inequalities
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inequalities open

I quote from http://www.artofproblemsolving.com/Forum/post-292977.html

[quote="darij grinberg"]Here is a nice one: http://www.mathlinks.ro/Forum/viewtopic.php?t=6138 .

[quote][color=blue]If a, b, c are any three positive reals, then prove the inequality

$\left(b^2c^2+c^2a^2+a^2b^2\right)\left(bc^2+cb^2+ca^2+ac^2+ab^2+ba^2\right)$
$\geq 2abc\left(b^2c^2+c^2a^2+a^2b^2+bc\left(b^2+c^2\right)+ca\left(c^2+a^2\right)+ab\left(a^2+b^2\right)\right)$.[/color][/quote]

Upon expanding, it becomes

$\left[4,3,0\right]+\left[3,2,2\right]\geq\left[4,2,1\right]+\left[3,3,1\right]$.

Try to bunch it out. I am not sure that it's impossible (with Muirhead, of course it is, but I mean Schur), since I am not really used with bunching in practice (well, you see that I didn't do the IMO one), but it would be a bit strange if it would work.

By the way, I look forward to hear Geoff's opinion on Vornicu-Schur next year. :D

  Darij[/quote]

So there are homogenous symmetrical ineqs like the above
\[ \left[4,3,0\right]+\left[3,2,2\right]\geq\left[4,2,1\right]+\left[3,3,1\right] \]
that cannot be bunched out with Muirhead but are correct anyway. (and thus have to be proved otherwise ;) ) The question is then, limiting it for the moment to two terms on either side: for [i]which pairs of partitions[/i] (two LHS, two RHS) is an ineq valid [b]even if Muirhead is 'violated'[/b], i.e. the LHS partitions don't dominate all the RHS ones??

Maybe this could lead to a new theorem more general than Muirhead???

I have an idea (I've discussed with mecrazywong but no result). I think that for a combination of symmetric sum where muirhead cannot be applied, a suitable factor could be found so that when this factor is multiplied, the product can be proved by muirhead. I have totally no idea how to ensure the existence of such factor and how to find it.

I have considered it. but only a thought slipped on my brain.not write it.
now I have a idea(May only a expression,though it may be useless to you,but I stand want to post it.
$\left[4,3,0\right]-\left[4,2,1\right]=\left[4,0,0 ; 1and2,0,0\right]=2\left[4,1,0 ; 1^2,0,0\right]$
$\left[3,3,1\right]-\left[3,2,2\right]=\left[3,1,1;1^2,0,0\right]$
and because $\left[4,1,0 \right] \geq \left[3,1,1 \right]$
so $\left[4,3,0\right]-\left[4,2,1\right] \geq \left[3,3,1\right]-\left[3,2,2\right]$
can you understand what I wrote.
for example:$\left[4,1,0 ; 1^2,0,0\right]$ means $\sum a^4b(b-c)^2$
$\left[4,0,0 ; 1and2,0,0\right]$ means $\sum a^4 (b-c)(b^2-c^2)$
$\left[3,1,1;1^2,0,0\right]$ means $\sum a^3bc(b-c)^2$
I think we can also prove  $\left[4,3,0\right]-\left[4,2,1\right] \geq 4(\left[3,3,1\right]-\left[3,2,2\right])$
can somebody understand me???? :)  :(

I think the above holds but I a not sure . For example
\[ \left[4,3,0\right]+\left[3,2,2\right]\geq\left[4,2,1\right]+\left[3,3,1\right]  \] (1)  is true.

So I make a conjecture when there are two <<parenthesis>> then the following holds

\[ \left[k,l,l-3\right]+\left[l,l-1,l-1\right]\geq\left[k,l-1,l-2\right]+\left[l,l,l-2\right]  \]

where $k\geq 4$ and $l\geq 3$ and $k=l+1$
. If you put $k=4$  and $l=3$ you will have the   (1).

This will not be hold for all for all $k,l$ but it will hold for some of them.
please cheack it thanks

Silouan, I think may be SOS works on you problem.
and another question:
$ \left[3,2,2\right]$ means $ a^3b^2c^2 \plus{} a^2b^3c^2 \plus{} a^2b^2c^3$ or $ 2(a^3b^2c^2 \plus{} a^2b^3c^2 \plus{} a^2b^2c^3)$?
thanks  :)

[quote="zhaobin"]Silouan, I think may be SOS works on you problem.
and another question:
$ \left[3,2,2\right]$ means $ a^3b^2c^2 \plus{} a^2b^3c^2 \plus{} a^2b^2c^3$ or $ 2(a^3b^2c^2 \plus{} a^2b^3c^2 \plus{} a^2b^2c^3)$?[/quote]

should be the second one. When we consider a symmetric sum we run over all the permutations.

I have an idea. It doesn't bring us terribly far, but here it comes for what it's worth. It's about "interpolating" the AM-GM. Let me first illustrate it with an example:
By AM-GM we have $[8,0]+[4,4]\geq 2[6,2].$
As well $[8,0]+[6,2]\geq 2[7,1]$ and $[6,2]+[4,4]\geq 2[5,3]$.
Adding up all three yields
$[8,0]+[4,4]\geq [7,1]+[5,3]$. See what I mean?
To put it more generally, let's introduce some convenient notation: for partitions $\alpha=(a_1,\ldots,a_k)$ and $\beta=(b_1,\ldots,b_k)$, define the component-wise sum $\alpha\oplus\beta: =(a_1+b_1,\ldots,a_k+b_k)$, the difference $\alpha\ominus\beta$ likewise (note that both depend of the order of the components), and the multiplication with a scalar $\lambda\alpha: =(\lambda a_1,\ldots,\lambda a_k)$.
Then we can write the AM-GM ineq for symmetric functions as
$\frac12([\alpha]+[\beta])\geq[\frac12(\alpha\oplus\beta)]$ and more generally, weigted with $0\leq t\leq1$:
$t[\alpha]+(1-t)[\beta]\geq[t\alpha\oplus(1-t)\beta]$.
Now it’s easy to formulate the following, a sort of [b]concavity [/b]:
[b] For partitions $\alpha,\beta\vdash n$ and $0\leq t\leq1$ we have $[\alpha]+[\beta]\geq [(1-t)\alpha\oplus t\beta] +[t\alpha\oplus(1-t)\beta]$. (*)[/b]
To prove that, we first claim the following : for partitions $\gamma$ and $\delta$ (not necessarily with the same sum) and natural $k>0$ such that all parts of $\delta\ominus k\gamma$ are $\geq 0$, we have
$[\delta\oplus k\gamma]+[ \delta\ominus k\gamma]\geq [\delta\oplus (k-1)\gamma]+[ \delta\ominus (k-1)\gamma]$. This follows easily by induction from AM-GM.
By choosing $\gamma=\frac1q( \alpha\ominus\beta)$ and $\delta=\frac1q( \alpha\oplus\beta)$ with arbitrarily large integer $q$, we can see that (*) is valid for all rational  $t$, and by continuity for all $0\leq t\leq1$.

EDIT: the above proof reveals a bit what happens, but I just noticed that (*) follows directly from the ineq mentioned before $t[\alpha]+(1-t)[\beta]\geq[t\alpha\oplus(1-t)\beta]$. Just interchange  $\alpha$ and $\beta$ and add them both. :)

[quote="silouan"]I think the above holds but I a not sure . For example
\[ [4,3,0] + [3,2,2]\geq[4,2,1] + [3,3,1]
\]
(1)  is true.

So I make a conjecture when there are two <<parenthesis>> then the following holds
\[ [k,l,l - 3] + [l,l - 1,l - 1]\geq[k,l - 1,l - 2] + [l,l,l - 2]
\]
where $ k\geq 4$ and $ l\geq 3$ and $ k = l + 1$
. If you put $ k = 4$  and $ l = 3$ you will have the   (1).

This will not be hold for all for all $ k,l$ but it will hold for some of them.
please cheack it thanks[/quote]

 $ [k,l,l - 3] + [l,l - 1,l - 1]\geq[k,l - 1,l - 2] + [l,l,l - 2]$'
equal to
$ \sum_{cyc}(a^kb^{l - 3}c^{l - 3}(b + c) - a^l b^{l - 2}c^{l - 2})(b - c)^2 \ge 0$
we may assume $ a \ge b \ge c$
then we get $ S_a = a^kb^{l - 3}c^{l - 3}(b + c) - a^l b^{l - 2}c^{l - 2} \ge 0$
and $ S_b = b^k c^{l - 3} a^{l - 3}(c + a) - b^l c^{l - 2}a^{l - 2} \ge b^k c^{l - 3} a^{l - 3} a - b^l c^{l - 2}a^{l - 2} \ge 0$
and also
$ S_b + S_c = (abc)^{l - 3}(b^4(a + c) - b^3ca + c^4(a + b) - c^3ab) \ge 0$
is it right?

what is going on here? I think you are still working on the proof of $ \left[4,3,0\right] \plus{} \left[3,2,2\right]\geq\left[4,2,1\right] \plus{} \left[3,3,1\right]$.
Or I haven't understood Silouan's conjecture. Because if you just multiply everything by $ (abc)^{l \minus{} 3}$ that doesn't change very much. And if we take a case $ k > l \plus{} 1$, we would lose homogenity and I'm not sure if the ineq holds for very small $ a,b,c.$  :?

Silouan, may you please clarify what you mean with your conjecture, why did you introduce $ k$ if you request later $ k \equal{} l \plus{} 1$ and what do you mean by the two <<parenthesis>>?? :?:

I introduced $k$   just for emphasis .

BTW, with parenthesis I meant $[...]$.   That's all.

Thanks Silouan. But so it seems to me that yours is not really a generalisation, just multiplying with $[l-3,l-3,l-3]$. :huh:

To Zhaobin: You say,[quote="zhaobin"]
I think may be SOS works on you problem.
[/quote]
Please tell me, who or what is SOS?
And any comments about my AM-GM thing (#10)? Maybe after all, that's too simple to bring anything substantially new.

[quote="zhaobin"]
 $[k,l,l-3]+[l,l-1,l-1]\geq[k,l-1,l-2]+[l,l,l-2]$'
equal to
$\sum_{cyc}(a^kb^{l-3}c^{l-3}(b+c)-a^l b^{l-2}c^{l-2})(b-c)^2 \ge 0$
we may assume $a \ge b \ge c$
then we get $S_a= a^kb^{l-3}c^{l-3}(b+c)-a^l b^{l-2}c^{l-2} \ge 0$
and $S_b=b^k c^{l-3} a^{l-3}(c+a)-b^l c^{l-2}a^{l-2} \ge b^k c^{l-3} a^{l-3} a-b^l c^{l-2}a^{l-2} \ge 0$
and also
$S_b+S_c =(abc)^{l-3}(b^4(a+c)-b^3ca+c^4(a+b)-c^3ab) \ge 0$
is it right?[/quote]

Yes . Zhaobin you are right and my conjecture is true for any $l$ .

[quote="spanferkel"]
Please tell me, who or what is SOS?
[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=65385 :)

An idea/conjecture, in fact I am not very sure about some part of it as well.

Define $a_1,...,a_n \in \mathbb{R}^{+}$,
$a = \left[ {\begin{array}{*{20}c} {a_1 } \\ {a_2 } \\ {...} \\ {a_n } \\ \end{array}} \right]$
and $v_1,...,v_k \in \mathbb{N}^n$, where the components of $v_i$ are $v_{i_1},...,v_{i_n}$.
Now we define $f_a(v_i) = \sum_{sym} a_1^{v_{i_1}}a_2^{v_{i_2}}...a_n^{v_{i_n}}$

[b]Muirhead[/b]: If $v_i \succ v_j$, $f_a (v_i) \ge f_a (v_j)$
[b]Conjecture[/b]:
Define $\succ_2$ on two sets $\{ u_1,...,u_k \}$ and $\{ v_1,...,v_k \}$.
If $\{ u_1,...,u_k \} \succ_2 \{ v_1,...,v_k \}$, we mean that
1.  $u_1 \succ u_2 \succ \cdots \succ u_k$,
2.  $v_1 \succ v_2 \succ \cdots \succ v_k$,
3.  $u_1 \succ v_1$, $u_1 + u_2 \succ v_1 + v_2$,..., $u_1 + ... + u_{k-1} \succ v_1 + ... +v_{k-1}$
4.  $u_1 + ... + u_n = v_1 + ... + v_n$

Then, $\sum f_a (u_i) \ge \sum f_a (v_i)$

This looks very promising, and if we can prove it, it would be a magnificent Christmas present :). Even for $k=2$ this would yield nice results.
The most intriguing thing to me is the fact that this seems to work even if the components of the partitions are not in non-increasing order. If this holds in general, we could obtain general Schur by it: $[m+2,0,0]+[m,1,1]\ge[m+1,1,0]+ [m+1,0,1]$.

For the conditions 1 & 2, it may be even sufficient to require them to be in (descending) alphabetical order.

[quote="Soarer"] I'm not sure if this can be dropped.[/quote]But what about my remark? And how then would you generalize that to a sum of $ k$ terms on each side? Should it be "majorize" for $ k\minus{}1$ of them and "minorize" for only one, as in the Suranyi case? That doesn't make one expect a very strong result then. And in fact, Suranyi is (relatively) quite weak.

Anyway I don't think it's too ambitiuos to maintain the general conjecture, even without all those $ k$ first conditions. [hide="I made a little typo, and I cannot edit it any more  :mad: "] Just to be complete, I forgot a factor but it doesn't really matter: it should be
$ [a](x,1,1)\plus{}[a^{\prime}] (x,1,1) \equal{} 120\sum_{i}(x^{a_{i}}\plus{}x^{a_{i}^{\prime}})\ge$ $ 120\sum_{i}(x^{b_{i}}\plus{}x^{b_{i}^{\prime}}) \equal{} [ b ](x,1,1)\plus{}[b^{\prime}] (x,1,1).$ [/hide]

We can write the general of your conjecture for $ n$ variables and $ k$ terms also as follows:

[b] Let $ A$ and $ B$ be stochastical $ k\times n$ matrices, meaning that each row is a  [hide="partition of 1."] [/b] i.e. $ a_{ij}\ge0$ and $ \sum_{j \equal{} 1}^{n}a_{ij}\equal{} 1$ for all $ i \equal{} 1,\dots,k$.

[size=75][b]Or if preferred, just require that all lines have the same sum.[/b][/size][/hide]
[b] Denoting the rows by $ \alpha_{i}$ and $ \beta_{i}$, we have [/b]
\[ \boxed{\sum_{i\equal{}1}^{k}[\alpha_{i}]\ge\sum_{i\equal{}1}^{k}[\beta_{i}]\Longleftrightarrow (\alpha_{1},\dots,\alpha_{k})\succ (\beta_{1},\dots,\beta_{k})}\]
(the $ k$-tuples are meant to be taken as vectors with length $ k\cdot n$)

Note that we can say "iff" because the "$ \Rightarrow$" part is trivially true. :)

And finally this conjecture seems very natural to me...

which remark?

By the way, if you allow arbitrary length of vectors (the general conjecture in my understanding), it is not true. The reason is because say
$ a\succ b$ and $ c\succ d$ we want to claim that $ [a]\plus{}[d]\ge [b]\plus{}[c]$ if $ (a,d)\succ (b,c)$. We know that $ [c]\ge [d]$. Pick one set of numbers so that the inequality is straight, and keep on adjoining zeros to the vector c and d (assume the entries of a,b,c,d are all positive), then we can see that in the conjecture we are proposing, the "weight" of the sym sum in $ [a]\plus{}[d]$ is larger and larger, so that the difference of $ [c]$ and $ [d]$ can be magnified arbitrarily.

I mean my remark about it being true if (n-1) of the variables are equal. (This remains valid for all $ n$ and $ k$. Doesn't it?)

I don't really understand  what you mean with "difference of [c] and [d], weight, straight" ?? All vectors must have the same length, so how can you add zeros only to c and d?
Can you give an example to illustrate what you mean and which makes the inequality false?

Oh I'm sorry, I thought your "general conjecture" allows vectors of arbitrary length which is false.

[quote="spanferkel"]Note that we can say "iff" because the "$ \Rightarrow$" part is trivially true. :)
[/quote]
Oops, this is wrong. I started wondering what happens for problems of type "find the best $ k$" if this $ k$ is irrational, like
"Hungkhtn ineq1" http://www.mathlinks.ro/Forum/viewtopic.php?t=23108  or
http://www.mathlinks.ro/Forum/viewtopic.php?t=77722.
Majorization alone would not allow to find the "best $ k$" if it is irrational.

In cases like these, there is in fact no majorization among the two 'big' arrays, but the ineq is true nevertheless. E.g. in the last example above, which can be written as $ [4,2,0]\plus{}k[2,2,2]\ge[3,2,1]\plus{}k[3,3,0]$, the majorization only holds for $ k\le\frac12$, but the best is $ k\equal{}\frac{\sqrt{3}}2$. [hide] (extending the notion of majorization to non-integer rational vector 'lengths' in the obvious way, i.e. multiplying everything with a common denominator) [/hide]

So the conjecture will 'only' stand as

\[ \boxed{\sum_{i \equal{} 1}^{k}[\alpha_{i}]\ge\sum_{i \equal{} 1}^{k}[\beta_{i}]\ \text{  if  }\ (\alpha_{1},\dots,\alpha_{k})\succ (\beta_{1},\dots,\beta_{k})}
\]

:wallbash: The conjecture is false for $ n\ge4$: :wallbash_red:
The ineq $ [4100]\plus{}2\cdot[2210]\ge2\cdot [3200]\plus{}[2111]$ is false (see $ \vec {x}\equal{}(1,1,0,0)$ ), in spite of $ (42222111)\succ(33222111)$ :(
 [hide] it is not the first time I've seen something of this shape serve as a counterexample... [/hide]
I still tend to think that it holds for three variables though. :roll: But in view of the counterexample above, I really don't see how to generalize one of the proofs of Muirhead, as given e.g. here: http://www.math.ust.hk/excalibur/v11_n1.pdf

Can you please write your conjecture (the $ n\equal{}3$ case) with details in one post? It seems to be scattered through all the topic, and I don't have the time to read it all now. Thanks!

  Darij

[quote="spanferkel"]OK.
(sorry for the delay, I've been rather busy these days)

[b] Let $ k,m\in\mathbb N$ and $ \alpha_{i}$ and $ \beta_{i}$ ($ i \equal{} 1,...,k$) partitions of $ m$ with three parts ($ \ge0$) each, then the conjecture says[/b]
\[ \boxed{\sum_{i \equal{} 1}^{k}[\alpha_{i}]\ge\sum_{i \equal{} 1}^{k}[\beta_{i}]\ \text{ if }\ (\alpha_{1},\dots,\alpha_{k})\succ (\beta_{1},\dots,\beta_{k})}
\]
(the $ k$-tuples are meant to be taken as vectors with length $ 3k$)[/quote]

And no more conditions? I would rather try setting $ k\leq 2$ than limiting myself to $ 3$-number partitions, because a counterexample is easily constructed for $ k \equal{} 3$ with $ 3$-number partitions:

Let $ \alpha_1 \equal{} \left(2,7,6\right)$, $ \alpha_2 \equal{} \left(9,5,1\right)$, $ \alpha_3 \equal{} \left(4,3,8\right)$, $ \beta_1 \equal{} \left(2,9,4\right)$, $ \beta_2 \equal{} \left(7,5,3\right)$, $ \beta_3 \equal{} \left(6,1,8\right)$.

These $ \alpha_i$ and $ \beta_i$ satisfy both $ \left(\alpha_1,\alpha_2,\alpha_3\right)\succ\left(\beta_1,\beta_2,\beta_3\right)$ and $ \left(\beta_1,\beta_2,\beta_3\right)\succ\left(\alpha_1,\alpha_2,\alpha_3\right)$. Hence, the conjecture would yield both $ \left[\alpha_1\right] \plus{} \left[\alpha_2\right] \plus{} \left[\alpha_3\right]\geq\left[\beta_1\right] \plus{} \left[\beta_2\right] \plus{} \left[\beta_3\right]$ and $ \left[\beta_1\right] \plus{} \left[\beta_2\right] \plus{} \left[\beta_3\right]\geq\left[\alpha_1\right] \plus{} \left[\alpha_2\right] \plus{} \left[\alpha_3\right]$, so that $ \left[\beta_1\right] \plus{} \left[\beta_2\right] \plus{} \left[\beta_3\right] \equal{} \left[\alpha_1\right] \plus{} \left[\alpha_2\right] \plus{} \left[\alpha_3\right]$, what is absurd.

How did I construct the above counterexample? Well, take a look at http://en.wikipedia.org/wiki/Magic_square .

Sorry for crushing a dream... I am wondering what happens if $ k \equal{} 2$, but I am more urgently wondering about an exercise on homology I have to do for tomorrow, so I am leaving the neat computations to my dear fellow MathLinkers. ;)

  darij

Dream crushed. :blush:
What a great idea of yours ... I would never have thought that magic squares might intervene in a problem of inequalities...  :D

[quote="Darij Grinberg"] And no more conditions? [/quote]
After some inspection: Darij, of course you are right, otherwise there are counterexamples even for $ k\equal{}2$:
$ [4,1,1]\plus{}[3,3,0]\ge[4,2,0]\plus{}[3,2,1]$ is false despite $ (43311)\succ(43221)$ :)

We need a condition concerning the 'leading' partitions on each side. (and maybe that can rescue the conjecture/dream :roll: ). Assume $ \alpha_1$ and $ \beta_1$ respectively are maximal (in the lexicographical order $ \succ_{lex}$) among the partitions on either side. Then obviously a necessary condition is $ \alpha_1\succ_{lex}\beta_1$.
Let's add this condition!

[hide] (or rather: sort of re-introduce it :blush: - see Soarer's post Sep 20, 2007.)
BTW, it would be so much easier to refer to it as post#32. Why have those numbers disappeared? :mad:

As the lexicographical order $ \succ_{lex}$ gives a weaker condition than $ \succ$, this would also include ineqs like
$ [9,3,0] \plus{}[6,6,0]\ge[8,2,2]\plus{}[7,4,1]$ (easy to prove by AM-GM :D )
 [/hide]
So we can state version 2.0.1 of the conjecture as follows:

[b] Let $ k,m\in\mathbb N$ and $ \alpha_{i}$ and $ \beta_{i}$ ($ i \equal{} 1,...,k$) partitions of $ m$ with three parts ($ \ge0$) each, where $ \alpha_1\succ_{lex}\beta_1$ (with the $ \alpha_i$'s and $ \beta_i$'s numbered in decreasing lexicographical order [/b] and assuming of course $ \alpha_1\neq\beta_1$).
[b] Then [/b]
\[ \boxed{\sum_{i \equal{} 1}^{k}[\alpha_{i}]\ge\sum_{i \equal{} 1}^{k}[\beta_{i}]\ \text{ if }\ (\alpha_{1},\dots,\alpha_{k})\succ (\beta_{1},\dots,\beta_{k})}
\]
(the $ k$-tuples are meant to be taken as vectors with length $ 3k$)

Then in your example, it would still be feasible to conjecture
$ [9,5,1]\plus{}[8,4,3]\plus{}[7,6,2]\geq$
$ [9,4,2]\plus{}[8,6,1]\plus{}[7,5,3]$. 	:)

[quote="spanferkel"]After some inspection: Darij, of course you are right, otherwise there are counterexamples even for $ k \equal{} 2$:
$ [4,1,1] \plus{} [3,3,0]\ge[4,2,0] \plus{} [3,2,1]$ is false despite $ (43311)\succ(43221)$ :)[/quote]

Thanks, this has escaped my attention.

[quote="spanferkel"]We need a condition concerning the 'leading' partitions on each side. (and maybe that can rescue the conjecture/dream :roll: ). Assume $ \alpha_1$ and $ \beta_1$ respectively are maximal (in the lexicographical order $ \succ_{lex}$) among the partitions on either side. Then obviously a necessary condition is $ \alpha_1\succ_{lex}\beta_1$.
Let's add this condition![/quote]

Are you sure it is of help? Either I am misunderstanding you or the condition $ \alpha_1\succ_{lex}\beta_1$ is generally worthless, because it can always be circumvented by adding an "alibi" partition to both sides which lexicographically dominates all other partitions appearing in the inequality. But maybe this condition [i]together with[/i] $ k \equal{} 2$ can result in something correct...

  darij

:blush: I hate writing this sort of posts :wallbash: happens just too often recently.  :wallbash_red:

Here is a genuine counterexample, based on Darij's idea:
Suppose the conjecture is true, yielding $ [9,4,2] \plus{} [8,6,1] \plus{} [7,5,3] \leq [9,5,1] \plus{} [8,4,3] \plus{} [7,6,2]$.
Adding $ \epsilon[10,5,0]$ to the LHS and $ \epsilon[10,4,1]$ to the RHS should flip the sign for any $ \epsilon\ge0$, as all conditions are fulfilled. But this is of course impossible if $ \epsilon$ is small enough. (My computer tells me that it is false for $ \epsilon\le .099876$.)

EDIT: just seen Darij's post. Mine is not an "alibi partition"!... I had circumvented this by [quote="spanferkel"]
 and assuming of course $ \alpha_1\neq\beta_1$).

:)[/quote]

Oh I haven't noticed your $ \alpha_1\neq\beta_1$ condition, but obviously the difference of corresponding elements of $ \alpha_1$ and $ \beta_1$ can be made arbitrarily small, so that condition can't be of much use either...

  darij

Good point. In fact, the conjecture doesn't hold for $ k\ge 3$. :(
Take $ [7,1,0]\plus{}[5,2,1]\plus{}[4,4,0]\ge 2[6,2,0]\plus{}[4,3,1]$. (Try $ x_3\equal{}0$ to see why this is wrong) :yankchain:

But I don't give up yet :) A stronger condition may work. What about the following, I'll call it
[b] version 2.1 [/b] :

[b] Let $ k,m\in\mathbb N$ and $ \alpha_{i}$ and $ \beta_{i}$ ($ i \equal{} 1,...,k$) partitions of $ m$ with three parts ($ \ge0$) each,
satisfying

$ (i)\ \ \alpha _1\succ\cdots\succ \alpha_k$
$ (ii)\ (\alpha_{1},\dots,\alpha_{k})\succ (\beta_{1},\dots,\beta_{k})$
(the $ k$-tuples are meant to be taken as vectors with length $ 3k$)

Then [/b] $ \sum_{i \equal{} 1}^{k}[\alpha_{i}]\ge\sum_{i \equal{} 1}^{k}[\beta_{i}].$

Note there is no condition $ \beta_1\succ\cdots\succ \beta_k$. Playing around with the majorization, I have the impression that (i) and (ii) together are strong enough :)
[hide] (until someone posts the next counterexample :maybe: ) [/hide]

For $ k\equal{}2$, It seems to me that the previous stronger version (2.0.1) may be proven by examining a finite number of cases (100 at most, according to the orders in which the partitions make up the parts of the two big vectors), each of which should be either trivial from Muirhead or a straightforward SOS. Maybe even a computer could do that – not of much interest. But there must be a more elegant method.


I want to improve in recursion, are there any books, pdf files, documents, or websites that I can go to in order to attain proficiency in recursion?

Learn about Binet's formula.

Work with fibonacci variants,

and read through

[url=http://en.wikipedia.org/wiki/Recurrence_relation#Example:_Fibonacci_numbers]THIS[/url]
and
[url=http://mathcircle.berkeley.edu/BMC3/Bjorn1/node4.html]THIS[/url]

Those are two very informative sites.


trigonometry
geometry
3D geometry
algebra
polynomial
conics
analytic geometry

[quote="brianb"]Hey yall here are some problems to solve:

Please explain and work fully:
(These are to be worked with no calculator, and in a method that can be feasibly used in a short amount of time, at least 2-3 minutes, preferably in 60 seconds or less)

ANY that can be worked will be of great help

Most will be somewhat formula based.

1) Find $ (.99)^{10}$ to the first three decimal places.

2)Find $ x^2 \plus{} y^2 \plus{} z^2 \equal{} a^2$ and $ x \plus{} y \plus{} z \equal{} x^3 \plus{} y^3 \plus{} z^3 \equal{} a$, write an equation expressing $ xyz$ in terms of $ a$.

3)Quadrilateral ABCD has right angles at B and D. If ABCD is kite shaped with AB-AD=20 and BC-CD=15, find the length of a radius of the circle inscribed in ABCD.

4) What kind of graph is $ r \equal{} 1 \plus{} 2\sin\theta$

5) Let S be the infinite sum of $ \dfrac{1}{2} \plus{} \dfrac{1}{2} \plus{} \dfrac{3}{8} \plus{} \dfrac{1}{4} \plus{} \dfrac{5}{32} \plus{} \dfrac{1}{8}...$
Let K be the sum of the squares of the roots of $ x^5 \minus{} 3x^4 \plus{} 6x^3 \minus{} 3x^2 \plus{} 1 \equal{} 0$
Find the product of S and K.

6) Find the remainder when $ 1! \plus{} 3! \plus{} 5!... \plus{} 49!$ is divided by $ 18$.

7)What is the $ 7th$ pentagonal number?

8) ABE is a right triangle with right angle E. C is a point on AE. D is a point on AB. CD forms a perpendicular line to AE. $ AC \equal{} 8, AD \minus{} 10, CD \minus{} 6,$ and  $ < BCD \equal{} 30 degrees.$ If BD= $ k(4\sqrt {3} \plus{} 3)$, find k.

9) The graph of the polar equation $ \cos\theta \plus{} r \equal{} \frac {1}{r}$ is a:

10)Evaluate: $ 2\arctan\frac {1}{3} \plus{} \arctan\frac {1}{7}$

11) Find the area of the triangle with vertices: (1,-3,-2) (5, -1, 2) (-1,1,2)

12) Given $ f(x) \equal{} x^5 \minus{} 5x^4 \plus{} 13x^3 \plus{} 121x^2 \minus{} 25x \plus{} 69$, find the sum of the cubes of its roots.

13) Tow poles are 50 feet apart on level ground. One pole is 10 feet taller than the other. A cable hangs form the top of the poles in the shape of a parabola. It's lowest point is 10 feet above a point on the ground 20 feet from the shorter pole. Find the height of the taller pole.

14) The phase shift of $ y \equal{} 5\sin3x \minus{} 12\cos3x$ is:

15)Solve for $ \theta$ where $ 0 < \theta < 90: \frac {\cos^2\theta \minus{} \sin^2\theta}{2\cos^2\theta\tan\theta} \equal{} 2 \minus{} \sqrt3$

16) $ (\sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt {1 \plus{} ...}}})(\sqrt {(7 \minus{} 3\sqrt5) \plus{} \sqrt {(7 \minus{} 3\sqrt5) \plus{} \sqrt {(7 \minus{} 3\sqrt5) \plus{} ...}}}) \equal{}$

17)$ \sum_{x \equal{} 2}^{50}\frac {3}{(x \minus{} 1)(x \plus{} 1)}$

18)How many values of x satisfy the equation: $ \log(x^2) \equal{} cos(x)$

19) Find the remainder when $ 19^{2001}$ is divided by 23.

Also, how would one go about simplifying something like : $ \sqrt {\sqrt2 \plus{} \sqrt5}$ or something like that to a single radical?

As soon as you get one, go ahead and post the solution.[/quote]

1: Taylor series (or binmial theorem)

2: Try to express the equations in term of the elementary symmetric polynomials $ \sigma_1\equal{}x\plus{}y\plus{}z,\sigma_2\equal{}xy\plus{}xz\plus{}zx,\sigma_3\equal{}xyz.$

6: From $ 6!$ on, every one of the term is divided by 18.

9: Conic

10:Use $ \tan(x\plus{}y)\equal{}\frac{\tan x \plus{} \tan y}{1\minus{}\tan x\tan y}.$

11: Shoelace.

12:Again,use elementary symmetric polynomials.

13: Put the system into a coordinate system with the lowest point in the origin.

14: Divide by $ \sqrt{5^2\plus{}12^2}.$

15: Show that the LHS equals $ \cot(2\theta).$

16: Try to find the limit of each root by solvin an equationlike $ x \equal{} \sqrt{1\plus{}x}$ etc.

17: Use $ \frac{2}{(x\plus{}1)(x\minus{}1)} \equal{} \frac{1}{x\minus{}1}\minus{}\frac{1}{x\plus{}1}.$

19: Euler-Fermat (or Totient thorem).

[quote="brianb"]Hey yall here are some problems to solve:

Please explain and work fully:
(These are to be worked with no calculator, and in a method that can be feasibly used in a short amount of time, at least 2-3 minutes, preferably in 60 seconds or less)

ANY that can be worked will be of great help

Most will be somewhat formula based.

1) Find $ (.99)^{10}$ to the first three decimal places.

2)Find $ x^2 \plus{} y^2 \plus{} z^2 \equal{} a^2$ and $ x \plus{} y \plus{} z \equal{} x^3 \plus{} y^3 \plus{} z^3 \equal{} a$, write an equation expressing $ xyz$ in terms of $ a$.

3)Quadrilateral ABCD has right angles at B and D. If ABCD is kite shaped with AB-AD=20 and BC-CD=15, find the length of a radius of the circle inscribed in ABCD.

4) What kind of graph is $ r \equal{} 1 \plus{} 2\sin\theta$

5) Let S be the infinite sum of $ \dfrac{1}{2} \plus{} \dfrac{1}{2} \plus{} \dfrac{3}{8} \plus{} \dfrac{1}{4} \plus{} \dfrac{5}{32} \plus{} \dfrac{1}{8}...$
Let K be the sum of the squares of the roots of $ x^5 \minus{} 3x^4 \plus{} 6x^3 \minus{} 3x^2 \plus{} 1 \equal{} 0$
Find the product of S and K.

6) Find the remainder when $ 1! \plus{} 3! \plus{} 5!... \plus{} 49!$ is divided by $ 18$.

7)What is the $ 7th$ pentagonal number?

8) ABE is a right triangle with right angle E. C is a point on AE. D is a point on AB. CD forms a perpendicular line to AE. $ AC \equal{} 8, AD \minus{} 10, CD \minus{} 6,$ and  $ < BCD \equal{} 30 degrees.$ If BD= $ k(4\sqrt {3} \plus{} 3)$, find k.

9) The graph of the polar equation $ \cos\theta \plus{} r \equal{} \frac {1}{r}$ is a:

10)Evaluate: $ 2\arctan\frac {1}{3} \plus{} \arctan\frac {1}{7}$

11) Find the area of the triangle with vertices: (1,-3,-2) (5, -1, 2) (-1,1,2)

12) Given $ f(x) \equal{} x^5 \minus{} 5x^4 \plus{} 13x^3 \plus{} 121x^2 \minus{} 25x \plus{} 69$, find the sum of the cubes of its roots.

13) Tow poles are 50 feet apart on level ground. One pole is 10 feet taller than the other. A cable hangs form the top of the poles in the shape of a parabola. It's lowest point is 10 feet above a point on the ground 20 feet from the shorter pole. Find the height of the taller pole.

14) The phase shift of $ y \equal{} 5\sin3x \minus{} 12\cos3x$ is:

15)Solve for $ \theta$ where $ 0 < \theta < 90: \frac {\cos^2\theta \minus{} \sin^2\theta}{2\cos^2\theta\tan\theta} \equal{} 2 \minus{} \sqrt3$

16) $ (\sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt {1 \plus{} ...}}})(\sqrt {(7 \minus{} 3\sqrt5) \plus{} \sqrt {(7 \minus{} 3\sqrt5) \plus{} \sqrt {(7 \minus{} 3\sqrt5) \plus{} ...}}}) \equal{}$

17)$ \sum_{x \equal{} 2}^{50}\frac {3}{(x \minus{} 1)(x \plus{} 1)}$

18)How many values of x satisfy the equation: $ \log(x^2) \equal{} cos(x)$

19) Find the remainder when $ 19^{2001}$ is divided by 23.

Also, how would one go about simplifying something like : $ \sqrt {\sqrt2 \plus{} \sqrt5}$ or something like that to a single radical?

As soon as you get one, go ahead and post the solution.[/quote]
for 5, is the sixth term really 1/8?
If it is 3/32 then,
 S $ \equal{}\dfrac{1}{2} \plus{} \dfrac{2}{4} \plus{} \dfrac{3}{8} \plus{} \dfrac{4}{16} \plus{} \dfrac{5}{32} \plus{} \dfrac{6}{64}...\equal{}2$
 K$ \equal{}3^2 \minus{}2*6\equal{}\minus{}3$
 the product is -6


If the digit $1$ is placed after a two digit number whose tens' digit is $t$, and units' digit is $u$, the new number is:

$\textbf{(A)}\ 10t+u+1 \qquad
\textbf{(B)}\ 100t+10u+1 \qquad
\textbf{(C)}\ 100t+10u+1\qquad
\textbf{(D)}\ t+u+1 \qquad
\textbf{(E)}\ \text{None of these answers}$

[hide="solution"]They have a 2 digit number already $ 10t \plus{} u$, but if we put a 1 after the you are going to have to multiply the $ 10t \plus{} u$, to keep them in their position, so the answer is:
$ 100t \plus{} 10u \plus{} 1$

B.
[/hide]

[hide="Solution"]

The new number looks like this:

$ \underline{t} \ \underline{u} \ \underline{1}$

This is same as $ 10^2 t \plus{} 10 u \plus{} 1$ or B. [/hide]

sorry to revive this really old thread, but aren't answers (b) and (c) exactly the same?

C is supposed to be $1000t+10u+1$: [[1950 AHSME Problems/Problem 7]]


The cost of holding a concert is the sum of the fixed cost, which is the same no matter how many people attend, and the variable cost, which depends on the number of people attending. If the total cost for a concert attended by 1000 people is $ \$75,000$ and the total cost of a concert attended by 1200 people is $ \$85,000$, what is the number of dollars in the fixed cost when holding a concert?

Let $ c$ be the fixed cost and $ p$ be the cost per person to attend.

$ c\plus{}1000p\equal{}75000$
$ c\plus{}1200p\equal{}85000$

$ 200p\equal{}10000$
$ p\equal{}50$

$ c\equal{}75000\minus{}50000\equal{}\boxed{25000}$ dollars.


ABCDEFGHIJKL is a regular dodecagon with side length 1.

1) Find the ratio of the area of eq. triangle AEI to that of the square ADGJ.

2) Find the area of the intersection of AEI and ADGJ.

1
let the center of dodecagon be $O$, $S_1$ be the are of circle and $S_2$ be the area of equilateral triangle.
then $OD=\frac{0.5}{\cos{\frac{11\pi}{24}}}=3.83\Rightarrow JD=2OD=7.66\Rightarrow JG=\frac{7.66}{\sqrt{2}}=5.41\Rightarrow S_1=29.337$
$\angle AOE=\frac{\pi}{3}\Rightarrow  \triangle AOE$ is equilateral
$S_2=\frac{\sqrt{3}}{4}(\frac{0.5}{\cos{\frac{11\pi}{24}}})^2=6.35$
now you can find $\frac{S_2}{S_1}$
my values are estimations but you can use actual values...

For #1, you might arrive at a more exact answer if you ignore the dodecagon's side length...
Anyone try #2?


superior algebra
superior algebra solved

need a little direction on this.  Let $H$ be a group of order 36. and H has element $a \in  H$ such that $a^{12} \neq e$ and $a^{18} \neq e$, show $H$ is cyclic. :)

  can i assume that $a \neq e$ and $\left|a\right|=36$ ?

Yes you can:
All possible orders for $a$ are the divisors of $36$, but all this numbers excepting $36$ itself are divisors of $12$ or $18$, and so $a$ cannot have their orders (because it would give $a^{12}=e$ or $a^{18}=e$), thus it must be $36$.


I believe it should be greater than or equal to (consider $n = 1$).
[hide]
$\frac{1}{2} = \sum_{i=1}^{n}\frac{1}{2n}$
do gaussian pairing and we see that $\frac{1}{2n} = \frac{1}{2n}$ and the rest are definitely greater because the smaller the denominator, the greater the number.
[/hide]

I think this can also be proven by induction.


function
real analysis
real analysis unsolved

y = f(f(f(x))); f(x) = 1/(1-x)...so at how many points is y not continuous?

Edit:

Source: http://answers.yahoo.com/question/index;_ylt=AvzaNbUWuWm1NK0876EulXDsy6IX;_ylv=3?qid=20080529204740AA68xLL&show=7#profile-info-ycee0IRqaa

This statement isn't solely directed at you, but I always wonder why so few posters provide background with their problems.  It only takes a few words to say whether you came up with a problem yourself, found it in a book, or had it on a test, and if you say what you've tried, what didn't work, what you got stuck on, etc., then it becomes an interesting mathematical story that everyone can participate in, a teaching moment rather than just a boring abstract exercise.

In any case, you can easily compute that $ y(x)\equal{}x$, which is continuous everywhere.  However, it's an odd question, since $ f(1)$ and $ f(f(0))$ do not exist, so $ y(0)$ and $ y(1)$ do not exist either.  So you need to have a notion of continuity at a point where a function is not necessarily defined.  I would need more information to understand the intention of the problem.

In general, when you compose continuous functions, you get a continuous function back.

Or, you can do it this way. You know $ f$ is discontinuous at $ x\equal{}1$ and therefor $ y$ is discontinuous  when $ f(f(x))\equal{}1$ and from here you get that there is no such $ x$.

But what do you mean that $ f$ is discontinuous at $ x\equal{}1$?  $ f$ does not even exist there.  The usual definition of continuity at a point does not make sense without the assumption that the function is defined at that point.

Hm... I may we wrong, but it makes sense if you look at it this way - it's not continuous! I mean, the condition for a function to be continuous at a point is to be defined in one, and if it's not continuous then it is discontinuous. So, not being defined at a point makes in not continuous - discontinuous... Hope I'm making some sense...

Well, if it's not defined, it's not a function at all!  You have to restrict yourself to the domain.


Determine all positive integer that are expressible in the form

$a^{2}+b^{2}+c^{2}+c$

where $a,b,c \in Z$

a) posted before.
b) bad title ( http://www.mathlinks.ro/Forum/viewtopic.php?t=135914 ).


$a,b,c$are positive integers.If $\prod_{cyc}(1+ab)$is a perfect square,show that $(1+ab),(1+bc),(1+ca)$are all perfect squares

Posted many times before...
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=4542


Find all prime number $ p$ satisfy: exit $ x,y\in N: p\equal{}x^2\plus{}y^2$

[quote="nktp"]Find all prime number $ p$ satisfy: exit $ x,y\in N: p \equal{} x^2 \plus{} y^2$[/quote]

Maybe $ 4n\plus{}1$??

Yes it is $ 4n\plus{}1$ but I can't find a nice solution!

Find all n can represent in the form $ x^2\plus{}2y^2$

If question is $ n\equal{}kx^2\plus{}ly^2$ where $ k,l$:constant then what is answer?

It is hard problem, considered by Gauss. Nesessary condition $ (k,l)\equal{}1, \ \binom{ \minus{} kl}p \equal{} 1$.

[quote="Rust"]It is hard problem, considered by Gauss. Nesessary condition $ (k,l) \equal{} 1, \ \binom{ \minus{} kl}p \equal{} 1$.[/quote]
Yes it is an open question
Find all $ n$ can represent in the form $ x^2\plus{}py^2$ where p is a prime.


Actually, I made this problem, but until now I could not find the answer.  :blush:

Let $ n$ be a positive integer.
For $ x_1 < x_2 < x_3 < \ldots < x_n$ and $ x_1,x_2,x_3,\ldots,x_n\in\Re^\plus{}$ ,prove that
$ \sum_{i \equal{} 1}^{n}\sqrt {\frac {x_i^2 \plus{} 1}{\prod_{j \equal{} 1}^{i}x_j}}\ > \ \sum_{k \equal{} \minus{} 1}^{n \minus{} 2}\sqrt {\frac {1}{n^k}}$

Are you sure this is true?  If you let all of the $ x_i$ be very close to some real number $ x$, then the left side equals

$ \sqrt{x^2\plus{}1}\sum\limits_{k\equal{}1}^{n}\sqrt{\frac{1}{x^k}}$

As $ n$ gets very large, this becomes very close to
$ \sqrt{x^2\plus{}1}\frac{\sqrt{x}}{\sqrt{x}\minus{}1}$
which is very close to a linear expression in $ x$ as $ x$ gets large.  But the expression on the RHS is linear in $ n$ as $ n$ gets large.  So let $ n$ be significantly larger than $ x$...


geometry
circumcircle
trigonometry
trig identities
Law of Cosines

Let $\Delta ABC$ an equilateral triangle and $P$ a point on the circumscribed circle of the triangle. If the circle radius is $r=1$ show that $PA^{2}+PB^{2}+PC^{2}=6$.

Since $P$ is on the circumcircle, say on minor arc AB, then $PACB$ is a cyclic quadrilateral.  Applying Ptolemy's theorem we have
\[PC=PA+PB\]
Also because $PACB$ is cyclic opposite angles are supplementary implying $\angle{APB}=\frac{2\pi}{3}$.  From special right triangles or otherwise we know $AB=\sqrt{3}$.  Using the law of cosines on triangle APB we have
\[3=PA^2+PAPB+PB^2\]
which implies
\[6=2PA^2+2PAPB+2PB^2=PA^2+PB^2+(PA+PB)^2=PA^2+PB^2+PC^2\]
as desired.


algebra
polynomial
Ring Theory
calculus
integration
function
domain

Find if there exists $R$ and $S$ two rings so that they are not isomorph but the polynomial rings $R[X]$ and $S[X]$ are isomorph.

I'm very curious about how these rings look like :).

That's interesting. I'll think about that (unless your rings are not unitary and commutative :P)

First some facts:

1. $|R[X]|=\left\{\begin{array}{c}|R| \text{ for } R \text{ infinite } \\ \aleph_0 \text{ for } R \text{ finite }\end{array}\right.$

2. $R[X]$ is a PID if and only if $R$ is a field. So, $R[X] \cong S[X]$ implies, that $R,S$ are both fields or both no fields.

3. If $R$ is an integral domain, then $R[X]^* = R^*$. In particular, $R[X] \cong S[X]$ implies $R^* \cong S^*$ when $R,S$ are integral domains.

4. $R$ is an integral domain if and only if $R[X]$ is an integral domain.

Now, if $R$ or $S$ is a field and $f : R[X] \to S[X]$ is an isomorphism, then $R$ and $S$ are fields by 2.. Restriction of $f$ yields an isomorphism $R^* \to S^*$. But we have also $f(0)=0$, so that $f$ yields a bijection $R \to S$, which is of course a homomorphism. So, $R$ and $S$ are isomorphic.

So, the answer is No, if $R$ or $S$ is a field.

I've posted here a lot of rubbish ... :blush: this remains:

Let $R,S$ be arbitrary rings s.t. $R[X] \cong S[X]$. Take a preimage $f \in R[X]$ of $X \in S[X]$.  Then we get
\[ R/(R \cap (f)) \hookrightarrow R[x]/(f) \cong S[X]/(X) \cong S  \]
Analogously, if $g \in S[X]$ is the image of $X \in R[X]$, then
\[ S/(S \cap (g)) \hookrightarrow S[x]/(g) \cong R[X]/(X) \cong R  \]

....

Since I didn't make any progress, I've started googling ;). The implication $R[X] \cong S[X] \Rightarrow R \cong S$ and the corresponding one for power series rings was intensively investigated in the seventies. Among other things, sufficient conditions for $R$ were found, and the first counterexample was found by Hochster in

M. Hochster, Non-uniqueness of coefficient rings in a polynomial rings, Proc. Amer. Math. Soc, 34 (1972), 81-82.

Unfortunately, I have not found yet a more detailed information as

[i]In his counterexample, he takes the coordinate ring A of the tangent bundle over the 2-sphere which is a finitely generated R-algebra and uses the geometric fact that the tangent bundle is stably trivial, but not trivial.[/i]

Besides, in another paper:

[i]there exists a Notherian ring $R$ with zero Jacobson radical which has a finitely generated nonfree module $P$ such that $P \oplus R \cong R^3$. Taking symmetric algebra of both sides yields $A[T]$ isomorphic to $R[X,Y,Z]$ where $T,\{X,Y,Z\}$ are indeterminantes over $A$ and $R$ respectively and $A$ is the symmetric algebra of $P$, and is not isomorphic to $R[X,Y]$.[/i]

You can find the example here:

http://www.math.lsa.umich.edu/~hochster/614/614.html

problem 6 in problem set #7


Math-M-Addicts is a math class held in Manhattan on Saturday mornings from 10 am to noon.  It is taught by four former IMO medalists.  The program is for students from 7th to 12th grade.

For more details, you can go to the [url=http://www.pushkinacademy.com/pamm]Math-M-Addicts web site[/url].  You can also download the attached document.

To apply for the program, you need to take a qualifying test, which is currently posted on the web site above.  Solutions are due Wednesday, October 4, 2006.

Please pass on the info to any interested students in the NYC area.

We at Math-M-Addicts received surprisingly few responses, and many students were confused about the program, so we decided to post on this forum and help answer some of the questions.

[u]Question 1[/u]: I think I am pretty good at math, but I am having problems with the test – why is that?
[u]Answer 1[/u]: The test was supposed to be hard, with the average score ranging between 1.5 and 2 questions.  The goal of the test was twofold – first, to decide on the scholarships and second, to give students a tangible measure of their progress.  However, it ended up intimidating people more than anything else.  Therefore, [b]we are abandoning the test[/b], and we hope to see all of you when Math-M-Addicts begins its second season.

[u]Question 2[/u]: How good should my math skills be in order to join the program?
[u]Answer 2[/u]: As we are planning to have two groups this year, we will be able to entertain a fairly broad range of skill sets; the only requirement is that one should be a math aficionado.  The fact that Math-M-Addicts is taught by former math Olympiad winners should not deter one from joining – all it means is that hopefully we are good at distilling information and have more fun in the process.

[u]Question 3[/u]: Math-M-Addicts is being done in collaboration with Pushkin Academy of Russian Heritage.  Should I be able to speak Russian or have Russian heritage in order to attend?
[u]Answer 3[/u]: Absolutely not!  While we are happy to collaborate with Pushkin Academy, our students are of various ethnic backgrounds.

Math-M-Addicts will start later than originally advertised, on November 4th.  For more information or with questions, please Email me at breydo@gmail.com.

Thanks,
Ruvim

Reminder: classes begin the Saturday.

Ruvim


Hello,

can someone give me some tips to solve this

0 = v * 0.17 + 1.232 - 240.1 / v^2

Does the 2nd degree formula apply to this (ax^2+bx+c=0) ? If so Howto handle the case where we have "a/x" and not ax?

Thanks.

You need to take the varible out of the denominator by multiplying both sides by it.
Also get rid of the decimal.

You'll get $170v^3+1234v^2+240100=0$

Just a suggestion: To type like above, learn LaTex through this forum. You'll see a link to it at the side of this page.


I was watching the movie Paycheck, and one of the scenes shows a computer screen saying "Error 41" followed by:

0x3172 0x1F21 0xF000 0xC132 0x5412
0x1357 0x94A1 0xFA77 0x1234 0x1A11

OK, so theres about a 0.0001% chance of this actually having any meaning at all but a) whoever wrote that must know at least a tiny bit of computing, as they wrote 0x, and b) if I was asked to give some error message for a computer in a movie that looked like a proper error message, I would make it a code :)

So if anyones bored one day..

You can start [url=http://www.monkeyfighter.com/code/tetris/tetris.h]here[/url]  ;)

[quote="Treething"]*dies*

What does that mean then :P ?[/quote]That is the question! :)

The person [url=http://www.drod.net/forum/viewtopic.php?TopicID=4672]here[/url] also has the same problem  :D

1. The movie paycheck was a very good action flick but the author didn't impress me as someone who knew much about science. I could be wrong, but if one looks around the curve of space back at one's self, he/she would be looking very far in the past; not the future. Thus, I presume the author found some fancy looking computer stuff like that on MonkeyFighter.com and added it to the movie because it looked impressive. In that case, the code mean nothing - or, anything you want, since false implies anything.

2. A common practice in computer debugging is to dump the stack when an error occurs.  This gives the programmer a trail of bread crumbs to follow to the error.  Modern computing systems don't usually print the hex dump. Java, for example, gives an extremely verbose stack dump when an uncaught exception occurs. Still, it is believable that the time telescope would dump hex - assuming it was coded as an embedded system.

3. If you want to know about that hex code: http://www.monkeyfighter.com/tetris.html


A man deposits 2000 dollars on June 1st each year for 20 years. If interest at 8% is compounded annually, what is the value of his account after 20 years?

Sorry if this is a basic problem, I just want to know how this problem can be solved using the formula $ S_{n} \equal{} \frac{x_{1} ( r^{n} \minus{} 1)}{r\minus{}1}$.

[hide="Discussion"] After his initial deposit, he has $ 2000$ dollars.

After his second deposit, he has $ 2000 \cdot 1.08 \plus{} 2000$ dollars (the first term is his money from the year before and the second term is his new deposit).

After his third deposit, he has $ (2000 \cdot 1.08 \plus{} 2000) \cdot 1.08 \plus{} 2000 \equal{} 2000 \cdot 1.08^2 \plus{} 2000 \cdot 1.08 \plus{} 2000$ dollars (same).

Hence after his $ 20^{th}$ deposit he will have

$ 2000 \cdot 1.08^{19} \plus{} 2000 \cdot 1.08^{18} \plus{} ... \plus{} 2000 \cdot 1.08 \plus{} 2000$

dollars.  Can you use the geometric series summation formula now?  :) [/hide]

A man pays off a \$30000 bet, interest and principal, in 10 equal installments made on Dec. 1st for 10 years. Find the amount of his annual payment if interest is charged at 12% per annum compounded annually.

I tried

$ 30000 \equal{} \frac {x( 1.12^{10} \minus{} 1)}{1.12 \minus{} 1}$

then solve for $ x$ but the answer appears to be wrong. I've tried small variations of this approach but no luck. Where's the screw-up?  :oops:

Also, I don't get this problem at all:

A fund is set up to provide two annual bursaries of \$1000 each for the next 20 years starting one year from now. How much money should be placed in the fund if interest is 10% per annum compounded annually?

Interest on a debt is not the same thing as interest on an investment.  In your calculation, the man accumulates wealth based on how much he has paid; what is actually happening is that his debt is [b]increasing.[/b]

I'm still stuck. I tried

$ 10x \equal{} 30000(1.12)^{9} \minus{} \Big( (1.12)^{9}x \plus{} (1.12)^{8}x \plus{} (1.12)^{7}x \plus{} \cdots \plus{} 1.12x \plus{} x \Big)$

$ 10x \equal{} 30000(1.12)^{9} \minus{} \frac {x (1.12^{9} \minus{} 1)}{1.12 \minus{} 1}$

But it appears to be wrong. And I'm still clueless on the bursary problem  :(

It looks right.  Why do you think it's wrong?

As for the second problem, you need to start with a principal $ P$ such that the $ 10\%$ interest combined with withdrawals of $ \$2000$ each year drains the fund exactly after $ 20$ years.

Well the way I did it I got an answer of \$3357 but using an algorithm I made on my calculator, I got an answer of about \$4510. I know the algorithm's right, it simply starts with $ 30000 \minus{} x$ where $ x$ is a guess of the annual payment. $ 30000\minus{}x$ is then multiplied by 1.12 then we subtract x again and this process is repeated 9 times. The goal is to input a guess for $ x$ such that the algorithm gives us a value close to 0, meaning we selected an amount that just over pays or under pays the dept by an infinitesimal amount. Seeing as 3357 is nowhere near the projected answer, I'm obviously doing something wrong in my calculation of the answer. Perhaps someone can give the problem a try and post their solution? I'd be extremely grateful  :D

Oops, I've been silly.  The issue is you're already including the $ 10$ payments in the subtraction on the right; the actual equation is just

$ 0 \equal{} 30000(1.2)^9 \minus{} x \cdot \frac {1.12^{10} \minus{} 1}{1.12 \minus{} 1}$

Edit: Okay I finally got it:

$ 0 \equal{} 30000(1.12)^{9} \minus{} \frac { x(1.12^{10} \minus{} 1)}{1.12 \minus{} 1}$

A subtlety is that you must raise the exponent by 1 in a geometric sum, which is why the exponent 10 is used. The final answer is about 4740 dollars.

This may be a year late, but since I just stumbled on it, what the heck.
[quote]A man pays off a \$30000 bet, interest and principal, in 10 equal installments made on Dec. 1st for 10 years. Find the amount of his annual payment if interest is charged at 12% per annum compounded annually.
[/quote]
If a man just incurred a \$30,000 debt from a bet, keeping in mind that I’m viewing this situation within the perspective of grim humor and worst scene scenario, I’m betting he won’t live long after making an arrangement with his mob bookie like paying it off in 10 equal installments over a period of 10 years.  If his bookie allows him to live, I’ll bet most of his worldly possessions, like his house and car, will be seized by his kind and benevolent bookie.  If his bookie agrees to his 10 year arrangement, it’s more than likely that he’ll end up paying something like a 100 times more than he owes.

Seriously though:

If the Dec. 1st your problem is referring to happens to be a year from now, then one way of solving your problem would be by the following procedure (the standard procedure I’m used to):
\[ \begin{gathered} 30,000 \equal{} x\left[ {1.12^{ \minus{} 1} \plus{} 1.12^{ \minus{} 2} \plus{} \cdot \cdot \cdot \plus{} 1.12^{ \minus{} 9} \plus{} 1.12^{ \minus{} 10} } \right] \hfill \\
\Leftrightarrow \hfill \\
30,000 \equal{} x\left[ {1.12^{ \minus{} 10} \plus{} 1.12^{ \minus{} 9} \plus{} \cdot \cdot \cdot \plus{} 1.12^{ \minus{} 2} \plus{} 1.12^{ \minus{} 1} } \right] \hfill \\
\Leftrightarrow \hfill \\
30,000 \equal{} \frac {{x[({\text{common ratio)(last term)}} \minus{} ({\text{first term)}}]}} {{({\text{common ratio)}} \minus{} 1}} \hfill \\
\Leftrightarrow \hfill \\
30,000 \equal{} \frac {{x[(1.12)(1.12)^{ \minus{} 1} \minus{} 1.12^{ \minus{} 10} ]}} {{1.12 \minus{} 1}} \hfill \\
\Leftrightarrow \hfill \\
30,000 \equal{} \frac {{x(1 \minus{} 1.12^{ \minus{} 10} )}} {{1.12 \minus{} 1}} \hfill \\
\Leftrightarrow \hfill \\
x \approx 5,309.524925 \hfill \\
\end{gathered}
\]
If the Dec. 1st your problem is referring to happens to be now, then another way of expressing your
\[ 0 \equal{} 30,000(1.12)^9 \minus{} \frac {{x(1.12^{10} \minus{} 1)}} {{1.12 \minus{} 1}} \Leftrightarrow x \approx 4,740.647254
\]
would be by the following procedure (again, the standard procedure I’m used to):
\[ \begin{gathered} 30,000 \equal{} x \plus{} x\left[ {1.12^{ \minus{} 1} \plus{} 1.12^{ \minus{} 2} \plus{} \cdot \cdot \cdot \plus{} 1.12^{ \minus{} 8} \plus{} 1.12^{ \minus{} 9} } \right] \hfill \\
\Leftrightarrow \hfill \\
30,000 \equal{} x \plus{} x\left[ {1.12^{ \minus{} 9} \plus{} 1.12^{ \minus{} 8} \plus{} \cdot \cdot \cdot \plus{} 1.12^{ \minus{} 2} \plus{} 1.12^{ \minus{} 1} } \right] \hfill \\
\Leftrightarrow \hfill \\
30,000 \equal{} x \plus{} \frac {{x[(1.12)(1.12)^{ \minus{} 1} \minus{} 1.12^{ \minus{} 9} ]}} {{1.12 \minus{} 1}} \hfill \\
\Leftrightarrow \hfill \\
30,000 \equal{} x \plus{} \frac {{x(1 \minus{} 1.12^{ \minus{} 9} )}} {{1.12 \minus{} 1}} \hfill \\
\Leftrightarrow \hfill \\
x \approx 4,740.647254 \hfill \\
\end{gathered}
\]
[quote]A fund is set up to provide two annual bursaries of \$1000 each for the next 20 years starting one year from now. How much money should be placed in the fund if interest is 10% per annum compounded annually?
[/quote]
My approach would be
\[ y \equal{} \frac {{(2*1,000)(1 \minus{} 1.10^{ \minus{} 20} )}} {{1.10 \minus{} 1}} \approx 17,027.12744
\]
I just can't get over how kind the man's bookie must be to agree to a 10 year plan  arrangement...

[url=http://www.mathisfunforum.com/viewtopic.php?id=10489]Serial Bonds[/url] anyone?


search
number theory unsolved
number theory

Solve in natural numbers $ x^y\plus{}y^z\equal{}z^x$

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Some of you may have seen this one before. It's a rather old problem, I think from Archimedes.

[img]http://www.princeton.edu/~mathclub/images/circles.gif[/img]

In the figure above the radius of the large half circle is 3, of the second largest is 2, and of the smallest is 1. Index the nested family of tangential circles counterclockwise by n = 1, 2,... Find a formula for radius of the nth nested circle.

Yes, this is one of the classic problems of inversion. One of the submitted designs for our ARML T-shirt in 2002 was about inversion and had this problem on it. I voted for it, but it didn't win.

Yes. This is a problem that lends itself very well to inversion. It can also be solved without using inversion (as I did when I first saw it in high school), but with a lot more difficulty.

Archimedes first studeid the figure with just two semicicles inscribed in a larger one (called an Arbelos, or "shoemakers knife"), and he may have also studied the figure with the chain of circles inscribed. But the first person to write down a solution was Pappus, and the figure is usually called a Pappus Chain. I don't know how Pappus solved the problem but it must have been without inversion, because that method wasn't discovered until the 19th century by Jacob Steiner.

Here are some interesting pages about it and how to solve it using inversion (spoiler):
http://mathworld.wolfram.com/Arbelos.html
http://mathworld.wolfram.com/PappusChain.html
http://jwilson.coe.uga.edu/EMT668/EMAT6680.2000/Simmons/Essay1/Essay1.html


function
integration
real analysis
real analysis solved

Find all continous functions $f:[0,1]\to[0,2]$ with the property that $\left(\int_{\frac1{n+1}}^{\frac1n}xf(x)dx\right)^2=\int_{\frac1{n+1}}^{\frac1n}x^2f(x)dx, \forall n\in\mathbb N^*$.

[i]Gabriel Marsanu, Andrei Nedelcu[/i]

We have $\left(\int_{\frac1{n+1}}^{\frac1n}xf(x)dx\right)^2\leq \int_{\frac1{n+1}}^{\frac1n} f(x)dx\cdot \int_{\frac1{n+1}}^{\frac1n}x^2f(x)dx$ and the equality iff $f=0$. But $\int_{\frac1{n+1}}^{\frac1n} f(x)dx\leq \frac{2}{n(n+1)}\leq 1$. It means that $f\equiv 0$.


geometry
geometric transformation
reflection
circumcircle
vector
geometry unsolved

Try this geometry problem from Mathematical Reflections:
Let ABC be an isoceles triangle, AB=AC. Point X on AC, point Y on BC such that XY is parallel to AB. Let O be the circumcenter of triangle CXY. Join BX, M is the midpoint of BX. Prove that AMC =90.

Thank you. :)

[quote="Nguyen Khanh"]Let ABC be an isoceles triangle, AB=AC. Point X on AC, point Y on BC such that XY is parallel to AB. Let O be the circumcenter of triangle CXY. Join BX, M is the midpoint of BX. Prove that AMC =90.[/quote]
[quote="erdugan"]Please correct the problem.We must prove that AMO=90[/quote]
They are impossible, because of $ \angle AMC > \angle AMO > \angle XMO \equal{} 90^{o}.$

[b][size=100]EDIT :[/size][/b]I am very sorry dear all my friends, for such that my blindness. I thought the point $ M,$ as the midpoint of $ YX$ instead of $ BX.$  :blush:

Kostas Vittas.

if ->AB means the vector AB, we say ->MO . -> AM = 0.
if we put the angle of x, ABC=x, we have:

->MO = ->MX + ->XO    (1)
->AM = ->AX + ->XO    (2)
(1),(2) => ->AM . ->MO = OX.AX.cos(A/2) + OX.XM.cos(x+(A/2)) - MX.AXcos(x+A) - 1

OX = CX * [sin(90-B)/sin(2B)]            (3)
CX/AX = [ sin(B-x)/sin(C) } * [ sin(A)/sin(x) ] =>
CX/AC = [ sin(B-x)sin(A) ] / [ sin(B-x)sin(A) + sin(C)sin(x) ] =>
CX = AC * [ sin(B-x)sin(A) ] / [ sin(B-x)sin(A) + sin(C)sin(x) ]    (4)

(3),(4) => OX = AC * [ sin(B-x)sin(A) ] / [ sin(B-x)sin(A) + sin(C)sin(x) ] * [sin(90-B)/sin(2B)] :oops:

AX = AB.(sin(x)/sin(x+A))

And we CAN find MX like this and now after all in the first equality we have every thing with sin(A) and sin(x). If we go ahead we see that the answer doesn't depend on x, and it is zero, and the problem will be solved. You can do it in less than 1 hour! :blush:  :lol:
 ( I think so!!!)

I WILL think about a better solution 14 hour later:D :idea:  :?:

if (AM,CY) = D:

BM=XM , AB || XD => AM=MD , AB=XD

now we want t say that AO=OD for this we say:

AB=XD=AC                    (1)
OX=OC                          (2)
OXY=(A/2)=OXC=OCX    (3)

(1),(2),(3) => AOC = DOX  => AO = OD

now we know AO=OD and AM=DM, so we can say AMO=90!!
I think this was a good and simple solution!!:D

sorry for the last difficult way :blush:  :blush:  :lol:

[quote="plane geometry"]Draw parallegram ABDX  △AOC≌△DOX
DO=OA
DONE[/quote]

That's brilliant!  :)


induction
number theory solved
number theory

Determine the set of all positive integers n for which n*2n-1 + 1 is a square !

Oh, this is a bulgarian probelm, actually. One may look up the solution in the Valentin's .pdf notes (I don't know the year offhand, though).

Cheers,
Thanos

n2^(n-1)+1=k^2 iff (k-1)(k+1)=n*2^(n-1). This means that k is odd. It's obvious that 2|k-1 and 2|k+1,m but 4 only divides one of them. n=1 is obviously not a soln, neither are n=2, n=3, so we can think of n as >3.

Case 1   2|k+1 and 4 does not divide it. It means that 2^(n-2)|k-1, so k-1>=2^(n-2) and k+1<=2n, so (k+1)-(k-1)=2<= 2n-2^(n-2), so 2^(n-2)+2<=2n, which is, by induction, false for any n except for a few cases: n=2, n=3, n=4, n=5. From these only n=5 is a soln, so we have the soln n=5 (5*16+1=81=9^2).

Case 2    4 does not divide k-1. This means that 2^(n-2)|k+1, so k+1>=2^(n-2) so k-1<=2n. This means that 2>=2^(n-2)-2n iff 1>=2^(n-3)-n in the set {4, 5}, which is proved by induction. 4 isn't a soln and we already found the soln n=5, so this is the only soln. Is it correct?


Find as much as possible about this recurrence:
$ a_{0} \equal{} 1, a_{n} \equal{} \frac {a_{n\minus{}1} \plus{} 2 a_{n\minus{}2} \plus{} 3 a_{n\minus{}3} \plus{} ... \plus{} n a_{0} } {n}$.
I don't know if it can be solved in closed form, so that's why I'm saying: find as much as possible about the recurrence.

Let $ c_0\equal{}0\equal{}c_1,c_2\equal{}1,a_n\equal{}c_{n\plus{}2}\minus{}2c_{n\plus{}1}\plus{}c_n$, then $ c_{n\plus{}2}\minus{}2c_{n\plus{}1}\plus{}c_n\equal{}\frac{c_{n\plus{}1}}{n}.$

$ na_n \equal{} a_{n\minus{}1} \plus{} 2a_{n\minus{}2} \plus{} ... \plus{} na_0$
$ (n\minus{}1)a_{n\minus{}1} \equal{} a_{n\minus{}2} \plus{} 2a_{n\minus{}3} \plus{} ... \plus{} (n\minus{}1) a_0$.

Subtracting gives

$ na_n \equal{} na_{n\minus{}1} \plus{} a_{n\minus{}2} \plus{} a_{n\minus{}3} \plus{} ... \plus{} a_0$.

Changing the index again gives

$ (n\minus{}1) a_{n\minus{}1} \equal{} (n\minus{}1) a_{n\minus{}2} \plus{} a_{n\minus{}3} \plus{} ... \plus{} a_0$.

Subtracting again gives

$ na_n \equal{} (2n\minus{}1) a_{n\minus{}1} \minus{} (n\minus{}2) a_{n\minus{}2}$.

When $ n$ is large, this is [i]approximately[/i] the recurrence $ a_n \equal{} 2a_{n\minus{}1} \minus{} a_{n\minus{}2}$, which has the general solution $ a_n \equal{} An \plus{} B$.

You are wrong. $ a_n$ increase as $ e^{2\sqrt n }$.

Exactly: If \[ A(x)\equal{}\sum_{i\equal{}0}^{\infty }a_nx^n\], then \[ A(x)\equiv e^{\frac{x}{1\minus{}x}}\equal{}e^{x\plus{}x^2\plus{}x^3\plus{}...}\equal{}1\plus{}x\plus{}\frac 32 x^2\plus{}....\]

$ a_n\equal{}a_0 \sum^{n}_{j\equal{}1}\binom{n\minus{}1}{j\minus{}1}\frac{1}{j!}\equal{}a_0  \,_1F_1(1\minus{}n,2,\minus{}1)$

But I cannot prove it directly (I think induction works though).

[quote="milin"]$ a_n \equal{} a_0 \sum^{n}_{j \equal{} 1}\binom{n \minus{} 1}{j \minus{} 1}\frac {1}{j!} \equal{} a_0 \,_1F_1(1 \minus{} n,2, \minus{} 1)$

But I cannot prove it directly (I think induction works though).[/quote]
It followed from $ A(x)\equal{}exp(\frac{x}{1\minus{}x})\to a_n\equal{}\sum_{j\equal{}1}^{n\minus{}1}C_{n\minus{}1}^{j\minus{}1}\frac{1}{j!}$, because $ (x\plus{}x^2\plus{}...)^j\equal{}\sum_n A_j(n)x^n$, were $ A_j(n)$ is number of solutions $ x_1\plus{}...\plus{}x_j\equal{}n, x_i\le 1$. It give $ A_j(n)\equal{}C_{n\minus{}1}^{j\minus{}1}$ and prove your result.


PRove $\sum_{k=0}^{n}\binom{n}{k}cos (k\theta)=(2cos \frac{\theta}{2})^{2}cos (\frac{n\theta}{2})$

P.S (Does theese problems come up ofthen in math olympiads :maybe: )

[quote="t0rajir0u"][hide="Solution"] This is just

$\Re \left( (e^{i\theta}+1)^{n}\right)$

Let $z = e^{ \frac{i\theta}{2}}$.  We can write this expression as

$\Re \left( z^{n}(z+\frac{1}{z})^{n}\right)$

Now, $z+\frac{1}{z}= 2 \cos \frac{\theta}{2}$ and $z^{n}= \cos \frac{n \theta}{2}+i \sin \frac{n \theta}{2}$, which produces our desired result. [/hide][/quote]

Could elaborate on the Re(z) part and how you got to it?

[quote="t0rajir0u"][hide="Discussion"] Recall that

$(x+1)^{n}= \sum_{k=0}^{n}{n \choose k}x^{k}$

The substitution of $\cos k \theta$ for $x^{k}$ suggests the substitution $x = \cos \theta+i \sin \theta$, and when this substitution is made we have

$\left( ( \cos \theta+i \sin \theta)+1 \right)^{n}= \sum_{k=0}^{n}{n \choose k}\left( \cos k \theta+i \sin k \theta \right)$

The real part of which is precisely the LHS of our given. [/hide][/quote]

Does this type of problem ofthen show up at math olympiads?


conics
ellipse
geometry
geometric transformation
reflection
analytic geometry
calculus

Suppose you have an ellipse-shaped pool table. You place the ball at one of the foci. You take the ball and shoot it. Prove that no matter where it bounces, it will always pass through the other focus.

I don't know the answer, but I have an idea...
[hide="Hint"]Let the point it firsts hits the ellipse be $X$. Let $O$ be the center of the ellipse. Also, let the point that intersects with the path of the ball and the major axis be $Y$. Draw a tangent line to the ellipse that passes through $X$. Now reflect the point $Y$ across this line. Now prove that $Y$ is the other focus by showing that $OY = \sqrt{a^{2}-b^{2}}$, where $2a$ is the length of the major axis, and $2b$ is the length of the minor axis. [/hide]

[hide="Another approach"]Take an arbitrary point on ellipse, say $C(x_{0},y_{0})$. Then find the lines $CF_{1}$, $CF_{2}$ and the tangent in $C$, and prove that $CF_{1}$ and $CF_{2}$ make the same angle with the tangent.[/hide]

[quote="vishalarul"]Farenhajt, how do you find the equation of the tangent line in C to the ellipse? (If I could do that, I would have used coordinate geometry :wink: )

I know how to take the tangent lines of circles, so your method would work if you had a circular pool table.[/quote]

It requires differentiation.  This isn't really a place for calculus problems, however, if you do want to differentiate it, I think the best wasy would be to express the ellipse parametrically as:

$x = a*\cos{t}$
$y = b*\sin{t}$

and use $\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Then, as you have the slope of the tangent line in terms of $t$, and you can find the slopes of the other lines to the point of tangency using standard analytic geometry tactics.  Once you have the slopes you can apply a bit of trig to get the result.

I think a geometric solution similar to what you thought of earlier might be more elegant and quite a bit slicker.


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_0^\frac{\pi}{4} \frac{x\plus{}\sin x}{1\plus{}\cos x}\ dx$.

[quote="kunny"]Evaluate $ \int_0^\frac {\pi}{4} \frac {x + \sin x}{1 + \cos x}\ dx$.[/quote]

$ 1+\cos x=2\cos^2\frac{x}{2},\frac{x+\sin x}{1+\cos x}=\frac{1}{2}(x\sec^2\frac{x}{2}+2\tan \frac{x}{2})$ the given integral equals to

$ \int_{0}^\frac{\pi}{4}\frac{1}{2}(x\sec^2\frac{x}{2}+2\tan \frac{x}{2})dx= \int_{0}^\frac{\pi}{4}xd\tan \frac{x}{2}dx+ \int_{0}^\frac{\pi}{4}\tan \frac{x}{2}dx
=x\tan \frac{x}{2}|_{0}^{\frac{\pi}{4}}=\frac{\pi(\sqrt{2}-1)}{8}$

You almost solved, but the answer is incorrect.
What's value of $ \tan \frac{\pi}{8}?$

[quote="kunny"]You almost solved, but the answer is incorrect.
What's value of $ \tan \frac {\pi}{8}?$[/quote]

let  $ x\equal{}\tan \frac{\pi}{8}$, we know $ \tan  \frac{\pi}{4}\equal{}\frac{2\tan \frac{\pi}{8}}{1\minus{}(\tan \frac{\pi}{8})^2}$ which is equivalent to $ 1\equal{}\frac{2x}{1\minus{}x^2}$

solve the equation,we obtain $ x\equal{}\minus{}1\pm\sqrt{2}$,but   $ x\equal{}\tan \frac{\pi}{8}>0$, we choose positive root, so  $ \tan \frac{\pi}{8}\equal{}\minus{}1\plus{}\sqrt{2}$

That's right.
The answer is $ \frac{\pi}{4}(\sqrt{2}\minus{}1)$.


[quote="Tung Lam"]Let a+b+c=1[/quote]

isn't that a bit strange start :?

how do you know a+b+c = 1 ?

you should prove it for all values... you forget that (a+b+c)^5 cn be smaller than 4(a+b+c)^3
(yes, it is simple, still you need to do it as a separate case...)

Dear Peter, I bet you have heard about homogenuous inequalities. This is one of them.

Peter, divide both sides by (a+b+c)^5, A=a/(a+b+c), ...

oh yes sorry :blush:

I misread the operation signs  :(


Is there any solution at the moment?  :blush:

[quote=hungkhtn]Let $ a,b,c,d$ be non-negative real numbers such that $ a^{2}+b^{2}+c^{2}+d^{2}=3$. Prove that
\[ ab+bc+cd+da+ac+bd \le a+b+c+d+2abcd.\][/quote]
If $d=0$ so we need to prove that $a+b+c\geq ab+ac+bc$, where $a^2+b^2+c^2=3$, which is
$(a+b+c)^2(a^2+b^2+c^2)\geq3(ab+ac+bc)^2$, which is obvious.
Let $abcd\neq0$ and $f(a,b,c,d)=a+b+c+d+2abcd-ab-ac-ad-bc-bd-cd+\lambda(a^2+b^2+c^2+d^2-3)$.
It's obvious that in a minimum point $\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=\frac{\partial f}{\partial d}=0$, which gives
$(a-b)(2\lambda+2cd-1)=(a-c)(2\lambda+2bd-1)=(a-d)(2\lambda+2cb-1)=(c-b)(2\lambda+2ad-1)=(d-b)(2\lambda+2ca-1)=(d-c)(2\lambda+2ab-1)=0$.
From here it remains to check two cases only:
1. $a=b$ and $d=c$, or $b=a$ and $d=c=\sqrt{\frac{3-2a^2}{2}}$ for $0< a<\sqrt{1.5}$;
2. $b=c=a$ and $d=\sqrt{3-3a^2}$, where $0<a<1$, which are true.

By Vasc's EV Theorem it's enough to check one case only: $b=c=a$.

[quote=arqady]
By [color=#f00]Vasc's EV Theorem[/color] [/quote]

What will say Vasc and what will say Hung ^^.


let $ \Delta ABC$ incribed a circle with centre $ (O)$.Let a point $ X$ inside triangle $ ABC$  such that $ \hat{XAB}\equal{} \hat{XBC}\equal{}a$.let $ P$ be a point such that $ PX$ perpendicular to $ OX$ ang $ \hat{XOP}\equal{}a$.And X;B;A and X;P;O are order in the counterclock wise.Prove that P lie on the fix line with every value of a.

$ \angle XAB \equal{} \angle XBC$ $ \Longrightarrow$ $ X$ is on a circle $ (K)$ tangent to $ BC$ at $ B$ and passing through $ A.$ $ OP$ cuts $ AB$ at $ Q.$ $ \angle XOP \equal{} \angle XAQ$ $ \Longrightarrow$ $ AXOQ$ is cyclic, let $ (S)$ be its circumcircle. Let $ R \in (S)$ be point diametrically opposite to $ O.$ The angles $ \angle OXR, \angle OXP$ are both right, $ R, X, P$ are collinear. The angle $ \angle OAR$ is also right, the line $ AR$ is fixed. Let $ AO$ cut $ (K)$ again at $ Z$ and let $ XB$ cut $ (S)$ again at $ Y.$ Then $ \angle XBZ \equal{} \angle XAZ \equal{} \angle XAO \equal{} \angle XYO$ and it follows that $ OY \parallel ZB,$ so that the line $ OY$ is also fixed. By Pascal theorem for the cyclic hexagon $ AXOYQR,$ the intersections $ B \equiv AQ \cap XY,$ $ D \equiv AR \cap OY,$ $ P \equiv XR \cap OQ$ are collinear. Lines $ AR, OY$ are fixed $ \Longrightarrow$ their intersection $ D$ is fixed. $ B$ is also fixed, therefore the collinearity line $ BPD$ is fixed. Q.E.D.

(When $ O$ is circumcenter of the $ \triangle ABC,$ $ BPD$ is the B-symmedian, which is totally irrelevant: If $ O$ is some other fixed point, $ BPD$ iis some other fixed line.)


a,b,c,d>0
Prove that:
[tex]\Large \sum \frac{a^3+b^3+c^3}{b+c+d} \ge a^2+b^2+c^2[/tex]

[quote="amanin"]Well,straightforward solution follows from Cauchy .
Between why did you post this problem here?[/quote]

How will you apply CS inequality????

very nice... so how are the turkish forums? i know a very little bit of it...


We want to find the inverse of $(n + 1)^2 \mod (2n + 1)^3$.
$(n + 1, 2n + 1) = 1$ so $(n + 1)^2$ is invertible.
Then $(n + 1)^2a + (2n + 1)^3b = 1$. Solving in $(a, b)$ we have that $a$ is the inverse.


algebra
polynomial
calculus
derivative
function
algebra open

IS it true?

If polynomial p(x) is given and deg p(x)=n
p has n positive roots then

Q(x) = p(x) - p' (x) has n positive roots too. p'(x) is a derivative of p(x)

Yes. Because a polynomial with no repeated roots has no roots in common with its derivative, none of the roots of $ Q$ are the roots of $ P$. Therefore it suffices to find the zeroes of $ 1 \minus{} P^\prime (x) / P(x)$, or $ 1 \minus{} \sum \frac{1}{x \minus{} r_i}$, where $ r_i$ are the roots of $ P$. All $ r_i$ are positive, so if $ x < 0$ then every term in the summation is negative, and $ 1$ minus a negative number can't be $ 0$.

You haven't actually shown that the roots of $ Q$ are real, although it's trivial to find $ n \minus{} 1$ of from the expansion you gave and then the last one can't have nonzero imaginary part.

Alternately, $ q(x) \equal{} e^x \frac {d}{dx} e^{ \minus{} x} p(x)$, which has $ n \minus{} 1$ positive roots by Rolle's theorem, and again the last root can't have nonzero imaginary part.

Only one step further is enough. We notice that $ 1\minus{}P'(x)/P(x)$ goes to $ \infty$ when $ x$ approaches $ r_i$ from left; goes to $ \minus{}\infty$ when $ x$ approaches $ r_i$ from right. Also the function goes to $ 1$ when $ x\rightarrow\infty$. This implies there is one root in each of the interval $ (r_i,r_{i\plus{}1})$ and one in $ (r_n,\infty)$.


function
inequalities proposed
inequalities


[quote="shortlist"]Given $ a_1,a_2....a_n$ in [0,1]
Find max , min : $ S \equal{} a_1(\frac {1}{2} \minus{} a_2) \plus{} a_2(\frac {1}{2} \minus{} a_3) \plus{} .... \plus{} a_n(\frac {1}{2} \minus{} a_1)$    :maybe:[/quote]$ S$ is linear in each $ a_i$, so we must have $ a_i\in\{0,1\}$ for en extremum.

For $ a_1,a_2....a_n$  such that $ S$ has a minimum, there must be a $ j$ such that $ a_j \equal{} 1$, but then we must also have $ a_{j \minus{} 1} \equal{} 1$. So $ a_i \equal{} 1\forall i$ and $ S_{min} \equal{} \minus{} \frac n2$.

For $ a_1,a_2....a_n$  such that $ S$ has a maximum, if there is a $ j$ such that $ a_j \equal{} 1$, then we must have $ a_{j \minus{} 1} \equal{} 0$ and $ a_{j \plus{} 1} \equal{} 0$.
If there is a $ k$ such that $ a_{k \minus{} 1} \equal{} 0$ and $ a_{k \plus{} 1} \equal{} 0$, then we must have $ a_k \equal{} 1$.
If there is a $ k$ such that $ \{a_{k \minus{} 1},a_{k \plus{} 1}\} \equal{} \{0,1\}$, then we must have $ a_k \equal{} 0$.

So the maximum is achieved for $ (0,1,0,1,...,0,1)$ or cyclic if $ n$ even and for $ (0,1,0,1,...,0,1,0)$ or cyclics if $ n$ odd. Thus $ S_{max} \equal{}\frac 12 \left[\frac n2\right]$. [edited] :)

acording me
$ S_{max}\equal{}\frac{\left[\frac{n}{2}\right]}{2}$
you can check again


what is the definition of polygon?
( may be obvious question but plz just answer the question,thanks.)

Polygon is a two dimensional closed figure in which a line segment is joined two other line segments and no line segments are intersecting each other.

Consult the mathematics bible.  :P

http://mathworld.wolfram.com/Polygon.html

[quote="prototypex24"]Can polygons be non-Euclidean (on a curved surface)?[/quote]

of course, why not?

[quote="prototypex24"]Can polygons be non-Euclidean (on a curved surface)?[/quote]

http://mathworld.wolfram.com/SphericalPolygon.html

http://mathworld.wolfram.com/SphericalTriangle.html


find five sets of functions f and g such that f[g(x)]=g[f(x)].  Naturally, f and g cannont be the same function...  can anyone help with this?

Simple.  There are an infinite number of solutions in the form f(x)=x and g(x) is any function or g(x)=x and f(x) is any function. :D

[hide]$f(x)=x, g(x)=x$
$f(x)=2x, g(x)=x$
$f(x)=3x, g(x)=x$
$f(x)=4x, g(x)=x$
$f(x)=5x, g(x)=x$ :D

You could also use functions like $f(x)=x+a, g(x)=x+b$. That's probably the most simple answer.[/hide]

[quote="kamikazetony"]Im still confused... is there any way you can expand your answer, and give me some examples?  Thanks![/quote]Using diophantus's solution.. let's say you have $f(x)=x$ and $g(x)=\sin^2(x-2)$. Then $f(g(x))=f(\sin^2(x-2))=\sin^2(x-2)$ and $g(f(x))=g(x)=\sin^2(x-2)$. This will work for whatever $g(x)$ is (over its domain).

Im still lost... what are five sets of the functions f and g such that f[g(x)]=g[f(x)]  I get the idea that you could add any number in there as he did above, but im not getting the reasoning behind it.

[quote="Gyan"]There are many others  apart from f(x)  or g(x) = x,

Eg f(x) = ax and g(x) = bx     (a and b can be any  number)
or f(x)  = x+ a    and g(x) = x+b
etc...[/quote]

So my answer would be like?
1. f(x)=2x and g(x)=2x
2. f(x)=3x and g(x)=10x
3. f(x)=6x and g(x)=4x
4. f(x)=4x and g(x)=2x
5. f(x)=8x and g(x)=7x

Sure but you can pick MANY others ... Sure putting socks first then the shoes is not the same thing as putting shoes first then the socks   ... but  substitute hat for shoes and you would be okay ...

You can even think $ f(x) =  x^a$   and   $g(x) = x^b$

or even  $  f(x) = 2x+2$    and $g(x) = 3x+4$

or $f(x) = ln(x)$  and  $ g(x) = exp(x) $ ...

list can go on ..

Use the inverse of a function.

If f(x) and g(x) are inverses of each other, then f(g(x)) = x = g(f(x)).

Say $f(x) = \frac{x}{x+1}$, then $g(x) = \frac{x}{1-x}$.

Okay, may not be the best question, but just for fun -

Can you suggest a function such as  f(x) (or (g(x)) is not x
or f(x) is not equal to g(x)
or f(x) is not inverse of g(x)
and it is none of the examples given above --
A little less obvious ... Just for fun

[hide]some thing like $f(x) = x^2+2$  and $g(x) =x^4+4x^2+6$
This does work.. BTW  [/hide]


When you rate a question, what is the 'criteria' for the difficulty of problem? MATHCOUNTS/AMC 8, AMC10 etc etc...

The criteria is how difficult you found the problem. If it's a problem that you were able to do in one second in your head, then it should be rated low. If it's a problem you had to really think about then rate it high.

But thats different for everyone isnt it?

If someone who took the IMO looked at a question that a kid from MATHCOUNTS did, wouldnt the results be a little different?

yes
which is why they ask everyone to rate it
that way, they can find out on average how hard a particular problem is to a wide range of people


calculus
derivative
real analysis
real analysis unsolved

$f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ is $C^{1}$ such that $\frac{df}{dx}=\frac{df}{dy}$, and $\forall x \in \mathbb{R}, f(x,0)>0$. Show that $\forall x,y \in \mathbb{R}, f(x,y)>0$

I could come up with a solution but I think $f$ only needs to be differentiable. Is that correct ?

I realized the proof requires $f$ to be $C^{1}$ actually.


trigonometry
real analysis
real analysis unsolved

hello, write the equation in the form
$ y'-\cos(x)\tan(y(x))=\sin(x)$. The solution of the homogenius part is
$ y(x)=\arcsin\left(Ce^\sin(x)\right)$.
Sonnhard.

That doesn't do much good, since the equation isn't linear. I don't expect an elementary solution.


trigonometry function geometry logarithms vector probability quadratics

here they are. deadline for answering is sunday sept 28, 6:00 PM EST. extra pts for generalizing, answer in [hide]spoiler[/hide]this is in black[hide]this is in spoiler[/hide]

1. Evaluate the product log (base 3) 16 * log (base 2) 27

2. Find the two smallest positive x (in radians) such that tan (:pi:/4+x)=9 tan (:pi:/4-x). Find the product of these x.

3. Right, you have a floor and a wall and two ladders. One of these ladders is twice the length of the other. Both touch the same point on the wall (same vertical height), and the shorter ladder makes an angle of :pi:/3 with the floor. What angle does the longer ladder make with the floor?

4. f(2x+1)=4x:^2:+14x. Find the sum of the roots of f(x)=0.

5. x:^2: + x cot A cos A - 1 =0 for all A between 0 and :pi:/2. What is the sum of the roots of this equation in terms of A?

Have fun!!! (don't worry, they'll get harder )

um. ok minor problem when i realize that most middle school peeps don't know trig. hold on for a while.

I hate trig.

[hide]

1. Obviously change of base, log16/2 * log27/log3 = 4*3 = 12

3. Put in the inverse function which is 1/2x-1 for x, ya get x:^2:+3x+10 = 0, so -3 is sum of the roots

[/hide]

[quote="ComplexZeta"]Why are you putting all these trig problems on a test for middle schoolers?[/quote]
Concured, I might have to retire if trigs and logs come into play...

on another note, does this mean round 1's over? It's not due till tomorrow if I recall correctly...

no round 1 is not over. ok, you can answer these questions if you want to, but they won't be part of the contest. sry about that folks. i'll get some easier questions. round 2 should be up later today or tomorrow.

alright...um...to continue the contest, i can get questions suitable for Alg. 1 students, geometry students, or alg. 2 students. which would y'all like?

any would be fine...

(spoiler)

[hide]is the first one 12???[/hide]

I don't really have a preference...

Algebra 1 and basic geometry would be nice because I haven't formally learned more than that.
I did study a bit of trig on my own, but that's about it.

...the last part (and especially the wink) makes me worried... :)
I mean, these questions are above and beyond MATHCOUNTS...maybe Getting Started, si? (meaning "may I move this?")

meaning no. i'll avoid trig, so don't worry.

um so would the geometry contest be ok then? (geometry means geometry and down, so it could include algebra 1 and whatever they teach before that)

yah sure. I wanted to review geom... and this is going to help

Well I'm not sure about the rest of the people, but most middle schoolers might not skip math and go into high school leveled math, so technically we should be sticking to only stuff we learn in middle school, meaning Algebra 1, Geometry, Basic and Introductory Number Theory, and Basic/Introductory proof.
That means Trigonometry, Logarithms (Exponents are OK), Vectors, Advanced Matrices, Advanced Probability, Sigma notation and other advanced sequences topics, and imaginary numbers as well as others are off limits.

The reason I'm writing this is that I got my State test result and went and saw what the state could and couldn't put in our State tests.

answers... well, it's really hints for how to get the answers since I'm too lazy to solve...

[hide] 1. Chain rule for logarithms.

2. Using tan(a+b) = (tan a + tan b) / (1-tan(a)tan(b)), the equation simplifies to the proportion(1+ tan x) / (1- tan x) = 9(1-tan x) / (1+ tan x).

3. The angle is arctan (sqrt (3) / 4).

4. Substitute (x-1)/2 for x in the function and solve for the roots.

5. The sum of the roots of any quadratic x^2 + ax + b is -a. [/hide]

yeah....I'm kinda in Algebra 1 this year so nothing too hard...so yeah...

[hide]1)change of base formula etc etc. 12[/hide]

Ok, on number 2, I got as far as [hide]tan x = 2, 1/2. [/hide] Where do i go from here?

Also, isn't number 5 just [hide] -cot(a)cos(a), -cos:^2:(a)/sin(a), whatever you wanna call it[/hide]?

Hmm, so is the answer in number 2

[hide]arctan(2)*arctan(1/2)[/hide]

[hide = s4]
You set $a = \frac{x-1}{2}$ to solve the problem. The function comes out to be $f(a) = x^2 + 5x - 6$. Since it is not a double root (redacted), the answer is simply $-b = -5$

[quote=fuzimiao2013]This thread is older than I am[/quote]

It's probably older than a lot of the people on this site

[quote=fuzimiao2013]This thread is older than I am[/quote]

same, I didn't realize how old aops forums was

[quote=fuzimiao2013]This thread is older than I am[/quote]

same lol

edit: @OP's last post was in 2011 LOL

I thought this was a newly posted thread and I was confused as to why the deadline was sep 28 o.0


will anyone please explain morphism,homothetics in detail?

Morphism of what¿ In arbitrary categories¿  :wink:
For homotetics, the same question: of affine (real) space¿


abstract algebra
induction
invariant
Ring Theory
vector
superior algebra
superior algebra solved

I'm sure I'm doing something wrong, but what is it? :?

I think it can be shown by indution. First of all, $m=n$ because the first group has dimension $m$ as a $\mathbb Z$-module, while the second one has dimension $n$. Now we can do induction after $n$. If $n=1$ it's trivial. Assume now we have shown it fo $1,2,\ldots,n-1$.

Since $a_m$ is a multiple of all the other $a_i$s, it means that all elements of $A$ (the first group) satisfy $a_nx=0$ and there is an element of order $a_n$. In the same manner we find that all the elements of $B$ (the second group) satisfy $b_nx=0$, and there is an element with order $b_n$. From this the fact that $a_n=b_n$ follows immediately.

Let $A,B$ be our two groups, and let $H$ be the group formed as the cartezian product of $n-1$ trivial groups and $\mathbb Z_{a_n}=\mathbb Z_{b_n}$. Then $\mathbb Z_{a_1}\times\ldots\times\mathbb Z_{a_{n-1}}$ is isomorphic to $A/H$, and $\mathbb Z_{b_1}\times\ldots\times\mathbb Z_{b_{n-1}}$ is isomorphic to $B/H$, and $A/H,\ B/H$ are isomorphic, so we can use the induction hypothesis.

What is wrong?

Isn't this the Invariant Factor Theorem  ;)  (finitely generated modules over PIDs  :P )
Elementary divisors should bring you back sweet memories, harazi :lol:

[quote="grobber"]
First of all, $m=n$ because the first group has dimension $m$ as a $\mathbb Z$-module, while the second one has dimension $n$.
[/quote]

What do you mean by dimension?
What do you say about $dim \mathbb{Z}_6$ and $dim \mathbb{Z}_2\times \mathbb{Z}_3$?

I think that the rest of the proof is corect and actually proves that $n=m$ without the use of the dimension argument.

You are excused  :cool:

there are some terms about modules that can be defined, such as the rank (the rank of the torsion free module that is the complement of the torsion submodule of M) and length (the supremum of lengths of composition series for M), but I don't remember about dimension. Does it exist? For vector space it is another name for rank, but what about general modules?

[quote="grobber"]Sorry about that... :blush:  :blush: Actually, because of this absurdity, the proof can't go on exactly as I wrote it up there, because I had $m=n$ and then induction after $n$. Now we have to do induction after one of $m,n$, but if we disregard details such as this one, I think and hope it's ok.

Sorry again. I'm really embarrassed, you know, but I'll force myself to smile now :).[/quote]

you can try an induction after $a_1\cdot a_2\cdot \ldots \cdot  a_n$. This does not depend on n or m and decreases after each factorisation you make.

Or induction after the smallest one among $m$ and $n$. I know it can be eaily repaired.


[quote="Rofler"]Given $ x,y,z,w \ge 0$
Prove that:
$ \frac {x \plus{} z}{(x \minus{} z)^2} \plus{} \frac {x \plus{} w}{(x \minus{} w)^2} \plus{} \frac {y \plus{} z}{(y \minus{} z)^2} \plus{} \frac {y \plus{} w}{(y \minus{} w)^2} \plus{} \sum_{cyc} \frac {1}{x} \ge \frac {8(1 \plus{} \sqrt {2})^2}{x \plus{} y \plus{} z \plus{} w}$

Also, find equality.[/quote]
Nice your inequality.

Using the Cauchy-Schwarz Inequality, we have
$ \begin{aligned} \frac {8}{x \plus{} y \plus{} z \plus{} w} & \equal{} \frac {16}{(x \plus{} z) \plus{} (y \plus{} z) \plus{} (x \plus{} w) \plus{} (y \plus{} w)} \\
& \le \frac {1}{x \plus{} z} \plus{} \frac {1}{y \plus{} z} \plus{} \frac {1}{x \plus{} w} \plus{} \frac {1}{y \plus{} w}. \end{aligned}$
Therefore, it suffices to prove that
$ \left[\frac {x \plus{} z}{(x \minus{} z)^2} \plus{} \frac {1}{2x} \plus{} \frac {1}{2z} \minus{} \frac {\left(1 \plus{} \sqrt {2}\right)^2}{x \plus{} z}\right] \plus{} \left[\frac {y \plus{} z}{(y \minus{} z)^2} \plus{} \frac {1}{2y} \plus{} \frac {1}{2z} \minus{} \frac {\left(1 \plus{} \sqrt {2}\right)^2}{y \plus{} z}\right] \plus{} \left[\frac {x \plus{} w}{(x \minus{} w)^2} \plus{} \frac {1}{2x} \plus{} \frac {1}{2w} \minus{} \frac {\left(1 \plus{} \sqrt {2}\right)^2}{x \plus{} w}\right] \plus{} \left[\frac {y \plus{} w}{(y \minus{} w)^2} \plus{} \frac {1}{2y} \plus{} \frac {1}{2w} \minus{} \frac {\left(1 \plus{} \sqrt {2}\right)^2}{y \plus{} w}\right] \ge 0.$
We shall prove that
$ \frac {x \plus{} z}{(x \minus{} z)^2} \plus{} \frac {1}{2x} \plus{} \frac {1}{2z} \minus{} \frac {\left(1 \plus{} \sqrt {2}\right)^2}{x \plus{} z} \ge 0,$
or
$ \frac {1}{(x \minus{} z)^2} \plus{} \frac {1}{2zx} \ge \frac {\left(1 \plus{} \sqrt {2}\right)^2}{(x \plus{} z)^2}.$
Indeed, using the Cauchy-Schwarz Inequality, we have
$ \frac {1}{(x \minus{} z)^2} \plus{} \frac {1}{2zx} \equal{} \frac {1}{(x \minus{} z)^2} \plus{} \frac {2}{4zx} \ge \frac {\left(1 \plus{} \sqrt {2}\right)^2}{(x \minus{} z)^2 \plus{} 4zx} \equal{} \frac {\left(1 \plus{} \sqrt {2}\right)^2}{(x \plus{} z)^2}.$

In the same manner, we also have
$ \frac {y \plus{} z}{(y \minus{} z)^2} \plus{} \frac {1}{2y} \plus{} \frac {1}{2z} \minus{} \frac {\left(1 \plus{} \sqrt {2}\right)^2}{y \plus{} z} \ge 0,$
$ \frac {x \plus{} w}{(x \minus{} w)^2} \plus{} \frac {1}{2x} \plus{} \frac {1}{2w} \minus{} \frac {\left(1 \plus{} \sqrt {2}\right)^2}{x \plus{} w} \ge 0,$
and
$ \frac {y \plus{} w}{(y \minus{} w)^2} \plus{} \frac {1}{2y} \plus{} \frac {1}{2w} \minus{} \frac {\left(1 \plus{} \sqrt {2}\right)^2}{y \plus{} w} \ge 0.$
By adding all these inequalities, the conclusion follows.

Equality holds if and only if $ x\equal{}y,$ $ z\equal{}w$ and $ x^2\plus{}z^2\equal{}2\left(1\plus{}\sqrt{2}\right)xz.$


When will the U.S. Physics Team 2006 be announced?
I heard it was supposed to be on April 17, but I still can't find it.

thats what i thought too.

maybe this week?

I don't think I'll make it anyways - the semifinal exam was way too hard (there is still a chance though, might be on the order of 10^-5, but it exists).  Maybe they already privately contacted the people who made it?

The finalists were notified by mail a week or so ago.

Yep, that's what I thought.  There is still next year.  Do you know about what percentage of the points do finalists usually have to earn in order to qualify?

Is it something like 50-60 percent or 85-90 percent?

hi.  i made it onto the team this year, and my coach is the academic director of the us physics team.

the score required to make the team can (and will) vary tremendously from year to year.  2006 was a very rough year compared to 2005, so the floor score was much lower.  exact statistics, unfortunately, are never released.

edit: just to confirm what durt said, i recieved my invitation around april 12 (give or take a day).

also, the team is not announced on any fixed date.  they post it only when they are confident that they know exactly who will attend the camp and not a minute earlier.

kenan diab, hawken school

while i don't know the exact score, i think i got around half the points.  unfortunately, i royally screwed up the capacitor question, i didn't get the thermo question, and i knew next to nothing about blackbody radiation, so b2 was basically 0.  the inductor question saved me.

[quote="shbmsc90"]Thanks [b]iostream.h[/b].
Do you know approx. how you did (if it's not confidential)?  Do you think you got half, two-thirds, or is it really just super-random what the floor would be?[/quote]

If I remember correctly, I got 122 on the first round (19 of the MC and 84 on the free response). I don't know if this is taken into consideration. On the semifinal, I feel like I got around 125 points, but it may be lower (if I'm overestimating because I'm a n00b at physics) or higher (if I'm used to harsh MOP grading and physics grading is actually easier).

I basically just read a good physics text - my class uses Halliday and Resnick.

There was this AOPSer Spoon last year who was so awesome amnd from U.S.A.I don't know where he is nowadays,he must've made the team.

Did you Mystic terminator?

[quote="shbmsc90"]Thanks; congrats on making the team!

It's pretty hard to get half the points.
I probably got like 10-20 percent of the points. :blush:
But I have almost no experince in physics problem solving.  This is the first year (11th grade) I have ever done physics (all I knew was F=ma before this year), and I'm in AP Physics B, so I try to patch up some of my knowledge from Calc BC (which I'm also taking only this year - pretty late looking at the biographies I read of previous finalists), so it's not like I was close or anything.  I didn't even take a basic diff. equations class.

What's the best way to practice for these exams?[/quote]it's nearly impossible to qualify for us physics team without ap physics c knowledge.  the semifinal always tests that.  the first thing you should do is learn that material.  you would also do well to learn thermodynamics w/ calculus (not covered in physics b or c) and more about waves/quantum/relativity.

in terms of math, bc calc is probably adequate.  i'm taking vector calc and linear algebra this year, and there was nothing on the prelim or semifinal exam that really required any of that.  diffeq's at this point are usually easy to solve just by separating variables or (maybe) using complex exponentials.  i'm not sure how much more math-heavy it gets at ipho level.

the best preparation for the exam is doing old semifinals.

I think I got 130 on both prelim and semifinal... for some bizarre reason on B2 I said that I * dt=esT^4 * dt=c * dT (s being a sigma, e being epsilon for emissivity because I don't know Tex) so I lost everything there, and I just didn't know how to do the last parts of A4. I think I got the rest though... tough test.

Errr... guys, are the questions from the semifinal available online? Could you pls tell where to find them, or just post them :)

By the way, I noticed today that if you go to the 2005 team page, click on the team, and then in the URL change 2005 to 2006, it'll show a page with this year's team.  However, there is not yet a link to it from the main page for the 2006 team, as it is not complete yet (all the names are there, but pics and bios are missing).

Or just click this link: http://www.aapt.org/olympiad2006/team.cfm

Heh, you can figure out from my username which one I am.


How many 4-digit integers can be formed with the set of digits {0, 1, 2, 3, 4, 5} such that no digit is repeated and the resulting integer is a multiple of 3?

[hide="Solution"]$ \begin{tabular}{|c|c|}\hline String & Permutations\\ \hline 0123 & 6\\ \hline 0135 & 6\\ \hline 0234 & 6\\ \hline 0345 & 6\\ \hline 1245 & 24\\ \hline \end{tabular}$

In all, there are $ 4\cdot6+24=48$.[/hide]

umm i like pie..there are 18 permutatiosn for those cases so its 18*4+24=96

I agree, 1023 for example hasn't been included there.

There are 18 permutations for those cases, because you can rearrange the 4 digits in any way except all those that begin with 0 (if you count those, e.g. 0123, then there are 24 permutations for these and the answer is 120). So, by starting with 0, you have 3 choices where to place it, then 3 choices again for any of the other digits you want to pick as second, then 2 choices, then one choice, so there are 3*3*2 = 18 permutations for those cases, and the answer is 18*4+24 = 96.


[hide="binomial expansions rock"]We want to determine $ 101^{50}\minus{}99^{50}\minus{}100^{50}$.

$ (100\plus{}1)^{50}\minus{}(100\minus{}1)^{50}\minus{}100^{50}$

Note that every other term in the first part cancels, leaving us with:

$ 2(100^{49}\cdot50\plus{}\text{blah})\minus{}100^{50}$

$ 2\text{blah}$

All of the terms in $ \text{blah}$ are positive, so $ 101^{50}>99^{50}\plus{}100^{50}$[/hide]

I figured out the error in my "proof" only about 2 hours ago. Well.

[quote="b-flat"]
Uhh, first of all, you don't look at the last few digits to compare two numbers.  To find which is larger, you compare starting at the most significant digit, which is the left-most digit.[/quote]

So $ 9.0>10$?  :wink:

Umm, no.

I was referring to this case only.  Obviously, if one has more digits, then it is larger.

Or, you could look at it as the most significant digit of 10 is 1, but the digit to which that would be compared would be a 0 (09=9).

[quote="b-flat"]Or, you could look at it as the most significant digit of 10 is 1, but the digit to which that would be compared would be a 0 (09=9).[/quote]

I intended it as a joke, but if you persist, why would 0 be the most significant digit of 09?

Ah. Serves me right. I actually selected the text in Mathematica and had it export it as TeX. I feel bad for not checking :(
I can edit it, but then all the posts after mine won't make any sense.