function
real analysis
real analysis unsolved

Find all strictly increasing functions $f: [0,+\infty)\to\mathbb{R}$ such that for every $x>y\ge0$ we have $f\left(\frac{x+y}{x-y}\right)=\frac{f(x)+f(y)}{f(x)-f(y)}.$
(V. Radchenko)

By setting $y=0$ we get $f(1)=\frac{f(x)+f(0)}{f(x)-f(0)}\Longrightarrow f(x)[1-f(1)]=f(0)f(1)+f(0)$ so $f(1)=1$ because if $f(1)\neq 1$ we get that $f$ is constant. Then $f(0)=0$.

By setting $x=ay \ \ a>1$ we get \[\frac{f(ay)+f(y)}{f(ay)-f(y)}=f\left( \frac{ay+y}{ay-y}\right)=f\left( \frac{a+1}{a-1}\right) = \frac{f(a)+1}{f(a)-1}\] .

So $[f(ay)+f(y)][f(a)-1]=[f(ay)-f(y)][f(a)+1] \ \Longrightarrow \ f(a)f(y)=f(ay)$ which is a Cauchy equation with the only solutions $f(x)=x^{c}$, and now it's easy to show that in this problem $c$ must be $1$.


Prove that, in the fibonacci sequence,

(u_n)^2 = (u_n-1)(u_n+1) - (-1)^(n-1)

For all n > 1

Using the strong form of mathematical induction. I have the solutuon of simple induction but can't seem to get the numbers to work out when trying the strong method.

Thanks for any help.

(also, small random question, is the null set a subset of itself? And does the statement that @ = @ prove this ... ie. does @=@ imply that the null set is a subset of the null set)

Well, yes, the empty set is a subset of itself (and, obviously, it's the only one!) :)

[quote="degreestudent"]Prove that, in the fibonacci sequence,

(u_n)^2 = (u_n-1)(u_n+1) - (-1)^(n-1)

For all n > 1

Using the [b]strong[/b] form of mathematical induction. I have the solutuon of [b]simple[/b] induction but can't seem to get the numbers to work out when trying the strong method.
[/quote]

I never really understood the difference between those two things. Isn't it just the same thing having 2 names?

But in fact weak induction is nothing but a special case of strong induction? If the integers of the set [0,n[ are enough to make sure the number $n+1$ is in the set, then the set {n} is enough too... So whenever you've proven something by weak induction like here, you've proven it by strong induction too. I agree that there's a small difference between the two things, but however they are so closely related that I'm still wondering why both of them are not just named 'induction'.

Well they are pretty much the same thing... only strong induction, when trying to show that P(r+1) is true, not only uses the truth of P(r) but can use P(r-1), P(r-1)... P(2) and P(1) if it wants to (assuming you have shown P(1) to be true). Since the nth Fibonacci number is dependant on the two (not just one) preceeding numbers this is useful I think to prove the formula of the nth number. I'm not sure that everything which can be proven by strong induction can also be proven by simple...

no I don't think that's true either. But it is true in the other direction: weak induction $\Rightarrow$ strong induction but not strong induction $\Rightarrow$ weak induction.
In this case you already have weak induction you said? Well then what's the problem?


Isn't this a simple problem?  Can any one help?  Common this is mathlinks after all.

[quote="Phydeaux"]I'm new here so I wasn't sure where to post this but, it does relate to this topic.  I've been trying to find the survival of a mutation, this is what one source says I should do:

"I then found the chance of survival to be [img]http://k.1asphost.com/AVAD/(1-e^-2SN).jpg[/img] where S is the selective value and N is the population size. Note that the larger N is, the smaller is this chance. The chance is always bigger than 2S, but when SN is large, the chance is close to 2S. When SN is 1 the chance is about 16% larger than 2S. When SN is 2 it's about 2% larger. When SN is 3 it's only about .25% larger." (Dr.Lee Spetner)

I don't understand fully could some one please explain?  Population size = 10,000 and, selective value = .1%.  Thanks for any help :).[/quote]

Well, I found how to do it.  The answer is 0.002000000004.

Well - you just plug in the numbers... right? OK, now I get your question :). It seems that you don't need any more help but if there's anything else you want to ask about that formula, feel free to do so ....

Oh, yeah, and by the way, I was wondering how could possibly a biologist come up with a formula like this (:P) so could you somebody please explain how does the mutation spread and what is its selective value? (I would like to try to derive it myself, it might be a very interesting and educational experience :). )


There are some differences between sets with related operations such as union , intersection and other important mathematical branches such as real numbers ,matrices ,vectors or complexes . For example real numbers with  sum is a group as well as n*n matrices with associated sum and etc. On the other hand sets with union or intersection are not groups .
  There is another difference (at least I think so).Zero or  neutral member of each branch is special ,e.g. zero 2*2 matrix can’t be summed to zero complex number .On the other hand in all discussions of sets ,there is only one empty set .According to axiom of specification ,having a special set ,the empty set can be yielded by a false Predicate . This is similar to  neutralization of a vector ,a matrix or a real number ,generally what is the current discussion .In these cases the original of zero is obvious but this is not regarded for empty set .
    We know that universe set doesn’t exist ,a set full of everything ,is it possible to have a set ,empty of anything ?!Are there any new set theories which regard some of these facts ?

[quote="christ"]On the other hand sets with union or intersection are not groups .[/quote]
But they do form a [url=http://en.wikipedia.org/wiki/Boolean_algebra_(structure)]Boolean algebra[/url], and the introduction of the symmetric difference gives you a [url=http://en.wikipedia.org/wiki/Boolean_ring]Boolean ring[/url] where symmetric difference, the addition, is an honest-to-god group operation.

As for your general question, there are [url=http://www.science.uva.nl/~seop/entries/settheory-alternative/]many, many models[/url] of set theory that use different collections of axioms.


Express P(c)=c^6+10c^4+25c^2 as a polynomial of least positive degree when c is a root of x^3+3x^2+4=0

There's only one root for the second equation, so it shouldn't be too hard... But I can't find the exact value, so if someone knows how...

[hide]This would probably make finding the exact value of $ P(c)$ harder, but divide $ P(c)$ by $ c^3\plus{}3c^2\plus{}4$:
$ P(c)\equal{}c^6\plus{}10c^4\plus{}25c^2\equal{}(c^3\plus{}3c^2\plus{}4)(c^3\minus{}3c^2\plus{}19c\minus{}61)\plus{}220c^2\minus{}76c\plus{}244$
Because $ c$ is a root of $ x^3\plus{}3x^2\plus{}4\equal{}0$, $ c^3\plus{}3c^2\plus{}4\equal{}0$.

Therefore $ P(c)\equal{}220c^2\minus{}76c\plus{}244$ when $ c$ is such a root.

This is the only way I made sense of "the polynomial of least positive degree."
[/hide]

[hide=A probably incorrect solution]$ \text{Factoring, we get }c^2(c^2 + 5)^2$

We also know that $ c^3 + 3c^2 = - 4\implies{c^2(c + 3) = - 4}\implies{c^2 = \frac { - 4}{c + 3}}$

Plugging back in we get:

$ \frac { - 4(c^2 + 5)^2}{c + 3}$

Which is of degree 3...idk where to go from here

Actually, $ P(c) \equal{} c^{2}(\textcolor{red}{c^{2}} \plus{} 5)^{2}$; furthermore, the degree has to be $ \in \mathbb{Z}^{\plus{}}$.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

I would like someone to explain how to go about finding out why a certain 3x3 matrix cannot be diagonalized. or maybe even a link that shows some examples would be great help. thanks.

Matrices on which field ? $ \mathbb{R}$ or $ \mathbb{C}$?

The main idea is: A matrix is diagonalizable if space $ F^n$ could be written as the direct sum of eigenspaces. If $ \lambda_1, \lambda_2, ..., \lambda_k$ are roots of its characteristic polynomial and $ \sum_{i\equal{}1}^{k} {\dim(\ker(M \minus{} \lambda_i I))} \equal{} n$ then $ M$ is diagonalizable.


We want to find all solutions to the equation (x^2)+(y^2)=(z^2) with gcd(x,y,z)=1

a) show that under the assumption gcd(x,y,z)=1, z must be odd and exactly one of the numbers x,y must be even. Withour loss of generality assume that y is even.

b) show that x+z and x-z are even, and gcd((x+z)/2, (x-z)/2)=1

c) Use part b) to show that x=(u^2)-(v^2), y=2uv, z=(u^2)+(v^2) for two coprime integers u,v in integer, one of which is even and the other is odd

Thank you

a)  $ x^2 \equiv 0, 1 \bmod 4$.  If $ z$ is even, then $ z^2 \equiv 0 \bmod 4$, so we must have $ x^2, y^2 \equiv 0 \bmod 4$; contradiction.  Hence $ z$ is odd, $ z^2 \equiv 1 \bmod 4$, and WLOG $ x^2 \equiv 1 \bmod 4, y^2 \equiv 0 \bmod 4$.

b)  $ x, z$ are both odd.  If a prime $ p$ divides $ \frac {x \plus{} z}{2}, \frac {x \minus{} z}{2}$ it divides their sum and difference (in other words, $ x, z$), contradiction.

c)  We have $ \left( \frac {y}{2} \right)^2 \equal{} \left( \frac {x \plus{} z}{2} \right) \left( \frac {x \minus{} z}{2} \right)$.  Because the two terms on the right are coprime, they are both squares.  Let the first be $ u^2$ and the second $ v^2$.  The conclusion follows.

A similar, but somewhat harder, question:  solve $ x^2 \plus{} y^2 \plus{} z^2 \equal{} w^2$ with $ \gcd(x, y, z, w) \equal{} 1$.


[hide="Question"]
If we were to balls and urns this, we'd have $ \binom{11}{3}\equal{}165$, which clearly overcounts by a lot.
However, $ \frac{165}{11}\equal{}15$, which is the right answer.
Is there a reason for this?
[/hide]

It's probably a coincidence. What could dividing by $ 11$ represent?

Maybe it was the $ 11$ used in balls and urns? :maybe: That was a wild stab in the dark.

Well, there's not much difference in something like partitioning $ 4$ into $ 2$ numbers. We get $ 1$ by dividing in that way, but it is obviously $ 3$. And there's nothing to say that it's going to be divisible.

I think sunehra is right... because both the cups and the balls are identical

Um, I provided a counter example in my last post...and there's still no proper reason to divide by anything.

well idk but the balls and urns methods overcounts because the four cups are at fixed places... but they are identical

And again, the logic behind the division is...??? There's some cases like $ (0,1,2,5)$ in which we have to divide by $ 4!\equal{}24$, but then other cases like $ (2,2,2,2)$ in which we don't need to divide. So why divide by $ 11$?

[quote="math154"]And again, the logic behind the division is...??? There's some cases like $ (0,1,2,5)$ in which we have to divide by $ 4! \equal{} 24$, but then other cases like $ (2,2,2,2)$ in which we don't need to divide. So why divide by $ 11$?[/quote]

oh yeah that makes sense... and it is a coincidence as you said because I tried another case that didn't work the same way.


Alex owns three times as many brown shoes as red shoes, twice as many black shoes as brown shoes, and four times as many white shoes as red shoes. What is the ratio of the number of white shoes to the number of black shoes he owns? Express your answer as a common fraction.

Number of brown shoes = a
Number of red shoes = b
Number of black shoes = c
Number of white shoes = d

It is given that $ 3b\equal{}a$, $ 2a\equal{}c$, and $ 4b\equal{}d$.  We wish to find $ \frac{d}{a}$.  Manipulating our first and third equations yields $ 12b\equal{}4a$ and $ 12b\equal{}3d$, so $ 4a\equal{}3d \implies \frac{d}{a}\equal{}\boxed{\frac{4}{3}}$.

brown=a
red=b
black=c
white=d
3a:b
2c:a
4d:b

3a:4d
3(2c):4d
6c:4d
3c:2d
[u]2white[/u]
3 black

2/3


A round oak pole has a cross-sectional area of 25 sq" and is 20' long.What is the weight of the pole in pounds ?
oak weighs 50 pds per cu'   or  0.029 pds per cu"

I have been going back over this & came up with 174 pounds. It is one of the possible answers.

Any time!  :D The message is too small. Please make the message longer before submitting.


12 minutes in an hour so it will lose $ 5\times 5\equal{}25$ seconds per hours and $ 25\times 24\equal{}600$ seconds which is 10 minutes.

Next time put all your questions on one topic!

[hide]24 hours is equal to 24*60=1440 minutes, which contains 1440/12=120 12-minute time periods. Since the clock loses 5 seconds every 12 minutes, it loses a total of 120*5=600 seconds in 24 hours. This is equal to 600/60=10 minutes.[/hide]


calculus
integration
trigonometry
logarithms
calculus computations

I think this is a standard enough integral. Just want to see a few different approaches to it.
$ \int{(cot(x))^{n}dx}$

[hide="Solution"]
Let $ n\geq 2$
$ I_{n}=\int\cot^{n}xdx=\int\cot^{n-2}x\cot^{2}xdx=\int\cot^{n-2}x\left(\frac{1}{\sin^{2}x}-1\right)dx=$

$ =\int\cot^{n-2}x\frac{1}{\sin^{2}x}-\int\cot^{n-2}xdx=\left\|\begin{array}{l}t=\cot x\\ dt=-\frac{1}{\sin^{2}x}\end{array}\right\|=$

$ =-\int t^{n-2}dt-I_{n-2}=-\frac{1}{n-1}t^{n-1}-I_{n-2}$, so:

$ I_{n}=\frac{1}{1-n}\cot^{n-1}x-I_{n-2}$
And if you compute for example $ I_{1}$ and $ I_{0}$ you will compute $ I_{n}$ for every $ n$.[/hide]


real analysis
limit
real analysis unsolved

1. let $ f(x) \equal{} \sum_{n\geq 0}3^{ \minus{} n}\sin(9^{n}\pi x)$ . Find  points at which f is not differentiable ?

2. let $ E\subset\mathbb{R}$ with finite measure , define $ f(x) \equal{} m(E\cap (E \plus{} x))$ where $ m$ is the Lebesgue measure on $ \mathbb{R}$  and $ E \plus{} x \equal{} \{ a \plus{} x : a\in E\}$.  Prove :
i) $ \lim_{n\to\infty}f(x) \equal{} 0$
ii) $ f$ is uniformly continuous on $ \mathbb{R}$

thanks in advance

2. (i) For any $ \epsilon>0$, we set $ M$ so large that $ m(E \symbol{92}(\minus{}M,M))<\epsilon$. Then it's straightforwrd to check $ f(x)<\epsilon$ whenver $ |x|>2M$.

(ii) Clearly $ f$ is continuous. Now consider a more generalized problem: if $ f: \mathbb{R}\to \mathbb{R}$ is continuous and $ \lim_{x\to \infty}f(x)$, $ \lim_{x\to \minus{}\infty} f(x)$ both exist (finite), then $ f$ is uniformly continuous. This is a standard excercise.


superior algebra
superior algebra unsolved

First of all $x=x^{2}=(-x)^{2}=-x$ hence $2x=0$. Now use $(x+y)^{2}=(x+y)$ to obtain the desired result.
Second idea: use lemma 2 in grobber's post: http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=ring+boolean&t=3041


Let $a,b,c,d$ be the positive numbers. Prove that $(a+b+c+d)^3\leq 4(a^3+b^3+c^3+d^3)+24(ab(c+d)+cd(a+b))$. ;)

Not a bad problem, in fact really nice one. First of all, observe that $\sum{a^3}-3\sum{abc}=(\sum{a})(\sum{a^2}-\sum{ab})$ (all sums being symmetric, that is $\sum{abc}$ has 4 terms and it is equal to $ ab(c+d)+cd(a+b)$). Also, observe that $2\sum{ab}=(\sum{a})^2-\sum{a^2}$. If I made well the computations, then the inequality reduces to $(\sum{a})(4\sum{a^2}-(\sum{a})^2)\leq 4\sum{a^3}$. This can be written as $4\frac{\sum{a^3}}{\sum{a}}+(\sum{a})^2\geq 4\sum{a^2}$. Apply AM-GM and then Cauchy-Schwarz and it's done. Really nice problem!

thank you very much Mr.harazi :oops:  :lol:  ;)  :D


[quote="dduclam"]Let $ a,b,c$ be nonnegative real numbers, prove that
\[ \sum_{cyc}\sqrt {\frac {bc}{(a \plus{} b)(a \plus{} c)}}\ge1 \plus{} \frac {9abc}{2(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}
\]
[/quote]
This problem loOks very nice but I think it is not true.Try $ b\equal{}c\equal{}999,a\equal{}10^{\minus{}18}$,we get $ LHS\minus{}RHS\equal{}\minus{}2.252252252$x$ 10^{\minus{}21}$

[quote="SUPERMAN2"][quote="dduclam"]Let $ a,b,c$ be nonnegative real numbers, prove that
\[ \sum_{cyc}\sqrt {\frac {bc}{(a \plus{} b)(a \plus{} c)}}\ge1 \plus{} \frac {9abc}{2(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}
\]
[/quote]
This problem loOks very nice but I think it is not true.Try $ b \equal{} c \equal{} 999,a \equal{} 10^{ \minus{} 18}$,we get $ LHS \minus{} RHS \equal{} \minus{} 2.252252252$x$ 10^{ \minus{} 21}$[/quote]

I don't think so, SUPERMAN2. If $ b\equal{}c$, WLOG we assum $ b\equal{}c\equal{}1$, the Inequality becoms
$ \frac1{a\plus{}1}\plus{}\sqrt{\frac {2a}{a\plus{}1}}\ge1\plus{}\frac{9a}{2(a\plus{}2)(2a\plus{}1)}$
$ \Leftrightarrow \sqrt{\frac2{a\plus{}1}}\ge\frac{\sqrt a}{a\plus{}1}\plus{}\frac{9\sqrt a}{2(a\plus{}2)(2a\plus{}1)}$

Now, please check again by your computer.  :wink:


[quote="felipesa"][quote="Spoon"]huh?[/quote]

what?[/quote]
you say "what?"
[quote="felipesa"]
non consider
Fall down TOMBAMENTO
mass tackles MASSA ROLDANAS
friction ATRITO [/quote]
what is this? i dont read portuguese

it just says that you don't have to consider friction and so?

Please note, when I said "huh?", the only thing there was a few words... NO PICTURE. Thus, it made no sense, there was not even a question asked. Hence "huh?".

[quote="Dr. No"][quote="felipesa"]non consider
Fall down TOMBAMENTO
mass tackles MASSA ROLDANAS
friction ATRITO

Find aceleration block A(4Kg),B(1Kg), C(2Kg) and D(6Kg)
F=100 N

If you have some problem to see the picture of problem  contact me[/quote]

The answer is 42.[/quote]

Your answer is wrong , you have to find the acelerations of blocks A,B,C,D

I think he was refering to the question of life, the universe, and everything ;)

is acceleration of A = $ \frac{1200}{341} $

[quote="Rushil"]hello? is my answer correct[/quote]

no is wrong A=x/489

x- you have to find

[quote="Rushil"]I give up. I cant do anything . I already spent a lot of time[/quote]

First, you have 7 equations and 9 acelerations

so you group and has 3 acelerations

Did you find ?

I ha 2 accelerations . Just give the sol. !!! :dry:

its too comlicated to get!!1 Why don't you post your solution


geometry
circumcircle
3D geometry
tetrahedron
sphere

Consider two non co-planar triangles in space that share a common edge. Let $ \ell$ be the line perpendicular to one triangle, which passes through that triangle's circumcenter, and $ m$ be the line perpendicular to the other triangle and passing through that triangle's circumcenter. Find the conditions under which $ \ell$ and $ m$ intersect.

Nobody yet?

[hide="Hint"]What do four non-coplanar points make?
[hide="Answer/Bigger Hint"]A tetrahedron![/hide][/hide]

Is there any, besides non co-planar?

[hide]
The four total points of the two triangles form a tetrahedron, which can be inscribed in a sphere.  Slicing this sphere along planes including the triangles (two faces of the tetrahedron) gives the circumcircles of the triangles.  $ l$ and $ m$ are thus axis of the sphere and will thus intersect at the center of the sphere.
[/hide]

Nope, that's it, and that solution was the motivation for this problem.

I was curious to see:
1. How quickly people would see that
2. If people came up with any other solutions

As an exercise, you might want to try proving that you can, in fact, always construct a circumsphere.


The function $ f(x)\equal{}x^3\plus{}3x$ is strictly increasing on the reals. Also $ f'(x)\ne 0$ so Implicit Function Theorem guarantees that it has inverse function at every point. Can we determine this inverse function explicitly?

Thanks

Uhhh, quite an unpleasant-looking function  :o
Thank you!


$u_n$ a sequence that tends to infinity, $v_n=u_{n+1}-u_n$ the sequence of its differences.
Suppose that $v_n \to  \infty$ and that $\{v_n\}$ is dense modulo $1$. Does it follow that $u_n$ is dense modulo $1$ ?

It doesn't matter if $v_n\to\infty$.It not follows that $u_n$ is dense in $[0,1]$. Youe claim is: The sequence of partial sums of a sequence which is dense in $[0,1]$ is dense in $[0,1]$.

Take $a_n$ a sequence dense in $[0,\frac 1 2]$ and all $a_n\in [0,\frac 1 2]$. The sequence $b_n=\begin{cases} a_{\frac n 2} \text { if } 2 \mid n \\ 1-a_{\frac{n-1} 2} \text{ else } \end{cases}$ is dense in $[0,1]$. and $\sum_{i=1}^n b_i$ has values of 1 and in $[0,\frac 1 2]$, hence not dense in $[0,1]$.

yes  :)  Nevertheless one can show for example that if $v_n \to  \infty$ and $v_{n+1}-v_n \to  0$ then $u_n$ is dense.


please start for me the solving of these 2 problems ,some hints about the first and the second one ,and the correct way i should start and what to do exactly .

thanks

[hide] Question 2)
a) is false put n=4 we get 16+4+1=21 which is nto prime
b) $ f(n)\equal{}n^2\plus{}n\plus{}1\equal{}n(n\plus{}1)\plus{}1$
as n is a positive integer  one out of n, (n+1) is even so $ f(n)$ is odd.[/hide]

[hide="1"]Let $ \Delta r$ be the small increment in $ r$ and $ \Delta V$ be the small increment in $ V$. Initially, the volume of the sphere is $ V\equal{}\frac{4\pi r^3}{3}$. After the inflation, the new volume is $ V\plus{}\Delta V\equal{}\frac{4\pi(r\plus{}\Delta r)^3}{3}$. You are given $ r\equal{}100\ \text{cm}$ and $ \Delta V\equal{}120,000\ \text{cm}^3$... you do the rest.[/hide]


Find the least value of $y=|x-a|+|x-b|+|x-c|$ where $a<b<c$.

[hide="Solution"]We'll discuss the intervals $(-\infty,a\rbrack,\lbrack a,b\rbrack,\lbrack b,c\rbrack,\lbrack c,+\infty)$

1. On the first the function is $y=-3x+a+b+c$ - decreasing linear function, hence it's minimum is in the right boundary: $y_{1min}=y(a)=b+c-2a$

2. On the second the function is $y=-x-a+b+c$ - decreasing linear function, hence it's minimum is in the right boundary: $y_{2min}=y(b)=c-a$

3. On the third the function is $y=x-a-b+c$ - increasing linear function, hence it's minimum is in the left boundary and is equal to $y_{2min}$

4. On the fourth the function is $y=3x-a-b-c$ - increasing linear function, hence it's minimum is in the left boundary: $y_{3min}=y(c)=2c-a-b$

Now $b-a>0\implies b-2a>-a\implies b+c-2a>c-a\implies y_{1min}>y_{2min}$ and $c-b>0\implies 2c-b>c\implies 2c-a-b>c-a\implies y_{3min}>y_{2min}.$

Therefore, $y_{min}=y_{2min}=y(b)=c-a.$[/hide]

Let's take three segments on the line of the x on the Gauss plane: the point x will obviously be between a and c. If we take a generic point different from a, b,c then we'll have that $|x-a|+|x-c|=a-c$ which is a invariant. So awe have to minimize $|x-b|$ which is at least $0$ so $x=b$

You can generalise this to n numbers quite easily...

$n$ friends live along a straight road. where should they meet such that the sum of their walking distances is at a minimum?


trigonometry function MATHCOUNTS calculus probability geometry integration

This was on the MCAS...what is the definition of "numerical difference"?

The actual question was along the lines of "What is the numerical difference between the median and the mode?" The median and mode could be clearly determined, 10 and 11, respectively. So is the answer -1 (when you subtract straight out) or 1 (assuming positive difference)? I asked around in school, there seems to be considerable numbers of those who said 1, as there are of those who said -1.

Personally, both should be considered valid, as the wording of the question was somewhat vague.

BTW is there any way to protest the MCAS if they only accept one of the answers?

dude you need to chillax about the MCAS

i mean in 8th grade i believe i used set notation and cited PIE on one of the open response questions, and used the law of sines to show AA similarity (which i later learned is BS) on another one. i see MCAS as something to have fun with.

but yeah i'm pretty sure it's 1

ZOMG IT'S SOOOOO TOTALLY AMBIGUOUS!!!!

My personal opinion IS that it can be either 1 or -1. I failed to find a definition of "numerical difference", so I figure we can consider each word separately. "numerical" simply means a measurement expressed in numbers. "difference" refers to the result of a subtraction problem.

And so it seems that "numerical" is rather redundant. What other sort of answer would one give?

And how do we know which subtraction to perform? 10-11 or 11-10? It really is uncertain.

If one makes the argument that "10 and 11" is different from "11 and 10": I believe that the word "AND" is commutative in almost all contexts, for example saying "shirts and pants" is the same as saying "pants and shirts". Therefore, we are given no idea as to which order to perform the subtraction.

So essentially we are being asked to perform the subtraction of two quantities without knowing the order. This is why I think that they should accept both 1 and -1 as correct answers.

OMG NASTOLGIA STOP BEING SUCH A HATER. WHERE IS THE <3 ?

Chillaxing? No problem...I'm already trying to reference random theorems obscure by MCAS standards. Makes MCAS actually fun...erm...

lol we should make a poll to see how many people were bored enough to say they differentiated that linear function to get the slope.

Ah...MCAS...those were the days...

Anyway it would shock me if it weren't 1, given MCAS stuff I have seen...

yeah but if i was walking down the street and i said something like "my stock rose -1 points" or whatever, it would be pretty weird. that was a pretty bad analogy.

I think by "numerical difference" they didn't want you putting in words the difference between what the word "median" means and what the word "mode" means.

If they just said "what is the difference" you could technically say "the median is the middle number in an ordered list of numbers while the mode is the number(s) that appear most frequently".  The should have asked for the "positive difference" or "positive numerical difference".

But I echo Catalyst.....chillax!  If you are even logged into this site you are almost guaranteed Advanced!

Yongyi, I think it is time to move past Nationals.  What happened, happened and dwelling on it can't help.

[quote="Yongyi781"]Unless something really bad happens to you like it does at Nationals. :|[/quote]

I did much worse than you did my first year, and to cap it off, I had two extremely disappointing years, the second much more so than the first even though I did much better.

I only had one disappointing year at nats :( Then I had an extremely disappointing one at states...Lol. I'm still baffled that I was able to think enough after that to beat 4 people in cd...

This is getting slightly off topic... Catalyst shouldn't be reminiscing his days of MathCounts, and I shouldn't have brought it up in the first place.

That said, does anyone remember the easiest question on MCAS ever? It was like, "What is..." but I forget. :)

That was definitely not a reminiscence, it was an effort to get you to stop talking about mathcounts (at least not in this thread).

I differentiated the linear thing too. actually, I managed calculus on half my open responses. Sooooooo bored.

Yeah, I've used calculus on MCAS open response before :D

I'm jealous... I never get the opportunity to use calculus on an open response. All of the 8th grade ones are like "What does this represent?" "What does that represent?" "What is the probability that..." "Which value goes next?" "blah blah blah", no area questions (integration!) or slopes (differentiation!). Yeah, pretty disappointing. :(

wait till 10th grade. it'll be pushing it, but forcing calc will be possible at times. I think half could get it forced, half were still too simple. Stupid MCAS. Oh, and physics sometimes has forceable calc.

I remember some math MCAS where I had like an average of 2/4 on all the open response...hmmm...

:huh: ...why has this thread been brought back from the dead?

Because of me! :ninja: Bahahahahahaha :D

what grade MCAS was this on?

the answer might change based on grade.

Personally I think it's -1 since they put 10 before 11 in the question.  :P


trigonometry
combinatorics proposed
combinatorics

we have $n+1$ points on a cricle with radius $1$.prove that there is exist two point such that the distance of them,maximum is $2sin(\frac{\pi}{n}$. :D

It would be nice if you would state understandable problems, at the last one I'm even not sure if you have read over what you wrote at least one time.
Did you mean:
When $n+1$ points are drawn on a circle with radius one, then there are two that have at most a distance of ${2\sin( \frac{\pi}{n} })$¿ Because then it's trivial and should be moved to some other forum in my oppinion.

sorry for that,because i was in a hurry.but i think it is good for this forum because it is a combinatoric problem. :?:

It's really easy, so I think PreOlympiad or so would be more appropriated...
By the way: you can change that $n+1$ into $n$.


ceiling function
number theory unsolved
number theory

WLOG, $a\leqslant b\leqslant c$.
$a+b|a^{2}+b^{2}\Rightarrow a+b|2a^{2}\Rightarrow 2a+2b|(2a)^{2}$.
\[p^{k}\parallel 2a+2b\Rightarrow p^{k}|(2a)^{2}\Rightarrow p^{\lceil\frac{k}{2}\rceil}|2a\Rightarrow p^{\lceil\frac{k}{2}\rceil}|2b\Rightarrow p^{\lceil\frac{k}{2}\rceil}|(2a,2b)=2(a,b)\Rightarrow 2a+2b|(2(a,b))^{2}\Rightarrow a+b|2(a,b)^{2}\Rightarrow 2a\leqslant a+b\leqslant 2(a,b)^{2}\]
Similarly $a+c|2(a,c)^{2}$. Hence
\[2a\leqslant a+c\leqslant 2(a,c)^{2}=2((a,b)\cdot\frac a{(a,b)},c)^{2}=2(\frac a{(a,b)},c)^{2}\leqslant 2(\frac a{(a,b)})^{2}\leqslant 2a \]
Hence $a=b=c=(a,b,c)=1$


limit
real analysis
real analysis unsolved

Let $x_0>0$ and for any positive  integer $n$ we define the sequence $(x_n)_{n\geq0}$ by considering $x_{n+1}=x_n+\frac{1}{\sqrt{x_n}}$. Find a).$\displaystyle \lim_{n\to\infty}x_n$  . b)$\displaystyle \lim_{n\to\infty}\frac{x_n^3}{n^2}$

(a) is easy. The sequence is increasing, and any limit must satisfy the equation $x=x+\frac1{\sqrt{x}},$ which has no real roots. Hence $\lim x_n=\infty.$

If done a little numerical fiddling to take a look at (b). The sequence $\frac{x_n^3}{n^2}$ decreases for $n>9,$ so it has a limit. Extrapolation suggests that the limit might be near $2.25.$

Added in edit: I can see an argument that says the limit must be exactly $\left(\frac32\right)^2=2.25.$ Details later.

Thanks, i've found the same answer as you, but i was a bit confused because i expected something more beautiful for the answer.
BTW, could you please write you proof for the asymptotic formula-i think our approaches are different

my proof:
first: we have $\lim_{n \ to +\infty} \frac{x_{n+1}}{x_n}$
then using stolze theorem:
we have $\lim_{n\to\infty}\frac{x_n^{3/2}}{n}= \lim_{n\to\infty}{x^{3/2}_{n+1}-x^{3/2}_n}$
then using   $\frac {3}{2} ({x_{n+1}-x_n}) \sqrt {x_n} \leq x^{3/2}_{n+1}-x^{3/2}_n \leq \frac {3}{2} ({x_{n+1}-x_n}) \sqrt {x_{n+1}}$
we have  $\lim_{n\to\infty}{x^{3/2}_{n+1}-x^{3/2}_n}=\frac{3}{2}$
and use this way,it is easy to prove what Moubinool  wrote :)


Find all $ n$ such that $ n^4\plus{}4$ is prime.

Thanks.


A coin is weighted so that it comes up heads with probability p and tails with probability 1-p, for some value p such that 0<p<1. I keep flipping the coin until head comes up. wat is the expected value of flips necessary?

[hide]
Let $N$ be the expected value we seek.  Then we get the equation:

$N = (1-p)(1+N)+p$

reason: Consider the first flip, if we get a tail, then we have to flip $N$ times more, plus the first flip.  if we get a head, we're done.

so solving we get $N = \frac{1}{p}$.
[/hide]
Wow...your first post. welcome to AoPS!!! :)

thanks for the reply =D

where did you get the +p though??

Think of that $+p$ as being $+p\cdot 1.$ With probability $p,$ $N=1.$

pkerichang has given the [b]conditional expectation[/b] argument for the expected value of a well-known probability distribution called the geometric distribution.

The key theorem of conditional expectation: If $X$ is a random variable, and $Y$ is another random variable that we choose to condition on, then \[E(X)=E(E(X|Y))\] The right hand side of what pkerichang has written is $E(E(N|Y))$ where $Y$ is the outcome of the first flip.

---

By the way, the two of you met on Saturday - or at least you were in the same room.

just think this way:
if the probability of head is 1 out of 50 (1/50) then obviously you would expect one to come up in 50 flips... so, it's 1/p


inequalities geometry inradius trigonometry

Let $ ABC$ be an obtuse triangle with $ \angle A>90^{\circ}$, and let $ r$ and $ R$ denote its inradius and circumradius.  Prove that

$ \frac{r}{R}\leq \frac{a\sin A}{a\plus{}b\plus{}c}$.

My solution is pretty long . . . I hope someone can come up with a better one.

(a+b+c)r=2Area=bc*sinA

$ bc*sinA \le R*a*sinA$

$ bc/a \le R$

$ 2bc/a \le 2R \equal{} a/sinA$

$ 2bc*sinA \le a^2$

4Area $ \le$ a^2

$ 4\sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)} \le a^2$ <--nvm

$ \sqrt {(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(a \minus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c)} \le a^2$ <--nvm

Edit: to show 4A is less than a^2, make a square with side length a, and place the triangle such that the triangle's side is lying on one side of the square, and the tip is facing inwards. Then Draw a semicircle internally on the side containing the triangle. From the condition <A is less than 90 degrees, the vertex A lies strictly inside the semicircle, so is less than a quarter of the area (it is quite obvious to see that the largest possible triangle would be 1/4 the area). Therefore, we are done. (And we can easily see it is strict).

Nice!
[hide]
It took me a lot longer to deduce $ a^2>4K$ - I drew the altitude $ h$ to $ a$, splitting $ a$ into $ m$ and $ n$.  Then

$ h^2\equal{}b^2\minus{}m^2\equal{}c^2\minus{}n^2$

Since $ a^2>b^2\plus{}c^2$,

$ b^2\minus{}m^2\plus{}c^2\minus{}n^2\equal{}2h^2<a^2\minus{}m^2\minus{}n^2$.

We know from AM-GM that since $ m\plus{}n\equal{}a$, $ a^2\minus{}m^2\minus{}n^2<\frac{a^2}{2}$.

So $ a^2>4h^2\Rightarrow a>2h$.  $ K\equal{}\frac{1}{2}ah$, so $ a^2>4K$.

Blah . . . I should have seen the easier method.

The rest of my solution was the same as yours.
[/hide]

r*(a+b+c)/2 <= a*R*sin(A)/2

Because the left hand side is the area of ABC, it can be replaced by b*c*sinA/2. Also

R= a/(2*sin(a))

With those replacements the new inequality becomes,

1)     2*b*c <= a^2

Because A is an obtuse angle, we get the relation.

a^2 > b^2 + c^2

adding and subtracting 2*b*c from the right hand side we get,

a^2 > 2*b*c + (b-c)^2

which of course implies 1.

Ah nice -- should've remembered the Pythagorean [i]In[/i]equality! :)

[u]Edit:[/u] Now that I notice, that's the same thing as my diagram, so I [i]really[/i] should've remembered it :(

[quote="E^(pi*i)=-1"] [color=darkred]Let $ ABC$ be a triangle with the circumradius $ R$ , the inradius $ r$ and $ A\ge 90^{\circ}$ .  Prove that $ \frac {r}{R}\le \frac {a\sin A}{a \plus{} b \plus{} c}$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Consider the semidisk $ S_d\left(O,\frac a2\right)$ with the diameter $ [BC]$ which contains the vertex $ A$ (its center $ O$ is the midpoint of the side $ [BC]$ ).

Denote the point $ D\in (BC)$ for which $ AD\perp BC$ . Observe that $ A\ge 90^{\circ}\Longleftrightarrow$ $ OA\le \frac a2$ . Since $ h_a\equiv AD\le OA$ obtain $ h_a\le \frac a2$ , i.e.

$ ah_a\le a\cdot \frac {a}{2}\ \Longleftrightarrow\ 2pr\le a\cdot R\sin A\ \Longleftrightarrow\ \frac {r}{R}\le \frac {a\sin A}{a \plus{} b \plus{} c}$ with equality iff $ A\equal{}90^{\circ}$ and $ AB\equal{}AC$ , i.e. $ \triangle\ ABC$ is $ A$ - right and isosceles.

[b]Remark.[/b] I used only the relations $ S\equal{}\frac {ah_a}{2}\equal{}pr$ and $ a\equal{}2R\sin A$ , where $ 2p\equal{}a\plus{}b\plus{}c$ .[/color]

[quote="The Zuton Force"]I did this sorta quickly, so it could probably be condensed to utilize fewer steps/substitutions, but whatever. If I'm not mistaken, we can make that a strict inequality, too.

[hide="Solution"]In triangle ABC extend side BA to some point D so that BDC is a right triangle with right angle at D and hypotenuse of length a. Now, consider the attached diagram. It's easy to see that:
$ 4[ABC] < 4 [BCD] \leq a^2$

Let $ K \equal{} [ABC]$ from hereon: $ 4K < a^2$

Multiply by $ \frac {bc}{4K}$ to get $ bc < \frac {a^2}{4K} bc$

Now use $ R \equal{} \frac {abc}{4K}$ to get: $ bc < Ra \implies bc \sin A < Ra \sin A$

Now, using $ K \equal{} \frac {1}{2} bc \sin A$, we get $ 2K < Ra \sin A$

Finally, note that $ K \equal{} rs \implies 2K \equal{} r(a \plus{} b \plus{} c)$. Therefore: $ r(a \plus{} b \plus{} c) < Ra \sin A$

$ \iff \frac {r}{R} < \frac {a\sin A}{a \plus{} b \plus{} c}$, as desired.[/hide][/quote]

Nice solution or we can simply do

$ S(ABC) < S(EBCF)$ ==>$ S(ABC) < a^2$

but we should prouve that $ a> max(b.c)$ and that clear because $ A>90>(B\plus{}C)$

we can say that [b]The Zuton Force[/b] has the best solution

[quote="E^(pi*i)=-1"] [color=darkred]Let $ ABC$ be a triangle with the circumradius $ R$ , the inradius $ r$ and $ A\ge 90^{\circ}$ .  Prove that $ \frac {r}{R}\le \frac {a\sin A}{a \plus{} b \plus{} c}$ .[/color] [/quote]

[color=darkblue][b][u]Proof I.[/u][/b] Consider the semidisk $ S_d\left(O,\frac a2\right)$ with the diameter $ [BC]$ which contains the vertex $ A$ (its center $ O$ is the midpoint of the side $ [BC]$ ).

Denote the point $ D\in (BC)$ for which $ AD\perp BC$ . Observe that $ A\ge 90^{\circ}\Longleftrightarrow$ $ OA\le \frac a2$ . Since $ h_a\equiv AD\le OA$ obtain $ h_a\le \frac a2$ , i.e.

$ ah_a\le a\cdot \frac {a}{2}\ \Longleftrightarrow\ 2pr\le a\cdot R\sin A\ \Longleftrightarrow\ \boxed {\ \frac {r}{R}\le \frac {a\sin A}{a \plus{} b \plus{} c}\ }$ with equality iff $ A \equal{} 90^{\circ}$ and $ AB \equal{} AC$ , i.e. $ \triangle\ ABC$ is $ A$ - right and isosceles.

[b][u]Proof II.[/u][/b] From the well-known characterization $ A\ge 90^{\circ}\ \stackrel {(*)}{\Longleftrightarrow}\ 2m_a\le a$ and the evident relation $ h_a\le m_a$ obtain $ 2h_a\le a$ $ \implies$

$ 4S\le a^2$ $ \implies$ $ 2pr\le a\cdot\frac a2$ $ \implies$ $ 2pr\le a\cdot R\sin A$ $ \implies$ $ \boxed {\ \frac {r}{R}\le \frac {a\sin A}{a \plus{} b \plus{} c}\ }$ with equality iff $ h_a\equal{}m_a\equal{}\frac a2$ , i.e. $ b\equal{}c$ and $ A\equal{}90^{\circ}$ .

[b]Remark.[/b] I used only the relations $ S \equal{} \frac {ah_a}{2} \equal{} pr$ , $ 4m_a^2\equal{}2\left(b^2\plus{}c^2\right)\minus{}a^2$ and $ a \equal{} 2R\sin A$ , where $ 2p \equal{} a \plus{} b \plus{} c$ .

$ (*)\ \ 2m_a\le a\ \Longleftrightarrow\ 4m_a^2\le a^2\ \Longleftrightarrow\ 2\left(b^2\plus{}c^2\right)\minus{}$ $ a^2\le a^2\ \Longleftrightarrow\ b^2\plus{}c^2\le a^2\ \Longleftrightarrow\ A\ge 90^{\circ}$ , i.e. $ \boxed {A\ge 90^{\circ}\ \Longleftrightarrow\ 2m_a\le a\ }$ .[/color]


function
number theory solved
number theory

Hello Freinds.
                    This problem is from Elimination Round of St.PBMO 2002.

 Given positive integers 'a' and 'b' , 2a+1 and 2b+1 are coprime. Find all possible values of greatest common divisors of  2^(2a+1)+2^(a+1)+1  and
 2^(2a+1)+2^(a+1)+1 .

 This is the beauty of  St.PBMO problems.  :D

[quote="Euler_India"]Given positive integers 'a' and 'b' , 2a+1 and 2b+1 are coprime. Find all possible values of greatest common divisors of  2^(2a+1)+2^(a+1)+1  and   2^(2a+1)+2^(a+1)+1 .[/quote]

Very beatiful: a statement in function of a alone for 2a+1 and 2b+1 coprime :D

I think you missed a little part ;)

I think the answer is $5$.
take $p$ dividing $2^{2a+1}+a^{a+1}+1$ and $2^{2b+1}+2^{b+1}+1$.
Then $(2^{a+1}+1)^2+1$ and its analogous for $b$ are divisible by $p$.
Let $k^2=-1 (mod..p)$ Then Either $2^{a+1}+1=2^{b+1}+1=k(mod..p)$ or
$2^{a+1}+1=k=-2^{b+1}-1$
Now  $2^{4a+4}=(k-1)^4=(-2k)^2=4(mod..p)$
Hence $2^{4a+2}=1(mod..p)$. Now if we denote $d$ the order of $2$ mod $p$ we deduce $d|4a+2$ and $d|4b+2$ which means $d=2$ since $(2a+1,2b+1)=1$ which implies $p=5$.
Hnce our divisor is a power of $5$.
Now if this power is at least $25$ we deduce
$2^{a+1}+1=7,18$ mod $25$
hence $2^{2a+1}, 2^{2b+1}=18,7$ mod $25$
Now since $2^x=18 $ (mod 5) if $x=15,5$ (mod 20$ we deduce $2a+1, 2b+1$ are not coprime
QED


Show that if the function: $f: R \to R$ is uniformly continuous, then the sequence of functions $f(x+1), f(x+\frac{1}{2}), f(x+\frac{1}{3}), \ldots f(x+\frac{1}{n}) \ldots$ is uniformly convergent.

I can't get this problem. I don't know where to start. I guess we can begin with defining $f$ has uniformly continuous, by saying for $p, q \in R$, $\exists \delta$ such that whenever $|p-q|< \delta$, we have $f(p)-f(q) < \epsilon$, for any fixed $\epsilon > 0$.

We can say a parallel definition for the convergence of functions, but I'm guessing that we need to note that $\frac{1}{n}$ gets smaller as $n$ gets larger. But what do I do next?

Thanks.

Ok, now let $q = x$, and let $p = x+\frac{1}{n}$.

Ok, so then $m$ be the smallest natural number such that $\frac{1}{m}\leq  \delta$.  Then, if $p-q < \frac{1}{m}\leq  \delta$, $|f(p)-f(q)| < \epsilon$.

Now, $p-q = (x+\frac{1}{n})-x = \frac{1}{n}$.

Thus, if $\frac{1}{n}< \frac{1}{m}$, then $|f(x+\frac{1}{n})-f(x)| < \epsilon$.

And $\frac{1}{n}< \frac{1}{m}$ $\iff$ $n > m$.

Therefore, $\forall \epsilon \in \mathbb{R}$, $\epsilon > 0$, $\exists m \in \mathbb{N}$ such that $\forall n \in \mathbb{N}$, $n \geq  m$, $|f(x+\frac{1}{n})-f(x)| < \epsilon$.


Donote :

$ p \equal{} a \plus{} b \plus{} c$      and     $ q \equal{} ab \plus{} ac \plus{} bc$   and    $ r \equal{} abc$    with     $ (a,b,c\,\geq\,0)$

by schur :     $ a^t(a \minus{} b)(a \minus{} c) \plus{} b^t(b \minus{} a)(b \minus{} c) \plus{} c^t(c \minus{} a)(c \minus{} b)\,\geq\,0$
with     $ t \equal{} 0$  we have :

$ (a \minus{} b)(a \minus{} c) \plus{} (b \minus{} a)(b \minus{} c) \plus{} (c \minus{} a)(c \minus{} b)\,\geq\,0$  $ \iff$   $ pq \minus{} 9r\,\iff\,0$    (1)

and with $ t \equal{} 1$  we have:

$ a(a \minus{} b)(a \minus{} c) \plus{} b(b \minus{} c)(b \minus{} a) \plus{} c(c \minus{} a)(c \minus{} b)\,\geq\,0$   $ \iff$   $ p^3 \minus{} 4pq \plus{} 9r\,\geq\,0$    (2)

and with $ t \equal{} 3$ we have :

$ a^2(a \minus{} b)(a \minus{} c) \plus{} b^2(b \minus{} a)(b \minus{} c) \plus{} c^2(c \minus{} a)(c \minus{} b)\,\geq\,0$ $ \iff$  $ p^4 \minus{} 5p^2q \plus{} 4q^ \plus{} 6pr\,\geq\,0$(3)

(1) and (2)   $ \iff$           $ \frac {4pq \minus{} p^3}{9}\,\preceq\,r\,\preceq\,\frac {pq}{9}$

and :                              $ \frac {9r}{p}\,\preceq\,q\,\preceq\,\frac {p^3 \plus{} 9r}{4p}$

[color=red]Formulas to know:[/color]

$ (a \plus{} b)(a \plus{} c)(b \plus{} c) \equal{} pq \minus{} r$

$ a^2 \plus{} b^2 \plus{} c^2 \equal{} p^2 \minus{} 2q$

$ a^3 \plus{} b^3 \plus{} c^3 \equal{} p^3 \minus{} 3pq \plus{} 3r$

$ a^4 \plus{} b^4 \plus{} c^4 \equal{} p^4 \minus{} 4p^2q \plus{} 2q^2 \plus{} 4pr$

$ a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2 \equal{} q^2 \minus{} 2pr$

$ a^3b^3 \plus{} a^3c^3 \plus{} b^3c^3 \equal{} q^3 \minus{} 3pqr \minus{} 3r^2$

$ ab(a \plus{} b) \plus{} ac(a \plus{} c) \plus{} bc(b \plus{} c) \equal{} pq \minus{} 3r$

$ ab(a^2 \plus{} b^2) \plus{} ac(a^2 \plus{} c^2) \plus{} bc(b^2 \plus{} c^2) \equal{} p^2q \minus{} 2q^2 \minus{} pr$

$ (a \plus{} 1)(b \plus{} 1)(c \plus{} 1) \equal{} p \plus{} q \plus{} r \plus{} 1$

$ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \equal{} p^2 \plus{} q^2 \plus{} r^2 \minus{} 2pr \minus{} 2q \plus{} 1$

$ (a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1) \equal{} p^3 \plus{} q^3 \minus{} 6pq \plus{} 8$

[color=red]Exemple :[/color]

$ a,b,c\,\geq\,0$  such that :  $ ab \plus{} ac \plus{} bc \plus{} abc \equal{} 4$.   Prove that :
\[ 3(a^2 \plus{} b^2 \plus{} c^2) \plus{} abc\,\geq\,10
\]
[color=green]Solution :[/color]

$ a^2 \plus{} b^2 \plus{} c^2 \equal{} p^2 \minus{} 2q$

and $ q \plus{} r \equal{} 4$
so:

$ 3(a^2 \plus{} b^2 \plus{} c^2) \plus{} abc\,\geq\,10$    $ \iff$   $ 3p^2 \minus{} 7q \minus{} 6\,\geq\,0$  (1)

by schur (2) we have with $ r \equal{} 4 \minus{} q$

$ q\,\preceq\,\frac {p^3 \plus{} 9r}{4p} \equal{} \frac {36 \plus{} p^3}{4p \plus{} 9}$

so:

(1)  $ \iff$    $ 3p^2 \minus{} \frac {7(36 \plus{} p^3)}{4p \plus{} 9} \minus{} 6\,\geq\,0$

$ \iff$   $ \frac {(p \minus{} 3)(5p^2 \plus{} 42p \plus{} 102)}{4p \plus{} 9}\,\geq\,0$

and since  $ 5p^2 \plus{} 42p \plus{} 102\,\geq\,0$  . so we have to show $ p \minus{} 3\,\geq\,0$

by  AM-GM  :   $ ab \plus{} ac \plus{} bc \plus{} abc \equal{} 4\,\geq\,4\sqrt [4]{(abc)^3}$ $ \iff$  $ 1\,\geq\,abc$

But     $ (a \plus{} b \plus{} c)^2\,\geq\,3(ab \plus{} ac \plus{} bc)$  $ \iff$  $ a \plus{} b \plus{} c\,\geq\,\sqrt {3(4 \minus{} abc}\,\geq\,3$

so ; $ p \minus{} 3\,\geq\,0$

[quote="DIMITRIS T"][quote="zaizai-hoang"]I think you copied from my article because the example is also my problem. Anyway, it is not important. But the example is very very easy.
\[ 3(a^2 \plus{} b^2 \plus{} c^2) \plus{} abc\ge 3(ab \plus{} bc \plus{} ca) \plus{} abc \equal{} 3(4 \minus{} abc) \plus{} abc \equal{} 12 \minus{} 2abc\ge 10
\]
which is always true by the AM-GM inequality we have $ abc\le 1$
In my opinion, this technique can solve almost of three-variables symmetric inequality.  :wink:[/quote]
Can u give a link for your article?
Thanks![/quote]
See here: http://batdangthuc.net/showthread.php?t=278

I don't understand why (1) <=> $ pq \minus{} 9r \geq 0$.

When I expand (a - b)(a - c) + .... I don't have the expanding of pq - 9r

See here: http://batdangthuc.net/showthread.php?t=278 <=== check this link for me (cai link die mat tieu goi >_< )[/quote]


ba salam khedmate bozorgan :
agar momken ast kasi mara dar hale in namosavi rahnamayi konad.
ba arze pozesh az nadashtane LAtex
agar a,b,c>o va bedanim a+b+c>abc sabet konid:
a^2 +b^2+c^2>abc

be ghole ramtin madani :
http://www.artofproblemsolving.com/Forum/files/latexhelp_140.pdf
rahnamaye latex dare ;)

sallam mohammad jan manzooram az soale sakht, to ye topic e digeh, in soal bood...

vay ageh bedooni che rahe halle zibayee dare,hal mikoni...

albatteh man rahe khode aghaye safa ro nakhoondam...

baba soalesh ghashange va baraye man sakht boodeh
khodamam hallesh nakardam
nemidoonam chera shomaha...
 :mad:

[quote="Omid Hatami"]Age eshtebah nakonam soale China West math Olympiad 2004 ast. Dar dorane javani hallesh kardam. Kheili masaleye ziba va moshkeliye.[/quote]
kodoomeshoon ooni ke $x,y,z$dare??
age mishe rahe halleto bezar ma ham estefade bebarim
ya rahnamai kon
merC :)

chon ebarat motagharen ast matavan farz kar $x>y>z$
hala be gaye$\sqrt{y^2 +z^2}$mishe gozasht$\sqrt2y$
be hamin tartib baraye $x,z$ pas masle tabdil mishe be
${\sqrt2+\frac{z}{\sqrt{x^2+y^2}}}>2$ ;)

$a=x^2$,$b=y^2$,$c=z^2$
$\Rightarrow \sqrt\frac{a}{b+c}+\sqrt\frac{b}{a+c}+\sqrt\frac{c}{a+b}>2$
darim:
$\sqrt{\frac{a}{b+c}}\geq\frac{2a}{a+b+c}$

$\sqrt{\frac{b}{a+c}}\geq\frac{2b}{a+b+c}$

$\sqrt{\frac{c}{a+b}}\geq\frac{2c}{a+b+c}$

hal in ravabet ra jam mizanim:
$\frac{2a+2b+2c}{a+b+c}=2$
va masale esbat mishavad ;)

[quote="fatehi"]latex lazem nadari mohamdhosein, bozorgan be man yad dadan ke agar "dota dollar sign avval va akhare ebarate riyazit begzari doroste" (ashegh)!

\[ a,b,c>o  \] va \[ a+b+c>abc  \] sabet konid
\[ a^2 +b^2+c^2>abc  \][/quote]
midanim:$(a^2+b^2+c^2)^2>abc(a+b+c)>(abc)^2$
$\Rightarrow (a^2+b^2+c^2)^2>(abc)^2$
$\Rightarrow a^2+b^2+c^2>abc$
va masale esbat mishavad ;)

ye soale tabe sakh vali jaleb:
$f(x)=x^3+x+1$
$f^{-1}(x)=?$

[quote="BaBaK Ghalebi"][quote="Ramtin Madani"]$f^{-1}(x)=\sqrt[3]{\frac{x-1}{2}+\sqrt{\frac{(x-1)^2}{4}+\frac{1}{27}}}+\sqrt[3]{\frac{x-1}{2}-\sqrt{\frac{(x-1)^2}{4}+\frac{1}{27}}}$[/quote]
wow!!
age mishe rahe halleto bezar
bad kasi ke io be man dad goft makoosesh bar hasbe tavabe mosalasati be dast miad
nazaret chie?? :?[/quote]

moaadele daraje sevom ro dar haalate koli mishe bar hasbe raadikaal bayaan kard.
man ham in formoolo az too majale didam. age  too moaadele gharaaresh bedi raahat saabet mishe ;)

ramtin rooye nesbataye mosalasatisham fekr kon ;)

neshan dahid ke:
$\frac{xyz}{x^3+y^3+xyz}+\frac{xyz}{y^3+z^3+xyz}+\frac{xyz}{z^3+x^3+xyz}\leq 1$
(baraye hameye adade haghighi va mosbate $x,y,z$ )

agar $a^2+b^2+c^2=3$
angah sabet konid:
$|a|+|b|+|c|-abc\leq 4$
( $a,b,c$ se adade haghighi hastand)

tamame tavabe haghighi rooye adade haghighi ra biabid be tori ke:
$f(x^2-y^2)=xf(x)-yf(y)$
(baraye har $x,y$ haghighi)

[quote="BaBaK Ghalebi"]tabe $f$ rooye tamae adade tabii pooshast va:
$\{\begin{array}{c}{1cr}f: N\to N\\forall m,n\in N;f(f(n)+f(m))=m+n\end{array}$
tamame maghadire momken baraye $f(1988)$ ra biabid[/quote]
sakhte??
ya arzeshe fekr kardan nadare??

[quote="Omid Hatami"]Be vozooh $f$ 1-1 ast. Dar zemn ba ezaye har $k\in \mathbb N\backslash\{1\}$ ye $k'$ hast ke $f(k')=k$. Pas $f(m)+f(n)=f(m+n-1)+f(1)$. Ager bezarim $n=2$ angah $f(m+1)-f(m)=f(2)-f(1)$. Pas chon har adade bozorgtar aaz $1$ dar borde tabe hast pas $f(2)-f(1)=1$. Hala $f(n+1)=f(n)+1=f(n-1)+2=\dots=f(1)+n$. Pas \[ m+n+4(f(1)-1)=f(m+n+2(f(1)-1))=f(f(m)+f(n))=m+n  \]
Yani $f(1)=1$ va $f(n)=n$ be ezaye har $n\in\mathbb N$[/quote]
merC :D

[quote="Babak Ghalebi"]$\sqrt{\frac{a}{b+c}}\geq\frac{2a}{a+b+c}$[/quote]
Chera?

mibakhshid agar soalam ahmaghaneast.Choon man sale avalam taze olympiado shoro kardam.

mitooni saadash koni va oon moghe be rahati mifahmi chera(yani khatiish koni)

[quote="Ramtin Madani"][quote="BaBaK Ghalebi"]If $a,b,c$ are positive reals such that $a+b+c=1$ , prove that
\[ 10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\geq 1  \]. :?:[/quote]

baayad saabet konim:
$10(a+b+c)^2(a^3+b^3+c^3)-9(a^5+b^5+c^5)\geq (a+b+c)^5$
agar hamichiz raa zarb koni va saade koni dar nahaayat bein naamosaavi miresi:
$a^4b+a^4c+b^4a+b^4c+c^4a+c^4b\geq 2(a^2b^2c+a^2bc^2+ab^2c^2)$
ke baa morhed dar miaad ;)[/quote]
merC
agha che ajab ma shoma ro didim
kam peidai
saret sholoogh shode dige be ma sary nemizani ;)  :D

[quote="amir2"]$n\in \mathbb{N}$ and |$a,b>0$ prove:
$(a+b)^2(a^2+b^2)^2\cdots(a^n+b^n)^2\ge (a^{n+1}+b^{n+1})^n$[/quote]
tebghe namosaviye kooshi darim
$(a^i+b^i)(a^{n+1-i}b^{n+1-i}) \ge (a^{\frac{n+1}{2}}+b^{\frac{n+1}{2}})^2$
az zarb in ravabet dar ham darim:
$(a+b)^2(a^2+b^2)^2...(a^n+b^n)^2 \ge (a^{\frac{n+1}2}+b^{\frac{n+1}2})^{2n}$
pas bayad sabet konim:
$(a^{\frac{n+1}2}+b^{\frac{n+1}2})^{2n} \ge (a^{n+1}+b^{n+1})^n$
dar natige:
$(a^{\frac{n+1}2}+b^{\frac{n+1}2})^2 \ge a^{n+1}+b^{n+1}$
ke age ino baz konim be rahati sanet mishe. :lol:

[quote="amir2"]$x,y,z>0$ prove:
$(x+y+z)^{x+y+z}\ge x^xy^yz^z$[/quote]
ya man joob zadam ya inke ... :lol:

[quote="amir2"]$\{a_i\}_{i=1}^n\ ,\ \{b_i\}_{i=1}^n$ be two sequence of nonegetive integer and $b_i\le a_i\ \ ,\ \ 1\le i\le n$
Prove:

$\prod_{i-1}^n\binom {a_i}{b_i}\le\binom {a_1+\cdots+a_n}{b_1+\cdots+b_n}$[/quote]
farz konim madresie $n$ kelas darad va kelase $i$ $a_i$nafar jamyiat darad$1 \le i \le n$
agar az kelase $i$ $b_i$ nafar ra entekhab konim $1 \le i \le n$ tedad rahhaye entekhab mishavad samte chape namosavi.
hal agar bekhahim az kole danesh amoozane madrese$b_1+b_2+...+b_n$nafar ra entekhab konim (ke in entekhab shaamele entekhabe ghable mishavad)mishavad samte raste tasavi pas hamishe samte rast bozorgtar mosaviye samte chap mishavad.
joob nazaram?... ;)


[hide]Unless there is a trick it is

$98*\frac{0.5}{0.7}=70$

$70$[/hide]

Are the containers equal?  I assume so.  But if the first container is exactly .5 mm wide, 49 mm long, and the length of the stick in height, than the 2nd container can hold 0 .7mm sticks of lead.

i fixed it, i always leave it the most important sentence >_>

[hide]The ratio of the volume of a .5mm lead stick to a .7 mm lead stick is $\frac{25}{49}$.  Call the volume of a .5mm lead stick $x$.  Then the volume of each container is $98x$.  Each .7 mm lead stick is $\frac{49}{25}x$ So there are $\frac{98x}{\frac{49}{25}x}$ or 50 lead sticks in the second container.[/hide]

[hide]The diameter is in a $\frac57$ ratio. Volume is cubed so $(\frac{5}{7})^{3}=\frac{125}{343}(98)=\frac{250}{7}$ sticks[/hide] I think I did something wrong.

[quote="Treething"]read the whole length thing[/quote]
I did. After looking at other's solution I notice that no one else did cubing. Don't you have to do cubing for volume?

No. Notice how I said that the lengths are the same =x

Not if one of the dimensions stay the same.

[quote="mathgeniuse^ln(x)"]Good Job Treething

Your Questions have tricked another.

I was about to consider volume, but i read it carefully, and got it right.[/quote]

But you only took into consideration one dimension.  Shouldn't you square?

I see why I am wrong. Also to math genius, your answer is wrong...

[hide]
The area of a cylinder of lead is:
Since the height $h$ is the same for both kinds of lead, we can cancel out the h's.
$.25^{2}\pi \cdot h=.0625 \pi h\Rightarrow.0625 \pi \cdot 98=6.125 \pi$

$.35^{2}\pi \cdot h=.1225 \pi h\Rightarrow.1225 \pi \cdot x=.1225x \pi$
So:
$.1225x \pi=6.125 \pi$
$1225x=61250$
$x=\boxed{50}$[/hide]

[quote="mathgeniuse^ln(x)"]Good Job Treething

Your Questions have tricked another.

I was about to consider volume, but i read it carefully, and got it right.[/quote]

Bold statement for someone who was tricked and got it wrong.

[quote="Treething"]Treething has two containers of lead. One contains 98 sticks of lead, and each stick of lead is 0.5 mm. in diameter. He has another container that contains sticks of lead that are 0.7 mm. in diameter and exactly as long. Assuming that each container contains the same amount of lead by volume, how many sticks of lead does this container have?[/quote]

[hide]50!
Calculating volume of the 0.5mm leads:
$0.0625h\pi$
Multiplying that by 98:
$6.125h\pi$

For the 0.7mm leads:
Each one is: $0.1225h\pi$
Since $h$ is the same for both leads you divide:

$\frac{6.125h\pi}{0.1225h\pi}=\boxed{50\text{ Lead Pieces}}$[/hide]

[size=75][color=darkred]Please hide your answers.
-nebula42[/color][/size]

[quote="SplashD"]isn't there space betweeen pieces of lead though? you can't fit [b]round[/b] pieces of lead perfectly into a container. are you taking that into account?[/quote]
The volume of the [b]LED[/b] is the same he said...

[quote="SplashD"]Well, my question would make for an interesting extension.[/quote]
It would be WAY too much work...


calculus
integration
trigonometry
algebra
partial fractions
calculus computations

I got a midterm next week, and I would really appreciate it if someone would check my answers. Sorry if it's a lot. Any bit of help would be much appreciated.

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

Problem: (Partial Fractions)
$\int{\frac{x^2+x-7}{x^3(x+2)} dx}$

My answer:
$\frac{5}{8}\ln|{\frac{x+2}{x}}|-\frac{9}{4x}+\frac{7}{4x^2}+C$

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

Problem: (Trig Sub)
$\int{\frac{\sqrt{x^2-4}}{x} dx}$

My answer:
$\sqrt{x^2-4}-2\sec{}^-{}^1{\frac{x}{2}}+C$

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

Problem: (Partial Fractions)
$\int{\frac{6}{x(x^2+9)}dx}$

My answer:
$\frac{2}{3}\ln|{x}|-\frac{1}{3}\ln|{x^2+9}|+C$

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

Problem: (Partial Fractions)
$\int{\frac{-5x}{(x^2+7)(x-2)}dx}$

My answer:
$\frac{5}{11}\ln|{x^2+7}|-\frac{10}{11}\ln|{x-2}|+\frac{35}{11}\int{\frac{1}{x^2+7}}$

I don't know how to integrate the last part : $\frac{35}{11}\int{\frac{1}{x^2+7}}$
I know I have to use arctan because the derivate of arctan is $\frac{1}{1+x^2}$ but I'm afraid to touch it.

Thanks in advance.

A formula: $\int\frac{du}{u^2+a^2}=\frac1a\tan^{-1}\left(\frac ua\right)+C.$

This can be verified simply by differentiating it (using the chain rule) to check it. It could be found in the first place using the "backwards" substitution $u=av,\,du=a\,dv,$ or with greater effort than necessary using the trig substitution $u=a\tan\theta.$

In your last example, that would make $a=\sqrt{7}.$

In general, since you have proposed answers to nearly all of your problems - have you tried differentiating your answers? Do that, and you may not need to ask us whether or not they are correct.


geometry
LaTeX
modular arithmetic
search
algorithm
induction

My solutions are all under a page. However, my solution to 4 is completely brute forced and my answer for 2 is a little off.

Here are mine. I used analytical geometry for #3 :blush:

don't talk to me about number four.  i am an idiot.

my solution to #3 is pretty.

#1,4,5 was pretty much the same as everyone elses

#2 I got 15 pretty fast using the computer, but I had difficulty proving it.

#3 Different than (and not as elegant as) the solutions so far

This is to make the post ten characters long...

EDIT: Well, in my first solution, I never caught that i had the following typo:

"If n is odd, and n=2k+1, then

[LaTeX formula]

If n is odd, and n=2k, then

[LaTeX formula]"

I hope they won't take off points for that...

[quote="not_trig"]calc rulz:

for #2, how do you know there are no solutions when

$n\equiv1\pmod{5^{2007}}$
and
$n\equiv0\pmod{2^{2007}}$

-or-

$n\equiv-1\pmod{5^{2007}}$
and
$n\equiv0\pmod{2^{2007}}$?

@azjps: And how did you write a program that handled such huge numbers?[/quote]

I said that there are (and there is in fact exactly one for each) solutions to that.

[quote="gregx007"]Lmao.  What sort of computer did you use?  And I'm assuming you used some sort of narrowed search algorithm:  otherwise, you have $10^{2007}$ computations to do, and assuming you finished within 1000 hours (about how long we have to work on the problems), thats still about $10^{2000}$ computations per second, which is way beyond the capability of any computer ever created.[/quote]

Mathematica.

What was the code fragment?  Obviously it did not end up doing a completely brute-force search.

[quote="ehehheehee"]Did you make it loop all the way from 1 to 10^2007 or did you narrow the search or solve the problem by hand first?

And to think that I was being cheap by using Visual Basic and finding the answers for the exponents from 1 to 7.[/quote]

I did that with my calculator.
It probably would've been a ton quicker with a computer.

[quote="not_trig"]
@azjps: And how did you write a program that handled such huge numbers?[/quote]

Sorry, I meant that I did something like in iostream.h’s #2 remark by calculating some of the smaller exponents (it’s fairly reasonable after seeing that smaller exponents return 15, that the answer for 10^2007 would also return 15).

[quote="K81o7"][quote="ehehheehee"]How long does it take for Mathematica to do that?  I never imagined there would be anything that powerful.[/quote]

Well it does $2^{10000000}$ in udner 10 seconds on my computer...[/quote]

Yes, but that is a logarithmic problem I believe.  It doesn't have to actually do $2^{10000000}$ calculations; it could run some code similar to the following:

$2^{a}= 2$, if $a=1$, $2*2^{a-1}$ otherwise.

A brute-force approach is linear in the search range, which is of course exponential overall.  This leads to some rather serious computation time issues, I believe.

Edit:

On second thought, I seem to be getting (quick) results with the following:

Reduce[Mod[n^3 - n, 10^2007] == 0, n, Integers].  I'd be curious to know the inner-workings of the Reduce command.  Also, do you think such a solution, without further justification, would get all 5 points credit?

here they are.

1.
Found S(n) when n was even and odd, then used cases.

2.
Found possibilities for n in mod $2^{2006}$ or $2^{2007}$, and $5^{2007}$. Used this to figure out how many soulutions there were.

3.
<sarcasm> I used geometric properties, to prove the statement and didn't assume something that trivializes the problem because I couldn't solve it. </sarcasm>

4.
Derived recursion,  found $T_{1}$, then used recursion to fing $T_{10}$.

5.
Used induction to find a formula for $A_{2n-1}$ then used it to find $A_{2007}$

min score: 55144 = 19
max score: 55255 = 22

I have no clue about commendations. I might get some.


What is the arithmetic mean of 14, 22 and 36?

(14+22+36)/3=72/3=24.
Note: Arithmetic mean= sum of numbers/amount of numbers

Once quickly notes the sum of 14 and 22 is 36, so its 36x2/3=2x12=24. I think that way as simpler arithmetic, especially if you arent as quick with factors of 72.

There might have been an error in the answer, so it would say 24 is incorrect.

Huh? I dont think anyone has posted yet saying 24 is wrong?

Sometimes FTW glitches and does not have the correct answers for the problems.

I don't think we should necessarily assume that though...


Lines can be drawn through the point (-4,3) so that the sum of the x and y intercepts of each line equals the slope of that line.  What is the sum of the slopes of such lines.
a) 0
b) -1
c) $\frac{1}{2}$
d)$\frac{-5}{4}$
e) None of these

Thank you!

[hide]The line goes through (-4,3) so equation is y=m(x+4)+3.
x=0: y-intercept 4m+3
y=0: m(x+4)+3=0 x-intercept -(3/m)-4
m=4m-(3/m)-1
Subtracting m and multiplying by m, 0=3m^2-m-3
The solutions have sum 1/3 [b](e)[/b].[/hide]


I have not as yet.  I would predict around mid-August for the problems.

[quote="mihail911"]this is my first year participation in the USAMTS...several people have told me that the first round is supposed to be sent somewhere in August.

has anyone received this round yet?[/quote]

Yeah, this is also my first year participating in USAMTS. I havent received the first round. I'm pretty sure its going to be around August 16th.

We'll be sending out the problems and posting them on the website in about 2 weeks.


Let $ a,b$ and $ c$ be positive real numbers. Prove that
\[ 1\plus{}\frac{3}{ab\plus{}bc\plus{}ca}\geq \frac{6}{a\plus{}b\plus{}c}.\]

AM-GM and we r done :
$ LHS \ge 2\sqrt{\frac{3}{\sum ab}}$
$ 2\sqrt{\frac{3}{\sum ab}} \ge 6/(\sum a) \leftrightarrow \sum a^2 \ge \sum ab$
which is true.


This is from BAMO 1999

#10 List x,y, and z in increasing order:

$x = \frac {a+b}{2}$

$y = \sqrt {ab}$

$z = \frac {2}{\frac {1}{a} + \frac {1}{b}}$

With $a \neq b$ and $a,b$ are both positive.

Here is what I did:

[hide="Um"]

x and y is AM-GM Inequality. So:

$x \geq y$

and $z = \frac {2ab}{a+b}$

When I compare it with $\frac {a+b}{2}$, I get:

$2ab \leq a^2+b^2$.

So, $x$ is greatest but I'm not sure for $y$ and $z$.

[EDIT] That inequality appeared on AM-GM too. So, does it mean it depends on $a$ and $b$ for $y$ and $z$?
What's the answer? :? [/hide]

what exactly is am-gm?  and how is it applied?  i keep hearing about it, but i have no idea what it is :?

AM-GM inequality says that the arithmetic mean is greater than or equal to the geometric mean.

In 2 variables, this is:

$\frac{a+b}{2} \geq \sqrt{ab}$

Recall that the geometric mean of numbers $a$ and $b$ is $x$ where $\frac{x}{a} = \frac{b}{x}$

I'm guessing it's usually applied to inequalities although I don't know how or have any specific examples.

[quote="joml88"][hide]Basic application of Power Mean Inequality: $AM\ge GM\ge HM$.

HM means harmonic mean... the harmonic mean of the set $\{a_1,a_2,\ldots,a_n\}$ is:

\[\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}.\][/hide]

Edit:

[hide]
Also, I will show you how to prove that $GM\ge HM$ for two variables.  Start with:

\[\sqrt{ab} ? \frac{2ab}{a+b}.\]

The '?' just means we don't know the relationship and the RHS is just a simplified form of the HM.  So now we square both sides:

\[ab ? \frac{4a^2b^2}{(a+b)^2}.\]

Cancel $ab$ and multiply through by $(a+b)^2$:

\[(a+b)^2 ?4ab.\]

Finally:

\[(a-b)^2 ? 0.\]

By the trivial inequality this is true with equality when $a=b$.  Now...for a formal proof you would need to reverse the steps.[/hide][/quote]

I don't like that whole question mark business.  Starting with $AM\geq GM$, make the substitutions $a=\frac {1}{x}$,$b=\frac {1}{y}$, we get:

$\frac {\frac {1}{x}+\frac{1}{y}}{2}\geq \frac {1}{\sqrt {xy}}$

Now if we flip it (and flip the inequality sign), we get $GM\geq HM$.

For the more advanced of us, this is the general inequality that gives us am-gm and a load others...
                      For all $x_i  \geq 0$,

If $p \geq q$ then $ \displaystyle ({\frac {x_i^{p}+x_2^{p}+....+x_n^{p}} {n}})^ {\frac {1}{p}} \geq ({\frac {x_i^{q}+x_2^{q}+....+x_n^{q}} {n}})^ {\frac {1}{q}}$

Where we define for p=o the GM;
$ \displaystyle {({x_1}{x_2}.....{x_n})}^{\frac {1}{n}}$ with equality in all if all the $x_i's$ are equal. Theres also a weighted version, but im not too sure on that...

[quote]For the more advanced of us, this is the general inequality that gives us am-gm and a load others...
[/quote]

It's known as the "Power Mean" inequality. (just thought I'd add that in.)

Thanks to all people.

blahblahblah, thanks a lot for showing it :) I got it now. :D


algorithm
number theory
Euclidean algorithm
relatively prime

Let $ n$ be a postive integer.

(1) Prove that $ n^2$ and $ 2n\plus{}1$ are coprime.

(2) Find the value of $ n$ such that $ n^2\plus{}2$ is a multiple of $ 2n\plus{}1.$

[hide="1"] Fromt the Euclidean Algorithm, $ gcf(n^2, 2n \plus{} 1) \equal{} gcf(n^2, (n \plus{} 1)^2)$.

However, two consecutive numbers are of different parities and thus relatively prime, so their squares are also relatively prime.[/hide]
[hide="2"]Using the Euclidean Algorithm, again, we eventually get to $ gcf(18, 4n \plus{} 2)$, so 4n+2 is 18 or 6, giving n=1 or n=4.[/hide]

For some reason I thought 4n+2 couldn't be equal to 6.

[hide="1"]
$ \gcd(n,2n\plus{}1)\equal{}1 \Rightarrow gcd(n^2, 2n\plus{}1)\equal{}1$
[/hide]

[hide="2"]
$ f(n)\equal{}4n^2\plus{}8$ is divisible by $ 2n\plus{}1$
$ \Rightarrow f(\minus{}\frac{1}{2})\equal{}9$ is divisible by $ 2n\plus{}1$
$ \Rightarrow n\equal{}1$ or $ 4$

After putting $ n\equal{}1$ and $ n\equal{}4$ into the original question, we see that both of them are solutions.
[/hide]

[/quote][hide="2"]
$ f(n) \equal{} 4n^2 \plus{} 8$ is divisible by $ 2n \plus{} 1$
$ \Rightarrow f( \minus{} \frac {1}{2}) \equal{} 9$ is divisible by $ 2n \plus{} 1$
$ \Rightarrow n \equal{} 1$ or $ 4$

After putting $ n \equal{} 1$ and $ n \equal{} 4$ into the original question, we see that both of them are solutions.
[/hide]

where did you get $ f(n) \equal{} 4n^2 \plus{} 8$?

Also
[hide="no. 2"]
$ n^2 \plus{} 2 \equal{} k(2n \plus{} 1)$

$ \therefore (n \minus{} k)^2 \equal{} k^2 \plus{} k \minus{} 2$

But, $ k^2 \plus{} k \minus{} 2 < k^2 \plus{} 2k \plus{} 1 \equal{} (k \plus{} 1)^2$   for positive $ k$

hence $ k \equal{} 2$ or $ 1$ and this gives $ n \equal{} 4,1$[/hide]

[quote="ocha"][hide="2"]
$ f(n) \equal{} 4n^2 \plus{} 8$ is divisible by $ 2n \plus{} 1$
$ \Rightarrow f( \minus{} \frac {1}{2}) \equal{} 9$ is divisible by $ 2n \plus{} 1$
$ \Rightarrow n \equal{} 1$ or $ 4$

After putting $ n \equal{} 1$ and $ n \equal{} 4$ into the original question, we see that both of them are solutions.
[/hide]

where did you get $ f(n) \equal{} 4n^2 \plus{} 8$?
[/hide][/quote]

$ 2n\plus{}1|n^2\plus{}2 \Leftrightarrow 2n\plus{}1|4n^2\plus{}8$, and I chose to multiply $ 2n\plus{}1$ by $ 4$ bacause I wanted to make $ f(\minus{}\frac{1}{2})$ an integer.

[quote="kunny"]Find the value of $ n$ such that $ n^2 \plus{} 2$ is a multiple of $ 2n \plus{} 1.$[/quote]
$ (n^2 \plus{} 2,2n \plus{} 1) \equal{} (2n^2 \plus{} 4,2n \plus{} 1) \equal{} (2n \plus{} 1,4 \minus{} n) \equal{} (4 \minus{} n,9)$ hence the answer can only possibly be $ 2n \plus{} 1 \equal{} 3\implies n \equal{} 1$ or $ 2n \plus{} 1 \equal{} 9\implies n \equal{} 4$ Check these all works.

@ocha: how did you deduce k=2,1?

@ stephencheng: okay cheers

[quote]
@ocha: how did you deduce k=2,1?[/quote]

We have $ (n \minus{} k)^2 \equal{} k^2 \minus{} k \plus{} 2$ which implies that $ k^2 \plus{} k \minus{} 2$ is square. however $ (k \minus{} 1)^2 < k^2 \plus{} k \minus{} 2 < (k \plus{} 1)^2$

Therefore we must have $ k^2 \minus{} k \plus{} 2 \equal{} k^2$ or $ k^2 \plus{} k \minus{} 2 \equal{} 0$ (we discard the negative root because k>0)[/quote]


calculus
integration
function
limit
ratio
real analysis
real analysis unsolved

Let $ f: \mathbb{R}_\plus{} \to \mathbb{R}_\plus{}$ be a decreasing function which satisfies $ \int_0^\infty f(x) \, dx < \infty$. Prove that $ \lim_{x \to \infty} xf(x) \equal{} 0$.

Sorry but I don't see how it should work :( Would you write something more?

Assume on the contrary: $ \lim_{x \to \infty}xf(x)\neq 0$. Therefore there exists $ \epsilon^{\prime}$, such that there does not exist $ x_{0}$, such that $ xf(x)<\epsilon^{\prime}$ for all $ x>x_0$. Therefore there exists arbitrarily great $ x$, such that $ f(x)\geq\frac{\epsilon^{\prime}}{x}$.

On the other hand, since $ \int_0^\infty f(x) \, dx < \infty$, we can find $ x_{0}$ for each $ \epsilon>0$, such that $ \int_{x_{0}}^\infty f(x) \, dx < \epsilon$.

Now for $ \epsilon^{\prime}$ we find $ x_{0}$, such that $ \int_{x_{0}}^\infty f(x) \, dx < \frac{\epsilon^{\prime}}{2}$. Further on we also find $ x_{1}\geq 2x_{0}$, such that $ f(x_1)\geq \frac{\epsilon^{\prime}}{x_1}$.

So:

$ \int_{x_{0}}^\infty f(x) \, dx = \int_{x_{0}}^{x_1} f(x) \, dx + \int_{x_{1}}^\infty f(x) \, dx \geq\\
\int_{x_{0}}^{x_1} f(x_1) \, dx + \int_{x_{1}}^\infty f(x) \, dx \geq
\int_{x_{0}}^{x_1}  \frac{\epsilon^{\prime}}{x_1} \, dx + \int_{x_{1}}^\infty f(x) \, dx \geq\\
\frac{\epsilon^{\prime}}{x_1}(x_1-x_0)+\int_{x_{1}}^\infty f(x) \, dx \geq \\
\frac{\epsilon^{\prime}}{2}.$

Contradiction with $ \int_{x_{0}}^\infty f(x) \, dx < \frac{\epsilon^{\prime}}{2}$.

t0rajir0u's implied proof is not at all obvious. In series terms, the theorem becomes this: if $ a_n$ is positive decreasing, and $ \sum a_n<\infty,$ then $ \lim na_n\equal{}0.$ The thing is that if $ na_n$ doesn't tend to zero, we can't conclude that $ a_n\ge \frac{c}{n}$ for all $ n.$

For a moment, I thought I'd found a counterexample - a descending staircase with very long steps - but upon redoing the computation, I realized that its integral did diverge. If fact, the computation somewhat resembles tmrfea's proof, which is sound.

A close relative of this has appeared a few times here on AoPS/Mathlinks: show that if $ a_n$ is decreasing and $ \sum a_n\equal{}\infty,$ then $ \sum\min\left(a_n,\frac1n\right)\equal{}\infty.$ I don't have time to look for the link right now.

I denote $ F(x)\equal{}\int_{0}^x f(t)dt$.

It's obvious that $ F$ is increasing and since from the hypothesis we know that it's bounded it follows that  $ \lim_{x \to \infty} F(x)$ exists and is finite.

Now I will use the following mean theorem:  Let $ f: [a,b] \to \mathbb{R}$ be a Riemann integrable function.Then there is $ \xi \in [a,b]$ such that $ f(\xi) \equal{} \frac{1}{b\minus{}a} \cdot \int_{a}^b f(t)dt$

$ \forall x \in \mathbb{R}_{\plus{}}$ there is $ \xi_x \in \left [ \frac{x}{2},x\right]$ such that $ f(\xi_x)\equal{} \frac{2}{x} \cdot \int_{\frac{x}{2}}^x f(t)dt$.And now we can deduce that

$ 0 \leq xf(x) \leq xf(\xi_x) \leq 2 \cdot  \int_{\frac{x}{2}}^x f(t)dt \equal{} 2(F(x) \minus{}F(x/2))$, so $ \lim_{x\to \infty} xf(x)\equal{}0$.

[quote="Kent Merryfield"]t0rajir0u's implied proof is not at all obvious. In series terms, the theorem becomes this: if $ a_n$ is positive decreasing, and $ \sum a_n < \infty,$ then $ \lim na_n \equal{} 0.$ The thing is that if $ na_n$ doesn't tend to zero, we can't conclude that $ a_n\ge \frac {c}{n}$ for all $ n.$[/quote]

Say $ \limsup xf(x) \geq \varepsilon > 0$.  Then there are points $ x_1,x_2,\ldots$ tending to $ \infty$ such that $ x_jf(x_j) \geq \varepsilon$.  It follows that $ \int f(x)dx \geq \sum (x_{j \plus{} 1} \minus{} x_j)f(x_{j \plus{} 1}) \geq \sum (x_{j \plus{} 1} \minus{} x_j)(\varepsilon/x_{j \plus{} 1}) \equal{} \sum \varepsilon \minus{} \varepsilon x_j/x_{j \plus{} 1}$

At this point it would be helpful to have control over the ratios $ x_j/x_{j \plus{} 1}$.  So pick the $ x_j$s inductively so that $ x_j \leq \frac {1}{2}x_{j \plus{} 1}$.  Then $ \sum \varepsilon \minus{} \varepsilon x_j/x_{j \plus{} 1} \geq \sum \varepsilon \minus{} \varepsilon/2 \equal{} \sum \varepsilon/2 \equal{} \infty$.

[quote="Kent Merryfield"]t0rajir0u's implied proof is not at all obvious. In series terms, the theorem becomes this: if $ a_n$ is positive decreasing, and $ \sum a_n < \infty,$ then $ \lim na_n \equal{} 0.$ ...[/quote]
http://www.mathlinks.ro/viewtopic.php?t=265059  question a.
Indeed here we can do the same things.
$ zf(z) \leq 2 \int_{z/2}^{z} f(x)dx \to 0$
and the last limit is true (for example ) by the Cauchy criterion
(Svejk in the some sense has used this idea.  )

[quote="Svejk"]
$ 0 \leq xf(x) \leq xf(\xi_x) \leq 2 \cdot \int_{\frac {x}{2}}^x f(t)dt \equal{} 2(F(x) \minus{} F(x/2))$, so $ \lim_{x\to \infty} xf(x) \equal{} 0$.[/quote]
Question: It is correct to use the Leibnitz-Newton Formula if we don't know for sure that $ f$ has antiderivates?

If $ f$ is integrable on $ [0, \infty]$, it is integrable on any subinterval.  In any case, monotonic functions are integrable.


That's not correct ! Take for ex. $a_1=...= a_n=2 ,\  k=1$ .....
:cool:

Simply use Cauchy.

\[\sum \frac{a_i^2}{a_{i + 1} - k} \sum (a_{i + 1} - k) \geq (\sum a_i)^2\]

so it suffices to prove that \[(\sum a_i)^2 \geq 4nk \sum (a_{i + 1} - k) = 4nk((\sum a_i) - nk)\]

which is equivalent to \[(\sum a_i - 2nk)^2 \geq 0.\]


[quote="13375P34K43V312"]What is the least integer N greater than 2004 for which the positive difference between the square of the number that is five more than N and the square of N is also a perfect square?[/quote]
[hide]$n^{2}-(n-5)^{2}=10x-25$ is a perfect square. The number is divisible by 5 clearly. If x is 2004 then we have 20040-25=20015. I don't have a calculator handy, but I know a trick, giving me the next perfect square to be $145^{2}$ in 10 seconds. The same trick tells me that this is equal to 21025, giving n=2105. But since I for some reason used n-5, our answer is actually n=2100.[/hide]
1.5 minutes


i'm having a hard time with this problem.

the only given information is the focus (-2,3) and the directrix x=2
using the information above write the standard equation

somebody help and explain please? :)

The vertex is (0,3), and it opens up to the left, and $ k \equal{} \minus{}2$.  Then
$ x \minus{} 0 \equal{} \frac {1}{4k}(y \minus{} 3)^2 \equal{} \boxed{\minus{}\frac {(y \minus{} 3)^2}{8}}$


real analysis
real analysis unsolved
combinatorics
combinatorial nullstellensatsz
combinatorics unsolved

Good evening

Let $c(a_1,...,a_n)=\frac{(a_1+...+a_n)!}{a_1!...a_n!}$. Prove that $c(a_1,...,a_n)$ is the constant term of
$\prod_{(i,j)\in [1,n]^2,i \neq j} (1-\frac{z_i}{z_j})^{a_j}$

I thought of a recursive method but was not able to complete it.

Sorry,maybe it is stupid, but what is constant term in this case ?

Of course not, I am unaware of the correct expression: I meant
for example in $a_0+a_1 X+a_2 X^2+...+a_n X^n$, $a_0$ is the constant term.

Hi,

in fact, it will follow from the how-can-one-think-of-such-an-ugly identity :
$\sum_{k_{ij}}(-1)^{\sum_{1 \leq i < j < n}k_{ij}}\left(\prod_{1 \leq i < j < n}{{a_{i}+a_{j}}\choose{a_{j}+k_{ij}}\right) \left(\prod_{1 \leq j < n}{{a_{j}+a_{n}}\choose{a_{n}+\sum_{i < j}k_{ij}-\sum_{i > j}k_{ji}}}\right) ={{a_{1}+\ldots+a_{n}}\choose{a_{1}, a_{2}, \ldots, a_{n}}}}$
 :ewpu:

(I'll try to find a reference, iff you're interested.)

How on earth can someone find something like this????

Indeed, this is what I cannot understand either..
The LHS of this identity was [i]conjectured[/i] by Freeman Dyson... (in the 60's - and it was proved not long after)

So, um, the question is, how was the guy at the oral examination (I suppose it's an oral examination, isn't it?) supposed to solve this problem?  :roll:

Answer: by using the Lagrange Interpolation Formula one can find a recurrence relation. Take it as a hint!

Yes, the proof of Dyson's constant term identity via Lagrange interpolation due to Good uses induction on $n$.

There are more ways to prove it.

One elegant proof due to F.V. Petrov and R.N. Karasev is again by using Lagrange interpolation and the Nullstellensatsz.

Yet another way (which I recently found in an exercise problem and it's hint in R.P. Stanley's book) is to use the (strange but not so difficult to prove) identity
$$\sum_{i=1}^{n} \prod_{1 \le j \ne i \le n} \left ( 1 - \frac{x_i}{x_j} \right ) ^{ -1} = 1$$
(One can use induction on $n$ to prove this identity. Or one can directly use Lagrange interpolation to the constant polynomial $f(x) \equiv 1$.)

Once proved this identity along with binomial theorem (and via induction on $n$) can be used to prove Dyson's constant term identity.


Since I consider it to be under the "what you read" category, I'm gonna post a thread here for you guys to post your Weblogs. If you don't know what this is, it's just like an online journal. Good examples are Xanga and LJ. By visiting any of the Weblogs listed below, we hope that you will get the idea and start your own.

Now, I guess I'll be the first to give out mine:

[url]http://www.xanga.com/MithysAlliance[/url]

To everyone who has a Xanga:

Join the Art of Problem Solving blogring by visiting [url]http://www.xanga.com/blogrings/blogring.asp?id=143909[/url]and clicking "join".

Ok, I'll post mine even though it's boring =D
http://www.magnusrex.tk

I have a [url=http://www.xanga.com/smoke_mirror]xanga[/url], username smoke_mirror.

wow...this is an old thread. I don't use my xanga any more. i now use livejournal, with which my username is mathfiend. u have to be in my friends ring to read my posts tho...

[quote="WarpedKlown1335"][url]http://www.xanga.com/WarpedKlown1335[/url][/quote]

[snipped]

  Darij

[PS. Oh, somebody seems to have deleted the swear-words!]

I just created one -[url]http://www.xanga.com/Demon_Slayer88[/url].

I frankly don't see how xangas and livejournals can qualify as journals or diaries.  The Internet is no place for your innermost thoughts, and something as personal as a TRUE journal/diary is not something to which one DIRECTS traffic.

They aren't supposed to be journals where to cast your deepest thoughts, just you daily mundane ramblings that you feel like to have to write down. Journals and diaries don't all have to be personal. The point of a journal or diary is to keep track of daily events, and you can elaborate on them by applying your own feelings and opinions, which is something you don't have to do in a public site.

Blog are nice and all, but I really don't see why anyone would really care about your life, unless your a celeberty or something.

Okay, here is it: http://www.mathlinks.ro/Forum/viewtopic.php?p=112006 (I had just to copy-paste it from my old olympiad notes and translate the few text into English). But in future, when you have a mathematical problem, please post it at MathLinks and in a form that would be more likely to attract people...

  Darij

Err yeah, I did before I think.  But err...please don't do that again though?  (Show my xanga posts on here.)  Thanks.

Also btw, nice proof :-).

[quote="WarpedKlown1335"]Umm...ok now that an excerpt from my xanga page has been showed, would anyone now like to prove Brahmagupta's theorem.  (And vectors aren't allowed.)[/quote]

Here you go (see attachment).  'Twas a good chance for me to learn how to use the picture environment.


arithmetic sequence
inequalities unsolved
inequalities

1) Find the smallest value of the following expression: $ P\equal{}(\frac{1}{cosa})^6\plus{}(\frac{1}{cosb})^6\plus{}(\frac{1}{cosc})^6$
    where a,b,c form an arithmetic sequence whose commen difference is $ \frac{\pi}{3}$
2) Find the smallest value of the following expression: $ Q\equal{}\frac{1}{(a\minus{}b)^2}\plus{}\frac{1}{(b\minus{}c)^2}\plus{}\frac{1}{(c\minus{}a)^2}$
    where $ a,b,c\in[0;2]$   (they are real numbers)

[quote="mr.danh"]1) Find the smallest value of the following expression: $ P \equal{} (\frac {1}{cosa})^6 \plus{} (\frac {1}{cosb})^6 \plus{} (\frac {1}{cosc})^6$
    where a,b,c form an arithmetic sequence whose commen difference is $ \frac {\pi}{3}$
2) Find the smallest value of the following expression: $ Q \equal{} \frac {1}{(a \minus{} b)^2} \plus{} \frac {1}{(b \minus{} c)^2} \plus{} \frac {1}{(c \minus{} a)^2}$
    where $ a,b,c\in[0;2]$   (they are real numbers)[/quote]
I know little English, so I can't understand the first, and here is my solution for the second
Without loss of generality, assume $ 2 \ge a>b>c \ge 0$, we have
\[ (a\minus{}c)^2 \le a^2 \le 4\]
\[ (a\minus{}b)^2 \le (2\minus{}b)^2\]
\[ (b\minus{}c)^2 \le b^2\]
hence
\[ Q \ge \frac{1}{4}\plus{}\frac{1}{b^2}\plus{}\frac{1}{(2\minus{}b)^2}\equal{}f(b)\]
we have
\[ f'(b)\equal{}\minus{}\frac{2}{b^3} \minus{}\frac{2}{(b\minus{}2)^3}\]
\[ f'(b)\equal{}0 \Leftrightarrow b\equal{}1\]
From now, we can easily check that
\[ Q \ge f(b) \ge f(1)\equal{}\frac{9}{4}\]
Equality holds for example at $ a\equal{}2,b\equal{}1,c\equal{}0$. :)

In the first problem, a,b,c such that $ a\minus{}b\equal{}b\minus{}c\equal{}\frac{\pi}{3}$. Do you understand ?
I hope we solve it soon.

[quote="mr.danh"]In the first problem, a,b,c such that $ a \minus{} b \equal{} b \minus{} c \equal{} \frac {\pi}{3}$. Do you understand ?
I hope we solve it soon.[/quote]
Now, I understand it and here is my solution :)


Hello everyone ^^, just wanted to send out a "Good luck." and more importantly "Have fun!" to everyone writing the Fryer, Galois and Hypatia tomorrow at their respective schools. I'll be writing the Galois so...hopefully I'll have fun with the questions and do well ^^ I'm aiming for Silver Standard so...yeah ^^''

-lZerol

Just in case: Don't discuss it until at least Sunday, since ICAE's taking it on Saturday.

[quote="un_o"]wasn't too difficult
took the hypatia btw[/quote]
ditto.  :D  :D  :D

lol, I bombed the fryer, misread the last question and got a tentative 31. T.T

Got 40 on Hypatia!!! (unofficially?) :D  :D

Even though many are likely to have the same...  :)

good job!
I didn't write any of them this year because of the fact that my math teacher didn't tell me when the contest was. Not being in high-school, I had absollutly no clue when it was. Oh well, I will write it next year.

The Fryer, Galois, and Hypatia contests/solutions are out... also the averages for gold, silver, and bronze standards are out... at [url]http://cemc.math.uwaterloo.ca/english/contests/index.shtml[/url]

On another note... it looks like a lot of people found the Hypatia insanely easy this year, as approximately 97 people got gold standard - 38, 39 or 40/40! compared to last year, where 9 people got 38+ ... Wow.  :D  :D

Ahh i fail, I got 39 on Hypatia (which I thought was good until I saw that gold standard was 38)... on 3c I did everything right and then wrote down a wrong answer at the end. :wallbash_red:

I got 38 on the Fryer, just like last year.
Only because I somehow thought that 4 threes use less numbers than 2 sixes, for the last fryer question. Argh.

i got 34 on fryer:
lost 2 points for questions's 1,2,3 each. :thumbup:

FGH Results (Gold Standard People) are up: [url]http://cemc.math.uwaterloo.ca/english/contests/past_result/2008/2008FGHResults.pdf[/url]

55 people got perfect on Hypatia this year... wow...  :maybe:  :maybe:


How many digits of pi do you know by heart? 3 would be the first digit.

I memorized the first 102 digits, but in a school contest, I only got up to digit #88 before someone said I had switched the 2 and the 8 in the ...34825... and I had said ...34285... because I was rushing. :|

But I won anyway! :D

isnt this kinda unnecessary for mathcounts?
you should put this on another forum

There's nowhere else to put it.

Besides, maybe Wickedjestr was asking fellow MATHCOUNTS students about this and wouldn't get an adequate response in The Round Table, for instance. And not exactly a lot of MATHCOUNTS people go to the Polls forum (although there's already a $ \pi$ poll there :) )

lol i know 32 cuz that's how many digits my calculator on my computer shows.

My [b]computer[/b] calculator shows up to "...33832795" but when I subtract that by 3.141592653589793238462643383279, I get "5.0288419716939931148196659300057e-31"

Beware, it's accurate only up to "5.028841971693993" and everything after that is baloney. :)

I know about 50, though I used to know 80ish and I forgot some.  :blush:

about 75... i have this book called "the joy of pi" and it has the first 1,000,000 digits of pi in it :rotfl:

Wait, what's the world record again? Isn't it like some guy from China who memorized 90000 digits? Hmm...i don't even see how thats possible

ELEVEN!!!!! YIPPEE!!!! :D

I know these:

3.1415926535.... (11 digits)

And that's more than my calculator can fit! :P


Find the number of $ (x,y)\in N$ satisfy
$ y^x \equal{} x^{100000}$

[hide]For $ x$=a factor of $ 100000$ , there is a corresponding natural number $ y$.[/hide]

Yes, but $ y^x \equal{} x^2$ has three solutions in $ \bf N$ even though 2 has only two factors.  You're not done yet.

[hide="unsure..help needed!"]Hm... Should I write $ x\equal{}a^n$ where $ n$ is a natural number and $ x$ is a natural number that is not a perfect square, cube, etc..?

So the question becomes finding the number of $ a^n$ such that $ \frac{2^5\times 5^5\times n}{a^n}$ is a natural number.

For $ a\equal{}2$,
$ n\equal{}6$ is an obvious solution.
When $ n$ increases by 1, the numerator needs one extra factor of $ 2$, which is impossible because of the distribution of $ 2^r$.
($ a\equal{}8$, $ n\equal{}2$ is the same as this)

For $ a\equal{}5$
$ n\equal{}6$ does not work and by the same logic $ n>6$ will for work.

For $ a\ne$ a factor of $ 100000$,
$ n\equal{}1$ already does not work.

So add $ 1$ to the number of factors of $ 100000$[/hide]


Each of the 143 friends has a different movie on his phone. They begin to transfer the movies they have on their phones. If two people can transfer all the movies in theor phones to one another in 10 minutes, how long does it take for everyone to have each of the 143 movies in their phones?

well, if pairs of friends can transfer at the same time, then
10min (143-1) = 1420 min = 23 hr 40 min

[quote="orangefronted"] can transfer [u]all[/u] the movies in theor phones to one another in 10 minutes[/quote]

(for example, $ 3$ people require $ 30$ minutes minimum, while $ 4$ people only require $ 20$)

Its too bad our phones don't transfer at that rate.

The correct answer should have been 90 min. I don't know why! :(

Note that by the $ n$th interval, a person may only have up to $ 2^n$ movies on his phone. Thus at time $ 70$, each person could have at most $ 128$ movies. In the next ten minutes, since there is an odd number of people, at least one person cannot transfer movies, so that person cannot get all $ 143$ movies. Hence the minimum is $ 90$. Now can you make a construction that does this?


geometry trapezoid symmetry rectangle AMC

darkquantum wrote:Isosceles trapezoid ABCD is circumscribed about cicle O. The longer base AB=16, and mABC=arcsin(0.8). Find the area of the trapezoid

[hide] One property of a quadrilateral circumscribed about a circle is that the sum of two opposite sides equals the sum of the other two opposite sides (equal tangents)

Let P be the foot of the perpendicular from C to AB. CP/BC=sin(PBC)=.8 => CP=.8BC. Pythagoras gives BP=.6BC. By symmetry, CD+1.2BC=AB=16 => CD=16-1.2BC. Now, by my little spiel at the beginning, AB+CD=2BC => 32-1.2*BC=2BC => BC=10 => CD=4. We know find that the area of the trapezoid is twice the area of PBC plus the area of the rectangle left over =6*8+4*8=80. Correct? no, probably not, i'm too prone to arithmetical errors...[/hide]

[quote]One property of a quadrilateral circumscribed about a circle is that the sum of two opposite sides equals the sum of the other two opposite sides (equal tangents) [/quote]

Waiit. How can you prove this?

Check out what happens when you draw in the points of tangency and compare lengths of tangents from the vertices.

Look at the bit in brackets. Draw a diagram..

You would just be able to state that as a fact though, its pretty well known..

Yupsters
80 is correct
though i did it using nasty trig...hmm, i really ought to look for nicer methods


We have:
$ A(n) \equal{} \left [\begin{array}{cc} n & \sum_{i \equal{} 1}^{n} x_i \\
\sum_{i \equal{} 1}^{n} x_i & \sum_{i \equal{} 1}^{n} {x_i}^2 \\
\end{array}\right]$
where $ x_n \equal{} 1 \minus{} \frac {1}{n}$

Prove:
\[ \frac {\det A(n \plus{} 1)}{\det A(n)} \equal{} \frac {n \plus{} 1}{n}
\]

there is probably sth wrong in the problem statement, since the determinants of the first three matrices calculate to $ 0,\frac{1}{4}{,\frac{13}{18}}$


Here is a nice but not hard problem for you. :)
Consider the triangle $ABC$. Let $D$ and $E$ points on $AB$ and $AC$ respectively such that $DE||BC$. $P$ is an arbitrary point in the interior. $F=PB\cap{DE}, G=PC\cap{DE}$. Let  $O_{1}$ and $O_{2}$ are the circumcenters of the triangles $PDG$  and $PFE$ respectively. Prove that $AP\perp{O_{1}O_{2}}$.

Well, in http://www.mathlinks.ro/Forum/viewtopic.php?t=24889 it was proven that, if Q denotes the point of intersection of the circumcircles of the triangles PDG and PFE different from the point P, then the points A, P, Q are collinear. Now, the common chord of two circles is always perpendicular to the line joining the centers of the circles; hence, the line PQ, being the common chord of the circumcircles of the triangles PDG and PFE, must be perpendicular to the line $O_1O_2$, which joins the centers $O_1$ and $O_2$ of these circles. But since the points A, P, Q are collinear, the line PQ is simply the line AP, and thus we get that the line AP is perpendicular to the line $O_1O_2$. Proof complete.

  darij

Ok, here is my proof for this one.
Let $X=AP\cap{DE}, Y=AP\cap{BC}$.$DE||BC\Rightarrow{\frac{DX}{BY}=\frac{XE}{YC}}, \frac{XF}{BY}=\frac{XG}{YC}$.From previous two equalities we obtain:$DX\cdot{XG}=FX\cdot{XE}$, thus $X$ is on the radical ax of  $(PDG)$ and $(PFE)$.


Find the remainder in the division by 7 of $A = (2243)^{325} + (1179)^{154}$.

:)

The first one is : http://www.mathlinks.ro/Forum/viewtopic.php?t=80464
Want more? See : http://www.mathlinks.ro/Forum/viewtopic.php?t=80466 .

Since $2243 \equiv 1179 \equiv 3$ (mod 7), the question reduces to:

$3^{325} +2^{154}$ (mod 7).  Considering the solution to your first problem, this is clearly:

3.


Bob and Meena play a two-person game which is won by the first person to accumulate 10 points. At each turn Bob gains a point with probability of $ \frac{1}{3}$ . If he doesn't get a point, then Meena gets a point. Meena is now ahead 9 to 8. What is the probability that Meena will win? Express your answer as a common fraction.

The probablility that Meena doesn't win is. $ \ (\frac {1}{3})^{2}$
or $ \frac {1}{9}$
So she wins with a probability of $ \frac {1}{9}$

The answer's 8/9.

~There's gotta be gypsies~
              <||;-)

Either Meena or Bob must win. Thus, we have $P(M)+P(B) = 1$
$P(B)$ is easy to calculate as Bob must get 2 points, while Meena gets none. This occurs with probability $(\frac{1}{3})^2 = \frac{1}{9}$
Thus, $P(M)+\frac{1}{9} = 1$ by plugging in $\frac{1}{9}$
The answer is $\boxed{\frac{8}{9}}


conics
ellipse
geometry proposed
geometry

Two ellipses intersect at $A,B,C,D$ with this order. Another ellipse is tangent to this ellipses at $M,N$ and $P,Q$.Prove that the lines $AC,BD,MN,PQ$ are concurrent.

Very nice problem, but I haven't a clue how to solve it.  Any hints?

I hope the following is correct :?.

Assume $A,B,C,D$ are all fixed, and perform the following "movement": vary the conic $\mathcal E$ tangent to the two initial ellipses $\mathcal E_1,\mathcal E_2$ in $M,N$ and $P,Q$ respectively, so that it stays tangent to $\mathcal E_1$ in $M,N$. Let $\ell=PQ$, and let $T=\ell\cap MN$. Also, let $P',Q'$ be the intersections between $\ell$ and $\mathcal E_1$.

The two points where $\mathcal E$ cuts $\ell$ determine an involution on $\ell$ (it's well-known that a pencil of conics determine an involution on a line). Let the two points be $P_{\mathcal E},Q_{\mathcal E}$ (since they depend on $\mathcal E$, which varies). Now, the conics passing through $A,B,C,D$ also determine an involution on $\ell$. The two involutions have two common corresponding pairs ($P',Q'$ and $P,Q$), which means that they must coincide, or, in other words, $A,B,C,D,P_{\mathcal E},Q_{\mathcal E}$ lie on a conic for all $\mathcal E$ which are tangent to $\mathcal E_1$ in $M,N$.

The conic $\mathcal E$ can degenerate to the double line $MN$, in which case $P_{\mathcal E}=Q_{\mathcal E}=T$, which means that in this case $A,B,C,D,T$ lie on a conic s.t. $T$ is a double point. This means that $T$ lies on two of the lines formed by $A,B,C,D$, and the conic in question actually consists of two secant lines.

I didn't get the actual order ($T=AC\cap BD$ and not $T=AD\cap BC$, for example), but those are just details, and it's just playing around with the fact that the conics involved are actually ellipses and not something else.


geometry
trigonometry
similar triangles
angle bisector

[quote="stergiu"]I'm sure you will enjoy this simple and classical problem:

   On the sides  $AB , BC$  of an equilateral  triangle  $ABC$  we take points $D , E$  such that  $BD = 2DA$  and  $CE = 2 EB$ . Prove that  $DE \perp BC$

 I gave it to my my pupils( class 10 ) but in 20 minutes nobody solved it !!!

[u] Babis[/u][/quote]

[hide]
Say $DB=CE=2x$ and $AD=EB=x$

Construct line $DF$ where $F$ is on $BC$ such that triangle $DBF$ is equilateral. So each side of the triangle is equal to $DB$ or $2x$. So $FB$ is also equal to $2x$, so we can conclude that $FE=EB=x$. So now since $DE$ is the median of triangle $DBF$, $DE \perp FB$, so $DE \perp BC$
[/hide]

[hide]Just use law of sines or law of cosines in $\triangle {BED}$ and the solution follows. [/hide]

[quote="Drunken_Math"][hide]
Say $DB=CE=2x$ and $AD=EB=x$

Construct line $DF$ where $F$ is on $BC$ such that triangle $DBF$ is equilateral. So each side of the triangle is equal to $DB$ or $2x$. So $FB$ is also equal to $2x$, so we can conclude that $FE=EB=x$. So now since $DE$ is the median of triangle $DBF$, $DE \perp FB$, so $DE \perp BC$
[/hide][/quote]Drunken_Math: how do you know $F$ will be on line $DF$ so that $DBF$ is equilateral? You can find a point $F$ on $DF$ that will make $DBF$ isosceles but you still have to prove that $DBF$ will be equilateral as well.

Jose: can you be more specific? I'm not seeing the obvious part that you claim  :(

How about this

[hide]Put a third point on BC, say, F such that BE=EF=FC

then, by similar triangles, triangle BDF is similar to ABC and therefore equilateral.

$\angle BFD = 60^o$

E is the midpoint of BF
$\therefore$ DE is the angle bisector of $\angle BDF$
$\therefore \angle EDF=\frac{1}{2}(60)=30^o$

$\angle DEF=180-60-30=90^o$

So DE is perpendicular to BC.[/hide]

Well ,

 the problem is solved. Nice solutions.The last solution in essentially the first with a litlle better wording.Congrats to all. I hope , I'll find some time to post a new geometry problem , but not difficult , as well.

   [u]babis[/u]


I need to find some applications or phenomenoms involve in arithmetic progression in real life.
Please help me!

This definitely smacks of a homework assignment, and we're supposed to be careful about answering posts that have that quality.

Having said that, though, I feel a responsibility to try to guide you.  Think about the nature of an arithmetic sequence of numbers.  Try to find the most trivial real-life example of one (and there is one that is most obvious) that would get you a grade of 'D' on your assignment...but once you've thought of that one, it will lead you to another, and then an even better example would come to you with continued thought.  Keep thinking of them and about them until you come up with one so swell that it would be sure to impress your instructor, and then use that one!

[hide="hint"]What's the most commonly sung song?[/hide]


linear algebra
matrix
vector
algebra
polynomial

If $x=e_{n}$ is the last element of the standard basis, $A^{k}x=e_{n-k}$.

This is a variation on a companion matrix; they're usually written with - signs by the coefficients, and those coefficients are either in the last row (with ones one step above the diagonal) or in the last column (with ones one step below the diagonal).
If you want a matrix with a specific characteristic or minimal polynomial, you use a companion matrix. They also come up naturally when converting high-order differential equations into systems.


The sum of two positive integers is 30 and their product is 216. Find the positive difference between the two numbers.

$ a\plus{}b\equal{}30,ab\equal{}16$
$ a\minus{}b\equal{}\sqrt{(a\plus{}b)^2\minus{}4ab}\equal{}\sqrt{836}$, so a and b aren't integers.

Suggestion: replace sum with difference and difference with sum in the problem. Solution would then be 42.

Product = 216 :wink:

So we get
\begin{align*} a - b & = \sqrt {(a + b)^2 - 4ab} \\
& = \sqrt {(30^2 - 4\cdot 216)} \\
& = \boxed{6} \\
\end{align*}

More quickly, as we know they are integers, we just find the factor pairs of 216.

1,216
2,108
3,72
4,54
6,36
8,27
9,24
12,18

So we want the ones with the sum of 30 which are 12,18.

Another easy way is to note that the average of the two numbers is 15, that is, 15-a,15+a. We have (15-a)(15+a)=225-a^2=216 so a=3, meaning the difference is 6.


geometry
circumcircle
inequalities
trigonometry
inequalities proposed

For any triangle of sides $a,b,c$ and circumradius $R$:

 $(a+b+c)R \ge \sqrt{a^2b^2+b^2c^2+c^2a^2}$.[/quote]

Putting $ a= x+y, b= y+z, c = x+z $ our inequality reduces to show :

$ \frac {x+y+z}{xyz}\geq \sum \frac {4}{(y+z)^2}$

Which  is true from the fact: $ \frac {1}{yz}\geq \frac {4}{(y+z)^2}$.

[quote="nickolas"]Isn't this a  particular  case  of  the  ineq : [tex] (x+y+z)^2R^2\geq xya^2+yzb^2+zxc^2  [/tex] , where  [tex] x,y,z>0[/tex]  ??[/quote]
can you post the solution? :)

I'm so gald for I have proved the inequality. :)
I find it is equal to $x^2+y^2+z^2 \geq -2xy \cos 2A-2yz \cos 2B- 2zx\cos 2C$
which is simple.


Compute $ i^{600} \plus{} i^{599} \plus{} \cdots \plus{} i \plus{} 1$.

Since $ i$ is a $ 600^{th}$ root of unity, $ i^{600} \equal{} 1\implies i^{600} \minus{} 1 \equal{} 0\implies i^{599} \plus{} i^{598} \plus{} \cdots \plus{} i \plus{} 1 \equal{} 0$

$ \implies i^{600} \plus{} i^{599} \plus{} \cdots \plus{} i \plus{} 1 \equal{} i^{600} \equal{} \boxed1$.

[quote="pubado"]Could you elaborate more on that please?[/quote]

His method is slightly advanced, but he is saying that since $i$ is a 600th root of unity, we have $i^{600}-1=0 \implies \frac{i^{600}-1}{i-1}=0 \implies i^{599}+i^{598}+\hdots+i^1+i^0=0 \implies i^{600}+i^{599}+\hdots+i+1=1$.