[hide]The total number of triangles that can be formed by 1989 points in a plane is a constant. To minimize the number of triangles with vertices in different groups we maximize the number of triangles with veritces in the same groups. This maximum is obtained when the number of  points in any one group is maximized. [/hide]

I think I agree with previous poster.  If each group must have a distinct number of items, put zero, 1, 2, 3, ..., 28 points in each of the first 29 groups.  Put the remaining 1583 points in the 30th group.

oops.  I meant maximized, which is, I think, a somewhat tougher problem.

I changed the question accordingly - sorry about that...

[hide]

Suppose we have the maximum configuration.

First I will forget about the distinct sizes bit to prove this little lemma thingy:

Consider any two of the groups, with sizes a and b. Let X be the number of ways of choosing two of the vertices from different groups and not from these two groups, and let Y be the number of ways of choosing just the single vertex outside of these two groups (so Y is actually 1989-a-b, but that doesn't matter).

Now there are X(a+b) + Y(ab) ways of choosing the triangles with at least one vertex from our chosen two groups. Suppose we swap one point over so the groups have size a-1 and b+1. Now there are X(a-1+b+1) + Y(a-1)(b+1) ways. Every other triangle not in these groups has been unchanged. So we must have Y(a-1)(b+1) <= Y(ab) since we originally chose the maximum configuration. This gives ab - b + a - 1 <= ab and thus a-b <= 1.

So what this tells us, is that in the maximum configuration, we are not able to find two groups with sizes a and b with a-b>=2 so that changing them to a-1 and b+1 will leave a valid configuration.

So this means that either all the group sizes are consecutive numbers, OR they are all consecutive apart from one "gap" of two points. (if there was more than one jump then we take the group below the first jump and above the second jump and apply that top lemma thing.)

So the first case: a + a+1 + .. + a+29 = 1989 gives a = 51.8, no. The second case gives, in the easiest way to write it, a + a+1 + .. + a+29 + a+30 - (a+k) =1989 for some k between 0 and 30. Solving this gives 30a - k = 1524. So the only k is 6 and a=51, giving the group sizes 51, 52, 53, 54, 55, 56, 58, 59, .., 81.

[/hide]


function
algebra
polynomial
combinatorics proposed
combinatorics

In how many ways can a 2x2x17 hole be filled with 1x1x2 blocks so that the top is even (nothing sticking out) and there are no "holes"?

[hide="Request"]
While I know that this can be solved fairly easily using recursion, it was mentioned to me that it could be done with generating functions. I would appreciate it if someone could show me a way more elegant than the recursion, be it generating functions or something else. Thanks. [/hide]

Could u explain how to do it using recursion?

You can call $A_{n}$ the number of ways to fill a $2\times2\times n$ hole. So, when you try to put a block in the bottom you have some posibilities. If you put the block filling two spaces in the bottom, you can call $B_{n}$ the number of ways to fill this new thing. If you put the block vertically, you can call $C_{n}$ the number of ways to fill this. Anyway, there will be a system of recursions, $A_{n}=4B_{n}+\cdots$, $B_{n}=\cdots$, and so on. I think there's a lot of posibilities, may be in $Z_{n}$ :P ...You will see how to solve the system of recursions.

Sorry if my English is so bad...

I have thought the generating function. The idea is to associate the hole to a polynomial. Each space, if we enumerate the rows, the columns and the others :P , will be $x^{l}y^{m}z^{n}$, where $l$ is the column, $m$ is the row and $n$ is the other. So, the entire hole is $P(x,y,z)=xyz(1+x)(1+y)(1+z+z^{2}+\cdots)$, or something like it. Each block has three ways to be putted, in the space $a$ and $ax$, or in $a$ and $ay$ or $a$ and $az$, the different rotations. So, $P(x,y,z)=(1+x)A(x,y,z)+(1+y)B(x,y,z)+(1+z)C(x,y,z)$, where $A,B,C$ are the polynomials of the initial spaces of the blocks. Now, we must see how many ways there are to write $P(x,y,z)$ in this way. I will think about it :P

Again, sorry if my English is so bad...

The problem was solved recursively [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=154481]here[/url].  Many simple recursions lead to simple generating functions, so in this sense the problem may be solved with generating functions.  I'm curious if Esteban's method actually leads anywhere. (Also, your English is pretty good  :).)


function
inequalities
inequalities unsolved

it's just the aplication of Popoviciu for convex function $ f(x)=\frac{1}{x}$ :)

Popoviciu would yield \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{9}{a+b+c}\geq 4\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right).\] This inequality is sharper.

Sorry.We just take the calculations.I"ll skip details it wiil yeld finally $3(\sum a)^{2}(\sum a^{2}b^{2})\geq 2abc(\sum a)^{2}+9a^{2}b^{2}c^{2}+2abc(\sum a)(\sum ab)$.Now we use $\sum a^{2}b^{2}\geq abc\sum a$ so we have to prove $(\sum a)(\sum a^{2})\geq 9abc$. :P

[quote="tranthanhnam"]let $a,b,c >0.$ Prove that :
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{3}{a+b+c}\geq 8*(\frac{1}{3(a+b)}+\frac{1}{3(b+c)}+\frac{1}{3(c+a)}$[/quote] (1)

$(1) \Leftrightarrow (a+b+c)(\sum \frac{1}{a}+\frac{3}{a+b+c})\geq \frac{8}{3}(a+b+c)(\sum \frac{1}{a+b})$
 $\Leftrightarrow 3+3+\sum \frac{a+b}{c}\geq \frac{8}{3}(3+\sum \frac{a}{b+c})$
 $\Leftrightarrow \sum \frac{a+b}{c}\geq \frac{8}{3}. \sum \frac{a}{b+c}+2$
This is easily proved by:
$\frac{a}{b}+\frac{a}{c}\geq \frac{4a}{b+c}$
so  $\sum \frac{a+b}{c}\geq 4. \sum \frac{a}{b+c}= \frac{8}{3}. \sum \frac{a}{b+c}+\frac{4}{3}.\sum \frac{a}{b+c}\geq \frac{8}{3}. \sum \frac{a}{b+c}+2$
(Because $\sum \frac{a}{b+c}\geq \frac{3}{2})$

The second problem is similar:

$(2) \Leftrightarrow (a+b+c)(\sum \frac{1}{a}+\frac{9}{a+b+c})\geq 4 (a+b+c)(\sum \frac{1}{a+b})$
 $\Leftrightarrow 3+9+\sum \frac{a+b}{c}\geq 4(3+\sum \frac{a}{b+c})$
 $\Leftrightarrow \sum \frac{a+b}{c}\geq 4. \sum \frac{a}{b+c}$
This is easily proved by:
$\frac{a}{b}+\frac{a}{c}\geq \frac{4a}{b+c}$

Bodan's post was edited 6785 times in total. :rotfl: There is definitely a bug :lol:

Nice solution detectivehien!

[quote="tranthanhnam"]Let $a,b,c>0$. Prove that: \[{1 \over a}+{1 \over b}+{1 \over c}+{3 \over a+b+c}\ge{8 \over 3}\left({1 \over a+b}+{1 \over b+c}+{1 \over c+a}\right).\] [/quote]
Wlog assume $a \ge b \ge c$. Consider sum $f(a,b,c)={1 \over a}+{1 \over b}+{1 \over c}+{3 \over a+b+c}\ge{8 \over 3}\left({1 \over a+b}+{1 \over b+c}+{1 \over c+a}\right)$, \[f(a,b,c)-f\left(a,{b+c \over 2},{b+c \over 2}\right)=(b-c)^{2}\left[{1 \over bc(b+c)}-{8 \over 3}\cdot{1 \over (a+b)(a+c)(2a+b+c)}\right]\] Clearing denominators, $S_{0}=6a^{3}+12abc+9a^{2}b+3ab^{2}+9a^{2}c+3ac^{2}-5bc(b+c)$.
Obviously, $S_{0}\ge 0$.
Let $t={b+c \over 2}\le a$, it remain to show that, \[{1 \over a}+{2 \over t}+{3 \over a+2t}-{8 \over 3}\left({1 \over 2t}+{2 \over a+t}\right) \ge 0 \\ \iff a^{3}+3t^{3}+a^{2}t-5at^{2}\ge 0 \\ \iff (a-t)(a^{2}+2at-3t^{2}) \ge 0.\] Done.


geometry
3D geometry
sphere
invariant
geometry unsolved

Let $ S$ be the sphere of radius $ 1$ in $ \mathbb{R}^3$. Let $ u_1,u_2$ be two inversions leaving $ S$ invariant. Can I write te composition $ u_1ou_2$ in the form $ fog$ where $ f$ is an isometry leaving $ S$ invariant and $ g$ is an  inversion leaving $ S$  invariant ? If yes then is the decomposition unique if we assume that all considered inversions have negative power ?

Uhm, I am probably being idiot right now, but wouldn't this imply (just by taking $ u_2\equal{}u_1$) that there exists an inversion which is an isometry?

  darij


What is the ratio (expressed as a common fraction) of the number of even factors to odd factors of 36?

[quote="FantasyLover"]the number of even factors of 36 is 6 (2,4,6,12,18,36)

and the number of odd factors of 36 is 3 (1,3,9)

answer :1/2[/quote]
Actually, it's $\boxed{\frac{2}{1}}$, there are $6$ even factors, and $3$ odd factors.  You just contradicted yourself.  :maybe:

How would you do a problem if you wanted to find the same ratio, except for the factors of like 1440?

[hide="Generalized"](See [url=http://www.artofproblemsolving.com/Wiki/index.php/Divisor]here[/url] for information on finding the number of factors of some positive integer $x$. This is necessary in understanding the following.)

The prime factorization of any integer $n$ is $2^{a} \times 3^{e_1} \times 5^{e_2} \times \cdots$. All odd factors cannot have an even factor. So all odd factors of $n$ are equal to the factors of $\dfrac{n}{2^a}$ which is, \[(e_1+1)(e_2+1) \cdots\]
All even factors must have $a > 0$, so all even factors of $n$ are equal to all factors of $n$ minus the ones that don't have even factors (which is above). We have, \[(a+1)(e_1+1)(e_2+1) \cdots - (e_1+1)(e_2+1) \cdots = a(e_1+1)(e_2+1)\cdots\]
We divide to get  \[ \dfrac{a(e_1+1)(e_2+1)\cdots}{(e_1+1)(e_2+1)\cdots}=\boxed{a} \][/hide]


I forgot how to prove the seven circles theorem. Can anyone tell me how?
The theorem is just basically this:

Draw an initial circle, and arrange six circles tangent to it such that they touch both the original circle and their two neighbors. Then the three lines joining opposite points of tangency are concurrent in a point. The figures above show several possible configurations(I'm sorry I could not provide an image, or it would have been much easier).
Anyway, could someone explain it to me?

Ehm... slightly wrong forum? ;)

http://www.mathlinks.ro/viewtopic.php?t=5190
http://www.mathpropress.com/stan/bibliography/sevenCircles.pdf

  darij


function
real analysis
real analysis unsolved

Please give any of non measurable function $ g$ such that: $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$.

You've already been given this answer, but I'll repeat it: there is no construction of such a function. In the standard model including the axiom of choice, they exist, but that's all we can say.


integration
calculus
real analysis
real analysis unsolved

Let $ (a_{n})$ be the sequence defined by

\[ a_{n}\equal{}\int\limits_{0}^{b}\frac{dx}{\sqrt{a\plus{}\sqrt{a\plus{}\cdots \plus{}\sqrt{x}}}},\]
where $ a\geq b>0$ and the number of the square roots under the integral sign is $ n$.

Prove that the sequence converges and find its limit.

$ f_n(x) \equal{} \sqrt {a \plus{} ... \plus{} \sqrt {x}}$ converges uniformly to the positive solution to $ a \plus{} c \equal{} c^2$ on $ [0, b]$ (I believe this follows from $ \sqrt {a \plus{} x}$ being Lipschitz), and since $ \frac {1}{x}$ is uniformly continuous in a neighborhood of $ c$ it follows that $ \frac {1}{f_n(x)}$ converges uniformly to $ \frac {1}{c}$, from which it follows that the original integral converges to $ \frac {b}{c}$.


inequalities
symmetry
calculus
logarithms
topology
triangle inequality
linear algebra

Definition 1:
$ (X,d)$ where $ X$ is a set. $ d: X \rightarrow \mathbb{R}$ satisfying:
(i)   $ d(x,y) \ge 0, \forall x,y \in X$
(ii)  $ d(x,y) \equal{} d(y,x)$
(iii) $ d(x,y) \equal{} 0 \Leftrightarrow x \equal{} y$
(iv) $ d(x,y) \le d(x,z) \plus{} d(z,y)$ for any $ z \in X$.

Definition 2:
$ (X,d)$ where $ X$ is a set. $ d: X \rightarrow \mathbb{R}$ satisfying:
(i)  $ d(x,y) \equal{} 0 \Leftrightarrow x \equal{} y$
(ii) $ d(x,y) \le d(x,z) \plus{} d(y,z)$ for any $ z \in X$

Observe that both definitions are equivalent.

Then why don't we define metric space according to Definition 2 in textbooks which requires a shorter list of properties.

Minimizing the number of axioms is never the be-all and end-all for the foundations of a subject. The whole thing has to make some kind of sense, and ease of memory is a very important consideration.

What you've written relies heavily on a [i]very[/i] specific way of entering the points into your axiom (ii). It would be hard to remember, and the slightest typo would ruin it.

Put yourself in the position of teaching the subject to a class.

Here's how the outline of the lecture would work with the standard definitions:

Present the axioms. Give them names, as appropriate. (Positive definite; symmetric, triangle inequality.)

Explain how these meet the one's common-sense assumptions about what "distance" means.

Immediately launch into presenting [i]many[/i] examples, verifying the axioms in each case:
$ \bullet\ \mathbb{R}$ with $ d(x,y)\equal{}|x\minus{}y|.$
$ \bullet\ \mathbb{R}^2$ with the 1-norm distance ("taxicab metric").
$ \bullet\ \mathbb{R}^2$ with the $ \infty$-norm distance.
$ \bullet\ \mathbb{R}^2$ with the Euclidean distance. (Do not verify the triangle inequality for this in this first lecture - postpone that until later.)
$ \bullet$ Bit strings of fixed length with the Hamming metric.
$ \bullet$ An example (draw picture) of a connected graph, with the usual graph-theoretic example.
$ \bullet$ And arbitrary set with $ d(x,y)\equal{}1$ for all $ x\ne y.$

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

Now contrast that lecture with the lecture given under your axioms.

Present axioms. (Unlike in the first presentation, in which I would rattle the axioms off of the top of my head without any written notes, for this one, I would have to have axiom (ii) very carefully written down in notes, and I would have to be very careful about transcribing it to the board.) Postpone any discussion of plausibility or common-sense assumptions.

Theorem 1: These axioms imply that $ d(x,y)\equal{}d(y,x).$ Present the proof of this.

Comment that after Theorem 1, we no longer have to be so careful about which order we write things in.

Theorem 2: These axioms imply that $ d(x,y)\ge 0.$ Present the proof of this.

Now, maybe we're ready to talk heuristics.

And then, maybe the time has come to present the list of examples - but in checking that these are metric spaces, do you really expect me to check your axiom (ii) directly? I don't think so.

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

I know which lecture I'd rather give. And note that the only thing I would [i]ever[/i] do with your axiom (ii) would be to use it to prove the standard axioms as theorems, and those are what I'd use afterwards.

It's interesting that these are equivalent formulations, and this might make a good problem on a first problem set dealing with metric spaces, but there are often reasons we choose one of several equivalent conditions as a "definition."  Here, the lack of common-sensicality of your second definition (which I just spent a fair amount of time trying to "explain" without relying on symmetry of $ d$) is really lacking.

To continue, while the two are equivalent, I look at your first definition and say, "Hmm, this kind of seems like a distance function."  And then I ask the professor "Is a metric in any way related to distance?"  And of course, the answer is "yes".  And then I say to myself, "Well now I don't really need to memorize these properties because they're just intuitive statements of distance."

I look at your second definition and I say, "So a metric is something obeying two properties.  What were they again...?  Which way did that inequality go?"

For me personally, the idea of compactness is similar.  I learned compactness to mean "closed and totally bounded".  Then of course we proved the other equivalent statements, such as finite subcovers and convergent sequences.  But to me, the closed+totally bounded definition made the most intuitive sense and certainly made it easier to remember.

In practice when you want to prove that something is a metric space, you usually only 'really' have to prove the triangle inequality since everything else is frequently more or less trivial.  So you want to make proving the triangle inequality as uncumbersome as possible.

Also it seems much more natural to write the inequality $ d(x,y) \leq d(x,z)\plus{}d(z,y)$ since the heuristic is that the inner $ z$'s 'cancel' in some sense.  Certainly it's much easier to keep track of things this way when you use the triangle inequality like $ d(a_0,a_n) \leq d(a_0,a_1)\plus{}\cdots\plus{}d(a_{n\minus{}1},a_n)$.

[quote="Kent Merryfield"][quote="blahblahblah"]In practice when you want to prove that something is a metric space, you usually only 'really' have to prove the triangle inequality since everything else is frequently more or less trivial.[/quote]
That would become apparent pretty quickly in the lecture I outline above as I go through the list of examples (some of you might have a slightly different list of examples). And note that I proposed delaying the the most difficult verification of the triangle inequality on that list - the Euclidean metric - to another lecture. Of course, that's the one case you already believe in, anyway.[/quote]

Out of curiosity, how would [i]you[/i] prove the triangle inequality for the euclidean metric on $ \mathbb{R}^d$?  Someone asked me how to do this the other day, and (trying to avoid invoking any 'powerful' inequalities), the best method I could come up with on the spur of the moment was to assume (by a linear change of variables) that $ z\equal{}0$, and then square both sides, cancel out various terms, and use the cauchy-schwartz inequality.  Is there a conspicuously quicker way?

Cauchy-Schwarz is the standard proof in an inner product space.  You can also prove [url=http://en.wikipedia.org/wiki/Minkowski_inequality]Minkowski's inequality[/url], which holds most generally in an $ L^p$ space, but its proof requires Holder, of which Cauchy-Schwarz is a special case.  In $ \mathbb{R}^d$ there's probably a Rearrangement proof, but offhand Cauchy-Schwarz is the most natural approach.

As far as the OP goes:  if we did things your way, we'd define finite fields to be finite alternative domains.

[quote="t0rajir0u"]Cauchy-Schwarz is the standard proof in an inner product space.  You can also prove [url=http://en.wikipedia.org/wiki/Minkowski_inequality]Minkowski's inequality[/url], which holds most generally in an $ L^p$ space, but its proof requires Holder, of which Cauchy-Schwarz is a special case.[/quote]

That's not really the question at hand.  The class Kent is talking about is an introductory analysis class for students who have had virtually no exposure to higher mathematics beyond the calculus sequence and possibly a linear algebra course, and the 'someone' who asked me this was a high school student.

I mean of course I could write some meaningless (to them) chain of equalities and inequalities like

\begin{align*}
\|x + y\|^2_2 &= \langle x + y,x + y\rangle \\
&= \|x\|_2^2 + \|y\|_2^2 + 2\langle x,y\rangle \\
&\leq \|x\|_2^2 + \|y\|_2^2 + 2\|x\|_2\|y\|_2 \\
&= (\|x\|_2 + \|y\|_2)^2
\end{align*}

but I was trying to find a method that's not so far from one that they might find on their own.

[quote="t0rajir0u"]Cauchy-Schwarz is the standard proof in an inner product space. You can also prove Minkowski's inequality, which holds most generally in an $ L^{p}$ space, but its proof requires Holder, of which Cauchy-Schwarz is a special case. In $ \mathbb{R}^{d}$ there's probably a Rearrangement proof, but offhand Cauchy-Schwarz is the most natural approach.[/quote]
And if I recall correctly, the proof of Holder requires Young's inequality, which I don't think is a trivial proof.

[quote="JRav"]
And if I recall correctly, the proof of Holder requires Young's inequality, which I don't think is a trivial proof.[/quote]

'young's inequality' is precisely equivalent to the statement that the mapping $ x\to \minus{}\log x$ is convex.  so yeah, it's more or less trivial.


probability
combinatorics proposed
combinatorics

A random walker in 2 dimensions walks inside a 30x30 box. From each point inside the box find the probability it will exit from sides. It has $ {n}$ courses.

Can you better explain the problem? I've got some questions...
[hide]
1)What do you mean "it will exit from sides"?
2)What are the n courses? He can step onto n different points?
3)From each point? Do you mean a random point or for every point?
4)In what directions can he move? Up, Down, Left, Right?
[/hide]


Hello guys this is my first problem on mathlinks

Find integer solutions (x, y) of the equation (x^2 + y)(x + y^2) = (x - y)^3.

[hide]We write the equation in the following form 2*y^2+x^2*y+3x^2+x=3xy and consider that this is a second degree equation on y.[/hide]

thanks for the hint [color=darkred]panos[/color], but i dont know what to do after that  :maybe: , could somebody help me with this problem?

[hide]As panos_lo suggested, use the quadratic formula on $ y$ in $ 2y^2 \plus{} (x^2 \minus{} 3x)y \plus{} (3x^2 \plus{} x) \equal{} 0$. What is necessary for $ y$ to be an integer?[/hide]

[hide="solution"] We claim that all solutions are $ (8, \minus{} 10), (9, \minus{} 6), (9, \minus{} 21), ( \minus{} 1, \minus{} 1),$ and $ (x, 0)$ where $ x$ is an integer.

It can be verified that the previous solutions work. We now will show these are the only solutions. To do so, we use casework based on the signs of x and y.

[hide="Case 1, at least one of x or y is zero"] Checking, $ (x, 0)$ works for all integers x. $ (0, y)$ does not work however, so our solutions from this case are $ (x, 0)$ for integers x.[/hide]

[hide="Case 2, both x and y are positive"] If $ x, y > 0$, then the LHS must be positive, meaning $ x > y$. However, if $ x > y$, the LHS is greater than $ (x^2)(x) \equal{} x^3$, while the RHS is less than $ x^3$. Thus, there are no solutions.[/hide]

[hide="Case 3, x>0, y<0"] Let $ y \equal{} \minus{} b$ so $ (x^2 \minus{} b)(b^2 \plus{} x) \equal{} (x \plus{} y)^3\implies{2b^2 \plus{} b( \minus{} x^2 \plus{} 3x) \plus{} (3x^2 \plus{} x)} \equal{} 0$. From this, we use the quadratic formula to get $ b \equal{} \frac {x^2 \minus{} 3x\pm\sqrt {x^4 \minus{} 6x^3 \minus{} 15x^2 \minus{} 8x}}{4}$. Now, $ x^4 \minus{} 6x^3 \minus{} 15x^2 \minus{} 8x \equal{} x(x \plus{} 1)^2(x \minus{} 8)$ must be a perfect square, so $ x(x \minus{} 8)$ must be a perfect square. $ x^2 \minus{} 8x < (x \minus{} 4)^2$, so we will attempt to use $ (x \minus{} 5)^2$ to bound $ x$. Clearly, $ x\ge{8}$. If $ x > 12$, $ x(x \minus{} 8) > (x \minus{} 5)^2$, so we check $ 8\le{x}\le{12}$. $ x \equal{} 8,9$ works, giving $ (x, y) \equal{} (8, \minus{} 10), (9, \minus{} 6),$ and $ (9, \minus{} 21)$.[/hide]

[hide="Case 4, x<0, y>0"] Let $ x \equal{} \minus{} a$ so $ (a^2 \plus{} y)(y^2 \minus{} a) \equal{} \minus{} 1(a \plus{} y)^3$ which expands and rearranges into $ 3a^2 \minus{} a \plus{} ay(3y \plus{} a) \equal{} 0$. $ ay(3y \plus{} a) > 0$ and $ 3a^2 > a$, so there are no solutions here.[/hide]

[hide="Case 5, x<0, y<0"] Let $ x \equal{} \minus{} a, y \equal{} \minus{} b$ so $ (a^2 \minus{} b)(b^2 \minus{} a) \equal{} (b \minus{} a)^3$ which expands and rearranges into $ 2b^2 \plus{} 3a^2 \minus{} 3ab \minus{} a^2b \minus{} a \equal{} 0$ or $ 2b \plus{} b( \minus{} a^2 \minus{} 3a) \plus{} (3a^2 \minus{} a) \equal{} 0$. Using the quadratic formula, $ b \equal{} \frac {a^2 \plus{} 3a\pm\sqrt {a^4 \plus{} 6a^3 \minus{} 15a^2 \plus{} 8a}}{4}$.

For this to be an integer, $ a^4 \plus{} 6a^3 \minus{} 15a^2 \plus{} 8a \equal{} a(a \minus{} 1)^2(a \plus{} 8)$ must be a perfect square. Thus, $ a(a \plus{} 8)$ must be a perfect square. $ a^2 \plus{} 8a < (a \plus{} 4)^2$, so $ a^2 \plus{} 8a \equal{} a^2 \plus{} 6a \plus{} 9, a^2 \plus{} 4a \plus{} 4, a^2 \plus{} 2a \plus{} 1,$ or $ a^2$.  Recalling that $ a$ must be positive, we find that $ a \equal{} 1$, giving $ b \equal{} 1$. Thus, our only solution is $ (x, y) \equal{} ( \minus{} 1, \minus{} 1)$ here.[/hide]

We have checked all cases, and have the solutions that we stated above.[/hide]

This took me forever...anybody have a faster method?

EDIT: [hide="alternatively"] Using the $ 2y^{2} \plus{} (x^{2} \minus{} 3x)y \plus{} (3x^{2} \plus{} x) \equal{} 0$ from what Victor posted, we may examine the discriminant and find $ x(x \minus{} 8)(x \plus{} 1)^2$ is a perfect square. Thus, $ (x \minus{} 4)^2 \minus{} 16 \equal{} y^2$. Using $ (x \plus{} y \minus{} 4)(x \minus{} y \minus{} 4) \equal{} 16$, we can examine all possibilities, and hopefully come up with the same answers.

Also, the problem becomes finding all x so $ x(x\minus{}8)$ is a perfect square, which probably uses bounding again, but I found it easier to separate into the above cases.[/hide]


calculus
integration
search
function
real analysis
real analysis theorems

i'm dead sure this has been done here before, but unfortunately i'm not able to find it using the search fn. so really sorry. the qn is this . let f be a continuous fn on [a,b] and also f is non-negative,not identically zero. show that $\int_{a}^{b}f(x)dx >0$ , its very clear intuitively, but i'm not able to prove it. ( i mean the darboux integeral here)any hints would be appreciated. thanks

Because $f$ is not identically zero and continuous there exists $[a_{0},b_{0}]\subset [a,b]$ such that $f(x)>0$ for every $x\in [a_{0},b_{0}]$. So $\int_{a_{0}}^{b_{0}}f(x)dx>0$.

$\int_{a}^{a_{0}}f(x)dx$ and $\int_{b_{0}}^{b}f(x)dx$ are $\geq0$ so $\int_{a}^{b}f(x)dx=\int_{a}^{a_{0}}f(x)dx+\int_{a_{0}}^{b_{0}}f(x)dx+\int_{a}^{a_{0}}f(x)dx>0$

Thor's argument isn't really enough; it assumes the nontrivial theorem linked to below. We need some interval $[a_{0},b_{0}]$ and a constant $c>0$ such that $f(x)\ge c$ for all $x\in[a_{0},b_{0}]$. This is easy to get by continuity; if $f(\alpha)>0$, we can find $\beta>0$ by continuity so that $|f(x)-f(\alpha)|<\frac{f(\alpha)}{2}$ when $|x-\alpha|\le\beta$.

A nice variation: If $f$ is Riemann integrable and strictly positive on an interval, $f$ has positive integral. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=37648]here[/url] for a proof.
The key is that we only need a single point of continuity (at a point where $f$ is positive); continuity everywhere is an excessively strong assumption.

[quote="kouboy"]thanks. why didnt i get this idea before? hmm i have to improve my analysis[/quote]
I like assigning images or catch phrases to certain arguments in analysis to help me remember. The key idea here (see what jmerry wrote for the details) is "stick a box under the graph."

Note that $f\ge0,$ $f$ Riemann integrable, $f$ not identically zero is not good enough to imply that the integral is positive; $f$ might be positive only points of discontinuity (for instance if $f$ is positive only at one point) and then you can't stick a box under the graph.

Another example for images or catchphrases. Prove that a monotone function on a bounded interval is Darboux (i.e., upper-lower sums) integrable, hence Riemann integrable. The key: "stack the boxes into one column."

[quote="jmerry"]Thor's argument isn't really enough; it assumes the nontrivial theorem linked to below. We need some interval $[a_{0},b_{0}]$ and a constant $c>0$ such that $f(x)\ge c$ for all $x\in[a_{0},b_{0}]$. This is easy to get by continuity; if $f(\alpha)>0$, we can find $\beta>0$ by continuity so that $|f(x)-f(\alpha)|<\frac{f(\alpha)}{2}$ when $|x-\alpha|\le\beta$.
[/quote]

I thought that theorem was well known.


Is there a triangle with sides of integer lengths such that the length of the shortest side is $ 2007$ and that the largest angle is twice the smallest?

Let $ a \le b \le c$ be the sides, $ a \equal{} 2007 \equal{} 3^2 \cdot 223,$ and $ \angle C \equal{} 2 \angle A.$ By the sine and cosine theorems,
$ \frac {c}{a} \equal{} \frac {\sin 2A}{\sin A} \equal{} 2 \cos A$

$ a^2 \equal{} b^2 \plus{} c^2 \minus{} \frac {bc^2}{a},\ \ c^2 (b \minus{} a) \equal{} a(b^2 \minus{} a^2) \equal{} a(b \plus{} a)(b \minus{} a)$

If $ b \equal{} a$ and $ \angle C \equal{} 2 \angle A \equal{} 2 \angle B,$ the $ \triangle ABC$ is isosceles right and the largest side $ c$ is not an integer. Thus $ a < b \le c,$ $ c^2 \equal{} a(a \plus{} b).$ But $ c^2 \equal{} a(a \plus{} b) \equal{} 3^2 \cdot 223 (a \plus{} b)$ implies $ c \equal{} 3 \cdot 223 \cdot n,\ n$ is an integer. Then $ 3^2 \cdot 223^2 \cdot n^2 \equal{} 3^2 \cdot 223 (3^2 \cdot 223 \plus{} b),$ $ b \equal{} 223 (n^2 \minus{} 3^2).$ Thus $ (a, b, c) \equal{} 223 \cdot (3^2, n^2 \minus{} 3^2, 3n)$ From $ c > a,$ we have $ 3n > 3^2,\ n \ge 4,$ and from $ \cos A \equal{} \frac {c}{2a} \equal{} \frac {3n}{2 \cdot 3^2} \equal{} \frac {n}{6},$ we have $ n \le 5.$ It follows that $ n^2 \minus{} 3^2 \equal{} 7$ or $ 16.$ The 1st possibility is excluded by $ b > a,\ n^2 \minus{} 3^2 > 9,$ the other one by $ b \le c,\ n^2 \minus{} 3^2 \le 3n \le 15.$

I've solved it the same way.Any other solutions?


Find all polynomials $ P(x)$ satisfying $ (x\minus{}16)P(2x)\equal{}16(x\minus{}1)P(x)$
for all real $ x$

Clearly $ x \equal{} 16$ and and $ x \equal{} 2$ are roots of $ P(x)$ (substitute $ x \equal{} 16$ and $ x \equal{} 1$ to make zero the LHS and RHS respectively), hence $ P(x) \equal{} (x \minus{} 2)(x \minus{} 16)Q(x)$, and $ 4(x \minus{} 16)(x \minus{} 1)(x \minus{} 8)Q(2x) \equal{} 16(x \minus{} 1)(x \minus{} 2)(x \minus{} 16)Q(x)$ for all real $ x$.  Similarly, $ x \equal{} 8$ and $ x \equal{} 4$ are also roots, or $ P(x) \equal{} (x \minus{} 2)(x \minus{} 4)(x \minus{} 8)(x \minus{} 16)R(x)$, where for all real $ x\notin\{2,4,8,16\}$, we have $ R(2x) \equal{} R(x)$.  Clearly, $ R(x) \minus{} R(3)$ has infinitely many roots $ x \equal{}3\cdot2^m$ for all positive integer $ m$, or $ R(x) \equal{} R(3)$ for all real $ x$.  Hence $ P(x) \equal{} A(x \minus{} 2)(x \minus{} 4)(x \minus{} 8)(x \minus{} 16)$ for any real constant $ A$.


After much persuasion from Orlando :), here are the links:

Problem 1: http://www.mathlinks.ro/viewtopic.php?p=18847
Problem 2: http://www.mathlinks.ro/viewtopic.php?p=18490
Problem 3: http://www.mathlinks.ro/viewtopic.php?p=18405
Problem 4: http://www.mathlinks.ro/viewtopic.php?p=18400
Problem 5: http://www.mathlinks.ro/viewtopic.php?p=18402
Problem 6: http://www.mathlinks.ro/viewtopic.php?p=18394

Why doesn't get the Singapore TST no link on the mathlinks front page ?

maybe cos its too easy?  :)

ok maybe it not the easiest around but compared to most countries ....
i messed up the second day pretty much, not being able to solve the easy problems

kennyworm


calculus
number theory unsolved
number theory

Need help for this 2 questions..  :blush:

1st Question:
[img]http://i3.photobucket.com/albums/y61/RawryNo0b/MathsQ7.jpg[/img]

2nd Question:
[img]http://i3.photobucket.com/albums/y61/RawryNo0b/MathsQ6.jpg[/img]

Thanks in advance!

Yay thanks for help! Smallest prime factor is 11. Thanks!

hmm i need help.......someone explain the method to me pls
either u post here or add me at normanyeo1@hotmail.com
thx


[hide]applying the pythagorean theorem, 16^2+12^2=400, so the car is 20 miles from its starting point.[/hide]

[hide]
use the pythagoran (spelling?) thereom.
one side is 12, one is 16 and the last is c^2
144+256=c^2
c becomes 20[/hide]

Theorem Pitagoras
12^2+16^2=20^2
so the ans is 20.

[quote="joejia"]A car travels on flat land 16 miles west and 12 miles south. How many miles is the car from its starting point?[/quote]

[hide]Applying the Pythagorean Theoreom, this will be:

$a^2+b^2=c^2$

$16^2+12^2=c^2$

$16^2 = 256$

$12^2 = 144$

$\sqrt{144+256} = 20$

The answer is 20.[/hide]


Hello, I've decided to study this topic a bit (the problems are very nice!!). Anyway, I was wondering if anybody has Paul Yiu's packet's on Mixtilinear Incircles? One of them is in AMM, but there are two others that I've see citations of in other papers. Also we can make this a thread to collect problems on Mixtilinear Circles.

Could you precise a bit more about these other two packets? Please quote the cites. I know, of course the packet in AMM, and a rapid search in the web page of Paul Yiu do not seems provide other sources....

In "On Mixtilinear Incircles and Excircles" by Khoa Lu Nguyen and Juan Carlos Salazar Forum Geometricorum Vol. 1 2006, the 5th reference is to Paul Yiu: Mixtilinear Incircles II, unpublished manuscript, 1998.

Also Darij references the second (and a third packet) by Paul Yiu here: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1040508572&t=24976. The links that he gives are broken.

At the books
Fukagawa-Pedoe:"Japanese Temple Geometry Problems" 1989 and
Fukagawa-Rigby:"Traditional Japanes Mathematics Problems of the 18th and 19th centuries" 2002,
there are several problems (not all solved, or with japanese references) about triangles and several circles of the type we are searching. Usually the proofs are metric and with a lot of computations.
Maybe other mathlinkers have some other references and/or more problems like these.


Is it possible for f(x) to be both increasing and decreasing everywhere on the interval (a,b)?

"Increasing" is not a property that a function has at a single point.  It's an interval property - a function can be said to be increasing on a certain interval. That has a concrete definition: $ f$ is strictly increasing on an interval $ I$ iff for all $ x,y\in I$ with $ x<y,$ we have $ f(x)<f(y).$

Just from that definition, a function cannot be both increasing and decreasing on the same interval. Of course a function can be both nondecreasing and nonincreasing on the same interval - that would make it constant on that interval.

It is possible to find a function, even a continuous function, for which there is no interval on which the function is increasing and no interval on which the function is decreasing. Any nowhere differentiable continuous function will do.

How about circle? I have thought about if f(x) is a circle, you could have f(x) increasing and decreasing on the same interval.

that is not a function, by definition a function must be well defined, which amounts to passing the vertical line test, which a circle does not


symmetry calculus integration trigonometry

1) There is a curved wire , 3/4th of a circle of radius a, which has a charge per unit length q = q' sin2@ , where @ is the angle from the center to any point on the wire. Calculate the electric field intensity at the center of the circle.

2) If the strength of the electric field which points downwards decreases with increase in altitude from the surface of the earth and assumes value of 150V/m at a distance 100m measured from the surface of the earth, calculate average net charge in 1 cubic metre of the air between the surface of the earth at the altitude of 100m measure from the ground.

In your charge per unit length, what is q'?  And your charge per unit length doesn't appear to have the correct units.

Take $ q' \equal{} q_0/l$ or something like that, its just a constant

1)
[hide][img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/e/c/a/eca64cdf3859c9722d07ed654e266b59ec27cf4b.gif[/img][/hide]

in 2) , is the question exactly the same , no info missing ?
which qn of aks is it ?

@ slogger, Your answer is wrong . I would recommend you do it fundamentally.

Anyway this is what I thought, by symmetry the components of the wire sub-tending angles $ [0,\frac {\pi}{2}]$ and $ [\pi , \frac {3 \pi}{2}]$ respectively cancel each other.So we only need to find the E due to the other quarter wire.

@ slogger: It is question 5.5, and the first one is 5.9.

btw , i must say , that i am myself sure the method i gave is wrong , it was just an attempt to get the answer after my initial method failed  :D

My initial view was exactly as yours and i got the value of
[img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/e/9/a/e9ac01742b692708ce47ff16e3e8d8ce5438e7b7.gif[/img]
integrate the above expression from pi/2 to pi , u'll get the answer ,
damn i was making a mistake in integration all the time uptil now  :blush:

Yes slogger this is exactly what I got,maybe lil diff, I am getting

Integration of [b]-2sin2@cos@[/b] from 0 to pi/4

did u get the answer ?
did u get my method ?
i took the quadrant from pi/2 to pi (whose E would not get cancelled )
then i considered an element of length dl making an angle theta from centre
then i found the field due to it and considered its component in the line joining centre and mid point of the considered quadrant as all the other component would get cancelled

The component you are talking of, are you sure it has to be integrated from pi/2 to pi.

See attached figure. (Srry I coudnt draw better )

Field due to A at centre is $ \frac {K q'sin(2(\frac {\pi}{2} \plus{} @))}{a}$ and due to symmetrical point B is $ \frac {K q'sin(2( \pi \minus{} @))}{a}$ , since both are equal ( $ \frac { \minus{} K q'sin2@ d@}{a}$)and by symmetry there resultant lies on the bisector of the quadrant, and the projection of this field due to both points on the bisector is $ \frac { \minus{} 2K q'sin2@ d@ cos(\frac {\pi}{4} \minus{} @)}{a}$. So now obviously I have to integrate it from 0 to pi/4.

@slogger the qns are from abhay singh :)

@pardesi , i do know that , i have only the first qn in my abhay kumar singh , secondly , do u get my method ?

Slogger, did you skip this step  :D

-$ \frac{\minus{}q'}{4 \pi \epsilon_o a} \int_{\frac{\pi}{2}}^{\pi} \sin 2(\frac{\pi}{2} \plus{} \theta) . \sin (45\minus{} \theta) d\theta$

 :maybe:

I am getting the ans as $ \frac{ \sqrt{2} q'}{6 \pi \epsilon_o a}$
  :maybe:


trigonometry
combinatorics unsolved
combinatorics

$ A \equal{} {a_1,a_2,...,a_{13}}$ , $ a_i \in R$ , $ a_i$ is not equal to $ a_j$. prove that there exist $ x,y$ $ \in$ $ A$ such that:

$ 0 < \frac{x\minus{}y}{1\plus{}xy} \leq 2\minus{}\sqrt{3}$

choose $ \pi \ge \alpha_i \ge 0$ such that $ \tan \alpha_i \equal{} a_i$
now we have
$ \tan( \alpha_i \minus{} \alpha_j) \equal{} \frac {\tan \alpha_i \minus{} \tan \alpha_j}{1 \plus{} \tan \alpha_i \tan \alpha_j} \equal{} \frac {a_i \minus{} a_j}{1 \plus{} a_ia_j}$
since $ \tan \beta \le 2 \minus{} \sqrt {3}$ iff (note that we assume $ \beta > 0$ ) $ \beta \le 15^\circ$ so we must find $ i,j$ such that $ |\alpha_i \minus{} \alpha_j| \le 15^\circ$
WLOG supposing $ \alpha_1 < \alpha _2 < ... < \alpha_{13}$
and suppose that such $ i,j$ do not exist then we have
$ \sum_{i \equal{} 1}^{12} (\alpha_{i \plus{} 1} \minus{} \alpha_{i}) \equal{} \alpha_{13} \minus{} \alpha_1 > 15^\circ \times 12 \equal{} 180^\circ$
which means $ \alpha_{13} > 180^\circ$ contradicting the first assumption.


function
integration
advanced fields
advanced fields unsolved

Let I(r)=\int_0^{2\pi} |f(re^{i\theta})|^pd\theta where . Show the monotonicity of  such that, prove  is subharmonic.( is harmonic function in unit disk)

Let $ I(r)\equal{}\int_0^{2\pi} |f(re^{i\theta})|^pd\theta$ where . Show the monotonicity of  such that, prove $ I(|z|)$  is subharmonic.( is harmonic function in unit disk)


Em mới tham gia diễn đàn mathlinks này và rất vui vì có rất nhiều mem đến từ Việt Nam, vì thế tại sao chúng ta lại không nhờ admin tạo nhóm thành viên Việt Nam nhỉ?

$\bigstar$ Ý bạn là sao. Nhóm thành viên Việt Nam thì để làm gì nhở  :roll: Hồi xưa ML còn có cái cờ tổ quốc để biết ai người nước nào. Từ ngày lên đời thì mất tiêu cái cờ ko bít đâu mà lần :P

Tại sao lại phải phân biệt rõ ràng nhỉ? Look at my signature!


Let $ ABC$ be a triangle and $ N$ be its nine-point center. Let $ N_a$, $ N_b$, $ N_c$ the nine-point centers of triangles $ NBC$, $ NCA$, $ NAB$, respectively. Prove that $ AN_a$, $ BN_b$ and $ CN_c$ are concurrent.

$ \mbox{If O circumcenter and H orthocenter of the triangle ABC, then the midpoint of [OH] \equal{} "nine \minus{} point center".}$
(view: [i]Nathan Altshiller-Court[/i]: [b]COLLEGE GEOMETRY[/b], sec.ed. p.104.)


What is the sum of the reciprocals of  the roots of the eqution

\[\frac{2003}{2004}x+1+\frac{1}{x}=0\]?

$\left(A\right)-\frac{2004}{2003}$
$\left(B\right)-1$
$\left(C\right) \frac{2003}{2004}$
$\left(D\right) 1$
$\left(E\right) \frac{2004}{2003}$

Multiply through by $x$ and by $2004$ to get a regular polynomial with integer coefficents, first:

$\frac{2003}{2004}x+1+\frac{1}{x}=0$
$\frac{2003}{2004}x^{2}+x+1=0$
$2003x^{2}+2004x+2004=0$

Now we have two methods.

[hide="Method 1"]Let the roots be $a$ and $b$.  We are looking for $\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$.  Using Vieta's formulae, $a+b=-\frac{2004}{2003}$, and $ab=\frac{2004}{2003}$.  Thus $\frac{a+b}{ab}=\boxed{-1}$[/hide]

[hide="Method 2"]It can be proven that if $P(X)$ is polynomial polynomial, then the polynomial with roots that are the reciprocals of $P(x)$ is simply $P(x)$ with its coefficients reversed.  So reversing the coefficients of our polynomial gives us:

$2004x^{2}+2004x+2003=0$

Now we can use Vieta's to find the sum of these roots.  We do $\frac{-b}{a}=\frac{-2004}{2004}=\boxed{-1}$

[/hide]

[hide="Vieta explained"]Vieta:

I'm not good at generalizing, and generalizations are hard to understand sometimes anyway, so I will give some examples.

If we have a polynomial like $ax^{2}+bx+c$ then we can quickly find the sum of its roots and product of its roots.

The sum is the negative of the second coefficient over the first coefficient, or $\frac{-b}{a}$.

The product is the negative or positive of the last coefficient over the first coefficient, or $\pm \frac{c}{a}$.

When dealing with the product, how do we know if it is positive or negative?  Well, if the highest power of x in our original polynomial is even, then it is positive.  If the highest power of x in our original polynomial is odd, then it is negative.

For example, in the polynomial $dx^{3}+ex^{2}+fx+g$ the sum of the roots is $\frac{-e}{d}$.  The product of the roots is $\frac{-g}{d}$.

Vieta's formulas are very important, and come up more than any other polynomial topic in math contests.[/hide]

[hide="More advanced but not needed for this problem"]If we have the polynomial $3x^{3}+4x^{2}+6x+9$, we can call the roots a, b, and c.  From above, we know that $a+b+c = \frac{-4}{3}$ and $abc=-3$.  However, we can also use Vieta's formulas to find the sum of the roots taken two at a time, or $ab+bc+ac$.  We just use the third coefficient over the first coefficient, or $\frac{6}{3}=2$.[/hide]

I learned this stuff last summer, and I was helped by a lot of the SC forum people in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81136]this[/url] post.  Check it out!

Do you recommend reading Mathematical Ideas by Adeel Khan?(from link, and just wondering if it helps)

You could use the quadratic formula, but the whole point of this problem is that you don't need to know the specific roots.

[quote="mysmartmouth"]Huh, I wonder if you know it's a website (PDF documents) of an AoPSer (chess64) not a book...

There's not much explanation, but I learned a lot by doing the problems, which was probably his intention.

However, he seems to have killed his website...I would instead read [url=http://tmp.mathideas.org/public/math/polynomials.pdf]this[/url] paper by him, it will teach you this kind of stuff about polynomials.[/quote]
OK thanks...I wasn't sure what it was...whether a website, a tutorial, a book, a... oh the possibilities are endless. Anyway, thanks.


Let $ABCD$ be a cyclic quadrilateral. $AD$ intersects $BC$ at $E$ and $AC$ intersects $BD$ at $F$. Show that $E$ is on the polar of $F$.

Lemma: The polar of $E = AD \cap BC$ is the line through $F = AC \cap BD$ and $G = AB \cap DC$.
On the other hand the polar of $E$ passes through $F$ iff the polar of $F$ passes through $E$ by La Hire's Theorem.

Yes, please show how to prove that lemma.

This is Theorem 7 in my note [i][url=http://www.cip.ifi.lmu.de/~grinberg/Butterfly.zip]On cyclic quadrilaterals and the butterfly theorem[/url][/i] as Theorem 9 [b]i)[/b] (for its proof, see the second proof of Theorem 6).

  darij

Or for a proof using Pascal's Theorem see the [url=http://www.math.ust.hk/excalibur/v11_n4.pdf]Mathematical Excalibur[/url].


Where could a middle schooler find information about chemistry. Is there a book for the chem olympiad similar to the AoPS? I have no experience with the subject but it facinates me.

You usually don't learn chemistry until high school, so there aren't many books on chemistry for middle school students. There are some very basic chemistry principles you might learn about in late middle school.

Can you give me some book/website names?
What grade are you in? You seem to post in the MC fourm yet you do chem, which you say is for high schoolers.

I'm entering 8th, but I read some chemistry for fun  :)

My textbook off Amazon is a Prentice Hall version, titled "Chemistry".

This was [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40570]discussed[/url] a while ago.


Let be $N=\{1,2,3,4,5,6,7,8\}$. I must write a number in each blank of the following chart such that:
*all numbers are different;
*all numbers are in $N$;
*there are NO consecutive numbers: nor vertical, nor horizontal, nor diagonal.

There is one solution to the puzzle (and possibly more):

···6·4···
·2·8·1·7
···5·3···

 :lol: I don't know if there is anything wrong with your proof. I just used trial and error.

Well, thanx, u are rite
I think I got it!

···b·c···
a·x·y··f
···d·e··

The trick I found is the following. Let's suppose $x\neq 1,8$.
Then it's easy to see that $f=x+1$. Then we have

···b·c··········
a·x·y··(x+1)
···d·e··········

And where is (x-1) in the chart ?It's easy to see that, in this way, there is NO room for (x-1)...

We deduce that, if a solution exits, $x=1$ or $x=8$.

Suppose $x=1$, then, we know that

···b·c···
a·1·y·2
···d·e···

But a similar idea tells us that $y=8$ (since it can't be $y=1$). Therefore we have

···b·c····
7·1·8··2
···d·e···

Now it's easy to finish.

If $b=3$, where is $4$ ? Then $d=4$ ( $e=4$ doesn't work).

···3·5····
7·1·8··2
···4·6···

Applying a symmetry

···4·6····
7·1·8··2
···3·5···

Your case can be found by simmetry too,

···6·4····
2·8·1··7
···5·3···

etc.


Let $ A, B$ be $ 3\times 3$ positive semidefinite Hermitian matrices, $ p_1 \plus{} p_2 \plus{} p_3 \equal{} 1$, where $ p_1, p_2, p_3 > 0$.
Prove or disprove that
$ \mid tr(A^{p_1}BA^{ p_2}BA^{p_3}B)\mid \le tr(AB^3).$
Without loss of generality, we may assume $ A$ is a diagonal matrix.

1) I don't understand the assumed condition $ n\equal{}3$.
The required result is equivalent to
If $ a,b>0$ and $ A,B\geq{0}$ then $ |tr(ABA^aBA^bB)|\leq{t}r(A^{1\plus{}a\plus{}b}B^3)$. Numerical experiments "show" that it's true.
Moreover it seems also to me that $ tr(ABA^aBA^bB)\geq{0}$.
2) The following results are true:
If $ A\geq{0},B$  are hermitian and if $ a>0$ then $ tr(ABA^aB)\leq{t}r(A^{1\plus{}a}B^2)$.
If $ A,B\geq{0}$ and if $ k$ is a positive integer then $ tr((AB)^k)\leq{t}r(A^kB^k))$ (Lieb-Thirring inequality).

[quote="loup blanc"]1) I don't understand the assumed condition $ n \equal{} 3$.
The required result is equivalent to
If $ a,b > 0$ and $ A,B\geq{0}$ then $ |tr(ABA^aBA^bB)|\leq{t}r(A^{1 \plus{} a \plus{} b}B^3)$. Numerical experiments "show" that it's true.
Moreover it seems also to me that $ tr(ABA^aBA^bB)\geq{0}$.
2) The following results are true:
If $ A\geq{0},B$  are hermitian and if $ a > 0$ then $ tr(ABA^aB)\leq{t}r(A^{1 \plus{} a}B^2)$.
If $ A,B\geq{0}$ and if $ k$ is a positive integer then $ tr((AB)^k)\leq{t}r(A^kB^k))$ (Lieb-Thirring inequality).[/quote]
Because for $ n\equal{}2$, the conclusion is true. The first step is to prove the case for $ n\equal{}3$. I know the statements of 2), thank you for sharing your knowledge.  :)


geometry
probability
ratio
MATHCOUNTS
number theory

For the state competition, get old national competitions, and learn everything that comes up in those.

Where are national competitions available?

[quote="led14pa2"]does any body know what are some math concepts and topics to study for the state competition. this is my 1st year in states and i really have no idea what to study. thanks in advance for the help.[/quote]

One thing that I believe helps is dividing up problems into categories.  This helps you focus on each concept at a time.  For instance, I view the major categories as aglebra, geometry, counting, number theory, and probability.

Once identified, you can go further and identity different kinds of problems within each category.  For instance probability problems could be grouped into counting problems (success outcome/total outcomes), geometry problems (finding ratios of regions like length, area, or volume), and algebra problems (finding algebraic relationships between different probabilities).

I hope that helps.

First, make sure you have a good understanding of how to solve all the common types of problems.  Many times, if you don't feel your solution is particularly effective, consult the local math person or AoPS, and you might get a more efficient solution.  An example is the "traveling from point A to point B on a grid" problems, where you can do it by brute force, but it takes longer.  This is especially important for the target round, where you can not afford to miss more than one problem if you want to make it to nationals (at least in competitive states like IL, CA, PA etc.).

Next, practice test taking skills.  Do practice tests or problems under time pressure.  If you're a clumsy person like me, try to have enough speed to have time to check your answers before going to the next question.  This often saves me 3-4 points on a sprint round.

Finally, aim high!  Don't feel that any goal is unrealistic; I improved my average state level practice test score by around 7 points my first year in one month!  I'm sure you can do better, since you have more than one month :)

Another thing that helps me alot when I am taking state level tests is to figure out which problems to try to awnswer first on the test. If you skip the harder problems and come back to them later, you have a better chance of making a good score. If you skipped three hard problems and finished the rest of the test but didn't answer the problems you skipped, then you get a better score than if you got those problems wrong and didn't finish the test.

Where can I purchase the previous National Competition Sets?

You can purchase old chapter, state and national competition set at the mathcounts.org web site.

Last year, my friend went to the chapter competion as a 7 and came in second for individual and first for countdown. He said that the proctors there gave out CDs with all the od national sets. He burned a copy for me and it has all the problems from 1983 till 2002. Everything too, got the masters, sprint, target, and the countdown.
It's pretty helpful but really hard too. :lol:

There's also a book called "the all-time greatest Mathcounts problems"

From the 1995 National Target #2
A dart board is designed to have two scoring areas, as shown. I an unlimited number is allowed, what is the largest finite score that cannot be attained?

I don't know how to make the picture so I'll describe it. It's a circle within  a circle. Hitting the inner circle gets you 15 points. And the outer circle gets you 7 points.

83. The book says the formula for the greatest finite score that can't be attained is the product of the two numbers minus the sum of the two numbers.
So it's 15(7) - 15 - 7


For $ x>0$, find the minimum value of $ \left(x\plus{}\frac{1}{x}\right)\left(2x\plus{}\frac{1}{2x}\right)$.

hello kunny, we have $ (2x^2\minus{}1)^2\geq0$ for all real $ x>0$, this is equivalent with
$ 4x^4\minus{}4x^2\plus{}1\geq0$
$ 4x^4\plus{}5x^2\plus{}1\geq9x^2$
$ 4x^4\plus{}4x^2\plus{}x^2\plus{}1\geq9x^2$
$ 2x^2\plus{}2\plus{}\frac{1}{2}\plus{}\frac{1}{2x^2}\geq\frac{9}{2}$
$ \left(x\plus{}\frac{1}{x}\right)\left(2x\plus{}\frac{1}{2x}\right)\geq\frac{9}{2}$.
The equal sign holds if and only if $ x\equal{}\frac{1}{\sqrt{2}}$.
Sonnhard.

[quote="kunny"]For $ x > 0$, find the minimum value of $ \left(x \plus{} \frac {1}{x}\right)\left(2x \plus{} \frac {1}{2x}\right)$.[/quote]

$ \left(x \plus{} \frac {1}{x}\right)\left(2x \plus{} \frac {1}{2x}\right) \equal{} 2x^2 \plus{} \frac {1}{2x^2} \plus{} 2.5$

Use AG now... so the minimum value is $ 4.5$


this problem from some russian olipiad
THERE IS 3 SUPERCOMPUTERS THEY ALL LOOK THE SAME AND THEY CAN NOT BE RECOGNIZEd. THE FIRST MADE IN USA AND IT ALWAYS TELL THE TRUTH THE SECOND ONE MADE IN CHINA AND IT ALWAYS LYING. THE THIRD ONE MADE IN RUSSIA AND IT GIVES RANDOM ANSWERS.EVERYONE OF THE COMPUTERS KNOW EVERYTHING  ABOUT EACH OTHER.YOU NEED TO FIND A QUESTION(THAT THE ANSWER ON IT IS YES OR NO) AND ASK ONE OF THE COMPUTERS  THE ANSWER ON IT HAVE TO ASSURE YOU THAT YOU NOT GONNA BUY THE CHINESE COMPUTER.

I think you should ask :
  If I'd ask you,are you the chinese computer ,what would you say. [im not sure if I understood the english]
If it is the chinnese computer ,It would be answered NO, but in this case it should be answer YES.So you buy another computer.
 ...it is the same trick if it is not the chinnese computer [at least I think soo]

Note : The negative of negative is postive :

EVERYONE OF THE COMPUTERS KNOW EVERYTHING ABOUT EACH OTHER.YOU NEED TO FIND A QUESTION(THAT THE ANSWER ON IT IS YES OR NO) AND ASK ONE OF THE COMPUTERS THE ANSWER ON IT HAVE TO ASSURE YOU THAT YOU NOT GONNA BUY THE CHINESE COMPUTER.

can u please elaborate on this? :)

I'm sorry for the unclarity!
you should ask one of the computers(you dont know who is who because they look the same) one question(the computers can only answer yes or no so you can not ask them question like "how do you fill?") can you find a question,  that can assure you, that you [b][i][u]not[/u][/i][/b] gonna pick the chinese computer?

notice: that the russian computer answers randomly "yes" or "no" so it very hard to recognize who is it by the answer

I can't find a one-question answer but I have a two-question one. Maybe someone can improve it:

[hide="Two questions."]
=====
Label the computers $ A,B,C$ left to right.

(1) Ask $ A$: "If I asked you yesterday is $ B$ Russia, what would you have said?"

If $ A$ is Russia, then $ B$ is not Russia. If $ A$ is USA or China and $ A$ says yes, then $ B$ is Russia and $ C$ is not Russia. If $ A$ says no, then $ B$ is not Russia and $ C$ is Russia.

(2) Using (1), we can find a non-Russia computer. Ask it: "Is 1+1=2?" If yes, it is USA so buy it. If no, then it is China so buy another.
=====
[/hide]

[quote="enndb0x"]I think you should ask :
  If I'd ask you,are you the chinese computer ,what would you say.[/quote]

What happens if the computer says "yes"? The computer could either be China or Russia, and so we can't determine which computer to take.

[quote="Brut3Forc3"][quote="jvenezuela716"][quote="Brut3Forc3"]I only know of a way to do this to ensure you don't pick the Russian (i.e., random answer) computer...[/quote]

Yeah, this is the first question in my post above.[/quote]
Oh, non-random computer can be done in one question. I'm still not sure how to do the non-liar, since if you ask the question to the random computer, you gain absolutely no information...[/quote]

In fact, because you gain no information, it is provable that you can't find the Chinese computer in one question.

And your first question is flawed; if it says yes, A might still be the Russian computer.

[quote="pythag011"]In fact, because you gain no information, it is provable that you can't find the Chinese computer in one question.

And your first question is flawed; if it says yes, A might still be the Russian computer.[/quote]
Really? Hmm.
Anyway, way to make sure you don't pick the random-picking computer:
[hide]Label the computers A,B,C in any order. Ask computer A if computer B tells the truth more often than C does. If A says yes, pick C. If A says no, pick B.
[hide="Reasoning"]We will never be able to tell if the computer we ask is the random computer or not (given one question only anyway), so we must accept that we will never pick the computer we ask. Because of this, we seek to somehow distinguish the two computers that we do not ask. For the question given, if A is a truth-teller, then the computer it indicates tells the truth more, so it is the random computer. If A is a liar, then the computer it indicates tells the truth less, so it is the the random computer. Either way, picking the other computer gets us what we want: a non-random computer. And of course, if A is the random computer, picking either of the other two will be okay.
Note that there in this process, there is no way to tell which computer is the truthteller or the liar...[/hide][/hide]

[quote="jvenezuela716"]I can assure both of you that my first question above works in finding a non-random computer.

@pythag011: If $ A$ is random, then no matter how $ A$ responds, we will be led to choose either $ B,C$, both of which are not random. Do you get it?[/quote]

No, explain please. How will we figure out if A is random from the first question? What if A is russia and says Yes? How will you be able to figure out its russia? What if B is russia and says no? How will we figure out its russia.

Asking the random computer cannot give you any information. And you cannot determine whether the computer you ask is random. So the only possible way is to ask a question that only the liar cannot answer. But I still have no idea on it.

Wait; are we allowed to ask questions that the computer cannot answer?

[hide="Proof if you cannot ask questions that a computer cannot answer"]
There is always a case that the computer you ask is the random computer. In this case, call the random computer A. Then either:

1. Computer B is a truth-teller and Computer C is a liar, or:
2. Computer B is a liar and Computer C is a truth-teller.

If you ask the random computer a question that isolates these two choices to 1 for "Yes" and 2 for "No", then you will gain no information because the random computer answers your question randomly, rather than according to the question you ask.

Now, if you ask the random computer a question such that both responses, "Yes" and "No" lead to choice 1, or both lead to choice 2, then you obviously will still not gain information.[/hide]

@Pythag011: If so, that would solve the problem with the following question: "Will the next word you say be 'Yes'?" Here, the liar cannot respond because if he chooses to say yes, he will say "No", after which he has to say "Yes", etc. Therefore, if the computer you ask cannot answer, then choose one of the other two. Otherwise, choose the computer that you asked.

The problem is not to take the chinese computer ,taking the russian or american is not the problem.

[quote="jvenezuela716"][quote="pythag011"]
No, explain please. How will we figure out if A is random from the first question? What if A is russia and says Yes? How will you be able to figure out its russia? What if B is russia and says no? How will we figure out its russia.[/quote]

[hide="Explanation."]
=====
The question is "If I asked you yesterday is $ B$ Russia, what would you have said?" If the response is yes, buy $ C$. If the response is no, buy $ B$. We claim that this will always result in the purchase of a non-Russia computer.

There are two cases for $ A$:

(i) If $ A$ is Russia, then no matter $ A$'s answer, we will buy $ B$ or $ C$, neither of which is Russia (because $ A$ is).

(ii) If $ A$ is USA or China, the construction of the question is such that their answers will be consistent and furthermore will answer truthfully the question "Is $ B$ Russia?". If the answer is yes, we will therefore buy $ C$, which is non-Russia. If the answer is no, we will therefore buy $ B$, which is non-random.
=====
[/hide]

Now as pythag011 said above, the original question regarding finding a non-China computer cannot be done in one question. So, to put this problem to rest, using any method discussed here (there are plenty), we can find a non-random computer with one question. Then, using a second, we can pose some trivial question that allows us to determine the identity of this non-random computer. For example, "Is 1+1=2?": if the output is yes, then we've found USA. Otherwise, it is China.[/quote]

Oh I see  :blush:

But still, we can't find the Chinese computer because we might ask the russian computer.

The question you should ask is:

Will the next word you say be "Yes"?

This way, you isolate the Chinese computer by the fact that it cannot answer the above question.


abstract algebra
number theory open
number theory

Solve in positive integers $x!+k=y^{2}$ where $k\ge2$ is arbitrary positive integer.

For $k>1$ it is simple to prove that the set of solutions are finitely.
Always there is a module $a$ that $x>a$ $x!+k\not\equiv y^{2}\bmod a$

[quote="juna"]
Always there is a module $a$ that $x>a$ $x!+k\not\equiv y^{2}\bmod a$[/quote] I think it is true if and only if k is not squre.

You are right!
If k is square, that is required abc hypothesis.
[url=http://mathworld.wolfram.com/BrocardsProblem.html]Brocards Problem[/url]


quadratics
inequalities
inequalities unsolved

Let $ 0 < x <y$ be real numbers. Let
$ H\equal{}\frac{2xy}{x\plus{}y}$ , $ G\equal{}\sqrt{xy}$ , $ A\equal{}\frac{x\plus{}y}{2}$ , $ Q\equal{}\sqrt{\frac{x^2\plus{}y^2}{2}}$
be the harmonic, geometric, arithmetic and root mean square (quadratic mean) of $ x$ and $ y$. As generally known $ H<G<A<Q$. Arrange the intervals $ [H,G]$ , $ [G,A]$ and $ [A,Q]$ in ascending order by their length.

[i]Lieber würde ich auf deutsch schreiben, aber hoffentlich verstehst du meine Ideen in englisch gut ...[/i]
__________________

The lengths of the intervals $ [H,G]$, $ [G,A]$ and $ [A,Q]$ are the numbers $ G\minus{}H$, $ A\minus{}G$ and $ Q\minus{}A$ respectively. The case $ x\equal{}y>0$ is very easy. Suppose that $ x\neq y$. Without loss of generality we may assume that $ x>y$. This implies that there exists the number $ k>1$ such that $ x\equal{}ky$. Hence
$ A\minus{}G\equal{}y\cdot\left (\frac{k\plus{}1}{2}\minus{}\sqrt{k}\right ),\\
Q\minus{}A\equal{}y\cdot\left (\sqrt{\frac{k^2\plus{}1}{2}}\minus{}\frac{k\plus{}1}{2}\right ),\\
G\minus{}H\equal{}y\cdot\left (\sqrt{k}\minus{}2\plus{}\frac{2}{k\plus{}1}\right ).$
You may also cancel the factor $ y>0$. Now prove the following inequalities ($ k>1$)
$ \frac{k\plus{}1}{2}\minus{}\sqrt{k}>\sqrt{\frac{k^2\plus{}1}{2}}\minus{}\frac{k\plus{}1}{2}\quad\Leftrightarrow\quad A\minus{}G>Q\minus{}A,\\
\sqrt{\frac{k^2\plus{}1}{2}}\minus{}\frac{k\plus{}1}{2}>\sqrt{k}\minus{}2\plus{}\frac{2}{k\plus{}1}\quad\Leftrightarrow\quad Q\minus{}A>G\minus{}H.$
Now it is obvious that $ A\minus{}G\ge Q\minus{}A\ge G\minus{}H$ for any pair $ (x,y)\in\mathbb{R}^\plus{}\times\mathbb{R}^\plus{}$.

[quote="M@ri@n"][i]Lieber würde ich auf deutsch schreiben, aber hoffentlich verstehst du meine Ideen in englisch gut ...[/i]
__________________

The lengths of the intervals $ [H,G]$, $ [G,A]$ and $ [A,Q]$ are the numbers $ G \minus{} H$, $ A \minus{} G$ and $ Q \minus{} A$ respectively. The case $ x \equal{} y > 0$ is very easy. Suppose that $ x\neq y$. Without loss of generality we may assume that $ x > y$. This implies that there exists the number $ k > 1$ such that $ x \equal{} ky$. Hence
$ A \minus{} G \equal{} y\cdot\left (\frac {k \plus{} 1}{2} \minus{} \sqrt {k}\right ), \\
Q \minus{} A \equal{} y\cdot\left (\sqrt {\frac {k^2 \plus{} 1}{2}} \minus{} \frac {k \plus{} 1}{2}\right ), \\
G \minus{} H \equal{} y\cdot\left (\sqrt {k} \minus{} 2 \plus{} \frac {2}{k \plus{} 1}\right ).$
You may also cancel the factor $ y > 0$. Now prove the following inequalities ($ k > 1$)
$ \frac {k \plus{} 1}{2} \minus{} \sqrt {k} > \sqrt {\frac {k^2 \plus{} 1}{2}} \minus{} \frac {k \plus{} 1}{2}\quad\Leftrightarrow\quad A \minus{} G > Q \minus{} A, \\
\sqrt {\frac {k^2 \plus{} 1}{2}} \minus{} \frac {k \plus{} 1}{2} > \sqrt {k} \minus{} 2 \plus{} \frac {2}{k \plus{} 1}\quad\Leftrightarrow\quad Q \minus{} A > G \minus{} H.$
Now it is obvious that $ A \minus{} G\ge Q \minus{} A\ge G \minus{} H$ for any pair $ (x,y)\in\mathbb{R}^ \plus{} \times\mathbb{R}^ \plus{}$.[/quote]
M@ri@n: You'll normaly write $ (x,y) \in \mathbb{R}^2_ \plus{}$. As far as I remember $ \mathbb{R}^ \plus{}$ is a group.

I'll provide the proofs and some other information :)
It's in fact very easy to "guess" the order of these intervals. Just plug in $ x \equal{} 1$ and $ y \equal{} 7$, and you will instantly know which order the intervals are!

First $ A \minus{} G \ge Q \minus{} A \iff 2A \ge G \plus{} Q$. After transfomation it is equivalent to:
$ x \plus{} y \ge \sqrt {\frac {x^2 \plus{} y^2}{2}} \plus{} \sqrt {xy}$.
But $ \sqrt {\frac {x^2 \plus{} y^2}{2}} \plus{} \sqrt {xy} \le \sqrt {x^2 \plus{} y^2 \plus{} 2xy} \equal{} x \plus{} y$ by AM-QM.
Then $ Q \minus{} A > G \minus{} H \iff Q \plus{} H \ge A \plus{} G$
$ \sqrt {\frac {x^2 \plus{} y^2}{2}} \plus{} \frac {2xy}{x \plus{} y} \ge \frac {x \plus{} y}{2} \plus{} \sqrt {xy} \iff$
$ \sqrt {\frac {x^2 \plus{} y^2}{2}} \minus{} \sqrt {xy} \plus{} (x \minus{} y)^2 \cdot \frac { \minus{} 1}{2(x \plus{} y)} \ge 0 \iff$
$ \frac {(x \minus{} y)^2}{2(\sqrt {\frac {x^2 \plus{} y^2}{2}} \plus{} \sqrt {xy})} \plus{} (x \minus{} y)^2 \cdot \frac { \minus{} 1}{2(x \plus{} y)} \ge 0 \iff$
$ (x \minus{} y)^2 \cdot \left ( \frac {1}{\sqrt {\frac {x^2 \plus{} y^2}{2}} \plus{} \sqrt {xy}} \minus{} \frac {1}{x \plus{} y} \right ) \ge 0 \iff$
$ (x \minus{} y)^2 \cdot \left ( \frac {x \plus{} y \minus{} \sqrt {\frac {x^2 \plus{} y^2}{2}} \minus{} \sqrt {xy}}{(\sqrt {\frac {x^2 \plus{} y^2}{2}} \plus{} \sqrt {xy})(x \plus{} y)} \right ) \ge 0 \iff$
$ (x \minus{} y)^2 \cdot \frac {2A \minus{} Q \minus{} G}{2A \cdot (Q \plus{} G)} \ge 0$. Which we have just proved, is true.
So $ A \minus{} G \ge Q \minus{} A \ge G \minus{} H$. QED

Just a fun observation:
$ 2A \minus{} G \minus{} Q \equal{} \frac {(x \minus{} y)^4}{8(2A \plus{} G \plus{} Q)(A^2 \plus{} GQ)}$ and $ Q \plus{} H \minus{} A \minus{} G \equal{} \frac {(x \minus{} y)^6}{16(2A \plus{} G \plus{} Q)(A^2 \plus{} GQ)A(Q \plus{} G)}$

Another fun thing :)
$ Q\minus{}A \equal{} \frac{(x\minus{}y)^2}{4(Q\plus{}A)}$
$ A\minus{}G \equal{} \frac{(x\minus{}y)^2}{4(A\plus{}G)}$
So now it is obvious that $ A\minus{}G \ge Q\minus{}A$.

Here is a beautiful solution.
[hide] The answer is $$d[G,A]>d[A,Q]>d[H,G],$$ where $d[x,y]$ denotes the length of the intervall $[x,y]$.
We will get this by proving the inequalities
\begin{align*}
&\text{(1)}  \; d[G,A] > d[H,G],
&\text{(2)}  \; d[G,A] > d[A,Q], \;\;\;
&\text{(3)}  \; d[A,Q] > d[H,G].
\end{align*}
We will use the following two "theorems" about the numbers $H$, $G$, $A$ and $Q$:
\begin{align*}
&\text{(I)} \; AH=G^2,
&\text{(II)} \; G^2+Q^2=2A^2,
\end{align*} or in other words,
[b](I)[/b] $G$ is the geometric mean of $A$ and $H$, and [b](II)[/b] $A$ is the quadratic mean of $G$ and $Q$.

The proofs of these "theorems" are trivial.

[b](1)[/b] Using this, we get $$G= GM[H,A] < AM[H,A],$$ because $H<A$, and we see that $AM[H,A]$ lies exactly between the numbers $H$ and $A$, hence $$d[G,A] > d[H,G].$$
[b](2)[/b] Furthermore, we get $$A= QM[G,Q] > AM[G,Q],$$ because $G<Q$, and we see that $AM[G,Q]$ lies exactly between the numbers $G$ and $Q$, hence $$d[G,A] > d[A,Q].$$
[b](3)[/b] Now we have to prove that $d[A,Q] > d[H,G]$.

To do so, we would need to proof that $d[G,Q] > d[H,A]$. This follows from
\begin{align*}
2A(Q-G) &= 2QM[Q,G](Q-G) \\ &> 2AM[Q,G](Q-G) \\ &= (Q+G)(Q-G)=Q^2-G^2 \\
&=(2A^2-G^2)-G^2=2A^2-2G^2 \\ &=2A^2-2AH=2A(A-H).
\end{align*}
Hence, the proof is done. [/hide]


I usually make mistakes when I have to fit the time of the verb in a given sentence. You know.... (which form of the past verb shuld I use  (present perfect,simle past , ect...)). What can you propose  me ?

ii) It is better to say "some friends" or "A few friends"

                                                                              Thankyou

Honestly, I have no idea which tense in which when I speak (but I'm a native speaker  :P ).  But when I learned Spanish and German, I tried to associate certain keywords are phrases with the particular tense, or recall examples that I could associate with the current situation.  The particular tenses usually have rules associated with them so try to remember those rules.

In response to your example, it depends.  If I'm talking casually, I'll say "I was with some friends..."  That's typically what I'd say.  I can't think of an example off the top of my head in which I would say "a few friends."  Maybe a formal conversation?

I would suggest you read a really good student grammar of English, because verb tense, modality, and aspect are a much bigger issue than anyone could address in one post.  I'd recommend this one:
[url]http://www.amazon.com/Students-Introduction-English-Grammar/dp/0521612888/ref=sr_1_1?ie=UTF8&s=books&qid=1241052783&sr=1-1[/url]

If there's anything specific you're wondering about, I'd be happy to answer.


inequalities
AMC
USA(J)MO
USAMO
inequalities open

you can find the solutions in:
1.www.mathlinks.ro/Forum/viewtopic.php?t=115680
2.www.kalva.demon.co.uk/usa/usoln/usol013.html


real analysis
function
vector
real analysis theorems

I'm working out of "Principles of Mathematical Analysis," by Rudin, and one of the theorems states:

The sequence of functions $ \{f_n\}$, defined on $ E$, converges uniformly on E if and only if for every $ \epsilon>0$ there exists an integer N such that $ m\geq$, $ n\geq N$, $ x\in E$ implies $ |f_n(x)\minus{}f_m(x)|<\epsilon$.

I must be reading something incorrectly, because to me this says that a sequence of functions is uniformly convergent if and only if it is a Cauchy sequence, which contradicts what I learned about Cauchy sequences earlier (that not all are convergent)

Is compactness or completeness supposed to be a condition on E? It is not stated in the book.

Any help would be appreciated. Thank you.

It is correct as stated. There is a pointwise limit function $ f(x)$. Given $ \varepsilon > 0$ chose $ N$ as in the statement. Passing $ m$ to infinity, we get that for every $ n \geq N$ and $ x \in E$ $ |f_n(x) \minus{} f(x)| \leq \varepsilon$

$ E$ does not need any metric structure at all, it has nothing to do with the statement. The point is that $ \mathbb{R}$ is complete.

So the functions are actually real valued functions, and that's why this theorem holds?

There is definitely a hidden assumption there- the functions are real-valued, or at least take values in a complete normed vector space.
The structure of $ E$ is irrelevant- it doesn't even need a topology.


$C= \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array}$

$[C^t]_{32}$

^What is that notation? I assume that the $_{32}$ is the scalar multiplier? What does the $t$ stand for?

First, when you see a 't' up there, it usually denotes the transpose of a matrix. Second, the 32 indicates the specified entry. It does not mean number '32', but it means row 3 column 2. So if you read the entire thing, it is indicating entry 3,2 of the transpose of $C$. Third, when you write a matrix, you can use \left( or \left[ for the big matrix enclosure and every \left command must end with a \right command.


geometry
AMC
AIME
email
USA(J)MO
USAMO
number theory

I'm going to apply and if I get in, I want to have a clear view of how the classes are going to be and what comes in them and how the camp is. So here are a couple questions:

1) What are the possible choices for classes?

2) How do you know what level to take?

3) Can you change from class? if so, Is there a time limit from which you can change to another class?

4) Are there any tests? If so, How often are they given?

5) What classes would you recommend?

1) 3 levels of Combinatorics, Number Theory, Algebra, Geometry.

That picture is clever. Lol, I get the 10 kinds of ppl: those who know binary and those who dont. FOR THOSE PEOPLE WHO DON'T KNOW, 10 IS 2 IN BINARY......sorry, but i had to mention it

[quote="shelly32494"]1) dont really know. however if its similar to last year... there's 3 levels of each number theory, geometry, alegbra & ummm. i forgot... sorry.
2) in the packet they give you [that also includes the list and description of classes] they'll also have recommendations [for example, AIME between 4&8 should take this class or something like that]; but you definitely dont need to follow that. read the descriptions & see what you know /dont know and decide whether you think you'd be good taking that class... dont worry bout it tho... becausee....
3) you can change classes! :) its really really easy. a lot of people do. i think they prefer it to be done in the first couple of days/first week. but i did know a guy who changed like, late second week/last week. -.- but since its really easy to change classes, you dont need to worry too much about your first selection [#2] of classes, because , after a couple of days, if you feel you're in the wrong level you can change =]]
4) yes, there are, however, its doesnt "count" [your parents arent going to see your test grades lol] its mostly a matter of pride & how much you've retained :P . its once /week. 'cept for the last week. so only twice during camp. pay attention in class hahahahah. ^_^ try not to goof off too much. or fall asleep . lol.
5) i really liked counting strategies & computational geometry [the two classes i took last year]. they were pretty challenging [esp computational geometry ] the teachers were great too :) i definitely recommend it :D[/quote]

I got my packet but I don't see anything about the classes  :?: .

[quote="blackbelt14253"][quote="ra5249"]6 to a room.
I would definitely recommend geometry proofs (3rd level geometry), and I found combinatorics 3rd level really helpful too.[/quote]

in case 6 to a room sounds scary, its not 3 to a bed, there's six beds. each room is like an apartment with 4 bedrooms (one for the counselor, 3 with 2 students each)

geometry proofs and nt level 3 are what i recommend. actually, i took level 2 of both geo and nt, since i was dumb back then, but whatever. just don't fall asleep during the lectures like some people. problem sessions are fun though.[/quote]

yeah and hey guys, go to the lectures/forums! they're really interesting and cover useful material!
notably, there was a forum on radical axes which i skipped. guess what #1 on the usamo this year was. darn.

notably, there was a forum on radical axes that i skipped as well. guess who got #1 on the usamo this year?  :o  lol i just want to make fun of THE RA XD

LOL I never liked to sleep with anyone in the same bed because what if he is kicking while he is sleeping. I will never sleep  :rotfl:

[quote="shelly32494"]really? ... thats weird.
email titu.[/quote]

i dont think the packets this year have anything about classes because mine didnt either

They probably send out the classes information after you send in your medical forms, etc.

yeah
well it would make more sense to send it together instead of having to mail two forms.

info about the classes should be online!


inequalities
Cauchy Inequality
inequalities proposed

Come on guys try to solve this.It looks ugly but it is very nice. :)

[quote="cezar lupu"]Vladut, at first glance, I proved that

                               \[ \sum\frac{a^{3}}{b^{3}+c^{3}}\geq\frac{1}{a^{4}+b^{4}+c^{4}+abc(a+b+c)}.  \][/quote]
Yes, this can be proved by a "first glance". :)
$\sum\frac{a^3}{b^3+c^3}=\sum\frac{a^4}{ab^3+ac^3}\geq\frac{(\sum a^2)^2}{\sum (ab^3+ac^3)}=\frac{1}{\sum_{sym}a^3b}\geq\frac{1}{a^4+b^4+c^4+abc(a+b+c)}$ by Cauchy-Schwarz and Schur inequalities respectively. :P

[quote="mateivld"]Let $a,b,c>0$.If $a^2+b^2+c^2=1$ prove that \[ \displaystyle\sum\frac{a^3}{b^3+c^3}\geq \frac{1}{2(1-\sum a^4-abc\sum a)}  \][/quote]
ok, :) It is due to leepakhin.
$\displaystyle\sum\frac{a^3}{b^3+c^3}\geq \frac{(\sum_{cyc} a^3)^2}{2\sum_{cyc}a^3b^3} \ge \frac{(\sum_{cyc} a^2)^2}{2\sum_{cyc}a^2b^2} = \frac{1}{2\sum_{cyc}a^2b^2}$
so we should only to prove:
$2(1-\sum a^4-abc\sum a) \ge 2(\sum_{cyc}a^2b^2) \iff 1-\sum a^4-abc\sum a \ge \sum_{cyc}a^2b^2 \iff 2 \sum_{cyc}a^2b^2-abc\sum a\ge \sum_{cyc}a^2b^2$
we have finished
for  $\frac{(\sum_{cyc} a^3)^2}{2\sum_{cyc}a^3b^3} \ge \frac{(\sum_{cyc} a^2)^2}{2\sum_{cyc}a^2b^2}$
please see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=67552]http://www.mathlinks.ro/Forum/viewtopic.php?t=67552[/url]

[quote="mateivld"]Let $a,b,c>0$.If $a^2+b^2+c^2=1$ prove that \[ \displaystyle\sum\frac{a^3}{b^3+c^3}\geq \frac{1}{2(1-\sum a^4-abc\sum a)}  \][/quote]

I will produce other solution( Because others have already posted the same solution as mine previous )
Because $a^2+b^2+c^2=1$ so we have:$1-( \sum a^4)-abc( \sum a )=( \sum a^2)^2-( \sum a^4 )-abc( \sum a )= \sum a^2(b^2+c^2-bc)= \sum a^2( \frac{(b^3+c^3)}{b+c} )$
So the given inequality is equivalent with:
$\sum \frac{a^3}{b^3+c^3} \geq \frac{1}{2 \sum a^2 \frac{(b^3+c^3)}{b+c} }$
Or in other form:
$2( \sum \frac{a^3}{(b^3+c^3)} )( \sum a^2 \frac{(b^3+c^3)}{b+c} ) \geq 1$
But due to Bunhiakowski inequality we have:
$LHS \geq 2( \sum \sqrt{ \frac{a^5}{b+c}} )^2$
So if we can prove : $\sum a^2 \sqrt{ \frac{a}{b+c}} \geq \frac{1}{ \sqrt{2}}$ the above inequality will also be proved..
By Cauchy inequality:
$F=\sqrt{ \frac{a}{b+c}}= \frac{a}{\sqrt{ab+ac}} \geq \frac{a}{\sqrt{ \frac{2a^2+b^2+c^2}{2}}}$
So $F= \sqrt{2} \frac{a}{ \sqrt{1+a^2}}= \sqrt{2} \frac{1}{ \sqrt{1+ \frac{1}{a^2}}}$
Finally: $F=2 \sqrt{2} \frac{1}{2 \sqrt{1+ \frac{1}{a^2}}}$
Again by Cauchy inequality $F \geq 2 \sqrt{2} \frac{1}{ \frac{4+1+\frac{1}{a^2}}{2}} =\frac{4\sqrt{2}a^2}{5a^2+1}$
So: $\sum \sqrt{ \frac{a^5}{b+c}} \geq \sum ( \frac{4 \sqrt{2}a^4}{5a^2+1})$
By putting $x=a^2$ $y=b^2$ $z=c^2$
The $RHS=f(x)+f(y)+f(z)$ with $f(x)= \frac{ 4\sqrt{2}x^2}{5x+1}$
But f(x) is convex for all x in [0,1] so by Jensen Inequality
$f(x)+f(y)+f(z) \geq 3f( \frac{x+y+z}{3})=3f( \frac{1}{3})= \frac{1}{ \sqrt{2}}$ Proved
Kummer
( Hope my solution is not too ugly :D )

I will give you my solution .Note that $\displaystyle \frac{a^3}{b^3+c^3}=\frac{a}{b+c}\cdot \frac{a^2}{b^2+c^2-bc}$ so we can aply Cebasev inequality ,and using that $\displaystyle\sum\frac{a}{b+c}\geq \frac{3}{2}$ we get $\displaystyle \sum\frac{a^3}{b^3+c^3}\geq \sum\frac{a^2}{2(b^2+c^2-bc)}\geq \frac{(\sum a^2)^2}{2(2\sum b^2c^2-abc\sum a)}$,from Cauchy. :)


(a) Let G = {M1, M2, ... , Mr} denote the set of n × n permutation matrices (those matrices which have only zeroes and ones as entries, and whose row and column sums are all one), and let A = (M1 + M2 + ... + Mr) / r denote the average of these matrices. Determine A.

(b) Let G = {M1, M2, ... , Mr} be any finite set of n × n matrices with real entries and such that G is a group under matrix multiplication. Let A = (M1 + M2 + ... + Mr) / r denote the average of these matrices. What can be said about A?

Your part (b) is wrong for the following reasons:
1) Not every group is cyclic (e.g., the symmetric group on more than 2 letters)
2) Just because $ M \neq I$ does not mean that $ M-I$ is invertible.  For example, consider the group $ G = \left\{ \left(\begin{array}{cc}0 & 1 \\ 1 & 0 \end{array}\right), I\right\}$ (or, indeed the symmetric group, as in part (a)).

This post belongs somewhere else -- I suspect it has appeared in either the Superior Algebra or Linear Algebra forums before, actually.


i was thinking of making some...any takers?

Sure, why not?  After next week no more exams or, really, school for me!

Count me in.  I need practice in not making mistakes :oops: .  From the looks of this thread, I don't think you will have any problems finding people willing to do them.

I will try as well. Thanks for making one

/me volunteers and promises to do it, after May 19.

[quote="Altheman"]yea...i have 4 ap tests so i don't expect to post anything before then...[/quote]

heh, me too... i'll probably take it after may 16th...

I'm in. The message is too small. Please make the message longer before submitting.

I'll take part. AP exams finish for me this wednesday so its all cool.

Wow I love how everyone just posts "I'll do it."

Okay, I'll do it too.

I'm in, all the way, in San Jose! (C to the A!!)

I'll do it if I have time. Or I'll make time if it looks really cool. Which it probably will, knowing you.

ap english is laying on the pain...huge paper that is like half of semester grade...

i have a bit of work to do, i am still playing around with time concerns/topic balancing...i will probably finish some time over the weekend

yet you still find time to post on aops  :D good luck!

i always find time to do math...haha...

so anyway, what do people think of the logistics for holding a contest...?

like i was thinking about people submitting answers...but i feel that is hard to have any time restriction/calculator restriction that way

i don't really have the time to do an aim session...so i guess i am leaning towards people just sending me answers for no particular reason

You could always post the problems at one particular time, then have answers due roughly 1 hour later, if people wanted their usernames to be in the top list.

It doesn't really matter if people cheat; the mock contests are made to practice, so you would only cheat yourself.

BTW, I'm interested  :D

just make it honor based i mean honestly why would someone cheat on a mock test

I don't think posting the questions and then submitting the answers an hour later would work. What if someone happens to be away from the computer at that particular time? It wouldn't really be fair to them.

I agree with noneoftheabove, just make it honor-based.

I'm in.

Or, you can just set up a definite time when they will be released, as an alternative to honor system. But this would have to be a time at least 2 days in advance.

I think I would want the problems, but may not "compete" per se. (or maybe I will.)


Harry Potter and the Half-Blood Prince!!!!!

Anyone very excited? the book is coming out July 16, and I heard that a main character dies...am I right?

Just pick up the book on the way and read it during camp. That's what I did with book 5. Except halfway through the camp, word spread and everyone asked if I could lend it.
My birthday happens to be July 17. It would be a great present.

[quote="LynnelleYe"]I'm excited, but it turns out that I'm leaving for CTY camp the day after it comes out. Since I want to enjoy it for as long as possible, and not finish it in a day, I'll probably just not read it until I come back. Too bad.
I didn't know it was 'official' that a main character would die, but with a war going on the book would be pretty unrealistic if nobody readers cared about died, so it's probably nearly a given.[/quote]
CTY camp? at where?

Does anyone know who dies in the story?

If that means Harry Potter dies... well that's not allowed in Harry Potter style books :P.  BTW at CTY no doubt one of the random prizes they give from time to time will be a trip to buy the Harry Potter book.

harry doesn't die yet. jk rowling said so herself.

AH I can't WAIT!

I'm a total bookworm...so hard to resist reading good books, hehehe.

What about Ron or Hermione?  I think that would be kind of unlikely...
My guess is that a character like [hide="spoilers of book 5"]Sirius[/hide] in the last book will be killed.

This person came by and told me the person casts a spell on Ron, and Ron tries to kill Hermionie, and Harry intervenes, but I don't believe it...but maybe it is, because hermionie is not "pure,"-referring to the title of the next book

ok...i don't think Harry or Ron or Hermione will die, they are the main charactors...remember there is stiil book seven.
And who will be the new Defense against Dark Arts teacher?? Lupin again?? Or Dumbledor's brother??
And what will happen to Lupin and Percy??

[quote="yif man12"]I hope Ron dies. Seriously, that kid gets on my nerves so much.[/quote]

Agreed. Definitely. If anyone goes, it should be Ron.

I'll read it in a day, even though I'm not that into Harry Potter anymore.

[quote="MathFiend"]woohoo! book comes out @ midnight! and i'm going!!!! hoorah! :)[/quote]
lol. Due to the differenct time zone, we will have the book at daytime in China.

LOL, did you read the story about the guy who took suicide because he learned a spoiler about harry potter? LOL, obviously a joke article but it was pretty funny.  :rotfl:

Yeah, I get it now!
I am very glad to see that it is much shorter than book 5.

[quote="SnowStorm"]I got it at midnight last night, fell asleep reading it and am about a third of the way through.  Good so far.[/quote]

Who dies?? what happens? ahhh I know I'm not going to buy the book but I want to know, lol.

[quote="mathclass"][quote="h_s_potter2002"]Yay, I finished the book![/quote]

aaaah I haven't gotten it. Is it good?[/quote]

All I want to say is that it was very suprising.  How the death hapenned, and what Harry planned to do later wasn't really something i expected.

let's not post ANYTHING of the contents of the book here....i'll start up a separate thread that can be used to discuss the book, that way you don't accidently come across something that spoils the book.

i got it yesterday night...i read for two hours and fell asleep...i read again this morning for an hour.

[quote]All I want to say is that it was very suprising. How the death hapenned, and what Harry planned to do later wasn't really something i expected.[/quote]

Really? I saw it coming, altough I expected her to not have it happen at the last second possible.

[quote="wobster109"]just my personal opinion, but

[color=red][b]IT SUCKED!!!!![/b][/color][/quote]

agreed, it didn't have enough action.....barely had any spell casting, too.

if the last part, the fight, had involved more detail and characters, the story would have been definitely better.

I liked the book - but it did have some elements that I thought were just... out of place. The whole boy-girl thing in the book - while it probably (some of it) should be in the book- was too much. There was probably about 2 chapters worth of paragraphs and sections just devoted to that, which doesnt have any importance in the book (didnt change much of anything, especially at the end when Harry..). Also, I feel that certain things that happened in the book were too cliched and weren't really original, which isnt something I expect from a Harry Potter book. There was a lot in the book to keep me interested, but there probably should have been a little more action, more spells. The fight scene could have been amazing if it was longer and more detailed. Last point here - did anyone else hate the Horcrux idea? It reminds me too much of video games where you have to go on all these little dangerous quests to get to the big boss...

^just my opinion


integration
calculus
trigonometry
calculus computations

how would i do this question using substitution method?

$\int xcosx\,dx$

You can't use a straightforward substitution, but you can use integration by parts.

[b]Hint[/b]
[hide][url]http://en.wikipedia.org/wiki/Integration_by_parts[/url][/hide]

[b]Solution[/b]
[hide]$I=\int x \cdot \cos x ~dx$

Let $u=x \iff du=dx$
Let $dv=\cos x ~dx \iff v=\sin x$

$I=x \cdot \sin x-\int \sin x ~dx+C_{0}$
$I=\boxed{x \cdot \sin x+\cos x+C}$[/hide]

ok i have not learnt that method yet but i shall learn now

thanks for your helps
bye

do i have to use integration by parts for $tan^{2}x$ also?

Not necessarily

Note that $\int\tan^{2}x~dx=\int(\sec^{2}x-1)~dx$, then it's done.

[quote="Hetidemek"][quote="skimnc"]
Let $u=x \iff du=dx$[/quote]

I think the $\Longleftarrow\ldots$ is not true...[/quote]
Yeah i know.  I just like the $\iff$ arrows..


A sequence of integers $a_{1}, a_{2}, a_{3}, \cdots$  is defined as follows: $a_{1}=1$, and for $n \ge 1$, $a_{n+1}$ is the smallest integer greater than $a_{n}$ such that $a_{i}+a_{j} \neq 3a_{k}$ for any $i, j, $ and $k$ in $\{1, 2, 3, \cdots, n+1 \}$, not necessarily distinct. Determine $a_{1998}$.

[quote="Peter"]A sequence of integers $ a_{1}, a_{2}, a_{3}, \cdots$  is defined as follows: $ a_{1} \equal{} 1$, and for $ n \ge 1$, $ a_{n \plus{} 1}$ is the smallest integer greater than $ a_{n}$ such that $ a_{i} \plus{} a_{j} \neq 3a_{k}$ for any $ i, j,$ and $ k$ in $ \{1, 2, 3, \cdots, n \plus{} 1 \}$, not necessarily distinct. Determine $ a_{1998}$.[/quote]
http://www.kalva.demon.co.uk/short/soln/sh98n4.html

[quote="chien than"]http://www.kalva.demon.co.uk/short/soln/sh98n4.html[/quote]
The link doesn't work in my computer because the link is wrong or maybe my computer is an ancient one. :)
So I will post the solution.

Initially, we can determine the first new values for $a_n$ are 1, 3, 4, 7, 10, 12, 13, 16, 19, 21, 22, 25. Since these are exactly the numbers of the forms $3k+1$ and $9k+3$, we conjecture that this is the general pattern. In fact, it is easy to see that the equation $x+y=3z$ has no solution in the set $K = \{ 3k+1, 9k+3 | k \in \mathbb{N} \} $ . We shall prove that the sequence $\{a_n\}$ is actually this set ordered increasingly.

Suppose $a_n>25$ is the first member not belonging to $K$ . We have several cases :

(i) $a_n=3r+2, r \in \mathbb{N}$. By the assumption, one of $r+1,r+2,r+3$ is of the form $3k+1$ (and smaller than $a_n$), and therefore is a member $a_i$ of the sequence. Then $3a_i$ equals $a_n+1$, $a_n+4$, or $a_n+7$, which is a contradiction because $1,4,7$ are in the sequence.

(ii) $a_n=9r$, $r \in \mathbb{N}$. Then $a_n+a_2=3(3r+1)$, although 3r+1 is in the sequence, a contradiction.

(iii) $a_n=9r+6$, $r \in \mathbb{N}$. Then one of the numbers $3r+3, 3r+6, 3r+9$ is a member $a_j$ of the sequence an thus $3a_j$ is equal to $a_n+3, a_n+12, a_n+21$, where $3,12,21$ are members of the sequence, again a contradiction.

Once we have revealed the scructure of the sequence, it is easy to compute $a_{1998}$. We have $1998=4 \times 499 +2$, which implies $a_{1998}=9\times 499+a_2=4494$.


If the sides of a triangle have lengths $ a, b, c$, such that $ a \plus{} b \minus{} c \equal{} 2$, and $ 2ab \minus{} c^{2} \equal{} 4$, prove that the triangle is equilateral.

[hide]$ (a\plus{}b\minus{}c)^2\equal{}2ab\minus{}c^2 \iff (a\minus{}c)^2 \plus{} (b\minus{}c)^2\equal{}0 \Rightarrow a\equal{}c, b\equal{}c$[/hide]

Yes, it's quite simple, though I didn't know how to solve it when I took the test (the 'simple' solutions tend to elude me). The idea of completing the square was somewhat novel to me, back then (i.e., May of this year). ;-_-

[quote="metafor"][color=darkred] If the sides of a triangle have lengths $ a, b, c$ such that $ a \plus{} b \minus{} c \equal{} 2$ and $ 2ab \minus{} c^{2} \equal{} 4$, then prove that the triangle is equilateral.[/color] [/quote]

[b][u]Proof.[/u] Otherwise[/b], $ 2ab \minus{} 4 \equal{} (a \plus{} b \minus{} 2)^2$ $ \implies$ $ a^2 \plus{} b^2 \minus{} 4a \minus{} 4b \plus{} 8 \equal{} 0$ $ \implies$ $ (a \minus{} 2)^2 \plus{} (b \minus{} 2)^2 \equal{} 0$ $ \implies$ $ a \equal{} b \equal{} 2 \equal{} c$ .


function
combinatorics solved
combinatorics

Let S be a set of n points in R^3. Consider the value of each of this points to be 1 or -1 such that for each line passing through at least 2 points in S the sum of the values of the points on that line is 0.
Prove that all the points are on the same line.

We do, of course, the same thing you (ral) did to solve the problem about the circles.

Take all the lines passing through a point (assume the points are $A_i,i\in \overline{1,n}$, and we consider all the lines through $A_i$). The sum of the sums of the functions of the points on these lines is $0=S+(k_i-1)f(A_i)$, where $S=\displaystyle \sum_{i=1}^nf(A_i)$ and $k_i$ is the number of lines passing through $A_i$. Unless $k_i=1\ \forall i\in \overline{1,n}$, we get $k_i>1\ \forall i\in \overline{1,n}$, so $f(A_i)=0\ \forall i\in \overline{1,n}$ (because $S$ and $f(A_i)$ have oppisite signs, so $S$ and $S$ have opposite signs, so $S=f(A_i)=0\ \forall i\in \overline{1,n}$), which is obviously false. This shows that $k_i=1\ \forall i\in \overline{1,n}$, and we're done.


Me too i have a solution do you want i post it ?
o
u
b
i

[quote="hxtung"]x1 >= x2 >= ... >= xn are real numbers such that x1k + x2k + ... +
xnk >= 0 for all positive integers k. Let d = max{ |x1|, ... , |xn| }.
Show that x1 = d and that (x - x1)(x - x2) ... (x - xn) <= xn - dn for all real x >= d.
I've solution, and you?[/quote]

S_k = x_1^k +...+x_n^k

nx_1 \geq S_1 \geq 0 => x_1 \geq 0

First case: x_n \geq 0 , d= x_1

Second case: x_n<0 suppose d=/=x_1

d=|x_n|

if |x_i|<|x_n|  for i\in{1;2;...;n-1}  \lim (x_i/x_n)^k =0
if |x_i|=|x_n| we have |x_1|<|x_n| => x_i=x_1 => \lim (x_i/x_k)^k=1

The limit of  S_k/x_n exist when k-->+\infty, this limit is L \geq 1

There exist k \in N such that S_(2k+1)/x_n > but x_n<0 we obtain a contradiction S_(2k+1) <0
So we must have d=x_1
-------------------------------------------------------------
Let x>x_1 if n=1 nothing  to prove, suppose n \geq 2

AM-GM

(x-x_2)....(x-x_n)  \leq ( x - 1/(n-1)(x_2+...+x_n))^(n-1)

S_1 \geq 0=> x_1 \geq -x_2-...-x_n

(x-x_2)....(x-x_n)  \leq ( x +x_1/(n-1))^(n-1)  \leq \sum_{j=0 up to n-1}
x_1^j.x^(n-1-j)

(x-x_1)...(x-x_n)  \leq (x-x_1)(\sum_{j=0 up to n-1}x_1^j.x^(n-1-j)=

x^n - x_1^n

Valentin where is the castle of Dracula ?
What is the real story of Dracula ?
Do you have any idea if he wrotes some mathematical papers ?


AMC
AIME
geometry
3D geometry
tetrahedron
USA(J)MO
USAMO

Seeing as how chess64 will be giving an AMC, it feels like a good time to have an AIME as well.  If anybody is interested, I can write up and AIME. I already have most of the test, since I have thought of various problems during the year. In order to prevent any conflict with chess64's contest, I would put up the test in the first week of july. Since the AIME is a longer test and we are all honest people, just pm your answers to me within 2 or 3 weeks after the test is posted (I will specify dates later).

I learned from my last AIME about difficulty and topic distribution. I will try to keep the geometry problems to a minimum. Although I have a few problems that look like pure geometry, but they are really another topic like number theory.

I'll take it when I get back from camp, which will be around July 17-ish.

I'll do it.

I need AIME practice to improve score.

Without loss of generality, I too shall participate :).

Bleh, I guess I'll enter, but I'm gonna get PWNED.
Oh well.

[quote="SplashD"]Do we sign up?

I'll take it.[/quote]

well, i suppose that i will send pm s to people who express interest

[quote="ShoeFactory"][quote="Altheman"]Although I have a few problems that look like pure geometry, but they are really another topic like number theory.[/quote]

That tetrahedron problem from last test![/quote]

:)

I will do it, if i have time that is. :lol:

Whoa, lots of mock tests. I'll take it  :)

There are only 2 mock tests, unless there are other ones that I'm not aware of.

I'll probably be in.... but anyone wanna write a mock USAMO?

Or like.. grade some of my solutions? lol

I think that the Mock AIME should be posted soon.

I'll take it, despite that I got owned at the AIME...

[quote="Altheman"][quote="pkerichang"]uh...when will it be out??
btw I'm taking it too (despite having posted before...)[/quote]

i am working hard on it, i am writing up solutions for every problem so that takes a while, while it is not the most time efficient method, it is my way of checking that the problems make sense and work out reasonably, i have work tomorrow morning, so i will post it some time later tomorrow[/quote]

You could just post it and write the solutions later, but then if there's a mistake, everybody would be mad at because they wasted so much time on it.

That's exactly why he's writing up solutions first, because the same issue arose with the mock AMC.

[quote="Altheman"][quote="mysmartmouth"][quote="asdf4"]Do we have to have a 3 hour time limit like last time?
otherwise, youd have to check every hour.[/quote]

Check what every hour?  His inbox?  Did I miss something?[/quote]

i don't know what he is taking about...just post your answers when you are done...we are going by the honor system[/quote]

Let's say you have a 3-hour time limit, and you can only turn in the answers in that 3 hour time slot. (happened once) Then, you ll have to check your inbox or this topic VERY often, otherwise, you will lose a lot of time to do them.
But, thats not happening so thats ok.

Almost done?

Also, I got 99.5 last year. That was my first time taking it.

[quote="nutz_for2.718281828"]

Also, I got 99.5 last year. That was my first time taking it.[/quote]

ok, I think you've said that like 20 times already. First of all, this is the AIME not AMC.  :wink:

Lol, Iversonfan. Trying to pick a fight with someone?   :rotfl:


ratio
superior algebra
superior algebra unsolved

Let $F_q$ be a finite field with odd number of elements $q=p^n$,$p>5$. Prove that if $q\neq3,5$ then there exists a point $(x,y)$, $x,y \in F^*_q$ on the "circle" $x^2+y^2=1$.

if $q=p^n$ , $q>5$ should do

it is a general theorem that in projective planes over a field with odd q elements, the curve
$x^2+y^2=z^2$ has exactly $q+1$ solutions

by now carefully looking at cases where $x=0$, or $z=0$,...

you exclude at most five solutions, so with $q>5$ there is still one that is good for yo

does this help?

If we know that the characteristic is at least 5, we can do this in the prime field with $p$ elements using ordinary number theory/simple algebra.

There are $\frac{p-1}2$ nonsquares, and $\frac{p-3}2$ squares other than $1$. Consider the pairs $(u,1-u)$ for each $u$ other than $0,1$. The ratio $\frac{1-u}u=\frac1u-1$ ranges over all values except $0$ and $-1$, so it takes either $k=\frac{p-1}2$ or $k=\frac{p-3}2$ square values (actually, whichever of these is odd, since squares other than 1 pair up). Whenever this ratio is a square, either both $u$ and $1-u$ are squares, or neither $u$ nor $1-u$ are squares. Whenever this ratio is not a square, exactly one of $u,1-u$ is a square.

Now, we count the three types of ordered pairs $(u,1-u)$. Let there be $a$ pairs of squares, $b$ pairs of nonsquares, and $c$ mixed pairs. Clearly, $c$ is the even number $p-2-k$. The total number of squares in the list is $p-3=2a+c$, using the fact that each square other than $0$ or $1$ appears twice, so $2a=p-3-c=k-1$. Since $k\ge\frac{7-3}{2}=2$, there is at least one pair of squares in the list, and we are done.
The total number of pairs of squares is in fact $\frac{p-5}4$ if $p\equiv1\mod 4$ and $\frac{p-3}4$  if $p\equiv3\mod 4$.

Note that fredbel's count counts each of these pairs as four solutions; that's what it is when looking at the equation $x^2+y^2=1$. The counts agree, since fredbel's method excludes six trivial solutions when $p\equiv1\mod 4$ and four trivial solutions when $p\equiv3\mod 4$. Those trivial solutions are $(0,1,1),(0,-1,1),(1,0,1),(-1,0,1)$ for all $p$ and $(1,i,0),(1,-i,0)$ where $i^2=1$ when $p\equiv1\mod 4$.

Actually, the entire proof above works for any finite field with an odd number of elements $q$ by replacing letters; we get the facts we need about squares from the fact that the multiplicative group is cyclic.