continuously; this constructs a path in $ SO_{n}$ between $ I$ and $ A$. Everything is in the path-component of $ I$, so $ SO_{n}$ is path-connected.

#2 Of course, that identification of Lie algebras comes with the bracket $ [A,B]\equal{}AB\minus{}BA$.
To get the Lie algebra of $ SO_{n}$, differentiate the relation $ X^{T}X\equal{}I$: we get $ X^{T}\cdot dX\plus{}(dX)^{T}\cdot X\equal{}0$, or equivalently that $ X^{T}\cdot dX$ is skew-symmetric. Choosing $ X\equal{}I$ as the point to base everything at (as we did for $ GL_{n}$), the subalgebra is identified with the skew-symmetric matrices.

[quote="jmerry"]#1
A constructive approach: choose a matrix $ A\in SO_{n}$. The eigenvalues come in complex-conjugate pairs on the unit circle except for some ones and an even number of $ \minus{}1$s, so we can find an orthonormal basis in which this matrix has the block-diagonal form $ \begin{bmatrix}S(\theta_{1}) & 0 &\cdots & 0 & 0\\ 0 & S(\theta_{2}) &\cdots & 0 & 0\\ \vdots &\vdots &\ddots &\vdots &\vdots\\ 0 & 0 &\cdots & S(\theta_{k}) & 0\\ 0 & 0 &\cdots & 0 & I\end{bmatrix}$, where $ S(\theta) \equal{}\begin{bmatrix}\cos(\theta) &\minus{}\sin(\theta)\\ \sin(\theta) &\cos(\theta)\end{bmatrix}$. Some of the $ \theta$s may be $ \pi$, of course.[/quote]
I know A can be writen in diagonol form,  but why the block-diagonal form ? where are those complex-conjugate pairs ?

Each $ S(\theta)$ represents the pair of eigenvalues $ e^{i\theta},e^{\minus{}i\theta}$.

As a general fact, if a real matrix is diagonalizable over $ \mathbb{C}$, it can be put into block-diagonal form over $ \mathbb{R}$, with each diagonal block either real or in complex number form $ \begin{bmatrix}a&\minus{}b\\b&a\end{bmatrix}$ corresponding to $ a\plus{}bi,a\minus{}bi$.


Let \omega(G) be set of orders of all elements of G and  G be Frobenius group. If H is also Frobenius group and |G|=|H| and \omega(G)=\omega(H), then whether or not G is isomorphic to H?


Find $ x\in R$ of this equation $ \boxed{x\ln\left (1 \plus{} \frac {1}{x} \right)^{1 \plus{} \frac {1}{x}} \minus{} x^3\ln\left (1 \minus{} \frac {1}{x^2} \right)^{1 \plus{} \frac {1}{x^2}} \equal{} 1 \minus{} x}$, With $ x > 0$.

hello, do you meant
$ \boxed{x\ln\left (1 \plus{} \frac {1}{x} \right)^{1 \plus{} \frac {1}{x}} \minus{} x^3\ln\left (1 \minus{} \frac {1}{x^2} \right)^{1 \minus{}\frac {1}{x^2}} \equal{} 1 \minus{} x}$
or
$ \boxed{x\ln\left (1 \plus{} \frac {1}{x} \right)^{1 \plus{} \frac {1}{x}} \minus{} x^3\ln\left (1 \minus{} \frac {1}{x^2} \right)^{1 \plus{} \frac {1}{x^2}} \equal{} 1 \minus{} x}$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, do you meant
$ \boxed{x\ln\left (1 \plus{} \frac {1}{x} \right)^{1 \plus{} \frac {1}{x}} \minus{} x^3\ln\left (1 \minus{} \frac {1}{x^2} \right)^{1 \minus{} \frac {1}{x^2}} \equal{} 1 \minus{} x}$
or
$ \boxed{x\ln\left (1 \plus{} \frac {1}{x} \right)^{1 \plus{} \frac {1}{x}} \minus{} x^3\ln\left (1 \minus{} \frac {1}{x^2} \right)^{1 \plus{} \frac {1}{x^2}} \equal{} 1 \minus{} x}$
Sonnhard.[/quote]

Dear Dr Sonnhard Graubner. It's $ \boxed{x\ln\left (1 \plus{} \frac {1}{x} \right)^{1 \plus{} \frac {1}{x}} \minus{} x^3\ln\left (1 \minus{} \frac {1}{x^2} \right)^{1 \plus{} \frac {1}{x^2}} \equal{} 1 \minus{} x}$

Thanhnam2902! :P

To make sure the second term on LHS is defined, we have $ x>1$, which implies $ LHS>0>RHS$. Hence, there is no solution.


geometry
calculus
derivative
trigonometry
perpendicular bisector

I can't seem to figure these two problems out - any help would greatly be appreciated!

If $a_1 = X$, $a_2 = X^{a_1}$, and, in general, $a_n = X^{a_{n-1}}$, calculate $a_{1,000,000}$ correctly to the nearest thousandth when $X = \left(\frac{4}{3}\right)^{\frac{3}{4}}$

A piece of paper in the shape of an equilateral triangle ABC has $AB = 8$. When $A$ is folded over to any point on $\overline{BC}$, a crease of length $X$ is formed which joins some point of $\overline{AC}$ to some point of $\overline{AB}$. As $A$ is moved along $\overline{BC}$ from $B$ to $C$, $x$ varies. If the greatest value of $x$ is $M$ and the least value of $x$ is $m$, find the ordered pair $(M, m)$.

[hide] For our purposes, the expression x^(x^(x^(...(x) with 1000000 x's can be considered infinite since we only need an approximation.

We can set x^(x^(x^(...(x)=y, then x^y=y, x=y^(1/y)

(4/3)^(3/4)=y^(1/y), so it's clear that y=4/3.

The answer would be 1.333

X can't be much less than 1, and since the answer is so small even with an infinite number of x's, we can assume that the rate of change is very slow, so our answer must be very close to the real answer.

[/hide]

[hide="2"]
Is the answer to the 2nd one:
$(M,m)=(4\sqrt3,4)$
and isnt this a calculus problem :huh:
[/hide]

[hide="solution"]
When $x$ is maximum, by guesswork we can figure out that $A$ coincides with $B$,
But that is the perpendicular bisector of $AB$ and so is $4\sqrt3$
The minumum, using guesswork again, would split the triangle into 4 congruent and equilateral triangles and there $x$ would be $4$
So arriving to my answer
[/hide]

Can you explain in more detail? Your answer just seems to use guesswork. Is there any other solution?

I think we can use differentiation here...
plot a graph $x-\text{distance of A as it moves in the line BC}$ then proceed

Yeah...if you want no calculus...
For 2, you can see that the locus of points for the midpoint of $AA'$ (A' is the point on BC) is the midsegment of the triangle, $DE$, such that $D$ is on $AB$ and $E$ on $AC$.  Hence, the segment with length $x$ (let it be $MN$ with $M$ on $AB$ and $N$ on $AC$), being the perpendicular bisector of $AA'$, must pass through this segment.  Let $MN$ intersect $DE$ at $F$.  Now we can calculate the value of $MN$.  Let $\angle{GAA'}$ have measure $\theta$, where $G$ is  the midpoint of $MN$.  Let $AA'$ intersect $MN$ at $H$.  Then, the length of $MN$ is \[ AH(\tan{(30^{\circ}-\theta)}+\tan{(30^{\circ}+\theta)})=AG\frac{1}{\cos{\theta}}(\tan{(30^{\circ}-\theta)}+\tan{(30^{\circ}+\theta)})=\frac{4}{\cos{\theta}}(\tan{(30^{\circ}-\theta)}+\tan{(30^{\circ}+\theta)})=\frac{4}{\cos{\theta}}(\frac{\sin{30^{\circ}}\cos{\theta}-\cos{30^{\circ}}\sin{\theta}}{\cos{30^{\circ}}\cos{\theta}+\sin{30^{\circ}}\sin{\theta}}+\frac{\sin{30^{\circ}}\cos{\theta}+\cos{30^{\circ}}\sin{\theta}}{\cos{30^{\circ}}\cos{\theta}-\sin{30^{\circ}}\sin{\theta}})=\frac{4}{\cos{\theta}}(\frac{\frac{1}{2}\cos{\theta}-\frac{\sqrt{3}}{2}\sin{\theta}}{\frac{\sqrt{3}}{2}\cos{\theta}+\frac{1}{2}\sin{\theta}}+\frac{\frac{1}{2}\cos{\theta}+\frac{\sqrt{3}}{2}\sin{\theta}}{\frac{\sqrt{3}}{2}\cos{\theta}-\frac{1}{2}\sin{\theta}}) \]
\[ =\frac{4}{\cos{\theta}}(\frac{\cos{\theta}-\sqrt{3}\sin{\theta}}{\sqrt{3}\cos{\theta}+\sin{\theta}}+\frac{\cos{\theta}+\sqrt{3}\sin{\theta}}{\sqrt{3}\cos{\theta}-\sin{\theta}} \]
\[ =\frac{4}{\cos{\theta}}(\frac{\sqrt{3}(\cos^2{\theta}+\sin^2{\theta})-4\cos{\theta}\sin{\theta}+\sqrt{3}(\cos^2{\theta}+\sin^2{\theta})+4\cos{\theta}\sin{\theta}}{3\cos^2{\theta}-\sin^2{\theta}}) \]
\[ =\frac{4}{\cos{\theta}}(\frac{2\sqrt{3}}{2\cos{2\theta}+1}) \]
Whew, that took a while to type out.  Anyways, we see that $0 \le \theta \le 30^{\circ}$  Thus, we maximize this expression with $\theta=30^{\circ}$, thus minimizing $\cos{\theta}$ and $\cos{2\theta}$.  This gives us $\frac{8\sqrt{3}}{\frac{\sqrt{3}}{2}*2}=8$, and minimize it by setting $\theta=0$, yielding $\frac{8}{\sqrt{3}}$.

Well I clearly messed this up somewhere along the way...hmm...meh