Find all triangles such that $a^2+b^2+c^2 = 8R^2$, where $R$ is the circumradius.

By sinelaw, $LHS = 4\sum R^2sin^2A$, whereby we need $2 = \sum sin^2A$.

This is the same thing as $1 = \sum cos^2(A)$.

--
difficult for me to proceed here..

ya, now we have $cosAcosBcosC=0$ which is easy to solve.

[quote="siuhochung"][quote="Singular"]can you explain how to get from $\sum cos^2A =  1$ to $\prod cosA = 0$ ?[/quote]
$\sum cos^2 A = 1-2\prod cosA$[/quote]
Does it mean: $\cos^2{A}+\cos^2{B}+\cos^2{C}=1-2\cos{A}\cos{B}\cos{C}$?

For any triangle $\triangle ABC$, the distance $OH$ between the circumcenter $O$ and the orthocenter $H$ is given by

$OH^2 = 9R^2 - (a^2 + b^2 + c^2)$

If $a^2 + b^2 + c^2 = 8R^2$, then $OH^2 = R^2,\ OH = R$, i.e., the orthocenter lies on the triangle circumcircle. For an acute angle triangle, both the circumcenter and orthocenter lie in the triangle interior, hence $OH < R$. For an obtuse angle triangle with the obtuse angle $\angle C > 90^o$, the circumcenter lies on the opposite side of the line $AB$ than the vertex $C$, while the orthocenter lies in the plane sector bounded by the rays opposite to the obtuse angle rays $CB, CA$, hence, $OH > R$. $OH = R$ is possible if and only if the triangle $\triangle ABC$ has a right angle.

Well, in my vain attempts to solve this, I did manage to find that it was equivalent to another interesting identity:
$\frac{\cos{A}}{b}+\frac{\cos{B}}{b}+\frac{\cos{C}}{c}=\frac{abc}{4A^2}$, with A being the area.

One of these days I'm going to have to sit down and learn all those damn triangle trigonometric relations.

Sorry for my typo earlier. I've edited it.

Sam, the identity being discussed was on one of our homework sheets for ARML! It was something like, bla bla bla about $\arccos(\alpha),\arccos(\beta),\arccos(\gamma)$ and then that led to $\alpha, \beta, \gamma$ are angles of a triangle and then from there it said prove: $\alpha^2 + \beta^2 + \gamma^2 = 1 - 2\alpha \beta \gamma$ which just so happens to be that identity.

[quote="warut_suk"]We have $\cos^2{A}+\cos^2{B}+\cos^2{C}=1-2\cos{A}\cos{B}\cos{C}$. Do we have something similar for $\sin^2{A}+\sin^2{B}+\sin^2{C}$?[/quote]

what are you looking for?
If you accept the existence of $cos{A}$, then why don't you directly consider $\sum sin^2 A = 3-\sum cos^2 A$?

I mean, can $\sum sin^2 A$ be written in form $k\prod sinA +$ something?

I'm too lazy to try to find it  :D


Hi everybody,

I do not know how to insert directly an image so here you can see my game

Your Download-Link #1: http://rapidshare.de/files/40580486/primemag1.bmp.html

In the first column you have the prime numbers a(j) 3,5,7,11 et 13

In the first line you have what I call the solutions a(i) 53,59,61,67,1543 (all the 5 numbers are prime too)

a(i,j)=a(i)mod(a(j))

So the goal of the game is to find the 5 numbers solutions such as :
- if you concatenate in the same line a(i,1) the number has to be prime
-all the numbers have to be primes 22111,34123,43543,94613,17929.

Can you find the solutions with a(j) = (17,19,23,29,31) instead of (3,5,7,11,13)?

Right!
Congratulations!!!
I'm trying to find a solution for both (lines and columns) primes.
It will not be easy.

5023, 39563, 63367, 220667, 315521

Actually, I don't find the solutions myself, computer programming helped me..  :wink:
So, I can find more solutions

I want to ask you a question..

Should the solutions be in increasing order?

P.S. I attach the initial image from http://rapidshare.de/files/40580486/primemag1.bmp.html, so that everyone can see it without downloading

Thank you for the picture.
I had the idea. I did it with increasing number but we could make solutions more harder by adding other constraints.
If someone can find one solution and only one such as the prime are consecutive (both line-column) or even a(i,j) itself prime or or ....
The game is open. I proposed a game with 5*5 board. Why not  bigger board (e.g. 10*10)? It depends on the density of the solutions.
I find another game more exciting with a lot of combination.
Soon I will post it.


[i] Please , does somebody have a simple solution to this problem ?[/i]

If function $f$ is continous on IR ,  $\int_{x}^{x+1}f(x)dx = f(x)$ for all reals $x$ , and $\lim_{x \to+\infty}f(x)$ exists , prove that $f$ is constant.

[u] Babis[/u]

There I have added a new and much simpler(and more elementary) solution to this problem.
 Babis

Mr Babis I find that $ f(R) \equal{}$ {a} so f is constant


algebra
polynomial
superior algebra
superior algebra unsolved

Prove that for $n\ge 3$,$x^{2^{n}}+x+1$ is reducible over $\mathbb Z_{2}$.

Let $ P(x)=x^{2^{n}}+x+1$.
1)Remark:  If $ n$ is odd ($ =2k+1$) then $ 2^{n}=4^{k}.2=2 mod 3$. Then $ x^{2}+x+1$ is a $ P$-divisor.
2) General case: Let $ Q(x)=x^{2^{2n}}-x$.
a) Lemma: $ P$ is a $ Q$-divisor.
Proof: Let $ K$ be the $ \mathbb{F}_{2}$-algebraic closure. If $ P(a)=0$ with $ a\in{K}$ then $ P'(a)=1$ and $ P$ has only single roots. Then it's sufficient to prove that $ Q(a)=0$: $ a^{2^{n}}=a+1$ then $ a^{2^{2n}}=a^{2^{n}}+1=a$ and $ Q(a)=0$.
b) $ Q$ is the product of all irreducible polynomials, on $ \mathbb{F}_{2}$, of degree $ p$ s.t. $ p$ is a $ 2n$-divisor. Thus $ P$ admits as divisor a polynomial on $ \mathbb{F}_{2}$, of degree $ \leq{2n}$, and therefore is reducible.


for image:
http://img692.imageshack.us/img692/9996/check.gif

Aston and Thomas are doing altitude measurements. Three of the same cone placed on the table. The size of each cone is 120 cm tall and 100 cm in diameter base. Then a sphere of radius 50 cm placed in the middle of the third cone. They want to know the high sphere from the surface of the table measured. The funny thing is, they both did not bring a ruler to measure the height of the sphere (height in question is the distance from the surface of the table to the center of the sphere). Can they get a definite answer? How 'measurement' is? (I put the three cone attached)

Please do not double post in two different threads  :wink:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=310652


This is not phrased in a way that makes for an easy application of induction.  I would suggest finding an equivalent statement that lends itself more easily to the method.  For example, "If the first person in line is a woman and no woman is directly in front of a man, then ...."

Another example is this beautiful idea of the late Bebe Panaitopol
[quote]Determine all sequences $ (a_n)_{n\geq 1}$ of positive real numbers with the property that $ \sum_{k\equal{}1}^n a_k^3 \equal{} \left ( \sum_{k\equal{}1}^n a_k \right )^2$ for all $ n\geq 1$.[/quote]
Prove what by induction? We need a predicate $ P(n)$. After trying low values of $ n$ we may conjecture $ P(n) : \equal{} "a_n \equal{} n"$, and prove it by induction.

[quote="absvalue"][quote]A group of people stand in line to purchase movie tickets. The first person in line is a woman and the last person in line is a man. Use proof by induction to show that somewhere in the line a woman is directly in front of a man.[/quote]

This seems so obvious... but I'm not sure how to prove it using induction. Could someone give me a hint?[/quote]

Induct on the number of people in the line. Base case would be $ n \equal{} 2$.

Assume that in a line of $ k$ people, there is a woman directly in front of a man. Now show that this same property holds in a line of $ k \plus{} 1$ people.

A proof by contradiction seems so much cleaner, though.

[quote="Lazarus"]Induct on the number of people in the line. Base case would be $ n \equal{} 2$.

Assume that in a line of $ k$ people, there is a woman directly in front of a man. Now show that this same property holds in a line of $ k \plus{} 1$ people.

A proof by contradiction seems so much cleaner, though.[/quote]

Thanks for your help! Using your hint, I've worked on it some more:

--------------------------------------
Let N represent the number of people in the line.
Basis step: Let N = 2. If there are only two people in the line, then clearly a woman is directly in front of a man. Our base case is confirmed and we can continue.
Induction hypothesis: Suppose that in a line of k people, a woman is directly in front of a man.
Next, we need to show that in a line of k+1 people, a woman is directly in front of a man.
Since the statement is true for k people, it is impossible to add one more person in such a way that would make the statement fals. Therefore, the statement is true for k + 1 people.
--------------------------------------

Is this all I need to do? Is there any way I can improve this?

Just want to make sure I understand how to tackle these induction questions. Thanks! :)

[quote="absvalue"]Since the statement is true for k people, it is impossible to add one more person in such a way that would make the statement fals. Therefore, the statement is true for k + 1 people. [/quote]  :|  As far as I can tell, this sentence is independent of any actual content of the problem.  Basically, you've done the following:
1) State premises.
2) Prove inductive hypothesis.
3) Say, "This sentence is the inductive step."
4) Conclude by induction.
The sentence I've quoted is step 3 -- there should be some content there, but there isn't.

Thanks. How would I go about improving step 3?

I've already given my advice: the statement as written is not very amenable to an inductive approach.  I would rewrite it in an equivalent form (and I gave a nudge in the direction of one good way of doing so) and then prove that statement by induction.

I think that continuing down the path you've chosen is a bad idea: you don't seem to be sufficiently comfortable with induction to tell the difference between a solid proof and a vague proof-idea that might be solidifiable.  At this level of familiarity, you should focus on proving statements using the cleanest possible structure, and the idea I've suggested leads to a much more natural inductive proof.

I think it's done by showing that lines satisfying JBL's proposition must have no man?

[quote="TZF"]I don't understand why the problem statement isn't conducive to induction.

[hide="Hint"]For a line of more than two people, the second to last person can either be a man or a woman. What can you say in each of these cases?[/hide][/quote]
Ah, that makes sense... thanks! :)

So, would something like this work?

Basis case: Let n=2. Clearly a woman is in front of a man in this case.

Induction hypothesis: Suppose that the statement is true for a line of k people.

Next we need to show that in a line of k+1 people, a woman is directly in front of a man. Consider the second to last position. If this position is held by a woman, then clearly a woman is directly in front of a man, and we are done. On the other hand, suppose this position is held by a man. Remove the last position and the line is identical to a line of k people. By our induction hypothesis, we know that a line of k people has a woman directly in front of a man, and we're done.


Given A={-2,-1,0,1,2,3}, B={1,2,3,4,5}

f is a  function from A to B  satisfying that  x+f(x)+xf(x)=2k+1,where k is an integer.

How many different kinds of such function are there?

Maybe it would help if we made $x+xf(x)+f(x)+1=2k+2$ or $(x+1)(f(x)+1)=2k+2$, so the product $(x+1)(f(x)+1)$ has to be even. So we only need to watch it modulo 2. Then A has the elements modulo 2: $A={0,1,0,1,0,1}$ and $B={1,0,1,0,1}$. Thus $A+1={1,0,1,0,1,0},B+1={0,1,0,1,0}$. Then the product $(A+1)(B+1)$ is even, when we have multiplication with $0$, and that happens exactly $3 \cdot 3 + 3 \cdot 5=9+15=24$ times.


Suppose that the functions $ f, g: \mathbb{Q}\rightarrow\mathbb{Q}$ are strictly increasing functions which attain every rational value. Is it necessarily true that the range of values of the function $ f\plus{}g$ is also the whole set $ \mathbb{Q}$?

No.

Define $ f$ and $ g$ as linear functions in $ \mathbb{Q} \bigcap (] \minus{} \infty, 0] \bigcup [5, \infty[)$ and

$ f(x) \equal{} 1 \minus{} \frac {1}{x \plus{} 1}$
$ g(x) \equal{} 1 \minus{} \frac {1}{x \plus{} 2}$

for $ 0 < x < 5$. Since $ \frac {1}{x}$ is a bijection between $ [a, b] \bigcap \mathbb{Q}$ and $ [\frac {1}{b}, \frac {1}{a}] \bigcap \mathbb{Q}$, both $ f$ and $ g$ can be bijections for convenient "linear extensions". Clearly they are increasing. However,

$ f(x) \plus{} g(x) \equal{} 2 \minus{} \frac {2x \plus{} 3}{x^2 \plus{} 3x \plus{} 2}$

for $ x \in [0, 5]$. Then

$ (f \plus{} g)(0) \equal{} \frac {1}{2}$
$ (f \plus{} g)(5) \equal{} \frac {71}{42}$

If the declaration were (was? were?) true, we would have $ (f \plus{} g)(x) \equal{} 1$ for some rational $ x \in [0, 5]$. Let us see...

$ 2x \plus{} 3 \equal{} x^2 \plus{} 3x \plus{} 2 \Longleftrightarrow x^2 \plus{} x \equal{} 1$.

Absurd!


limit
logarithms
LaTeX
real analysis
real analysis unsolved

explore lower & upper bounds; find also sup and inf of a set:

$ \lbrace \frac{1}{n}\minus{}\frac{1}{n\plus{}1}\plus{}...\plus{}\frac{(\minus{}1)^{k\minus{}1}}{n\plus{}k\minus{}1}\rbrace$        -  $ n, k\in N$

[hide="maybe"]$ S_{n,k}\equal{}(\minus{}1)^n\sum_{j\equal{}n}^{k\plus{}n} \frac{(\minus{}1)^j}{j}$

we want upper and lower bounds of $ X\equal{}\cup_{n\equal{}1}^\infty \cup_{k\equal{}0}^\infty \{ S_{n,k} \}$

so we notice.. for a fixed n

$ S_{n,0} > S_{n,2} > \dots > S_{n,2m} > \dots \ge \lim_{m \to \infty} S_{n,2m} \equal{} (\minus{}1)^{n\minus{}1}[\ln 2\minus{} S_{1,n\minus{}1}]$

$ S_{n,1}<S_{n,2m\minus{}1}<S_{n,2m\plus{}1} < \dots \le \lim_{m \to \infty} S_{n,2m\plus{}1} \equal{} (\minus{}1)^{n\minus{}1}[\ln 2\minus{} S_{1,n\minus{}1}]$

gives us $ S_{n,0} > S_{n,k} > S_{n,1}$ which is nice.

$ \sup X \equal{} \sup_{n \in N} S_{n,0} \equal{} \sup_{n \in N} \frac{1}{n}\equal{} 1$

$ \inf X \equal{} \inf_{n \in N} S_{n,1} \equal{} \inf_{n \in N} \frac{1}{n}\minus{}\frac{1}{n\plus{}1} \equal{} 0$[/hide]

[quote="kenn4000"]$ X \equal{} \cup_{n \equal{} 1}^\infty \cup_{k \equal{} 0}^\infty \{ S_{n,k} \}$[/quote]

Latex suggestion: use \bigcup instead of \cup when you want super- or subscripts.  (This works for a number of symbols: \bigcap, \bigwedge, etc.)

[quote="kenn4000"]gives us $ S_{n,0} > S_{n,k} > S_{n,1}$ which is nice. [/quote]  Would be nice, but doesn't appear to actually be a consequence of what you've written: as far as I can tell, you've shown neither $ S_{n, 2m} > S_{n,1}$ nor $ S_{n, 2m \plus{} 1} < S_{n, 0}$.  (Your answers are correct, though.)


ratio
limit
function
calculus
calculus computations

$ u_0$>0,$ u_1$>0  :
the sequence $ u_n$ with
$ u_{n\plus{}2}$ = 1+$ \frac{u_{n\plus{}1}}{u_n}$
determine all $ u_0$,$ u_1$ such that the sequence is convergen!!!

Numerical evidence strongly suggests that this sequence converges to $ 2$ for all positive $ u_0,u_1.$

However, for the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=232920]other topic[/url] you posted with the ratio in the definition inverted, it appears that this only tends to $ 2$ if it is constantly $ 2,$ which happens if $ u_0 \equal{} u_1 \equal{} 2.$ Otherwise, it seems to tend towards a stable 2-cycle.

You can play exactly the same game as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=229371]here[/url], only now the trapping intervals are of the form $ [a,b]$ with $ a<b$, $ (a\minus{}1)(b\minus{}1)\equal{}1$. Again, you have to stay near the endpoints if you want to preserve the interval. If you have $ u_{n}/u_{n\plus{}1}$, this is feasible and you get $ a,b,a,b,\dots$ in the limit most of the time, but $ u_{n\plus{}1}/u_n$ gives $ a,b,b$ and then a shorter interval. I leave it to somebody else to turn this into a formal argument.

The problem in the other thread, $ u_{n \plus{} 2} \equal{} \frac {u_n}{u_{n \plus{} 1}}$, seems to be the more interesting one.

In that case the set of initial conditions $ u_0,u_1$ for which the limit $ \lim_{n\to \infty}u_n$ exists is nontrivial.  By introducing the function $ f(u_0,u_1) \equal{} \lim_{n\to \infty}u_{2n} \minus{} u_{2n \plus{} 1}$ and proving that $ f$ is continuous, I believe we can show that $ S \equal{} \{ x\in \mathbb{R}^2: f(x) \equal{} 0\}$ is actually a one-dimensional subset of $ \mathbb{R}^2$, not just the singleton $ \{(2,2)\}$.

could you write full solution with this problem??


Let $g_i(x)$ and $h(x)$ be the real polynomials with real coefficients and deg(g)=<4,deg(h)=<2.Find the maximum number of real roots of $\sum\sqrt{g_i(x)}+h(x)=0$.

[quote="Rushil"]Can you give any hint???? :spider:[/quote]
If I were able to solve it, then it would not be in the Open Questions Forum.


probability
AMC
AIME
USAMTS
geometry
geometric transformation
homothety

[quote="mchoi815"]Since no one's doing this...

Pick your Master's round winner.  1/228 chance youll make it right..
Name your competitors too if you feel like it.[/quote]
o_O masters is hard to predict because not everyone gets to do it.

That said, if Wenyu Cao is in Masters, he's gonna win.

Yeah.  1 in 228 chance.  youre the first one to pick.  i have only seen a national masters round. so i'm not very sure who will do well, which is why i didnt pick

[quote="ZhangPeijin"]What's the masters round? :huh:[/quote]
It's a round in which 4 people (top 2 written and cd) are given a difficult topic to answer questions on and give a presentation answering the proposed question.

I think masters round is so hard to predict because it's not really about who knows more.  It's really about who gives the best presentation.  At least that's how I'd judge it.  Last year, basically everyone's presentation sucked.

Wenyu Cao will win it, easily. I won't be surprised at all if he makes it. If he does not, Yuan will most likely win, because they are like the studliest MCers around. But I can assure you all that I will not win. However, if I do not make the masters round, I will have a slightly higher chance of winning than if I actually do make the masters round.

dude that could turn into a conditional probability problem...

I think if I somehow randomly make it, I can explain things very well and make it interesting to boot (crazy kid + humor XD) IF I understand how to do the problem.

Which I probably won't.

ok... lol
why is no one saying kevin chen? but yeah wenyu cao and allen yuan are definite possibilities...

(Shouldn't be talking cause everyone here can pwn me badly)
I'm not sure but I don't know if Kevin Chen can compare with Wenyu or Allen at this type of stuff. What did you get on the AMC12/AIME?

im going with an upset, but i think that 6th grader from ca will win (seriously, no joke)

jorian

[quote="diophantient"][quote="rd5493"]was last years the one about the light buls.  it took me like forever to figure out the formula.  was it?  i predict that either kevin will win (hmmm... every1 seems to know who im talking about) or that annoying yaun kid from michigan who enjoys editing Wikipedia to make it look like he won already this year.  No kidding.  he really edited Wikipedia, but he's changed it back since then.[/quote]
Last year had to do with homothecy.[/quote]

To be more specific or if anybody doesn't know what homothecy is, it was about investigating segments drawn from a certain point to every point on another figure like a square and looking at what the midpoints form when connected.

I think the light bulb one was from All-Time Greatest.  1991?

[quote="rd5493"]was last years the one about the light buls.  it took me like forever to figure out the formula.  was it?  i predict that either kevin will win (hmmm... every1 seems to know who im talking about) or that annoying yaun kid from michigan who enjoys editing Wikipedia to make it look like he won already this year.  No kidding.  he really edited Wikipedia, but he's changed it back since then.[/quote]

Thanks for the respect there...did you know that mentioning that in your post in fact makes you more suspicious than I?

Last year, Daniel Li already studied the subject, so there was a bit of an advantage there.  Anyways...I'd be ecstatic to make Masters.

[quote="13375P34K43V312"]um if you think allen yuan is annoying then you are a sad sad person
first of all anyone can edit the wiki, and you don't have much evidence that it was indeed him

and second toward jhredsox you have absolutely no basis for your prediction, it seems pretty arbitrary to me.

Wenyu Cao had 144 AMC 12 and unknown AIME, I have yet to find out, he had a [b]100[/b] on USAMTS, made the elite eight of the itest ToC (lost to Arnav), and is, all in all, a beast at upper level math, which is what the masters round (somewhat) stresses. His USAMTS scores and four commendations show that he is an extremely effective communicator of mathematics, and unless he has a really quiet or really annoying voice then he will own everyone there too.[/quote]

Towards the first two paragraphs:  Jeez, lighten up!
 :|

[quote="13375P34K43V312"][quote="bpms"][quote="13375P34K43V312"]I actually have three Masters round ideas in mind, I might be able to write up at least two full ones by Nationals, if anyone's interested.[/quote]
lolz, how would we give presentations?
or the people who made nationals could do a practice somewhere hosted by you, where we give our presentations.
That actually is a good idea, we could meet up and do presentations or something there.[/quote]

dude, i hate to say it, but that's a pretty good idea
then one that i wrote last night (in place of attending my own math jam), is actually quite difficult, but Ubemaya would kill it, because it was the basis of a class we both took at MathPath. Except he's not going to MC so yeah.[/quote]

The hypercube master's round was also based on a class at Mathpath, yet it is widely acclaimed as great. The problem was that none of the presenters had gone to Mathpath. :)

[quote="13375P34K43V312"]1995 I believe

Last year's was pretty boring in my opinion, Mathcounts was Mathcounts and gave it a lame name ("midshape"), even though they should have been studs and called it homothety or even loci.
2005's was much more interesting (hypercubes).

I'm hoping for combinatorics this year, something like counting SETs.[/quote]

When I thought about the SET thing like a week after last years nats, I realized just how simple it was. I won't say anything, just in case you do make one on it.

Not really, at least for the case of only SETs. It's pretty easily shown that any two cards can be a member of exactly one SET, and then it's pretty trivial.


function
number theory unsolved
number theory

Let for every natural $ n$ function $ \varphi (n)$ is Eulers`s function, and is define as:
\[ \varphi (n)=n(1-\frac{1}{p_{1}})(1-\frac{1}{p_{2}})...(1-\frac{1}{p_{m}}),\]
where $ p_{1},p_{2},...,p_{m}$-all prime divisors of $ n$. For example,
\[ \varphi(1)=\varphi(2)=1, \varphi(3)=\varphi(4)=2.\]
Find all $ n\leq 1000,$for which $ \varphi(n)=\varphi(n+1).$

Excluding your examples, it should not be hard to notice that WLOG $ n$ is even and that any odd divisor of $ n$ divides it only once.  So, $ n=2^{a}p_{1}p_{2}\cdots p_{k}$ where the $ p_{i}$ are distinct odd primes.  So, we have reduced to $ \phi(n+1)=\phi(n)=2^{a-1}\prod_{i=1}^{k}(p_{i}-1)=2^{a-1+k}\prod_{i=1}^{k}(p_{i}-1)/2$. Since $ n+1$ is odd, it must have at least $ a-1+k$ odd prime divisors.  As $ 3^{7}>1000$, there can only be at most $ 6$ prime divisors of $ n+1$ implying that $ a-1+k\le 6$.  This should help narrow down the search.  :)

[quote="baz"]Excluding your examples, it should not be hard to notice that WLOG $ n$ is even [...].[/quote]
Hi, baz. :) Actually, sorry but you're wrong on this point. For example, $ \varphi(104) = \varphi(105) = 48$.

[quote="baz"][...] So we have reduced to $ \phi(n+1)=\phi(n)=2^{a-1}\prod_{i=1}^{k}(p_{i}-1)=2^{a-1+k}\prod_{i=1}^{k}(p_{i}-1)/2$. Since $ n+1$ is odd, it must have at least $ a-1+k$ odd prime divisors.[/quote]
But in fact you can't exclude $ v_{2}(p_{i}-1) \ge 2$, can you?! :maybe: Thus this isn't correct too.

--Alright, technically WLOG $ n$ is even then we are leftover with $ n+1$ (or maybe $ n-1$).  NOTE: I was not trying to supply a proof.  Instead, I was just trying to help sieve through the 1000 values of $ n$ by eliminating some.

[quote="s.tringali"][quote="baz"]Excluding your examples, it should not be hard to notice that WLOG $ n$ is even [...].[/quote]
Hi, baz. :) Actually, sorry but you're wrong on this point. For example, $ \varphi(15) = \varphi(16) = 8$ and $ \varphi(255) = \varphi(266) = 128$. [/quote]
--15 is odd and 16 is even

[quote="s.tringali"][quote="baz"][...] So we have reduced to $ \phi(n+1)=\phi(n)=2^{a-1}\prod_{i=1}^{k}(p_{i}-1)=2^{a-1+k}\prod_{i=1}^{k}(p_{i}-1)/2$. Since $ n+1$ is odd, it must have at least $ a-1+k$ odd prime divisors [b]which are $ \equiv 3\pmod{4}$-but useless[/b].[/quote]
But in fact you can't exclude $ v_{2}(p_{i}-1) \ge 2$, can you?! :maybe: Thus this isn't correct too.[/quote]
--you are right.  Thanks for catching that error.   :)

[quote="baz"]
[quote="s.tringali"][quote="baz"]Excluding your examples, it should not be hard to notice that WLOG $ n$ is even [...].[/quote]
Actually, sorry but you're wrong on this point. For example, $ \varphi(15) = \varphi(16) = 8$ and $ \varphi(255) = \varphi(266) = 128$. [/quote]
--15 is odd and 16 is even[/quote]
Weariness seems to play a nasty trick on everyone. :oops: I read "[i]even[/i]", but in fact I wrote about the [i]odd[/i] case. :roll: Anyway, $ \varphi(104) = \varphi(105)$, so... :wink: I'm going to edit my previous post, thanks.

[quote="baz"][...] it should not be hard to notice that any odd divisor of $ n$ divides it only once.[/quote]
Not really! For example, $ 975 = 3 \cdot 5^{2}\cdot 13$, but $ \varphi(975) = \varphi(976) = 480$.

the phi function phi(n) also represents the number of positive integers less than or equal to n that are relatively prime to n.

[quote="Yuriy Solovyov"]So, what is the answer??? :ninja:[/quote]

I came up with $ n=1,3,15,104,164,194,255,495,584,975$.
How does this sound?

[quote="baz"][quote="s.tringali"][quote="baz"][...] it should not be hard to notice that any odd divisor of $ n$ divides it only once.[/quote]
Not really! For example, $ 975 = 3 \cdot 5^{2}\cdot 13$, but $ \varphi(975) = \varphi(976) = 480$.[/quote]
Notice that my claim was true here.  I claimed this statement for $ n$ when $ n$ was even.  So, $ 976=2^{4}\cdot 61$ is alright.[/quote]

But n is not even! Taking n=976, the claim would be $ \varphi(976)=\varphi(977)$!

[quote="jb05"][/quote][quote="baz"][quote="s.tringali"][quote="baz"][...] it should not be hard to notice that any odd divisor of $ n$ divides it only once.[/quote]
Not really! For example, $ 975 = 3 \cdot 5^{2}\cdot 13$, but $ \varphi(975) = \varphi(976) = 480$.[/quote]
Notice that my claim was true here.  I claimed this statement for $ n$ when $ n$ was even.  So, $ 976=2^{4}\cdot 61$ is alright.[/quote][quote="jb05"]

But $ n$ is not even! Taking $ n=976$, the claim would be $ \varphi(976)=\varphi(977)$![/quote]

jbo5,
Not that it is monumental, but if you look back in the posts, I was dealing with $ n$ even and then $ n+1$ [b]or [/b]$ n-1$.  So, yes $ n=976$ and $ n-1=975$.
Either way this problem is most easily done with the aid of a computer.  :)


[quote="cooley"]How can we prove that if (an) is a sequence of positive real numbers with lim an+1/an is L, then lim an^(1/n) is also L?[/quote]
take  $\epsilon > 0$ such tath  $\epsilon < L$
$n_{0}\in IN$ such that
$L-\epsilon \leq \frac{u_{n+1}}{u_{n}}\leq L+\epsilon$ for all $n \geq n_{0}$
we have
$(L-\epsilon)^{n-n_{0}}\leq \frac{u_{n}}{u_{n_{0}}}\leq (L+\epsilon)^{n-n_{0}}$
it's easy now  :wink:


I'm not an IMO guy but... do you think it could be possible that they could be going onto a career not related to contest math?

Bit off topic, so it's hidden...
[hide]
[quote="fredbel6"]
Anyway, I just went on to study to become an engineer, and switched to mathematics after one year.
That is my biggest mistake in my life and I hope I never make a bigger one  :( .[/quote]
Wow!?! ... Could you explain it, plz  :) ? It sounds like your math-study is terribly boring or dreadful  ;)  ... and I hope maths isn't this way ^^.
(If you've already talked about just give me a link...  :D )[/hide]

In order to avoid confusion : starting engineering was the mistake, going to maths was the rescue plan!

[quote="conceive"][quote="José"]IMO people just study math at college[/quote]

Sure they do but my queston is about what will they do with their MO knowlwdge? Most of it cannot be used in college..

On the other hand I would wonder if some IMO person never returned to MO math..[/quote]

yes, one of the China team members last year quitted MO stuff. (he didn't even join China MO)

But this doesn't answer my question. I think only Davron posted reasonable post.

My question is here: http://www.mathlinks.ro/Forum/viewtopic.php?p=513063#p513063

[quote="Davron"]We have to give to others our own knowledge, what we learned from the others. My advice help to people it will
be great. Other's will love you, respect and never forget you. I wish you good luck !!!

Davron[/quote]

I totally agree! You know, the best help I got preparing for the IMO this year was from ex-contestants, young enthousiasts who are trying to help the "new generation"... I hope that I'll be able to do the same thing in years to come.

[quote="blahblahblah"]For the vast majority of IMO participants, I'm sure the answer is 'none of the above', and moreover, 'nothing like any of the above'.[/quote]

And if "nothing like the above", then what replied they should say, I mean the vast majority?

Anyway I can name at least four IMO participant (and not only IMO participant..) who are children of some old IMO participants...

[quote="José"][quote="Davron"]We have to give to others our own knowledge, what we learned from the others. My advice help to people it will
be great. Other's will love you, respect and never forget you. I wish you good luck !!!

Davron[/quote]

Are you going to IMO this year Davron?[/quote]

Sorry, JosA I didn't see this post, sorry again !
yes i am going to IMO ... :-)

davron

I just post something to make this topic alive.

Well I never went to IMO (or even USAMO) but I'll answer anyway b/c I did do high school contests and I post in this forum. ;) I plan to teach math! :D :D :D :D (Maybe work in finance for a time as well.)

[quote="conceive"]I have a question dedicated to IMO participants.

What will you do after you finish your IMO career? (atfer you will be 20 or more years old)..

=============

There are many possibilities what to do. Anyway yesterday I have met one imo gold medalist and he said to me, that he is not doing MO math since he won his gold  :)[/quote]

They do whatever they want to do next besides MO.
Just like athletes do whatever they want to do next after they turn 30.

Well, but they have so much knowledge, much of it cannot be use anywhere.. and only in MO.

What can they do? To become a poster here for the rest of their lives??

Personally I've found this topic very interesting.

Yes, they can post here. There are other math forums too. They can write books on olympiad problem solving. They can coach. They can propose problems for olympiads. These are all uses for MO skills that don't have an upper limit for age. :)

Or they work at McDonald's for the rest of their lives.

I wish some of them did because that one lady keeps getting my price wrong! arg!

[quote="fleeting_guest"][quote="Xevarion"] They can write books on olympiad problem solving. They can coach [olympiads]. They can propose problems for olympiads. [/quote]

It is not really post-olympiad if the job description has "olympiad" in it.[/quote]

Is a schoolteacher not a "post-school" job?

[quote="Klebian"][quote="fleeting_guest"]It is not really post-olympiad if the job description has "olympiad" in it.[/quote]

Is a schoolteacher not a "post-school" job?[/quote]

As notions such as "those who can't, teach" indicate, olympiads are not the only context where it is relevant to ask what comes afterward.

[quote="manuel"]It is also nice to generalize olympiad style results and ideas into big research...[/quote]

I agree. Maybe "innocent" IMO problem can be generalized into important things, and maybe useful

Actually I think people after IMO career do not do MO math any more... :lol:

At least not MO math.

conceive

[quote="José"]IMO people just study math at college[/quote]

Wow!! Do they? Well I did theorethical physics so I guess not very far from math.

Ursula

Even though I have never been to IMO, my opinion is they get some career in Maths and/or share their knowledge with younger generations by either teaching only several ppl(maybe like one's kids) or writing olympiad books and coaching.
This is one of the best math-related round table topics, thank you conceive for starting it. This topic made me want to start another IMO-related topic, which I think is also quite interesting.


I was wondering if people know what AoPS is programmed with. obviously php is involved (which can be concluded from the .php tags). Javascript also from the source code. I wonder what the classroom is programmed with. Is this site backed by a RDBMS like oracle?

The classroom is written in java.
The forum and all the rest of the functionality was created with a modded phpbb script - http://www.phpbb.com/.


I saw these in a Russian problem book a long time ago, also attributed to Ramanujan.  There are equalities between some/all of the above and they are not hard (but tedious) to prove by direct calculation.   How he discovered the formulas, I don't know.


calculus
integration
calculus computations

Determine whether the following improper integral converges or diverges without evaluating it:

$\int_{0}^{1-}\frac{dx}{\sqrt{1-x^{4}}}$

[quote="julien_santini"]$\sqrt{1-x^4} \sim \sqrt{4(1-x)}$ at point $x=1^{-}$. That is, the integral is convergent.[/quote]
Could you please explain that a bit more?  I don't see exactly why that means it's convergent.  Thank you very much.   :)

julien is using a version of the limit comparison test. He's comparing the given integral to

$\int_0^{1^-}\frac{dx}{\sqrt{4(1-x)}}=\frac12(-2\sqrt{1-x})|_0^{1^-}=1$

Since that integral converges, and since the one we are given behaves in essentially the same way in the limit as $x\to1^-,$ then the given integral converges by comparison.

Only more likely, julien never actually computed that integral, and made what amouts to a quick mental substitution, effectively the same thing as $u=1-x,$ to say, "It behaves locally like one over the square root, and that converges."

You have to say also $\int_{0}^{1-}\frac{dx}{\sqrt{1-x^{4}}}>0$. But this is trivial since integrand is positive as $x\in [0,1[$.


arithmetic sequence
number theory proposed
number theory

Let $\tau(m)$ be the number of divisors of the integer $m$.

Prove/Disprove the following statement:

Given positive integers $n, d > 1$, there exist $n$ consecutive positive integers $x_1, x_2, ..., x_n$ such that $\tau(x_i) \neq d \ \forall i \in \{1,2,...,n\}$

Claim: there is an $x$ such that all, $x,x+1,x+2,...,x+n-1$, have more than $d$ divisors.

Prove:
Take $n$ different primes $p_i$ ($0 \leq i <n$), then take an $x \equiv -i \mod p_i ^d$ by the Chinese remainder theorem.
Then $x+i$ has at least the $d+1$ divisors $1,p_i,p_i^2,...,p_i^d$

Well done Darij  :D

Here is a followup question.

Given an infinite arithmetic progression $x_t = a + tb$ of positive integers, prove/disprove that we can find n consecutive members of the the arithmetic progression, each of which does not have exactly m divisors.

1.: I'm [b][u]not[/u][/b] Darij  :!: (as the very last line of my signature says)

2.: Do the same:
Claim: there is an $x=a+bt$ such that all, $x,x+b,x+2b,...,x+(n-1)b$, have more than $d$ divisors.

Prove:
Take $n$ different primes $p_i$ ($0 \leq i <n$) not dividing $b$, then take an (sufficently large) $x \equiv -i \cdot b \mod p_i ^d$ and $x\equiv a \mod b$ by the Chinese remainder theorem.
Then $x+i\cdot b$ has at least the $d+1$ divisors $1,p_i,p_i^2,...,p_i^d$ and is in that sequence.


differential equation
trigonometry
logarithms
calculus
derivative
topology
geometry

$ \exp: \mathbb{C}\rightarrow \mathbb{C}$ is the only differentiable function with properties $ f'\equal{}f$ and $ f(0)\equal{}1$.

Can you please give me a hint :)

Not quite into complex analysis methods, but I'll try to help (probably not the best way, though): If it were $ \exp: \mathbb{R}\rightarrow \mathbb{R}$, you would just solve the differential equation $ f'\equal{}f$ with initial condition $ f(0)\equal{}1$ and that would be quite fast (just separation of variables). Is there a way to generalize it to $ \mathbb{C}$?

Idd, if we work with $ \mathbb{R}$ it is easy, but I could not prove it in our case.

$ f(z) \equal{} u(x,y) \plus{} iv(x,y)$ where $ z \equal{} x \plus{} iy$ and $ u,v$ are real functions.  Since $ f'(z) \equal{} \frac {\partial u}{\partial x} \plus{} i\frac {\partial v}{\partial x} \equal{} \frac {\partial v}{\partial y} \minus{} i\frac {\partial u}{\partial y}$ the differential equation $ f' \equal{} f$ yields the PDEs
\[ \begin{cases} \frac {\partial u}{\partial x} \equal{} \frac {\partial v}{\partial y} \equal{} u \\
\frac {\partial v}{\partial x} \equal{} \minus{} \frac {\partial u}{\partial y} \equal{} v \end{cases}
\]
Solving each in variable $ x$ yields
\[ \begin{cases} u \equal{} \frac {\partial v}{\partial y} \equal{} e^{x}\phi_1(y) \\
v \equal{} \minus{} \frac {\partial u}{\partial y} \equal{} e^{x}\phi_2(y) \end{cases}
\]
Substituting then gives a system of ODEs with initial conditions $ \phi_1(0) \equal{} 1$, $ \phi_2(0) \equal{} 0$ (ICs corresponding to $ f(0)\equal{}1$)
\[ \begin{cases} \minus{} \phi_1'(y) \equal{} \phi_2(y) \\
\phi_2'(y) \equal{} \phi_1(y) \end{cases}
\]
the solution to this well known system is easily seen to be $ \phi_1(y) \equal{} \cos (y)$, $ \phi_2(y) \equal{} \sin (y)$, so
\[ \begin{cases} u(x,y) \equal{} e^x\cos y \\
v(x,y) \equal{} e^x\sin y \end{cases}
\]
and hence
\[ f(z) \equal{} e^x(\cos y \plus{} i\sin y) \equal{} e^z
\]

How much theory are you assuming?  Since a solution directly from Cauchy-Riemann has been posted, I'll do a few oblique ones...

Here's a thought: such a function must be equal to $ e^x$ on the real line, right?  Now it's true in general that if two analytic functions agree on a set with an accumulation point, they agree everywhere.

Another thought: $ f$ and $ f'$ have unique power series.  Equate coefficients--done! (though in a subtle way, this relies on the first observation)

Another one: If $ f$ is nonzero in some neighborhood, then $ \log f \equal{} x\plus{}C$, where $ \log$ is some local branch of the logarithm.  Then $ e^C\equal{}1$ so $ f\equal{}e^z$.  Patch together the neighborhoods!

Or: Consider $ g(z)\equal{}f(z)\minus{}e^z$.  Then $ g(0)\equal{}0$ and $ g'\equal{}g$.  So all of the derivatives of $ g$ are $ 0$ at $ 0$.  Show that this implies that $ g$ must be $ 0$ everywhere.

Another way to look at it:

First, $ f(x)\equal{}e^x$ is the solution to $ f'(x)\equal{}f(x),f(0)\equal{}1$ on the real line, using the usual ODE uniqueness theorem.
Next, we get the differential equation $ g'(y)\equal{}ig(y), g(0)\equal{}e^x$ for the function $ g(y)\equal{}f(x\plus{}iy)$ on the vertical line at $ x$. By the ODE uniqueness theorem, the only solution is $ g(y)\equal{}e^x\cdot e^{iy}$.

It should also be possible to prove existence and uniqueness theorems on $ \mathbb{C}$ analogously to those on $ \mathbb{R}$.

Thanks guys, but I don't know stuff like ODE uniqueness, partial derivatives etc
I should not use those stuff.

There is one essential theorem in a differential equations class: [url=http://en.wikipedia.org/wiki/Picard-Lindel%C3%B6f_theorem]existence-uniqueness[/url]. We're using the uniqueness side- I outlined a proof [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=865325#865325]here[/url].

There are tricks using the Mean Value Theorem for specific simple differential equations. They're not really much easier than the general uniqueness proof.

If you aren't familiar with partial derivatives, how do you define or work with complex differentiability?  It seems partial differentiation is at the heart of the very definition of complex differentiability, so if you aren't familiar with it, then you probably don't have an understanding of the original question that you're asking.  Any first course in complex analysis must absolutely require multivariable calculus as a prerequisite; if you haven't had a course in multivariable calculus, start with that.

As someone who studied complex analysis before multivariate calculus, I have to disagree.  Complex differentiation should certainly not be defined using partial derivatives, but by analogy with the real derivative of one variable.

There are many different levels at which material can be approached.  Rigorous proofs of the key theorems in complex analysis require some background about the topology of metric spaces, for example, but a discussion can get pretty far on geometry alone.  It's not fair to say that the original poster doesn't really understand.

[quote="LydianRain"] ... but by analogy with the real derivative of one variable.
....
[/quote]

Exactly, I must solve it like I only know that the derivative is the limiet ...

You really need the real-variable theory here. The mean value theorem and related arguments are essential, and you can't prove any uniqueness results without them. Fortunately, you can take it one dimension at a time- you don't actually need complex analysis unless you go with the power series argument.

An essential piece of my argument: what does the function look like on a vertical line? That doesn't have to actually be a partial derivative, since we're just restricting the usual derivative limit to approach on a line, but it's very much the partial derivative mindset.


Prove that $ \frac {8}{27}\ge ab \plus{} bc \plus{} ca \minus{} abc$ if $ a,b,c\ge0$ and $ a \plus{} b \plus{} c \equal{} 1$

Note that $ (1 \minus{} a)(1 \minus{} b)(1 \minus{} c) \equal{} ab \plus{} bc \plus{} ca \minus{} abc$. Hence we need to show that
\[ \frac {8}{27}\ge (a \plus{} b)(b \plus{} c)(c \plus{} a)
\]
From AM-GM,
\[ \left(\frac {a \plus{} b \plus{} b \plus{} c \plus{} c \plus{} a}{3}\right)^3\ge (a \plus{} b)(b \plus{} c)(c \plus{} a)
\]
Hence we are done.


Rules are [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=222906]here[/url].

Will need 8 or 12 to start.

[quote="KING1"]I'll join, but when we challenge someone..which gun do we use?[/quote]

You use the gun which you haven't given on yourself.

Hmm, I haven't played one of these in a while.  I'm in.

I think I'll give this another go.

By the way this should be in Games, not FF. :P

$ E\equal{}mc^2$

So I am in....... :wink:

Decline, i.e. I wind up with B.

I'm annoyed; to win I have to live through this AND get maybach to accept my gun w/ 6 bullets.

[hide="Aaaah!  My brain hurts!"]Okay, so after ernie added a new, confuzzling rule, james4l challenged none other than the psychologist, perhaps in an attempt to succeed him?  The psychologist declined, then his apprentice declined, and then that's when ernie decided this rule was stupid and that his brain screwed.  But then, Mewto55555 helped ernie to realize that he declined the gun with only 1 bullet!  james4l contemplated his chances.  "Well, I have a 5/6 chance of surviving, but I already outscrewed the odds once, so what if I do it again?"  He not-so-confidently put the gun to his head and clicked the trigger.  BANG!  Nothing.  "Hmmm, now I must go study the frequencies of maybach, but what if Mewto55555 stays alive?  Oh dear, oh dear..."  Then, Mewto55555 slowly put his gun to his head, realizing that the math nerd would take his place as the psychologist.  BANG!  He slowly drifted away, saying, "I decline the gun, decline, decline..."

[b][color=green]james4l: Lives;[/color][color=red] Mewto55555: Dies[/color][/b]

Teh Alive:

james4l
[b]maybach- C'mon, against a MATH NERD?!?!?  Not cool.[/b][/hide]

Meh, I'm 4/5 in choosing. 7/9 if you count the last one I played in. Everyone always targets me =(

The drama is intensifying with every second james4l thinks and contemplates on whether to take or switch the gun.  The audience is getting sleepy, the puppet annoyed.

[hide]james4l, after about an hour of contemplative thinking, chose to switch the gun.  Once again, it didn't really matter, because each gun had 3 bullets.  "Hmmm... so I have a 1/2 chance of surviving, but with my luck, I could probably bump it up to 65/128."  maybach, annoyed, responded, "Shut up, you nerd!  I'm about to shoot myself!  Literally!"  And so he did.  But nothing happened.  "Sigh, maybe I should've shot myself 6 times."  james4l nervously put the gun to his head, [hide="and slowly pulled the trigger."]The bullet wasn't so slow, however.

[b][color=green]maybach: Lives;[/color] [color=red]james4l: Dies[/color][/b]

Congrats, maybach!  You are the winner!  But I still have [hide="one more challenge for you..."]"You see that table over there, maybach?  You know, the one with 6 guns?" ernie politely asked.  "Yeah, I do.  Whaddya want me to do, shoot james4l with them?  I'd be glad to do so."  ernie responded, "Uhhh... no.  There is one last challenge for you.  Each one of those guns has a certain number of bullets inside them.  You must choose one, and [b]shoot yourself with it[/b]."  maybach, quite confused, stuttered, "WH-WH-WHAAAAAAAAAA?!?!?  I DON'T REMEMBER THIS PART OF THE GAME!!!"  ernie responded, "Well, that's my twist!  So which gun will you choose?"[/hide][/hide][/hide]

Um........Gun #2????:(

:gulp:  This isn't going to be good!

[quote="ernie"]The swagger of maybach suddenly went into the negatives.  But what he should've done (to perhaps cheat) is to look at my last post in HSB. :D  He wasn't thinking straight, as :gulp: is not an emoticon.  :P  Slowly, realizing that this could be the end, maybach out the gun to his head.  The crowd gasped.  Finally, he clicked the trigger.  [hide="BANG!"]What?  Confetti?  It appears as if maybach has won the whole challenge!  Our winner!  The only survivor!  Congratulations, maybach!  (He could have just looked at my last HSB post, as the sequence of numbers was posted there. :P)  Bullet analysis will be posted later.[/hide][/quote]

That's great! But I didn't look at HSB.  Most of the math there is too hard  :blush: .  Whew.

[hide="Gun Choices"]KING1: I forget, lol.
Bugi: 4/2
cf249: 2/4
dragon96: 4/2
maybach: 5/1; 3/3
spacguy524: 4/2
cgyao15: 1/5
james4l: 5/1; 1/5
Mewto55555: 6/0; 0/6
game.guy: 3/3
Xaenir: 3/3

Average for Round 1: 3.37/2.63
Average for Round 2: 1.33/4.67
Average for Game: 2.93/3.07[/hide]

[hide="Player Challenging + Accepting/Declining"]
cgyao15: challenged with B (lived); declined (died)
KING1: accepted (died)
spaceguy524: challenged with B (died)
dragon96: declined (died)
Mewto55555: challenged with B (lived); accepted (lived); accepted (lived)
maybach: challenged with A (lived); accepted (lived)
Bugi: accepted (died)
Xaenir: challenged with B (died)
game.guy: challenged with B (lived); modkilled
james4l: challenged with B (lived); accepted (lived)
cf249: challenged with A (died)

james4l: challenged with A, accepted (lived); declined, declined (died)
Mewto55555: accepted, declined (died)
maybach: challenged with A, accepted (lived)

For stats, I will only use Round 1.  Brackets show Round 2, if you are interested.

Challenger: B preferred over A, 6-2 [A over B, 2-0]
Challengee: Accepted over Declined, 6-2 [Declined over Accepted, 3-2]
Death of Challengee if Gun Accepted: 2/6
Death of Challengee if Gun Declined: 2/2
Average Gun Distribution of Challenger/Challengee During Shootout: 2.125/3.875 [2/4][/hide]

lol..I put 6 in one, and 0 in the other :P

Yeah, but it would help if you told me which one was which.


x + 2/x = 5 + 2/5
x = ?

Do I convert 5 + 2/5 to 27/5  but not sure how to reduce x + 2/x?

instead of converting anything
you should notice that $x=5$
the right hand side is the exact same thing as the left hand side,
just with $x$ as $5$

Did notice, just would like to know the steps used in reaching x.

I figure:

   x +  2/x = 5 + 2/5
5(x + 2/x) = x(5 + 2/5)
  5x + 10   =  5x + 2x
   5x - 7x   =  -10
            x  =   5

Is this method correct?

um, you just assumed that x = 5 in the second step of the solution.

[hide]You would convert 5 + 2/5 to 27/5. Then multiply through by x: x^2 + 2 = 27x/5; subtract 27x/5 from both sides and we are left with the quadratic x^2 -27x/5 + 2 = 0

To factor, we can multiply the whole thing by 5; 5x^2 - 27x + 10 = 0; (5x-2)(x-5) = 0, so x = [b]5 or 2/5.[/b] (You guys missed the last solution as well)[/hide]

[quote="learningfish"]I used 5 from multiplying by the opposite sides denominator.
You explained it well thanks[/quote]

Don't ever do that. It just so happened that one of the solutions for $x$ in your equation was the denominator of the other side, it will very infrequently happen that way.

Let me show you something:

$\frac{1}{4}= \frac{2}{8}$

True statement, obviously. Now let's do what you did and multiply by the other side's denominator:

$\frac{8}{4}= \frac{2*4}{8}$

$2 = 1$

That's why you can't multiply by the other side's denominator. Most of the time you will get a false statement. Remember the golden rule of algebra: What you do to one side of the equation you must do to the other.

So if you multiplied BOTH sides by BOTH denominators, it would be fine:

$\frac{1}{4}= \frac{2}{8}$

$\frac{4*8}{4}= \frac{2*4*8}{8}$

$8 = 8$.


calculus
integration
logarithms
derivative
function
search
calculus computations

I am told to prove that this is correct:

$\int\frac{1}{x+x^4}dx=\frac{1}{3}\ln\left(\frac{x^3}{1+x^3}\right)$

How do i prove that?

Presumably, up to a constant.  It is enough to show that the derivative of the RHS is equal to the integrand on the LHS.

Ypu can find a derivitate of $\frac{1}{3}\ln\left(\frac{x^3}{1+x^3}\right)$  :)  :)

If you were told [color=red]to compute [/color]it then  you had to write :

  $\int \frac{1}{x+x^4}dx = \int \frac{(x^3 + 1) -x^3}{x(1+x^3)}dx$=
 =  $\int( \frac{1}{x} - \frac{x^2}{x^3+1})dx = lnx - \frac{1}{3} ln(x^3+1) +c$

 which easily leads to the same answer.

[u]babis[/u]

A bit too hard algrebra for me though :) Hum, but ok, let's presume i am allowed to do it the other way around. But how do you derivitate this? I get some really nasty numbers.

Use this $(c \cdot f(g(x)))' =c \cdot g'(x) \cdot f'(g(x))$ and this $( \ln x )' = \frac{1}{x}$ .  :)

[quote="JBL"]You take the derivative just like you would take the derivative of anything else, in this case with applications of the chain and quotient rule.  Then you simplyify, simplify, simplify.  It might be a little nicer if you re-write $\frac{x^3}{x^3 + 1} = 1 - \frac1{x^3 + 1}$ (no quotient rule).[/quote]

Nope . It is nicer if he writes $\frac{1}{3}\ln\left(\frac{x^3}{1+x^3}\right)= \frac{1}{3} \left( \ln (x^3) -\ln (x^3+1) \right)$

[quote="Sporally"]I am not sure about what to do about the $f'(g(x))$, but here is what i get.

$\frac{1}{3}\cdot \frac{1}{x^3\cdot (1+x^3)}\cdot \frac{3x^2}{(1+x^3)^2}$

Does that sound right to you?[/quote]

Almost . It is $f(x)= \ln x$ therefore $f'(x) = \frac{1}{x} \Rightarrow f'(g(x))= \frac{1}{\frac{x^3}{x^3 + 1}} = \frac{x^3 + 1}{x^3}$

 :)

This is wrong : $\frac{1}{\frac{2^3}{2^3 + 1}}=0.0139$ . This is correct $\frac{1}{\frac{x^3}{x^3 + 1}} = \frac{x^3 + 1}{x^3}$

Generally $\frac{1}{a}$ is the inverse of $a$ and the inverse of $\frac{x^3}{x^3 + 1}$ is $\frac{1}{\frac{x^3}{x^3 + 1}}$ .

But the inverse of $\frac{x^3}{x^3 + 1}$ is also $\frac{x^3 + 1}{x^3}$ .

That is how we get the rule $\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}$ .

 :)

Yes. Ramanujan, that is much nicer.  Actually, my brother (who is not particularly strong in math) got into an argument with his math teacher on just this topic, but with the derivative of $\ln \frac x{x+1}$ -- I was very proud of him for being right!

Sporally: $\frac{a}{\frac bc} \neq \frac{\frac ab}c$.  You are using a calculator (and failing to properly use parentheses) instead of thinking about what the symbols mean.

[quote="JBL"]  Actually, my brother (who is not particularly strong in math) got into an argument with his math teacher on just this topic, but with the derivative of $\ln \frac x{x+1}$ -- I was very proud of him for being right!
[/quote]

 :lol:

OK, now i see it, finally. I've got to work with the lower fraction first of and then i can take the inverted of the result.

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

And now i got to the bottom finally - thx a lot for your help.

I have a couple of other questions though - mostly yes or no answers:

$g(x)=x\cdot \sqrt{x}$, $x: [0;4]$

And i've got this formula to calculate the lenght of the function:

$L=\int_a^b\sqrt{1+(g'(x))^2}dx$

Two questions for this formula:

1) Is this a formula that always applies for the lenght of a function? I was told it wasn't.

2) If the answer for 1) is a yes, can a TI-84 plus tell me the lenght of the function like it easily can tell me the integral of the function?

If not, can you plz verify this for me:

$L=\int_0^4\sqrt{1+(g'(x))^2}dx=9.07$

1) Yes look the pdf file for a brief description .
    You can search for arc length at google .

2) [quote="Sporally"]can a TI-84 plus tell me[/quote]   :?  What is this ???

Anyway the result is correct .  :)

[quote="Sporally"]A calculator from Texas Instruments called TI-84 plus - was wondering if it could tell me the lenght of a function when i tell it the function...[/quote]

I dont know .... but the above integral gives the length of the function.  :)

Ok, thx for your help through it all. Hope someone else can answer the question about the arc lenght on a calculator ;)

I never used TI-84, but I'm not aware of such an option on my TI-89. But if it can do integrals for you, you can easily type it in yourself, or write a program. I can't remember how to prompt for a function in TI calculator language, though. Is it the same as for anything else?

> prompt("Enter your function, f(x) = ", f)
> prompt("Enter the right bound: ", a)
> prompt("Enter the left bound: ", b)
> disp(int($\sqrt$ (diff(f, x)^2 + 1), x, a, b))

Doesn't that do what you want? I might have gotten the syntax a little bit wrong; it's been ages since I did any calculator programming. But anyway, considering how short it is, it seems to me that you may as well just type in the formula. It's a good think to remember anyway.

Here is what i do

"clicking the 'PRGM' button"
"choosing 'NEW' and assigning a name for the program"
"click 'PGRM', go to 'I/O' and pick 'Prompt', which is then moved to the display below the name of the program"
" i then write the function (is this the particular function i need in this case)"
"adding another two 'Prompt' and assigning 'a' and 'b' which in this case are '0' and '4'.
"adding 'Disp' which is found where 'Prompt' is found and it is displayed below the three 'Prompt's".

PROGRAM:LENGTH
:Prompt $x\cdot \sqrt{}(x)$
:Prompt $0$
:Prompt $4$
:Disp $int(\sqrt{}(1+$

I guess you want me to replace the $0$ with $a$, the $4$ with $b$ and the $x\cdot \sqrt{}(x)$ with something else - but i am pretty sure you don't want me to write "f(x)=" in letters, do you?

How do i write the "diff(f, x)". Not in letters, right?

What about this ?? [url]http://www2.ohlone.edu/people2/joconnell/ti/[/url]   :)

Forgot about this thread. Will take a look at it - your link seems to explain how to do it.


For what value of $ n$ is $ 5 \times 8 \times 2 \times n \equal{} 7!$?

knowing factorials up to about $ 8!$ is helpful. $ 7! \equal{} 5,040$. the left side is $ (2)(8)(5)n \equal{} 80n$. dividing, we get $ \frac{5,040}{80} \equal{} 63$

[hide=my sol]We get the equation $5\cdot8\cdot2\cdot n=7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1$. Simplified, we get $n=7\cdot3\cdot3$ which makes $n=\boxed{63}$. This sol is simpler than the one above because we don't need to know $7!$[/hide][hide=for people that want to report my bump]I'm allowed to bump as long as I don't violate the ToS[/hide]

[quote=eduD_looC]What is the ToS?[/quote]

Terms of Service

[hide=Neat lil question]$7!$ $=$ $7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1$
This is equal to $5\cdot8\cdot2\cdot{n}$

Instead of working out $7!$ and dividing by $5\cdot8\cdot2$, we can cancel out numbers from both sides of the equals sign.

After cancelling, we get $7\cdot6\cdot3$ $=$ $2n$

$2n$ $=$ $126$

$n$ $=$ $\boxed{63}$
[/hide]


AMC
USA(J)MO
USAMO
function
inequalities open
inequalities

Let $f$ be a function, let us call it "very convex" if...
$\frac{f(x) + f(y)}{2} \ge f\left(\frac{x+y}{2}\right) + |x-y|$
prove that no $f$ exists.

I have half of a solution. It was #1 on day 1 from 2000.

It's from the 2000 USAMO:
[url]http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=27&year=2000[/url]
http://www.mathlinks.ro/Forum/viewtopic.php?t=47401

MYM has a very nice open question from this problem. :lol:


Given a,b,c>0 Prove that:
$ \frac {a^2 \plus{} kbc}{(a\plus{}b)(c\plus{}a)} \plus{} \frac {b^2 \plus{} kca}{(a\plus{}b)(b\plus{}c)} \plus{} \frac {c^2 \plus{} kab}{(b\plus{}c)(c\plus{}a)} \geq \frac {3(1 \plus{} k)}{4}$
($ k \geq \minus{} 1$)

[quote="dragon1"]Given a,b,c>0 Prove that:
$ \frac {a^2 \plus{} kbc}{(a \plus{} b)(c \plus{} a)} \plus{} \frac {b^2 \plus{} kca}{(a \plus{} b)(b \plus{} c)} \plus{} \frac {c^2 \plus{} kab}{(b \plus{} c)(c \plus{} a)} \geq \frac {3(1 \plus{} k)}{4}$
($ k \geq \minus{} 1$)[/quote]
It equivalent to:
$ 4(k\plus{}1)(a^2(b\plus{}c)\plus{}b^2(c\plus{}a)\plus{}c^2(a\plus{}b)) \ge 3(k\plus{}1)(a\plus{}b)(b\plus{}c)(c\plus{}a)$
$ <\equal{}> a^2(b\plus{}c)\plus{}b^2(c\plus{}a)\plus{}c^2(a\plus{}b) \ge 6abc$ (because ($ k \ge \minus{}1$ )
Which is obvious true by AM-AM :wink:  :)

[quote="dragon1"]Given a,b,c>0 Prove that:
$ \frac {a^2 \plus{} kbc}{(a \plus{} b)(c \plus{} a)} \plus{} \frac {b^2 \plus{} kca}{(a \plus{} b)(b \plus{} c)} \plus{} \frac {c^2 \plus{} kab}{(b \plus{} c)(c \plus{} a)} \geq \frac {3(1 \plus{} k)}{4}$
($ k \geq \minus{} 1$)[/quote]

We have the old:

$ (1) \sum\frac {a^2}{(a \plus{} b)(a \plus{} c)}\geq \frac {3}{4}$

$ (2) \sum\frac {kbc}{(a \plus{} b)(a \plus{} c)} \equal{} k \sum\frac {bc}{(a \plus{} b)(a \plus{} c)}\geq \frac {k3}{4}$

since (1) and (2) we have:

$ \sum\frac {a^2 \plus{} kbc}{(a \plus{} b)(a \plus{} c)}\geq \frac {3(1 \plus{} k)}{4}$
$ (k \geq \minus{} 1)$ :)

Opps! Linguras, K can be negative so your second inequality might inverse the sign.


algebra
polynomial
geometry unsolved
geometry

The side lengths of a triangle are the roots of a cubic polynomial with rational coefficients. Prove that the altitudes of this triangle are roots of a polynomial of sixth degree with rational coefficients.

$S_{\triangle ABC}^{2}=p(p-a)(p-b)(p-c)$ is rational because p is rational and polynomial $f(x)=(x-a)(x-b)(x-c)\in\mathbb{Q}[x]$. Since $h_{a}^{2}=\frac{4S^{2}}{a^{2}}$,etc. the statement is obvious(polynomial $(x^{2}-h_{a}^{2})(x^{2}-h_{b}^{2})(x^{2}-h_{c}^{2})$ is with rational coeff.)


Given Problem:
$ \frac{\binom{n}{r}\plus{}5\binom{n}{r\plus{}1}\plus{}10\binom{n}{r\plus{}2}\plus{}\binom{n}{r\plus{}3}\plus{}5\binom{n}{r\plus{}4}\plus{}\binom{n}{r\plus{}5}}{\binom{n}{r}\plus{}4\binom{n}{r\plus{}1}\plus{}6\binom{n}{r\plus{}2}\plus{}4\binom{n}{r\plus{}3}\plus{}\binom{n}{r\plus{}4}}\equal{}\frac{n\plus{}k}{r\plus{}k}$

I think there is a missing $ 10$ in the numerator...
Revised Problem:
$ \boxed{\frac{\binom{n}{r}\plus{}5\binom{n}{r\plus{}1}\plus{}10\binom{n}{r\plus{}2}\plus{}10\binom{n}{r\plus{}3}\plus{}5\binom{n}{r\plus{}4}\plus{}\binom{n}{r\plus{}5}}{\binom{n}{r}\plus{}4\binom{n}{r\plus{}1}\plus{}6\binom{n}{r\plus{}2}\plus{}4\binom{n}{r\plus{}3}\plus{}\binom{n}{r\plus{}4}}\equal{}\frac{n\plus{}k}{r\plus{}k}}$

[hide="Hint"]
Use Vandermonde's Identity: $ \sum_{k\equal{}0}^{r}\binom{m}{k}\binom{n}{r\minus{}k}\equal{}\binom{m\plus{}n}{r}$.
[/hide]


modular arithmetic
number theory proposed
number theory

[quote="k.l.l4ever"]Let $ a,b,c,d$ be positive integers such that:
$ a^2 \plus{} b^2 \plus{} ab \equal{} c^2 \plus{} d^2 \plus{} cd$
Prove that $ a \plus{} b \plus{} b \plus{} d$ is composite.[/quote]

Wrong : $ a\equal{}2,b\equal{}1,c\equal{}2,d\equal{}1$ $ \implies$ $ a^2 \plus{} b^2 \plus{} ab \equal{} c^2 \plus{} d^2 \plus{} cd\equal{}7$ while $ a \plus{} b \plus{} b \plus{} d\equal{}5$ is prime

Sorry,I've edited the statement of this problem.

this problem is well-known and maybe posted before,
suppose that $ a\plus{}b\plus{}c\plus{}d\equal{}p$
then we have $ (a\plus{}b)^2\minus{}ab\equal{}(c\plus{}d)^2\minus{}cd \implies (a\plus{}b)^2\minus{}ab \equiv (c\plus{}d)^2\minus{}cd \pmod {p}$ but we have $ a\plus{}b \equiv \minus{}c\minus{}d \pmod {p}$ so we have $ ab \equiv cd \pmod {p}$
and we get $ a^2\plus{}b^2 \equiv c^2\plus{}d^2 \pmod {p}$
so we have $ a^2\minus{}c^2 \equiv d^2\minus{}b^2 \pmod {p}$
but $ a\plus{}c \equiv \minus{}(b\plus{}d) \pmod {p}$
from the last congruency we get
$ \minus{}(a\minus{}c) \equiv d\minus{}b \pmod {p} \implies p | d\minus{}b\plus{}a\minus{}c \implies a\plus{}b\plus{}c\plus{}d |d\minus{}b\plus{}a\minus{}c$
which is a contradiction!unless we have $ d\minus{}b\equal{}c\minus{}a  \implies d\plus{}a\equal{}c\plus{}b \implies 2 | a\plus{}b\plus{}c\plus{}d\equal{}p \implies p\equal{}2\equal{}a\plus{}b\plus{}c\plus{}d$
so at least one of the variables $ a,b,c,d$ must be $ 0$
we r done!    :)

[quote="k.l.l4ever"]Let $ a,b,c,d$ be positive integers such that:
$ a^2 \plus{} b^2 \plus{} ab \equal{} c^2 \plus{} d^2 \plus{} cd$
Prove that $ a \plus{} b \plus{} c \plus{} d$ is composite.[/quote]
$ a^2 \plus{} b^2 \plus{} ab \equal{} c^2 \plus{} d^2 \plus{} cd$ $ \implies$ $ (a \plus{} b)^2 \minus{} ab \equal{} (c \plus{} d)^2 \minus{} cd$ $ \implies$ $ (a \plus{} b \minus{} c \minus{} d)(a \plus{} b \plus{} c \plus{} d) \equal{} ab \minus{} cd$
$ a^2 \plus{} b^2 \plus{} ab \equal{} c^2 \plus{} d^2 \plus{} cd$ $ \implies$ $ (a \minus{} b)^2 \plus{} ab \equal{} (c \minus{} d)^2 \plus{} cd$ $ \implies$ $ (a \minus{} b \plus{} c \minus{} d)( \minus{} a \plus{} b \plus{} c \minus{} d) \equal{} ab \minus{} cd$

So $ (a \plus{} b \minus{} c \minus{} d)(a \plus{} b \plus{} c \plus{} d) \equal{} (a \minus{} b \plus{} c \minus{} d)( \minus{} a \plus{} b \plus{} c \minus{} d)$

So $ a \plus{} b \plus{} c \plus{} d$ prime would imply :
either $ a \plus{} b \plus{} c \plus{} d$ divides $ |a \minus{} b \plus{} c \minus{} d|$ and so $ a \plus{} b \plus{} c \plus{} d\le |a \minus{} b \plus{} c \minus{} d|$ which is impossible since $ a,b,c,d > 0$
either $ a \plus{} b \plus{} c \plus{} d$ divides $ |\minus{} a \plus{} b \plus{} c \minus{} d|$ and so $ a \plus{} b \plus{} c \plus{} d\le |a \minus{} b \minus{} c \plus{} d|$ which is impossible since $ a,b,c,d > 0$

Hence the result : $ a \plus{} b \plus{} c \plus{} d$ is composite.

[u]Edit [/u]: did not see shoki already answered, sorry.

[quote="pco"]$ a^2 \plus{} b^2 \plus{} ab \equal{} c^2 \plus{} d^2 \plus{} cd$ $ \implies$ [/quote]$ (a \minus{} b)^2 \plus{} 3ab \equal{} (c \minus{} d)^2 \plus{} 3cd$

See also : http://www.artofproblemsolving.com/Forum/viewtopic.php?t=345528


geometry
inequalities
trigonometry
analytic geometry
graphing lines
slope
symmetry

Region $ R$ contains all the points $ (x,y)$ such that $ x^{2}+y^{2}\leq 100$
and $ sin(x+y)\geq  0$. Find the area of region R.

[hide]
Lets just look at the second inequality for now

In the first cycle of the sine, $ x+y$ must be between $ 0$ and $ \pi$
this happens between the lines $ y =-x$ and $ y=-x+\pi$

In the second cycle, $ x+y$ must be between $ 2\pi$ and $ 3\pi$ corresponding to the region between $ y =-x+2\pi$ and $ y =-x+3\pi$

the overall graph becomes a series of parallel lines with slope -1 with alternating shading.

now we go back to the first graph, a circle centered at the origin with radius $ 10$.
You will notice that for each intersecting shaded trapezoidy thing in the 1st quadrant, there is an equal area unshaded trapezoidy thing in the 3rd quadrant. Because of this symmetry, the total shaded area is exactly half of the circle.

$ A =\frac12\pi (10)^{2}= 50\pi$
[/hide]


I am not sure how to solve this.

Find the number of times the two given curves intersect at the origin.

x^3 + x + y = 0 and y^3 – 3 * x^2 * y – 2 * x^3 = 0

More than direct answer I would more appreciate method / logic.

Thanks for any help

Can you explain what you are asking?

At the origin, times of intersect can only be 1 or 0
when x=0, y=0 ,then two equations =0
So the ans. is  1 , right?

Or can you further explain your ques.?

Is this your equation ?

$x^3+x+y=0$

$y^3-3x^2y-2x^3=0$
 :)

[hide]

Substitute $y=-x^3-x$ into the second equation you will get
$x^7(x^2+3)=0$ after some simplifying job . So you can see that $x^7=0$ which impies it intersects seven times at the origin .  :) [/hide]

:?: But , what is the ans.?

I think it's 1! :lol:


anyone have probabilistic proofs that 2-d brownian motion never hits a fixed point but that it does get arbitrarily close?  i know the standard log/harmonic approach and a complex analysis approach but I'd like to see a straight prob one.

May be try to approximate the Brownian motion by a scaled random walk?


Yesterday, I created some inequalities, one of them is the following:

Let $a,b,c$ be nonnegative real numbers such that $a^2+b^2+c^2=1$, prove that
\[ 1\leq\frac a{ab+1}+\frac b{bc+1}+\frac c{ca+1}\leq\frac{3\sqrt3}4. \]

I have NOT solved it. Is it true and new?

Well, the left inequality can be sharpened :)
See here : http://www.mathlinks.ro/Forum/viewtopic.php?t=75146

I think mine is different from that.

Yesterday, I reduced the second inequality to proving that
\[ 32b^2-24\sqrt3b(1+a)+27a^2+22a+11\geq0,  \]
with equality when $a=\tfrac13$, and $b=\tfrac2{\sqrt3}$.

I had assumed that $z=\max\{x,y,z\}$, and put $a=xy$, $b=x+y$.

How should I proceed with the proof?

Nobody is interested in this nice inequality?


In an international meeting of $n \geq 3$ participants, 14 languages are spoken. We know that:

- Any 3 participants speak a common language.

- No language is spoken more that by the half of the participants.

What is the least value of $n$?

I bet $10$ dollars that the answer will be $8$! :oops:

case 1: $n$ is even, say $n=2k$, then we have that every language kills at most $\binom{k}{3}$ trios, so we must have: $14\binom{k}{3}$ is at least $\binom{2k}{3}$ and this only happens for $k>3$, so the smallest possible value is $8$, Igor told me this is the correct answer so i guess there is a nice configuration with it!

case 2: $n$ is odd we get no better bound!

[quote="remo"]It is easy to construct the required sets using a 7 point projective plane + an extra point. Languages are spoken by lines + the extra point or by complements of lines.[/quote]

Can you please explain? This sounds like a very cool method, but being a novice, I have no idea what you are talking about.

Anyways, I did trial and error and found: 1234, 5678, 1256, 3478, 1278,3456, 1357,2468,1368,2457,1458,2367,1467, 2358. (1-8 denote the 8 people, and each of the 14 "clusters" gives the people who speak this language.) So indeed, the answer is 8.

Draw a triangle and mark it's vertices with 1,2,3. Draw its inscribed circle and mark the points where it touches the sides of triangle with 5,6,7 (so that 5 is on the side 2,3 and 6 on the side 1,3). Also mark the incenter with 4 and add one more point to that plane and mark it 8. Then read again the explanation and you'll see your example :)

I must be misunderstanding something. Isn't the only possible value of $n$ 8?  Draw table with the 14 languages on the horizontal axis and the $n$ participants on the vertical axis. Put a marker in a cell if the participant speaks the corresponding language. The number of makrs is at least $n$ choose $3$, and at most $0.5*14n=7n$. This inequality is equivalent to $n \le 8$. What did I miss???

Wherefore have you infered the number of markers must be at least $\binom{n}{3}$ ? To better see this is not true, imagine there is no limitation on the number of persons speaking a same language, and have each person speaking all $14$ languages; then obviously any $3$ persons speak a common language, since they all three speak all languages. Then you have precisely $14n$ markers, and your inequality with $\binom{n}{3}$ is clearly misplaced.

Thusly I have herefore been proven wrong whence mavropnevma asked wherefore have I inferred that lower bound for the number of markers. I reach the obvious conclusion that I am thusforth incorrect insofar as the lowerbound so henceforth ignore my hitherto comment. Good day.


$x$ and $y$ are positive integers.

prove that the equation $x^{2}-y^{3}=7$ does not have solution.

we have, (mod.4), that $y \equiv 1 (mod.4)$, and that $(y+2)(y^{2}-2y+4) = x^{2}+1$!

but $y+2$ will have one prime factor of form $4k+3$, and $x^{2}+1$ no!


Find the remainder when the polynomial:

$x + x^3 + x^9 + x^{27} + x^{81} + x^{243}$

Is divided by:

(a) $x - 1$
(b) $x ^2 - 1$

Rep, I'm a bit confused as to why P(1)=6 for (a). Could you perhaps explain it another way?

Oh, duh! I didn't even think about the original function. That's quite obvious. Thanks, Rep. :)

Is the remainder to the new question 5x?


Simple AIME problem but strange wording that I have a hard time understanding it.

[b]Source: AIME 2003 #3[/b]

Let the $S = {(1,2,3,5,8,13,21,34)}$.  Susan makes a list as follows: for each two-element subset of $S$, she writes on her list the greater of the set's two elements.  Find the sum of the numbers on the list.

[hide="Answer"]The number of two-element subsets is ${8\choose 2}=8\cdot 7=56$. All $7$ subsets that include $34$ will have $34$ as the larger number, six of the seven subsets that include $21$ will have $21$ as the larger number, etc., so the answer will be $34\cdot 7+21\cdot 6+13\cdot 5+8\cdot 4+5\cdot 3+3\cdot 2+2=484$.[/hide]

That's the way i did it as well.
Also note that you have terms of a fibonacci sequence.  I don't know if that fact can be used...

I noticed that too, but couldn't find any correlation. I don't know if there's any point to it. Maybe it's supposed to throw you off on a tangent. :P

There isn't any significance in Fibonacci occuring (none that I can see at least).  The writers of the AIME problems probably just felt like using Fibonacci.

I think it can help with the addition. For example, I had $34\cdot 7+21\cdot 6+13\cdot 5+8\cdot 4+5\cdot 3+3\cdot 2+2\cdot 1$, which can be simplified to $(2+3)+(3+5)+2(5+8)+2(8+13)+3(13+21)+3(21+34)+4\cdot 34$. From there, we have $(2+3)+(3+5)=(5+8) \Rightarrow 3(5+8)+2(8+13)=13+2\cdot 34$, and $3(13+21)+3(21+34)=3(34)+3(55)=3\cdot 89$, so the total sum is $13+6\cdot 34+3\cdot 89=484$, which is easier to multiply out. I don't know if it's any easier overall, after factoring in the extra work, but it might be.


The area of a square plot of land is 325 square meters. What is the perimeter of the square, in meters? Express your answer in simplest radical form.

The side length is $ \sqrt{325}$. Simplifying this, we get $ 5\sqrt{14}$. THe perimeter is this multiplied by 4, which is $ \boxed{20\sqrt{14}}$.


has anyone downloaded internet explorer 7?

I downloaded it just now.
I think it looks quite cool, though I'm not convinced by it overall yet: unless I'm much mistaken, pages are now taking longer to load than they did on IE 6...

a little but just a [size=59]little tiny bit[/size]

plus, you can have more than one hompage

yes, i have it.
it has the following features.

Tabs like firefore and Phishing Filter plus beautiful outlook,but i have heard that it needs be bought otherwise it will damage the windows.

It emits a squeal so loud and so high-pitched that it breaks all windows in the area.

If you were IE and you saw firefox, you'd squeal pretty loudly too.


algebra
polynomial
search
number theory
prime numbers
algebra solved

Let $f(x)$ be polynomial with integer coefficients satisfy: $\gcd (f(x);f'(x))=1$. Prove that: there exists positive integer $m$ and prime number $p$ such that $p$ divides $f(m)$ but $p^2$ does not divide $f(m)$.

I had sent the proble with the same idea of this problem.Do the search
Easy problem...  :D  :D

the condition means, that f is the product of dictinct linear factors in the complex numbers. the result is that there is a prime p such that in $F_p$(the integers mod p), f has a linear factor which is not multiple. if the factor is multiple, then two complex roots are the same in $F_p$, that means their difference is divisible by p, by their difference is algebraic, so if it is divisible by p, then the constant term of the polynomial of which their difference is a zero must be divisible by p, thus the difference is divisible through only a finite number of primes, thus there are only finitely many primes where a linear factor is multiple. if the result is wrong, then there are only finitely many primes through which f(m) is divisible as m runs through all integers. now let the product of all these primes be P, the constant term of f be $a_0$. look at $f(Pa_0)$. factor out $a_0$ to get a term which is a multiple of P plus 1. thus, the second factor is not divisible through any of the primes, contradiction.
maybe there's a more elementary and nicer solution.

Peter

I didn't read what Peter wrote, but I bet it's Ok :).

Anyway, here's what I did (I hope it's not exactly the same thing :)):

For all $m\in \mathbb N$ and $p$ prime s.t. $p|f(m)$, $p$ doesn't divide $f'(m)$. This is because there are polynomials $P,Q\in \mathbb Z[x]$ s.t. $P(x)f(x)+Q(x)f'(x)=1$.

Now assume $p|f(m)$. Then $p|f(m+kp),\ k\in \overline {1,p-1}$. We assume that $p^2|f(m),\ f(m+kp)$. But $f(m+kp)=f(m)+kpf'(m)+M_{p^2}$ (we expand the polynomial in a Taylor series). This means that $p^2|kpf'(m)\Rightarrow p|kf'(m)\Rightarrow p|f'(m)\Rightarrow$ contradiction.

your argument is much nicer...it does also show a much stronger theorem, namely that there exists an m for every prime p.

Peter

Hmm Peter. Nice solution for a 16-year old student!
Anyway, there is a simpler one:
a) let g(x)=f'(x). Then since gcd(f(x),g(x))=1 then 1=m(x)g(x)-n(x)f(x) and if p divides f(x) then p doesn't divide g(x)
b) consider p|f(t). Then f(p+t)-f(t)=p*g(t) mod p^2( easily check)
So by a) one of f(t) or f(p+t) is not divisible by p^2
QED

but in this case too much theory didn't help to find a nice solution :D

[quote="harazi"]I remember to have solved some time ago a beautiful problem from Iran: if f is a polynomial without double roots, then for any n there exists a number k such that in f(k) at least n prime numbers appear with exponent 1. I think this is a little bit harder.[/quote]
Is it on the forum?

Yes, it is the topic Just beautiful by Omid Hatami.

If you found it then just paste the link here.

[quote="grobber"]For all $m\in \mathbb N$ and $p$ prime s.t. $p|f(m)$, $p$ doesn't divide $f'(m)$. This is because there are polynomials $P,Q\in \mathbb Z[x]$ s.t. $P(x)f(x)+Q(x)f'(x)=1$.[/quote]

Just for the record, this is not true.

Yeah, it's bogus. There are, however, integer polynomials $P,Q$ such that $Pf+Qf'$ is some constant integer $t$. Now take $p$ to be one of the (infinitely many) primes which divide some $f(m)$ but not $t$.

Yes, the correction was pretty obvious.
[quote="Myth"]My follower :)[/quote]
I do hope you are kidding...


I'm back with another undergraduate math dispute:

Carnegie Mellon vs. McGill!!!

Help!! I was just settling down today after a trip to C.Mellon. The school made a very, very good impression on me: I had my heart set on going there. My mom was ready to send the deposit. All of a sudden I get an e-mail from McGill saying that I had been admitted off the wait list!
I am planning on majoring in math/physics. Carnegie offers an honors program to graduate with a bachelor's and master's in math in just four years. Seniors at CMU get job offers in October of their senior year from prestigious companies and firms, not to mention the social scene and campus at CMU fit me very well. I did visit McGill, but all I saw was the campus (which was massive); I didn't get a feel for the quality of the math dept. or the social climate. Of course, the biggest advantage of McGill is the price: it's heads and shoulders below CMU's, which offered me little in the way of financial aid or scholarships (still in review).

I need some quick advice! Is McGill good/cheap enough for me to forget CMU, a comfortable fit and a great school, in a good city?

I'm going to summer programs at both, oddly enough (the summer school held by the Center for Nonlinear Analysis at CMU and the Canadian Undergraduate Mathematics Conference held at McGill).  I'd probably say that CMU is a bit better, but McGill is a nicer place and the differences for an undergraduate in quality of education probably won't be that significant.  In terms of quality of peers, they placed around the same on the Putnam this year I think, for whatever that's worth.

Sent a PM to you, and for WIW  short answer - CMU - My opinion only.


[quote="José"]Find all pairs of positive integrers $(a;b)$ such that $a+b \le 100$ and $\frac{a+\frac{1}{b}}{\frac{1}{a}+b}=13$.[/quote]

[hide]Simplify to $\frac ab=13$ which has 7 solutions.[/hide]

[quote="SplashD"]
[hide]Simplify to $\frac{a}{b}=13$ which has 7 solutions.[/hide][/quote] :wink: :D

[hide]Simplify the rational expression.[/hide]

[hide]$\text{Simplify to}$ $\frac{a}{b}$

$\text{Then, we have: }$
$\left\lfloor \frac{100}{13}\right\rfloor = \boxed{7}$ $\text{solutions}$

:juggle:[/hide]


Yeah, but can anybody say why? Here...

[hide]Say length is l and width is w. So for the first rectangle, the area is l*w, obvoiusly. If the length is increased by 20%, it becomes 1.2*l in the new rectangle. Using this logic, the new width is 1.25*w. So the area of the new rectangle is 1.2*l*1.25*w or 1.5*l*w. Obviously there is a 50% increase from the area of the old rectangle to the area of the new rectangle.[/hide]

[hide]I just plugged in values like this:

5x4 rectangle turns into
6x5 rectangle
20
30
1/2=50%[/hide]

[hide] Call the length and width L and W. Thus, the new rectangle's area is:
1.2L*1.25W=1.5LW.

It is unmistakable than there is a 50% increase in area. [/hide]

[hide]Let the area be x. Since the length is 20% longer and the width is 25% longer, than the new area is (1+.2)(1+.25)x=1.5x, so the increase is [b]50%[/b].[/hide]

[quote="Treething"][quote="mathgeek2006"]If the length of a rectangle were increased by 20% and the width by 25%, by what percent would its area increase?[/quote]

[hide="5 second answer"]
Increased by a factor of 5/4 and 6/5.

5/4 * 6/5 = 6/4 = 1.5
[b]50%[/b]
[/hide][/quote]

Thanks, Treething, that was helpful.  It would save a few seconds on Countdown.  I alway just multiply it out with a 10x10, or something easy to multiply, but your way is much easier.


I am so confused.  I think the operative word is "directly" that is throwing me off.

In cellular respiration, most ATP is produced directly as a result of _____.

A.  the Krebs cycle
B.  glycolysis
C.  fermentation
D.  the diffusion of H+ ions across the inner mitochondrial membrane
E.  the electron transport chain

Both the Krebs Cycle AND glycolysis produce ATP.  But there can't be two answers.  This is driving me nuts.

If the question uses the word "most" wouldn't the answer be the Kreb's cycle? I might be wrong since I'm going from memory- and it's been a summer since bio.
I think the question is asking which process produces more molecules of ATP.

Uh, I'm going with the D. the diffusion of H+ ions across the inner mitochondrial membrane.

In both the Krebs cycle and Glycolysis, only a small number of molecules of ATP are formed (2 in Glycolysis, 4 in the actual Krebs Cycle). The real use of these processes comes in the storing of energy in NADH and FADH2 which release their hydrogens when the electron transport chain steals their electrons. The hydrogens cause a proton gradient which creates potential energy until a hydrogen diffuses across the inner mitochondrial membrane through the use of enzyme ATP Synthetase which releases enough energy to bond a phosphate group to ADP, creating ATP. With the great number of NADH and FADH2 created in Glycolysis and the Krebs Cycle, this gives many H+ ions to the process, creating the greatest amount of ATP out of the process (approx. 2 ATP per FADH2 and 3 ATP per NADH, creating 28 ATP).

Wind Slicer you're right!  Thank you so much!


The hexagon with the R is colored red.  Each hexagon is colored either red, yellow or green, such that no two hexagons with a common side are
colored the same color.  In how many
different ways can the figure be colored?

[asy]path a=(0,0)--(10,0)--(15,8.7)--(10,17.3)--(0,17.3)--(-5,8.7)--cycle;
draw(a);
draw(shift((15,8.7))*a);
draw(shift((15,-8.7))*a);
draw(shift((30,17.3))*a);
draw(shift((30,0))*a);
draw(shift((30,-17.3))*a);
draw(shift((45,26))*a);
draw(shift((45,8.7))*a);
draw(shift((45,-8.7))*a);
draw(shift((60,17.3))*a);
draw(shift((60,0))*a);
draw(shift((60,-17.3))*a);
draw(shift((75,8.7))*a);
draw(shift((75,-8.7))*a);
label("$R$",(5,10),S);[/asy]

There are $ \boxed2$ arrangements for the second column (YG and GY), after which the rest is fixed.

I still don't get it. How come the olors are fixed?

shouldn't it be 8 tho? can someone show me the cases?

The hexagons in the second column to the left must be yellow and green, but either color can be on top: 2 possibilities. With either possibility, the rest of the figure is easily colored in, with the color of every other hexagon being forced. (ex. In the third column, the middle hexagon must be red, and the top and bottom must be the opposite of the top and bottom in the second column) Thus, there are only $\boxed{2}$ possibilities.

AoPS's answer

please don't post in old threads ( >1 month or so) unless you have something constructive to say. In your case somebody else already posted the same solution.

You know how there's a number before the problem? I wonder what number 1 is.


geometry
3D geometry
pyramid
combinatorial geometry
geometry unsolved

Let E be a finite set of points in the 3D-space with the following properties:

[b]a.)[/b] E is not coplanar.
[b]b.)[/b] No three points of E are collinear.

Prove that either there are 5 points in E which are edges of a convex pyramid in whose interior there is no further point of E, or there is a plane which contains exactly three points of E.

The convex hull of the set of points is a polyhedron $\mathcal P$. Assume not all its faces are triangular, and take $A,B,C,D$ to be four consecutive points of one of the non-triangular faces. Let $X$ be a point not in the plane $(ABCD)$ whose distance to this plane is minimal. The pyramid $XABCD$ contains no point from our set inside it.

So, you just proved that all faces are triangular (and a point inside every face). It's not all, I think.

Oh, sorry, I forgot about the case when there are points inside the triangular faces.

Ok, here's another try, which just a modification of the well-known solution to [url=http://mathworld.wolfram.com/SylvestersLineProblem.html]Sylvester's problem[/url]:

Assume that on each plane determined by three points in our configuration, there's at least one other point. Let $A,B,C,D$ be such that the distance between $A$ and the plane $BCD$ is minimal. We may also assume that all the points in our set which belong to $(BCD)$, other than $B,C,D$, lie inside $BCD$. Let $X$ be such a point, and let $P_a$ be the projection of $A$ on $(BCD)$.

$X$ lies in one of the triangles $P_aBC,P_aCD,P_aDB$. Assume it's the first one, and let $P$ be the projection of $P_a$ on $(ABC)$. The distacne from $X$ to $(ABC)$ is at most $P_aP$, which is less than $AP_a$, and this contradicts the minimality of the distance between $A$ and $(BCD)$.
I hope it's Ok this time.

Now it seems OK. By the way, is not it true that for any finite not-complanar set in $\mathbb{R}^3$ there exists a plane, which contains exactly three points of it?

It's Darij posting right now.

I also find Grobber's solution correct, and I would like to add that the problem is exercise 34 in Arthur Engel, [i]Problem-Solving Strategies[/i], New York 1998, chapter 3.

As for Fedor's question:

[quote="Fedor Petrov"]is not it true that for any finite not-complanar set in $\mathbb{R}^3$ there exists a plane, which contains exactly three points of it?[/quote]

I remember Peter Scholze giving a counterexample to this at Rostock. Unfortunately, I don't remember the exact counterexample. Peter, could you find it again?

  Thanks,
  Darij

[quote="Darij"]
...

I remember Peter Scholze giving a counterexample to this at Rostock. Unfortunately, I don't remember the exact counterexample. Peter, could you find it again?

...
[/quote]

Yes, could you? :) (or anyone else)


calculus integration function limit trigonometry logarithms algebra

Hi, can anyone show me how to evaluate this integral.  Thanks again.

Question: $ \displaystyle\int_{\minus{}\infty}^{\infty}\frac{e^{2x}}{ae^{3x}\plus{}b}dx$ where $ a,b>0$

If you make the substitution $ u\equal{}e^x,$ this becomes
\[ \int_0^{\infty}\frac{u}{au^3\plus{}b}\,du\]
This puts it into the category of a rational function; factoring the denominator and applying partial fractions offers a method, although it will take some effort.

Would it be possible to use residues?  If we let
$ f(z)\equal{}\frac{z}{az^{3}\plus{}b}$, then there are poles at $ z\equal{}\sqrt[3]{\frac{\minus{}b}{a}}$.  As long as $ b\neq 0$, then the pole won't occur on the real line, so we get
$ \oint_{\Gamma}f(z)\ dz\equal{}2\pi\sum_{i}\text{Res}(f,z_{i})\equal{}\lim_{r\rightarrow\infty}\int_{0}^{r}f(z)\ dz\plus{}\int_{C_{r}^{\plus{}}}f(z)\ dz$,
where $ C_{r}^{\plus{}}$ is the semi-circle of radius $ r$ closing the path.  I believe there is some variation of Jordan's lemma which says that the second contour integral goes to 0 since the degree of the denominator exceeds the degree of the numerator by 2.  So I guess the hard part would  be computing the residues at the poles enclosed by the path.

I'm not sure if this actually works, I'm just throwing it out there.

[hide]Define $ I \equal{} \int_{\minus{} \infty}^{\plus{} \infty} \frac{e^{2x}}{ae^{3x} \plus{} b}\,dx$. Let's consider the change of variables

$ e^{3x} \equal{} \frac{b}{a} \tan^2u$,

which implies that

$ x \equal{} \frac23 \ln \left(\sqrt{\frac{b}{a}} \tan u\right)$ and $ dx \equal{} \frac23 \frac{\sec^2u}{\tan u} \,du$.

Therefore, with this substitution we have

$ I \equal{} \int_0^{\pi/2} \frac{(b/a)^{2/3} (\tan u)^{4/3}}{b(1 \plus{} \tan^2u)} \cdot \frac23 \frac{\sec^2u}{\tan u} \,du \equal{} \frac{2}{3 a^{2/3} b^{1/3}} \int_0^{\pi/2} (\tan u)^{1/3}\,du$.

I will leave as an exercise the proof that

$ \int_0^{\pi/2} (\tan u)^{p}\,du \equal{} \frac{\pi}{2} \sec\left(\frac{p \pi}{2}\right)$, with $ 0 < p < 1$.

In the present case we have $ p \equal{} \frac13$, and so we have

$ I \equal{} \frac{2}{3 a^{2/3} b^{1/3}} \cdot \frac{\pi}{2} \sec\left(\frac{\pi}{6}\right)$

$ \equal{}> I \equal{} \frac{2 \pi}{3 \sqrt{3} a^{2/3} b^{1/3}}$.[/hide]

Very nice solution Carcul!  Thanks!
[quote="Carcul"]I will leave as an exercise the proof that

$ \int_0^{\pi/2} (\tan u)^{p}\,du \equal{} \frac {\pi}{2} \sec\left(\frac {p \pi}{2}\right)$, with $ 0 < p < 1$.[/quote]
$ \displaystyle I \equal{} \int_0^{\frac {\pi}{2}}\tan^p\theta d\theta$, where $ 0 < p < 1$.
Let $ y \equal{} \sin\theta$ then $ \displaystyle dy \equal{} \cos\theta d\theta$, and when $ \theta \equal{} 0$, $ y \equal{} 0$ and when $ \theta \equal{} \frac {\pi}{2}$, $ y \equal{} 1$
Thus, $ \displaystyle I \equal{} \int_0^1 y^p(1 \minus{} y^2)^{ \minus{} \frac {p}{2} \minus{} \frac {1}{2}}dy$.
Let $ t \equal{} y^2\Rightarrow dt \equal{} 2ydy$, and the limits of integration remains the same.
Therefore, we obtain $ \displaystyle I \equal{} \frac {1}{2}\int_0^1t^{\frac {p \minus{} 1}{2}}(1 \minus{} t)^{ \minus{} \frac {p}{2} \minus{} \frac {1}{2}}dt \equal{} \frac {1}{2}\text{B}\left(\frac {p \plus{} 1}{2},\frac { \minus{} p \plus{} 1}{2}\right) \equal{} \frac {\pi}{2\cos\left(\frac {p\pi}{2}\right)}$, where $ 0 < p < 1$.


When you sketch a figure of an adult male, it is recommended that the head be $ \frac{1}{8}$ of
the total height. If the full drawn height of a 6-foot man is 10 inches, how many
inches should the drawing of his head be from top to bottom? Express your answer as
a decimal to the nearest hundredth.

We simply multiply.

$ \frac{1}{8} \times 10 \equal{} 1.25$

Our answer is $ \boxed{1.25}$.

[hide]multiply.

$ \frac {1}{8} \times 10 \equal{} 1.25$

or if u really wanted to, convert $ \frac {1}{8}$ into a decimal: $ 0.125$ then multiply

Our answer is $ \boxed{1.25}$.  [/hide]
BTW how do you do multiple boxes around an answer?

For example: \boxed{\boxed{\boxed{5}}} would turn out $ \boxed{\boxed{\boxed{5}}}$.

[quote="ernie"]Um... just a question for you myyellowducky82, but why do you have to post the same answer with virtually the same to get to the solution?[/quote]
well when i wuz typing it up i realized that isabella got the same thing xD... and i just figured out how to write fractions in LATEX...kinda wanted to practice my skillz then.
also i did the thing where u could convert 1/8 into a decimal...

Why does everyone think the world revolves around Microsoft and Windows?!?!?! Ever heard of Fedora? Linux? Open office? Writer? Is it just me and #H34N1?

well it's the most word-processing program used nowadays.

Well...that's only because when you buy computers a lot already have Windows installed so it's easy peasy lemon squeezy. You have to install Fedora, but it's free, and plus they give out the source code.

meh.

I've heard of practically all of those, but Microsoft is definitely the most common.  Not too much into those programs though.

thats true.
btw guyz this is gettin' a tiny bit spammy...


function
real analysis
real analysis unsolved

Suppose $ f$ is an extended value function on $ ( \minus{} \infty, \infty)$, there exists a countable set $ D$ with the following property. For each $ t$, there exists $ t_n$ in $ D$, $ t_n \to t$ such that $ f(t)\to f(t_n)$ as $ n \to \infty$.

Of course the problem is to prove this statement. I believe the key is to find an appropriate metric, but given my lack of imagination, I would like to ask in this forum. There's a hint telling us to consider the graph $ (t, f(t))$ and introduce a metric.

Thank you.
T.Y.M.

Edited: Now it should look better  :)

The graph $ G\equal{}\{(t,f(t))\colon t\in\mathbb R\}$ is a subset of $ \mathbb R^2$, and therefore it inherits a metric from $ \mathbb R^2$. Since $ \mathbb R^2$ is separable, so is $ G$. In other words, there is a countable set $ D\subset\mathbb R$ such that the set $ \{(t,f(t))\colon t\in D\}$ is dense in $ G$, and this is what the statement asks for.

I appreciate your help. Your argument is simple yet elegant.
I simply overlooked the fact that $ R^2$ is separable.

Thank you.


number theory
greatest common divisor
number theory unsolved

Find all non-negative integers $ x,y,z$ such that $ 5^x \plus{} 7^y \equal{} 2^z$.
 :lol:
([i]Daniel Kohen, University of Buenos Aires - Buenos Aires,Argentina[/i])

I dont know whether i am anywhere close to the solution, but i got using modulo some powers of two
that x is of the from 8l+2 whereas y is of the form 8l+1

Any help will be appreciated

Thx,
Befuddlers

After more than 2 years, I finally got a solution: [url=http://bboyjordan.wordpress.com/2010/07/25/list-of-funny-problems-2nd/]Problem 17- A sum of prime powers is not a power of 2[/url].

Btw, I copied it below:

[i]Solution.[/i] We will prove that if $(x,y,z)$ is a solution that $(x,y,z)\in S:=\{(0,0,1), (0,1,3), (2,1,5)\}$.

[i]1.[/i] If $x=y=0$ then $z=1$, so that $(0,0,1) \in S$.

[i]2[/i].If $x=0$ and $y\ge 1$ then $7^y+1=2^z$. If $2\mid y$ then in $\mathbb{Z}/4\mathbb{Z}$ we have $2^z=(-1)^y+1=2$, so that $y=1$, but $2=7^y+1\ge 7^1+1$ is absurd. Otherwise, $2\nmid y$, and in $\mathbb{Z}/16\mathbb{Z}$ we have $2^z=7^y+1=7\cdot 49^{\frac{y-1}{2}}+1=7+1$: this is enough to find the other trivial solution $(0,1,3)\in S$.

[i]3. [/i]If $x\ge 1$ and $y=0$ then $5^x+1=2^z$. Since $5^x+1\ge 6$ we have that $z\ge 2$, and taking $\mathbb{Z}/4\mathbb{Z}$ we obtain $0=2^z=5^x+1=2$, absurd.

[i]4.[/i] From now, we can suppose $\min\{x,y\}\ge 1$, so that $2^z=5^x+7^y\ge 5+7=12$ implies that $z\ge 4$.

[i]5.[/i] In $\mathbb{Z}/3\mathbb{Z}$ we have that $(-1)^z=2^z=5^x+7^y=(-1)^x+1$, so $2\mid x$ and $2\nmid z$.

[i]6.[/i] In $\mathbb{Z}/4\mathbb{Z}$ we have that $0=2^z=5^x+7^y=1+(-1)^y$, so $2\nmid y$.

[i]7.[/i] In $\mathbb{Z}/16\mathbb{Z}$ we have that $0=2^z=5^x+7^y=25^{\frac{x}{2}}+7\cdot 49^{\frac{y-1}{2}}=$ $9^{\frac{x}{2}}+7=3^x+7$ but residues of $3^i$ are exactly $[3,9,11,1]$, so that we can conclude that $4\mid x-2$.

[i]8.[/i] To sum up all observations in points 5,6,7 , we can reformulate the problem as "find all $(a,b,c)\in \mathbb{N}^3$ such that $5^{4a+2}+7^{2b+1}=2^{2c+1}$ for some $c\ge 2$.

[i]9. [/i]In $\mathbb{Z}/25\mathbb{Z}$ we have that $0=5^{4a+2}=2^{2c+1}-7^{2b+1}=2^{2c+1}-7\cdot (-1)^b$; and since $\text{gcd}(13,25)=1$ then also $0=13(2^{2c+1}-7\cdot (-1)^b)=4^c-16\cdot (-1)^b$. Now residues of $4^i$ are exacty $[4,16,14,6,24;21,9,11,19,1]$, and we are searching $16\cdot (-1)^b \in \{9,16\}$, that is equivalent to $5\mid c-2$.

[i]10.[/i] To sum up all observations in points 8,9 , we can reformulate the problem as "find all $(a,b,d)\in \mathbb{N}^3$ such that $5^{4a+2}+7^{2b+1}=2^{5(2d+1)}$. We divide the rest of the solution in two cases:

[i]PART 1.[/i]

[i]11.[/i] Let's find all solutions in the case $b=0$, that is $5^{4a+2}+7=2^{10d+5}$ in non-negative integers.

[i]12.[/i] The equation in point 11 is equivalent to $5^2\cdot (5^{4a}-1)=2^5\cdot (2^{10d}-1)$, and we can see the trivial solution with $a=d=0$ that is $(x,y,z)=(2,1,5) \in S$. Let's show that no other integer solutions exist.

[i]13.[/i] Consider the 2-adic valuations: it is true that $5=\upsilon_2(2^5(2^{10d}-1))=\upsilon_2(5^2(5^{4a}-1))=$ $\upsilon_2(5^{4a}-1)$. And by the lifting lemma it is equal to $\upsilon_2(\frac{5^2-1}{2})+\upsilon_2(4a)$, that implies $5=2+\upsilon_2(4a)$, so $2||a$.

[i]14.[/i] Consider the equation in $\mathbb{Z}/13\mathbb{Z}$ (after observing that $13\mid 5^4-1$), and we obtain: $6\cdot 10^d=2^5\cdot 2^{10d}=5^2\cdot 5^{4a}+7=5^2+7=6$ that is equivalent to $6=\text{ord}_{13}(10)\mid d$

[i]15.[/i] To sum up points 13,14, the equation becomes equivalent to $5^2\cdot (5^{2^3\cdot (2h-1)}-1)=2^5(2^{2^2\cdot 3\cdot 5k}-1)$ for some $h,k$ positive integers.

[i]16.[/i] After observing that $17\mid \text{gcd}(5^8+1,2^4+1)$ (that is equivalent to $\text{ord}_{17}(5)=2\text{ord}_{17}(2)=16$ ), consider the equation of point 15 in $\mathbb{Z}/17\mathbb{Z}$: we have that $5^2(5^{2^3\cdot (2h-1)}-1)=5^2((-1)^{2h-1}-1)=-2\cdot 5^2=1$ and $2^5(2^{2^2\cdot 3\cdot 5k}-1)=15((-1)^k-1) \in \{0,4\}$. This is a contradiction.

[i]PART 2.[/i]

[i]17.[/i] Now let's show that in the case $b\ge 1$, the equation $5^{4a+2}+7^{2b+1}=2^{10d+5}$ has no solutions.

[i]18. [/i] In $\mathbb{Z}/49\mathbb{Z}$ (the information in this point makes the difference from the previous part) we have $37^a=50\cdot 37^a=2\cdot(5^2\cdot 5^{4a})=2(5^{4a+2}+7^{2b+1})$ $=2(2^{10d+5})=64\cdot 2^{10d}=15\cdot 44^d=37^{18+16d}$. Now, since $\text{ord}_{49}(37)=21$, we conclude that $21\mid (18+16d)-a$, that is equivalent to $3\mid a-d$ and $7\mid (2d+4)-a$.

[i]19.[/i] Observe that $7^6-1=2^4\cdot 3^2\cdot 19\cdot 43$ and take the equation in $\mathbb{Z}/9\mathbb{Z}$, we have: $4^a+4^b=4\cdot (7\cdot 4^a+7\cdot 4^b)=4\cdot (25\cdot 5^{4a}+7\cdot 49^b)$ $=4\cdot (2^5\cdot 2^{10d})=2^7\cdot 2^{10d}=2\cdot 4^{2d}$. We can see directly by hand that residues of $4^i$ are exactly $[4,7,1]$ and residues of $2\cdot 4^{2i}$ are $[5,8,2]$. And we can see that $(a,b,d)$ is a solution if and only if $3\mid a+b-d$.

[i]20.[/i] Since we know that $3\mid a-d$ for point 18, then it is also true that $3\mid (a+b-d)-(a-d)=b$. In particular if $d$ is a divisor of $7^6-1$ then $7^{2b+1}$ is constant in $\mathbb{Z}/d\mathbb{Z}$.

[i]21.[/i] In $\mathbb{Z}/43\mathbb{Z}$ we have $25\cdot 23^a+7=5^{4a+2}+7^{2b+1}=2^{10d+5}=32\cdot 35^d$. Now $\text{ord}_{43}(23)=21$ and $\text{ord}_{43}(35)=7$: complete residues of LHS are $[23,31,0,18,2,21,28; 17,22,8,30,20,5,4; 24,11,13,16,42,38,32]$, and complete residues of RHS are $[2,27,42,8,22,39,32]$.

[i]22.[/i] Looking previous residues we can see that if $(a,d)$ is a solution modulo $7$ ( we can say that because $7=\text{ord}_{43}(35)\mid \text{ord}_{43}(23)\mid \varphi(43)$ ) then $(a,d)\in\{(5,1),(5,3),(3,4),(2,5),(7,7)\}$.

[i]23. [/i]In noone of previous couples $(a,d)$ modulo $7$ we verify that $7\mid (2d+4)-a$, as requested in point 18. This is a contradiction. []

Paolo Leonetti


[quote="hien"]Let $a,b,c>0$ satified $a+b+c=1$. Find minimum of expression:
$A=\frac{1}{a^{3}+2b^{3}+3c^{3}}+\frac{1}{b^{3}+2c^{3}+3a^{3}}+\frac{1}{c^{3}+2a^{3}+3b^{3}}$[/quote]
I think $\inf A=\frac{11}{6},$ but my proof is very ugly.

This problem is difficult because coeffients isn't symmetry. The problem due to me thinking when i taught inequality to my student


geometry
3D geometry
algebra solved
algebra

Can every integer be written as the sum of five perfect cubes?
[The perfect cubes are 1, 8, 27, ... but also 0, -1, -8, -27, ...]

You've got it :)

There is another technique:

6n = (n - 1)3 + (n + 1)3 - n3 - n3

and then notice that

6n + k = (n - 1)3 + (n + 1)3 - n3 - n3 + k = (n - 1)3 + (n + 1)3 - n3 - n3 + k3

modulo 6.

>> 6n = (n - 1)^3 + (n + 1)^3 - n^3 - n^3 : actually I started from there, but then I missed the mod 6 trick  :(


Compute: $\textstyle \sqrt{2\frac{1}{4}} + \sqrt{1\frac{7}{9}}$.  Express your answer as a mixed numeral.

Your solution was going well, but then you slipped up on the final step.

Your LaTeX was impressive for a first effort.  To add a line break, you can type either \\ or \newline.  For an extra touch, you might also want to add an equals sign between expressions, and to have expressions aligned under each other.  There is a LaTeX environment called eqnarray* that will do the job.  For example, you can write
\begin{eqnarray*}
221
  & = & 225 - 4 \\
  & = & 15^2 - 2^2 \\
  & = & (15 - 2)(15 + 2) \\
  & = & 13 \cdot 17.
\end{eqnarray*}

(You can see how I did that by putting your cursor over the display or double-clicking on it.)

Doh!

[hide]So $2\frac{5}{6}$.

Wow, how did I do that?[/hide]

I'm fixing my previous "solution" to eqnarray*, but leaving mistakes in.[/hide]

It can be written in-forum. Click on the equation and/or press quote to see how.

Yes, you don't need any software to use LaTeX on this forum.  Just start typing within dollar signs.  For example, you can type $a^2 + b^2 = c^2$ to get the Pythagorean theorem.

(I disabled LaTeX in this post so you can see what to type.)