Howdy peoples. Im looking for an explanation of why this doesnt work:

When one is looking for the surf area of a sphere, why can t one integrate the circumferences of the infinite amount of circles making up the sphere perpin to the x-axis with respect to x?

Thanx in advance!

The simple answer is that by integrating, or "summing", the circumferences of circles like that you're approximating segments of the sphere by cylindrical surfaces. Well at first this might seem like a good approximation, but think about how good of an approximation it is near the poles (or endpoints of your integration). You can see that it's not very good - because the sphere's surface gets very steep near the poles points while the cylinder stays flat. Hence the sphere has more surface area there than the cyclinder is able to approximate - no matter how small your segments get.

Another way to see it is to think about how your method would work for arc lengths of a curve. If you approximated a curve by your method you would be approximating them by flat pieces, and disregarding the curves steepness. Integrating this would just give you the length of the interval that the curve covers. That clearly isn't a very good approximation - no matter how many segments you take.


when is a function moving left? lets say the function is x(t), is it moving left when i do a sign chart for v(t) ?

a function x(t) is moving left when the velocity function v(t) is negative--because the v(t) is the derivative of x(t).
a function x(t) is moving right when the velocity function v(t) is positive.

By the way, this does not belong here--it belongs in the calculus forum.


calculus
function
geometry
derivative
integration
logarithms
calculus computations

my work is hidden
please check my work / help me with those I don't know.

1. Find the vol. of a solid generated by revolving the region bounded by $ y \equal{} 0$ and $ y \equal{} \frac {1}{\sqrt {x}}$ for $ 1 \le x \le 2$ about the line $ y \equal{} \minus{} 1$
2. Find the volume of the solid generated by revolving the region bounded by the lines $ x \equal{} 0$, $ y \equal{} 0$, $ x \equal{} 1$, and the curve $ y \equal{} e^{x^{2}}$. about the $ y$ - axis
3. SOLVED
4. Solved
[hide]True or false: If $ f(x)$ is continuous and $ 0 \le f(x) \le 1$ for all $ x$ in the interval $ [0,1]$, then for some number $ x$, $ f(x) \equal{} x$. [/hide]
5. SOLVED
6. Can you write the func. $ f(x) \equal{} x$ as the product of two differentiable functions ($ g(x)$ and $ h(x)$) if $ g(0) \equal{} h(0) \equal{} 0$
[hide]Yes. But I don't know how to show this. It's true for $ x^{\frac {1}{2}}$ and $ x^{\frac {1}{2}}$ or $ x^{\frac {1}{3}}$ and $ x^{\frac {2}{3}}$.[/hide]
7. Show that if
$ \frac {a_{0}}{1} \plus{} \frac {a_{1}}{2} \plus{} ... \plus{} \frac {a_{n}}{n \plus{} 1}$
then
$ a_{0} \plus{} a_{1}x \plus{} ... \plus{} a_{n}x^{n} \equal{} 0$
for some x in $ [0,1]$
8. Solved
[hide]A gas pipeline is to be constructed  from a storage tank to a house. A road runs straight from the tank, passing at its closest $ 300$ feet away from the house at a point $ 600$ feet from the tank. Pipe laid along the road costs $ \$8$ per foot, while pipe laid off the road costs $ \$10$ per foot. What is the minimum cost for which this pipeline can be built? [/hide]
9. Find the area bounded by $ y \equal{} \sqrt {x}$, $ \frac {1}{x}$, $ \frac {1}{x^{2}}$
10. The region in the first quadrant bounded by $ x \equal{} 2$ and $ y \equal{} 4$ and $ x^{2} \equal{} 4y$ is revolved around the $ y$-axis. Find the volume.

#4: When you say two curves "must cross", you are implicitly invoking the Intermediate Value Theorem. Try to manipulate things to make that explicit.

#5: It is that simple.

#6. No. $ x^{\frac13}$ and $ x^{\frac23}$ are not differentiable on the real line- in particular, they're not differentiable at zero. Consider the product rule...

#3 . There are only 4 functions . No more
#4.  Use fixed point theorem
# 7 Use the Intermediate value theorem and derivative formula
Do you mean there is $ x\in [0,1]$ st . $ a_{0} \plus{} a_{1}x \plus{} ... \plus{} a_{n}x^{n} \equal{} 0$? :lol:

I have removed 3 and 5 because they have been solved.
I have corrected the problem listed for 7.

Seven I don't understand at all.

For 4 I can't see how to do this with intermediate value. An attempt of a proof by counter (or whatever you call it) using the squeeze theorem has now been included. Does that work?

I'm a little confused by 6. I'm guessing that because of the product rule, we have troubles when x=0, because g(x)*h(x) = f(x) means that solving for h(x) or g(x) always excludes x=0.

1,2,9,10 I need some help solving, I never was any good at those.

8. Seems to me like I'd be calculating something similar to a calculation of light /electricity bending / taking the path of least resistance. I hate that light can do more calculus than me. I need some help on figuring this out.

It's extremely impolite to remove questions once they've already been discussed -- don't do it any more.

from now on problems that are solved will be in hidden sections.

#1 Hint: [hide]Just break the volume into cylinders.[/hide]
And I got: [hide]$ \pi(\ln{2} \plus{} 4\sqrt {2} \minus{} 3)$ [/hide]

That answer doesn't make sense, is there supposed to be a ")"

[quote]#6: [hide]
I believe it is impossible to have a g(x) and h(x) that meet those standards.
Consider the equation: $g(x)h(x)=x$ . Then $g'(x)h(x)+g(x)h'(x)=1$ . If you set x=0, then you arrive at the conclusion that 1=0, which is false. Therefore, it is impossible. This result implies that either h(x), g(x), or both must not be differentiable at x=0 for any h(x) and g(x). Note: it is possible to have an h(x) and g(x) such that h(x)g(x) =x, but one or both of these two functions themselves will not be differentiable at x=0, therefore not meeting the requirements of the problem.
[/hide]
[/quote]


let $g_{n}$ = geometrical mean of ${n\choose 0},\dots,{n\choose n}$

I tried to find $\lim g_{n}$ but I gived up :(

hint :
show that $g_{n+1}^{n+2}\geq 2^{n}g_{n}^{n+1}$
so $g_{n+1}\geq 2^{\frac{n(n+1)}{2(n+2)}}$ so the limit is $\infty$  :wink:

It's easy.
More : it can be shown that  $g_{n+1}^{n+2}\geq 2^{n}(n+1)g_{n}^{n+1}$


My solution is exactly same as yours, but the beginners don't notice that, Farenhajt. :wink:

Take the logarithm with base $2$ of both sides.


Say $ x_1, x_2,....x_k$ are independently exponentially distributed random variables with rates $ \lambda_1,\lambda_2....\lambda_k$. How do you derive that $ x_1\plus{}x_2......x_n$ is a gamma distribution (1,k)?

if you can do it for only $ 2$ variables, it is a simple induction. Can you ?

The sum is only a gamma distribution if $ \lambda_1 \equal{} \lambda_2 \equal{} \cdots$ ; in other words, if the exponential distributions are identical. The most straightforward way to show that would be by using a transform method: moment generating functions or characteristic functions. That requires the up-front effort in setting up the transform method.


limit
number theory theorems
number theory

I read that the statement:
(1) $ \sum_{n=1}^{\infty}\frac{\mu(n)}{n}= 0$
implies the prime number theorem. But it is well known that:

(2) $ \sum_{n=1}^{\infty}\frac{\mu(n)}{n^{s}}=\frac{1}{\zeta(s)}$

doesn't (1) follow from (2) like this:

$ \lim_{s\to 1}{\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{s}}}=\lim_{s\to 1}{\frac{1}{\zeta(s)}= 0}$

i know there's something wrong with this, otherwise PNT wouldve been proven long before Hadamard and de la Vallée Poussin proved it  :P can you tell me what's wrong with this? thanks.

[quote="winter"]I read that the statement:
(1) $ \sum_{n = 1}^{\infty}\frac{\mu(n)}{n}= 0$
implies the prime number theorem. But it is well known that:

(2) $ \sum_{n = 1}^{\infty}\frac{\mu(n)}{n^{s}}=\frac{1}{\zeta(s)}$

doesn't (1) follow from (2) like this:

$ \lim_{s\to 1}{\sum_{n = 1}^{\infty}\frac{\mu(n)}{n^{s}}}=\lim_{s\to 1}{\frac{1}{\zeta(s)}= 0}$

i know there's something wrong with this, otherwise PNT wouldve been proven long before Hadamard and de la Vallée Poussin proved it  :P can you tell me what's wrong with this? thanks.[/quote]

Why is $ \sum_{n = 1}^{\infty}\frac{\mu(n)}{n}=\lim_{s\to 1}{\sum_{n = 1}^{\infty}\frac{\mu(n)}{n^{s}}}$? :wink:


LaTeX
trigonometry
algebra
system of equations

1) If x=cy+bz,y=a+cx,z=bx+ay where x,y,z are not all zero,then prove that $a^2+b^2+c^2+2abc=1$.

2)Let $\"l\"$,$\"m\"$ and $g$,$t$ be the roots of $ax^2+bx+c=0$ and $px^2+qx+r=0$ respectively.If the system of equations ly+mz=0 and gy+tz=0 has a non trivial solution,then prove that  $\frac{b^2}{q^2}=\frac{ac}{pr}$.

3)If one root of the equation $ax^2+bx+c=0$ is equal to the nth power of the other,then show that.....this is a well known property and i don't know the latex to write it down :D Someone plaese prove it.

4)a-bx-cx=0
    -ay+b-cy=0
    -az-bz+c=0

Prove that $xy+yz+zx+2xyz=1$

Isn't there an easier method for the first one?And the others are to be solved to :)

No one getting the sums :|   Can't be....

[quote="shobber"][quote="shyong"][hide="1"]
first of all i think you left out something , it shoudl be $y=az+cx$

Consider a triangle $ABC$ with side length $x,y,z$ at the opposite angle of $A,B,C$ respectively . Then we have

$x=y\cos C+z\cos B$
$y=z\cos A+x\cos C$
$z=x\cos B+y\cos A$

which is equivalent to the given equation where $a,b,c$ are $\cos A,\cos B,\cos C$ respectively .

We wish to prove that $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$ which is a well-known result  ;) [/hide][/quote]

Why can you make this substitution? How do you know that $x, y, z$ can form a triangle? Besides, the problem did not say that they are positive numbers.[/quote]

The 1st and 4th ones are confusing me :huh:

[quote="shadysaysurspammed"]Solutions......?[/quote]
I think #4 is incomplete in some way, because assuming that $a,b,c$ can be any real numbers, then I can create $x,y,z$ that don't satisfy the given equation.


if a cyclic heptagon (seven sides that is) has three angles of $120^{\circ}$, then it has two equal sides.

[quote="riddler"]I think you are being a bit unclear because the $3$ angles can be anywhere and the equal sides can also be anywhere.[/quote]

you're right. the problem asks you to handle all these cases.

[quote="riddler"]The question is not a question it is just a statement, do you have to prove something????
and whats the difference between a cyclic and a simple heptagon????[/quote]

well, you have to prove the statement given.

you are given a heptagon with all the 7 vertices on a circle. moreover, three of its angles are equal to $120^{\circ}$.

you must prove that it has two equal sides.

that's all.

first off, at least 2 of the 120 degree angles has to have 1 other angle or less inbetween them

this is because if you have more than that between the 120 angles, you could have an angle at 1,4,7 but the 1 and 7 are right next to eachother

this leaves 2 cases 1) 2 120 degree angles are next to eachother
2) 2 120 degree angles have 1 space in between them.

we can say that each side subtends a certain arc label them $1,2,3...7$

drawing it would help to understand
case 1
let the first 120 degree angle be between 7 and 6 and it is an inscribed angle, so $1+2+3+4+5$ equal $2*120 = 240 degrees$
likewise then next angle (between 6 and 5) would be $1+2+3+4+7=240$
subtracting those two yields $5=7$ (the arcs 5 and 7) so since 2 arcs are congruent, then 2 chords are congruent

case 2
(2 120 degree angles are 1 apart and the 3rd is 2 spaces from one and 1 space from the other)
let the first 120 degree angle be between 7 and 6 and it is an inscribed angle, so $1+2+3+4+5 = 240 degrees$

then the next 120 degree angle would be be between 5 and 4 so $1+2+3+6+7 = 240$

subtracting those leads us to 4+5=6+7
those equal 120 since $1+2+3+6+7 = 240$ and there are 360 degrees in a circle so $360-240 = 120 = 4+5$

then the other has to be 2 away from one of those so it would be between 2 and 3 (or 1 and 2 but that is not a separate case because the figure could be flipped)
so $1+4+5+6+7 = 240 $
substitute again $120 +120 = arc 1 = 240$
$arc 1 = 0 degrees$
but that is a contradiction since that would make the figure a hexagon, therefore the 1st case is the only valid case were 2 sides are congruent

An alternative presentation of the solution:

Let $A_{1},A_{2},\ldots,A_{7}$ be the cyclic (convex, non-degenerate) heptagon.

[i]Case 1:[/i] $\angle A_{1}= \angle A_{2}$.

Let $d$ the perpendicular bisector of $A_{1}A_{2}$. Then, $A_{2}A_{3}$ and $A_{1}A_{7}$ are symmetric wrt $d$. Hence, $A_{2}A_{3}= A_{1}A_{7}$.

[i]Case 2:[/i] $\angle A_{1}= \angle A_{3}= \angle A_{5}$.

We have $\angle A_{2}A_{5}A_{4}= 180^\circ-\angle A_{4}A_{3}A_{2}= 60^\circ$ and $\angle A_{7}A_{5}A_{3}= 180^\circ-\angle A_{2}A_{1}A_{7}= 60^\circ$. Thus, $\angle A_{7}A_{5}A_{4}= \angle A_{6}A_{5}A_{4}$, so $A_{6}= A_{7}$, i.e. our heptagon is degenerate, contradiction.


modular arithmetic
number theory
greatest common divisor

How many positive integer values of x less than 1000 satisfy x is congruent to 4 mod 9 and x is congruent to 5 mod 12?

[hide="Solution(?)"]$ x\equiv 4\pmod 9\implies x\equal{}9k\plus{}4,\; k\in\mathbb{Z}$.
$ x\equiv 5\pmod{12}\implies x\equal{}12j\plus{}5,\;j\in\mathbb{Z}$.
$ \therefore 9k\plus{}4\equal{}12j\plus{}5\implies 9k\minus{}12j\equal{}1$.

We now have a linear Diophantine equation. However, $ \text{gcd}(9,12)\equal{}3$, which means there are $ \boxed{\text{no solutions}}$.

Using substitutions, we end up with $ 9k\equiv 1\pmod{12}$, but (again, because $ \text{gcd}(9,12)\neq 1$), there is no modular multiplicative inverse for 9 in modulo 12, so there is no way to simplify further.[/hide]

Checking, we have all the odds (which $ x$ must be) to be congruent to either $ 1$ or $ 7$.

umm here is a straightforward way to prove that there are no solutions
[hide]
$ x \equiv 4 \pmod{9} \implies x \equiv 1 \pmod{3}$
$ x \equiv 5 \pmod{12} \implies x \equiv 2 \pmod{3}$
contradiction
[/hide]

side note to ernie
what mod base?


Is there a way that you can force a win in checkers?  After i saw that connect 4 approach(which would take a long time to learn) I thought that since connect 4 seems much more complicated then checkers, maybe checkers has a therom to win.

Not with 2 rational players playing their best.


Does anybody know if the expansion will carry through to this year as well, or was that just for last year?

Yo, according to mah sourcez, that will carry through to this year, yo.


[quote="horselvr101"]Does anyone have an idea about when the Intermediate Trigonometry and Complex Numbers class will be offered again?[/quote]

Our plan is to offer it this Fall.  The Fall schedule will be up soon.


Suppose that $n$ people each know exactly one piece of information, and all $n$ pieces are different. Every time person $A$ phones person $B$, $A$ tells $B$ everything that $A$ knows, while $B$ tells $A$ nothing. What is the minimum number of phone calls between pairs of people needed for everyone to know everything? Prove your answer is a minimum.

$2n-2$.
We can see that this can always be done by induction:
with $n=1$, we need zero calls. WOLOG, let A be the first caller, and let B be the last reciever (A=B is possible). For each new person added, we need only two additional calls: the new person calls A on the first call, and B calls the new person as the last call.

To see that this is optimal: it is clear that with $n>1$, each person needs to recieve a call at least once. Thus a minimum of $n$ calls is necessary. WOLOG, let A be the first reciever, B be the second, and so on. It is clear that with $n>2$, A must recieve at least one additional call. With $n>3$, B must recieve at least one additional call, etc. Therefore we have $n$ original calls, plus $n-2$ additional calls. $n+n-2=2n-2$


We have just revamped the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Guide.php]Internet Math Resources Guide[/url] at AoPS.

New to our resources is the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_MathScholarships.php]Math Scholarships page[/url].

If you know of any scholarship resources that we could add to that page, please email me at crawford@artofproblemsolving.com.

Thanks for posting the link to the [url=http://www.ditdservices.org/Articles.aspx?ArticleID=24&NavID=0_0]Davidson Young Scholars[/url] program on the math scholarships page. All the United States AoPSers should give that program a look, if they are still under the age of 16.


geometry
rectangle
power of a point
Pythagorean Theorem
AMC

In circle O, perpendicular chords AB and CD intersect at E. AE=2, EB=12, CE=4. Find the radius of circle O.

[hide]By Power of a Point, DE = 6. By Pythagorean, CD = 4Sqrt(10) and BD = 6Sqrt(5). Then we can form triangle CBD. [CBD] = (1/2)(10)(12) = 60. But by the formula Area = abc/4R, [CBD] = (10*4Sqrt(10)*6Sqrt(5))/(4R) = 300Sqrt(2)/R. Solving 300Sqrt(2)/R = 60, R = 5Sqrt(2).[/hide]

yes fiery
thats how did it on the contest

Another method.  From O, draw OF perpendicular to CD and OG perpendicular to AB.  F must be the midpoint of CD.  Thus FE = 1.  OFEG is a rectangle, so OG = 1.  G is the midpoint of AB, so BG = 7.  Thus, by the Pythagorean Theorem, R = 5:sqrt2:


Okay, let us see if anyone out here   is interested in doing  some math
Source:  Based on an Old mathbook

"Did your teacher give  you this problem, or is it from MNO?" I asked "It looks   rather hard."

"Nahin Uncle", said Chintoo , "Its called NMO, and actually I made it up. Its an equation, and my age (in years) is one of the root"

"Then  it can't be that hard", I chuckled, " Polynomial equation, all coefficients are integer, your  root is an integer.. let me try x=7 .. No that gives 77"

"Do I look 7 yeas old"., Chintoo  pouted?

"Okay baba, lets us try a larger  integer,.. no that does not give a zero either, it gives 85"

"You KNOW  I am older than that... Not funny." STOP Kidding."

Ok so we stop kidding... but how old is Chintoo?

f(7)=77 f(k)=85 f(m)=0     let m be the age

k>7 and m>k

  If f is a polynomial with integer coefficients for any 2 integers a and b

     (a-b) | { f(a)-f(b) }

 So (k-7) | { f(k)-f(7) }    so k-7 | 8 . But the only factors of 8 are 1,2,4,8

 so k = 8 or 9 or 11 or 15.

  (m-7) | { f(m) - f(7) }  Since m>7 ,  (m-7) | 77 . 77 has 2 factors 7,11.
  m = 14 or 18  ........ (1)

  also (m-k) | { f(m) -f(k) } Since m>k so m-k | 85 . 85 has 2 factors 5,17

if k=8 ,this gives  m=13 or 25 contradicts eqn 1
   k=9  this gives  m=14 or 26 , this could be a soln
   k=11                m= 16 or 28 contradicts eq 1
   k=15                m= 20 or 32 contradicts eq 1

so k=9 and m=14

Correct, of course,[hide] though a minor nitpick -
[quote] (m-7) | { f(m) - f(7) } Since m>7 , (m-7) | 77 . 77 has 2 factors 7,11.
m = 14 or 18 ........ (1) [/quote]
There are 4 factors of 77, two other  factors of 77, (other than  7,11)  are  (1,77). They do not work (for example m can not be 8 because m>k>7 etc) so it makes no difference in final answer.
(siilary two other factors of 85 (namely 1 and 85) do not make any difference too)
 [/hide]


probability
limit
integration
topology
probability and stats

Let $ N_k(t), t\geq 0, k\in \mathbb{N}$ be independent Poisson processes with intensities $ \lambda_k, k\in \mathbb{N}$, and the total intensity $ \Lambda: \equal{}\sum_{k}\lambda_k\equal{}\plus{}\infty$. Put $ X(t)\equal{}\gamma t\minus{}\sum_{k\equal{}1}^\infty c_kN_k(t)$ where $ \gamma$ and $ c_k,k\in \mathbb{N}$ are positive and $ \sum_{k}c_k\lambda_k<\plus{}\infty$. Denote $ \tau\equal{}\inf\{t: X(t)<0\}$. Is it true that $ P(\tau\equal{}0)\equal{}1$?

Does this helps ?

http://www.emis.de/journals/EJP-ECP/_ejpecp/include/getdoc2fa8.pdf

Otherwise you can take a look at the book of Kyprianou "Introductory Lectures on Fluctuations Theory of Lévy Processes with Application

I think there is proposition about small time behaviour of lévy processes other than Coumpound processes

my two cents

I do not think the linked paper would help because the `ruin time' $ \tau$ seems to be a more delicate thing than $ \rho: \equal{}\lim_{t\to 0\plus{}}P(X_t<0)$. In the paper, they relate $ \rho$ with some 'trajectory-wise' objects, but these objects are of a 'time-averaging' nature and I do not see how one can deduce anything about the first exit time from this.
 I don't have  an immediate access to the book by Kyprianou, but I'll try to get it. Thank you for the references!

About the question from the initial post: I do not know the answer; I think that
$ P(\tau>0)\equal{}1$ in every case, but this is a guess only, provided by some very non-rigorous reasoning  :maybe:

Alex-k, am I right in saying that you try to see if point "0" is a regular point with respect to the open set $ \R^{ - *}$ for your process.

Here is a result about regular point for Lévy process found in the mentioned reference page 142 Theorem 6.5

Theorem:

For any Lévy process $ X$, other than a compound Poisson process, the point "0" is regular for the set $ B = \R^{ - *}$  if and only if :

$ \int_{0}^{1}\frac {1}{t}P(X_t)dt = \infty$

and the latter holds iff one of the 3 following conditions hold :

1-$ X$ is a process of unbounded variation (your case :))
2-$ X$ is a process of bounded variation and drift is positive
3-$ X$ is a process of bounded variation , dirft is null and : the follwoing hold : $ \int_0^1\frac {x\Pi(dx)}{\int_0^x\Pi( - \infty, - y)dy} = \infty$
(with $ \Pi$ the Lévy measure of the pure jump process)
So you are clearly in case 1 so your intuition sounds good to me.

The author gives three references and does not prove the theorem here they are :

Rogozin "the local behavior of processes with independent increments"1968 in russian (which should not be a problem for you Iguess  :) )
Shtatland "On local properties of processes with independent increment" 1965
Bertoin "Regularity of the Half-line for Lévy processes"  Bull. Sci. Math. 121 (1997)

Hope this helps

[b]TheBridgeTwo[/b], thanks a lot!!! It's great that I should not invent this bycicle once again  :roll: One comment: in the case under consideration,  the variation is bounded because $ \int|u|\Pi(du)\equal{}\sum_kc_k\lambda_k<\plus{}\infty$. But this should be covered by the case 2). Once again, thanks!

You are welcome

But this result looks more like a "Harley Davidson" than a bycicle to me  :)


function
inequalities unsolved
inequalities

FIND MAX
A=$ \frac{1}{sinx\plus{}4}\minus{}\frac{1}{cosx\minus{}4}$

hello, differentiate your function with respect to $ x$.
Sonnhard.


function
limit
integration
calculus
derivative
real analysis
real analysis solved

let $f:R^{+}->R^{+}$ be a continuous function such that
$\lim_{x->\infty} f(x)\int_{0}^{x}f^{2}(t)dt = 1$
Find an equivalent of f.

Very easy problem. Let $ F(x)=\int_{0}^{x} {f^2(t)dt}$. Then we have that the derivative of $ F^3$ is equivalent with 3, then by Hospital rule that $ F$ is equivalent with $ (3x)^{\frac{1}{3}}$ and thus $f$ is equivalent with $ (3x)^{-\frac{1}{3}}$. Quite similar to a problem from Vranceanu Procopiu, this year,

that's also my soln. What 's Vranceanu Procopiu ? Could you post some interesting pb from here ?

A romanian inter-county contest, which is quite nice (especially this year). Unfortunately, there are just 2 problems at each class and I think they were already posted.


Η παρακάτω άσκηση πιστεύω ότι είναι πάρα πολύ δύσκολη .Είναι πιστεύω ένα θέμα αντίστοιχο της δυσκολίας των ΙΜΟ pr 6 .
Ελπίζω να σας αρέσει.....

<<Lattice point >> ονομάζεται ένα σημείο στο καρτεσιανό επίπεδο που και οι δύο συντεταγμένες του είναι ακέραιοι .
Ένα <<ακέραιο ορθογώνιο σύνολο>> (αντίστοιχα ένα <<ακέραιο τετράγωνο σύνολο>>) είναι το σύνολο των <<Lattice point>> που βρίσκονται στο εσωτερικό ή στην περίμετρο ενός ορθογωνίου (παραλληλογράμμου) (αντίστοιχα  ενός τετραγώνου) με κορυφές <<Lattice point>> .
Αντιστοιχίζουμε σε κάθε <<Lattice point>> έναν πραγματικό αριθμό έτσι ώστε  το άθροισμα όλων των αριθμών σε κάθε  <<ακέραιο τετράγωνο σύνολο>>  να είναι κατά απόλυτο μικρότερο του 1 .
Αποδείξτε ότι το άθροισμα όλων των αριθμών σε κάθε <<ακέραιο ορθογώνιο σύνολο>> είναι κατά απολύτο μικρότερο του 4 .

Καλή διασκέδαση  :wink:

Σιλουαν εισαι σιγουρος για τις τρεις τελευταιες σειρες? δεν καταλαβαινω τι εννοεις οταν λες το αθροισμα ολων των αριθμων.

παρα πολυ δύσκολη είναι πρέπει να δωθεί αναλυτική λήση όταν αυτό γίνει για να γίνει κατανοητό από όλους μας
ευχαριστώ

[quote="kostas"]ρε παιδια καταλαβαινει κανεις τι εννοει ο Σιλουαν οταν λεει οτι το αθροισμα ολων των αριθμων σε καθε ακεραιο τετραγωνο συνολο ειναι 1.?[/quote]

Ένα παράδειγμα σε τετράγωνο $2 \times 2$ .Πρέπει

$\\ \left|\frac{2}{5}-\frac{1}{7}-\frac{1}{12}-\frac{1}{3}\right|<1 \,\, , \left|\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{3}\right|<1 \\ \\ \\ \left|\frac{3}{5}-\frac{1}{7}-\frac{3}{12}-\frac{1}{3}\right|<1 \,\, , \left|\frac{3}{5}-\frac{4}{7}+\frac{1}{2}-\frac{1}{3}\right|<1$

και

$\\ \left|\frac{2}{5}-\frac{1}{7}-\frac{1}{12}-\frac{1}{3}+\frac{3}{5}-\frac{4}{7}-\frac{3}{12}+\frac{1}{2}-\frac{1}{6}\right|<1$

Οι αριθμοί μπορεί να είναι λάθος όμως γιατί αυτό πρέπει να
ισχύει σε κάθε τετράγωνο .  :roll:


Show that if p is an odd prime, then there are exactly (p - 1)/2 quadratic residues of p.

Just see that a^2 = (-a)^2 = (p-a)^2 (mod p).

That shows that there can't be more than (p-1)/2 quadratic residues (mod p). Here's how to show that there are (p-1)/2:

Let a and b share a quadratic residue m (mod p), and let 1<=a<=b<p.

a2=kp+m.
b2=lp+m.
a2-b2=p(k-l).

Thus either a-b or a+b must divide p. Neither a-b nor a+b can be as large as 2p, so either a-b=0 or a+b=p. Thus the only numbers (mod p) that share a quadratic residue (mod p) are a and p-a for a=1,2,...,(p-1)/2. (I suppose there's a reason why 0 shouldn't be called a quadratic residue? Otherwise it wold be (p+1)/2.)


Let $ G$ Abelian locally compact group. Prove that :

(a) If group $ G$ is discrete then $ \hat{G}$ is compact.

(b) If group $ G$ is compact, then $ \hat{G}$ is discrete.

[hide="Remarks"]There are hints how to prove this, for (a) we have to notice that $ L_1(G)$ is unital Banach algebra, what is trivial to show, but I don't know how to continue from that.
For (b) we have to notice that $ \{\chi\in \hat{G}: |\mu(\chi)| > \frac {1}{2}\}$ is an open set, and that if we have a normed Haar measure of G, then the dual group $ \hat{G}$ is an orthonormal set in $ L_2(G,\mu)$. Unfortunately I don't know how to prove any of that, I can't say that it's easy to see that $ \chi\rightarrow \mu(\chi)$ is a continuous function ( or it is? ).[/hide]

Any hints are welcome.

Can anyone point out some hints on how to show that the set $ \{\chi\in \hat{G}: |\mu(\chi)| > \frac {1}{2}\}$ is open?  :help:


hi guys,
let f(x)=1/(1+2*int(x)-x)
prove that f^(2^n)(n)=f(f(f(...f(n))))=n+1 for every integer n>=0.
any idea ???

what is int(x) .... is it by any change [x] = integer part of x ?

I think that induction after "n" should work! :D
have you tried that?

cheers! :D :D

Induction looks kind of nasty to me. Remember that when you go from n to n+1 not only do you have to compose with f 2^n more times, but you also change the argument of f, from n to n+1. It could work, but it doesn't look that good, does it? :D

Indeed grobber, like Myth I think once said: The easiest idea is usually the messiest and hardest perhaps! :D

cheers!

Oh, so induction DOES work, and it's not even messy! :D  :D  :D


Apodeikste tin anisotita gia olous tous thetikous pragmatikous arithmous:

    $ \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\plus{}\frac{1}{2\sqrt[3]{abc}}\geq\frac{(a\plus{}b\plus{}c\plus{}\sqrt[3]{abc})^2}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$

Cauchy it !  :wink:
Έχουμε
 $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {1}{2\sqrt [3]{abc}}\equal{}\sum (\frac{c^2}{c^2(a\plus{}b)})\plus{}\frac{(\sqrt[3]{abc})^2}{2abc}$

και από Cauchy-Swartz
$ \sum (\frac{c^2}{c^2(a\plus{}b)})\plus{}\frac{(\sqrt[3]{abc})^2}{2abc}\geq\frac{(a\plus{}b\plus{}c\plus{}\sqrt[3]{abc})^2}{c^2(a\plus{}b)\plus{}b^2(a\plus{}c)\plus{}a^2(b\plus{}c)\plus{}2abc}$

και το μόνο που μένει πλέον είναι να παρατηρήσούμε ότι $ c^2(a\plus{}b)\plus{}b^2(a\plus{}c)\plus{}a^2(b\plus{}c)\plus{}2abc\equal{}(a\plus{}b)(b\plus{}c)(c\plus{}a)$

[quote="silouan"]Cauchy it !  :wink:
[...]
και το μόνο που μένει πλέον είναι να παρατηρήσούμε ότι $ c^2(a \plus{} b) \plus{} b^2(a \plus{} c) \plus{} a^2(b \plus{} c) \plus{} 2abc \equal{} (a \plus{} b)(b \plus{} c)(c \plus{} a)$[/quote]

Oooh...

Damn. Good. Catch. :coolspeak:

Cheerio,

Durandal 1707

[quote="silouan"]Cauchy it !  :wink:
Έχουμε
 $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {1}{2\sqrt [3]{abc}} \equal{} \sum (\frac {c^2}{c^2(a \plus{} b)}) \plus{} \frac {(\sqrt [3]{abc})^2}{2abc}$

και από Cauchy-Swartz
$ \sum (\frac {c^2}{c^2(a \plus{} b)}) \plus{} \frac {(\sqrt [3]{abc})^2}{2abc}\geq\frac {(a \plus{} b \plus{} c \plus{} \sqrt [3]{abc})^2}{c^2(a \plus{} b) \plus{} b^2(a \plus{} c) \plus{} a^2(b \plus{} c) \plus{} 2abc}$

και το μόνο που μένει πλέον είναι να παρατηρήσούμε ότι $ c^2(a \plus{} b) \plus{} b^2(a \plus{} c) \plus{} a^2(b \plus{} c) \plus{} 2abc \equal{} (a \plus{} b)(b \plus{} c)(c \plus{} a)$[/quote]
ωραίος!!! δηλ το φερνεις σε καταλληλη μορφη για να εφαρμόσεις την andreescu...

Είναι 1/(α+β)+1/(β+γ)+1/(γ+α)>= (α+β+γ)^2 Αυτό βγαίνει εύκολα με ομώνυμα
Επίσης 1/2(αβγ)^(3/2) είναι >= [(αβγ)^(1/3)]/(α+β)(β+γ)(γ+α) Αυτό βγαίνει με χιαστί και ανάπτυξη του (α+β)(β+γ)(γ+α).Τώρα εύκολα απο ανισότητα Α βγαίναι το ζητούμενο.[/quote]

Κάτι δεν καταλαβαίνω... Βάλτην αναλυτικά γιατί μάλλον κάνεις κάποιο λάθος. :wink: [/quote]

η ανισοτητα ειναι ομογενης, γεγονος που μας επιτρεπει να υποθεσουμε οτι $ abc\equal{}1$

Tωρα γραφεται ισοδυναμα ως
$ \sum (a\plus{}b)(b\plus{}c)\plus{} \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{2} \geq (a\plus{}b\plus{}c)^2\plus{}1\plus{}2(ab\plus{}bc\plus{}ca)$

ομως ισχυει

$ \sum (a\plus{}b)(b\plus{}c) \plus{} \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{2}$

$ \equal{} \sum a^2\plus{} 3\sum ab \plus{} \frac{\sum a^2b \plus{} \sum a^2c \plus{} 2abc}{2}$

$ \geq (a\plus{}b\plus{}c)^2 \plus{} \sum ab \plus{} \sum a^2c \plus{} 1 \geq (a\plus{}b\plus{}c)^2 \plus{} 1 \plus{} 2(a\plus{}b\plus{}c)$

διοτι,
$ \sum a^2b \geq \sum a^2c$,αληθης απο αναδιαταξη

$ \sum ab \plus{} \sum a^2c \geq 2 \sum a$

που ισχυει απο την ανισοτητα αριθμητικου γεωμετρικου μεσου ως εξης:

$ \sum ab \plus{} \sum a^2c \plus{} \sum a \geq \sum 3a(\sqrt[3]{abc}) \equal{} 3\sum a$

$ \leftrightarrow \sum ab\plus{} \sum a^2c \geq 2 \sum a$

Ηλίας

[quote="Athinaios"]
διοτι,
$ \sum a^2b \geq \sum a^2c$,αληθης απο αναδιαταξη
Hλίας[/quote]
Μπορείς να το εξηγήσεις αυτο λίγο? Δεν νομίζω ότι ισχύει πάντα...! :wink:

[quote="new_member"]
...............................
 Xoris vlavi tis genikotitas mporoume na theorisoume oti $ a\geq b\geq c\geq0$  opote i parapano anisotita einai alithis.

   I lysi mou itan paromoia me ayti tou Silouan..Hlia den katalavaino giati $ abc \equal{} 1$...Pote leme oti mia anisotita einai omogenis?
   Mporei kapoios na eksigisei?Prospathisa na vro kati online alla mataia..

~Kostas[/quote]

Αν ήταν αναδιάταξη, γιατί  να μην ισχύει και η $ a^2c \plus{} b^2a \plus{} c^2b \geq a^2b \plus{} b^2c \plus{} c^2a$ ;

 Επομένως δεν είναι αναδιάταξη. Αλλά ούτε με τον παραπάνω τρόπο λύνεται, διότι ως μη συμμετρική δεν μπορούμε να διατάξουμε και τις τρεις μεταβλητές  :) .

[u]Μπάμπης[/u]
 [color=red][b](Α! Για τις ομογενείς ανισότητες και τις συμμετρικές , δες το βιβλίο : Στεργίου - Σκομπρής : Αλγεβρικές ανισότητες, εκδόσεις Σαββάλα. Δανίσου το από κάποιον φίλο σου που το έχει.)[/b][/color]

[quote="new_member"]Gia tin anisotita pou egrapse o Hlias oti einai alithis,nomizo pos isxyei...mias kai:

       $ a^2b \plus{} b^2c \plus{} c^2a\geq a^2c \plus{} b^2a \plus{} c^2b$  kai isodynama exoume
       $ ab(a \minus{} b) \plus{} bc(b \minus{} c) \plus{} ca(c \minus{} a)\geq0$   or
       $ (a \minus{} b)(b \minus{} c)(a \minus{} c)\geq0$

 Xoris vlavi tis genikotitas mporoume na theorisoume oti $ a\geq b\geq c\geq0$  opote i parapano anisotita einai alithis.

   I lysi mou itan paromoia me ayti tou Silouan..Hlia den katalavaino giati $ abc \equal{} 1$...Pote leme oti mia anisotita einai omogenis?
   Mporei kapoios na eksigisei?Prospathisa na vro kati online alla mataia..

~Kostas[/quote]

Ομογενης ονομαζεται η ανισοτητα στην οποια αν αντικαταστησεις τις μεταβλητες με τις ιδιες πολλαπλασιασμενες με εναν πραγματικο αριθμο κ, τα κ απλοποιουνται και μενει παλι η ιδια ανισοτητα. Ετσι μπορεις χωρις να βλαπτεται η γενικοτητα να αντικαταστησεις τις μεταβλητες με τις ιδιες πολλαπλασιασμενες με εναν αριθμο που να απλοποιει την ανισοτητα. Εδω πχ ο Ηλιας πολλαπλασιαζει με $ k \equal{} \frac {1}{\sqrt [3]{abc}}$ οποτε προκειπτει η ιδια ανισοτητα με μεταβλητες με γινομενο 1, συνηθιζεται ομως οταν χρησημοποιουμε την ομογενοια μιας ανισοτητας να ονομαζουμε τις νεες μεταβλητες που προκειπτουν οπως τις αρχικες, οποτε λεμε απλα: λογο ομογενοιας εστω οτι αβγ=1 η α+β+γ=14 κλπ

Για να βαλεις τωρα τις μεταβλητες μιας ανισοτητας σε καποια διαταξη (ειτε για να χρησημοποιησεις την ανισοτητα αναδιαταξης ειτε για οτιδηποτε αλλο), πρεπει η ανισοτητα να ειναι συμμετρικη. Συμμετρικη ειναι η ανισοτητα η οποια μενει ιδια αν αλλαξουμε δυο μεταβλητες της (οπως πχ η $ a^2 \plus{} b^2 \plus{} c^2 \geq ab \plus{} bc \plus{} ca$). Η ανισοτητα που γραφεις πιο πανω ειναι κυκλικη, δεν ειναι ομως συμμετρικη (προσπαθησε να αλλαξεις 2 μεταβλητες της και δες τι γινεται), αρα δεν μπορεις να υποθεσεις οτι α>β>γ.

Η λυση του Ηλια ομως ειναι σωστη διοτι η αρχικη ανισοτητα ειναι συμμετρικη, οποτε μπορει απο την αρχη να υποθεσει οτι α>=β>=γ. Γι αυτο και μεσα στην παρασταση του εμφανιζονται και οι δυο παραστασεις ($ a^2b \plus{} b^2c \plus{} c^2a$ και $ a^2c \plus{} b^2a \plus{} c^2b$), οποτε μπορει να φτασει στη λυση εφαρμοζοντας ανισοτητα αναδιαταξης σε μια απο τις 2 αυτες παραστασεις, διακρινοντας περιπτωσεις για τη διαταξη των μεταβλητων (εδω μαλιστα δεν χρειαζονται περισσοτερες απο 2 περιπτωσεις διοτι λογο κυκλικοτητας των παραστασεων μπορει να θεωρησει οτι: a=max{a,b,c}, η οτι a=min{a,b,c}, πραγμα που επιτρεπεται σε καθε κυκλικη αλγεβρικη παρασταση και διαφερει απο το να θεωρησεις οτι α>=β>=γ.)

Ελπιζω να βοηθησα

[quote="NickNafplio"]
Η λυση του Ηλια ομως ειναι σωστη διοτι η αρχικη ανισοτητα ειναι συμμετρικη, οποτε μπορει απο την αρχη να υποθεσει οτι α>=β>=γ. Γι αυτο και μεσα στην παρασταση του εμφανιζονται και οι δυο παραστασεις ($ a^2b \plus{} b^2c \plus{} c^2a$ και $ a^2c \plus{} b^2a \plus{} c^2b$), οποτε μπορει να φτασει στη λυση εφαρμοζοντας ανισοτητα αναδιαταξης σε μια απο τις 2 αυτες παραστασεις, διακρινοντας περιπτωσεις για τη διαταξη των μεταβλητων (εδω μαλιστα δεν χρειαζονται περισσοτερες απο 2 περιπτωσεις διοτι λογο κυκλικοτητας των παραστασεων μπορει να θεωρησει οτι: a=max{a,b,c}, η οτι a=min{a,b,c}, πραγμα που επιτρεπεται σε καθε κυκλικη αλγεβρικη παρασταση και διαφερει απο το να θεωρησεις οτι α>=β>=γ.)
Ελπιζω να βοηθησα[/quote]

Αρκιβώς! Απλά ήθελα να τονίσω ότι αυτό που λέει ο Ηλίας δεν προκύπτει από αναδιάταξη! Ετσι όπως το λέει το NickNafplio είναι μια χαρά :wink:

yep παιδια συγνωμη για το μπερδεμα που προκαλεσα.
Τωρα αντι για την σχεση:
$ \sum a^2b \geq \sum a^2c$

μπορουμε να κανουμε το εξης(ισως πιο απλο..)
$ \sum (a\plus{}b)(b\plus{}c) \plus{} \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{2}$
$ \equal{} \sum a^2\plus{} 3\sum ab \plus{} \frac{\sum a^2b \plus{} \sum a^2c \plus{} 2abc}{2}$
$ \equal{}(a\plus{}b\plus{}c)^2 \plus{} \sum ab \plus{} \frac{\sum a^2c \plus{} \sum a^2b}{2} \plus{} 1 \geq (a\plus{}b\plus{}c)^2 \plus{} 1 \plus{} 2(a\plus{}b\plus{}c)$

διοτι,

$ ab \plus{} a^2b \plus{} a^2c \plus{} ac \geq 4a\sqrt[4]{(abc)^2}\equal{}4a$
,απο ανισοτητα αμ-γμ
κυκλικα , προσθεση κατα μελη και τελειωσαμε.

Many thanks NickNafplio...thanks pou eksigises tin enoia tis omogeneias kai tis symmetrias...katanoisa to lathos mou.


function
number theory unsolved
number theory

[quote="every"]Find all function $ f: R^{ \plus{} } \minus{} > R^{ \plus{} }$ such that
$ f(f(x)) \plus{} f(x) \equal{} (2006^{3^{2008}} \plus{} (2006^{3^{2008}})^2)x$[/quote]
 Mods: Please lock this topic, this is a problem from M&Y, issue 3,,2008, and we can't discuss it now


I need to list all the Stack-Sortable Permutations for n=4.  I know it will be 14 as it moves with the Catalan numbers, but I really don't understand the definition.

Cn is the number of stack-sortable permutations of {1, ..., n}. A permutation w is called stack-sortable if S(w) = (1, ..., n), where S(w) is defined recursively as follows: write w = unv where n is the largest element in w and u and v are shorter sequences, and set S(w) = S(u)S(v)n, with S being the identity for one-element sequences.

If anyone can help me out, that would be great.

Thanks!

I can't make heads or tails of that definition, but an equivalent definition is that of $ 231$-[i]avoidance[/i]: a permutation $ w \equal{} w_1\cdots w_n$ [i]contains[/i] $ 231$ if there are $ i < j < k$ such that $ w_k < w_i < w_j$; otherwise, $ w$ [i]avoids[/i] $ 231$.  In terms seemingly closer to the definition you gave, a permutation is stack-sortable (equiv. $ 231$-avoiding) if it has the form $ u \, n \, v$ where every entry of $ u$ is smaller than every entry of $ v$ and $ u, v$ are both stack-sortable.

How's that?

It's a very poort definition but figured what needed to be figured out, thanks!


search
combinatorics unsolved
combinatorics

Each positive integer shall be coloured red or green such that it satisfies the following properties:

- The sum of three not necessarily distinct red numbers is a red number.
- The sum of three not necessarily distinct green numbers is a green number.
- There are red and green numbers.

Find all such colorations!

The number $ 1$ must be red$ (R)$ or green$ (G)$.
Suppose $ 1\in R$.

We have that $ 1\plus{}1\plus{}1\equal{}3, 3\plus{}1\plus{}1\equal{}5, 5\plus{}1\plus{}1\equal{}7,....$ if we go on we find that $ (1,3,5,....,2\alpha \plus{}1)\in R$.

Now we show that $ R$ can't have an even number.
If an even number $ \beta$ is in $ R$ then all the even numbers $ >\beta$ must be in $ R$
We know that $ (\beta \minus{}2)\in G$ and must be $ (3\cdot(\beta \minus{}2))\in G$. But $ 3\cdot(\beta \minus{}2)>\beta$, and $ 3\cdot(\beta \minus{}2)\in R$. contradiction.

So all the even numbers are in $ G$. Obviously each sum of three even numbers is an even number.

The case $ 1\in G$ is totally equivalent, so we have finished.

Not a big mistake and can be easily corrected... If $ \beta \equal{} 2$ ?

[quote="Kirill"]If $ \beta \equal{} 2$ ?[/quote]

if $ \beta\equal{}2$ then $ R\equal{}(1,2,3,4,5.....)$ and $ G\equal{}\emptyset$. But $ G\neq\emptyset$ because

[quote="orl"]- There are red and green numbers. [/quote] :wink:

This one has appeared before, too:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=477003617&t=146599


logarithms
linear algebra
matrix
algebra unsolved
algebra

Let $ (a,b,c,d)\in{\mathbb{R_{+}^{*}}-{1}}$.

we note that:

$ \lambda$=$ \ln{\frac{a}{c^{2}}.\ln{\frac{a}{d^{3}}-(\ln{b})^{2}}}$

with $ \lambda\neq0$

solve the following system :

$ \left\{\begin{matrix}a^{x}b^{y}=c^{2x-1}+c^{2x+1}&\\b^{x}a^{y}=d^{3y-1}+d^{3y+1}&\end{matrix}\right.$

thanx.

dividing the first equation by $ c^{2x}$ transforms it into $ \left(\frac{a}{c^2}\right)^xb^y\equal{}\frac{1}{c}\plus{}c$. Similarly, the second equation transforms into $ \left(\frac{a}{d^3}\right)^yb^x\equal{}\frac{1}{d}\plus{}d$. For simplicity, let $ u\equal{}\frac{a}{c^2}, v\equal{}\frac{a}{d^3}, A\equal{}\frac{1}{c}\plus{}c, B\equal{}\frac{1}{d}\plus{}d$. So our equations are

\[ u^xb^y\equal{}A
b^xv^y\equal{}B\]

Taking $ \ln$'s,

\[ x\ln u \plus{} y\ln b \equal{} \ln A
x\ln b \plus{} y\ln v \equal{} \ln B\]

Subtracting $ \ln u$ times the second equation from $ \ln b$ times the first, we get $ \minus{}\lambda y \equal{} \ln A \ln b\minus{}\ln B \ln u$, so $ y\equal{}\frac{\minus{}1}{\lambda}(\ln A \ln b\minus{}\ln B \ln u)$.

Subtracting $ \ln b$ times the second equation from $ \ln v$ times the first, we get $ \lambda x \equal{} \ln A \ln v \minus{} \ln B \ln b$, so $ x \equal{}\frac{1}{\lambda}(\ln A \ln v\minus{} \ln B \ln b)$.


When five standard six-sided dice are rolled sequentially there are $ 6^5 \equal{} 7776$ possible outcomes. For how many outcomes is the sum of the five rolled numbers exactly 27?

First, what possible outcomes give you 27?  For convenience, let's list the possibilities in descending order:

One would be 6,6,6,6,3.  Now we can break one off a unit from somewhere to pass to lower numbers like this:  6,6,6,5,4 then 6,6,5,5,5 and now notice that moving a unit around would not give us anything new.

Now since the problem gives us the number of [u]sequences length 6[/u], we have to take into account that the sequence 6,6,6,6,3 is going to be different from 3,6,6,6,6 and so on.

For 6,6,6,6,3: this is easy... how many ways can you place 3?  5 total.
For 6,6,6,5,4: There are 5C3=10 ways to place the 6's... then you can write 4 before 5 or 5 before 4 in each one, so that accounts for 20 total patterns.
For 6,6,5,5,5:  There are 5C3=10 different ways to place the 5's and the other slots get filled with 6.  So 10 different patterns.

Then there are 5+20+10=35 total patterns, so $ \frac{35}{7776}$ is our answer.  Of course it's reduced because 35 is only divisible by 5 and 7, but 7776 is only divisible by 2 and 3.

Hope this helps, and hope I didn't make a silly error :)

to get 27 we have

6,6,6,6,3 -> 5 ways to arrange
6,6,6,5,4 -> 20 ways to arrange
6,6,5,5,5 -> 10 ways to arrange
________________________

35 total ways, so our answer is 35.

@Rightannihilator: Look out! The question asks for the number of ways to get 27,

not the probability of getting 27!


From Cauchy-Schwarz inequality,
\begin{align*}\sum_{i\neq j}\frac {x_i^2}{x_ix_j}&\ge\frac {\left(\sum_{i\neq j}x_i\right)^2}{\sum_{i\neq j}x_ix_j}\\
& = \frac {(n - 1)^2(x_1 + x_2 + \dots + x_n)^2}{(x_1 + x_2 + \dots + x_n)^2 - (x_1^2 + x_2^2 + \dots + x_n^2)}\\
& = \frac {(n - 1)^2t^2}{t^2 - t} = \frac {(n - 1)^2t}{t - 1}\end{align*}
P.S. May I know where you found this problem?

Thanks a lot Nayel for your solution. I think it is a team selection question of Turkey. But I am not sure.


inequalities
n-variable inequality
algebra
IMO Shortlist

Let $ a_1\geq \cdots \geq a_n \geq a_{n \plus{} 1} \equal{} 0$ be real numbers. Show that
\[ \sqrt {\sum_{k \equal{} 1}^n a_k} \leq \sum_{k \equal{} 1}^n \sqrt k (\sqrt {a_k} \minus{} \sqrt {a_{k \plus{} 1}}).
\]
[i]Proposed by Romania[/i]

l don't know what telescopic sum is...it's a classical sum !
i'll give you the solution so maybe it will help.

let $x_k=\sqrt{a_k}-\sqrt{a_{k=1}}$ for $k=1,\ldots,n$ so $a_1=(x_1+\cdots+x_n)^2, a_2=(x_2+\cdots+x_n)^2,\ldots,a_n=x_n^2$

after having developped the squares, we get \[  \sum_ {k=1}^n a_k=\sum_{k=1}^n k x_k^2+2 \sum_{1\leq k<\leq n} k x_k x_l   (1)\]

$x_k x_l$'s coefficient ($k<l$) in the last sum is equal to $k$ because this product appears exactly one time in each expression of $a_1,\ldots,a_n$ but isn't present in $a_{k+1},\ldots,a_n$

the square on the right member of the inequality equals \[   \left( \sum_{k=1}^n \sqrt k x_k \right)^2= \sum_{k=1}^n k x_k^2+2\sum_{1 \leq k<l \leq n} \sqrt{kl} x_k x_l.    (2)\]

since the value of (1) isn't $\geq$ than the value of (2), the result follows.

-   $\displaystyle \sum_{k=1}^n (\sqrt{a_k}-\sqrt{a_{k+1}})=(\sqrt{a_1}-\sqrt{a_2})+(\sqrt{a_2}-\sqrt{a_3})+\cdots+(\sqrt{a_n}-\sqrt{a_{n+1}})$

-   $a_1\geq a_2 \geq \ldots \geq a_2 \geq a_{n+1}=0$

that's all what l can tell you. howewer, the fact that $a_{n+1}=0$ doesn't intervene in the solution  written above.

now, if you really meant it that way, then what i've written first is a solution! we have $\sum^n_{k=1} (\sqrt{a_k}-\sqrt{a_{k+1}})=\sqrt{a_1}-\sqrt{a_2}+\sqrt{a_2}\pm...-\sqrt{a_{n+1}}$, so all terms except $+\sqrt{a_1}-\sqrt{a_{n+1}}$ cancel, so the right side of your inequality is just $\sqrt{a_1}$. on the left side we get the square root of $a_1$ plus some other nonnegative terms, so the inequality is true.

Peter

[Edit: wrong index]

l gave the official solution but yours is shorter and smarter, peter : congratulations.
therefore, this solution is very easy, and the problem is an imo problem ! (they have become harder in the future)

Lemma: If $k \geq l$, then $\sqrt{k+d} - \sqrt{k} \leq \sqrt{l+d} - \sqrt{l}$.

Proof of lemma: Just expand it lol

Now, rearranging the inequality gives $\sqrt{a_1+a_2+\ldots+a_n} - \sqrt{a_1} \leq \sqrt{2a_2}-\sqrt{a_2}+\sqrt{3a_3}-\sqrt{2a_3}+\ldots+\sqrt{ka_k}-\sqrt{(k-1)a_k}$. Now, repeatedly applying the lemma gives $\sqrt{2a_2}-\sqrt{a_2} \geq \sqrt{a_1+a_2}-\sqrt{a_1}$, $\sqrt{3a_3}-\sqrt{2a_3} \geq \sqrt{a_1+a_2+a_3}-\sqrt{a_1+a_2}$, and so on. The sum telescopes, giving the desired inequality.


Function is convex in every variable (actually it's 0, so it's linear in every variable, but eh) so it obtains it's maximum when each variable is at the endpoints, and so on.

The function can be rewritten as

\[ S = (x_1+x_2+\ldots+x_{100})-(x_1x_2+x_2x_3+\ldots+x_{100}x_1) \]

Let $k$ be the number of variables that are $1$, and $100-k$ are the number of variables that are $0$. Clearly, $x_1+x_2+\ldots+x_{100}=k$. For the other one, want to minimize the number of variables that aren't $0$, since we want to maximize the entire function. We will show by contradiction that this is when the solution set is

\[ (x_1,x_2,\ldots,x_{100})=(1,0,1,0\ldots,0) \]

Suppose this wasn't the maximal solution set, $S$. Let's change any of the $1's$ to a zero; there are still no pairs of consecutive ones, and since all of our variables are consecutive pairs, then there must be a consecutive pair for them to be $1$; so our pairwise terms are still a sum of $0$. However, we minimize our $\sum x_i$ by $1$, so our new solution set $S'$ is less than our original solution set $S$, and thus $S$ is greater than any solution set containing less than $50$ ones.

Suppose we add a $1$ anywhere. Our $\sum x_i$ is increased by $1$, clearly. However, say we had $(x_{i-1},x_i,x_{i+1})=(1,0,1)$ before and now we have it $(1,1,1)$. We have two new consecutive pairs of consecutive $1's$, so we have TWO of the pairwise products change from $0$ to $1$; thus the pairwise sum increases by $2$, and the overall sum $S'$ is decreased by $1$. Again, this shows that $S$ is maximal. Thus, our given solution set is maximal, with a sum of $\fbox{50}$.


calculus
derivative
function
integration
calculus computations

1. Express 1/(1-x)^2 as a power series by differentiating 1/(1-x). What is the radius of convergence? My book says:
1/(1-x)=1+x+x^2+x^3....=Sigma n=0...infinity x^n. Then it differentiating each side, get: 1/(1-x)^2=1+2x+3x^2+....Sigma n=1...infinity nx^(n-1). However, Here I have a problem, I think the derivative (1-x)^(-1) is -1(1-x)^(-2), so there should be a negative sign there.
2. Find a power series representation for ln(1-x) and its radius of convergence. My book says"We notice that, except for a factor of -1, the derivative of this function is 1/(1-x).  Then it integrates both sides: -ln(1-x)=integral sign 1/(1-x) dx=...
Here I don't understand where did that -1 come from?

Thanks for help.

Please try to write by Latex. Sometimes peoples will  not read your post.

Your differentiation for (1) is incorrect since you have not used the chain rule correctly.  Notice that you also need to multiply it by $ \frac{d}{dx}(1\minus{}x)\equal{}\minus{}1$.


geometry
3D geometry
quadratics
perimeter
trapezoid
trigonometry
number theory

1.

Call the common difference $ a$. Then, the angles are

$ 50, 50\plus{}a, 50\plus{}2a, 50\plus{}3a.$

Since $ \angle 3<90$, $ a<20 \implies 3a<60 \implies 50\plus{}3a<110$

Therefore, the answer is $ 109$

Comments and Feedback:

This was a fairly easy question, but many people assumed that

$ a\equal{}19$, and got $ 107$. Should you get $ a<20$ and then

realize that you have to manipulate it, it is very easy.

Personally, I assumed $ a\equal{}19$ and thought that $ 50\plus{}3 \cdot 19\equal{}109$. LOL.

2. The prime factorization of the number is $ p_1^3$ or

$ p_2p_3$, for some primes $ p_1$, $ p_2$, $ p_3$.

Checking $ n$ where $ n$ is no more then $ 3$ away then

perfect cubes, we get $ 341$. Using the second factorization,

we can easily see that the smallest is $ \boxed{33}$.

$ 33\equal{}3 \cdot 11, 34\equal{}2 \cdot 17, 35\equal{}5 \cdot 7$.

Comments and Feedback:

This question was bashable, which was probably the way most

people did it. Personally, I brute forced up from $ 30$, got

the hit $ 33$, and checked everything below it.

I don't really see how you can get this without testing at least

some numbers.

3.

We know that $ a^2\plus{}b^2\equal{}(a\plus{}b)^2\minus{}2ab$.

Therefore, using the $ 2$ quadratic equations that we have,

we can easily get the sum and the product of the roots of both

equation, and the result follows.

Comments and Feedback:

The sum of squares factorization helps a lot,

but this was easily brute forceable by finding the roots.

Personally, I found this very easy.

4.

Let the smaller side be $ x$. Then, the larger side is $ 2x$.

The perimeter is obviously $ x\plus{}x\plus{}x\plus{}2x\plus{}2x\plus{}2x\equal{}9x$.

Then, extend the sides to make a big equilateral triangle.

Similar triangles gives

$ \frac{3x}{4x}\equal{}\frac{8}{4x} \implies 3x\equal{}8 \implies 9x\equal{}24$.

Comments and Feedback:

This was very easy if you saw to extend the sides.

Even if you didn't, there are other ways to do it.

In order of easiest to hardest, in my opinion,

Extending, Using the trapezoid, Drawing radii.

Those are the $ 4$ that I got. I won 't post solutions for the

others, because, well, I don't know them  :lol:

8. Using Arithmetic sequences, we get S_k = k(3k-1)/2. Since k and 3k-1 are relatively prime, one is a perfect square and one is 2 times a perfect square. 3k-1 cannot be a perfect square because -1 is not a quadratic residue mod 3. Assume that k is even. Since k is a perfect square, it can be expressed in the form $ 2^2n o$, where o is odd. We get that the largest power of 2 that divides S_k is 2^2n-1. Since 2n-1 is odd, s_k is not a perfect square. Now brute force using odd squares.

6. Use cos x= sin(pi/2-x) to get sinx+siny=pi/2. square to get sinx*siny, then turn tanx +cot x into 1/(sinx siny).

For number 6 I think I got the opposite of the actualy answer, but adding the three numbers came out to be right

So $ \sin(2\sin x) \equal{} \cos(2\cos x)$

When the sine of an angle is the cosine of another angle are equal, those angles are complementary, so

$ 2\sin x \plus{} 2\cos x \equal{} \frac\pi2$

$ \sin x \plus{} \cos x \equal{} \frac\pi4$, squaring

$ 1 \plus{} 2\sin x\cos x \equal{} \frac {\pi^2}{16}$

$ \sin x\cos x \equal{} \frac {16 \minus{} \pi^2}{32}$

$ \tan x \plus{} \cot x \equal{} \frac1{\sin x\cos x} \equal{} \frac {32}{16 \minus{} \pi^2}$

so $ 32 \plus{} 16 \plus{} 2 \equal{} 50$

I got the right answer in the end, but where did I go wrong?

[quote="Twiz"][quote="nat mc"]$ 1 \plus{} 2\sin x\cos x \equal{} \frac {\pi^2}{16}$

$ \sin x\cos x \equal{} \frac {16 \minus{} \pi^2}{32}$[/quote]
Messed up rearrangement.[/quote]
well he messed with the sign on purpose to make the answer turn out in the form we need
i got the same thing, but if we keep it be pi squared-16, we'd get a negative value for tan + cot, but given the range of x, it should be positive
hmm

[quote="nat mc"]I got the right answer in the end, but where did I go wrong?[/quote]
Right here :
[quote]When the sine of an angle is the cosine of another angle are equal, [b]those angles are complementary[/b][/quote]

You assumed
\[ \sin a \equal{} \cos b \implies a \plus{} b \equal{} \pi/2,
\]
where $ (a,b) \equal{} (2\sin x, 2\cos x)$.  But in fact, $ a$ and $ b$ are in $ [0,2]$, so $ \sin a \equal{} \cos b$ only implies
\[ a \pm b \equal{} \pm \pi/2 .
\]
For instance, in general, we could have $ a \equal{} \frac {3\pi}{4}$ and $ b \equal{} \pi/4$.

In this case, we in fact have $ a \minus{} b \equal{} \pm \pi/2$, because of the restrictions that junggi pointed out.  The sign of $ \pi/2$ is, of course, irrelevant.

2.  Either one or two of the numbers are even.  If two of them are even, one is a multiple of $ 4$, hence must be $ 8$; this doesn't work, so $ n \plus{} 1$ is even; furthermore, it's twice a prime.

Either $ n$ or $ n\plus{}2$ must be divisible by $ 3$, so the other is a product of two primes greater than $ 3$.  This first occurs at $ 35 \equal{} 5 \cdot 7$, and we find that $ 33, 34, 35$ works.

4.  The part of the hexagon to the left of AD is five congruent equilateral triangles with side length $ \frac{8}{3}$.  The rest is eight more congruent equilateral triangles with the same side length.

6.  [b]Sum to product identity:[/b]  $ \sin a \minus{} \sin b \equal{} 2 \sin \left( \frac{a\minus{}b}{2} \right) \cos \left( \frac{a\plus{}b}{2} \right)$.

Write as $ \sin(2 \sin x) \equal{} \sin( \frac{\pi}{2} \plus{} 2 \cos x)$.  The rest is casework.  It turns out that the case that $ 2 \sin x \equal{} \frac{\pi}{2} \plus{} 2 \cos x$ gives us an answer in the right form.

8.  $ k(3k\minus{}1) \equal{} 2n^2 \Leftrightarrow (6k\minus{}1)^2 \minus{} 6(2u)^2 \equal{} 1$.  This is a [url=http://en.wikipedia.org/wiki/Pell%27s_equation]Pell's equation[/url] whose solutions are described by the coefficients of $ (5 \plus{} 2 \sqrt{6})^n$.  In fact, $ (5 \plus{} 2 \sqrt{6})^3 \equal{} 481 \plus{} 198 \sqrt{6}$, which gives $ k \equal{} 81$.


modular arithmetic
number theory unsolved
number theory

find x,y,z are positive integer such that z is not divided by 3, y is a prime number.
x^3 -  y^3 =  z^2

$ (x \minus{} y)(x^2 \plus{} y^2 \plus{} xy) \equal{} z^2$
if $ \gcd(x \minus{} y,x^2 \plus{} y^2 \plus{} xy) \equal{} d$ then $ d | 3$
if $ d \equal{} 3$ then $ 3 | z$ which is a contradiction with the assumption.
so we get $ d \equal{} 1$ and then
$ x \minus{} y \equal{} u^2 \ \ \ , \ \ \ x^2 \plus{} y^2 \plus{} xy \equal{} v^2$
which implies
$ 4x^2 \plus{} 4xy \plus{} 4y^2 \equal{} 4v^2 \implies (2x \plus{} y)^2 \plus{} 3y^2 \equal{} (2v)^2 \implies$

$ 3y^2 \equal{} (2v \minus{} 2x \minus{} y)(2v \plus{} 2x \plus{} y)$
note that $ y$ is a prime so we have to check 3 cases:

1-$ 2v \minus{} 2x \minus{} y \equal{} y,2x \plus{} y \plus{} 2v \equal{} 3y \implies x \equal{} 0,v \equal{} y$ impossible,

2-$ 2v \minus{} 2x \minus{} y \equal{} 1,2x \plus{} y \plus{} 2v \equal{} 3y^2$
$ \implies 4u^2 \equal{} 3y^2 \minus{} 6y \minus{} 1 \implies u^2 \equiv \minus{} 1 \pmod {3}$ impossible,

3-$ 2v \minus{} 2x \minus{} y \equal{} 3,2x \plus{} y \plus{} 2v \equal{} y^2 \implies (y \minus{} 3)^2 \minus{} (2u)^2 \equal{} 12$ we'll get $ y \equal{} 7,u \equal{} 1 \implies x \equal{} 8,z \equal{} 13$
so the only solution is $ (x,y,z) \equal{} (8,7,13)$.


limit
logarithms
inequalities
calculus
calculus computations

Evaluate the product $\frac{e \sqrt[3]{e} \sqrt[5]{e} ... }{ \sqrt{e} \sqrt[4]{e} \sqrt[6]{e} ... }$

Well, the problem statement has the square root of $e$ twice in the denominator, but let's assume that's a typo. The other thing to be careful of is that you can make the sum come out differently by choosing a different pattern for the partial products. RaduSorici's choice is the most logical one, though, and I agree with his result.

[quote="Xevarion"]Well, the problem statement has the square root of $e$ twice in the denominator[/quote]

You are right, that was a mistake  :blush:

However, if I define the product very explicitly:

$\lim_{n \to \infty} \frac{e \sqrt[3]{e} \sqrt[5]{e} ... \sqrt[2n-1]{e} }{ \sqrt{e} \sqrt[4]{e} \sqrt[6]{e} ... \sqrt[2n]{e} }$

Then it does tend to one exact value (without using $\gamma$!) because the product is equivalent to

$e^{ 1 - \frac{1}{2} + \frac{1}{3} -+... }$

And the exponent converges.   ;)

[b]Lemma 1[/b] ([color=red]the [u]Catalan's identity[/u][/color]) : $\boxed {\ \sum^n_{k=1} \frac{1}{n+k}=\sum^{2n}_{k=1} (-1)^{k-1}\cdot \frac 1k\ }\ .$
[b]Proof.[/b] $\sum^n_{k=1}\frac{1}{n+k}=$ $\sum^{2n}_{k=1}\frac 1k-\sum^n_{k=1}\frac 1k=$ $\sum^n_{k=1} \frac{1}{2k-1}+\sum^n_{k=1}\frac{1}{2k}-\sum^n_{k=1}\frac 1k=$ $\sum^n_{k=1}\frac{1}{2k-1}-\sum^n_{k=1}\frac{1}{2k}=$ $\sum^{2n}_{k=1} (-1)^{k-1}\frac 1k\ .$

[b]Lemma 2.[/b] $\boxed {\ L_n\equiv\lim_{n\to\infty} \sum^n_{k=1}\frac{1}{n+k}=\ln 2\ }\ .$
[b]Proof.[/b]From a well-known remarkable inequality $(*)$ for $n\in N$ and $\left(\forall\right)k\in \overline {1,n}\ ,\ \frac{1}{n+k+1}<\ln (n+k+1)-\ln (n+k)<\frac{1}{n+k}\Longrightarrow$
$\sum^n_{k=1}\frac{1}{n+k+1}<\ln (2n+1)-\ln (n+1)<\sum^n_{k=1}\frac{1}{n+k}\Longrightarrow$
$L_n+\frac{1}{2n+1}-\frac{1}{n+1}<\ln\frac{2n+1}{n+1}<L_n\Longrightarrow$
$\ln\frac{2n+1}{n+1}<L_n<\ln\frac{2n+1}{n+1}+\frac{1}{n+1}-\frac{1}{2n+1}\ .$
For $n\to\infty$ in the previous bilateral inequality we obtain $\lim_{n\to\infty} L_n=\ln 2\ .$

$(*)$ [b]Remark.[/b] For $k,n\in N^*\ : \ \left(1+\frac 1n\right)^n<e<\left(1+\frac 1n\right)^{n+1}\Longrightarrow$
$\frac{1}{n+1}<\ln (n+1)-\ln n<\frac 1n\Longrightarrow$
$\frac{1}{n+k+1}<\ln (n+k+1)-\ln (n+k)<\frac{1}{n+k}\ .$


In square ABCD, N is on side AD and M is on side AB, such that segments CM and CN cut the square into 3 regions with equal areas.  What is the length of CN?

Are we able to calculate CN without knowing the length of the side of the square?

[hide]I got CN= a :sqrt: 13 /3. a is the side length of the square.[/hide]

Yes you are correct.

And no, you can't find the exact numerical value of CN because the ratio of the areas of the three regions can remain constant regardless of the length of the side of ABCD. So, that changes CN based upon, as you said, the side length.

I totally agree with cj.
[hide]
Let $a$ = side of the square.

Right triangle NDC has an area of $\frac{1}{3}a^2.$

Its base is $a.$
Let it height be $h = ND.$

Then: $\frac{1}{2}(a)(h) = \frac{1}{3}a^2\;\;\Rightarrow\;\;h = \frac{2}{3}a.$

Hence: $NC^2 \;= \;a^2 + \left(\frac{2}{3}a\right)^2 \;= \;\frac{13a^2}{9}$

Therefore: $NC \;= \;\sqrt{\frac{13a^2}{9}} \;= \;\frac{\sqrt{13}}{3}\;a$[/hide]

Heh, yep I left that little detail out.  Suppose the side length is 3.


Source: AHSME 1973 #12

The average (arithmetic mean) age of a group consisting of doctors and lawyers is 40.  If the doctors average 35 and the lawyers 50 years old, then the ratio of the number of doctors to the number of lawyers is
$\text{(A)} \ 3: 2 \qquad \text{(B)} \ 3: 1 \qquad \text{(C)} \ 2: 3 \qquad \text{(D)} \ 2: 1 \qquad \text{(E)} \ 1: 2$

[hide]

x is the amount of doctors and y the amount of lawyers.

  35x + 50y
  -------------   = 40
     x + y

35x + 50y = 40x + 40y

10y = 5x

Thus the ratio is 2 : 1

[/hide]

[hide]
let $x$ be the number of doctors
let $y$ be the number of lawyers

The total age of the doctors is $35x$ and the total age of the lawyers is $50y$
We also know the total age of the whole group is $40(x+y)$
So set up an equation :

$40 (x + y) = 35x + 50y \\ 8(x+y) = 7x + 10y\\ x = 2y\\ \frac{x}{y} = 2\\$

So the ratio of the number of doctors to lawyers is $\boxed {(D)2 : 1}$
[/hide]


function
combinatorics open
combinatorics

Given a set of size n, how many ways are there to partition it, and then order each partition?  For example, there is one way to do it when n=1:

{1}

3 ways to do it when n=2:

{1,2}
{2,1}
{1},{2}

and 13 ways to do it when n=3:

{1,2,3}
{1,3,2}
{2,1,3}
{2,3,1}
{3,1,2}
{3,2,1}
{1},{2,3}
{1},{3,2}
{2},{1,3}
{2},{3,1}
{3},{1,2}
{3},{2,1}
{1},{2},{3}

Thus far, the fastest solution I can come up with (I'm not certain it is correct) is to sum over all k, where k is the number of partitions:

$ \sum_{k\equal{}1}^{n}\dfrac{n!S(n,k)}{k!}$

Where S(n,k) is the Stirling number of the second kind.  I would prefer something faster--a closed form if possible.  I'm eventually going to need this for a computer program where computations like this one will be performed many times.  However, even if there's no faster solution, I'd still like to know whether my solution is correct.  It's been a while since I've done combinatorics.

If I'm not mistaken, this is [url=http://www.research.att.com/~njas/sequences/A000262]A000262[/url], the number of "sets of lists."  Its exponential generating function $ e^{\frac {x}{1 \minus{} x} }$ is obtained by composing the generating function for partitions with the generating function for non-empty ordered sets, and there are a few formulae available on the OEIS entry, in particular a recursion that should allow for reasonably fast computations.

Interesting--I never knew there was an entire online database for mathematical sequences.  But that page is definitely what I'm looking for.  Thanks!

The [url=http://www.research.att.com/~njas/sequences/index.html]OEIS[/url] is an unspeakably useful resource in solving enumerative problems.  It provides a very quick check for the question, "Should this problem have a nice answer?" and only [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=260825]occasionally[/url] leads me astray.


inequalities
calculus
integration
inequalities proposed

Edit: Ignore this post please... I posted a wrong inequality...

How do you use Holder on this one, yptsoi? (I mean the original question)  :? I see only a solution with means which is pretty obvious.

The following inequality holds for $a_1,a_2,...,a_n$ positive reals:
$a_1+a_2+...+a_n \geq (n-1)\sqrt[n]{a_1a_2...a_n}+\sqrt \frac{a_1^2+a_2^2+...+a_n^2}{n}$.

More general, for any positive integral $k$ we have
$n^{k-2}(a_1+a_2+...+a_n) \geq (n^{k-1}-1)\sqrt[n]{a_1a_2...a_n} + \sqrt[k]{\frac{a_1^k+a_2^k+...+a_n^k}{n}}$.

[quote="Vasc"]The following inequality holds for $a_1,a_2,...,a_n$ positive reals:
$a_1+a_2+...+a_n \geq (n-1)\sqrt[n]{a_1a_2...a_n}+\sqrt \frac{a_1^2+a_2^2+...+a_n^2}{n}$.

[/quote]
Let $a_1+a_2+...+a_n=nu,$ $a_1a_2+a_1a_3+...+a_{n-1}a_n=\frac{n(n-1)}{2}v^2$ and
$a_1a_2\cdot...\cdot a_n=w^n.$ $( v\geq0, w\geq0 )$
Hence, our inequality equivalent to $nu-(n-1)w\geq\sqrt{\frac{n^2u^2-n(n-1)v^2}{n}}\Leftrightarrow$
$\Leftrightarrow nu^2-2nuw+(n-1)w^2+v^2\geq0.$ But $v\geq w.$ Hence remain to prove that
$nu^2-2nuw+nw^2\geq0,$ which is true.
It is interesting, that $v^{2n-2}\geq u^{n-2}w^n$ by Maclaren inequality.Therefore your inequality can be strengthened for $n>2.$

For example: let $a,$ $b,$ $c$ and $d$ are non-negative real numbers. Prove that
$a+b+c+d\geq\frac{8}{3}\cdot\sqrt[4]{abcd}+\frac{4}{3}\cdot\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}.$

[quote="arqady"]$a\geq0,$ $b\geq0,$ $c\geq0.$ Prove that
$3(a+b+c)\geq 7\sqrt[3]{abc} +2 \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$
This is stronger.  ;)[/quote]

How do you prove this?

[quote="arqady"]
Let $a_1+a_2+...+a_n=nu,$ $a_1a_2+a_1a_3+...+a_{n-1}a_n=\frac{n(n-1)}{2}v^2$ and
$a_1a_2\cdot...\cdot a_n=w^n.$ $( v\geq0, w\geq0 )$
[/quote]

Maybe a something like this ? ;)

Nice your proof, Arqady. But does it work for any positive integer $k$?

I think that $ 3(a\plus{}b\plus{}c) \ge \frac{13}{2}\sqrt[3]{abc}\plus{}\frac{5}{2}\sqrt[3]{\frac{a^3\plus{}b^3\plus{}c^3}{3}}$ holds true too.  :maybe:

[quote="caubetoanhoc94"][quote="arqady"]$ a\geq0,$ $ b\geq0,$ $ c\geq0.$ Prove that
$ 3(a \plus{} b \plus{} c)\geq 7\sqrt [3]{abc} \plus{} 2 \sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}$
This is stronger.  ;)[/quote]

We also have:
$ \frac {11}{3}(a \plus{} b \plus{} c) \ge 8 \sqrt [3]{abc} \plus{} 3 \sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}$  :P

This is stronger  :lol:[/quote]
No, the ineq which arqady posted (the author is can_hang) is stronger than mine.
But the hint in my old proof also solved it  :)
You can see that proof here: (#183)
http://www.maths.vn/forums/showthread.php?t=9244&page=19

[quote=Yimin Ge]For positive reals  $a,b,c$, prove:
\[ 3(a+b+c)\geq 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}} \]

When does equality hold?[/quote]
Let $ a,$ $ b$ and $ c$ are non-negative numbers. [url=http://www.artofproblemsolving.com/community/c6h228831p1268135]Prove that:[/url]
\[ 3(a\plus{}b\plus{}c) \geq 7\sqrt[3]{abc}\plus{}2\sqrt[3]{\frac{a^3\plus{}b^3\plus{}c^3}{3}}\]


When corresponding terms of two different arithmetic sequences are multiplied, the new sequence below is formed.  Find the 8th term of this new sequence.
1440, 1716, 1848, ...

how would you solve this?
is this an aime question?

Let the first arithmetic sequence be : $p,p+b,p+2b,...,p+7b,...$ and the second sequence be $k,k+n,k+2n,...,k+7n,...$
The new sequence is $pk,(p+b)(k+n),(p+2b)(k+2n),...,(p+7b)(k+7n),...$

From the given information :
$pk=1440$
$(p+b)(k+n)=1716 \rightarrow pk+bk+pn+bn=1716 \rightarrow bk+pn+bn=276$
$(p+2b)(k+2n)=1848 \rightarrow pk+2(bk+pn+bn)+2bn=1848 \rightarrow bn=-72$

$(p+7b)(k+7n)=pk+7(bk+pn+bn)+42bn$
                    $=1440+7(276)+42(-72)$
                    $=348$

Therefore 8th term is $348$


ABCD (clockwise from top left) is a rectangle. CEF is an equilateral triangle, where E is on DA and F is on AB. Prove that the area of triangle EAF is equal to the area of triangle CBF plus the area of triangle CDE.

Let angle DCE = a. Then CED = 90 - a, AFE = 60 - a, BFC = 60 + a, and BCF = 30 - a.

Area of triangle AEF = 0.5sin(60-a)cos(60-a) = ((sqrt3)/8) cos 2a + (1/8) sin 2a.

Triangle BCF = ((sqrt3)/8) cos 2a - (1/8) sin 2a and triangle CED - (1/4) sin 2a.

Therefore, triangle AEF = triangle BCF + triangle CED.


Let ABC be a triangle, and let X be a point on the side AB that is not A or B. Let P be the incentre of the triangle ACX, Q be the incentre of the triangle BCX and M be the mid-point of the segmant PQ. Show that $|MC|>|MX|$

This is the one problem i couldnt do on the IrMO paper two,  :( but however many else i got wrong i do not know... :)

Cool problem.  Here's a sketch of the solution.  By calculating the medians of triangles $CPQ$ and $XPQ$ (in terms of the sides) we see that it suffices to prove $XP^2+XQ^2 > CP^2 + CQ^2$.  A few applications of Pythagoras and calculations reduces this to

\[(XC)(AC-AX) + (XC)(BC-XB) > 0 \iff AC+ BC > AX+ BX = AB,\]

which is obvious from the triangle inequality.

what IRMO is that?
where did you get the Qs?

The Irish Mathematical Olympiad, I'd imagine.

oh i c... i thought it was Iranian Math olympiad... :)

the diagrame looks terrific when I drew it,
where do you get these problems form,I am going to give it to my class and award any student solving it


Suppose that $ a\star b\equal{}ab\minus{}b$ for all integers $ a$ and $ b$. What is the value of $ 3\star (\minus{}2)$?

Substituting into the equation, we have a=3, b=-2.

3(-2)--2
-6+2
$ \boxed{ \minus{} 4}$


geometry
circumcircle
induction
parallelogram
geometric transformation
reflection
conics

8.In the plane are given 4 points  $A,B,C,D$. The points $A_1,B_1,C_1,D_1$ are orthocenters of the triangles $BCD,CDA,DAB,ABC$.$A_2,B_2,C_2,D_2$ are orthocenters of the triangles $B_1C_1D_1,C_1D_1A_1,D_1A_1B_1,A_1B_1C_1$ and...   .Prove that the circumcircles  of all  triangles formed by middelpoints of  this triangles are passing through the same point.$\clubsuit$

If we could show that the 9-point circles of the first four triangles $\triangle ABC,\ \triangle BCD,\ \triangle CDA,\ \triangle DAB$ pass through the same point Z and that the 9-point circles of the next four triangles $\triangle A_1B_1C_1,\ \triangle B_1C_1D_1,\ \triangle C_1D_1A_1,\ \triangle D_1A_1B_1$ also pass through the same point Z as the 9-point circles of the first four triangles, it would follow by induction that the 9-point circles of the last four triangles $\triangle A_2B_2C_2,\ \triangle B_2C_2D_2,\ \triangle C_2D_2A_2,\ \triangle D_2A_2B_2$ pass through the point Z as well.

Let P, Q, R, S be the midpoints of the quadrilateral sides AB, BC, CD, DA, forming a parallelogram PQRS. Let U, V be the midpoints of the quadrilateral diagonals AC, BD and X the intersection of the parallelogram diagonals PR, QS. $PU \parallel BC$ is a midline of the triangle $\triangle ABC,$ $RV \parallel BC$ is a midline of the triangle $\triangle BCD,$ $UR \parallel DA$ is a midline of the triangle $\triangle CDA,$ $VP \parallel DA$ is a midline of the triangle $\triangle DAB.$ Hence $PU \parallel RV,\ UR \parallel VP,$ PURV is a parallelogram and its diagonals PR, UV cut each other at half at the midpoint X of the diagonal PR of the parallelogram PQRS. Thus X is a midpoint of UV and V is a reflection of U in X. Consequently, the triangles $\triangle SUP \cong \triangle QVR$ are congruent and the angles $\angle SUP = \angle QVR$ are equal.  Let (K), (L), (M), (N) be the 9-point circles of the triangles $\triangle ABC,\ \triangle BCD,\ \triangle CDA,\ \triangle DAB,$ respectively, i.e., circumcircles of their medial triangles $\triangle PQU,\ \triangle QRV,\ \triangle RUS,\ \triangle SVP.$ The circles (K), (M) meet at U and at another point Z.

$\angle QZR = 360^\circ - (\angle UZQ + \angle UZR) =$

$= 360^\circ - [(180^\circ - \angle UPQ) + (180^\circ - \angle USR)] = \angle UPQ + \angle USR =$

$= 180^\circ - (\angle USP + \angle UPS) = \angle SUP = \angle QVR$

It follows that the quadrilateral QRZV is cyclic, i.e., the circumcircle (L) of the triangle $\triangle QVR$ passes through the second intersection Z of the circles (K), (M). In exactly the same way, we can show that the circle (N) also passes through the point Z.

Let O be the circumcenter of the triangle $\triangle ABC$ and D* the isogonal conjugate of D with respect to the triangle $\triangle ABC.$ Let the line OD* meet the circumncircle (O) at X, Y. The isogonal conjugate of the line $XY \equiv OD^*$ is a hyperbola $\mathcal H$ passing through the triangle vertices A, B, C, the asymptotes of which are the Simson lines with the poles X, Y. Since X, Y are diametrically opposite points of (O), these Simson lines are perpendicular to each other. They meet at the orthopole Z of the line XY, which lies on the 9-point circle (K) of the triangle $\triangle ABC.$ Thus the hyperbola $\mathcal H$ is normal, with perpendicular asymptotes intersecting at $Z \in (K).$ Since the orthocenter $D_1$ of the triangle $\triangle ABC$ is the isogonal conjugate of the circumcenter $O \in XY,$ $D_1 \in \mathcal H.$ Since D is the isogonal conjugate of $D^* \in XY \equiv OD^*,$ $D \in \mathcal H.$ Thus all 4 points A, B, C, D lie on the normal hyperbola $\mathcal H$ and consequently, the remaining ortocenters $A_1, B_1, C_1$ of the triangles $\triangle BCD,\ \triangle CDA,\ \triangle DAB$ also lie on $\mathcal H$ (see the lemma below). But then the 9-point circles (L), (M), (N) pass through the intersection Z of the hyperbola asymptotes. Moreover, since the orthocenters $D_1, A_1, B_1, C_1$ of the first four triangles all lie of $\mathcal H,$ the orthocenters $D_2, A_2, B_2, C_2$ of the next four triangles $\triangle A_1B_1C_1,\ \triangle B_1C_1D_1,\ \triangle C_1D_1A_1,\ \triangle D_1A_1B_1$ also lie on $\mathcal H$ and consequently, their 9-point circles $(K_1),\ (L_1),\ (M_1),\ (N_1)$ pass through the same intersection Z of the hyperbola asymptotes.

[b]Lemma:[/b] Let the vertices of a triangle $\triangle ABC$ lie on a normal hyperbola. Then the orthocenter H of this triangle lies on the same normal hyperbola.

Choose the coordinate origin at the midpoint of AB and WLOG, let the coordinates of A, B be (-1, 0), (+1, 0). The general equation of a normal hyperbola passing through the points A, B is

$y^2 - x^2 + pxy + qy + 1 = 0$

$y = f(x) = \frac{-(px + q) \pm \sqrt{(px + q)^2 + 4(x^2 - 1)}}{2}$

Since $CH \perp AB$, the x-coordinates of H, C are equal, $x_H = x_C.$ Assuming that the vertex C lies on this hyperbola, equations of the lines $BC,\ AH \perp BC$ are

$BC: \ \ y = m(x - 1) = \frac{f(x_C)}{x_C - 1}\ (x - 1) = \frac{f(x_H)}{x_H - 1}\ (x - 1)$

$AH: \ \ y = -\frac 1 m\ (x + 1) = -\frac{x_H - 1}{f(x_H)}\ (x + 1)$

The coordinates $x_H, y_H$ of the orthocenter $H \in AH$ then obey

$y_H = -\frac{x_H - 1}{f(x_H)}\ (x_H + 1) = \frac{1 - x_H^2}{f(x_H)}$

Substituting the hyperbola equation for $f(x_H)$ and dropping the index H from the x-, y-coordinates of H,

$1 - x^2 = y f(x) = \frac{-y(px + q) \pm y \sqrt{(px + q)^2 + 4(x^2 - 1)}}{2}$

$y (px + q) - 2(x^2 - 1) = \pm y \sqrt{(px + q)^2 + 4(x^2 - 1)}$

Squaring, subtracting equal terms from both sides, and reducing by $4(x^2 - 1),$

$(x^2 - 1) - y (px + q) = y^2$

$y^2 - x^2 + pxy + qy + 1 = 0$

which means that H lies on the same normal hyperbola as the one passing through the triangle vertices A, B, C


function
limit
calculus
integration
probability and stats

Find the value of $ k$ such that

$ F(x,y)\equal{}kxy(2x\plus{}3y)$ , $ x\plus{}y\leq 1$ , $ x>0$ , $ y>0$

will be a joint distribution function.

I know there is something to do with $ \lim_{x\rightarrow\infty}\lim_{y\rightarrow\infty}F(x,y)\equal{}1$, but how should we compute the limit with the restriction $ x\plus{}y\leq 1$ ? Thanks

The restriction tells you what your region of integration is.

So , how does that helps in the problem?

The question as stated doesn't make any sense. Are you sure this isn't supposed to be a joint density function? Then it would make sense, with the additional assumption that the function is zero outside of the stated domain.

I could see specifying a distribution function on a certain set with the tacit assumption that it is either $ 1$ or $ 0$ outside that set, as needed. But you can't do that here. Note in particular that the formula you give is not constant on $ x\plus{}y\equal{}1.$


Right... so this is where we write stupid poems in chinese that don't have to make sense.

Example (yes I actually wrote these):
我们今天吃鸡蛋，
炒的，煮的，和跟     饭。 (it got kind of blurred is supposed to be he gen fan as in with rice)
蘸点酱油撒点盐，
最后吃了真不甜。

从前有个绿青蛙。
它的朋友是王八。
她们每天不洗澡，
只在游泳池里爬。

我有一只 鸡，
一点不美丽。
它总爱  睡  觉，
也爱吃大米。

This is in simplified.
Anyways, here is a program to write in chinese: http://www.njstar.com
you can download chinese-writing software for free here

[i]stupid[/i] chinese poems? someone could find this offensive...like me...im watching u... :D

[quote="Treething"]pi gu[/quote]

You are sick.

haha
this is so pointless
but i like ur poem ra
very pointless but good

你的中文挺好的.我可写不出这种诗.

Great job.

It is kind of pointless though.

yeah thanks! =D
but for some reason it doesn't show up for me anymore...

My Japanese is better than my chineese :(

[quote="alkjash"]You know both? That's weird.

Hey, my friend knows both.[/quote]

i know a little japanese and a little chinese (i forgot half the stuff i learned at chinese school)

我不讲英语。
我总告诉谎言。
很好, 我现在懂英语。
You wouldn't believe me eitherwise.

[quote="1=2"]uuhh... lie.[/quote]

Haha. It took me a moment to get that. (sorry, I'm not very smart)

[quote="ra5249"]dude wtf
im the one who wrote the poems if you didnt notice
if i can write chinese but not speak chinese?!? thats really stupid[/quote]

i think what you mean is that you have the skills to go to a language translator and copy and paste...you can't really write chinese

Maybe he meant you were lying because you can speak English too? I dunno.

kaoticity

when you grow up

name your kid xiao

and laugh at him (xiao wang lol lol lol)

Woh... tsi ee guh dah buhn dahn!!!!

Dway ah! Dway Ah!

Woh.. Tsi ee guh dah  buhn dahn

Nee yeh tsi!!!

Dway!

Shien zhai kai tsi yao pee guu!!!!

Shee shoo wong xing ping wing ding xu bu wu poo hong wang bang shugaxan wongtakan

[quote="footballrocks41237"][quote="Treething"]ni hao
pi gu

zao gao

wei ba.[/quote]

you win.  it doesn't make a sentence.[/quote]

poems dont have to make a sentence, do you know anything about poetry? i personally liked the rhythm and meter of his poem, it reminded me of the communist poem the red wheelbarrow

so much depends
upon

the red wheel
barrow

beside the white
chickens.

nice [color=white]The message is too small. Please make the message longer before submitting.[/color]

check out my haiku

once there was a fox
it liked eating diet lox
it wore fluffy socks

this isnt my haiku but heres one:
haikus are easy
sometimes they dont make sense
refrigerator

hmm nice one noneoftheabove
except ur second line only has 6 sylables -_-

I've seen this one before.  Should be [quote="noneoftheabove"]haikus are easy
[b]BUT[/b] sometimes they dont make sense
refrigerator[/quote]

hmm...
i like tacos too
they smell like your yellow shoe
i own a ship's crew

fine
wo shuo zhong wen hua
ni hui bu hui shuo zhong wen
xian zai shuo zai jian

http://www.rinkworks.com/poetry/p/herring.haiku.shtml

This is a haiku.
This is the middle of it.
And this is the end.

umm...
doesnt sound very chinese

bonjour, les amis.
je m'appelle jacques vonnegut.
je joue au basket

ha! in french!

(btw thats not my real name -_-)


induction
search
number theory unsolved
number theory

if $ R$ is a number of $ 81$ digits such that all of them are $ 1$, show that $ R$ is divisible by $ 81$....Oops!

I just wonder to ask Is it always that any number the sum of its digits is $ a$ , then it must be divisible by $ a$?

[quote="Hidden Scofield"]I just wonder to ask Is it always that any number the sum of its digits is  , then it must be divisible by ?[/quote]

No, $ 73$ is not multiple of $ 10$.

[quote="Hidden Scofield"]That's the answer when I divide the number $ R$ on $ 81$.

But I need the proof technique ....without using calculator .

Nemion , I meant all digits of $ a$ is $ 1$. I dropped it out from my question    :P[/quote]
if $ a \equal{} 3^n$ it is true!and you can prove it by induction on $ n$
[hide="hint"]note that if $ R_n$ is a number of $ 3^n$ digits such that all of them are $ 1$ then:
$ R_n \equal{} \frac {10^{3^n} \minus{} 1}{9}$ [/hide]

The $ R_n$ form is easy to obtain:

Notice that $ 10^n\minus{}1\equal{}999...999$ where 9 appears $ n$ times. As $ \frac{999...999}{9}\equal{}111...111$, you've got your form.

Second part: Aren't all even numbers counterexamples? And if the next question is: does it hold for odd numbers, there are always numbers $ 5, 7, 11$ as counterexamples.

So I'll ask a "converse", are there any numbers not in form $ 3^n$ which satisfy the claim?

[quote="Hidden Scofield"]O.k.

how do you get the form for $ R_n$ ?  :maybe:

do you have an counter example if $ a\ne 3^n$.

 :roll:[/quote]
I think it is obviously
$ R_n$is a number of$ 3^n$  digits such that all of them are$ 1$ therefore $ 9R_n$ s a number of$ 3^n$  digits such that all of them are$ 9$!
and $ 9R_n \plus{} 1 \equal{} 10^n$
sorry,hsiljak give the same solution before me

ali666's technique is generalized [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=275968159&t=205928]here[/url].  As for the second question...

We want to find integers $ k$ such that $ k | \frac{10^k \minus{} 1}{9}$.  Clearly $ \gcd(k, 10) \equal{} 1$.  Consider the smallest prime factor $ p$ of $ k$ that isn't $ 3$ and write $ k \equal{} mp$ where $ m$ is divisible by $ 3$ or primes larger than $ p$.  We want $ p | 10^{mp} \minus{} 1$ and we know that $ 10^{p} \equiv 10 \bmod p$, so $ p | 10^m \minus{} 1$.  It follows that $ p \equiv 1 \bmod m$, but this is impossible if $ m$ is only divisible by primes larger than $ p$, so it follows that $ m$ is a power of $ 3$.

So the only numbers that could work that aren't powers of $ 3$ are of the form $ 3^n \cdot p$ where $ p \equiv 1 \bmod 3^n$ is prime.  The problem here is that $ \lambda(3^n \cdot p)$ is always even, so Carmichael's theorem isn't enough to guarantee divisibility and we need to explicitly find a prime with the property that the order of $ 10$ is odd.

[quote="t0rajir0u"] $ p | 10^m \minus{} 1$.  It follows that $ p \equiv 1 \bmod m$
[/quote]
I dont understand this step :maybe:
for second question we have infainite $ a$ with this property.note that:
if $ b|\frac {10^{3^n} \minus{} 1}{9} ,(b,3) \equal{} 1$ then $ a \equal{} b\times3^n$ work.

[quote="hsiljak"][quote="ali666"]
for second question we have infainite $ a$ with this property.note that:
if $ b|\frac {10^{3^n} \minus{} 1}{9} ,(b,3) \equal{} 1$ then $ a \equal{} b\times3^n$ work.[/quote]

Can we find one such $ b$? The claim that we have an infinite number of $ a$s is valid if we find at least one $ b$ ($ a$).[/quote]
yes...actually to prove the claim we need find infinite $ b$...
$ b \equal{} \frac {10^{3^n} \minus{} 1}{9\times3^n}$ for all $ n$ work...
$ (b,3) \equal{} 1$ because $ 3^{n \plus{} 1}$ not divide  $ \frac {10^{3^n} \minus{} 1}{9}$...
and to prove it you can use very useful t0rajirou's generalization...or use induction!

Of course it does. Now I see it.

[hide="Remark"]Hmm.. this could make a nice NT problem for a local competition ;)[/hide]

[quote="ali666"][quote="t0rajir0u"] $ p | 10^m \minus{} 1$.  It follows that $ p \equiv 1 \bmod m$
[/quote]
I dont understand this step :maybe: [/quote]
Sorry, ignore that.   :|


AMC
AIME
geometry
3D geometry
LaTeX
complex numbers

Suppose that the sum of the squares of two complex numbers  'x' and 'y' is  7 and the sum of the cubes is 10. What is the largest real value
(x + y) can have?

Can somebody give me a hint, so that I can try and start it?

this is what I started off doing;

x^2 + y^2 = 7
x^3 + y^3 = 10

I called '(x + y)' z.

then I had

z^2 = 7 + 2xy

$ \text{\LaTeX}$ bashing :D

[hide="Click for solution"]

Let $ x + y = S$ and $ xy = P$. Note $ x^2 + y^2 = 7$ and $ x^3 + y^3 = 10$. Now,
\begin{eqnarray*}x^2 + y^2 & = & (x + y)^2 - 2xy \\
x^3 + y^3 & = & (x + y)(x^2 + y^2) - xy(x + y)\end{eqnarray*}
Thus,
\begin{eqnarray*}S^2 - 2P & = & 7 \\
7S - P\cdot S & = & 10\end{eqnarray*}
From the first equation, $ P = \frac {S^2 - 7}{2}$. Substituting into the second equation,
\[ S^3 - 21S + 20 = 0.
\]
But $ S^3 - 21S + 20 = (S - 1)(S + 5)(S - 4)$, so the largest real value of $ x + y$ is $ 4$.

[/hide]

EDIT: 1996th post!  Guess what year i was born in :D

But S^3 - 21S + 20 = (S-1)(S+5)(S-4), so the largest real value of x+y is 4.

I am sorry, but I do not understand why this means the answer is 4.

P.S. - Hmm, were you born in 1991?

[quote="Wickedestjr"]But S^3 - 21S + 20 = (S-1)(S+5)(S-4), so the largest real value of x+y is 4.

I am sorry, but I do not understand why this means the answer is 4.
[/quote]

Remember from your factoring lessons that if $ (S\minus{}1)(S\plus{}5)(S\minus{}4)\equal{}0$, then one of the factors $ (S\minus{}1),(S\plus{}5),(S\minus{}4)$ must be 0. These values of $ S$ are $ 1, \minus{}5$, and $ 4$, respectively. Which is the largest out of those three?

Wickedestjr, you are being sarcastic, right?
I've been getting that impression for a while...

Yes I am being sarcastic, but that is completely irrelevant to this thread.

@Yongyi781 - Sorry about that, I think I got (-4) instead of 4. I don't know what I did wrong.

@Yongyi871 and AIME15 - thanks for the help

I just read about factorizations in Volume 1, and now I think I know how to solve problems like these ones.

It's the complex number part that threw you off, eh?

The factorization part really isn't that hard.

What do you mean by "complex number part"?

[quote="Wickedestjr"]Suppose that the sum of the squares of two complex numbers  'x' and 'y' is  7 and the sum of the cubes is 10. What is the largest real value
(x + y) can have? [/quote]

THEYRE COMPLEX NUMBERS.

Oh, sorry. :fool:

I already knew what complex numbers were though. They are any numbers in the form a + bi. I think that I assumed that it was imaginary, because I forgot that 'b' could be zero.


Three small circles are congruent and mutualy tangent and the large circle is tangent to all the small circles.The point P is on the large circle and PA, PB, y  PC are tangents to the three smal circles.Show that PA+PB=PC

leónsotelo

The three small circles are inscribed in the large circle.

leonsotelo

The original problem:

 http://it.stlawu.edu/~mbos/Problem_of_the_Week/fall_1998/problem4.html

leonsotelo

[b]Lemma:[/b] Given $ (O,R)$ and $ (O',R')$ are internal tangent at $ X (R > R'). P$ lies on $ (O), PY$ is a tangency of $ (O')$ then $ \frac {PY}{PX} \equal{} \sqrt {\frac {R \minus{} R'}{R}}$
[b]Proof: [/b]Let $ P'$ be a point lie on $ (O) (P\not\equiv P). P'B'$ is a tangency of $ (O').$
$ PX \cap (O') \equal{} F, P'X\cap (O') \equal{} \{F'\}$
It's easy to prove that $ FF'//PP'$
We have $ \frac {PY^2}{FX^2} \equal{} \frac {PF.PX}{FX^2} \equal{} \frac {P'F'.P'X}{F'X.F'X} \equal{} \frac {P'Y'^2}{F'X^2}$
$ \Rightarrow \frac {PY}{FX} \equal{} \frac {P'Y'}{F'X} \Rightarrow \frac {PY}{P'Y'} \equal{} \frac {FX}{F'X} \equal{} \frac {PX}{P'X}$
$ \Rightarrow \frac {FY}{PX} \equal{} \frac {F'Y'}{P'X}$
Now let $ XZ$ be a diameter of $ (O).ZZ'$ is a tangency of $ (O')$
Put $ P'\equiv Z$ we have $ \frac {PY}{PX} \equal{} \frac {ZZ'}{ZX} \equal{} \sqrt {\frac {R \minus{} R'}{R}}$
[b]Return to our problem:[/b]
Denote by $ X,Y,Z$ the points contact of $ (O_a), (O_b), (O_c)$ to $ (O)$ then triangle $ XYZ$ are equilateral.
Applying the lemma above we have $ \frac {PA}{PX} \equal{} \frac {PB}{PY} \equal{} \frac {PC}{PZ} ( \equal{} \sqrt {\frac {R \minus{} R'}{R}})$
$ \equal{} \frac {PA \plus{} PB}{PX \plus{} PY} \equal{} \frac {PA \plus{} PB}{PZ}$
$ \Rightarrow PC \equal{} PA \plus{} PB$ (QED)

It's a particular case of [url=http://en.wikipedia.org/wiki/Casey%27s_theorem]Casey's theorem[/url].


Hello all.

I am in Algebra II, but I am self studying Math Analysis(Precalculus).  I want to be the best math student  in my grade, however I am plagued by slow work and stupid mistakes.  I understand all the concepts, but I never execute to perfection and almost always work very slow.  I normally am one of the last to finish.

I average a 93 in Algebra II, but I want to reach perfection at 100 percent.

How can I execute my math skills to perfection?

In other words how can I stop making stupid mistakes and work faster without EVER sacrificing accuracy.

I appreciate any advice you may give me

---Potential

You will never achieve [i]perfection[/i]. Everyone makes dumb mistakes, even IMO qualifiers. There is no way you can eliminate them entirely; however, you can reduce them greatly. First, read the article on this site ([url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php[/url]) and follow it. Next, practice. This will help you with both mistakes and speed. Get more sleep, too.

[quote="Potential"]Thanks.

You are probably right.  It is just that I have become so tired of trailing behind other students who have no intellectual advantage over me.

But you are right in saying that perfection is impossible.[/quote]

Dude if this is class work, speed doesnt really matter that much because on tests they usually give you like more than enough time.  But always check your work, this isnt like the AMCs or mathcounts where time is actually important so turning your test in first doesnt matter.

[quote="Potential"]Thanks.

You are probably right.  It is just that I have become so tired of trailing behind other students who have no intellectual advantage over me.

But you are right in saying that perfection is impossible.[/quote]

Well how are the other students better than you? I'm one of the top students in my math class but due to procrastination and laziness I too am one of the last ones to finish (not that I care?). It could be that your math class is the elite of the grade because my class, no offense, is very slow. Try reading your Algebra II textbook more than doing precalc. Seriously. I spend 2 weeks with calculus instead of Algebra II (that's what I'm taking now) and I was seriously messed up for the unit exam (sigh...a 91%  :( ). Do practice problems for the next unit you'll be doing so that you'll always be ahead. Don't do other problems from future chapters until you are 100% positive that you know the material for the next chapter.

Another thing: where applicible, do some word problems relating to the concepts. For example, we are doing inverse, direct, and joint relationships in math class today, and I wizzed by the problems but I slowed down a little when I got to the word problems.

@Art of Owna: Yeah for tests you need to take your time. But, when your teacher gives you practice classwork problems, then it's time to race  :wink: .


Find all the functions $ f: \mathbb{R}\rightarrow \mathbb{R}$ such that :

$ f(x^3\plus{}y^3)\equal{}xf(x^2)\plus{}yf(y^2)\ ,\ (\forall)x,y\in \mathbb{R}$

The idea is similar to post#19 of http://www.mathlinks.ro/viewtopic.php?t=54046.

BTW, a typo down there is
$ f(bx\plus{}c)\equal{}bf(x)$ should be corrected to $ f(bx\plus{}c)\equal{}bf(x)\plus{}f(c)$.


Alexa has a mixture consisting of 8 ounces of red jelly beans and
13 ounces of green jelly beans. What is the minimum number of
ounces of white jelly beans she must buy so that, when added to the
old mixture, her new mixture is an integer number of pounds?

1 pound=16 ounces (or $ \$1.6$ if my currency exchange is accurate :lol: )

13+8=21
16*2=32

32-21=11

Since there are 16 ounces in a pound and we have a total of only 21 we must add 11 to get to the next pound which is 32.

the answer is 11.


[quote="aidan"]You have a sheet of paper and a pair of scissors. You cut the paper into either 8 or 12 pieces. Then you take one of these pieces and cut it into either 8 or 12 pieces, and so on. After 2 cuts you may have 15, 19, or 23 pieces (by making two '8-cuts', one '8-cut' and one '12-cut', or two '12-cuts').

There is an integer $ N$ such that any integer larger than $ N$ can be the number of pieces of paper you have after enough cuts. Find $ N$.[/quote]
[hide]You start off with 1 and you add 7 or 11. Normally your max would be 59 (by residue classes), but since you start with 1 $ \boxed{N\equal{}60}$[/hide]

[quote="aidan"]What are residue classes? And can the Chicken McNugget theorem be used here? Anyhow, correct answer  :)[/quote]You can use Chicken McNugget (it's derived from residue classes anyhow)


inequalities
search
function
invariant
rearrangement inequality

it says:  $\displaystyle \sum_{a,b,c}\frac{a}{b+c}\heq\frac{3}{2}$

The above post is missing the $\ge$ sign.

For positive reals $a, b, c$:

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$.

Yeah, and this inequality has at least 6 solutions. ;)

Can someone give me those 6 solutions. I have tried very hard but unsuccessful. :huh:

For now, because i'm lazy and i'm busy with other things, i'll post only one solution. Here it goes:

 It is very easy to see that from AM-HM it follows that $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geq\frac{9}{2(a+b+c)}$. So this inequality is equvalent with Nesbitt like this: $\sum\frac{a+b+c}{b+c}\geq\frac{9}{2}$ and by making computations we get this famous inequality. ;)

  Is it clear? :D

Another, more complicated, proof goes like this:

apply Cauchy-Schwartz to $x_1 = \sqrt{a(b + c)}, x_2 = \sqrt{b(a + c)}, x_3 = \sqrt{c(a + b)}$ and $y_1 = \sqrt{\frac{a}{b+c}}, y_2 = \sqrt{\frac{b}{a+c}}, y_3 = \sqrt{\frac{c}{a+b}}, $ to get

$\left(a(b + c) + b(a + c) + c(a + b)\right)\left(\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} \right) \geq (a + b + c)^2$

$\Rightarrow \frac{a}{b+c} + \frac{b}{a + c} + \frac{c}{a+b} \geq \frac{a^2 + b^2 + c^2 + 2(ab + bc + ac)}{2(ab + bc + ac)}$

$\Rightarrow \frac{a}{b+c} + \frac{b}{a + c} + \frac{c}{a+b} \geq 1 + \frac{1}{2}\frac{a^2 + b^2 + c^2}{ab + bc + ac} \geq \frac{3}{2}$

because we know that $a^2 + b^2 + c^2 \geq ab + bc + ac$

I hope this makes sense.

[hide="solution 3"]$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$
Without loss of generality, let $a+b+c=1$, then we have to prove
$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\geq\frac{3}{2}$
$\Leftrightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\geq\frac{9}{2}$
By Cauchy-Schwarz Inequality,
$(1-a+1-b+1-c)(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c})\geq\ (1+1+1)^2=9$
The conclusion follows.[/hide]
[hide="solution 4"]$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$
$\Leftrightarrow 2a(a+b)(a+c)+2b(b+c)(b+a)+2c(c+a)(c+b)-3(a+b)(b+c)(c+a)\geq 0$
After expansion, this is equivalent to $2a^3+2b^3+2c^3-a^2b-a^2c-b^2c-b^2a-c^2a-c^2b\geq 0$
or $\sum_{sym}a^3\geq\sum_{sym}a^2b$
This is true by Muirhead Inequality since $(3,0,0)\succ (2,1,0)$[/hide]

Although it's a bit gratuitous (you might want to use the search function next time) here are a couple more solutions.

If $a\geq b\geq c$, then $\frac {1}{b+c}\geq \frac {1}{a+c}\geq \frac {1}{a+b}$

By the rearrangement inequality, this implies

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

and

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

Adding those inequalities together, we get that

$2\left(\frac {a}{b+c}+\frac {b}{c+a}+\frac {c}{a+b}\right)\geq 3$

or

$\frac {a}{b+c}+\frac {b}{c+a}+\frac {c}{a+b}\geq \frac {3}{2}$

The second proof uses Jensen's inequality, and is my personal favorite.  Let $a+b+c=1$.  $f(x)=1/x$ is convex on $(0,\infty)$, so that

$\frac {a}{b+c}+\frac {b}{c+a}+\frac {c}{a+b}\geq \frac {1}{a(b+c)+b(c+a)+c(a+b)}$

and since $2(ab+bc+ca)\leq \frac {2}{3}(a+b+c)^2$ (prove this!), we're done.

[quote="varun"]Well, thx for giving the proofs.
I couldn't understand most of the part, except the first one. Well I think I will like to study them some day.
For now, can someone explain me the rearrangement inequality?

THX again. :P[/quote]

If we have $a_1 \geq a_2 \geq \ldots \geq a_n$ and $b_1 \geq b_2 \geq \ldots \geq b_n$ and $f(i)$ is some permutation of numbers from 1 to n, then we have that

$a_1b_1+\ldots+a_nb_n \geq a_1b_{f(1)}+\ldots+a_nb_{f(n)} \geq a_1b_n+...+a_nb_1$

For example, if you have $5>4>3$ and $100>50>20$ (for example, if you need to take as much money as you can but you have to take 5 times some amount, then 4 times some other amount..), it's obvious that $5*100+4*50+3*20$ is maximum amount, and $5*20+4*50+3*100$ is minimum amount you can take.

Try to prove that $a^3+b^3+c^3 \geq a^2 b +b^2 c+c^2 b$, if a, b and c are positive numbers.

[hide]Since inequality is invariant to cyclic permutations we may assume that $a \geq b \geq c$ or $a \geq c \geq b$. Let's solve the first case. If $a \geq b \geq c$ then $a^2 \geq b^2 \geq c^2$. So we have that

$a \geq b \geq c$
$a^2 \geq b^2 \geq c^2$

and by rearrangement we get that $a^3+b^3+c^3 \geq a^2 b +b^2 c+c^2 b$, what we wanted.[/hide]

This proof is from Arthur Engel ' s book

Suppose that \[ f(a,b,c)= \frac{a+b+c}{b+c} + \frac{a+b+c}{a+c} + \frac{a+b+c}{a+b}-3 \] then \[ f(a,b,c) =(a+b+c) \left ( \frac{1}{a+b}+ \frac{1}{a+c}+ \frac{1}{b+c} \right ) -3  \]
or \[ \frac{1}{2} [(a+b)+(b+c)+(c+a)] \left ( \frac{1}{a+b}+ \frac{1}{a+c}+ \frac{1}{b+c} \right ) -3 \]
set $a+b=x$, $b+c=y$, $a+c=z$ and yields
\[ 2f(a,b,c) = (x+y+z) \left ( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right ) -6 \]
\[ = \frac{x}{y} + \frac{y}{x} + \frac{x}{z} + \frac{z}{x} + \frac{y}{z} + \frac{z}{y} -3 \geq 3 \] you can get it by doing AM-GM in every two term

In the Jensen proof, all we need to do is to say that:

$f(x)=\frac{x}{1-x}$ Homogenizing:

$\frac{f(a)+f(b)+f(c)}{3} \ge f(\frac{a+b+c}{3}). $

The RHS reduces to $f(\frac{1}{3})=\frac{1}{2}$ and we're done.

[quote="idioteque"]
If we have a_1 \geq a_2 \geq \ldots \geq a_n and b_1 \geq b_2 \geq \ldots \geq b_n and f(i) is some permutation of numbers from 1 to n, then we have that

a_1b_1+\ldots+a_nb_n \geq a_1b_{f(1)}+\ldots+a_nb_{f(n)} \geq a_1b_n+...+a_nb_1
[/quote]
What do you mean by the second part of the inequality, is it
a(i)b(j), where i and j are from 1 to n.


Let $ X$ and $ Y$ hilbert spaces and let $ A\in \mathcal{L}(X,Y)$. Let $ A^*$ the adjoint of $ A$. Prove that $ A^*$ is injective if and only if $ R(A)$ is dense in $ Y$  , with $ R(A)\equal{}\{v\in Y: \exists u\in X s.t A(u)\equal{}v\}$


calculus
integration
function
limit
trigonometry
real analysis
real analysis unsolved

Let $f: [0,\pi]\to \Re$ be a continous function with the second dervivative continous. Compute:
$\lim_{n\to\infty}\int_{0}^{\pi}\frac{\sin \left ( n+\frac{1}{2}\right ) t}{\sin\frac{t}{2}}f(t)dt$

do you know Fourier series ? If so, write it as the expansion of $f$ ..

I don't think he does... Anyway, see [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=riemann&t=69329]here[/url] for other proofs (I never read them thoroughly, I just bumped into that topic while searching for the Riemann-Lebesgue Lemma).


function
geometry
ratio
geometry unsolved

let $ ABCD$ a convex quadrilateral, and let $ A'$,$ B'$,$ C'$ and$ D'$  are points such that $ A$ ,$ B$,$ C$ and $ D$ are the middle of $ [DD']$,$ [AA']$,$ [BB']$ and $ [CC']$.
find the the surface of the quadrilateral $ A'B'C'D'$ in function of the surface of $ ABCD$
from moroccan TST 2000

[quote="mathangel"]Please clarify these:

$ A, B, C, D$ are points in the space or in a plane ?

Does "middle" means "midpoint" ?[/quote]
I think he means points in a plane where "middle" means "midpoint," so yes.
[hide="Solution"]
Connect $ B'D$, $ D'B$, $ A'C$, $ C'B$, $ AC$ and $ BD$.  Now, by area ratios, we have that $ x \equal{} [ACD] \equal{} [C'AD] \equal{} [AC'D']$, $ y \equal{} [BCD] \equal{} [B'CD] \equal{} [C'DB']$, $ w \equal{} [ADB] \equal{} [D'AB] \equal{} [D'BA']$, and $ z \equal{} [ABC] \equal{} [A'BC] \equal{} [A'CB]$.  Thus, $ [A'B'C'D'] \equal{} [ABCD] \plus{} [B'C'C] \plus{} [C'D'D] \plus{} [AA'D'] \plus{} [ABB'] \equal{} (\frac {[ABC] \plus{} [ACD] \plus{} [BCD] \plus{} [ABD]}{2}) \plus{} 2x \plus{} 2y \plus{} 2z \plus{} 2w \equal{} \frac {x \plus{} y \plus{} z \plus{} w}{2} \plus{} 2x \plus{} 2y \plus{} 2z \plus{} 2w \equal{} \frac{5x \plus{} 5y \plus{} 5z \plus{} 5w}{2} \equal{} \boxed{5[ABCD]}$. [/hide]