I was trying to find a general solution in positive integers to simultaneous equations of the form

x+y-z=n
(x^2)+(y^2)-(z^2)=n

I won't post the steps in my solution because I don't know latex and it will look a mess

I got (x-n)(y-n)=0.5n(n-1) which I know is correct.

So (x,y) = (a+n , b+n)  where (a,b) is a positive factor pair of 0.5n(n-1)

However, I am unsure what to do with the case where (x-n), (y-n) are negative, but x, y are positive. For example, when n=6 we get (x-6)(y-6)=15.

This could be (x-6)= -3 , (y-6) = -5, which gives (x,y)=(3,1)

In this particular case, the value of z is negative so the solution is invalid, but is there some way that I can prove that the factors cannot be negative, or find the criteria for which negative factors give positive solutions for the general case?

Any help would be appreciated, and thanks in advance.

[b]EDIT: I just found out that a couple of BMO questions have actually been based on equations like this. Q2 of BMO1 2007/2008 was to solve these equations where n=12. And in BMO1 1998/1999 Q1 can be solved once the youngest child's age has been found by fiddling around to get: b+c-a=5 and (b^2)+(c^2)-(a^2)=5[/b]

[hide="Solution"]If we have
\[ x \plus{} y \minus{} z \equal{} n\]
and
\[ x^2 \plus{} y^2 \minus{} z^2 \equal{} n,\]
then WLOG we have $ x \equal{} x^2, y \equal{} y^2,$ and $ z \equal{} z^2$. But the only time this happens is when $ (x,y,z) \equal{} (1,1,1)$. $ \Box$[/hide]

[quote="pytheagle"]then WLOG [/quote]  Similarly, since $ 1 \plus{} 4 \equal{} 5$ and $ 2 \plus{} 3 \equal{} 5$ we have $ 1 \equal{} 2$ and $ 4 \equal{} 3$   :roll:

To answer the OP's question: $ x \plus{} y \minus{}z \equal{} n$ and $ z > 0$ gives $ n < x \plus{} y$.  From this we have that one of $ x, y$ must be at least $ \frac {n}{2}$.  Without loss of generality ( :wink: ), say $ x \geq \frac {n}{2}$.  If $ x < n$ then $ |x \minus{} n| \leq \frac {n}{2}$.  Also if $ 0 < y < n$ then $ |y \minus{} n| < n$.  Thus $ |(x \minus{} n)(y \minus{} n)| < \frac {n}{2} \cdot n < \frac {n(n \plus{} 1)}{2}$.  So there are no solutions in this case.

@pytheagle No. There are many solutions that are not (1,1,1). Example for n=4.

(x-4)(y-4)=6

(x,y)=(10,5),(5,10),(6,7),(7,6)

Plugging these into x+y-z=4 gives

(x,y,z)=(10,5,11),(5,10,11),(6,7,9),(7,6,9)

And these are only the solutions for n=4. There are many solutions for all values of n.

My question was whether negative factors (x-n),(y-n) could ever yield positive integer solutions, and if so, when?

Thanks JBL  :)

I had a feeling that negative factors would never give solutions, but you can never be sure with these things, and I wasn't sure how to prove it.

EDIT: Wait, how does proving that (x-n)(y-n) < n(n+1)/2 prove anything? I only just realised that you have a + rather than a -. It should be (x-n)(y-n) < n(n-1)/2 we are trying to prove shouldn't it?

Sorry, I solved something slightly different, but there's enough wiggle room in my estimates to fix it:

From $ n < x \plus{} y$ we may assume $ x > \frac {n}{2}$ and so $ |x \minus{} n| < \frac {n}{2}$, while from $ 1 < y < n$ we have $ |y \minus{} n| \leq n \minus{} 1$.  Thus $ |(x \minus{} n)(y \minus{} n)| < \frac {n}{2} \cdot (n \minus{} 1) \equal{} \frac {n(n \minus{} 1)}{2}$.

By the way, LaTeX is very easy to learn -- you're already using some of its notations (e.g., the carrot for exponentiation).  Enclose code in dollar signs, and you can click on other people's code to see how they do what they do.  See also http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 and [[LaTeX:Symbols]] and [[LaTeX:Commands]].

TYVM for the clear explanation, and I'm definitely going to start using LaTeX now since it's so much easier to read.


algebra
polynomial
complex numbers
algebra proposed

Let $ f(x)$and $ g(y)$ be two monic polynomials of degree=$ n$ having complex coefficients.
We know that there exist complex numbers $ a_i,b_i,c_i \forall 1\le i \le n$, such that
$ f(x)\minus{}g(y)\equal{}\prod_{i\equal{}1}^n{(a_ix\plus{}b_iy\plus{}c_i)}$.
Prove that there exists $ a,b,c\in\mathbb{C}$ such that
$ f(x)\equal{}(x\plus{}a)^n\plus{}c\text{ and }g(y)\equal{}(y\plus{}b)^n\plus{}c$.

this is not a hard problem if my solution is correct.
are u there in the training camp. how is it going on. please post all the problems

$ f(x) - g(y) = \prod_{i = 1}^n{(a_ix + b_iy + c_i)}$.
We know $ \prod_{i = 1}^{n} a_i = 1$ as $ f(x)$ is monic. Factoring out the $ a_i$'s, and writing $ b_i/a_i = d_i, \; \; c_i/a_i = e_i$, we get
$ f(x) - g(y) = \prod_{i = 1}^n{(x + d_iy + e_i)} \Rightarrow \prod_{i = 1}^{n} d_i = - 1$.
Comparing co-efficients of $ x^{n - i}y^i$ for $ 1\leq i \leq n - 1$ on both sides, we get
$ \sum_{1\leq j_1\leq j_2...\leq j_i \leq n} d_{j_1}d_{j_2}\cdots d_{j_i} = 0$.
In other words, $ d_1,d_2,\cdots,d_n$ are the roots of the equation $ X^n - (-1)^n = 0$.
Without loss of generality $ d_i = - e^{2\pi(i - 1)/n} \hspace{0.05in} \text{ for } 1\leq i \leq n$.
We write $ \omega = e^{2\pi/n}$. Hence, $ d_i = - \omega^{i - 1}$.
$ f(x) - g(y) = \prod_{i = 1}^n{(x - \omega^{i - 1}y + e_i)}$---------(*).
Substituting, $ x = \omega^{i - 1}y - e_i$ in (*), we get
$ f(\omega^{i - 1}y - e_i) - g(y) = 0 \;\; \text{ for } 1\leq i \leq n$------------(**).
Substituting, $ y = 0, \;x = \omega^{i - 1}y - e_i$ in (*), we get
$ f(\omega^{i - 1}y - e_i) - g(0) = \;\;\prod_{j = 1}^n{(\omega^{i - 1}y - e_i + e_j)} \text{ for } 1\leq i \leq n$-------------(***).
Using (**) in (***), we get
$ g(y) - g(0) = \;\; \prod_{j = 1}^n{(\omega^{i - 1}y - e_i + e_j)} \text{ for } 1\leq i \leq n$
Simplifying the RHS, we get
$ g(y) - g(0) = \;\; \prod_{j = 1}^n{\left(y - \frac {(e_i - e_j)}{\omega^{i - 1}}\right)} \text{ for } 1\leq i \leq n \;\;(\because (\omega^{i - 1})^{n} = 1)$
Sum of the roots of the polynomial $ g(y) - g(0)$ is $ \sum_{j = 1}^{n} \frac {(e_i - e_j)}{\omega^{i - 1}} = \frac {ne_i - \sum_{j = 1}^{n} e_j}{\omega^{i - 1}} \text{ for } 1\leq i \leq n$
In other words, ${ \frac {ne_i - \sum_{j = 1}^{n} e_j}{\omega^{i - 1}} = ne_1 - \sum_{j = 1}^{n} e_j} \text{ for } 1\leq i \leq n$. Writing $ e_i$ in terms of $ e_1$, we get
$ e_i = \frac {1}{n}\sum_{j = 1}^{n} e_j - \left( \frac {1}{n}\sum_{j = 1}^{n} e_j - e_1 \right)\omega^{i - 1} \text{ for } 1\leq i \leq n$.
We note that $ \frac {1}{n}\sum_{j = 1}^{n} e_j$ and $ \frac {1}{n}\sum_{j = 1}^{n} e_j - e_1$ are independent of $ i$. Hence, setting $ a = \frac {1}{n}\sum_{j = 1}^{n} e_j$ and $ b = \frac {1}{n}\sum_{j = 1}^{n} e_j - e_1$, we get
$ e_i = a - b\omega^{i - 1} \text{ for } 1\leq i \leq n$.
Substituting the above expression for $ e_i$ in (*), we get
$ f(x) - g(y) = \prod_{i = 1}^n{\left(x - \omega^{i - 1}y + a - b\omega^{i - 1}\right)} = \prod_{i = 1}^n{\left(\left(x + a\right) - \omega^{i - 1}\left(y + b\right)\right)}$
$ \Rightarrow f(x) - g(y) = \left(x + a\right)^n - \left(y + b\right)^n \;\;(\because \prod_{i = 1}^n{\left(X - \omega^{i - 1}Y\right)} = X^n - Y^n)$.
$ \Rightarrow f(x) = (x + a)^n + c, \;\;g(y) = (y + b)^n + c$.


[quote="Ignite168"]Co-winners!

Oh and fireemblem, I can prove 1+1=41, want to see?

$1+1=41_{\frac{1}{4}}$[/quote]

Are you Chuck Norris?

[quote="mathgeniuse^ln(x)"][quote="Ignite168"]Co-winners!

Oh and fireemblem, I can prove 1+1=41, want to see?

$1+1=41_{\frac{1}{4}}$[/quote]

You must be proud of yourself.
Unfortunately you can't use a number larger than that number when you are in that base.

Like base 2, you can only use 0,1
Base n, you can only use 0,1,2,...,n-1

No exceptions.

[color=white]Chuck Norris can prove that he failed Math.[/color][/quote]

How do you write numbers in bases less than 1 then? :?

[quote="mathgeniuse^ln(x)"][quote="Ignite168"]Co-winners!

Oh and fireemblem, I can prove 1+1=41, want to see?

$1+1=41_{\frac{1}{4}}$[/quote]

You must be proud of yourself.
Unfortunately you can't use a number larger than that number when you are in that base.

Like base 2, you can only use 0,1
Base n, you can only use 0,1,2,...,n-1

No exceptions.

[color=white]Chuck Norris can prove that he failed Math.[/color][/quote]

Or you could just consider it unsimplified.

[quote="Ignite168"][quote="SplashD"][quote="mathgeniuse^ln(x)"]Let's just say you can't.  But I really don't know the answer to your question, SplashD.[/quote]

Would you numbers like $\frac{1}{5}\frac{1}{6}_{\frac{1}{4}}$?[/quote]

It's much easier to express it as $\frac{13}{60}_{10}$ though.[/quote]

Not if you're trying to prove 1+1=41

[quote="SplashD"][/quote][quote="Ignite168"][quote="SplashD"][quote="mathgeniuse^ln(x)"]Let's just say you can't.  But I really don't know the answer to your question, SplashD.[/quote]

Would you numbers like $\frac{1}{5}\frac{1}{6}_{\frac{1}{4}}$?[/quote]

It's much easier to express it as $\frac{13}{60}_{10}$ though.[/quote][quote="SplashD"]

Not if you're trying to prove 1+1=41[/quote]

Where would $\frac{1}{5}\frac{1}{6}_{\frac{1}{4}}$ come in in that?

Anyway, I think you can have complex bases (not sure though) so I'm pretty sure you can have fractional bases.

[quote="Ignite168"][/quote][quote="SplashD"][/quote][quote="Ignite168"][quote="SplashD"][quote="mathgeniuse^ln(x)"]Let's just say you can't.  But I really don't know the answer to your question, SplashD.[/quote]

Would you numbers like $\frac{1}{5}\frac{1}{6}_{\frac{1}{4}}$?[/quote]

It's much easier to express it as $\frac{13}{60}_{10}$ though.[/quote][quote="SplashD"]

Not if you're trying to prove 1+1=41[/quote][quote="Ignite168"]

Where would $\frac{1}{5}\frac{1}{6}_{\frac{1}{4}}$ come in in that?

Anyway, I think you can have complex bases (not sure though) so I'm pretty sure you can have fractional bases.[/quote]

I was just giving a random example.

[quote="biffanddoc"]1+1 cannot =41 in any base because it 41 has 2 digits and 1 is 1 digit. That means that 1 is at least half of the base we are working in, and so the base has to be <2. But we cant have a 4 in base 2 or less, so yeah $1+1\neq41$[/quote]

Well, Chuck Norris can still probably prove that 1+1=41.

FINE. but I cant prove it. only chuck norris can.

If base -1/4 existed, then 1+1=41. But thats kinda weird.

[quote="SplashD"][/quote][quote="Ignite168"][/quote][quote="SplashD"][/quote][quote="Ignite168"][quote="SplashD"][quote="mathgeniuse^ln(x)"]Let's just say you can't.  But I really don't know the answer to your question, SplashD.[/quote]

Would you numbers like $\frac{1}{5}\frac{1}{6}_{\frac{1}{4}}$?[/quote]

It's much easier to express it as $\frac{13}{60}_{10}$ though.[/quote][quote="SplashD"]

Not if you're trying to prove 1+1=41[/quote][quote="Ignite168"]

Where would $\frac{1}{5}\frac{1}{6}_{\frac{1}{4}}$ come in in that?

Anyway, I think you can have complex bases (not sure though) so I'm pretty sure you can have fractional bases.[/quote][quote="SplashD"]

I was just giving a random example.[/quote]

Indeed, you can have fractional bases. Fractional bases are quite a bit of fun, and are quite useful also. My specialty presently is fractional bases, and from what I know, I'm the only person to investigate fractional bases in depth since their discovery by A. J. Kempner in 1936.

Yes, you may also have negative bases, or negative fractional bases, irrational bases, or complex bases. I don't know too much about complex bases; I recommend investigating Donald Knuth's work on irrational bases. I don't know a source for complex bases.

Neal, there are such exceptions, as I outlined above. If you could use only 1 and 0 in an improper fractional base, every integer would have an infinitely long representation in every improper fractional base. But an integer should be able to be represented without a decimal point in any rational base. Thus we use up to the larger of the denominator or numerator of the fractinal base as digits, as shown above.

In fact, 1+1=41 in base 1/4. For a fractional base $\frac pq$, we use for digits $0,1,2,\cdots,max(p,q)-1$. In base 3/2 we use 0,1,2 as digits; in base 1/4 we use 0,1,2,3. (Or, as in the case at hand, we may use, instead of 0, $max(p,q)$, so for base 3/2, we'd use 1,2,3 as digits, and for 1/4 we'd use $1,2,3,4$. )

I am kind of hazy on proper fractional bases - I'm an expert more in improper fractional bases - but anyway, in that variation of base 1/4, indeed 1+1=41. (Or, if you want to, 1+1=2 in base 1/4. The rules are kind of loose for proper fractional bases. In bases less than 1, or irrational bases, a number does not have a unique representation. For instance, base 1/4: 4 in base 10 is the same as .1 or 4 or ...333 (extending infinitely to the left) in base 1/4.)

Once, I wanted to prove that 1+1=3, but I came across a problem:

1+1 = 3 in base x
(3*1)+(0*x) = 2
0x = -1
1+1 = 3 in base -1/0

Is this possible?

Also - how do I use fractions in latex?

Is it possible?  Not unless you're Chuck Norris.

And to do fractions, you can use \frac{numerator}{denominator}.


Let $ W$ be a subspace of an inner product space $ V$.  Suppose $ \exists$ a projection $ T$ on $ W$ along $ W^{\perp}$ such that $ N(T)\equal{}W^{\perp}$.  Prove that $ \parallel{}T(x)\parallel{}\leq \parallel{}x\parallel{}$ $ \forall x\in V$.

A a function $ T: V\rightarrow V$ is called a projection on $ W_{1}$ along $ W_{2}$ if, for $ x\equal{}x_{1}\plus{}x_{2}$ with $ x_{1}\in W_{1}$ and $ x_{2}\in W_{2}$, we have $ T(x)\equal{}x_{1}$.

Thank you.

??$ \parallel{}x_1\parallel{}^2 \equal{} \parallel{}x\parallel{}^2 \minus{} \parallel{}x_2\parallel{}^2 \leq \parallel{}x\parallel{}^2$

Thank you.  Do you by any chance know how to prove Parseval's Identity/Equality?

If $ \{e_1,...,e_n\}$ be an orthonormal basis for $ V$ then $ x\equal{}\sum_{i\equal{}1}^n \left\langle x,e_i\right\rangle e_i \Longrightarrow \forall y \in V$ we have $ \left\langle x,y\right\rangle \equal{}\sum_{i\equal{}1}^n \left\langle x,e_i\right\rangle \overline{\left\langle y,e_i\right\rangle}$.

I know what it says, I need a proof of it.

If your version of parseval is the one ehsan2004 wrote above, then just take inner product with y in the equality about x.

This is what I have-please correct if wrong:
$ \left\langle x,y\right\rangle\equal{}\left\langle\sum_{i\equal{}1}^{n}\left\langle x,v_{i}\right\rangle v_{i},y\right\rangle\equal{}\sum_{i\equal{}1}^{n}\left\langle\left\langle x,v_{i}\right\rangle v_{i},y\right\rangle\equal{}\sum_{i\equal{}1}^{n}v_{i}\left\langle\left\langle x,v_{i}\right\rangle,y\right\rangle$

How do I now get that last inner product in there?


algebra
polynomial
number theory unsolved
number theory

Prove:
$ \exists a \in \mathbb{N^*} | \frac{a^{29} \minus{} 1}{a\minus{}1}$ has at least 2007 distinct prime factors.

(in English: prove that exists a positive integer $ a$ such that $ \frac{a^{29} \minus{} 1}{a\minus{}1}$ has at least 2007 distinct prime factors.)

Thanks in advance.

If $ p|\frac{a^{29}-1}}{a-1}, p>29$, then $ p=1\mod 29$ and for all prime $ p=1\mod 29$ we have at least one $ a_p\not =1mod p$, suth that $ p|\frac{(a_p+lp)^{29}-1}{a_p+lp-1}$. Thtrefore by chinese theorem we can choose a, suth that $ a\equiv a_{p_i}\mod p_i, i=1,2,...,n\ge 2007$.

Obivicious $ f(a)\equal{}\frac{a^{29}\minus{}1}{a\minus{}1}$ is a polynomial of a.


1. An arrow is spun once and lands on one of three equally likely regions, numbered 1,2, and 3. (Please imagine a circle, and divided into three regions that says 1,2,3 on it) List all eight possible events for one spin of the arrow. [color=darkred]I don't think there are eight outcomes. I don't really know the rules for spinning the arrow though. But how can one arrow points two numbers at the same time and even three numbers at the same time? [/color]The answers on my book is { },{1}, {2},{3},{1,2}, {1,3},{2,3},{1,2,3}. [color=darkred]I don't understand, hope you can tell me.[/color]
2. If two cards are drawn from an ordinary deck without replacement, what is the probability that the cards form a pair? [color=darkred]Since i don't really know how to play cards, so i don't understand what it means by 'form a pair'? [/color]
3. I am confused with the counting principle which is to multiplicate and the disjoint sets in a problem, like i don't know whether i need to use counting principle to time everything together or use disjoint sets to plus?
Thanks so much for helping me out.[/b]

[hide="Number 2"]
A pair refers to two cards that have the same face value (e.g. Two Kings or two fives is a pair).
An ordinary deck of cards has 52 cards with 4 cards of each face value. If two cards are drawn the first card can be any card and the second card must be one of the 3 remaining cards that share that same face value.  Since they are drawn without replacement the second card is drawn from 51 cards. Thus the probability in question is $\frac{3}{51}$.[/hide]

Number 3 - I'm not quite sure what you're asking.  Two sets are disjoint if they have no elements in common.  Since you're asking about probability, lets look at it from a probability standpoint.
Say we have two events $A$ and $B$.  By definition $P(A\cup B) = P(A)+P(B)-P(A \cap B)$.  But if $A$ and $B$ are disjoint then $P(A \cap B ) = 0$. Thus for disjoint events $A$ and $B$, $P(A \cup B) = P(A)+P(B)$.
For your question about multiplication:
In probability, probabilities of events are multiplied when the events are independent.
By definition two events $A$ and $B$ are independent if:
$P(A \cap B) = P(A)\cdot P(B)$

Number 1 appears to be worded incorrectly

if the arrow can only land in one of the three regions, then it can't be in more than one of them
Are you sure you copied the question word for word from the book?

[quote="DiscreetFourierTransform"][quote="archimedes1"]Number 1:
Well, if you ripped off the arrow and threw it at the circle, that would work.  If you missed the circle completely, you'd get the empty set.  Likewise, if the arrow landed across all three regions, you'd get {1,2,3}.

Keep in mind these 8 events are not equally likely.[/quote]

That really made my day!

That would be the funniest probability question to see on an AMC:
Billy is playing a game with a spinner and two dice.
He gets fed up, and decides to throw the pointy spinner at his friend Bobby.
Assume that there is equal probability of the spinner landing in either of Bobby's eyes, nostrils, and mouth (and that it will land nowhere else).
There is a 1/2 chance he misses Bobby completely, because Bobby is a champion dodgeball player.

If Billy hits Bobby in the eye, then he gets to roll the two dice. If the sum of the two dice is exactly 8, then Billy wins all of the fruit in the refrigerator.

What are the odds in favor of Billy getting to eat fruit?[/quote]

The chance that Bobby gets hit in the eye is $\frac{1}{6}$, since there is a 1/3 chance that he hits Bobby in the eye, if he hits him at all.

The chance that the sum of the two dice is exactly 8 is $\frac{5}{36}$ because there are:

$(2,6)$
$(3,5)$
$(4,4)$
$(5,3)$
$(6,2)$.

Thus the resulting probablility is $\frac{1}{6}\cdot \frac{5}{36}= \boxed{\frac{5}{216}}$.

 :)

[quote="DiscreetFourierTransform"]Close! Each eye has it's own probability of 1/16, not the eyes being treated as one entity.
So it's really 1/8 that he gets hit in an eye.
So it's $\frac{1}{8}\cdot \frac{5}{36}= \frac{5}{288}$

The odds in favor of that are $5: 283$.. or $0.017667844522968197879858657243816$ to $1$[/quote]

How about the nostrils? Shouldn't they count seperately too if the eyes do?  :lol:

oooo I got pwned ;-)
They should..

Whatever, the idea is the same.
I like these kind of silly problems on math tests.

Or maybe... we should take the average area of the nostrils, eyes, and mouth and apply it to the problem.

Hahaa...look for this as #14 on the 2009 AMC 12.  :P


for given  irrational A ,positive real E,show there exists natural no. n,
s.t. (n^2*A)<E,where (G) =min of fractional part of G and 1-G. ;)


geometry
3D geometry
rotation
binomial coefficients

An 11 x 11 x 11 wooden cube is formed by gluing together $11^3$ unit cubes.  What is the greatest number of unit cubes that can be seen from a single point?
(A) 328
(B) 329
(C) 330
(D) 331
(E) 332

[hide]
It seems to me that 3 faces will be visible so just the surface area of 3 faces, but there the edge cubes are visible on 2 faces and 1 vertex will be visible on three faces, so its:
$3(11*11) - (11+10+10)$
So E 32 ?
[/hide]

Close.  You're on the right track with the "grouping" -- can you make it even easier on yourself?

[hide]There are at most 3 facees to see. One of the faces has 11*11 cubes; the ther face has 10*11 cubes; the third has 10*10 cubes. The total is 331. D) [/hide]

hatsup,

You are correct, but there is a simpler solution.  You are thinking in 2 dimensions.  Is there a more efficient way to think about this problem?

[hide]If you can look at 3 faces at once, then you can see 121 unique cubes on one side, 110 on a second side and 110-10=100 on the third, making 331. However, if you harness the power of a mirror, you can now see 100 cubes on a forth side and another 91 cubes on the fifth, meaning you can see a total of 522 cubes.[/hide]

Ha -- no mirrors allowed.

Still, this is also a 2-D counting solution.  You did however mention that you can see three faces at once...

Well if you harness the power of mirrors,  (but NO x-rays) you can see all the sides.. (Same question can be asked in the following form too -   you can rotate the cube  - or if the outer layer is colored, how  many cubes are colored etc..)

The hint Mathew is giving, it  will answer the original, as well as this question, in an easy  way...
[hide]Ask yourself how many cubes you can not see?[/hide]

I think I got the 3D solution.
[hide]If you can see the outside 3 faces, then what you cant see is a $10\times 10\times10$ cube. The outside three faces "cap" the $10\times 10\times10$ cube to make the $11\times 11\times 11$ cube!
Therefore we can see $11^3 -10^3 =331$ cubes.[/hide]

Nicely done.

There are a number of great relationships in this problem including the fact that the digits of the answer reveal binomial coefficients:
\[ (10 + 1)^3 = {3 \choose 3}10^3 + {3 \choose 2}10^2 + {3 \choose 1}10^1 + {3 \choose 0}10^0, \]
so
\[ (10 + 1)^3 - 10^3 = {3 \choose 2}10^2 + {3 \choose 1}10^1 + {3 \choose 0}10^0 = 3 \cdot 10^2 + 3 \cdot 10^1 + 1 \cdot 10^0 = 331. \]

[hide]BTW - Similary the answer to the other problem would be   $11^3 - 9^3$ [/hide]


Prove that for $a,b,c>0$ we have $\displaystyle\sum\frac{a}{b+c}\geq \frac{21}{16}+\frac{27(\sum a^3)}{16(\sum a)^3}$

Take $a+b+c=3$, the inequality is equal to
$\frac{a}{3-a}+\frac{b}{3-b}+\frac{c}{3-c} \geq \frac{21}{16}+\frac{a^3+b^3+c^3}{16}$
Let prove that $\frac{a}{3-a} \geq \frac{-2}{16}+\frac{9a+a^3}{16}$
This one is equal to $(a-1)^2(a^2-a+6) \geq 0$
Obtain the inequality for b,c too then add them up to have what we need  :) .


Sorry if this is in the wrong section

If $\sin\theta+\cos\theta=.8$ then what does $\sin2\theta$ equal?


function
integration
real analysis
real analysis unsolved

What is the necessary and sufficient condition for the continous functions $ A,B$ in order to exist two solutions $ f_1, f_2$ of the equation $ y\"+ Ay'+By=0$ such that $ f_1 f_2=1$?

Suppose that $y$ and $1/y$ are solutions:

*If $y$ and $1/y$ are independant solutions, they form
a basis of solution. The wronskian is $\frac{-2y'}{y}$ so
we can conclude that $y'(t) \neq 0$ for all $t \in R$

We will express $A(t)$ and $B(t)$ as
functions of $y$. We know that $Ay'+By=-y\"$ and $-Ay'y+By^2=y\"y-2y'$
(because $y$ and $1/y$ are solutions)
This is a Cramer be cause the determinant is $2y'y \neq 0$ for all $t$.
A few calculation leads to $B(t)=\frac{-y'}{y^2}$.

So, since $y$ and $1/y$ are solution we can assume that $y$ increasing
Hence $B(t)<0$
But we can also assume that $y$ is decreasing.
hence $B(t)>0$, contradiction.

So the $y$ and $1/y$ do not form a basis: $y=\frac{k}{y}$ where
$k \neq 0$ is a constant. Thus $y \neq 0$ is constant wich leads to
$B(t)=0$

This is why I think that $B(t)=0$ is a necessary and sufficient condition.
Is it correct ?

[quote="alekk"]This is why I think that $B(t)=0$ is a necessary and sufficient condition.
Is it correct ?[/quote]
I dont think so !  I found  $\displaystyle B(t)e^{\int 2A(t)} = K (constant)$ (négative if you want only real solutions).

 :cool:  Am I right ?   :cool:

[quote="alekk"]I think it should be more interesting to post proof rather that affirmations like this  :)
It is possible that my sln is false since there were some ugly calculation  :([/quote]
It's not possible , it's sure  :D: take for ex. $A=0,B=-1$
Anyway, in fact you made a mistake because  if $ \displaystyle y,\frac{1}{y}$ are solutions then a   few calculation leads to $\displaystyle B(t)=\frac{-y'^2}{y^2}$ (and not to $\displaystyle \frac{-y'}{y^2}$ :D )
Now it's easy : $\displaystyle y = Ce^{ \pm{\int \sqrt{-B(t)}}}$ . Put it in the differential equation and finally you should find $\displaystyle B(t)e^{\int 2A(t)} = K (constant)$
Reciprocally from $\displaystyle B(t)e^{\int 2A(t)} = K (constant)$ you can easily check  that $\ \displaystyle y_1=e^{\int \sqrt{-B(t)}},y_2=\frac{1}{y_1}$ are solutions of the  differential equation .
 :cool:  Do you follow me ?  :cool:


Let x, y be integers different from -1 such that
\[ \frac{{x^4  \minus{} 1}}{{y \plus{} 1}} \plus{} \frac{{y^4  \minus{} 1}}{{x \plus{} 1}}
\]  is an integer. prove that  \[ x^4 y^{44}  \minus{} 1\] is divisible by [b]x+1[/b]

It's have been posted before but i have forgotten the link.
Hint:From the hypothesis easy to see that $ x\plus{}1|y^4\minus{}1$ then $ x^4y^{44}\minus{}1\equal{}y^{44}(x^4\minus{}1)\plus{}y^{44} \minus{}1\vdots x\plus{}1$


[color=purple] Let two triangles $ A_1A_2A_3$ and $ B_1B_2B_3$ in plane which have three altitudes and bisectors $ l_i,h_i$ and $ l'_i,h'_i$ respectively. with $ i\equal{}1,2,3$ and two areas $ S$ and $ S'$.
  Prove that $ a^2x^2\plus{}b^2y^2\plus{}c^2z^2\ge 16SS'max\lbrace\frac{l_il'_j}{h_ih'_j}\rbrace$
In which $ j\equal{}1,2,3$[/color]

[quote="evarist"][color=purple] Let two triangles $ A_1A_2A_3$ and $ B_1B_2B_3$ in plane which have three altitudes and bisectors $ l_i,h_i$ and $ l'_i,h'_i$ respectively. with $ i \equal{} 1,2,3$ and two areas $ S$ and $ S'$.
  Prove that $ a^2x^2 \plus{} b^2y^2 \plus{} c^2z^2\ge 16SS'max\lbrace\frac {l_il'_j}{h_ih'_j}\rbrace$
In which $ j \equal{} 1,2,3$[/color][/quote]
 [color=purple] Sorry because my mistake  :( in the above inequality $ a,b,c$ are three sides length of triangle $ A_1A_2A_3$ and $ x,y,z$ are three sides length of triangle $ B_1B_2B_3$ :) [/color]


Ring Theory
superior algebra
superior algebra solved

Can someone please give me an example of an uncountable proper subring of $\mathbb{R}$?

Thanks. :)

The union of $\mathbb Q$ and all [url=http://en.wikipedia.org/wiki/Liouville_number]Liouville numbers[/url].

Edit: You'd better check if this indeed works.  :|

The transcendence degree of $\mathbb R$ over $\mathbb Q$ is $2^{\aleph_{0}}$, so pick a transcendence basis $(e_\alpha)_\alpha$ and consider the field obtained by appending to $\mathbb Q$ all $e_\alpha$ except for one.


superior algebra
superior algebra solved

Prove, that polynom x^(8k+4) + x^(2k+1) + 1 isn't reducible in the field Z2("zet-two).

Excuse me, I made mistake. My question is:  for which k it is true?

:?: Stop! But our polynom ends on x^6, although must ends on x^7

More exactly: on [b]x^6+1[/b] instead [b]x^7+1[/b]


I've been trying to dig up study materials or links to study materials for the International Physics Olympiad and it's chemistry counterpart, but I can't seem to find anything nicely organized like the AoPS list or [url=http://olympiads.win.tue.nl/ioi/study/books.html]IOI list.[/url]
EDIT: I also think it is very easy to enter the communities of olympiad mathematics and iformatics. It is as simple as logging on to a website. But I don't think there is a AoPS equivilant for physics or chem.

Can anyone help shead some light on what I should be studying? I've been working through the free MIT OCW physics video lectures and the book at http://www.ochem4free.com, but I know those aren't going to be teaching the AoPSy techniques I will need.

Also, [url=http://www.math.upenn.edu/%7Ewilf/DownldGF.html]generatingfunctionology[/url]is available free.

Well, worry about the prelim levels first. I know that up to USNCO all you need is raw theory and some lab skills. Once you get good at that then I recommend doing past IPhO/IChO problems (that's my strategy at least)


inequalities
geometry
incenter
angle bisector
geometry proposed

Let $ ABCD$ be a circumscribed quadrilateral in which $ AB\cap CD\equal{}\{M\}$ and $ AC\cap BD\equal{}\{N\}.$ Knowing that $ AD>\max\{AB,BC,CD\},$ prove that:

\[ 90^{\circ}<m(\angle{AND})<90^{\circ}\plus{}\frac{m(\angle{AMD})}{2}.
\]

[b][u]Solution.[/u][/b]

$ \bullet$ The first inequality: $ m(\angle{AND}) > 90^{\circ}.$

One of the angles $ \angle{AND}$ and $ \angle{ANB}$ has the measure greater (or equal) than (to) $ 90^{\circ}.$ Suppose that $ m(\angle{ANB})\geq 90^{\circ}.$

Then:

$ \Rightarrow\left |\begin{array}{lcl} AB^2\geq NB^2 + NA^2 \\
 \\
CD^2\geq NC^2 + ND^2\end{array}\right.\Longrightarrow\boxed{AB^2 + CD^2\geq NA^2 + NB^2 + NC^2 + ND^2} \dots\dots\dots (1)$

We also have that $ m(\angle{AND})\leq 90^{\circ}.$ So:

$ \Rightarrow\left |\begin{array}{lcl} AD^2\leq NA^2 + ND^2 \\
 \\
BC^2\leq NB^2 + NC^2\end{array}\right.\Longrightarrow\boxed{AD^2 + BC^2\leq NA^2 + NB^2 + NC^2 + ND^2} \dots\dots\dots (2)$

Since $ ABCD$ is a circumscribed quadrilateral and $ AD > \max\{AB,BC,CD\}$ we deduce from [b][color=blue]Pithot's theorem[/color][/b] that:

$ AD + BC = AB + CD \ |^2 \Longleftrightarrow AD^2 + BC^2 + 2AD\cdot BC = AB^2 + CD^2 + 2AB\cdot CD.$

Observe that:

$ AD\cdot BC < AB\cdot CD \Longleftrightarrow AD(AB + CD - AD) < AB\cdot CD\Longleftrightarrow$$ (AD - AB)(AD - CD) > 0,$

which is obviously true. Combining the last two relations we obtain that: $ \boxed{AD^2 + BC^2 > AB^2 + CD^2}\dots\dots\dots (3)$

But from the relations $ (1)$ and $ (2)$ respectively we get that: $ AD^2 + BC^2\leq AB^2 + CD^2,$ which contradicts $ (3).$ So our assumption is wrong.

$ \bullet$ The second inequality: $ m(\angle{AND}) < 90^{\circ} + \frac {m(\angle{AMD})}{2}.$

Firstly, let us define some new points. Let $ I$ be the incenter of $ \triangle{MAD}$ (i.e. incenter of $ ABCD$), $ K,R$ and $ S$ be the midpoints of segments $ [AD], \ [AC]$ and $ [BD].$

We have: $ m(\angle{AID}) = 90^{\circ} + \frac {m(\angle{AMD})}{2}.$ So we have to show that $ m(\angle{AND}) < m(\angle{AID})\Longleftrightarrow I\in Int\left (\mathcal{C}\left (K;\frac {AD}{2}\right )\right ).$

From [b][url=http://agutie.homestead.com/FiLes/NewtonTheorem.htm]Newton's theorem[/url][/b] we know that $ I\in RS.$ Moreover we'll prove that $ \boxed{I\in (RS)}$

For the begining let's prove that $ R\in (NA) \ \wedge \ S\in (ND).$ The parallel line to $ AD$ passing through $ B$ intersects $ AC$ in $ X$ and $ MD$ in $ Y.$ It follows that:

$ \frac {BN}{ND} = \frac {BX}{AD} < \frac {BY}{AD} < 1 \ \Rightarrow ND > BN\Rightarrow S\in (ND).$ Similarly we can prove that $ R\in (NA).$

Now suppose that $ I\in RS\backslash (RS).$ Let $ E\in AI\cap BD$ and $ F\in DI\cap AC.$ Applying the angle bisector theorem we'll have:

$ \frac {EB}{ED} = \frac {AB}{AD} < 1\Rightarrow BE < DE\Rightarrow E\in (BS).$ If $ E\in (NS)$ then $ AE$ intersects the segment $ (RS),$ which consists a contradiction of our supposition. Hence $ E\in (NB).$

Analogicaly $ F\in (NC).$ Then lines $ AE$ and $ DF$ meet "above" line $ RS$ (these lines cannot meet on $ RS$). Thus $ \boxed{I\in (RS)}$

We have:

$ \left |\begin{array}{lcl}KR = \displaystyle\frac {CD}{2} < \frac {AD}{2}\Longrightarrow \boxed{R\in Int\left (\mathcal{C}\left (K;\frac {AD}{2}\right )\right )} \\
 \\
KS = \displaystyle\frac {AB}{2} < \frac {AD}{2}\Longrightarrow \boxed{S\in Int\left (\mathcal{C}\left (K;\frac {AD}{2}\right )\right )}\end{array}\right.$ $ \Longrightarrow$ Any point that lies on $ (RS)$ is in interior of $ \mathcal{C}\left (K;\frac {AD}{2}\right ),$ because the circle is a convex set. We conclude that ${ I\in Int\left (\mathcal{C}\left (K;\frac {AD}{2}\right )\right )}.$


Quadrilateral $ ABCD$ is congruent to
quadrilateral $ WXYZ$. We are given that
$ AB \equal{} 5$ cm, $ BC \equal{} 7$ cm, $ YZ \equal{} 6$ cm and
$ ZW \equal{} 4$ cm. What is the perimeter of
quadrilateral $ ABCD$?

[asy]unitsize(0.5inch);
path a=(0,0)--(5,-2)--(6,8)--(-4,6)--cycle;
draw(a);
draw(rotate(180,(3,10))*a);
label("$A$",(0,0),S);
label("$W$",(6,20),N);
label("$B$",(5,-2),S);
label("$X$",(1,22),N);
label("$C$",(6,8),E);
label("$Y$",(0,12),W);
label("$D$",(-4,6),W);
label("$Z$",(10,14),E);[/asy]

substitute YZ for DC and the other one on WXYZ they say for the only segment on ABCD that doesnt have a length

4+5+6+7=22

ummm, no offense, but if you cant do this stuff, especially THIS particular one, you shouldnt be on this site...

see. thats what i thought! but then i thought also that Bots couldnt post messages. i'm sorry, any mods/admins for my stupidity

For confirmation, check GameBot's email.  It is
g a m e @ a r t o f p r o b l e m s o l v i n g . c o m
And I'm sure AoPS wouldn't give out free emails to anyone.   :P

Since WXYZ is congruent to ABCD, the perimeter of ABCD is the perimeter of WXYZ. Therefore, the perimeter is 5+7+6+4=22

[quote="mz94"]

ummm, no offense, but if you cant do this stuff, especially THIS particular one, you shouldnt be on this site...[/quote]

Please don't say things like that next time. The person who clicked "Ask on Forum" might be offended, and anyway, when I first started, I was terrible. My rating was horrible and I could only solve maybe, 2 questions a game. But it's been 4 months now, and my rating is now usually between the high 1300s and low 1400s. With lots of practice, I got tons better. This person didn't know how to solve this problem, so he/she asked it on the forum. That's what this forum is for - so we can all learn from each other.

is it just me or is all the needed LaTeX/pictures are missing?

[quote="ernie"]Let's just say that [b]many[/b] of the pictures are missing.[/quote]

$ \text{many}\implies\text{almost all, if not all}$  :P  :wink:

Actually, I've seen a picture on about 2 problems.  :P

And sometimes I get pictures and no one else does.  :huh:

But we're getting off topic, so let's stop discussing this no-picture-syndrome please.  :)


geometry
perimeter
rectangle
analytic geometry
conics
parabola
quadratics

A circle circumscribes the rectangle ABCD as shown. Vertex A is located at the origin of an (x,y) - coordinate system, vertex B is on the positive x-axis, vertex C is on the parabola y=x^2 and vertex D is on the positive y-axis. If the perimeter of the rectangle is 24 then what is the area of the circle? Please explain how you get your answer.

[hide="Solution"]Let the vertex $ C$ have coordinates $ (x,x^2)$. Thus $ AB \equal{} x$ and $ AD \equal{} x^2$, and the perimeter is $ 2x^2 \plus{} 2x \equal{} 24$. Solving this quadratic yields $ x \equal{} 3$. Thus the diagonal of the rectangle, which is also the diameter of the circle, is $ \sqrt {AB^2 \plus{} AD^2} \equal{} \sqrt {3^2 \plus{} (3^2)^2} \equal{} \sqrt {90} \equal{} 3\sqrt {10}$, and the radius is $ \frac {3\sqrt {10}}{2}$ Thus the area is $ \pi\left(\frac{3\sqrt{10}}{2}\right)^2 \equal{}\boxed{\frac{45\pi}{2}}$.[/hide]

You are brilliant, leoxnlin! Thanks a lot for the reasonable answer.

Where do you get the pictures of the problems?


geometry
inradius
circumcircle
inequalities
inequalities unsolved

I'm calling this a conjecture since I didn't have really enough time to look for a proof. Anyway, it seems very plausible from my dynamic sketches.

Let ABC be a triangle with altitudes $h_a$, $h_b$, $h_c$, angle bisectors $l_a$, $l_b$, $l_c$, exradii $r_a$, $r_b$, $r_c$, inradius r and circumradius R. Prove or disprove the inequality

$8 < \frac{h_ar_a+h_br_b+h_cr_c}{Rr} \leq \frac{l_ar_a+l_br_b+l_cr_c}{Rr} \leq \frac{27}{2}$.

  Darij

The inequality $ \displaystyle 8 < \frac{ h_{a}r_{a}+ h_{b}r_{b}+h_{c}r_{c}}{Rr}$ is equivalent to: $\displaystyle \frac{ (4R+r)^{2}+p^{2}}{2R^{2}} > 8$ which is obvious.

 :D  :D

[quote="Lagrangia"]The inequality $ \displaystyle 8 < \frac{ h_{a}r_{a}+ h_{b}r_{b}+h_{c}r_{c}}{Rr}$ is equivalent to: $\displaystyle \frac{ (4R+r)^{2}+p^{2}}{2R^{2}} > 8$ which is obvious. [/quote]

Well, guess we all have our opinion of what's obvious, don't we?  :D

These look like two very non-trivial steps to me, could you explain a bit please?

Well.. here is the full solution of the identity:

$\displaystyle \frac{ \sum h_{a}r_{a}}{Rr} = \frac{2r_{a}r_{b}r_{c}}{Rr}  \left( \sum \frac{1}{r_{b}+r_{c}} \right).$ This is because: $\displaystyle h_{a} = \frac{2r_{b}r_{c}}{r_{b}+r_{c}}.$

Then $\displaystyle \frac{2r_{a}r_{b}r_{c}}{Rr}  \left( \sum \frac{1}{r_{b}+r_{c}} \right)  = \frac{2r_{a}r_{b}r_{c}}{Rr} \left( \frac{ \sum (r_{a}+r_{b})(r_{a}+r_{c})}{ \prod (r_{a}+r_{b})} \right).$

Now we use the fact that: $r_{a}r_{b}r_{c} = p^{2}r,$ $\displaystyle \prod (r_{a}+r_{b}) = 4p^{2}R,$ $\displaystyle \sum (r_{a}+r_{b})(r_{a}+r_{c}) = \sum r_{a}^{2} + 3\sum r_{a}r_{b} = (r_{a} + r_{b}+r_{c})^{2} + 2\sum r_{a}r_{b}.$

Also $\displaystyle \sum r_{a} = 4R+r$ and $\sum r_{a}r_{b} = p^{2}.$

We put all these relations head to head and it follows that $\displaystyle \frac{ \sum h_{a}r_{a} }{Rr} = \frac{(4R+r)^{2}+p^{2}}{2R^{2}}.$

Thus we have $8Rr + r^{2}+p^{2} >0,$ which is obvious.

The second inequality is trivial, since $l_a\geq h_a$ and so on. And for the third, you can use the fact that $\frac {p^2-l_a^2}{2} \geq l_ar_a$, then it's well known that $2l_a^2 = \frac {a^2b^2+a^2c^2}{(b+c)^2} + b^2+c^2$,  $Rr = \frac {abc}{2(a+b+c)}$ and you have an inequality in terms of the sides of the triangle... i don't know if it's true because i didn't make the calculations, but it seems to be  :lol:


Vieta
superior algebra
superior algebra solved

Just want to confirm 2 questions on my mid term.  :?

[size=150][b]QUESTION 1:[/b][/size]
Solve the equation 12x3 - 4x2  - 5x + 2 = 0 given that it has a repeated root.

[size=150][b]QUESTION 2:[/b][/size]
If :alpha:, :beta:, :gamma: are roots of the equation ax3 + bx2 + cx + d = 0, determine the equation whose roots are :alpha:2 , :beta:2 , and :gamma:2

Hopefully you guys can help, appreciate the help  :)

edit: (sorry about  posting in the wrong forum)

All right, I will answer to you, fotia.

Question 1:$12x^3-4x^2-5x+2=0\Longleftrightarrow (2x-1)^2(3x+2)=0$,so $x=-\frac{2}{3},\frac{1}{2}$  (multiple root).

Question 2:In using vieta's formula,we have $\alpha+\beta+\gamma =-\frac{b}{a},\alpha\beta+\beta\gamma+\gamma\alpha =\frac{c}{a},\alpha\beta\gamma =-\frac{d}{a}$.

Therefore ${\alpha}^2+{\beta}^2+{\gamma}^2=(\alpha+\beta+\gamma )^2-2(\alpha\beta+\beta\gamma+\gamma\alpha) =\frac{b^2-2ca}{a^2}$, ${\alpha}^2{\beta}^2+{\beta}^2{\gamma}^2+{\gamma}^2{\alpha}^2=(\alpha\beta)^2+({\beta\gamma)^2+(\gamma\alpha)}^2-2\alpha\beta\gamma (\alpha+\beta+\gamma )=\frac{c^2-2bd}{a^2}$, ${\alpha}^2{\beta}^2{\gamma}^2=(\alpha\beta\gamma)^2=\frac{d^2}{a^2}$.

Since the desired equation is given by the following equation.

$(x-{\alpha}^2)(x-{\beta}^2)(x-{\gamma}^2)=0$
${\Longleftrightarrow x^3-(\alpha}^2+{\beta}^2+{\gamma}^2)x^2+({\alpha}^2{\beta}^2+{\beta}^2{\gamma}^2+{\gamma}^2{\alpha}^2)x-{\alpha}^2{\beta}^2{\gamma}^2=0$,

consequently the answer is $a^2x^3-(b^2-2ca)x^2+(c^2-2bd)x-d^2=0$.

kunny

[quote="kunny"]
Question 2:In using vieta's formula,we have $\alpha+\beta+\gamma =-\frac{b}{a},\alpha\beta+\beta\gamma+\gamma\alpha =\frac{c}{a},\alpha\beta\gamma =-\frac{d}{a}$.
...

kunny[/quote]

ok, i worked out how you got your solution but just curious how you went on implementing vieta's theory into this.  How did u get $\alpha+\beta+\gamma =-\frac{b}{a},\alpha\beta+\beta\gamma+\gamma\alpha =\frac{c}{a},\alpha\beta\gamma =-\frac{d}{a}$.

-sorry again :? just not familiar with the formula

All right,fotia.

Let $\alpha,\beta$ and $\gamma$ be roots of the cubic equation $ax^3+bx^2+cx+d=0$.

We can write $ax^3+bx^2+cx+d=a(x-\alpha)(x-\beta)(x-\gamma)$.

Comparing the coefficients of $x^3,x^2,x$,and constant term of both sides of  the equation,you will obtain our three equations which is Vieta's formula.In Japan we call it [b]the relation of root and coefficient for cubic equation[/b].

kunny


From Utah State Math Contest, Senior 2002

31. sec t + tan t = ?

(A) cot(-t/2 + :pi:/4)
(B) cot(t/2 - :pi:/4)
(C) 1/2*cot(-t + :pi:/2)
(D) 1/2*cot(t-:pi:/2)
(E) tan(-t/2)

35.  In a rhombus, one angle has twice the measure of another angle.  The shorter diagonal is 10 cm.  What is the perimeter of the rhombus?

(A) 10
(B) 20
(C) 40
(D) 20:rt3:
(E) 40:rt3:

37. If logb2 = A and logb 5 = c where b > 0 with b =/ 1, then logb 500 is equal to which of the following?

(A) 3A + 2c
(B) 10Ac
(C) A^2 + c^3
(D) 2A + 3c
(E) None of the above

cool utah i took the junior exam check out the results (GO CHURCHILJRHIGH they took 1st in 7th grade) too bad old winston is dead

i'm too lazy to write the solutions for them (i have done these in school like 4 times already)

[hide]cot(-.5t+Pi/4)

40

2A+3c[/hide]

_______________________

A word from your friendly moderator: In the future, please post your answers in spoiler so that the next person who comes along gets to do the problem without seeing the answer.  This spoils all the fun, hence the word 'spoiler'.   Also, in the future, don't just post answers.  Show how you got your answers, so that we can see several different ways of doing a problem.

- mathfanatic

P.S.  Don't worry - I didn't delete your solution.  Just highlight the blank space above and voila!


function
ceiling function
algebra
domain
complex analysis
complex analysis unsolved

What's the range of $ z^{ \minus{} 1}*cos(z^{ \minus{} 1})$, $ 0 < |z| < \delta$, where $ \delta$ is any positive number.

Denote the function by $ f(z)$.  Clearly, $ f\left(\frac {1}{2N\pi}\right) \equal{} 2N\pi$, while $ f\left(\frac {1}{(2N \plus{} 1)\pi}\right) \equal{} \minus{} (2N \plus{} 1)\pi$ for any integer $ N$.  For sufficiently large $ N$, $ \frac {1}{2N\pi} < \delta$.  For any positive real $ M > 0$, take $ N \equal{} \lceil\max\{M,\frac {1}{\delta}\}\rceil$, and taking $ z \equal{} \frac {1}{2N\pi}$ and $ z \equal{} \frac {1}{(2N \plus{} 1)\pi}$, you have two numbers whose images by $ f$ are respectively positive and negative, and both larger than $ M$ in absolute value.  Clearly this means that $ f$ is not bounded.  But $ f$ is continuous, hence by the intermediate value theorem, the range of $ f$ is $ \mathbb{R}$.

 :blush: Sorry jmjerry, you're absolutely right.  Totally missed that this is the $ \mathbb{C}$ section...   :(

You just answered the wrong question, daniel73. Take note of where this is posted- the domain is a punctured disk in $ \mathbb{C}$, rather than a punctured interval in $ \mathbb{R}$.

Yes.

jmerry, you are absolutely right. Do you have any idea?

the problem is a corollary of theorem named Casorati-Weierstrass theorem.

let $ A\equal{}\{f(z) : 0<|z|<\delta\}$ where $ f(z)\equal{}\frac{1}{z}cos(\frac{1}{z})$

The Casorati-Weierstrass theorem tells that $ \overline{A}\equal{}\mathbb{C}$.

We know that $ A\subset \mathbb{C}$
need to prove $ \mathbb{C}\equal{}\overline{A}\subset A$
suppose not, there is a point $ z\in\mathbb{C}\equal{}\overline{A}$ st $ z\not\in A$, using the definition of limit point $ z$ and do some calculations you will have a contradiction

1234567a,

You are right. By the problem is that how to find the solutions of $ \frac {1}{z}cos(\frac {1}{z}) \equal{} w$, for any $ w\in\mathbb{C}$


1. Whats the radius of Albatrosss' tuna can if the volume is 150 :pi: cm^3 and the height is 6 cm.

2. Whats the volume of Eustines box if the height is 10 the length is 5 times the height and the width is half the length minus 15

3. Whats the numaeric value of the surface area minus the volume of a sphere with radii 5

4. Boberts fish was killed by Treethings Cat poison. If the base of the tank was 10 * 15 in how many in^3 of water needs to be pumped out to get the water level down 3 in

Formulas:
Cylinder Volume: B*H or r^2 * :pi: * h
Box Volume: l*w*h
Sphere Volume: 4/3 :pi: r^3
Sphere Surface area: 4r^2 :pi:

This should be easier.   :D

[quote="A+MATH"]1. Whats the radius of Albatrosss' tuna can if the volume is 150 :pi: cm^3 and the height is 6 cm.

2. Whats the volume of Eustines box if the height is 10 the length is 5 times the height and the width is half the length minus 15

3. Whats the numaeric value of the surface area minus the volume of a sphere with radii 5

4. Boberts fish was killed by Treethings Cat poison. If the base of the tank was 10 * 15 in how many in^3 of water needs to be pumped out to get the water level down 3 in[/quote]

Where can I find how high the water level in Boberts fish tank was?

[hide="Solution"]Solving $150\pi=6r^2\pi$, we get $r=5$.

The dimensions of Eustine's box are 10x50x10 or 10x50x17.5, depending on how the problem is interpreted. So the answer is either 5000 or 8750.

The surface area of a sphere with radii 5 is 100pi. The volume is (500/3)*pi. So the answer is (300/3-500/3)pi=-200pi/3[/hide]

You don't need it. The problem more or less says: "how much water should be pumped out in order to lower the water level by three inches?"

Yeah you don't need the height. The answers are correct.

Oops :blush: , I read the problem as "... water level down to 3 in".

[hide="Beginning of Answer"]10*15*x=150x

10*15*(x-3)=150x-450[/hide]

[hide="End of Answer"]150x-(150x-450)=450[/hide]

[quote="A+MATH"]1. Whats the radius of Albatrosss' tuna can if the volume is 150 :pi: cm^3 and the height is 6 cm.
[/quote]

[hide]$6*pi*r^2=150pi$
$r^2=25$
$r=5 cm$[/hide]

[quote="A+MATH"]2. Whats the volume of Eustines box if the height is 10 the length is 5 times the height and the width is half the length minus 15
[/quote]

[hide]height=10
length=50
width=10
10*10*15=1500[/hide]

[quote="A+MATH"]
3. Whats the numaeric value of the surface area minus the volume of a sphere with radii 5

Sphere Volume: 4/3 :pi: r^3
Sphere Surface area: 4r^2 :pi:
[/quote]

[hide="my solution"]area=100pi
volume=500pi/3
300pi/3-500pi/3=-200pi/3[/hide]

[quote="A+MATH"]4. Boberts fish was killed by Treethings Cat poison. If the base of the tank was 10 * 15 in how many in^3 of water needs to be pumped out to get the water level down 3 in [/quote]

[hide="my solution"]10*15*3=450[/hide]

[color=darkred]Posts combined[/color]

I combined the quadruple post. Please don't do that again ;) .

Nice Problems...

Those kinds of problems , I bet it often comes out in AMC8...I took it before....So.. i know..

I have a request that students stop discussing problems about cat poison.

I think the value of the shock is not nearly so high as the value of membership to the community.


[quote="parrot"]Howś can I solve it

X1 + X2 + 4X3 + 2X4 = 5
−3X1 − X2 − 6X4 = 3
−X1 − X2 + 2X3 − 2X4 = 1[/quote]Please develop your $\LaTeX$ skills. It makes your statements more 'readable' fashion.

[quote="parrot"]How can I solve it (for integers? positive integers?)

$x_{1}+x_{2}+4x_{3}+2x_{4}= 5$
$-3x_{1}-x_{2}-6x_{4}= 3$
$-x1-x_{2}+2x_{3}-2x_{4}= 1$[/quote]


[quote="ch1n353ch3s54a1l"][quote="ch*yeuk"]How would you do this algebraically? =) I know the answer, but I don't know how to get to it.

A line segment has endpoints Y(7,2) and Z(-5,-7). Determine the coordinates of the points that divide YZ into three equal parts[/quote]
[hide]
well you know that the distance between the poitns on the x axis is 12. 12/3=4. distance between 2 and -7 is 9. 9/3=3. so the points are, if you add 4 to x and 3 to y, (-1,-4), (3,-1)[/hide][/quote]

Ah thanks! Seems so simple now =/ ehh. What I did was find the midpoint of YZ.. then the midpoint of that midpoint and Y and .. yeah. Thanks a lot =D


induction
number theory unsolved
number theory

[quote="usamo"]for a given positive integer $n$, define A(n) to be the highest power of 2 that divides n. (example: A(40)=8)
let $S_n=\sum_{i=1}^{2^{i-1}}A(2i)$.
find $n_{max}<1000$, s.t. $S_n$ is perfect square.[/quote]
If $S_n=\sum_{i=1}^{2^{n-1}}A(2i)$
By induction: $S_n=(n+1)2^{n-1}$
Then n_{max}=899


Can you please explain the Well-ordering principle and Archimedean property?I have trouble in understanding it.Thank you.

[url=http://en.wikipedia.org/wiki/Well_ordering_principle]Well-ordering principle[/url]

[url=http://en.wikipedia.org/wiki/Archimedean_property]Archimedean property[/url]

As far as I know, the Well-ordering principle more or less states that every nonempty set made up of positive integers has a smallest element. Additionally, I believe it is equivalent to the Axiom of Choice.


cincodemayo5590 wrote:Compute the summation:

By the way, this isn't the same sum as at HMMT last year (and unfortunately it requires calculus, too)...

[hide]

by PFD. Canceling alot of terms out, we are left with the equivalent sum (work this out yourself!)

Call the summation S. Then we know that  is the power series of  centered at . It converges for (I believe) , so we can plug in  to get , so we have . Plugging this back in, our final answer is

[/hide]


$ CH_3COOCH_3$ - does it respond to haloform test?  :maybe:

In the mechanism of the haloform reaction there is the formation of an enolate from the methyl ketone and hydroxide. Esthers are less acidic than ketones, so they should not undergo this reaction. Also, and perhaps more revealing, since hydroxide is used in the haloform reaction, saponification of the esther is the easiest path.


symmetry
function
number theory
relatively prime
prime numbers
greatest common divisor
number theory unsolved

Let $ m,n$ be unequal positive integers. If $ m^{n}\equal{}n^{m}$,find all possible ordered pairs $ (m,n)$.

Hello. First we check the case m=n. This gives us infinitely many solution. All (x,x) pairs of positive integers satisfies the condition. Now assume that they are unequal. Since the equation is symmetric, w.l.o.g. we may assume that m is greater than n. Let the greatest common divisor of m and n be d. m=d.a and n=d.b (here a and b are relatively prime numbers.) Turn to the equation.                         d^n . a^n = d^m . b^m. Since m is greater than n we may rewrite this equation as a^n = d^(m-n) . b^m. We can easily see that b^m should divide a^n. But they are relatively prime numbers so b=1. Now d^(m-n) = a^n. Putting m=d.a and n=d we get          d^(da-d)=a^d. If we take d. root of this equation (they are all positive numbers) we get d^(a-1) = a. Now start solving this equation. a cannot be 1 because we assumed that m and n are unequal numbers. So a should be greater than or equal to 2. If d=1 we get a=1 which is a contradiction. If d=2 we get 2^(a-1)=a. Here a=2 gives us a solution which is (m,n)=(4,2). By the symmetry (2,4) is also a solution. But if a is greater than 3 one can show that 2^(a-1) reaches a. (by induction.) Now we can say that d^(a-1) is greater than 2^(a-1) so we don't have any solution except (4,2) and (2,4) :P

The other way to look at this problem is to write it as

$ m^{\frac{1}{m}} \equal{} n^{\frac{1}{n}}$

The function $ y \equal{} x^{\frac{1}{x}}$ is increasing for $ 0 < x < e$ and decreasing for $ e < x$ (just take logs and differentiate.)

at positve integer values of x it takes values $ 1$, $ \sqrt{2}$, $ 3^{\frac{1}{3}}$,  $ \sqrt{2}$, ...
for $ x > 4$, $ y$ decreases and is larger than $ 1$, so $ 1 < y <  \sqrt{2}$
So for $ y > 4$ it cannot take on any previous value at positive integer x.
Therefore the only solutions for $ n \ne m$ are (2,4) and (4,2).

[quote="PhilG"]The other way to look at this problem is to write it as

$ m^{\frac {1}{m}} \equal{} n^{\frac {1}{n}}$

The function $ y \equal{} x^{\frac {1}{x}}$ is increasing for $ 0 < x < e$ and decreasing for $ e < x$ (just take logs and differentiate.)

at positve integer values of x it takes values $ 1$, $ \sqrt {2}$, $ 3^{\frac {1}{3}}$,  $ \sqrt {2}$, ...
for $ x > 4$, $ y$ decreases and is larger than $ 1$, so $ 1 < y < \sqrt {2}$
So for $ y > 4$ it cannot take on any previous value at positive integer x.
Therefore the only solutions for $ n \ne m$ are (2,4) and (4,2).[/quote]

let $ m<n$.
As the function  $ y \equal{} x^{\frac {1}{x}}$  is increasing for  $ 0 < x < e$  and decreasing for  $ e < x,\implies 0<m<e$ and   $ e<n$
$ \implies m\equal{}1$  or  $ 2$
note because of the nature of the function, correspond to each $ m$ there is one and only one $ n \implies n\equal{}1$  or  $ 4$  but  $ m\ne n\implies (m,n)\equal{}(2,4) .$
Therefore the only solutions for $ m \ne n$ are $ (2,4)$ and $ (4,2)$


Find the largest real a for which one can find a function f defined on the positive integers with values in the same set such that for all n:
   f(n+1)>=f([(f(n))^a])+f(n), where [] is the integer part.
  Harazi

Almost 3 years...

My answer is $a=\frac{1}{4}$ and function is $f(n)=n^{2}$.

I have manage to prove that if $a>\frac{1}{4}$ then it is possible to show that $f(n)\geq n^{k}$ for any $k$ for sufficiently large $n$, but it contradicts with natural condition $f^{a}(n)\leq n+1$.


inequalities
quadratics
algebra
polynomial
inequalities proposed

(a) Let $x,y,z$ be positive numbers such that $x+x^3+y+y^3+z+z^3=6$. Prove that
$x^3y+y^3z+z^3x \le 3$.

(b) What happens for $x,y,z$ real numbers ?

can we prove that $x^4+y^4+z^4+xy^3+yz^3+zx^3 \geq 2(x^3y+y^3z+z^3x)$?
I'm not sure.

is it an implication of your very nice ineq, vasc?
\[3(a^3b+b^3c+c^3a) \le (a^2+b^2+c^2)^2 \]
(if i didn't remember it wrongly)

[quote="siuhochung"]
$\[3(a^3b+b^3c+c^3a) \le (a^2+b^2+c^2)^2$
[/quote]
can you post the solution?

i lost the link of the thread.
I think it is a very hard inequality.
Vasc had a very very beautiful, incredile, amazing one-line solution.. :D

[quote="sbos"]Using famous $(x^2+y^2+z^2)^2\geq3(x^3y+y^3z+z^3x)$ and Cauchy-Schwarz we get
\[3(x^3y+y^3z+z^3x)\leq(x+y+z)(x^3+y^3+z^3)=a(6-a)\leq9.\]
Everything is true even if some of  $x,y,z$ is negative.
It is really a cosequence of that Vasc’s inequality as siuhochung supposed.[/quote]
The ineq. $(x^2+y^2+z^2)^2 \le (x+y+z)(x^3+y^3+z^3)$ is not true for any real numbers x, y, z (see le case $x+y+z=0$).
Then why does everything is true even if some of  $x,y,z$ is negative ?

Excuse me, I'm sorry for the nonsense I've written.

Here's a proof for $(a^2+b^2+c^2)^2\geq3(a^3b+b^3c+c^3a)$.  Put \[a=x,\; b=x+y,\; c=x+y+z.\]
The inequality becomes
\[x^2(yz+y^2+z^2)+x(7yz^2+6y^2z+y^3+z^3)+y^4+z^4+5y^3z+8y^2z^2+4yz^3\geq0.\]
Its quadratic on $x$ and $yz+y^2+z^2\geq0$ so its enough to prove that discriminant is negative.
The inequality $D\leq0$ is equivalent to \[y^6+z^6+4y^5z+2yz^5+2y^4z^2\geq3y^2z^4+6y^3z^3.\]
Put \[t=\frac{y}{z}\,.\] an we get
\[t^6+4t^5+2t^4-6t^3-3t^2+2t+1\geq0,\] or
\[(t^3+2t^2-t-1)^2\geq0,\] which is obviously true.\\
The inequality also follows from the identity
\[ (a^3+b^3+c^3-5a^2b-5b^2c-5c^2a+4ab^2+4bc^2+4ca^2)^2+\]
\[+3(a^3+b^3+c^3-a^2b-b^2c-c^2a-2ab^2-2bc^2-2ca^2+6abc)^2=\]
\[=4(a^2+b^2+c^2-ab-bc-ca)((a^2+b^2+c^2)^2-3(a^3b+b^3c+c^3a)).\]
This looks impossible but in fact its natural. The quadratic polynomial above is always positive which means
that it can be written as a sum of squares. That's the way how we get this strange identity.

Indeed Sbos, the inequality $3(x^3y+y^3z+z^3x)\leq(x+y+z)(x^3+y^3+z^3)$ is true FOR ANY REAL $x,y,z$.

[quote="Vasc"]There is an implication of my another inequality, which was posted just in this topic. ;)[/quote]
This ineq. is above "my another inequality". In fact, this ineq. is much more interesting for $xyz <0$. I considere these inequalities,  $3(x^3y+y^3z+z^3x)\leq(x+y+z)(x^3+y^3+z^3)$ and $3(x^3y+y^3z+z^3x)\leq(x^2+y^2+z^2)^2$, to be twin, since I discovered them at the same time.

[quote="sbos"]The inequality $3(x^3y+y^3z+z^3x)\leq(x+y+z)(x^3+y^3+z^3)$ can be proved FOR ANY REAL $x,y,z$ using the same method as above (for $3(x^3y+y^3z+z^3x)\leq(x^2+y^2+z^2)^2$). I think it is not neccessary to write all the details, everything is just the same. So that's the reason why the inequlity holds for any real $x,y,z$.[/quote]
You are right:
 $(x+y+z)(x^3+y^3+z^3)-3(x^3y+y^3z+z^3x)=\frac{1}{2}\sum(x^2-y^2-xy+yz)^2.$ ;)

[quote="hungkhtn"]Oh, I enconter to see that it's same as my problem post in VietNam forum for the new year 2006. :
See here :

http://diendantoanhoc.net/forum/index.php?showtopic=9873

In this topic, there's an other hard problem which still has no solution :

Find the best constant $k$ for which the following ineq's always true :

\[ a^4+b^4+c^4+k(ab+bc+ca)^2 \ge (1+3k)(ab^3+bc^3+ca^3)  \]

[hide]Hi, soarer, I remember that I had send to you this problem before[/hide][/quote]

I don't think so.... I think you have only sent me an inequality that's posted by vasc before (something like $a^4+b^4+c^4+ab^3+bc^3+ca^3 \ge 2(a^3b+b^3c+c^3a)$)

[quote="Soarer"][quote="hungkhtn"]Oh? I don't see that It's posted since 2005? Great problem and I'm sorry. :blush:[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?t=6026&postorder=asc&start=40

very beautiful proof Mr Stefan V has there..[/quote]

Because there of them have the same way for it.

[quote="Vasc"][quote="hungkhtn"]
Find the best constant $k$ for which the following ineq's always true :

\[ a^4+b^4+c^4+k(ab+bc+ca)^2 \ge (1+3k)(ab^3+bc^3+ca^3)  \][/quote]
We have two values almost the best of k such that

$3(a^4+b^4+c^4)+(ab+bc+ca)^2 \geq 6(a^3b+b^3c+c^3a)$

and
$a^4+b^4+c^4+a^3b+b^3c+c^3a \geq \frac{2}{3}(ab+bc+ca)^2$.[/quote]
How did you prove it?Again One-line solution...

Indeed, one-line solution.
The more general inequality
$\sum a^4+(3r^2-1)\sum a^2b^2+3r(1-r)abc\sum a \geq 3r\sum a^3b$
is equivalent to
$\sum(2a^2-b^2-c^2-3rab+3rbc)^2 \geq 0$.


inequalities
geometry
parallelogram
function
inequalities unsolved

[quote="Soarer"]Denote $a = (a_{1},...,a_{n})$, $a_{i}\in \mathbb{R}^{+}$, $\sum_{sym}x_{1}^{a_{1}}...x_{n}^{a_{n}}$ by $[a]$, $>$ for majorization, then

is it true that
$a > b$, $c > d$, $a > c$, $a+b = c+d$ implies $[a]+[b] \ge [c]+[d]$?[/quote]That depends if "majorization" includes ordering of the $a_{i}$. Normally the definition of it doesn't.  :maybe: Remember [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=397958#397958]this one[/url] which can be written as $[4211]+[2222]\ge[3311]+[3122]$ if you want all the $a_{i}$ strictly positive. Take $x=(1,1,1,\epsilon)$ :maybe:

sorry, i forgot to write that a,b,c,d are all ordered. And yes, in fact it is an attempt to prove my conjecture 8 months ago.  :blush:

I wanted to find out what the conditions mean geometrically. It is straightforward (for three variables, that is) but maybe helpful.
Let $N$ be the total degree of the symmetric sums. In the $\mathbb R^{3}$ with axis corresponding to $a_{1},a_{2},a_{3}$, everything happens in the plane spanned by $(N,0,0), (0,N,0), (0,0,N)$ with homogenous coordinates. As $a_{1}\ge a_{2}\ge a_{3}$, it is further restricted to the red triangle $LOK$ with angles $30,60,90$. $L(N,0,0), K(\frac{N}{2},\frac{N}{2},0),O(\frac{N}{3},\frac{N}{3},\frac{N}{3})$.
If we choose inside this triangle a partition (point) $a$, the region majorized by $a$ is the light blue quadrilateral $aH_{1}OH_{2}$ with right angles at $H_{1},H_{2}$. Once we havechosen $b\prec a$, we find that $c$ and $d$ are then restricted to the two dark blue parallelograms. Because of $a+b=c+d$, they are symmetrical with repect to the point $M=\frac{a+b}2$. Note the dark blue parallelograms may be truncated by the triangle sides $OK$ and/or $OL$, but I think that the conjecture holds for the whole of them. On the contrary, the condition $c\succ d$ cannot be omitted, as shows the counterexample $[420]+[321]\ge[411]+[330]$, which is false for $(x,y,z)=(2,2,1)$.

as I have to submit some reports for math courses i'm taking, I think I won't work on this for some time. I hope somebody find my idea useful:
I found that the concept of majorization can be extended to any partially ordered ring, not necessarily totally ordered but satisfies $x > 0 \wedge y > 0 \Rightarrow xy > 0$ and $x > y \Rightarrow x+c > y+c$. Of course, the old majorization inequality won't work here because rings can be discrete, but symmetric sum function is well defined since product and addition is defined in ring. My guess is, for a suitably-chosen ring and order, we can have some ineq better than muirhead. In some sense, this was how I formulated my conjecture.

@ soarer: wish you good luck for your work :) And I would never have thought of Muirhead et.al. in terms of [i]rings [/i]- that might become rather abstract...

As to the conjecture/question, I think more conditions need to be added  :( e.g.,
$[620]+[332]\ge[530]+[422]$ holds, but
$[530]+[332]\ge[440]+[422]$ is false :wallbash:

Writing it in the form $[a]-[c]\ge[d]-[b]$, we would need some way of 'measuring the difference' of two symmetric functions.


inequalities
greatest common divisor
number theory unsolved
number theory

Let m,n be two not null natural numbers, their greatest common divisor beeing 1 and 4|m.
Demonstrate that if [code] \frac{m}{n}>\sqrt{2}[/code] then we have the "stronger" inequality
[code]\dfrac{m}{n}-\dfrac{7}{mn}>\sqrt{2}[/code]

The stronger inequality is true even when m is 2 mod 4. 2|m is sufficient.
[hide]If $ \dfrac{m}{n} > \sqrt {2}$, then $ \dfrac{m^2}{n^2} > 2$, so $ m^2 \minus{} 2n^2 > 0$. Since $ m^2 \ne 2n^2$, $ m^2 \minus{} 2n^2$ is at least the greatest common divisor of $ m^2$ and $ 2n^2$. That is, $ m^2 \minus{} 2n^2 \ge \gcd (m^2, 2n^2) \equal{} \gcd (m^2, 2) \equal{} 2$.
$ m^2 \minus{} 2n^2 \minus{} 14 \ge \minus{} 12$
$ m^2 \ge 4$ so
$ m^4 \minus{} 2m^2n^2 \minus{} 14m^2 \ge \minus{} 48 > \minus{} 49$.
$ m^4 \minus{} 14m^2 \plus{} 49 > 2m^2n^2$
$ (m^2 \minus{} 7)^2 > 2m^2n^2$
$ \left(\dfrac{m^2 \minus{} 7}{mn}\right)^2 > 2$
$ \dfrac{m^2 \minus{} 7}{mn} > \sqrt {2}$
$ \dfrac{m}{n} \minus{} \dfrac{7}{mn} > \sqrt {2}$, as desired.[/hide]


Let $ E$ be the intersection of the diagonals of the convex quadrilateral $ ABCD$. The perimeters of $ \triangle AEB$, $ \triangle BEC$, $ \triangle CED$, and $ \triangle DEA$ are all same. If inradii of $ \triangle AEB$, $ \triangle BEC$, $ \triangle CED$ are $ 3,4,6$, respectively, then inradius of $ \triangle DEA$ will be ?

$\textbf{(A)}\  \frac {9}{2} \qquad\textbf{(B)}\  \frac {7}{2} \qquad\textbf{(C)}\  \frac {13}{3} \qquad\textbf{(D)}\  5 \qquad\textbf{(E)}\ \text{None}$


Many people including myself want a list of all programs and contests out there. If you could post ones that you know here that would help everybody.

AoPS already started a list [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php]here[/url].  It's not complete, but if you know a contest that isn't on that list, email Mr. Crawford, and it'll be one step closer to complete!  When the AoPSWiki is up, this list will be something that everyone can edit themself.


solve equation!
$ x^2\plus{}81\equal{} 3^y \plus{}   1993^2$  x,y are positive integer!

You can substitute "x" by 1993 and "y" by 4 and in that way you can get the answer to the equality

1993²+81=81+1993²


[hide]
Let $a=\sqrt[3]{1+x}$ ,$b=\sqrt[3]{1-x}$ , the equation can be reduced to
$(a-2b)^2=0$ which is $a=2b$ . Solve it and get $x=\frac{7}{9}$ ;)
[/hide]


inequalities
calculus
derivative
function
inequalities unsolved

Prove for $ 0 < k \leq 1$ and $ a_i \in \mathbb{R}^\plus{},$ $ i \equal{} 1,2 \ldots, n$ the following inequality holds:

\[ \left( \frac{a_1}{a_2 \plus{} \ldots \plus{} a_n} \right)^k \plus{} \ldots \plus{} \left( \frac{a_n}{a_1 \plus{} \ldots \plus{} a_{n\minus{}1}} \right)^k \geq \frac{n}{(n\minus{}1)^k}.\]

[hide]By Power Mean Inequality
\[ LHS\ge n\left(\frac {\frac {a_1}{a_2 \plus{} \ldots \plus{} a_n} \plus{} \ldots \plus{} \frac {a_n}{a_1 \plus{} \ldots \plus{} a_{n \minus{} 1}}}n\right)^k.
\]
We claim that $ \frac {a_1}{a_2 \plus{} \ldots \plus{} a_n} \plus{} \ldots \plus{} \frac {a_n}{a_1 \plus{} \ldots \plus{} a_{n \minus{} 1}}\ge\frac {n}{n \minus{} 1}$. Since the inequality is homogeneous we may assume without loss of generality that $ a_1\plus{}\ldots\plus{}a_n\equal{}n$. Consider $ f(x)\equal{}\frac{x}{n\minus{}x}$ in the interval $ (0,n)$. The second derivative of this function is $ f''(x)\equal{}\frac{2n}{(n\minus{}x)^3}$ which is positive. Hence the function is convex. Since $ (a_1,a_2,\ldots,a_n)$ majorizes $ (1,1,\ldots,1)$, by Karamata's inequality we have $ f(a_1)\plus{}f(a_2)\plus{}\ldots\plus{}f(a_n)\ge f(1)\plus{}f(1)\plus{}\ldots\plus{}f(1)\equal{}\frac{n}{n\minus{}1}$. Hence our claim holds.

Now we have
\[ n\left(\frac {\frac {a_1}{a_2 \plus{} \ldots \plus{} a_n} \plus{} \ldots \plus{} \frac {a_n}{a_1 \plus{} \ldots \plus{} a_{n \minus{} 1}}}n\right)^k\ge\frac{n}{(n\minus{}1)^k},
\]
as desired.[/hide]

[quote="joh"]By [b]Power Mean Inequality[/b]
\[ LHS\ge n\left(\frac {\frac {a_1}{a_2 \plus{} \ldots \plus{} a_n} \plus{} \ldots \plus{} \frac {a_n}{a_1 \plus{} \ldots \plus{} a_{n \minus{} 1}}}n\right)^k
\]
[/quote]

It's wrong. By [b]Power Mean Inequality[/b] we get that:

\[ LHS\leq n\left(\frac {\frac {a_1}{a_2 \plus{} \ldots \plus{} a_n} \plus{} \ldots \plus{} \frac {a_n}{a_1 \plus{} \ldots \plus{} a_{n \minus{} 1}}}n\right)^k
\]
because $ \boxed{k\leq 1}$

However, this inequality is wrong. For example, if $ n \equal{} 3 \ \wedge\ k \equal{} \frac {1}{2}$ and $ a_{1} \equal{} \frac {1}{15}\ \wedge \ a_{2} \equal{} a_{3} \equal{} 1,$ we obtain a contradiction. For $ k\geq 1$ the inequality is true (see [b]joh's post[/b]).


[quote="ayush_2008"]Q1 Find probability of getting 3 consecutive sixes in 10 throws of a die.

Q2 A die is thrown again and again until we obtain 2 consecutive sixes are obtained .find the probabiltiy of getting 2 consecutive sixes[/quote]
I know the answer to the first question[which] $ 1/216 [/is]

The second question needs to have some restrictions placed on it. The question as it stands would have an answer of 1, as continuously rolling the die will eventually result in two sixes in a row. In any given two throws, the probability that both are sixes is 1/36. If some other limitation was implied, please state it.

[quote="rachitgoel"][quote="ayush_2008"]Q1 Find probability of getting 3 consecutive sixes in 10 throws of a die.

Q2 A die is thrown again and again until we obtain 2 consecutive sixes are obtained .find the probabiltiy of getting 2 consecutive sixes[/quote]
I know the answer to the first question[which] $ 1/216$ [/is][/quote]

This is clearly wrong, as it actually computes the probability of getting $3$ out of $3$ sixes in the first $3$ throws, which is different from what the problem asks. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=498&t=552917]here[/url] a similar(but easier) problem.

Consider the $n$-tuples in which each element is a whole number between $1$ and $6$. Let $p(n,k)$ be the number of such $n$-tuples which don't contain a triple six, and in which the final $k$ elements, but no more, are sixes. Let $q(n)$ be the number of $n$-tuples which contain a triple six anywhere.

Start with $p(3,0)=180,\ p(3,1)=30,\ p(3,2)=5,$ and $q(3)=1$. Then $p(n+1,0)=5(p(n,0)+p(n,1)+p(n,2)),$ $p(n+1,1)=p(n,0),$ $p(n+1,2)=p(n,1),$ and $q(n+1)=6q(n)+p(n,2)$. This can be tabulated as \[\begin{array}{c|cccc}n&p(n,0)&p(n,1)&p(n,2)&q(n)\\\hline 3&180&30&5&1\\4&1075&180&30&11\\5&6425&1075&180&96\\\vdots&\vdots&\vdots&\vdots&\vdots\end{array}\] in which each row sums to $6^n$. The $n=10$ row gives us $q(10)=1903176$, so the answer is $\frac{q(10)}{6^{10}}=\frac{979}{31104}\approx 3.1475\%$.


Set $ {1,2,..,N}$ is coloured by three colours.Each colour is coloured to at least $ \frac{N}{4}$
Prove that there exist three numbers $ x,y,z$ of three colours such that $ x\equal{} y\plus{}z$

If N=4 with {1;3} blue , {2} green and {4} red it seems to be impossible !

There might need to be restriction on $ N$. For $ N \equal{} 5$, you can't even get at least $ N/4$ colors for all three. I can't find a counterexample for $ N \equal{} 6$. Perhaps the problem is for $ N\geq 6$?

OK, It's $ N>5$ and replace "at least $ \frac{N}{4}$ " to " $ >\frac{N}{4}$ . Sorry

Sorry but for N=8 , at least 3 elements for every colors is impossible  :oops:


floor function
superior algebra
superior algebra unsolved

Let $n\ge2$ and $\forall k\;\; 3\leq k\leq n+1,\; t_{k}=(1\;k-1)(2\;k-2)\dots(\lfloor\frac{k}{2}\rfloor k-\lfloor\frac{k}{2}\rfloor)$.
Prove that $S_{n}=<t_{k}|3\leq k\leq n+1>$ and that each element of $S_{n}$ can be regarded as a product of at most $2n-3$ of $t_{k}$s.

I think this problem is somehow related, any idea?: Prove that $(1 2 \dots n)$ cannot be written as a product of less than $n-1$ transpositions.


summer program
MathPath
probability
AwesomeMath

If I go to mathpath, this would be my first time at a sleep-away camp, and I'm not sure if I want to be away from home for the whole 4 weeks. I know the 2-week option is new this year, so obviously no one's tried it yet. But if you had had an option, would you have chosen 2 weeks?

I was 11 last year (and the youngest in the camp), and I had no problems doing 4 full weeks.

And I hadn't been away from home before.

I was at MathPath last year, and I didn't think that four weeks was too long.  (If anything, I wanted MathPath to be longer!)  I think that you won't really get the true MathPath experience in two weeks.

I was at Mathpath last year, and all I can say is that real fun starts after 2 weeks...

Agreed. The first two weeks are fun too, but the second two are better.

I was wondering about you guys. Do you guys usually take a friend or another mathlete with you so you have someone you know?

No, usually you don't; some kids know other people going though. Certainly encourage your classmates and friends to apply!

There were multiple kids from certain cities/schools. For me, there were a few kids I knew from MC nats, and four more from my city, one of which I knew extremely well (MathPath 2009 people would know why). :P

You'll make friends there! Don't worry about that.

I would encourage you to go to the 4 week period over the 2 week period because as mathq said, most of the fun happens only after the 2 weeks. By the end of the 2 weeks, you finally get to know most of the camp and really start feeling more comfortable being away from home and enjoying the classes. Choosing the 2 week option is essentially Being taken out just when you're really beginning to enjoy yourself. I was concerned when I first decided to come, but my homesickness quickly wore off.

[quote="Eplsilon"]Has anyone tried the 2 week option? I'm going, and I think 2 weeks is probably better, Since I've never been away from home before. But I just wanted to hear what somebody who's done it thinks.[/quote]
MathPath was my first time away from home as well. However, I applied for four weeks and it was awesome. I wasn't homesick in the least, except for one day when I was sick, and that was one week through.

EDIT: Most years, nobody gets sick at MathPath. Last year, I was the only one. Do not worry about being sick. The probability that you will be sick if you go to MathPath is about 0.5%. Even if you do get sick, adults will regularly check in on you, and will do their best to make you feel well as soon as possible.

I went to mathpath this year, for the whole time, and I was glad I did. The only time that I got homesick was when the two-weekers left. However, after 4 weeks, I was really glad that I had not missed out on all the stuff in the third and fourth weeks. If you choose the2-week option, try to leave the 4-week option open.

I am applying for this summer, but I have never been to a sleep-away camp. As a 14 year old, this is my last year of eligibility, so I don't want to miss out on anything. However, I am a little bit hesitant about staying the full 4 weeks. Professor Maurer wrote to me that they are trying not to have any 2-week campers, so I think that I should try the full camp.

But I have a few questions:
1) Are you allowed to have a cell phone?
2) Is there access to computers/email?
3) How is the food? If it is not great, are we allowed to bring/keep some of our own? (I am a very picky eater, unfortunately.)

Thanks.

Last I checked, cell phones are permitted

[quote="goldenboy1.618"]
1) Are you allowed to have a cell phone?
2) Is there access to computers/email?
3) How is the food? If it is not great, are we allowed to bring/keep some of our own? (I am a very picky eater, unfortunately.)[/quote]

1) Yes, you are allowed to have a cell phone, though it is not required.
2) Yes, you have access to computers/email pretty much whenever you are not in class.
3) The food, last year, was top-of-the-line cafeteria food. It was really good. Of course, mathpath will be in a different place this year, but I expect the food will still be pretty good.

It won't help you to bring food, because first the food will spoil after a while, and second, you can't bring food for a whole month.

[quote="mathq"]Yeah, and I want to mention one more thing, I was grade 8 when I was enrolled in MathPath last year, now that I am grade 9, I look back and figured that MathPath isn't worth going for. If you are just planning to go for learning, I suggest choosing somewhere else. If you are aiming to find friends who are also interested in math as you do, go for it. I found myself friends who I still call, and we are even planning to go to same university.
I'm not saying that MathPath is not educational, there is too much recreation activity which is not worth 4500 bucks. I understand that this camp is aiming for middle school kids, but I INSIST that you focus more on making friends. They are gonna be your life long friends.[/quote]

I completely disagree.
Making friends is very important at MathPath, but you are missing a great opportunity if you don't learn things from it. You are guaranteed to learn a whole lot from MathPath if you want to learn from it.

For concerns about picky eating, keep in mind that you will be eating at a college style dining hall. There will be a variety of choices of hot food made, but there will usually be a soup / salad line there for you to choose from if you don't like the main choices. Your local university is probably a good indicator of food quality.

To address Eric's and mathq's argument, I would say that Mathpath offers a more enriching and horizon-broadening environment for learning mathematics. If your goal is to hone down your problem solving / bag of tricks for competitions, you might want to consider other camps, such as AwesomeMath which are more geared towards those sort of competitions. However, if you want a summer camp that exposes you to a variety of mathematics that chances are you've never heard of before, Mathpath would be a better choice. Speaking as a Mathpath alumni, I can say for certain that you'll make friends that have the same interest in mathematics as you do, which you can use to expand and explore your natural curiosity, a trait which I am sure exists at other places as well. Mathpath is one of the best experiences I have ever had, exposing me to worlds of mathematics and introducing me to a variety of characters that I will stay friends with for quite some time.

At MathPath this year, there were kids who did a lot of math (like me) and others who just socialized the whole time.  It's a choice.  I suggest you do a lot of math, challenge yourself, and always do your homework, but get exercise and do activities, especially on weekends, when there is no math to be done.  Because there isn't that much homework, you'll still have time for a lot of socializing in addition to exercise and math.

Revival......please try not to revive.

I went to MathPath in 2008 when I was 11, one of the youngest there, and I had no trouble with it. As hm29168 said, when 2 weeks are over you're just getting to know everyone there.


inequalities
function
inequalities unsolved

For the first inequality,if we can prove
[quote="chidat"]
$\frac{a}{b^{2}+c^{2}}+\frac{b}{a^{2}+c^{2}}+\frac{c}{b^{2}+c^{2}}\geq\frac{3\sqrt{3}}{2\sqrt{a^{2}+b^{2}+c^{2}}}$[/quote]
the inequality will be solved easily

1.Wehave that is equivalent to $\displaystyle \sum \frac{a}{b+c}+\sum \frac{\sqrt{a(a+b+c)}}{b+c}\geq \frac{3+3\sqrt 3}{2}$.So we just need to prove $\displaystyle \left(\sum \frac{\sqrt a}{b+c}\right)(\sqrt{a+b+c})\geq \frac{3\sqrt 3}{2}$,equivalent to $\displaystyle \left(\sum \frac{\sqrt a}{b+c}\right)(\sum \sqrt a)\geq \frac{9}{2}$.Now we use Chebasev inequality so we have $3\displaystyle\sum \frac{a}{b+c}\geq \frac{9}{2}$,obvious. :)

The first was posted again and I gave a solution
http://www.mathlinks.ro/Forum/viewtopic.php?t=55240

No one wants to solve the second ineq ???
But I think it's very nice.

Hi anybody!
I've just found the solution of fist problem
This problem is my teacher's and I found it on internet
[url=http://journals.cms.math.ca/cgi-bin/vault/public/view/CRUXv17n5/body/HTML/152?template=CRUX]CLICK HERE[/url]
                                              BYE! :D

[quote="mateivld"].Now we use Chebasev inequality so we have $3\displaystyle\sum \frac{a}{b+c}\geq \frac{9}{2}$,obvious. :)[/quote]
 the sign of Tchebucheff was reversed!

Is really necessary all this? If I don´t mistake in my calculations this functions has a negative second derivate, so it´s just a Jensen´s Inequalitie!

[quote="Nameless"]Let  $x,y,z$ be positive numbers such that $x+y+z=1$. Prove that
$\sum\frac{1+\sqrt{x}}{1-x}\geq \frac{9+3\sqrt{3}}{2}$.[/quote]
$\sum\frac{1}{1-\sqrt{x}}\geq\sum\frac{3}{4}+\left(\frac{9}{4}+\frac{3\sqrt{3}}{2}\right)x=\frac{9+3\sqrt{3}}{2}$.
[hide="Proof."]
$\frac{1}{1-\sqrt{x}}-\frac{3}{4}-\left(\frac{9}{4}+\frac{3\sqrt{3}}{2}\right)x=\frac{(2+\sqrt{3})(2-\sqrt{3}+\sqrt{x})(1-\sqrt{3x})^{2}}{4(1-\sqrt{x})}\geq 0$.
[/hide]


algebra
polynomial
binomial theorem
algebra unsolved

Find all reals $ \alpha$ for which there is a nonzero polynomial $P$ with real coefficients such that
\[\frac{P(1)+P(3)+P(5)+\cdots+P(2n-1)}{n}=\alpha P(n) \quad \forall n \in \mathbb N,\]
and find all such polynomials for $ \alpha=2$.

[hide]
For $ n\equal{}1$, we get $ P(1)\equal{}\alpha P(1)$, so $ \alpha \equal{}1$ or $ P(1)\equal{}0$. When $ \alpha \equal{}1$, the constant $ P(x)\equal{}1$ is a solution. Otherwise, let $ P(x)\equal{}(x\minus{}1)Q(x)$.

If the given equation holds for $ n$ and $ n\plus{}1$, then we have
\[ \alpha (n\plus{}1)P(n\plus{}1)\equal{}\sum_{k\equal{}1}^{n\plus{}1}P(2k\minus{}1)\]
\[ \alpha nP(n)\equal{}\sum_{k\equal{}1}^n P(2k\minus{}1)\]
Subtracting,
\[ \alpha ((n\plus{}1)P(n\plus{}1)\minus{}nP(n))\equal{}P(2n\plus{}1)\]
for all positive integers $ n$. So the above polynomials are equal for all positive integers, making them identical. We then have
\[ \alpha ((x\plus{}1)P(x\plus{}1)\minus{}xP(x))\equal{}P(2x\plus{}1)\]
Let the degree of $ P$ be $ d$, the leading coefficient be $ L$, and the next coefficient be $ M$. Equating $ x^d$ term gives
\[ \alpha (L\plus{}(dL\plus{}M)\minus{}M)\equal{}2^dL\Rightarrow \alpha \equal{}\frac{2^d}{d\plus{}1}\]
Note that $ d\neq 0$ (else $ P(x)\equal{}P(1)\equal{}0$) so $ d$ is positive. So $ \alpha$ must equal $ \frac{2^d}{d\plus{}1}$ for some positive integer $ d$.

Now consider $ \alpha \equal{}2$, so $ 2^{d\minus{}1}\equal{}d\plus{}1\Rightarrow d\equal{}3$. The above, after substituting for $ Q$, becomes
\[ (x\plus{}1)Q(x\plus{}1)\minus{}(x\minus{}1)Q(x)\equal{}Q(2x\plus{}1)\]
Plug in $ x\equal{}\minus{}1$ to get $ Q(\minus{}1)\equal{}0$. Then let $ Q(x)\equal{}(x\plus{}1)R(x)$ so that $ x\plus{}1$ cancels, leaving
\[ (x\plus{}2)R(x\plus{}1)\minus{}(x\minus{}1)R(x)\equal{}2R(2x\plus{}1)\]
Finally, plug in $ x\equal{}\minus{}1$ again to get $ R(0)\equal{}0$, but also $ d\equal{}3$ implies the degree of $ R$ is 1. So $ R(x)\equal{}Cx$, making $ P(x)\equal{}Cx(x\minus{}1)(x\plus{}1)$ for $ C\neq 0$.
[/hide]

[quote="pleurestique"][quote="scorpius119"]
$ \alpha (L \plus{} (dL \plus{} M) \minus{} M) \equal{} 2^dL\Rightarrow \alpha \equal{} \frac {2^d}{d \plus{} 1}$
[/quote]

everything appears right up until there. how do you get a $ dL$ in the coefficient if $ d$ is the degree. you weren't differentiating[/quote]

It's the binomial theorem: $ dL\plus{}M$ is the coefficient of $ x^{d\minus{}1}$ in $ P(x\plus{}1)\equal{}L(x\plus{}1)^d\plus{}M(x\plus{}1)^{d\minus{}1}\plus{}\cdots$ (the rest of the terms are insignificant).


[quote="anthony19891019"]$a+b+c=3$
Proof:$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq ab+bc+ac$[/quote]
$a\geq0,$ $b\geq0$ and $c\geq0.$
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35839
the soarer's and mamat's posts.

[quote="anthony19891019"]Can anyone use mixing variables to solve it? :)[/quote]
The gift for you, anthony19891019. :)
Let $a=\min\{a,b,c\}$ and $F(a,b,c)=\sqrt{a}+\sqrt{b}+\sqrt{c}-ab-bc-ac.$
Then $b+c\geq2,$ and
 $F(a,b,c)-F\left(a,\frac{b+c}{2},\frac{b+c}{2}\right)=\sqrt{b}+\sqrt{c}-\sqrt{2(b+c)}-bc+\left(\frac{b+c}{2}\right)^{2}=$
$=\frac{(b-c)^{2}}{4}-\frac{(\sqrt{b}-\sqrt{c})^{2}}{\sqrt{b}+\sqrt{c}+\sqrt{2(b+c)}}=\frac{(\sqrt{b}-\sqrt{c})^{2}((\sqrt{b}+\sqrt{c})^{2}(\sqrt{b}+\sqrt{c}+\sqrt{2(b+c)})-4)}{4(\sqrt{b}+\sqrt{c}+\sqrt{2(b+c)})}.$
But $(\sqrt{b}+\sqrt{c})^{2}(\sqrt{b}+\sqrt{c}+\sqrt{2(b+c)})-4\geq(b+c)\sqrt{2(b+c)}-4\geq0.$
Hence, $F(a,b,c)-F\left(a,\frac{b+c}{2},\frac{b+c}{2}\right)\geq0.$
Remain to prove that $F(a,b,b)\geq0,$ where $a=3-2b$ and $b\in[1,1.5]$
$F(a,b,b)=\sqrt{3-2b}+2\sqrt{b}-2b\sqrt{3-2b}-b^{2}.$ Try to finish.

It is not apparent to me how the line after "But..." follows...
By the way, this post was linked as "difficult to understand", but I think it is really elementary if this one step is clearly explained...


Let $ a,b,c>0.$ Prove that:
$ \sum \frac{a^3}{2a^2\plus{}b^2}\geq \frac{a\plus{}b\plus{}c}{3}$

Clearing denominators leaves

\begin{align*} 9(a^3 + b^3 + c^3) \ge& 3 ( a^3 + b^3 + c^3 ) \\
& + 3 ( a^2b + b^2c + c^2a ) \\
& + 3 ( a^2c + b^2a + c^2b ) \end{align*}

Which is true by Muirhead/rearrangement

[quote="No Reason"]Let $ a,b,c > 0.$ Prove that:
$ \sum \frac {a^3}{2a^2 \plus{} b^2}\geq \frac {a \plus{} b \plus{} c}{3}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22937

[quote="No Reason"]Let $ a,b,c > 0.$ Prove that:
$ A = \sum \frac {a^3}{2a^2 + b^2}\geq \frac {a + b + c}{3}$[/quote]
We have:
$ A = \sum\frac {a^4}{a(2a^2 + b^2)}$
By CBS:
$ A\geq \frac {(a^2 + b^2 + c^2)^2}{2a^3 + ab^2 + 2b^3 + bc^2 + 2c^3 + ca^2} = B$
We only prove that : $ B\geq \frac {a + b + c}{3} (1)$
That right!
$ (1)\Rightarrow (\sum a^4) + 5(\sum a^2b^2)\geq 3(ab^3 + bc^3 + ca^3) + 2(a^3b + b^3c + c^3a) + abc(a + b + c)$
$ \Leftrightarrow \sum(a^4 + 3a^2b^2 + 2c^2a^2 - a^2bc - 2a^3b - 3a^3c)(2)$
Beside:
Suppose $ f(x) = a^2 + 3b^2 + 2c^2 - bc - 2ab - 3ca = a^2 - (2b + 3c)a + 3b^2 - bc + 2c^2$
 $ \Delta_a = - 8b^2 + 16cb + c^2$
 $ \Delta_b = 288c^2\geq 0$
 $ \Leftrightarrow \Delta_a\leq 0 (because - 8 < 0)$
 $ \Longleftrightarrow f(x)\geq 0 (because 1 > 0)$
$ \Longrightarrow (2) right \Rightarrow (1) right\Rightarrow$  that inequalities proved

[quote="No Reason"]
$ A\geq \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{2a^3 \plus{} ab^2 \plus{} 2b^3 \plus{} bc^2 \plus{} 2c^3 \plus{} ca^2} \equal{} B$
We only prove that : $ B\geq \frac {a \plus{} b \plus{} c}{3} (1)$
That right!
[/quote]
I don't think so! Try $ a\equal{}3$ and $ b\equal{}c\equal{}2.$ :wink:


number theory
greatest common divisor
algebra
polynomial
number theory unsolved

[quote="stvs_f"]what is gcd(2^561 - 2,3^561 - 3,...,561^561 - 561)[/quote]
Let $ p$ be a prime, then
$ p \mid \gcd(2^{561} \minus{} 2,3^{561} \minus{} 3,...,561^{561} \minus{} 561)$ $ \iff$
$ n^{560} \equiv 1 \bmod p$ or $ p \mid n$, $ \forall n \in \{1,2,3,..,561\}$
So if $ p \minus{} 1 \mid 560$ it is obvious that $ p \mid \gcd(2^{561} \minus{} 2,3^{561} \minus{} 3,...,561^{561} \minus{} 561)$.

(Btw. from now on $ G \equal{} \gcd(2^{561} \minus{} 2,3^{561} \minus{} 3,...,561^{561} \minus{} 561)$)

Notice $ 560 \equal{} 2^4 \cdot 5 \cdot 7$. So $ 560$ has $ (4 \plus{} 1)(1 \plus{} 1)(1 \plus{} 1) \equal{} 20$ divisors. These are:
$ 1$, $ 2$, $ 4$, $ 8$, $ 16$,
$ 5$, $ 10$, $ 20$, $ 40$, $ 80$,
$ 7$, $ 14$, $ 28$, $ 56$, $ 108$,
$ 35$, $ 70$, $ 140$, $ 280$, $ 560$.
So $ p \minus{} 1 \mid 560$ $ \iff$ $ p \in \{2,3,5,9,17,6,11,21,41,81,8,15,29,57,109,3671,141,281,561\}$. But only some of these are primes, so
$ p \in \{2,3,5,11,17,29,41,71,109,281\}$.
This motivates me to guess that:
$ G \equal{} 2 \cdot 3 \cdot 5 \cdot 11 \cdot 17 \cdot 29 \cdot 41 \cdot 71 \cdot 109 \cdot 281$.
It's obvious that if $ p \le 561$ then $ p^2 \nmid G$ since then would $ p^2 \mid p^{561} \minus{} p$, which is absurd.

Assume $ p \minus{} 1 \nmid 560$, $ p \le 561$ and $ p \mid G$. Then $ 2^{560} \equiv 3^{560} \equiv ... \equiv (p \minus{} 1)^{560} \equiv 1 \bmod p$. Let $ a \equal{} (p \minus{} 1,560)$. Then $ 2^a \equiv 3^a \equiv ... \equiv (p \minus{} 1)^a \equiv 1 \bmod p$ where $ a < p \minus{} 1$. It should then be "well known" that it is not possible, but can anyone give a proof?

Assume $ p > 561$ and $ p \mid G$. Let $ a \equal{} (p\minus{}1,560)$. Then $ 2^a \equiv 3^a \equiv ... \equiv 561^a \equiv 1 \bmod p$. In other words the polynomia $ p(x) \equal{} x^a \minus{} 1$ have the ability $ p \mid p(2), ..., p(561)$. If $ a < 560$ it will have $ 561$ roots $ \mathbb{Z}/p\mathbb{Z}$ in but a degree less than $ 560$, which is impossible, so assume $ a \equal{} 560$, and therefore $ p \equal{} 560k\plus{}1$.
On the other hand, we know that $ x^{560}\minus{}1 \equiv (x\minus{}2)(x\minus{}3)...(x\minus{}561)$, and by comparing coefficients: $ 2\plus{}3\plus{}...\plus{}561 \equiv 0$. Then $ p \mid 561 \cdot 281 \minus{} 1$, which isn't possible when $ p > 561$ and $ p \equiv 1 \bmod 560$.


hi,
theres a wording problem i cannot solve, Im not sure how to write out the equation from the problem.

this is the question:
Bus company charges a half of its normal far for concession card holders. If company takes $ 2600 and carries 900 passengers, then how many of those passengers were concession card holders if the normal far was$4.

All i know so far is to
Let X be number of concession card holders. Can anyone tell me what is the equation?been workin on it for a while. I will know what to do if i know the equation. I know its suppose to be an easy question but i feel so dumb not knowing it.

thank u

THE QUESTION WAS NOT THAT CLEAR.
BUT IF I UNDERSTOOD OT CORRECTLY, EQUATION IS
        LET X BE NO. OF PERSONS NOT HAVING MEMBERSHIP CARD.
  4X + (900-X)2 =2600

THIS WOULD GIVE CARD HOLDERS =900-X = 500 :wink:


Hallo! Ich soll als Vertreterin des Pentakratischen |Vorstands|  :D die Liste der Bundessieger 2004 hier posten.

[b]Richard Bamler*****[/b], Gilching
[b]Anne Fernengel[/b], Kleinostheim    :-)
[b]Friedrich Feuerstein***[/b], Heidelberg
[b]Alexander Fromm[/b], Berlin
[b]Robert Graf**[/b], Augsburg
[b]Darij Grinberg**[/b], Karlsruhe
[b]Asar Hage-Ali[/b], Ahrensfelde
[b]Daniel Harrer[/b], Regensburg
[b]Paul-Fiete Hartmann[/b], Göttingen
[b]Annika Heckel[/b], Immenstadt
[b]Christian Hercher**[/b], Kahla
[b]Alexander Ivanov[/b], Rottweil
[b]Stefan Kahler[/b], Bayreuth
[b]Xin Li**[/b], Münster, Westf.
[b]Benjamin Matschke[/b], Cottbus
[b]Momsen Reincke[/b], München
[b]Peter Scholze**[/b], Berlin
[b]Jewgeni Strekalowski***[/b], Weimar
[b]Dominik Wrazidlo[/b], Wiesloch

Für Fehler entschuldige ich mich schonmal im Voraus, ich habe bestimmt welche gemacht oder jemanden vergessen oder... *sehr besorgt sei*. Auf jeden Fall an alle von mir herzlichen Glückwunsch!!!


algebra
polynomial
calculus
integration
number theory
relatively prime
Rational Root Theorem

[quote="Peter"]Let $ f(x)=x^{n}+5x^{n-1}+3$, where $ n>1$ is an integer. Prove that $ f(x)$ cannot be expressed as the product of two nonconstant polynomials with integer coefficients.[/quote]

First, it's an easy job to show that $ f(x)$ has no linear factor because we have a good weapon:

 [b]Rational Root Theorem[/b] Let $ P(x)=p_{n}x^{n}+\cdots+p_{0}\in Z[x]$ and $ p_{n}\neq 0$. If $ r$ and $ s \neq 0$ are relatively prime integers such that $ P \left( \frac{r}{s}\right)=0$, then we have $ r \vert p_{0}$ and $ s \vert p_{n}$.

If $ f(x)=x^{n}+5x^{n-1}+3$ has a linear factor in $ Z[x]$, then $ f$ has a rational root. By applying [b]RRT[/b], we find that $ 1$, $ -1$, $ 3$, and $ -3$ are only candidates. However, one can easily check that none of $ f(1)$, $ f(1)$, $ f(3)$ and $ f(-3)$ can be zero. This is a contradiction.

Now, we establish that $ f$ has no factor in $ Z[x]$ with arbitrary degrees $ \geq 1$. Here, we present two different solutions using standard techniques:

[b][First Solution][/b] Assume that $ f(x)$ can be expressed as the product of two nonconstant polynomials with integer coefficients:
\[ f(x)= \left( a_{k}x^{k}+\cdots+a_{0}\right)\left( b_{m}x^{m}+\cdots+b_{0}\right), \]
where $ 1 \leq k,m<n$, $ k+m=n$,  $ (a_{0},b_{0})=(\pm 1, \pm 3)$, and $ (a_{k},b_{m})=(\pm 1, \pm 1)$. Let $ B=\{ j \in \{0, \cdots, m\}\; \vert \; b_{j}\not\equiv 0 \; (mod \; 3) \}$. The main trick is to think of $ l=Min B$, which is well-defined because $ m \in B$. Since $ b_{0}= \pm 3$ or $ 0 \not\in B$, we get $ 1 \leq l \leq m$. Since $ b_{0}\equiv \cdots \equiv b_{l-1}\equiv 0 (mod \; 3)$, after reading the coefficient of $ x^{l}$ modulo $ 3$ in
\[ x^{n}+5x^{n-1}+3=\left( a_{k}x^{k}+\cdots+a_{0}\right)\left( b_{m}x^{m}+\cdots+b_{0}\right), \]
we find that the coefficient of $ x^{l}$ in $ x^{n}+5x^{n-1}+3$ is not divisible by $ 3$. Indeed, we get
\[ a_{l}b_{0}+\cdots+a_{1}b_{l-1}+a_{0}b_{l}\equiv a_{0}b_{l}\not \equiv 0 (mod \; 3). \]
Since, in $ x^{n}+5x^{n-1}+3$, only two terms $ x^{n}$ and $ x^{n-1}$ have the coefficient not divisible by $ 3$, we see that $ l=n$ or $ l=n-1$. It now follows that $ k=n-m \leq n-l \leq 1$. So, we get $ k=1$ and $ a_{1}=a_{k}= \pm 1$.  Now, look at the polynomial factorization again:
\[ x^{n}+5x^{n-1}+3=\left( a_{1}x+a_{0}\right)\left( b_{m}x^{m}+\cdots+b_{0}\right). \]
Since $ a_{1}=\pm 1$ and $ a_{0}= \pm 1$, we find that $ a_{1}x+a_{0}=\pm x \pm 1$ so that $ f(x)=x^{n}+5x^{n-1}+3$ is divisible by $ x+1$ or $ x-1$.  In other words, $ f(1)=0$ or $ f(-1)=0$. However, this is a contriadiction because $ f(1)=9$ and $ f(-1)=3+4(-1)^{n-1}$. This completes the proof.

[b][Second Solution][/b] Assume $ f(x)=F(x)G(x)$, where $ F(x), G(x) \in Z[x]$ and $ 1 \leq deg F < n$. Since $ 3=f(0)=F(0)G(0)$, we may also assume that $ (F(0), G(0))=(1,3)$ or $ (-1,-3)$. The key idea is to empoly complex numbers. By the fundamental theorem of algebra, we can write
\[ F(x)=A(x-z_{1}) \cdots (x-z_{k}) \;\; \text{and}\;\; G(x)=B(x-z_{k+1}) \cdots (x-z_{n}), \]
where $ z_{1}, \cdots, z_{n}\in C$ and $ A=B=\pm 1$. After looking at $ F(0)$, we find that
\[ \pm 1 =F(0) = \pm A z_{1}\cdots z_{k}\;\; \text{or}\;\; \left\vert \prod_{1 \leq i \leq k}z_{i}\right\vert =1. \]
The key observation is that $ z_{1}$, $ \cdots$, $ z_{k}$ are also zeros of $ f=FG$. It follows that
\[ {z_{i}}^{n}+5{z_{i}}^{n-1}+3=0 \;\; \text{or}\;\;{z_{i}}^{n-1}(z_{i}+5)=-3 \]
for all $ i \in \{1, \cdots, k \}$. Hence, we obtain
\[ 3^{k}= \left\vert \prod_{1 \leq i \leq k}{z_{i}}^{n-1}(z_{i}+5) \right\vert = \left\vert \prod_{1 \leq i \leq k}(z_{i}+5) \right\vert =\vert F(-5) \vert. \]
Now, looking at $ f(-5)$, we deduce that
\[ 3 = \vert f(-5) \vert = \vert F(-5) G(-5) \vert=3^{k}\vert G(-5) \vert. \]
Since $ G(-5) \in Z$, this means that $ k=1$ or $ deg F=1$. Hence, $ f(x)=x^{n}+5x^{n-1}+3$ has a linear factor $ F(x)$ with integral coefficients. Since $ F(0)=\pm 1$ and $ A=\pm 1$, we have $ F(x)= \pm x \pm 1$ so that $ F(x)$ is divisible by $ x+1$ or $ x-1$. This is impossible as  in the first solution.

Rather unpretty solutions were posted above.I would like to write down another.I am deeply convinced that Extended Eisenstein criterion was already discussed here,that's why I will leave this theorem without proof.
[b]Extended Eisenstein criterion[/b]:
Suppose that polynomial $ P(x)\in\mathbb{Z}[x]$ has the following form:
\[ P(x) \equal{} a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} \dots \plus{} a_1x \plus{} a_0
\]
And there exist prime number $ p$,such that $ p^2\not{|} a_0$ and $ p|a_0,a_1,\dots,a_k$,but $ p\not{|} a_{k \plus{} 1},\dots,a_n$,then $ P(x)$ has an irreducible factor of a degree greater than $ k$.As a particular case if we take $ k \equal{} n \minus{} 1$ then $ P$ is irreducible.
Let's return to the initial problem.
By the above mentioned theorem,$ f$ has an irreducible factor of a degree $ n \minus{} 1$,therefore,it is enough to show that  $ f$ has no integer roots.
Suppose to the contrary that $ f$ has an integer root,say $ r$.Then
$ r^{n} \plus{} 5r^{n \minus{} 1} \plus{} 3 \equal{} 0$.
This equation implies that $ |r|$ is either $ 3$ or $ 1$.It is not hard to check the case $ |r| \equal{} 1$ and to observe that if $ n > 2$ then $ 9|3$ what is impossible.
It means that either $ n \equal{} 2$ and $ |r| \equal{} 3$ or $ f$ has no roots.
But equation $ r^2 \plus{} 5r \plus{} 3$ has no integer roots,contradiction.
$ \blacksquare$.

[quote=Peter]Let $f(x)=x^{n}+5x^{n-1}+3$, where $n>1$ is an integer. Prove that $f(x)$ cannot be expressed as the product of two nonconstant polynomials with integer coefficients.[/quote]

Remark: Follows directly by [url=https://en.m.wikipedia.org/wiki/Perron%27s_irreducibility_criterion]Perron's Criterion [/url].


Well, yeah. It can't be a coincedence that mathgenius123 posted the homework we got [i]and[/i] the number of the Mandelbrot Individual rounds was 8875, which is the number in the url of the homework.

@mathgenius123: You really should stop posting homework problems. That's not good for you.

To not spam, I'll... tell you to multiply out the product for #2 on the first team play  and to plug in numbers for #1 on the same round. See, I didn't spam :lol:


geometry
geometric transformation
reflection
geometry unsolved

Consider triangle ABC with angle bisectors AA1, BB1, CC1. Denote by U the intersection of AA1 and B1C1. Let V be the projection from U onto BC. Let W be the intersection of the angle bisectors of
∠BC1V and ∠CB1V . Prove that A, V,W are collinear.

We only need to prove that:
\[ \frac{C_1V}{C_1A} \equal{} \frac{B_1 V}{B_1A}\]
This is equivalent to
\[ \frac{VC_1}{VB_1} \equal{} \frac{AC_1}{AB_1} \equal{} \frac{UC_1}{UB_1}\]
 which happens only if $ VU$ is the bisector of $ \angle{C_1VB_1}$. This is a well-known result.

[quote="mathangel"]We only need to prove that:
\[ \frac {C_1V}{C_1A} \equal{} \frac {B_1 V}{B_1A}
\]
[/quote]
Why?I don't understand.Can you help me?

Suppose that $ W_1, W_2$ are intersections of the two bisectors with the line $ AV$. Proving that $ A, V, W$ are collinear is the same as proving $ W_1$ concides with $ W_2$. And this happens if and only if
\[ \frac{W_1 A}{W_1 V} \equal{} \frac{W_2 A}{W_2 V}\]
which turns out to be equivalent to that forementioned condition.


[quote="moldlee"]Show that $ {{2007} \choose {91}} \equiv 5 \pmod{91}$.[/quote]Have you tried writing out what $ {{2007} \choose {91}}$ is using the formula $ {{n} \choose {r}} \equal{} \frac { n!}{ (n \minus{} r)! r!}$?

(PS: that formula has been written from memory, which isn't particularly great!)

EDIT: Or (after some thought) could you not remove multiples of 91  from $ {{2007} \choose {91}}$  first and then use the formula? (i.e.2007 -91 =1916, and 91-91 =0 so we have $ {{1916} \choose {0}}\pmod {91}$  and so on). I realise it is unreasonable to use the formula with such large number

Since 91=13*7, could you possibly prove that it's 5 mod 7 and 5 mod 13, then etc.?


If $a,b,c>0$ and the denominators of the fractions(which follow) are strictly positive, then prove that
$\sum\frac{a^2+b^2}{8c^2-ab}\ge\frac67$

My edition of hungkhtn's post :D

We are to prove $\sum \frac{a^2+b^2}{8c^2-ab} \ge \frac{6}{7}$, which is equivalent to
\[\sum\left(\frac{a^2+b^2}{8c^2-ab}+2\right)=\sum \frac{(a-b)^2+16c^2}{8c^2-ab}\ge \frac{48}{7}.\]

We will prove that $\sum \frac{c^2}{8c^2-ab}\ge \frac{3}{7}.$

Indeed,
\[\sum \frac{c^2}{8c^2-ab}=\sum \frac{a^2b^2c^2}{8c^2a^2b^2-a^3b^3}\ge a^2b^2c^2\cdot \frac{9}{24c^2a^2b^2-a^3b^3-b^3c^3-c^3a^3} ,\]
but $a^3b^3+b^3c^3+c^3a^3 \ge 3a^2b^2c^2$, so we have result.


On a circular dartboard with concentric scoring areas, it is possible to score 15, 16, 17, 18, or 19 points with one dart.  What is the fewest number of darts necessary to score exactly 100 points on this dart board?

Try it first without a calculator.

[hide="I really shouldnt answer mc problems anymore but..."]We can express 100 as 15*6+10. We can express 19 as 15*1+4. We know that if we use 6 numbers, they must add up to 15*6+10. We can do this by adding 19+19+17+15+15+15.

If we can make n of the numbers add up to (n+1)*15 or greater, we can use less then 6 numbers. If we use 4 19's, we get 5*15+1. Now to use less then 6 numbers, we must use 1 number to get from 5*15+1 to 6*15+10. But we cannot use 25, so we cannot sum to 100 with those numbers with less then 6 numbers. Therefore the answer is 6.[/hide]

[hide]
The lowest I got is 15+16+17+18+19+15=100, there are 6 numbers, so the answer is 6.[/hide]


inequalities
trigonometry
function
number theory

If 3n+1 iis a perfect square then n+1 can be written as sum of squares of three numbers.

If A+B+C = pi
then show that SinA + SinB -CosC less than equal to 3/2.

Please make separate posts for your problems, and please use LaTeX.

For Number Theory:
[hide]As $3n+1$ can't be the square of a multiple of $3$, we know that $3n+1 = (3k\pm1)^2$ for some integer $k$. In the case $3n+1 = (3k+1)^2$, we get that $n=3k^2+2k$, so $n+1 = 3k^2+2k+1 = k^2+k^2+(k+1)^2$. In the case $3n+1 = (3k+1)^2$, we get $n+1 = k^2+k^2+(k-1)^2$, so we're done.[/hide]

Try to generalise this: it works for any prime $p$ instead of 3 :)

If $p$ is a prime and $pn + 1$ is a perfect square, then $n + 1$ is the sum of $p$ perfect squares.

Well, the generalization can be solved exactly as [b]Kurt Gödel[/b] did :
We have : $pn+1 = (pk+1)^2$, then :
$pn+1 =(pk)^2 + 2pk + 1$
$\iff n= pk^2 +2k$.
Hence :
$n+1 = pk^2 + 2k + 1$
$\iff n+1 = \underbrace{k^2 + k^2 + \ldots + k^2}_{p \ times} + 2k + 1$
$\iff n+1 = \underbrace{k^2 + k^2 + \ldots + k^2}_{(p-1) \ times} + (k+1)^2$
So, $n + 1$ is indeed the sum of $p$ perfect squares.

:)

[quote="Arne"]How can you be sure that $pn + 1$ is the square of a number of the form $pk + 1$?

Your proof is not complete![/quote]
Ok, sorry. :oops:

I think the following one works ;) :
We have : $pn+1=a^2$, hence : $n = \frac{a^2-1}p$.
From that we get that $p$ either divides $a+1$ or it divides $a-1$.
So, we have : $a = kp+r$, with $|r|=1$.
$n = \frac{(kp+r)^2-1}p = k^2p+2kr$
$\iff n+1 = k^2 + k^2 + \ldots + k^2 + (k^2+2kr+1) = k^2+k^2+ \ldots + k^2 + (k^2 + 2kr + r^2) = k^2 + k^2 + \ldots + k^2 + (k+r)^2$
So $n$ is the sum of $p$ perfect squares. :)

Indeed. And that's why $p$ needs to be prime.

[color=darkblue][b]The trigonometrical inequality:[/b] $\sin A+\sin B-\cos C\le \frac 32\ \ (1).$
[b][u]Proof.[/u][/b] From the well-known conditioned trigonometrical inequality $\{x,y,z\}\subset \left[-\frac{\pi }{2},\frac{\pi}{2}\right]\ ,\ x+y+z=\pi\Longrightarrow \cos x+\cos y+\cos z\le \frac 32\ \ (2)\ .$
for $x: =\frac{\pi}{2}-A\ ,\ y: =\frac{\pi}{2}-B\ ,\ z: =\pi -C$ results $(1)$.
[u]Remark.[/u] The inequality $(2)$ results from the concavity of the function $\cos\ : \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\rightarrow [0,1]$.[/color]