Problem:

Alan and Ben run at constant speeds, but in opposite directions, around a track.  They meet every 20 seconds.  Alan requires 45 seconds to complete a lap.  How many seconds does Ben require?

Thanks.

Hi,

This problem is asking for how many seconds it takes Ben to complete a lap. We can just deal with the first lap as every lap after that will take the same amount of time since he runs at a constant speed. We are given that they meet every 20 seconds. So, let's look at the first lap. They both start and run in opposite directions, so their distances after the 20 seconds must add up to that of the whole circle. Since it takes Alan 45 seconds to complete the whole course, he has covered $ \frac{20}{45} \implies \frac{4}{9}$th of the lap in 20 seconds. Thus, Ben has completed the remaining $ \frac{5}{9}$th in 20 seconds. Therefore, it would take Ben $ 20(9/5) \equal{} 36$ (as $ \frac{5}{9} \cdot \frac{9}{5} \equal{} 1$ (1 representing the whole course being run)) seconds to cover the whole course once.

Does this problem assume that both people start at the same place, or can the answer be found regardless?

Well, I really haven't considered that. For example, say they start halfway(opposite the field of each other), then, you can't really do it the way I did it. So, I basically assumed they both started at the same spot and went in opposite directions. This problem doesn't state that, so I guess it's OK to assume that :P


inequalities
function
trigonometry
inequalities proposed

Let $\alpha,\beta,\gamma$ be real numbers such that $0<\alpha<\frac{\pi}{2},\ 0<\beta<\frac{\pi}{2},\ 0<\gamma<\frac{\pi}{2}$ and $\cos 2\alpha+\cos 2\beta+\cos 2\gamma=\frac{1}{2}.$

Prove the following inequality.

\[ \tan \alpha \tan \beta+\tan \beta \tan \gamma+\tan \gamma \tan \alpha \leq \frac{3}{2}.  \]

[quote="kunny"]Let $\alpha,\beta,\gamma$ be real  numbers such that $\cos 2\alpha+\cos 2\beta+\cos 2\gamma=\frac{1}{2}.$

Prove the following inequality.

\[ \tan \alpha \tan \beta+\tan \beta \tan \gamma+\tan \gamma \tan \alpha \leq \frac{3}{2}.  \][/quote]
Let $a=\cos^2 \alpha,$ $b=\cos^2\beta$, and $c=\cos^2\gamma$. So that we need to prove that
\[ \frac{\sqrt{(1-a^2)(1-b^2)}}{ab}+\frac{\sqrt{(1-b^2)(1-c^2)}}{bc}+\frac{\sqrt{(1-a^2)(1-c^2)}}{ac}\leq \frac32, \]
where $a^2+b^2+c^2=\frac74$.

[quote="kunny"]Let $\alpha,\beta,\gamma$ be real  numbers such that $\cos 2\alpha+\cos 2\beta+\cos 2\gamma=\frac{1}{2}.$
Prove the following inequality.
\[ \tan \alpha \tan \beta+\tan \beta \tan \gamma+\tan \gamma \tan \alpha \leq \frac{3}{2}.  \][/quote]
If $\alpha=\beta=\gamma=arctan\sqrt\frac{5}{7}$ then
$cos2\alpha=cos2\beta=cos2\gamma=\frac{1}{6},$ but
$tan\alpha tan\beta+tan\alpha tan\gamma+tan\beta tan\gamma=3\cdot\frac{5}{7}>\frac{3}{2}.$ ;)

Perhaps they must be angles of a triangle?

[quote="kunny"]Sorry that, Assume that  $0<\alpha<\frac{\pi}{2},0<\beta<\frac{\pi}{2},0<\gamma<\frac{\pi}{2}.$

kunny[/quote]
but the numbers $arqady$ give us are satisfy your condition...


Find the value of [size=150] [b]i^i[/b][/size]

$ e^{i\pi} \equal{} \minus{} 1\Longrightarrow ( \minus{} 1)^{ \minus{} i} \equal{} e^{\pi}.$

its[size=200] [b]i^i[/b][/size] and not [b](-1)^i[/b].
any by the way there r a few options
[b]1) pie/2

2) -3*pie/2

3) 2007pie/2

4) 2009pie/2[/b]

I do know that, just sent an approach.

Please do $ Latex$. :wink:

The answer i am getting doesn't seem to be in any of your options  :(

[hide="My Approach"]
$ i = 0 + 1i = cos\frac{\pi}{2} + i sin\frac{\pi}{2} = e^{i\frac{\pi}{2}} \\
\\
So, \ i^i = e^{i^2\frac{\pi}{2}} = e^{-\frac{\pi}{2}}$

and e power nothing is in your options  :!:
or did you mention the wrong options here??

P.S. instead of the argument [b]pie/2[/b] for i,
u can also take [b]3pie/2[/b]

[/hide]

There's [u]infinitely many[/u] values for $ i^i$:

$ i = e^{i\left({\pi\over 2} - 2k\pi\right)}\implies i^i = e^{ - {\pi\over 2} + 2k\pi},k\in\mathbb{Z}$

Provided the given answers should be of the form $ e^{\mathrm{given\,value}}$, among them only for (3) we can find an integer $ k$, namely $ {2007\pi\over 2} = - {\pi\over 2} + 2\cdot 502\pi$. Therefore the answer is $ e^{2007\pi/2}$.


Find all $u,v$ are positive integers such that |$3^{u}-5^{v}$| <= $3^{u/2}$
The answer is (3,2)

I think it has a great chance to be an open problem.  :?

What is your initial problem?

I don't think it's his lemma. I know it appeared  in an exponent Diophantine problem, somewhere in 1987 AMM, Elementary prob.

[quote="Myth"]Do you mean it is an elementary problem? :?[/quote]

Nope! I mean editors placed it in Elementary Section problem!


let $ f(x)\equal{}x^k\plus{}5x^{k\minus{}1}\plus{}3$ for integer $ k>1$.  Show that $ f$ can not be written as product of two polynomials with integer coefficients (with non-zero, positive, degree of course).

This is a trivial consequence of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=605327559&t=150]Perron's criterion[/url].


If $ a,b,c >0$ such that $ a\plus{}b\plus{}c \equal{} 1$, then
$ \frac{1}{a\plus{}a^2\plus{}a^3\plus{}a^4}\plus{}\frac{1}{b\plus{}b^2\plus{}b^3\plus{}b^4}\plus{}\frac{1}{c\plus{}c^2\plus{}c^3\plus{}c^4} \geq \frac{729}{40}(a^2\plus{}b^2\plus{}c^2)$

[quote="Gibbenergy"]If $ a,b,c > 0$ such that $ a + b + c = 1$, then
$ \frac {1}{a + a^2 + a^3 + a^4} + \frac {1}{b + b^2 + b^3 + b^4} + \frac {1}{c + c^2 + c^3 + c^4} \geq \frac {729}{40}(a^2 + b^2 + c^2)$[/quote]
This inequality actually follows from the known inequality
\[ \frac {1}{a} + \frac {1}{b} + \frac {1}{c} + 48(ab + bc + ca) \ge 25
\]
Indeed, notice that the original inequality is equivalent to
\[ \sum \frac {1}{a(a + 1)(a^2 + 1)} \ge \frac {729}{40}(a^2 + b^2 + c^2)
\]
Moreover, it is easy to check that
\[ \frac {1}{(x + 1)(x^2 + 1)} \ge \frac {27(29 - 27x)}{800} \quad \for all x \in [0,1]
\]
Applyint this inequality, it suffices to show that
\[ \sum \frac {29 - 27a}{a} \ge 540(a^2 + b^2 + c^2)
\]

\[ 29\left( \frac {1}{a} + \frac {1}{b} + \frac {1}{c} - 9\right) \ge 180[3(a^2 + b^2 + c^2) - 1]
\]

\[ \frac {1}{a} + \frac {1}{b} + \frac {1}{c} - 9 \ge \frac {180}{29} [3(a^2 + b^2 + c^2) - 1]
\]
Since $ \frac {180}{29} < 8$, it suffices to prove
\[ \frac {1}{a} + \frac {1}{b} + \frac {1}{c} - 9 \ge 8[3(a^2 + b^2 + c^2) - 1]
\]

\[ \frac {1}{a} + \frac {1}{b} + \frac {1}{c} + 48(ab + bc + ca) \ge 25
\]
Thus, all we have to do is to prove the last inequality, which is not hard to prove. :)

I wonder if your method still works with

$ \sum \frac {1}{a \plus{} a^2 \plus{} a^3 \plus{} a^4 \plus{} a^5}  \geq \frac {2187}{121}(a^2 \plus{} b^2 \plus{} c^2)$

Provided that $ a,b,c$ are positive and $ a \plus{} b \plus{} c \equal{} 1$


how do you solve

e 1-5x =3 ½

log10x-2 = log10(x-2)

lnx^4 = 4lnx

I'm assuming you mean

$ e^{1\minus{}5x}\equal{}3^{\frac{1}{2}}$
take the natural log of both sides to get
$ 1\minus{}5x \equal{} \frac{1}{2}\ln{3}$, so
$ x \equal{} \frac{1\minus{}\frac{1}{2}\ln{3}}{5}$

I don't know what you mean by the second question. Sorry

[quote="iniquitus"]I don't know what you mean by the second question. Sorry[/quote]
$ \log_{10}x\minus{}2\equal{}\log_{10}(x\minus{}2)$

$ \ln^4x\equal{}4\ln x$


Hello folks, I have a couple of queries.

[b] Query 1 [/b]

I observed (and this is also universally acknowledged) that almost everything is logical in science, (Science , in this context is - Math, Physics, Computer Science, these are the only sciences that I have dealt with enthusiastically) but, I feel some aspects (or concepts) of these "sciences" are very difficult to comprehend or assimilate,so my query is - When everything is logical in these "sciences", then, why do I (which includes myself and many people) have great difficulty in understanding these concepts/theories? (Don't ask me for an answer, I simply don't know it, so I have asked you people)

[b] Query 2 [/b]

Is there any way by which I can comprehend difficult scientific theories or concepts? (Say- By making analogies etc, "comprehend" in this context can have other connotations too, say like reducing/simplifying an abstruse concept/theory into something tangible)

I would appreciate if people found some time to answer these queries of mine.

Thanks!

[b] P.S.: [/b] The title "A Philosophical Query" might be a misnomer, but I couldn't concoct a much better/appropriate title, so do bare with it. :lol:

[quote="Gen8"]why do I (which includes myself and many people) have great difficulty in understanding these concepts/theories? [/quote]
Either the concept has never been explained well to you or you have not put enough effort into understanding it.  What more is there?

[quote="Gen8"]Is there any way by which I can comprehend difficult scientific theories or concepts?[/quote]
Learn them from multiple sources.  I find this a good way to combine different perspectives and get a broader picture of various concepts.

This is a strangely worded question.  Learning is learning.  You're probably either setting too high a standard for yourself or trying to learn too much too quickly.

I interpreted the first question slightly differently, along the lines of "if the sciences are based entirely on logic, why are they not completely obvious?"

I think the problem with this reasoning is that logic itself is not something built in to the human brain and must be acquired through a long, difficult process called learning, especially more difficult or advanced ideas. Also, while the sciences are based on logic, this does not mean that they follow immediately from logic; many ideas require very lengthy, tricky logical reasoning that is not at all obvious. For example, many brilliant math solutions are based on some different insight that is logical in itself but not at all a logical consequence of the problem.

Sorry if any of this is incoherent, I am still trying to understand what I'm saying myself.

On a fundamental level, you do have to give time for your neurons to rewire or make new connections or whatever it is they do that embodies learning. Since this is an entirely biological process, it's going to have to take some time.

Of course, I'm no neuroscientist, but it makes sense to me to have our biology itself be an inherent limiting factor. The reason some are more apt to learning than others is because their meta-circuitry (however that is embodied) enables quicker connections, guides neuron connections, etc. In essence, some people have learned to learn better. Whether this is a result of environment or genetics is debatable, and the standard answer here is both.

[b] @ stuck [/b]

I guess your reply is only relevant to a certain degree, I was seeking a reply which was more philosophical (or at least sounds philosophical) in content. Your reply was more biological (or rather neurological). You also had been slightly tangential.

My first query certainly is philosophical in nature.

My second query was and I paraphrase it - Is there any scientific/logical way by which I can understand (at least to a small extent) abstruse scientific theories/principles?

In response to this query, some people have suggested (not on this forum) - make analogies etc, but making analogies doesn't always suffice.

Are there any efficient/logical ways by which I can understand complex concepts/theories much easily? (Say- By citing examples, counter-examples etc)

I'm having problems in comprehending difficult scientific text.

Please suggest some ways.

It would help if you told us what topic you were trying to understand, instead of being extremely vague. Then we could probably offer more directed suggestions.

The best strategies for learning "science" in general are to sit on it and think about it, learn from multiple sources (including asking people/professors, multiple texts, etc), and where applicable, toy around with the concepts and/or do practice problems, as t0rajir0u suggested.

In many scientific subjects, it helps (for me, anyway) to begin a new topic by trying to understand why, qualitatively, it may make sense. Then, go through the math. Then ask yourself why the math "agrees" with the qualitative understanding. Once you've done that, ask yourself what specific intuitive facts the mathematical model offers beyond the qualitative model.

[quote="Gen8"]My first query certainly is philosophical in nature.[/quote]
Philosophically, you're skirting around a more important question:  why do you expect physics to be logical in the first place?  This is an assumption we make about the physical world - that it can be described by patterns that we can understand.  Isn't this remarkable?  Isn't this, in hindsight, not obvious at all?  It shouldn't be easy to comprehend the nature of the universe!

[quote="Gen8"]Is there any scientific/logical way by which I can understand (at least to a small extent) abstruse scientific theories/principles? [/quote]
This is such a strangely worded question.  How does anyone understand abstruse scientific theories and principles?  By learning the simpler ones and developing their intuitions, conversing with their colleagues and mentors, developing heuristics and a general perspective...

[quote="Gen8"]make analogies etc, but making analogies doesn't always suffice. [/quote]
Analogies are extremely powerful.  I would argue that really the only way we understand anything is through a sufficiently sophisticated analogy or metaphor, even if we aren't consciously aware of it.  Fundamentally, I like stuck's answer; what we can understand is essentially limited by what our brains evolved to need to understand to survive.  Thus we understand Euclidean geometry because we needed to process visual information to hunt and navigate, etc.  I don't know why you don't like his reply; it's quite philosophical.  One implication of this biological connection is the assertion that [b]mathematics is a uniquely human endeavor.[/b]

[quote="Gen8"]I'm having problems in comprehending difficult scientific text. [/quote]
There are a multitude of reasons you could be having trouble.  The text could be too difficult for your level of understanding; it could be poorly written; you could be attempting a text you simply don't have the background for; you may not be reading carefully enough; you may be lacking in some mathematical or conceptual foundation that you must find and re-learn... this is a difficult question to answer without knowing what, exactly, you're having difficulty with.

[quote="t0rajir0u"]Analogies are extremely powerful. I would argue that really the only way we understand anything is through a sufficiently sophisticated analogy or metaphor, even if we aren't consciously aware of it. Fundamentally, I like stuck's answer; what we can understand is essentially limited by what our brains evolved to need to understand to survive. Thus we understand Euclidean geometry because we needed to process visual information to hunt and navigate, etc. I don't know why you don't like his reply; it's quite philosophical. One implication of this biological connection is the assertion that mathematics is a uniquely human endeavor.[/quote]
This is very helpful when you get into advanced fields, as you're trying to do.  As an example where I think analogies are really important, topology.  Many statements in topology seem really "mathy" and hence hard to understand but if you actually try to visualize what the statement is, it makes more sense.  For example, here's a problem I had on my final exam: Prove that every nonvanishing vector field on $ B^{n\plus{}1}$ points directly outward at some point of $ S^{n}$, and directly inward at some point of $ S^{n}$.  I admit, if a previous professor hadn't made the joke in another unrelated class, I wouldn't have understood it, but the joke is that you can't comb all of the hairs on a tennis ball.  That is the analogy for this statement.  I think it helps understand just what is going on.

Also, if you're going to try to learn advanced material, you're going to need some sort of background-simple as that.  You can't attempt to understand string theory (the deep mathematical theory) without some pretty good knowledge of quantum mechanics.  And you probably won't understand a course on how to perform brain surgery unless you have very advanced knowledge of neurology.

[quote="t0rajir0u"]Philosophically, you're skirting around a more important question: why do you expect physics to be logical in the first place? This is an assumption we make about the physical world - that it can be described by patterns that we can understand. Isn't this remarkable? Isn't this, in hindsight, not obvious at all? It shouldn't be easy to comprehend the nature of the universe![/quote]

This is something that has been intriguing me for a long time.
[b]How are we able to explain so many things in nature logically? [/b]
More importantly, [b][u]Why must there be a reason for everything?[/u][/b]
[b]Have we(mankind), at any point of time, assumed that there must be a reason for all that we observe (or predict)?[/b]

[quote="t0rajir0u"]
Analogies are extremely powerful. I would argue that really the only way we understand anything is through a sufficiently sophisticated analogy or metaphor, even if we aren't consciously aware of it. [/quote]

How come analogies work so well in explaining several things?
Is there a reason behind this?

On analogies:  I assume analogies work so well because they take something abstract or hard to understand, and liken it to something that everyone/most people can.  They simplify things by comparing them to more understandable material
(Random thought: A Brief History of Time has many of these analogies that helped me a bunch in understanding many of the material)

[quote="keyree10"][b][u]Why must there be a reason for everything?[/u][/b][/quote]
This isn't exactly your question, but you might be interested in [url=http://www.dartmouth.edu/~matc/MathDrama/reading/Wigner.html]this essay[/url].

[quote="keyree10"]How come analogies work so well in explaining several things?
Is there a reason behind this?[/quote]
Analogies allow us to [b]compress[/b] the understanding of many related situations by restricting our attention to the similarities between those situations.  This means that we don't need to store as much information to gain as much understanding.  The flip side, of course, is that one must recognize when analogies break down and think about the unique properties of various situations as appropriate.

I second [b] Quickster94 and JRav's[/b] posts, they had actually (in a way) summarized what I wanted to know.

Apart from the requisite knowledge/expertise in  scientific disciplines (I agree, these are of paramount importance when I strives to understand obscure scientific theories/principles), and 'analogies',[b] does there exist any other efficient/relevant techniques to streamline my understanding of difficult scientific concepts.[/b]

Say something like a well-constructed, coherent analogy that explains something that's unknown/complex by representing into something known, thereby allowing us to compare the similarities between the known and the unknown (Not necessarily as described, but [b] something that simplifies a complex scientific theory/tenet into a tangible form).[/b]

I'm facing difficulty in comprehending certain aspects of Computer Science, and Mathematics. (That is combinatorics)

If you need help with combinatorics, t0r and JBL are probably your two best bets on this forum.

[quote="Gen8"][b] does there exist any other efficient/relevant techniques to streamline my understanding of difficult scientific concepts.[/b] [/quote]
Different techniques will work for different people and learning what works for you is part of your learning process.

[quote="Gen8"]a well-constructed, coherent analogy that explains something that's unknown/complex by representing into something known[/quote]
You mean [b]generically[/b]?  Good luck with that.  If such a thing existed we'd have taught it to everyone by now.


Let a convex quadrilateral ABCD inscribed the circle (O). AC meets BD at K . A circle $ (\omega_{BC})$ which touches KB,KC and intouches (O) at E. Denote M,N,P be the incenters of triangles KBC,ABC,DBC. Prove that EM is the bisector of angle NEP.

We know E,M,F,C are concyclic   B,E,M,G are concyclic
∠BMC=90+1/2∠BKC  ∠GFC=90+1/2∠BKC   =>  C,M,P,F are concyclic
=> E,M,P,F,C are concyclic   => ∠PEC=∠PMC
Analogously,  ∠BEN=∠BMN
So ∠PEC=∠BEN
From http://www.mathlinks.ro/viewtopic.php?t=233718 we know  ∠BEZ=∠CEZ
Thus we aquire   ∠NEM=∠PEM


modular arithmetic
invariant
induction
combinatorics proposed
combinatorics

Let $A_1A_2\ldots A_n$ be a convex polygon, which is triangulated (i.e. divided up into triangles, any two of which do not intersect, intersect in a vertex, or have a common side). Suppose each vertex of the polygon is adjacent to an even number of triangle edges, except for $A_{i_1},\ldots,A_{i_k}$ (with $i_1<i_2<\ldots<i_k$). Prove that:

(a) $k$ is even;

(b) $n\equiv i_1-i_2+i_3-i_4+\ldots+i_{k-1}-i_k\pmod 3$.

Part (a) is easily proved for example considering the graph of triangulation and using that sum of the degrees is always even.
For part (b):
First note that number of triangles in a triangulation is always $n-2$ (easily proved by calculating the sum of angles of the polygon in two ways)
For each triangle consider the number of it's edges that belong to the polygon. Sum of these numbers is always $n$ because every edge of the polygon is exactly in one of the triangles. There are $n-2$ such numbers, so one of them is greater than $1$. So the correspondig triangle has at least two edges of the polygon so it is $A_iA_{i+1}A_{i+2}$ for sum $i$. (Just here we assume that $n+1=1$ and $n+2=2$)

Lemma: The requested property is invariant under rotating the numbering of vertices.
Proof : It is sufficient to show that if for numbering $1,2,\ldots,n$ the property holds then so does for the numbering $2,3,\ldots,n,1$. (The general property will be proven by using this assertion $k$ times.)
Consider two cases :
* $i_k\neq n$ : In this case the sum is invariant by adding one two each $i_j$ so the property holds.
* $i_k=n$ : In this case the new sum would be
\[ (i_k+1-n)-(i_1+1)+(i_2+1)-(i_3+1)+\ldots-(i_{n-1}+1)=-n-(i_1-i_2+i_3-\ldots-a_n)=-n-n=n (mod 3) \]

So we may assume that the triangle we mentioned earlier is $A_{n-1}A_nA_1$.
Now we use induction for the polygon $A_1A_2\ldots A_{n-1}$.
Suppose $j_1<j_2<\ldots<j_t$ are the numbers of the vertices for the polygon $A_1A_2\ldots A_{n-1}$ that have odd degree.
Then we have four cases :
* $j_1=1$ and $j_t=n-1$ : The sum for the $A_1A_2\ldots A_n$ will be
\[ j_2-j_3+j_4-\ldots-j_{t-1}=(j_1-j_t)-(j_1-j_2+j_3-\ldots-j_t)=(1-(n-1))-(n-1)=1+(n-1)=n (mod 3) \]
* $j_1=1$ and $j_t\neq n-1$ : The sum for $A_1A_2\ldots A_n$ will be
\[ j_2-j_3+j_4-\ldots+j_t-(n-1)=(j_1-(n-1))-(j_1-j_2+\ldots-j_t)=2-n-(n-1)=n (mod 3) \]
* $j_1\neq 1$ and $j_t=n-1$ : The sum will be
\[ 1-j_1+j_2-\ldots-j_{t-1}=(1-j_t)-(j_1-j_2+\ldots-j_t)=(1-(n-1))-(n-1)=n (mod 3) \]
* $j_1\neq 1$ and $j_t\neq n-1$ : The sum will be
\[ 1-j_1+j_2-\ldots+j_t-(n-1)=(1-(n-1))-(j_1-j_2+\ldots-j_t)=(1-(n-1))-(n-1)=n (mod 3) \]


I'm going to be there next Saturday. You all will be there, right?

[quote="numberdance"]*grin* Hopefully I do better next week than I did at OSSM yesterday.[/quote]
You went to OSSM too? I couldn't go. Darn.. I really wanted to see you in person.. I guess I will have to wait or do I know you already?

I don't think we know each other.

Yes, I was there. 2nd place 8th grade girl Division 4, behind the girl from Summit Middle School. I must admit, it was pretty awesome to see a girl beat all the 8th grade boys in that division. But I wasn't in the division's top 5.

Waiting until Saturday probably isn't that long, even though it's going to seem like forever just because I want to get Saturday over with...


Prove that: With all arbitrary real numbers $x,y,z\in[1,2]$, we always have following inequality:
$(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq6\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)$
When equality was occured?

i just saw this in the inequalities proposed problems section
the same problem...
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84210


Can someone please explain to me (in mathematical context) what a module or mod is, where it is used and how it is used? thank you

Well, basically modulo=remainder.  The wiki can explain it better than I can.

[url]http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic[/url]

Just so you know, a module and a mod are two completely different things. A module is a concept of abstract algebra (which you can look up on wikipedia) and a mod is a modulus which is what the previous poster talked about.


May I give a hint ?
[hide]
 $ B = \left[\begin {array} {LLL} I & 0 & 0 \\
0 & A^3 & 0 \\
0 & 0 & A^6 \end {array} \right].\left[\begin {array} {LLL} a_{1}I & b_{1}I & c_{1}I \\
a_{2}I & b_{2}I & c_{2}I \\
a_{3}I & b_{3}I & c_{3}I \end {array} \right].\left[\begin {array} {LLL} A & 0 & 0 \\
0 & A^2 & 0 \\
0 & 0 & A^3 \end {array} \right]$
The determinant of the left and the right matrix is obviously not equal to zero, then we just need to prove that the determinant of the matrix in the middle is not equal to zero.
If $ m_i$ is the eigenvalue(s) of $ A$ with eigenvector $ [x_i,y_i,z_i]$, then we can prove that $ m_i$ is also the eigenvalue(s) of $ A$ with eigenvector $ [x_i,0,0,y_i,0,0,z_i,0,0]$,$ [0,x_i,0,0,y_i,0,0,z_i,0]$ and $ [0,0,x_i,0,0,y_i,0,0,z_i]$. From the description above we can know that if $ A$ has nonzero eigenvalue, then the matrix in the middle has nonzero eigenvalue (also).
Sorry if I am wrong.
[/hide]


Now that I can actually make LaTeX documents (thanks Mr. Rusczyk!) I have two questions:

(1) PICTURES - How do I make graphs and other pictures myself to put inside my documents?  I could print them out and draw them in myself, but that's not great.  Note I do not have a scanner.

(2) PARENTHESES - How do I make parentheses that go around a whole fraction?  Like, I want to avoid this:

[tex]\displaystyle(\frac{1}{2})(\frac{1}{5})[/tex]

for the parentheses, there are 2 ways you can do it:

using \left( and \right) will let LaTeX fit the right size for you:
[tex]\displaystyle \left( \frac{2}{5} \right)[/tex].

OR, you can tell it your own size if you want it to be different from what it gives automatically:

[tex]\displaystyle \Bigg( \frac{2}{5} \Bigg)[/tex]

the size specifications can be found in the AoPS Latex manual.

For the first question, there is an entire section in the AoPS manual on importing graphics. Make sure you have the Go2PDF thing, though.

Thanks for the reply to (2)!  For (1) I meant not how to put them in PDFs, I mean how to make the images themselves.

OH, I see. You can use any program you want... word, paint, geometer's sketchpad (this is ideal, but not necessary if you don't have it), photoshop, ... the possibilities are endless!


trigonometry
function
trig identities
Pythagorean Theorem
geometry

A trigonometry problem:

Let cotA = -4 and four ninths (where A is in QII).  Find secA.

sorry, I know this is a dumb question, but I'm honestly clueless.

Use $ \cot^2A \plus{} 1 \equal{} 1/\sin^2A$ identity then, convert $ \sin A$ into $ \cos A$, and then to $ \sec A$

$ \cot{A}\equal{}\minus{}\frac{40}{9}$ Because $ \cot{A}\equal{}\frac{1}{\tan{A}}$, $ \tan{A}\equal{}\minus{}\frac{9}{40}$. Thus, $ \tan^2{A}\equal{}\frac{81}{1600}$. By Pythagorean Identities, $ \tan^2{A}\plus{}1\equal{}\sec^2{A}$. So, $ \sec^2{A}\equal{}\frac{1681}{1600}$. Taking the square root, $ \sec{A}\equal{}\frac{41}{40}$.

thanks for the replies, they're really helpful.

[quote="Textangle"]Taking the square root, $ \sec{A} \equal{} \frac {41}{40}$.[/quote]

Before undoing a squaring procedure, you need to determine which sign to use.

In quadrant 2, cosine is negative so it's reciprocal, secant, is also negative. You want $ \sec A \equal{} \minus{} \frac{41}{40}$

Oh sorry :blush: I forgot. Sorry!
I have to remember that, thanks.

It's always helpful to learn the trig identities because of problems like this. When you find the value of any trig function of X, you can find all the others.

You could also draw a right triangle in the second quadrant like in the attached picture:

$ \cot{A} \equal{} \frac{x}{y} \equal{} \minus{}\frac{40}{9}$

Because we are in the second quadrant, $ x$ must be negative and $ y$ must be positive. One set of values of $ x$ and $ y$ that would work are $ x\equal{}\minus{}40$ and $ y\equal{}9$. (or you could use the general case $ x\equal{}\minus{}40k, y\equal{}9k, k>0$)

Then by pythagorean theorem, $ r\equal{}\sqrt{x^2\plus{}y^2}\equal{}\sqrt{(\minus{}40)^2\plus{}9^2}\equal{}41$

Therefore $ \sec{A}\equal{}\frac{r}{x}\equal{}\frac{41}{\minus{}40}\equal{}\minus{}\frac{41}{40}$


How can I put cyrilic in math mode.
I use

\newcommand{\cir}[1]{\fontencoding{OT2}\fontfamily{cmr}\selectfont#1}

to swith to cyrilic, but it doesn't work in math mode.

example:

\[ 5+2=7 \cir{oliver} \]

you'll need to different command, one in text mode, and one in math mode.  The one to use in math mode is:

\usepackage[X2,T1]{fontenc}
\DeclareMathAlphabet{\mathcir}{X2}{cmr}{m}{n}

But in the latex companion, they say that the "OT2" encoding is too old.  You should use X2, that's why I used X2.  It looks different from OT2 but I just don't know, it all look russian to me... :?

is cyrilic a language?


[quote="ityaa"]let x,y,z>0 prove
$3(x^{2}y+y^{2}z+z^{2}x) \geq xy^{2}+yz^{2}+zx^{2}+6xyz$[/quote]

i think this does not hold for y=1, z=4, x->0.

 :)


geometry
circumcircle
incenter
perpendicular bisector
power of a point
radical axis
geometry unsolved

Figure I，O respectively are the circumcenter and incenter of △ ABC，AO intersects circumcircle at D，BI intersects AC、circumcircle at the E、F，CI intersects AB、circumcircle at the K、L； P、 M respectively are the  midpoint of CL、 IL，Q、N respectively are the  midpoint of  BF、IF，the Perpendicular bisectors of  IL, IF intersect BD、 CD at S、T，the circumcircle of △ SPK、△ TQE respectively  intersect  LS、FT at U、V
Verify: U，I，V are collinear，UV / / EK，LS / BS = FT / CT

no bady solved it？
Actually it's not much difficult if you try！

Let line EK intersect (CBA) at points X and Y
Let (IBK) intersect KE at points K and F
Let (BFE ) intersect CA at points G and E Angle IFE = IBK = CBI , so IF intersects BC at point D and D is on (BFE)
Easy to see that angle EBL = 90 - A/2 = KIB = KFB  , so Power of point L wrt. (BFE) = BL*BL
Easy to see that angle EGF = EBF IKF , so CKFG is cyclic and Power of point L wrt. (CKF) = KL*CL = BL*BL , so L is on Radical line of (CKF) and (BFE) , so L is on GF
Let (EBA) intersect KE at points H and E
Angle BHK = BAC = BLK , so KBHL is cyclic , so angle KHL = KBL = ACL = LFK , so LF = HL
Let Z is foot of perpendicular from L on HF , easy to see that HZ = ZF and triangle BHF ~ BLI
So if M is midpoint of IL then angle BMI = BZF
We have proved that SL is perpendicular to KE
Return to our problem
Let U' is on SL and angle SUI = 90
Let prove that PSU'K is cyclic
SL*UL = IL*ML = IL*IL/2 = Cl*KL/2 = PL*KL , so done and U' = U . Problem done

Thank you，skytin ！
For a long time no baby do it，I have forgot this problem，Next I 'll give my solution.

Dear Lym, Skytin and mathlinkers.
Sorry to revive this old thread, But I would like to add my solution for this amazing problem.

Let The circumcircle of the triangle $\triangle{AKC}$ intersect the line $KE$ at $X$. The perpendicular from $F$ to $KE$ intersect $CD$ at $T_1$. Let $XA$ intersect $FT_1$ and $FL$ at $Y$ and $Z$ respectively.
First, It's obvious that $FL$ is the perpendicular bisector of $AI$ since $F$ and $L$ circumcenters of $\triangle{AIC}$ and $\triangle{AIB}$ respectively. Now, remark that:
$\angle{FYA}=90-\angle{AXK}=90-\angle{AFL}=\angle{FIA}$, So points $A$ , $F$ , $Y$ and $I$ are concyclic.
we have also:
$\angle{ZXE}=\angle{ACK}=\angle{AFZ}=\angle{AFE}$
and $\angle{CXE}=\angle{CAK}=\angle{CFE}$
Combining those two results, we deduce that the pentagon $XFZEC$ is cyclic.
So, $\angle{ZCE}=\angle{ZXE}=\angle{ACI}$ therefor: $\angle{ZCI}=\angle{ACB}=\angle{AYI}$ but we have also: $\angle{ZCT_1}=90+\angle{ACI}=90+\angle{YXE}=\angle{ZYT_1}$. Combining these last results we deduce that, the pentagon $YZIT_1C$ is also cyclic. It follow that:
$\angle{YT_1I}=\angle{AZI}=180-2\angle{YFI}$ from which follow that $T_1$ is equidistant from $I$ and $F$ and so $T_1 \equiv T$ and then $EK\perp FT$.
Now, let $R$ be the midpoint of $AF$. We can easly prove that: $AQER$ is cyclic, Since $TV$ , $QE$ and $AR$ are concurrent, we must have ; by Radical Axis Theorem, that $ARVT$ is a cyclic quadrilateral. It's Obvious that $AIRN$ is cyclic. So, let it circumcircle intersect the circumcircle of $EQTV$ at $R_1$ and $R_2$. By Radical Axis theorem again we have $F$, $R_1$ and $R_2$ are collinear. Another use of the Radical Axis Theorem on circles $\odot(INT)$ , $\odot(R_1R_2Q)$ and $\odot(INR)$ implies that point $V$ lies on $\odot(INT)$ from which follow that $IV \perp FT$.
We conclude then that $IV \parallel EK$, similary we have $UI \parallel EK$ and hence, points $U$ , $V$ and $I$ are collinear on a line parallel with $EK$.

Dear lym.
I'm really interested on a different proof for this beautifull problem, so please, it will be very nice if you could add it here.


Michael tells the following riddle about his favorite number: ''My number is the sum of 126 plus one-fourth my number minus one-eighth my number.'' What is Michael's favorite number?


Find the number of surjections from A = {a, b, c, d} to B = {1, 2, 3}


If $ a,b,c\geq0$  prove that

$ \sqrt{a(b\plus{}c)}\plus{}\sqrt{b(c\plus{}a)}\plus{}\sqrt{c(a\plus{}b)}\leq\sqrt{2}(a\plus{}b\plus{}c)$ :)

[hide]First notice that we can allow $ a\plus{}b\plus{}c\equal{}1$, for the inequality is homogeneous.

Therefore, the inequality that we have to prove is

$ \sqrt{a\minus{}a^{2}}\plus{}\sqrt{b\minus{}b^{2}}\plus{}\sqrt{c\minus{}c^{2}} \le \sqrt{2}$

Now, we notice that $ f(x)\equal{}\sqrt{x\minus{}x^{2}}$ is a convex function going downwards, so we use Jensen's Inequality.

$ \frac{1}{3}(f(a)\plus{}f(b)\plus{}f(c)) \le f(\frac{a\plus{}b\plus{}c}{3})$.

Then our desired result follows.[/hide]

$ \sqrt{(a\plus{}a)(b\plus{}c)}\le \frac{1}{2}\cdot (a\plus{}a\plus{}b\plus{}c)$ sum these to finish the problem

[quote="galois01"]If $ a,b,c\geq0$  prove that

$ \sqrt {a(b \plus{} c)} \plus{} \sqrt {b(c \plus{} a)} \plus{} \sqrt {c(a \plus{} b)}\leq\sqrt {2}(a \plus{} b \plus{} c)$ :)[/quote]
[hide="Solution"]
By Cauchy, \[ [a\plus{}b\plus{}c][(\sqrt{b\plus{}c})^2\plus{}(\sqrt{a\plus{}c})^2\plus{}(\sqrt{a\plus{}b})^2]\ge (\sqrt{a(b\plus{}c)}\plus{}\sqrt{b(a\plus{}c)}\plus{}\sqrt{c(a\plus{}b)})^2\] \[ 2(a\plus{}b\plus{}c)^2\ge (\sqrt{a(b\plus{}c)}\plus{}\sqrt{b(a\plus{}c)}\plus{}\sqrt{c(a\plus{}b)})^2\] \[ \sqrt{2}(a\plus{}b\plus{}c)\ge (\sqrt{a(b\plus{}c)}\plus{}\sqrt{b(a\plus{}c)}\plus{}\sqrt{c(a\plus{}b)})\] and we are done.  [/hide]

[quote="galois01"]If $ a,b,c\geq0$  prove that

$ \sqrt {a(b \plus{} c)} \plus{} \sqrt {b(c \plus{} a)} \plus{} \sqrt {c(a \plus{} b)}\leq\sqrt {2}(a \plus{} b \plus{} c)$ :)[/quote]

\[ \sum\sqrt{a(b\plus{}c)}\le\sqrt{\left(\sum a\right)\left(\sum(b\plus{}c)\right)}\equal{}\sqrt 2(a\plus{}b\plus{}c)\]

The following inequality is stronger and easier. :lol:
Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that
\[ \sqrt {a(b \plus{} c)} \plus{} \sqrt {b(c \plus{} a)} \plus{} \sqrt {c(a \plus{} b)}\leq\sqrt {6(ab\plus{}ac\plus{}bc )}\]

[quote="arqady"]The following inequality is stronger and easier. :lol:
Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that
\[ \sqrt {a(b \plus{} c)} \plus{} \sqrt {b(c \plus{} a)} \plus{} \sqrt {c(a \plus{} b)}\leq\sqrt {6(ab \plus{} ac \plus{} bc )}
\]
[/quote]

[hide]Let $ ab \plus{} ca \equal{} x, bc \plus{} ab \equal{} y, ca \plus{} bc \equal{} z$. Then it's equivalent to
\[ \sqrt x \plus{} \sqrt y \plus{} \sqrt z\le\sqrt {3(x \plus{} y \plus{} z)}\iff \sum(\sqrt x \minus{} \sqrt y)^2\ge 0
\]
[/hide]


If $V$ is connected, must $\overline{V}$ be connected? Must $\mathring{V}$ be connected?

Suppose that there exist open sets $A, \, B$ with $\overline V \subseteq A \cup B$ and $\overline V \cap A \cap B = \emptyset$ .

We have  $V \subseteq \overline V \subseteq A \cup B$ and $V \cap A \cap B \subseteq \overline V \cap A \cap B = \emptyset$

Since $V$ is connected is $V \cap A = \emptyset$ or  $V \cap B = \emptyset$ . Suppose that $V \cap A = \emptyset$ .

Then $V \subseteq (X \setminus A)$ and because $(X \setminus A)$ is closed $\overline V \subseteq (X \setminus A)$  therefore $\overline V \cap A = \emptyset$.

This proves that $\overline V$ is connected.

What about the converse ??

 :)

in R^2, consider 2 disjoint balls with a line that connects them. It is path connected but not its interior.


When written in base two a multiple of seventeen cotains exactly three of the digits 1.

Prove that this number must contain at least six of the digits 0, and if it contains exactly seven of the digits 0 then it must be even.

We can write
$ 17k \equal{} 2^{a} \plus{} 2^{b} \plus{} 2^{c}$ , where $ a,b,c$ are distinct positive integers and $ a > b > c$.
To prove the first statement we only need to prove that $ a \geq 9$.
We may take out a common term to obtain
$ 2^{c}(1 \plus{} 2^{a \minus{} c} \plus{} 2^{b \minus{} c})$, and therefore
$ (1 \plus{} 2^{a \minus{} c} \plus{} 2^{b \minus{} c})$ is a multiple of 17. To simplify,
$ 1 \plus{} 2^{n} \plus{} 2^{m} \equal{} 17p$ , where $ 8 \geq n > m$. Note that p is odd.
However, the largest possible value for the LHS is $ 1 \plus{} 2^{8} \plus{} 2^{7} < 17 \times 23$.
So we may test for $ p \equal{} 1,3,...,21$, to find that there is no solution.

For the second statement, prove by contradiction.
Suppose there exists such an odd integer. Then;
Since the number has three 1's and seven 0's,
we can write
$ 17k \equal{} 2^{9} \plus{} 2^{0} \plus{} 2^{c}$, where $ 0 < c < 9$
Testing the 8 values for c in between, we find that none work.
Hence we only need to find one even example.
A quick check finds that $ 17 \times 34 \equal{} 2^{9} \plus{} 2^{6} \plus{} 2^{1}$
is the desired solution


calculus
integration
combinatorics proposed
combinatorics

Yes, a famous result due to Schinzel states that for each $n$ there is a circle of the plane which goes through exactly $n$ integer points.

Pierre.

If I'm not wrong, one of the circles you are looking for is the one with center $(\frac 1 3 , 0 )$ and radius $r = \frac {25} 3.$

Pierre.

[quote="indybar"]Is there a circle which passes through five points with integer co-ordinates?[/quote]

Yes, even if you don't believe.  :)  See http://www.mathlinks.ro/Forum/viewtopic.php?t=5556 .

  Darij

From the mathworld link http://mathworld.wolfram.com/SchinzelsTheorem.html
I don't understand how the number of integral solutions to the equation $x^2+y^2=n$, given by $r(n)=
4(d_1-d_3)$ where $d_1$ is the number of divisors of $n$ of the form $4k+1$ and $d_3$ is the number of divisors of the form $4k+3$ identifies circles
$(x-\frac{1}{2})^2+y^2 = \frac{5^{k-1}}{4}$ for $n=2k$ and $(x-\frac{1}{3})^2+y^2= \frac{5^{2k}}{9}$ for $n=2k+1$
Can anyone help?


Since $6a^2=(2a)*(3a)$
         $-3=(-1)*(3)$
We are tring to get $7a$(the middle term) with the stuff below so we can write
                          $3a$---------------$(-1)$
                          $2a$---------------$(3)$
and we multiple the terms lying on the diaganol, then add them up
                         $3*3a+(-1)*2a=9a-2a=7a$
Here we get the middle term, so we get
                         $6a^2+7a-3=(3a-1)(2a+3)$

Here's a slightly different method -

First, take the coefficiants of the first and last terms, multiply them, and them find find two factors that have add up to the second term, and multiply out the the product i mentioned in the first sentence.

So you would do 6*-3 = -18, and find two factors that have a sum of 7, and multiply out to -18, which are -2 and 9.

Then you would write $6a^2-2a+9a-3$, whcih has the same value as the first expression.  You would then do group facotring (I think that's what it's called) and get 2a(3a-1)+3(3a-1) = (2a+3)(3a-1).

[b]
Also you can find the roots of the equation $ax^2+bx+c=0$ Then the answer is $ax^2+bx+c=a(x-x_1)(x-x_2)$[/b]

Usually finding the roots of an equations takes much longer than factoring, and time is of the essence in competiotions.

[quote="OnionJuice"]Im workin on rational expressions and applications.

Okay, how does this: [b]6a2+7a-3[/b] become this: [b](3a-1)(2a+3)[/b] ???

Anybody??[/quote]By the distributive property (applied twice),
$(3a-1)(2a+3)$
$=(3a-1)(2a)+(3a-1)(3)$
$=(3a)(2a)+(-1)(2a)+(3a)(3)+(-1)(3)$
$=6a^2-2a+9a-3$
$=6a^2+7a-3$.

You're probably asking how to factor though. That's a different matter  :D.


Hi!!
I'm interested in getting an introductory physics text for high school, that does not use calculus.
Rather, uses more algebra 1 and 2.
If you know of any please suggest them to me.

I used Serway and Faughn, but I think they skipped some of the more important conceptual points.

My school uses PHYSICS by Giancoli sixth edition.
It's an okay book.  It has clearly been watered down for high school students.


I have been reading combinatorics but I can reach an answer.
I need to count all possible combinations of combining
two elements of n +
three elements of n +
four elements of n +
...
n-1 elements of n

Where the order and is not relevant and without replacemente.

Ej with four elements: A B C D

1. A-B,C,D
2. A-C,B,D
3. A-D,B,C
4. B-C,A,D
5. B-D,A,C
6. C-D,A,B
7. A-B,C-D
8. A-C,B-D
9. A-D,B-C
10. A-B-C,D
11. A-B-D,C
12. A-C-D,B
13. B-D-C,A

I think I didnt miss any. Can anyone help me in how to calculate this number of possible combinations?

Thank you in advance

you seem to want to calculate this:

$ \sum^{n\minus{}1}_{k\equal{}2} \binom {n} {k} \equal{}\minus{}\left[\binom{n}{n}\plus{}\binom{n}{0}\plus{}\binom{n}{1}\right]\plus{}\sum^{n}_{k\equal{}0} \binom {n} {k} \equal{}$

$ \minus{}(2\plus{}n)\plus{}\sum^{n}_{k\equal{}0} \left[\binom {n} {k}1^{n\minus{}k}1^k\right]\equal{}\minus{}(2\plus{}n)\plus{}\left(1\plus{}1\right)^n\equal{}2^n\minus{}(2\plus{}n)$

so n=4 gives us 13. a little less exhausting than counting them all  :D

Can you give me hints of what rules are you applying in the different steps or a book or webpage.

Thank you

(2^n)-(2+n)= 10 not 13 :(.

But I will to try to find the solution with what you wrote.

Thank you

You appear to be looking at [url=http://www.research.att.com/~njas/sequences/A000110]Bell numbers[/url] minus 2 (since you're excluding the "all singletons partition" and the "one part partition").

[quote="varelse"]the proposed solution do not takes into account the options:

7. A-B,C-D
8. A-C,B-D
9. A-D,B-C

Thank you[/quote]

ah, so maybe we are calculating different things.  my sum was for all combinations of single groups of a given size.  if you are accounting for A-B,C-D as different than A-B,C,D .. you will miss all groups of this type with my method.

slightly different question.. and i dont think i can tweak my method in any obvious way without making it way ugly.. hmm

JBL you got it right

Kenn4000 your coments have been usefull too.


group theory
superior algebra
superior algebra unsolved

Show that for $n \geq 3$, $Z(S_n) = \{ \epsilon \}$ where $Z(S_n)$ is the center of the symmetric group of $n$ symbols, and $ \epsilon $ is the identity element in $S_n$.

I have an intuitive idea of why this is so, but I can't figure out how I might write it out as a proof.

An element $x$ is central iff it is unchanged by conjugation by every element $g$: that is, $x^g = g^{-1}xg = x$.  In the symmetric group, if you write a permutation $\pi$ out in cycle form $(a_1 a_2 a_3 \ldots)(b_1 b_2 b_3 \ldots) \cdots$ then the conjugate of $\pi$ by a permutation $\sigma$ has cycle form $(a_1^\sigma a_2^\sigma a_3^\sigma \ldots) \cdots$: in other words, you just apply $\sigma$ to the entries in the cycles.  Verify this and then see which permutations are unaffected if you apply any permutation to their cycles.


geometry
geometric transformation
dilation

Propose a method for making the following conversion (this problem features an organic synthesis technique known as [i]functional group transposition[/i]):

$ C_6H_5\minus{}COCH_2CH_3 \longrightarrow C_6H_5\minus{}CH_2COCH_3$.

Is this viable? ( I still think aq. NaOH can only convert X(halide) into OH)
1. $ \alpha$ chlorination
2. aq. NaOH
3. Mosingo redn
4. Oppenauer oxidation
 :)

There are several problems with that:

1) Polychlorination can occur

2) An alpha,beta-unsaturated ketone can form (which in this case would be highly favourable)

3) A quasi-Favorskii rearrangement could also occur.

[quote="Pardesi"]...there aren't any groups affected due toa cid other than carbonyl ofcourse[/quote]

Of course... and what about chlorine? Isn't it reduced?

oh so there is amalgamated Zn how did i forget this...
well u can use thioacetal reduction then

[quote="pardesi"]i would suggest
the same but with some extra aditions
1)chlorination in acidic media so that it stops at after a single substitution
2)clemmensen should do ...there aren't any groups affected due toa cid other than carbonyl ofcourse
2)use aq.$ \ce{KOH}$ to prevent rearrangement
3)oppenaur oxidation
or if any alkene gets formed use Wacker process[/quote]

Thats why i said Mozingo  :D

well here's a completely new synthesis
1)MVP reduction
2)dehdration
3)hydroboration oxidation
4)oppenaur oxidation

Only one doubt does HBO give that required alcohol? (Don't try to prove by anti-markownikoff rule give a correct explanation)

why not...
the boron attacks the beta carbon atom to the phenyl group...since the TS can accomadate the positive charge on the alpha carbon atom because of resonance stabilization...
is this ur problem or there is something else that is worrying u...

i think pardesi's method is absolutely correct.
this is because in hydroboration oxidation--[b]OH[/b] group is attached to less hindered side of the less hindered carbon atom.
So,pardesi is right.

@pardesi-good method

I also think the hydroboration will not be selective. Also, trying to oxidise alcohols like Ph-CH2-CHOH-CH3 is problematic, since most oxidizing agents are also somewhat acidic. That's why these transposition reactions are never made through alcohols oxidation.

Mind I just ask. What is the meaning of transposition in Organic Chemistry?

carcul is their any acidic thing(probably except water) in oppenaur oxidation.....
also as far as i know(refer Morrison) HBO prefers least hindered site at the same time tries to add in such a way that the beta carbon tho where B gets attached gets stabilized due to incumbent positive charge. :)

[quote="BanishedTraitor"]What is the meaning of transposition [/quote]

Transposition is the "transportation" of a functional group from one carbon to another in a molecule. For example, -CH2CO- -----> -COCH2-.

[quote="Pardesi"]carcul is their any acidic thing(probably except water) in oppenaur oxidation.....[/quote]

A trimethoxialuminium is a very good Lewis acid.

Carcul any suggestion...
what is ur method :)

Here's one of several methods.

[hide="Click here"]1. TsNHNH2
2. BuLi (2 eq.)
3. MeSSMe
4. BuLi (1 eq)
5. HgCl2, H2O, MeCN[/hide]

I will leave with you the interpretation of these reactions.

Reduction to alcohol(LiAlH4)
Dehydration to get conjugated product
Foll by treatment with MMPP(epoxidation) + acidification.

Acidification with what? And what would be the product expected?

acidification with dil. acid should do. well, since epoxide ring breaks and a benzylic carbocation is formed adjacent to OH. so H- migrates and you get a ketone(pinacol-pinacolone)

With dilute acid a glycol can form. And shouldn't methyl migrate instead/also?

even if a glycol forms pinacol rearrangement will take place, wont it. also isnt the migratory aptitude H>aryl>alkyl.

probably a better syntsesis wud be reaction of epoxide wid Hcl foll by treatment wid Ag+.


algorithm
combinatorics
Inequality
partition
IMO Shortlist

Let $ A_0 \equal{} (a_1,\dots,a_n)$ be a finite sequence of real numbers. For each $ k\geq 0$, from the sequence $ A_k \equal{} (x_1,\dots,x_k)$ we construct a new sequence $ A_{k \plus{} 1}$ in the following way.
1. We choose a partition $ \{1,\dots,n\} \equal{} I\cup J$, where $ I$ and $ J$ are two disjoint sets, such that the expression
\[ \left|\sum_{i\in I}x_i \minus{} \sum_{j\in J}x_j\right|
\]
attains the smallest value. (We allow $ I$ or $ J$ to be empty; in this case the corresponding sum is 0.) If there are several such partitions, one is chosen arbitrarily.
2. We set $ A_{k \plus{} 1} \equal{} (y_1,\dots,y_n)$ where $ y_i \equal{} x_i \plus{} 1$ if $ i\in I$, and $ y_i \equal{} x_i \minus{} 1$ if $ i\in J$.
Prove that for some $ k$, the sequence $ A_k$ contains an element $ x$ such that $ |x|\geq\frac n2$.

[i]Author: Omid Hatami, Iran[/i]

Considering the convex function $ f$:$ f(x_1,x_2,...,x_n)\equal{}\sum_{k\equal{}1}^{n}x_k^2$ and assume the opposite assertion.We will get a contradiction quickly :)

[quote="hxy09"]Considering the convex function $ f$:$ f(x_1,x_2,...,x_n) \equal{} \sum_{k \equal{} 1}^{n}x_k^2$ and assume the opposite assertion.We will get a contradiction quickly :)[/quote]
Could you elaborate?

Let the funktion $ f(A_k)$ be $ f(A_k)\equal{}\sum x_k^2$. f(x) is strict increasing. Assume not so $ f(A_k)\leq(A_k\plus{}1)$ we have $ \sum (x_i\plus{}1)^2\plus{}\sum (x_j\minus{}1)^2\leq\sum x_k^2$. Hence $ n \leq2*(\sum x_j\minus{}\sum x_i)$ and $ \frac{n}{2} \leq\sum x_j\minus{}\sum x_i$, where $ |\sum x_j\minus{}\sum x_i|$ is the smallest possible. Now if every $ |x_k|$ is less than $ \frac{n}{2}$, then let $ |z|$ be the greatest number from $ |x_1|,|x_2|...|x_k|$ and we put $ z$ into set $ I$ and every other number one by one if positive into that set of I and J that has the less sum than the other and if negative into the set of I and J that has the greater sum than the other. It's easy to see that $ |\sum x_i\minus{}\sum x_j|$ is always less than |z| which is less than $ \frac{n}{2}$, contradiuction to $ \frac{n}{2} \leq\sum x_j\minus{}\sum x_i$ is the smallest. Hence f(x) is strict increasing and so it's clear that there is a k so that $ A_k$ satisfield the condiction.

Assume all numbers belong in $(-n/2,n/2)$ always. Given a sequence $A_k=(x_1,...,x_n)$ define $f(A_k)=x_1^2+...+x_k^2$, and notice $f(A_{k+1}) = \sum_{i \in I} (x_i+1)^2 + \sum_{i \in J} (x_i-1)^2 = f(A_k)+n-2(I-J)$, where $I,J$ denote the sums of the sets $I$ and $J$ for convenience. Notice that each individual number has only $n$ "possibilities", since the fractional part is constant. Therefore eventually some configuration will repeat, and therefore $f(A_k)=f(A_{k+m})$ for some $k,m$. This means that $I-J$ has an average value of $n/2$ when it goes from $A_k$ to $A_{k+m}$.

Therefore there exists some $I-J \ge n/2$. However, when this happens, if $0 < a \in I$, which must exist, then define $I'=I-a$, $J'=J+a$ and since $a<n/2$, we get $|I'-J'| < |I-J|$, a contradiction.

Hence, done.

Let $\{ m\}$ denote the fractional part of a real number $m$. Also, for $A_k=(x_1,..., x_n)$ let $s_k=\sum_{i=1}^n x_i^2$. Consider the values of $\{ \sum_{i\in I} a_i - \sum_{j\in J} a_j\}$ where $I\cup J$ is a partition of the first $n$ integers. If $M$ is the minimal value, then we claim that as long as $A_k$ contains no element $x$ with $|x|\ge n/2$, we have $s_{k+1}\ge s_k + 2M$. Note that the fractional part does not change if the $a_i$ are replaced by $x_i$, as $a_i-x_i$ is an integer for all $i$. Suppose no element of $A_k$ has absolute value at least $n/2$. Then let $A_{k+1}=(y_1,...,y_n)$. We have $$\sum_{i=1}^n y_i^2=\sum_{i\in I} (x_i+1)^2+\sum_{j\in J} (x_j - 1)^2$$ $$s_{k+1}=s_k + n + 2\left(\sum_{i\in I} x_i - \sum_{j\in J} x_j\right)$$ We claim that $$n + 2\left(\sum_{i\in I} x_i - \sum_{j\in J} x_j\right)>0$$ which implies that $$n + 2\left(\sum_{i\in I} x_i - \sum_{j\in J} x_j\right)\ge 2\{ \sum_{i\in I} x_i - \sum_{j\in J} x_j\} = 2\{ \sum_{i\in I} a_i - \sum_{j\in J} a_j\} \ge 2M$$ To prove that the sum is always positive, suppose that $I\cup J$ was the optimal partition and $n + 2\left(\sum_{i\in I} x_i - \sum_{j\in J} x_j\right)\le 0$. Then it follows $S=\sum_{i\in I} x_i - \sum_{j\in J} x_j \le -n/2$. But then if we add a positive element to $I$ from $J$ (or remove a negative element from $I$ and add it to $J$), say $x_k$, then the sum increases by $2|x_k|$. But then note $2|x_k| < n$ by definition. If $|S|$ is optimal, then $|S|\le |S+2|x_k||$. Since $S$ is negative, it follows $S+2|x_k| \ge -S$. But then this implies $|x_k|\ge -S \ge n/2$, contradiction! Thus if $I\cup J$ is the optimal partition, then $n + 2\left(\sum_{i\in I} x_i - \sum_{j\in J} x_j\right)>0$. Thus if $A_k$ has no element $x$ with absolute value at least $n/2$, then $s_{k+1}\ge s_k + 2M$. However, this means that eventually $s_k\ge n^3/4$, which solves the problem as $\sum_{i=1}^n x_i^2\ge n^3/4$ implies for some $x_j$ we have $x_j^2\ge n^2/4$, or $|x_j|\ge n/2$. $\square$

Suppose for the sake of contradiction that $|x|\le n/2$ for all $x\in A_k$ for all $k$. The key idea is to consider \[f(A) = \sum_{x\in A}x^2.\] Indeed, note that \[f(A_{k+1})-f(A_k) = n  + 2\left(\sum_{i\in I}x_i - \sum_{j\in J}x_j\right)\] where $A_k=(x_1,\ldots,x_n)$.

[b]Lemma:[/b]  We have \[\left|\sum_{i\in I}x_i - \sum_{j\in J}x_j\right|<\frac{n}{2}.\]

[i]Proof:[/i]  Start with $I=\emptyset$, and add on one element from $A_k$ at a time. We see that \[\sum_{i\in I}x_i - \sum_{j\in J}x_j\] changes by $2x_i<n$ at a time, so at some point, it is less than $n/2$ away from $0$. This shows that there is a partition with difference between the split less than $n/2$, and as $I\cup J$ is the optimal partition, the lemma follows. $\blacksquare$

This implies that \[f(A_{k+1})-f(A_k)>0.\qquad\qquad(\star)\] Now, note that \[f(A_{k+1})-f(A_k)\equiv 2\left(\sum_{i\in I}a_i - \sum_{j\in J}a_j\right)\pmod{1},\] and since there are finitely many partitions $I\cup J$, we see that there is some $\delta\in(0,1)$ such that \[\{f(A_{k+1}-A_k)\}\ge \delta,\] or is an integer. In either case, $(\star)$ implies that \[f(A_{k+1})-f(A_k)\ge\delta.\] Therefore, $f(A_k)$ grows arbitrarily large, in contradiction to the fact that the $|x_i|$s are always bounded. This completes the solution.

[hide=lol]Assume not.

Let $\Delta$ be the quantity $\sum_{x \in I} x - \sum_{y \in J} y$. Note that by a greedy arguement (order them, and consider the prefix sum and consider the moment it exceed half of the total sum) we have $\Delta \leq M$, where $M$ is the maximum of these numbers. Also, note that there are only finitely many possibe values, so if it's never $\frac{n}{2}$, then there's some $c > 0$ such that it don't exceed $\frac{n}{2} - c$.

Then consider the sum of squares of the numbers. It increases by $n + 2 \Delta \geq n - 2 (\frac{n}{2} - k) = k$ at each step, so at some point it will be larger than $n^3$. Then some element would be larger than $n$, a contradiction. [/hide]

Let $A_k=(x_{1k},x_{2k},...,x_{nk})$, where $x_{1k}\leq x_{2k}\leq...\leq x_{nk}$.

[color=#f00][b]Lemma.[/b][/color] If $|a_1|\leq...\leq |a_n|$ and denote $A=\{a_1,...,a_n\}$ then there exists some $I\in A$ such that
$$|\sum_{i\in I}a_i-\sum_{i\notin I}a_i|\leq |a_n|\hspace{20pt}(1)$$
[u]Proof. [/u]
We proceed by induction on $n$. If $n=1$ it is trivial, suppose $n=k$ is true, then for $n=k+1$, there exists some $I\in \{a_1,...,a_{n-1}\}$ satisfying $(1)$. Then if $\sum_{i\in I}a_i\leq \sum_{i\notin I}a_i$ we can just put $a_n$ in $I$, which will imply
$$|\sum_{i\in I}a_i-\sum_{i\notin I}a_i|=|a_n|-|\sum_{i\in I}a_i-\sum_{i\notin I}a_i-a_n|\geq a_n$$
the other case is similar. $\blacksquare$

Let $s_k=\sum_{i=1}^nx_{ik}^2$, we claim that if $|x_{nk}|<\frac{x}{2}$ then $s_k<s_{k+1}$. Indeed, we have(writing $x_{ik}=x_i$)
$$s_{k+1}-s_k=\sum_{i\in I}(x_i+1)^2-x_i^2+\sum_{i\in J}(x_i-1)^2-x_i^2=2(\sum_{i\in I}x_i-\sum_{j\in I}x_i+n>2(-\frac{n}{2})+n=0$$
as desired.
Hence if such $k$ does not exist then $s_k$ is unbounded, contradiction.

Assume the opposite, so that for any $k$ and any $x \in A_k$, we have $|x| < \frac{n}{2}$.
Now we claim that for $A_k = \{x_1 , x_2 , \dots , x_n\}$ and $A_{k+1} = \{y_1 , y_2 , \dots , y_n\},$ $\sum_{i=1}^{n} x_i^2 < \sum_{i=1}^{n} y_i^2,$ if we can prove this claim then we will obviously be done.

Note that after constructing $A_{k+1},$ $\sum_{i=1}^{n} x_i^2$ changes by $n + 2(\sum_{i\in I}x_i - \sum_{j\in J}x_j)$.
Now we greedily construct a partition of $I' \cup J' = \{1,2, \dots , n\}, I' \cap J' = \emptyset,$ so that $|\sum_{i \in I'}x_i - \sum_{j \in J'} x_j| < \frac{n}{2}$.
For this, first start with $x_1$ and put it in $I'$. Then for any $t \geq 2$, put $x_t$ in $I'$ or $J'$ such that $|\sum_{i \in I'}x_i - \sum_{j \in J'} x_j|$ is minimal. Since $|x_i| < \frac{n}{2}$ for any $i$, this greedy construction always keeps the inequality $|\sum_{i \in I'}x_i - \sum_{j \in J'} x_j| < \frac{n}{2}$.
So, since $I$ and $J$ were optimal sets, we must have $|\sum_{i \in I}x_i - \sum_{j \in J} x_j| < \frac{n}{2}$ as well.
So, $\sum_{i=1}^{n} x_i^2$ changes by $n + 2(\sum_{i \in I}x_i - \sum_{j \in J} x_j) > 0$, and we are done. $\square$

Solved with [b]gabrupro[/b]. We spent quite some time on this only because of my brilliant assumption that $(-1)^2 = -1$ :gleam:

Suppose not. Let $S$ denote the sum of squares of the terms of $A_k$, we claim this is always increasing, which would be a contradiction.

If $S_i$ and $S_j$ are the sums of numbers in $I$ and $J$, then see that $S$ increases by $2(S_i - S_j) + |I| + |J| = n + 2(S_i - S_j)$. So it suffices to show $S_i - S_j < \frac{n}{2}$, but this is easy to see since otherwise, because all elements in the set are $< \frac{n}{2}$, we can move over a term from $I$ to $J$ while decreasing the sum, which was supposedly minimal. Therefore, $S$ is indeed increasing so we are done. $\blacksquare$

Assume the contrary. The key claim is the following:
[b][color=#f00]Claim:[/color][/b] $\sum_{i=1}^n x_i^2$ always increases after every new sequence constructed.
[i]Proof.[/i] Note that we have \begin{align*} \sum_{i=1}^n y_i^2 &= \sum_{i\in I} (x_i+1)^2 +  \sum_{i\in I} (x_j+1)^2 \\ &= (x_i^2+2x_i+1) +  \sum_{i\in I} (x_j^2-2x_j+1) \\ &= n + \sum_{i=1}^n x_i^2 + 2\left(\sum_{i\in I} x_i - \sum_{j \in J} x_j \right).\end{align*} It suffices to prove that $\sum_{i\in I} x_i - \sum_{j \in J} x_j < \frac{n}{2}.$ Suppose that this is not true; then moving [i]any[/i] element $i\in \{1, \ldots, n\}$ to $J$ changes $\sum_{i\in I} x_i - \sum_{j \in J} x_j$ by an amount $-2x_i < -2\cdot \frac{n}{2} = -n$. However, this would imply that this new partitioning creates a lower $\left|\sum_{i\in I} x_i - \sum_{j \in J} x_j\right|$, contradiction. $\blacksquare$

Now notice that each individual $x_i$ can be transformed into finitely many numbers, as it is bounded between $-\frac{n}{2}$ and $\frac{n}{2}$ and must be an integer amount away from it's starting value. This implies that there are finitely many values of $\sum_{i=1}^n x_i^2$ attained, which is a contradiction by the claim.

Assume that all $|x| < \frac{n}{2}$, always. Then, note that
\[\sum_{i=1}^n x_{i,k+1}^2 = \sum_{i\in I} (x_{i,k}+1)^2 + \sum_{i\in J} (x_{i,k}-1)^2 = \sum_{i=1}^n x_{i,k}^2 + 2\cdot (\sum_{i\in I} x_{i,k} - \sum_{j\in J} x_j) + n \geq \sum_{i=1}^n x_{i,k}^2 - 2\cdot |\sum_{i\in I} x_{i,k} - \sum_{j\in J} x_j| + n\]
Note that  $|\sum_{i\in I} x_{i,k} - \sum_{j\in J} x_j| \leq \max |x_i|$. This is because one can sort all of the $x_i$ values, and consider beginning with the largest and if the running total is positive, subtract, and if the running total is negative, add. Note that the set of $|\sum_{i\in I} x_{i,k} - \sum_{j\in J} x_j| $ values is finite and they are also all $<\frac{n}{2}$. Thus, under our initial assumption, $- 2\cdot |\sum_{i\in I} x_{i,k} - \sum_{j\in J} x_j| + n> \epsilon$, always, for some constant value of $\epsilon$. Thus, $\sum_{i=1}^n x_{i,k}^2$ increases by a constant amount at each iteration, so eventually it is at least $n\cdot \frac{n}{2}^2$ and the initial assumption is incorrect, so there must eventually be some $|x| \geq \frac{n}{2}$. $\blacksquare$.


In case people haven't heard, PUMaC (Princeton University Mathematics Competition) is holding an off-site portion this year.  If you are in a math circle or math enrichment group, or you go to a good school for math, you might consider bugging your math teacher to participate.  If you can't find a team at your school, a MathLinks/AoPS team (consisting of 8 or fewer people on this forum) would also be acceptable, though perhaps more difficult to coordinate.

Information is here:  http://cgi.math.princeton.edu/mathclub/index.php/Power_Test_2007

Register online at http://webscript.princeton.edu/~mathclub/pumacos/index.php by October 5 (sorry for the somewhat short notice).  The registration fee is $ \$$40 due by October 19.

The tests will be sent by email to teams on Friday, November 2, at 9:00 AM, and they will be due on Saturday, November 10, at 9:00 AM through e-mail (encouraged) or mail.

I've been semi-exaggerating about how smart the Canadian math community is for the past two years, so register and show the Americans what Canada is made of!

Questions? PM me or Xevarion.

Thanks for the information!

I will try to make a team.
EDIT: On second thought, I think I will be really busy at the time of the competition because of university stuff so I probably will not participate.

I would like to participate...but I think I could only do so on an AoPS team.


Find $ n\in$N and $ n\ge2$ such that
 $ (x\plus{}1)^2\plus{}y^2_1\equal{}...\equal{}(x\plus{}n)^2\plus{}y^2_n$ has roots $ (x;y_1;...;y_n)\in$Z


The Problemo:

$n>3$ is a perfect square. $a_{i}$ ( $1 \leq i \leq n$ ) are positive integers and $a_{1}> a_{2}> \cdots > a_{n}$ . Prove that:

 $6\sum^{n}_{i=1}( 1+\sqrt{a_{n}}) \geq 5\sqrt{n}+4( \sqrt{n})^{3}+3( \sqrt{n})^{2}$ :ninja:

Common guys, :)
I will post my solution after the first is posted...


MATHCOUNTS
AMC
AIME
logarithms
complex numbers

I don't want the answers or any questions and I don't need the topics.

also what usually apears on the mathcounts tests?

If you're looking for a study sheet specifically for MathCounts, the Toolbox is great: [url]http://www.mathcounts.org/webarticles/articlefiles/105-toolbox.pdf[/url] Most problems in the MathCounts tests, including the School Round, are built up from these topics. Hope it helps!

Try to avoid triple posting ;)

Buy AOPS volume 1 (found on this site)

You've got to be kidding me, biffan. There is no way you need AoPS volume 1 for [b] school [/b] round. I'd recommend the site easyas recommended, and going through warm-ups and work-outs.

You definetly don't need all of AoPS volume 1 for school round, but the first couple chapters and the ones on permuatations/combinations is real useful. Even if you only look at those chapters for the school round, if you make chapter/state/national the rest will come in handy.

You may not need it, but if you have that mastered, as far as I know you will have no problem whatsoever with both school and chapter. It's like practicing AIME problems to do better on the AMC. It isn't necessary, but it definitely works.

I would also practice any problems posted in this forum, even if they're a harder difficulty than the school round (which all of them are that I've seen). Make sure that either you can solve them on your own (preferable), or that you understand the solutions of others (only if you can't solve it on your own).

Hi David :D  :D  :D

Try the [url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=181&z=46]go figure thing[/url] on the mathcounts website.  It's fun when you're really bored, though the solutions they give you are pretty bad.  It doesn't work with Firefox, so you'll have to use IE or something.

Good luck on the sprint next week :P  You'll do well on it, I'm sure.

You have to be able to do the problems, but speed is also important. Even if the problems are easy if it takes you forever to compute you can't do very well.
The best way I know to improve speed is just do lots of problems, and come up with shortcuts.

this might not help, but for mathcounts I never really studied, but still got high scores. mathcounts competition isn't like the SAT or etc; it's more like the olympiads and stuff; you don't really need to study except what you should already have been doing (schoolwork and mathclub etc work relating with mathcounts).

PS good luck on your mathcounts.

[quote="biffanddoc"]I REALLY disagree with that statement. You should study  a lot of old problems and AoPS volume 1 to succeed at not only school round or just MATHCOUNTS, but in other competitions.[/quote]
that was just for me, not for everyone. I get lots of perfects by studying/not studying so either way works for me. Depends on his mathematical skills too. I can't concentrate on 1 contest like that because I have too many things to do. :(  :starwars:

A little algebra from book 1 helps, theres absolutely no geometry.  The understanding of a few topics is good enough.

it probably would help but you should also try to remember some tricks where you look at it and know the answer
i personally am not good at mathcounts cuz im kinda slow
if ur jus having problems on the problems they give u
and jus practice the problems that ur teacher gives you (if he/she does)
cuz normally i think those are what are usually like are the type of problems he competetion will use
so once you get used to those problems then it shud be easier

If I remember correctly (and i maybe not), AoPS volume one had nothing to do with permutations or combinations.. it was logarithms (nothing to do with MC), then imaginary numbers (not sure on that one), and then a useful chapter (but not for school round) on solving equations or something to that effect.

No, it's volume one, I have it right in front of me:

1. Exponents and Logarithms
2. Complex Numbers
3. Linear Equations.
4. Proportions.
5. Using the Integers.

Wasn't counting chapter 20 something? (Volume 1)

Oh, my mistake. I was thinking of Vol. II, since that begins with logs as well.

I'm not sure why Vol. I includes logs and complex numbers, as those aren't normally introduced until Algebra II. Anyway, just skip over those chapters if you're studying for MathCounts.

Also, does it have only those five chapters? I can't find my copy of it at the moment to check.

5 out of about 29.. didn't want to list em all lol.

[quote]I'm not sure why Vol. I includes logs and complex numbers, as those aren't normally introduced until Algebra II. Anyway, just skip over those chapters if you're studying for MathCounts. [/quote]

you'll probably need the exponents part of ch.1 for mathcounts( but not for school round, I don't think)
[url=http://www.artofproblemsolving.com/Books/AoPS_B_TOC.php?page_id=1]chapters[/url]


limit
real analysis
real analysis unsolved

We choose $m$ $n$-tuples $u_{i}=(a_{i1},a_{i2},\cdots, a_{in})\in\mathbb{N}^{n}$ and let $\|u_{i}\|: =\sum_{j=1}^{n}a_{ij}$

Find the $\lim_{m\to\infty}\frac{\sum_{i=1}^{m}\|u_{i}\|}{m}$

Alexandros

From Cesaro-Stoltz : $\lim_{m\to\infty}\frac{\sum_{i=1}^{m}\parallel u_{i}\parallel }{m}=\lim_{m\to\infty}\frac{\sum_{i=1}^{m}\sum_{j=1}^{n}a_{ij}}{m}= \lim_{m\to\infty}\frac{\sum_{i=1}^{m+1}\sum_{j=1}^{n}a_{ij}-\sum_{i=1}^{m}\sum_{j=1}^{n}a_{ij}}{m+1-m}$ so ${\lim_{m\to\infty}\frac{\sum_{i=1}^{m}\parallel u_{i}\parallel }{m}=\lim_{m\to\infty}\parallel u_{m+1}\parallel }$

The problem is terribly incomplete. Of course, what spix said is correct but to make this observation would not require knowing what $u_{j}$ are at all. Please, correct the problem statement so that it would make sense. One possible correction would be to add the word distinct before $n$-tuples.


For how many positive integers $a<100$ will $a^3+23$ be divisible by 24?
(A) 4
(B) 5
(C) 9
(D) 10

MCrawford wrote:For how many positive integers  will  be divisible by 24?
(A) 4
(B) 5
(C) 9
(D) 10

Solution in spoiler:

[hide]

The desired equivalence relation is a^3  1 mod 24.

Consider integers 0  b < 24.  It should be clear that b^3 mod 24  (some multiple of GCD(b,24)).  For example, 6^3 mod 24 must be equivalent to a multiple of 2.  Thus, any b that are multiples of 2 or 3 cannot satisfy the equivalence relation b^3  1 mod 24.  So we need to only test the values b = 1, 5, 7, 11, 13, 17, 19, 23.

1: 1^3  1 mod 24

5: 5^2  25  1 mod 24, so 5^3  5 mod 24

7: 7^2  49  1 mod 24, so 7^3  7 mod 24

11: 11^2  121  1 mod 24, so 11^3  11 mod 24

13: 13^2  (-11)^2  121  1 mod 24, so 13^3  13 mod 24

17: 17^2  (-7)^2  49  1 mod 24, so 17^3  17 mod 24

19: 19^2  (-5)^2  25  1 mod 24, so 19^3  19 mod 24

23: 23^3  (-1)^3  -1 mod 24.

Only b = 1 satisfies b^3  1 mod 24.  Thus a = 1, 25, 49, 73, 97 all satisfy a^3  1 mod 24.  The answer is then 5, (B).[/hide]


Jill baked a rectangular cake for her brothers, John and Jerry. She planned to cut the cake in half and send an equal amount home with each brother. However, when she took the cake from the fridge, she saw that her son, Joe, had cut out a rectangular chunk and eaten it. How can she make a single straight cut through the cake that will guarantte each brother an equal amount? Jill has some string on hand, but no measuring tools. Also, she may not cut the cake through the middle, parallel to the cake's top surface. Each brother should recieve an equal amount of the frosted top.

Source: Optional problem my teacher gave us.

depends on where the rectangle is cut and if the rectangle cut is similar to the original cake... the problem is not clear enough

Yes I agree. I don't really understand this question either.

Was there a picture...?

And if a string could help, then how could the cake possibly be cut?

no there wasn't a picture. and jill had a knife to cut.

but how are we supposed to solve the problem without a picture/knowing where the rectangle is?

if we can pretend that the rectangle is parallel to the side of the cake, then she can use the string to roughly measure and cut it parallel to the side and, ok confusing....

but we still need some knowing where the rectangle is cut or some picture or something

I think this problem is complete. Different answer on different situation.

Don't you mean "incomplete"?...

This problem is far from perfect.

[quote="aanjohri"]Jill baked a rectangular cake [/quote]

I just realized the cake is rectangular.    :)

Wait never mind that doesn't help much, sorry.  :blush:

There was a picture of a triangular piece of cake on a plate with a candle. However, it looked like a clip art picture and I think it was not meant as a diagram, but just a little accompanying picture.

ALSO, this was at the bottom of the problem:
"This problem is tricky! You may not solve it successfully, but you must show/explain your attempt in order to receive full credit."

EDIT: i think i was the only one who didn't know it was a rectangular cake.  :(

Then why was it triangular?!?!?![/quote]

That was just a picture from clip art and i think it was just put there to make the problem look "pretty."
It had nothing to do with the question.

I think you have to cut through the diagonal and use the string to help you cut straight.

Which diagonal? Cutting through a diagonal works in these two cases:

1) The slice is cut out of a corner, and is similar to the whole cake.

2) The slice is (somehow) taken out of the inside of the cake, and the diagonal coincides with the cake's diagonal.

In either case, you can cut along the diagonal of the cake, which coincides with the diagonal of the slice.

I will get the solution to this problem in late august/ early september, so i will post up the answer then.

Also, I asked a friend this question and he said that you could try balancing the cake on the string so that on each side of the string, there is an equal amount of cake.

However, I am not sure if that would work. How would you balance a cake on a string?

Okay, imagine the cake is on a coordinate plane. The coordinates of the cake are (0,0), (10,10), (0,10), and (10,0). You place the string with one end on (0,0) and the other end on (10,10). Then you cut the cake across the line with the equation Y=(1)(X) + 10.

anyway, @wickedestjr
if there is a rectangular chunk taken out, and does not fall directly in line with ur cut, one person would be getting less

I understand the problem but it is just missing a block of  information. You could cut the cake diagonally if the the rectangular piece of cake is small and is cut diagonally from a corner than it is possible that she will give an equal amount of cake to her brothers.

It's a very strange problem, but there is probably a good solution to it that would lower all this confusion.

take the centers of the cake and the rectangle cut out. cut a straight line through the center of the cake and the center of the rectangle. since any cut through the center of a rectangle will cut it into 2 equal peices.
it  doesnt really mattter where the rectangle is as long as the part eaten was a rectangle.

[quote="aanjohri"] Also, I asked a friend this question and he said that you could try balancing the cake on the string so that on each side of the string, there is an equal amount of cake.
quote]

I like that solution! ha! i bet it might actually work

How would you even get the cake on the string if you were to try it. You wouldn't be able to balance it without moving your hands at least a little bit. It would also probably break into two un equal pieces.

Would if you had a balance beam or something, then you could cut the cake into a lot of pieces and keep moving the pieces around from side to side across the beam until it balanced perfectly then you'd have the two halves. I think that might be the solution.

but unfortunately, you don't have a balance beam

you only have string

Odd...No one remarks on abacadea's post... His solution is [b]clearly[/b] correct

( I am pretty sure this problem meant to imply (this is my assumption) that the rectangular piece cut out had all four sides parallel/perpendicular to the sides of the whole cake; so what I mean is that if my aformentioned assumption is true, then abadacadaea is right. If my assumption is wrong, then this might be a rather hard (I am not really sure, maybe there is an easy but slick solution) construction problem, especially since it is 3D, and probably does not belong in the elementary school forums. So, if my assumption is wrong, maybe reading up on the [b]Pancake theorem[/b] will help. This theorem states that with two figures (of any crazy shape/size) in one plane, there always exists one line that can split both figures in half). Hm, I am getting off topic.

  Also, to add to his solution, I will say that to find the centers of a rectangle with a piece of string, one can take the string lay it across one diagonal, and then if it is long enough (I will assume the string the string in this problem is "long enough" since the length was unspecified) lay it also across the other diagonal (If you are not sure what I mean, imagine the ends of the string at two adjacent corners and the middle of the string connects the other two adjacent corners - its liike 75% of a figure 8 made with the string).

Doing this with both rectangles (it can be done simultaneously if the string is long enough) will construct the centers of the two rectangles (the center is the intersection of the two diagonals)

Then what would you be using the string for? I don't think the string was just a piece of irrevelant information.

TO show where the line is conencting the mipoints.

@ hurdler: No, even if the sides are not all parallel, it would still work because any line passing through the center of a rectangle divides it in half.

Oh, you're right fishy.

I forgot about that....

For example (time to test my new asy skillz!)

[asy]draw((-10,0)--(-10,10)--(10,10)--(10,0)--cycle);
fill((-10,0)--(-10,10)--(10,10)--(10,0)--cycle,orange);
fill((1,3)--(5,7)--(7,5)--(3,1)--cycle,white);
draw((-10,7.5)--(10,2.5),blue+linewidth(1));
dot((0,5));
dot((4,4));[/asy]

As you can see, the regions seperated by the blue line are equal in area, so abacadaea was correct.


How much credit will we get for drawing a precise drawing and measuring lengths and angles in a geometry problem?

I can't say I know for sure, but I wouldn't expect more than 1 (I doubt you'd get 0).

[quote="13375P34K43V312"]I believe you do get credit for getting the correct answer, regardless of how you get it.[/quote]

Well, I remember last year, I got the correct answer to a problem and wrote two pages of my analysis to the problem (not exactly a solution), and I only got a 1.

At best, performing a measurement will give you a 1.  Unless you can prove that your diagram and measurement were perfect (IE:  Construction, which is a valid technique), you haven't actually "solved" the problem.  Being able to prove that your answer is correct is probably both the most difficult and most helpful portion of the USAMTS, so a measurement/approximation basically defeats the purpose of the contest.


Determine whether the following statements about integers is true (T) or false (F):

1)  $2^4-4^2$ is an integer.

2)  $\frac{-\pi}{\pi}$ is not an integer.

3)  $\frac{3}{5}$ is not an integer.

4)  $\frac{1}{\frac{1}{2}}$ is an integer.

5)*  $-10^{1992}$ is not an integer.

6)**  $\frac12+\frac13+\frac16$ is not an integer.

Enjoy.

[hide="Hint"]no matter its positive or negative, if it's just a normal number with those signs, they are integers...

Decimals( I'm not sure though ) or Fractions are not integers
[color=darkred]Decimals and fractions are not integers. Correct.[/color][/hide]

"integers----for kbabe" hmm..this should be interesting...

k...

1. True

2. False (cuz it would equal to -1, woudn't it?)

3. True

4. True

5. False

6. False cuz it equals to 1...(i didn't do it right the first time....)

mm...did i do it right?

[quote="LuCy4EvA"][quote="KBabe"][quote="LuCy4EvA"]check 2 and 6[/quote]

i just did.[/quote]

I mean that it's wrong[/quote]

didn't u mean that my ORIGINAL answers were wrong? becuz i just edited it when i said "i just checked them"

[quote="KBabe"][/quote][quote="LuCy4EvA"][quote="KBabe"][quote="LuCy4EvA"]check 2 and 6[/quote]

i just did.[/quote]

I mean that it's wrong[/quote][quote="KBabe"]

didn't u mean that my ORIGINAL answers were wrong? becuz i just edited it when i said "i just checked them"[/quote]

kk

I'm glad you understand what integers are. Very good job, KBabe. Keep it up.


AMC
AIME
MATHCOUNTS
geometry
number theory

As a book, Introduction to Number Theory is good. If you've finished the number theory in Vol 1 and Vol 2 then it will probably just serve as a review. I also recommend the Intermediate Number Theory Seminar, and doing as many AMC 10/12 and AIME number theory problems as you can find.

[quote="batteredbutnotdefeated"]If I have completed the number theory in AoPS volume 1&2, what would be the best way to improve even further in Number Theory? I find it to be my weak point...[/quote]

Just find harder problems to do like isabella said, it'll help your skills go to a higher level.

Actually, isabella, the vol.1 book does not go in depth on number theory, so the Introduction to Number Theory is a good start along with intermediate number theory.

Doing AMC 10/12 questions, they have great number theory questions, buy some of the books.
AOPs has some reccomendations, if you go to their resources site.

By the way, are you in middle school focusing on MATHCOUNTS or highschool to the AMC 10/12/AIME level?

I'm in middle school focusing on getting into a good high school and making USAMO. :P

The only number theory I read was from Volume 1 and Volume 2. I've never even seen Intro to Number Theory, so I can't comment on that. I think Volume 1 and Volume 2 give a really good starting point, and then it's just doing problems. This pretty much applies to everything, but in my opinion, even more so for number theory than anything else (except maybe geometry?).

 I feel that for a lot of number theory problems, you need to have done something at least somewhat similar to it; the same thing for geometry, as in a lot of them, if you can draw in another line that may not be obvious, it helps immensely. So basically, start doing lots of number theory problems. After you get to the point where you're working on olympiad problems, PEN can be really helpful.

By the way, I'm not a big fan of making working through books your #1 priority. I don't really enjoy going through books; I'd rather just work on anything I find interesting.

People recommend doing PEN. Before that I would make sure you know how to do linear congruences and basic contradiction techniques.

[quote="xpmath"]
By the way, I'm not a big fan of making working through books your #1 priority. I don't really enjoy going through books; I'd rather just work on anything I find interesting.[/quote]

I actually enjoy going through books, unless the material taught is tedious or way too advanced for me. :)

[url=http://www.artofproblemsolving.com/Forum/resources.php?c=200]Problems In Elementary Number Theory.[/url]


Find all the functions $ f: \mathbb{N}^* \rightarrow \mathbb{N}^*$ such that $ (\forall)n\ge 1\ ,\ f(1) \plus{} f(2) \plus{} ... \plus{} f(n) \equal{} k^3\ ,\ k\in \mathbb{N}$ and $ k^3 \le n^3$ .

For $ n\equal{}1$, the condition $ k^3<n^3$ implies $ k\equal{}0$, but then $ f(1)$ isn't positive.

[quote="alex2008"]I edited . Please when the typo is obvious or i missed a case solve the problem without consider that .[/quote]

A typo isn't always obvious. It could as well have been $ f: \mathbb{N}^*\to\mathbb{N}$, for example. It's the duty of the thread starter to make sure there are no errors and/or omissions in the statement.

I'm not sure if this is right..

We know that $ f(1)\le 1$ and that $ f(1)=k^3$ of which the only value that works is $ f(1)=1$ we have $ f(1)+f(2)\le 8\Leftrightarrow 1+f(2)\le 8\rightarrow f(2)=7$, because $ f(1)+f(2)$ has to be a perfect cube. Following this you see that $ f(1)+f(2)+\cdots+f(n)=n^3$.

Now some equations:

\begin{eqnarray*}
f(1)+f(2)+\cdots+f(n)+f(n+1)&=&(n+1)^3\\
f(1)+f(2)+\cdots+f(n)&=&n^3\\
n^3+f(n+1)&=&(n+1)^3\\
f(n+1)&=&(n+1)^3-n^3\\
f(n+1)&=&(n+1)^3-((n+1)-1)^3\\
f(n)&=&n^3-(n-1)^3\\
f(n)&=&\boxed{3n^2-3n+1}
\end{eqnarray*}


calculus
integration
trigonometry
logarithms
calculus computations

Prove the following inequality.

\[ 1\leq \int_0^{\frac{\pi}{2}} \sqrt{1-\sin ^ 3 x} \ dx\leq \frac{1}{2}\{\sqrt{2}+\ln (1+\sqrt{2}\ ) \} \]


arithmetic sequence
inequalities proposed
inequalities

If $\min |a_i - a_j| = 0$ then we are done. Otherwise due to homogeneity we may assume wolog that $\min |a_i - a_j| = 1$. Then $\sum a_i^2$ becomes minimum only when $\{a\}$ (when sorted) is an arithmetic sequence with common difference 1. Then
\[
\sum_{i=1}^n a_i^2 = \sum_{i=1}^n (i-\alpha)^2 = n\alpha^2- 2\alpha\sum_{i=1}^n i +  \sum_{i=1}^n i^2
\]
for some $\alpha$. We can then find the $\alpha$ that minimizes the above expression, and rest is just computation.

Please justify why $\sum{{a_{i}}^{2}}$ aproaches its minimum when $(a_{i})$(when sorted) is an arithmetic progression.I have doubts in the case when $\max(a_{i})<0$.

Okay, suppose that $\sum a^2$ is minimum and some two consecutive terms, $a_i, a_{i+1}$ differ by more than 1. If $|a_i| > |a_{i+1}|$, then we may move $a_i$ towards $a_{i+1}$ and thus decreasing $\sum a^2$ while preserving $\min|a_i - a_j| = 1$. Similarly if $|a_i| \leq |a_{i+1}|$


hello, i have
$ \int\tan(x)(1\plus{}\sec(x))\,dx\equal{}\ln|\sec(x)|\plus{}\sec(x)\plus{}C$.
Sonnhard.


geometry parallelogram ratio inequalities calculus

Let ABCD and A'B'C'D' be  parallelograms. Let M,N,P,Q be midpoints of AA', BB', CC', DD'. Prove that MNPQ is a parallelogram.

Welcome to the site. I will move this challenge problem to the Intermediate forum for the usual reasons.

eum... tokenadult, this basic 9th grade geometry... what is left for beginners section geometry then... and even as a challenge.

Yes, Peter, it is very simple. Maybe Tokenadult had another proof in mind, but there is a very easy one.

A kind of generalization (not so easy anymore):

Let ABCD and A'B'C'D' be parallelograms. Let M, N, P, Q be four points on the lines AA', BB', CC', DD' which divide the segments AA', BB', CC', DD' in one and the same ratio (all internally or all externally). Prove that the quadrilateral MNPQ is a parallelogram.

  Darij

Well in fact that one is easy as well: just apply the theorem above enough times :D:D

Not in the case when the ratio is irrational...

  Darij

it is.

any irrational can be arbitrarily approximised by rationals ;)

I really didn't think you would do this... although I admit that the proof I just gave in the "Sequence Inequality" thread is yet more brutal... it seems that we are competing in applying calculus to completely trivial elementary facts...

  Darij

A littly competition is healthy :D

Nah, I didn't really try for it. Anyway you don't even need calculus:

Set the ratio in binary notation (like 101001,01101001...)

we choose the first digit, 1 or 0. We choose the second, 1 or 0. And so on...

If you think a little about it, that _exactly_ the same as getting any ratio in this way :D

Peter, could you please stop complaining every time a moderator promotes a post?  I mean it seems like more than half the time a mod (usually me or tokenadult to be precise) moves a post you have say something like:  this is elementary,  this only takes a few steps,  this is trivial...   I don't see any point to it.  We are just trying to keep the getting started forum at the right level.  If you want to complain about its intended level take that up with one of the site administrators.  All we are doing is moving the post, not deleting it.

Thank you

Joseph L.

jolm, I only complained twice, twice about the same thing. So it's not necessary to flame me for "always complaining".

As for the rest, read my topic in the Q&S.


function calculus derivative algebra polynomial

Given real numbers $a,b,c$ prove that

$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ always has at least one real solution for $x$.

[hide]
Trivial if a,b,c aren't distinct.
Or else let LHS be $f(x)$
WLOG assume $a > b > c$
$f(a) = (a-b)(a-c) > 0$
$f(b) = (b-a)(b-c) < 0$

Then use the intermediate value theorem.
[/hide]

[quote="t0rajir0u"][hide="Fun solution"] This is the derivative of the polynomial

$g(x) = (x-a)(x-b)(x-c)$

Which obviously has extrema.  QED. [/hide][/quote]

this is nice; i think we should be careful about being concise, if two of the roots are distinct, then we use a [i]direct[/i] application of rolle's theorem at x=a, x=b, if all the roots are the same, then the derivative has the solution x=a=b=c


Using logs to base ten, find all real values of x for which
$ x^{log x} \equal{} 100x$.

I did guess and check and came up with:

[hide] 1/10 and 100. [/hide]

Could someone teach me the actual method for doing this? Thanks.

Write it in exponential form.

$ a^b \equal{} c \rightarrow \log_a c \equal{} b$

Applying this,

$ x^{\log x} \equal{} 100x \rightarrow \log_x 100x \equal{} \log x$

$ \log_x x \plus{} \log_x 100 \equal{} \log x$

$ 1 \plus{} \frac {\log 100}{\log x} \equal{} \log x$

$ \log x \plus{} 2 \equal{} (\log x)^2$

Let $ u \equal{} \log x$, then

$ u^2 \minus{} u \minus{} 2 \equal{} 0$

$ u \equal{} \frac {1 \pm 3}{2} \equal{} 2$ or $ \minus{} 1$.

$ \log x \equal{} 2$

$ \boxed{x \equal{} 100}$

$ \log x \equal{} \minus{} 1$

$ \boxed{x \equal{} \frac {1}{10}}$

I'm assuming you know the basic properties of logarithms, about sums, subtractions,change of base, etc.

You are given equation $ x^{\log x} \equal{} 100x$. Apply $ \log$ to both sides to get $ \log(x^{\log x}) \equal{} (\log x)^2$ and $ \log(100x) \equal{} \log 100 \plus{} \log x \equal{} 2 \plus{} \log x$. Together, denoting $ \log x \equal{} \lambda$, they give $ \lambda^2 \equal{} \lambda \plus{} 2$, with roots $ \lambda \equal{} \minus{}1$ and $ \lambda \equal{} 2$, so $ x \equal{} 1/10$ and $ x\equal{}100$.


Thinking about going to AMSP this summer? To apply, you'll need to take an admission test, and the first one has just been posted (a day early, so you can get a headstart.)

Also, the early application deadline is approaching--Feb. 11. If you think you might go to AMSP this summer, you should apply early since students admitted early receive a 10% discount on the tuition. [b]Early applicants must take this admission test.[/b]

Visit [url]http://www.awesomemath.org[/url] to see the test and get all the application information.

Here's a direct link to the test: [url]http://www.awesomemath.org/tests/TestA_2008.pdf[/url]

Regardless of whether you apply early, if you decide to submit solutions to this admission test they must be postmarked by Feb. 11.

Thanks for the quickness! If only it were not due the day before the AMC...

Anyone have any idea how many problems are needed (usually) to pass?

Just do as many of the problems as you can. Don't worry about that.

May we discuss the problems for the A test after Feb 11th?


calculus
integration
function
calculus computations

Hey,

[b]1. The problem statement, all variables and given/known data.[/b]
Given,
$ {dy} \equal{} {{{{f}^{\prime}}{(x)}}{dx}}$

Where $ {y}$ is found by integrating both sides of the above equation, which of the following is correct,
$ {y} \equal{} {{F(x)}\plus{}{C}}$
or
$ {y} \equal{} {F(x)}$

[b]2. Relevant equations.[/b]
Knowledge of Differential Equations.

[b]3. The attempt at a solution.[/b]
[hide="My Attempt at a Solution."]
I have always understood the standard/correct convention to be,
$ {{\int_{}^{}}{{{f}^{\prime}}{(x)}}{dx}} \equal{} {{F(x)}\plus{}{C}}$

Recently though, when solving differential equations I have noticed that the convention used is,
$ {{\int_{}^{}}{{{f}^{\prime}}{(x)}}{dx}} \equal{} {F(x)}$

Where, it is understood that the function $ {F(x)}$ already includes the constant $ {C}$, or (as I have heard it phrased) it is understood that the function $ {F(x)}$ has absorbed the constant $ {C}$.

Which is correct?
[/hide]
Thanks,

-[u][b][url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url][/b][/u]

Neither means anything until you've specified what $ F$ is and what $ y$ is supposed to be.  If $ F$ is a generic antiderivative and $ y$ is a [b]particular[/b] solution, you're free to use the second convention; if you've explicitly computed $ F$ and $ y$ is the [b]general[/b] solution, you have to use the first.


calculus
integration
logarithms
trigonometry
real analysis
real analysis unsolved

I will be very (and I mean [b]veeeery[/b]) surprised if it turns out that this expression has an elementary antiderivative. Where did you get it from?

I found it on a italian site :[url=http://www.matefilia.it/scolerivolftp/problemaaprile2004.htm]here[/url].But there isn't the solution.

I'm afraid it was their idea of an April joke :lol:. But it seems you are not alone to be fooled: see [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26004[/url].


x,y and z are non-negative real numbers which are NOT the sides of a triangle

Prove x/(y+z-x) +y/(z+x-y) + z/(x+y-z)  <= -1


The cargo area of a truck is a rectangular prism that measures 8 feet by 21 feet by 11 feet. Each television set measures 1.5 feet by 1 and two-thirds feet, by 1 and one-third feet. How many TV sets can be loaded into the truck?

P.S.- If you are focused on being selected as the best answer, be sure to show your work. Thanks.

[hide]Let's get some bounds!
Upper: $8*21*11=1848$ cubic feet. Each television set is $(3/2)*(5/3)*(4/3)=10/3$, for a maximum of $1848*3/10=554.4$ televisions.
Lower: If we arrange them to form a giant rectangular prism, we see that $21$ works best with $3/2$, $8$ works best with $4/3$, leaving $5/3$ for $11$, which isn't too shabby, as the smallest empty space we could get would be $1/3$ feet while here we have $1$ foot. This gives us a lower bound of $504$.
I don't feel like calculating all the myriad diagonal ways right now.[/hide]


parameterization
conics
ellipse
trigonometry
calculus
calculus computations

I'm really lost on how to find parametrizations with positive, etc. orientations. In particular:

1. Parametrize the boundary curve $C = dS$ of the surface $S: \frac{x^{2}}{9}+\frac{y^{2}}{9}+\frac{z^{2}}{16}= 1$. Z is less than or equal to 0.
2. Find a counterclockwise parametrization of the ellipse $x^{2}+\frac{y^{2}}{4}= 1$

Seeing the process would be very handy since i'm pretty shaky on this...

Do you know how to parametrize a circle? The first curve is a circle, and the second is a stretched circle.

for an ellipse, it is easy:

you have $\frac{x^{2}}{a^{2}}\;+\;\frac{y^{2}}{b^{2}}\;=\;1$

so, you essentially want to find something that implies $\mathbb{A}^{2}+\mathbb{B}^{2}=1$.

and what better than to use the trigonometric relation $\cos^{2}{\theta}+\sin^{2}{\theta}=1$,
or the hyperbolic relation $\sec h^{2}\;{\theta}+\tan h^{2}\;{\theta}=1$,
or even the algebraic relation  $\left( \frac{2\theta}{1+\theta^{2}}\right)^{2}+\left( \frac{1-\theta^{2}}{1+\theta^{2}}\right)^{2}\;=1$ ?

you yourself should pick an [u]easy [/u]one, i.e. use the trig relation:
$x=\cos\theta,\;\;\;y=2\;\sin\theta\;\;\;\quad \;\;\;where\;\;\;0\;\leq\;\theta\;\leq\;2\pi$

this is, of-course, for a [b]counterclockwise [/b]parametrization, where
$\theta$ will increase from $0$ all the way to $2\pi$.

use a similar reasoning for the ellipsoid $\frac{x^{2}}{a^{2}}\;+\;\frac{y^{2}}{b^{2}}\;+\;\frac{z^{2}}{c^{2}}\;=\;1\;\;\;$ to get

$x=3\cos\theta\;\sin\phi,\;\;\;y=3\sin\theta\;\sin\phi,\;\;\;z=4\cos\phi,\;\;\;\quad \;\;\;$
$where\;\;\;0\;\leq\;\theta\;\leq\;2\pi,\;\;\;\quad \;\;\;and\;\;\;\quad 0\;\leq\;\phi\;\leq\;\pi$

[ note: if $z\;\;\leq\;\;0$, then $\frac{\pi}{2}\;\;\leq\;\;\phi\;\;\leq\;\;\pi\;\;$ ]

i myself will tend to use the parametrization:
$x=\frac{6\;\theta}{1+\theta^{2}+\phi^{2}},\;\;\;y=\frac{6\;\phi}{1+\theta^{2}+\phi^{2}},\;\;\; z=\frac{4\;{(1-\theta^{2}-\phi^{2})}}{{1+\theta^{2}+\phi^{2}}}$

but then again, i am crazy!   :D  :rotfl:

[ PS: given all of this, i am sure you can now find a way for parametrization
of a boundary curve for an ellipsoid ]


the number N divides the number 171 with a remainder of 6. List all the numbers N can be.

I sorta know how to do this one. I know that N has to divide 177.

[hide="my answer"]
11,15,33,55,165

A way to solve this problem

$171-6 =165$
then prime factorize 165
$3*5*11$
find all the possible combinations of multiplications that are bigger than 6
so the answers are

$11,15,33,55,165$[/hide]

[quote="bryantacademy"]I know that N has to divide 177.[/quote]

No, you have it backwards. $N$ has to divide $165$, since $165+6=171$. ;) It should be pretty simply from there.

[hide]
For it to have a remainder of 6 when dividing 171, it must be a factor of 165.
They are
1,3,5,11,15,33,55,165
the 2 digit ones are
11,15,33,55
[/hide]

this is a two-way post...

biffanddoc, the numbers must be two digit

jesusfreaksomenumbersomenumbersomenumber, fyi i do that a lot... not good... is there a way i can remember tht?

First of all, why do the numbers have to be two-digit? Is that a requirement which you forgot to list?

As for my part, I'm not exactly sure. I guess practice works. If you have the time when you're not taking a competition or something, reread every question carefully to practice, and eventually you'll start doing it all the time.

Also, don't bother adding the numbers at the end of my username. ;)

[hide]first it has to be divisible by 165 because something divided by x has a remainder of 6... then it has to be a factor of 165 (prime factor) 3*5*11. so there are 2+2+2 total factors . Ones that don't work are 3,5 because well... the remainder is 6. So there are 4 that work. Is that what the problems is asking :?:   :| [/hide]

Sorry I meant
[hide]11, 15, 33, 55[/hide]


complex numbers
superior algebra
superior algebra unsolved

Is the ideal $ I=\bigg( \frac{3+\sqrt{-47}}{2}, 2 \bigg)$ principal in $ R=\mathbb{Z}\bigg[\frac{1+\sqrt{-47}}{2}\bigg]$ (i.e. the ring of algebraic integers in $ \mathbb{Q}(\sqrt{-47})$)?  Prove it.

(I edited this b/c the question originally asked is it principal)

I will attempt to answer my own problem.  [b]I would greatly appreciate someone letting me know if this is alright.  [/b]

Let $ \gamma=( 1+\sqrt{-47})/2$, and suppose $ I$ is principal, i.e. $ I=(x+y\gamma)$ for some integers $ x,y$.  So, $ 2 \in I=(x+y\gamma)$ implies $ (x+y\gamma)\mid 2$.  Taking norms, we get $ (x^{2}+47y^{2})/4$ divides 4.  Multiplying through by 4, it follows that $ x=1,2$ or 4.  Now, $ 1+\gamma=( 3+\sqrt{-47})/2 \in I=(x)$ yields $ 1+\gamma = x (a+b\gamma)$ for some integers $ a,b$.  Hence, $ x=\pm 1$ after comparing real parts.  But, this is impossible since 1 is not in $ I$, because if it were then there would exist integers $ a,b,c,d$ such that (note: $ \gamma^{2}=\gamma-12$)
$ 1=( (3+\sqrt{-47})/2 ) (a+b\gamma)+2(c+d\gamma)=(1+\gamma)(a+b\gamma)+2c+2d\gamma$
$ ~= (a-12b+2c)+(a+2b+2d)\gamma$.
Adding real and imaginary part yields $ 2a-10b+2c+2d=1$ which is the desired contradiction.


This is an easy problem and can be done by various methods.
could you please post diffrent methods to approach this problem

Q) What is the SUM of
       5/2 +9/4 + 17/8 + ....................... till n terms

Thanking You,
mittal :roll:

@mittal , the Most Basic Approach that i can suggest is :->


I went to my favorite taco stand with some friends. We ordered a total of 8 tacos and 3 burritos. John handed the cashier 3 dollars to help pay for the burritos, Tim gave her 2 dollars to help pay for the tacos, and then I gave her the rest of the money for the burritos and the rest of the money for the tacos. As I did this, I noticed that the ratio of the amount of money I paid for burritos to the amount of money I paid for tacos was 2:5. Later we went back to the same taco stand. This time we bought 10 tacos and 4 burritos. I paid for the whole thing by myself, and this time I spent 18 dollars more than I had last time. How much money was that?

let $ t$ be cost of one taco,
$ b$ be cost of burrito
\[ \frac {3b - 3}{8t - 2} = \frac25 \\
\\
10t + 4b = 18 + (3b - 3) + (8t - 2)\]
from first eqution we get
\[ 15b - 15 = 16t - 4 \\
15b - 16t = 11\]
from second equation we get
\[ 10t + 4b = 18 + 3b + 8t - 5 \\
2t + b = 13 \\
16t + 8b = 104\]
adding these gives
\[ 23b = 115 \\
b = 5\]
from
\[ 2t + b = 13\]
we get
\[ 2t = 8 \\
t = 4\]
now cost of 10 tacos and 4 burritos is
\begin{align*}
&10\cdot 4 + 4\cdot 5 \\
=& 40 + 20 \\
=& 60\end{align*}