Hello everyone!
I know that most of us here in the pre-olympiad section work a lot to get ready for the olympiad. But I know that a very important thing is to study special books dedicated to this topic. That is why I want to try to gather the most important things in one place and to make it available for every one.
A good idea would be if everybody who finds out about a special article, book etc. posts its title here, and if somebody who sees the title knows where it can be found( internet, book-shop etc.) tells the rest.
That way it would get very easy to find the materials we need for preparation.
For example I search now the following books:
International Mathematical Olympiad 1959 - 1999 (Hardcover)
by Istvan Reiman (Editor)
International Mathematical Olympiad Problems (Problem Solving) (Paperback)
by Derek Holton (Author)
Count Down: The Race for Beautiful Solutions at the International Mathematical Olympiad (Paperback)
by Steve Olson (Author)
International Mathematics Olympiad Volume III: 1991-2004 (Anthem Science, Technology & Medicine) (Paperback)
by Istvan Reiman (Author)
International Mathematical Olympiad Volume I: 1959-1975 (Anthem Science, Technology & Medicine) (Paperback)
by Istvan Reiman (Author)
International Mathematics Olympiad Volume II: 1976-1990 (Anthem Science, Technology & Medicine) (Hardcover)
by Istvan Reiman (Author)
International Mathematical Olympiads, 1959-1977 (New Mathematical Library) (New Mathematical Library) (Paperback)
by Samuel L. Greitzer (Editor)
Challenging Mathematical Problems With Elementary Solutions, Vol. 1 and vol.2(Paperback)
by A. M. Yaglom (Author), I. M. Yaglom (Author)
The second All-Russian Olympiad in mathematics (Unknown Binding)
by Dmitri Thoro (Author)
Research Problems in Discrete Geometry (Hardcover)
by Peter Brass (Author), William O.J. Moser (Author), János Pach (Author)

Most of them are very rare, and in my country they can not be found.
And if they exist, they are not translated very correct.
So if someone konws if these books, for example are availabe in a public web library, may live the link etc.

The idea is to help each other to find the resorces we need to study.
By having more options to find a material, we will reduce the search time and we will chose the most efficient way of finding the papers we need.

Nice idea, but I think this would be better suited [url=http://www.artofproblemsolving.com/Forum/index.php?f=319]here[/url].

yes, you are right, I will try to move the topic there as soon as possible!


function
geometry
3D geometry
combinatorics proposed
combinatorics

Let $ X$ be a finite set of positive integers and $ A$ a subset of $ X$. Prove that there exists a subset $ B$ of $ X$ such that $ A$ equals the set of elements of $ X$ which divide an odd number of elements of $ B$

i think this generalizes to any (finite) poset

Yes. Any rank function will do the same job as divisibility.

the idea goes something like this... consider the Hasse diagram of the poset. start by putting the elements in $ A$ that are maximal in $ A$, in $ B$. then traverse the lower cones of these elements downward, adjusting elements along the way and only adjusting an element once higher-level elements have been adjusted. more specifically, do the following: (i) when you reach an element in $ A$ and it's $ \leq$ an odd number of elements of $ B$ so far, leave it alone, otherwise add it to B; (ii) when you reach an element not in $ A$ and it's $ \leq$ an odd number of elements of $ B$ so far, add it to $ B$, otherwise leave it alone.


vector
algebra
polynomial
linear algebra
linear algebra unsolved

Given $f$ is an endomorphism of a complex vector space finite dimensional $E$. Denote:
\[ F: L(E)\to L(E) \]
\[ F(g)=f\circ g-g\circ f \]
Prove that $f$ is diagonalizable iff $F$ is diagonalizable

Hmmm.....I am kind of new to Linear Algebra and all that stuff.....but I think this should work....
Consider all this in a basis in which $f$ is diagonalisable.....(I think the other way around should also work...but it's a bit more work...)
Basicly....Consider matrices $B_{i,j}$ which have $0$ everywhere and just a $1$ at the intersection of the $i^{th}$ line with the $j^{th}$ column....
Now....Look how $F$ acts on these matrices.....and notice that $B_{i,j}$ is actually an eigenvector for $F$ with eigenvalue $\lambda_{i}-\lambda{j}$ where the $\lambda$'s are the corresponding eigenvalues of $f$....So the conclusion should be clear..The other way around should also work...I hope...
Can somebody tell me if this is at least partially understandable?
Thanks :)
This is the way I wrote it since it was a problem in one of my problem sets (due last week :) ) and it was about matrices in general...but I hope it's the same thing?

I think it's been already posted  :)
Anyway, here how one can solve this:
consider $G: g \to  fg$ and $D: g\to gf$.

Then if $P$ is a polynomial, $P(G)(g)=P(f)g$. So the minimal polynomial of $G$ is equal to the minimal polynomial of $f$. Anf by the same argument, the minimal polynomial of $D$ is also the same that $f$.
If $f$ is diagonalizable, so are $G$ and $D$. But $G$ and $D$ commute so $D-G$ is also diagonalizable.

Take now a polynomial $P$. Then $P(F)(g)=P(f)g-gP(f)$. Hence the minimal polynomial of $F$ is equal to the minimal polynomial of $f$...

Sorry....why is $P(F)(g)=P(f)g-gP(f)$? I mean if you compose $F$ twice you will not get $f\circ f \circ g + g\circ f \circ f$

ok , forget what I wrote for the second part.. I was in a hurry  :)

[quote="DusT"]Hi!
Sorry to be annoying, but can anyone tell me if what I said is correct ?
I mean, I have shown that the vectors (that also form a basis) $B_{i,j}$ are all eigenvectors...and thus $F$ should be diagonalizable ?
Thanks,
Simion[/quote]
If there is no wrong in your post, you have only done necessary condition

$u \in L(E)$, $T(u): L(E)\to L(E)$, $T(u)(v)=uv-vu$.
One has proved that if $u$ is diagonalizable, so is $T(u)$. Exactly the same proof shows that if $u$ is nilpotent,
$T(u)$ is nilpotent.

Now we will use the well known fact that on a complex vector space, for every $u \in L(E)$, there is a unique decomposition $u=d+n$ where $d$ is diagonalizable and $n$ nilpotent and $nd=dn$.

Suppose now that $u=n+d$ and $T(u)=T(n)+T(d)$ is diagonalizable. Then, (since $T(d)$ is diagonalisable and $T(n)$ nilpotent),  $T(u)=T(n)+T(d)$ is precisely the decomposition we were talking about. So if $T(u)$ is diagonalisable, by the uniqueness od the decompositon, $T(n)=0$. This means that $n$ commutes with all endomorphisms. So $n$ is an homothethy. But remember that $n$ is nilpotent ! So $n$ is nul. This is why $u=d$, wich proves that $u$ is diagonalisable.


Hi.

I was wondering what year I should buy from MathCounts. I'm trying to prepare for the State. What's the hardest year, in your opinions?

Thank you all, F. Turner

He's talking about nats...
Why in the world would you buy states when you can get it online?

Actually, you can't any states before 2000. So maybe he's talking about <2000 states and any nats.

Well, the older states are really too easy if you're preparing for states.

I got all the states from 1990 to 2000.
I could scan them for you guys... or something.

um, have you ever heard of copyright laws?

[quote="ZhangPeijin"]Oh...
But it educational.
Don't copyright laws not apply to educational material?
or academic?[/quote]

DOH! Okay, I think I'll just go to the library, borrow halo, and copy it, cuz they say it increases hand-eye coordination.

That wouldn't be a very great law, what defines something as academic? Also, then why would people make academic books in the first place if they didn't make money?

Actually dude, here is an example. A music teacher buys a copy of a score for each part. Then makes a copy per student for the cellos, then the violins, etc. FOR EDUCATIONAL purposes, people are allowed to make one copy per student.

Of course, in this case, this would be better described as DISTRIBUTION, since zhangpeijin probably is not georgiamathdude's teacher.... or is he?  :ninja:

I've already posted this at numerous other topics, but look at it again.

I sent a letter to MATHCOUNTS asking if it was legal.

Here is their response.

Hello,

Thank you for contacting MATHCOUNTS. We publish and sell old MATHCOUNTS
Competition Booklets, so as long as we have published the competition for
public use you may receive a copy of the test. Hopefully this answers your
question, if not please feel free to contact me to discuss further.

Best regards,
Janice

I think that it's legal to share old tests.

Does that mean only state tests and lower are allowed to be shared?

I'm not sure. I comprehended it in which we could distribute any tests that have been either sold, or given to the public.

No dude. That means that if we can't get it online for FREE we can't get it from the copying machine. They have not published it for public use.

Wait guys, nevermind, I don't have the state tests to distribute. My scanner doesn't seem to be able to scan anymore.
 :(

I don't have the link, but click on the topic titled "MATHCOUNTS resources HERE!" or something like that..... To answer the original question, if you want to get competition sets, get 07 cause it was the hardests... If you want to buy just nats competitions, buy 2006 cause if you were in MC last year you will remember the probelems..... I suggest called Sports Awards


Quadrilateral $ ABCD$ is circumscribed about a circle. Three sides of the quadrilateral are 9,  17, and 12. Find the length of side $ \overline{AD}$.

[asy]import olympiad;

pair A = (153,-29);
pair B = (67,-173);
pair C = (252,-173);
pair D = (237,-62);
pair O = (179,-109);

draw(Circle(O, 64), linewidth(2));
draw(A--B--C--D--cycle, linewidth(2));

label("$A$", A, N);
label("$D$", D, NE);
label("$C$", C, SE);
label("$B$", B, SW);
label("12", (A+B)/2, NW);
label("17", (B+C)/2, S);
label("9", (D+C)/2, ENE);[/asy]

The two segments containing the points of tangency of the tangents from any point to a circle are equal. Thus, we can let the segments going out from $ A$ be $ a$, from $ B$ be $ b$, etc. to get
$ a\plus{}b\equal{}12$ (1),
$ b\plus{}c\equal{}17$ (2),
$ c\plus{}d\equal{}9$ (3),
and $ a\plus{}d\equal{}AD$.

We can easily find this value from (1)+(3)-(2): $ (a\plus{}b)\plus{}(c\plus{}d)\minus{}(b\plus{}c)\equal{}12\plus{}9\minus{}17\equal{}\boxed4$.

Is it true that the sum of one pair of opposite sides equals the sum of the other pair?

Yes; both pairs sum up to $ a\plus{}b\plus{}c\plus{}d$. :)

No, but does $ a\plus{}c \equal{} b\plus{}d$?

Uh..that's not your original question. And the answer is no. (Look at what $ a,b,c,d$ are defined as).

*sigh*

Is $ AD \plus{} BC \equal{} AB \plus{} CD$?

I already said that $ AD\plus{}BC\equal{}AB\plus{}CD\equal{}a\plus{}b\plus{}c\plus{}d$... :wink:  :P

the answer on ftw is $5$ but I also think it's $4$.


(copied from '93 Nats, #4) What will be the diameter in inches of a sphere with twice the volume of a sphere of radius 9 inches? Express your answer in simplest radical form.

[hide="hint"]the volume of a sphere is $\frac{4r^{3}\pi}{3}$ where r=radius[/hide]

[hide] We know that twice the volume of the small sphere is $\frac{4}{3}(\pi9^{3})(2)$.  Now, this is equal to the volume of the sphere, which has a radius of r.  Thus,
\[\frac{4}{3}(\pi(r^{3}))=\frac{4}{3}(\pi9^{3})(2) \]
We can cancel a lot of stuff of to get that $r^{3}=2*9^{3}\Rightarrow r=9\sqrt[3]{2}$.  Thus, the diameter is $2r=\boxed{18\sqrt[3]{2}}$ [/hide]

^ too complicated method
i guessed
 18(2)^(1/3) and got it right! :!:


probability
quadratics
calculus
function
geometry
3D geometry
integration

$ax^{2}+bx+c = 0$
What is the probability that the roots of the above quadratic equation would be real given that $0 \le (a, b, c) \le 1$ i.e. $a, b, c$are any real numbers between $0$ and $1$?

[quote="Farenhajt"]I'm not sure if this belongs to Intermediate, since I can see only the solution using a double integral... But if anyone has a simpler idea, go ahead :)

[hide="Serious calculus in here"]We must find the probability of $b^{2}-4ac\geqslant 0$ when $a,b,c\in[0,1]$. Observe the function $b(a,c)=2\sqrt{ac}$. By the geometric probability principle, we must find the volume of the body lying both above the surface $b(a,c)=2\sqrt{ac}$ and within the cube $a,b,c\in[0,1]$. Since it's easier to calculate the complementary volume, we'll do just that - the desired volume is the difference between the volume of the said cube, which is $1$, and the following:

\begin{eqnarray*}V &=& \int_{0}^{1}\int_{0}^{1}2\sqrt{ac}\,\text{d}a\,\text{d}c\\ &=& \int_{0}^{1}\text{d}c\int_{0}^{1}2\sqrt{ac}\,\text{d}a\\ &=& \int_{0}^{1}2\sqrt{c}\left(\frac{a^{3/2}}{3/2}\bigg|_{0}^{1}\right)\text{d}c\\ &=& \int_{0}^{1}{4\over 3}\sqrt{c}\,\text{d}c\\ &=&{4\over 3}\cdot\frac{c^{3/2}}{3/2}\bigg|_{0}^{1}\\ &=&{8\over 9}\end{eqnarray*}

Therefore, both the desired volume and the probability are ${1\over 9}$[/hide][/quote]

you have just calculated only one part of the complementary volumes.... Other parts are there ... Just give it another thought .... There may be many more integrals coming your way....

Actually, you're right, since there's a part of the volume that lies outside the cube. Hence from the calculated volume we must subtract the following:

\begin{eqnarray*}V_{1}&=& \int_{1/4}^{1}\int_{1/(4a)}^{1}2\sqrt{ac}\,\mathrm{d}c\,\mathrm{d}a\\ &=& \int_{1/4}^{1}{4\over 3}\left(\sqrt{a}-\frac{1}{8a}\right)\mathrm{d}a\\ &=& \frac{7-3\ln 2}{9}\end{eqnarray*}

because $ac={1\over 4}$ is the intersection of $b=2\sqrt{ac}$ and $b=1$, but then we must add back the following:

$V_{2}=\int_{1/4}^{1}\left(1-\frac{1}{4a}\right)\mathrm{d}a=\frac{3-2\ln 2}{4}$

which is the volume of the unit-high cilindroid that remains within the cube.

Therefore, the complementary volume is $V={8\over 9}-\frac{7-3\ln 2}{9}+\frac{3-2\ln 2}{4}=\frac{31-6\ln 2}{36}$ and the probability is $P=1-V=\frac{5+6\ln 2}{36}$

Hope it's right this time. And thanks :)


how do we prove that a cauchy sequence is bounded

It's only been three days since the last time I answered [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=cauchy+sequence+bounded&t=74539]this[/url]; you could try searching before you post.


geometry
geometric transformation
reflection
angle bisector

$AD$, $AE$ and $AF$ are the altitude, internal angle bisector and median respectively from $A$ of $\triangle ABC$. If these three lines tetrasect $\angle A$, and $\angle B \geq \angle C$, determine the angles of the $\triangle$.

[hide="Big Hint"]
The line from a vertex to the circumcentre is the reflection of the altitude from that vertex in the angle bisector from that vertex. ie. $\angle IAH = \angle IAO$.
[/hide]


geometry circumcircle inradius trigonometry

If you have a regular n-gon.  Is there a simple formula that gives you the circumradius and inradius?

The inradius of any regular n-gon is equal to the apothem. If you know the area of that n-gon then just use the formula $\frac {lp}{2} = A$ where $l$ is the apothem and $p$ is the perimeter. Or if you know the length of the side you can find the apothem by trig using $\frac {x}{2 \tan \frac {180}{n}}$.

The circumradius can be found using $\frac {x}{2 \sin \frac {180}{n}}$.


inequalities
function
calculus
derivative
integration
real analysis
real analysis unsolved

Pg 295, #10 of The Way of Analysis by Robert Strichartz:

If f is analytic on (a,b) prove that for every x_0 in (a,b) there exists a neighborhood of x_0 and constants M and r such that the k-th derivative of f, |f^(k)(x)|<=M(k!)(r^k) for all k and x in the neighborhood.

Are we allowed to extend $f$ into complex plane and use [url=http://en.wikipedia.org/wiki/Cauchy_integral_formula]Cauchy's integral formula[/url]? Perhaps not since the question is posted in "Real Analysis". But I can't think of any other proof. :(

Since $f$ is analytic at $x_{0}$, there is some interval $(x_{0}-c,x_{0}+c)$ on which $f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}}{n!}(x_{0})(x-x_{0})^{n}= \sum_{n=0}^{\infty}a_{n}(x-x_{0})^{n}$. Choose $R<c$: by uniform convergence of the series on $[x-R,x+R]$, $|a_{n}|<AR^{-n}$ for some constant $A$.

Now, we shift the series. For any $x\in (x_{0}-R,x_{0}+R)$, $|y-x|\le R-|x-x_{0}|$, write $f(y)=\sum_{n=0}^{\infty}a_{n}(y-x+x-x_{0})^{n}$
$f(y)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_{n}\binom{n}{k}(y-x)^{m}(x-x_{0})^{n-k}$
$f(y)=\sum_{k=0}^{\infty}(y-x)^{k}\sum_{j=0}^{\infty}a_{k+j}\binom{k+j}{k}(x-x_{0})^{j}$ (using absolute convergence)
$f(y)=\sum_{k=0}^\infty c_{k}(x-x_{0})^{k}$, where $c_{k}=\sum_{j=0}^{\infty}a_{k+j}\binom{k+j}{k}(x-x_{0})^{j}$. We must have $c_{k}=\frac{f^{(k)}(x)}{k!}$, so we estimate the size of $c_{k}$.

$|c_{k}|\le \sum_{j=0}^{\infty}|a_{k+j}|\binom{k+j}{k}|x-x_{0}|^{j}$
$|c_{k}|\le \sum_{j=0}^{\infty}AR^{-k-j}\binom{k+j}{k}|x-x_{0}|^{j}$
$|c_{k}|\le AR^{-k}\sum_{j=0}^{\infty}\binom{k+j}{k}\left(\frac{|x-x_{0}|}{R}\right)^{j}= AR^{-k}\left(1-\frac{|x-x_{0}|}{R}\right)^{-k-1}$
$c_{k}\le AR\left(R-|x-x_{0}|\right)^{-k-1}= \frac{AR}{R-|x-x_{0}|}\left(R-|x-x_{0}|\right)^{-k}$
The power series for $f$ centered at $x$ thus converges to $f$ on $\{y: |y-x|\le R-|x-x_{0}|\}$. Considering all possible $R$, its radius of convergence (to $f$) is at least $c-|x-x_{0}|$.

If we further restrict $|x-x_{0}|\le s<R$, we can set $M=\frac{AR}{R-s}$ and $r=\frac1{R-s}$.

We can derive most of the nice properties of functions defined by power series with algebra. It's just a lot nicer to use complex analysis.


yep thats right, sorry forgot to include that

[quote="sen"]let $a=\alpha_1\cdot 10^{n-1} + \alpha_2\cdot 10^{n-2}+\alpha_3\cdot 10^{n-3}+\cdots \alpha_n$

Find all possible values of $\frac{10^n + 1}{a}$ if it is an integer[/quote]Let $b=\frac{10^n+1}{a}$ be positive WLOG. Since $b$ must be an integer, $1<b<10$. Write $10^n+1=ab$. Since $ab$ is odd, $b$ must be odd. Also, $ab\equiv1\pmod{10}$, $b\ne5$. Observe that $ab$ is not divisible by 3, so $b=7$ is the only possible value. This happens when $n\equiv3\pmod{6}$. If you are including negative $b$, then the answer is $b=\pm7$.


algebra
polynomial
function
calculus
derivative
calculus computations

Let $ \{p_{n}(x)\}$ be a sequence of polynomials, and let $ p_{n}(x)$ converge to $ f(x)$ uniformly on the real line.  Prove that $ f(x)$ is a polynomial.

  :)

For large values of $ n$, $ p_{n+1}-p_{n}$ is a bounded polynomial.  :wink:

If the degree of $ p_{n}(x)$ was not bounded then it would not converge uniformly, so Let $ N$ be the bound. Obviously for every $ p_{n}(x)$ and $ m>N$ the derivate of order $ m$ at $ 0$ is $ 0$ so, by the uniformly convergence the same happens for $ f(x)$ so $ f(x)$ is a polynomial. QED

I reasoned like mlok.  Since the $ p_{n}(x)$ converge uniformly, we must have $ \forall \epsilon >0$ an $ N$ such that for all $ n\geq N$ we have $ |p_{n}(x)-p_{n+1}(x)|<\epsilon$ $ \forall x \in \mathbb{R}$. This means you eventually have a bounded polynomial on the real line (a constant), so the limit function must be a polynomial also.

For Afonso's solution, I must nitpick.  We know nothing about $ f$ except that it is the uniform limit of the $ p_{n}(x)$ which implies it is continuous, but not necessarily $ C^{k}$ for arbitrarily high k; so taking derivatives of $ f$ might be out of order.

See also : http://mathlinks.ro/Forum/viewtopic.php?t=138048


quadratics
inequalities
algebra
quadratic formula

Hi,

I have been using the AoPS Vol 1: the Basics (you guys should already know that by now :P), and I stumbled upon this problem.

EXCERCISE 6-10. Find all real values of $ x$ which satisfy $ \sqrt {x^2 + 1} + x^2 + 1 = 90$

I did this problem the "not so great way". This is what I did:

$ \sqrt {x^2 + 1} + x^2 + 1 = 90 \Rightarrow \sqrt {x^2 + 1} + x^2 - 89 = 0 \Rightarrow \sqrt {x^2 + 1} = 89 - x^2 \Rightarrow (\sqrt {x^2 + 1})^2 = (89 - x^2)^2 \Rightarrow x^2 + 1 = x^4 - 178x^2 + 7921 \Rightarrow x^4 - 179x^2 + 7290 = 0$ (I subtracted the $ x^2 + 1$ from both sides)

Let $ y = x^2.$ So I now have:
$ y^2 - 179y + 7920 = 0.$ And I decided to use the quadratic formula to find the roots.

$ \frac { - b\pm\sqrt {b^2 - 4ac}}{2a}$
That gives me: $ \frac {179\pm\sqrt {32041 - 31680}}{2(1)} \Rightarrow \frac {179\pm\sqrt {361}}{2} \Rightarrow \frac {179\pm 19}{2}$
Therefore, the two roots are:
$ y = \frac {179 + 19}{2} = \frac {198}{2} = \boxed{99}$ and,
$ y = \frac {179 - 19}{2} = \frac {160}{2} = \boxed{80}$
Hence, we have $ y = 99,80$ or $ x^2 = 80,99;$ Therefore the two roots of x can be: $ x = \sqrt {80} \Rightarrow \boxed{4\sqrt {5}}$ and $ x = \sqrt {99} \Rightarrow \boxed{3\sqrt {11}}$

However, the way the book did it was a bit different. They let $ y = \sqrt {x^2 + 1},$ which was very smart. So, their quadratic was $ y^2 + y - 90 = 0$ (very clean). So the solutions that they got were: $ y = 9, - 10,$ hence, $ \sqrt {x^2 + 1} = - 10,$ or $ \sqrt {x^2 + 1} = 9.$ They discard the first one because the square root cannot be a negative unless otherwise specified, and for the second one they got $ x = 4\sqrt {5},$ which is what I got.

So, their answer was $ x = \pm 4\sqrt {5}$, but if they solved the second one, they would have also gotten $ x = 3\sqrt {11},$ which they discarded because of the $ - 10.$

My question is if this problem was solved the way I did, how would I know that $ 3\sqrt {11}$ wouldn't be a valid root?

Thank you for your time. (I might have made a typo...too many things to type :P)!

Regards,
limac

[quote="limac"]

My question is if this problem was solved the way I did, how would I know that $ 3\sqrt {11}$ wouldn't be a valid root?

[/quote]

$ 3\sqrt{11}$ isnt a solution to the equation. plug it in and see  :wink:

Ahhh nuts...I forgot the most important part...Check your solutions. **Headdesk**

Thank you for pointing that out mz94! :)

Regards,
limac

hello, when you square both sides of the equation $ \sqrt {x^2 \plus{} 1} \equal{} 89 \minus{} x^2$ you must observe that the inequality $ 89 \minus{} x^2\geq0$ must hold, because both sides of your equation must be nonnegative, in the other case your result is not equivalent to your original equation.
Sonnhard.

Oh, that also makes sense. But that arises a question: We are afterall solving this equation for $ x$, so how come both the solutions don't work. As in shouldn't both $ \frac{179\plus{}19}{2}$ and $ \frac{179\minus{}19}{2}$ satisfy the equation, as they are both the solutions to $ x$. And can this be analogous to something like $ (x\minus{}3)(x\plus{}4)\equal{}0$, where both the values of $ x$ satisfies the equation?

not in this case. limac, in the case you are talking about, all your x terms have integer coefficients. in this problem, however, we have an $ \sqrt{x^2\plus{}1}$ term, which yields 2 solutions, whereas usually only one of them is correct

Ah...I see. When we get $ \sqrt{x^2\plus{}1}\plus{}x^2\plus{}1\equal{}90 \Rightarrow \sqrt{99\plus{}1}\plus{}100\equal{}90,$ we get $ 10\plus{}100\equal{}90,$ which is not true. However, if we would have taken the negative square root, i.e. $ \sqrt{100} \equal{} \minus{}10,$ we would have $ \minus{}10\plus{}100\equal{}90,$ which is true. But when they says $ \sqrt{a},$ they imply the positive square root, therefore, we rule out the possibility of $ x^2\equal{}99,$ or $ x\equal{}3\sqrt{11}$

Thank you guys very much your your help! :)

Regards,
limac


Driving his wife to work one morning, Sam encounters heavy traffic, and his average speed is only 20 mph. After dropping his wife at her office, he returns home along the same route at an average speed of 40 mph. What was his average speed for the entire trip? Express your answer as a mixed number

Using harmonic mean, we find $ \frac{2}{\frac{1}{20} \plus{} \frac{1}{40}} \equal{} \boxed{26\frac{2}{3}}$.

Let's say the distance each way is 100 miles.  Then on the way to her office, it takes him $ t\equal{}5$ hours and on the way back, it takes, $ 2.5$ hours.  Therefore, it takes him $ 7.5$ hours to travel 200 miles, an average speed of $ 26\frac{2}{3}$ mph.

Since the distance is the same, with and without traffic, and we want to find the average, harmonic mean can come into play.
$2ab/a+b=2(20)(40)/[20+40]=$ $26$ $2/3$
Thus, $26$ $2/3$ is your answer.


[quote="N.T.TUAN"]Show that the equation $3y^{2}= x^{4}+x$ has no solutions in positive integers.[/quote]

$3y^{2}= x(x+1)(x^{2}-x+1)$ $\Rightarrow $ either $x=0$ (mod 3), either $x=-1$ (mod 3)

1) $x=3u$ : $y^{2}= u(3u+1)(9u^{2}-3u+1)$. But $u$, $3u+1$ and $9u^{2}-3u+1=(3u+1)^{2}-3(3u+1)+3$ are coprimes, so each is a perfect square. Then $9u^{2}-3u+1=a^{2}$ and $(2a)^{2}=(6u-1)^{2}+3$. But equation $A^{2}-B^{2}=3$ implies $B=1$ or $B=-1$ and so $u=0$ and $x=0$

2) $x+1=3u$. We can write $y=3^{p}z$ and it is easy to see that $x^{2}-x+1=(x+1)^{2}-3(x+1)+3$ is divisible by 3 but not by 9. So, $3^{2p}$ divides $x+1$ and we have :
$z^{2}=x\ \frac{x+1}{3^{2p}}\ \frac{x^{2}-x+1}{3}$ and these 3 factors are coprime, hence $x$ and $x+1$ are perfect squares, hence $x=0$ or $x=-1$

So the equation $3y^{2}= x^{4}+x$ has no solutions in positive integers.

--
Patrick


twleve less than a # is the same as 6 , decreased by 8 times the # .find the #

$ x \minus{} 12 \equal{} 6 \minus{} 8x$

$ 9x \minus{} 12 \equal{} 6$

$ 9x \equal{} 18$

$ x \equal{} \boxed{2}$

You can also use the elimination method.

Elimination is used when you need to solve more than one equation. There is no need in this problem to take the trouble of creating more variables. The solution that ernie posted is the quickest and most direct.


algebra
function
domain
calculus
integration
number theory
complex numbers

Find integer pairs (a,b) satisfying:

$ a^2 \plus{} b^2 \equal{} (1^2 \plus{} 2^2)(3^2 \plus{} 4^2)(5^2 \plus{} 6^2)$

Its very easy , doesnt require any prerequisite number theory theory  :P .

Here's my solution
Divide both sides by 25.
Then a^2/25 + b^2/25 = 305

So, (a/5)^2 + (b/5)^2 = 305

Now 305 is congruent to 1 mod 4.
So one of the squares is divisible by 4.
Implies one of the above numbers(a/5 0r b/5) is an even number less than 16.(none of the squares can cross 17)
Checking we get it works for 16 and 4.
For 16 other number is 7 and for 4 other number is 17.
So,
solutions are 80 and 35
0r 85 and 20.

[quote="Conjurer"]There are many more.I am not looking into your solution right now since its late , but i will see to it later.[/quote]

I dont see how you can have more solutions, unless you want negative integers as well, which would add only 6 more to the ones mentioned by befuddlers.

Yes, positive negative both. Is your solution same as befuddlers's?

befuddlers assumed a/5 and b/5 were integers.
There's a lot of solutions.

$ (a^2 \plus{} b^2)(c^2 \plus{} d^2) \equal{} (ac\pm bd)^2 \plus{} (ad\mp bc)^2$

rewrite the equation as
$ (1^2 \plus{} 2^2)^3(5^2 \plus{} 6^2)$

$ (1^2 \plus{} 2^2)^2$ can be written in 2 different ways as sum of two squares ($ a \equal{} c \equal{} 1,b \equal{} d \equal{} 2$ in the identitity)   $ (3^2 \plus{} 4^2)$ and $ (5^2 \plus{} 0^2)$

$ (3^2 \plus{} 4^2)(1^2 \plus{} 2^2)$ can be written as sum of 2 squares by the identity($ a \equal{} 1,b \equal{} 4,c \equal{} 1,d \equal{} 2$) in two ways

$ ($one of them$ )(5^2 \plus{} 6^2)$ which is the RHS can be written as sum of 2 squares in two ways

$ ($other one of them$ )(5^2 \plus{} 6^2)$ which is the RHS can be written as sum of 2 squares in two ways

$ (5^2 \plus{} 0^2)(1^2 \plus{} 2^2)$ can be written as sum of 2 squares by the identity($ a \equal{} 1,b \equal{} 4,c \equal{} 1,d \equal{} 2$) in two ways

$ ($one of them$ )(5^2 \plus{} 6^2)$ which is the RHS can be written as sum of 2 squares in two ways

$ ($other one of them$ )(5^2 \plus{} 6^2)$ which is the RHS can be written as sum of 2 squares in two ways

So I think there re $ 8$ ways.

The numbers can be found by the identity, but it's too much work for me...

OK I am posting an amazing solution to this amazing problem:

Define $ z1 \equal{} \pm 1 \pm 2i$

$ z2 \equal{} \pm 3 \pm 4i$

$ z3 \equal{} \pm 5 \pm 6i$

Let $ z \equal{} a \plus{} bi$

According to given condition

$ |z|^2 \equal{} |z1|^2|z2|^2|z3|^2 \equal{} |z1 z2 z3|^2$

$ (z1 z2 z3)$ can be found easily. Rest is comparing.

[quote="Conjurer"]$ |z|^2 \equal{} |z_1|^2|z_2|^2|z_3|^2 \equal{} |z_1 z_2 z_3|^2$[/quote]
Yes that's my solution also. To show that there are no other solutions, it's best to write it as $ |1 \plus{} 2i|^6|5 \plus{} 6i|^2$. state that the norm of [url=http://en.wikipedia.org/wiki/Gaussian_integers]Gaussian integers[/url] is closed under multiplication or just show the identity I gave to justify this step. since both $ 1^2 \plus{} 2^2$ and $ 5^2 \plus{} 6^2$ are primes,  $ (1 \plus{} 2i)$ and $ (5 \plus{} 6i)$ are Gaussian primes. And a few more sentences complete the proof.

P.S. I was waiting for a solution that "doesn't require any prerequisite number theory theory", and you disappointed me.

[quote="Conjurer"]O'cmmon , it was a complex no. question hitherto you introduced gaussian n all.[/quote]

I agree.  :D

[quote="Conjurer"]O'cmmon , it was a complex no. question hitherto you introduced gaussian n all.[/quote]
Don't call "gaussian n all". It's only a name given to those complex numbers. Just like you introduced "complex n all", stating the word "complex numbers" were not necessary, I just gave the same idea in the prev. post without saying complex. and the identity I gave is merely a proof of $ |z_1|\cdot |z_2| \equal{} |z_1\cdot z_2|$

I introduced "gaussian n all" because you are waiving past important parts of the solution. I was trying to fill the gaps using the theory of Gaussian integers. Your solution is very unrigorous.

[quote="Conjurer"]$ (z_1 z_2 z_3)$ can be found easily.[/quote]
to give only one representation as sum of 2 squares. How does that help? Or if you represent it as different ways, it still remains to prove that there aren't any other.

To find all, one needs to find all complex $ z \equal{} a \plus{} bi$ satisfying $ |z| \equal{} |z_1z_2z_3|$ such that $ a,b$ are integers. To do that one needs:

one or more of these kind of results:

1) Gaussian integers form a unique factorization domain with units $ 1, \minus{} 1, i,$ and $ \minus{} i$

2)The ring of Gaussian integers is the integral closure of $ \mathbb Z$ in the field of Gaussian rationals $ Q(i)$ consisting of the complex numbers whose real and imaginary part are both rational.

3)A Gaussian integer $ a \plus{} bi$ is prime if and only if:

    * one of $ a,b$ is zero and the other is a prime of the form $ 4n \plus{} 3$ or its negative $ \minus{} (4n \plus{} 3)$
    * or both are nonzero and $ a^2 \plus{} b^2$ is prime.

[b]Of course it might be possible to do in a simpler way, without introducing all these. [color=red]Then give a detailed rigorous step by step solution of the problem[/color], not that outline style.[/b]

Thnx for completing the proof rigorously but doesnt it logically prove itself that there cannot be any other solutions. I mean we are find the point(s) z in the Argand plane , which satisfies |z|^2 = |z1 z2 z3|^2, where z1,z2,z3 are constants, isnt it just like a simple complex equation, equating real and imaginary part type?

[quote="Conjurer"]doesnt it logically prove itself that there cannot be any other solutions. I mean we are find the point(s) z in the Argand plane , which satisfies |z|^2 = |z1 z2 z3|^2, where z1,z2,z3 are constants, isnt it just like a simple complex equation, equating real and imaginary part type?[/quote]
No it doesn't. There are infinitely many complex numbers $ z$ that satisfy that equation, which lie on the circle with the origin at the center and radius $ |z_1 z_2 z_3|$. Or I could have misunderstood it as well, then show how to do that...

But there are finitely many integer points on that circle, which we need to find.

(integer points mean those complex numbers which have both real and imaginary part as integers, or that's the very definition of gaussian integers itself)

As fur as I understand, this is the whole point of rigorousness, avoiding one's intuitions and proving something by justifying each step.

I havn't completely written the proof rigorously yet, only said it can be done in that way.

If anyone is interested: Here is my solution with Complex Numbers:

Gaussian integers are not really something advanced. Check out the definition of it in the link I gave.

[hide]Let $ N(z) \equal{} |z|^2$ denote the norm of a gaussian integer $ z$.

Let $ a \plus{} bi \equal{} z$. We need to find all Gaussian integers $ z$ satisfying $ N(z) \equal{} (N(1 \plus{} 2i))^3N(5 \plus{} 6i)$

Since Gaussian integers form a unique factorization domain, there is a unique factorization of $ z$ into gaussian primes.

Let $ z \equal{} z_1z_2\cdots z_n$, where $ z_k$ are gaussian primes not necessarily distinct for $ k \equal{} 1,2,\cdots ,n$

Because the norm of Gaussian integers is multiplicative:

$ N(z) \equal{} N(z_1z_2\cdots z_n) \equal{} N(z_1)N(z_2)\cdots N(z_n)$

$ N(z_1)N(z_2)\cdots N(z_n) \equal{} (N(1 \plus{} 2i))^3N(5 \plus{} 6i)\cdots (*)$

by point (3) in the previous post:

If $ z$ is a gaussian prime, then $ N(z)$ is prime.

but since $ N(1 \plus{} 2i)$ and $ N(5 \plus{} 6i)$ are primes,

we have two prime factorizations of a same number in both $ RHS$ and $ LHS$ of $ (*)$. Hence they should be the same.

Without loss of generality:
$ N(z_1) \equal{} N(z_2) \equal{} N(z_3) \equal{} N(1 \plus{} 2i), N(z_4) \equal{} N(5 \plus{} 6i),n \equal{} 4$

If $ p,q$ are two gaussian primes such that $ N(p) \equal{} N(q)$ then: $ \{|re(p)|,|im(p)|\} \equal{} \{|re(q)|,|im(q)|\}$

Now we can find all possible $ z_1,z_2,z_3,z_4$ and each such quadruple corresponds to one $ z \equal{} z_1z_2z_3z_4$, and each such $ z$ corresponds to one pair $ a \equal{} re(z),b \equal{} im(z)$

maybe this much details wasn't needed...[/hide]


Ross Mathematics Program
function
number theory proposed
number theory

Let $A = \left\{ n + \pi ( n - 1 ) | n \in \mathbb{N} \right\}$ and $B = \left\{ n + p_n | n \in \mathbb{N} \right\}$.
Show that
\[ A \cup B =\mathbb{N} \mbox{ and } A \cap B = \{ \}  \]

something should be clarified... what is $\pi(n)$?? number of partitions of some sort of $n$...?  and what about $p_n$?

$\pi(n)$: number of primes $\leq n$
$p_n$: $n$-th prime

[quote="pleurestique"]so... more generally, if $g(n)$ is a sequence of natural numbers and $f(n) =$ the number of elements in the sequence $g$ that are $\leq n$, then the sequences $n+g(n)$ and $n+f(n)$ are *complementary*... a nice proof is presented in Ross Honsberger's Mathematical Gems III.[/quote]

The proof is relatively straightforward, but I will modify your statement to be a little more precise.

Let $s_n$ denote a [i]monotonically increasing[/i] sequence of positive integers such that $s_1 \ge 1$.  Let $\omega(n)$ be defined as the number of members of $s_k \le n$.  Hence, $s_{\omega(n)} = n$ and $\omega(s_k) = k$.  Additionally, $\omega(0) = 0$.

[b]Lemma[/b]:  The sets $A = \{ n + \omega(n-1) | n \in \mathbb{N} \}$ and $B = \{ n + s_n | n \in \mathbb{N} \}$ are disjoint, and their union is the naturals.

Proof:

Consider the values of $a_n = n + \omega(n-1)$ at two consecutive terms $n, n+1$ such that $n \neq s_k$ for any $k$.  Then $\omega(n-1) = \omega(n)$ and so the two terms are consecutive.

Consider the values of the sequence $a_n = n + \omega(n-1)$ as $n$ approaches a value of $s_k$.  At $n = s_k$ we have $a_n = s_k + \omega(s_k - 1) = s_k + k - 1$.  Additionally, $a_{n+1} = s_k + 1 + \omega(s_k) = s_k + k + 1$.

We see that $a_n$ is a sequence that produces consecutive integers "most of the time," but skips integers of the form $s_k + k$, precisely the set $B$.  Moreover, $a_1 = 1$, so we are guaranteed that $A \cup B = \mathbb{N}$.

The original problem can be solved by putting $s_n = p_n$ and $\omega(n) = \pi(n)$.

Q.E.D.

Note that setting $s_n + n = n^2$ or $s_n + n = \frac{n(n+1)}{2}$ produces a function $n + \omega(n)$ that produces an integer sequence that skips precisely the squares or the triangulars.  Both of these problems were proposed and solved in Engel, but without this generalization.

[quote="t0rajir0u"]
Note that setting $s_n + n = n^2$ or $s_n + n = \frac{n(n+1)}{2}$ produces a function $n + \omega(n)$ that produces an integer sequence that skips precisely the squares or the triangulars.[/quote]

Very nice ! :)


In the product $ B2 \times 7B\equal{}6396$, $ B$ is a digit.  The value of $ B$ is

\[ \textbf{(A)}\ 3 \qquad
\textbf{(B)}\ 5 \qquad
\textbf{(C)}\ 6 \qquad
\textbf{(D)}\ 7 \qquad
\textbf{(E)}\ 8
\]

[hide="Click for alternate solution"]
If $ B \times 2$ ends in $ 6$, then $ B$ is $ 3$ or $ 8$.  Since the product exceeds $ 6000$, $ B$ must be $ 8$, or answer $ \boxed{\textbf{(E)}}$.
[/hide]


analytic geometry
trigonometry
integration
geometry
angle bisector
geometry unsolved

Consider triangle ABC,the continues charges distribution lies on the side of BC with the linear charge density such as m,prove that the direction of electric force in the point A lies on the bisector of the viewing angle BAC

Put the coordinate origin in $ A$ and the positive x-axis along the bisector $ AK$ of the angle $ \alpha = \angle CAB.$ Let $ \lambda$ be the uniform charge density along the segment $ a = |BC|$ and $ h_a = |AD|$ the A-altitude of the $ \triangle ABC.$ Let a point $ P \in |BC|$ be arbitrary and denote $ \varrho = |AP|,$ $ \varphi = \measuredangle PAK,$ $ \psi = \measuredangle PAD.$ Using Coulomb law and $ \cos \psi = \frac {h_a}{\varrho},$ the electric field component $ E_y$ at $ A$ in the y-axis direction (perpendicular to the angle bisector $ AK$) is

${ E_y = \int_{0}}^{a} \sin \varphi\ \frac {\lambda\ \text da}{4 \pi \varepsilon \varrho^2} = \frac {\lambda}{4 \pi \varepsilon } \int_{ - \frac {\alpha}{2}}^{ + \frac {\alpha}{2}} \sin \varphi\ \frac {\varrho\ \text d\varphi}{\varrho^2 \cos \psi} =$

$ = \frac {\lambda}{4 \pi \varepsilon} \int_{ - \frac {\alpha}{2}}^{ + \frac {\alpha}{2}}\sin \varphi\ \frac {\varrho^2\ \text d\varphi}{\varrho^2 h_a} = \frac {\lambda}{4 \pi \varepsilon h_a} \int_{ - \frac {\alpha}{2}}^{ + \frac {\alpha}{2}} \sin \varphi\ \text d\varphi = 0$


function
limit
inequalities
number theory solved
number theory

Solve the equations :
 a) $\phi (x^2)=x^2-x$
 b) $\phi (nx)= \phi(x) , n >1$

 $\phi$ is Euler's function

a) $\phi(x^2) = x \cdot \phi(x)$, for any $x\in\mathbb{Z}^+$. So solve $\phi(x) = x-1$ in positive integers, which is done iff $x$ is a prime.

Write $n=a\cdot b$ where $a$ is the maximal divisor of $n$  that is coprime with $x$.
Then $\phi (nx) = \phi (abx) = \phi (a) \phi(bx) = b \phi(a)\phi(x)$, where the last step comes from standard properties of $\phi$ since all prime divisors of $b$ divide $x$.
So we need $b \phi(a) = 1 \implies b=1, \phi(a)=1 \implies a=n=2$, so $x$ is odd and $n=2$.

b) You need $\gcd(n,x) = 1$, if $\phi(nx) = \phi(x)$. In fact $\phi(nx) = \phi(x)$ only if $v_p(\phi(nx)) = v_p(\phi(x))$, where $v_p(\cdot)$ is the $p$-adic valuation of its argument. So letting $p\in\mathfrak{P}$ divide both $n$ and $x$, you'd have $v_p(\phi(x)) \leq 1 + v_p(\phi(nx))$, absurd! Hence by multiplicativity $\phi(nx) = \phi(x)$ iff $\phi(n) = 1$ and $\gcd(n,x) = 1$, that's to say iff $n = 2$ and $x$ is odd.

Oh, sorry... It seems I was too slow, ZetaX. No matter, I believe two is better than one (even if similar arguments were used). :lol:

In addition to that :
$\text{let } n \in \mathbb N^*$

Show that :
c) $\lim_{n \to +\infty} \ \phi(n) = +\infty$
d) $\phi(n) > ln(n)$
e) $\phi (n) > \sqrt{n} - 1$

 :gleam:

All of yours, mathmanman, follow from Kendall-Osborn's inequality: "for any integer $n > 6$, $\phi(n) \geq \sqrt{n}$".

Hmm... if you say so.. but, can you prove it (or them)? :)

Well, I suppose you're asking for a proof of Kendall-Osborn's. Ok, here is! Letting $n > 6$, you have $\phi(n) \geq \sqrt{n}$ iff $\phi(n) = \prod_{p \mid n} \left(1 - \frac{1}{p}\right) \geq \frac{1}{\sqrt{n}}$, where LHS product is over all positive rational primes dividing $n$. Moreover remind that $\phi(2n) = \phi(n)$, if $n$ is odd. Hence you're actually reduced to check it just for squarefree even numbers.

Therefore let $n = 2p_1 p_2 \ldots p_r$, where $r\in\mathbb{Z}^+$ and $p_1, p_2, \ldots, p_r$ are pairwise distinct rational (positive) primes. If $p\in\mathfrak{P}$ is odd, then $\phi(p) > \sqrt{p}$, and particularly $\phi(p) > \sqrt{2p}$, if $p \geq 5$. So trivially $\phi(n) \geq \sqrt{n}$, if $p_r \geq 5$.

Unfortunately this argument fails if $n > 6$ doesn't contain any prime factor $\geq 5$. Anyway, letting $n = 2^r \cdot 3^s$, with $r, s \in \mathbb{N}$ and $n > 6$, you handle your counts to find out everything works. And that's all folks...


conics
parabola
ellipse
geometry
parallelogram
geometry proposed

$M$ is midpoint of $BC$.$P$ is an arbitary point on $BC$.
$C_{1}$ is tangent to big circle.Suppose radius of $C_{1}$ is $r_{1}$
Radius of $C_{4}$ is equal to radius of $C_{1}$ and $C_{4}$ is tangent to $BC$ at P.
$C_{2}$ and $C_{3}$ are tangent to big circle and line $BC$ and circle $C_{4}$.
[img]http://aycu01.webshots.com/image/4120/2005120338156776027_rs.jpg[/img]
Prove : \[r_{1}+r_{2}+r_{3}=R\] ($R$ radius of big circle)

Hey, people  don't you see this beautiful problem. :?  :?  :?

I think nobody is interested in this interesting problem.
I'll send the solution soon.

I don't think I'll get into all the details.

Let $O_i,O$ be the centers of circles $i,\ i\in\overline {1,4}$ and the large circle, respectively. Let $\ell$ be the line parallel to $BC$ through $O_1$, and let $X_i,X$ be the projection of $O_i,O$ on $\ell$, respectively.

Because circles $2,3$ are tangent to $BC$ and circle $4$, $O_2,O_3$ lie on the parabola having $O_4$ as focal point and $\ell$ as directrice (call the parabola $\mathcal P$). At the same time, the circles $2,3$ are tangent to the big circle and to circle $4$ , so the $O_2,O_3$ also lie on the ellipse having $O_4,O$ as foci and large axis $r_1+R$ (call the ellipse $\mathcal E$). It's easy to see that $d(O_4,\ell)+d(O,\ell)=r_1+R$, and in these conditions, it's also easy to see that if a point $P$ belongs to $\mathcal P\cap \mathcal E$, then so does its symmetric wrt the center of $\mathcal E$. $O_2,O_3\in\mathcal P\cap \mathcal E$, so the symmetric of $O_2$ wrt the midpoint of $O_4O$ also belongs to $\mathcal P\cap \mathcal E$. On the other hand, the vertex of $\mathcal P$, which is $M$, lies inside $\mathcal E$, so $\mathcal P\cap\mathcal E$ has at most two points, meaning that $O_2OO_3O_4$ is a parallelogram, so $OO_3=O_2O_4$, and since $OO_3=R-r_3,\ O_2O_4=r_2+r_1$, we get the conclusion.

I hope that it's Ok, and that  people can understand it :).

This seems something like Sangaku problem in Edo era. :lol:

[quote="gollywog"]i guess inversion and go... but i'm too lazy to do it :D[/quote]

i used inversion,with the intersection of circle$C(1)$,and arc$BC$, as the

center,

really i dont have enough time to post the solution,but im not lazy :wink:


Let $ y_i\equal{}\frac{1}{1\minus{}x_i}$, then $ x_i\equal{}1\minus{}\frac{1}{y_i}$, therefore $ y_i$ are roots of
$ \frac{(1\minus{}\frac{1}{y})^{n\plus{}1}\minus{}1}{1\minus{}\frac 1y \minus{}1}\equal{}0$ or $ P(y)\equal{}(y\minus{}1)^{n\plus{}1}\minus{}y^{n\plus{}1}\equal{}0$.
Therefore $ \sigma_1(y)\equal{}\sum_iy_i\equal{}\frac{C_{n\plus{}1}^2}{C_{n\plus{}1}^1}\equal{}\frac{n}{2}.$


function
real analysis
real analysis unsolved

If $ f$ is derivable on $ [a,b]$ and $ f^{'}(a)\equal{}f^{'}(b)$ prove that exist $ c\in{(a,b)}$ so that $ f(c)\minus{}f(a)\equal{}(c\minus{}a)f'(c)$.
I find a resolvation using the function $ h$ , $ h(x)\equal{}\frac{f(x)\minus{}f(a)}{x\minus{}a}$ for $ x\in{(a,b]}$ and $ h(x)\equal{}f'(a)$  for $ x\equal{}a$. This function is continuous on $ [a,b]$, derivable on $ (a,b)$ . But I can't prove that $ h(a)\equal{}h(b)$ . Because then from Rolle's theorem follows the result

why not
\[ h\left( x \right) \equal{} \frac{{f\left( b \right) \minus{} f\left( a \right)}}
{{b \minus{} a}}
\]at $ x \equal{} a$?

I think that for $ h(x) \equal{} \frac {f(b) \minus{} f(a)}{b \minus{} a} , at  x \equal{} a$ the function $ h$ isn't continuous

Isn't this a problem from GM?Because I found one just like this, and in the hypothesis it was $ f(a) \equal{} f(b)$, not $ f^\prime (a) \equal{} f^\prime (b)$. Are you sure this is the right hypothesis?

The problem isn't from GM. Where I found it the condition is $ f'(a)\equal{}f'(b)$.


Using only the digits 7, 8, and 9, how many seven-digit palindromes can you make?

Please explain this to me. Thanks.  :)

a palindrome is a number of the form ABCDCBA (same backwards and forwards)

Looking at this

3 possibilities for A times 3 for B times 3 for C times 3 for D

81

A palindrome is as mewto said ABCBA.
You only have to do 3*3*3 (3 for every choice 7, 8 or 9) because the last two digits must be the same as the first two. Actually it is 3*3*3*1*1
 :) Still, mewto had a good explination

But there are 7 digits.  Therefore, Mewto's explanation stands since the palindrome would be in the form "ABCDCBA", allowing for $3^4 = 81$ possibilities for ABCD.


The Goldbach conjecture states that every even number greater than or equal to four is a Goldbach number, a number that can be expressed as the sum of two primes.
So, here is the amazing demostration you've been waiting:
[hide][color=blue][size=200]Happy April Fools[/size][/color] :rotfl:[/hide]

well,  that was  a  bad prank.  was  so excited to hear that


AMC
USA(J)MO
USAMO
search
algebra unsolved
algebra

Let: $x_{1}, x_{2}, ... , x_{n}$, and $y_{1}, y_{2}, ... , y_{n}$ are non-negative reals. Show that $\sum(min(xixj, yiyj)) \leq  \sum(min(xiyj, xjyi))$, where each sum is taken over all $n^{2}$ pairs $(i, j)$

use search button !  :mad: it has been posted many times


AMC
AIME
geometry
3D geometry
prism
number theory
least common multiple

#11 on the 1994 AIME reads:

Given 94 identical bricks, each 4 x 10 x 19, how many different heights of tower can be built (assuming each brick adds 4, 10 or 19 to the height)?

Can somebody give me a good solution to this?  I can't really follow Kalva's method.

I assume you don't, because if you did the answer would just be 1.... :P

Err, not quite... since each brick can have height 4, 10, or 19. I think you do use all the bricks. But every time I see this problem, I never remember how to solve it. And since it's on the 1994 AIME, which was really hard... yeah. I can't follow Kalva's solution either; the only part I get is the first part about 5*10 = 2*19 + 3*4...

[quote="krustyteklown"]do you need to use all 94 bricks?[/quote]
Yes, you do - and there isn't just one height, because each brick can either use it's short, medium, or long end to add height to the tower.

Call the heights S, M, and L. Then find out how many different amounts of S, M, and L you can have in 94 slots. After you do this find least common multiples and find the overlap and subtract from the previous total. I say this but I haven't solved the problem yet. This is what I would do at first glance. I'll tell you if I come up with a solution  :P

[quote="krustyteklown"]What it meant was there could only be 0-4 "true" bricks adding 10 on[/quote]

Because if you had 5+ bricks of 10, you could substitute 2 19's and 3 4's and you would get an equal height. So you just assume you have <5 bricks of 10 so you dont count equivalent heights.

Classic problem.

Not really that hard ... come on now guys!

Isn't it something to do with subtracting $LCM(4,10)+LCM(4,19)+LCM(10,19)$, then adding $LCM(4,10,19)$? Or am I mixing it up with a different kind of problem?

I think you're thinking of the problem about "how many blocks does a diagonal in a axbxc dimension rectangular prism cross"

Anyways, that's just the inclusion exclusion principle.

Yeah, I thought so.

Well, I'm going to have a go at a solution, although I may not have time to finish it right now.

[hide="Answer"]The smallest possible height is all 94 blocks with height 4. Since 5 blocks of height 10 can be replaced with 3 blocks of height 4 and 2 blocks with height 19, we have 4 more possible heights by switching 4's for 10's. Then, you can start replacing each 4 with a 19 for 94 more heights, then take one with all 19's and switch some with 10's for four more possibilities; so far we have $1+4+94+4=103$. Next, we take all 93 combinations of 4's and 19's and switch out a certain number for 10's, which (I think) gives us $93\cdot 4=372 \Rightarrow 372+103=\boxed{475}$.[/hide]

I'm pretty sure I messed up in there somewhere... :?

I don't see where LCM comes into play in this problem. It's really just a matter of swapping stuff out (and remember you can run out of bricks toward the upper limit)

Treat the bricks as heights 0, 6, 15, for convenience. Clearly you can only get multiples of 3. The question is which multiples of 3 you can't get. That's pretty much it.

Ah, nice, that makes it much simpler.

[hide="Answer"]Essentially, we have the equation $15a+6b+c=n$, where $a+b+c=94$. We cannot make $3$ and $9$, but we can make every other multiple of $3$ up to the point where we have too many $15$'s to get all the multiples of $3$ in-between. $1389$ is the first of these, since all numbers of the form $15n+9$ require 4 $6$'s to make, but $1389=91\cdot 15+4\cdot 6$, which is not possible. We also have $1398$, $1404$, and $1407$ before we reach $1410=94\cdot 15$. So, we have $6$ multiples of $3$ out of $470$ that are impossible, but we must also add $0$ as a possible value, so our answer is $470-6+1=\boxed{465}$.[/hide]

Wow, I was surprisingly close to the answer last time... I wonder what I did wrong in reaching that.

Would the tower have to be a rectangular prism, or could it be something like a lego pyramid?

It doesn't have to be a rectangular prism, but it can't be a Lego pyramid or something. They all have to be stacked on top of each other.


geometry
3D geometry
function
articles
limit

[hide="My book"]
Fundamentals of physics by Halliday, Resnik, Walker
6th edition, reduced price edition for developing countries
Page $ 431,432$
[/hide]
[hide="First doubt"]
[quote]
$ \cdots$

[size=150]Linear Expansion[/size]

If the temperature of a metal rod of length $ L$ is raised by an amount $ \Delta T$, its length is found to increase by an amount
\[ \Delta L \equal{} L\alpha \Delta T
\]
in which $ \alpha$ is a constant called the [b]coefficient of linear expansion[/b].

$ \cdots$
[/quote]

I was wondering about the following scenario:

A rod of length $ L$ undergoes a raise in temperature first by $ \Delta T_1$ and then ny $ \Delta T_2$

I shall calculate the final length in two ways

$ \boxed{1}$
Total temperature increase is $ \Delta T_1 \plus{} \Delta T_2$. Thus
Final length =$ L \plus{} \Delta L \equal{} L \plus{} L\alpha (\Delta T_1 \plus{} \Delta T_2) \equal{}$
$ \boxed{L(1 \plus{} \alpha(\Delta T_1 \plus{} \Delta T_2))}$

$ \boxed{2}$
After the first increase $ \Delta T_1$ the length will be
$ L \plus{} \Delta L_1 \equal{} L \plus{} L\alpha \Delta T_1 \equal{} L(1 \plus{} \alpha \Delta T_1)$

Now considering $ L(1 \plus{} \alpha \Delta T_1)$ as the initial length and implying the second change $ \Delta T_2$ the final length will be
$ L(1 \plus{} \alpha \Delta T_1) \plus{} \Delta L_2 \equal{} L(1 \plus{} \alpha \Delta T_1) \plus{} L(1 \plus{} \alpha \Delta T_1)\alpha \Delta T_2 \equal{} \boxed{L(1 \plus{} \alpha \Delta T_1)(1 \plus{} \alpha \Delta T_2)}$

But the two results are not identity. Equating them would imply $ L \equal{} 0\text{ or }\alpha \equal{} 0\text{ or }T_1 \equal{} 0\text{ or }T_2 \equal{} 0$ all of which are untrue.
[/hide]
[hide="Second doubt"]
[quote]
$ \cdots$

If the temperature of a solid or liquid whose volume is $ V$ is increased by an amount $ \Delta T$, the increase in volume is found to be
\[ \Delta V \equal{} V\beta \Delta T
\]
where $ \beta$ is the [b]coefficient of volume expansion[/b] of the liquid or the solid

$ \cdots$
[/quote]
OK, for one thing the first doubt is here too.

Besides, I was wondering that why wouldn't the following work for a solid?

Each dimension of a cube undergoes linear expansion. Thus If $ V_i,V_f,L_i,L_f$ are respectively the initial and final volume,the initial and final length of a side of the cube then
\[ \frac {V_f}{V_i} \equal{} \left(\frac {L_f}{L_i}\right)^3
\]
[/hide]
[hide="Question"]
[quote]
$ \cdots$

The coefficient of volume expansion and linear expansion for a solid are related by
\[ \beta \equal{} 3\alpha
\]
$ \cdots$
[/quote]
Why?
[/hide]

Please help me [b]Pradesi, JRav, Carcul, hsiljak, sludgethrower[/b] or anyone else.

I [b]don't have any teacher[/b] to clear my questions. I only read my book and try to understand things.

So please help me to understand even if my questions are too stupid.

Hmmm. I have also encountered the first dilemma in my high school days, but I was explained that the trick is in the fact that $ \alpha$ is a temperature-dependent function, so this kind of calculation never gives uniform results (if you split your thermal interval in subintervals).
This may have another type of explanation, but this is the first I've remembered.

As far as doubt no. 2 is concerned: I'll be brave and say that for small temperature changes, your formula in fact holds:

it is equivalent to $ (3\alpha \Delta T \plus{} 1)^3 \equal{} 3\alpha\Delta T \plus{} 1$ which is true for small temperature changes.

Again, we encounter the calculation problem in the case of bigger thermal intervals, as the change of coefficient isn't taken into account.

As far as the last question is concerned:

The logic of the proof is exactly the same you applied in your "conjecture" about the cube expanding due to heating:

$ \alpha_V \equal{} \frac {1}{V} \frac {\partial V}{\partial T} \equal{} \frac {1}{L^3} \frac {\partial L^3}{\partial T} \equal{} \frac {1}{L^3}\left(\frac {\partial L^3}{\partial L} \cdot \frac {\partial L}{\partial T}\right) \equal{} \frac {1}{L^3}\left(3L^2 \frac {\partial L}{\partial T}\right) \equal{} 3 \cdot \frac {1}{L}\frac {\partial L}{\partial T} \equal{} 3\alpha_L$

and here's the approval (from the Wiki article http://en.wikipedia.org/wiki/Coefficient_of_thermal_expansion)

[quote]This ratio arises because volume is composed of three mutually orthogonal directions. Thus, in an isotropic material, one-third of the volumetric expansion is in a single axis (a very close approximation for small differential changes). Note that the partial derivative of volume with respect to length as shown in the above equation is exact, however, in practice it is important to note that the differential change in volume is only valid for small changes in volume (i.e., the expression is not linear). As the change in temperature increases, and as the value for the linear coefficient of thermal expansion increases, the error in this formula also increases. For non-negligible changes in volume:

    $ ({L \plus{} }{\Delta L})^3 \equal{} {L^3 \plus{} 3L^2}{\Delta L} \plus{} {3L}{\Delta L}^2 \plus{} {\Delta L}^3 \,$

Note that this equation contains the main term, $ 3L^2$ , but also shows a secondary term that scales as $ 3L{\Delta L}^2 \equal{} {3L^3}{\alpha}^2{\Delta T}^2\,$, which shows that a large change in temperature can overshadow a small value for the linear coefficient of thermal expansion. Although the coefficient of linear thermal expansion can be quite small, when combined with a large change in temperature the differential change in length can become large enough that this factor needs to be considered. The last term, $ {\Delta L}^3$ is vanishingly small, and is almost universally ignored.[/quote]

This reminds me of the old simple interest dilemma..... (this was taught to us for a long time without any mention that compound interest existed :!: )..... I used to think that I would never keep my money deposited for long in a place, and reinvest it anually. :P

OK. I mostly understood what you said. Only a few things to say....

Does all of it mean that if we use a appropriate constant and don't split the interval, is the formula quoted from my book totally accurate for small or large changes alike? :maybe: Or is there a better formula but for the sake of simplicity this one is accepted? (like the simple interest in intermediate school level)

[quote="hsiljak"]$ \alpha$ is a temperature-dependent function[/quote]
In the quoted formula, $ \alpha$ is dependent on which temperature, initial temperature?

[quote="hsiljak"]so this kind of calculation never gives uniform results (if you split your thermal interval in subintervals).[/quote]
Does this mean that if we don't split the interval, the formula is more accurate? I was thinking that the formula would be most accurate if we split the interval into differentially small subintervals?
[quote="hsiljak"]Again, we encounter the calculation problem in the case of bigger thermal intervals, as the change of coefficient isn't taken into account. [/quote]
But this time the interval needn't split, so there shouldn't be a problem with changing $ \alpha_L$. We can just use the appropriate $ \alpha_L$ according to the corresponding temperature?

[quote="hsiljak"]As far as doubt no. 2 is concerned: I'll be brave and say that for small temperature changes, your formula in fact holds:

it is equivalent to $ (3\alpha \Delta T \plus{} 1)^3 \equal{} 3\alpha\Delta T \plus{} 1$ which is true for small temperature changes.[/quote]
Lets forget about the other formulas. Can you physically explain why my formula doesn't hold for large $ \Delta T$? I mean my reasoning is simple, each edge gets increased by a factor $ \left(\frac {L_f}{L_i}\right)$(whatever formula determines $ L_f$), no problem with that, then the volume must increase by a factor of $ \left(\frac {L_f}{L_i}\right)^3$?

Actually the formula $ \Delta L \equal{} L\alpha\Delta T$ holds for very small temperature change, i.e. $ \Delta T$ tends to $ 0$. If it is not considered as very small temperature change, we cannot consider the $ \Delta L$ part to be very small, which will generate problems.
Actually, when you heat a rod or something, at every instant you have to change the initial length of the rod in your formula, unless the increment $ \Delta L$ is very small. Else, you will face some of these severe problems:

i)Total temperature increase is $ \Delta T$
Thus final length=$ L \plus{} \Delta L \equal{} L \plus{} L\alpha\Delta T \equal{} \boxed {L(1 \plus{} \alpha\Delta T)}$

ii)Total temperature increase is $ \Delta T \equal{} \frac {\Delta T}{2} \plus{} \frac {\Delta T}{2}$
Thus after $ \frac {\Delta T}{2}$ increase in temperature, length=$ L\left(1 \plus{} \alpha\frac {\Delta T}{2}\right)$

Final length=$ \boxed{ L\left(1 \plus{} \alpha\frac {\Delta T}{2}\right)^2}$

You will find other equations as $ \boxed{ L\left(1 \plus{} \alpha\frac {\Delta T}{3}\right)^3}$, or $ \boxed{ L\left(1 \plus{} \alpha\frac {\Delta T}{4}\right)^4}$ from similar process. But all these equations are equivalent to the first one if and only if $ \Delta T$ is very very small. I hope you understand.
 :)

Yes I was wondering about exactly this in the first place, but I thought the book would have something to say regarding this, especially when We have a not so complicated formula:

$ L_f \equal{} \lim_{n\to \infty} L_i(1 \plus{} \alpha \frac {\Delta T}{n})^n \equal{} \boxed{L_ie^{\alpha \Delta T}}$
($ L_i,L_f$ respectively the initial and final temperatures)

This works fine for small or large $ \Delta T$ and it's not contradictory within itself.

So why isn't this given anywhere? Surely there is some problem? Please someone put light...

[quote="Akashnil"]
Does all of it mean that if we use a appropriate constant and don't split the interval, is the formula quoted from my book totally accurate for small or large changes alike? :maybe: Or is there a better formula but for the sake of simplicity this one is accepted? (like the simple interest in intermediate school level)

[quote="hsiljak"]$ \alpha$ is a temperature-dependent function[/quote]
In the quoted formula, $ \alpha$ is dependent on which temperature, initial temperature?

[/quote]

These two questions are pretty much connected. $ \alpha$ depends on the initial temperature. Standard $ \alpha$ is given for $ 20^0C$ for instance, and it would be different for some other temperatures. Most of the coefficients given in textbooks and physics problems are some sort of average values for the expected heating range, so the results that can be obtained using them are pretty accurate (i.e. the coefficients are most of the time "tweaked" for your basic$ L(1 \plus{} \alpha T)$ formula). Even for engineers (working in limited heating bounds), your formula is precise enough, with the right selection of coefficients.
Now, back to the first question: the formula you have and the coefficient given will give accurate results. Splitting the interval would be appropriate if you could have access to precise info about the coefficients at certain temperatures. Note that the coefficients don't change linearly with the change of temperature!
Interesting synchronicity: last night I had a brief conversation on this subject with an older engineer (to make sure my reasoning is correct) and he said: "You want to ask whether there is an advanced formula which can be used for thermal expansion (like composite interest formula)? It may be obtainable, but not necessary."

[quote="Akashnil"]
[quote="hsiljak"]so this kind of calculation never gives uniform results (if you split your thermal interval in subintervals).[/quote]
Does this mean that if we don't split the interval, the formula is more accurate? I was thinking that the formula would be most accurate if we split the interval into differentially small subintervals?
[/quote]
Yes, but you should then know $ \alpha$ for each subinterval.

[quote="Akashnil"]
[quote="hsiljak"]Again, we encounter the calculation problem in the case of bigger thermal intervals, as the change of coefficient isn't taken into account. [/quote]
But this time the interval needn't split, so there shouldn't be a problem with changing $ \alpha_L$. We can just use the appropriate $ \alpha_L$ according to the corresponding temperature?

[quote="hsiljak"]As far as doubt no. 2 is concerned: I'll be brave and say that for small temperature changes, your formula in fact holds:

it is equivalent to $ (3\alpha \Delta T \plus{} 1)^3 \equal{} 3\alpha\Delta T \plus{} 1$ which is true for small temperature changes.[/quote]
Lets forget about the other formulas. Can you physically explain why my formula doesn't hold for large $ \Delta T$? I mean my reasoning is simple, each edge gets increased by a factor $ \left(\frac {L_f}{L_i}\right)$(whatever formula determines $ L_f$), no problem with that, then the volume must increase by a factor of $ \left(\frac {L_f}{L_i}\right)^3$?[/quote]
[/quote]

Honestly, I don't think that there exist a physical explanation (I hope someone will correct me if I'm wrong). Your formula, in my opinion holds for all temperature ranges (if there hadn't been thermal deformations), but it is hard to obtain same results using this reasoning and the derived formula, due to the fact that $ \alpha$ is constantly changing, and its averaging for practical purposes won't make it totally accurate, and the nonlinearity appearing after expanding your formula kind of distorts the calculation. But the formula you derived is absolutely correct for all ranges.
Here's more about it: http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/ThermProps.htm (See the Volume Expansion paragraph)

BTW, I'll try to find out more about these formulae.

P. S. I really don't see a reason why one couldn't use the $ Le^{\alpha T}$ formula-it is pretty logical, going straight form the $ \alpha\equal{}\frac{1}{L}\frac{dL}{dT}$... But still, why it isn't in textbooks?  :maybe:

The exponential formula seems OK: see http://www.agu.org/reference/minphys/6_fei.pdf (formula 5)

P. S. This paper is interesting because of its approach to the thermal coefficient change (as a non-linear function with certain constant coefficients).

Thanks a lot for everything. :)

I am not actually interested about finding more accurate formulas, at least at this moment :oops: . The book formula is very appropriate, at least when solving a problem at a test :roll: . I was worried that my conception was somewhere unclear. However the thread helped me very much.

P.S. BTW, the paper was quite interesting.


A polynomial $f(x) \in \mathbb{Z}[x]$ satisfies $f(0) \equiv f(1) \equiv 1 \pmod{2}$

Show that $f(a) \neq 0$ for all integer values $a$.


geometry
3D geometry
sphere
dodecahedron
counting
distinguishability
geometric transformation

Five identical cubes, each having a volume of 1 cubic inch, are fused together to form a solid object. Adjoining cubes are fused together in such a way that their contiguous faces cover one another entirely without any overlap. No cube is connected to the rest of the object at nly a finite number of points or lines. How many different shapes could these 5-cubic-inch objects have?

[hide="answer"]29[/hide]

What is the maximum number of bounded volumes that may be formed by five interpenetrating spheres (e.g., soap bubbles), considering only the surfaces of the spheres as bounds and counting only volumes that are not further subdivided?

[hide="answer"]29[/hide]

Each side of a "regular" (i.e., perfectly symmetrical) dodecahedron is painted a uniform shade of red or blue, so that any two sides painted red are mutually indistinguishable, as are any two sides painted blue. Two clolor patterns for the entire dodecahedron are mutually indistinguishable if and only if the colors on all twelves sides of one can be made to coincide with identical colors on all twelves sides of the other by suitable rigid rotations of either pattern as a whole without disturbing the arrangement of its colors. If any four sides are painted red and the other eight blue, how many distinguishable color patterns can the dodecahedron have, including all twelves sides in each pattern?

[hide="answer"]12[/hide]


What are "bad boxes" and how does one avoid them? I always get dozens of them per document even when I have zero errors after compiling.

Assistance appreciated!

I cannot be 100% sure of what bad boxes are, but I have noticed that whenever I compile a document like this:
[code]
This is a line
\\
This is another line
[/code]
I get a bad box. It seems to me that an empty line makes a bad box. You can fix this by changing the code to:
[code]
This is a line\\

This is another line
[/code]
However, I have gotten into the habit of using the first method :), and my documents come out fine.

If you want extra space between your paragraphs automatically, set the \parskip in the preamble.  Also, do a search for "bad box" on this forum to find what the ever-useful stevem has said about them.  (And why, in fact, would you ever post a question in this forum without doing a search for the key word, I wonder?)


Here is a simple, but very beautiful problem: a polynomial of degree 5 and rational coefficients has a multiple root. Prove that it has a rational root. Are there any larger numbers than 5 with this property?

Since any rational polynomial with multiple roots is reducible, if we assume that our polynomial has no linear rational factors, it means that it's of the form $(x-\alpha)(x-\bar\alpha)\cdot (x-\alpha)(x-\bar\alpha)(x-\beta)$, where $\alpha$ is not real, and $(x-\alpha)(x-\bar\alpha)$ is rational. This means that $\beta$ is rational, contradiction.

As for the second part, we can simply take polynomials $P=(x-\alpha)^2(x-\bar\alpha)^2Q$, where $\alpha$ is not real, $Q$ has no real roots (and has degree at least $2$), and both $(x-\alpha)(x-\bar\alpha)$ and $Q$ have rational coefficients.


trigonometry
function
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Hi, have problem with these questions to.

solve these equations:
π =pi
1. cos (π/4-2x)=1/(2^1/2)

2. sin(π /6-3x)=-1/2

Some different task now:

given that cos x = 2/(5^1/2) and that 3π/2 < x < 2π. solve the exact functionvalue for

1. sin(3x-3π/2)

2. cos (4x-7π/4)

antoher task:

solve the equation:

1. 4 cos^2(x) + 8sinx + 1 = 0
2. 4sin^2(x) - 8 cosx + 1 = 0

[hide="HINT the first two"]
$\frac{1}{\sqrt{2}}$ and $-\frac{1}{2}$ both yeild a special triangle angle when in the $\cos^{-1}$ and $\sin^{-1}$ functions.
[/hide]
[hide="HINT the last two"]
Use the Pythagorean Identities, especially $cos^{2}x+sin^{2}x = 1$, to get to one trigonometric function in the equation.
[/hide]

thanks for the help but i still cant solve the problem. could u be more specific?

$\LaTeX$ed:
solve these equations:
1. $\cos{(\frac{\pi}{4-2x})}=\frac{1}{\sqrt{2}}$
2. $\sin{(\frac{\pi}{6-3x})}=\frac{-1}{2}$

given that $\cos{x}=\frac{2}{\sqrt{5}}$ and that $\frac{3\pi}{2}< x < 2\pi$. solve the exact functionvalue for
3. $\sin{(3x-\frac{3\pi}{2})}$
4. $\cos{(4x-\frac{7\pi}{4})}$

solve the equations:
5.$4\cos^{2}(x)+8\sin{x}+1 = 0$
6.$4\sin^{2}(x)-8 \cos{x}+1 = 0$

[hide="1 and 2"]
AstroPhys means:
Do $\cos^{-1}$ on both sides for #1.
The rhs will be a function of a special right triangle.

Do $\sin^{-1}$ on both sides for #2.
Again, the rhs will be a function of a special right triangle.
[/hide]

thanks for the answers on 3 and 5 but still got problem with 4 and 6

Elaborating on my previous hints for the last two
[hide="HINTS"]
$\cos^{2}x = 1-\sin^{2}x$
$\sin^{2}x = 1-\cos^{2}x$
[/hide]

[hide="Solution to the last two"]
1.
So we have $\cos^{2}x = 1-\sin^{2}x$.
Make that substitution into number one, and for ease of typing, I'll call $\sin x = a$.
Now we have
$4(1-a^{2})+8a+1 =0$
Solve that for $a$ with the quadratic formula, get that
$-4a^{2}+8a+5=0$
$a = 2.5$ or $a=-0.5$

Now we have $\sin x = 2.5$ or $\sin x =-0.5$
And from special triangles, we know that this only works for $x = 0.5$, which is $\boxed{\frac{-\pi}{6}}$

The other solution follows by the same pattern, except you substitute so that it's a quadratic in cosine instead of sine.

[/hide]

[quote="DiscreetFourierTransform"]
Now we have $\sin x = 2.5$ or $\sin x =-0.5$
And from special triangles, we know that this only works for $x = 0.5$, which is $\boxed{\frac{-\pi}{6}}$
[/quote]

Actually, the range of $\sin x$ is $[-1,1]$, so you don't need special triangles.

Sorry. I combined two statements into one.
1. We know this only works for $x = 0.5$.
2. From special triangles, this is $\frac{-\pi}{6}$

They don't imply each other.

And it just so happens that $\sin^{-1}2.5 = \frac{\pi}{2}-\frac{\ln\left(\frac{5\sqrt{21}+23}{2}\right)}{2}\cdot i$

Here we go:
Given that $\cos{x}=\frac{2}{\sqrt{5}}$ and that $\frac{3\pi}{2}< x < 2\pi$. Solve the exact functionvalue for:
4. $\cos{(4x-\frac{7\pi}{4})}$

Note that:

$\frac{3\pi}{2}< x < 2\pi \Rightarrow \sin(x) =-\sqrt{1-cos^{2}(x)}$ (because sine is negative in third and fourth quadrant)

$\cos{x}=\frac{2}{\sqrt{5}}\Rightarrow \sin^{2}(x) = \frac{1}{5}$

$\cos(\frac{7\pi}{4}) = \frac{\sqrt{2}}{2}$
$\sin(\frac{7\pi}{4}) = \frac{-\sqrt{2}}{2}$

$\cos(3x)$
$= \cos(2x+x)$
$= \cos(2x)\cos(x)-\sin(2x)\sin(x)$
$= (2\cos{^{2}}(x)-1)\cos(x)-2\sin(x)\cos(x)\sin(x)$
$= (2\cos{^{2}}(x)-1)\cos(x)-2\cos(x)(1-cos^{2}(x))$
$= 4\cos^{3}(x)-3\cos(x)$

$\sin(3x) = \sin(2x+x) = \sin(2x)\cos(x)+\sin(x)\cos(2x)$
$= 2\sin(x)\cos(x)\cos(x)+\sin(x)(2cos^{2}(x)-1)$
$= 2\sin(x)\cos^{2}(x)+\sin(x)2\cos^{2}(x)-\sin(x)$
$= 4\sin(x)\cos^{2}(x)-\sin(x) = (4\cos^{2}(x)-1)\sin(x)$

Now we have:
$\cos{(4x-\frac{7\pi}{4})}$
$= \cos(4x)\cos(\frac{7\pi}{4})+\sin(4x)\sin(\frac{7\pi}{4})$
$= \frac{\sqrt{2}}{2}[\cos(4x)-\sin(4x)]$
$= \frac{\sqrt{2}}{2}[\cos(3x+x)-\sin(3x+x)]$
$= \frac{\sqrt{2}}{2}[(\cos(3x)\cos(x)-\sin(3x)\sin(x))-(\sin(3x)\cos(x)+\cos(3x)\sin(x))]$
$= \frac{\sqrt{2}}{2}[((4\cos^{3}x-3\cosh)\cosh-\sinh(4\cos^{2}x-1)\sinh)-((4\cos^{2}x-1)\sinh\cosh+(4\cos^{3}x-3\cosh)\sinh)]$
$= \frac{\sqrt{2}}{2}[((4\cos^{4}x-3\cos^{2}x)-\sin^{2}x(4\cos^{2}x-1))-(-\sqrt{1-cos^{2}x}\cosh(4\cos^{2}x-1)-\sqrt{1-cos^{2}x}(4\cos^{3}x-3\cosh))]$

From this line you should plug in the value for cosine given at the begining of the post. I might have made some errors so check each line and ask to make sure you undersand.

some small errors but overall good (so small that even i noticed them), thanks to everybody for the help...


First:
$ x^{3}\plus{}2x\plus{}1\equiv x\plus{}2x\plus{}1 \equiv 1 \ (mod2)$ and $ RHS\equiv (\minus{}1)^{n} \ (mod2)$
thus $ n\equal{}2k$
$ x^{3}\plus{}2x\plus{}1\equal{}4^{k}$x must be odd, so
$ x\equiv 1 \vee 3 \ (mod4)$
Now we check that $ k\equal{}1$ is a solution and for $ k\geq 2$ we have:
$ RHS\equiv 0 \ (mod16)$ so we must have $ x^{3}\plus{}2x\plus{}1\equiv 0 \ (mod16)$
which is impossible (just make a table mod16) thus we have no roots for $ k\geq 2$.
So the unique solution is $ x\equal{}1 \ n\equal{}2$ :wink:

[quote="polskimisiek"]$ x^{3} \plus{} 2x \plus{} 1\equiv 0 \ (mod16)$
which is impossible (just make a table mod16) [/quote]
Check $ x\equiv 13 \mod 16$.


Να εξεταστούν ως προς την ομοιόμορφη σύγκλιση οι παρακάτω σειρές

1.  $ \sum_{k\equal{}0}^{\infty}(1\minus{}x^{2})x^{3k}$      για      $ \minus{}1<x<1$

2.  $ \sum_{k\equal{}0}^{\infty}(1\minus{}x^{2})x^{3k}$      για   $ \minus{}1\plus{}d\leq{x}\leq1\minus{}d$ με $ 0<d<1$

[quote="spinos"]Να εξεταστούν ως προς την ομοιόμορφη σύγκλιση οι παρακάτω σειρές

1.  $ \sum_{k \equal{} 0}^{\infty}(1 \minus{} x^{2})x^{3k}$      για      $ \minus{} 1 < x < 1$

2.  $ \sum_{k \equal{} 0}^{\infty}(1 \minus{} x^{2})x^{3k}$      για   $ \minus{} 1 \plus{} d\leq{x}\leq1 \minus{} d$ με $ 0 < d < 1$[/quote]

Αρκεί να θυμηθείς την ακτίνα σύγκλισης της γεωμετρικής σειράς.

Cheerio,

Durandal 1707


[quote="m4467"]$ (a \plus{} 1)(b \plus{} 1)(c \plus{} 1) \equal{} 2abc$[/quote]
It is equavalent to $ (1\plus{}\frac 1a)(1\plus{}\frac 1b)(1\plus{}\frac 1c)\equal{}2$. Let $ a\le b\le c$, then $ a\equal{}2$ or $ a\equal{}3$.
1. $ a\equal{}2$ give $ (b\minus{}3)(c\minus{}3)\equal{}12$ solutions are $ b\equal{}4,c\equal{}15$, $ b\equal{}5,c\equal{}9,$ $ b\equal{}6,c\equal{}7$.
2. $ a\equal{}3$ give $ (b\minus{}2)(c\minus{}2)\equal{}6$, solutions are $ b\equal{}3,c\equal{}8$, $ b\equal{}4,c\equal{}5$.


vector
calculus
derivative
abstract algebra
geometry
geometric transformation
group theory

$ H$ is sub lie group of lie group $ G$. $ g_{e}$ is lie algebra of $ G$ and $ h_{e}$ is lie algebra of $ H$.
$ g_{e}=h_{e}+m$, where $ m$ is some predetermined supplent of $ he$ in (finite dimentional) vector space $ g_{e}$.
So $ m$ can be canonically identified as $ g_{e}/h_{e}$. ($ m$ is determined by theory of symmetric homogeneous space).

So $ G$ can be considered as a principal bundle over $ G/H$ with structure group $ H$, with $ \pi: G \to  G/H$
as the canonical projection. The tangent linear mapping (i.e derivative) of $ \pi$ at $ e$ is $ T(\pi): T_{e}(G) \to  T_{\pi (e)}(G/H)$.
The kernel of this mapping is $ he$. So $ T_{\pi (e)}(G/H)$ can be canonicaly identified as $ g_{e}/h_{e}$.

For any $ s, x \in G$,  consider left translation $ x \mapsto s \bullet x$. The derivative of this mapping is $ u \mapsto s \bullet u$,
a diffeomorphism from $ g_{e}$ onto $ T_{s}(G)$.
Similairly, $ Ad(s) \bullet u=s \bullet u \bullet s^{-1}$ is a automorphism of $ g_{e}$.

For any $ t \in H$, $ u \in g_{e}$, clearly we have $ t \bullet u \in T_{t}(G)$. So $ T(\pi) (t \bullet u) \in T_{\pi (t)}(G/H)=T_{\pi (e)}(G/H)=g_{e}/h_{e}=m$.

And consider $ t \in H$, $ u \in m \subset g_{e}$, $ Ad(t) \bullet u$  also $ \in m$ by theory of
symmetric homogeneous space.

Now how can I show:

$ T(\pi) (t \bullet u) =Ad(t) \bullet u$ in $ m$?

Clearly $ T(\pi) (t \bullet u) =T(\pi) (t \bullet u \bullet t^{-1}\bullet t)$

But $ \pi$ and $ \bullet t$ is commutative for any $ t \in H$, (right translation by $ t$ doesn't change
left cosets of $ H$ in $ G$).

so $ T(\pi) (t \bullet u)=(T(\pi) (t \bullet u \bullet t^{-1})) \bullet t =T(\pi) (t \bullet u \bullet t^{-1})$
since $ \bullet t = 1_{G/H}$.

Now  by above mentioned indentification (The kernel of $ T(\pi)$ is $ he$ so $ m$ can be canonically identified as $ g_{e}/h_{e}$), we have $ T(\pi) = 1_{m}$.
And the fact for any $ u \in m$, $ t \bullet u \bullet t^{-1}=Ad(t) \bullet u \in m$:

 $ T(\pi) (Ad(t) \bullet u)= (Ad(t) \bullet u)$.


How many number can be created it from this set $ x \equal{} \left\{ {0,1,2,3,4,5,6,7,8,9} \right\}$ that is composed from $ 3$ even digits and its tens field (I mean 42[color=red]1[/color]0 by tens) is odd. And the numbers must be less than $ 5234$.
Have Fun! :surf:

Hello everyone
  We have $ 5$ even numbers (considerating $ 0$ as even) and $ 5$ odd numbers ,but the number should be at most $ 5234$ , so the number of numbers  :lol:
is $ 2\cdot 5 \cdot 5 \cdot 5 \cdot 5 \equal{}1250$

[quote="enndb0x"]Hello everyone
  We have $ 5$ even numbers (considerating $ 0$ as even) and $ 5$ odd numbers ,but the number should be at most $ 5234$ , so the number of numbers  :lol:
is $ 2\cdot 5 \cdot 5 \cdot 5 \cdot 5 \equal{} 1250$[/quote]

close, but you forget combinations like 5210. Not only that, your variation is for 5 digit numbers. these must be 4 digit numbers ($ <5234$) with 4 even digits, but we had a restriction taht the tens digit must be odd. this cannot be true, and we have a contradiction. therefore, we have, in actuality, a grand, whopping total of....

[hide="click here to reveal monster #"]$ \boxed{0}$ numbers[/hide]

So great solution.  :rotfl:
Is the best way here to write them and count it.  :lol:

[quote="tomwolf2008"]Although if you really count it you'll get 0[/quote]
 :huh:
What do you mean?
I may have for example $ 4652$  :|

[quote="Bugi"]You said it has 4 even digits, and the tens digit is odd, so it has at least 5 digits, contradiciton.[/quote]
It was a typo its $ 3$ even digits plus the tens digit is odd.

[quote="Bugi"]First digit: 2 ways
Second digit: 5 ways
Third digit: 5 ways
Fourth digit: 5 ways
250?[/quote]
I don't know  :(


1. Determine all pairs (x, y) of integers such that

2^x + 2^(2x+1) + 1 = y^2.

You means that

$ 2^x\plus{}2^{2x\plus{}1}\plus{}1\equal{}y^2$

Just for the beauty of LaTeX... :maybe:

http://www.mathlinks.ro/viewtopic.php?p=572815#572815

Additionally, it is a number theory problem.

http://www.mathlinks.ro/Forum/viewtopic.php?p=572815#572815

This is an IMO problem! Please try searching before posting, especially when you are posting well known problems

Daniel

[quote="mstoenescu"]You means that

$ 2^x \plus{} 2^{2x \plus{} 1} \plus{} 1 \equal{} y^2$

Just for the beauty of LaTeX... :maybe:[/quote]

Yes , this was an IMO problem(I think was at 2006)

$ 2^x \plus{} 2^{2x \plus{} 1} \plus{} 1 \equal{} y^2$  we add $ \minus{} 1$ both sides and we have : $ 2^x(1 \plus{} 2^{x \plus{} 1}) \equal{} (y \minus{} 1)(y \plus{} 1)$
 cause $ (y \plus{} 1) \minus{} (y \minus{} 1) \equal{} 2\implies(1 \plus{} 2^{x \plus{} 1}) \minus{} 2^x \equal{} 2\implies 2^x \equal{} 1\implies x \equal{} 0$ and we easy can find $ y \equal{} 2$ and $ y \equal{} \minus{} 2$

[i]Thanks god, we can edit LaTex [/i]


Now the dollar signs for currencies can be made simpler on the board. Use a backslash to "escape" the dollar sign.

So

[code]\$1,000[/code]

will generate \$1,000

The \\$ still works in the regular $ \LaTeX$ code as expected.

Great! It would be also nice to update the "odd number of dollar signs" error message accordingly (it still contains the recommendation to use $\$ $ for the currency sign)

I like it. My trial run on the test forum has drawn a couple of responses.

Maybe change the following message a little?

[quote]
You have an odd number of LaTeX tags $ in your document and BBCode is enabled. You might have forgotten to close a tag of a LaTeX formula, so please check your document. If you want to use the symbol $ to represent currency (dollars) either disable BBCode (not recommended) or
use the code $\$$ to create a image of the dollar sign (so that your post can be safely quoted afterwards).

Remember that it's best to use the Preview function so you can spot mistakes in your LaTeX syntax.
[/quote]

[quote="fedja"]Great! It would be also nice to update the "odd number of dollar signs" error message accordingly (it still contains the recommendation to use $ \$$ for the currency sign)[/quote]I missed this :) You have a point, that message is no longer useful in its current form.

How did fedja write \$\\$ \$ when quoting his post gives $ \$$? :maybe:

With BBCode disabled (this is a bad option because of the quoting problem, but if you post pure text and don't expect a mathematical discussion in the thread, you can use it too)

Dojo, use the [code tag to show code, don't disable BBCode. As a matter of fact in the very near future, the option of disabling BBCode won't be available anymore anyway.


function
integration
trigonometry
calculus
limit
real analysis
real analysis unsolved

I need to evaluate asymptotics for the following function $ f(t) \equal{}\int_{1}^{\infty}\frac{1\minus{}\cos{tx}}{x^{1\plus{}\alpha}}dx$, where $ 0 <\alpha < 2$, more precisely i want to have $ f(t) \equal{}ct^{\alpha}\plus{}O(t^{2})$. As i understand, we can not differentiate here, so i tried to decompose integral into 2 with limits depending on t, but it didnt help.

One thing jumps out immediately. Taking the limit under the integral (justified by the Dominated Convergence Theorem), we see that $ \lim_{t\to0}f(t)\equal{}0.$ Hence that leading term of $ 1$ that you wrote down cannot be correct.

Try this: make the substitution $ u\equal{}tx.$ We get that

$ f(t)\equal{}t^{\alpha}\int_{t}^{\infty}\frac{1\minus{}\cos u}{u^{1\plus{}\alpha}}\,du.$

Now, let $ A\equal{}\int_{0}^{\infty}\frac{1\minus{}\cos u}{u^{1\plus{}\alpha}}\,du.$ (Whatever that number turns out to be.) We have

$ f(t)\equal{}At^{\alpha}\minus{}t^{\alpha}\int_{0}^{t}\frac{1\minus{}\cos u}{u^{1\plus{}\alpha}}\,du.$

For the latter integral, one should be able to productively expand the integrand as a power series in $ u.$

[b]Kent Merryfield[/b], thanks for the helpful advice, it works.


If you take a random 4 digit number, what is probability that all of its digits are even?

Well, digits are : 0,1,2,3,4,5,6,7,8,9 -> 5 even numbers (including 0)
Consequently, for the first digit : 5/10=1/2, same goes for the three other digits, so, that makes :
$\frac {1}{2^4}$

:)

[hide]
there are 5 possible digits for all the digits of the number
there are 10 possible digits
5/10=1/2
$\left(\frac{1}{2}\right)^4=\frac{1}{16}$[/hide]

[quote="1234567890"]There is a problem to above solutions:
Can 0 be the first digit?[/quote]

If 0 was the first digit, then it wouldn't be a 4 digit number...  ;)

[hide]
There are 9 choices for the first digit, 10 for the second, 10 for the third, and 10 for the fourth.
There are 4 even digits possible for the first digit, and 5 for the second, third, and fourth.
$\frac{4\cdot5\cdot5\cdot5}{9\cdot10\cdot10\cdot10}=\frac{4}{72}=\boxed{\frac{1}{18}}$[/hide]

[hide]
ooops, okay so since zero can not be the first digit then there are 4 choices for the first and 5 for the rest.  There are 9 total choices for the first and 10 for the rest. $\frac{4 \cdot 5^3}{9 \cdot 10^3}=\frac{1}{18}$[/hide]


linear algebra
matrix
vector
linear algebra unsolved

Hello, I'm stuck with the following problem:

Let A be the matrix [1, 1 ; 0,1] and consider the following
differential equation:

dy/ dt = A*y

Now here's the problem:

By considering the solution of dy/dt = A*y
Show the matrix A is not diagonalizable.

Ok so far I found the solution is given by y(t)=C_1* e^t + C_2* t*e^(t)

where C_1, C_2 are constants vectors in R^2.

Then I have no clue how to proceed, any ideas?

Try to solve a system $\frac{dY}{dt}=MY$ where $M$ is a diagonal matrix. You will easily see that solutions are not of the form of the one you have found (no term with $te^t$).

Hi alekk , thanks for your reply. That's what I tried but got stuck, didn't know where to use the fact that M is a diagonal matrix.
Care to provide another hint ?

Thanks in advance

weird that you show  that it is not diagonisable by differential equations, usually you work the other way around

i would just say, if it diagonisable, it would have $A=C^{-1}DC$ with C invertible
this however would imply that D is identity and thus $A=C^{-1} C=I$

and that is impossible, A is not identity


First of all, anthranilic acid exists as a zwitter ion, so you cannot basically apply the ortho effect here.
Secondly in o-anthranilic acid, the H attached to N(with + charge), is hydrogen bonded to the oxgen of COO-, which makes it a weaker acid, compared to para isomer where there is no intramolecular H bonding.
 :D

heve a look at [url=http://www.mathlinks.ro/viewtopic.php?t=181815]this[/url] interesting discussion building on between carcul and banishedtraitor


[quote="alex2008"]Let be $ x,y,z\ge 0$ such that $ \frac {4}{x} \plus{} \frac {3}{y} \plus{} \frac {2}{z} \equal{} 6$ . Show that :

$ 2xy \plus{} 3zx \plus{} 4yz \plus{} 5xyz\ge 33$[/quote]
Yes it is easy
[hide="solution"]
$ 2xy\plus{}3xz\plus{}4yz\equal{}6xyz$
So we have to prove that:
$ xyz \ge 3$.
From AM-GM $ \frac {4}{x} \plus{} \frac {3}{y} \plus{} \frac {2}{z} \ge 3\sqrt[3]{\frac{24}{xyz}} \Longleftrightarrow xyz \ge 3$.
[/hide]


Princeton
college
real analysis
complex analysis
IMC
college contests

And where is it. Let's start the topic now :)

Blagoevgrad, Bulgaria according to the website.

Princeton sends a team again. We haven't decided the people yet. Possibly I will go again. :)

[quote="DIMITRIS T"]What math knowleges I must have for this competition?
Is it possible for a freshman to have succes in this competition?
can  you give me any related links?

Thanks in advance :)[/quote]
I will come to IMC 2008 !! I am from Airlangga University, SUrabaya, Indonesia.
The materials you need for IMC 2008 is Real Analysis, Complex Analysis, COmbinatorics and Abstract Algebra  :lol:

The material you mentioned above is for preparing in IMC in general not only this year's one right???????

I might come to visit for a couple of days, after the IMO. I am not sure yet though ... it's kind of far from Bucharest.

Princeton team is basically decided now.

Returning:

Simion Filip
Aleksandar Lishkov
Marius Pachitariu
Nathan Savir (:D)

New members:

Adam Hesterberg
Quoc Ho
Eric Kang
Charles Li
Anton Malyshev
Lily Shen

As usual, although Princeton is (one of?) the only American university attending, our team is actually representing a variety of countries!
Simion: Moldova/Romania
Alex: Bulgaria
Marius: Romania
Me: United Kingdom
Adam: USA
Quoc: Vietnam
Eric: New Zealand
Charles: Australia
Anton: USA
Lily: China

I hope to see some other Mathlinkers again at the IMC this year! Does anyone want to go to Greece with me during grading?  :wink:

I am in the Sharif University team, but maybe I can not come to Bulgaria at that time.

Here is the current team of Sharif University of Technology:

Omid Hatami
Naser Talebizadeh
Jalil Kazemitabar
Behzad Mehrdad
Jaber Zarezadeh
Nima Ahmadipour anari
Mostafa Eynollah zadeh Samadi
Javad Moghadamzadeh

[quote="Mr.Doe"]I'm coming!!!! :D  :D  :D  I'm so happy, this is my first IMO![/quote]
but the topic is about IMC! :)

[quote="SimonM"]Does anyone know which UK universities send teams in?

(Not that it will be a major factor affecting my choice of university)[/quote]IMO I'm telling you this [b]should NOT[/b] be your major factor affecting the choice of your university. The contest is nothing compared to the IMO (both in terms of organization and the medal/prizes worthiness.) You can always participate like an individual.

Yeah, I said it wouldn't be a factor affecting it, I was just curious to see if any did (I'm fairly decided where I want to go anyway)

Oh, I read "Note that it will be" instead of "Not that it will be" :D

Team of Kyiv Taras Shevchenko National University:

Ivan FESHCHENKO  (gold medal at SEEMOUS-2008, 1st prize at IMC-2007)
Sergii SLOBODIANIUK  (silver medal at Iran ISMO-2008, 1st prize at IMC-2007, 2nd prize at IMC-2006, gold medal at IMO-2005)
Oleksandr KRAVETS  (silver medal at Iran ISMO-2008, 1st prize at IMC-2007, 2nd prize at IMC-2006, bronze medal at IMO-2005)
Maksym TANTSIURA (silver medal at SEEMOUS-2008)
Ivan IURCHENKO  (gold medal at SEEMOUS-2008, 1st prize at IMC-2007)
Danylo RADCHENKO (gold medal at SEEMOUS-2008, gold medal at IMO-2007, bronze medal at IMO-2006)
Oleksii KLURMAN  (3rd prize at IMC-2006, 2nd prize at IMC-2005)
Larysa KHOMENKO  (bronze medal at Iran ISMO-2008, 2nd prize at IMC-2007, 2nd prize at IMC-2006)
Taras TYMOSHKEVYCH  (3rd prize at IMC-2007, 1st prize at IMC-2006, 2nd prize at IMC-2005)
Ivan LIVINSKYI  (2nd prize at IMC-2007, 2nd prize at IMC-2006)
Andriy ARMAN (silver medal at SEEMOUS-2008)

Team Leader: Dr. Andriy BONDARENKO

There should be also team from Kyiv Polytechnic Institute (1 student) and team from Odessa I.I.Mechnikov National University (3 students, 1 team leader).

[quote="Xevarion"]Princeton team is basically decided now. [/quote]

A few changes due to visa problems...

Simion Filip
Aleksandar Lishkov (hopefully)
Marius Pachitariu
Nathan Savir
Adam Hesterberg
Eric Kang
Anton Malyshev
Lily Shen
Peter Diao

We are lucky to have Peter Varju for our team leader. Some of you may know him from past IMCs (that's when I first met him).

I'm coming :) Jagiellonian University Team:

Krzysiek Dorobisz
Maciek Gawron
Tomek Kobos
Przemyslaw Mazur
Sylwester Zając

I'm not coming back, I've had it with blagoevgrad, twice was enough. What's the point of international competitions if they're always on the same place?

[quote="dienbk"]Hi all
 I'm a new member of 4rum and I don't know where to find IMC 2008 problems and solutions. Have you ever post them in 4rum and the link to download them. Thanks[/quote]

http://www.imc-math.org.uk/index.php?year=2007&item=problems
http://www.imc-math.org.uk/index.php?year=2006&item=problems
http://www.imc-math.org.uk/index.php?year=2005&item=problems
etc.  :wink:

Please, can you post the results of IMC 2008?

They're announcing the final results right now at the closing ceremony, which I was warned is boring enough to warrant skipping. They should be up on http://imc-math.org/ shortly, I suppose.

I know the website but in the previous years they posted them very late, almost a week after the closing ceremony. Please, may be they give them by e-mail or something like that and you can put them here.

[quote="AdrianDD"]Mai sa fie, din cata lume a fost acolo, chiar nu are nimeni rezultatele?[/quote]

English ?

Do you have the results of the IMC2008? There were so many people there, I can't belive nobody has them.

I know that all of our team members except one got First Prize (I was not that one).

Great. Congratulations! An the other participants?

I know some information about MSU team:
1. Alexander Efimov 220  :first:
4. Vasily Astakhov 213
Omid, what you know about prize cut-offs and how many points participants of your team have?

Yeah, I saw Alexander after the second day's exam. He seemed very happy. He told me he solved 5 problems in first day and all of the problems of second day. Nice job.  :)

I don't know really, because I was in Bulgaria only for two days of exam. They told me the result after that.

I did not do well at the exam though. In the first day I misunderstood one of the easy problems in the exam, and in the second day I missed two.

Yes but how can  we find out about the results?


inequalities
induction
inequalities solved

I saw this before but I don't remember where exactly ...


Use the Cauchy-Schwarz inequality to solve the following max/min problem: If the (long) diagonal of a rectangular box has length c, what is the greatest sum of the length, width, and height of the box can be?  For what box shape does the maximum occur?

[hide] Let $ u\equal{}(x,y,z)$ be the vector of lenght, width and height of the box. By Cauchy-Schwarz inequality, taking $ v\equal{}(1,1,1)$ you have
$ |(u,v)| \leq \sqrt{(u,u)(v,v)}$ where $ (u,v)$ represent the inner product of $ u$ and $ v$, and in this case it become $ x\plus{}y\plus{}z \leq \sqrt{3}\sqrt{x^2\plus{}y^2\plus{}z^2}\equal{}\sqrt{3}c$. Equality holds when $ x\equal{}y\equal{}z\equal{}\frac{1}{\sqrt{3}}c$.[/hide]


well friends,
the problem is -
"In a list of 200 numbers, everyone(except the end ones)is equal to the sum of the two adjacent number in the list.The sum of all the numbers is equal to the sum of the first 100 of them.find that sum if the 35th number in the list is 6."

i got the sum of all the numbers in the list as 9.is it correct? :?:  :?:
i can post a solution if needed :D

feel free to post ur solution..... :D

See here[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=236132[/url]

Go to post number #15 for an excellent solution by miles.


find the product of the solutions of

\[ x^{%Error. "logx" is a bad command.
}=\frac{1000}{x}\]

i got it down to $ %Error. "logx" is a bad command.
+%Error. "logx" is a bad command.
(%Error. "logx" is a bad command.
)=3$
but now im stuck (assuming the above is right in the first place)

Eh, what does this have to do with logs? Did you typo in typing the problem or something?

It should be $ x^{\log x} \equal{} \frac {1000}x$.

[b]DaReaper:[/b] You essentially got it down to $ (\log x)^2 \plus{} \log x \minus{} 3 \equal{} 0$. It's a quadratic, so factor it.
[b]
Edit:[/b] Sorry, ignore this post.

yea it was a massive typo, but math 154 is exactly got to exactly where i was stuck

so
\[ \log x \equal{} \frac { \minus{} 1\pm\sqrt {13}}{2}
\]
?

then what?

My apologies, I blanked out and thought it could be factored nicely.

You can note that $ \log x \equal{} \frac { \minus{} 1\pm\sqrt {13}}2\implies x \equal{} 10^{\frac { \minus{} 1\pm\sqrt {13}}2}$. Let $ x_1 \equal{} 10^{\frac { \minus{} 1 \plus{} \sqrt {13}}2}$ and $ x_2 \equal{} 10^{\frac { \minus{} 1 \minus{} \sqrt {13}}2}$. What is their product?

You can also skip this step by noting that the sum of the roots of the quadratic is $ \minus{} 1$ by Vieta's Formulas. Try this approach as well.

Or...
[hide="Better"]
Get to $ (\log x)^2\plus{}\log x\minus{}3\equal{}0$. From Vieta's, the sum of the roots is $ \minus{}1$.
Then the product of the roots of $ x$ is $ 10^{\minus{}1}\equal{}\boxed{\frac{1}{10}}$.
[/hide]

Edit: Oh, didn't see that last line of math154's post.


1) A man walks from Dublin to Cork in Ireland. He doesn't pass a single pub (pubs, or bars, are very common in Ireland). How does he manage this?

2) at a party, the king suddenly killed the queen. However, despite people knowing that it happened, no-one really cared, and no-one payed much attention. Why not?

3) A man takes a vacation with his family. even though he always took the elevator down to the lobby, he always took the stairs when he went back to the room. why didn't he take the elevator at night, when he came back?

[quote="SoxFan34"]2) at a party, the king suddenly killed the queen. However, despite people knowing that it happened, no-one really cared, and no-one payed much attention. Why not?[/quote]
[hide]maybe they were cards or chess pieces.  not an actual king and queen.[/hide]

[hide="#3"]This one is pretty common.  :)  The man is very short, so he can reach the low numbered buttons (such as the lobby), so he can take the elevator down in the morning. However, he can't reach the button for his own room, so he has to take the stairs in the evening.[/hide]
Wow, #1 seems pretty hard.  :maybe:

For #1, maybe he plans his route specially, or maybe every single pub in Ireland was blown up by terrorists.

Nebula42, you are WRONG!

So are you, mustafa. The IRA protects their Guiness very much.

However, perfect number is right ; it's a chess game

[hide="#1"]the guy goes into every bar so he doent [b]pass[/b] any of them[/hide]

wow u must be desperate for replies :rotfl:  :rotfl:  :rotfl:

[hide]he walked through the sewers[/hide]
That's for number one I believe

[hide="good guess?"]He walked through the woods.[/hide]

[hide="guess #2"]Since pubs are common in Ireland, each one should be on every map. He just avoids the streets that have pubs.[/hide]

[hide="#1"]

He passes many pubs - not just a single pub.

[/hide]


If $a,b,c > 0$ then show that:

$\sum \left( \frac{a}{a+b}+\frac{2\sqrt{a}}{\sqrt{a}+\sqrt{c}}\right) \geq \frac{9}{2}$

Be careful: it's not $\frac{a}{b+c}$ as one might expect, but $\frac{a}{a+b}$ instead.

[quote="hendrata01"]If $a,b,c > 0$ then show that:

$\sum \left( \frac{a}{a+b}+\frac{2\sqrt{a}}{\sqrt{a}+\sqrt{c}}\right) \geq \frac{9}{2}$
[/quote]
I think, it's wrong. Try $a=1,$ $b=\epsilon$ and $c=\epsilon^{2},$ where $\epsilon\rightarrow0^{+}.$ :wink:

I'm sorry about that, I seemed to miss one important condition in the problem.

So this is the original problem:
If $x,y,z > 1$, prove that
$\sum \frac{1}{1+x}+2\sum \frac{1}{1+\sqrt{xy}}\geq \frac{3}{2}\sum \frac{1}{1+\sqrt[3]{x^{2}y}}$
(So the last term is a symmetric summation, it has 6 terms in it).

This is the explanation of the mistake:
In order to "mask" the trick I used to create the problem, I substituted $x = a/b, y = b/c, z = c/a$ and voila, you get my first problem. But apparently this inequality may not hold if $xyz=1$, or if any of the $x,y,z$ is less than one.

Again sorry about that. This new problem should be much easier.


quadratics
modular arithmetic
number theory theorems
number theory

can you help me prove this?

 A perferct square is always a quadratic residue of an odd prime

thank you :)

Suppose this is not true for an $ a^2<p$ where $ p>2$ is a prime and $ a\in \mathbb{N}$.

Then, take $ m>a$ such that $ p\vert m\minus{}a$.

You get $ m^2\equiv a^2 \pmod{p}$, which is true because $ m\equiv a \pmod{p}$.


Compute the smallest prime number $ p$ such that $ (p^3 \plus{} 2)(p^2 \plus{} p)$ has exactly $ 42$ positive factors.

I don't think it is possible. We have p(p+1)(p^3+2), all three factors which are relatively prime. Thus since the number has 42 factors, it is of the form $ a^6\cdot b^2\cdot c$, where a,b,and c are prime. So p=c. p+1 then must equal $ a^6$ or $ b^2$ , so p must equal either (b+1)(b-1), where b would have to be 2 if it was prime, but this doesn't work out, or $ (a^3\plus{}1)(a^3\minus{}1)$, which isn't prime.