show that every integer can be represented as sum of 4 squares.

Every _nonnegative_ integer ;)

Steps for a solution:

[b]a)[/b] Show that if $x$ and $y$ are sums of four squares, so is $x \cdot y$.
[hide="Big hint"]\[(a^{2}+b^{2}+c^{2}+d^{2}) \cdot (e^{2}+f^{2}+g^{2}+h^{2}) = \\ = (ae-bf-cg-dh)^{2}+(ae+bf+ch-dg)^{2}+(ag-bh+ce+df)^{2}+(ah+bg-cf+de)^{2}\][/hide]
We thus only need to prove the theorem for odd primes $p$.
[b]
b)[/b] Let $p$ by an odd prime, show there are integers $x,y$ such that $x^{2}+y^{2}\equiv 1 \mod p$.
[hide="Hint"]Consider the sets $A= \left\{ x^{2}+1 \ | \ x= 0,1,...\frac{p-1}^{2}\right\}$ and $B= \left\{-y^{2}\ | \ y= 0,1,...\frac{p-1}^{2}\right\}$.[/hide]

[b]c)[/b] Let $k>0$ be minimal such that $kp$ is a sum of four squares. Show that $k<p$.
[hide="Hint"]Use b), in fact use that you can assume $|x| , |y| < \frac{p}{2}$[/hide]

[b]d)[/b] Show that this minimal $k$ (from b) ) is odd and assume it to be $>1$ from now (we want to get a contradiction).

[b]e)[/b] Let $kp = a^{2}+b^{2}+c^{2}+d^{2}$. Find $x \equiv a , \ y \equiv b , \ z \equiv c , \ w \equiv d \mod k$ such that $x^{2}+y^{2}+z^{2}+w^{2}= mk$ and $m<k$. As $1<k<p$, show that $m>0$, too.
[hide="Hint"]See the hint of c).[/hide]

[b]f)[/b] Write $pmk^{2}=(kp) \cdot (mk)=s^{2}+t^{2}+u^{2}+v^{2}$ as sum of four squares using the representations you already know and using the hint of a). Show that $k| s,t,u,v$ and conclude that $np$ is a sum of four squares.

[b]g)[/b] See that the results of e) and f) are contradictory, thus $p$ is a sum of four squares and we are finished.

PS: If you want a lot more to do, show that there are exactly \[\mathrm{d}_{4}(n) = 8 \cdot \sum_{ 4 \nmid k | n}k\] ways of writing $n$ as sum of four squares of integers (permutations and different signs are not counted only once).

here's a paper I had written a long time ago on this.
at the end the proof of Jacobi's theorem is incomplete, but the proof of this theorem, should be satisfying.
What Zetax said is basically the proof shown here. although i have seen other, more elegant, ways of proving this result; but this is the classic solution.

I have a question regarding th proof of this fact using hurwitz numbers. Why doest the proof using ussual quarternions doesnt work, i cant see it.

A geometric reason is that the lattice $\mathbb Z[i,j,k]=\{a+bi+cj+dk|a,b,c,d \in \mathbb Z\}$ is not 'dense' enough, i.e. there are points $z \in \mathbb Q[i,j,k]$ such that there is no $z^\prime  \in \mathbb Z[i,j,k]$ such that $|z-z^\prime|<1$; actually, the only exceptions are the remaing Hurwitz numbers, where we get $|z-z^\prime|=1$.
An algebraic reason is that $\mathbb Z[i,j,k]$ is not "integrally closed" (if adapted to quaternions), a property necessary for unique factorization and usefull even if not used directly.

I cannot give more insights as I don't know the specific proof, but I can take a look if you show/describe it.

There was a whole NYSML Power Round (2008) on representations of integers as a sum of squares. The last question was fun, and uses this result. Prove that if an integer is at least 170, then it can be expressed as the sum of five positive squares.

See here [url]http://www.jstor.org/stable/10.4169/0025570X.83.3.210[/url] a stronger result.


A positive integer is equal to the sum of the squares of its first four positive divisors. What is the largest prime which divides this positive integer?

By first four, do you mean starting from 1 and going up?

First note that the number cannot be odd (because the sum of $ 4$ odd squares is even). So it is even and the smallest divisors are $ 1,2$. Now the next smallest is either $ 4$ or a prime number $ p$. If it is $ 4$ then the next divisor must be odd. But if $ n$ is the next divisor and it is odd we have:

$ n^{2}\equiv 1 (mod 4)$ and therefore $ 1^{2}+2^{2}+4^{2}+n^{2}\equiv 2 (mod 4)$ which is a contradiction. So the next divisor must be a prime $ p$ and the one after that must be even so it is $ 2p$.

So our number is $ 1^{2}+2^{2}+p^{2}+(2p)^{2}= 5(1+p^{2})$. Since $ p$ does not divide $ p^{2}+1$ so $ p=5$ and our number is $ 130 = 1 \times 2 \times 5 \times 13$. So the answer is $ 13$.

This problem is equivalent to BMO 1989/1.


Let's consider conected graphs which can be obtained by hex-comb!
When this graph has perfect matching?
This graph is bipartite, and if we color vertices black-white, we must have equal number of white and black vertices! But, are there some conditions that ensures existance of perfect matching?

it has a perfect matching if it satisfies Halls condition


Euler
function
abstract algebra
group theory
induction
number theory

prove that any group $G$ that has $n$ elements and such that $n$ and $\phi(n)$ are coprime is cyclic.
($\phi(n)$ is the euler function )

Alekk, do you have another solution than the one from the article of Richard Antetomaso? I'd be glad to see one. Once I tried to find another solution, but didn't manage.

Is it a nasty solution? (please don't give any hints :)) I gave it some thought last night, but didn't manage to do much.. :(

I didn't find it easy at all. It is a difficult, but nice solution.

Gallian wrote a paper on the following result"If n & phi n are relatively prime then $Z_n$ is the only group of order n" The article is in American Mathematical Monthly.I forgot which edition.I ll check & tell you.

Am I allowed to use the Feit-Thompson Theorem, stating that a simple finite non-abelian group must have even order? :)

I'm just a bit reluctant when it comes to using theorems with proofs spanning something like $250$ pages, which I will never read or understand :). But, I guess there's no harm in posting a proof using that strong result, just until we find a decent proof :).

The order of the group is of the form $p_1\ldots p_k$, here $p_i$ are distinct and odd primes (they are odd if we assume $k>1$, which we do, to get the simple case $k=1$ out of the way). Then we use the mentioned theorem to conclude that the group is not simple, and thus has a non-trivial normal subgroup $H$ with $|H|=p_1\ldots p_t$ (after a renumbering of the primes) for some $1\le t<k$.

We can now use induction on $k$. By the induction hypothesis, $H$ is cyclic, and so is $G/H$. For an element $x\in G$, let $\bar x$ be its image through the projection from $G$ to $G/H$. Given elements $g_1,\ldots,g_k$ of orders $p_1,\ldots,p_k$ respectively with $g_1,\ldots,g_t\in H$, if we assume that $g_1^{a_1}\ldots g_k^{a_k}=g_1^{b_1}\ldots g_k^{b_k}\ (*)$ for $a_i,b_i\in\overline{0,p_i-1}$, then the same relation holds if we replace each $g_i$ with $\bar g_i$, and from here and the fact that $H,G/H$ are cyclic we easily deduce the fact that $a_i=b_i,\ \forall i$, meaning that $G$ consists precisely of the products of the form $(*)$, so, in particular, it's generated by the $g_i$'s. This means that all we need to do in order to show that $G$ is abelian is prove that $g_j,g_i$ commute, for all $j>t,i\le t$.

According to the above, fix $j>t,i\le t$. We have $g_jg_ig_j^{-1}=g_i^a,\ a\in\overline{1,p_i-1}$, so $g_i=g_j^{p_j}g_ig_j^{-p_j}=g_i^{a^{p_j}}$, so $g_i^{a^{p_j}-1}=1$, meaning that $p_i|a^{p_j}-1$, which is impossible unless $a=1$, given the fact that $p_j\not|\ p_i-1$ (implied by the hypothesis). We have thus obtained the fact that $g_i,g_j$ commute. Since this is true for any $j>t,i\le t$, it means, according to the paragraph above, that $G$ is abelian, and from here it's easy: with the notations above, $g_1\ldots g_k$ has order $p_1\ldots p_k=|G|$.

Unfortunately no & i dont have much hope in ever finding it out.

I think I managed to show by completely elementary means that $G$ cannot be simple, completing grobber's solution. Keeping in mind the induction hypothesis, all we need to do is to show that if $G$ is non-abelian and simple, then it must have a proper, non-cyclic subgroup (and thus cannot be our group). That is, it suffices to show that a minimal non-cyclic group cannot be simple. We'll do more than that: we'll show that a minimal non-abelian group cannot be simple!

[b]Claim. [/b] Let $G$ be a non-abelian, simple group. Then, $G$ has a proper, non-abelian subgroup. In other words, a minimal non-abelian group cannot be simple.

[i]Proof. [/i] Assume to the contrary that $G$ is simple, but all of its proper subgroups are abelian. Consider any two different maximal subgroups $U,V < G$. Clearly, $U$ is a subgroup of its normalizer $N_G(U)$; on the other hand, by the simplicity of $G$, $N_G(U)$ is properly contained in $G$ (else $U$ would be normal), and the maximality of $U$ implies that $U = N_G(U)$. Hence, every maximal subgroup of $G$ coincides with its normalizer.

Next, we show that $U \cap V = \langle 1 \rangle$, i.e. any two distinct maximal subgroups of $G$ have a trivial intersection. Assume to the contrary that $x \neq 1$ is an element of both $U$ and $V$. Since, by assumption, $U$ is abelian, one has $C_G(x) \supseteq U$, while the simplicity of $G$ implies that $G$ has a trivial center, and hence $C_G(x)$ is contained properly in $G$. As $U$ is maximal, we immediately get $C_G(x) = U$. Likewise, $C_G(x) = V$, which is already a contradiction with $U \neq V$. Hence, $U \cap V = \langle 1 \rangle$.

Then it is easy. If $U$ is any maximal subgroup, then by what we showed any two distinct conjugates of $U$ intersect trivially. As the conjugacy classes are precisely $n/|U|$, we get that the conjugates of $U$ contain altogether $(|U|-1)n/|U|$ non-$1$ elements. This, of course, is strictly $< n-1$, so there is an element $y \in G$ not contained in any of the conjugates of $U$. Choose a maximal group $V$ containing $y$. Analogously, the conjugates of $V$ contain altogether $(|V|-1)n/|V|$ non-$1$ elements. But the conjugates of $U$ and $V$ are all maximal groups and thus intersect trivially (pairwise), so we have already found
\[
(|U|-1)n/|U| + (|V|-1)n/|V| = n \Big(  2 - \frac{1}{|U|} - \frac{1}{|V|} \Big) \geq n
\]
non-$1$ different elements of the $n$-order group $G$. Contradiction. Proof concluded.  :)

Combining this with grobber's arguments, we have an elementary solution to the initial problem.

--Vesselin

Actually, it can be shown that if $G$ has square-free order, then it is the semidirect product of two cyclic groups (a delightful result). It is obvious from here: if $G \cong A \ltimes B$ is the semidirect product of the non-trivial cyclic groups $A \cong \mathbb{Z}_a$ and $B \cong \mathbb{Z}_b$ ($ab=n$, $a,b > 1$) with respect to the homomorphism $\phi: B \to \mathrm{Aut} \, A$, then $\phi$ is trivial [i]unless[/i] $b = |B|$ and $\phi(a) = |\mathrm{Aut} \, A|$ are not co-prime. But the conditions $(n,\phi(n)) = 1$ and $ab=n$ obviously entail $(b,\phi(a)) = 1$, so $\phi$ is necessarily trivial, i.e. $G \cong \mathbb{Z}_a \times \mathbb{Z}_b \cong \mathbb{Z}_{ab}$ is the direct product of $\mathbb{Z}_a$ and $\mathbb{Z}_b$, as claimed.

--Vesselin


It becomes from a school test

Find all functions such as

$ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ and $ f'(0)\equal{}2$

This is the famous Cauchy's Equation, and with the assumption that $ f$ is continuous, we have $ f(x) \equal{} kx$ for constant $ k$. Since $ f'(0) \equal{} 2$, it follows that $ f(x) \equal{} 2x$.

[quote="giannis"]It becomes from a school test

Find all functions such as

$ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ and $ f'(0) \equal{} 2$[/quote]
$ \forall x\in \mathbb{R}: \ f'(x)\equal{}\lim_{h\to 0}\frac{f(x\plus{}h)\minus{}f(x)}{h}\equal{}\lim_{h\to 0}\frac{f(h)}{h}\equal{}f'(0)\equal{}2$
then $ f(x)\equal{}f(0)\plus{}\int_{0}^{x}f'(x)\equal{}0\plus{}2x$

[quote="not_trig"][quote="aviateurpilot"] $ \lim_{h\to 0}\frac {f(h)}{h} \equal{} f'(0) \equal{} 2$
[/quote]

Why is this true? Does it have to do with additivity?[/quote]

Note that since $ f(0 \plus{} 0) \equal{} f(0) \plus{} f(0)$ we must have $ f(0) \equal{} 0$.  Then

$ f'(0) \equal{} \lim_{h \to 0} \frac {f(0 \plus{} h) \minus{} f(0)}{h} \equal{} \lim_{h \to 0} \frac {f(h)}{h}$ by definition.


Let $ a;b;c \ge 0;ab\plus{}bc\plus{}ac\plus{}6abc\equal{}9$.Prove that:
$ a\plus{}b\plus{}c\plus{}3abc \ge 6$
  :roll:

[quote="caubetoanhoc94"]Let $ a;b;c \ge 0;ab \plus{} bc \plus{} ac \plus{} 6abc \equal{} 9$.Prove that:
$ a \plus{} b \plus{} c \plus{} 3abc \ge 6$
  :roll:[/quote]
The inequality is equivalent to $ 2\sum a \geq \sum ab \plus{} 3 < \equal{} > \sum (a \plus{} 2)^2 \geq (\sum a)^2 \plus{} 18$
but : $ \sum (a \plus{} 2)^2 \geq (\sum a \plus{} 6)^2/3$
so it's suffice to prove that $ (\sum a \plus{} 6)^2 \geq 3(\sum a)^2 \plus{} 54$ ,which is equivalent to $ (\sum a \minus{} 3)^2 \geq 0$ (True):D

sr .......................wrong again >"< mod plz delete this stupid post :(

[quote="long14893"][quote="caubetoanhoc94"]Let $ a;b;c \ge 0;ab \plus{} bc \plus{} ac \plus{} 6abc \equal{} 9$.Prove that:
$ a \plus{} b \plus{} c \plus{} 3abc \ge 6$
  :roll:[/quote]
The inequality is equivalent to $ 2\sum a \geq \sum ab \plus{} 3 < \equal{} > \sum (a \plus{} 2)^2 \geq (\sum a)^2 \plus{} 18$
but : $ \sum (a \plus{} 2)^2 \geq (\sum a \plus{} 6)^2/3$
so it's suffice to prove that $ (\sum a \plus{} 6)^2 \geq 3(\sum a)^2 \plus{} 54$ ,which is equivalent to $ (\sum a \minus{} 3)^2 \geq 0$ (True):D[/quote]

You're wrong  :P


algebra
polynomial
limit
function
binomial theorem
continued fraction

Well since both $\pi$ and $e$ are trancendental, that is they cannot be solutions to a polynomial equation with rational coefficients. So no matter how you manipulate these, you'll have difficulty proving you've got $\pi$ or $e$.

However, we can at least define $e$ as $e=\lim_{x\to\infty}(1+\frac{1}{x})^{x}$. I tried manipulating this a bit and didn't get anywhere, though someone else perhaps could.

But I did make an interesting discovery about $e$ in the process by using Newton's Binomial Theorem:

$e=\lim_{x\to\infty}(1+\frac{1}{x})^{x}$

$={x\choose{0}}1+{x\choose{1}}\frac{1}{x}+{x\choose{2}}\frac{1}{x^{2}}+\dots$

$=\sum_{n=0}^{\infty}{x\choose{n}}\frac{1}{x^{n}}$

$=1+\frac{x}{1}\frac{1}{x}+\frac{x(x-1)}{2}\frac{1}{x^{2}}+\frac{x(x-1)(x-2)}{3!}\frac{1}{x^{3}}+\dots$

Since $x\to\infty$, all the non-dominant terms can be neglected, so we get:

$e=1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\dots$

$e=\sum_{n=0}^{\infty}{\frac{1}{n!}}$

Which doesn't answer the given question at all, but makes me happy.

Somehow I doubt that these are Intermediate-level problems.

Edit:

[quote="Wikipedia"]In 1748 Euler published a very important theorem showing that a particular kind of continued fraction is equivalent to a certain very general infinite series.[/quote]

I am tempted to say that such an equivalence is necessary for these problems.

Edit 2:

http://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula#The_exponential_function
http://en.wikipedia.org/wiki/Generalized_continued_fraction#The_equivalence_transformation


AMC AIME USA(J)MO USAMO Putnam trigonometry geometry

I'm kinda stuck here...I'm a senior in high school, graduating this summer (class of '08).  I recently took the AIME and I got a huge surge of confidence because I got a score of 6 (also managed to answer problems 9 and 12)...was one careless mistake away from getting a 7 (but I don't make that many careless mistakes! i like that)  I finally felt like qualifying for the USAMO wasn't too far away...except that I'm a senior now.

The problem is, the Putnam is way too hard right now.  The AIME seems to be just right.  I'd feel very accomplished if I could "qualify" for the USAMO...  Is it a total waste of time practicing for the AMC12/AIME at this stage?  I'm having a good time with the problems, and I think I can improve a lot.  Can I still take the AMC12/AIME and list myself as grade 13 or something...?

And to the people who are getting 10+ or near perfects on the AIME.  I'm not a genius I guess, but I'm decent at math...I got 800 on the SATs...is there this invisible barrier that you suddenly hit when you want to get better, or do you think a decent but definitely not extraordinary person like me can get there?  Right now, I feel like I'm really lacking in the stuff that comes with experience (like using modular arithmetic to solve equations faster...and using trig in geometry problems, i know like nothing about complex numbers and algebraic manipulation..)  I want to take the AMC12 adn AIME problem solving Aops courses over summer to prepare...

If you have fun doing math, it doesn't have to be in a competition setting.  Learn about the concepts behind AIME-level problems if they interest you and take interesting college classes (there are problem-solving classes, too).  You can't officially take the AMC/AIME as a college student, so unless you're planning on not going to college...

As a graduating senior, you should ask yourself what you plan on doing in math beyond competing because there is a lot of real math to do.  Are you planning on majoring in math?

If you've decided what college you want to go to, definitely email their math departments. It can't possibly hurt and you might be very pleased with the results. I'm actually going to my state college (largely for finance reasons, partly for stupid application decisions such as only seriously applying to 4 schools, etc.) but, me being in Illinois, the math department there [i]isn't[/i] sub-par; it's just not MIT.
Regardless, I emailed their math department and found out that I can actually get into Real Analysis as a first semester freshman! MANY colleges have these opportunities and you should definitely check them out.

If competitive math really interests you, then try find out if your campus holds any local contests... if not the Putnam is probably your best shot, although it's a pretty big leap from the AIME admittedly.

One other option is looking into a summer program like PROMYS. I'm not sure how many places actually take graduating seniors, but they are out there (PROMYS being one). I personally went to Hampshire (which I know does not take graduating seniors unfortunately) and I can honestly say it was by far the best math experience I've ever had. I learned everything from number theory to group theory to topology to analysis. If nothing else, programs like that definitely help you figure out where your mathematical interests are (for example, I'm not a huge fan of number theory but I love algebra and analysis).

Lastly, you could also look into local research opportunities. They take some work to dig up ones that you would actually be able to get into, but I know people who have done them as incoming freshman, and they are 100% worth the time.

[quote="mathemonster"]kops why did u only apply to 4 schools??[/quote]
Uh like I said I was really dumb about the application process... basically I liked U of I so I decided I wasn't going to do much research and only apply to schools I really wanted to go to (since U of I was basically a guarantee to get in)... so that's what I did. I sorta regret it now cause I didn't get into MIT and I could have maybe gotten into Stanford for example... but oh well. Like I said I arranged things with the math department at U of I so it should work out ;)

[quote="kops723"]
One other option is looking into a summer program like PROMYS. I'm not sure how many places actually take graduating seniors, but they are out there (PROMYS being one). I personally went to Hampshire (which I know does not take graduating seniors unfortunately) and I can honestly say it was by far the best math experience I've ever had. I learned everything from number theory to group theory to topology to analysis. If nothing else, programs like that definitely help you figure out where your mathematical interests are (for example, I'm not a huge fan of number theory but I love algebra and analysis).
[/quote]

[url=http://mathcamp.org/]Mathcamp[/url] also takes graduating seniors.


conics
parabola
geometry unsolved
geometry

Let $ ABC$ is triangle and let $ d$ is a line. Find point $ M$ lie on $ d$ for $ MA^2\plus{}3MB^2$  is minimum.

Is there any point of having point C? What if you just say segment AB?

Anyways, I think this is the solution [hide]
Let the perp from B to d be Q, the perp from A to D be R. Let BQ=r, AR=n, QR=x. It is clear that the best point is between Q and R. Let the distance between a point you choose on d (call it Y) and Q be p. Then QR=x-p. So by Pythag, we want to minimize 3(r^2+p^2)+(x-p)^2+n^2. Everything is constant except for 4p^2-2xp. So this expression is minimized at the vertex, when p=x/4. [/hide]

Hi! dgreenb801. This problem is one of work.

It's problem 1.
This is problem 2:
Find pint $ N$ for $ NA^2\plus{}2NB^2\plus{}5NC^2$ is minimum.

We solve it similarly using coordinates.
Let C be (0,0), B be (a,0), and A be (b,c). Then, for any  point (x,y), the expression you gave is $ (x \minus{} b)^2 \plus{} (y \minus{} c)^2 \plus{} 2(x \minus{} a)^2 \plus{} 2y^2 \plus{} 5x^2 \plus{} 5y^2$. The non-constant part of this is equal to $ (8x^2 \minus{} x(4a \plus{} 2b)) \plus{} (8y^2 \minus{} 2cy)$. So we take the vertex of each of these parabolas for x and y, so the point is (a/4+b/8,c/8).

We solve general problem .
$ x_1,..,x_n > 0$ and $ n$ point $ A_1,..,A_n$ in the plane .
Find point N such that $ \sum_{i \equal{} 1}^n x_i NA_i^2$
Here is my solution
Call G is the point such that
$ \sum_{i \equal{} 1}^n x_i\vec{GA_i} \equal{} \vec{0}$
Then we have
$ \sum_{i \equal{} 1}^n x_i(NA_i^2 \minus{} GA_i^2) \equal{} \sum_{i \equal{} 1}^n x_i(\vec{NA_i} \minus{} \vec{GA_i})(\vec{NA_i} \plus{} \vec{GA_i}) \equal{} \sum_{i \equal{} 1}^n \vec{NG}\sum_{i \equal{} 1}^n x_i (\vec{NA_i} \plus{} \vec{GA_i}) \equal{} \frac {\vec{NG}^2}{\sum_{i \equal{} 1}^n x_i}\geq 0$.
So we have $ \sum_{i \equal{} 1}^n x_i NA_i^2\geq \sum_{i \equal{} 1}^n x_iGA_i^2$
Equality hold if and only if $ N\equiv G$
Note : Only need $ \sum_{i \equal{} 1}^n x_i > 0$


linear algebra
matrix
vector
algebra
polynomial
linear algebra unsolved

Let $ A$,$ B$  2 matrix with real elements.If
$ BA^2\equal{}B^2$ and $ 4AB\plus{}I_{n}\equal{}3BA$ prove that $ AB\equal{}BA$

We can assume that $ A,B$ are $ (n,n)$ complex matrices.
The key is: $ 4AB\plus{}I\equal{}3BA$ implies  $ B$ is an invertible matrix.
Proof: if $ B$ isn't invertible then there exists a vector $ x\not\equal{}0$ s.t. $ Bx\equal{}0$; thus $ 3BAx\equal{}x$ and $ 1/3\in{spectrum(BA)}$.
Moreover $ spectrum(AB)\equal{}spectrum(BA)\equal{}\{\lambda_1,\cdots,\lambda_n\}$ and the 2 following sets are the same: $ \{4\lambda_1\plus{}1,\cdots,4\lambda_n\plus{}1\}\equal{}\{3\lambda_1,\cdots,3\lambda_n\}$. Thus if $ u$ is a $ BA$-eigenvalue then $ (3u\minus{}1)/4$ is another one (or eventualy the same) . The sequence $ u_{0}\equal{}1/3, u_{k\plus{}1}\equal{}(3u_{k}\minus{}1)/4$ is infinite, what is a contradiction.
What follows is easy: $ A^2\equal{}B,AB\equal{}BA,A^3\plus{}I\equal{}0$.  QED

Let $ x\in ker(B)$

$ Bx\equal{}0$

$ 4ABx\plus{}x\equal{}3BAx$  => $ x\equal{}3BAx$  (1)

multiply to the right by $ A$ the second relation of the statement, and apply the vector $ x$

$ 4ABAx\plus{}Ax\equal{}3BA^2x\equal{}3B^2x\equal{}0$ => $ Ax\equal{}\minus{}4ABAx$  (2)

(1) => $ Ax\equal{}3ABAx$ (3)

(2) and (3) give $ Ax\equal{}0$  again with (1) $ x\equal{}3BAx\equal{}0$

We have proved   $ ker(B)\equal{}{0}$

$ B$ is invertible

The first relation of statement gives $ A^2\equal{}B$ plug in the second relation
you have $ A^3 \plus{}I\equal{}0$

We can generalize as follows: if $ 4AB\plus{}I\equal{}3AB$ and $ B^{p}A^k\equal{}B^{p\plus{}1}{Q(A)}$ where $ k,p$ are non zero natural  integers and $ Q(x)\in\mathbb{C}[x]$ then $ B$ is a polynomial in $ A$.
Proof: we have seen that $ B$ (and $ A$ in a similar way) are invertible. Then $ Q(A)$ is invertible and $ B\equal{}A^k{Q(A)^{\minus{}1}}$. We notice that $ Q(A)^{\minus{}1}$ is a polynomial in $ A$.  QED.


Given a four digit number $n=\overline{abcd}$ where $n$ is in base 10 and all the prime factor,$p_i$, of $n$ satisfy the condition $p_i\geq 7$ . Prove or Disprove that we can always find a number $k$ such that

$n|\underbrace{aa\cdots aa}_{\text{k term}}\underbrace{bb\cdots bb}_{\text{k term}}\underbrace{cc\cdots cc}_{\text{k term}}\underbrace{dd\cdots dd}_{\text{k term}}$

\[ Let \ \underbrace{aa\cdots aa}_{\text{k term}}\underbrace{bb\cdots bb}_{\text{k term}}\underbrace{cc\cdots cc}_{\text{k term}}\underbrace{dd\cdots dd}_{\text{k term}}=A \]
$A=\frac{10^k-1}9 (a*10^{3k}+b*10^{2k}+c*10^k+d). \ Let \ n=p_1^{a_1}*p_2^{a_2}\dots p_s^{a_s}, \ p_i|n.$ $If \ p_i\not=2,5, then \ exist \ k_i: \ p_i^{a_i}|B=(10^{k_i}-1)/9,$
for example $k_i=a_i \ if \ p_i=3, \ else \ k_i=(p_i-1)*p_i^{a_i-1}. \ If \ k_i|k \ for \ all \ i, \ then \ n|B|A.$


[hide]Okay, factor out the $x^{2}$ to get $x^{2}(x-6)$.  As you can see, the roots are 0, 0, and 6.  Two of the roots aren't distinct.  Since the coefficient is positive, we can just add any positive value of $p$ for the roots to be distinct.  [/hide]

Edit: I just realized a flaw in my solution, I didn't consider an upper bound on this.  I found the real answer [hide]$0<x<32$[/hide]but it uses some calculus which I think isn't allowed on this forum.

[quote="seamusoboyle"]Alternately, one of the roots of this equation is:

$\frac{1}{2} \sqrt[3] {64-4p+4 \sqrt{p^2-32p}}+ \frac {8}{ \sqrt[3]{ 64-4p+4 \sqrt{p^2-32p}}}+2$

 :D
So p has to satisfy what you got from calculus!!! :D[/quote]
Did you use formulas to find the roots??

pathetic attempt
[hide][quote="Euler1001"]If f(x) = x^3 - 6x 2 + p , where P is an arbitrary constant.  For what values of p does f(x) have 3 distinct real roots?[/quote]
if f(x) has real roots, then f(x)=(x+a)(x+b)(x+c) where a b and c are real numbers. since
ab+bc+ac=-6
and a+b+c=0
find all real values for abc
so thats
ab+c(a+b)=-6
a+b=-c
so ab+-a^2-2ab-b^2=-6
so -a^2-ab-b^2=-6
so -(a^2+ab+b^2)=-6
so -(a^3+b^3)=-6(a-b)
a^3+b^3=6b-6a
with this we can set bounds on a b and -a-b=c
so with these bounds we can solve for it[/hide]


$ PQ$ and $ RS$ are two diametrically opposite tangents to a circle of radius $ r$ where $ P$ and $ R$ are the end-points of the said diameter.
$ PS$ and $ RQ$ intersect at $ X$ which happens to be on the circumference of the circle. Hence find the value of $ 2r$.

$ A)$ $ \sqrt {PQ \cdot RS}$

$ B)$ $ \frac {PQ \plus{} RS}{2}$

$ C)$ $ \frac {2PQ \cdot RS}{PQ \plus{} RS}$

$ D)$ $ \sqrt {\frac {PQ^2 \plus{} RS^2}{2}}$

[hide="My approach:"]
I tried doing it by using similarity of the different right angles formed and the Theorem on Equal Ratios. However I was unsuccessful, perhaps maybe because I obtained too many equations and it all ended up in a tangle and I was caught in a vicious circle where I returned to the same equality all over again.
[hide="The Answer:"]
Correct Option is: $ A$.
[/hide]
[/hide]

if PQ is a diameter, how come it be a tangent :huh:  :huh:  :huh:

correct question==
PR is diameter

then it is a pure geometric approach with similarity argument($ \angle PXR\equal{}\pi/2$). to show that diameter is geometric mean

Hello
instead of $ Q$ it should be $ R$ .
Solution:
angle $ PQR$=$ \alpha$
$ 2r \equal{} \frac {PQ}{cot\alpha}$
$ 2r \equal{} \frac {RS}{tan\alpha}$
multiply both d equation u will get the result.
Thank u

[quote="kabi"]Hello
instead of $ Q$ it should be $ R$ .
Solution:
angle $ PXR$=$ \alpha$
$ 2r \equal{} \frac {PQ}{cot\alpha}$
$ 2r \equal{} \frac {RS}{tan\alpha}$
multiply both d equation u will get the result.
Thank u[/quote]
no need for $ \alpha$ though :)  :wink:

[quote="kabi"]Hello
instead of $ Q$ it should be $ R$ .
Solution:
angle $ PXR$=$ \alpha$

$ 2r \equal{} \frac {PQ}{cot\alpha}$

$ 2r \equal{} \frac {RS}{tan\alpha}$
[/quote]

$ \angle PXR \equal{} 90^0.$ Then, you get $ 2r\equal{}\infty$ or $ 2r\equal{}0$. Multiplying will give an indeterminate form.

Isn't it so?

[quote="kabi"]hello
I m sorry it is not $ PXR$ but $ PQR$
I have edited the previous post.[/quote]

Ok got the answer. Thank You.
$ \tan \alpha = \frac{2r}{PQ} \\
\\
\tan \alpha = \frac{RS}{2r}$

Equating, $ 2r = \sqrt {PQ \cdot RS}$

Why are you introducing $ \tan \alpha$ and stuff like that?
Just note that $ \triangle SRP \sim \triangle RPQ \Rightarrow \frac{RS}{PR} \equal{} \frac{PR}{PQ} \iff 2r \equal{} \sqrt{RS \cdot PQ}$


complex numbers
topology
advanced fields
advanced fields unsolved

Let $X$ and $Y$ be two topological spaces. How to find a covering map $p: (X,x_{0})\to (Y,y_{0})$ and two paths $\alpha,\beta\in \Omega(Y,y_{0})$ s.t. $\tilde{\alpha}(1)=\tilde{\beta}(1)$ but $\alpha\not \sim \beta$.

just a brief sketch of the thing, annoying (and boring) details are marked by a star...

consider the two-folds covering $p: (S^{1}, 1) \to (S^{1}, 1), z \mapsto z^{2}$, (where $S^{1}$ is seen as the subset of complex numbers of modulus $1$), let $\gamma$ be a generator of the fundamental group, and consider the two paths $\gamma^{2j}: [0,1] \to S^{1}$, $j=0,1$.
clearly, the two paths are non-homotopic, and it's * to see that they share the same endpoint in the given covering.

I understand that $X$ and $Y$ are not given: you are asked to come up with $X$, $Y$, $\alpha$, etc. Otherwise $Y$ could be contractible, in which case $\alpha$ and $\beta$ are homotopic.

Still, the problem is poorly designed: you can just take $X$ and $Y$ to be the same space, with $p$ being the identity map. Then the problem reduces to finding two non-homotopic paths with the same endpoints.


algebra
polynomial
combinatorics proposed
combinatorics

Find all value of $ n$ for which there are nonzero real numbers $ a, b, c, d$ such that after expanding and collecting similar terms, the polynomial $ (ax \plus{} b)^{100} \minus{} (cx \plus{} d)^{100}$ has exactly $ n$ nonzero coefficients.

The coefficient of $ x^k$ cancels out if $ a^k b^{100\minus{}k} \equal{} c^k d^{100\minus{}k}$, that is $ \left( \frac{ad}{bc} \right)^k \equal{} \left( \frac{d}{b} \right)^{100}$.
Note that for fixed $ a,b,c,d$, the value of $ k$ is uniquely defined by the above equation, unless $ a \equal{} c$ and $ d\equal{}b$ (in which case $ k$ can be arbitrary).
Therefore, the answer is $ n\equal{}0$, $ 100$, or $ 101$.

What about $ (x\plus{}1)^{100}\minus{}(x\minus{}1)^{100}$? Note that f $ a^{k}/b^{k}\equal{}c^{k}/d^{k}$ when $ k$ is even we can possibly have $ a/b\equal{}\minus{}c/d$.

You're right. That gives one more solution: $ n\equal{}50$.

[hide="Solution"]

Note that

$(ax+b)^{100}-(cx+d)^{100}=\sum_{i=0}^{100}{\binom{100}{i}x^i(a^ib^{100-i}-c^id^{100-i})}$.

The coefficient of $x^i$ is thus $0$ iff $a^ib^{100-i}-c^id^{100-i}=0$.

It is easy to construct exactly 1 coefficient being $0$ (e.g. $a=c, b \neq \pm d$) - e.g. $n=100$. We now assume that more than 1 coefficient is $0$.

Therefore $a^ib^{100-i}=c^id^{100-i}, a^jb^{100-j}=c^jd^{100-j}$ for some $i,j$. Dividing these gives

$a^{i-j}b^{j-i}=c^{i-j}d^{j-i}$

or

$(\frac{a}{b})^{i-j}=(\frac{c}{d})^{i-j}$.

Therefore, $\frac{a}{b}=\pm \frac{c}{d}$, with the negative case happening only when $i-j$ is even.

If $\frac{a}{b}=\frac{c}{d}$, either all of the coefficients are 0 or none are, since
$a^{100}-c^{100}=\frac{a}{b}(a^{99}b)-\frac{c}{d}(c^{99}d)=\frac{a}{b}(a^{99}b-c^{99}d)$
$=\frac{a^k}{b^k}(a^{100-k}b^k-c^{100-k}d^k)=\frac{a^{100}}{b^{100}}(b^{100}-d^{100})$. This gives $n=0,101$.

If $\frac{a}{b}=-\frac{c}{d}$, then note none of $i=1,3,\hdots,99$ can be $0$, and either all or none of $i=0,2,\hdots,100$ are $0$, by similar logic to above. This gives $n=101$ (again) or $n=50$.

Our solutions are thus $n=0, 50, 100, 101$.[/hide]

I agree with these solutions, but there is mistake in translation of this problem. In original version (on webpage of Olympiad) the exponents are 1000 not 100.


yea base change works, and when you do, it simplifies and gives you an integer answer

This is called chain rule for logarithrm.

Chain rule states that $(\log_a b)(\log_b c) = \log_a c$

Applying this:

$(\log_2 3)(\log_3 5) = \log_2 5$

Multiplying by $\log_5 8$, we get $\log_2 8$ which equals 3.

I'll leave a proof of a chain rule for other people.

I believe Holier already proved it. Or at least outlined the proof.

We are taught to prove it with exponets.

[quote="236factorial"]What is $(\log_23)(\log_35)(\log_58)$? Something to do with the base change formula?[/quote]

Yes, you do use the change of base formula...thus:

$log 3/log 2(log 5/log 3)(log 8/log 5)$ And using magic we can cancel out $log 3 and log 5$ leaving us with with $log 8/log 2$, then we reverse the change of base and get log $log 2 of 8$ so that's $3$.


trigonometry
Gauss
algebra proposed
algebra

1)
let $z_1$ = $ x_1 + iy_1 $
and $z_2$ = $ x_2 + iy_2 $

Thus , the given relations imply,

$ (x_1 +x_2)^2 + (y_1 + y_2)^2 =1 $
$ (1+x_1)^2 = (x_1)^2 $
and $ (1+x_2)^2 = (x_2)^2 $
Thus , we have
$ z_1 = -0.5 + ai $ and $ z_2 = -0.5 -ai $
Thus, we can easily get the required realtion(i'm a bit lazy!!!  :no: )

2) Let $z = cis( \alpha) $
thus,$ \frac{1}{z} = cis (- \alpha) $
and$ \frac{1}{z^2} = cis ( -2 \alpha)$

acc. to question, this implies

$ 2\cos{ \alpha} + \cos { 2 \alpha} < - \frac{1}{8} $
and $ \sin {2 \alpha} = 0 $

Thus, $ \alpha $ (smallest) = $ \pi $

From which the rquired result follows. Even in the general case this is true...

From $|1+z_i|=|z_i|\ (i=1,2)\Longleftrightarrow z_i=-\frac{1}{2}$.Let $z_3=-1$, we have $z_1+z_2=z_3$
and $z_1,z_2$ are conjugate on the unit circle for Gauss plane,yielding $|z_1+z_2|=1$.
Drawing graph gives $z_1+z_2=-1$.Thus $(1+z_1)(1+z_2)=1+z_1+z_2+z_1z_2=z_1z_2$.Q.E.D.

good geomtric interpretation. I still have to do this aspect of complex nos. But effectively, aren't the two solutions the same..

Also, is my second solution correct??? :)  :(

$z\in \Im,\ |1+z|=|z|\nrightarrow z=-\frac 12.$
$z\in \Im,\ |1+z|=|z|\Longleftrightarrow Re\ z=- \frac 12.$

EDIT : sorry, the only thnig we can coclude is that the real part of the no.s is $ \frac{1}{2} $ Right???

I don't understand. Be more clearly. Write more pliant.

I was just saying what you have written in your edited post now!!!


Let $a,b,c$ be positive reals s.t. $a^2+b^2+c^2=1$. Prove the following inequality \[ \sum \frac{a}{a^3+bc} >3 . \]

[i]Proposed by A. Khrabrov[/i]

I know one member in this forum can answer it very well. I saw him sovle this problem .
His solution is really great, natural and really nice.
He is Ng.T.TU. Hey TU write your solution right now!!!!!!
I want to post that solution but It is not mine. So I don't want the others think I stole the solution from him.

I am a new member. Hi everyone!!!!!!!!!! :D

N.T.TU will never post his proof unless you say you hate it.

Somehow I don't really get the point  :D

[quote="Alexander Khrabrov"][quote="perfect_radio"]

PS: A solution was given by Iura in http://www.mathlinks.ro/Forum/viewtopic.php?t=44430 (which is probably correct, but I'm afraid to check it) .  I'm wondering whether there is an elegant solution :?

 :D[/quote]

Iura's solution is true, but contain a lot of inessential mistakes.
Canonical solution is very short, five lines only.[/quote]

Glad to hear that Iura's soln is a correct soln.

And what do you mean by "canonical"? Anyway, if it is that short, I'd like to see it  :D

[quote="nttu"]$ c < c^3 +ab $[/quote]
why is this true?

[quote="nttu"]If $ a \geq a^3+bc ; b \geq b^3+ac $ and $ c \geq c^3 +bc $ , the ineq is obviously true .[/quote]
thanks! Your solution is beautiful.  :lol:

At the first line with using Couchy we get
 $ \frac{1}{a^2+\frac{bc}{a}} + \frac{1}{b^2+\frac{ac}{b}} \geq \frac{4}{a^2+b^2+\frac{bc}{a} + \frac{ac}{b}} $
I do not understand your solution

For $p,q,x,y>0$, you can use Cauchy-Schwalz's extended version $\frac{x^2}{p}+\frac{y^2}{q}\geq \frac{(x+y)^2}{p+q}$ :)

[quote="silouan"]I know very well this but why $ a^2+b^2=1 $
The problem says $ a^2+b^2+c^2=1 $[/quote]

we increase the denominator from $a^2+b^2$ to $a^2+b^2+c^2$. thus, we get smth smaller so the ineq is still true, right?

[quote="silouan"]Oh yes! I understand it. i make a thought. We suppose that a<=a^3+bc
b<=b^3+ac.............. etc. But If the same time we have that   a<=a^3+bc but
b>b^3+ac and c>c^3+ab. What happens?[/quote]

   No , I don't suppose that  :o  . We only need one in three ineqs .

[quote="nttu"]This problem can be solved by Cauchy's ineq :
 $ \frac{1}{a^2+\frac{bc}{a}} + \frac{1}{b^2+\frac{ac}{b}} \geq \frac{4}{1+\frac{bc}{a} + \frac{ac}{b}} $
 Furthermore , $ \frac{4}{1+\frac{bc}{a} + \frac{ac}{b}} + \frac{bc}{a} + \frac{ac}{b} \geq 1 $
 So it's enough to show : $ \frac{c}{c^3+ab} \geq \frac{bc}{a} + \frac{ac}{b} $
 The last ineq is true for $ c < c^3 +ab $[/quote]

your solution is correct, and very short also :) i doubted that such soln exists, but....

Now I have to say it is the most un-natural soln I've seen on ML (there may be other, but i'm not very old on the forum). I mean how on earth did you think of those ineqs  :?

[quote="silouan"]Sure! It is a perfect solution but a solution very strange. Has anyone another solution to this problem? Maybe the official?[/quote]

It's the official solution. I don't know another solutions except offitial and Iura's one.

I have a different solution:
1/(a^2+(bc)/a) + 1/(b^2+(ac)/b) + 1/(c^2+(ba)/a)>=3^2/(a^2+b^2+c^2+ab+bc+ca);
the last is equivalent to 2>=ab/c+bc/a+ca/b;then we have to prove that 2abc>=a^2*b^2+c^2*b^2+a^2*c^2
=> we have to prove that a^4+b^4+c^4 + 4abc>=1=a^2 + b^2 + c^2 and we have that a^3 + bc>a and we sum with a,b,c and we have what we had to prove.

[quote="nttu"] $ \frac {c}{c^3 \plus{} ab} \geq \frac {bc}{a} \plus{} \frac {ac}{b}$
 The last ineq is true for $ c < c^3 \plus{} ab$[/quote]

Why is  $ \frac {c}{c^3 \plus{} ab} \geq \frac {bc}{a} \plus{} \frac {ac}{b}$ true for $ c < c^3 \plus{} ab$?  :maybe:

Clear the denominators and you'll see why :!:

[quote="nttu"]If $ a \geq a^3+bc ; b \geq b^3+ac $ and $ c \geq c^3 +bc $ , the ineq is obviously true .[/quote]
your solution is not true!


geometric transformation
geometry proposed
geometry

we are given a circle segment .A cincle inscribed in tis circle segment is tangent at A and B.
prove that AB passes through a fixed point , independently of the inscribed circle. :D

$\Gamma$ is the fixed circle and $C,D$ are two fixed points on it. Denote $\omega$ the variable circle internally tangent to $\Gamma$ at $A$ and tangent to $\overline{CD}$ at $B.$ Lines $AC,AD$ cut $\omega$ again at  $C',D'.$ Since $A$ is the exsimilicenter of $\omega \sim \Gamma,$ it follows that $CD \parallel C'D'$ $\Longrightarrow$ Arcs $BC'$ and $BD'$ of $\omega$ are equal $\Longrightarrow$ $AB$ bisects $\angle CAD$ internally. Hence, $AB$ always passes through the midpoint of the arc $CD$ of $\Gamma$ not containing $A.$

i proved that AB passes through the midpoint of arc CD using a dilatation with center A,
carrying the inscribed circle at the "big" circle. I found it very cool, despite that it is easy.

AB passes through the midpoint of arc CD..easy by angle chase


Let ABCD be a convex quadrilateral, and let M and N be the midpoints of its sides AD and BC, respectively. Assume that the points A, B, M, N are concyclic, and the circumcircle of triangle BMC touches the line AB. Show that the circumcircle of triangle AND touches the line AB, too.

  Darij

A nice problem.

The given condition is equivalent to $\angle BMC=\angle AMN\iff AMN,BMC$ are similar, iff $\frac{AM}{AN}=\frac{BM}{2BN}$. This is equivalent to $\frac{AN}{BN}=\frac{2AM}{BM}$, which, in turn, is equivalent to what we want to prove ($AND,BNM$ are similar).

Oh. Now I really don't understand why my proof took me 1 page.  :blush:

Anyway, the condition that the circumcircle of triangle BMC touches the line AB and the condition that the circumcircle of triangle AND touches the line AB are both equivalent to the condition that the quadrilateral AMNB is harmonic.

  Darij

Yes, this follows easily: we have $AN\cdot BM=2AM\cdot BN$, and if we also take Ptolemy's Theorem into consideration, we find that the products of the opposite sides are equal for the quadrilateral $ABMN$.

Let $ABCD$ be a convex quadrilateral, and let $M$ and $N$ be the midpoints of its sides $AD$ and $BC$, respectively. Assume that the points $A, B, M, N$ are concyclic, and the circumcircle of triangle $BMC $ touches the line $AB$. Show that the circumcircle of triangle $AND$ touches the line $AB$.


limit
logarithms
function
real analysis
real analysis unsolved

show :D

$ \textbf{(i)}\lim_{n\to\infty}\frac{1}{\ln p_{n}}\prod_{i=1}^{n}\frac{p_{i}}{p_{i}-1}=\boxed{e^\gamma}$

$ \textbf{(ii)}\lim_{n\to\infty}\ln p_{n}\prod_{i=1}^{n}\frac{p_{i}}{p_{i}+1}=\boxed{\frac{\zeta{(2)}}{e^\gamma}}$

sorry, i should have mentioned above $ p_{i}$ refers to the $ i^{th}$ prime number. :|

anyhow, these 2 are collectively known as [b]merten[/b]'s [i]prime number[/i] theorems
(well, the first one only, i think) [ according to hardy & wright's text that is ]

it suffices to say if you can prove $ \textbf{(i)}$, then $ \textbf{(ii)}$ follows, because we have
$ 1-\frac{1}{p_{i}^{2}}\;=\;(1+\frac{1}{p_{i}})\;(1-\frac{1}{p_{i}})$

and so,
$ \mathbb L\;=\;\lim_{n\to\infty}\ln p_{n}\prod_{i=1}^{n}\frac{p_{i}}{p_{i}+1}\;=\;\lim_{n\to\infty}\ln p_{n}\;\;\frac{1}{\prod_{i=1}^{n}(1+\frac{1}{p_{i}})}$

$ =\;\lim_{n\to\infty}\ln p_{n}\;\;\frac{\;\;\frac{1}{\prod_{i=1}^{n}(1-\frac{1}{p_{i}^{2}}})\;\;}{\;\;\frac{1}{\prod_{i=1}^{n}(1-\frac{1}{p_{i}}})\;\;}$

$ =\frac{\lim_{n\to\infty}\frac{1}{\prod_{i=1}^{n}(1-\frac{1}{p_{i}^{2}}})}{\lim_{n\to\infty}\frac{1}{\ln p_{n}\prod_{i=1}^{n}(1-\frac{1}{p_{i}}})}$

now remembering euler's product formula for the zeta function, i.e.
$ \zeta{(s)}\;=\;\lim_{n\to\infty}\frac{1}{\prod_{i=1}^{n}(1-\frac{1}{p_{i}^{s}})}$

we have $ \mathbb L\;=\;\frac{\zeta{(2)}}{e^\gamma}$
[ this follows because $ \textbf{(i)}$ already asserts ${ \lim_{n\to\infty}\frac{1}{\ln p_{n}\prod_{i=1}^{n}(1-\frac{1}{p_{i}}})}$

$ =\lim_{n\to\infty}\frac{1}{\ln p_{n}}\prod_{i=1}^{n}\frac{p_{i}}{p_{i}-1}\;=\;e^\gamma$ ]

therefore, we now [b]only [/b]need to show $ \textbf{(i)}$  :D

hint: of-course, the answer $ \boxed{e^\gamma}$ immediately suggests to us
we take $ \ln$ of ${ \frac{1}{\ln p_{n}\prod_{i=1}^{n}(1-\frac{1}{p_{i}}})}$
and then show $ \lim_{m,n\to\infty}\;[\sum_{i=1}^{n}\;\sum_{j=1}^{m}\frac{1}{j\;p_{i}^{j}}\;-\;\ln \ln p_{n}]\;=\;\boxed{\gamma}$

[ PS: i know how to show it, but i want to see if someone here
posts a [b]different [/b]method...  :maybe: ]


calculus
integration
algebra
function
domain
quadratics
Ring Theory

prove that there are infinitely many gaussian primes.

then see if you can prove that there are infinitely many prime elements in the integral domain whose elements are in the form a+b*sqrt(d). a and b are rational and d does not contain perfect squares. d may be negative or positive.

There are infinitely many Gaussian primes because if there were a finite number, you could take their product and add 1, thereby creating a number relatively prime to all existing primes (remember, primes are not units, and there do exist primes).  We can use a norm argument to show that this product is not a unit because the norm of $ a+bi$ is $ a^{2}+b^{2}$ which is $ 1$ only if one of $ a, b$ is $ 0$.

However, $ \mathbb{Q}[\sqrt{d}]$ is a field, and therefore has no primes (suppose $ a^{2}-db^{2}=N$, then the reciprocal of $ a+b\sqrt{d}$ is $ \frac{a-b\sqrt{d}}{N}$).  Perhaps you meant $ \mathbb{Z}[\sqrt{d}]$?  In that case, you can make the same argument for $ d<0$ (and you always know the set of primes is nonempty, because not everything is a unit (take $ \sqrt{d}$), and the norms are naturals and therefore well-ordered).

However, all of this is from the point of view of the ring itself.  If, instead, you make the association to the equation $ a^{2}-db^{2}=p$, there's a very nice way to see it.  If $ a+b\sqrt{d}$ has norm $ p$, $ p$ a prime in the integers, then it is prime.  Therefore, it suffices to show there are infinitely many $ p$ for which there are solutions to $ a^{2}-db^{2}=p$ in the integers.  However, using a well-known argument with Minkowski's Theorem (if you haven't seen it I can outline it), we know that if there is a nonzero solution to the congruence $ a^{2}-db^{2}\equiv 0\mod{p}$, then there is a solution to the above equation.  We know that there is a solution to this equation as long as $ \left(\frac{d}{p}\right)=1$ (ie, d is a quadratic residue modulo p), and we know that there are infinitely many primes $ p$ for which this holds: if $ p$ works, then $ p+4|d|$ has to work, by an application of Quadratic Reciprocity (in fact we can use QR to determine the general form of $ p$).  This is overkill for the problem at hand, but it's nicely generalizable.

Well, here's my take on this:

We will prove by contradiction.
Let pn be the largest prime in Q[sqrt(d)] and call it an+bn*sqrt(d).

We assume that there are finitely many primes in Q[sqrt(d)].
When we multiply these n primes together, we have (a1+b1i)(a2+b2*sqrt(d))(a3+b3*sqrt(d))...(an+bn*sqrt(d)).
Note that multiplying elements of Q[sqrt(d)] produces another element of Q[sqrt(d)]: (p+q*sqrt(d))(r+s*sqrt(d)) = pr+qr*sqrt(d)+ps*sqrt(d)-qs = (pr-qs)+(qr+ps)*sqrt(d).

Thus, multiplying all of the primes returns a composite number of the form a'+b'*sqrt(d).
Adding 1 gives us a'+1+b'*sqrt(d). This new number is also an integer, according to Theorem 8.2, because it is the sum of two integers.
1 is a unit of any quadratic field, so we know that none of our finite number of primes equals 1. This means that a'+1+b'*sqrt(d) is 1 mod p, where p is any prime.
Yet because a'+1+b'*sqrt(d) is an integer, it has a prime factorization. From above, we know that this factorization contains none of the predetermined primes, which means it contains new primes.

This contradicts our idea that there are finitely many primes, and thus we have proved that there are infinitely many primes in Q[sqrt(d)].

Does adding the 1 work, or do I need something else?

Theorem 8.2: If A and B are integers in Q[sqrt(d)], then A-B is also an integer.

[quote="sylow_theory"][quote="Riders on the Storm"]prove that there are infinitely many gaussian primes.
[/quote]

A Gaussian integer which is [b]real[/b] is prime if and only if it is an odd prime so that $ p\equiv 3(\bmod 4)$. Since there are infinitely many such prime the rest is trivial ...[/quote]

A prime in a ring is defined as an element whichis such that when it is expressed as the product of two things, exactly one of the two things is a unit.

[quote="Riders on the Storm"]This contradicts our idea that there are finitely many primes, and thus we have proved that there are infinitely many primes in Q[sqrt(d)].

Does adding the 1 work, or do I need something else?[/quote]

You DO indeed need something else, something very key.  You need to prove that the set of primes in that ring is nonempty.  So for each of these proofs, you need to give some prime as an example.  But this particular ring is a field: see my post above.

[quote="sylow_theory"][quote="K81o7"]
A prime in a ring is defined as an element whichis such that when it is expressed as the product of two things, exactly one of the two things is a unit.
[/quote]
I dont mean to be insulting, but I think you might be wrong. (Maybe we use different definitions).

[b]Definition:[/b] An [b]irreducible[/b] non-zero non-unit element $ p$ in a ring $ R$ is such that if $ p=ab$ then $ a$ or $ b$ is a unit.

[b]Definition:[/b] A [b]Principal Ideal Domain[/b] (PID) is a domain $ D$ such that all ideals $ N\triangleleft D$ are principle, i.e. $ N=\left< a \right>$ for some $ a\in D$.

[b]Definition:[/b] A [b]prime[/b] non-unit non-zero element $ p\in R$ is prime iff $ p|ab$ then $ p|a$ or $ p|b$.

[b]Theorem:[/b] If $ p$ is prime then it is irreducible.

[b]Theorem:[/b] If $ D$ is a PID and $ p\in D$ is irreducible then $ p$ is prime.

Note: The converse is only true for PID's. In fact I can give you a conterexample to what you said.[/quote]

I think that by "prime", I meant "irreducible" :blush: : I guess I don't have much experience using these terms.  But in a Euclidean ring, I guess we can use them interchangeably.  (Thanks for letting me know this btw: I never knew there was a difference between the two terms :) )

[quote="K81o7"] But in a Euclidean ring, I guess we can use them interchangeably.  (Thanks for letting me know this btw: I never knew there was a difference between the two terms :))[/quote]
Your welcome.

The reason why you can use these terms interchangeably in an Euclidean Domain is that an Euclidean Domain is a Principal Ideal Domain and so the theorem applies.

Btw,

does anyone know how to find congruence classes in a quadratic field?

[quote="Riders on the Storm"]K81o7--

Well, isn't it impossible for the set of primes in a quadratic field to be empty? If a field Q[sqrt(d)] (I think you said Z[sqrt(d)] is the correct term, and you might be right, but the original book I got this problem from said Q) has integers, then each integer can be factored down into primes, right? This shows that primes exist.

Maybe I'm missing something from your first post; if so, please fill me in.  :)

[/quote]
Well in the case of $ \{a+b\sqrt{d}\parallel a, b \in \mathbb{Q}\}$, where $ \mathbb{Q}$ is the set of rationals, every nonzero element has a multiplicative inverse: ie, every element is a unit.  Then by the definition of a prime, there are no primes, because a prime is by definition not a unit.  Notice also that every nonzero element divides any element of the ring.  This holds in any field (because in a field, every nonzero element is, by definition, a unit)

However, I have a feeling I'm completely misunderstanding what you're saying, because you say "if this field has integers."  From the terminology I've seen, a quadratic integer is of the form $ \frac{a+b\sqrt{d}}{2}$, where $ d$ is squarefree and $ a, b \in \mathbb{Z}$ (ie, the solution to a monic quadratic equation).  Is this what was meant?  Perhaps there's a convention I don't know...I've always seen $ \mathbb{Q}$ to represent the rationals, but maybe here it means "quadratic"...

Though now that I think about it, I have seen a field defined as follows:

$ R_{D}=$
\[ \{a+b\omega\parallel a, b \in \mathbb{Z}, \omega=\sqrt{D}, D \equiv 2, 3 \mod{4}\}\]

\[ \{a+b\omega\parallel a, b\in \mathbb{Z}, \omega=\frac{-1+\sqrt{D}}{2}, d \equiv 1\mod{4}\}\]
where of course the ring depends upon $ D$.  Well, as long as the distributive law holds (which it does, of course), you use the product+1 argument, because no prime is a unit...

Edit: In response to your last post: what exactly do you mean?  As in find the size of the congruence class?  The nature?  We can always define congruence the same way as in the integers...I know things behave nicely in the Gaussian integers, but past that, I don't know.  Try relating size of the ring of residues to norm...

sorry for my failure to phrase stuff correctly. your second paragraph is entirely correct. It's just my book (origin of problem) uses Q instead of Z. (no explanantion, I'm assuming Q stands for quadratic.

[quote="Riders on the Storm"]then each integer can be factored down into primes, right? This shows that primes exist.[/quote]
No, you cannot factor the integers down to primes, you can only factor them down to irreducible elements (number rings are very likely to be not UFD or even PID). Please do not neglect the difference.

And what do you mean by finding congruence classes¿

And I think it would be better to discuss such things in the future not here but in the [url=http://www.mathlinks.ro/Forum/index.php?f=8]Superior Algebra Forum[/url] .

well, when we examine Z, we have congruence classes for modular arithmetic, right?

since the problem states it, I suppose there is some kind of analog of this for quadratic fields.
I'm not entirely sure what that means.

Here's the problem itself--maybe seeing the wording might help.

"find the congruence classes (mod (3+sqrt(3))/2) in Q[sqrt(-3)]"

Z[sqrt(-3)] is what is meant, I think, if K81o7 is right.

[quote="K81o7"]using a well-known argument with Minkowski's Theorem (if you haven't seen it I can outline it), we know that if there is a nonzero solution to the congruence $ a^{2}-db^{2}\equiv 0\mod{p}$, then there is a solution to the above equation.[/quote]
Hasse-Minkowski's theorem requires nontrivial local solutions for all primes, in other words a nonzero solution $ \mod$ any integer (or any prime power).
And additionally, it only gives rational solutions then.

Counterexample if only solutions $ \mod p$ are given:
$ d=71$, $ p=7$

Counterexample for the rational solution(s) being an integer solution:
Take $ d=34$, $ p=-1$ (here $ p$ isn't prime, but you never used any property of $ p$).

Hmm....though can't you take that nontrivial solution and construct a lattice...oh okay I see the problem now.  I was thinking you could do the same thing as with $ a^{2}+b^{2}=p$ in a lattice, but I see now it doesn't hold...

But what if $ d<0$?  Despite the fact that you have an unnatural lattice, it only seems like a "stretched out" version of what you'd use for $ a^{2}+b^{2}=p$...and then it appears there's nothing wrong: you just take an ellipse as your convex region instead of a circle.


If A and B are the sets A = $ \{x\epsilon\mathbb{R}/\minus{}3\leq x\leq2\}$ and B = $ \{x\epsilon\mathbb{R}/\minus{}2<x<4\}$  then A - B is the interval:

a) [-2;4]
b) [-3;-2]
c) [2;4]
d) [-2;2]
e) [-3;4]

That's correct, but how to do ?
What I need to resolve ?
I'm lost in A-B .  :(

$ A \minus{} B$ means the things that are in $ A$ but not in $ B$, that is $ \minus{}3 \leq x \leq \minus{}2$, that is $ [\minus{}3, \minus{}2]$.

In the same way, $ B \minus{} A \equal{} (2, 4)$, that is $ 2 < x < 4$.

Have you tried drawing some number lines: I think they might help?

If you still don't get it, ask again and I'll see what I can do.

I think I understand now, do you know others questions only for training ?
(I need examples)

Thanks.

Sorry, luiseduardo, I haven't seen questions exactly like this before. If I find any similar questions, I'll let you know.

Hi luiseduardo, I've come across some similar questions in "Set Theory and Related Topics" by Seymour Lipschutz which can be found at [url]http://www.mcgraw-hill.co.uk/html/0070381593.html[/url], pages 40 to 44.

Thank you very much Andrew. But do you know other simple question ?

I think I understand, but I know other kind to do:

A = [-3,2] = -3,-2,-1,0,1,2
B = ]-2,4[ = -1,0,1,2,3

Cutting values:

The remains of A are: -3 and -2. The anwer :)


How? \include{usamts33.pdf} isn't working for me (usamts33.pdf is in the same folder as the file I'm working on)

with \include, latex "copy and paste" the content of the file included.  If it's a pdf, latex can't.  you need to include a text file but the best file to include is a .tex

Hmm.... I have an OpenOffice file with one of my USAMTS solutions in it, which I could convert to PDF, but how do I use that file in my Latex file?

you want to include your solutions in a latex file, that's right?  and the solutions are in a file like word (it's not openoffice, but the same kind of document), right?

Exactly. I'd reallllly like to not have to retype it all!

there has been some discussions about a word latex convertor, but I think you gonna have to retype it all.  Sorry


number theory
prime numbers
prime factorization

I mean natural factors and m,n are natural numbers

There are, I think, eight cases as stated.  Do you mean the case where $gcd(m, n) = 1$?  That's the easy one.

no, i didn't mean a special case!I meant any case

[quote="alex__ib"]Honestly i don't know  :mad: .I actually thought of something like the factors of n are:1,x,y,z,n and for m there are:1,a,b,m
I was thinking of the cases like x=a or x=b ,but i am stuck! :huh:[/quote]

Please somebody correct me if I'm wrong.

As far as I can remember, a number $x^{a}$ has $a+1$ factors. Suppose that a number $y^{b}$ has $b+1$ factors. if $x^{a}$ and $y^{b}$ are multiplied, then you simply multiply the number of their factors. So the number of factors is $(a+1)(b+1)$. This means, that you just multiply the number of factors of each number.

But in this case, $x$ and $y$ should be prime numbers raise to a positive exponent. Now, if you did not define $n$ and $m$, then I think it is impossible to find a precise number of factors of $m$ multiplied by $n$ since $n$ and $m$, can have the same factors.

A number $n$ has $5$ factors and a number $m$ has $4$ factors

If $n$ has $5$ factors, then it must be in form $a^{4}$, where $a$ is prime.
and $m$ can be:
1. $b*c$, where $b$, $c$ are prime, one of them may equal to $a$, but not necessary. But both of them can't be $a$.
2. $b^{3}$, where $b$ is prime, $b$ may equal to $a$, but not necessary.

Let's look at the first case..
If $b$ and $c$ are not equal to $a$, then $mn=a^{4}*b*c$, $mn$ has $(4+1)*(1+1)*(1+1)=\boxed{20}$ factors
If one of them is equal to $a$, wlog, assume $b=a$, then $mn=a^{5}*c$, $mn$ has $(5+1)*(1+1)=\boxed{12}$ factors.

Let's look at the second case
If $b$ is not equal to $a$, then $mn=a^{4}*b^{3}$, $mn$ has $(4+1)*(3+1)=\boxed{20}$ factors.
If $b=a$, $mn=a^{7}$, $mn$ has $\boxed{8}$ factors.

Therefore, mn can have 8, 12 or 20 factors..

[quote="guile"]But in this case, $x$ and $y$ should be prime numbers raise to a positive exponent.[/quote]

[b]Distinct[/b] prime powers.  Two relevant lemmas here.  First, a definition.

[b]Definition:[/b]  Let $d(n)$ denote the number of positive divisors of $n$.

Then the claims.

[b]Lemma 1:[/b]  Let the prime factorization of $n$ be $\prod_{i=1}^{n}p_{i}^{e_{i}}$.  Then

$d(n) = \prod_{i=1}^{n}(e_{i}+1)$

[b]Lemma 2:[/b]  $gcd(a, b) = 1 \implies d(ab) = d(a) d(b)$


$ ABC$ is a triangle for which $ BE$, $ CF$ are medians. If $ BE \equal{} 15 units$, $ CF \equal{} 36 units$ and $ BE$ perpendicular to $ CF$, then what is the area of triangle $ ABC$ [/code]

[hide="ans"]360 sq.units[/hide]

[hide="proof"]remeber medians divide the triangle into 6 triangle of equal area so. the area of the triangle is $ 3.\frac{1}{2}(10)(24)$[/hide]


What is the sum of all odd integers between $ \minus{}10$ and 12?

[hide="solution"]
The integers we are summing range from $ \minus{}9$ to $ 11.$ We see the integers from $ \minus{}9$ to $ 9$ cancel and we are only left with $ \boxed{11}.$[/hide]

What happened to 10?

Nvr mind... 4got to read odd nums


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let Ge be a group and H<G. Suppose |G/H|=2. Prove that aH=Ha for all a in G.

left-cosets partition $ G$, as well as right-cosets. thus $ aH \equal{} G \backslash H \equal{} Ha$ for all $ a \in G \backslash H$.

I am dealing with this problem, and I found this post. How about the general problem: Let $ G$ be a group and $ H<G$. Suppose $ |G/H|\equal{}p$, where p is a prime. Prove that $ H$ is normal in $ G$. I have not found any solution yet. Was this problem posted in the forum?

As far as I know, if $ G$ be a [b]finite[/b] group containing a subgroup $ H$ with $ [G: H]\equal{}p$, where $ p$ is the [b]smallest prime divisor[/b] of $ G$ ,then $ H$ is normal in $ G$.

[quote="ehsan2004"]As far as I know, if $ G$ be a [b]finite[/b] group containing a subgroup $ H$ with $ [G: H] \equal{} p$, where $ p$ is the [b]smallest prime divisor[/b] of $ G$ ,then $ H$ is normal in $ G$.[/quote][hide="Proof"]The kernel $ K$ of the group homomorphism
\[ \Phi: \left\{\begin{array}{ccc} G & \rightarrow & Bij(G/H) \\
g & \mapsto & (g'H\mapsto gg'H) \end{array}\right .
\]
is a normal subgroup of $ G$ contained in $ H$. $ [G: K]$ divides $ |Bij(G/H)| \equal{} p!$ which implies that $ [H: K]$ divides $ (p \minus{} 1)!$ (*).

If $ H$ is not normal, then let $ q > 1$ be a prime divisor of $ [H: K]$. Then $ q < p$, by (*). But $ q$ divides $ |G|$, hence $ p\le q$, by the minimality of $ p$. Contradiction.

EDIT: It is shown that $ \Phi$ induces an injection $ G/H\hookrightarrow Bij(G/H)$.[/hide]

[quote="MegaRockman"]How about the general problem: Let $ G$ be a group and $ H < G$. Suppose $ |G/H| \equal{} p$, where p is a prime. Prove that $ H$ is normal in $ G$.[/quote]
[hide="counterexample for p > 2"]$ G \equal{} D_{p}$ and $ H \equal{} \langle s\rangle$, where $ D_{p}$ is the dihedral group of order $ 2p$, and $ s\in D_{p}$ is one of reflections.[/hide]


integration
real analysis
real analysis unsolved

Let $ f, g: [0,1]\to\mathbb{R}$ be two integrable fucntions and $ \phi: [0,1]\to\mathbb{R}$ is nondecreasing, continuous in $ 0$ and $ \phi(0)\equal{}0$. Prove that there exists $ c\in (0,1)$ such that

 \[ \displaystyle\int_0^1f(x)dx\int_0^cg(x)\phi(x)dx\equal{}\int_0^1g(x)dx\int_0^cf(x)\phi(x)dx.\]


Could you tell me what is the combinatorical interpretation of the formula:
$\sum_{k=0}^{n} (-1)^k \binom{n}{k}=0$

I think: $\left|\coprod\limits_{i=1}^n A_k\right|=\sum\limits_{i=1}^n\left|A_i\right|-\sum\limits_{1\le i<j\le n}\left|A_i\cap A_j\right|+\sum\limits_{1\le i<j<k\le n}\left|A_i\cap A_j\cap A_k\right|-\ldots$

First, the original formula is true only for $n \ge 1$.  (It fails for $n = 0$.)

In words, the formula says that the number of even subsets of $1..n$ is equal to the number of odd subsets of $1..n$.  We will show that the two numbers are equal by constructing a bijection from the first set to the second set.

First, from the universe $1..n$, let's fix one number, say $1$.  Given an even subset $S$ of $1..n$, let's map it to $S \oplus \{ 1 \}$, the set obtained by switching the status of $1$.  Note that this new subset will have an odd number of elements.  It's easy to check that this map is a bijection.

Here is the general result:
http://www.mathlinks.ro/Forum/topic-41800.html


geometry
conics
ellipse
geometry unsolved

Hi all,

I must start by saying I really am not a mathematician and it has been 30 years since I studied math.

The problem is that I am trying to compute how to plot intersecting points on a standard ellipse, and I am trying to find the formulae to calculate where the points should be plotted.

I know the length of the major and minor axis and the focal point, given the focal point is x=0 and y=0, the elipse minor axis is y=40 and y=-40 and the major axis is x=10 and x=-10.

The big problem I have is that given any number of points I need to plot them evenly around the elipse so they intersect the line.

Is there a formulae that enables me to pass in the focal point co-ordinates, the co-ordinates of the x and y axis plus the number of points I need plotted and the result being an x,y co-ordinate for each of the points I need plotted.

Apologies if I am making this sound confusing.

Your help would be gratefully appreciated.

Mike

Seems more like Algebra... I think a moderator should move this to that forum.


Find all solutions $(x,y,z)$ to the following system of equations.
\[ \\\log 2xy=\log x\log y\\\log yz=\log y\log z\\\log 2xz=\log x\log z \]

Masoud Zargar

I don't understand, if $2xy=x+y, 2yz=y+z, 2xz=x+z$, then $xy+xz+yz=y+z+x$, we can create an infinite number of polynomials with the roots $x,y,z$... :o

Shobhit, I think you need to go back and learn your logarithms ;)

For example, $\log 2xy = \log x \log y \Leftrightarrow \log 2 + \log x + \log y = \log x \log y$.

This is a pretty cool problem, btw.

ughhh..what is wrong with me today :wallbash:

[hide="hint"]
Simplify with the usual log identities and then let $a=\log x$, etc.
[/hide]

[quote="rem"][quote="boxedexe"][hide="HINT"]
Try Factoring
[/hide][/quote]

Why and how would you factor? :huh:[/quote]

After substituting $a = \log{x}$, $b = \log{y}$, and $c = \log{z}$, we can use the Simon's favorite trick factoring, basicaly $ab \pm a \pm b + 1 = (1\pm a)(1 \pm b)$

[quote="rem"]log2xy = log2 + logx + logy = logxlogy
logyz   =            logy + logz = logylogz
log2yz = log2 + logx + logz = logxlogz
Now, let logx = a, logy = b, logz = c, log2 = m. Get a system:
m + a + b = ab
b + c = bc
m + a + c = ac.
Substracting first from third,get:
c - b = a (c - b)
(1 - a)(c - b) = 0
If a = 1, get m + 1 + b = b, which is impossible.
So , c - b = 0, or c = b.
b + c = bc.
2b = b^2, so b = 0 , 2.
When b = 0, c = 0, a = - m. logx = - log2, which has no solutions for x.
When b = 2, c = 2, a = 2 + m
logx = 2 + log2
x = 10^(2 + log2).
logy = logz = 2, so z = y = 100.
How is this a FUN system? :huh:[/quote]

It is fun if you get the right idea.  You also didn't find all solutions.  And just as the previous post says, it can be factored that way.
[hide="One more hint"]
There are 2 solutions.
[/hide]

P.S. Please use Latex. :)
EDIT: I just saw somethig. You have $x=10^{2+\log 2}$ which is equal to $x=10^{\log 200}=200$.  A solutio is $(x,y,z)=(200,100,100)$.
Masoud Zargar

In my solution I wrote that logx = - log2 has no solutions for x.:wallbash_red:
Sorry. This yields x = 10 ^ (- log 2) = 1/2.
Then logy = logz = 0, so y =  z = 1.
The other solution is (1/2, 1, 1). :)

Well done; however, there is a solution that is way simpler.
[b][i]Solution:[/i][/b]

$\\\log 2xy+1=\log 20+\log x+\log y=\log x\log y+1\\ \log^2 20=(\log x-1)^2(\log y-1)^2(\log z-1)^2\\ \pm\log 20=(\log x-1)(\log y-1)(\log z-1)\\\implies\pm\log 20=\log x-1\implies x=200, \frac{1}{2}\\\\ (\log 200-1)(\log y-1)=\log 20\implies \log y-1=1\implies y=100\\ (\log\frac{1}{2}-1)(\log y-1)=\log 20\implies \log y-1=-1\implies y=1\\\\ (\log 100-1)(\log z-1)=1\implies \log z-1=1\implies z=100\\ (\log 1-1)(\log z-1)=1\implies \log z-1=-1\implies z=1\\\\ \therefore (x,y,z)=(200,100,100),(\frac{1}{2},1,1)$

Masoud Zargar


calculus
integration
trigonometry
calculus computations

sec^2 (t/4)             a=0, b=pie

let u= sec

du/dx= sec tan

du= sec tan dx

i got stuck or messed up right here

U^2 (t/4)

now i know i'm supposed to antidefferentiate but the t/4 is confusing is it a constant or do i differentiate it

Use u= t/4

Then du= dt/4, dt= 4 du
$\int \sec^{2}(t/4) dt = 4 \int \sec^{2}u du= 4 \tan u+C = 4 \tan (t/4)+C$

So with the bounds of [itex]\pi[/itex] and 0, the definite integral is equal to 4.


inequalities
function
search
inequalities proposed

Let $a,b,c,d$ be nonnegative real numbers such that $a+b+c+d=1$. Prove that
\[1+32abcd\geq 3(ab+bc+cd+da+ac+bd). \]
Determine equality case(s).

If it's true then it's easily killable with Lagrange multiplyers as on the boundary (when one of the variables is equal to 0) inequality is trivial and for point inside we get ($a=b$ or $cd=\frac{3}{32}$) and alike symetrically (changed place of $a,b,c,d$) so it's just enough to check some cases (not so many).
If one looks for a nicer solution then probably mixing variables works as the problem is very similar to IMO SL 26/1993

It is TRUE and solved. What Megus mentioned is not for high school mathematics. I did not use mixing variable.

I know that you may use Convex function, Thuan. Rewrite it in form of $ab$ and $cd$ and notice that $0\le ab\le\frac{(a+b)^{2}}{4}, 0\le cd\le \frac{(c+d)^{2}}{4}$. However, the general problem as follow

\[(n-1)(a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2})+na_{1}a_{2}...a_{n}\ge n^{2}\]

can be solved without mixing variable, too. Anyway, this problem and its generalization are old (I think they belong to VASC) so you can search it in this forum.

Since, $2(ab+bc+cd+da+ac+bd)=1-a^{2}-b^{2}-c^{2}-d^{2}$, we can rewrite our inequality as
$3(a^{2}+b^{2}+c^{2}+d^{2})+64abcd\geq 1$ <=> $3(a^{2}+b^{2}+c^{2}+d^{2})(a+b+c+d)^{2}+64abcd\geq (a+b+c+d)^{4}$, since the last inequality
is homogenous and symmetric $WLOG$ let $d=1$, $a\leq b\leq c\leq d\leq 1$ $\Rightarrow$  $a+b+c\leq 3$, $abc\leq 1$.
<=> $3(a^{2}+b^{2}+c^{2}+1)(a+b+c+1)^{2}+64abc\geq (a+b+c+1)^{4}$. Let $3\geq a+b+c=x$, $1\geq abc=z$, $ab+bc+ca=y$ $\Rightarrow$ $9xz\leq 27z$
<=> $3(x^{2}-2y+1)(x+1)^{2}+64z\geq (x+1)^{4}$
<=> $x^{4}+x^{3}+x+1+32z\geq 3y(x+1)^{2}$

<=> $x^{4}+x^{3}+x+1+32z\geq 3(x^{2}y+2xy+y)$

using the inequalities
$x^{3}+9z\geq 4xy$, $x^{2}\geq 3y$, $x^{4}+27z\geq x^{4}+9xz\geq 4x^{2}y$ we will have ..................

In fact, I've heard that this method is called "$Brute$ $Force$" method, but I couldn't find any information about this method
in the internet  :( . I think this technics is very powerful, for instance in the Vasc's book there are many problems that can be
solved with this method, and for some of them it's the shortest way to do it.

However, are there any articles, or do you know any other information about "$Brute$ $Force$" ?

p.s. Thanks pvthuan, I think this inequality is really nice.  :)


$\bigstar$ Find the best k constant of the following inequality:
\[\frac{1}{2+\sqrt[k]{a}}+\frac{1}{2+\sqrt[k]{b}}+\frac{1}{2+\sqrt[k]{c}}\ge 1\]
If $a,b,c$ are positive real numbers such that $a+b+c=3$

For $ k>1$ the inequality is correct, what do you mean find the best k constant???!!!

for $ k>1$ and $ a,b,c$ positive reals, we have $ \frac{a+b+c}{3}\leq\sqrt[k]{\frac{a^{k}+b^{k}+c^{k}}{3}}$

[hide="If you ask why here is a famous inequality!"]
If $ a_{1},a_{2},\cdots,a_{n},w_{1},w_{2},\cdots,w_{n}$ are poritive reals and $ r,s$ are nonzero reals such that $ r<s$, then we have:

$ \left(\frac{\sum^{n}_{i=1}w_{i}a^{r}_{i}}{\sum^{n}_{i=1}w_{i}}\right)^{\frac{1}{r}}\leq\left(\frac{\sum^{n}_{i=1}w_{i}a^{s}_{i}}{\sum^{n}_{i=1}w_{i}}\right)^{\frac{1}{s}}$

now put $ n=3,w_{i}=1,r=1,s=k$ then we have the inequality said above...!
EDITED!
[/hide]

now by Cauchy we have:

$ \left(\frac{1}{2+\sqrt[k]{a}}+\frac{1}{2+\sqrt[k]{b}}+\frac{1}{2+\sqrt[k]{c}}\right)\left(2+\sqrt[k]{a}+2+\sqrt[k]{b}+2+\sqrt[k]{c}\right)\geq9$

and:

$ \left(\frac{1}{2+\sqrt[k]{a}}+\frac{1}{2+\sqrt[k]{b}}+\frac{1}{2+\sqrt[k]{c}}\right)\geq\frac{9}{6+\sqrt[k]{a}+\sqrt[k]{b}+\sqrt[k]{c}}$

but because $ k>1$:

${ \frac{\sqrt[k]{a}+\sqrt[k]{b}+\sqrt[k]{c}}{3}\leq\sqrt[k]{\frac{a+b+c}{3}}=1\Longrightarrow\sqrt[k]{a}+\sqrt[k]{b}+\sqrt[k]{c}}\leq3$
and
$ \frac{1}{2+\sqrt[k]{a}}+\frac{1}{2+\sqrt[k]{b}}+\frac{1}{2+\sqrt[k]{c}}\geq1$


geometry
perimeter
search
circumcircle
IMO Shortlist
geometry proposed

prove that in the all of equailateral triangles that the vertex of this triangles lies on $ AB,AC,BC$ the $ F_1F_2F_3$ has the less perimeter that $ F_1$ is the projection of conjugate of $ F$ that $ F$ is the Fermat point in the triangle $ ABC$ (and $ F_2,F_3$ like that)  (Fermat point is a piont that sees the sides of triangle with equal angles)

MJ GEO, please give your posts meaningful subject. The subject has to say what the problem is about in best way possible. This is in fact a requirement for being able to search old problems. As for this problem check issue #13 of  IMO Shortlist 93 for a proof and further results ([url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80880[/url]).

thank for your refrence,but i cant find anything between too problems.now see my solution.first its easy to prove that $ F_1F_2F_3$ is equailateral and then prove that $ F_1F_2$ is per perpendicular to $ AF$ and the others like that ( $ F_1$ is on $ AB$ and $ F_2$ is on $ AC$ ) now if $ F$ be the Fermat point then $ S(ABC)\equal{}S(AF_1FF_2)\plus{}S(BF_1FF_3)\plus{}S(CF_3FF_2)$ so $ S(ABC)\equal{}(AF\plus{}BF\plus{}CF)(x)$ that $ x$ is $ F_1F_2,F_2F_3,F_1F_3$ and its trivial that $ AF\plus{}BF \plus{}CF<AK\plus{}BK\plus{}CK$ that $ K$ is an arbitray point in triangle and the problem is solve.oh im sorry too much because of writing wrong problem,we must prove $ F\minus{}1F_2F_3$ has the less surface :oops:

Let $ \triangle A'B'C'$ be an arbitrary equilateral triangle circumscribed in $ \triangle ABC.$ Circumcircles $ (X),$ $(Y),$ $(Z)$ of $ \triangle A'CB,$ $\triangle B'AC,$ $\triangle C'BA$ concur at the Miquel point of $ \triangle A'B'C' \cup ABC,$ i.e. the 1st Fermat point $ F$ of $ \triangle ABC.$ Lines $ FA,FB,FC$ are pairwise radical axes of the circles $ (X),(Y),(Z).$ Then $ \triangle XYZ$ is equilateral whose sidelines are the perpendicular bisectors of $ FA,FB,FC.$ Let $ M,N$ be the projections of $ Z,Y$ on $ B'C'$ and $ Z'$ the projection of $ Y$ on $ ZM.$ If $ Z \not\equiv Z',$ then $ YZ \ge YZ' \equal{} MN \equal{} \frac {_1}{^2}B'C'$ and we get similar expressions cyclically. Thus, $ \triangle A'B'C'$ attains its maximum area (perimeter) if and only if $ \triangle A'B'C'$ is centrally similar to $ \triangle XYZ.$ That is, if $ \triangle A'B'C'$ is centrally similar to the pedal triangle $\triangle F_1F_2F_3$ of the 1st Isodynamic point $ F'$ of  $\triangle ABC.$

For each equilateral triangle $ \triangle DEF$ inscribed in $ \triangle ABC$  there exists another equilateral triangle $ \triangle D'E'F'$  circumscribed in $ \triangle ABC$  and  centrally similar to $ \triangle DEF.$ By Gergonne-Ann theorem we have $ [ABC]^2 \equal{} [DEF] \cdot [D'E'F']$ $\Longrightarrow$ $ [DEF]$ attains its minimum iff $ [D'E'F']$ attains its maximum, i.e. if $ \triangle D'E'F'$ is centrally similar to the pedal triangle of $ F'$ WRT $ \triangle ABC.$ Then $ \triangle DEF$ is identical with $ \triangle F_1F_2F_3.$


For every real numbers $x,y(y\ge 0)$ we denote $S(x,y)=max(1-2x+y ,1+2x+y,x^2+y^2)$
Find the minimum value of $S(x,y)$

$S(x,y) \ge 1-2x+y$
$S(x,y) \ge 1+2x+y$
then
$S(x,y) \ge 1+y \ge 1$
let $x=y=0$
we can get
$S(x,y) =1$
so the answer is 1.
am I wrong?

Remark:It is also can do if replace $y \in R$ to $y \ge 0$  :)

then it is equal to find the minimum of
$F(y)={1+y,y^2}$
you can see it will happen when $1+y=y^2,y<0$
that means $y=\frac{1-\sqrt 5}{2}$,the answer is $y=\frac{3-\sqrt 5}{2}$
this is because:
if $y \ge \frac{1-\sqrt 5}{2},1+y \ge \frac{3-\sqrt 5}{2}$
if $y \le \frac{1-\sqrt 5}{2},y^2 \ge \frac{3-\sqrt 5}{2}$
we are done :)


Prove $(x+y+z)^3-x^3-y^3-z^3=3(x+y)(y+z)(z+x)$ geometrically.
i.e. using geometry.

You can imagine a cube with side x+y+z and divide it up into 27 rectangular parallelepipeds (with sides x,y,z) and "see" it in drawed picture, but it's not much easier then just multiplying out both sides :)

BTW: do you know any much more simplier solution?

Sorry... even i do not have a simple solution for this problem other than the algebraic one. Happy New year all!  :)


analytic geometry
search
probability and stats

I hope this is the right forum for this; it's definitely college-level, but there didn't seem to be a combinatorics board in College Playground; this Statistics board is the closest I could get. I trust a moderator will move it somewhere more appropriate if I chose wrongly ;)

This is a very interesting problem proposed by John Conway.

Consider an integer plane $ \mathbb Z\times\mathbb Z$ as representing an infinite checker board. We start the game by putting at most one checker on any given square where $ (a,b)$ where $ b\leq0$ (that is, the lower half of the plane). Checkers may be jumped over one another either horizontally or vertically as in the diagram below:
[img]http://komrades.org/checkersweb.png[/img]
The goal is to find the smallest possible configuration of starting checkers (with all $ b\leq0$) such that you can reach a given target square $ (0,g)$ using only the moves described above. For instance, for $ g \equal{} 1$, the configuration shown in the example above would work, with the two checkers shown being at $ (0,0)$ and $ (0, \minus{} 1)$. The configuration for $ g \equal{} 2$ looks like
[img]http://komrades.org/g2.png[/img]

There are a number of interesting problems we can ask ourselves about this game. Most of these were proposed by my Combinatorics prof (well, to us, at least. I'm sure he isn't the first to ask these questions :))

1) What is the minimum number of checkers needed to win for $ g \equal{} 1,2,3,4$?
2) Prove that it is not possible to win for $ g \equal{} 5$ regardless of how many checkers you place down.
3) What if we are allowed to jump diagonally, as well as horizontally/vertically? Find a new $ g$ such that it is not possible to win.
4) Extend this game in the obvious way to $ n$ dimensions. Show that for any $ n$ we play with, there is some $ g$ so that the goal $ g$ is impossible to win, regardless of how many checkers are placed down.

Hope you guys enjoy this problem as much as I have!

I've talked to Conway about this, so I'm not going to spoil things for everyone by just reciting what he told me, but here's a

[hide="Hint"]Assign a value to each coordinate in such a way that a move does not increase the sum of the values of all occupied coordinates. Then the value of the final position is at least as large as the value of the starting position that gets you there. [/hide]

Grobber posted a solution to a part of this a long time ago:
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=760797933&t=21914

The idea is also presented in Engel, I think.

For a solution to the first two parts, my combinatorics prof posted this: http://www.math.uwaterloo.ca/~dgwagner/MATH249/checkers.pdf

For a solution to the second two parts, I wrote this: http://komrades.org/personal/Ass03_half.pdf

I found these things really enlightening to do, I hope others enjoy them similarly :)

[b]EDIT:[/b] And, um, oops, I just discovered the Combinatorics board in the olympiad section. Could a mod please move this topic there? It's much more fitting.


i got another way and simpler way of doing 2nd prob
take $(a+b)(b+c)(c+a)+abc$
let $s=a+b+c$
then i can write above as $(s-a)(s-b)(s-c)+abc$ which can be written as $s^{3}-s^{2}(a+b+c)+s(ab+bc+ac)-abc+abc$
now it can written as $s^{3}-s^{3}+(a+b+c)(ab+bc+ac)$
which is $(a+b+c)(ab+bc+ac)$
hence
$(a+b)(b+c)(c+a)+abc = (a+b+c)(ab+bc+ac)$
use abc = 1
$(a+b)(b+c)(c+a)+1 = (a+b+c)(1/c+1/b+1/c)$
hence proved....


Pentru cei (încå) necunoscåtori o initiere si câteva exemple interesante [url=http://www.math.ust.hk/excalibur/v11_n1.pdf]aici[/url].

vezi ca se supara Cezar cand te aude cu Muirheadu :))

Nu inteleg de ce Mu$*$rhead e asa banned subject  :)  am pus post-ul pentru ca n-am gasit nici un topic explicit (des)p(r)e tema asta  :arrow:

Despre $\ Inegalitatea\ lui\ Muirhead$ poti gasi mai multe si in sectiunea $\ Inequalities\ Theorems\ and\ Formulas$ ;) si bineinteles in inegalitatile propuse de mai marii forumului acolo.

materialu ala e util, parerea mea. stiu ca este metoda nu chiar eleganta de a rezolva o inegalitate, dar ies foarte multe asa. la juniori daca o stii e super tare ca majoritatea inegalitatilor sunt simetrice. (aici fac observatia ca daca inegalitatea nu e simetrica, atunci poti sa abandonezi din start metoda. daca e simetrica, ai mari sanse de reusita). la seniori se dau inegalitati mai "smechere" dar uite ca anul acesta am avut o inegalitate destul de greu de abordat prin metode elegante, si majoritatea celor de la baraj au rezolvat`o (si eu ma numar aici) cu omogenizare, desfacut paranteze, si apoi inegalitatea mediilor sau muirhead aplicata de mai multe ori. si acum imi aduc aminte, erau 243 de termeni in fiecare membru, si nici nu iesea imediat. trebuia sa aplici medii de vreo 3-4-5 ori. totusi, imediat cum am citit inegalitatea, m`am apucat de calcule, spunandu`mi in gand ca "nu se poate sa nu iasa asa". consider ca am procedat bine, deoarece dupa o ora aveam o problema redactata complet, si moralu sus ca sa zic asa :)

[quote="stancioiu sorin"]Cred ca majoritatea de pe ML stiu <snipped>, ca sa-l citez pe Cezar.
 :)[/quote]Ai grija la exprimare!

Eu l-am citat pe Cezar, Valentin.Acolo unde a scris si el asa de ce n-ai cenzurat??Oricum, imi cer scuze pentru exprimare. :oops:

Da.Am vazut si unde era postul ala al lui Cezar cu Inegalitatea lui Muirhead.Uitati-va pe $\ Inegalitatea\ lui\ Finsler$, pe sectiunea asta.Gasiti voi. :) Era doar o gluma totusi.

[quote="sorin"]Acolo unde a scris si el asa de ce n-ai cenzurat??[/quote]
D-l student Cezar Lupu e moderator si stie foarte bine ce vorbeste. Bine ca iti ceri scuze macar  :)

Din putinele mele cunostinte, Cezar Lupu nu e profesor, ci student in anul II(sper sa fie corect) de facultate la Universitatea din Bucuresti. :) Oricum, mi-am cerut scuze si daca e nevoie, mai imi cer.Sa zicem asa:
$\ {Imi\ cer\ scuze}^{2006(n+2006)}$, $\ n\geq\ 2006$.

Off-topic: Am corectat post-ul, cred ca l-am confundat cu altcineva :oops: . O $\infty$ de scuze!
On-topic: Intr-adevar, si eu consider cå Muirhead e cea mai bunå metodå "calculatorie" în cazul în care avem sume simetrice, fiind de folos la JBMO etc.. Am dat link-ul deoarece (no offense) topic-ul de la sectiunea sus-numitå era de fapt mai mult o definitie si nu då exemple (din câte îmi amintesc) de cazuri de aplicabilitate sau alte relatii.

Ok,...Sa stiti ca Inegalitatea lui Muirhead este foarte utila.Sa zicem urmatoarele probleme, date la etapa regionala a olimpiadelor rusesti:
   [b] Problema 1[/b]
 Sa se demonstreze ca  pentru orice numere reale $\ x,y$ este adevarata inegalitatea $\ 2x^4+2y^4\geq\ xy(x+y)^2$.
   [b]Problema 2[/b]
  Sa se arate ca daca $\ a,b\geq\ 0$, atunci avem inegalitatea
$\ a^3(b+1)+b^3(a+1)\geq\ a^2(b+b^2)+b^2(a+a^2)$.

   Cele doua inegalitati sunt simple, dar pot fi solutionate usor folosind Inegalitatea lui Muirhead.

[b]Problema 1[/b].
$(4,0)\succ (3,1)$=>$x^4+y^4 \ge x^3y+xy^3$
$(4,0)\succ (2,2)$=>$x^4+y^4 \ge 2x^2y^2$
[b]Problema 2[/b]
$(3,1)\succ (2,2)$=>$a^3b+b^3a \ge a^2b^2+b^2a^2$
$(3,0)\succ (2,1)$=>$a^3+b^3 \ge a^2b+b^2a$
Insumând obtinem cele 2 relatii cerute, si uite rezolvarea unei probleme de la RRO  :)  E adevårat cå multe inegalitati mai putin "smechere" de la regionale pot fi solutionate astfel (a se vedea problemele mai vechi de la faza regionala din India).

Solutiile sunt corecte.La inegalitatea a doua, mai pot zice doar ca putea fi scrisa sub forma $\ (a-b)^2(a+ab+b)\geq\ 0$, care e adevarata in conditiile problemei.

pentru "juniorii" forumului si cei care nu stiu aceste probleme...
1)
$a,b,c>0, \sum a=3$.
demonstrati ca $\sum\frac{1}{a^{2}}\geq\sum a^{2}$
baraj 3 - 2006
2)
$x,y,z>0, xyz\geq 1$
demonstrati ca $\sum \frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}\geq 0$
OIM 2005

insa, apropo de metoda, eu nu m`as risca sa fac o problema la judeteana sa zicem, cu muirhead...

Problema a doua e prea lunga pentru a o mai posta. :( Solutie frumoasa si eleganta a gasit Iurie Boreico pentru aceasta problema.

[/quote]Din putinele mele cunostinte, Cezar Lupu nu e profesor, ci student in anul II(sper sa fie corect) de facultate la Universitatea din Bucuresti. Smile Oricum, mi-am cerut scuze si daca e nevoie, mai imi cer.[quote]

* observatzia era doar ca se zice "imi mai cer " :)  :blush: [/quote]

scuzatzi probleme tehnice din mesaju anterior :)

Am si eu o intrebare in legatura cu inegalitatea lui Muirhead.

E valabil si pt $ a^3b\plus{}b^3c\plus{}c^3a\ge a^2bc\plus{}b^2ca\plus{}c^2ab$?


Ethanenitrile from calcium carbide in 5 steps.

Where have you posted . Here ? Please give the link.


number theory
prime numbers
number theory unsolved

Let $ p$ is prime numbers and $ p > 3$ . Find all pairs of integers $ (a,b)$ that satisfy this equation $ a^2 \plus{} 3ab \plus{} 2p(a\plus{}b) \plus{} p^2 \equal{} 0$,and $ a,b \ne 0$

analyse the discriminant wrt $ a$...
u'll get $ \Delta\equal{}9b^2\plus{}4pb\equal{}b(9b\plus{}4p)$ which must be a perfect square.
then analyse the different cases for $ \gcd(b,9b\plus{}4p)$.

$ \Delta\equal{}9b^{2}\plus{}4pb\equal{}q^{2}$  $ \implies$
$ 9b^{2}\plus{}4pb\minus{}q^{2}\equal{}0$  $ \Delta\equal{}16p^{2}\plus{}4.9q^{2}\equal{}4(4p^{2}\plus{}9q^{2})$  $ \implies$
$ (4p^{2}\plus{}9q^{2})$  must be perfect square  $ \implies$ $ 4p^{2}\plus{}9q^{2}\equal{}s^{2}$  $ \implies$
 $ 4p^{2}\equal{}(s\minus{}3q)(s\plus{}3q)$  $ \implies$  $ (s\minus{}3q,s\plus{}3q)\equal{}d$  $ \implies$  $ 2|d$  $ \implies$
$ i)s\minus{}3q\equal{}s\plus{}3q\equal{}2p$ $ \implies$ $ q\equal{}0$  $ \implies$ $ 9b\plus{}4p\equal{}0$  $ \implies$ $ 9|p$ contradiction!
$ ii)s\minus{}3q\equal{}2,s\plus{}3q\equal{}2p^{2}$

[quote="rajelaz"][quote="shoki"]
then analyse the different cases for $ \gcd(b,9b \plus{} 4p)$.[/quote]

How to divide cases???
for example , i've got problem
"Determine all $ n$ for $ n \in \mathbb{N_0}$ and $ (n \plus{} 3)(n^2 \plus{} 3)$ is perfect square.."
i'm so sorry... i m just newbie here  :oops:[/quote]
i don' t know how u got such problem( :| ) but anyway see that $ \gcd(b,9b \plus{} 4p) \in \{1,2,4,p,2p,4p\}$.
case 1-$ b\equal{}x^2,9b\plus{}4p\equal{}z^2 \implies z^2\minus{}(3x)^2\equal{}4p$.the rest is very trivial.
case 2-$ b\equal{}2x^2,9 \times 2x^2\plus{}4p\equal{}2z^2 \implies 2p\equal{}z^2\minus{}(3x)^2$.the rest is very trivial.
case 3-$ b\equal{}4x^2,9 \times 4x^2\plus{}4p\equal{}4z^2$.the rest is very trivial.
all other cases are trivial too.
if u don't know how to solve them do the same thing that cnyd did:i.e. for case 1 ,for example we have $ 4p\equal{}(z\minus{}3x)(z\plus{}3x)$
which leads to these cases:
$ 1\minus{} \pm4p\equal{}(z\minus{}3x),(z\plus{}3x)\equal{}\pm1,2\minus{}\pm2p\equal{}(z\minus{}3x),(z\plus{}3x)\equal{}\pm2,3\minus{}\pm p\equal{}(z\minus{}3x),(z\plus{}3x)\equal{}\pm4,...,6\minus{}\pm1\equal{}(z\minus{}3x),(z\plus{}3x)\equal{}\pm4p$


geometry
congruent triangles
geometry unsolved

How many non-congruent triangles can you fit on a 5x5 geoboard.  Non-congruent meaning not the same triangle in a different position but a totally different triangle.  I know this problem can be solved by drawing all of the possible triangles out I was just wondering if there was a faster mathematical way of doing it.  Any help is appreciated! :)

Not really, I think...

$%Error. "comb" is a bad command.
{25}{3}-12\times%Error. "comb" is a bad command.
{5}{3}-4%Error. "comb" is a bad command.
{5}{4}-4$ gives the number of (non-congruent) triangles. um... I think you just have to draw them


number theory
greatest common divisor
least common multiple

$ a_1, a_2,..., a_n$ are positive integers such that:
1) $ a_i < 2008$ for all i=1,2,...,n
2) The least common multiple of any pair of these numbers is greater 2008.
Show that $ \sum_{i \equal{} 1}^n \frac {1}{a_i} < 2$
:)

Edited!

Shouldn't "greatest common divisor" be "least common multiple", since 2 numbers < 2008 can't have a gcd that's > 2008.

@[b]Carbon57[/b]: I edited my problem! Do you have a solution for the exact problem?


integration
limit
function
calculus
logarithms
real analysis
real analysis unsolved

Let $0<c_1<\varphi(x)<c_2$ for positive $x$. Define $I_{\alpha}(T)=\int_0^T \varphi(t) t^{\alpha}dt$. If $\lim_{T\rightarrow \infty} I_{-1}(T)/\log(T)= C$, does it imply that $\lim I_{\alpha}(T)/T^{\alpha+1}=C/(\alpha+1)$?

No, it doesn't. Just consider the function $\varphi$ that is $c_1$ most of the time but, on a very sparse set of intervals $I_k=[a_k,e a_k]$ equals $c_2$. Then, for each $T$, the integral $I_{-1}(T)$ is between $c_1\log T$ and $c_1\log T$ plus $(c_2-c_1)$ times the number of intervals $I_k$ intersecting $[0,T]$, which can grow as slow as we want. On the other hand, when integrating a power bigger than $-1$, these intervals wil matter a lot. So, some additional restrictions on $\varphi$ are needed.


What integer is closest to the value of $ \sqrt[5]{1000}$?

we 1st find the # closest to 1000 so we start with 2^5=32 putting into chart looks like

2^5=32
4^5=1024

AH HAH! we found the number closest to the 1000 so the answer is 4

[size=200][b][i][u]PLEASE RATE MY ANSWER THANK YOU SO MUCH !!![/u][/i][/b][/size]

 :pilot:  :pilot:  :pilot:  :pilot:

:gathering:  :gathering:  :gathering:  :gathering:
wow


vector
topology
advanced fields
advanced fields unsolved

Hello all could somebody please help me to figure out how to prove this:

[quote]For a Riemannian manifold $ (M,g)$, define $ \Xi \equal{}\{V \in \mathfrak{X}(M) : g([V,W],U) \plus{} g(W, [V,U])\equal{}0 \; \;\forall \;  W,U \in \mathfrak{X}(M)\}.$    Prove that $ \Xi$ is a finite dimensional subspace of $ \mathfrak{X}(M)$.   In addition, show that if  $ V$ and $ V'$ are in $ \Xi$ so is $ [V,V'].$[/quote]

Any help will be greatly appreciated.

I guess proving $ \Xi$ is a subspace is not hard and the fact it is finite dimensional follows by taking the standard vector fields.

Ive been having troubles with the second part though.  I guess it follows from some relation between $ g$ and the Lie bracket but I am not sure what that is and if that is the way to prove this problem.

Any hint will be greatly appreciated (so by that I mean I dont want you to solve the problem but give me some ideas)

Thank you.

the first point seems to be really a standard check, as you said.

regarding its dimension, i'd say that a vector field in $ \Xi$ could be determined by its value at a point, but i haven't written a line about it.

the last point seems to be a consequence of the jacobi equation, that is
$ [[X,Y],Z]\plus{}[[Y,Z],X]\plus{}[[Z,X],Y]\equal{}0$ for any choice of $ X,Y,Z$
but, again, i made no actual computations so i could be completely wrong.


find x,y,z such that  x 2 +y 2 +z 2 =xyz
                              x^3+y^3+z^3=(xyz) 2  :(  :lol:


If $x$ and $y$ are really small and $x<y$ the minimum can be as small as possible.

Yes, it's problematic.   For example, if $y = 2x$ then we have

$A = x-\frac{1}{2x^{3}}$

And as $x \to 0^{+}$ we have $A \to-\infty$.

[quote="chien than"]Let $x;y$ be positive real numbers satisfying $x \neq y$ .Find min:
$A=x+\frac{1}{xy(x-y)}$[/quote]
I don't understand? I'll consider finding min of $|A|$.
Set $x$ constant. We must maximize $y(x-y)$; clearly $x<y$ for minimum in the original so we want to minimize \[y(y-x)=y^{2}-xy=\left(y-\frac{x}{2}\right)^{2}-\frac{x^{2}}{4}\geq-\frac{x^{2}}{4}\]
Thus \[A\geq x+\frac{-4}{x^{3}}\] with equality at $2y=x$ in the original.
To minimize $x-\frac{4}{x^{3}}$ is impossible since it approaches infinity... ok I see the issue now: $0<x<y$ doesn't coincide with $2y=x$

---

Again fix $x$ and find the derivative of $A$, i.e. \[\frac{\partial A}{\partial y}=\frac{1}{x}\cdot \frac{\partial A}{\partial y}\left(yx-y^{2}\right)^{-1}=\frac{2y-x}{x(y(x-y))^{2}}\] with the bottom strictly larger than zero so the only relative extrema occurs at $2y=x$ and \[A=2y+\frac{1}{2y^{3}}\]. The min of the abs. value must occur at a relative extrema or absolute extreme except any infinities can be ignored. The minimum of the absolute value of \[A=2y+\frac{1}{2y^{3}}\] is the minimum of $A$ ($y$ positive) which is at $4y^{4}=3$ or at $y=\sqrt[4]{\frac{3}{4}}$ where \[A=\frac{4y^{4}+1}{2y^{3}}=\frac{2}{y^{3}}=\frac{2}{\sqrt[4/3]{\frac{3}{4}}}\]

[quote="chien than"]Let $x;y$ be positive real numbers satisfying $x \neq y$ .Find min:
$A=x+\frac{1}{xy(x-y)}$[/quote]
$1,5A=(x-y)+\frac{1}{2}x+y+\frac{1,5}{xy(x-y)}\geq 4\sqrt[4]{(x-y)\frac{1}{2}xy\frac{1,5}{xy(x-y)}}=4\sqrt[4]{\frac{3}{4}}$


LaTeX
inequalities
inequalities unsolved

let$a,b, c$  positive reals  such that :$abc \leq 1$
 prove that :
  $a/c +b/a +c/b \geq a+b+c$

[quote]Why don't you type $\LaTeX$ clearer, it would be more readable:

[quote="tranthanhnam"]Let $a,b,c$ be positive reals such that: $abc \le 1$. Prove that:
\[ \frac{a}{c}+\frac{b}{a}+\frac{c}{b} \ge a+b+c  \][/quote][/quote]
Inequality:
$\frac{a}{c}+\frac{b}{a}+\frac{c}{b} \ge a+b+c$
$\iff a^2b+b^2c+c^2a \ge a+b+c$ from $abc \le 1$
By Muirhead's:
$\frac{a^2b+b^2c+c^2a}{\sqrt[3]{(abc)^2}} \ge a+b+c$
...

ah`, i think that  solutionyour is wrong
    because  $abc \leq 1$  , not $abc \geq 1$  .
  hence , i want to  have a new solution
 can you help me ?

hmm... tranthanhnam, I think it's true by Muirhead's:
\[ \frac{a^2b+b^2c+c^2a}{abc} \ge \frac{a^2b+b^2c+c^2a}{\sqrt[3]{(abc)^2}} \ge a+b+c  \]
$abc \le \sqrt[3]{(abc)^2}$ follows from $abc \le 1$.

I have a question.If $\ abc$<$\ 1$, can we say that $\ a+b+c$>$\ ab+bc+ca$ ?Because if we prove this one, the problem can be solved using Cauchy in Engel Form(or Berstrom, it's the same thing).

We presume that a=>b=>c.
Using the re-arangement inequality for numbers a=>b=>c and 1/c=>1/b=>1/a we have
\[ \frac{a}{c}+\frac{b}{a}+\frac{c}{b}>\frac{a}{a}+\frac{b}{b}+\frac{c}{c}=3 \]
but c<=b<=a<=1(a,b,c are positive and abc<1)
\[ a+b+c<3 \]
Q.E.D
if a=b=c=1 we have both members equal

[quote="luke999"]We presume that a=>b=>c.[/quote]

The inequality is not symmetric so your proof is flawed :)

Indeed, but the power of the rearrangement ability is exactly that (a,b,c),(1/a,1/b,1/c) are always opposedly sorted, so it is very easy to fix. :)

The rest of his proof is also flawed  :D

Indeed, the other part is completely wrong. :)

Anyway, I'm afraid to say something very stupid now, but I've been wondering this all along... what's the problem with inequalities of which 1 side is symmetric and the other is not (but is cyclic on 3 variables)?

It is obvious that $\frac ac +\frac ba +\frac cb \geq a+b+c$ is true $\forall a,b,c\in\mathbb{R}^+$ if and only if $\frac ab +\frac bc +\frac ca \geq a+b+c$ is true $\forall a,b,c\in\mathbb{R}^+$, so it is sufficient to prove $\frac ac +\frac ba +\frac cb + \frac ab +\frac bc +\frac ca \geq 2(a+b+c)$, which is symmetric... ?

That's an idea I've had in the past as well. It's not completely rigorous... so I don't know...

Why/where is it not rigorous? The selection of a random triplet $(a,b,c)$ is identical to the selection of a random triplet $(a,c,b)$... if it is true for all triplets $(a,b,c)$ then it is by definition true for all triplets $(a,c,b)$.

If it is correct, I don't understand how such problems still pose a challenge to people. But I don't really see the flaw...

This reasoning is not rigorous as it is wrong. The following gives a counterexample:
Consider the statement
$\forall a,b,c \in \mathbb{R}^+: \frac a b + \frac b c + \frac c a \ge \frac 1 2 \big( \frac a b + \frac b c + \frac c a + \frac b a + \frac c b+ \frac a c \big)$.
It is equivalent to
$\forall a,b,c \in \mathbb{R}^+: \frac b a + \frac c b + \frac a c \ge \frac 1 2 \big( \frac a b + \frac b c + \frac c a + \frac b a + \frac c b + \frac a c \big)$.
Now consider the symmetrization of the first statement
$\forall a,b,c \in \mathbb{R}^+: \frac a b + \frac b c + \frac c a + \frac b a + \frac c b + \frac a c \ge \frac a b + \frac b c + \frac c a + \frac b a + \frac c b + \frac a c$,
which is obviously true. This however does NOT prove the first statement, which in fact is wrong (e.g. $a=1$, $b=3$, $c = 2$).


MATHCOUNTS
algebra
difference of squares
special factorizations

What is the sum of the prime factors of 2^16 -1.

     Any hints whould be helpful.

                             Thanks :lol:

[hide="hint"]  Try applying the difference of squares factoring trick. [/hide]

[hide="Further more"]  Try applying it more than once! [/hide]

[quote="prototypex24"]Following G-units hints,
[hide]
you'll get $(2^2-1)(2^2+1)(2^4+1)(2^8+1)=(3)(5)(17)(33)$

The prime factors are 3, 5,11,17
Sum:  36
[/hide][/quote]

prototypex24, isn't $(2^8+1)=257?$ Also, where did you get the 11 from? I think the answer should be [hide]257 + 17 + 5 + 3 = 282???[/hide]

I agree with plokoon51. $2^{16}-1$ is not divisible by $11$...


let us start to list the qualifiers for the 2007 state comprehensive contest

[list]

Name                  @                                                     Qualified Site
John Berman @ UNC -Wilmington
Jenny Chen         @                                                    UNC -Wilmington
Bryce Taylor          @UNC-Charlotte
Brendan Fletcher       @UNC-Charlotte

You can find many more from the threads for each site on this forum, though I'm not sure what purpose this list is really supposed to serve.

I would be much more interested to know which of last year's ARML participants didn't qualify. There are always a few, and I know of two already (Damien and Lizi).

[quote="jb05"]I would be much more interested to know which of last year's ARML participants didn't qualify. There are always a few, and I know of two already (Damien and Lizi).[/quote]

Last year, while students took the test, coaches had a meeting. During the meeting, Mr. John Goebel said they would invite students who did not qualify through the regionals, but made the USAMO to take the State Comprehensive Final.

At first, he said some people suggested that let students who made the AIME Honor Roll be invited to take the final. But they changed since too many students made the AIME Honor Roll last year in NC.

So if Damien and Lizi made the USAMO, they may still get invited to take the final.

Really? YES!!!!!!! I really hope so! :blush:

Thanks to Mr. Goebel!

I'm pretty sure Lizi didn't make USAMO - his AMC12 score was 120ish, but his AIME score may not have been high enough.

On that subject, I don't know if Vivek made it - he didn't do very well on AIME either.  I know he did worse than last year, and I think he scored about a 7.


Let be $ n,k\in \mathbb{N}$ such that $ k<n$ , and let be $ x_1,x_2,...,x_n\in \mathbb{R}_\plus{}$ such that $ \prod_{i\equal{}1}^nx_i\ge 1$ . Show that :

$ \frac{x_1^3\plus{}x_2^3}{x_1^2\plus{}kx_1x_2\plus{}x_2^2}\plus{}\frac{x_2^3\plus{}x_3^3}{x_2^2\plus{}kx_2x_2\plus{}x_3^2}\plus{}...\plus{}\frac{x_n^3\plus{}x_1^3}{x_n^2\plus{}kx_nx_1\plus{}x_1^2}\ge \frac{2n}{k\plus{}2}$

[quote="alex2008"]Let be $ n,k\in \mathbb{N}$ such that $ k < n$ , and let be $ x_1,x_2,...,x_n\in \mathbb{R}_ \plus{}$ such that $ \prod_{i \equal{} 1}^nx_i\ge 1$ . Show that :

$ \frac {x_1^3 \plus{} x_2^3}{x_1^2 \plus{} kx_1x_2 \plus{} x_2^2} \plus{} \frac {x_2^3 \plus{} x_3^3}{x_2^2 \plus{} kx_2x_2 \plus{} x_3^2} \plus{} ... \plus{} \frac {x_n^3 \plus{} x_1^3}{x_n^2 \plus{} kx_nx_1 \plus{} x_1^2}\ge \frac {2n}{k \plus{} 2}$[/quote]
Use $ \frac{a^3\plus{}b^3}{a^2\plus{}kab\plus{}b^2} \ge \frac{a\plus{}b}{k\plus{}2}$ and AM-GM inequality :)