monovariant
invariant
combinatorics unsolved
combinatorics

As the graph, a pond is divided into 2n (n $\geq$ 5) parts. Two parts are called neighborhood if they have a common side or arc. Thus every part has three neighborhoods. Now there are 4n+1 frogs at the pond. If there are three or more frogs at one part, then three of the frogs of the part will jump to the three neighborhoods repsectively. Prove that for some time later, the frogs at the pond will uniformily distribute. That is, for any part either there are frogs at the part or there are frogs at the each of its neighborhoods.

[img]http://www.mathlinks.ro/Forum/files/china2005_2_214.gif[/img]

All three problems are extremely easy  ... :?  Is this really the final round?

For this problem (easy exercise on monovariants), here is an outlined solution:

1. Show that every part will be visited by a frog at some point;
2. Consider the number $N$ of pairs of adjecent parts of the pond, none of which is occupied by a frog. Observe that no move can increase this quantity (it is a [i]monovariant[/i]). Thus, as a non-increasing sequence of non-negative integers, this quantity stabilizes at some point. The problem is equivalent to showing that it stabilizes to $0$. Assume to the contrary that it stabilizes to some $k \geq 1$: that is, from some point onwards, there are precisely $k$ pairs of adjecent parts of the pond, none of which is occupied by a frog.
3. Let $A,B$ be adjecent parts of the pond, none of which is occupied by a frog. Notice that $N$ will be strictly decreased if a frog visits $A$.
4. Derive a contradiction from 1), 2) and 3).

Or, did I misunderstand the problem?

--Vesselin

That proof looks pretty solid and nice  :) . A generalisation of part 1 can be found here: http://www.mathlinks.ro/Forum/viewtopic.php?t=24406 (most likely as both were written by vess, it appears to have sprung out of this thread?)

Unbelievable!
Problem from CMO, which can be solved in 5 minutes!

I think, it is not CMO.

And the other two 'problems' are no less trivial ...  :?

[quote="Myth"]Unbelievable! Problem from CMO, which can be solved in 5 minutes! [/quote]

If there are only $4n$ frogs, does every vertex get exactly 2 frogs eventually?

Do you ask whether we will reach the moment s.t. all vertex will have exactly 2 frogs?
Consider $K_4$ and the distribution $4,2,2,0$. We will obtain an infinite cycle, so answer for this question is no.

[quote="Myth"]Do you ask whether we will reach the moment s.t. all vertex will have exactly 2 frogs? [/quote]
Yes.  This type of problem, for any graph, is known as "chip-firing game". It is allowed to move the frogs one vertex at a time and for a small enough number of frogs the final configuration is independent of all choices.  Unfortunately I don't remember the details of the theorem.

[quote] Consider $K_4$ and the distribution $4,2,2,0$. We will obtain an infinite cycle, so answer for this question is no.[/quote] What if $n > 2$?

[quote="grobber"]I found [url=http://mathforum.org/epigone/sci.math.research/kinshumskun]a link[/url]. Apparently, if we start with fewer frogs than the number of edges in the graph, the process always ends in a configuration which depends only on the initial configuration.[/quote]

There are many other interesting things about the game.
There is a way to use chip-firing configurations to introduce the structure of a commutative group on the set of spanning trees of the graph.  I think it is a finer invariant than the number of spanning trees; there are two graphs with same number of spanning trees but different group structures.  Also, the number of moves depends only on the initial configuration, it is polynomially bounded for undirected graphs and exponential for directed graphs.
If I can find a reference I will post it.

If I did'nt make a mistake this problems is similar to a problem in Poland 1994 .
The problem is like this.
n girls sitting around a table. At first one of them holds k cards, at one step if there are some girls holding 2 cards then one of them will give the 2 cards two the girls besides her. Find k so that the game will end.

It was also shortlisted, I think. If $n$ is odd, then the game ends for all $k\le n$, while if $k$ is even, the game ends for $k\le n-1$, I think.

I used conex parts which should stabilize from some point and obviously can't.
But why was this problem given with $4n+1$ as long as it works with $2(2n-4)+3$.
Am I wrong?

Well, you can have 2 frogs on each of (2n-1) squares - the last one will never be visited.

yes but that means that the $2n$-th part's neighbours all have at least one frog. As I understand the problem, you don't need to have every part visited for a distribution to be uniform.

[quote="ikap"]why was this problem given with $4n+1$ as long as it works with $2(2n-4)+3$. Am I wrong?[/quote]
Maybe to make the problem easier.   It is a plausible strategy, but incomparably harder, to try to characterize the final configuration of frogs.  If the number of frogs is $4n+1$, the process is non-terminating, so that method cannot be used.  If the number of frogs is less than $3n$, one has the terminating case of the chip-firing game and there is an elaborate theory about the final configuration.


I understand that this changes every year, but what is the usual minimum USAMO score necessary to make Red MOP? Or, since USAMO is really focusing on #1 and 4 from now on, around what score would be the cutoff in those two problems?

[quote="andersonw"]Usually, the cutoff is around 9-10 (1 problem fully solved, partial credit on 1-2 other problems), but now that they're only going to use 2 problems, it will probably be 7-8 (or even lower), because some people taking USAMO may solve, say, problem 2 and not problem 1, but it wouldn't count toward red MOP. Also, if you solve 1 problem, you can only get additional partial credit on 1 other problem, unlike when you could get partial credit on 5 other problems.

For example, if only 1 and 4 were counted toward red MOP on last year's USAMO, the cutoff score would have most likely been 7 or 6 instead of 8, because lots of people got 2 free points on problem 2, which wouldn't count under the new system.
(unfortunately, I got a 7 last year :( )[/quote]
You sure? I was under the impression that all of the USAMO problems counted for red MOP.

The only difference was that the USAJMO participants could only use the last 3 and 6 questions.

Really? Then what's the point of taking the whole test if only 2 problems count? It would be pretty sad to get points on other problems but those 2..

Also for tiebreaker, in case some 2 people tied on #1 and 4 scores.

[quote="Bugi"]To get to Blue MOP, Black MOP?[/quote]

Unfortunately solving 6 problems on USAJMO doesn't get you into anything but Red.

So basically, your USAJMO doesn't give you anything? No international competitions, nothing more than red MOP?

Why is it introduced then?

According to the AMC Director, it is created to give a chance to beginners in olympiad who do not have much experience so that they won't get to intimidated by the difficulty of USAMO. But I don't see how we would really get frightened because of USAMO if we at least tried...

I know multiple people at my school who are just almost good enough to qualify for USAMO. They are definitely good enough for USAJMO, and would much rather take it than the USAMO, which is very intimidating. Remember, the set of people on AoPS is exactly the set of people who would be motivated by a really difficult contest - for others who are different, it's a good idea to have the intermediate level. AIME to USAMO is a gigantic jump, especially for people unfamiliar with proofs.

This year:

Red MOP: 8+
Blue MOP: 20+
Black MOP: 27+

Blue was 18 this year.

To get to red, you should just be comfortable with #1 and #4.

what if someone gets a 9 on USAMO but solves problem 2 and get partial credit on problem 1?
(this scenario might be possible because the ninth grader could have first solved 2 and not had enough time to do 1 properly)

would they still qualify for red MOP assuming the cutoff is about 7-8?

[quote="andersonw"]Under the current rules, no.
(I think this is pretty clear based on what was already said in this thread, and it's also not that hard to search "USAJMO" to find more details on it ;) )[/quote] Are you sure? Doesn't it say that if you're in the USAMO you can do as many problems as you want, while the USAJMO is just limited to those two?

Then that would be unfair because USAMO would have 6 problems that count toward qualification, while USAJMO would only have 2.

1. That the entire USAMO score counts for red MOP is a frequently suggested proposal on this forum, frequent enough to (I think) mislead people here. I have not seen AMCDirector say anything about changing their original policy of counting only J3/J6 and 1/4.

2. To the above: It's not necessarily unfair. It is harder to make the USAMO than the USAJMO.

Just wondering what would a sophemore get from taking the USAJMO (since that would only help you get to Red MOP only)?

Absolutely nothing, other than possibly the honor of being a USAJMO winner (sort of like MVP of the AAA baseball world series?)

You'd get the experience of solving olympiad questions in an actual test environment, hopefully it would help you do better the next year in the USAMO.


Hi!

Starting about 2 days ago, whenever I click the link for hidden text

([hide="this"]nothing[/hide])

it doesn't display the text, it just takes me back to the top of the page.

Is there any way to fix this?

Thanks

You probably have a bad copy of the script that expands hidden text. Try clearing your cache.


indeed the solution is 13/29 R.

Both the top half and bottom half equal 13/8.

So 1/Req = 8/13 + 8/13 + 1

yes the answer is $\frac{13}{29}R$
in the figure below:
$AM+MN=2R\Rightarrow 2R+AN=\frac{2}{3}R\Rightarrow \frac{2}{3}R+ND=\frac{5}{3}R\Rightarrow\frac{5}{3}R+AO=\frac{5}{8}R\Rightarrow\frac{5}{8}R+R=\frac{13}{8}R$
so top and bottom are both $\frac{13}{8}R$ the resistance between $A$ and $B$ is, as if you had three resistors in parallel two of each with $\frac{13}{8}R$  and one with $R$ so
$\frac{1}{R_{A,B}}=\frac{1}{\frac{13}{8}R}+\frac{1}{\frac{13}{8}R}+\frac{1}{R}$
$R_{A,B}=\frac{13}{29}R$

[quote="spx2"]in the following drawing we have a electricity problem
a octogon.any line on it represents a resistor of resistance
$R$.It asks to give the resistance between points $A$ and $B$
namely $R_{AB}$.
i tried to do it and it gave me $\frac{7}{17}\cdot R$.
but in the book ofcourse like every problem i solve i have to
get it wrong.
book says $\frac{13}{29}\cdot R$ or something like that.[/quote]

Let's consider just the network above $AB$. (the network below $AB$ is symmetric with it, and so will have the same net resistance)

The outer triangle is $(1+1)||1 = 2||1 = \frac{2}{3}$

Added to the next resistor to which this combination is joined in series $\frac{2}{3}+1 =\frac{5}{3}$

The $\frac{5}{3}$ is in parallel with the next resistor $\frac{5}{3}||1 =\frac{5}{8}$

This combination is in series with 1, $\frac{5}{8}+1 = \frac{13}{8}$

Finally, there is a parallel network of three resistors the 13/8 above $AB$, 1 between $AB$, and 13/8 below $AB$

$R_{AB} =(\frac{8}{13}+1+\frac{8}{13})^{-1}=\frac{13}{29}R$

Well, this one is easy, it can be done with paralell/series combinations .. but check this out ..

I've written this program a few months ago, its use is intuitive, just feed it with a file named rin.txt with description of a graph = scheme. For your scheme it is

8 13
0 1
0 2
0 3
0 7
0 4
1 2
2 3
3 7
4 7
4 5
5 6
5 7
6 7

Enjoy!

{sorry, i couldn't post it as a .cpp file so here it is:}

[code]#include <stdio.h>
#include <stdlib.h>
#define MAX 100

int g[MAX][MAX];
int p[MAX];
int N,M;
double c[MAX][MAX];

void gread()
{
  FILE *f;
  int i,j,k;
  f=fopen("rin.txt","rt");
  fscanf(f,"%d%d",&N,&M);
  for(i=0; i<=N; i++)
    {
    p[i]=0;
    for(j=0; j<=N; j++)
      {g[i][j]=0;}
    }
  for(i=0; i<M; i++)
    {
    fscanf(f,"%d%d",&j,&k);
    p[j]++; p[k]++;
    g[j][k]++; g[k][j]++;
    }
  fclose(f);
}

double rcalc()
{
  int i,j,k;
  double q,ii;
  for(i=1; i<N; i++)
    for(j=1; j<N; j++) {c[i][j]=g[i][j];}
  for(i=1; i<N; i++)
    {
    c[i][i]=-p[i];
    c[i][0]=-g[i][0];
    }

  for(i=1; i<N-1; i++)
    {
    for(j=i+1; j<N-1; j++)
      {
      q=-c[j][i]/c[i][i];
      for(k=0; k<N-1; k++) {c[j][k]+=c[i][k]*q;}
      }
    }

  for(i=N-2; i>0; i--)
    {
    c[i][0]/=c[i][i];
    for(j=1; j<i; j++){c[j][0]-=c[i][0]*c[j][i];}
    }

  ii=0;
  c[N-1][0]=0;
  for(i=1; i<N; i++){ii+=(1.0-c[i][0])*g[0][i];}
  return 1.0/ii;
}

int main()
{
  double rr;
  gread();

  rr=rcalc();
  printf("%.15lf\n",rr);

  while (1);

  return 0;
}
[/code]

[quote="spx2"]to AMIRTHULSA

in the figure below:
$AM+MN=2R\Rightarrow 2R+AN=\frac{2}{3}R\Rightarrow \frac{2}{3}R+NO$

i agree until here
now $AO$$ is in parallel with what has been calculated until now so the formula
for parallel must be aplicated not series.[/quote]
 :huuh:
THAT IS THE FORMULA FOR PARALLEL


A rectangular array has 9 rows and 2004 columns. In the 9 * 2004 cells of the table we place the numbers from 1 to 2004, each 9 times. And we do this in such a way that two numbers, which stand in exactly the same column in and differ around at most by 3. Find the smallest possible sum of all numbers in the first row.

nobody has idea about this problem? I think the smallest possible sum of all numbers in the first row occur when the first row is:
"1,1,1,2,3,...,1999,2000,2001,2001" but I can't prove :(

The problem was too hard  :?:  or anyone didn't think about :!:


[hide]
The faster train overtook the slower train 4 hours after it departed from the station.  That means,[b] relative to the slow train[/b], the fast train went 100 miles.  Then that means those 100 miles came from the slow train's 2 hour head start.  So the slow train's speed is $50$ mph.
[/hide]

[quote="sharkman"]A train leaves a station at 9am traveling at a constant rate of speed.  Two hours later a second train, traveling at an average rate of 25 miles per hour more than the first train, leaves the same station, going in the same direction, on a parallel track.  If the faster train overtakes the first train at 3pm, on the same day, what is the average rate of speed of the first train?[/quote]

[hide]The first train traveled 6 hours and the second train traveled 4 hours.  So if we call the speed of the first train $s$, then

$d=6s$
$d=4(s+25)$
$6s=4s+100$
$s=50$

The answer is [b]50[/b].[/hide]


$a,b,c\ge 0$. Prove or disprove:
$\sum\sqrt[4]{b^4+c^4+14b^3c}\ge 2(a+b+c)$

[quote="Gibbenergy"][quote="chidat"]$a,b,c\ge 0$. Prove or disprove:
$\sqrt[4]{b^4+c^4+14b^3c}\ge 2(a+b+c)$[/quote]
are u sure it is right? Where is a?[/quote]
I'm sorry. I've edited it

[quote="Nameless"][quote="zhaobin"]see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=68174]http://www.mathlinks.ro/Forum/viewtopic.php?t=68174[/url][/quote]
No, they are different. This ineq is stronger than my ineq. And this ineq can't be proved by my ineq's way[/quote]
you are right :blush:

Why noone solve this? I think it's very nice and hard. But I can't solve it :blush:


If you are have a 30% chance of achieving a positive outcome in something 1 time, what are the odds for you doing this 26 times in a row.

Also if there is a 10% chance of you achieving a positive outcome in something 1 time, what are the odds of doing this 5 times in a row.

Thx!

jommie

[hide="first problem"]$(\frac{3}{10})^{26}$, whoever wants to calculate that.[/hide]

[hide="second"]$(\frac{1}{10})^5=\frac{1}{100000}$.[/hide]

when you divide that first one out i get:
0.00000000000002541865828329


It might help if you posted the problem.

lol sry, but I havent gotten it back yet

Heres what i can remember

|f(x)| + g(x) = 4x ; x <= -2
|f(x)| + g(x) = -2x2  ; -2 < x <= 0
|f(x)| + g(x) = h(x) ; x > 0

What is the least possible value of f(10)?

Can't you have f(x) = -[huge gigantic number]?  Then it will be arbitrary how small you can make f(10)...  :?

i honestly dont know, sry :blush: :(

there may be more to this problem

il be sure (or someone else) to post the full problem tomorrow, when i get it back

$g(x), f(x), h(x)$ are polynomials

--what it said at the beginning of the problem

and that's the complete problem

He only forgot the condition that f, g, and h are all polynomial functions. That's why the minimum is bounded.

What you do is add what they give you. The |f(x)| cancel out, since f(x) has different signs on those intervals.
g(x)=-x^2-2x
f(x)= :pm: (x^2+2x) (f(x) can be either, not both)

Therefore the minimum of f(10) = -120.

this "math league" was the mandelbrot, was it not?

I think not.  This is the Math League of [url]http://www.mathleague.org[/url].


function
real analysis
real analysis unsolved

Could you please help me to solve the following exercise:

Let $ f: [a, b]\rightarrow\mathbb{R}$ be a continuous function that is differentiable on $ (a,b)$ and such that $ f(a)\neq f(b)$. Show that there exists $ c\in (a,b)$ such that \[ \frac{a f(b)\minus{}b f(a)}{f(b)\minus{}f(a)}\equal{}c\minus{}\frac{f(c)}{f'(c)}.\]

are you sure the function is not restricted by condition $ f(x) \neq 0, \forall x \in [a,b]$ ?
if yes, you can apply Cauchy mean value theorem to functions: $ \frac {x}{f(x)}$ and $ \frac {1}{f(x)}$

a counter exemple for the case when f can be 0 is:
$ f : [0,1] \minus{} > R$, $ f(x) \equal{} (x \minus{} p)^2, x \in [0,p]$, $ f(x) \equal{} 0, x \in (p,1]$, $ 0 < p < 1$


Haven't seen it, but I don't plan to.  Fornicating marionettes isn't my personal idea of quality entertainment, but to each his own.

Hey, it was made by the creators of South Park, so I've got to give it a shot.

Incredibly hilarious.

Yes, there were inappropriate scences such as fornicating puppets.. but man, it was hilarious.

I highly recommend it to anyone who finds Southpark funny.

I plan to see it soon.

Omang, the soundtrack to the movie is hilarious.

i saw it and it was awesome ans hilarious
even though it isa couple of days after i saw it i still have the theme song running through my head

I haven't seen it yet..

but it seems really hilarious...

I think I will like that movie...

by the fact that it includes Kim Jung Ill....


A person is riding in a hot air balloon. For the first hour and a half the wind current is a constant 22.0 miles per hour in the direction N 37.5 E. The the wind current changes to 18.5 miles per and heads the balloon in the direction S 52.5 E. If this continues for another 2 hours, what is the bearing of the balloon from its starting point?

The movement of the balloon is determined by its velocity vector, in this case, the wind.  You can think of a vector as being an arrow.  So in the first hour and a half, the balloon travels $ 22\times 1.5\equal{}33$ miles.  But this is just the length of the vector.  You need to now draw a Cartesian plane, and draw an arrow of length 33 with angle N 37.5 E from the origin.  Now from the end of this arrow, draw another one with length $ 18.5\times 2\equal{}37$ miles and angle S 52.5 E (why?).  Be careful here.  Then once that is done, draw your resultant vector, which connects the origin to the end of the second vector.  That is the vector you need to work with.


algebra polynomial probability percent induction inequalities Euler

We choose random a unitary polynomial of degree $n$ and coefficients in the set $1,2,...,n!$. Prove that the probability for this polynomial to be special is between $0.71$ and $0.75$, where a polynomial $g$ is called special if for every $k>1$ in the sequence $f(1), f(2), f(3),...$ there are infinitely many numbers relatively prime with $k$.

Well, actually this is the only problem from the USA TST 2005 that I know. ;)

Its leading coefficient is 1. That is, a polynomial of the form $ X^n+a_{n-1} X^{n-1}+...$.  :)

can you post your solution harazi?this problem is really interesting

Probability P(n) equal to $P(n)=\prod_{p\le n}(1-\frac{1}{p^{p}})$, were p primes less or equal n. For example P(1)=1,P(2)=0.75, $P(3)=\frac{13}{18}$, e.tc.

Here's a solution. The latter part is a bit messy though.
Suppose $ P(x)$ is a non-special polynomial. Then for some $ k$, we have $ (P(x), k) > 1$, for all integer $ x$, with finitely many exceptions. Let $ p_{1}, p_{2} ... p_{t}$ be the prime factors of $ k$. Then, for any $ x$, at least one of the following is true: $ p_{1} | P(x) ; p_{2} | P(x) ; ... p_{t} | P(x)$. Consider the equation $ P(x) \equiv 0 (mod p_{i})$. The solution set of this equation is: $ x \equiv i_{0}, i_{1} ... , i_{r} (mod p_{i})$, for some set $ S_{i} = {i_{1} ... i_{r}}$. Suppose for each prime $ p_{i}$, this solution set does not encompass all elements of $ Z_{p_{i}}$. Then for each prime $ p_{i}$, assign to it a number $ q_{i}$, such that $ q_{i}$ is not in $ S_{i}$. Then $ (P(q_{i}), p_{i}) = 1$. Consider the solution to the set of equations $ a \equiv q_{i} (mod p_{i})$, for $ i = 1,2 ... t$ (which exists due to the Chinese Remainder Theorem). Then, we have   $ (P(a), k) = 1$. Consider the infinite family $ a_{i} \equiv a (mod k)$. Then $ k | a_{i} - a | P(a_{i}) - P(a)$, so $ (P(a_{i}), k) = 1$, so we can find infinitely many integers $ n$ such that $ (P(n), k) = 1$, which contradicts our choice of $ k$.
Thus we have proven that, for any non-special polynomial $ P(x)$, we can associate with it a prime $ p$, so that $ P(x) \equiv 0 (mod p)$, for all integers $ x$.

Lemma 1: If $ p$ is a prime, then $ 1^k + 2^k + ... + (p - 1)^k$ is divisible by $ p$, iff $ p - 1 | k$.
Proof: If $ p - 1|k$, we have $ i^k \equiv 1 (mod p)$ for $ i = 1, 2 ... p - 1$, so $ 1^k + ... + (p - 1)^k \equiv p - 1 (mod p)$, so it is not divisible by $ p$.
Now suppose $ k$ is not divisible by $ p - 1$. Pick a generator $ g$ in $ Z_{p}$. We know that the sets $ {g^1 ... g^{p - 1}}$ and $ {1, 2... p - 1}$ coincide in $ Z_{p}$, so the sets $ (g^k)^1, (g^k)^2 ... (g^k)^{p - 1}$ and $ 1^k, 2^k ... (p - 1)^k$ also coincide in $ Z_{p}$, so $ 1^k + ... + (p - 1)^k \equiv (g^k) ((g^k)^0 + (g^k)^1 + ... + (g^k)^{p - 2}) \equiv (g^k) \frac {(g^k)^{p - 1} - 1}{g^k - 1} \equiv 0 (mod p)$, since $ g^k - 1$ is not $ 0$ in $ Z_{p}$, since $ k$ is not divisible by $ p$. Thus lemma is proven.
Lemma 2: Suppose $ Q(x)$ is an integer polynomial of degree $ p - 2$, where $ p$ is an odd prime. If $ p | Q(x)$ for all $ x$ such that $ (p,x) = 1$, then all coefficients of $ Q(x)$ are divisible by $ p$.
Proof: Let $ Q(x) = a_{0} + a_{1} x + ... + a_{p - 2} x^{p - 2}$ Working in $ Z_{p}$, we have $ p | a_{0} + u a_{1} + u^2 a_{2} + ... + u^{p - 2} a_{p - 2}$, for $ u = 1, 2, ... p - 1$. Call this equation (U). To prove that $ p | a_{i}$, it suffices to prove $ p | v a_{i}$ for some $ v$ such that $ (p,v) = 1$. Thus it would be sufficient, in order to prove $ p | a_{k}$, to find some integers $ t_{1,k}, t_{2,k} ... t_{p - 1,k}$, so that $ t_{1,k} * (1) + t_{2,k} * (2) + ... + t_{p - 1,k}* (P - 1) \equiv v a_{i} (mod p)$, for some $ v$ such that $ (v,p) = 1$. (Recall that (1), (2) .. (P-1) are equations of the form p | ... ). If this is true, since $ 0 \equiv t_{1,k} * (1) + t_{2,k} * (2) + ... + t_{p - 1,k}* (P - 1) \equiv v a_{i} (mod p)$ , it would follow that $ p | a_{i}$. Now it remains to find $ t_{i,k}$ so that $ t_{1,k} * (1) + t_{2,k} * (2) + ... + t_{p - 1,k}* (P - 1) \equiv v a_{i} (mod p)$  - but this is equivalent to find $ t_{i,k}$ so that $ p | 1^s t_{1,k} + 2^s t_{2,k} + ... + (p - 1)^s t_{p - 1,k}$, for all $ s = 0,1,2 ... p - 2$ except $ k$. From Lemma 1, it's not hard to see that $ t_{i,k} = i^{p - 1 - k}$ satisfies all these conditions: so we have found our numbers $ t_{i,j}$, and consequently proved that $ p | a_{i}$, which in turns prove Lemma 2.

Let $ P(x) = x^n + a_{n - 1} x^{n - 1} + ... + a_{1} x + a_{0}$, and $ P(x)$ is not special, with respect to a prime $ p$. Since $ P(p) \equiv 0 (mod p)$, we have $ a_{0} \equiv 0 (mod p)$. Now choose $ x$, so that $ (p,x) = 1$. By Fermat's Little Theorem, $ x^{r(p - 1) + s} \equiv x^{s} (mod p)$, for integer $ r$. So we have $ P(x) \equiv Q(x) (mod p)$, where $ Q(x) = b_{0} + b_{1} x + ... + b_{p - 2} x^{p - 2}$, and $ b_{0} = a_{0} + a_{p - 1} + a_{2p - 2} + ...$, $ b_{1} = a_{1} + a_{p} + ..., .... b_{p - 2} = a_{p - 2} + a_{2p - 3} + ...$, where $ a_{n} = 1$. From Lemma 2, we have: $ p | b_{0}, b_{1} ... b_{p - 2}$, so $ p | a_{0} + a_{p - 1} + ...$, $ p | a_{1} + a_{p} + ...$, ... $ p |a_{p - 2} + a_{2p - 3} + ...$ and also, $ p | a_{0}$ from before. By backtracking our steps, we can see that these $ p$ conditions are equivalent to the polynomial $ P(x)$ being non-special.

Lets do the easy part first: the upper bound. We shall prove that exactly $ 25$ percent of all such polynomials are 'non-special', with respect to the prime $ 2$. The condition for a prime being special with respect to $ 2$ is fairly simple: $ 2 | a_{0}$, $ 2 | a_{1} + a_{2} + ... + a_{n - 1} + 1$ (since coefficient of $ x^n$ is $ 1$). To prove that $ 25%
$ of all sets $ {a_{0} ... a_{n - 1}}$, with $ 1 \leq a_{i} \leq n!$, first note since $ n!$ is even, exactly half of all sets satisfy the condition that $ 2|a_{0}$. Of these sets, we can easily set up a bijection between between the sets satisfying $ 2 | a_{1} + a_{2} ... + a_{n - 1} + 1$ and those satisfying $ 2 | a_{1} + .. + a_{n - 1}$: biject any set $ {x_{1} ... x_{n - 1)}}$ to the set $ {y_{1} ... x_{n - 1}}$, where for each $ x_{1}$ we find the number $ y_{1}$ so that $ (x_{1}, y_{1})$ is one of the following sets: $ (1,2), (3,4) ... (n! - 1, n!)$.

Lemma 3: Suppose we select $ c$ numbers from the set $ {1, 2 ... tp + k}$ where $ 0 \leq k \leq p - 1$. The probability that the sum of these numbers is equal to $ d (mod p)$, for some integer $ d$, is at most $ \frac {t + 1}{tp + k}$.
Proof: Let $ f(c, d)$ be the number of ways of selecting $ c$ numbers (one number may be used more than once), from the set $ {1,2 .. tp + k}$, which have a sum equal to $ d (mod p)$. We want to prove that $ f(c,d) \leq (t + 1)(tp + k)^{c - 1}$ (since in total there are $ (tp + k)^{c}$ such sets). Let's prove the result by induction on $ c$.
For $ c = 1$, it's easy to see there's at most $ t + 1$ different ways of selecting $ 1$ number that is equal to $ d (mod p)$. Suppose it is true for $ k = c - 1$. The number of ways of having a set with $ c$ numbers, with the first one being $ r (mod p)$, so that they sum to $ d (mod p)$, is equal to $ f(c - 1, d - r)$. Since there are $ t + 1$ numbers in $ {1,2 ... tp + k}$ equaling $ i (mod p)$, for $ 1 \leq i \leq k$, and $ t$ such numbers equaling $ j (mod p)$ for $ k + 1 \leq j \leq p$, we have the recurrence
 $ f(c,d) = (t + 1) f(c - 1, d - 1) + (t + 1) f(c - 1, d - 2) + ... + (t + 1) f(c - 1, d - k) + t f(c - 1, d - k - 1) + ... + t f (c - 1,d)$,
 from which we immediately get our inequality $ f(c,d) \leq (t + 1)(tp + k)^{c - 1}$, using the induction assumption. This proves Lemma 3.

Continuing from before, $ P(x)$ is not special with respect to prime $ p$ iff: $ p | a_{0} + a_{p - 1} + ...$, $ p | a_{1} + a_{p} + ...$, ... $ p |a_{p - 2} + a_{2p - 3} + ...$ and $ p | a_{0}$. If $ p \leq n$, then $ p | n!$; if we let $ Pr(X) =$ probability of an event $ X$ occuring (not to confuse it with the polynomial $ P(x)$), we can see that the probability of any of the $ p$ conditions necessary for $ P(x)$ not to be special, with respect to $ p$, is the probability of a set of some $ c$ integers from $ {1,2 ... n!}$ having sum equal to $ 0 (mod p)$; by using the same bijection method with which we proved that $ 25$ percent of all our polynomials are not special with respect to the prime $ 2$, we can see that the probability of any of the $ p$ conditions to be true, is exactly $ \frac {1}{p}$. Thus the probability, that all of those $ p$ conditions are true, is equal to $ \frac {1}{p^p}$, for a prime $ p \leq n$ (since the conditions are mutually indepedent, we can multiply the probabilities).
For a prime $ p$, with $ n < p < n!$, the same method fails to work, since the probability maybe slightly more than $ \frac {1}{p}$ for some of the individual conditions to work (since $ p$ no longer divides $ n!$, the same bijection method fails to work).
So instead, we now rely on the crude estimate that Lemma 3 provides. We now have, that the probability of any of the $ p$ conditions is true, is at most $ \frac {t_{p} + 1}{tp + k}$, where $ t_{p}p + k = n!$ (since we seek to find to a set, that has sum of elements equal to $ 0 (mod p)$, or $ - 1 (mod p)$ - in the case where the term $ a_{n} = 1$ occurs, so Lemma 3 is directly applicable). However, the probability of the first condition, $ p |a_{0}$ occurs is clearly less than $ \frac {1}{p}$, so the probability of $ P(x)$ being not-special with respect to $ p$, is at most $ \frac {(t_{p} + 1)^{p - 1}}{p(n!)^{p - 1}}$.
Clearly, the probability of a polynomial being not-special, is at most the sum of the probabilities that the polynomial is not special with respect to each prime, that is smaller than $ n!$. It suffices to prove this number is less than $ 0.29$, i.e.
$ \sum_{p\le n} \frac {1}{p^p} + \sum_{p > n} \frac {(t_{p} + 1)^{p - 1}}{p(n!)^{p - 1}} < 0.29$.
For $ n = 2$, we just write out the $ 4$ possible polynomials and find out $ x^2 + x + 2$ is the only non-special one; for $ n = 3$ we verify the above inequality directly (and it is incredibly tight); for $ n = 4$ a method very similar to the below works, and for $ n \ge 5$ the following method works to kill that last inequality:
If $ p > n$, since $ tp + k = n!$, and $ 0 \le k \le p - 1$, we have $ t + 1 \le (n - 1)!$, so $ \frac {t + 1}{n!} < \frac {1}{n}$. So:
$ \sum_{p > n} \frac {(t_{p} + 1)^{p - 1}}{p(n!)^{p - 1}} < \sum_{p > n} \frac {1}{p n^{p - 1}} < \sum_{n + 1\le m\le n!} \frac {1}{(n + 1)n^{m - 1}} < \frac {1}{(n + 1)n^n} \sum_{m \ge 1} (\frac {1}{n})^m = \frac {1}{(n + 1)n^n} \frac {1}{1 - \frac {1}{n}} < \frac {1}{n^n}$,
so since $ n \ge 5$,
$ \sum_{p\le n} \frac {1}{p^p} + \sum_{p > n} \frac {(t_{p} + 1)^{p - 1}}{p(n!)^{p - 1}} < \sum_{p \le n} \frac {1}{p^p} + \frac {1}{n^n} < \frac {1}{2^2} + \frac {1}{3^3} + \frac {1}{5^5} + \sum_{n \ge 5} \frac {1}{n^5} < \frac {1}{2^2} + \frac {1}{3^3} + \frac {1}{5^5} + 0.002 < 0.29$.
Note that $ \sum_{n \ge 5} \frac {1}{n^5} < 0.02$, using the inequality $ \frac {1}{n^5} < \frac {1}{20 n^2} - \frac {1}{20 (n + 1)^2}$ (which can be proven easily by computation), and using telescoping sums.

epitomy01's solution is very long and rather hard to read... I've compiled a more readable and shorter solution.

Define a polynomial to be $p$[i]-special[/i] if $f$ does not vanish modulo $p$. Then a polynomial is special iff it is special with respect to every prime.

[b]Lemma 1:[/b] If $\deg f < p$, then $f$ does not vanish modulo $p$.
[b]Proof:[/b] This is trivial and well-known. $f$ has at most $\deg f$ roots over $\mathbb{F}_p$ as $\mathbb{F}_p$ is a field, and the result follows. $\square$

Hence it follows for the selection of polynomials in the problem, it suffices to consider whether or not polynomials are $p$-special where $p$ is a prime $\le n$.

[b]Lemma 2:[/b] Given a random polynomial specified in the problem, the probability $f$ is $p$-special is $1 - \frac{1}{p^p}$.
[b]Proof:[/b] Given a random polynomial $f$, reduce it modulo $p$ and then modulo $x^p - x$. Clearly this does not alter whether or not the polynomial vanishes modulo $p$. Hence it suffices to consider the $p^{p}$ polynomials $a_0 + a_1x + ... + a_{p-1}x^{p-1}$ where $0 \le a_i \le p-1$ (remark each of them corresponds to $\frac{n!^{n!}}{p^p}$ polynomials before reducing them). Only $1$ of them vanishes modulo $p$ , hence it follows the probability a polynomial vanishes modulo $p$ is $1 - \frac{1}{p^p}$ and we are done. $\square$

Remark that a polynomial being $p$-special and $q$-special are independent events when $p,q$ are distinct primes by CRT. Hence it follows the probability a random polynomial is special is $\prod_{p \le n} \left (1 - \frac{1}{p^p} \right )$.

Now we prove the theorem. As $1 - \frac{1}{2^2} = 0.75$, it follows $\prod_{p \le n} \left (1 - \frac{1}{p^p} \right ) \le 0.75$. Thus it suffices to show $\prod_{p \le n} \left (1 - \frac{1}{p^p} \right ) \ge 0.71$, or $\prod_{p} \left (1 - \frac{1}{p^p} \right ) \ge 0.71$.

Remark that for all $k$, $\prod_{k \le p} \left (1 - \frac{1}{p^p} \right ) > \frac{1}{\zeta(k)}$ by the Euler Product Formula.
Now, observe:
$\prod_{p} \left (1 - \frac{1}{p^p} \right ) > \frac{\left ( 1 - \frac{1}{2^2} \right ) \left ( 1 - \frac{1}{3^3} \right ) \left ( 1 - \frac{1}{5^5} \right ) \left ( 1 - \frac{1}{7^7} \right )}{\zeta(8)} > \frac{0.72}{\zeta(8)}$

But:
$\zeta(8) = (-1)^9 \cdot \frac{(2\pi)^8B_8}{2 \cdot 16!} = \frac{\pi^8}{9450}$ because by a little algebra $B_8 = \frac{-1}{30}$
It's not hard to estimate then that $\zeta(8) < 1.005$, hence $\prod_{p} \left (1 - \frac{1}{p^p} \right ) > \frac{0.72}{1.05} > 0.71$, as desired so we are done. $\square$

(Bounding $\zeta(n)$ without using the exact value while still getting a good enough value is possible, but this route for the last few steps of bounding is pretty mindless and just number crunching horror)

Sorry to bump this, but I was going back through old TSTs to practice for the upcoming USAMO, and I randomly picked the 2005 one to do. Unfortunately, the problem statement here is just unclear enough that I can't actually do the problem without some clarification. Specifically, the statement is clearly false if $n = 1$, so is the restriction on $n$ that $n \ge 2$, or something else? Additionally, the definition of special starts with a polynomial $g$ and then references a polynomial $f$; I assume these are supposed to be the same? Finally, I know someone clarified this in the comments, but a rewording of this problem that avoids the word "unitary," instead using "monic" or even "with leading coefficient 1" would be great. If anyone has access to the original statement of this problem, please clarify these things and maybe even repost it so other people who use this for practice don't have the same problem.

EDIT: Also, hide tags for the solutions. Those would be nice.

Rebump. I realize it's possible nobody has the old TST, but I'm hoping somebody will give me a usable version of the problem soon so I can get to work on it.

Suppose $n \geq 2$. By CRT it suffices to look at primes only, by periodicity it suffices to find a single integer value $m$ such that $p \nmid f(m)$. In general, by Lagrange interpolation there is a single polynomial of degree at most $p-1$ which is always $0 \pmod{p}$, namely the zero polynomial, so we can ignore $p>n$.

For primes $p \leq n$, arbitrarily select the coefficients of every term with degree at least $p$, leaving $p$ undetermined coefficients, and call this polynomial $f_0(x)$. Then by Lagrange interpolation, in $\mathbb{F}_p$ there always exists exactly one polynomial $f_1(x)$ with degree at most $p-1$ such that $f_0(x)+f_1(x)=0$ for all $x \in \mathbb{F}_p$, so the proportion of the remaining $(n!)^{p}$ choices for the rest of the coefficients that results in a polynomial not being special is exactly $p^{-p}$.

By only looking at $p=2$ we easily get the upper bound. For the lower bound, summing over all $p$ we have a probability of at most $\sum_{p \text{ prime}} p^{-p}$ that a polynomial is not special. However,
$$\sum_{p\text{ prime}} p^{-p} \leq -1-4^{-4}+\sum_{n=1}^\infty n^{-n} \approx -1-4^{-4}+1.291286=.28737975,$$
which is small enough. $\blacksquare$


function
calculus
derivative
analytic geometry
graphing lines
slope
LaTeX

how do you determine the interval with which each function is increasing or decreasing?

example: the graph of y=2e^x

[u]One Way is to Use Calculus[/u]

Taking the derivative of $ y \equal{} 2e^x$,
$ y' \equal{} 2e^x$ (notice how the derivative of this function equals itself!)
From this new function, which tells the exact slope of the original curve at any value of $ x$,
it is clear to see that the slope will always be positive no matter what $ x$ is. Therefore,
the original function $ y \equal{} 2e^x$ is increasing for all values of $ x$ ($ \minus{} \infty\le{x}\le{\infty}$)

I'm not sure if this proof is rigorous, but I believe it is.
Let $ k$ be an any positive real number.

Then, we have

$ 2e^{x\plus{}k}\minus{}2e^{x}$
$ \equal{}2e^{x}e^{k}\minus{}2e^{x}$
$ \equal{}2e^{x}(e^{k}\minus{}1) > 2e^{x}(e^{0}\minus{}1) \equal{} 0$

So it is always increasing

It is also worth mentioning that $ 2e^x>0$ for any real $ x$.

this example is simple...
strictly saying:
function is increasing if and only if for every $ a>b \rightarrow f(a)>f(b)$
and function is decreasing if and only if for every $ a>b \rightarrow f(a)<f(b)$
so in most cases picking 2 different x's and seeing how function changes will determine if function is increasing or decreasing

PS: sorry im not very good in latex


prove that the number $\binom{2^n-1}{i},i=0,1,...,2^{n-1}-1$,has pairwise different residius modulo $2^n$. :)  :)  :)

Also from Sanct-Peterburg Math Olympyad 2003.


analytic geometry
geometry
3D geometry
sphere
inequalities
conics
ellipse

Did anyone find an elegant solution to problm 3? I think I found the answer, but my solution isn't very ingenious.

$r-t \geq 0$
also
$2u \geq 0$
$r-t+2u \geq 0$
Note that
$4(r+s+t+u)+(r-t+2u)$
$=5r+4s+3t+6u$
$=100$
Therefore
$100-4(r+s+t+u)=(r-t+2u) \geq 0$
$100-4(r+s+t+u) \geq 0$
$25 \geq r+s+t+u$

Hence the maximum value of $r+s+t+u=25$, equality holds r=s=t=25/3, u=0
$s-u \geq 0$
$2t \geq 0$
$s-u+2t \geq 0$
Note that
$5(r+s+t+u)-(s-u+2t)$
$=5r+4s+3t+6u$
$=100$
Therefore
$5(r+s+t+u)-100=s-u+2t \geq 0$
$5(r+s+t+u)-100 \geq 0$
$r+s+t+u \geq 20$

Thus the minimum value of $r+s+t+u=20$, equality holds r=20, s=t=u=0
[edit]---- looks too long before..... no need to double space..... lol

Yeah, nice proof beta, looks like mine ;-).
I think for max case, you mean r = s = t = 25/3, u = 0.

yeah....... this round seems easy...... maybe Round 2 will be harder??

yeah, last year the rounds got significantly harder (at least for me). round 2 seemed hardest last year.

Interesting solution... Mine took a lot longer than that, I used 4 dimensional inequalities. Yours is much quicker, congratulations.

Here's mine:

Since r >= s >= t >= u >= 0, the numbers r-s, s-t, t-u, and u are non-negative. To find the lower bound, we calculate as follows:

$r+s+t+u$

$=(r-s)+2(s-t)+3(t-u)+4u$

$\ge (r-s)+\tfrac{9}{5}(s-t)+\tfrac{12}{5}(t-u)+\tfrac{18}{5}u$

$=\tfrac{1}{5}\big(5r+4s+3t+6u\big)$

$=\tfrac{1}{5}\big(100\big)$

$=20$.

The minimum of 20 can be achieved when (r,s,t,u) = (20,0,0,0). We similarly find the upper bound:

$r+s+t+u$

$=(r-s)+2(s-t)+3(t-u)+4u$

$\le \tfrac54(r-s)+\tfrac{9}{4}(s-t)+3(t-u)+\tfrac{9}{2}u$

$=\tfrac{1}{4}\big(5r+4s+3t+6u\big)$

$=\tfrac{1}{4}\big(100\big)$

$=25$.

This maximum is attained when (r,s,t,u)=(25/3,25/3,25/3,0). Thus, the minimum and maximum values are 20 and 25, respectively.

how many solutions did you guys get for number 2?

If each variable in the equations was instead to represent a component of a coordinate squared, then we would be left with a 4-dimensional sphere-like shape, and a series of inequalities equations.  By finding the closest point on the sphere-like equation to the origin and the farthest point from the origin that obeys the inequalities, we can find the smallest and greatest sums of $r+s+t+u$.
Geometrically, as a point moves along a sphere equation on one plane, the distance to the origin will change the least if it moves along the plane with distances from the origin at the axes closest to each other in value.
Define:
$r$ to be $w^2$ in a 4-dimensional coordinate system;
$s$ to be $x^2$ in a 4-dimensional coordinate system;
$t$ to be $y^2$ in a 4-dimensional coordinate system;
$u$ to be $z^2$ in a 4-dimensional coordinate system;
Through substitution we obtain the following equations:
$5w^2+4x^2+3y^2+6z^2=10^2$;
$w^2 \geq x^2 \geq y^2 \geq z^2 \geq 0$
The first can be recognized as an equation for a 4-dimensional ellipse, which will from now on be referred to as a hyper-sphere.  The second equation can be reduced to
$|w| \geq |x| \geq |y| \geq |z| \geq 0$
which can be further broken down to the six equations:
$|w| \geq |x|$, $|w| \geq |y|$, $|w| \geq |z|$, $|x| \geq |y|$, $|x| \geq |z|$, $|y| \geq |z|$
From the coefficients of the variables we can determine that the minimum distance from the origin exists at the point $(0,0,0,\frac{10}{\sqrt{6}})$, where $z^2$ is at its greatest and all other components are zero.  This however, does not satisfy the inequalities.  To find the next point that does satisfy the inequalities, we must move the point along the $w/z$ plane, for it is the plane whose other pole is the next closest to the origin, so will cause the distance to the origin to move the least. Because moving along this plane means that both $x^2$ and $y^2$ must remain at zero, we know that the next viable point along this line is that at $(2\sqrt{5},0,0,0)$, where $w^2$ is at its greatest and all other components are zero.

Therefore, the coordinates of the closest point to the origin are given by $w=2\sqrt{5}, x=y=z=0$, and the distance is given by $r=20, s=t=u=0$ or $20$.

Following this same logic for the farthest point, we can see that the farthest point is at $(0,0,\frac{10}{\sqrt{3}},0)$, where $y^2$ is at its greatest and all other components are zero.  This however, does not satisfy the inequalities. From the inequalities, we know that at the $r$ and $s$ must be at least as great as $t$, so the closest viable point to $(0,0,\frac{10}{\sqrt{3}},0)$ must be that in which $r$, $s$, and $t$ are equal and $u$ is 0.

Therefore, the coordinates of the furthest point from the origin are given by $w=x=y=\frac{5}{\sqrt{3}}, u=0$, and the distance is given by $r=s=t=\frac{25}{3}, u=0$ or $25$.

If I had to rank them, I'd give them all 2's or 3's, all done on first day.  P1 took a grand total of 5 minutes, so it gets a 1. P5 stymied me for a while, then I remembered a property of centroids and it fell into place.  My only mistake I found, (unfortunately on tuesday), was that I had one of the solutions to P2 in my work, but not in my solutions box. I hope they get harder...

[quote="beta"][quote="ZennyK"]Good thinking, Zach... nice and short.  :)
DavinciCoder, can you post yours? It sounds cool. I'll try to put mine up later today if I have a break in my homework. ;)

How challenging would you say the problems from round 1 were, on a scale of 1-10, where 1 = way to easy and 10 = extremely difficult?[/quote]

1)--3
2)--4
3)--5
4)--3
5)--2[/quote]I probably did something wrong on problem 4 because that one seemed by far the easiest to me. I just solved 172 + (172 - 4) + (172 - 2*4) + ... + (172 - (n-1)*4) = 180(n - 2) to get n = 12. Try not to laugh if there's a glaring error in my solution that I'm missing  :blush: .

[quote="AntonioMainenti"]I probably did something wrong on problem 4 because that one seemed by far the easiest to me. I just solved 172 + (172 - 4) + (172 - 2*4) + ... + (172 - (n-1)*4) = 180(n - 2) to get n = 12. Try not to laugh if there's a glaring error in my solution that I'm missing  :blush: .[/quote]

Yeah... I got 12 for that one too...... but i think #5 is even easier if you know law of cosines

As an engineer, I recognized this problem as a "Linear Programming" problem.  Linear Programming is applicable when every constraint is a linear function of the variables and the function to be maximized (or minimized) is also a linear function of the variables.

One of the theorems of Linear Programming states that the maximum (or minimum) will occur at a vertex of the n-dimensional object formed by the constraints.  The conditions $r\ge s$, $s\ge t$, $t\ge u$, and $u\ge 0$ gives us a 4-dimensional pyramid, apex at the origin, open at the top.  The condition, $5r+4s+3t+6u=100$, gives us a 3-dimensional tetrahedron.  Thus, there are 4 points to be tested.

Attached is an illustration of a simpler 2-variable problem.  The shaded areas are excluded by the constraints, $r\ge s\ge 0$.  The constraint, $5r+4s=100$, reduces the allowable figure to a line segment.  Whatever linear function of r and s you seek to maximize or minimize, a solution will exist at one of the points shown in green.

I'm a first-timer, and I'm worried that my solutions seem to be a lot longer than everyone else's (except #1 and #4 which seem to be uniformly short).  Is there a disadvantage to this?  Also, I made the same error as the other people who got only 8 solutions for #2 by ignoring $x^2+y^2+11=0$ in my list of solutions.  However, I did mention in my proof that $(\pm6,\pm5)$ is the solution to this equation, so they are in there somewhere.  On #3, I used what seems to be a somewhat unique approach, but it got me the right answer.  Opinions?

Edit: as soon as I posted this, it started working. *sigh*

Disclaimer: I may be totally wrong, and probably am.

Regarding your question dhuang: I think here is a bit of a problem. if you have s+2t+3u, in the context of your proof I agree you must minimize it. Th problem is, if the minimum is zero, it generates a contradiction. If s+2t+3u=0, then s and t and u all equal zero. Hence, f=r. In the context of the maximum, r=100/3. However, when you plug this back into the original equation, I think you find
5(100/3) + 4(0) + 3(0) + 6 (0) =100
500/3=100
A contradiction. Hence the minimum of that expression cannot be zero given the context of the original problem. Similarly, the minimum does not work.

[quote="dhuang614"]I have a question for #3.  I did the problem a different way, but I still don't understand why I got a wrong answer.  Here is my solution.[/quote]

sran thinks you have a contradiction. If you had completed the necessary second half of your proof, you would have found a contradiction that would have pointed out your error to you. Unfortunately, it lookes like you didn't even realize that the second half was necessary.

You said, "Let $f=r+s+t+u.$" and worked your way down to

[quote="dhuang614"]$f\leq\frac {100}{3}- \frac {1}{3}(s+2t+3u).$

Note that in order to maximize $f$, we must minimize $\frac {1}{3}(s+2t+3u)$.  Because the minimum of $\frac {1}{3}(s+2t+3u)$ is 0, $\max{f}=\frac{100}{3}$.[/quote]

That last line is your error. What you have really proven is that
$\max(f)\leq \frac{100}{3}$. You have not found the maximum, you have found an upper bound. The part of the proof that you skipped is proving that the value if $f$ can go all that way up to your upper bound. Which it cannot, because $f\leq 25$.

Erin Schram
USAMTS grader

For questions number 2 i used mostly logic.  :)  That didn't cost me a lot of work.[/list]

I'm sorry i mean that i used mostly logic for questions number 3

For number 2, i got 12 solutions. Did anyone else get that many?

I solved number 1 nearly the exact same way Davincicode :)

There's an edit feature for your posts so you don't have to repeat.  12 solutions was the correct number for #2.  Check the Math Jam transcript for all the right answers: [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=52]here[/url].

Hi!  I am a first timer here as well.  I believe that I have solved all the problems correctly - my answers match the other posted answers and I can't find any flaws in my proofs.

Here's the thing: my proofs (for the most part) were considerably longer than the other proofs posted here (my solution to #2 took 7 pages).

From looking at other solutions, I see that I provided proofs for many assertions that others assumed were obvious.  For instance, in the context of #1, the following is a typical statement regarding the minimum sum of the numbers on the vertices of the pentagon (thoobik, I hope you don't mind my quoting you):

"Clearly, the smallest possible value for $s$ is $1+2+3+4+5=15$."

While this does seem to be a rather obvious fact, I was not sure how much detail was required in the solutions.  That is, could I make such an assertion or should I prove it.  I decided to err on the side of caution, and provided two lemmas which formalize a (very) generalized version of that assertion and applied them to prove it.  Is this the right approach? Is there anything wrong with this approach?  Should I have stuck with "Clearly...?"

Also, on other problems (#4 for example) I am not sure I was explicit enough (particularly with the proof of Lemma 1).

All feedback on my solutions will be greatly appreciated.  Thanks!

-Eric

PS: I have attached my solutions (whopping 15 pages)

First, a bit of general advice.  If a statement is ridiculously obvious (such as the five-member subset of (1,2,3,4,5,6,7,8,9,10) with the smallest sum is (1,2,3,4,5)), or if it's very well known (the sum of the interior angles of an n-gon is (n-2)180 degrees), you don't have to prove it.  In general, if you are refering to a theorem, a good general guideline is, if you can name it, or it's extremely well-known, you don't have to prove it (but if even if you can name an obscure theorem, you should probably state what it is, just in case the grader doesn't know; on another forum, a USAMTS grader said that there was once some confusion about what "Stewart's theorem" was).

For #3, you should definitely see if you can find a shorter solution.  The general guidelines are that solutions probably shouldn't exceed 3 pages.

For #4, see the first paragraph.

In general, the shorter you can make your solutions without taking away from their substance, the better off you are.

By the way, here are my solutions:

I thank dts for the feedback, but still welcome other comments.

Also, do the graders have a preference between a clear solution involving a minimum amount of english and a solution with lots of text?

[quote="paniag"]Also, do the graders have a preference between a clear solution involving a minimum amount of english and a solution with lots of text?[/quote]

We graders prefer clear solutions. The amount of text varies. Some solutions are best described in equations and some solutions are best described in words.

[quote="paniag"]Hi!  I am a first timer here as well.  I believe that I have solved all the problems correctly - my answers match the other posted answers and I can't find any flaws in my proofs.

Here's the thing: my proofs (for the most part) were considerably longer than the other proofs posted here (my solution to #2 took 7 pages).

From looking at other solutions, I see that I provided proofs for many assertions that others assumed were obvious.  For instance, in the context of #1, the following is a typical statement regarding the minimum sum of the numbers on the vertices of the pentagon (thoobik, I hope you don't mind my quoting you):

"Clearly, the smallest possible value for $s$ is $1+2+3+4+5=15$."

While this does seem to be a rather obvious fact, I was not sure how much detail was required in the solutions.  That is, could I make such an assertion or should I prove it.  I decided to err on the side of caution, and provided two lemmas which formalize a (very) generalized version of that assertion and applied them to prove it.  Is this the right approach? Is there anything wrong with this approach?  Should I have stuck with "Clearly...?"
[/quote]

Even thoobik's solution for problem 1/1/16 is longer than I like. Dividing his steps into lemmas simply meant that he broke the flow of his argument and had to repeat definitions for $n$ and $s$. I do think it is clear and obvious that the five smallest numbers in a set of numbers is the five-element subset with the smallest sum, so he was right not to bother with a proof.

There are problems with a long and detailed style:

(1) We graders get tired of reading so much. In problem 1/1/16, you prove general lemmas for a specific problem. Of course, by proving the general case, you had to switch to variables, summation notation, and function notation. That notation is more difficult to read that just using the numbers from 1 to 10.

(2) Sometimes important details are mislaid in the mass of other details. For example, in Lemma 2 in problem 1/1/16 you forget to mention that set $A$ has to have a fixed size. Because you forgot that, the statement of the lemma is false. Furthermore, you don't prove that your sum $cS(A)+S(B)$ is smaller than the sum for all other sets, you prove that it is smaller for all sets $A'$ that differ from $A$ by one element. You could chain together inequalities of sets differing in one element to prove it for all sets, but you forgot to do so in your proof. These are unimportant details in an unimportant lemma, so you won't lose points for it, but an important detail could be overlooked almost as easily.

(3) Conciseness of a proof counts toward getting a commendation or for being selected as an official solution. It is difficult to be elegent when bogged down in proofs of the obvious. One way to earn a commendation is to find an elegant generalition of the result, but we mean generalize the [i]result[/i], not the intermediate details.

Looking at your solution for problem 2/1/16, I see you used the Fermat's Last Theorem shortcut, which we did expect some participants to discover. You gave a footnote for the proof of Fermat's Last Theorem. That explains to me why your solutions are overly wordy. You are mimicking the style in mathematics journals. In this case, that is writing for the wrong audience.

Articles in journals are cutting edge work that needs a lot of support because it is new. New techniques have to be defined carefully. New discoveries have to be tested carefully. Sources have to be documented carefully.

On the other hand, I have taken articles from journals and turned them into seminar talks. This involves a great deal of rewriting, because flow and clarity are important in a talk, even more important than being careful. The listening audience must understand the concept the first time because there is no way for them to rehash the material. I have to emphasis the key points by skipping over other details. It is a different style.

Solutions for a contest, such as the USAMTS, are written for an audience that is already familiar with the material. Even if a grader could not solve a USAMTS problem himself (it happens often), he or she has probably seen hundreds of solutions of that problem by the time he or she grades yours. The goal of your writeup is not to cover every detail of the solution but to demonstrate that you understand the problem, its solution, and the vital nuances of the solution.

To spend time rigorously proving an obvious property means that you do not understand why rigor is important. For example, earlier in this thread I mentioned to dhuang614 that in problem 3/1/16 that though he had found an upper bound for $r+s+t+u$, he had not proven that the sum ever reached his upper bound, which it doesn't. His lack of rigor left him with a false answer. That instance of rigor was important. If I simply told him he needed more rigor, and he had chosen to give the derivation of his upper bound in even more detail and rigor, it would not have helped him. That would be focussing on unimportant rigor.

By the way, looking at your solution for 3/1/16, I see terrible flaws of style. You divide the proof of the minimum into a logical set of cases, and you do proceed with rigor. But in case 1, you find r=20. In case 2, you assume that r still equals 20, even though you used s=0 in case 1 to find the value of r, and case 2 has s>0. Instead, you use $r-\epsilon_0$ in the place of r to suggest that the value you want for r is not really 20. What you really should have said was, "Let $r=20-\epsilon_0$," so that you could still use the variable r to mean r and then use the number 20 to mean the value of r from case 1. As for your proof of the maximum, that is too long. I will wait until I am actually grading your submission to finish reading it.

[quote="paniag"]Also, on other problems (#4 for example) I am not sure I was explicit enough (particularly with the proof of Lemma 1). [/quote]

Both Lemma 1 and Lemma 2 in your solution of problem 4/1/16 are well-known results. You do not have to prove them in your solution and we graders will skip reading your proofs of those lemmas in the interest of saving time.

Erin Schram
USAMTS grader

Well, as long as people are posting their solutions, here's mine.

Erin Schram,

I greatly appreciate your comments.  I will present Round 2 solutions differently.

Thank you,
-Eric


Ross Mathematics Program
probability
linear algebra
matrix
function
expected value
conditional probability

Two teams, A and B will play a series of games. The probability of A winning each game is p. The overall winner is the first team to have won two more games than the other.

a) Find the propbability that team A is the overall winner

b)Find the expected number of games played.

Ah, the Wimbledon championships ended today. (Not that this is a good model for tennis, anyway - in tennis, $ p$ alternates between two values depending on who is serving.)

Let $ P$ be the conditional probability that $ A$ eventually wins, given that they are tied. (Hence the overall probability of $ A$ winning starting from the beginning will be $ P.$) Let $ P_1$ be the probability that $ A$ wins, given that $ A$ has the advantage, and let $ P_2$ be the probability that $ A$ wins, given that $ B$ has the advantage.

From the tied state, we move to the state of $ A$ having the advantage with probability $ p$ and to the state of $ B$ having the advantage with probability $ 1 - p.$ That gives us
\[ P = pP_1 + (1 - p)P_2\]
Similar consideration of what happens from the advantage-$ A$ state and the advantage-$ B$ state gives rise to two more equations:
\begin{align*}P_1 & = p + (1 - p)P \\
P_2 & = pP + (1 - p)\cdot 0\end{align*}
In matrix form, we have this system:
\[ \left[\begin{array}{ccc|c}1 & - p & p - 1 & 0 \\
p - 1 & 1 & 0 & p \\
- p & 0 & 1 & 0\end{array}\right]\]
Solving it (off stage), we get that $ P = \frac {p^2}{1 - 2p + 2p^2},$ $ P_1 = \frac {p - p^2 + p^3}{1 - 2p+p^2}$ and $ P_2 = \frac {p^3}{1 - 2p + 2p^2}.$

$ P$ is the answer we want. We say that $ A$ wins the match with probability $ \frac {p^2}{1 - 2p + 2p^2}.$

If we put the values $ p = 0,p = 1,p = \frac12$ into this expression, we get the results we should expect. A graph of this function will be sigmoidal, steepest in the middle, and approximately doubling the advantage in a neighborhood of the middle. That is, if $ p$ is $ 51\%,$ then the probability of $ A$ winning the match will be nearly $ 52\%.$

We can solve for the expected time in the same way. Let $ T$ be the conditional expectation of the remaining time, given a tied state, and let $ T_1$ and $ T_2$ be the expected remaining time from the states of advantage-$ A$ and advantage-$ B,$ respectively. We get the following system of equations:
\begin{align*}T & = 1 + pT_1 + (1 - p)T_2 \\
T_1 & = p + (1 - p)(1 + T) \\
T_2 & = p(1 + T) + (1 - p)\end{align*}
That gives us $ T = \frac {2}{1 - 2p + 2p^2}$ and also $ T_1 = \frac {3 - 4p + 2p^2}{1 - 2p + 2p^2}$ and $ T_2 = \frac {1 + 2p^2}{1 - 2p + 2p^2}.$

$ T$ is the answer we want. Starting at the beginning, the expected length of the match is $ \frac {2}{1 - 2p + 2p^2}$ games. This has a minimum of $ 2$ games when $ p = 0$ or $ p = 1$ (which makes sense) and a maximum value of $ 4$ games when $ p = \frac12.$


Given the vertex $A$ and the centroid $M$ of a triangle $ABC.$ Find the locus of all vertices $B$ such that the angles $A,$ $B,$ $C$ of triangle $ABC$ simultaneously satisfy the following three inequalities:

$\begin{equation*}
40 \leq A\leq 70 ;\;\;\;\;\;\;\;\;\;\; 40 \leq B\leq 70
;\;\;\;\;\;\;\;\;\;\; 40 \leq C\leq 70.
\end{equation*}$

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


Find the number of different sequences $ a_{1}$, $ a_{2}$,...,$ a_{8}$ that satisfy following conditions:

(i) $ a_{1}=31$

(ii) for each $ i>1$ there exists index $ j<i$ such that $ 4$ l$ {a_{i}-a_{j}}$l $ \leq 9-i$

(iii) There exists index $ i$ and $ j$ such that $ a_{i}-a_{j} \geq 4$

[quote="delegat"]Find the number of different sequences $ a_{1}$, $ a_{2}$,...,$ a_{8}$ that satisfy following conditions:

(i) $ a_{1} = 31$

(ii) for each $ i > 1$ there exists index $ j < i$ such that

(iii) There exists index $ i$ and $ j$ such that $ a_{i} - a_{j} \geq 4$[/quote]
  Firstly, I must point that it should be integer or natural number sequence, because if not than the problem is trivial and answer is infinite :)
 Secondly, the condition
$ 4$ l$ {a_{i} - a_{j}}$l $ \leq 9 - i$ means that for $ i > 5$ there is a number $ a_{j}$ with $ j < i$ such that $ a_{i} - a_{j} = 0$ and discussion about  $ a_{2}$ means that $ a_{2} = ({32,31,30})$ $ a_{3} = ({33,32,31,30,29})$ $ a_{4} = ({34,33,32,31,30,29,28})$ $ a_{5} = ({35,34,33,32,31,30,29,28,27})$
and because of iii)   $ a_{5} = 35$ or  $ a_{5} = 27$ and later discussion is only trivial counting....  :)

[quote] Firstly, I must point that it should be integer or natural number sequence, because if not than the problem is trivial and answer is infinite[/quote]

I am sorry for this but it is obvious I think.

[b] vedran[/b]  Your solution is very nice

In fact, we notice that $ a_{6}$ , $ a_{7}$ and $ a_{8}$ are equal to some of $ a_{i}$ , where $ 1 \leq i \leq 5$.

Because of third condition all  $ a_{i}$'s, $ 1 \leq i \leq 5$., are different  (difference of greatest and smallest of them is less than 4, and due to third condition it must be at least 4).

So we can choose $ a_{i}$ in two ways: $ a_{i}\equal{}\max\{a_{j} ,j<i\}\plus{}1$ or $ a_{i}\equal{}\min \{a_{j}, j<i\}\minus{}1$.

We have in total $ 2^{4}$ different ways to choose first 5 elements of sequence and for each of them $ 5^{3}$ ways to choose last 3 terms of sequence so in total we have: $ 2^{4}.5^{3}\equal{}2000$ such sequences.


Very nice inequality, since equality occurs again for $ x_1 \equal{} ... \equal{} x_{n \minus{} 1} \equal{} \frac 1{\sqrt[4] {n \minus{} 1}}$ and $ x_n \equal{}  \sqrt[4] {n \minus{} 1}$. :lol:
Of course, EV-Theorem gives the shortest solution. :wink:

[quote="can_hang2007"]Let $ a_1,a_2,\ldots, a_n$ be positive real numbers such that $ a^4_1 \plus{} a^4_2 \plus{} \cdots \plus{} a^4_n \equal{} n.$ Prove the inequality
$ \left( n \plus{} \sqrt {n \minus{} 1}\right) (a^2_1 \plus{} a^2_2 \plus{} \cdots \plus{} a^2_n \minus{} n) \ge (a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)^2 \minus{} n^2.$
:)[/quote]
I found in my computer the following:
Let $ a_1,a_2,\ldots, a_n$ be positive real numbers such that $ a^4_1 \plus{} a^4_2 \plus{} \cdots \plus{} a^4_n \equal{} n.$
If $ n\le k\le n\plus{}\sqrt{n\minus{}1}$, then
$ k(a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2\minus{}n)\ge (a_1\plus{}a_2\plus{}...\plus{}a_n)^2\minus{}n^2$.

The following related inequality holds, too.
Let $ a_1,a_2,\ldots, a_n$ be positive real numbers such that $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equal{} n.$ Prove the inequality
$ (a^2_1 \plus{} a^2_2 \plus{} \cdots \plus{} a^2_n)^2 \minus{} n^2 \ge \frac {n(n \minus{} 1)}{n^2 \minus{} n \plus{} 1}(a^4_1 \plus{} a^4_2 \plus{} \cdots \plus{} a^4_n \minus{} n).$

[quote="Vasc"][quote="can_hang2007"]Let $ a_1,a_2,\ldots, a_n$ be positive real numbers such that $ a^4_1 \plus{} a^4_2 \plus{} \cdots \plus{} a^4_n \equal{} n.$ Prove the inequality
$ \left( n \plus{} \sqrt {n \minus{} 1}\right) (a^2_1 \plus{} a^2_2 \plus{} \cdots \plus{} a^2_n \minus{} n) \ge (a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)^2 \minus{} n^2.$
:)[/quote]
I found in my computer the following:
Let $ a_1,a_2,\ldots, a_n$ be positive real numbers such that $ a^4_1 \plus{} a^4_2 \plus{} \cdots \plus{} a^4_n \equal{} n.$
If $ n\le k\le n \plus{} \sqrt {n \minus{} 1}$, then
$ k(a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2 \minus{} n)\ge (a_1 \plus{} a_2 \plus{} ... \plus{} a_n)^2 \minus{} n^2$.
[/quote]
Yes, I found you posted a particular case here: http://www.mathlinks.ro/viewtopic.php?t=236754. And I just thought of the general and proved it.  :blush: :)

P/s: This inequality holds for all $ k \le n\plus{}\sqrt{n\minus{}1},$ no need to have $ k \ge n$. ;) :)

[quote="can_hang2007"] P/s: This inequality holds for all $ k \le n \plus{} \sqrt {n \minus{} 1},$ no need to have $ k \ge n$. ;) :)[/quote]
You are right, this condition are not necessary.


trigonometry
inequalities
inequalities unsolved

if $\alpha ,\beta, \gamma$ are angels of triangle
prove that
$\sin\alpha+\sin\beta+\sin\gamma\leq \sin({\alpha+\frac{\beta}{2}})+\sin({\beta+\frac{\gamma}{2}})+\sin({\gamma+\frac{\alpha}{2}})$

I rename the angles into A, B, C, so that the problem rewrites as follows:

[color=blue][b]Problem.[/b] If A, B, C are the angles of a triangle, prove that

$\sin A+\sin B+\sin C<\sin\left(A+\frac{B}{2}\right)+\sin\left(B+\frac{C}{2}\right)+\sin\left(C+\frac{A}{2}\right)$.[/color]

The [i]solution[/i] is pretty simple:

After the sum of angles in triangle ABC, we have A + B + C = 180°. Thus, B + C < 180°, and consequently, $\frac{B+C}{2}<\frac{180^{\circ}}{2}=90^{\circ}$, so that $\sin\frac{B+C}{2}<1$.

By a well-known formula from trigonometry,

$\sin B+\sin C=2\sin\frac{B+C}{2}\cos\frac{B-C}{2}$.

Since $\sin\frac{B+C}{2}< 1$ and $\cos\frac{B-C}{2}> 0$ (since $-90^{\circ}<\frac{B-C}{2}<90^{\circ}$), this implies that

$\sin B+\sin C<2\cos\frac{B-C}{2}$.

But since A + B + C = 180°, we have $\frac{A+B+C}{2}=\frac{180^{\circ}}{2}=90^{\circ}$, and thus

$\frac{B-C}{2}=\frac{A+B+C}{2}-\left(C+\frac{A}{2}\right)=90^{\circ}-\left(C+\frac{A}{2}\right)$,

and thus, $\cos\frac{B-C}{2}=\sin\left(C+\frac{A}{2}\right)$, so that our inequality $\sin B+\sin C<2\cos\frac{B-C}{2}$ becomes

$\sin B+\sin C<2\sin\left(C+\frac{A}{2}\right)$.

Similarly, we can find

$\sin C+\sin A<2\sin\left(A+\frac{B}{2}\right)$;
$\sin A+\sin B<2\sin\left(B+\frac{C}{2}\right)$.

Summing up these three inequalities, we get

$\left(\sin B+\sin C\right)+\left(\sin C+\sin A\right)+\left(\sin A+\sin B\right)$
$<2\sin\left(C+\frac{A}{2}\right)+2\sin\left(A+\frac{B}{2}\right)+2\sin\left(B+\frac{C}{2}\right)$;

in other words,

$2\left(\sin A+\sin B+\sin C\right)<2\left(\sin\left(A+\frac{B}{2}\right)+\sin\left(B+\frac{C}{2}\right)+\sin\left(C+\frac{A}{2}\right)\right)$.

This yields

$\sin A+\sin B+\sin C<\sin\left(A+\frac{B}{2}\right)+\sin\left(B+\frac{C}{2}\right)+\sin\left(C+\frac{A}{2}\right)$,

and thus, your inequality is proven.

  darij

You are right, equality can never hold. I went wrong about this in my post #2; now I have edited my post and corrected the mistake.

  Darij


trigonometry calculus integration function logarithms algebra polynomial

The integral is as follows:

$\int{\frac{{5}}{(x-1)\sqrt{x^{2}-2x-24}}dx}$

I understand that a u-substitution and some inverse trig substitutions are involved, but I'm wondering how to deal with the $(x-1)$ - should I bring it into the square root, or somehow complete the square or maybe something else? Any hints or help would be appreciated.

This is the second integral:

$\int{\frac{{2x^{2}+4x+22}}{x^{2}+2x+10}}dx$

This one gives me an extremely peculiar answer when done out on the TI-89 Titanium, and no terms really cancel out when factored, so I'm wondering if there is a u-substituion involved/inverse trig (particularly arctan) involved. Someone told me this was simplified by Integration by Parts?

Well - any help is welcome

On the first one: note that completing the square is going to give you a term of $(x-1)^{2}.$ Looking ahead to that, make the substitution $u=x-1.$ $dx$ becomes $du.$ $x-1$ becomes $u.$ and $x^{2}-2x-24=(x-1)^{2}-25=u^{2}-25.$ So now you have

$\int\frac{5\,du}{u\sqrt{u^{2}-25}}.$

Now that it's "cleaned up" a little, look at it as if it were a new problem. Perhaps a trig substitution (although that's not the only idea)?

On the second one: this is a rational function. All rational functions have antiderivatives that are linear combinations of rational functions, logarithms, and arctangents.

The first step will be to do long division, getting a polynomial (well, in this case, that's just the constant $2$) plus a rational function that is a proper fraction.

Now - can the denominator be factored? In this case, if you're staying within the real numbers, the answer is no. Your denominator is an irreducible quadratic. Complete the square in that, and make another "little" substitution - in this case, $u=x+1.$ Now look at what you have - can you see how to proceed?

Write the second integral as \[2\int\frac{x^{2}\ dx}{(x-1-3i)(x-1+3i)}+4\int \frac{x \ dx}{(x-1-3i)(x-1+3i)}+22\int \frac{dx}{(x-1-3i)(x-1+3i)}\] and use partial fractions and integration by parts.

Edit:  As for the wierd output on your calculator, note the identity

$\tan^{-1}z = \frac{i}{2}\ln\left(\frac{i+z}{i-z}\right)$, where here $\ln$ represents the complex log given by $\ln z = \ln|z|+iarg(z)$.

Edit #2: Here's a much easier way (I skimmed over what Kent said and this may be what he mentioned): \[\int\frac{2x^{2}+4x+22}{x^{2}+2x+10}\ dx = 2\int{\frac{x^{2}+2x+10+1}{x^{2}+2x+10}\ dx = 2\int \left(1+\frac{1}{x^{2}+2x+10}\right) \ dx}\]

[quote="Kent Merryfield"]On the first one: note that completing the square is going to give you a term of $(x-1)^{2}.$ Looking ahead to that, make the substitution $u=x-1.$ $dx$ becomes $du.$ $x-1$ becomes $u.$ and $x^{2}-2x-24=(x-1)^{2}-25=u^{2}-25.$ So now you have

$\int\frac{5\,du}{u\sqrt{u^{2}-25}}.$

Now that it's "cleaned up" a little, look at it as if it were a new problem. Perhaps a trig substitution (although that's not the only idea)?[/quote]

Hmm, well now that its cleaned up, I suppose we can do the following:

$5\int\frac{du}{u\sqrt{u^{2}-25}}$

and since $\int\frac{du}{u\sqrt{u^{2}-a^{2}}}= \frac{1}{a}\sec^{-1}\frac{|u|}{a}+C$, we let $a=5$ and $u=x-1$.

Therefore, our the answer to our integral is equivalent to:

$\sec^{-1}|(x-1)|+C$

Is this right? Doesn't seem to be - but I dunno...

For the second one,

$\int\frac{2x^{2}+4x+22}{x^{2}+2x+10}\ dx = 2\int{\frac{x^{2}+2x+10+1}{x^{2}+2x+10}\ dx = 2\int \left(1+\frac{1}{x^{2}+2x+10}\right) \ dx = 2\int dx+2\int{\frac{1}{x^{2}+2x+10}}dx}$

So at this point, would I integrate the first part as is and then the second part using u-substitution? Or do we use a u-substitution on the entire integral? I know that $2\int dx = 2x$. But for the second integral, would I now use u-substitution, where $u = x+1$? So:

$2\int{\frac{1}{x^{2}+2x+10}}dx = 2\int{\frac{du}{(u+1)^{2}+3^{2}}}= \frac{2\tan^{-1}\left(\frac{u}{3}\right)}{3}+C$

Substituting $x+1$ for $u$, we get $\frac{2\tan^{-1}\left(\frac{x+1}{3}\right)}{3}+C$, and combining it with the first integral we get $2x+\frac{2\tan^{-1}\left(\frac{x+1}{3}\right)}{3}+C$

The second one is correct. The first one should be

$\frac15\sec^{-1}\left|\frac{x-1}5\right|+C,$

for $x>6$ or $x<-4.$

Wait - where does the absolute value sign go then? The formula is:

$\int\frac{du}{u\sqrt{u^{2}-a^{2}}}= \frac{1}{a}\sec^{-1}\frac{|u|}{a}+C$

Which means that the absolute value needs to be placed somewhere right?

[quote="Kent Merryfield"]On second thought, it does work to have the absolute value there. I'll go back and edit.[/quote]

Two final questions - does the absolute value have to go over the entire fraction or just over $u$. Also, when you say "for
$x > 6$ or $x <-4$, do I have to include this in the solution? Or would it be okay to leave this part out.

[quote]does the absolute value have to go over the entire fraction or just over $u.$[/quote]
Come on - just look at that and think for yourself. Exactly how much difference does it make?

[quote]do I have to include this in the solution? Or would it be okay to leave this part out.[/quote]
I think you're asking whether I would deduct points for omitting that on a test paper. Probably not (it is, after all, implicit in the domain of the arcsecant), but it is something you should be aware of. I think you should focus your attention less on doing things in some officially approved manner and more on getting at what things mean.

Let's put this another way. Suppose you are asked the following:

$\int_{-10}^{26}{\frac{{5}}{(x-1)\sqrt{x^{2}-2x-24}}dx}$

How are you going to approach it? Are you going to grab your pencil and start scratching away, trying to get the formulas right? Or are you going to back away and see that the question is complete nonsense, as the function is not defined on a rather large chunk in the middle of the interval $[-10,26]$?

That antiderivative formula with the absolute value is really two entirely different formulas, one for positive $u,$ one for negative $u.$ The absolute value is a minor convenience that allows us to write these two different formulas on one line, but that doesn't change that fact that it's still two formulas and that separate parts of the domain require separate treatment.


The ozonolysis of the substance $ \text{C}_{10} \text{H}_{16} \text{O}_{4}$(A) leads to the sole product $ \text{C}_5\text{H}_8\text{O}_3$(B).The hydrolysis of B yields an acid(C) and an alcohol(D).On hetaing C we get $ \text{CO}_2$ and the alcohol D gives positive iodoform test.if the compound A is subected to hydrolysis an acid E will result which on prolonged hesting gives $ \text{C}_6\text{H}_6 \text{O}_3$(F).The compound F can react with $ \text{H}_2\text{O}$ to form an acid G which is an isomer of E but with different Melting Point.Identify all the compounds from $ \text{A} \rightarrow \text{F}$ with necessary reactions and explainations.
Also what is the product of acid G with cold Alk. $ \text{KMnO}_4$.

This is a very nice and instructive problem. What is the source? Do you have more problems like this one?

Source:A test
Yes will be getting to solve many such problems .Surely will post them here

Finding the structure of A seems to be the easy part. Isn't the hydrolysis of A in acidic medium?

what i couldn't get is if there is only one product then how come that has $ 3$ $ \text{O}$ because the cleavage should have been symmetric
also i don't know if the medium is acidic this was the data provided

[quote="Pardesi"]how come that has $ 3$  $ \text{O}$ because the cleavage should have been symmetric[/quote]

Hint: what happens in an ozonolysis reaction?

OMG that was the most stupid doubt i fussed up this question yesterday because of this  :rotfl:  :rotfl:  :rotfl:

Now answer the following three questions:

1) The hydrolysis of B gives an acid and an alcohol. What does this tell us about B?

2) On heating acid C carbon dioxide is released. What does this mean about the structure of C?

3) Alcohol D gives a + iodoform test. So what must be the general structure of D?

1 is an ester 3 has methyl group at alpha position
but what about 2?

There are two principal types of acids that loose easily carbon dioxide on heating (at about 100ºC): beta-ceto acids and glycidic acids.


i have similar question but a bit complicated looking

find all functions which are R->R
and satisfy    f( xf(x) + f(y) ) =[f(x)]^2  + y

here read a^n as ''a raised to the power n''

We want to find all $ f: \mathbb{R}\rightarrow\mathbb{R}$ with $ f(xf(x)\plus{}f(y))\equal{}f(x)^2\plus{}y$.
This should not be hard.

If we fix $ x$, we see $ f$ is bijective. So there exists $ f(a)\equal{}0$.
Now we compare $ P(0,y)$ and $ P(a,y)$, we will get $ f(0)^2\plus{}y\equal{}0\plus{}y\Rightarrow f(0)\equal{}0$.

$ P(0,y): f(f(y))\equal{}y$
We have $ zf(z)\equal{}yf(y)$, where $ z\equal{}f(y)$.

$ P(x,0): f(xf(x))\equal{}f(x)^2$.
$ af(a)\equal{}bf(b)\Rightarrow f(a)\equal{}\pm f(b)$.

Hence, we have for $ z\equal{}f(y)$,
$ f(y)\equal{}\pm f(z)\equal{}\pm y$

Suppose we have $ f(a)\equal{}a$ and $ f(b)\equal{}\minus{}b$ for $ a,b\ne0$ simultaneously.
$ P(a,b): f(a^2\minus{}b)\equal{}a^2\plus{}b\Rightarrow a^2\minus{}b\equal{}\pm(a^2\plus{}b)$.
Then we either have $ a\equal{}0$ or $ b\equal{}0$. Contradiction!

Hence, $ f(x)\equal{}x$ or $ f(x)\equal{}\minus{}x$.

[quote="xxp2000"]We want to find all $ f: \mathbb{R}\rightarrow\mathbb{R}$ with $ f(xf(x) \plus{} f(y)) \equal{} f(x)^2 \plus{} y$.
This should not be hard.

If we fix $ x$, we see $ f$ is bijective. So there exists $ f(a) \equal{} 0$.
Now we compare $ P(0,y)$ and $ P(a,y)$, we will get $ f(0)^2 \plus{} y \equal{} 0 \plus{} y\Rightarrow f(0) \equal{} 0$.

$ P(0,y): f(f(y)) \equal{} y$
We have $ zf(z) \equal{} yf(y)$, where $ z \equal{} f(y)$.

$ P(x,0): f(xf(x)) \equal{} f(x)^2$.
$ af(a) \equal{} bf(b)\Rightarrow f(a) \equal{} \pm f(b)$.

Hence, we have for $ z \equal{} f(y)$,
$ f(y) \equal{} \pm f(z) \equal{} \pm y$

Suppose we have $ f(a) \equal{} a$ and $ f(b) \equal{} \minus{} b$ for $ a,b\ne0$ simultaneously.
$ P(a,b): f(a^2 \minus{} b) \equal{} a^2 \plus{} b\Rightarrow a^2 \minus{} b \equal{} \pm(a^2 \plus{} b)$.
Then we either have $ a \equal{} 0$ or $ b \equal{} 0$. Contradiction!

Hence, $ f(x) \equal{} x$ or $ f(x) \equal{} \minus{} x$.[/quote]

But how do you know it's bijective? And what is $ b$?

Let $ x\equal{}0$ and $ a\equal{}f(0)$, we have
$ f(f(y))\equal{}y\plus{}a^2$.
Looking at the RHS, $ f$ can take any real value and is surjective.
If $ f(b)\equal{}f(c)$, we will have
$ f(f(b))\equal{}f(f(c))\equal{}b\plus{}a^2\equal{}c\plus{}a^2\Rightarrow b\equal{}c$. So $ f$ is injective.

Every time you see something like $ f(f(y))\equal{}y$, you can try if $ f$ is bijective.

For your second question, you might get confused on the notation.

I first let $ a$ solve $ f(a)\equal{}0$, from which we derive $ f(0)\equal{}0$.

Then $ a$ and $ b$ are arbitrary values. (This might be the confusing part)
For example, we show, if $ af(a)\equal{}bf(b)$ for some $ a,b$, we can prove $ f(a)\equal{}\pm f(b)$.

I wish this helps.


geometry
circumcircle
parallelogram
geometry proposed

A line $\ell$ is tangent to the excircle of the triangle $ABC$ at $B$. Let $T$ be the foot of perpendicular from the orthocenter of the triangle $ABC$ onto line $\ell$. Let $M$ be the midpoint of the side $AC$. Prove that the triangle $BTM$ is isosceles

I'm sure you meant circumcircle instead of excircle.

The projection of $N$ (the nine-point center) on $\ell$ is the midpoint of $BT$ (because $N$ is the midpoint of $OH$), and we have to prove the same for $M$, so the problem is equivalent to showing that $MN\perp BT\iff MN\|OB$. Let $X$ be the midpoint of $HB$. From the fact that $XBOM$ is a parallelogram, the conclusion follows.

That's a real quickie. BTW, it was already posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=3875 and solved by... well, guess whom.

  Darij


function
real analysis
algebra
domain
real analysis unsolved

How to find a functions $f_{i},f\in L^{p}(\mathbb{R}$, $1\leq p<\infty$ such that $f_{i}\to f$ uniformly but $(f_{i})$ does not converge in $L^{p}$? (with respect to Lebesgue measure space)

You need to take advantage of the infinite measure of the domain, as you wouldn't be able to do this in a space of finite measure. So something has to go on as you go to infinity.

My suggestion: let $f_{n}(x)=\begin{cases}c_{n}&0\le x\le n\\ 0&\text{otherwise}\end{cases}$

The idea is that we're going to have $f_{n}$ converge uniformly to zero. Now I'll let you fiddle with $c_{n}$ until it becomes the example you want.


What is the value of $a$ for wich $P(x)=x^3-x^2-x+a$ , $a$ form $R$ has :
A) a double root that is integer x1=x2
B) a triple root that is integer x1=x3=x3

in general if a polynomial $f$ has a root $r$ of multiplicity $m$, then $r$ is a root of multiplicity $m-1$ of $f'$

B) $P'(x) = 3x^2-2x-1 = (3x+1)(x-1)$ has roots $-1/3$ and $1$: no roots of multiplicity 2. so there is no such $a$ in this case.
A)  since we're concerned with integer roots the only possibility is $P(x) = (x-1)^2(x+a)$. expanding this we get

$(x^2-2x+1)(x+a) = x^3+(a-2)x^2+(-2a+1)x+a$,

and by equating coefficients, $a = 1$


search
pigeonhole principle
number theory proposed
number theory

Find the smallest value of n for which the following statement is true:

Out of every set of n positive integers one can always choose seven numbers whose sum is divisible by 7.

By a theorem of Erdos, among $ 2n\minus{}1$ positive integers, we can always choose $ n$ of them whose sum is divisible by $ n$. So the answer is at most 13. I think it's not difficult to prove that 12 is not enough.

Can you tell me a link for this theorem of Erdos please  (and perhaps a demonstration) ?

This theorem is called the [url=http://planetmath.org/encyclopedia/EGZTheorem.html]Erdős-Ginzburg-Ziv[/url], or [url=http://en.wikipedia.org/wiki/Zero-sum_problems]EGZ[/url] [url=http://www.mathlinks.ro/viewtopic.php?t=14797]theorem[/url]. $ 12$ integers (or $ n\minus{}1$ for the more general case) are not enough because we can take $ \{\overbrace{1, \dots, 1}^6,\overbrace{7, \dots, 7}^6\}$ where clearly no sum of $ 7$ elements is divisible by $ 7$.

We can also use this theorem to solve [url=http://www.mathlinks.ro/viewtopic.php?search_id=717150361&t=247568]some[/url] [url=http://www.mathlinks.ro/viewtopic.php?search_id=799938950&t=54868]Pigeonhole-looking[/url] [url=http://www.mathlinks.ro/viewtopic.php?search_id=475462728&t=57546]problems[/url]. It has an [url=http://www.mathlinks.ro/viewtopic.php?search_id=470764114&t=19496]extension[/url]. It is actually very strong when applicable.


I don't actually read or speak Spanish, but I have a calendar published by Universidad Autónoma del Estado de Morelos that has a problem for every day of the year. Some of them are pretty good. I'll drop by every once in a while to post one of them.

En un tablero de ajedrez de $8\times8$, el rey se encuentra en el cuadro que está el la esquina inferior izquierda. ¿De cuántas formas puede llegar a la esquina superior izquierda en exactamente 7 movimentos? (Recuerda que el rey mueve de uno en uno.)

De alguna manera este problema me recuerda al problema A5 de la Putnam del 2003.

En fin, me parece que se puede reformular el problema como encontrar el numero de soluciones de la ecuación $x_1+x_2+x_3+x_4+x_5+x_6+x_7=0$ donde $x_i\in\{-1,0,1\}$ y para todo $i$ se cumple que $x_1+x_2+\cdot+x_i\geq0$.

Pense que era mucho mas sencillo, pero, salvo por el ataque por fuerza bruta, no se me ocurre ningún procedimiento rápido. Combi nunca ha sido mi fuerte :blush: .

Por solicitud del mismo Kent Merryfield, el problema ha sido movido a la seccion de Olimpiadas de Matematica.

Primero veamos que es posible hallar la cantidad de caminos en el caso dado por un conteo manual refinado: escribimos $1$ en la casilla inicial, y para cada fila $i$, en la fila $i+1$, escribimos el número de $(k, i)$ en $(k-1, i+1); (k, i+1); (k+1, i+1)$ salvo que $k-1$ sea cero. Como en cada casilla de la fila $i+1$ puede haber escritos varios números, si esto ocurre los reemplazamos por la suma. Repetimos hasta llegar a la fila 8, pero es asqueroso. El procedimiento da como resultado 127, si no tuve errores de aritmética. Para el caso de $n$ algunas ideas:

Consideramos caminos que van de la casilla $(1, 1)$ a $(1, n+1)$ en $n$ movimientos, donde el tablero tiene un límite a la izquiera de $(1, 1)$ pero puede ser infinito verticalmente y hacia la derecha.

Decimos que un camino de n movimientos es "de tipo k" cuando pasa por exactamente k casillas de la columna 1, sin contar $(1, 1)$ y $(1, n+1)$ (o sea que un camino puede ser de tipo $0, 1, ..., n-1$). Veamos que entonces un camino de tipo k es una unión de k+1 caminos de tipo 0. Además, un camino de tipo 0 y $i$ movimientos es un camino legal de $i-2$ movimientos. Sea $a(n, k)$ la cantidad de caminos de $n$ movimientos de tipo $k$, y sea $a(n)$ la cantidad de caminos de $n$ movimientos. Entonces la suma que queremos hallar es,

\[a(n) = a(n, 0)+a(n, 1)+...+a(n, n-1) \\
a(n) = a(n-2)+a(n-1, 0)a(1, 0)+a(n-2, 0)a(2, 0)+...+a(1, 0)a(n-1, 0)+a(n, 2)+...+a(n, n-1) \\
a(n)=a(n-2)+a(n-3)0+a(n-4)1+a(n-5)a(1)+...+a(n-3)0+a(n, 2)+...+a(n, n-1)\]

que puede seguir descomponiéndose recursivamente, pero no parece dar una fórmula linda...


are there left-handed at Scuola Normale Superiore Pisa?

mmmh...
interesting question...
Are you a believer of the theory which connects mathematical capabilities with being left-handed?
I don't know much about, but according to me it is only a thing which involves some popular folklore.
I can answer you about Sissa (or Isas); not a lot of people are left-handed, 3 or 4 in a group of 40 people. It is in the average considering the fact that after the 60s the concept of "the left hand is the hand of the devil' is collapsed and in the primary/secondary school no one was forced to use the right hand even if he (or she) was left-handed.

By the way, an interesting thing is that, according to research, LEFT-handed people tend to use more of their RIGHT brain, which is the one devoted to artistical creation, whereas the left part of the brain (used mainly by right-handed people) is the one devoted to logical reasoning.

But I think that's not really imperative: I've known much left-handed people with no mathematical abilities at all and right-handed people painting and drawing like professionals.

[quote="mattilgale"]i know that 3/6 that the italian olympic team team at the 2005 IMO were left-handed...

sorry for my english[/quote]

i was been at IMO2005 with my team but i haven't seen left-handed in italian team... but i'm not sure  :D   who are the olimpic contestants in Italy who used left hand?

I think left-handed are very very  good in maths.. my theory is based on my experience. :)

I know a lot of people who are not... much more than who are... (i can assure you that i know a large amount of people from everywhere in the world, so my statistic is well-based)

i've 7 lefthanded people in my classroom and i'm ambidextrous...i can say that none of those lefthanded is good at maths

but who of italian team is left-handed??
who of scuola normale superiore students is left-handed?

I was been at IMO2005 in the italian team and I'm left-handed... I'm not very sure, but I think that the other left-handed are Luca Barbieri (29 points, silver medal) and Maria Colombo (19 points, bronze medal, sixth girl in the world). I don't remember about any russian member...did you see us?

I don't know if it is true or it is folklore...
but I'm sure that at the stage senior 2005 there are a lot of left-handed, more than the normal %

Being left handed or right handed doesn't make much of a difference in your mathamatical capabilites. But it is true that a person uses the left side of his/her brain when doing math.  :coolspeak:


A is an (m x n) matrix. B is an (n x m) matrix and n < m. Show that C=AB is not invertible.

the columns of C will be combinations of the n columns of A
thus they aren't m linear independent columns
thus C is not  invertible

Another way of saying this: $\text{rank}(AB)\le\min(\text{rank}(A),\text{rank}(B)).$

If $n<m,$ then the rank of either an $n\times m$ or an $m\times n$ matrix cannot exceed $n.$ Thus $\text{rank}(AB)\le n<m.$ But the rank of an invertible $m\times m$ matrix must be $m.$

The language may look different, but this is really the same thing that fredbel6 said.


induction
geometric sequence
algebra proposed
algebra

How can you find it , Kunny ? It's very nice !

@Romano.
[quote="kunny"]Solving the recursive formula without using induction, $x_n=1+\frac{2}{3^{n+1}-1}$[/quote] From $x_{n+1}=\frac{2x_n+1}{x_n+2}$ we obtain $\frac{x_{n+1}+1}{x_{n+1}-1}=3\cdot \frac{x_n+1}{x_n-1}$. Hence the result.

[quote="romano"]Nice solution , Kunny and Myth !
  Did you guess it ? How can you guess it ?[/quote]
Suppose we have $x_{n+1}=\frac {ax_n+b}{cx_n+d}$ and let $u,v$ be the roots of the equation $cx^2+(d-a)x-b=0$. Furthermore, suppose $u,v$ are real and distinct numbers.
Then, if $y_n=\frac {x_n-u}{x_n-v}$ we have
\[
y_{n+1}=\frac {x_{n+1}-u}{x_{n+1}-v}=\frac {(ax_n+b)-u(cx_n+d)}{(ax_n+b)-v(cx_n+d)}=
\]
\[
\frac {(a-uc)x_n+b-ud}{(a-vc)x_n+b-vd}=\frac {(a-uc)(x_n-u)}{(a-vc)(x_n-v)}=\frac{(a-uc)}{(a-vc)}y_n.
\]
Thus, $(y_n)$ is a geometric sequence and the rest is easy.
Note that this works if $x_0\ne u,v$.

[quote="romano"]Nice solution , Kunny and Myth !
  Did you guess it ? How can you guess it ?[/quote]

$x_{n+1}-1=\frac{2x_n+1}{x_n+2}-1=\frac{x_n-1}{x_n+2}$,so

$\frac{1}{x_{n+1}-1}=\frac{x_n+2}{x_n-1}=1+\frac{3}{x_n-1}$

Set $y_n=\frac{1}{x_n-1}$, we have $y_{n+1}=3y_{n}+1$


function
calculus
derivative
trigonometry
probability
topology
calculus computations

An interesting question:

[color=blue]Problem[/color] Do we have a real function like $f$ that is continuous on interval $I$ but its not differentiable at any point in that interval.

Edit: Strange, I just saw the other topic with the same name, Sorry for posting this.
http://www.mathlinks.ro/Forum/viewtopic.php?t=122547
How come two people think of a same thing at the same time ? :)

Suppose we have set $A$ which is a subset of set $B.$ In general, how can we prove that $A\ne B?$ There are two main methods of proof: by explicit construction or by size. To execute this by explicit construction, you would need to build an element of $B\setminus A.$

Now, what about size? A proof by size shows that $B$ is in some sense "larger" than $A$ and hence cannot be the same set. I can think of three different notions of size: (1) cardinality, (2) Baire category, or (3) measure.

As an illustration, consider $A$ to be the set of real algebraic numbers and $B=\mathbb{R}.$ To ask if $A$ is not all of $B$ is to ask whether transcendental numbers exists.

You can prove this by explicit construction: can you show that $\sum_{n=0}^{\infty}10^{-n!}$ is transcendental? Or (much, much more difficult), can we prove that either $e$ or $\pi$ is transcendental?

Or you can prove this by size - in this case, by cardinality. The set $A$ of algebraic numbers is countable, while the set $B$ of real numbers is uncountable. The proof by size fails to produce a single specific instance of a transcendental number but it does show that "most" real numbers are transcendental.

---

Now, what about the case at hand? $B$ is the set of continuous functions. (We can make that the continuous functions on some interval if we wish.) $A$ is the set of continuous functions that have a derivative at at least one point. How can we prove that $A$ is not $B?$

There is explicit construction. My favorite for this is

$\sum_{n=1}^{\infty}\left(\frac34\right)^{n}\cos\left(4^{n}x\right)$

The example is essentially due to Weierstrass. Others may have other favorite examples, using the same basic ideas.

But what about size? Can we prove this by size?

Cardinality is a non-starter. The sets $A$ and $B$ have the same cardinality. (Note that $A$ contains the set of all antiderivatives of elements of $B.$) But that leaves us two other versions of size to try, and they both work.

It is possible to show that in the complete metric space $C[a,b],$ the set of functions with even a one-sided derivative at any point is the countable union of nowhere dense subsets. So we can use Baire category. (I don't know this proof myself, but I do know that it appears in some textbooks.)

What about measure? The argument here uses Brownian motion. You can think of Brownian motion as a way to assign a probability measure to the set of continuous functions. (Continuous functions with a particular initial condition, but we'll have no problem making that general.) In that probability space, the functions with any derivatives anywhere are a subset of probability zero.

Both "size" proofs then show that in some sense, "most" continuous functions are nowhere differentible.


I would like to change the font size of a certain text fragment in my document, without changing the font size of my entire document. I know the commands \large, \Huge etc., but I wondered if I couldn't make it even larger (or arbitrarly large/small). Normally, there should be way to do this, because I've seen a few documents with really HUGE titles  :D

Thanks!

Posted many times:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=74117[/url]

 :)

I'm sorry, but [quote="ssn116"]Hi,
If you want to change the font size of the chapter/section headings (I couldn't make out exactly what it was that you needed), an easy way to do it is to use the titlesec package. For example, for Large section titles, use:
[code]
\titleformat{\section}[hang]{\center\Large}{\Large\thesection}{12 pt}{}{}
[/code]
[/quote]
isn't really what I was looking for. I don't want to change the font of all my chapter/section/subsection titles, but only a few words. You know, first page, really huge title. But only that title, not any other in the document. Isn't there a command, like, dunno, \fontsize[30pt]{Mathlinks} ?

Did you read the hole topic, 2 last replays? :huh:

Tu es fénéant je vois! :wink:

Non, je n'suis pas fénéant  :D

I've read the last two replies as well, but this is what the extsizes-read-me says: [quote]How to use extsizes.
Your documentclass command should look like:
	\documentclass[14pt]{extreport}
or
	\documentclass[9pt]{extarticle}
The sizes available are 8pt, 9pt, 10pt, 11pt, 12pt, 14pt, 17pt, and 20pt.
There should be no need to change any other part of your document.
[/quote]

Therefore, I assume that, once again, it's just for formatting the WHOLE document...

Désolé pour le retard...

Try with the scalfnt package. To use it have a look at [url=http://www.mackichan.com/index.html?techtalk/517.htm~mainFrame]this[/url] (Again something to rad :D )

Bon courage!


Find all triles $ (x,y,z)$ of integers such that $ x^2(y\minus{}z)\plus{}y^2(z\minus{}x)\plus{}z^2(x\minus{}y)\equal{}2$

Just the main hint (the other part is only a boring casework
[hide]Notice that $ x^2(y \minus{} z) \plus{} y^2(z \minus{} x) \plus{} z^2(x \minus{} y) \equal{} (x \minus{} y)(x \minus{} z)(y \minus{} z)$[/hide]

I would like to continue what silouna has left. So here we are:
Observe that $ (x\minus{}y)\plus{}(y\minus{}z)\equal{}x\minus{}z$. On ther other hand, $ 2$ can be written as a product of three distinct integers. So we have three posibilities:

$ x\minus{}y\equal{}1$
$ x\minus{}z\equal{}2$
$ y\minus{}z\equal{}1$   Thus we have $ (x,y,z)\equal{}(k\plus{}1,k,k\minus{}1)$ for some integer $ k$,

$ x\minus{}y\equal{}\minus{}2$
$ x\minus{}z\equal{}\minus{}1$
$ y\minus{}z\equal{}1$  From this we have $ (x,y,z)\equal{}(k\minus{}1,k\plus{}1,k)$ for some integer $ k$,

$ x\minus{}y\equal{}1$
$ x\minus{}z\equal{}\minus{}1$
$ y\minus{}z\equal{}\minus{}2$  And finally we have $ (x,y,z)\equal{}(k,k\minus{}1,k\plus{}1)$ for some integer $ k$.

[i]Very goog thought silouan. Bravo[/i]!! :)


[quote="Mathias_DK"][quote="geniusbliss"]What is the validity of the assumption - say for 3 varibles a,b,c .. assuming that $ a\ge b\ge c$ for symmetric inequalities.I know it that it sounds correct logically but is there a rigourous logical proof? ... also watever the proof is why does it make the same assumption invalid in cyclic inequalities.. and wat is the proof for assuming a certain variable to be the maximal or minimal element or middle..?

i hope someone gives a rigourous proof(which i hope exists..)[/quote]
Assume that $ f(x,y,z)$ is symmetric.
Then $ f(x,y,z) \equal{} f(x,z,y) \equal{} f(y,x,z) \equal{} f(y,z,x) \equal{} f(z,x,y) \equal{} f(z,y,x)$.
Now let $ (a',b',c')$ be any permutation of $ (a,b,c)$ such that $ a' \ge b' \ge c'$, then $ f(a',b',c') \equal{} f(a,b,c)$.
So if we need to prove $ f(a,b,c) \ge 0$ it is equivalent to proving $ f(a',b',c') \ge 0$ for all $ a' \ge b' \ge c'$, i.e. $ f(a,b,c) \ge 0 \forall a,b,c \iff f(a,b,c) \ge 0 \forall a \ge b \ge c$.
If $ g(x,y,z)$ is cyclic then we only know that $ g(x,y,z) \equal{} g(z,x,y) \equal{} g(y,z,x)$, so we cannot use the same argument.
If we need to prove $ g(a,b,c) \ge 0$, then consider $ (a',b',c') \equal{} (b,c,a)$ and $ (a'',b'',c'') \equal{} (c,a,b)$:
Then $ g(a,b,c) \equal{} g(a',b',c') \equal{} g(a'',b'',c'')$
But we know that: $ a \ge b,c$ or $ b \ge a,c$ or $ c \ge a,b$ which is equivalent to:
$ a \ge b,c$ or $ a' \ge b',c'$ or $ a'' \ge b'',c''$.

So in order to prove $ g(a,b,c) \ge 0 \forall a,b,c$ it is sufficient to prove that one of these hold:
$ g(a,b,c) \ge 0$ or $ g(a',b',c') \ge 0$ or $ g(a'',b'',c'') \ge 0$.
Since either $ a \ge b,c$ or $ a' \ge b',c'$ or $ a'' \ge b'',c''$ it is enough to prove $ g(a,b,c) \ge 0 \forall a \ge b,c$.[/quote]
thanks a lot mathias !


2^200-(2^192)*31+2^n is a perfect square. Find value of n. :?:

euclid, please avoid typos!  The first time I read through this, I thought it was somehow wrong... anyway, revised:

[quote="Euclid_great"]$ 2^{200} - 2^{192}\cdot 31 + 2^n = 2^{192}(2^8 - 31) + 2^n$

=$ 2^{192}(256 - 31) + 2^n$

=$ 2^{192}\cdot 225 + 2^n$

$ \therefore$ for some m $ \in \mathbb N$

$ 2^n = m^2 - 2^{192}\cdot 225$

$ \Rightarrow 2^n = (m - 2^{96}\cdot 15)(m + 2^{96}\cdot 15)$

$ m - 2^{96}\cdot 15 = 2^{\alpha}$ and $ m + 2^{96}\cdot 15 = 2^{\alpha + \beta}$(For some non negative integers $ \alpha,\beta)$

Hence $ 2^{97}\cdot 15 = 2^{\alpha + \beta} - 2^{\alpha} = 2^{\alpha}(2^{\beta} - 1)$

$ \Rightarrow 2^{\alpha} = 2^{97}\ and\ 2^{\beta} - 1 = 15$

$ \alpha = 97$ and $ \beta = 4$

$ \therefore n = 2\alpha + \beta = 198$

DONE!!![/quote]

[quote="Euclid_great"]I am not  good at typing.  :(  :(

So the correct answer:

$ 2^{200} - 2^{192}\cdot 31 + 2^n = 2^{192}(2^8 - 31) + 2^n$

=$ 2^{192}(256 - 31) + 2^n$

=$ 2^{192}\cdot 225 + 2^n$

$ \therefore$ for some m $ \in \mathbb N$

$ 2^n = m^2 - 2^{192}\cdot 225$

$ \Rightarrow 2^n = (m - 2^{96}\cdot 15)(m + 2^{96}\cdot 15)$

$ m - 2^{96}\cdot 15 = 2^{\alpha}$ and $ m + 2^{96}\cdot 15 = 2^{\alpha + \beta}$(For some non negative integers $ \alpha,\beta)$

Hence $ 2^{97}\cdot 15 = 2^{\alpha + \beta} - 2^{\alpha} = 2^{\alpha}(2^{\beta} - 1)$

$ \Rightarrow 2^{\alpha} = 2^{97}\ and\ 2^{\beta} - 1 = 15$

$ \alpha = 97$ and $ \beta = 4$

$ \therefore n = 2\alpha + \beta = 198$

DONE!!![/quote]

An objection!!!!!!!
You haven't proved that $ 2^{\alpha} = m^2 - 2^{96}\cdot 15$
and the other one also.

Lakshya:

$ m \minus{} 2^{96}15 \in \mathbb{Z}$ and $ m \plus{} 2^{96}15 \in \mathbb{Z}$, so both are integral divisors of $ 2^n$.  Now what form could an integral divisor of $ 2^n$ take on?  How about $ 2^a$, with $ a \le n$?