pigeonhole principle
quadratics
calculus
integration
topology
calculus computations

What exactly is a dense set?
and how would you show that a set is dense?

I am asking this since i am working throught ACoPS and got to questions that says, "[b][color=orange]Show that the set of real numbers that are zeros of a quadradic equations with integer coefficients is dnse[/color][/b]" and "[b][color=green]Let $D$ be the set of dyadic rationals, the rational numbers whose denominators are powers of two. Show that $D$ is dense[/color].[/b]"

The set $S$ is dense in $\mathbb{R}$ iff for every $a<b,$ there exists $s\in S$ such that $a<s<b.$

For instance, the set of rational numbers is dense in $\mathbb{R},$ as is also the set of irrational numbers. But the set of integers is not dense in $\mathbb{R}$ as there are no integers between $2.3$ and $2.8.$

More generally, if $X$ is any metric space and $S\subset X,$ then we say that $S$ is dense in $X$ iff for every $x\in X$ and $\epsilon>0,$ there exists $s\in S$ such that $d(s,x)<\epsilon.$

Said more quickly: $S$ is dense in $X$ iff the closure of $S$ equals $X.$

thank.
so how would you show that a set is dense?

For the first question: Consider $\{q^{2}(x-p/q)^{2}: p,q\in\mathbb{Z}\}$

For the second question:  Given $\epsilon>0$, choose $n$ so large that $2^{-n}<\epsilon$.  Then apply the pigeonhole principle to the set $\{m2^{-n}: m\in\mathbb{Z}\}$.

For every $x\in\mathbb{R}$ and every $\epsilon>0$, you want to find, in the first case, a quadratic with integral coefficients having a root $r$ with $|x-r|<\epsilon$.  For the second you want to find $n,m\in\mathbb{Z}$ such that $|x-m2^{-n}|<\epsilon$.

exactly what do you mean by "apply pigeon hole principle"?

i know what the pigeon hole principle is but i am not sure how you would apply it to this case.

You can prove it without pigeons:
1) a subset of $\mathbb R$ is dense iff its complement does not contain any nonempty open intervals;
2) the complement of the set $\{k/2^{n}\colon k\in\mathbb Z\}$ does not contain open intervals of length more than $1/2^{n}$. Therefore,
3) the complement of the set of all dyadic rationals does not contain nonempty open intervals.


inequalities
calculus
derivative
linear algebra
matrix
inequalities proposed

if $x,y,z$ are positive numbers such that $xy+yz+zx=xyz.$ Prove that

                                  $ x+y+z+6.sqrt(3)-9\geq2.sqrt(xyz). $

First homogenize: $x+y+z+(6\sqrt{3}-9)\frac{xyz}{xy+yz+zx}\geq2\sqrt{xy+yz+zx}$. Now apply to the new homogeneous ineq the condition $xy+yz+zx=1$ to get $x+y+z+(6\sqrt{3}-9)xyz\geq2$. The last one can easily be proven by Langange multipliers.

Can you please post the solution by Lagrange's multipliers?

Consider $f(x,y,z,l)=x+y+z+(6\sqrt{3}-9)xyz+l(xy+yz+zx-1)$
Taking the partial derivatives equal to zero we get the equations:
$1+(6\sqrt{3}-9)yz+l(y+z)=0$  (1)
$1+(6\sqrt{3}-9)xz+l(x+z)=0$  (2)
$1+(6\sqrt{3}-9)xy+l(x+y)=0$  (3)
$xy+yz+zx=1$                          (4)
Substracting (2) from (1) we get $(x-y)[(6\sqrt{3}-9)z+l]=0$ and similarly $(y-z)[6\sqrt{3}-9)x+l]=0$, $(z-x)[(6\sqrt{3}-9)y+l]=0$.
If $x=y=z$ from (4) we get that the common value is $\frac{\sqrt{3}}{3}$. Obviously we cannot have simultaneously $x\not=y,y\not=z,x\not=z$ so suppose wlog $x=y$. Then, in order not to slip into the case $x=y=z$, we must have $x=y=-\frac{l}{6\sqrt{3}-9}$. Plugging these values into (3) we get $l=-\sqrt{6\sqrt{3}-9}$(we end up in a trivial eqn of second degree and we choose the negative value of l because $x,y$ positive). So $x=y=\frac{1}{\sqrt{6\sqrt{3}-9}}$ Plugging these values into (4) we get that $z=\frac{3\sqrt{3}-5}{\sqrt{6\sqrt{3}-9}}$ and re-plugging all values into $f$ we will get an expression which is equal to 2,03.
 Now for $x=y=z=\frac{\sqrt{3}}{3}$ we get that $f=2$. If we take the Hessian matrix of $f$ it should not be big deal to prove that it is defined positively for the point $(\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3})$(everything vanishes except for the term $(6\sqrt{3}-9)xyz$ in some cases). Thus the value 2,03 can be discarded anyway and finally we get the inequality wanted (well, i stilll owe the Hessian of $f$ but....... too lazy for this!)


How many way can you tile a $3$x$n$ rectangle with $2$x$1$ dominios?

I haven't had time to think about it much, but take the following case into consideration:
[hide="Hint"]If $n\equiv 1\pmod 2$, it is not possible to tile it using $2\times 1$ dominos; the proof is very basic (you cannot divide an odd number by 2 and get remainder 0).[/hide]
Masoud Zargar

Sorry for reviving an old thread.  The problem has been solved here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126738


hope my solution is right:
we may assume:$a=\max \{a,b,c\}$
case1:
$a \ge b \ge c$
we have:
$2a+b \ge 2b+c$ and $2a+b \ge 2c+a$
thus:
\[ \frac{\sqrt[3]c-\sqrt[3]a}{2a+b}+\frac{\sqrt[3]a-\sqrt[3]b}{2b+c}+\frac{\sqrt[3]b-\sqrt[3]c}{2c+a} \ge \frac{\sqrt[3]c-\sqrt[3]a}{2a+b}+\frac{\sqrt[3]a-\sqrt[3]b}{2a+b}+\frac{\sqrt[3]b-\sqrt[3]c}{2a+b}=0  \]
case2:
$a \ge c \ge b$
we have:
$2b+c \le 2c+a$,$2b+c \le 2a+b$,
thus:
\[ \frac{\sqrt[3]c-\sqrt[3]a}{2a+b}+\frac{\sqrt[3]a-\sqrt[3]b}{2b+c}+\frac{\sqrt[3]b-\sqrt[3]c}{2c+a}\ge \frac{\sqrt[3]c-\sqrt[3]a}{2b+c}+\frac{\sqrt[3]a-\sqrt[3]b}{2b+c}+\frac{\sqrt[3]b-\sqrt[3]c}{2b+c}=0  \].
thus,the problem is proved.

[quote="manocanh"][quote="zhaobin"]hope my solution is right:
we may assume:$a=\max \{a,b,c\}$
$a \ge b \ge c$
we have: $2a+b \le 2c+a$
[/quote]
[quote]\[ \frac{\sqrt[3]c-\sqrt[3]a}{2a+b}+\frac{\sqrt[3]a-\sqrt[3]b}{2b+c}+\frac{\sqrt[3]b-\sqrt[3]c}{2c+a} \fbox{= ?} \frac{\sqrt[3]c-\sqrt[3]a}{2b+c}+\frac{\sqrt[3]a-\sqrt[3]b}{2b+c}+\frac{\sqrt[3]b-\sqrt[3]c}{2b+c}=0  \].[/quote]
a small mistake !!!!!!!!!  :D :D[/quote]
sorry,thanks for you careful reading.


abstract algebra
category theory
geometry
induction
advanced fields
advanced fields unsolved

$ E$ is a linear space with dimension $ n$ and $ \alpha \in \wedge^{n\minus{}1}(E)$.Show that $ \exists \alpha_{1},...,\alpha_{n\minus{}1} \in E^{*}$ such that:
$ \alpha \equal{}\alpha_{1}\wedge ..\wedge \alpha_{n\minus{}1}$
(Note:$ E^{*}$ is the dual space of $ E$)

how should the right side yield (canonical) an element of $ \wedge^{n\minus{}1} E$?

[quote="-oo-"]how should the right side yield (canonical) an element of $ \wedge^{n \minus{} 1} E$?[/quote]
Since $ \alpha \equal{}\alpha_{1}\wedge ..\wedge \alpha_{n\minus{}1}$ are functionals their wedge is an element of  $ \wedge^{n \minus{} 1} E$.(By definition of wedge)

???
The problem is true if we replace $ E^{\ast}$ with $ E$ (I think I even showed it for something more general than linear spaces - probably, free modules over Euclidean rings or maybe PID's).

  darij

This is true or not?
$ E^{*}\equal{}\wedge^{1}(E)$ !!

[quote="payman_pm"]This is true or not?
$ E^{*} \equal{} \wedge^{1}(E)$ !![/quote]

No. $ E\equal{}\wedge^1 E$.

  darij

I'm confused. Here is how we built the definition:

Take $ \mathcal{T}_s^r(V)$ to be the tensors of covariant order $ r$ and contravariant order $ s$.

So if you take $ s \equal{} 0$ and $ r \equal{} 1$, you have $ \mathcal{T}^1(V) \equal{} \{\phi : V\to\mathbb{R}\}$ where $ \phi$ is linear and all that. i.e. $ \mathcal{T}^1$ consists of all linear functionals, i.e. $ \mathcal{T}^1 \equal{} V^*$.

OK. Now to get to $ \wedge^r(V)$ we define the $ \mathcal{A}$ to by the alternating map and take $ \mathcal{A}(\mathcal{T}^r(V)) \equal{} \wedge^r(V)$.

Take $ r \equal{} 1$ again. It seems to me that since there is nothing to alternate (a covariant order being $ 1$), so $ \wedge^1(V) \equal{} \mathcal{A}(\mathcal{T}^1(V)) \equal{} \mathcal{T}^1(V) \equal{} V^*$.

Sorry to intrude on your question, but I was under the same impression. Where is the error in reasoning here? That $ \mathcal{A}$ is everything. Then it would just make $ \wedge^1(V)$ a subset of $ V^*$.

I have been used to defining $ \wedge^k V$ as $ \left(\mathcal{A}\left(\mathcal{T}^k V\right)\right)^*$. This way $ \wedge^k V$ is of a similar nature as $ \otimes^k V$. Or do you define $ \otimes^1 V$ to be $ V^*$ as well?

  darij

I haven't done much with this, but notationally isn't $ \otimes^k(V)\equal{}\mathcal(T)^k$? In that case, yes, $ \otimes^1(V)\equal{}V^*$ since it consists of all linear functionals $ \phi : V\to\mathbb{R}$.

I've only seen this in differential geometry out of Boothby's book. Maybe it is more standard in other branches of math to do it the other way?

I have never read analysis books on this matter, but the notations in algebra books definitely imply $ \otimes^1 V \equal{} V$.

Anyway, in this particular problem it is irrelevant since we can take any arbitrary symmetric bilinear form on $ E$ and use it to identify $ E$ with $ E^*$.

  darij

Do it as if it is college algebra, the only thing you need for the problem is force.  First write down some basis $ \{e_i\}$.  All involving terms would be of the form $ f_ie_1\wedge e_2\wedge\ldots\wedge\hat{e_i}\wedge\ldots\wedge e_n$.

To combine 2 terms are fairly easy, e.g.   $ f_1e_1\wedge e_3\wedge\ldots\wedge e_n + f_2e_2\wedge e_3\wedge\ldots\wedge e_n = (f_1e_1 + f_2e_2)\wedge e_3\wedge\ldots\wedge e_n$.

To combine more, e.g., $ (f_1e_1 + f_2e_2)\wedge e_3\wedge\ldots\wedge e_n + f_3e_1\wedge e_2\wedge e_4\ldots\wedge e_n$, you noticed that $ f_3e_1\wedge e_2\wedge e_4\ldots\wedge e_n=\frac{f_3}{f_1}(f_1e_1 + f_2e_2)\wedge e_2\wedge e_4\wedge\ldots\wedge e_n$ and thus the above term is combined into $ (f_1e_1+f_2e_2)\wedge(\frac{f_3}{f_1}e_2+e_3)\wedge e_4\wedge\ldots\wedge e_n (*)$.

The rest is the same story.  Inductively, if you look at $ (*)$, you easily see as you have combined $ k$ terms to get $ b_1\wedge\ldots\wedge b_{k-1}\wedge e_{k+1}\ldots\wedge e_n$ and want to proceed one more step, then for $ i\leq k-1$, the ${ b_i}$ term is a linear combination of $ e_i, e_{i+1}$.  Because of this, note that we are able to let the next term $ f_{k+1}e_1\ldots\hat{e_{k+1}}\wedge\ldots e_n=\tilde{f_{k+1}}b_1\wedge\ldots\wedge b_{k-1}\wedge e_{k}\wedge e_{k+2}\ldots\wedge e_n$ so the induction goes on.  When you and anywhere you have $ f_i=0$ you just don't need to do anything, so the factions won't cause any problem.


analytic geometry
quadratics
geometry
rhombus
trigonometry
algebra
quadratic formula

Given equilateral triangle $ABC$, where $A=(0,0)$ and $B=(3\sqrt{2},3\sqrt{2})$, there are two possible locations in the $xy$=plane for the point C. Find the sum of all possible $y$-coordinates of $C$.

[hide]The distance from $A$ to $B$ is 6.

$\sqrt{\left(y-0\right)^{2}+\left(x-0\right)^{2}}=6$ and $\sqrt{\left(y-3\sqrt{2}\right)^{2}+\left(x-3\sqrt{2}\right)^{2}}=6$

The first equation can be simplified to $x^{2}+y^{2}=36.$

The second can be simplified to $x^{2}-6x\sqrt{2}+y^{2}-6y\sqrt{2}+36=36,$ or $x^{2}+y^{2}=6x\sqrt{2}+6y\sqrt{2}.$

Now we know that $6x\sqrt{2}+6y\sqrt{2}=36,$ or $x\sqrt{2}+y\sqrt{2}=6.$

Isolating $x$ gives $x=\frac{6-y\sqrt{2}}{\sqrt{2}},$ or $x=3\sqrt{2}-y.$

Substitute this back into $x^{2}+y^{2}=36$ to get $\left(3\sqrt{2}-y\right)^{2}+y^{2}=36,$ or $y^{2}-6y\sqrt{2}+18+y^{2}=36.$

This can be further simplified to $2y^{2}-6y\sqrt{2}-18=0.$

Using the quadratic formula, $y=\frac{6\sqrt{2}\pm\sqrt{72+72}}{2},$ or $y=3\sqrt{2}\pm6.$

$3\sqrt{2}+6+3\sqrt{2}-6=\boxed{6\sqrt{2}}$[/hide]

I think you made a mistake expanding $(x-3\sqrt{2})^{2}$.  Here's my solution:

[hide]The two possible locations for $C$ (let them be $C_{1},C_{2}$) are on opposite sides of $AB$.  Since $AC_{1}BC_{2}$ is a rhombus, the midpoint of $C_{1}C_{2}$ is also the midpoint of $AB$.  The midpoint of $AB$ has $y$-coordinate $\frac{3\sqrt{2}}{2}$.  Then $\frac{3\sqrt{2}}{2}$ is the average of the $y$-coordinates of the possible locations for $C$, so the sum is $3\sqrt{2}$.[/hide]

[hide]
Here's another method, using complex numbers. If you actually wanted to find the points, you could set $C = 3e^{i\frac{\pi}{4}}$ from which we get the two points as

$Ce^{i\frac{\pi}{6}}$ and $Ce^{-i\frac{\pi}{6}}$

and evaluate the ugly trig functions.[/hide]

[quote="matt276eagles"]I think you made a mistake expanding $(x-3\sqrt{2})^{2}$.[/quote]
I did, but my answer is still different from yours. :huh:


Aratati ca  $A=2^{x}+3^{x}+4^{x}+15^{x}-5^{x}-9^{x}-10^{x}$ este pozitiv pentru orice x real

Probabil ca ar treb si justificat de ce expresiile din paranteze sunt pozitive... :maybe: E o chestie de rutina bine inteles (x<1;x>1;x=1) dar vroiam sa precizez :P


Hello all! Today, i'll introduce an old problem of sequence, but i'm finding some full solutions. I think it's hard for a lot of mathlinkers! :P

Let a, b are 2 real numbers such that $ 0<a,b<4$. We conside the sequence $ a_n$: $ a_0\equal{}a; a_1\equal{}b; a_{n\plus{}2}\equal{}\frac {2(a_{n\plus{}1}\plus{}a_{n})}{\sqrt{a_{n\plus{}1}}\plus{}\sqrt{a_{n\plus{}1}}}$.
Show that this sequence converges, find $ lim a_n$

Please lock this topic. It's running problem of MYM


symmetry
number theory
number theory unsolved

Hi all,

I have unfortunately very little background in number theory, and was wondering if anyone could point me in the right direction regarding this question:

Is there any sort of general underlying rule as to when a given sum of positive integers is prime?

Clearly, if they all share a common factor, the sum cannot be prime, but it is also easy to see that even if they don't share a common factor, then the sum can still be composite. Is there a more complex theory underlying such a question, perhaps for certain special cases, that I can read up on?

Thanks for your help
Sev

As stated, your question is far too general.  Of course there are many necessary conditions for a sum of positive integers to be prime (there must be an odd number of odd ones, and similar conditions modulo other small primes $ p$), but these same conditions apply to a single positive integer.  In other words, this problem is exactly as hard as summing the positive integers and determining whether the resulting sum is prime.

Is this question motivated by a competition problem?  If so, there may be extra symmetry in the sum one can take advantage of.  For example, the sum $ 1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} n$ is clearly composite because it is always divisible by (a factor of) $ n \plus{} 1$ by Gaussian pairing.


Let  $S(n)$ be the sum of digits of $n$. Prove that $S(2n^2+3)$ is never a perfect square.

We know that for every integer number $S(n)\equiv n (mod 9)$ ,and we know $n^2\equiv 0,1,4,7(mod 9)$ but $2n^2+3\equiv 2,3,5,8(mod 9)$. ;)


modular arithmetic
number theory
prime numbers
combinatorics proposed
combinatorics

Let $ p_1,\dots,p_k$ be prime numbers, and let $ S$ be the set of those integers whose all prime divisors are among $ p_1,\dots,p_k$. For a finite subset $ A$ of the integers let us denote by $ \mathcal G(A)$ the graph whose vertices are the elements of $ A$, and the edges are those pairs $ a,b\in A$ for which $ a \minus{} b\in S$. Does there exist for all $ m\geq 3$ an $ m$-element subset $ A$ of the integers such that
(i) $ \mathcal G(A)$ is complete?
(ii) $ \mathcal G(A)$ is connected, but all vertices have degree at most 2?

For part (i), the answer is no. The reason for this is, if we consider a prime $ p$ which is greater then every $ p_k$, then $ a_i \not \equiv a_j \pmod{p},\forall 1\leq i < j \leq m$. Thus, $ |\{a_i - a_j|1\leq i < j \leq m\}| < \infty$, yieding the statement.

For part (ii), we will construct $ A$. Consider a prime number $ p$ different from $ p_1,p_2,...,p_k$. Put $ h = \text{ord}_p (p_1)$ then $ p_1^{\alpha + th}\equiv p_1^{\alpha} \pmod{p},\forall t\in\mathbb{N} \& ^*$. Now randomly take a positive integer $ \alpha_1$ and put $ \alpha_t = \alpha_1 + (t - 1)h$. $ A = \{a_0 = n,a_i = n + p_1^{\alpha_i}p_2p_3...p_k|i = 2,3,...,m - 1\}$. It is clearly that $ a_i - a_0 = p_1^{\alpha_i}p_2...p_k\in S$. This yields that $ \mathcal{G}(A)$ is connected (since every vertex of its is connected to $ a_0$). But for $ i > j > 1$ then $ a_i - a_j = p_2p_3...p_kp_1^{\alpha_j}(p_1^{(i - j)h} - 1)$, which is a multiple of $ p$, so $ a_i - a_j \not\in S$ so $ \text{deg}(a_i) = 1,\forall i = 1,2,...,m - 1$. Hence, $ A$ satisfies the condition.

[b]Mashimaru:[/b] But then $\deg(a_0)>2$, so $A$ does not work.

However, if we let $P=p_1p_2\cdots p_k$, then taking $a_i=1+P+\cdots+P^{i-1}=\frac{P^i-1}{P-1}$ for all $i\ge1$,
\[P^i\left(\frac{P^j-1}{P-1}\right)=a_{i+j}-a_i\in S\Longleftrightarrow j=1\](if $j>0$), so $A=\{a_i\}_{i\ge1}$ satisfies (ii).


linear algebra
matrix
algebra
polynomial
function
parameterization
quadratics

How many eigenvalues to the matrix
$\left[ \begin {array}{cccc} 4&1&0&4\\\noalign{\medskip}1&5&4&0 \\\noalign{\medskip}0&4&5&0\\\noalign{\medskip}4&0&0&5\end {array} \right]$
are larger than 3?

I used some elemntary transformation and diagonalized the matrix to having $(-11,-1,1,1)$ on the diagonal. Are these the eigenvalues of the matrix?

No, although that is interesting. How did you diagonalize it?

Adding and substracting columns and rows.... in the worst case scenario just write Hamilton-Cayley`s relation and solve the equation :roll:

So what you did was essentially to show that the matrix is invertible? That doesn't help much...  The characteristic polynomial can't be solved, that's the whole point of the question.

I`ve given more thought to the problem:
1)notice that the matrix is symmetric that means that it has only real eigenvalues
2)this is it were it get`s messy write down the equation you get from Hamilton-Cayley, a 4th degree  equation, use some derivates to find out in which intervals it annulates.....

Note that what spix said about the eigenvalues in his first post has to be wrong. The sum of the eigenvalues is the trace of the matrix, which in this case is $19.$

Look at $A-3I.$ This matrix has a negative determinant.

-----

Oops, I'd better correct that, the product of a bad calculation. $A-3I$ has a positive determinant. That means that the number of eigenvalues greater then $3$ is an even number. Since the trace of $A-3I$ is still positive, that means it the eigenvalues can't all be less than $3.$ Thus the number of eigenvalues greater than $3$ must be two or four.

But if all four eigenvalues were greater than $3$, we would have $\det A>81.$ Since $\det A=11,$ we can't have this. So we conclude that $A$ has exactly two eigenvalues greater than $3.$

[Edited after I saw jmerry's post below.]

Why would an odd number number of eigenvalues be less than 3? It has to be wrong, since there are 2. I still have no idea how to do this problem.

spix: The problem is not meant to be done like that.

I know that my original post is totally incorrect and I apologise for that. Anyway Kalle could you please point out the way in which the problem should be solved? For me the best approach remains studying the function given by the caracteristic polynomial  knowing that it has real roots  :P

It's two.

We look at the matrix $A-3I$, and compare it to the matrix $B=\begin{bmatrix}0&0&0&4\\0&0&4&0\\0&4&0&0\\4&0&0&0\end{bmatrix}$. $B$ has two eigenvalues equal to $4$, and two equal to $-4$. Clearly, $\|A-3I-B\|\le 3$ in any matrix norm, so the eigenvalues of $A-3I$ differ from corresponding eigenvalues of $B$ by no more than 3. That gives two negative eigenvalues (between $-7$ and $-1$) and two positive eigenvalues (between $1$ and $7$).

Think of this as a generalization of the "strictly diagonally dominant" idea- the signature is the same as that of the matrix with just the big elements.

Unfortunately I don't know what a matrix norm is, so I don't understand your solution. Is there a way to understand the the eigenvalues differ at most by 3 using only very basic theory?

It's a variant on the Gershgorin circle theorem- see this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=32062]thread[/url] for a statement of the original version. In addition, grobber's post there can be adapted to our situation- as long as we keep the matrix symmetric, the eigenvalues stay real.

It's not exactly "very elementary", but it's a very nice theorem.

[quote="Kent Merryfield"]$A-3I$ has a positive determinant. That means that the number of eigenvalues greater then $3$ is an even number. [/quote]
How  does this conclusion work? Is it because you know that the characteristic polynomial is a continuous function and that it tends to infinity as the parameter grows large? What if the polynomial has a zero without becoming negative?

Remember that in this case $A$ (and thus $A-3I$ as well) is a real symmetric matrix. Thus it has (counting multiplicities) four real eigenvalues.

If $\lambda_j,\,1\le j\le 4$ are the eigenvalues of $A,$ then the eigenvalues of $A-3I$ are $\lambda_j-3.$

The determinant of $A-3I$ is the product of its eigenvalues. That product is positive iff an even number of the factors is negative. (Which, since there are four, is the same as an even number being positive.)

You do have to count multiplicities. Hence when you said, "What if the polynomial has a zero without becoming negative?" we'll have to count that as two eigenvalues.

Tomorrow I have a lecture in linear algebra that among other things covers things about "positive definite" quadratic forms. As I understand it, by considering $A - 3I$ as the matrix for a quadratic form you can complete the square to find out how many positive eigenvalues it has. But if there is such an easy solution, why hasn't anyone suggested it? I'll guess I see after the lecture; I've only skimmed through the text quickly.

Yes, by looking at the quadratic form for $A - 3E$ and completing the square you find that two of the eigenvalues will be positive. Furthermore, since the two positive coefficients in the diagonalization you get by completing the square are different from each other, you know that the original matrix has two eigenvalues larger than three, not counting multiplicities.


Find more than two consecutive integers that have a sum of 11.

[hide="click to see stuff"]
5+6=11
and
this one took me a while but:
-4+-3+-2+-1+0+1+2+3+4+5+6=11
[/hide]
Use the hide thing people.


calculus
integration
function
trigonometry
limit
real analysis
Support

Suppose $f$ is a real function with complex values which is piecewise continuous and piecewise of class $C^{1}$, such that $\int_{R}{ |f(t)|dt}$ converges.  Prove that
    $\frac{1}{2\cdot\pi}\cdot\int_{R}{ g(u)e^{ixu}du}=\frac{f(x^{+})+f(x^{-})}{2}$ where $ f(x^{+})$ is the limit at right in $x$ and $ g(x)=\int_{R}{ f(t)e^{-itx}dt}$.

If $f$ has jump discontinuities, which you are allowing, then $g(u)$ will behave like $\frac{C}u$ for large $u.$ Hence we cannot assume that $g$ is absolutely integrable and must turn to computing the integral $\int_{\mathbb{R}}g(u)e^{iux}du$ as an improper integral. In fact, what I prove will concern the Cauchy prinicpal value of that integral, which is an even more restrictive intepretation.

Let $S_M(x)=\frac1{2\pi}\int_{-M}^Mg(u)e^{iux}du$

$=\frac1{2\pi}\int_{-M}^M(\int_{-\infty}^{\infty}f(y)e^{-iuy}dy)e^{iux}du$

$=\frac1{2\pi}\int_{-\infty}^{\infty}f(y)(\int_{-M}^Me^{iu(x-y)}du)dy$

$=\frac1{\pi}\int_{-\infty}^{\infty}f(y)\frac{\sin M(x-y)}{x-y}dy= \int_{-\infty}^{\infty}f(x-y)\frac{\sin My}{\pi y}dy$

Let $H(x)$ be the function such that $H(0)=0$ and $H'(x)=\frac{\sin x}{\pi x}.$ $H(x)$ is odd, smooth (given by a power series with an infinite radius of convergence), bounded, and we know that $\lim_{x\to\infty}H(x)=\frac12$ and $\lim_{x\to -\infty}H(x)=-\frac12.$ Note that $\frac{d}{dx}H(Mx)=\frac{\sin Mx}{\pi x}.$

We split our integral into two pieces and integrate by parts, differentiating $f$. Since $f$ may have have finitely many jump discontinuities, this must actually be performed as a Stieltjes integral, and the terms which I will write as integrals may also include a finite number of point evaluations. This will not change the substance of the computation.

$\int_0^{\infty}f(x-y)\frac{\sin My}{\pi y}dy= f(x-y)(H(My)- \frac12)|_0^{\infty}+\int_0^{\infty}f'(x-y)(H(My)-\frac12)\,dy$

Using the fact that $f$ must be bounded and that $\lim_{y\to 0^+}(H(My)- \frac12)=-\frac12$ this becomes

$\frac{f(x^-)}2+\int_0^{\infty}f'(x-y)(H(My)-\frac12)\,dy$

Similarly,

$\int_{-\infty}^0f(x-y)\frac{\sin My}{\pi y}dy= f(x-y)(H(My)+ \frac12)|_{-\infty}^0+\int_{-\infty}^0f'(x-y)(H(My)+\frac12)\,dy$

$=\frac{f(x^+)}2+\int_{-\infty}^0f'(x-y)(H(My)+\frac12)\,dy$

So $S_M(x)=\frac{f(x^-)+f(x^+)}2+\int_{-\infty}^{\infty}f'(x-y)(H(My)- \frac12\text{sgn}(y))\,dy$

That last integral may also need to include a finite number of point evaluations. I would like to claim that $|f'|$ is absolutely integrable. We have what is just as good: that $df(x)$ is a finite measure.

The function $(H(My)- \frac12\text{sgn}(y))$ is uniformly (for all $M$) bounded and tends pointwise to zero as $M\to\infty$. Therefore, by the Dominated Convergence Theorem,

$\lim_{M\to\infty}\int_{-\infty}^{\infty}f'(x-y)(H(My)- \frac12\text{sgn}(y))\,dy =0.$

Hence, $\lim_{M\to\infty}S_M(x)=\frac{f(x^-)+f(x^+)}2.$

I have to admit that there are some functions that ruin my argument. For instance, let $f(x)=\frac1{x^2}\sin(x^3)$ for $x>1.$ Then I can't claim that $f'$ is absolutely integrable and I need another argument besides the DCT for interchanging limit and integral in the last step.

[quote="Kent Merryfield"]
Let $S_M(x)=\frac1{2\pi}\int_{-M}^Mg(u)e^{iux}du$

$=\frac1{\pi}\int_{-\infty}^{\infty}f(y)\frac{\sin M(x-y)}{x-y}dy= \int_{-\infty}^{\infty}f(x-y)\frac{\sin My}{\pi y}dy$
[/quote]
This may be continued as
$=\int_{0}^{\infty}(f(x+y)+f(x-y))\frac{\sin My}{\pi y}dy$
Assume now that $f(x+)+f(x-)=0$. Then the function $\frac{f(x+y)+f(x-y)}{\pi y}$ is in $L^1(0,+\infty)$ as a function of $y$ (this easily follows from $f$ being simultaneously piecewise $C^1$ and integrable). Then the Riemann-Lebesgue lemma allows to conclude that $S_M(x)\to 0$ as $M\to\infty$. To reduce the general case to this particular one, it suffices to subtract from $f$ some nice function for which convergence is easy to demonstrate (for instance, an infinitely smooth function with compact support or, in view of Kent's argument, any function such that $f,f'\in L^1$). What I did here is the classical proof of [b]Dini's criterion[/b] for convergence of  Fourier series or integrals. The criterion, as you may guess, is the following: if $f\in L^1$ and $\frac {f(x+y)+f(x-y)-2A}y$ is integrable near $0$ with respect to $y$, then $S_M(x)\to A$.


If bobby uses only hammer, he can build 6 houses in 7 years. If he uses only a saw, he can build 9 houses in 10 years. If he uses a hammer and a saw at the same time, how long will it take him, to the nearest year, to build 100 houses?

With a hammer, he builds $ \frac{6}{7}$ houses a year. With a saw, he builds $ \frac{9}{10}$ houses per year.
With both, he builds $ \frac{6}{7}\plus{}\frac{9}{10}\equal{}\frac{60}{70}\plus{}\frac{63}{70}\equal{}\frac{123}{70}$ houses per year.

$ \frac{100}{\frac{123}{70}}\approx57$ years.

P.S. Isn't it kinda hard to use both at the same time, though?

Bah no the answer is $ \infty$ He hammers one hand into nothingness and saws off the other. Oops...


A mass is connected to one end of a massless string,the other end of which is connected to a very thin frictionless vertical pole.The string is initially wound completely around the pole(in a very large number of little horizontal circles),with the mass touching the pole.The mass is released,and the string gradually unwinds.What angle does the string make with the pole at the moment it becomes completely unwound?

I also couldn't understand the problem.That's why I wrote it here. :(


calculus
integration
trigonometry
calculus computations

It doesn't converge.. $ \tan$ is undefined in an infinitive number of points in $ [0,\infty]$.

That isn't true, it converges to $ \frac{\pi}{2}$

I have basically no background in residue theory as I have not yet picked up a complex methods book but i know some basics so here it goes...

To integrate \[ \int_{\minus{}\infty}^{\infty} \frac{\sin{x}}{x} dx\] we find the poles as follows, \[ P \int_{\minus{}\infty}^{\infty} \frac{e^{\imath x}}{x} dx \equal{} \imath \pi \sum_{y\equal{}0} \mathbf{Res}(\frac{e^{\imath z}}{z}) \equal{} \pi \imath  \] Equating real and imaginary parts we get
\[ \int_{\minus{}\infty}^{\infty} \frac{\cos{x}}{x} dx \equal{} 0\] and
\[ \int_{\minus{}\infty}^{\infty} \frac{\sin{x}}{x} dx \equal{} \pi \].

...and i can't get past this unfortunately. i don't know if i'm on the right track or not.

While both of those are correct results, they don't have anything to do with the original problem - they don't help. And milin is right: the integral diverges.

Or if you're going to try to extract a number from it through some limiting process (related to, but more complicated than, a principal value), then you'd better explain precisely what that limiting process is; the meaning is not plain from the original notation.


integration
calculus
real analysis
real analysis unsolved

I thought this question was routine on first glance...but maybe im not approaching it right.

$\int{\int{e^{-y^{2}}dx}dy}$.
over triangular region bounded by (0,0), (1,0) and (0,1).

is there any solution the wouldnt require using a taylor expansion or the change of variable theorem?

Thanks in advance

Maybe that's not the right triangle? If the vertices were $(0,0),(1,1),(0,1)$, this would be routine. As it is, only numerical methods can work.

how would i go about determining what
${\sum_{n=0}^{\infty}{\frac{(-1)^{n}}{{n!(2n+1)}}}}$.

converges to exactly?

You wouldn't. The only thing you can do is estimate it numerically, and the partial sums of that series work very well for that purpose.

ah ok. i assumed there was some method for reducing series to some simpler (non summation or integral apprximation) expression.


function
number theory solved
number theory

Is it known that p_n ~ n.lnn
It is known a better equivalent, with mor terms ?

10x

It is known [1] that :
n(ln(n) + ln(ln(n)) - 3/2) < p_n < n(ln(n) + ln(ln(n)) - 1/2), for n  \geq  21.

More generally [2], for every positive number A, there exists a constant N such that, for n  \geq  N :
nln(n) + nln(ln(n)) - n - An < p_n < nln(n) + nln(ln(n)) - n + An.

According to Ribenboim [3], we have :
p_n = nln(n) + nln(ln(n)) - n + o( nln(ln(n))/ln(n) ).

References :
[1] J.B.Rosser, L.Schoenfeld, 'Approximate formulaes for some functions of prime numbers', Illinois J.Math., 6 (1962), p.64-94.
[2] J.B.Rosser, 'Explicit bounds for some functions of prime numbers', Amer. J.Math, 63 (1941), p. 211-232.
[3] P.Ribenboim, Nombres premiers : mystères et records, P.U.F., p.167.

Pierre.


find all of the fuction $f: R...R$ such that
$f(\frac{1+x}{1-x}\ ) = af(x)$
a is constant real number

Denote $g(x) = \frac{1+x}{1-x}$.  Then

$\frac{1+x}{1-x} = y$
$1 + x = y - yx$
$(1+y)x = y-1$
$x = \frac{y-1}{y+1}$

So $g^{-1}(x) = - \frac{1-x}{1+x} = - \frac{1}{g(x)}$.  From this we derive $g(g(x)) = - \frac{1}{x}$ and $g^{(4)} (x) = x$.

Rewrite as $f(g(x)) = a f(x)$.  The above expression gives $f(g^{(4)} (x)) = f(x) = a^4 f(x)$ and hence $a = 1, -1$ or $f(x) = 0$.

And, err, I'm stuck  :|   Can't seem to get much more out of that.


calculus
articles
MATHCOUNTS
AMC
AMC 8
AIME
ratio

i'll win nationals, obviously  :D  :D  :D

[hide="For all of you who do not understand sarcasm"]i'm kidding[/hide]

The first place person from CA will dominate you all (not me). just kidding, but I think she has a pretty good chance.

I dunno, I've heard rumors about this kid from Ohio who nobody knows about that finished second at states but is a top contender for the winner at nationals. His name is Brian something?... :lol:

Seriously I have no idea as I've never seen any of these top people.

Sam Keller! He is gonna own you all!!! Whoops, mistake.  :blush:  :blush:  :P [/u]

Have no doubt I do that he is own we all. To max limit.

Wow. I sound like Yoda.

[quote="Press alt f4"]Mark Zhang![/quote]

Drew, Mark Zhang started this post...and he's mocking me (James Zhou)

[quote="tarquin"]Jeffrey Chan.. HE'S AMAZING![/quote]

i thought he was a freshman now???

[quote="solafidefarms"];)  ;) Me! :lol:  :lol:  :P  :P

Billy[/quote]

I second that

[quote="09husbri"]Yes - the MathCounts people know what they are doing, and a fair system means as much to us as it does to them. If there were a fairer system that is expedient, efficient, and watchable as the current one is, they would find it and use it. Granted, the current system isn't perfect, but it works okay. I think this goes back to people refusing to accept Neal Wu as national champion. Get over it! ;)[/quote]

I'm willing to accept that, but the unfair part is 2nd place gets $2000 more than 3rd place, yet they never faced each other, same for 4th place... now tell me that's fair

No one can, Bob.  This forum is for writing, not speaking. :roll:  :D  :P

after the top 4 they should have a tournament where everyone matches against everyone else once the person with most wins wins and for tiebreaker between 2 people they find who won between the two who are tied

Actually, that is the best alteration idea I've seen yet - a round robin with the top four competing for different levels of prize money!

I think what they should at least do is have some sort of seperate, actual question tiebreaker for the people who tied to make the countdown. As someone who got tiebreakered out I would like to see that rule in place.

And I could care less that tiebreakered isn't a word. . . it is now.

To some extent, I feel your pain; I got tiebreakered out of 57th place (I got 58th, there was a tie at 24 points from 57th to 65th). I wanted to make that list so badly!

[quote="sirorange"]I know that. You definetly haven't said that before :roll:. But I complain about getting 14th too much so. . .   But still, yeah that does stink, but making the list and making the countdown are two seperate things, and I'm not sure they should have one to make the list. But in all actuality, you did make the top 57.[/quote]

I bet I complain about getting 41st more... Though I should be happy with my rank... I didn't even expect to make top 25% with my awful awful WORST score ever of 3 on target... I usually get 6-8 on practices... my brain completely froze on competition day... I could've easily made a 7 on that target and a 25 on that sprint any other day... I guess I'm glad that I made top 25% but I'm still disappointed that my score was so low...

Yeah. See, I was sick that day, and the water spilling, I bet you've heard that story before about me. Anyways, I could have easily gotten a 25 on the sprint, and I got my worst target score ever. (I had gotten perfects on like all of the practices except two.) But there are 214 people who would have loved to have been where I was, so I have to be semi-happy.

[quote="bob123"]I bet I complain about getting 41st more... Though I should be happy with my rank... I didn't even expect to make top 25% with my awful awful WORST score ever of 3 on target... I usually get 6-8 on practices... my brain completely froze on competition day... I could've easily made a 7 on that target and a 25 on that sprint any other day... I guess I'm glad that I made top 25% but I'm still disappointed that my score was so low...[/quote]

It's worse for me. I got 51st with the same score. You were on the top of the tiebreaker and I was on the bottom.  :cursing:

Exact same thing for me last year. While I couldn't have come close to Greg's score, I could have gotten as high as a 38 without stupid mistakes, which, with a tiebreaker, could have put me in countdown. :dry:

When I took last year's in practice for the first time, I got a 45. Which is kinda annoying, because those were all my type of questions, and this year, was definetly not.

I found that there were a lot of problems that I couldn't solve in the time but almost immediately understood afterwards. I blame this on the pressure.


logarithms
limit
real analysis
real analysis unsolved

$ a_k \ge 0, b_k \equal{} \minus{}\frac{\ln a_k}{\ln k}$

$ \lim \inf b_k > 1 \implies \sum_{k\equal{}1}^\infty a_k < \infty$

but what about converse?

If $ a_k\equal{}\frac1{k\ln^2k}$ (at least for $ k\ge 2$) then $ \lim b_k\equal{}\lim\frac{\ln k\plus{}2\ln\ln k}{\ln k}\equal{}1$ but the series converges.

is there a closed form for this $ \sum a_k$ ?

or the generalized $ \sum_{k\equal{}1}^\infty \frac{1}{k\ln^p k}$

There's no good reason to expect there to be a closed form. Such series often make good examples to use when the issue is convergence versus divergence (of positive term series).


[quote="E^(pi*i)=-1"]Let $ ABC$ be a triangle, and let $ H$ and $ \omega$ be its orthocenter and circumcircle respectively.  Let $ BD$ be a diameter of $ \omega$.  Prove that $ AHCD$ is a parallelogram.[/quote]
[hide="Solution"]
Since $ H$ is the orthocenter of $ \triangle ABC$, we have that $ AH\perp BC$ and $ CH\perp AB$.  Also, since $ BD$ is a diamter of the circle, $ DC\perp BC$ and $ AD\perp AB$.  Hence, $ AD\parallel CH$ and $ CD\parallel AH$, which means that $ AHCD$ is a parallelogram. [/hide]

[b]Another proof[/b]
Let $ O$ be the center of the circle $ \omega$.Trace $ OA_1 \bot BC$ , $ A_1 \in (BC)$.
$ \{\begin{array}{cccc} OA_1=\frac{DC}{2} \\
\\
OA_1=\frac{AH}{2} \end{array}$ $ \implies$ $ \{\begin{array}{cccc} DC=AH\\
\\
AH \parallel{} DC \end{array}$ $ \implies$ $ AHCD$ paralelogram

Dear Mathlinkers,
we use only the fact that AH is 2 time the distance from O to BC (Carnot relation).

Sincerely
Jean-Louis


Find all positive integers which can be written as $ \frac{a\plus{}b}{c}\plus{}\frac{b\plus{}c}{a}\plus{}\frac{c\plus{}a}{b}$ for some positive integers $ a,b,c$, where $ GCD(a,b)\equal{}GCD(b,c)\equal{}GCD(c,a)\equal{}1$.

it was posted here :http://www.mathlinks.ro/viewtopic.php?t=220438


geometry
rectangle
analytic geometry
perimeter

A rectangular corral is to be built using 160 feet of fence. One side of the corral will be part of a straight 100-foot wall of an adjacent building. What is the maximum number of square feet possible for the area of the corral?

1600sqft. if it is correct, i will add a sol.  :)

I don't know if you're correct... just post the solution.

I'd say that he's not correct. Here's my solution:

[hide]
what we are given is that we have to make a rectangular area on the side of the building, but we don't have to fence along the wall. So, if we draw a diagram with a line on the bottom and a rectangle on top of that, lable the side of the rectangle across from the line (wall) x, and the other two sides y. We have $x+y+y = 160$, or $x + 2y = 160$. What we want to find the maximum value to is $xy$. We can do this in a couple ways. Linear programming (hate it, takes forever, and i'm not going to explain it), and using basic algebra.

Isolate x in the first equation:

$x = 160 - 2y$

Then substitute that into the area formula:

$A = (160 - 2y)(y)$
$A = -2y^2 + 160y$

now, if we where to graph this, the maximum value would be the vertex. The x-coordinate of the vertex (in this case the value of y) is $\frac{-b}{2a} = \frac{-160}{-4} = 40$

So, $y = 40$ and $x = 80$, making the area $\boxed{3200 \text{ ft}^2}$
[/hide]

do we have to use the entire wall as a side? I don't think your dimensions are a rectangle anymore.

[quote="236factorial"]A rectangular corral is to be built using 160 feet of fence. One side of the corral will be [b]part[/b] of a straight 100-foot wall of an adjacent building. What is the maximum number of square feet possible for the area of the corral?[/quote]

no we don't have to use the [b]entire[/b] wall as a side. And as for my dimensions, it's an 80x40 rectangle. What's wrong with that?

[quote="236factorial"]okay, then, why can't it just be a square?  :?[/quote]

Because 160 is not divisible by 3  :P

If it were a square, the side would be $\frac{160}{3}$

Squaring that gives $\frac{25600}9$ or $2844\frac 49$ which is less than 3200 :P

wait a second guys. One of the sides are given as 100 ft. so it is 100*30=3000.  :P

actually, one of the sides is given as PART of 100 feet. Therefore, my solution still stands as the maximum at 3200 :D

i agree with hamster 1800. 80x40 is the largest.  :blush:

Sorry, I was just a little stupid  :blush:  (didn't realize perimeter differences for wall).


Let your first number you choose be 1 and create a set of numbers  w where {w} is an element of N. Using the fact that 2n+1 is always odd, we can conclude that no matter what # we choose, X is = or grt than Y.

Sorry, but I don't quite understand what you're trying to say...

He didn't actually say anything.  His response looks like a joke.

I don't this is problem is right at all. If $a_{1}=3$ and $a_{2}=3$, then $X=4$ and $Y=5$.  Also, if we let $a_{3}=3$, then $X=14$ and $Y=13$.  So is there something else we need to know?

[quote="barasawala"]Let $a_{1},...,a_{k}$ be natural numbers. We are to choose integers $n_{1},n_{2},...,n_{k}$ such that $1\leq n_{i}\leq a_{i}$ for all $i=1,2,...,k$. Let $X$ be the number of ways to choose such that $n_{1}+...+n_{k}$ is odd. Let $Y$ be the number of ways to choose such that $n_{1}+...+n_{k}$ is even. Show that $X\geq Y$[/quote]

I'm sorry. I'm quite sure this one is correct now. (Note that there are two corrections: the $0$ and the $\leq$)

Let $a_{1},...,a_{k}$ be natural numbers. We are to choose integers $n_{1},n_{2},...,n_{k}$ such that $0\leq n_{i}\leq a_{i}$ for all $i=1,2,...,k$. Let $X$ be the number of ways to choose such that $n_{1}+...+n_{k}$ is odd. Let $Y$ be the number of ways to choose such that $n_{1}+...+n_{k}$ is even. Show that $X\leq Y$

[quote="zgorkster"]Proceed by induction on $k$.  If $n_{k}$ is odd, then $X=Y$, and if $n_{k}$ is even, you can use the inductive step to show that $X$ is still at most $Y$.[/quote]

Yeah...  :)


algebra
polynomial
function
number theory unsolved
number theory

Let f(x(1), ..... ,x(n)) be a polynominal with integer coefficients of degree less then. Prove that if N is the number of n tuples (x(1), .... ,x(n))
with 0<=x(i)<13 and f(x(1), ..... ,x(n))=0 (mod 13), then N is divisible by 13.
i can't understand anything can someone explain me

It is not true. Let n=1 $f(x)=x^{12}-1$. Number of solutions is 12 (not divisible to 13).

what is the meaning of f(x(1),........,x(n)) polynominal

All function $Z_{p}\to Z_{p}$ can be given as polinomial.

I can't understand. what is the meaning of the notation f(x(1),......,x(n))


modular arithmetic
function
number theory
relatively prime

Prove that
\[\binom{2p}{p}\equiv 2 \pmod{p^{3}}\]
(p>3)
A combinatorial proof would be especially impressive.

[hide="A start..."]
\[\dbinom{2p}{p}= 2^{p}\cdot (2p-1)(2p-3)\ldots (p+2)(p-2)(p-4) \ldots (3)(1) \cdot \frac{1}{(p-1)!}\] \[\equiv 2^{p}\cdot (p^{2}-4)(p^{2}-16)\ldots (p^{2}-(p-1)^{2}) (-1)^{-1}\equiv 2^{p}\cdot \left( \prod_{k=1}^{\frac{p-1}{2}}(p^{2}-4k^{2}) \right) \cdot (-1) \]
Now, since everything with a power of $p$ $>$ 2 will be divisibile by $p^{3}$, we have that the expression is \[\equiv-2^{p}\cdot \left( (-1)^{\frac{p-1}{2}}\cdot \prod_{k=1}^{\frac{p-1}{2}}(4k^{2})+(-1)^{\frac{p-3}{2}}\sum_{k=1}^{\frac{p-1}{2}}(4k^{2})\right)\] which is just nasty. I'm hoping there's a better way to go about this...  :P[/hide]

[b]Solution.[/b] We are to prove that $p^{3}|\binom{2p}{p}-2$.  Notice that
\[S=\binom{2p}{p}-2=\sum_{k=1}^{p-1}\binom{p}{k}^{2}=p^{2}\sum_{k=1}^{p-1}k^{-2}\binom{p-1}{k-1}^{2}.\]
Since $\gcd(p,a)=1\forall a\in[1,p-1]\cap\mathbb{Z},$ all congruency classes modulo $p$ have a unique inverse,i.e. $\mathbb{Z}/p\mathbb{Z}$ is a field.  Hence,
\[S\equiv p^{2}\sum_{k=1}^{p-1}k^{-2}\left(\prod_{j=1}^{k-1}j^{-1}(p-j)\right)^{2}\equiv p^{2}\sum_{k=1}^{p-1}k^{-2}\equiv p^{2}\sum_{k=1}^{p-1}k^{2}\equiv 0\pmod{p^{3}},\]
as required. $\Box$

[quote="boxedexe"]
Since $\gcd(p,a)=1\forall a\in[1,p-1]\cap\mathbb{Z},$ all congruency classes modulo $p$ have a unique inverse,i.e. $\mathbb{Z}/p\mathbb{Z}$ is a field.  Hence,
\[S\equiv p^{2}\sum_{k=1}^{p-1}k^{-2}\left(\prod_{j=1}^{k-1}j^{-1}(p-j)\right)^{2}\equiv p^{2}\sum_{k=1}^{p-1}k^{-2}\equiv p^{2}\sum_{k=1}^{p-1}k^{2}\equiv 0\pmod{p^{3}}, \]
as required. $\Box$[/quote]

How do you know that
\[\left(\prod_{j=1}^{k-1}j^{-1}(p-j) \right)^{2}\equiv 1 \pmod{p}\]
?

Note that the last step, we have that $\sum_{k=1}^{p-1}k^{2}=\frac{(p-1)(p)(2p-1)}{6}$

This is where the $p\ne 2,3$ comes from. Since $p>3$, we have that the $p$ in the numerator does not cancel.

[quote="me@home"]I don't think thats correct... take $i=2$ then supposedly $p^{3}|p(p+1)/2$ which is absurd.[/quote]

Oh snap... i'll fix it later.. gonna go to sleep now.

[quote="Karth"]Note that $\binom{2p}{p}= \sum_{i=0}^{p}\binom{p}{i}.$ Note that, for $i = 2, 3, 4, \cdots, p-1,$ $p^{3}| \binom{p}{i}.$ So, we have $\binom{p}{0}+\binom{p}{p}\equiv 2 \bmod{p^{3}}.$[/quote]

First off, it should read $\binom{2p}{p}= \sum_{i=0}^{p}\binom{p}{i}^{2}$. Secondly, I'm pretty sure this method can only show the congruence modulo $p^{2}$, and not $p^{3}$.

Indeed $p \parallel \binom{p}{i}$ so the method would not work. Boxedexe provides an excellent method for strengthening it to $p^{3}$.

Is $p$ supposed to be a prime? Also, why does this work:
\[p^{2}\sum_{k=1}^{p-1}k^{-2}\equiv p^{2}\sum_{k=1}^{p-1}k^{2}\pmod{p^{3}}\]
?

I believe you meant to ask why $p^{2}\sum_{k=1}^{p-1}k^{-2}\equiv p^{2}\sum_{k=1}^{p-1}k^{2}\pmod{p^{3}}$.  This is due to the fact that $\sum_{k=1}^{p-1}k^{-2}\equiv \sum_{k=1}^{p-1}k^{2}\pmod p$.  This fact follows since every $k$ here is relatively prime to $p$ and so each one has a unique inverse modulo $p$.

At first, I thought that was confusing because $k^{2}$ for $k=0,1,...,p-1$ does not form a residue class. But the idea that he is using is:

$\sum_{j=0}^{p-1}f( j^{-1})\equiv \sum_{j=0}^{p-1}f(j)\mod p$

where $f(x)$ is a function,$p$ is a prime, and $y^{-1}$ denotes the inverse of $y$ modulo $p$. In this case, he took $f(x)=x^{2}$.


dar mored soal 4 be nazar shoma ba donbale fibonatchi be dast  nemiad
be ezae har zoji az donbaleh &adad 1 hokm sadegh ast. [/b]

ali joon man ke behet goftam ke.khob farz mikonim baraye hardo ozve fiboonatchi doroste .bayad esbat koni mishe az har adadi ba in donbale be yek ozve donbaleye fiboonatchi resid.(ke shayad esbatesh kheili sakht bashe)
ama age ba adade fard beri kar kheili rahate.

man ye topic daghighan ba hamin onvan va dar hamin zamine dadam,fekr konam mitoonesti too oonja in soal ro beporsi. :D

albate to hanooz ba in site zid ashena nisti,dar ayande bishtar ashena khahi shod.


Solve for $ x$: $ \left(\minus{}\frac{1}{3}\right)(\minus{}4 \minus{}3x)\equal{}\frac{1}{2}$

hello, we have
$ \minus{}\frac{1}{3}(\minus{}4\minus{}3x)\equal{}\frac{1}{2}$
$ \frac{1}{3}(4\plus{}3x)\equal{}\frac{1}{2}|\cdot3$
$ 4\plus{}3x\equal{}\frac{3}{2}|\minus{}4$
$ 3x\equal{}\frac{3}{2}\minus{}\frac{8}{2}\equal{}\minus{}\frac{5}{2}$
$ x\equal{}\minus{}\frac{5}{6}$
Sonnhard.

I don't understand Dr. Graubner's notation.

$ \minus{}\frac{1}{3}(\minus{}4\minus{}3x)\equal{}\frac{1}{2} \implies$
$ \minus{}(\minus{}4\minus{}3x)\equal{}\frac{3}{2} \implies$
$ 4\plus{}3x\equal{}\frac{3}{2} \implies$
$ 3x\equal{}\minus{}\frac{5}{2} \implies$
$ x\equal{}\minus{}\frac{5}{6}$.

The answer shows up 5/6. It's wrong right?

Yes, I believe so. You can report the error.


calculus
integration
limit
function
real analysis
absolute value
real analysis unsolved

Here is my solution to this nice but easy problem. I think it is in my book of analisys. Anyway, here it goes:

 [color=blue] [b] Problem. [/b]

   Let $f: [0,1]\to\mathbb{R}$, $f\in C [0,1]$. Prove that \[ \lim_{n\to\infty} n\int_{0}^{1} x^{n}f(x) dx=f(1).  \]

 [/color]

  [i] Solution. [/i]

    Integrating par parts, we have \[ n\int_{0}^{1} x^{n} f(x) dx=f(1)-\int_{0}^{1} x^{n} f(x)dx -\int_{0}^{1} x^{n+1} f'(x) dx.  \]

     But, \[ 0\leq | \int_{0}^{1} x^{n} f(x) dx | \leq \int_{0}^{1} x^{n} |f(x)| dx\leq\frac{1}{n+1}\cdot\sup_{x\in [0,1]} f(x)dx,  \]
     so $\lim_{n\to\infty}\int_{0}^{1} x^{n} f(x) dx =0$. Analogues $\lim_{n\to\infty}\int_{0}^{1} x^{n+1} f'(x)dx=0$. So the conclusion follows. :)  ;)

Here is more interresting problem:

  [color=blue] [b] Problem. [/b]

     Let $f: [0,1]\to\mathbb{R}$ be a continuous function, and $(a_{n})_{n\geq 1}$ a sequence such that

                                                  \[ a_{n}=n\int_{0}^{1} x^{n} f(x^{n}) dx.  \]

      Find $\lim_{n\to\infty} a_{n}$.

     [/color]

Yes, Moubi!!! That's the answer. :)  It would be nicer if you could post a full solution. ;)

[quote="Cezar Lupu"]

     Let $f: [0,1]\to\mathbb{R}$ be a continuous function, and $(a_{n})_{n\geq 1}$ a sequence such that

                                                  \[ a_{n}=n\int_{0}^{1} x^{n} f(x^{n}) dx.  \]

      Find $\lim_{n\to\infty} a_{n}$.

     [/quote]

With the change variable $y=x^n$, $a(n)=\int_{0}^{1}f(y)y^{1/n}dy$ apply
Lebesgue's dominated convergence

Wow I have never thought to apply Lebesgue theorem although I'm taking a measure theory course this year.
 Moubi, I would be grateful if you could complete the solution (in detail, please :D )
 Anyway, my solution is quite long. :)

[quote="Cezar Lupu"]

    Integrating par parts, we have \[ n\int_{0}^{1} x^{n} f(x) dx=f(1)-\int_{0}^{1} x^{n} f(x)dx -\int_{0}^{1} x^{n+1} f'(x) dx.  \]

[/quote]

Cezar Lupu, why you think that $f'$ is exist?
And even if $f'$ exist...
[quote]
 Analogues $\lim_{n\to\infty}\int_{0}^{1} x^{n+1} f'(x)dx=0$
[/quote]
why do you think that $f'$ is bounded???
 :?:  :?:

Write the integral as $\int_0^\alpha nx^nf(x)\mbox dx+\int_\alpha^1nx^nf(x)\mbox dx\ (*)$ for $\alpha\in(0,1)$. Fix $\alpha$ so close to $1$ that $|f(x)-f(1)|<\varepsilon,\ \forall x\in[\alpha,1]$. For large $n$, the left part of $(*)$ will be $<\varepsilon$ in absolute value, so we'll have $|\int_0^1nx^nf(x)\mbox dx-\frac{n(1-\alpha^{n+1})}{n+1}f(1)|\le\varepsilon(1+\frac{n(1-\alpha^{n+1})}{n+1})$. The conclusion follows.

you  need f just to be almost everywhere continuous and the limit wil be
f(0-)
just apply the theorem that have told you Moubi ,
we have int(nx^nf(x),0,1)=int(x^(1/n)f(x^(1/n)),0,1)
let ( V_n) the sequence of functions :[0,1] ---> IR,x --> x^(1/n)f(x^(1/n))
(1): ( V_n) converge pointwise to V:[0,1] ---> IR,x --> f(o-) if x=/=0,1 and V(0)=0, V(1)=f(1)
(2) for  M=||f||°°  then ,for all n and all x ||V_n(x)|| <=M
 and we apply directly the theorem ...

This was also discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=402145#p402145]here[/url]

[quote="flip2004"][b] If  $f\in C[0,1]$ prove   $\lim_{n\to \infty} n\int\limits_{0}^{1}x^{n}f(x) \; dx = f(1) .$  [/b]
[size=75] [color=red] [b] Generalization ? [/b] [/color] [/size][/quote]

[size=75][color=blue][b] Generalization [/b][/color]  [/size]
[hide][b] Consider $h\in C[0,1] , h([0,1]) \subseteq [0,1]$ and $K_n\in C[0,1] , \; K_n\ge 0$ on $[0,1]$.   If
i)  $\lim\limits_{n\to \infty}\int\limits_{0}^{1} K_n(t) =1$ ,
ii) there exists   $z_0 , z_0\in [0,1]$ ,  such that   $\lim\limits_{n\to \infty}\int\limits_{0}^{1} K_n(t)|h(t)|^j =z_0^j$   for  $j\in \{1,2\},$  then
\[ \lim\limits_{n\to \infty}\int\limits_{0}^{1} K_n(t)f\left(h(t)\right) =f(z_0) \; \;,\; \; \forall f\in C[0,1] .  \]
[/b]

[MORE INTERESTING ]  Let us introduce the monomials $e_0(t): =1\; , \; e_j(t)=t^j , \; j\in {\mathbb N}^* .$
[color=blue][b]
Let $J_n: C[0,1] \to {\mathbb R} \; ,\; n\in {\mathbb N}^* ,$ be a sequence of linear positive functionals  such that
\[ \lim\limits_{n\to \infty} \left| J_n(e_j) -e_j(\lambda) \right| = 0 \; \; \mbox{for}\; \; j\in \{0,1,2\}, \lambda \in [0,1] .  \]
Then  $\lim\limits_{n\to \infty} \left| J_n(f) - f(\lambda) \right| = 0 \; \; \mbox{for all }\; \; f\in C[0,1] .$
[/b][/color]
Discovered by Tiberiu Popoviciu (around 1950), Harald Bohman (around 1951) and by P.P.Korovkin ($\approx$1951/1952) .
[/hide]

turning back to the original problem, couldn't it be reduced to a problem of convergence of measures? after all, it's the convergence of $\mu_n = g \mathcal{L}^1$, where $g = (n+1)x^n$. it comes out that $\mu_n$ is a measure on $[0,1]$ converging to dirac delta (*) in $1$, and you should get your result, since $\int f d\mu_n = \int f d\mu = f(1)$.

part (*) is easy to prove, since repartition function $F_n(x) : = \int_0^x g_n(t)d\mathcal{L}^1(t) \rightarrow 0 = \int_0^x d\delta_1 =: F(x)$, and $F_n(1) : = \int_0^1 g_n(t)d\mathcal{L}^1(t) \rightarrow 1 = \int_0^1 d\delta_1 =: F(1)$, and so we have that $F_n \rightarrow F$ pointwise, and convergence is shown.

actually, i didn't check if this kind of job would be working on generalization, but at least it suggest another kind of generalization... (maybe the same)


[quote="mathVNpro"]Let $ ABC$ be a triangle with $ (I)$ is its incricle. Let $ A_0$ be the tangency point of $ (I)$ with $ BC$. Let $ A_2$ be the the reflected point of $ A_0$ with respect to $ I$. Let $ A_1$ be the intersection of $ AA_2$ and $ BC$. $ A_3$ is the second intersection of $ AA_2$ and $ (I)$. Let $ M_a$ be the midpoint of $ BC$. Prove that $ M_aA_3IA_0$ are concyclic.[/quote]
We use this lemma: [i]Let the incircle of triangle ABC touch side BC at D, and let DE be a diameter of the circle. If line AE meets BC at F, then BD = CF[/i].
By this lemma we have $ M_a$ is midpoint of  $ A_0A_1$, then $ MA_3\equal{}MA_0$ because that $ A_0A_3A_1$is right triangle. So $ M_aA_3I\equal{}MA_0I\equal{}90$then $ M_aA_3IA_0$ are concyclic

By homothetic  A1 is the tangent point of exincircle toward angleA
Thus BA1 CA0  Ma is the midpoint of A1A notice A0A2 is the diameter
MaA3=MaA1=MaA0    => MaA3 is tangent to incircle
done!

Dear Mathlinkers,
1. let (1) be the incircle of ABC, I the center of (1),
(2) the circle with diameter A0A1 and center Ma.
2. According to the Morley's circle (see http://perso.orange.fr/jl.ayme vol. 2, Les circles de Morley…), A0. A1, I, Ma and A3 are cocyclic.
Sincerely
Jean-Louis


function
calculus
derivative
real analysis
real analysis solved

We have the equation: x^(2n)-nx^2+1=0 for any n>=3 natural!
a)prove that the equation has four real solutions an,bn,cn,dn which verify the relation:
an<-1<bn<0<cn<1<dn.
b)Compute the limits of an,bn,cd and dn when n-->oo.

bye bye!

Let f(y) be the fonction : f(y) = y^n - n*y + 1, defined in [0,+oo[
(f(0)=1 , f(1)=2-n < 0 , f(4)=4^n-4*n+1 > 0) ==> f has two roots U(n),V(n) such that : 0 < U(n) < 1 < V(n) < 4
The derivative f'(y)=n*(y^(n-1)-1) has only one positive root so the positive roots for f(y) are only U(n),V(n) .

So the initial equation has only four real roots :
-2 < a(n)=-sqrt(V(n)) < -1 < b(n)=-sqrt(U(n)) < 0 <  c(n)=sqrt(U(n) < 1 < d(n)=sqrt(V(n)) < 2

U(n)^n-n*U(n)+1=0  ==> 0 < U(n)=(1+U(n)^n)/n  < 2/n  ==>  U(n) --> 0  ==>  b(n)-->0 and c(n)-->0
V(n)^n-n*V(n)+1=0 ==> 1 < V(n)=(n*V(n)-1)^(1/n) < (4*n-1)^(1/n) ==> V(n)-->1 ==> a(n)-->-1 and d(n)-->1

See you ..


Find the fuction $f: (0\to+\infty) \to (0\to+\infty)$ sastifying the relation:
$(x+y)f(f(x)y)=x^{2}f(f(x)+f(y))$
 :blush:

Let's give a solution.We have $(x+y)f(f(x)f(y))=x^{2}f(f(x)+f(f(y)))=y^{2}f(f(y)+f(f(x)))$.Now we have that f is injective,we just fix y and if we have $f(x)=f(x_{0})$ we obtain that $x^{2}=x_{0}^{2}$ so we have $x=x_{0}$.Now we make $y=x^{2}-x,x>1$.We have that $f(f(x)(x^{2}-x))=f(f(x)+f(x^{2}-x))$ and from injectivity we have $f(x)(x^{2}-x-1)=f(x^{2}-x)$.Now we observe that we can make that we can make $x^{2}-x-1$ negative and we are done. :lol:

[quote="mateivld"]Let's give a solution.We have $(x+y)f(f(x)f(y))=x^{2}f(f(x)+f(f(y)))=y^{2}f(f(y)+f(f(x)))$.[/quote]

you are wrong here it must be $(x+f(y))f(f(x)f(y))...$

Sorry for that.We obtain that $\frac{x^{2}f(f(x)+f(f(y)))}{x+f(y)}=\frac{y^{2}f(f(y)+f(f(x)))}{y+f(x)}$.Now assume that we have $f(x)=f(y)$ we obtain that $x^{2}y+x^{2}f(x)=y^{2}x+y^{2}f(y)$ so we have $(x-y)(xy+(x+y)f(x))=0$ so necessarly we have $x=y$ so f is injective.The proof follows as in my last post. :lol:

f is injective
Let $x=\frac{1+\sqrt{5}}{2},\ y=1$
$\rightarrow f\left(f\left(\frac{1+\sqrt{5}}{2}\right)\right)=f\left(f(1)+f\left(\frac{1+\sqrt{5}}{2}\right)\right)$
$\rightarrow f\left(\frac{1+\sqrt{5}}{2}\right)=f\left(\frac{1+\sqrt{5}}{2}\right)+f(1)\rightarrow f(1)=0$ unreasonableness
Not exist function f


vector
geometry
3D geometry
pyramid
USAMTS
HMMT
similar triangles

On a 4 x 4 x 3 rectangular parallelepiped, vertices $A,B,$ and $C$ are adjacent to vertex $D$. The perpendicular distance from $D$ to the plane containing $A,B,$ and $C$ is closest to

$\text{(A)} \ 1.6 \qquad \text{(B)} \ 1.9 \qquad \text{(C)} \ 2.1 \qquad \text{(D)} \ 2.7 \qquad \text{(E)} \ 2.9$

Too lazy to do it, but:

[hide]Put it in a 3-space grid, take cross-product of  and , then you'll have normal to plane, so you can get equation to plane; then just do point to plane formula.[/hide]

[hide]AB and AC have length 5, BC has length 4¬/2. Let E be the midpoint of BC. Then BE = 2¬/2, and by pythagorean, AE = ¬/17. ADE is another right triangle and its altitude should be the desired answer.

DE = ¬/8, AE = ¬/17, AD = 3. Let the foot of the altitude be X. By similar triangles, DX/DA = DE/EA. (short/hypotenuse) so

$\frac{DX}3 = \frac{\sqrt8}{\sqrt{17}}$
$\frac{DX^2}9 = \frac{8}{17}$
$DX = \sqrt{\frac{72}{17}}$
which is approx equal to 2.057. The answer is C[/hide]

Can anyone explain the more elegant solution?

Elemennop wrote:Using the picture in the OP's post:

[hide] The bottom left corner closer to you is . Let , , , . Let , and . Take cross product of  and , which will be ; this vector is also the normal to the plane. The equation of the plane  can then be found to be . We can then use the formula for point to plane, which is . Plugging this all in, we get the distance from D to the plane be equal to , which is approximately equal to , so the answer is .[/hide]

Why not let D be the origin? It'd be slightly easier.

[quote="WarpedKlown1335"]How about expressing the volume in two different ways.  That makes this problem really easy.[/quote]
[hide]Nice:
 [color=darkred]b[/color][color=orange]h[/color]=[color=darkred]b[/color][color=orange]h[/color]
[color=darkred](4)(4)(1/2)[/color][color=orange](3)[/color]=[color=darkred](1/2)sqrt(32)sqrt(17)[/color][color=orange]h[/color][/hide]

Solve for h.

find the area of ABC (it is an isosceles triangle..., and use the volume of the pyramid with that base and height, and the volume you know is 8

[quote="WarpedKlown1335"]How about expressing the volume in two different ways.  That makes this problem really easy.[/quote]
yea you can also use de Gua's thm.

Check out USAMTS Round 2 #5 this year...

A problem on HMMT last year was similar to this but it asked for finding the expression of the area of the face opposite the vertex in terms of the areas of the side faces (it was a right angle at the vertex).


Euler
graph theory
combinatorics open
combinatorics

If we want to describe three numbers different from each other, we use three symbol $\neq$.
 $a \neq b \neq c \neq a$
How many symbols $\neq$ we need to describe n numbers different from each other( at least)
-- :blush:

obviously $ n \choose 2 $... if for some pair, $a, b$, we *don't* say $ a \neq b$, then there's still the possibility that $a = b$, no matter between which *other* pairs we put $\neq$ signs.

unless me lad meant in a row ..........but he didn't specify it, me lad, oh me lad.

you lad... you may be right... i may be wrong!  you lad

$n(n-1)/2$ for odd $n$ and $n^2/2-1$ for even $n$. If $n$ is odd, than the complete graph $K_n$ with $n$ vertices is Euler graph (i.e. degrees of its vertices are even), hence it has an Euler cycle, which uses each edge esactly once. If $n$ is even, note that each number must appear at least $n/2$ times in our line, hence we have at least $n^2/2$ numbers and $n^2/2-1$ = signs. An example may be constructed in a following way: consider $K_n$ and add additionaly double edges $1-2,3-4,\dots,(2n-2)-(2n-1)$.
Now we have $n^2/2-1$ edges, and all degrees except two are even. So, the exists an Euler cycle.

Oh.That's my solution,too.
but I can't fing whathappen when n is even

[quote="Vihang"]For three numbers isn't $a \neq b \neq c$ enough because trasitivity ensures $a \neq c$  :?:

If this is true then only n-1 signs are required.
$a_1 \neq a_2 \neq ... a_{n-1} \neq a_n$ ensures all the numbers are different.  :huh:[/quote]

Not quite...  Suppose you have $a=c=3$, $b=5$, for example.


There are 4 representatives from each of 4 companies at a convention. At the start of the convention, every person shakes hands once with every other person except the other representatives from their company. How many handshakes are there?

Each of the 16 people shake hands with 12 other people. Thus, teh answer is 16*12 / 2 = 96, since we overcount each shake once.

16 people need to shake 12 other people 16*12=192 BUT WAIT THERES MORE! WE OVERCOUNTED SO WE NEED TO DIVIDE BY 2 SO WE CAN TAKE OUT THE PART WE OVER COUNTED 0/\0


Prove if a,b,c>0 then
$ \frac{a^3\plus{}b^3\plus{}c^3}{\sqrt{3abc}}>\equal{}\sum\frac{a^2}{\sqrt{b\plus{}c}}>\equal{}\frac{\sqrt[6]{abc}(a\plus{}b\plus{}c)}{\sqrt{2}}$

Try $ a\equal{}b\equal{}c\equal{}0,01$  ^^

[quote="tranquoc"]Prove if a,b,c>0 then
$ \frac {a^3 \plus{} b^3 \plus{} c^3}{\sqrt {3abc}} > \equal{} \sum\frac {a^2}{\sqrt {b \plus{} c}} > \equal{} \frac {\sqrt [6]{abc}(a \plus{} b \plus{} c)}{\sqrt {2}}$[/quote]
It's wrong. Let $ a\equal{}b\equal{}c\equal{}1$  :)

Assume that :$ abc=1$
$ \bigstar$ :
$ \sum a^3+b^3+c^3 \ge \sum  a^2\sqrt[4]{a} =\sum \frac{a^2}{\sqrt[4]{bc}} \ge \sum \frac{\sqrt{2}a^2 }{\sqrt{b+c}}$
Done
$ \bigstar$
${ 3 \sum \frac{a^2}{\sqrt{b+c}} \ge (\sum a^2})(\sum \frac{1}{\sqrt{b+c}} ) \ge  (a+b+c)^2 \frac{3\sqrt{3}}{\sqrt{2(a+b+c)}} \ge ..$
Done ;)


Consider N closed intervals, having the the extremities natural numbers between 1 and L. Every number $i$ from the interval [1, L] has associated a ponder $c_{i}$.
We name a cover a set of natural numbers between 1 and L with the property that every interval contains at least an element of the set. The cost of a cover is equal with the sum of the ponders from the cover.
For a set of given intervals find the minim cost of a cover.

INPUT

The input file contains two numbers N and L. On the next line there are L  numbers $c_{1}$, $c_{2}$, ..., $c_{i}$, representing the ponders of the numbers from the interval [1, L]. The next N lines contain each contain two numbers a, b (1<=a<=b<=L) representing the extremities of the intervals.

OUTPUT

In the output file there will be a single number representing the minim cost of a cover.

Restrictions

1<=N<=60000
1<=L<= 1000000
0<=$c_{i}$<=1024

SAMPLE INPUT

4 10
1 3 6 4 5 1 0 1 3 2
1 3
3 5
6 9
4 4

SAMPLE OUTPUT

5

Time limit: 0.5 seconds

Smells like Flow or Mincut. But cant make the graph!

mahbub, aren't the bounds on N and L a bit [i]too[/i] big for a flow implementation to handle?

Anyway, this can be done with a $\mathcal O (L+N \log N)$ DP. Is it just me or are most Romanian problems generally pretty interesting? :)

Here are this year's problems at the nationals http://oni2007.around25.ro/index.php?uid=5
I can translate if needed.

How about... all of them?

Heh, just kidding. How about you just choose a couple that look interesting?


Prove that the hypotenuse (side across from the right angle) of a right triangle inscribed in a circle is the diameter of the circle.

[geogebra]5297328df71e94f40293b44fe19fd3569d7ddf14[/geogebra]

(I am not sure of the difficulty level, so it might not belong here)

Let's say you have the right triangle $ ABC$ inscribed in a circle such that $ \angle CAB$ measures $ 90^{\circ}$. Then by a [url=http://en.wikipedia.org/wiki/Inscribed_angle#Property]property[/url] of inscribed angles in a circle, the arc that is opened is equal to twice the inscribed angle which is $ 90^{\circ}$, and since by definition if the arc equals $ 180^{\circ}$ then the segment from the two points the make up the arc is a diameter of the circumscribed circle.


Given a triangle ABC.M is an arbitrary point in space.Prove the ineq
$ aMB\plus{}bMC\plus{}cMA$≥$ \sqrt{a^2b^2\plus{}b^2c^2\plus{}c^2a^2}$

[quote="SUPERMAN2"]Given a triangle ABC.M is an arbitrary point in space.Prove the ineq
$ aMB \plus{} bMC \plus{} cMA$≥$ \sqrt {a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}$[/quote]
Anyone helps me?I really need a solution.


Is this law true?

when a sphere is circumsribed by a cylinder:

the ratio of the volume of the sphere to the volume of the cylinder circumscribing the sphere is 2/3.

Yes, the ratio is 2/3. Since the volume of sphere is $\frac{4}{3}\pi*r^{3}$, and the volume of the cylinder circumcribing it is $2r*\pi*r^{2}$, (the height is 2r, since it's twice the radius of the sphere. If you simplify the ratio $\frac{\frac{4}{3}\pi*r^{3}}{2r*\pi*r^{2}}$, you get $\frac{4/3}{2}$, which simplifies to $2/3$

Yeah! I proved it on the bus once! Who was the inventor of this law? Archimedes?

Idk.. I just proved it when I saw the post you made...

Proving stuff can be fun and frustrating  :)


This has nothing to do with math, but...

anybody knows where I can find the paper and solutions to the 2005 CAP exam? its not on the CAP website...

They say on their site that you must e-mail them to obtain a password. I am not sure if they give the passwords to students looking to write the exam. I dont see why not unless they do something lame like reuse questions.

Nope, they haven't uploaded the 2005 paper yet. Though, the older papers may be found here http://www.physics.ubc.ca/~outreach/CAPexams/oldexams_e.htm

The password is only for the solution PDFs. You can find out the password by emailing to the address indicated, and you'll receive the password in a couple days. This is done to discourage students from simply looking at the solutions without really trying the problems.


I don't get it. You want to find $x^2+y^2=0(\mod n)$.

But you can just take $x=y$  :?

I dont know if this is what you are asking, but $N$ is a sum of squares if and only if every prime of the form $4k+3$ dividing $N$ has even order.

[quote="hotkirstin"]I know that only primes of form 4k+1 can be written as sums of two squares... but what primes can be of the form x^2 + y^2 = 0 (mod p), or mod p^a, or just any number in general???[/quote]

If $x^2+y^2=\Lambda p^a$  then every prime in $\Lambda p^a$ which is $1 \ mod \ 4$ must have an even power as Pascual2005 told b4.

Best regards.

The Crazy Math


Prove that
$ p^{2}\mid \binom{2p}{p} \minus{} 2$
for every prime $ p$

It is from Wolstenholme theorem.
$ p^2|2(1...(p\minus{}1)(p\plus{}1)...(2p\minus{}1)\minus{}(p\minus{}1)!^2)$


hi,
i am trying to prove that the perpendicular bisectors of a triangle are concurrent...how do i go about doing that??

NO SOLUTION PLS! gime a hint or tip or something ok? tnx

tnx and God bless

i have an idea...i thought about it, and i was like, why not prove that if two perpendicular bisectors intersect, then their intersection is equidistant from the vertices, and then i extend a line from the midpoint of the third side to the intersection, which then forms two triangles....I can prove that triangles congruent, and that means the adjacent angles are congruent, and if 2 lines intersect to form congruent adjacent angles, then they're perpendicular. ....by definition the line i extended is a perpendicular bisector, and it intersects the other perpendicular bisectors at the same point, and therefore by definition of concurrent lines....the 3 perpendicular bisectors of a triangle are concurrent.

wow, this is awesome. i just had a eureka moment

Are the perpendicular bisectors always cevians? :wink:

math154 is right, I am sorry, I automatically thought of heights, now I read the text and saw.
1) Prove that they cut
2) Where do they usually cut?


Prove that there exists a term in the fibonacci sequence that ends with 6 zeroes. (divisible by a million in other words)

This problem is too fun to leave unanswered. Note that knowing any two consecutive terms determines the entire Fibonacci sequence [i]in both directions[/i]. Also, adding or subtracting the same two numbers (mod 10^6) will always leave the same answer. Since we must eventually get two consecutive terms of the Fibonacci sequence that are the same as two consecutives before since there are only (10^6)^2<infinity ways to choose a and b distinctly (mod 10^6), the Fibonacci sequence must cycle (mod 10^6) with some period. Since F_0=0:equiv:0 (mod 10^6), there must be (infinitely many) other terms divisible by 10^6. The same argument will hold for any number, not just 10^6.

ComplexZeta's proof is perfect, but there is a constructive proof based on Yuri Matijasevich (*) Lemma (very nice challenge btw) :

Fibonacci number Fm is a multiple of (Fn)^2 if and only if m is a multiple of nFn.

It means that if a | F(n0) then a^2 | F(n0*F(n0)). It's then possible to determine N such a a^k | F(N) as soon as we have a | F(n0).

For instance :

F(15) = 610, so 100 | F(15*610)=N1, 10000 | F(N1*F(N1))=N2, 100000000 | F(N2*F(N2)), ....

(*) genius who solved Hilbert tenth problem, cf http://www.informatik.uni-stuttgart.de/fmi/ti/personen/Matiyasevich/Julia for instance.

Even better than that, it can be proven rather easily with Pigeonhole that the Fibonacci Sequence will cycle in [i]any[/i] mod, meaning that you can find a term in the sequence with exactly n zeroes, where n is some positive integer.

That's false. Let L_n denote the nth Lucas number. (That is, L_1=1, L_2=3, L_n=L_(n-1)+L_(n-2) for n>=3.) Find a term of the Lucas sequence divisible by 5.

But I was talking about Fibonacci? It's different than Lucas...

Yes, I know, but how would you use Pigeonhole when the same thing for Lucas numbers doesn't work?

The only reason it works is that there is a 0 term in the loop somewhere. With Lucas' sequence we can't guarantee this. However what darkquantum said about the fibonacci sequence is true. For any positive integer we can find a term divisible by it in the sequence.

I agree with that. I do not agree that we can prove this result by the pigeonhole principle, however.

I was under the impression that pigeonhole was being used to show there must be a pair that repeats. Which is exactly what it is really, just much easier to see.

Yes. For pigeonhole to work in Fibonacci, we just take the residues of pairs of consecutive terms in teh sequence. Since there are finitely many possible pairs of such residues in any mod, then you are guaranteed by pigeonhole for one pair to repeat... Uh-oh, forgot the rest of the proof, hafta dig it back up....

I'll post later...

O yea, the rest of the proof:

since there are only finitely many pairs with different residue 0 to n-1, then we know that at least one pair will appear twice by pigeonhole (well, duh). Now, if one pair shows up again, that means that the entire sequence will cycle (do you see why?), meaning taht (1,1) will show up again. Now, in order for a (1,1) to show up, there must be a 0 in the pair before, meanign that there is at least one term that is divisible by your mod.


see above.
the other birthday threadgot overspammed. im doing this to try to see that if we get 23 posts, two people will have the same birthday

ui got the explanation of 23C2/365; i still dont believe it, cheifly because 43C2 is greater than 365 and if we have a group of 43 people, its definitely not going to have 2 ppl be the same bday

edit: forgot my own birthday
8/19/92

May 9, 1990
It sometimes coincides with Mother's Day

December 21, 1984, although my first high school ID card said 11.22.18 instead of 12.21.84.

[quote="ARTI"][quote="jli"]march 1 1993[/quote]

February,18 1989.
I wouldn't believe that there are kids here of an age as yours [i]jli[/i] but with a considerable number of posts. :?
Artiona[/quote]

January 18,1992

I believe Silverfalcon's also around this age.

March 24th, 1988

Bah haha ha, to be 12 again, lol.  Those were the days...lots more time for math....

I thought Silverfalcon was a freshman in high school.  Wouldn't that make him my age (around 14 or 15)?

[quote="Scrambled"]ui got the explanation of 23C2/365; i still dont believe it, cheifly because 43C2 is greater than 365 and if we have a group of 43 people, its definitely not going to have 2 ppl be the same bday[/quote]

btw, the ${23}\choose{2}$ is just a rough guide. the actual calculation is $P(k+1) = \prod_{i=1}^k \frac{364-k}{365}$ and so $k \geq 22$ implies $P \leq 0.5$, where $P(k)$ is probability of k people all different birthdays. (PS what is latex for implies?) This means that the probability of two people having the first birthday would be more than 0.5.

I am from planet Planet.
I was born on Enuj 25, 9002.
and my birthday is not [hide=Puzzled spoilers]June 25 on Earth. Try and figure it out! :P [/hide]

[quote=barry.yyc]12/15/2009[/quote]

I'm a bit older than him/her.

[quote=adriancheung]June 26 2011[/quote]
what you're not even a 10-year-old?
how did you know this site at all?


This is easy but may be someone wants to solve it.Here it is...

A rope with mass m and length L is hanging from the ceiling.Find the tension force at the bottom point.

I think that you need an additional variable -- either the engle between the end of the rope and the ceiling (easier case), or the horizontal distance between the hanging points (harder case).

You can take that angle as $\frac{\pi}{2}$. :)

Then the tension is zero. Just look at the forces acting on one half of the rope: the reaction from the ceiling (vertical), the weight (vertical), and the reaction from the other part (horizontal)...


geometry
area of a triangle
Heron\u0027s formula

[hide="Answer"]A regular hexagon is composed of $6$ equilateral triangles; the length of each side in the given hexagon is $r$, so the area of the hexagon is $6\cdot \frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{2}$.[/hide]

or you could use the formula $\frac 12 ap$ where $a$ is the apothem and $p$ is the perimeter.  jf'197s way seems much faster though. :roll:

I'm wondering why the given side of the hexagon is r.

The length of the side of the hexagon is r because if you draw the hexagon in the circle and connect the center of the circle to the vertices of the hexagon, you form six equilateral triangles. (60 degree angle, two congruent radii as sides) Since all of the sides of an equilateral triangle are congruent, the side length is the radius, r.

Sorry, that should've been obvious to me  :blush:

Heron's formula would maybe faster here: [tex]A~=~\sqrt{3/2*1/2*1/2*1/2} ~ = ~\sqrt{3/16} ~ = ~ \sqrt{3}/4[/tex]
And then times six of course.

If you're talking about the area of each equilateral triangle (which you seem to be, since you're multiplying it by 6), it's far easier to just use $\frac{s^2\sqrt{3}}{4}$, which is what I did.

[quote="236factorial"]A regular hexagon is inscribed inside a unit circle. What is the area of the hexagon? Express your answer as a fraction in simplest radical form.[/quote]

[hide]
p=6
a=(sqrt3)/2
ap/2=3(sqrt3)/2
[/hide]


I am a sophomore currently attending high school, and I love math and I believe I am better than the average high school student at math. I'm willing to spend a lot of time studying math, however, I've been led to believe my entire life that only geniuses can understand the math at the level of the AIME and the USAMO and succeed to that level. This may seem like an incredibly dumb question, but I was wondering if I try enough, would I be able to score well on the AMC 12 and move on to the AIME and in some off chance, possibly the USAMO?

That's a difficult question to answer. It really depends on you. I would certainly try, as problem solving skills in general are applicable to pretty much all ocupations. If you have a natural aptitude for math, even if just a small one, then try to go all of the way. I wish you luck.

Winning USAMo or some other huge math contest gives you soooo many benefits(i.e. scholarships, etc.), that I think you should keep trying. :)

No way, only SUPERGENIUSES can ever make the AIME, not to mention USAMO. Actually, nah I think just about everybody (and certainly those with > average math aptitude) can make AIME after some studying. As for USAMO, you should definitely try, because a lot of people have the potential to solve around 8 problems on the AIME, but of course, you'll have to try those problems yourself to assess your particular situation. I would say that everyone has the potential to improve their AMC by 20 points and their AIME by 3 points, and this is beyond your natural improvement as you grow older.

IMO, genius is just being insightful at the right times. And in AIME's, with some practice, and practice is enough to improve your score considerably, you can get into USAMO. Though, I don't think anybody can change my belief that it's mostly geniuses, contrary to what I said above, that can make it past USAMO into MOSP. So erm, be content with your USAMO score... I guess that's what I'm saying. :(

I won USAMO a couple of times, it gets boring.


function
logarithms
algebra unsolved
algebra

Find all continuous functions $ f$ $ :$ $ R$ $ \longrightarrow$ $ R$  that
$ \forall$ $ x,y$ $ \in$ $ R$ $ :$    $ f$ $ ($ $ x$ $ f(x)$+ $ y$ $ f(y)$ $ )$ $ \equal{}$ $ 2$ $ f(x)$ $ \times$ $ f(y)$

We will prove that f(x)=0 or f(x)=\frac12.

Case 1) suppose there exists x such that $ f(x)\equal{}0$.
Then $ f(xf(x))\equal{}0$ for all $ x$. Particularly, $ f(0)\equal{}0$. We claim $ f(x)\equal{}0$. Suppose it is not true. Then we have $ f(a)\equal{}b\neq0$. By continuity of $ f(x)$, we can always assume $ |b|<0.5$. Since $ f(x)$ and $ xf(x)$ are all continuous. Equation $ xf(x)\equal{}t$ always has solution when $ t$ is between 0 and $ ab$. So $ f(t)\equal{}0$ for any t between $ 0$ and $ ab$.

We also have $ f(2xf(x))\equal{}2f(x)^2$. Apply this to $ f(a)\equal{}b$ again and again, we will have $ f((2b)^{2^n\minus{}1}a)\equal{}b(2b)^{2^n\minus{}1}\neq0$. We can always find big enough $ n$ such that $ (2b)^{2^n\minus{}1}a$ falls in between $ 0$ and $ ab$. Here we get contradiction.

Case 2) there is no x such that f(x)=0.
Obviously $ f(0)\equal{}\frac12$. We claim $ f(x)\equal{}\frac12$. Suppose it is not true. By the continuity of $ f(x)$ and $ \log_{\frac12}^x$, we can find $ f(x)\equal{}a\equal{}(\frac12)^r$, where $ r>0.5, r\neq1$ and is rational. Since $ f(2xf(x))\equal{}2f(x)^2$. We have $ f(2ax)\equal{}2a^2\neq a$. Applying $ f(xf(x))\equal{}f(x)$ again and again, we will have $ f(a^nx)\equal{}a$ and $ f((2a^2)^m2ax)\equal{}2a^2$ for any non-negative integer $ n,m$. But we will get contradiction if we can find $ n,m$ such that $ a^n\equal{}(2a^2)^m2a$, or $ a\equal{}(\frac12)^{\frac{m\plus{}1}{2m\plus{}1\minus{}n}}$, or $ r\equal{}\frac{m\plus{}1}{2m\plus{}1\minus{}n}$. The latter equation is apparently solvable by the definition of $ r$.

Hence, we have either $ f(x)\equal{}0$ or $ f(x)\equal{}\frac12$.


function
calculus
derivative
integration
logarithms
calculus computations

the graph of the function y is a logistic curve. Show that the rate of growth is maximum at the point of inflection, and that this occurs when y= L/2

I'll let your logistic function $ y\equal{}\frac{L}{1\plus{}e^{\minus{}t}}$. The rate of growth is given by $ r\equal{}\frac{dy}{dt}\equal{}\frac{e^{\minus{}t}L}{(1\plus{}e^{\minus{}t})^2}$.

$ \frac{dr}{dt}\equal{}\minus{}\frac{e^{\minus{}2t}(e^t\minus{}1)}{(1\plus{}e^{\minus{}t})^3}L$. So when $ t<0$, $ \frac{dr}{dt}>0$ and when $ t>0$, $ \frac{dr}{dt}<0$. It means that $ r$ increases when $ t<0$, and decreases when $ t>0$, hence $ r$ has its maximum value when $ t\equal{}0$. And since $ r\equal{}\frac{d^2y}{dt^2}$ and changes sign before and behind $ t\equal{}0$, the point $ (t, y)\equal{}\left(0,\frac{L}{2}\right)$ is the point of inflection.

I am sorry! I forget you should use the dy/dt=y*(L-y) directly and without first turn this derivative to the integral form( which is the logistic function).

[quote="J.Y.Choi"]I don't know what you mean exactly. Do you want to solve the differential equation $ \frac {dy}{dt} \equal{} y(L \minus{} y)$??

tip: You must learn LaTex. The following website may help you a lot.
http://www.codecogs.com/components/equationeditor/equationeditor.php[/quote]
yes, but so can you also show me that the rate of growth is maximum at the point of inflection, and that this occurs when y= L/2.

Seperate variables and do integral in each hand side, we get $ \int\frac{dy}{y(L\minus{}y)}\equal{}\int dt$.

Using the fraction $ \frac{1}{y(L\minus{}y)}\equal{}\frac{1}{L}\left(\frac{1}{y}\plus{}\frac{1}{L\minus{}y}\right)$, the right hand side is $ \frac{1}{L}\ln\left(\frac{L}{L\minus{}y}\right)$.
So, the general solution is $ y\equal{}\frac{L}{1\plus{}Ae^{\minus{}Lt}}$, $ A(>0)$ is constant. The next work is what I said in the previous post.

$ y'\equal{}y(L\minus{}y)$

$ y''\equal{}y'(L\minus{}y)\minus{}y'y\equal{}y'(L\minus{}2y)\equal{}y(L\minus{}y)(L\minus{}2y)$

$ y''\equal{}0 \to y\equal{}0, \frac{L}{2}, L$

$ y'\equal{}0$ when $ y\equal{}0, L$; $ y'\equal{}\frac{L^2}{2}>0$ when $ y\equal{}\frac{L}{2}$

since its the greatest of the critical points, its the local max for y'

is this what you wanted?


taken from http://www.mfdabbs.pwp.blueyonder.co.uk/Maths_Pages/SketchPad_Files/Japanese_Temple_Geometry_Problems/Japanese_Temple_Geometry.html

1.

Given chord AB with midpoint P of a circle centered at O, suppose lines from A and B intersect OP produced at Q.  Suppose OP intersects the circle again from the same side as AB at R.  Let AQ intersect the circle at A and A'.  Suppose the section A'QB is inscribed with a circle of radius r.  Prove that:

1/PQ + 1/QR = 1/r

2.

Suppose A" B" and C" are circles with centers A B C such that B" and C" are tangent to A" at P and Q.  Let d be the length of the exterior tangent between B" and C".  Prove PQ^2 = ([ad]^2)/[(a+b)(a+c)]

3.

In an equilateral triangle ABC of side 1, cevians are drawn from B and C meeting the triangle again at P and Q respectively and intersecting at O.  APOQ and BOC are both inscribed with the same radius circle; find the radius.

4.

A point A is on a circle of center O and radius 1 with diameter RS such that AOS = 90deg; take a second circle with diameter AO, suppose tangents to this circle from R and S intersect at T and intersect the first circle at R' and S', then suppose the triangularish section TR'AS' is inscribed with a circle of radius r.  Find r.

5.

ABCD is a square of sidelength 1,  M is the midpoint of CD, suppose AC intersects BM at P, find the radius of the incircle of ABP.

Solution to problem number 5:

We know that triangles ABP and CMP are similar. So $2=\frac{|AB|}{|CM|}=\frac{|AP|}{|CP|}$

Now, draw line paralel to MB, passing through C and intersectiong line AB in N. We now know that triangles ANC and ABP are similar with coeficient $\frac{|AP|+|CP|}{|AP|}=\frac{3}{2}$. So the radiuses of incircles of ANC and ABP are also in ration 3/2.

Now we know that if triangle X,Y,Z has sides x,y,z, radius of incenter r and are S, then S=(x+y+z)r/2. From this we can find out that radius of incentre of ANC is $\frac{3}{2(\sqrt{2}+\frac{\sqrt{5}}{2}+\frac{3}{2})}$ and from it we conclude that radius of incentre of APB is $\frac{1}{\sqrt{2}+\frac{\sqrt{5}}{2}+\frac{3}{2}}$

I hope there's no mistake. :D


AMC
USA(J)MO
USAMO
inequalities solved
inequalities

Isn't it USAMO 2003?The solution key is to take x+y+z=3,and first term  <= 4/3+4x/3,also adding other such values we get the desired result.

Yeah,  it is really usamo problem. You can find the solution in this section. View the last topic ;)


The graph of $ x^2\minus{}4y^2\equal{}0$:

$ \textbf{(A)}\ \text{is a hyperbola intersecting only the }x\text{ \minus{}axis} \\
\textbf{(B)}\ \text{is a hyperbola intersecting only the }y\text{ \minus{}axis} \\
\textbf{(C)}\ \text{is a hyperbola intersecting neither axis} \\
\textbf{(D)}\ \text{is a pair of straight lines} \\
\textbf{(E)}\ \text{does not exist}$

[quote="AIME15"]The graph of $ x^2 \minus{} 4y^2 \equal{} 0$:

$ \textbf{(A)}\ \text{is a hyperbola intersecting only the }x\text{ \minus{} axis} \\
\textbf{(B)}\ \text{is a hyperbola intersecting only the }y\text{ \minus{} axis} \\
\textbf{(C)}\ \text{is a hyperbola intersecting neither axis} \\
\textbf{(D)}\ \text{is a pair of straight lines} \\
\textbf{(E)}\ \text{does not exist}$[/quote]
[hide]$ x^{2}\equal{}4y^{2}$
$ x\equal{}2y$
$ x\equal{}\minus{}2y$
This is a degenerate hyperbola or a pair of straight lines. D[/hide]


geometry
geometric transformation
rotation
number theory

John is examining the houses in his neighborhood. There are 12 houses with each having one light on their porch. How many different lighting patterns could be created by these 12 lights?

How would you do this problem?

[hide="Hint 1"] Either a given house has its light on, or it doesn't. [/hide] [hide="Hint 2"] Binary. [/hide] [hide="Hint 3"] Say $1$ means a light is on and $0$ means a light is off.  How many different strings of twelve $1$s and $0$s are there? [/hide]

[b]Follow-up:[/b]  John lives in a cul-de-sac.  The twelve houses are arranged in a circle (in other words, patterns that are the same upon rotation don't count as different).  Now how many patterns are there?

The first house either has its light on or it doesn't. There are 2 possible combinations. The second light also has two possible combinations. The total number of combinations are:
2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2^12 = [b]4096[/b]

(I think) :(

Kevin K's solution looks right.

The follow-up is quite a bit trickier especially since 12 is not a prime number. If you have trouble with it, try it with a prime number of houses first, and think about why you need to subtract 2 before dividing by the number of houses. (One of the reasons is that it won't divide evenly until you subtract 2; this is in fact a combinatorical proof for Fermat's Little Theorem--think about the combinatorical argument)

That looks correct. By the way, the BINARY hint was because you could think of the problem as a binary system with 12 digits, then solving for the total amount of numbers in the binary system (2^12).


inequalities
algebra
polynomial
inequalities proposed

[quote=zdyzhj][quote=Ji Chen]If $ x,y$ are nonnegative real numbers, then

$ 5x^2y^4\plus{}3x^5\plus{}3y^5\plus{}5x^3y\plus{}4\geq 5x^2y^3\plus{}5x^4\plus{}5y^4\plus{}5x^2y,$

with equality if and only if $ x\equal{}y\equal{}1.$[/quote]

easy to prove[/quote]

 If each one of us we would have said "easy to prove" any inequality, we would have had more posts than arqady, because it would be absolutely unfair!

[quote=pco][quote=oldbeginner] If each one of us we would have said "easy to prove" any inequality, we would have had more posts than arqady, because it would be absolutely unfair![/quote]
Dont bother with zdyzhj's posts. He/she never produced any useful post. Maybe it is just an automatic tool used for just trolling forums.
Here are, for example, his/her full sixteen last posts (parenthesis are added google translate :) ) :
1) "easy to prove"
2) "这些无聊不等式老是编有什么用" (These boring inequalities are always used for what)
3) "we needn't UVW at all."
4) "easy to prove"
5) "easy to prove with BW"
6) "two lines down"
7) "陈胜利谁阿不认识" (Chen Shengli who did not know)
8) "去年都同你讲了这类问题我基本已系统解决了" (Last year you told me this kind of problem I have basically solved the system)
9) "easy to prove"
10) "easy to prove"
11) "easy to prove"
12) "can be proved by BW"
13) "炸弹一个" (A bomb one)
14) "无需差分只需一行 one line down" (There is no need for a difference)
15) "I finally found a nice poof wiht our former tools for the two inequality."
16) "I proved it by BW"
...[/quote]

We cannot know anything about zdyzhj's abilities. I think we shouldn't judge a book by its cover. Maybe he doesn't have time to show his solution. :P

Maybe the margin is too small for his proofs?

[quote=KaiRain][quote=dragonheart6][quote=adityaguharoy]I knew that Buffalo way was so called as it sweeps everything that comes along like a buffalo. But how is it that the name SHED technique comes ?[/quote]
Interesting about the name of Buffalo Way.[/quote]
I feel a little inconvenient when you said "SHED" in stead of "Buffalo Way". What does it stand for, dragonheart6 ? :)[/quote]

I saw it in the link below, line 16, "The SHED technique (buffalo way)":
https://artofproblemsolving.com/community/c1902h1042832_links_to_inequalities_theorems_and_notes

[quote=dragonheart6]
I saw it in the link below, line 16, "The SHED technique (buffalo way)":
https://artofproblemsolving.com/community/c1902h1042832_links_to_inequalities_theorems_and_notes[/quote]
I see, thank you  :-D

[quote=SBM][quote=dragonheart6][quote=Ji Chen]If $ x,y$ are nonnegative real numbers, then

$ 5x^2y^4\plus{}3x^5\plus{}3y^5\plus{}5x^3y\plus{}4\geq 5x^2y^3\plus{}5x^4\plus{}5y^4\plus{}5x^2y,$

with equality if and only if $ x\equal{}y\equal{}1.$[/quote]

After homogenization (letting $x = \frac{a}{c}, \ y = \frac{b}{c}$), the SHED technique (Buffalo Way) works (need to check several cases $a\ge b\ge c$, or $a\ge c\ge b$ etc.).[/quote]

Why we only need to check two cases $a\ge b\ge c$, or $a\ge c\ge b$?[/quote]

Actually, I said "need to check several cases ... etc."

[quote=SBM][quote=dragonheart6]
Actually, I said "need to check several cases ... etc."[/quote]
Ok, so we need to check how many cases? What's it?
P/s: Sorry for my bad English![/quote]

Actually, it is not a short solution.

By letting $x = \frac{a}{c}, y = \frac{b}{c}$ with $a, b\ge 0$ and $c > 0$, the inequality is written as (after clearing the denominators)
$$f(a, b, c) = 3 a^5 c-5 a^4 c^2+5 a^3 b c^2+5 a^2 b^4-5 a^2 b^3 c-5 a^2 b c^3+3 b^5 c-5 b^4 c^2+4 c^6 \ge 0.$$

We need to check six cases.

1) $a \ge b \ge c$: Let $b = c + s, a = c + s + t$ for $s, t\ge 0$. We have
$f(c+s+t, c+s, c)$ is a polynomial in $c, s, t$ with non-negative coefficients. The inequality is true.

2) $b \ge a \ge c$: Similar to 1).

3) $c \ge a \ge b$: Similar to 1).

4) $c \ge b \ge a$: Similar to 1).

5) $a \ge c \ge b$: Let $c = b + s, a = b + s + t$ for $s, t \ge 0$. We have
$f(b+s+t, b, b+s) = q_4 b^4 + q_3b^3 + q_2b^2 + q_1b + q_0 \ge 0$
where
\begin{align}
q_4 &= 15 s^2-15 s t+10 t^2, \\
q_3 &= 35 s^3-30 s^2 t+25 s t^2+15 t^3, \\
q_2 &= 30 s^4-30 s^3 t+30 s^2 t^2+40 s t^3+10 t^4, \\
q_1 &= 12 s^5-20 s^4 t+10 s^3 t^2+35 s^2 t^3+20 s t^4+3 t^5, \\
q_0 &= 2 s^6-5 s^5 t+10 s^3 t^3+10 s^2 t^4+3 s t^5.
\end{align}
It is easy to prove that $q_4, q_3, q_2, q_1, q_0 \ge 0$.
For example, $q_0 \ge 2 s^6-5 s^5 t+10 s^3 t^3
= s^6 + s^6 + 10s^3t^3 - 5s^5t \ge 3\sqrt[3]{s^6 \cdot s^6 \cdot 10s^3t^3} - 5s^5t
= 3\sqrt[3]{10} s^5 t - 5s^5 t \ge 0$.

6) $b \ge c \ge a$: Similar to 5).

[quote=SBM][quote=dragonheart6]
Actually, I said "need to check several cases ... etc."[/quote]
Ok, so we need to check how many cases? What's it?
P/s: Sorry for my bad English![/quote]

By the way, can you get SOS by using your argument?

I can also get an SOS (Sum of Squares):
$$\mathrm{LHS} - \mathrm{RHS}  = \frac{1}{3}x(x-1)^2(3x-4)^2 + \frac{1}{3}y(y-1)^2(3y-4)^2 +
\frac{1}{6}z^T Q z
$$
where $z = [xy^2, xy, y, x, 1, y^2, x^2]^T$ and ($Q$ is positive semi-definite)
$$Q = \left(\begin{array}{ccccccc} 30 & -15 & -14 & -10 & 9 & 0 & 0\\ -15 & 24 & -20 & -23 & 5 & 14 & 15\\ -14 & -20 & 120 & -5 & -16 & -73 & 8\\ -10 & -23 & -5 & 116 & -16 & 11 & -73\\ 9 & 5 & -16 & -16 & 24 & -4 & -2\\ 0 & 14 & -73 & 11 & -4 & 54 & -2\\ 0 & 15 & 8 & -73 & -2 & -2 & 54 \end{array}\right).
$$

[quote=Hangdaj]That's good a real hit!
https://artofproblemsolving.com/community/c6h205593p2401384
I need to the hint, thank you, Mr. Dragon Heart[/quote]

There are some solutions there.

[quote=SBM][quote=dragonheart6][quote=Hangdaj]I just need to your solution![/quote]

I gave an SOS solution for that.[/quote]

Very nice! Very impressive! How did you get it?[/quote]

Did you try your argument?

[quote=dragonheart6][quote=SBM][quote=dragonheart6]

Did you try your argument?[/quote]

Argument ...?[/quote]

I mean your method to derive SOS. See
https://artofproblemsolving.com/community/forum-user-search/393259/6[/quote]

Oh, I was try to use [DRIVE!sos], but I did not geting the results :(


geometry
rectangle
MATHCOUNTS
number theory
greatest common divisor

Is there a formula for finding the number of squares the diagonal of a square passes through?

Also..which book out all the Introductory Set do you think will help the most for Mathcounts?

Yes there is...

For a rectangle with the sides $ a$ and $ b$ the formula is $ a\plus{}b \minus{} gcd(a,b)$

[quote="batteredbutnotdefeated"]Master AoPS volume 1. Do lots of problems. Make nats. Understand the enigma that is life.(the last one doesn't count)

I've got the first 2 parts down, only 1 more to go :lol:  :wink:[/quote]

Huh?

He mastered volume 1 and did lots of problems.
He only needs to make nats

[quote="How this relates to the topic"]
Squares the diagonal of a rectangle passes through..
Is there a formula?

--------------------------------------------------------------------------------

Is there a formula for finding the number of squares the diagonal of a square passes through?

[b]Also..which book out all the Introductory Set do you think will help the most for Mathcounts?[/b]

[/quote]

Intro NT or Intro CP, depending on your weakness.  Algebra and geometry are taught in school, so if you can buy only one, buy one that covers your weakest area.

[quote="SuperNerd123"]
For a rectangle with the sides $ a$ and $ b$ the formula is $ a \plus{} b \minus{} gcd(a,b)$[/quote]

But that doesnt work for squares, right?

Lets take a $ 8$ by $ 8$ square for example.......using your formula we get 8 squares, but the diagonal passes through 16 squares!

No, it passes through 8 squares. Draw it for a 3 x 3 square.

[quote="gm wrote"]
[b]that doesnt work for squares, right?[/b]

Lets take a  by  square for example.......using your formula we get 8 squares, but the diagonal passes through 16 squares!
[/quote]

All squares are rectangles but not all rectangles are squares

Passing through the corner of a square does not count as actually passing through the square.

Yeah and if the diagonal passed through 16 squares that would mean it passed through $ \frac14$ of the the figure

[quote="SuperNerd123"][quote="gm"]
[b]that doesnt work for squares, right?[/b]

Lets take a  by  square for example.......using your formula we get 8 squares, but the diagonal passes through 16 squares!
[/quote]

All squares are rectangles but not all rectangles are squares[/quote]

Of course, if the formula doesn't work for a square, it didn't work for some rectangle which means the formula would have been false.

[hide="proof of formula"]
The diagonal will pass through $ a$ squares going out the top of the square (assuming that $ a$ is the vertical dimension), and $ b$ squares going out the side of the square, giving us $ a\plus{}b$ squares.

Of course, that's probably not what the formula is.  There are times when the diagonal passes through a vertex of the square, in which case we've double counted it.  This happens $ \text{gcd}(a,b)$ times.  The rest follows.

If anyone wants to fix this up/make it more clear (trust me, there's a lot of room for that), go ahead.
[/hide]


The parabola $ 2y \equal{} 2x^2 \minus{} 5$ and the line $ 3x \plus{} 2y \equal{} 0$ intersect in two points. If the two intersection points are named $ (a,b)$ and $ (c,d)$ find the ordered pair $ (a \plus{} c, b \plus{} d)$, giving your answer in decimal form.

EDIT: $ 2y$

[hide="interesting problem names you have here"]
I'm assuming you mean $ 2y\equal{}2x^2\minus{}5$.
Substitute for 2y. $ 2x^2\minus{}5\plus{}3x\equal{}0$.
Simplify.

QED

[/hide]

[quote="CircleSquared"][hide="interesting problem names you have here"]
I'm assuming you mean $ 2y \equal{} 2x^2 \minus{} 5$.
Substitute for 2y. $ 2x^2 \minus{} 5 \plus{} 3x \equal{} 0$.
Simplify.

QED

[/hide][/quote]

Or the typo could have been for "hyperbola" rather than "parabola" ;)

Yes, I did mean $ 2y$ (sorry!).

I should have look for the similarities before diving into isolating $ y$ of each side and set both equations equal. Neat! thanks!

01 means problem #1
01 means contest #1 (there are 5 each year)
04 is just the year

easy reference for me I guess.


Let $ 0<x<y<z.$ Prove that $ |xz\minus{}y^2|<y(z\minus{}x).$

We have to prove that $ xy-yz<xz-y^2<yz-yx \Leftrightarrow y(x+y)<z(x+y) \  \text inforumla{and} \  x(z+y)<y(z+y)$, which is true.


[b]Problem.[/b] [i]There are $127$ countries on a planet such that each sends $4$ or $5$ of their leading mathematicians to an international meeting on algebraic topology.  Everyone shakes hands with all possible mathematicians from [u]other[/u] countries. If there are a total of $587$ participants at this meeting, what is the sum of the number of hands shook by each mathematician at this meeting.[/i]

[quote="E^(pi*i)=-1"][hide="This solution seems too easy."]
$127*4=508$, so $587-508=79$ of the countries sent $5$ mathematicians, and $127-79=48$ of the countries sent $4$ mathematicians.  Any mathematician from a country that sent $5$ mathematicians shakes everyone's hand except for the $5$ people from his own country - that is, $587-5=582$ hands.  Likewise, each mathematician from a country that sent $4$ mathematicians will shake $583$ hands.  Thus the total number of hands shaken is

$48*4*583+79*5*582=111936+229890=341826$.

I doubt that this kind of multiplication is necessary in the correct solution.
[/hide][/quote]

[hide="I think..."]
This is the best way to do it; it's not necessarily pretty but it's straightforward.  The only thing is that you counted each handshake twice, so the answer should be 341826/2 = 170913.

[/hide]

uhh...perhaps he had something else in mind? this is pretty trivial...and if you read the wording of the problem carefully, it is a bit unclear, but i don't think you divide by two...like each individual has a sum of handshakes, then add those

you do not divide by two.  read 'total number of hands shaken'.  divide by two to count number of handshakes.  multiply by two because IT TAKES TWO HANDS TO SHAKE HANDS!  nothing changes in the end.