trigonometry
parameterization
algebra
polynomial
function
inequalities proposed
inequalities

Let $\varphi = \frac{1+\sqrt{5}}{2}$ and  $f(x)= 1+\varphi \cos{x}+A\cos{2x}+B\cos{3x}\;$  with real parameters $A$ and $B$.
It is known that  $f(x) \ge 0$ for all  $x \in [0,\pi]$ . Find $A$ and $B.$

Here what I think for this one

$f(x)$  is even trigonometric polynomial so for any $x \in [-\pi;\pi]$ $f(x) \geq 0$

$f$ is $2\pi$ periodic function so $f(x)$  is positive for any $x\in \mathbb R$

Now a problem posted by harazi said  $f$ a trigonometric polynomial is postive on $\mathbb R$ then there exist $g$ such that $f(x)=|g(e^{ix})|^2$


geometry
parallelogram
geometry unsolved

Let $ABCD$ be a convex cuadrilateral, such that $AB=CD$, but it hasn't got any opposite parallel sides. Let $M$ and $N$ be the midpoints of $AD$ and $BC$. Prove that the angle between $MN$ and $AB$ and the angle between $MN$ and $CD$ are equal.

draw the parallelograms $AMB'B$ and $DMC'C$. then since $BB'$ is parallel and equal in length to $CC'$ (because they're both parallel and equal to $AM$ and $MD$), and since $BN = NC$, it follows that $B', N, C'$ are collinear and $B'N = NC'$. finally, since $MB' = MC'$, triangles $MB'N$ and $MC'N$ are then congruent, so $\angle B'MN = \angle C'MN$, and the result follows.


Find $ n\in Z$ for $ z\equal{}(1\plus{}i)^n\minus{}(1\minus{}i)^n$ is a complex number.

A) $ n\equal{}2k, k \in Z$

B) $ n\equal{}4k, k \in Z$

C) $ n \neq 4k, k \in Z$

D) $ n \in Z$

I'm pretty sure there is a missing keyword in the problem for otherwise the problem would be extremely trivial, according to definitions... :ninja:

[quote="thanhnam2902"]Find $ n\in Z$ for $ z \equal{} (1 \plus{} i)^n \minus{} (1 \minus{} i)^n$ is a complex number.

A) $ n \equal{} 2k, k \in Z$

B) $ n \equal{} 4k, k \in Z$

C) $ n \neq 4k, k \in Z$

D) $ n \in Z$[/quote]

if $ n\equal{}4k , k \in Z$ then $ (1\minus{}i)^{n} \minus{}(1\minus{}i)^{n}$ is not a cmplex number. so $ n \neq 4k ,k\in Z$ or $ \boxed C$

[quote="enndb0x"][quote="thanhnam2902"]Find $ n\in Z$ for $ z \equal{} (1 \plus{} i)^n \minus{} (1 \minus{} i)^n$ is a complex number.

A) $ n \equal{} 2k, k \in Z$

B) $ n \equal{} 4k, k \in Z$

C) $ n \neq 4k, k \in Z$

D) $ n \in Z$[/quote]

if $ n \equal{} 4k , k \in Z$ then $ (1 \minus{} i)^{n} \minus{} (1 \minus{} i)^{n}$ is not a cmplex number. so $ n \neq 4k ,k\in Z$ or $ \boxed C$[/quote]How is that not a complex number? If it is according to this 'definition', the fundamental theorem of algebra would be false...

Sounds like there may be some language barriers here.

In the American(/English?) math system:
A real number is your standard number line, with $ 0$, $ \minus{}1$, $ \pi$, and everything else.
A complex number is any number of the form $ a\plus{}bi$ for $ a,b$ real numbers. This [i]includes[/i] real and imaginary numbers.

10000th user's remark is that $ (1\plus{}i)^n \minus{} (1\minus{}i)^n$ is always complex, which is true. It might be real or it might not be, but that doesn't change whether it is complex or not.

Maybe the problem meant "not real", but that doesn't make a lot of sense.

Question can be fixed: Find $ n\in Z$ when $ z$ is a non-real complex number.

Can't we just substitute 1 for n and 4 for n? This is clearly enough to weed out the other answers, leaving C. :maybe:


could use some help with these... :oops:

1)

set up the letters A,F,K,N,O,T alphabetically and for every combination set a nummber it represents

a) what nummber is ANKFOT ?
b) what does nr 711 spell?

afterwards ill come with more

I'm assuming that you mean the first permutation alphabetically is 1, the second is 2, and so on:

[hide]

a) The first permutation is AFKNOT. There are 24 words where F is the second letter and 24 where is is the third letter (ahead of ANKFOT, i.e. starting with A).  ANKFOT is the 7th with A first and F fourth.  So it is number 55.

b) There are 720 total permutations.  So 711 very near the end.  Clearly it must start with TO.  There are 6 permutations of the last four with N first and 6 with K, so K is next.  It is the third of these 6, so F is next (the middle letter).  It is the first of these , so A is next and then N.  The word is TOKFAN.

[/hide]


logarithms
calculus
calculus computations

compute this limit:
$ u_n\equal{} (n\plus{}1)^{1\plus{}\frac{1}{n\plus{}1}}\minus{} n^{1\plus{}\frac{1}{n}}$
Here's what i did:
$ n^{1\plus{}\frac{1}{n}}\equal{}n.e^{\frac{ln(n)}{n}}$
$ \equal{}n.(1\plus{}\frac{ln(n)}{n}\plus{}o(\frac{ln(n)}{n}))$
$ \equal{}n\plus{}ln(n)\plus{}no(\frac{ln(n)}{n}))$
de meme
$ (n\plus{}1)^{1\plus{}\frac{1}{n\plus{}1}}\equal{}n\plus{}1 \plus{} ln(n\plus{}1) \plus{}(n\plus{}1)o(\frac{ln(n\plus{}1)}{n\plus{}1}))$
The difference gives:
$ u_n\equal{}1 \plus{} (n\plus{}1).o(\frac{ln(n\plus{}1)}{n\plus{}1}))  \minus{} n.o(\frac{ln(n)}{n}))
     \equal{}1 \plus{} no(1) \plus{} o(1)$

wich dont give the limit.
Thanks.
NB:The limit is 1

We can use that $ e^x\equal{}1\plus{}x\plus{}O(x^2)$ as $ x\to 0.$ We don't need to limit ourselves to saying $ o(x)$ when we have $ O(x^2).$

Note also that $ \frac{\ln(n\plus{}1)}{n\plus{}1}$ is asymptotically the same as $ \frac{\ln n}{n}.$ Lump those together in your $ O$ calculations.

$ (n\plus{}1)\left(1\plus{}\frac{\ln(n\plus{}1)}{n\plus{}1}\plus{}O\left(\frac{\ln^2n}{n^2}\right)\right)\minus{}n\left(1\plus{}\frac{\ln n}{n}\plus{}O\left(\frac{\ln^2n}{n^2}\right)\right)$

$ \equal{}1\plus{}\ln(n\plus{}1)\minus{}\ln n\plus{}O\left(\frac{\ln^2n}{n}\right)$

Since $ \ln(n\plus{}1)\minus{}\ln n\equal{}\ln\left(1\plus{}\frac1n\right)\equal{}O\left(\frac1n\right),$ we have everything except the $ 1$ tending to zero and the limit is $ 1.$


The natural numbers x,y satisfy 2x^2 + x = 3y^2 + y.
Prove that x-y, 2x+2y+1 and 3x+3y+1 are squares !

2xx+x=3yy+y
x-y+2xx-2yy=yy
(x-y)(2x+2y+1)=yy
(x-y, 2x+2y+1) =1
Therefore x-y, 2x+2y+1 are perfect squares.
x-y+3xx-3yy=xx
(x-y)(3x+3y+1)=xx
Therefore 3x+3y+1 is a perfect square too.


Two players [b]A [/b]and [b]B[/b] are playing the final of a championship which consists of a series of matches.  Probability of [b]A[/b] winning the match is [b]2/3 [/b]and that of [b]B[/b] is [b]1/3[/b].  The winner will be the one who is ahead by [b]2 [/b]games as compared to the other player and wins at least [b]6[/b] games.  It is given that [b]B[/b] has won the first four games, find the probability that [b]A [/b]wins the championship.

This is a little messy and involved.  Also, most of the arithmetic is omitted.

Let us denote by $p_{1}$ the probability that $A$ wins if $A$ is ahead by 1 and can win on her next turn (so, has at least 5 wins); by $p_{0}$ the probability that $A$ wins if $A$ and $B$ are tied (and each has at least 4 wins, so can win with 2 more wins), and by $p_{-1}$ the probability that $A$ wins if $B$ is ahead by 1 and $B$ can win on his next turn (so, has at least 5 wins).  Then we have $p_{1}= \frac{2}{3}+\frac{1}{3}p_{0}$, $p_{0}= \frac{2}{3}p_{-1}+\frac{1}{3}p_{1}$ and $p_{-1}= \frac{2}{3}p_{0}$.  We can solve these equations to get $p_{0}= \frac{4}{5}$, $p_{-1}= \frac{8}{15}$ and $p_{1}= \frac{14}{15}$.

Now, let $p(m, n)$ be the probability that $A$ wins given that the current state of the game is that $A$ has $m$ wins and $B$ has $n$.  We wish to compute $p(0, 4)$.  Note that $p(m, n) = \frac{2}{3}p(m+1, n)+\frac{1}{3}p(m, n+1)$ and that $p(k, k) = p_{0}$, $p(k+1, k) = p_{1}$ and $p(k, k+1) = p_{-1}$ for $k \geq 4$.  This allows us to work backwards to find that the answer is $p(0, 4) = \frac{1088}{3645}$.

Incidentally, starting the game by giving $B$ a head-start of 3 makes it very close to even (slightly to $A$'s advantage).


limit
logarithms
real analysis
real analysis unsolved

[color=darkred]Prove that for any $ n\in N^{*}$ the equation $ x^{n}\plus{}nx\equal{}1$ has only one positive root $ x_{n}$ and $ x_{n}\searrow 0$, $ \lim_{n\to\infty}\left(1\plus{}x_{n}\right)^{n}\equal{}e$.[/color]

[quote="Virgil Nicula"][color=darkred]Prove that for any $ n\in N^{*}$ the equation $ x^{n}\plus{}nx \equal{} 1$ has only one positive root $ x_{n}$ and $ x_{n}\searrow 0$, $ \lim_{n\to\infty}\left(1\plus{}x_{n}\right)^{n}\equal{} e$.[/color][/quote]
let $ n\to\plus{}oo$ therefore
we have $ x^{n}\plus{}nx \equal{} 1$ therefore we can easly show that $ x\to 0$.
ok then $ (1\minus{}nx)^{\frac{1}{n}}\equal{} x\to 0$ therefore we can say that
$ x$=$ \frac{1}{n}\plus{}\alpha_{n}$ where $ (\alpha_{n})^{\frac{1}{n}}\equal{}o(\frac{1}{n})$
therefore
$ (1\plus{}x)^{n}\equal{} (1\plus{}\frac{1}{n}\plus{}o(\frac{1}{n}))^{n}\equal{} e^{n(\ln (1\plus{}\frac{1}{n}\plus{}o(\frac{1}{n})))}\equal{} e^{n(\frac{1}{n}\plus{}o(\frac{1}{n}))}\equal{} e*e^{o(1)}\to e$ when $ n\to\plus{}oo$

[quote="Virgil Nicula"][color=darkred]Prove that for any $ n\in N^{*}$ the equation $ x^{n}\plus{}nx \equal{} 1$ has only one positive root $ x_{n}$ and $ x_{n}\searrow 0$, $ \lim_{n\to\infty}\left(1\plus{}x_{n}\right)^{n}\equal{} e$.[/color][/quote]

You also posted here (pb 7)
http://www.mathlinks.ro/viewtopic.php?t=70659


How many distinct values are possible for the expression $ 8 \plus{} 4 \times 6 \div 2$ if one set of parentheses may be placed in the expression?

Well first, let's see how many ways we can place the parentheses and THEN solve each one.
$ (8 \plus{} 4)x6/div2$
$ 8 \plus{} (4x6)/div2$
$ 8 \plus{} 4x(6/div2)$
$ (8 \plus{} 4x6)/div2$
$ 8 \plus{} (4x6/div2)$
$ (8 \plus{} 4x6/div2)$

So now we see that there are 6 ways that the parentheses can be put. Now let's solve them to find each distinct answer.

$ (8 \plus{} 4)x6/div2$ Solve in parentheses first
$ 12x6/div2$ Now solve from left to right
$ 72/div2$ Divide
$ 36$

Now we have ONE distinct answer, 36. Now let's solve the second one.

$ 8 \plus{} (4x6)/div2$ Solve in parentheses first
$ 8 \plus{} 24/div2$ Divide according to order of operations
$ 8 \plus{} 12$ Add
$ 20$

Now we have a second distinct answer, 20. Currently we have two different answers. Now to solve the third one.

$ 8 \plus{} 4x(6/div2)$ Solve in parentheses first
$ 8 \plus{} 4x3$ Multiply
$ 8 \plus{} 12$ Add
$ 20$

Now we have a third answer... which apparently is NOT distinct, it's a repeat from the last equation; so we still have two different answers.

$ (8 \plus{} 4x6)/div2$ Solve in parentheses first
$ (8 \plus{} 24)/div2$ Add
$ 32/div2$ Divide
$ 16$

Now we have third distinct answer, 16. Now for the fourth one (currently we have 3 different answers).

$ 8 \plus{} (4x6/div2)$ Solve in parentheses first
$ 8 \plus{} (24/div2)$ Divide
$ 8 \plus{} (12)$ Add
$ 20$

Now we see ANOTHER repeat, 20.

$ (8 \plus{} 4x6/div2)$ Solve in parentheses first; multiply
$ (8 \plus{} 24/div2)$ Divide
$ (8 \plus{} 12)/div$ Add
$ 20$

We see 20 again. We have four distinct solutions, therefore the answer is $ 4$

There are $6$ ways to place parenthesis around $2$ or more terms in the expression.

$(8+4\times 6\div 2)=20$

$ (8+4\times 6)\div 2=10$

$ 8+(4\times 6\div 2)=20$

$(8+4)\times 6\div 2=36$

$ 8+(4\times 6)\div 2=20$

$ 8+4\times (6\div 2)=20$

Thus there are only $\boxed {3}$ distinct solutions.


integration
limit
calculus
derivative
calculus computations

$f$ continous on $\mathbb{R}$ and $k \in \mathbb{R}$ such as
$\int_{0}^{x}f(t) dt = \frac{x}{2} ( f(x)+k ) $ for every $x \in \mathbb{R}$
how much is $f(0)$ ?

Divide both sides by $x$ and let $x\to 0$: you'll get $f(0)=\frac12(f(0)+k)$.

How do you show
that $\int_{0}^{x}\frac{1}{x}f(t) dt=f(0)$?

He just skipped a couple steps.  Use L'hopital's rule:

[hide]$\int_{0}^{x}f(t) dt = \frac{x}{2}[f(x)+k )]$
$\frac{\int_{0}^{x}f(t) dt}{x}= \frac{1}{2}[f(x)+k]$; Divide by $x$
$\lim_{x \rightarrow 0}\frac{\int_{0}^{x}f(t) dt}{x}=\lim_{x \rightarrow 0}\frac{1}{2}[f(x)+k]$; Take the limit of both sides as $x \rightarrow 0$.  On the left we have $\frac{0}{0}$ so we may proceed to take a derivative of both the top and the bottom by L'hopital's rule.
$\lim_{x \rightarrow 0}\frac{f(x)}{1}=\frac{1}{2}[f(0)+k]$
$f(0)=\frac{1}{2}[f(0)+k]$
$2f(0)=f(0)+k$
$\boxed{f(0)=k}$[/hide]

skimnc - quoting L'Hôpital is circular reasoning. To use the rule requires one to know the derivative - but notice that if $F(0)=0,$ then $\lim_{x\to 0}\frac{F(x)}x$ is the very definition of $F'(0).$

In this case $F(x)=\int_{0}^{x}f(t)\,dt.$ The reason we know its derivative at zero is the Fundamental Theorem of Calculus. The hypothesis "$f$ continous on $\mathbb{R}$" - excuse me, fixing the spelling, [b]continuous[/b] on $\mathbb{R}$ - is there because it's needed for the FTC. But that's also a little stronger than you need - you only need assume that $f$ is continuous at $0$ and integrable on an interval containing zero.

Well yes the problem stated it was continuous at zero so I assumed he knew that.  If it weren't then the right side would not be valid either.  I was just branching off what mlok had done.


In the link CONTEST we can find only the current round of the contest 7. Where I can find the others contest (the problems).

I'm slowly rebuilding the contest page. However it will take some time for that to happen.


Putnam
function
number theory
greatest common divisor
search
calculus
calculus computations

Find all continuously differentiable real  functions f such that for every rational number q, the number
f(q) is rational and has the same denominator as q.

(The denominator of a rational number q is the unique
positive integer b such that q = a/b for some integer a
with gcd(a, b) = 1.) (Note: gcd means greatest common
divisor.)

This was B5, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244054]posted here[/url].

oh!  thanks  , din knew the location before.

can u give me the links of other 2008 putnam q too ?

Just search for Putnam 2008. All 12 problems were posted by Dr. Merryfield, with proper naming "Putnam 2008 problem XY".

http://www.mathlinks.ro/search.php is your friend. :lol:


In short: Positron KE=$2.000$ MeV collides with electron at rest. The two are annihilated. Two photons produced. One in direction of positron, other in opposite direction. Find energies of photons.

My solution: Conservation of Energy yields $E_{0e^{+}}+KE_{e^{+}}+E_{0e^{-}}=E_{p_{1}}+E_{p_{2}}$ The entire left side is known and the energies of the photons on the right are solely kinetic (since they are photons).

Now conservation of momentum yields $p_{e^{+}}=p_{p_{1}}+p_{p_{2}}$. Here is where it is shaky. I said that because we knew KE of all three of those, we can say momentum is $\frac{KE}{c}$ (where we have our KE in MeV), so $\frac{KE_{e^{+}}}{c}=\frac{KE_{p_{1}}}{c}-\frac{KE_{p_{2}}}{c}$. Should I use a relativistic conversion or something? My final answers are $0.511 ~MeV$ for the one travelling opposite direction and $2.511 ~MeV$ for the one travelling in the same direction.

well, here's what i would do to get rid of your admittedly wishy-washy step:

you know the momentum of the positron because you can calculate velocity from the relativistic expression for kinetic energy ($K = mc^{2}(\gamma-1)$) and plug into the relativistic expression for momentum ($p = \gamma mv$).  then, photons have no rest mass, so their momentum is just $\frac{E}{c}$.  so we have 2 equations, 2 unknowns like before, so you can solve.

idk if that changes anything, compute it yourself.

Thanks, that is actually how I did the problem the first time. It does give a different answer. So this raises a couple of questions. Why does the first way not work? Guess: It doesn't take into consideration rest energy/mass. (Meaning for a particle with rest mass $p\neq \frac{KE}{c}$).

Question 2: So let's take into consideration all the energy. We have $E^{2}=(mc^{2})^{2}+p^{2}c^{2}$ and so momentum is $p=\sqrt{\frac{E^{2}}{c^{2}}-m^{2}c^{2}}$ of which all the values are known. This yields yet another answer. Of course, I am notoriously bad at typing things into a calculator  :oops: so that could be the problem.

your guess as to why your first way doesn't work is correct.  that can be easily verified by comparing your faulty expression $p = \frac{E}{c}$ to the general equation $E^{2}= (pc)^{2}+(mc^{2})^{2}$.

your second way ought to work, but it looks like it would just generate irritating algebra, since you'd have to substitute $E = mc^{2}+K$ and expand it, simplify, whatever.

Problems of this kind often lead to irritating equations... That's why 4-vectors are the way to go! :)

4-vectors are four dimensional vectors -- for example, $(ct, x, y, z)$, where $t$ is the time, $c$ the speed of light and $x, y, z$ are the coordinates. They make certain relativity problems much easier because they have some neat properties (their scalar product is invariant under Lorentz transformations). Try googling them for more information, I don't have the time to elaborate right now :maybe:

I've used four-vectors in relativity. Is it possible to apply this in Compton Effect/Pair-production type problems? I'd be really interested to see considering how much easier relativity becomes with them.

Of course you can. For example, the formula for the Compton effect can be derived in a few lines using 4-vectors :)


superior algebra
superior algebra unsolved

Let $ A$ be a ring such that $ 1 \plus{} 1 \neq 0$ and $ |U(A)| \equal{} n$. If the product of the elements of $ U\left(A\right)$ in some specific order is $ 1$ (e. g. if there exists an order in which the product of the elements of $ U\left(A\right)$ equals $ 1$), then prove that $ 4\mid n$.

Sorry, what does "in a specific order" mean?

The ring isn't supposed to be commutative, so "the product of the elements in U(A)" isn't welldefined without defining in which order the elements shall be multiplied.

A friend of mine got this one.

We prove the more general statement that if $ G$ is a finite group containing an element of order 2 such that the product of the elements of $ G$ in some order is $ 1$, then $ 4$ divides $ |G|$.  The result follows, since the conditions imply that $ \minus{}1$ is an element of order 2.

We assume for contradiction that $ |G|\equiv 2\ (mod\ 4)$. (clearly $ |G|$ is even)

Let $ G$ act on itself by left multiplication, then each $ g\in G$ can be written naturally as a permutation in $ S_n$.  Let $ \phi(g)$ be the sign of this permutation.  Let $ x$ be an element of order 2, and note that the permutation assigned to $ x$ is the product of $ |G|/2$ transpositions, so it must be odd, hence $ \phi(x)\equal{}\minus{}1$.  It follows that $ \phi(xy)\equal{}\minus{}\phi(y)$ for all $ y$, so that $ |\phi^{\minus{}1}(1)| \equal{} |\phi^{\minus{}1}(\minus{}1)|$.

Applying $ \phi$ to the equation $ x_1 x_2 ... x_n \equal{} 1$ gives $ (\minus{}1)^k \equal{} 1$, where $ k\equal{}|\phi^{\minus{}1}(\minus{}1)|\equal{}|G|/2$, hence $ |G|/2$ is even, so $ 4$ divides $ |G|$, contradiction.  We conclude that $ 4$ divides $ |G|$.


Got a question, just can't figure it. Fairly easy I'm sure but I'm just not understanding it.

A right Triangle the hypotenuse is 10 meters. One leg(base) is 2 meters longer than the other. What are the lengths of the legs? The drawing shows the hypotenuse as "10m", the base as "x + 2m" and the other leg as "x". I'm just not seeing the way of Factoring to figure out what the legs will be. Thanks for any help in advance!!!

huh? isnt this trivial? anyway, solving \[ x^2+(x+2)^2=100 \] gives (x+8)(x-6)=0 from which u get that one of the sides has length 6 and hence the other has 8. from now on, please post your questions in the appropriate forum. thanks


integration
logarithms
real analysis
real analysis solved

Evaluate the following definite integrals.

(1)$\int^e_1 5^{\ln x}dx$

(2)$\int^1_0 \frac{x+1}{(x^2+1)^2}dx$

for the first one it is easy to see that $5^{\ln x}=x^{\ln 5}$...

for the second one we have
$\displaystyle\int\frac{x+1}{(x^2+1)^2}dx=-\frac12\int\left(\frac1{x^2+1}\right)'dx+\text{arctan}x+\frac12\int x\left(\frac1{x^2+1}\right)'dx=-\frac12\frac1{1+x^2}+\text{arctan}x+\frac12\frac x{x^2+1}-\frac12\text{arctan}x+C=\frac{x-1}{2(1+x^2)}+\frac12\text{arctan}x+C$.
Now evaluate it in $0$ and $1$.


inequalities
induction
inequalities open

[quote="tranvinhphuctk14"]Disprove or Prove that:
If $a,b,c>0$ and $k \in N$ then:

$\large \sqrt[2k]{\frac{a}{b+c}}+\sqrt[2k]{\frac{b}{a+c}}+\sqrt[2k]{\frac{c}{a+b}}> 2.$[/quote]

More generally, for $k\in\mathbb{R},k\geq 2$ we have \[\sqrt[k]{\frac{a}{b+c}}+\sqrt[k]{\frac{b}{c+a}}+\sqrt[k]{\frac{c}{a+b}}>2.\]

do you mean $k\ge 2$ because the right side isn't satisfied when $k=1$

if you do, then the right side can be easily proven using induction

Right side easily proved with induction? The RHS is 2... do you mean the greater than side? Because this doesn't follow from induction either... we prove the opposite:
We want $\sqrt[k+1]{\frac{a}{b+c}}+\sqrt[k+1]{\frac{b}{c+a}}+\sqrt[k+1]{\frac{c}{a+b}}\geq \sqrt[k]{\frac{a}{b+c}}+\sqrt[k]{\frac{b}{c+a}}+\sqrt[k]{\frac{c}{a+b}}$
But we have the exact opposite is true because $\sqrt[k+1]a\leq \sqrt[k]a$

nvm. there used to be a two part inequality but he editted it out

oh what did it say?
(I notice Bodan has a knack for editting, look at his sig. then look at the 4 edits on a rather small post...)

you can have a look [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76277]here[/url]


hi,

we can prove using the following function f that the number of N is the same of the number of pair numbers  :D .

f:$ N\to P$
  $ n\to 2n$

P = pair numbers.

f is a bijetion .

don't you find that a little bit illogical :D?

Possibly.  That's why we have the notion of [url=http://en.wikipedia.org/wiki/Schnirelmann_density]Schnirelmann density[/url].


inequalities
calculus
inequalities solved

Show that any real values satisfying x+y+z=5 and xy+yz+zx=3
lies between -1 and 13/3.

First of all, x^2+y^2+z^2=19.

For the >=-1 part:

Let's assume x<-1 => y+z>6 => xy+xz=x(y+z)<-6 => yz>9 => y^2+z^2>18 => x^2<1 => false since x<-1, so |x|>1, so we have a contradiction.

For the <=13/3 part:

Let's assume x>13/3 => 0<y+z<2/3 => yz<1/9 => x(y+z)>26/9 => y+z>26/9x => 5>x+26/9x => 9x^2-45x+26<0 => 9(x-13/3)(x-2/3)<0 => x is in (2/3,13/3), contradiction because x>13/3.

We can find examples of values -1 and 13/3:

the triplets (x,y,z) = (-1,3,3) and (13/3,1/3,1/3) are both solutions.

I hope it's correct (I might have made some mistakes on the way).

Here is my solution
We know that (x+y) 2 \geq 4xy
But x+y=5-z and 4xy=4z(5-z) and after some more calculs we have -3z 2 +10z+13 \geq 0
  That means that -1 \leq z \leq 13/3
  Idem for x,y

  Hope my solution is corect! :)  :)  :)


Square $ ABCD$ has side length 2. A semicircle with diameter $ AB$ is constructed inside the square, and the tangent to semicircle from $ C$ intersects side $ AD$ at $ E$. What is the length of $ CE$?

[img]http://mathprob.com/_private/get.php/c6c5992d7d4c0d51827a56e3dff9f310/da_square.PNG[/img]

[hide]From graphing I got $ CE \equal{} \frac{5}{2}$[/hide]

[hide]
Let DE = x.  Then $ EA \equal{} 2\minus{}x$.  Next, let T be the point of tangency between the semicircle and segment CE.  Then $ CE^2 \equal{} DE^2 \plus{} DC^2 \equal{} x^2 \plus{} 4 \equal{} (CT \plus{} TE)^2$.  But CT = CB = 2, and TE = EA = 2-x, so

$ (2\plus{}2\minus{}x)^2 \equal{} x^2 \plus{} 4$,

from which it easily follows that x = 3/2, and CE = 5/2.
[/hide]


Let $ x,y,z,$ be the natural numbers satisfying equation $ x^2 \plus{} y^2\equal{}z^2$

Show that for every x,y,z numbers, $ (z\minus{}x)(z\minus{}y)\equal{}2n^2$ where $ n$ is a natural number too.

[quote="cenx"]Let $ x,y,z,$ be the natural numbers satisfying equation $ x^2 \plus{} y^2 \equal{} z^2$

Show that for every x,y,z numbers, $ (z \minus{} x)(z \minus{} y) \equal{} 2n^2$ where $ n$ is a natural number too.[/quote]

$ (z \minus{} x)(z \minus{} y)\equal{}z^2\minus{}zx\minus{}zy\plus{}xy\equal{}\frac{x^2\plus{}y^2\plus{}z^2}{2}\minus{}zx\minus{}zy\plus{}xy\equal{}2 \left(\frac{x\plus{}y\minus{}z}{2}\right)^2$

It's easy to check by modulo 4 that if $ 2|z$ then $ 2|x$ and $ 2|y$ and if $ 2$ doesn't divide $ z$ then $ 2$ only divides one of $ x,y$.

[quote="cenx"]Let $ x,y,z,$ be the natural numbers satisfying equation $ x^2 \plus{} y^2 \equal{} z^2$

Show that for every x,y,z numbers, $ (z \minus{} x)(z \minus{} y) \equal{} 2n^2$ where $ n$ is a natural number too.[/quote]

Since general solution of $ x^2\plus{}y^2\equal{}z^2$ are $ (x,y,z)\equal{}(2abc,a^2c\minus{}b^2c,a^2c\plus{}b^2c)$ and $ (x,y,z)\equal{}(a^2c\minus{}b^2c,2abc,a^2c\plus{}b^2c)$, we get :

$ (z \minus{} x)(z \minus{} y) \equal{}(a^2c\plus{}b^2c\minus{}2abc)(a^2c\plus{}b^2c\minus{}a^2c\plus{}b^2c)$ $ \equal{}2b^2c^2(a\minus{}b)^2$

Q.E.D.


How would I write

{y|y is a former president}

in $\LaTeX$? Or on this forum?

Is this what you want?

$\{y|y \text{ is a former president}\}$

[code]\{y|y \text{ is a former president}\}[/code]

I think, it is better way to use "\mid" command for "|".
\[\{y\mid y \text{ is a former president}\}.\]

I did not know about \mid.  Thanks for sharing   :)


Prove that for $n\geq 2$, \[\underbrace{2^{2^{\cdots^{2}}}}_{n\text{ terms}}\equiv \underbrace{2^{2^{\cdots^{2}}}}_{n-1\text{ terms}}\; \pmod{n}.\]

Taking a closer look at my proof in http://www.mathlinks.ro/viewtopic.php?p=849418.

[quote=lazizbek42]Induction[/quote]

Yes, I solved it that way but the induction isn't easy.

For completeness' sake, I will simply copy my recent reply on [url]https://artofproblemsolving.com/community/c146h150506p849417[/url].

We will show that with $a_1 = 2, a_2 = 2^2, a_3 = 2^{2^{2}}, \ldots$, we have that $a_i \equiv a_j \pmod{n}$ for all $i,j > \log_2(n)$. First note that if $a_k$ is congruent to $a_{k+1} = 2^{a_k}$ modulo $n$ for some $k$, then $a_k \equiv {a_j} \pmod{n}$, for all $j > k$ by induction. This means that we are done as soon as we can find two consecutive terms which are congruent modulo $n$. Now assume by induction that $a_k \equiv a_{k+1} \pmod{n}$ for all $n < 2^k$, which is trivially true for $k = 1$. We will prove that $a_{k+1} \equiv a_{k+2} \pmod{n}$ holds for all $n < 2^{k+1}$. So assume $2^k \le n < 2^{k+1}$ and write $n = l2^m$ with $l$ odd. \\

First we will deal with the even case; assume that $m \ge 1$. We actually intend to show that already $a_k \equiv a_{k+1} \pmod{n}$ in this case. First it is clear that $a_i \equiv a_j \equiv 0 \pmod{2^m}$ for all $i,j \ge k \ge m$, so it is sufficient to show $a_k \equiv a_{k+1} \pmod{l}$. But since $l < 2^k$, this already follows from the induction hypothesis.

So now let's move on to the odd case and assume $m = 0$. Since $\varphi(n)$ is even, we obtain (either by what we just proved or the induction hypothesis) that $a_{k+1} \equiv a_k \pmod{\varphi(n)}$. But by Euler's Theorem this implies $a_{k+1} = 2^{a_{k}} \equiv 2^{a_{k+1}} = a_{k+2} \pmod{n}$.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $a_{ij}$ denote a real symetric matrix with coefficients in $[-1,1]$. Define the huge polynomial
   $p(x)=\sum_{A\in \{1,2,...,n\}}{X^{|A|}\cdot\prod_{i\in A}{\prod_{j\in A^{c}}{a_{ij}}}}$. Here $A^c$ is the complementary set of $A$ (the numbers that are not in $A$). Prove that this polynomial has all its roots on the unit circle.

No activity in linear algebra forum? Any idea on this one? The hint given was to consider the polynomial in $n$ variables $X_1,...,X_n$ obtained replacing $X^{|J|}$ with $\prod_{j\in J}{X_j}$, but even with this "hint" I have no idea.


calculus
integration
logarithms
real analysis
real analysis unsolved

Okay, so I am modeling a rocket, and for simplicity sake, let's say it is governed by

$ \frac {dv}{dt} \equal{} v$

So seperation of variables -> $ \frac {dv}{v} \equal{} dt.$

Take the definite integral of both sides,

$ \int^{v_f}_{v_i} \frac {dv}{v} \equal{} \int^{t}_{0} dt$

$ \ln|v_f| \minus{} \ln|v_i| \equal{} t$
$ \ln|\frac {v_f}{v_i}| \equal{} t$
$ |\frac {v_f}{v_i}| \equal{} e^t$

Okay, this is where I am getting confused.

So is it $ |v_f| \equal{} |v_i|e^t$ ?

But that doesn't really make sense since that would essentially be two equations:

$ v_f \equal{} v_i(e^t)$ and $ v_f \equal{} \minus{} v_i(e^t)$

Obviously it can't be both. Do I need some type of extra information to see which solution is to the original differential equation is the one that actually models my rocket? That doesn't make since to me since this is just a first order system, it should only require one initial condition. I graphed both, and it was clear that $ v_f \equal{} v_i(e^t)$ was the solution that made the most sense, but then that was like we never used absolute values. So when is it necessary to use the absolute values when integrating $ \frac {1}{x}$.

Thanks for the help!

I've never particularly liked the $ \ln |x|$ formula- it gives the false impression that you can cross the origin. How would I put it? The most general antiderivative of $ \frac1x$ is $ \ln(cx)$ for some nonzero constant $ c$. On the real line, we use a negative $ c$ if $ x$ is negative, and a positive $ c$ if $ x$ is positive; we can't have both, because $ \frac1x$ is not integrable on any interval containing the origin.
If $ x$ has units, we put the reciprocal of those units into $ c$; we can only take logarithms of unitless numbers.

How does that apply here? We want $ t\equal{}0$ when $ v\equal{}v_i$, so we use $ \ln\left(\frac{v}{v_i}\right)$ as the antiderivative. Solving $ \ln\left(\frac{v_f}{v_i}\right)\equal{}t$, we get $ v_f\equal{}v_i\cdot e^t$.

A short version of what I think:
$ \ln(x)$ is an antiderivative of $ \frac1x$ on $ (0,\infty)$.
$ \ln(\minus{}x)$ is an antiderivative of $ \frac1x$ on $ (\minus{}\infty,0)$.
$ \ln|x|$ is not an antiderivative of $ \frac1x$ on $ (\minus{}\infty,\infty)$.

That makes a lot of sense, thanks.

Would there ever be a case in some type of other problem that the final result would be $ y \equal{} \minus{} a(e^t)$ instead of $ y \equal{} a(e^t)$ ?

Yes- that goes with the initial condition $ y(0)\equal{}\minus{}a$.


What is the least common multiple of the first ten positive integers?

You can just do 1*5*7*8*9 because you need all prime numbers so 5 and 7, but 8 covers for 2 and 9 covers for 3. This gives you 2520.


$ \forall a,b,c>0$,distinct from $ 1$, such that $ a\plus{}b\plus{}c\equal{}1$,prove that:

$ \frac{8a^3}{b\plus{}c}\plus{} \frac{8b^3}{c\plus{}a}\plus{} \frac{8c^3}{a\plus{}b} \ge (1\minus{}a)^2\plus{}(1\minus{}b)^2\plus{}(1\minus{}c)^2$

We have

$ LHS \ge 4(a^2\plus{}b^2\plus{}c^2) \ge RHS$

It is done here.

[hide="@Obel1x"]
Te kom tregu qe esht veshtir me kriju mosbarazime t'zorshme  :P

[/hide]


geometry
geometric transformation
dilation

The hydrogenation of an alkyne $ R\minus{}C \equiv C\minus{}R'$ using lithium ($ Li$) and liquid ammonia ($ NH_3$) results in an alkene and the addition is anti-stereochemistry.

How can we explain this phenomenon?

That phenomenon is explained by the mechanism of the reaction. First, note that when an alkali metal is dissolved in anhydrous liquid ammonia, one obtains a deep blue solution (when dilute, but with a bronze appearence when we have concentrated solutions) that behaves much like a solution of sodium ions and solvated electrons - in fact, it is experimentally found that the electrons are delocalised all over the solution, as they would be in a metal. These solutions are very usefull reducing agents, and the reactions that use them are called [i]dissolving metal reductions[/i] - another very important reaction of this type is the Birch reduction.
In the mechanism that follows, I will represent the single electron transfers as if they occur from lithium atoms. In the first step, the lithium atom transfers one electron to one of the carbons of the triple bond, producing the radical-anion I. Next, a proton transfer from ammonia gives the vinylic radical II.
Note that, although not shown, the species I and II exist as an equilibrium between both possible conformations: the one shown and another with both R groups in the same side - this is similar to the cis-trans interconversion of oximes (RR'C=N-OH).
In the next step, a second electron transfer from a lithium atom gives the trans-vinylic anion III: now, it is this step that determines the anti stereochemistry of the reduction, as the trans anion is preferably formed in order to minimize steric repulsions between the R groups.
Finally, a second proton transfer from ammonia completes the mechanism to produce the final product, the trans-alkene IV.

Ah, that denotes the mechanism in a much clearer way than in the book thanks!


Three circles are arranged in a row so that each is tangent to the circles next to it.  The radii of the two cirlces at the two ends are 5 and 3.  What is the length of the line segment AB that passes through the center of each circle? (hint: endpoint A is on the larger circle and endpoint B is on smaller circle)

[hide] my guess was that it had to be 16 + something (16 is the sum of the diameters of the two outer circles) , but i can figure out what[/hide]

Your problem is ambiguous; there are no restrictions on the size of the middle circle. All you can say is that $ AB\ge16$.


Guess the game from the screenshot.

Try to guess the game from the screenshot (name & platform). The correct guesser posts a new screenshot afterwards.

Rules:
-No unreleased games
-If you get the image online, try to reupload the image onto something like http://www.imageshack.us (or just attach it here) with [b]a different filename[/b] so it isn't immediately obvious what the game is.
-If no one figures it out after three days, a new screenshot may be posted.
-Only one guess per post, and no double posting.

First one:
[img]http://img292.imageshack.us/img292/3677/snshotop1.png[/img]

actually it's from the gba but close enough  :lol:

your turn.

[quote="junggi"][img]http://me.yogurttree.com/junggi/asdgasdg.jpg[/img]
possibly the best game ever created for <insert game console here>.[/quote]
SSB:M

Someone else can post the pictar....

[quote="ragnarok23"][quote="junggi"][img]http://me.yogurttree.com/junggi/asdgasdg.jpg[/img]
[/quote]
 :omighty:  :omighty: :omighty: [/quote]
I second that.

To keep it going.
[img]http://img53.imageshack.us/img53/2601/screenshotofyu8.png[/img]

[quote]looks like pokemon[/quote]
Yeah, that's right.

Stepmania, PC (you only use C400?)

[img]http://i9.photobucket.com/albums/a72/randomdragoon/screenshotofavg.png[/img]

PRESS R OR Z TWICE TO DO A BARREL ROLL

and yeah, i only use c400

i should move up, but i'm too lazy to figure out how to use c mods other than 400 (yeah, that's right)

[quote="junggi"][quote="randomdragoon"]<image of starfox64 for nintendo64>[/quote]
or something like that

[img]http://me.yogurttree.com/junggi/asdgdsahah.jpg[/img][/quote]
it's Playstation! (my best guess lol)
okay it's DBZ rite?

here's a nice text-based game:
[img]http://img235.imageshack.us/img235/3488/nznnskldjflkdjlksfjcp4.png[/img]

It's not Star Fox 64. It's Star Fox for the SNES.

Also, don't post another screenshot if your answer is a total guess or obviously wrong.

[quote="ProtestanT"][quote="junggi"][quote="randomdragoon"]It's not Star Fox 64. It's Star Fox for the SNES.
[/quote]
doh how could I be so stupid, i mean just look at the graphics, ewwww[/quote]
junggi is my guess close?[/quote]
not at all, my friend, not at all.
If it was DBZ, they would be flying, firing a bunch of weird fireball thingies and have huge yellow hair.

[quote="randomdragoon"]I'm guessing it's some version of Tekken.[/quote]
yup, tekken 5.

[quote="ProtestanT"]
it's Playstation! (my best guess lol)
okay it's DBZ rite?

here's a nice text-based game:
[img]http://img235.imageshack.us/img235/3488/nznnskldjflkdjlksfjcp4.png[/img][/quote]
I believe this game is Nuke Zone, a browser-based game for the PC.
EDIT: Yeah, visited their website and wikipedia entry, definately is.

Next:
[img]http://i9.photobucket.com/albums/a72/randomdragoon/screenshotofavg2.jpg[/img]

[quote="ProtestanT"]i was wonderin how u figured it out...then i googled "command centre missile silo research lab shipyard airfield war factory barracks power plant advanced turret headquarter torpedo launcher sam site machinegun", and most of the top links were about nukezone... i think the SAM site gave it away[/quote]
Actually, I just went on wikipedia for a list of browser-based MMORPGs. It was sorted by category, so I could immediately skip all of the medieval/fantasy/etc. categories, and go straight to the proper category, which had about eight games in it. It didn't take much longer to realize that the wikipedia screenshot for Nuke Zone looked extremely similar.

Here's one while you wait:

It isn't actually me playing - its from a friend.

[img]http://www.garia.net/g6/anewdawn.jpg[/img]

sweet game really.

Eve Online, game for PC.

This one is a really fun game:
[img]http://img329.imageshack.us/img329/9396/screenshotoa3.jpg[/img]

kyy's picture looked like a tree trunk, and random's looked like crabs

Haha, search up some more ships from Eve.  That one isn't even the ugliest...Try "Eve Ships Mimitar," they have trash cans...fly swatters...whatnot.

[img]http://games.kikizo.com/media/halo2_june04/06c.jpg[/img]

the best game =)

HALO 2
http://img120.imageshack.us/my.php?image=picture52tj3.png

Its a great game, either this or original would be OK.[/code]


Could anyone give me any hints for this AOPS Vol 2 Logarithm Problem?

If log 36 = a and log 125 = b, express log(1/12) in terms of a and b.

[quote="Dr Sonnhard Graubner"]hello, what base has your logarithm?
Sonnhard.[/quote]
When there isn't a base written it is assumed to be in base 10  :wink:


ratio
real analysis
real analysis unsolved

Show that there is real a   s.t
                           1/3≤{a^n }≤2/3
holds for all positive integers n. (where {x} is fractional part of x)
to view all the Some problems from the Uzbekistan universities olympiads visit
http://www.mathlinks.ro/Forum/topic-81099.html
(alekk if you have solution , please send)

Use the golden ratio, or some number like that. What are they called again? [url=http://mathworld.wolfram.com/PisotNumber.html]Pisot Numbers.[/url]


abstract algebra
Ring Theory
superior algebra
superior algebra solved

Let $R$ be a P.I.D. and $F$ its field of fractions. Suppose $S$ is a ring with $R \subset S \subset F$.

(a) Show that all elements $\alpha \in S$ can be written as $a/b$, where $a,b \in R$ and $1/b \in S$.

(b) Show that $S$ is a P.I.D.

(c) Show that if $S$ is finitely generated as an $R$-module, then $S = R$.

Thanks!

Are you sure about (c)?  Consider $R = Z[\sqrt{-3}]$ and $S = Z\left[\frac{1+\sqrt{-3}}{2}\right]$.

Let $T=\{x\in R|\frac{1}{x}\in S\}$. Then merely by the ring axioms for $S$, $T$ is multiplicatively closed. Obivously, $T^{-1}R$ naturally injects in $S$.

(a) Let $\alpha\in S$. Then $\alpha=\frac{a}{b}$ for some coprime $a,b\in R$.
There exist $c,d\in R$ such that $ac+bd=1$. We have $\frac{ac+bd}b=c\alpha+d\in S$, so $\frac{1}{b}\in S$.
It is now clear that $S\simeq T^{-1}R$.
For this reason, (b) should now be purely theoretic: a localisation of a PID is PID.

(c) $R$ is noetherian. $S$ is finitely generated as an $R$ module, so it is a noetherian $R$-module. Assume it is different from $R$. Then it contains some $\frac{1}{b}$ for some $b\in R$ that is not a unit in this ring. Consider the sub $R$-modules $R/b^{n}$ for $n\in\mathbb N^{*}$. These form an increasing sequence of sub-modules that never becomes stationary.
----------------------------------------------------------------------------
rgep's example is not appropriate for this problem's context because $\mathbb Z[\sqrt{-3}]$ is not PID. I think it is not even factorial.


the vertices of a triangle are thepoints of intersection of the lines y=-x-1, x=2, and y=1/5x+13/5. find an equation of the circle passing through all three vertices

Hint: Since the AMC contest is done you can go look at the solution for problem number 21 and compare it to this one...

[quote="Leeyuhang"]when they give you an equation and they want you to find the area how DO you go about doing it?
i have no clue at all. one example is the contest #2 in the getting started section. how do you solve a problem like that?[/quote]

I presume you are talking about finding the area of a circle given its equation.. well, to find the area, you have to find the radius. You can rearrange the formula for a circle in such a way that tells you the radius.

have you taken algebra II leeyuhang? its in there about equations of circles

no, i don't but i'll look in my dad's collection of textbooks.
isn't algebra 2 also called intermediate algebra or something like that?

well basically the circle equation is of the form (x-h)^2+(y-i)^2=r^2

where (x,y) is a point on the circle, and (h,i) is the origin, and r is the radius. dont get mad if i messed that up, i think...im floating

origin as in 0,0 or the circle? does h and i stand for something?
floating?

yeah, as i said...im floating aka delirious


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let A and B be two subgroups of a group G. If |A|=p, a prime integer, then show that either A⋂B={e} or A⊆B.

thank you.... :)

just guess the amount of subgroups of A ;)

[hide="Umm.."]Let A and B be subsets of a group G, and let |A|=p where p is prime.
Suppose $ A \cap B \not= \{ e \}$. Since we know that $ A \cap B \not= \emptyset$ ,there is $ g \in {A \cap B}$ such that $ g \not= e$. Since $ o(g)=p$, so $ |<g>|=p$. That is $ < g > = A$. But we know that ${ <g> \subseteq B}$. Thus $ A \subseteq B$.[/hide]


inequalities
trigonometry
geometry
trig identities
Law of Sines
geometry proposed

[quote="Virgil Nicula"]$ \triangle\ ABC$ $ \implies$ $ \max\ \{b,c\} < \sqrt {R(a \plus{} 2m_a)}\ (*)\ ,$ where $ m_a$ is the length of the $ A$- median.
[b]Remark.[/b] For $ a \equal{} b$ and $ c\rightarrow 0$ , we"ll have $ m_a\rightarrow \frac b2$ , $ R\rightarrow\frac b2$ and $ (*)$ becomes "[b]equality[/b]" ![/quote]
[hide="Solution"]
Without loss of generality, allow us to say that $ b\ge c$.  Now, we see that $ m_a\plus{}\frac{a}{2}> b\implies 2m_a\plus{}a> 2b\implies \sqrt{2m_a\plus{}a}>\sqrt{2b}$.  The Law of Sines gives us that \[ R\equal{}\frac{b}{2\sin \angle ABC}\implies \sqrt{R(2m_a\plus{}a)}>\sqrt{2bR}\equal{}\sqrt{2b(\frac{b}{2\sin \angle ABC})}\equal{}\frac{b}{\sqrt{\sin \angle ABC}}>b\] and we are done. [/hide]


videos
function
algorithm
AMC
AIME
probability
expected value

I used to be a beast at minesweeper, but I"m out of practice now. What are your best times?

beginner:4, average around 9
intermediate: 36, average around 45
expert: 114, average around 140

not as good as you... but here are my best and average times:

Beginner: Best-6, Average-10
Intermediate: Best-40, Average-51
Expert: Best:152, Average-170

Beginner: 6
Intermediate: 42
Expert: 999  :P I'm too lazy to beat it

Those are my best times.

(Best times:)

Beginner: 7
Intermediate: ... 42?
Expert: 128

I stopped playing when I cramped my hand too much clicking.  :P

[quote="t0rajir0u"]

I stopped playing when I cramped my hand too much clicking.  :P[/quote]

Same. I used to be able to play for 2 hours straight, then I realized I had played for 2 hours, then I immediately began to feel the fatigue.

YES!  After about two months, I finally got a new Intermediate best time: 40 seconds!  :D

[quote="SplashD"]You know there's a lot of luck involved in minesweeper as well.[/quote]
my friend just clicked randomly and got 1 second twice in a row. :o

Expert always comes down to luck and I always get unlucky.

[quote="sq53e"]I play a lot of "customized" minesweeper games (30*24 with 200 mines). I did beat it once, with 432 seconds.[/quote]

Has anyone every beat customized with a 24*30 board and 667 mines?

[quote="SplashD"]Has anyone every beat customized with a 24*30 board and 667 mines?[/quote]

That's like impossible... unless you're really, really lucky

[quote="sq53e"]It's actually possible to beat it if you know the cheat codes.[/quote]

Cheat codes? What cheat codes?

For me:
Beginner: 2 seconds
Intermediate: 39 seconds
Expert: 89 seconds

[quote="louiethechewy123"][quote="moogra"]It's not really reviving when it's not even on the bottom half of the first page. Reduced my expert time to 229[/quote]

What? This is the 7th page...[/quote]

[url=http://www.artofproblemsolving.com/Forum/index.php?f=360]This page?[/url]

I wouldn't ever revive stuff on the 7th page, but as it was in the upper middle of the first page, it's okay.

I got 54 in intermediate today, 6 in beginner (average 15ish). I need to get better beginner average-wise.

1 second custom (snapshot is attached  :rotfl: )
78 seconds beginner

HOW can you do 10 clicks in one second?
 :o

This is what I did:

Custom --> Height 24 --> Width 30 --> Mines 10 --> OK

Click a random square. A long as you don't click a mine or number, pretty much the whole screen goes out. Repeat this process and eventually you will get a one click win!

I used to play all three levels on my old Windows XP computer but ever since I switched to my current computer two years ago, I’ve been playing only the Advanced level on the Windows Vista version.

[list][img]http://i638.photobucket.com/albums/uu109/susannah_dingley/Minesweeper.jpg[/img][/list]
So my best time at Advanced level on my new computer is 198 seconds – and I win between two and three out of every ten games. I’ve forgotten my best times on my old computer, but I most probably have solved Advanced Minesweeper in much less than 198 seconds before. I was younger and my reflexes were quicker then. :P

I beat the 26*30, 668 mines game.
The screen print is included.

What I find strange about the new WINDOWS XP's is that they always allow you to repeat the one you just did.  :huh:

I beat a 500-mine game once, but I tried ... let's see ... 109 times before completing it.
EDIT: It took me 2 hours. xD

I just got a new computer, so I don't remember my best times..... but since this computer is a touchpad and I haven't gotten around to getting a mouse, I probably won't be getting good times anytime soon. :P

I've played a bit more and have gotten some better times:
Flagging:
Beginner 1 (3bv = 1 lol)
Intermediate 25? was on old computer
Expert 88

Non-flagging
Beginner: 4.71 (3bv=14) (minesweeper clone)
Intermediate: 30.20 (3bv=65) (hi-games.net)
Expert: 126 (3bv = 174) (hi-games.net)

3bv/s:
Beginner: 3.28 (arbiter)
Intermediate: 2.15
Expert: something crappy

IOE:
Beginner: 1.00 (clone)
Intermediate: 0.905 (clone)
Expert: something crappy

My best times:
Beginner: 10
Intermediate: 34
Expert: 88

My average times:
Beginner and Intermediate: I haven't played that again recently...
Expert: 120

Expert: -52 minutes (end_time - start_time)
[negative time run]


How many ways can you take 4 numbers of the set {$ 1,2,3,4,5,6,7,8$} to make a sum of $ 18$?

This can also maybe be done by generating a fn  :maybe:  :)

[hide="Well..."] If the numbers [i]aren't[/i] distinct (and if order matters), we want the coefficient of $ x^{18}$ in $ (x \plus{} x^2 \plus{} ... \plus{} x^{8})^4$, to which we can apply the [url=http://en.wikipedia.org/wiki/Multinomial_theorem]Multinomial theorem[/url], although for this problem size it doesn't help much.  This approach becomes much more helpful if $ 4$ is replaced with a much larger number.  [/hide]

[quote="LightEntropy"]is there a way to do this using counting principles? i.e nCr? Sorry if this doesn't make sense, i'm new to competitive, harder, math and am trying to understand a faster way of doing it rather the slow way of trial and error.[/quote]

[hide="Yes."](You may want to read chapters 13 and 15 of [url=http://www.maths.mq.edu.au/~wchen/lndmfolder/lndm.html]this[/url] to understand what I am doing, although I try to explain the method decently.)

If the numbers don't need to be distinct and if order matters, this approach works:

Set it up like a linear equation:

$ a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \equal{} 18$

We're trying to find the number of solutions for that equation when $ a_1,a_2,a_3,a_4 \in \{1,2,3,4,5,6,7,8\}$.

The number of solutions of that equation when $ a_1,a_2,a_3,a_4 \in \{1,2,3,4,5...\}$ is the same as the number of solutions of $ b_1 \plus{} b_2 \plus{} b_3 \plus{} b_4 \equal{} 14$ when $ b_1,b_2,b_3,b_4 \in \{0,1,2,3,4,5...\}$ because $ (a_1 \minus{} 1) \plus{} (a_2 \minus{} 1) \plus{} (a_3 \minus{} 1) \plus{} (a_4 \minus{} 1) \equal{} 14$. Basically, we dropped them all down one and it still works the same way.

The number of solutions to $ a_1 \plus{} a_2 \plus{} a_3 \plus{} \cdots \plus{} a_k \equal{} n$ without any restrictions is $ {n \plus{} k \minus{} 1 \choose n}$. (That's $ _{n \plus{} k \minus{} 1}C_n$, but the other way, which is called a binomial coefficient, is more convenient).

So without restrictions, we have:

$ {14 \plus{} 4 \minus{} 1 \choose 14} \equal{} {17 \choose 14} \equal{} 680$.

Now we want to get rid of all of the cases where any $ b_i > 7$ (remember, we dropped everything down one in order to make the first number in those sets be zero, so we're now using 7 instead of 8). So for every $ i \equal{} 1,2,3,4$:

$ S_i \in \{(b_1,...b_4): b_i > 7\}$

And from the 680 solutions from before, we'll subtract $ |S_1 \cup S_2 \cup S_3 \cup S_ 4|$; that is, we're subtracting every case where $ b_i > 7$.

$ |S_1| \equal{} |S_2| \equal{} |S_3| \equal{} |S_4| \equal{} {14 \plus{} 4 \minus{} 8 \minus{} 1 \choose 14 \minus{} 8} \equal{} {9 \choose 6}$.

Note that this is equal to the number of solutions in $ (b_1 \plus{} 8) \plus{} b_2 \plus{} b_3 \plus{} b_4 \equal{} 14$ We got the 8 from the fact that the limit on the other set is 7, and then we add one.

There are also 4 sets that we counted equal to the binomial coefficient up above, so later, we'll multiply that by 4.

For every $ |S_n \cap S_m|$ when $ s,n \equal{} 1,2,3,4$, the number of solutions to

$ (b_1 \plus{} 8) \plus{} (b_2 \plus{} 8) \plus{} b_3 \plus{} b_4 \equal{} 14$
$ b_1 \plus{} b_2 \plus{} b_3 \plus{} b_4 \equal{} \minus{} 2$

is zero. Note that its corresponding binomial coefficient

{14+4-16-1 \choose 14-16} = {1 \choose -2}

is meaningless. So that leaves us with

$ |S_1 \cup S_2 \cup S_3 \cup S_ 4| \equal{} 4{9 \choose 6} \equal{} 336$

The solution is then

$ {17 \choose 14} \minus{} 4{9 \choose 6} \equal{} 680 \minus{} 336 \equal{} \boxed{344}$.

When you get used to it, it's a lot quicker than it seems at first. When you keep using this method, you'll soon see a pattern that'll be easily identifiable for you personally.[/hide]

[b]@ LordCarbo:[/b] [i]I think order doesn't matter ;) [/i]

[hide="Solution"]Let's subtract 1 from each of the numbers, so that we want to pick numbers from 0, 1, ... 7 that add to 14.
Note that these numbers sum to 7*8/2 = 28, so the other four numbers will add to 28-14 = 14 as well.

Thus, we want to place the numbers into the format _+_+_+_=_+_+_+_ where numbers on the LHS were chosen and numbers on the RHS were not chosen

We know that sets such that if a is in the subset then 7-a is also in the subset will automatically satisfy this. This way, if we mark the numbers from 0 to 7 which are chosen, they will be symmetric, and it suffices to chose two of the four numbers 0 1 2 3. There are 4C2 = 6 ways of doing this.

The only other possibility is if on one side we chose _XX_ and on the other side we chose X__X. There are two ways of picking the sides, so there are 2 ways here.

This gives a total of 8 ways.[/hide]

Sorry for the confusion guys, order does not matter and the terms must be distinct.


Sequence
algebra
Inequality
IMO Shortlist

Let $c_1, \ldots, c_n \in \mathbb{R}$ with $n \geq 2$ such that \[ 0 \leq \sum^n_{i=1} c_i \leq n. \] Show that we can find integers $k_1, \ldots, k_n$ such that \[ \sum^n_{i=1} k_i = 0 \] and \[ 1-n \leq c_i + n \cdot k_i \leq n \] for every $i = 1, \ldots, n.$

[hide="Another formulation:"]
Let $x_1, \ldots, x_n,$ with $n \geq 2$ be real numbers such that \[ |x_1 + \ldots + x_n| \leq n. \] Show that there exist integers $k_1, \ldots, k_n$ such that \[ |k_1 + \ldots + k_n| = 0. \] and \[ |x_i + 2 \cdot n \cdot k_i| \leq 2 \cdot n -1 \] for every $i = 1, \ldots, n.$ In order to prove this, denote $c_i = \frac{1+x_i}{2}$ for $i = 1, \ldots, n,$ etc.
[/hide]

Please be so kind and write a full solution.  :)

Exactly, let $\frac{-c_i}{n}=m_i+x_i,m_i=[\frac{-c_i}{n}],l_0=0;l_i=[1+\sum_{j=1}^i x_j],k_i=m_i+l_i-l_{i-1}$.
We have $\sum_j k_j=\sum_j m_i +1+[\sum_j x_j] \Longrightarrow \sum_j \frac{-c_j}{n}<\sum_i k_i \le 1+\sum_j \frac{-c_j}{n}.$
Therefore if $0<\sum_j c_j \le n$, then $\sum_j k_j=0.$
We have $\frac{-c_i}{n}-1<m_i\le \frac{-c_i}{n} \Longrightarrow -n<nm_i+c_i\le 0$.
It give $-n<nk_i +c_i\le n$.


I think I know what modulus (remainder) math is, but how can you use for problems?????  :?:

Modulos have applications in number theory.  For example, finding answers to problems such as
"What are the last 2 digits of $2^{1993}$?"  can be solved much more easily with modulos.  For a more detailed explanation, search AoPS for "Fermat's Little Theorem."

I'll do out an example of using modular arithmetic, other than the more common "what is the remainder of" some huge number type of problem.

Say you have several 13 cent stamps and several 17 cent stamps that total 206 cents.

You can write one equation 13x + 17y = 206

Using modular arithmetic, you can divide each term by 13 or 17 and leave only the remainder behind.  I will divide by 13 giving me a remainder equation of :

0x + 4y = 11 (mod 13) meaning those are the remainders when you divide by 13

so 4y = 11, or 4y = 11 + some number of 13s

4y = 11, 4y =11+13, 4y = 24(this gives a whole # for y)

if 4y = 24 then y = 6  if y = 6, then
13x + 17(6) = 206
13x + 102 = 206
13x = 104
x = 8

So you buy 8 of the 13 cent stamps and 6 of the 17 cent stamps.

Now sometimes there can be more than one answer to a problem like this.  For example, if the stamps add up to 664 cents, you have this equation:

13x + 17y = 664

Dividing by 13 gives you :

0x + 4y = 1 (mod 13)
4y = 1, or 4y = 14, or 4y = 27, or 4y = 40 (finally one with whole # solution)

4y = 40 so y = 10

13x + 17y = 664
13x + 17(10) =664
13x + 170 = 664
13x = 494
x = 38

So you can buy 38 of the 13 cent stamps and 10 of the 17 cent stamps.

However, since y = 10 (mod 13), y could also equal 10+13, or 10+13+13

If y = 23, then you can solve for x = 21
If y = 36, then you can solve for x = 4

This example had 3 solutions, so it would depend on the problem which was the one your looking for, or maybe the answer is that there is 3 solutions!!!

This is just one example of using modular arithmetic, read up on it for more :lol:

Great example!  Here is a more complex concept that requires mods:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79917

Yeah, that helped me out a lot.  I knew it had to do with remainder and such, but now I understand more about what it is and how it can be used...


geometry
analytic geometry
graphing lines
slope
combinatorics proposed
combinatorics

if n>=3 points are given in plane s.t. they all dont lie on a single line,then they determine atleast n-1 different slopes ,where equality is possible only if n is odd and n>=5 :welcome:

You can find this in the Chapter 10 of [u]Proofs from the book[/u] written by Martin Aigner and Gunter M Ziegler : [url=http://books.google.com/books?vid=ISBN3540404600&id=KvQr9l0wgf8C&pg=PA59&lpg=PA59&dq=the+slope+problem&sig=QGXdWe9ogcbJG294Zi19elJHKKU]The slope problem.[/url]


Does anyone know if I can get a few (3-5) sample WOOT problems? My dad is reluctant to spend $ \$$695 and wants to see if I am ready.

We haven't published answers -- you can ask the problems on the message board (or search for them, some have probably been discussed before).

Also, I think one of the wonderful things about taking an AoPS course is that you can register for the class, and if its too difficult, you can still quit before 2 weeks (I think) after the course's start.

So how much should I be able to score on AIME before WOOT? I want to make sure I take it at the right time.

Look here:
http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php
Specifically, this is the question to your answer:
"Students who are not capable of consistently scoring 5 or higher on the American Invitational Mathematics Exam should strongly consider waiting until they have more experience before joining WOOT."

Also see [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=257]here[/url]. You might also want to attend the math jam on September 8.

[quote="rrusczyk"]Take a look at the PDF on [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOTSchedule.php]the schedule page[/url].  You'll find some sample problems there.[/quote]

Just one question, those sample problems, are they easier, aligned or harder than the program itself?

They're fairly representative of the spread.  You can focus on one end or the other in the course of WOOT, depending on your interest and ability.

No, no, no I never considered taking WOOT :P

Just wanted to know at what difficulty is it worked there, gathering info on everything :)  :roll:


Um... lets say that there is a clock that when the hour hand makes one revolution around the center, the minute hand makes 16, and the second hand makes 36. Yah, wierd... so the question is not counting when they meet at the beginning and end, how many times do all 3 hands meet during one revolution of the hour hand?

Easier problems for people who couldn't solve the original question (these feature the same concept):

Question 1 wrote:It is 4'o clock now. When is the next time the hour hand and the minute hand meet?
Hint:[hide]Think about how many degrees/radian the hands move/hour[/hide]
If you apply it to different angle differences between the hands...
Question 2 wrote:The hour hand and the minute hand is together now. (The answer to Question 1) When will the hands make a right angle?

Remember that [hide]Question 2's answer must be later than Question 1's answer[/hide].

I think you'll get how to solve Chinaboy's question now.

Ok.

[hide]Hour hand - (moves 6 degrees every 12 minutes),(moves .5 degrees every 1 minute).

Minute hand - (moves 6 degrees every minute)

Between the minute hand and the hour hand, the amount of degrees is 120.  We want to know the time when the difference of degrees between the hour hand and the minute hand is zero.

Here is the equation...

120+.5m-6m=0.

Solving for m, we get m=21.8[/hide]

Are there any better ways to solve this?

I do it that way, too. The only difference is 120 + .5x = 6x which is the same thing as yours... but then there always might be easier shortcuts...

Barnacle, to help with Tare's problem here is a little more detail

[hide]
Since the hour hand moves 360 degrees in 12 hours, the hour hand moves 30 degrees per hour (30h) So for example at 1:00 the hands are 30(1)  degrees apart as the minute hand is back in it's original place.

Now, since the minute hand of a normal clock moves 360 in 60 minutes, it moves 6 degrees per minute (6m).  But since after 1:00 the minute hand is actuallu catching up to the hour hand(until it passes it), we can call that minus 6m.

Finally, each minuts the hour hand moves 30/60 = 1/2 a degree (further away from the minute hand...again until they have crossed) so call that plus .5m

We have the distance between the hands then as 30h - 6m + .5m = 30h - 5.5m

If you want to know when the hands are in the same place, set that equal to zero

30h - 5.5m = 0

Since this problem asked for the first time they cross after 4:00, substitute 4 for h giving you

30(4) - 5.5m = 0  or 120 - 5.5m = 0 which is what everyone else gave as the formula.

m = 21.8181...

Be careful when using the 30h - 5.5m formula as once the minute hand passes the hour hand you can get negative answers, so the answer is actually the absolute value of the formula.  Also for times like 11:02, you will get values well over 180 degrees, in these cases subtract your value from 360 to get the actual angel(non-reflex)

Hope that helps.  Try using this logic to do the original problem with the crazy clock :)
[/hide]


Can somebody give me any information about ARML 2009 for WA states, I'm in Camas.

Just saw this ...

Yes WA will have teams for 2009 ARML.  We are not one of those better organized states.  Typically this kicks off in Spring (March / April).  There seems to be higher interests now and hopefully there will be more teams from WA this year.  Note that new teams would have to start with Division B (currently our Gold team competes in Division A, and Red team competes in Division B).

I am also running the 2008-2009 ARML Power Contest, this power team is made up by a selected group of students from local schools (Bellevue, Redmond, Garfield, Lakeside, TJ, Interlake, Nathan Hale, etc.).  Due to the collaborative problem solving nature it is not feasible to includes students far away from puget sound area.  With November Round 1, we are currently tied for 4th place.  Round 2 is scheduled Feb 7.

More to come after Mu Alpha Theta State competition.

In order to get selected to the 2009 WA ARML team does one have to participate in the 2008-2009 power contest first?

I was also a little curious on the criteria for getting selected.. what counts?

No, but the opposite might hold some truth, let me explain.

Mr. Norris (of TJHS) runs WA ARML team, I organize the practices (including mini-ARML / mockup with Oregon team last year) and recommend team members.  I don't see any particular reason to restrict the number of teams, but each team should have 15 students (the relay format requires students break into 5 sub-teams of 3 each).  If we are considering more than two teams which is what WA has in the last couple of years, we need to get multiple of 15 committed students.  The cost is about $400 or so per student ($75 fee, plus air fair to/from Las Vegas, some meals, and misc expenses.).  Math enthusiastic students who are up to ARML level challenge are all encouraged to join, in my view.  In 2008 MIC and Fall Classic, some parents and coaches talked to me about ARML, I believe with the interest level we are likely to get more than 2 teams this year, but let me check with Mr. Norris first.  Please note that currently Gold team competes in Division A, Red team competes in Division B.  New team will start in Division B.

ARML Power Contest is a write-in 2 round contest (Nov and Feb).  Each round is 45min, questions are similar to ARML Contest Power Round.  I am the sponsor and organizer for WA ARML Power Contest (2007-2008, and 2008-2009).  Team size is smaller (around 10 or so), team members are typically among the top math students and contest winners in WA.  AMC-12, AIME scores are key indicators (e.g. if you are USAMO qualifier and you are local to Puget Sound area, you will be invited).  Local competitions are looked into as well (e.g. MIC, Mu Alpha Theta).  Typically Power team becomes the core of our Gold team.  Puget Sound is the primary target area due to practical consideration of team practices and taking the tests together and this area has the highest concentration of talents in WA.   If you are interested for next year's Power Contest, drop me a PM.

To answer your question: you don't have to be in ARML Power team to participate ARML.  If you are  contributing member of WA ARML Power Team, chances are you will be placed in WA Gold team.

Please check ARML site (http://www.arml.com) to learn more.

Thanks for the clarification and your insights, minghu.

I will send you a PM; check your inbox


In triangle $ABC$, point $D$ divides $BC$ in the ratio 1:2, and $F$ divides $AD$ in the ratio 1:2. Line $BF$ intersects line $AC$ in point $E$. What is the ratio of the area of figure $EFDC$ to the area of  triangle $ABF$?

[quote="lotrgreengrapes7926"]Is it $2AF=FD$ and $2BD=DC$? Or the other way around? :?[/quote]
Yes, it is $2AF = FD$ and $2BD = DC$. I am not sure how to include a diagram in PDF inline in a message. Can anybody say how to include a PDF diagram inline in a message? Is the only way to include a diagram is including it as an attachment?

[hide] :idea: ABC=BFD;  AEF=EFD
ABF+AEF= BFD+EFD  =>   ABE=BED and BED=EDC
=>  ABE=BED=EDC=1/3ABC
   Therefore :   ADC-AEB=FECD-ABF
=>   1/2ABC-1/3ABC=FED+BED-BFD
=>    1/6ABC=2FED
=>     1/12ABC=FED
    BFD=BED-EFD=1/3ABC-1/12ABC=1/4ABC
    EFDC=DEC+FED=1/3ABC+1/12ABC=5/12ABC
So that:   BFD : EFDC=(1/4ABC):(5/12ABC)=3/5[/hide]

[quote="TRAN THAI HUNG"] ABC=BFD;  AEF=EFD[/quote]
I think you mean $ABF$ for the first one. And I don't think this is right. I got $2ABF=BFD$ and $2AEF=EFD$.  :?


function
number theory unsolved
number theory

Find all functions  $ f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that \[ f(x\plus{}y) \plus{} f(xy) \equal{} f(x) \minus{} f(y) \plus{}1,\] for all $x,y \in \mathbb{Z}$.

there have to be some mistake here... I mean, plug in it $ x \equal{} y \equal{} 0$ and you get $ f(0\plus{}0)\plus{}f(0.0) \equal{} f(0)\minus{}f(0)\plus{}1 \equal{} 1$, that is $ 2f(0) \equal{} 1$ and $ f(0) \equal{} \dfrac{1}{2}$ which is not in $ \mathbb{Z}$, contradiction!.

[quote="Lousin Garckz"]there have to be some mistake here... I mean, plug in it $ x \equal{} y \equal{} 0$ and you get $ f(0 \plus{} 0) \plus{} f(0.0) \equal{} f(0) \minus{} f(0) \plus{} 1 \equal{} 1$, that is $ 2f(0) \equal{} 1$ and $ f(0) \equal{} \dfrac{1}{2}$ which is not in $ \mathbb{Z}$, contradiction!.[/quote]

Sorry, this the correct one
$ f(x\plus{}y) \plus{} f(xy) \equal{} f(x)f(y) \plus{}1$, $ f : \mathbb{Z} \rightarrow \mathbb{Z}$

[quote="rajelaz"] Sorry, this the correct one
$ f(x \plus{} y) \plus{} f(xy) \equal{} f(x)f(y) \plus{} 1$, $ f : \mathbb{Z} \rightarrow \mathbb{Z}$[/quote]

Let $ P(x,y)$ be the assertion $ f(x \plus{} y) \plus{} f(xy) \equal{} f(x)f(y) \plus{} 1$
Let $ f(1)\equal{}a$

If $ a\equal{}1$, $ P(x\minus{}1,1)$ $ \implies$ $ f(x)\plus{}f(x\minus{}1)\equal{}f(x\minus{}1)\plus{}1$ and so $ f(x)\equal{}1$ $ \forall x$

Consider now $ a\neq 1$
$ P(0,0)$ $ \implies$ $ (f(0)\minus{}1)^2\equal{}0$ and so $ f(0)\equal{}1$
$ P(\minus{}1,1)$ $ \implies$ $ (a\minus{}1)f(\minus{}1)\equal{}0$ and so $ f(\minus{}1)\equal{}0$
$ P(\minus{}2,1)$ $ \implies$ $ 1 \equal{} (1\minus{}a)f(\minus{}2)$ and so $ 1\minus{}a|1$ and so $ a\in\{0,2\}$

If $ a\equal{}2$ $ P(x,1)$ $ \implies$ $ f(x\plus{}1)\equal{}f(x)\plus{}1$ and we got the solution $ f(x)\equal{}x\plus{}1$ $ \forall x$

If $ a\equal{}0$ $ P(x,1)$ $ \implies$ $ f(x\plus{}1)\equal{}1\minus{}f(x)$ and so $ f(x\plus{}2)\equal{}f(x)$ and we got the solution $ f(x)\equal{}0$ if $ x$ odd and $ f(x)\equal{}1$ if $ x$ even.

So three solutions :

$ f(x)\equal{}x\plus{}1$ $ \forall x$
$ f(x)\equal{}1$ $ \forall x$
$ f(x)\equal{}0$ if $ x$ odd and $ f(x)\equal{}1$ if $ x$ even


factorial
real analysis
real analysis solved

i've been doin crap with factorials and the harmonic series and i've made a mistake that i cant figure out.  the end result is 1 = ( n! ) ^ 2, which is reminiscent of the 3=1 or 2=1 puzzles, so its like one of those puzzles, but i dont think it breaks a basic rule

ln(a/b) = ln(a) - ln(b)

also:

ln(a) + ln(b) = ab

(all rheiman sums from 1 to n)

[SUM] ln(1/ i) = ln(1/1) + ln(1/2) + ln(1/3) ... ln(1/n)

or

[SUM] ln( 1 / i ) = ln ( 1/1 * 1/2 * 1/3 ... 1/n )

[SUM] ln( 1 / i ) = ln ( 1/n! )

[SUM] ln( i ) = ln ( 1 * 2 * 3 ... n )
                   = ln( n! )

now the application (and my problem) :

ln( 1 / i ) = ln(1) - ln(i)

these are identical values just different representations, so:

[SUM] ln( 1 / i ) = [SUM] [ ln(1) - ln( i ) ]

[SUM] ln( 1 / i ) = [SUM] ln( i )          // ln(1) = 0

ln( 1 / i! ) = ln( i! )   // each raised to a power of e

1 / i! = i!  //which doesnt make any sense or even more nonsense

1 = ( i! )^2

i cant see what i did wrong, help

[quote="oooooh ya"][SUM] ln( 1 / i ) = [SUM] [ ln(1) - ln( i ) ]

[SUM] ln( 1 / i ) = [SUM] ln( i ) // ln(1) = 0
[/quote]
That transition is pure nonsense.
I don't even know what you mean by //, but this has to be nonsense.

[quote="oooooh ya"][SUM] ln( 1 / i ) = [SUM] [ ln(1) - ln( i ) ]
[/quote]
Correct
[quote]
[SUM] ln( 1 / i ) = [SUM] ln( i )          // ln(1) = 0
[/quote]
sign error, should be

[SUM] ln( 1 / i ) = [b]-[/b] [SUM] ln( i )


:?: find all polynomials with integer coeff.P,s.t.P(x)=2^n,has integer sol.for all naturals n.
any  :idea:

does 0 belong to natural interger?if so,I guess I've got a solution,if not
I can just solve the situation when the constant A0 is odd
define S(a,b)={x|P(x)=2^a(mod2^(a+1)),x belong to{1,2.....2^b}}   (a=<b)
T(a,b)={x|P(x)=0(mod2^(a+1)),x belong to{1,2.....2^b}}     (a=<b)
we can prove by inductive methord that |S(a,b)|=|T(a,b)|=2^(b-a) :huh:
so S(a,a)=T(a,a)=1

let Xa and Ya be the only element in S(a,a) and T(a,a),respectively
we can also prove that {Ya,Ya+2^(a+1)}={X(a+1),Y(a+1)} :huh:

let's suppose when x>t,P(x) is strickly increase,
consider an interger "r" s.t. the solution of P(x)=2^r is larger than t,
so when m>r,there is only one solution s.t.P(x)=2^s,
define the solution with Gm,then {Gm(m>r)}is strickly increase,
and Gm=Xm(mod2^(m+1))
from the fact that {Ya,Ya+2^(a+1)}={X(a+1),Y(a+1)}
it's easy to prove that there exsist infinite "c">m
s.t.X(c+1)>=X(c)+2^c :huh:

for each of the c,P(Gc)=2^c,P(Gc+2^c)=<P(G(c+1))=2^(c+1)
if deg(P(X))>=2,we can prove when c is large enough,
P(Gc+2^c)>=2^(c+1), :huh: which is a contrariety

as for deg(P(X))=1 or 0,it's easy :P
  that I didn't provethe sentenses end with :huh:.  but I think to prove them is not hard

:lol: ,nice approach
 0 isnt natural no.,but i think yours approach may work :D
 will tell if get :roll:


calculus
integration
trigonometry
logarithms
LaTeX
calculus computations

I know the solution from the back of the book but i cant seem to work it out right.

Int from pi to 0 : tan[x/2]

using the natural log to int.  The solution the book gives is ln2.  How do I work this out?
Thank you!!![/b]

$\int_0^{\pi}\tan\left(\frac x2\right)\,dx=\infty.$

What, exactly, is the problem you're looking at?

yeah thats pretty much it only its supposed to be

$ \int_0^{\frac{\pi}2}\tan\left(\frac x2\right)\,dx.$

[moderator edit]

from 0 to pi/2

I was working it out using

tan(u)du
and I wind up with
2ln sec|u|  evaluated from 0 to pi

or as an alternative i get

-ln cos|u| evauated from 0 to pi

I am sorry I'm new and I am not familiar with using the displaystyle

Let's just push the substitution through. You're clearly making the substitution

$u=\frac x2,\,x=2u,\,dx=2du.$

Just remember now that although $x$ goes from $0$ to $\frac{\pi}2,$ $u$ goes from $0$ to $\frac{\pi}4.$ That's $\frac{\pi}4,$ not $\pi.$ So we now have

$2\int_0^{\frac{\pi}4}\tan u\,du.$

Finish it from there.

Tell me: what is $\cos\left(\frac{\pi}4\right)$ ? The exact value, not the decimal. This is one of the values you must know.

Second issue: is there a way of rewriting $\ln\left(\sqrt{u}\right)$ so that it doesn't have any square root symbols? And can you simplify $\ln\left(\frac1u\right)$?

We just started going over natural logs in class so I'm a little shaky with some of the specifics of simpilfying and finding dy/dx and ints, but this is how I approached it.

cos(pi/4) =  :rt2: / 2

then: -2ln( :rt2: /2)
= -2ln( :rt2: ) - -ln(2)
= -2ln(2^(1/2)) - -ln2
= -ln2 - -ln2 ?
I must be doing something wrong.
As for the ln(1/u) cant that be written as
ln1-ln(u)
Is that what you meant?

[quote="illcrowflu"]That helps a lot. Thank You

I plug in  -2ln(cos(pi/4)) -  -2ln(cos(0))
and I get .693147... which is equal to ln2

But what I dont understand is why the book writes the solution as ln2
is there something that I'm missing.  Can that evaluation be simplified to avoid getting the decimal?

Thanks again!![/quote]Just enjoying this--
The right hand gives you zero--from the cos $\rightarrow$ 1; ln 0[b].[/b]
The left, a multiple of ln(cos(:pi:/4)) --by -2 -- can be rewritten with that [b]multiplier in the exponent[/b] ([b]inside the log[/b]).  ~Square 1/ (1/:sqrt:2)...  Tada.    But I'm not familiar to this forum, so I might well be clueless here.
        :downto:

P.S.  Yeah, I hope I wasn't giving anything too much away.

[quote]As for the ln(1/u) cant that be written as
ln1-ln(u) [/quote]
I'd never let a calculus student of mine get away with leaving $\ln1$ in an answer. Come on, now, $\ln1=0.$ What I wanted you to say was that $\ln\left(\frac1u\right)=\ln(u^{-1})=-\ln(u).$

$-2\ln\left(\frac{\sqrt{2}}2\right)=-2\ln\left(\frac1{\sqrt{2}}\right) = -2\ln(2^{-1/2}) = -2\left(-\frac12\right)\ln2=\ln2.$

fedwini, I'd edit your bad LaTeX code to make it readable if I could figure out what you were trying to say, but I can't. It looks like you're trying to use LaTeX interchangably with the emoticons just as a collection of symbols. That's not a good way of doing it. Try writing expressions entirely in LaTex. For instance, (dollar)\frac{\sqrt{2}}{2}(dollar) becomes $\frac{\sqrt{2}}{2}$.

Thanks a lot Kent you were a big help - I appreciate your time.
BTW I am new to the forum but I plan to keep posting as it has been so helpful.

For everyone else's sake how do I write the equations out like you were rather than using regular ascii characters?

[quote]For everyone else's sake how do I write the equations out like you were rather than using regular ascii characters?[/quote]
It's called LaTeX. There's some information posted over on the side, as one of the site guides, and there's also a LaTeX forum for questions and problem.

Go look at your own #3 post on this topic, which I edited, to get an idea.

A dollar sign puts you into LaTeX mode, and another dollar sign gets you out. Within that, a backslash introduces a named object, with the names mostly being mnomonic: \int for integral, \infty for infinity, \frac for fraction, and so on. ^ is for superscript, _ for subscript. Curly braces {} delineate and enclose the scope of each object. Thus \frac{x^2+1}{2x+e^{x-1}} becomes $\frac{x^2+1}{2x+e^{x-1}}.$

[quote="Kent Merryfield"][quote]$\ldots$
fedwini, I'd edit your bad LaTeX code to make it readable if I could figure out what you were trying to say, but I can't. It looks like you're trying to use LaTeX interchangably with the emoticons just as a collection of symbols. That's not a good way of doing it. Try writing expressions entirely in LaTex. For instance, (dollar)\frac{\sqrt{2}}{2}(dollar) becomes $\frac{\sqrt{2}}{2}$.[/quote][/quote]I've got that fixed now--and I do need to use better LaTeX.  There is one thing that I wanted to mention, however: (illcrowflu,) Hovering over any LaTeX picture shows the coding, but ~ If you cannot get that to work for you with long strings, quoting someone else's message will always show you what they wrote--though you'd intend to cancel out of the Reply.

[quote="jmerry"]Actually, selecting and copying that part of the post should work. If you paste that in, you'll get the code, without the dollar signs.[/quote]Good deal--I mean good to know.  :lol:


function
floor function
algebra proposed
algebra

Let $ [x]$ denote the integer part of $ x$, i.e., the greatest integer not exceeding $ x$. If $ n$ is a positive integer, express as a simple function of $ n$ the sum

$ \left[\frac{n\plus{}1}{2}\right] \plus{} \left[\frac{n\plus{}2}{4}\right]\plus{}...\plus{}\left[\frac{n\plus{}2^k}{2^{k\plus{}1}}\right]\plus{}...$

Big hint (try to solve it w/o):

[hide]$ \lfloor{x\plus{}\frac{1}{2}}\rfloor \equal{} \lfloor{2x}\rfloor\minus{}\lfloor{x}\rfloor$[/hide]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125439
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34668

There's also at least one other thread where this problem was posted in which I gave a slightly different solution.

I also think it would be very instructive, although overkill, to give a proof by [b]generating functions.[/b]


$ \begin {cases} x^2\plus{}y^2\equal{}z^2\\xz\equal{}y^2\\xy\equal{}10\end {cases}$

[quote="thanhnam2902"]$ \begin {cases} x^2 \plus{} y^2 \equal{} z^2 \\
xz \equal{} y^2 \\
xy \equal{} 10\end {cases}$[/quote]

See that $ x,y,z$ are not equal to zero. Then
\[ \Rightarrow x^2  \plus{} xz \equal{} z^2
\]
\[ \Leftrightarrow t^2  \plus{} t \minus{} 1 \equal{} 0,t \equal{} \frac{x}{z}
\]
\[ \Rightarrow \left[ \begin{array}{l}
 t \equal{} \frac{{ \minus{} 1 \plus{} \sqrt 5 }}{2} \\
 t \equal{} \frac{{ \minus{} 1 \minus{} \sqrt 5 }}{2} \\
 \end{array} \right.
\]

[quote="Honey_S"][quote="thanhnam2902"]$ \begin {cases} x^2 \plus{} y^2 \equal{} z^2 \\
xz \equal{} y^2 \\
xy \equal{} 10\end {cases}$[/quote]

See that $ x,y,z$ are not equal to zero. Then
\[ \Rightarrow x^2 \plus{} xz \equal{} z^2
\]

\[ \Leftrightarrow t^2 \plus{} t \minus{} 1 \equal{} 0,t \equal{} \frac {x}{z}
\]

\[ \Rightarrow \left[ \begin{array}{l} t \equal{} \frac {{ \minus{} 1 \plus{} \sqrt 5 }}{2} \\
t \equal{} \frac {{ \minus{} 1 \minus{} \sqrt 5 }}{2} \\
\end{array} \right.
\]
[/quote]
Thank you. Oh! This system is easy.


probability
function
expected value
probability and stats

Brief interruptions to my work occur  on average every ten minutes and the number of interruptions in any given time period has a Poisson distribution. Given that an interruption has just occurred, find the probability that I will have less than t minutes to work before the next interruption. If the random variable T is the time I have to work before the next interruption, find the probability density function of T.

I need an uninterrupted half hour to finish an important paper . Show that the expected  number of interruption before my first uninterrupted period of half an hour or more is $ e^3 \minus{}1$. Find also the expected length of time between interruptions that are less than half an hour apart. Hence write down the expected wait before my first uninterrupted period of half an hour or more.

if you have already heard about Poisson processes, you immediately know that the time length between two consecutive interruptions is exponentially distributed with mean $ 10$ minutes. From here, everything is straightforward.
1: $ T$ has density $ \lambda \exp(\minus{}\lambda t) \, dt$ on $ (0,\infty)$ with $ \lambda \equal{} \frac{1}{10 \text{ min}}$
2: properties of the geometric distribution gives you the expected number of interruption before the first uninterrupted period of half an hour or more, etc ...

Is this OK ?

I'm not familiar with the geometric distribution, but I solved it like this:

$ p \equal{} P(T > 30) \equal{} e^{ \minus{} 3}$

Expected number of interruptions before the first uninterrupted half hour:

$ 0\cdot p \plus{} 1\cdot (1 \minus{} p) p \plus{} 2 \cdot (1 \minus{} p)^2 \cdot p \plus{} 3 \cdot (1 \minus{} p)^3 \cdot p \plus{}  \cdots \equal{} p (\frac {1}{p^2} \minus{} \frac {1}{p}) \equal{} e^3 \minus{} 1$

For the expected time to pass before the first uninterrupted half hour I calculated about 2h 40min. What's your solution?
Are you studying in Cambridge?

1: yes, your solution is essentially what I had in mind (http://en.wikipedia.org/wiki/Geometric_distribution)
2: For the expected time to pass before the first uninterrupted half hour it is the same idea:
\[ \equal{} E[T|T < 30] \cdot p\plus{}2\cdot E[T|T < 30] \cdot (1\minus{}p) p\plus{}3 \cdot E[T|T < 30] \cdot (1\minus{}p)^{2}\cdot p\plus{}... \]


I'm not sure if this is the right forum.Anyway here's the question:-
In a convex quadrileteral ABCD the angle bisectors of angle ABC and angle CAB meet at point [b]P[/b].The angle bisectors of angle BAD and angle DBA meet at [b]Q[/b].Prove that if A,B,C,D are concyclic then so is A,B,P,Q.

Since $ ABCD$ is cyclic, $ \angle DAC \equal{} \angle DBC \equal{} x$. Then, letting $ \angle CAB \equal{} y$, $ \angle DBA \equal{} z$, we see through simple angle chasing that

$ \angle BPA \equal{} 180 \minus{} \frac{\alpha\plus{}\beta\plus{}\gamma}{2} \equal{} \angle BQA$. So $ ABPQ$ is cyclic.


Find all functions $f(n)$ such that $f(2n) = f(n+1)^2 - f(n-1)^2$

1)  $f : \mathbb{N} \rightarrow \mathbb{N}$
2)  $f : \mathbb{R} \rightarrow \mathbb{R}$

Is it really an olympiad problem??  :maybe:

I'm getting a lot of weird stuff, yet I still can't manage to conclude.

For the 1st one, I think there are none, but I can't prove it. (I can tell you what I have so far)

For the 2nd one, I guess there are infinitely many, but I don't think there's an easy characterization. (I can show you a few examples)


LOL, you could do homework with all that time. :)

well, i only play on weekends, so if you're implying me doing extra work upon extra work... :P

any other arguements?

Its an addicting free online mmorpg much like runescape, though i've never played runescape. Its 2D sidescroll, and may sound shabby at first, but most definitely isn't. I would emphasize the addicting part.

So I'm guessing that your avatar is your Maplestory character?

not my character. just a random avatar i found. i find that ms character make good avatars...  :D

[quote="Stratigizer"]Lets talk about this game of extremes.
Either you know about it, or dont, and love it, or hate it.
Post igns, character info, and try to convince me to quit.
 ;)[/quote]

Oh this game... I tried to start it but it wont get on my computer... I think that its pretty good... a little like runescape... I did make an account though i think. mabey on someone else's com.

its on http://www.maplestory.com

actually, that site is just a portal.

goto http://www.mapleglobal.com for the site, download, and a bit of info.

its completely free, which adds to the awesomeness.

[quote="Spencer"][quote="MathFiend"]is there any way to play without downloading it? it would take nearly 5 hours on just the .exe file alone on my computer.[/quote]

yea. it says 56 hours for my com.[/quote]

Im trying to download right now...

Hey, if I download the thing on the bottom, then the main game loads for only likke 1 hour.

I used to play this...
Got to be pretty boring though
I mean killing the same things over and over....
What the point??
My opinion
[hide]Quit before it takes over ur life!!![/hide]
I got some people at my school who talk about it for hours
It like the only thing they do
This is better than Maplestory
[hide] :ddr: [/hide]

If trying to download the game,
Do the parts
It is way faster
It takes 4 hours on my comp to download all of it
but only 2 hours with parts

Well, you only have pretty much 1 choice for armor.
use the two-handed swords, scimitar and lionheart.
tho they are more expensive, they look cooler and are the most balanced.

anyways, i always liked magic claw better, as the terrain is no obstacle for attacking. and regular [i]claws[/i] are for theives, yes, the throwing star casters...
anyways, im still 46, doing all these awesome new quests!
[/i]

umm... spam???

what is wrong with your posts?

WOAA!! I've seen plenty of double posts, but never a quadruple post :oops_sign:  :w00tb:

Hehe I'm a hardcore RS fan.

Maplestory has only one skill: BEATING CUTE FUZZY THINGS TILL THEY YELL  :help:

Seems a bit too simple for me. On the other hand RS has 21 skills.

Farmers for life.

LVL 57 finally ice-lightning if u guys forgot
I got 3 anakamoon
If anyone is on scania, I may give u guys one.
my name is [i]yomama789[/i]
Do any of u guys have BC ores, I need 50 of them and I onli get 20 right now

[quote="Stratigizer"]Well, you only have pretty much 1 choice for armor.
use the two-handed swords, scimitar and lionheart.
tho they are more expensive, they look cooler and are the most balanced.
[/quote]

I'm 31, blunt weapon page.
And you say I have only 1 choice, but I'm still not too knowing of everything MapleStory, could you tell me which armor I should get? Lvl requirement 30 or 35, stuff I'd keep until 40.
And I have lvl 25 armor, if there's any you think would be worth keeping until 40, please advise!

um, yeah, i have no clue what happened there...sorry about that!

im not sure about the stat requirements, but at lv 30, your top is the green hwarang shirt or w/e you call them, black martial arts pants, the shoe you get from nella's last quest, and it doesn't really matter the glove.

lv 35 you should get brown/blue helm, forgot the name, and white/black chainmail, which is overall. keep your lv 30 shoes, and you should have lv 35 gloves from john's sleepywood jump quest.

If you have an orihalcon burgernet helm, keep it til 35, and if you have venon, or perhaps the lv 25 glove, keep until 35. however, i dont think you should keep anything til lv 40. stat-wise, its not beneficial compared to cost, and it just doesn't look as cool!

goto http://www.hidden-street.net to look at the stats and requirements, but also, to everyone who doesn't know, goto http://www.happymapling.net for a look at equips up to lv 100!

ok, enough of that...

Ludi's coming out soon
I'm so prepared for it
if no one believes me go to mapleglobal to look

It's okay, I want to try the pq though.
I uninstalled MapleStory because it was distracting me from mathcounts -_-

pq is what, 35-50?
but it seems like the other pq, it just takes too long to get in ><

I guess state hasn't happened yet? Anyways, good luck! and I'm from Texas too, so go Texas ;)

i love maple lol

the best part of maple is calculating profits regarding scrolling and junk, i'm soo rich from just doing nothing but scrolling.
15dex 6 slots left sauna robe
11 att gloves
etc XD

yeah, i too just discovered my lucky hand in scrolling. my overall robe is +6, with 3 10% def scrolls. i also have +12 jump +4speed shoes, with like 2 10% jump scrolls...
also, my friend borrowed a fish spear and tried 3 10% for spear att, and they all worked.

yeah, seems like its gonna be all 10% for me =)


algebra
polynomial
superior algebra
superior algebra unsolved

Prove that if [b]K[/b] is a field and [i]x_1, x_2, ... x_n[/i] are elements of algebric closer of [b]K[/b], then [b]K[/b]([i]x_1, x_2, ... x_n[/i]) is a collection of all the expressions of the form [b]f[/b]([i]x_1, x_2, ... x_n[/i]) in wich [b]f[/b] beloning to [b]K[X_1,...X_n][/b] is a polynomial such that [i]deg[/i](X_i) [b]f[/b] < [b][K([i]x_i[/i]):K][/b]


The rules of this game are as follows:

1. Each player gets to pick 5 things they would have with them if they were stranded on a barren wasteland. These items cannot be larger than a house.(e.g. a tent would be a good idea)
2. players can trade items by typing the username, the item that they want, and the item they are willing to trade from the person they want to trade with. The person can then decide whether to trade or not.
3. every week or so I will post a problem. Each player will PM me their solutions to me, and the players who get the problem wrong must give up an item.
4. The players with insufficient items for survival will be eliminated.(I will be the one to decide who has insufficient items)
5. There will be a max of 5 players per game.
6. The winner of this game may host the next one.

[hide=sign ups][/hide]

sign up

[hide="signups"]
Bugi
westiepaw
[/hide]

sign up

[hide="signups"]
Bugi
westiepaw
spaceguy524
[/hide]

Sign up.

[hide="signups"]
Bugi
westiepaw
spaceguy524
cyberspace
[/hide]

One more...

WAIT! Never mind. There's already 5 people.  :P

o.k.... The competition begins.

You 5... Pick your 5 items, and be smart about them. If you're not smart about them, you might just die when the game starts. :)

What does the barren island have on it?  Is there any hope of escape, or should I choose items for long-term survival, rather than escape?  Thanks!

Ok, i will sign up since Bugi dropped out.


[quote="alanchou"]uh, triangle???[/quote]

oh heh that was a typo, a freudian slip to be exact  :rotfl:

[quote="CircleSquared"][quote="123456789"]Let a circle be inscribed in an isosceles trapezoid with base lengths $ m$ and $ n$. Express the diameter $ d$ of the circle in terms of $ m$ and $ n$.[/quote]

[hide]
Draw the segments from the center of the circle to each vertex of the trapezoid, and the perpendiculars to each edge.
It's easy to see that the lateral sides must be $ \frac {m \plus{} n}{2}$ by congruencies.
Draw the height from the upper left vertex.  Use the pythagorean theorem.
$ \frac {m \minus{} n}{2}^2 \plus{} h^2 \equal{} \frac {m \plus{} n}{2}^2$
$ \frac {m^2 \plus{} n^2}{4} \minus{} \frac {mn}{2} \plus{} h^2 \equal{} \frac {m^2 \plus{} n^2}{4} \plus{} \frac {mn}{2}$
cancel, combine like terms
$ h^2 \equal{} mn$
or in other words
$ (2r)^2 \equal{} mn$
$ r \equal{} \frac {\sqrt {mn}}{2}$
[/hide]

too simple..someone tell me what i did wrong

edit: yes alan i meant circle[/quote]It's meant to be simple. It's High School Basics. And I was asking for $ d$, but $ r$ is just as good.


$ 1)$ Given $ a, b, c \geq\ 0$ and $ a\plus{}b\plus{}c\equal{}3$.
Prove that: $ \frac {7a^2 \plus{} 9}{b^2 \plus{} c^2} \plus{} \frac {7b^2 \plus{} 9}{c^2 \plus{} a^2} \plus{} \frac {7c^2 \plus{} 9}{a^2 \plus{} b^2}\ge 24$
$ 1)$ Given $ a, b, c \geq\ 0$ and $ ab\plus{}bc\plus{}ca\equal{}3$.
Prove that: $ \frac {4a^2 \plus{} 3}{b^2 \plus{} c^2} \plus{} \frac {4b^2 \plus{} 3}{c^2 \plus{} a^2} \plus{} \frac {4c^2 \plus{} 3}{a^2 \plus{} b^2}\ge \frac {21}{2}$

I see some long proofs in here: http://diendantoanhoc.net/forum/index.php?showtopic=45718
Indeed, both of them can be solve easily by CS and Schur ineq  :)

I'm also a fan of Vasc's problems , note that the first and the second one are equivalents so I only write the solution of the frst one:

it's equivalent to:

$ 7\sum{ \frac{a^2}{b^2+c^2}} +( \sum{a})^2 \sum{\frac{1}{b^2+c^2}} \geq 24$

using Cauchy it follows that :

$ 7\sum{ \frac{a^2}{b^2+c^2}} \geq  7\frac{(\sum{a^2})^2}{2\sum{(ab)^2}}$

and $ \sum{ \frac{1}{b^2+c^2} }  \geq  \frac{ 3+ \frac{(\sum{a^2})^2}{2\sum{(ab)^2}}{\sum{a^2}}$

WLOG $ a+b+c=1$  it's equivalent to :
$ f(r)= 4-23q+23q^2+20q^3+42r-96rq \geq 0$  and thus f is increasing
and it's easy to check that :  if $ q\leq \frac{1}{4}$ $ f(0 )\geq 0$  and if $ q\geq \frac{1}{4}$  $ f(\frac{4q-1}{9}) \geq 0$


Are given 25 numbers . The sum of each 4 of them is positive prove that the sum of all those numbers is positive .

can't we just add all the "4-possibilities" up? :maybe:
because then each number would appear $ \binom{24}{3}$ times,I mean let $ x_1,x_2,\ldots ,x_{25}$ be our 25 numbers,and we had:

$ x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \equal{} a_1$
.
.
.
$ x_{22} \plus{} x_{23} \plus{} x_{24} \plus{} x_{25} \equal{} a_{\binom{25}{4}}$

where $ a_i$ are positive numbers for $ i \equal{} 1,2,\ldots ,\binom{25}{4}$

now add all the equalities we had above,we would end up with:

$ \binom{25}{3}\sum_{i \equal{} 1}^{25} x_i \equal{} \sum_{i \equal{} 1}^{\binom{25}{4}} a_i > 0$

$ \Rightarrow \sum_{i \equal{} 1}^{25} x_i > 0$
 :maybe:


inequalities
geometry
perimeter
rectangle
triangle inequality
Pythagorean Theorem

A convex polygon does not extend outside a square side 1. Prove that the sum of the squares of its sides is at most 4.

[hide="Hint"]
Divide into right triangles. Then use the fact that $ x^{2}\leq x \ \forall x \leq 1$.[/hide]

[hide="Interesting"]Consider the orthagonal components of each side respective to the square. By the triangle inequality, $ a+b \geq \sqrt{a^{2}+b^{2}}$ for component lengths $ a$ and $ b$. Note that the sum of the lengths of these components equals the perimeter of a binding rectangle of that polygon.

Let us orient our components such that the achieved binding rectangle has edges parallel to those of the square. This rectangle must be fully contained within the square, and hence no component can have length greater than 1. It follows that the square of the length of a component cannot exceed the length of that component.

Now, note that the sum of the squares of the side lengths equals the sum of the squares of the lengths of the components (Pythagorean Theorem), which does not exceed the sum of the lengths of the components, which is equivalent to the perimeter of the binding rectangle.

Since neither the length nor width of this binding rectangle can exceed 1, the perimeter of this rectangle cannot exceed 4. Transitively, the sum of the squares of the side-lengths of the polygon cannot exceed 4.

Testing the unit square case gives that this bound is indeed achievable, so 4 is the best upper bound. [/hide]

...
Bah my computer stalled. Anyway, it turns out that this is basically what cinco suggested.


f(x+y)=f(x)f(y) and f(a)=1990. Prove that f is injectiv.

I can't see why you can't take $ f(x) \equal{} e^{g(x)}$, where $ g(x)$ is a pathological solution of the Cauchy equation with some of the Hamel basis elements (that don't intersect with $ a$) set to $ 0$ by $ g$.

You can set that only if this funciton is monotone or injectiv.

Why? The function $ f: \mathbb R\to\mathbb R$ can be shown to have values in $ \mathbb R^\plus{}$ only such that MellowMelon's reasoning does work. Can you provide a proof?


geometry
3D geometry
combinatorics unsolved
combinatorics

A cube is divided into n^3 little cubes, prove that we can assign a DIFFERENT integer to each little cube so that the sum of each of the 3n^2 rows parallel to an edge is zero.

edit: it means the sum of numbers in the n little cubes in all 3n^2 rows is zero

for $ n\equal{}2$ it's obviously umtrue  :wink: since $ a_{1,1,1} \plus{} a_{1,1,2} \equal{} 0$ and also $ a_{1,1,1} \plus{} a_{1,2,1} \equal{}0$ so it must be
$ a_{1,2,1} \equal{} a_{1,1,2}$ which is a contraddiction since the numbers were supposed to be different.


In general does $ \sum a\plus{}b$ mean $ (\sum a)\plus{}b$ or $ \sum (a\plus{}b)$?

Umm well is this a question that's attempting to get an answer for one of the USAMTS problems?

If so, you just have to look it up on your own.

Typically, $ \sum a \plus{} b \equal{} (\sum a) \plus{} b$. Heh, I see what problem you're looking at... I don't think this counts as cheating though. :)

[quote="Brut3Forc3"]Typically, $ \sum a + b = (\sum a) + b$. Heh, I see what problem you're looking at... I don't think this counts as cheating though. :)[/quote]
If it's not considered cheating for telling this, I believe it's $ \sum{(a + b)}$ cuz in $ \text{\LaTeX}$, when you just type a+b in the brackets for sum, your exact original reading shows up.

Brackets or parenthesis? Brackets usually disappear on everything...

[quote="ProtestanT"][quote="pythag011"][quote="gauss1181"]I meant these brackets: { }[/quote]

Yeah, if you type a latex expression and use brackets as parenthesis, they disappear. The summation notation in latex does not require brackets...

usually I type \displaystyle\sum_{i=1}^5 x^2 rather than usually I type \displaystyle\sum_{i=1}^5{x^2}

and it does the exact same thing.[/quote]
However, when one of the \mod things enclose something in brackets, it's also in parenthesis. I think it's \pmod. 4\pmod{3 3k4} shows up like this: $ 4\pmod{3 3k4}$[/quote]

\pmod is not a symbol. \sum is.

Just assume one and solve the problem with it.

[quote="t0rajir0u"]In context, this depends on whether $ b$ has a term that contains the index of the summation.  This notation is ambiguous, though; to be precise, you should add parentheses appropriately.[/quote]

I totally agree.  Just use your common sense and knowledge of summation notation. :wink:


If I have a contour diagram of:

[url=http://imageshack.us][img]http://img118.imageshack.us/img118/9608/graphdu7.jpg[/img][/url]

then what would the graph of:

f(-x,y) = f(x,y)

and

f(-x,y) = -f(x,y)

look like??

my guess is that the first one is the same as the original one translated with the y-axis and the second one translated with the x-axis. Is this right?

[quote="JBL"][quote="EquinoX"]f(-x,y) = f(x,y)
...
translated with the y-axis[/quote] Hmm?[/quote]

Why a Hmmm??

Because neither of the lines I quoted make sense ;)

What I meant translated with the y-axis is pretend that the y-axis is a mirror, so therefore everything to the right of the y-axis will be reflected to the left of the y-axis. That's what I assume
 f(-x,y) = f(x,y) looks like and I just want to confirm whether this is correct or not?

Your notation is wrong. You are talking about f(-x,y)=0, not f(-x, y)=f(x,y).. There is a big difference.. Suppose you have f(x,y) = x+y. Then f(-x,y)=-x+y. So, if you plot x+y=0 you get y = -x, and plotting -x+y=0 you get y=x. This goes by your assumption. But f(x,y) = f(-x,y) is x+y = -x+y => x=0, and this has nothing to do with what you are talking about.

Yes what I am trying to say is to find what the function f(-x,y) looks like if f(x,y) is on the picture above...