let $x,y,z \geq 0 ,x+y+z=1$ and $k \geq 1 $ then
$ \sqrt \frac{x}{k+y} +\sqrt \frac{y}{k+z} + \sqrt \frac{z}{k+x} \leq \frac{3}{\sqrt{3k+1}} $
[b][color=red][size=150]and what is left side(unsolved[/size])[/color]?[/b]

[quote="zhaobin"][quote="siuhochung"]you reversed the sign of jensen.[/quote]
what do you mean?[/quote]
I was talking about boll's wrong solution before he removed it.

after a long time,I find this still unsolve and no discuss.

[quote="ityaa"]let $x,y,z \geq 0 ,x+y+z=1$ and $k \geq 1$ then
$\sqrt \frac{x}{k+y} +\sqrt \frac{y}{k+z} + \sqrt \frac{z}{k+x} \leq \frac{3}{\sqrt{3k+1}}$
[b][color=red][size=150]and what is left side(unsolved[/size])[/color]?[/b][/quote]
is it unsolved?NO,it is just unsovled for 2005.but not for 2006 ;) .because I sloved it yesterday night or today morning(I can't rememeber clearly).May it will be a present for ityaa and me(Or all the mathlinks who interest in it) :)
PROBLEM:let $x,y,z \geq 0 ,x+y+z=1$ and $k \geq 1$ then
$\sqrt \frac{x}{k+y} +\sqrt \frac{y}{k+z} + \sqrt \frac{z}{k+x} \leq \frac{3}{\sqrt{3k+1}}$ $(1)$

Here goes my solution:
$Lemma 1$  $a,b,c \ge 0$,then:
$(a^2+2)(b^2+2)(c^2+2) \ge 3(a+b+c)^2$ $(2)$
proof of $lemma 1$:because two of $a^2-1,b^2-1,c^2-1$ must have the same sign.
without loss of general we assume $(a^2-1)(b^2-1) \ge 0$
so $(a^2+2)(b^2+2) \ge 3(a^2+b^2+3)$
thus $(a^2+2)(b^2+2)(c^2+2) \ge 3(a^2+b^2+1)(1+1+c^2) \ge 3(a+b+c)^2$
$(2)$ is proved.
$Lemma 2$  $a+b+c=3,a,b,c \ge 0$ then $4(a^2+b^2+c^2)+a^2b^2c^2 \ge 13$
proof of $lemma 2$:let $f(a,b,c)=4(a^2+b^2+c^2)+a^2b^2c^2$
I will prove $f(a,b,c)\ge f(\frac{a+b}2,\frac{a+b}2,c)$ $(3)$
in fact $(3) \iff 4(a^2+b^2-2(\frac{a+b}2)^2 \ge c^2((\frac{a+b}2)^4-a^2b^2)$
$\iff 32(a-b)^2 \ge c^2 ((a+b)^2+4ab)(a-b)^2 \iff 32 \ge c^2((a+b)^2+4ab)$
but we have $c^2((a+b)^2+4ab) \le 2(a+b)^2c^2 \le 2(\frac{a+b+c}2)^4 <32$
so  $(3)$ is true.we get  $f(a,b,c)$ get it's minimum when $a=b=c=1$
so  $4(a^2+b^2+c^2)+a^2b^2c^2 \ge 13$.
lemma 2 is proved.
for the original problem:
if there exist:$x,y,z$ ,satisfy $\sqrt \frac{x}{k+y} +\sqrt \frac{y}{k+z} + \sqrt \frac{z}{k+x} > \frac{3}{\sqrt{3k+1}}$
let $a= \sqrt \frac{x}{k+y} ,b=\sqrt \frac{y}{k+z},c=\sqrt \frac{z}{k+x}$,
(easy to see $0 \le a,b,c \le 1,abc <1$
then we have:$a+b+c > \frac{3}{\sqrt{3k+1}}$(and let $m=a+b+c > \frac{3}{\sqrt{3k+1}}$)
with calaculation we can get:
$x=\frac{k(a^2+a^2b^2+a^2b^2c^2)}{1-a^2b^2c^2}$
$y=\frac{k(b^2+b^2c^2+a^2b^2c^2)}{1-a^2b^2c^2}$
$x=\frac{k(c^2+c^2a^2+a^2b^2c^2)}{1-a^2b^2c^2}$
I will prove that $x+y+z>1$ $(4)$(then get the contradiction)
in fact:
$(4) \iff \frac{k(a^2+a^2b^2+a^2b^2c^2)}{1-a^2b^2c^2}+\frac{k(b^2+b^2c^2+a^2b^2c^2)}{1-a^2b^2c^2}+\frac{k(c^2+c^2a^2+a^2b^2c^2)}{1-a^2b^2c^2} >1$
$\iff k\sum_{cyc}a^2+ k\sum_{cyc}a^2b^2+3ka^2b^2c^2 > 1-a^2b^2c^2$
$\iff k\sum_{cyc}a^2+ k\sum_{cyc}a^2b^2+(3k+1)a^2b^2c^2 > 1$ $(5)$
let $u=\frac{3a}{m},v=\frac{3b}{m},w=\frac{3c}{m}$
$(5) \iff k \frac{m^2}{3^2}\sum_{cyc}u^2+ k\frac{m^4}{3^4}\sum_{cyc}u^2v^2+(3k+1)\frac{m^6}{3^6}u^2v^2w^2 > 1$   $(6)$
because $m=a+b+c > \frac{3}{\sqrt{3k+1}}$)
so we have just to prove
$\frac{k}{3k+1}\sum_{cyc}u^2+ \frac{k}{(3k+1)^2}\sum_{cyc}u^2v^2+\frac{1}{(3k+1)^2}u^2v^2w^2 \ge 1$
$\iff 2k(3k+1)\sum_{cyc}u^2+2k\sum_{cyc}u^2v^2+2u^2v^2w^2 \ge 2(3k+1)^2$ $(7)$
but we have:
$2k(3k+1)\sum_{cyc}u^2+2k\sum_{cyc}u^2v^2+2u^2v^2w^2$
$=(6k^2-2k)\sum_{cyc}u^2+k(4\sum_{cyc} u^2+2\sum_{cyc}u^2v^2+u^2v^2w^2 +1)+(2-k)u^2v^2w^2-k$
$=(6k^2-2k)\sum_{cyc}u^2+k(u^2+2)(v^2+2)(w^2+2)+(2-k)u^2v^2w^2-k$
$=(\frac{3k^2-k}2)(4\sum_{cyc}u^2+u^2v^2w^2)+k(u^2+2)(v^2+2)(w^2+2)+(2-k-\frac{3k^2-k}2)u^2v^2w^2-k$
$\ge 2(3k+1)^2$
(the last inequality is because lemma 1,2,$2-k-\frac{3k^2-k}2 \le 0,uvw \le 1$
so we prove $(7)$ is ture.
which means $x+y+z >1$
$contradiction$

$Remark$
1.I think $a= \sqrt \frac{x}{k+y} ,b=\sqrt \frac{y}{k+z},c=\sqrt \frac{z}{k+x}$ is the pivotal step.
2.the nice proof for lemma 1 is wuhao.
3.I think 1 is not the best constant in $k\ge 1$

[quote="ityaa"]let $x,y,z \geq 0 ,x+y+z=1$ and $k \geq 1$ then
$\sqrt \frac{x}{k+y} +\sqrt \frac{y}{k+z} + \sqrt \frac{z}{k+x} \leq \frac{3}{\sqrt{3k+1}}$
[b][color=red][size=150]and what is left side(unsolved[/size])[/color]?[/b][/quote]
and now goes to the left side.
and sure the left side is $\sqrt \frac{x}{k+y} +\sqrt \frac{y}{k+z} + \sqrt \frac{z}{k+x} \geq \frac{1}{\sqrt{k}}$
PROOF:(the same approach to the right side)
If there exist $x,y,z$,such that  $\sqrt \frac{x}{k+y} +\sqrt \frac{y}{k+z} + \sqrt \frac{z}{k+x} < \frac{1}{\sqrt{k}}$
similar:
we will  prove:
$\frac{k(a^2+a^2b^2+a^2b^2c^2)}{1-a^2b^2c^2}+\frac{k(b^2+b^2c^2+a^2b^2c^2)}{1-a^2b^2c^2}+\frac{k(c^2+c^2a^2+a^2b^2c^2)}{1-a^2b^2c^2} <1$(a,b,c is defined in the last post,and here we have $a+b+c< \frac{1}{\sqrt{k}}$,and let $l=a+b+c$)
then let:$u=\frac{a}{l},v=\frac{b}{l},w=\frac{c}{l}$,then $u+v+w=1$
similar to the right side:we should prove

$\sum_{cyc}u^2+\frac 1 k \sum_{cyc}u^2v^2+\frac{3k+1}{k^3}u^2v^2w^2 \le 1$
but
$\sum_{cyc}u^2+\frac 1 k \sum_{cyc}u^2v^2+\frac{3k+1}{k^3}u^2v^2w^2 \le \sum_{cyc}u^2+ \sum_{cyc}u^2v^2+4u^2v^2w^2$  (because $k \ge 1$)
let prove the stronger one

$\sum_{cyc}u^2+ \sum_{cyc}u^2v^2+4u^2v^2w^2 \le 1$
but we have: $\sum_{cyc}u^2v^2 \le (\sum_{cyc}uv)^2 \le \sum_{cyc}uv$ (notice $u+v+w=1$)
and $4u^2v^2w^2 \le \sum_{cyc}uv$ easy to know. ;)
so the last inequality is true.
wo compelete the whole problem

ityaa,is it ok now?
PS: the equality can hold when $x=y=0,z=1$ or change the odds


Let $\frac{n+\sqrt{n}+\sqrt[3]{n}+\dots+\sqrt[n]{n}}{n}$.Is decrease the series, if n>4 ?

[quote="csabafarkas"]Let $f(n) = \frac{n+\sqrt{n}+\sqrt[3]{n}+\dots+\sqrt[n]{n}}{n}$.Is decrease the series, if n>4 ?[/quote]I guess you are asking if the sequence $f(n)_{n\ge1}$ is decreasing for $n>4$. The answer is no - $f(13)$ is maximal, and $f(n)$ is decreasing for $n > 14$. The limit of this sequence was found in
http://www.mathlinks.ro/Forum/viewtopic.php?t=76909 .

OK!How are you prove that the series is decrease if n>14? :roll:


Let positive real numbers $ a,b$ satisfy $ b \minus{} a > 2.$ Prove that for any two distinct integers $ m,n$ belonging to $ [a,b),$ there always exists non-empty set $ S$ consisting of certain integers belonging to $ [ab,(a \plus{} 1)(b \plus{} 1))$ such that $ \frac {\displaystyle\prod_{x\in S}}{mn}$ is square of a rational number.

Well, i can't remember detail proof.
This is what my teacher taught me.

let $ T \equal{} \{x|a\le x \le [b] \plus{} 1\}$
and if $ x,y \in T$ satisfy above condition(in problem).
Then we connect $ x$ and $ y$
Clain) these $ x,y$s set consists connected graph

But it is not difficult to prove it.
and then, we are done

Any full solution for this problem, please !!!

I think, the hint, though pretty unclear, could be carried out. The idea is to show existence of a chain of integers $m=r_0, r_1,\ldots, r_j=n$, with $r_j\in [a,b)$ and $r_i r_{i+1}\in[ab,(a+1)(b+1))\,, i=0,\ldots,i-1$.
For example, the plan of proving it might be carried like that: Assume $m<n$ and start with $r_0=m$. If $r_0 n \in [ab,(a+1)(b+1)) $ we are done, else set $r_1$ to be the closest to $n$ integer in $[a,b)$ that keep $r_0r_1$ in this interval. Then proceed with $r_1$ instead with $r_0$ and so on. Eventually we should arrive at $n$.
Finally there is one more obstacle: some of the products $r_i r_{i+1}$ may be the same. In that case we omit all of the equal ones if their number is even or leave just one of them if their number is odd.

[quote=simplyconnected43]Let $m \sim n$ be the relation $ \frac {\displaystyle\prod_{x\in S} x}{mn}$ is a rational square. I claim that $\sim$ is an equivalence relation. Denote $[ab, (a+1)(b+1))$ as $\textit{the big interval}$ henceforth, and $[a, b)$ as $\textit{the little interval}$ henceforth.

Symmetry is immediate, and we can evidently prove reflexiveness by showing that there is at least one perfect square in $\textit{the big interval}$. Observe that there exists a real $0 \le r < 1$ such that $\sqrt{ab} + r$ is an integer. I claim that this, when squared, is less than $(a+1)(b+1)$. This computation is simple, observe that \[ (\sqrt{ab} + r)^2 = ab + r + 2r\sqrt{ab} < ab + 1 + 2\sqrt{ab} < ab + a + b + 1 = (a+1)(b+1) \] The next step is to show transitivity, which is slightly more nontrivial. Assume $a \sim b$ and $b \sim c$. Then, let $S_1$ be the associated set with the first one and $S_2$ be the associated set with the second one. Then, it is clear $S_1 \oplus S_2$ (where $\oplus$ denotes symmetric difference) is a working subset. (if the symmetric difference is empty, we have $ac$ perfect square, so just take the perfect square in $\textit{the big interval}$.

Now that we have proven $\sim$ is simply an equivalence relation, it remains to show that all elements are equivalent to each other. Fortunately, this turns out to be not that hard. I claim that $x \sim (x+1)$ for all integers $x$ in the little interval, which will show the result by definition. This proof turns out to be rather easy; observe that $x \sim y$ if $xy$ is in $\textit{the big interval}$, for obvious reasons. So all we need to do is find an integer $y$ such that \[ ab \le xy \le (x+1)(y) < (a+1)(b+1) \] and we will be done since $x \sim y \sim (x+1)$. This process is relatively easy. Note that we can lose here only if we have some $y$ for which
\begin{align*}
xy &< ab \\
(x+1)(y + 1) &\ge (a+1)(b+1)
\end{align*}
This is of course ludicrous; subtracting the equations gives $x + y > a + b$, but $a \le x \le y < b$, so this is impossible.[/quote]

Hi,
The idea of showing an equivalence relation is indeed excellent. But isn't there a mistake in your last step? if you are trying to show that there is a $y$ in $[a,b)$ such that both $xy$ and $(x+1)y$ are in $[ab,ab+a+b+1)$, your last step is not sufficient, as it doesn't cover the case that only one of them is in the interval.


algebra
polynomial
inequalities
trigonometry
induction
algebra unsolved

[hide="hint"]Let $ v_n \equal{} u_n \sqrt {6} \minus{} 1$, then $ v_1 \equal{} 2$ and $ v_{n \plus{} 1} \equal{} 4v_n^3 \plus{} 3v_n\equal{}iT_3(iv_n)$ where $ T_3$ is the third [url=http://en.wikipedia.org/wiki/Chebyshev_polynomials]Chebyshev polynomial[/url] of the first kind.[/hide]

I don't understand. Can you explain for me?Thanks

Help me this exercise:
Find $ u_n:$
   $ u_1\equal{}2$
   $ u_{n\plus{}1}\equal{}u^3_n\plus{}3u^2_n\minus{}3$

Same general idea: try looking at the sequence $ v_n \equal{} \minus{} \frac {u_n \plus{} 1}{2}$ and relate it again to $ T_3$.

How does this help?  Well the $ k$th Chebyshev polynomial $ T_k$ satisfies $ T_k(x) \equal{} \cos (k\arccos x)$, so if a sequence satisfies $ a_{n \plus{} 1} \equal{} T_k(a_n)$ then it follows easily by induction that $ a_n \equal{} \cos \left(k^{n \minus{} 1}\arccos a_1\right)\equal{}T_{k^{n\minus{}1}}(a_1)$
[hide="in this case"]Using the above method the answer comes out to be
\[ u_n \equal{} \minus{} 1 \minus{} 2\cos \left(3^{n \minus{} 1}\arccos \left( \minus{} \frac {3}{2}\right)\right) \equal{} \minus{} 1 \plus{} 2T_{3^{n \minus{} 1}}\left(\frac {3}{2}\right)
\]
[/hide]

[quote="karis"]Help me this exercise:
Find $ u_n:$
   $ u_1 \equal{} 2$
   $ u_{n \plus{} 1} \equal{} u^3_n \plus{} 3u^2_n \minus{} 3$[/quote]
Let $ K_n\equal{}U_n\plus{}1$
So we have
$ K_1\equal{}3\equal{}a\plus{}\frac{1}{a}$
$ K_{n\plus{}1}\equal{}K_n^3\minus{}3K_n$
By induction we have
$ K_{n\plus{}1}\equal{}a^{3^n}\plus{}\frac{1}{a^{3^n}}$
:D


According to an industry report, 26 percent of all households have at least one cell phone. Further, of those that do have a cell phone, the mean monthly bill is 55.90 dollars with a standard deviation equal to 9.60 dollars. Recently, a random sample of 400 households was selected. Of these households, 88 indicated that they had cell phones. The mean bill for these 88 households was 59.00 dollars. What is the probability of getting a mean bill of 57.00 dollars or more if the numbers provided by the industry report are correct?

Can anyone help me figure out why the answer approximately 0.1401?

Thanks in advance.

hint[hide]Use z-score and the z-score table.[/hide]

Right, but would I just use the latter info in the question, or should I be using all the info given?
Would I just do: =(57-55.9)/(9.6/sqrt(88))?

Thanks.


Putnam
integration
calculus
inequalities
function
college contests

A5) Find the maximum value of $\int_{0}^{y}\sqrt{x^{4}+(y-y^{2})^{2}}dx$ for $0\leq y\leq 1$.

I don't have a solution for this yet. I figure this may be useful: Let the integral be denoted $f(y)$, then according to the [url=http://mathworld.wolfram.com/LeibnizIntegralRule.html]Leibniz Integral Rule[/url] we have

$\frac{df}{dy}=\int_{0}^{y}\frac{y(1-y)(1-2y)}{\sqrt{x^{4}+(y-y^{2})^{2}}}dx+\sqrt{y^{4}+(y-y^{2})^{2}}$

Now what?

Your way might work, but here is a simpler way:  Use the inequality $\sqrt{a+b}\le \sqrt{a}+\sqrt{b}$.

[quote="Jimmy"]$g(y) = \int_{0}^{y}\sqrt{x^{4}+(y-y^{2})^{2}}\ dx \le \int_{0}^{y}\sqrt{x^{4}}\ dx+\int_{0}^{y}\sqrt{(y-y^{2})^{2}}\ dx =$ $y|y^{2}-y|+\frac{y^{3}}{3}$

Now for $0\le y < 1$, we have $g(y) = y|y^{2}-y|+\frac{y^{3}}{3}=-y(y^{2}-y)+\frac{y^{3}}{3}= y^{2}-\frac{2y^{3}}{3}$

$g'(y) = 2y-y^{2}\ge 0 \ \forall y\in[0,1]$

So, $g$ is increasing on the interval and thus reaches a maximum at $g(1) = \frac{1}{3}$.[/quote]

I think there is something missing here: we do not have

$g(y) = y|y^{2}-y|+\frac{y^{3}}{3}=-y(y^{2}-y)+\frac{y^{3}}{3}= y^{2}-\frac{2y^{3}}{3}$

we only have

$g(y) \leq y|y^{2}-y|+\frac{y^{3}}{3}$

and a function is not necessarily increasing on some interval because it is bounded above by such a function on that interval, but I do see your point (perhaps) that since the maximum of the square-root of a function is occurs at the same place as the maximum of the function itself, it works-out despite the integral which isn't being squared?

[quote="Ravi B"]What Jimmy meant is that \[g(y) \le h(y)\] where \[h(y) = y |y^{2}-y|+\frac{y^{3}}{3}\, .\] As Jimmy showed, $h$ is increasing on $[0, 1]$.  Thus \[g(y) \le h(y) \le h(1) = \frac{1}{3}\] for $0 \le y \le 1$.  Because $g(1) = \frac{1}{3}$, we see that the maximum of $g(y)$ (over $0 \le y \le 1$) is $\frac{1}{3}$.[/quote]

Thank you! I needed to see that to understand.

-Ben


Oh, this is a smart way of solving it. Thanks!

[quote=El1te_Postfarmer][hide=hint]I'm a postfarmer so i will post a useless hint to a prealgebra problem. [/hide][/quote]

stop postfarming.

[quote=GameBot]Compute: $ \frac{6!\plus{}5!}{5!}$.[/quote]

This is very simple $6!$ is $6 \cdot 5!$ so we can rewrite the problem as

$\frac{6 \cdot 5! + 1 \cdot 5!}{5!}$
$\frac{7(5!)}{5!}$
$7$

Sincearly
Jason


One mass, $ m_{1}$ = 0.215 kg, of an ideal Atwood machine rests on the floor 1.10 m below the other mass, $ m_{2}$ = 0.255 kg.

a. If the masses are released from rest, how long does it take $ m_{2}$ to reach the floor?
b. How high will mass $ m_{1}$ ascend from the floor? [Hint: When $ m_{2}$ hits the floor, $ m_{1}$ continues to move upward.]

Atwood Machine:
[img]http://upload.wikimedia.org/wikipedia/commons/3/38/Atwoodmachine.gif[/img]

My question deals with (b.)

Work for (a.). The acceleration of the two blocks is:

$ a \equal{} \frac{g(m_{2}\minus{}m_{1})}{m_{1}\plus{}m_{2}}$, so the acceleration is 0.834 $ m/s^{2}$.

$ x\equal{}x_{0} \plus{} v_{0}t\plus{}\frac{1}{2}at^{2}$
$ 0\equal{}1.10\plus{}\frac{1}{2}(\minus{}0.834)t^{2}$ (-0.834 because falling)
So, $ t\equal{}1.624$ seconds.

Assuming my work is right for a, how do you do Part b?

Your work for a is correct.
When mass 2 reaches the floor, mass 1 has ascended 1.10 m. From this precise moment onwards, the tension is zero and the only force applied on mass 1 is its weight (and so its acceleration is g). The distance traveled by mass 1 from that point onwards can be calculated from the expression

$ v_F^2 \equal{} v_o^2 \minus{} 2gd$

where $ v_F$ is the final speed (=0), $ v_o$ is the speed with which mass 1 reaches the height 1.10 m, g is the acceleration of gravity and d the distance traveled (the minus sign comes from the fact that the velocity and acceleration vectors have opposite senses). To compute $ v_o$ we use a similar expression for the 1st part of the motion (from rest to height 1.10m):

$ v^2 \equal{} v_o^2 \plus{} 2ad$

where now $ v_o \equal{} 0$, $ v \equal{} v_F$, a = 0.834 m/s^2, and d = 1.10 m. Note that, the 1st part of the motion both acceleration and speed have the same sense. Finally, the total height H attained by mass 1 will be H = 1.10 + d.

[quote="1/(ln x)"]One mass, $ m_{1}$ = 0.215 kg, of an ideal Atwood machine rests on the floor 1.10 m below the other mass, $ m_{2}$ = 0.255 kg.

a. If the masses are released from rest, how long does it take $ m_{2}$ to reach the floor?
b. How high will mass $ m_{1}$ ascend from the floor? [Hint: When $ m_{2}$ hits the floor, $ m_{1}$ continues to move upward.]

Atwood Machine:
[img]http://upload.wikimedia.org/wikipedia/commons/3/38/Atwoodmachine.gif[/img]

[/quote]

I would like to ask a question regarding this system....

if the distance betwwen the 2 masses are not mentioned......if suppose the system is released from rest, and after 1 second the mass $ m_{2}$ is stopped momentarily, after how much time will the tension in the string above mass $ m_{1}$ be regained?

won't mass $ m_{1}$ be acted upon by only gravity? if so, how will it decelerate?

What if the mass $ m_{2}$ is stopped for 1 second , 1 second after the system is set into motion?


rotation
geometry
geometric transformation
complex numbers
geometry unsolved

Let $ ABC$ be a triangle and $ M$ the midpoint of $ BC$. We construct externally the isoceles right triangles $ ABP$ and $ ACQ$ with bases at sides of triangle $ ABC$. Prove that $ PMQ$ is also a isoceles right triangle.

[hide]An application of complex numbers does it quickly:

Let $ A\equal{}0$.  Then,

$ M\equal{}\frac{1}{2}(B\plus{}C)$
$ Q\equal{}\frac{1}{\sqrt{2}}C(\frac{1}{\sqrt{2}}\minus{}\frac{i}{\sqrt{2}})$
$ P\equal{}\frac{1}{\sqrt{2}}B(\frac{1}{\sqrt{2}}\plus{}\frac{i}{\sqrt{2}})$

Rotate $ MQ$ about $ M$ by $ \frac{\minus{}\pi}{2}$to get

$ MQ^\prime\equal{}\frac{1}{\sqrt{2}}C(\frac{1}{\sqrt{2}}\minus{}\frac{i}{\sqrt{2}})(\minus{}i)\minus{}\frac{1}{2}(B\plus{}C)(\minus{}i)$

$ \equal{}\frac{\minus{}i}{2}C\minus{}\frac{1}{2}C\plus{}\frac{i}{2}B\plus{}\frac{i}{2}C$

$ \equal{}\minus{}\frac{1}{2}C\plus{}\frac{i}{2}B$

$ \equal{}\frac{1}{\sqrt{2}}B(\frac{1}{\sqrt{2}}\plus{}\frac{i}{\sqrt{2}})\minus{}\frac{1}{2}(B\plus{}C)$

$ \equal{}MP$

Thus, we have that $ MPQ$ is an iscoceles right triangle with vertex $ M$.
[/hide]

[hide="Solution"]
Consider $ N$ external to $ \triangle ABC$ with $ BN \equal{} NC$ and $ \angle BNC \equal{} 90$.  Then, since $ M$ is the midpoint of $ BC$, we have that $ MN \equal{} BM$ and $ \angle BMN \equal{} 90$, so $ \triangle BMN\sim \triangle BPA$, so a spiral similarity maps $ \triangle BPM\sim \triangle BAN$.  Thus, $ \frac {PM}{AN} \equal{} \frac {BP}{AB} \equal{} \frac {1}{\sqrt {2}}$ since $ \triangle BPA$ is a right isosceles triangle.  It follows that $ PM \equal{} \frac {AN}{\sqrt {2}}$.  Similarly, $ QM \equal{} \frac {AN}{\sqrt {2}} \equal{} PM$.  Furthermore, the similarities imply that\[ \angle PMQ \equal{} 180 \minus{} \angle PMB \minus{} \angle QMC \equal{} 180 \minus{} \angle ANB \minus{} \angle ANC \equal{} 180 \minus{} \angle BNC \equal{} 90\]from which we conclude that $ \triangle PMQ$ is right and isosceles. [/hide]

Dear Mathlinkers,
ABGF and ACDE are the wo external squares construct on ABC.
1. The traingles AFB and ACE are A-isosceles.
2. By the rotation (center A, angle 90°), BE = CF and BE orthogonal to CF.
3. Accordling to the midline theorem applies to the triangle BCE, (BE) // (MQ)  and  BE = 2.MQ.
4. Mutatis mutandis,	(CF) // (MP) and CF = 2.MP.
…
Sincerely
Jean-Louis


Consider two column vectors $ x$ and $ y$ that are orthogonal, and an arbitrary projection matrix $ P\equal{}A(A^TA)^{\minus{}1}A^T$.  Is $ Px$ and $ Py$ necessarily orthogonal?

My first thought is to just let the matrix notation work for us. We note that $ P^T\equal{}P$ and $ P^TP\equal{}P^2\equal{}P.$ Now, suppose $ x^Ty\equal{}0.$ Look at:

$ (Px)^T(Py)\equal{}x^TP^TPy\equal{}x^TPy.$

That doesn't look like it should necessarily be zero. So look for the counterexample.

Let $ P\equal{}\begin{bmatrix}1&0\\0&0\end{bmatrix},\ x\equal{}\begin{bmatrix}1\\1\end{bmatrix},\ y\equal{}\begin{bmatrix}1\\\minus{}1\end{bmatrix}.$

Then $ x^Ty\equal{}0$ but $ x^TPy\equal{}1.$

$ Px$ and $ Py$ are not necessarily orthogonal.

Note: we can get that $ P$ as $ A(A^TA)^{\minus{}1}A^T$ for $ A\equal{}\begin{bmatrix}1\\0\end{bmatrix}.$


function
vector
algebra unsolved
algebra

Find all functions $f: \mathbb R \to \mathbb R$ such that \[f(x+f(x)+y)=f(y)+2x\] holds for all $x,y \in \mathbb R$.

[quote="Leonhard Euler"]Find all functions f:R->R such that $ f(x+f(x)+y)=f(y)+2x$[/quote]

We have the property $ P(x,y)$ : $ f(x+f(x)+y)=f(y)+2x$ $ \forall x,y$

$ 1).$ $ f(x)$ is bijective
$ P(\frac{a-f(0)}{2},0)$ : $ f(\ldots)=a$ and $ f(x)$ is surjective
$ P(x,-f(x))$ : $ f(x)=f(-f(x))+2x$ and so $ f(x_{1})=f(x_{2})$ implies $ x_{1}=x_{2}$ and $ f(x)$ is injective
Q.E.D.

$ 2.)$ $ f(0)=0$
$ P(0,x)$ : $ f(x+f(0))=f(x)$ and, since $ f(x)$ is injective, $ f(0)=0$
Q.E.D.

$ 3.)$ $ x+f(x)$ is bijective
$ P(x,0)$ : $ f(x+f(x))=2x$ and so $ x+f(x)=f^{-1}(2x)$ and $ x+f(x)$ is bijective.
Q.E.D.

$ 4).$ $ f(x+y)=f(x)+f(y)$
Since $ f(x+f(x))=2x$, $ P(x,y)$ may be written $ f(x+f(x)+y)=f(y)+f(x+f(x))$ and since $ x+f(x)$ is surjective, for any $ z$ it exists $ x$ such that $ x+f(x)=z$ and so $ f(z+y)=f(y)+f(z)$.
Q.E.D.

$ 5).$ the only continuous solutions are $ f(x)=x$ and $ f(x)=-2x$
The only continuous solutions of this Cauchy's equation are $ f(x)=ax$ and so we have $ ax+a^{2}x+ay=ay+2x$ and so $ a^{2}+a-2=0$ and so $ a=1$ or $ a=-2$
Q.E.D.

$ 6).$ With axiom of Choice, it exists non continuous solutions.
Using (classical way) a base $ \{x_{i}\}$ for the $ \mathbb{Q}$-vector space $ \mathbb{R}$, it is immediate to see that $ f(x_{i})=x_{i}$ or $ f(x_{i})=-2x_{i}$.
So we can build two $ \mathbb{Q}$-vector space $ A$ and $ B$ such that $ A\cap B=\{0\}$ and $ A+B=\mathbb{R}$ ($ A$ using the part of the base such that $ f(x_{i})=x_{i}$ and $ B$ using the part of the base such that $ f(x_{i})=-2x_{i})$.
Then, if $ a(x)$ and $ b(x)$ are the projections of a real $ x$ on $ A$ and $ B$ (such that there is a unique decomposition $ x=a(x)+b(x)$ with $ a(x)\in A$ and $ b(x)\in B$, then $ f(x)=a(x)-2b(x)$

So the general solution of the original equation is the following one :

Let $ A$ and $ B$  be two $ \mathbb{Q}$-vector space $ A$ and $ B$ such that $ A\cap B=\{0\}$ and $ A+B=\mathbb{R}$
Let $ a(x)$ and $ b(x)$ be the projections of a real $ x$ on $ A$ and $ B$ (such that there is a unique decomposition $ x=a(x)+b(x)$ with $ a(x)\in A$ and $ b(x)\in B$

Then $ f(x)=a(x)-2b(x)$ is a solution (and all solutions are of this form).

If $ A=\mathbb{R}$ and $ B=\{0\}$, then we have the continuous solution $ f(x)=x$
If $ A=\{0\}$ and $ B=\mathbb{R}$, then we have the continuous solution $ f(x)=-2x$
With axiom of Choice, a lot of other pairs $ (A,B)$ exist and so a lot of non continuous solutions.


trigonometry
function
limit
calculus
calculus computations

For which $ x \in \mathbb{R}$ will the $ \sum_{n\equal{}0}^{\infty} \sin nx$ infinite series be convergent?

If the terms don't go to zero, the series doesn't converge. The terms only go to zero if they're all zero- $ x$ is a multiple of $ \pi$.

On the other hand, $ \sum_{n\equal{}0}^{\infty} f(n)\sin nx$ converges for all $ x$ whenever $ f$ is a decreasing function with $ \lim_{n\to\infty}f(n)\equal{}0$.


I think there is no a soluiton except working  with a calcultator!  :wink:

It is a problem from a competition in US and it actually has a a very nice solution but I cannot figure out. I think 12 is just a random number.

I hope anyone can try this interesting problem.

[quote="thamtu"]A competitor number is a positive integer that has two more prime factors than its predecessor. For example, 8 is a competitor number because it has three prime factors(2.2.2) but 7 has only 1(7).
Of the first twelve competitor numbers, how many of them are divisible by 8?[/quote]
If $ n>1$ is not prime, then $ n$ had at least two prime factors. If I undestand your condition all composite $ n>1$ you call competitor. Because number of primes had dtnsity 0, numbers $ n$,  suth that $ 8|n$ had density $ \frac 18$.

I don't understand your solution clearly, Rust. But from what I have done, among 12 first competition numbers(I think the largest number is 88), there are 5 numbers that are divisible by 8. However, I do not have a good solution for this problem. :(
Can anyone try to come up with a good solution?

actually, there are four numbers that are divisible by 8 among the first twelve math competition numbers. However, can anyone have good idea to solve this problem?
Thank you.


Using the following method: prove (a x b) x c = (a•c)b – (b•c)a

Let I be a unit vector in the direction of a and let J be a unit vector perpendicular to I and parallel to the plane of a and b. Let K = I x J.

a) Show how:
a = a1I
b = b1I + b2J
c = c1I + c2J + c3K

b) Show that: (a x b) x c = -a1b2c2I + a1b2c1J

c) Substitute for I and J in terms of a and b.

Would it not be easier to just use matrices and show the final matrix being equal to the RHS?

nope...cant use matrices.

need to do it algebraically etc


induction
combinatorics unsolved
combinatorics

The [i]Lucas numbers $ l_{0}, l_{1}, l_{2},...,l_{n}...$ [/i] are defined on the same recurrence relation defining the Fibonacci numbers, but the Lucas numbers posses different initial conditions.

$ l_{n}\equal{}l_{n\minus{}1} \plus{} l_{n\minus{}2},$ (n≥2)$ l_{0}\equal{}2, l_{1} \equal{} 1$

(a)$ l_{n} \equal{} f_{n\minus{}1} \plus{} f_{n\plus{}1}$for n≥1.

I've tried to do induction, but I'm just having a hard time to get it to work.

So Let n=1
$l_{1}=f_{0}+f_{2}$
1 = 0 +1

Check

Assume $ l_{n} \equal{} f_{n \minus{} 1} \plus{} f_{n \plus{} 1}$

$ l_{n \plus{} 1}$=$ l_{n} \plus{} l_{n \minus{} 1}$ stated in our first assumption

$ l_{n} \equal{} f_{n \minus{} 1} \plus{} f_{n \plus{} 1}$ and *$ l_{n \minus{} 1} \equal{} f_{n \minus{} 1} \plus{} f_{n \plus{} 1}$*

$ l_{n \plus{} 1} \equal{} f_{n \minus{} 1} \plus{} f_{n \plus{} 1} \plus{} f_{n \minus{} 1} \plus{} f_{n \plus{} 1}$

then by simplifying with Fibonacci Identities I get

$ l_{n\plus{}1} \equal{} f_{n} \plus{} f_{n \plus{} 2}$ for n≥1.

My question now is with the part between the **, can I make that assumption based on my induction assumption since that is n-1 and my induction assumption is n.

I think you meant!

[i][b]*$ l_{n \minus{} 1} \equal{} f_{n \minus{} 2} \plus{} f_{n }$*

$ l_{n \plus{} 1} \equal{} f_{n \minus{} 1} \plus{} f_{n \plus{} 1} \plus{} f_{n \minus{} 2} \plus{} f_{n }$
[/b][/i]

That is continuous induction. You can assume all from 1-n to hold good and then prove for n+1, in which case you can still backtrack (as you do for normal induction) and prove the result.


The positive number $ a$ is chosen such that the terms $ 20, a, \frac{5}{4}$ are the first, second and third terms,
respectively, of a geometric sequence. What is the value of $ a$?

We know that $ \frac{20}{a}\equal{}\frac{a}{\frac{5}{4}}$.  Cross multiplying yields
\[ a^2\equal{}25\\
a\equal{}\pm 5\]
Since $ a$ is positive, our answer is simply $ \boxed{5}$


logarithms
function
algebra unsolved
algebra

[quote="linker"]Solve the equation :

$ (x \plus{} a)^{\log_ab} \plus{} a \equal{} (x \plus{} b)^{\log_ba} \plus{} b$ , where $ a \neq b$ and $ a, b > 1$ .

Posted [url=http://www.mathlinks.ro/viewtopic.php?t=182890]here[/url] before , but nobody  posted a solution .[/quote]

At least, we have the trivial solution $ x\equal{}0$.

Are you looking for "all" solutions ?

[quote="linker"]yes , of course .
When I said "solve the equation" I tough about finding all solutions . Or maybe to prove that $ x \equal{} 0$ is the only solution .[/quote]

OK.

WLOG say $ a>b>1$

Let $ f(x)\equal{}(x\plus{}a)^{\log_a(b)}\plus{}a$. We have $ 1>\log_a(b)>0$ and so $ f(x)$ is concave.
Let $ g(x)\equal{}(x\plus{}b)^{\log_b(a)}\plus{}b$. We have $ \log_b(a)>1$ and so $ g(x)$ is convex.

So $ f(x)\equal{}g(x)$ has at most one solution on $ [\minus{}b,\plus{}\infty)$

Since $ f(0)\equal{}g(0)$, the unique solution of this equation is 0.

[quote="greentreeroad"]so where is the second intersection of two function?[/quote]

There is no second intersection if we consider that $ f(x)$ is only defined for $ x\geq \minus{}a$ and that $ g(x)$ is only defined for $ x\geq \minus{}b$

Obviously, in some pecular cases (for example $ \frac{\ln(a)}{\ln(b)}\in\mathbb{N}$), $ g(x)$ may be defined for $ x<\minus{}b$ and $ f(x)$ may also be defined (if $ \frac{\ln(a)}{\ln(b)}$ is an odd integer) for $ x\leq \minus{}a$. In these pecular cases, other solutions may appear.

But, in general case, (for example $ \frac{\ln(a)}{\ln(b)}$ non rational), 0 is the unique solution.


How many rectangles is in this image? Is there a quicker way of solving this - dont realli think counting it is the only way...

There is a nice combinatorial approach.
[hide][asy]import cse5; for(int i=0; i<4; ++i) D((0,i)--(4,i)); for(int i=0; i<5; ++i) D((i,0)--(i,3)); for(int i=0; i<3; ++i) D((4,i)--(5,i)); D((5,0)--(5,2));[/asy]

Ignore the last right two boxes for now, and focus on the $ 3\times 4$ rectangle. If we pick two vertical and two horizontal lines (where the lines are not the same), we will create a rectangle. Since is a one-to-one correspondence, all we need to do is find the number of ways that we can pick two different vertical and two different horizontal lines. This is easy; there are 5 vertical and 4 horizontal lines, so
$ \binom{5}{2}\cdot\binom{4}{2}= 10\cdot 6 = 60$

We're not done yet; we still have to account for the last two boxes. I'll let you do this; there are $ 3\cdot 5 = 15$ ways (I think... tell me if you think you have a different answer).

So the grand total is $ 60+15 = \boxed{75}$.

This is a lot easier than counting them all... (unless you're very organized :P ).[/hide]

Thanks vishalarul- You are a great help...

A slightly different approach:
[hide]Ignore the box that is "missing", pretend there is a box there for now.  So we have a complete 3x5 rectangle.  We'll count all the rectangles in this rectangle, and then subtract the rectangles that include the upper right box.
$ \binom{6}{2}\binom{4}{2}=90$
This is how many rectangles there are in the full 3x5.
Now we count the number of rectangles which involve the upper right box:
$ 3\cdot 5 = 15$
So our total is
$ 90-15=\boxed{75}$.[/hide]

Yeah, there are definitely 15 rectangles that contain the last 2 boxes because $ \binom{3}{2}\cdot\binom{5}{1}=15$. You calculate $ \binom{3}{2}$ because of the bottom three horizontal lines. You would calculate $ \binom{5}{1}$ because you know that in order for a rectangle to contain the last 2 boxes, it must contain the right-most vertical line. Thus, you're only allowed to choose 1 of the 5 remaining vertical lines in this diagram.

I think it should be
[hide] [b]$ \dbinom{3}{1}$[/b]$ *\dbinom{5}{1}= \boxed{15}$ (same answer, different equation), instead because you are also choosing only 1 line from the bottom three horizontal lines, since you must choose the top one to include the box.[/hide]

solve for rectangles of diferent areas 1*1 2*1 1*2 and so on. its just like solving for squares.....almost

Hmm... why would you do that?
It's way easier to do what the other people said...

[quote="lokesh"]Actually, you don't have to choose the top one to include at least one of the two leftover rectangles. You could choose both of the bottom-most horizontal lines and still include the bottom-most leftover rectangle. Also, by your reasoning it would actually be $ \binom{2}{1}$, not $ \binom{3}{1}$.[/quote]
why do you say "the bottom-most leftover rectangle"? There is only one left-over rectangle (right?), and that's at the upper right corner. There are still 3 lines under it, and you have to choose the top, so by my logic it is \dbinom{3}{1}.

I was talking about calculating the number of rectangles that include the 2 rectangles (the "leftover rectangles") that are the bottom right corner of the diagram. Based on your last post, it seems like you're talking about a different rectangle. We're talking about two completely different (but both valid) ways of solving this problem. You're doing it the $ 90-15=75$ way (which seems like complimentary counting to me) and I'm doing it the $ 60+15=75$ way (casework).

ooh, I didn't know you were doing it that way. I thought you were referring to archimedes's post (the post above yours), which stated that there were 15 rectangles that included the missing rectangle, without stating a fast way of finding that, which I thought you adding on. sorry  :) .


algebra
polynomial
function
topology
complex analysis
complex analysis unsolved

Let $ f$ be analytic in a convex open set $ U$. Suppose that for all $ w, z\in U$ there exists $ \eta$ on the straight line connecting  $ w$ and $ z$ such that $ \frac{f(w)\minus{}f(z)}{w\minus{}z}\equal{}f'(\eta)$. Show that $ f$ is a polynomial of degree at most 2. (Conclude that not all real variable theorems carry over to complex variables.)

All the first year grad students here turned in solutions, but when we compared afterward we found fault with all of each other's solutions. Now I'm very curious if there is some nice way to do this that none of us could think of.

If none of you guys could do it this may be flawed too. Don't have time to check things carefully (in fact never have time to solve non-trivial problems on mathlinks, and regret it when I do anyway).

If $ f'' \equal{} 0$ in $ U$ there is nothing to prove. Otherwise, let the zeros of $ f''$ be $ (a_n)$.

Fix $ w \in U\setminus\{a_n\}$. Take a ball $ B$ around $ w$ contained in $ U\setminus\{a_n\}$.
Then
$ f(z) \equal{} f(w) \plus{} (z \minus{} w)f'(z\eta(z) \plus{} w(1 \minus{} \eta(z)))$ for $ z$ in $ B$, and $ 0 < \eta(z) < 1$
Since $ f''(z) \neq 0$ in $ B$ we can apply the inverse function theorem to $ f'$ to conclude that $ \eta(z)$ is analytic.
But then $ \eta(z) \equal{} K$, $ 0 < K < 1$

So in $ B$, $ f(z) \minus{} f(w) \equal{} (z \minus{} w)f'(w \plus{} K(z \minus{} w))$
Now, through whatever method, this does imply that $ f'''(w) \equal{} 0$

So, $ f$ is a polynomial of degree at most 2.


linear algebra
matrix
linear algebra unsolved

[quote="smurfy"]Hello,

I'm new to linear algebra and unsure how to handle a problem.  Here is the scenario: I have 10 lists of words chosen at random from a list of 5000 possible words (the word lists vary in size).  I want to represent a simplified view of the similarity of these lists on a 2d plot (i.e. each list represented by a spot, and the distance of each spot from the other spots being indicative of the overlap between the list).  I have a matrix made from the ten lists in which each number in the matrix corresponds to the % of shared words (relative to the smaller sized list of the two being compared).

I imagine that these could be compared with principal components analysis, but I suspect that there is a simpler solution.  Could someone please point me in the right direction?

Thanks![/quote]
Try PCA together with projection on the leading two components or [url=http://www.statsoft.com/textbook/stmulsca.html]multidimensional scaling[/url]. Both have their advantages and disadvantages.


probability
function
calculus
algebra
domain
expected value

what is the average amount of times a coin should be tossed such that the number of heads and tails are the same?

Err.. does that mean the average number of flips it takes to get equal number of heads and tails?

The best interpretation I have: flip the coin repeatedly, noting how many heads and how many tails have been accumulated. When the totals are equal, stop and record how many coins we have flipped. It is possible that we will never stop, but we can show that that has probability zero.

This therefore gives a meaningful probability distribution, from which we can attempt to find an expected value.

The expected value of this distribution does not exist- it is infinite.

A solution, using generating functions and calculus:
[hide="Solution"]WLOG, the first coin is a head. After an even number of flips, the heads either exceed the number of tails or are equal. If they are equal, we stop and the last coin was a tail. The number of ways to fit $2n$ flips between these two known ones, without the number of tails in this subsequence ever exceeding the number of heads, is the well-known sequence of Catalan numbers $C_n.$
Some facts about the Catalan numbers:
$C_n=\frac1{n+1}{2n\choose n}$
$C_{n+1}=C_0C_n+C_1C_{n-1}+\dots+C_{n-1}C_1+C_nC_0$
$\sum_{n=0}^\infty C_nx^n=\frac{1-\sqrt{1-4x}}{2x}=\frac{2}{1-\sqrt{1-4x}}$
The probability that our experiment lasts $2n+2$ coin flips is $\frac{C_n}{2}\left(\frac14\right)^n.$

Define $f(x)=\sum_{n=0}^\infty C_nx^n=\frac{2}{1-\sqrt{1-4x}}$ for $-\frac14\le x\le\frac14$
The probability that our experiment terminates is $\frac12f\left(\frac14\right)=1.$
Now the expected value is $\sum_{n=0}^\infty (n+1)C_n\left(\frac14\right)^n.$ Replacing $\frac14$ by $x,$ we have that $\sum_{n=0}^\infty (n+1)C_nx^n=(xf(x))'=f(x)+xf'(x).$
Evaluating this as $x\to\frac14,$ $f'(x)\to\infty$ and the expected value is infinite.

Actually, I skipped the justification for some of that- evaluating the power series is fine inside the radius of convergence, but the point we care about is on the boundary. Abel's theorem saves us there by saying that if a power series converges at a point on the boundary of its domain, it converges to the limit of the function from inside. By the contrapositive, our series for the expected value must diverge. For the probability series, we show that the series converges by noting that the partial sums must all be less than 1 and all of the terms are positive.

It can be shown that $4^{-n}C_n=O\left(n^{-\frac32}\right).$ This makes the convergence and divergence of the two series clear.
[/hide]

This problem seems rather hard and advanced for this forum.

I think there should be a slightly easier way of approaching the problem:
[hide]
Suppose the answer is finite, and equals n.
The first toss is WLOG a head. If the next toss is a tail, we are done in two turns. Otherwise its a head and we have to wait on average n tosses (including that head) until we are back to a position with one more head than tails. Again theres a 50/50 chance of us finishing, otherwise we wait another n turns, etc etc. You can get a big formula out of that.
[/hide]

[quote="TripleM"]You can get a big formula out of that.
[/quote]which reduces to $1=0.$

On the other hand, how are you going to convince yourself that the experiment terminates almost surely but the expected value is still infinite?
There's still reason to do it the hard way.

[quote="jmerry"][quote="TripleM"]You can get a big formula out of that.
[/quote]which reduces to $1=0.$
[/quote]
Well no, it reduces to show that the expected value is infinite..
[quote="jmerry"]
On the other hand, how are you going to convince yourself that the experiment terminates almost surely but the expected value is still infinite?
[/quote]
I'm not sure what you mean..  both our formulas show the same thing, that the expected value is infinite.

There may well be a flaw somewhere in what I had, but I can't see it off the top of my head.


Find all  integers such that

  $ phi(phi(n^2)) \equal{} n$.

Obviosly $ n\equal{}1$ solution. If n>1 odd and p prime divisor (odd), then $ p|\phi(n^2)\to 2|p\minus{}1|\phi (\phi (n^2))$ contradition.
Let n is even, then $ n\equal{}2^kp_1^{k_1}...p_m^{k_m}, k\ge 1$, $ p_i$ - odd primes. It give \[ v_2(\phi (n))\equal{}\sum_{i\equal{}1}^m v_2(p_i \minus{}1) \plus{}k\minus{}1\ge m\plus{}k\minus{}1\].
Therefore $ v_2(n^2)\ge m\plus{}k\minus{}1$ and $ v_2(\phi (\phi (n^2))\ge (m\plus{}k\minus{}2)\plus{}m>k$ if n had more the one odd prime divisors.
It give $ n\equal{}2^kp^{k_1}$. If $ k_1\equal{}0$, then $ \phi(n^2)\equal{}2^{2k\minus{}1}, \phi(\phi(n^2))\equal{}2^{2k\minus{}2}\equal{}2^k\to k\equal{}2$.
It give second solution $ n\equal{}4$.
Let $ k_1>0$, then $ \phi(n^2)\equal{}2^{2k\minus{}1}(p\minus{}1)p^{2k_1\minus{}1},v_p(\phi (\phi(n^2))\equal{}2k_1\minus{}2\equal{}k_1\to k_1\equal{}2$.
If $ q|p\minus{}1$ for odd prime q, then $ q|\phi(\phi(n^2))$, therefore $ p\minus{}1\equal{}2^l$ and
$ v_2(\phi(n^2))\equal{}2k\minus{}1\plus{}l, \phi(\phi(n^2))\equal{}2^{2k\minus{}2\plus{}2l}p^2\equal{}2^kp^2\to k\plus{}2l\minus{}2\equal{}0$ contradition.

[quote="marko avila"]¿how do you define $ v_2(n)$?[/quote]

$ u_p(n)$ is the greater power of $ p$ that divides $ n$. So $ u_p(n)\equal{}k \Longleftrightarrow p^k|n$ AND  $ p^{k\plus{}1}\nmid  n$.

Alexandros


[quote="Michael Niland"]Prove that        $ \(7^{\frac{1}{2}}$ + $ \(7^{\frac{1}{3}}$ + $ \(7^{\frac{1}{4}}$  $ <$ 7      and

$ \(4^{\frac{1}{2}}$ + $ \(4^{\frac{1}{3}}$ + $ \(4^{\frac{1}{4}}$ $ >$ 4 ,      without using calculator[/quote]
As a matter of fact, the second inequality can be strengthened greatly, I think that the LHS>5.

The second problem easily follows from the bernoulli inequality...

$ (x\plus{}1)^{6}\plus{}(x\plus{}1)^{4}\plus{}(x\plus{}1)^{3}\ge 6x\plus{}4x\plus{}3x\plus{}3>4\iff 13x>1$ where $ x\equal{}4^{\frac{1}{12}}\minus{}1$.

Or, another ripoff of arqady's solution:
$ 4^{1/2}\equal{} 2$
$ 4^{1/3}> 1$
$ 4^{1/4}> 1$
hence the LHS is greater than 4.


topology
advanced fields
advanced fields unsolved

Suppose we have a group $ G$ acting on a differentiable manifold $ M$ as diffeomorphisms, such all isotropy groups $ Gx\equal{} \{ g \in G : gx \equal{} x \}$ are finite, every orbit $ Gx$ is closed and discrete.

Then i want to prove that $ G$ is discrete and acts properly discontinuously on $ M$.

Here is what i've done : i think discreteness of $ G$ can be done as follows : suppose $ G$ is not discrete => we can take the sequence of distinct elements $ g_n \to id$ in $ G$, then
$ g_n(s) \to s$ for every $ s \in M$ (*), but the fact that orbits are dense means that for every $ s$ in $ M$, there exists a neighborhood $ U$ of it s.t. the only point of $ Gx$ in $ U$ is $ x$.
Hence (*) would imply that $ g_n(s) \equal{} s$ for large enough $ n$, but this contradicts to the finiteness of isotropy groups and our choice of distinct $ g_n$'s.

Could you please help me with discontinuity of action: i need to prove that there is a neighborhood $ U$ of every point $ x$ in $ M$ such that the $ card \{ gU \bigcap U \neq \emptyset \}$ is finite.

Thanks.

[quote="jmerry"]A [url=http://en.wikipedia.org/wiki/Properly_discontinuous_action]Wikipedia link[/url] for the relevant definition. Confusingly enough, it would also be correct to call the action continuous.

Actually, one of the definitions on that page is obvious directly from your hypotheses.[/quote]

Still, i don't see the proof. Sorry, but i don't understand what you are getting at .
How can i prove that the fact that $ G_x$ is finite and orbits are closed and discrete, then  $ card \{ gU \bigcap U \neq \emptyset \}$ is finite ?

[quote="jmerry"]I was talking about the defintion with "the intersection of an orbit and a compact set is finite". That's immediate from your hypotheses- what you want is the equivalence of the definitions.[/quote]

Ok, i also see that we are done with this particular definition.
Trying to prove the equivalence, suppose we have that $ Gx$ is dense and $ G_x$ is finite, then given any $ x$ we can find its neighborhood $ U$ which doesn't contain any points from $ Gx$, except $ x$.
The the question is to find a neighborhood $ V$ of $ U$, such that as long as $ hV \cap V \neq \emptyset$ we must have $ hx \in V$. It would be possible if we had some kind of a uniform continuity of all the maps $ h : M \to M$, then could have taken a neighborhood of $ x$, such that all the images $ hV, h \in H$ have uniformly bounded diameter.

But we don't seem to have something like this, do you have any ideas how to do that ?


Determine $ m$ so that $ 4x^2\minus{}6x\plus{}m$ is divisible by $ x\minus{}3$. The obtained value, $ m$, is an exact divisor of:
$ \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 48 \qquad\textbf{(E)}\ 64$

[hide="Solution"]
Synthetic division shows that $ m\plus{}18$ must be $ 0$, so $ m\equal{}\minus{}18$, which is a divisor of $ \boxed{\textbf{(C)}\ 36}$.
[/hide]

[hide="Similar"]3 must be a root of the quadratic, ie $ 4 \cdot 3^2 \minus{} 6 \cdot 3 \plus{} m \equal{} 0 \implies m \equal{} \minus{}18 \implies \boxed{\textbf{(C)} \; 36}$[/hide]


Show that the equation

\[ r^4 \minus{} s^4 \equal{} 3^n(2r^2s^2)\]

has no solutions $ (r,s,n)$ on the positive integers.

I think it's not difficult at all
Let $ (r,s)\equal{}d$
Hence $ r\equal{}r_{1}.d$ and $ s\equal{}s_{1}.d$ with $ (r_{1},s_{1})\equal{}1$
Form now the equation is equivalent to
$ r_{1}^{4}\minus{}s_{1}^{4}\equal{}3^{n}(2r_{1}^{2}s_{1}^{2})$
We obtain $ r_{1}\equal{}s_{1}\equal{}0$ when if it's not, then $ gcd(r_{1}^{4}\minus{}s_{1}^{4},r_{1}^{2})\equal{}1$ which lead to a contradiction.
So the equation has no positive solution


Moe uses a mower to cut his rectangular $ 90$-foot by $ 150$-foot lawn. The swath he cuts is $ 28$ inches wide, but he overlaps each cut by $ 4$ inches to make sure that no grass is missed. He walks at the rate of $ 5000$ feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow his lawn?

$ \textbf{(A)}\ 0.75 \qquad
\textbf{(B)}\ 0.8 \qquad
\textbf{(C)}\ 1.35 \qquad
\textbf{(D)}\ 1.5 \qquad
\textbf{(E)}\ 3$

[hide="Click for solution"]
The area of the lawn is $ 90 \times 150\equal{}13500$ square feet.  Moe cuts about two square feet for each foot he pushes the mower forward, so he cuts $ 2(5000)\equal{}10000$ feet per hour.  Therefore, it takes approximately $ \frac{13500}{10000}\equal{}1.35$ hours, which corresponds to answer choice $ \boxed{\textbf{C}}$.
[/hide]

[quote="varunrocks"]May I know how got that he mows 2 square feet?[/quote]

28 inch swath-4 inch overlap=24in=2ft


[b]An eight-bit binary word is a sequence of eight digits each of whcih is either 0 or 1. Find the number of eight-bit binary words.[/b]

[hide="i got"]128[/hide]

[hide="they got"]256[/hide]

:?:
[hide]Each spot has 2 possibilities.  There are 8 spots, so the total combinations is $ 2^{8}\equal{}256$.[/hide]

[quote="mihail911"]yeah, i understand that. the two possibilities are 1 and 0. however the first spot can't be 0, otherwise it would really be a 7-bit number.  :wink: so there is only 1 possibility for the number in first digits place.

that's where my problem lies.[/quote]
Sorry...
 :oops: Still, I don't think you're supposed to assume that.  Otherwise, they would have said 8-digit binary number instead of "eight-bit binary word."

$ 00000000$ is considered a word.

I must commend you for being able to think ahead, and assuming a 1 must be in the first place.

hmm..i guess that is the case. thanks fraenkel and archimedes!


How many positive counting numbers are there between 2 and 90, inclusive?

well...um... you just subtract 2 from 90... and add 1 to include the extra number... you get 89.

hope this helped!  :D


geometry
power of a point
radical axis
angle bisector
AoPS Books

This is a very well known fact. The intersection point is called the radical centre.
How I would prove it:

All points on the line $AD$ have equal powers with respect to circles through $ADB$ and $ADC$.
Similar case with all points on the straight line $CF$.

The intersection point of $AD$ and $CF$ has equal powers with respect to all three circles, so it lies on the third radical axis, i.e: $BE$

The point is called the radical centre in China. Is the name Radical Point used more frequently in Western countries?

[quote="Jan"]All points on the line $AD$ have equal powers with respect to circles[/quote]
Could you explain what you mean by equal powers

It's the same way you prove concurrency for angle bisectors of a triangle; it's some property of sets(commutivity). Bisector of A is the points equidistant from B and C, B is equal from A and C, so the point where they meet is equidistant from A,B and C, and sansifies the requirements of being on the angle bisector of C, equidistance from A and B.

That's also why there is an incircle, because you can make a circle with that common distance.

I don't think that proof is rigourous, I'm having trouble remembering it. It is in the AoPS books somewhere.

[quote="PenguinIntegral"]I don't think that proof is rigourous.[/quote]
Of course the proof is rigourous. It is a great theorem that the three radical axises of three circles are concurrent or parallel.


integration
calculus
calculus computations

Prove:
$\int\limits_0^a {x^3 f(x^2 )dx = \frac{1}{2}\int\limits_0^{a^2 } {xf(x)dx} }$

This is just a change of variables. Change $y=x^2$ in the first equation and you'll get the second (except with $y$ instead of $x$ as your variable, but that doesn't matter). [/code]


conics ellipse geometry circumcircle incenter

Let $\mathcal E$ be an ellipse with foci $B,C$, center $O$ and large axis $AD$, s.t. $AB=BO=OC=CD$. Show that The tangent to $\mathcal E$ in the point $M$ is parallel to the line formed by the incircle and circumcircle of $MBC$.

I wrote it as it was proposed. Basicly, it has nothing to do with conic sections, but I guess it was formulated like this because the author wanted it to be solved by 11’th graders, and there’s no synthetic geometry in the 11’th grade (in Romania).

Note that the normal line bisects the angle which is formed by the two segments joining the two foci.

the problem can be rewritten as follows.

In triangle $ABC$, $b+c=2a$. The circumcenter is $O$ and the incenter is $I$. Prove that $OI$ is perpendicular to $AI$.

Without loss of generality, suppose the projection of $I$ on $BC$ is nearer to $C$.

In triangle $ABC$, $b+c=2a$. The incenter is $I$. Line $\ell$ passes through $AI$ and $\ell$ is perpendicular to $AI$. Suppose $\ell_1$ is the perpendicular bisector of $BC$ and $\ell_2$ is the perpendicular bisector of $AB$. We need to prove that the intersections of $\ell_1, \ell_2$ and $\ell$ are the same, say $P, Q$, are the same. We can prove that $IP=IQ$.

Suppose $A=2\alpha$ and $C=2\gamma$.

It is not difficult to figure out that [tex]IP = \frac{\frac{a}{2} - r\cot \gamma}{\sin (\alpha + C)}[/tex] and [tex]IQ = \frac{\frac{c}{2} - r\cot \alpha}{\sin \alpha}[/tex].

\[IP=IQ \Longleftrightarrow \frac{\frac{c}{2} - r\cot \alpha}{\frac{a}{2} - r\cot \gamma} = \frac{\sin \alpha}{\sin(\alpha + C)}\]

Extend $AI$ to meet with $BC$ at $A'$, then it is not hard to find out $BA'=\frac{1}{2}c$ thus $\frac{\sin \alpha}{\sin(\alpha + C)} = \frac{BA'}{AB} = \frac{1}{2}$

so we are going to prove
\[\frac{\frac{c}{2} - r\cot \alpha}{\frac{a}{2} - r\cot \gamma} = \frac{1}{2}\]

Notice that $b=2a-c=r\cot \alpha + r\cot \gamma$,
we are going to prove
\[c-2r\cot\alpha = \frac{a}{2}-r\cot\gamma =\frac{a}{2}-(2a-c-r\cot\alpha)\]
or
\[\frac{a}{2} = r\cot\alpha\]
or
\[\frac{a^2}{4} = r^2\cot^2\alpha = \frac{4S^2}{(a+b+c)^2}\frac{\cos^2\alpha}{1-\cos^2\alpha} = \frac{4S^2}{9a^2}\frac{\cos^2\alpha}{1-\cos^2\alpha}\]

From cosine's law, $2\cos^2\alpha - 1 = \cos 2\alpha = \cos 2A = \frac{b^2+c^2-a^2}{2bc}$, we have $\cos^2 = \frac{(b+c)^2-a^2}{4bc} = \frac{3a^2}{4bc}$.

Substitute this and $S^2=p(p-a)(p-b)(p-c) = \frac{3}{2}a^2(\frac{3}{2}-b)(\frac{3} {2}-c)$ into above, we finally know the equality holds.

I will do the same as Liyi, but without the trig bash...

[quote="grobber"]Let $\mathcal E$ be an ellipse with foci $B,C$, center $O$ and large axis $AD$, s.t. $AB=BO=OC=CD$. Show that The tangent to $\mathcal E$ in the point $M$ is parallel to the line formed by the incircle and circumcircle of $MBC$.[/quote]

I think you wanted to say:

[i]Let $\mathcal E$ be an ellipse with foci B and C, center O and large axis AD, such that AB = BO = OC = CD. Let M be a point on $\mathcal E$. Show that the tangent to $\mathcal E$ in the point M is parallel to the line formed by the incenter and the circumcenter of triangle MBC.[/i]

In fact, since B and C are the focii of the ellipse $\mathcal E$, and M is a point on it, a well-known fact shows that the tangent to $\mathcal E$ in the point M is the external angle bisector of the angle BMC. Thus, we have to show that the external angle bisector of the angle BMC is parallel to the line formed by the incenter and the circumcenter of triangle MBC.

On the other hand, we know that CM + MB = 2 BC. This is because CM + MB = CA + AB (by the definition of the ellipse), but CA = BC + AB, thus CM + MB = CA + AB = (BC + AB) + AB = BC + 2 AB = BC + 2 BO (since AB = BO). On the other hand, since BO = OC, we have 2 BO = BC, and thus CM + MB = BC + 2 BO = BC + BC = 2 BC.

Now I will slightly change the notations. I will use the following lemma:

[b]Lemma 1.[/b] Let ABC be a triangle with CA + AB = 2 BC. Then the external angle bisector of the angle CAB is parallel to the line formed by the incenter and the circumcenter of triangle ABC.

Of course, applying Lemma 1 to the triangle BMC, which satisfies CM + MB = 2 BC, we get the assertion of the problem. So it just remains to prove Lemma 1.

In order to prove Lemma 1, I will use another lemma:

[b]Lemma 2.[/b] Let ABC be an arbitrary triangle, and X, Y, Z be the points where the external angle bisectors of the angles CAB, ABC, BCA meet the sidelines BC, CA, AB. Further, let O be the circumcenter and I the incenter of triangle ABC. Then, the points X, Y, Z lie on one line, and this line is perpendicular to the line OI.

One possible proof of this lemma was given in paragraph 3 of [url=http://de.geocities.com/darij_grinberg/Schroeder/Schroeder.html#2]posting [Schroeder2] of my Schrder Points Database[/url].

From lemma 2, it follows that the lines YZ and OI are perpendicular. In order to prove Lemma 1, we must show that the external angle bisector of the angle CAB is parallel to the line OI. Hence, this is equivalent to showing that the external angle bisector of the angle CAB is perpendicular to the line YZ.

If AB = BC, from CA + AB = 2 BC it also follows that also CA = BC, so that AB = BC = CA, and the triangle ABC is equilateral. I leave this case to the interested reader  :D ...

So we can assume $AB \neq BC$. Let WLOG be AB < BC; then, from CA + AB = 2 BC it also follows that CA > BC, so that AB < BC < CA. This chain of inequalities shows us where the points Y, Z lie: For instance, since BC < CA and $\displaystyle \frac{BZ}{AZ} = \frac{BC}{CA}$, we have BZ < AZ, and hence the point Z lies on the extension of the segment AB beyound B. Similarly, the point Y lies on the extension of the segment CA beyound A. Therefore, the external angle bisector of the angle CAB is the internal angle bisector of the angle YAZ. Hence, instead of proving that the external angle bisector of the angle CAB is perpendicular to the line YZ, it will be enough to show that the internal angle bisector of the angle YAZ is perpendicular to the line YZ. But this will obviously follow once it will be shown that the triangle YAZ is isosceles, i. e. that AY = AZ.

So it remains to prove AY = AZ. But we have $\displaystyle \frac{BZ}{AZ} = \frac{BC}{CA}$, so $BZ \cdot CA = AZ \cdot BC$. Using BZ = AZ - AB, we get $\left( AZ - AB \right) \cdot CA = AZ \cdot BC$, or equivalently $\left( AZ - c \right) \cdot b = AZ \cdot a$. This leads to $\displaystyle AZ = \frac{bc}{b-a}$. Similarly, $\displaystyle AY = \frac{bc}{a-c}$. But b - a = a - c, since b + c = 2a (just because CA + AB = 2 BC). Hence, AZ = AY, and we can claim victory.

  Darij

Here's what I did (let's rename everything :)).

Given triangle $ABC$ such that $CA+AB=2BC$, show that the external bisector of $\angle BAC$ is parallel to $OI$.

Let $A'$ be the point where $AI$ (the bisector of $\angle BAC$) cuts the circumcircle of $ABC$. It's well-known that $A'B=A'C=A'I=t$. We use Ptolemy in $ABA'C$ to get $\displaystyle tCA+tAB=AA'\cdot BC=AA'\cdot\frac{CA+AB}2$. This means that $AA'=2t=2A'I$, so $I$ is the midpoint of $AA'$, meaning that $OI$ is the perpendicular bisector of $AA'$, giving $OI\perp AI$. From here, the conclusion easily follows (the two bisectors of $\angle BAC$ are perpendicular).

Another proof (of: In triangle ABC, b+c=2*a. Let O, I be circumcentre, incentre. Prove AI is perpendicular to OI):
First using the following result (proved in http://www.mathlinks.ro/Forum/viewtopic.php?t=14655, the general version):
Let K be the point on BC s.t. BK=BC and L on AC s.t. LB=BC. then KL is perpendicular to OI.
it suffices to prove KL is parellel to AI. AI is jsut the angle bisector of <BAC. Also, AK=AL since a-c=b-a, so AKL is iscoceles and <AKL=a/2=<BAI. So KL is parellel to AI as required.


Hi,
Does anybody know the results and problems of CWMO 2007?Or the official web page?
Thank you.

To all members from China and for all people,who interested in this olympiad:
i've got a problems of the first day,but i don't know,may i post it here?
Maybe it is forbidden...
Thank you.

There's no reason why it should be forbidden. Go ahead.  :)

[quote="watchcatno1"]Well there have been a series of posts intended at cheating - Indian and Vietnamese just to name a few. Erken's apprehension is justified.[/quote]
What do you mean?Could you give me links of this posts,of course if they are still here.Thank you.

A Chinese website has the Chinese version of the problems:
http://www.aoshoo.com/bbs1/dispbbs.asp?boardID=15&ID=10880&page=1

Maybe some of us can translate it to English. Had I have time, I would do.

I have translate the problems into English for those friends who can not read Chinese.

if anyone has solutions for cwmo 2007 please post here:)

[quote="Erken"]May i post all of them here?[/quote]

Yes, you may do it in seperate posts and ask then a mod to link them on the contest section.

edit: so it was already done

Pay attention to the dates of posts SCP, Erken asked that in 2007, and has long since posted the problems for the Contests Section..

Basil11, discussions of the problems can be found here: http://www.artofproblemsolving.com/Forum/resources.php?c=37&cid=63&year=2007&sid=c92902d3a29e02c3513d6739958339f2


Help me this exercise!
Given ABC is an acute triangle. prove that
$ \frac{sinA\plus{}sinB\plus{}sinC}{cosA\plus{}cosB\plus{}cosC} \leq \frac{tanA\plus{}tanB\plus{}tanC}{3}$

Converting the right hand side in terms of sine and cosine, we have:

$ \frac {\tan{A} \plus{} \tan{B} \plus{} \tan{C}}{3} \equal{} \frac {\frac {\sin{A}}{\cos{A}} \plus{} \frac {\sin{B}}{\cos{B}} \plus{} \frac {\sin{C}}{\cos{C}}}{3} \equal{} \frac {\sin{A} \plus{} \sin{B} \plus{} \sin{C}}{3(\cos{A} \plus{} \cos{B} \plus{} \cos{C})}$

Now since ABC is an acute triangle, the sines and cosines of its angles are all positive. 1/3 of a positive number is always less than the original positive number, therefore the inequality

$ \frac {\sin{A} \plus{} \sin{B} \plus{} \sin{C}}{\cos{A} \plus{} \cos{B} \plus{} \cos{C}} \le \frac {\tan{A} \plus{} \tan{B} \plus{} \tan{C}}{3}$

is true.

I'm not really sure if this is a legit explanation or not...

[quote]$ \frac {\frac {\sin{A}}{\cos{A}} \plus{} \frac {\sin{B}}{\cos{B}} \plus{} \frac {\sin{C}}{\cos{C}}}{3} \equal{} \frac {\sin{A} \plus{} \sin{B} \plus{} \sin{C}}{3(\cos{A} \plus{} \cos{B} \plus{} \cos{C})}$[/quote]

That's not correct.

This is [[Chebyshev's Inequality]] on similarly sorted $ \{\sin A, \sin B, \sin C\}, \{\sec A, \sec B, \sec C\}$.

@azjps: Chebyshev's wouldn't work in this case.  For one thing, the side being multiplied by 3 is the one which, expanded, equals $ \tan{A} \plus{} \tan{B} \plus{} \tan{C}$.  Also, it would be required that secA+secB+secC=1/(cosA+cosB+cosC), too.

And I have some serious doubts about the truth of this inequality.  When multiplied out and rearranged, the inequality should be:
$ 2\sin{A} \plus{} 2\sin{B} \plus{} 2\sin(A \plus{} B) \plus{} \tan{A}\cos(A \plus{} B) \plus{} \tan{B}\cos(A \plus{} B) \plus{} \tan(A \plus{} B)\cos{A} \plus{} \tan(A \plus{} B)\cos{B} \text{(should be)} \leq \tan{A}\cos{B} \plus{} \tan{B}\cos{A}$.
If we look at the last two terms of the LHS and the two terms on the RHS, see that, if we make $ A \plus{} B$ extremely close to $ 90^{\circ}$, we can easily outmatch both the terms on the RHS with $ \tan{A}$ and $ \tan{B}$.

Oh wow..Sure enough, trying angles of 45, 46, and 89, we get that (sin45+sin46+sin89)/(cos45+cos46+cos89)=4.330...$ >$1.64...=(tan45+tan46+tan89)/3

In the future, it would be much appreciated if the given inequalities are known to be correct...

My apologies.  It turns out I was using radians, and I apologize for being a bit rash...

And Chebyshev's does indeed work.  I had just thought azjps had meant a direct application.

Chebyshev works. We get

$ \frac{\tan A \plus{} \tan B \plus{} \tan C}{3} \geq (1/9)(\sin A \plus{} \sin B \plus{} \sin C)(\sec A \plus{} \sec B \plus{} \sec C)$. We now wish to show that $ \frac{\sec A \plus{} \sec B \plus{} \sec C}{9} \geq \frac{1}{\cos A \plus{} \cos B \plus{} \cos C}$, i.e., $ \frac{\sec A \plus{} \sec B \plus{} \sec C}{3} \geq \frac{3}{\frac{1}{\sec A} \plus{} \frac{1}{\sec B} \plus{} \frac{1}{\sec C}}$, but this is just AM-HM, so we're done.

$$\sqrt{3}\le \frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}<2$$
[url]https://artofproblemsolving.com/community/c4h2859064p25374615[/url]
$$\displaystyle{\frac{\cos A+\cos B - \cos C}{\sin A+\sin B - \sin C}} \geq -\frac{\sqrt{3}}{3}$$
[url]https://math.stackexchange.com/questions/124466/prove-that-displaystyle-frac-cos-a-cos-b-cos-c-sin-a-sin-b-sin-c[/url]
n any acute triangle ,we have:
$$\sin A + \sin B + \sin C >\sqrt{2}(\cos  A + \cos B + \cos C).$$
[url]https://artofproblemsolving.com/community/c6h17035p15727576[/url]
[url]https://artofproblemsolving.com/community/c6h1191507p5813666[/url]


Try this problem!

Find all polynomials f(x) such that f(x)+f(x+2)=2f(x+1)
for all real x.

I don't find it difficult, nevertheless interesting!  :)  :D

I'm really bad at this type of problem.

[hide="I found 1"]

$f(x) = x$ definitely works because $f(x) = x, f(x+2) = x+2, f(x) + f(x+2) = x+x+2 = 2x+2$. And $2f(x+1) = 2(x+1) = 2x+2$.[/hide]

I'll have to think to find the other ones..  :huh:


Try this one. I think it is very nice but I am not sure if it is strict or not:
$a,b,c \geq 0$ Prove:
$5(a^4+b^4+c^4)+17(a^2b^2+b^2c^2+c^2a^2) \geq 4abc(a+b+c)+18(a^3b+b^3c+c^3a)$
I am sure it is correct 100%  :D
Have fun with this

There is a very effective way to solve homogeneous ineqs . We change the problem into $S= A(b-c)^2+B(a-c)^2+C(a-b)^2 \geq 0$
 In this problem : $A= 14b^2+5bc+4a^2-4c^2$
  We always have $A+B \geq 0$ ,$A+C \geq 0$ , $C+B \geq 0$
 If $(b-a)(b-c) \leq 0$ then $B \geq 0$
 So $S \geq 0$


function
algebra
domain
trigonometry
search
functional equation
real analysis

find all reals fonction :
$ f(x \plus{} y) \plus{} f(x \minus{} y) \equal{} 2f(x)f(y)$

Hey, if we substitute to our functional equations $ x\equal{}y\equal{}0$ then we get:

          $ 2f(0)\equal{}2f^2(0)$ or $ f(0)\equal{}0$ or $ f(0)\equal{}1$.

 [u][b]Case 1[/b][/u]: $ f(0)\equal{}0$

  Then by substituting $ y\equal{}0$ in our functional equation we get:

    $ 2f(x)\equal{}2f(x)f(0)\equal{}0$ hence $ f(x)\equal{}0$ and that's the only function that satisfies the condition.

          [u][b]Case 2[/b][/u]: $ f(0)\equal{}1$

  Then by substituting $ x\equal{}0$ we receive:   $ f(y)\plus{}f(\minus{}y)\equal{}2f(0)f(y)\equal{}2f(y)$ or $ f(y)\equal{}f(\minus{}y)$ and
hence all the even functions satisfy the functional equation.

[quote="new_member"]Then by substituting $ x \equal{} 0$ we receive:   $ f(y) \plus{} f( \minus{} y) \equal{} 2f(0)f(y) \equal{} 2f(y)$ or $ f(y) \equal{} f( \minus{} y)$ and hence all the even functions satisfy the functional equation.[/quote]No, this only implies that any solution of the functional equation is necessarily even.

This is known as D'Alembert's functional equation.  If the domain is all of $ \mathbb{R}$ some solutions are $ \cos (cx)$ and $ \cosh (cx)$ for arbitrary $ c$.

[b]Joeblow[/b],thanks for pointing out my mistake.Could you please provide any links for D'Alembert's functional equation?
A quich search on wiki didn't give me anything.

If you have access to JSTOR or can find this in your library:

[url=http://www.jstor.org/pss/2323650]The D'Alembert Functional Equation[/url], F. J. Papp, The American Mathematical Monthly, Vol. 92, No. 4 (Apr., 1985), pp. 273-275.

Edit: Google also gives [url=http://books.google.com/books?id=8EWnEh18rVgC&source=gbs_navlinks_s]this E-book[/url], pages 103-105 contain a proof and are available free.  Seems they build the solution on a discrete set and then use a continuity argument, so this does not resolve the general question of non-continuous solutions.

Thank you,[b]JoeBlow[/b].I appreciate you're sharing your information.


USAMO
inequalities
function
inequalities proposed

Here's another method:[hide]Let the numbers be denoted by xi.  Let s be the sum of the numbers, and t be the sum of their squares.  Because n :ge: 4, either s :ge: 2n or s :le: 3n - 4.  We will divide the proof into those two cases.

Assume that s :ge: 2n.  Then the average of the numbers is at least 2, which means that the maximum number is at least 2, so we are done.

Assume that s :le: 3n - 4.  Consider the sum of (xi - 2)(xj - 2) over all i <> j.  We will show that this sum is nonpositive.  If we can do that, then at least one term of this sum must be nonpositive, which means that some xi is at least 2, so we would be done.

Okay, let's analyze the sum in the previous paragraph.  A little algebraic manipulation shows that this sum is equal to (s - 2(n-1))2 - t + 4n - 4.  Because s is between n and 3n - 4, the term (s - 2(n-1))2 is at most (n - 2)2.  Also the problem says that t :ge: n2.  Plugging those inequalities back shows that the sum from the previous paragraph is indeed nonpositive.  Thus we are done.  [/hide]

Here's yet another method: [hide]Again, let the numbers be denoted by xi, let s be the sum of the numbers, and let t be the sum of their squares.  Consider the sum of (xi - 2)(xi - 2 + n) over all i.  A little algebra shows that this sum is equal to t + (n - 4)s - 2(n - 2)n.  Because t :ge: n2, s :ge: n, and n :ge: 4, we see that the sum is at least 0.  Hence at least one term of the sum is nonnegative.  That means for some i, either xi :ge: 2 or xi :le: 2 - n.  The first case means we are done.  In the second case, the sum of the other n - 1 numbers must be at least 2n - 2, which means that one of the other numbers is at least 2, and we are again done.[/hide]

Ravi B's sum :Sigma:  = :Sigma: (xi - 2)(xi - 2 + n) made the problem look so simple.

[hide]Assume that for each i, xi < 2. Then

s - xi < 2(n-1), where s =  :Sigma: xk  :ge: n

that is, xi > s - 2(n-1)  :ge:  n - 2(n-1) = 2 - n

or xi-2+n > 0

So (xi - 2)(xi - 2 + n) < 0 for each i.

That is,  the sum :Sigma:  is negative.

On the other hand the sum :Sigma:  is  nonnegative as demonstrated above. A contradiction!

[/hide]

The Kalva solution uses something called the smoothing principle, which ends up being fairly important in inequalities...

Here was an idea I had when working on this problem...

Proof is by contradiction ---

For each i, define

[tex] x_i = 2 - a_i [/tex].

Our assumption is that all the xi are non-negative.

Then

[tex] \sum a_i \ge n \Rightarrow \sum x_i \le n[/tex]

[tex] \sum a_i^2 \ge n^2 \\
\Rightarrow
4n - 4 \sum x_i + \sum x_i^2 \ge n^2 \\
\Rightarrow
n^2 - 4n \le \sum x_i^2 - 4\sum x_i < (\sum x_i)^2 - 4\sum x_i (*)\\
\le n^2 - 4n (**).
[/tex]

(*) Follows since all the xi are nonnegative.

(**) Follows since n :ge: 4 and [tex] \Sum x_i \le n [/tex].

Contradiction.

Assuming my proof is right... I have found quite a few problems that can be solved using this neat substitution of assigning variable quantities to the distance each variable is from where you want it to be ... its a cool trick, when it works.

Nice solution!  I think you are right that working with positive variables is easier than working with variables less than say 2.  That way, inequalities like  :Sigma: xi2 < (:Sigma: xi)2 become more apparent.

Yet another way!

I'll first need a (relatively simple) [b]lemma[/b]: Given non-negative reals [tex]p_1,\ldots,p_t[/tex] with [tex]\sum p_k^2=q[/tex], we have [tex]\sum p_k\ge\sqrt{q}[/tex].

[b]Proof[/b]: The proof is rather simple. We have [tex]\left(\sum p_k\right)^2 = p_1^2+\cdots+p_t^2+(\text{other non-negative terms})\ge q[/tex], which establishes the result. The equality case, though easy to find, is unnecessary for this proof. [tex]\blacksquare[/tex]

Let [tex]x_1, ... , x_n[/tex] be a set of reals each less than 2, with [tex]\sum x_k^2\ge n^2[/tex]. They can't all be positive, since then we would have [tex]\sum x_k^2 < \sum 2^2 = 4n \le n^2[/tex], a contradiction. Thus, we can divide them into two classes: the positive ones [tex]a_1, \ldots, a_m[/tex] (each less than 2) and the non-positive ones [tex]b_1,\ldots,b_{n-m}[/tex] where m is an integer less than n. Notice that

[tex]\displaystyle 4m+\sum_{k=1}^{n-m}b_k^2 > \sum_{k=1}^{m} a_k^2 + \sum_{k=1}^{n-m}b_k^2 = \sum_{k=1}^n x_k^2 \ge n^2[/tex], i.e.

[tex]\displaystyle \sum_{k=1}^{n-m} b_k^2 \ge n^2-4m[/tex].

We can now apply the lemma to [tex]-b_1,\ldots, -b_{n-m}[/tex] to find that

[tex]\displaystyle \sum_{k=1}^{n-m}(-b_k) \ge \sqrt{\sum_{k=1}^{n-m}b_k^2} = \sqrt{n^2-4m} [/tex].

Finally, [tex]\displaystyle \sum_{k=1}^n x_k = \sum_{k=1}^m a_k - \sum_{k=1}^{n-m}b_k < 2m -\sqrt{n^2-4m}[/tex].

Now, the function [tex]f(m)=2m-\sqrt{n^2-4m}[/tex] is an increasing function of m, and so it is maximized when m is maximized, i.e. when m=n-1. Thus,

[tex]f(m)\le f(n-1) = 2n-2-\sqrt{n^2-4n+4} = n[/tex].

Therefore, we see that [tex]\sum x_k < 2m -\sqrt{n^2-4m} \le n[/tex], and so both inequalities in the problem cannot be simultaneously satisfied if all the x_i are less than 2.

[quote="Nukular"]
Assuming my proof is right... I have found quite a few problems that can be solved...[/quote]

Can you share the problems? Thanks!

The one im thinking of in particular was on the first MOP test in Blue this year... I should probably wait until after mop ends to post it...

Another solution, which is pretty fast (drilling for USAMO):

The proof is easy for n=4, so I won't put it up.

For n>4, assume all a_i<2.  Then sum(a_i)<2n.           (1)

Assume sum(a_i)>=m (We know m is at least n.)
Then 2sum(a_i)+2sum{j not equal to i}(a_i*a_j)=
              sum[a_i(2+sum{j not equal to i}(a_j))]>=
              sum(a_i*m)>=
              m^2.                                                               (2)

By (1) and (2), 2sum{j not equal to i}(a_i*a_j)>=m^2-4n, so
sum(a_i^2)+2sum{j not equal to i}(a_i*a_j)=[sum(a_i)]^2>=m^2 + n^2-4n.
But then sum(a_i)>=sqrt(m^2 + n^2 - 4n), which is greater than m for all n>4.  We may then repeat the algebra done after (2), substituting sqrt(m^2 + n^2 - 4n) for m.  But then sum(a_i) has no lower bound.  Contradiction.

Daniel Litt

I'm confused, why can't all the numbers be 1?

Then the sum-of-squares condition would not be satisfied.

[hide="another method?"]
WLOG, $ a_n \ge a_{n\minus{}1}\ge...\ge a_1$. Thus,
$ na_n \ge n$
$ a_n \ge 1$
And
$ na_n^2 \ge n^2$
$ a_n^2 \ge n$
Since n>3 and n is an integer, $ n \ge 4$. Thus,
$ a_n^2 \ge 4$
$ a_n \ge 2$. Since n is positive. [/hide]

How do you show that $ n a_n^2 \ge n^2$?

The second equation...$ \sum a_i^2 \ge n^2$

How do you know that $ a_n^2 \ge a_i^2$?

This is tricky for a #4, IMHO.

Say we have $ i$ negative terms and $ n\minus{}i$ nonnegative terms among $ a_{1},a_{2},\ldots,a_{n}$. WLOG, let the negative ones be $ a_{1},a_{2},\ldots,a_{i}$. Assume that $ a_{1},\ldots,a_{n}<2$. Note that $ i\ge1$, or else $ a_{x}^{2}<4$ for all $ x$ (as $ 0\le a_{x}<2$), giving $ a_{1}^{2}\plus{}a_{2}^{2}\plus{}\cdots\plus{}a_{n}^{2}<4n\le n^{2}$ for $ n\ge4$, contradiction. Also, we obviously have $ n\ge i$.

We have $ a_{1}\plus{}a_{2}\plus{}\cdots\plus{}a_{n}\ge n$, and $ a_{i\plus{}1}\plus{}a_{i\plus{}2}\plus{}\cdots\plus{}a_{n}<2(n\minus{}i)$ by assumption. Thus, $ a_{1}\plus{}a_{2}\plus{}\cdots\plus{}a_{i}>n\minus{}2i$. Note that

$ a_{1}^{2}\plus{}a_{2}^{2}\plus{}\cdots\plus{}a_{i}^{2}<(a_{1}\plus{}\cdots\plus{}a_{i})^{2}$

since $ a_{x}<0$ for all $ 1\le x\le i$; this is clear upon expanding the RHS. Now,

$ a_{1}^{2}\plus{}a_{2}^{2}\plus{}\cdots\plus{}a_{n}^{2}$
$ \equal{}(a_{1}\plus{}\cdots\plus{}a_{i})^{2}\plus{}4(n\minus{}i)$
$ <(n\minus{}2i)^{2}\plus{}4(n\minus{}i)$
$ <n^{2}\minus{}4(i\minus{}1)(n\minus{}i)$
$ \le n^{2}$,

contradiction.

Hmmm, maybe my method seems super easy and stuff...

I don't see anything WRONG, though. There probably is something. Can you guys review this? :)

We proceed by contradiction. Assume that all of the terms are less than 2. We then have

$ a_1^2 \plus{} a_2^2 \plus{} \dots \plus{} a_n^2 < 4n$

because the maximum value for each term is 4.

However, we are given
$ a_1^2 \plus{} a_2^2 \plus{} \dots \plus{} a_n^2 \geq n^2$, and this is not always true with the previous inequality.

Therefore, not all of the terms are less than 2, i.e. the highest term must be greater than or equal to two.

Right?  :huh:

The terms can be negative. Some can be less than $ \minus{}2$, and thus your first inequality doesn't hold. We can even have an equality case: $ n\minus{}1$ copies of $ 2$ and one of $ \minus{}n\plus{}2$.

I'm pretty sure I got this problem right when I took it live. Of course, that was ten years ago and I don't remember details.

[hide]Assume all the $a_i$ are less than 2. Let $\min(a_1, \dots, a_n) = a_n.$ It follows that $a_n > 2-n.$ If we have $a_1 = a_2 = \dots = a_{n-1} = 2, a_n = 2-n,$ we have
\[ a_1^2 + \dots + a_n^2 = 4(n-1) + (2-n)^2 = n^2, \] which is the maximum sum we can get from the numbers. But since, $a_1, \dots, a_{n-1} < 2,$ and $a_n > 2-n,$ which is clearly negative, we have that \[ a_1^2 + \dots + a_n^2 < n^2, \] which is a contradiction. [/hide]

[quote]If we have $a_1 = a_2 = \dots = a_{n-1} = 2, a_n = 2-n,$ we have
\[ a_1^2 + \dots + a_n^2 = 4(n-1) + (2-n)^2 = n^2, \] which is the maximum sum we can get from the numbers. [/quote]
While this statement is actually true, the justification for this is not strong enough. For instance, you could have $a_1 < -2$; in this case, $a_1^2 > 4$.

[hide="My solution"]Suppose for the sake of contradiction that $\max(a_1, a_2, \ldots, a_n) < 2$. If we let $b_i = 2 - a_i$, we have $b_1, b_2, \ldots, b_n > 0$.

Suppose without loss of generality that $b_1 \geq b_2 \geq \cdots \geq b_n$. Then . $a_1 + a_2 + \cdots + a_n \geq n$ is equivalent to $b_1 + b_2 + \cdots + b_n \leq n$, and $a_1^2 + a_2^2 + \cdots + a_n^2 \geq n^2$ is equivalent to $(2-b_1)^2 + (2-b_2)^2 + \cdots + (2-b_n)^2 \geq n^2$.

Let $f(x) = (2-x)^2$. $f$ is convex, and $(b_1 + b_2 + \cdots + b_n, \underbrace{0, 0, \ldots, 0}_{n-1})$ majorizes $(b_1, b_2, \ldots, b_n)$, so by Karamata's inequality,
$f(b_1) + f(b_2) + \cdots + f(b_n) \leq f(b_1 + b_2 + \cdots + b_n) + (n-1)f(0).$

Observe that $0 \leq b_1 + b_2 + \cdots + b_n \leq n$. Since $f$ is convex, $f(b_1 + b_2 + \cdots + b_n) \leq \max(f(0), f(n))$. $f(0) = 4 \leq n^2 - 4n + 4 = f(n)$ since $n \geq 4$, so $f(b_1 + b_2 + \cdots + b_n) \leq f(n) + (n-1)f(0) = n^2 - 4n + 4 + 4n - 4 = n^2$, with equality holding if and only if $(b_1 + b_2 + \cdots + b_n, \underbrace{0, 0, \ldots, 0}_{n-1}) = (b_1, b_2, \ldots, b_n)$, which cannot happen since $b_n > 0$. Therefore, $n^2 \leq f(b_1) + f(b_2) + \cdots + f(b_n) < n^2$, which is a contradiction.

(I am aware that a solution nearly identical to this (the use of Karamata's inequality was instead called "smoothing"), though the solution is written using the deprecated [tex][/tex] tags, rendering it very difficult to read.) [/hide]

I am not quite sure of what smoothing is. Is this solution right because it seems very easy because the problem seems very weak.
[hide="Solution"]
Assume for the sake of contradiction that all of the $a_i$'s are less than $2$. Therefore, we can make $a_i=2-b_i$ where $b_i$ is positive. The first inequality the becomes $ \sum b_{i} < n$. We then apply this to the second inequality and we get that $ \sum b_i^{2}-4b_i \ge n^2 - 4n$. Now we have that $b_i<n$ because the $b_i$'s are positive and have sum $< n$. Therefore we get that $\sum b_i^{2}-4b_i = \sum b_i(b_i-4) < \sum b_i (n-4) < n(n-4) = n^2-4n$. Contradiction, therefore, the max is $>2$.
[/hide]

Excuse me if I am wrong, but is this trivialized by $n-1$-EV?

It suffices to show that $a_i<2\rightarrow \sum a_i^2<n^2$.  If all are positive, the sum of squares is $<4n\le n^2$ and we are done.  If there are two negatives $x,y$, then $(x+y)^2+0^2$ yields a larger sum of squares.  So WLOG there is one negative.  Taking this negative $a$ and a positive $b<2$, then $(a-2+b)^2+2^2$ yields a larger sum of squares (it is equivalent to $(2-a)(2-b)>0$.  So the absolute max is $n-1$ copies of $2$ and $1$ copy of $-n+2$, which yields a sum of squares of $n^2$.  However equality can't be achieved so we are done.

Is this rigorous enough:

Suppose on the contrary that $max(a_1, a_2, ..., a_n) < 2$. Let us partition the set of {$a_i$} into {$b_i$} of nonnegative integers and {$c_i$} of negative integers. First, we prove that {$c_i$} is nonempty.

If it was empty, then $\sum a_i^2 < 4n \leq n^2$, contradiction.

WLOG, $a_1 \geq a_2 \geq \dots \geq a_n$.

We now perform a series of steps; each increasing one of $s_1 = a_1 + a_2 + \dots + a_n$ and $s_2 = a_1^2 + a_2^2 + \dots + a_n^2 \geq n^2$ without breaking any bounds, until we get a maximal sum that leads to a contradiction.

This works, since a maximal sum that didn't break any of our bounds contradicting the max $a_i \geq 2$ statement implies that none of the smaller  {$a_i$} series work, since each can be "smoothed out" to each other.

Now for our steps:

$\textbf{Step 1}$: We can increase $b_i$ to 2-$\epsilon$ for some infinitesimally small $\epsilon$. This increases $s_1$ and $s_2$.

$\textbf{Step 2}$: Since $\bigg(\sum c_i\bigg)^2 \geq \sum c_i^2$, we can collect all the $c_i$ terms into $a_n$ without changing $s_1$ and increasing $s_2$. This adds zeros to $b_i$.

$\textbf{Step 3}$: Repeat step 1 for the zero's.

We now have n-1 $2-\epsilon$'s and a large negative number. We can now perform the proof by contradiction:

From the first inequality we have:

$$ \sum a_i \geq n \rightarrow (n-1) \cdot (2 - \epsilon) + a_n \geq n \rightarrow n-2 > -a_n \rightarrow a_n > 2-n$$
From this, we also have:

$$ \sum a_i^2 < 4 \cdot (n-1) + a_n^2 < 4\cdot (n-1) + (n-2)^2 = n^2$$
Which contradicts the second inequality. Hence, $max(a_1, a_2, \dots a_n) \geq 2$.

Similar to above, but arrives at finish [i]slightly[/i] differently.
Assume for sake of contradiction that $\max(a_i)<2$. WLOG let $a_1\leq a_2 \leq \cdots \leq a_n$. Furthermore denote $s=a_1+a_2+\cdots+a_n$. We have that $n\leq s<2n$. If $a_1\geq 0$, we have that.
$$4n>a_1^2+a_2^2+...+a_n^2\geq n^2 \implies 4n>n^2\implies n<4$$But $n>3$ and $n$ is an integer, so we have a contradiction, and we are done. Hence, we can assume $a_1<0$. Now observe that $x^2$ is convex, hence we can unsmooth by replacing $a_1$ and $a_i$ with $a_1+a_i-2$ and $2$, for all $i\in{2,3\cdots n}$. We end up with
$$\sum_{k=1}^{n} a_k^2<(s-2(n-1))^2+(n-1)\cdot 2^2=a_1^2+4(n-1)$$
Note that $a_1+2(n-1)\leq n \implies a_1\geq 2-n$. Since $a_1<0$, we have that $a_1^2\leq (2-n)^2=(n-2)^2$. Thus
$$\sum_{k=1}^{n} a_k^2=a_1^2+4(n-1)\leq (n-2)^2+4(n-1)=n^2$$ which is a contradiction, so we win. $\square$

If anyone had a solution with $n-1$ EV to share, that would be great.

[Hide = Proofread for my friend please]
We have $max(a_i^2) \geq n$, else the second condition does not hold. WLOG let $min(a_i) = a_1$ and $max(a_i) = a_n$ Also, let $a_1 = -(\sqrt{n} + j)$ for positive real $j$.

We have:
$-(\sqrt{n} + j) + a_2 + ... + a_n \geq n$
$a_2 + ... + a_n \geq n + \sqrt{n} + j$
$a_n(n - 1) \geq a_2 + ... + a_n \geq n + \sqrt{n} + j$
$a_n(\sqrt{n} + 1)(\sqrt{n} - 1) \geq \sqrt{n}(\sqrt{n} + 1) + j$
$a_n \geq \frac{\sqrt{n}}{\sqrt{n} - 1} + \frac{j}{n - 1}$
$a_n \geq 1 + \frac{1}{\sqrt{n} - 1} + \frac{j}{n - 1}$

Case 1: $\frac{1}{\sqrt{n} - 1} + \frac{j}{n - 1} \geq 1$
Which immediately gives us $a_n \geq 2$

Case 2: $\frac{1}{\sqrt{n} - 1} + \frac{j}{n - 1} < 1$
$\frac{j}{(\sqrt{n} + 1)(\sqrt{n} - 1)} < \frac{\sqrt{n} - 2}{\sqrt{n} - 1}$
$j < n - \sqrt{n} - 2$ -----$(1)$

By second equation, we have:
$(n + j^2 + 2j\sqrt{n}) + a_2^2 + ... + a_n^2 \geq n^2$
$(n - 1)a_n^2 \geq n(n - 1) - j(j - 2\sqrt{n})$
$a_n^2 \geq n - \frac{j(j - 2\sqrt{n})}{(n - 1)}$
$a_n^2 \geq n - \frac{(n - \sqrt{n} - 2)(n + \sqrt{n} - 2)}{(n - 1)}$ (From $(1)$)
$a_n^2 \geq n - \frac{n^2 - 5n + 4}{n - 1} = n - \frac{(n - 4)(n - 1)}{(n - 1)} = n - n + 4 = 4$
$a_n \geq 2$

Hence proved. The positive real case is trivial to prove

I am sleepy now, so I am not sure, but what's wrong with the below solution?
After seeing all these solutions, I am feeling like I have made a silly her, bit can't find it
WLOG, Assume $a_n\geq a_{n-1}\geq\ldots\geq a_1$
Then, $na_n\geq a_1+a_2+\ldots+a_n\geq n\implies a_n\geq 1>0$.
And, $na_n^2\geq a_1^2+a_2^2+\ldots+a_n^2\geq n^2\implies na_n^2\geq n^2$.
As $a_n$ greater than 0, $a_n\geq\sqrt{n}$. As $n\geq 4$, $a_n\geq 2$

BTW, this is also a PUTNAM question iirc

Umm, can someone plz check th solution, I am really in a confused state

[quote=tetrahedr0n]Let $a_{1}, a_{2}, \dots, a_{n}$ ($n > 3$) be real numbers such that \[ a_{1} + a_{2} + \cdots + a_{n} \geq n \qquad \mbox{and} \qquad a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2} \geq n^{2}.  \] Prove that $\max(a_{1}, a_{2}, \dots, a_{n}) \geq 2$.[/quote]

This is a beautiful problem

[quote=tetrahedr0n]Let $a_{1}, a_{2}, \dots, a_{n}$ ($n > 3$) be real numbers such that \[ a_{1} + a_{2} + \cdots + a_{n} \geq n \qquad \mbox{and} \qquad a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2} \geq n^{2}.  \] Prove that $\max(a_{1}, a_{2}, \dots, a_{n}) \geq 2$.[/quote]

[hide=Hint]Those squares are saying something! Use Cauchy Swharz. ;)[/hide]

[quote=mihaig][quote=Math-wiz]I Then, $na_n\geq a_1+a_2+\ldots+a_n\geq n\implies a_n\geq 1>0$.
And, $na_n^2\geq a_1^2+a_2^2+\ldots+a_n^2\geq n^2\implies na_n^2\geq n^2$.[/quote]
Try $2,2,2,2,-3$ . Then $na_n^2< a_1^2+a_2^2+\ldots+a_n^2=n^2$ .[/quote]

Ohh! Yes, got your point

Does this work?
[hide = my attempt]
WLOG $a_n\geq a_{n-1}\geq \cdots a_1$.
Now for any integer $i$, with $1\leq i\leq n-1$, we define \[p_{ni}:= \frac{(a_n-a_i)a_i}{a_n+a_i}\;\;q_{ni}:=\frac{(a_n-a_i)a_n}{a_n+a_i}\]
Notice that since $a_n\geq a_i$ by assumption, $p_{ni}\leq q_{ni}$.
We the replace $(a_n,a_i)$ with $(a_n-p_{ni},a_i+q_{ni})$ which evidently preserves the fact that $\sum a_i\geq n$.\\
On the other hand since \[(a_n-p_{ni})^2+(a_i+q_{ni})^2 = a_n^2+p_{ni}^2+a_i^2+q_{ni}^2-2\frac{a_na_i(a_n-a_i)}{a_n+a_i}+2\frac{a_na_i(a_n-a_i)}{a_n+a_i} = a_n^2+a_i^2+p_{ni}^2+q_{ni}^2\geq a_n^2+a_i^2\]
thus it also preserves $\sum a_i^2 \geq n^2$.
Note further that if we keep doing this operation to any $a_k\geq a_l$, we would eventually arrive at a state where $a_n = a_{n-1} = \cdots =a_1$, for which since \[na_n^2\geq n^2\Rightarrow a_n\geq \sqrt{n}\geq 2\] we are done
[/hide]

For the sake of contradiction assume that $\max{(a_i)}<2$.
Suppose we have no negative reals, then $$4n>a_{1}^{2} + a_{2}^{2} + \ldots + a_{n}^{2} \geq n^{2}\implies 4>n,$$which contradicts $n\geq 4$.

Let $i<n$ denote the number of positive reals, $a_1,\ldots,a_i$. The rest, $a_{i+1},\ldots,a_n$, are negative. Thus,
$$a_{i+1}+\ldots+a_n> n-2i\Longleftrightarrow 2i-n>\vert a_{i+1}\vert +\ldots+ \vert a_{n}\vert $$
and $$a_{i+1}^2+\ldots+a_n^2> n^2-4i.$$
Thus, $$(2i-n)^2>(\vert a_{i+1}\vert +\ldots+ \vert a_{n}\vert )^2\geq a_{i+1}^2+\ldots+a_n^2> n^2-4i.$$
This is equivalent to $$4i^2-4ni>-4i\Longleftrightarrow i>n-1.$$
This arrive at a contradiction. We conclude that $\max{(a_i)}\geq 2$.

This problem dies fairly easily to smoothing. Suppose for the sake of contradiction that such a sequence existed with all $a_i < 2$. The point is that we can actually make most of the variables $2-\varepsilon$ and one of the variables $-n + 2 + (n-1) \varepsilon$ by fixing some $a_i$, and pushing $a_i$ and $a_j$ apart for each $j \neq i$ such that $a_j = 2 -\varepsilon$. Under this process the value of $\sum a_i^2$ increases while $\sum a_i$ stays constant.

So it suffices to show the problem cannot work in that case. Fortunately, this is obvious, as $$\sum_{i=1}^{n} a_i^2 = (n-2)^2 - 2n(n-1) \varepsilon + (n-1)^2 \varepsilon^2 + 4(n-1) - 2(n-1) \varepsilon = n^2 - (n-1)^2 \varepsilon < n^2.$$

Solved with otis walkthrough.

Suppose there existed a sequence with $a_i < 2$ for all $i$, and the rest of the problem conditions were true. If everything was positive, then the sum of the squares is less than $n\cdot (2^2) = 4n \le n^2$, contradiction. Now we assume there is at least one negative term. For any two $i\ne j$ with $a_i$ and $a_j$ both negative, we can replace $a_i$ and $a_i$ with $a_i + a_j$ and $0$ and the conditions still must hold true. For any $i$ with $a_i \ge 0$, we can replace $a_i$ with $2 - \epsilon$ for some arbitrarily small positive constant $\epsilon$. Now we can assume there is only one negative term, and the rest of the terms are equal to $2 - \epsilon$. Let the negative term be equal to $-d$.

We have $n \le \sum a_i \le 2(n-1) - d$, so $d\le n - 2$. Hence $\sum a_i^2 <  2^2 \cdot (n-1) + (n-2)^2 = n^2$, contradiction. Therefore, there exists a term in the sequence that is at least $2$.

Assume for contradiction $b_i = 2-a_i > 0$ for each $i$. Let $S = \sum b_i$. Note that
\[S = \sum b_i = \sum (2-a_i) = 2n - \sum a_i\le 2n - n = n.\]
We also have
\[ n^2\le \sum a_i^2 = \sum (2-b_i)^2 = 4n - 4S + \sum b_i^2\implies \sum b_i^2\ge n^2 - 4n + 4S.\]
But note that $\sum b_i^2 < S^2$ as all $b_i$ are positive. Thus we have
\[ S^2 > n^2 - 4n + 4S\implies (S-2)^2 > (n-2)^2\implies S < 4-n \text{ or }S > n.\]
We have $4-n \le 0$ so we cannot have $S < 4-n$. We have also shown that $S\le n$, so $S > n$ is impossible as well. Thus we have arrived at a contradiction.


$ABC$ is a triangle. a line parallel to $BC$, intersects $AB,AC$ in points $L,K$.

a line parallel to $AC$, intersects $BC,AC$, in points $N,M$.

a line parallel to $AB$, intersects $BC,AC$, in points $P,Q$.

prove that:

the intersection of $AB,KN$.

the intersection of $BC,MQ$.

the intersection of $CA,PL$.

are colinear...

:rotfl:  i dint know the problem is too dificult...

[quote="amir2"]I know who is vasheghani , I have meet him once.
He is so good at geometry[/quote]

ya he is really good in geometry...

know hi is learning for imo 2006...

A classical, nice and very easy problem !

[hide="Touch here and see the my solution"]
I note the intersections $X\in BC\cap MQ$, $Y\in CA\cap LP$, $Z\in AB\cap KN$. From the Menelaus' theorem applied to the transversals $\overline {XMQ}$, $\overline {YLP}$, $\overline {ZNK}$ for the triangle $ABC$ result the following relations (with orientated segments):

$\frac{XB}{XC}\cdot \frac{QC}{QA}\cdot \frac{MA}{MB}=1;$

$\frac{YC}{YA}\cdot \frac{LA}{LB}\cdot \frac{PB}{PC}=1;$

$\frac{ZA}{ZB}\cdot \frac{NB}{NC}\cdot \frac{KC}{KA}=1$.

But $\frac{KA}{KC}=\frac{LA}{LB},\ \frac{MB}{MA}=\frac{NB}{NC},\ \frac{PC}{PB}=\frac{QC}{QA}.$

From the product of the all above relations we obtain the relation

$\frac{XB}{XC}\cdot \frac{YC}{YA}\cdot \frac{ZA}{ZB}=1$, i.e. the points $X$,
$Y$, $Z$ are collinearly.

[u]Remark.[/u] Here is a generalization of this problem (with orientated segments):

$\triangle ABC$, $\{A_1,A_2\}\subset BC$, $\{B_1,B_2\}\subset CA$, $\{C_1,C_2\}\subset AB$,

$X\in BC\cap B_1C_2$, $Y\in CA\cap C_1A_2$, $Z\in AB\cap A_1B_2.$

$\left(\frac{BA_1}{A_1C}\cdot \frac{CB_1}{B_1A}\cdot \frac{AC_1}{C_1B}\right)\bullet\left(\frac{BA_2}{A_2C}\cdot \frac{CB_2}{B_2A}\cdot \frac{AC_2}{C_2B}\right)=1\Longrightarrow Z\in XY.$

You can analyse and another particular cases of the last general problem !
[/hide]
\[ \overline {\underline {\left|\ Merry\ Christmas\ !\ \right| }}  \]


function
calculus
derivative
real analysis
real analysis unsolved

Hi,

what convergence theorems are there for
functions
$ L^1([\minus{}\pi,\pi]) \ni f: [ \minus{} \pi,\pi] \longrightarrow \mathbb{C}$
which are $ \emph{only}$ assumed to be differentiable
throughout $ [\minus{}\pi,\pi]$ but nothing more? In particular,
it is not assumed that $ f'$ is continuous? Are there
examples of such a function for which there is at least
one point in $ [\minus{}\pi,\pi]$ for which the Fourier partial sums
do not converge to the function value? After all, there
continuous functions which exhibit such behaviour. What
can be said in the case of differentiable functions having
a derivative which is not continuous? Something $ \emph{must}$
be worse than in the continuously-differentiable case. Does
anyone know illustrative examples?

Thank you,

.,.

One can have a function which is differentiable everywhere but which has a very badly behaved derivative. For instance:

$ f(x)\equal{}x^2\sin(x^{\minus{}10}),$ extended to be zero at zero. That's differentiable everywhere, but the derivative is pretty convincingly not locally integrable around the origin.

That's not directly an example for Fourier series, but because of examples like that, I strongly suspect that one cannot state any positive theorems about convergence for everywhere differentiable functions that one couldn't also state for continuous functions.

On the other hand, weak derivatives are more interesting and useful than pointwise classical derivatives, anyway.

Suppose the weak derivative of $ f$ on $ \mathbb{T}$ is an $ L^1$ function? That is the same as $ f$ being absolutely continuous.

Or even weaker: suppose the weak derivative of $ f$ on $ \mathbb{T}$ is a bounded (signed) measure? That is the same as $ f$ being of bounded variation. Note that absolute continuity implies bounded variation, so this second condition is more general.

One may directly prove that if $ f$ is of bounded variation, then $ S_n(f)$ converges pointwise everywhere to the average of the right and left hand limits of $ f.$ (Functions of bounded variation can have at worst countably many jump discontinuities.) It follows from this that if $ f$ is absolutely continuous, then $ S_nf\to f$ everywhere.

Well, the standard Dini criterion for convergence ($ S_nf(x)\to f(x)$ whenever $ \frac{f(x\plus{}t)\plus{}f(x\minus{}t)\minus{}2f(x)}{t}\in L^1$ as a function of $ t$) certainly ensures that pointwise differentiability suffices for pointwise convergence. A good (and not too hard) exercise is to construct an example for which the convergence is not uniform, which shows that something is, indeed, worse ($ C^1$ is enough for absolute uniform convergence). At last, the interesting question I'll leave unanswered is whether it is possible to fix some rate of convergence and ensure that the Fourier series converges at that rate a.e. (for $ C^1$, the answer is "Yes" even for absolute convergence (just use Cauchy-Schwarz on the tail of the series; for $ C$, it is "No"; the Fourier series can converge as slowly as it wants everywhere).


snowflakes make this forum a much better looking...
but this occupies to much system resources...
iexplore.exe occupies > 80%, even up to 97% of CPU Time...
the system really slows down...

I agree. I've had a really hard time posting and stuff a few minutes ago. I'm in an Icafe and the computers here have lousy CPU's  :D

sorry for that. I will remove them. on my comp there is only a 25% usage of CPU, and everything's checked out ok.  :(

Hey Valentin, don't be so lazy.  :D And get some Christmas mode/style establised very soon!  :)

[quote="leepakhin"]What snowflakes are you talking about? :?[/quote]

This is an insider issue of long-time ML/AoPS users. This .php forum was extended by certain features last year, e.g. emoticons having a Christmas cap and snowflakes falling from top to the bottom of the page! Comeback wanted!

Nothing against snowflakes, but as an additional skin with an option to turn off! Currently I can only access ML from an old computer which really suxx0rz (the internet connection at home is broken), and while it currently hangs when 5 windows are opened, I guess any additional style effects would make it hang with 1 window opened, thus making work completely impossible. Besides, at our IMO training camps in Rostock and Bad Homburg we have even worse computers and we cannot do anything about that. And this is only about Germany, I guess in Vietnam it is a bit worse...

Guys, please do remember that not everybody has the newest computer, the quickest internet connection, the best antivirus, Linux, Firefox etc. This is why I have, e. g., pleaded for a (more-or-less) text-only skin for ML some time ago. This skin was then removed when ML moved to a faster server, but I still think it would be useful to have such a skin again, rather than wasting traffic and processor time for unnecessary decorations!

  Darij

Yeah they were all in plan, but unexpected matters appeared :( They will come eventually :)


abstract algebra
group theory
superior algebra
superior algebra unsolved

[quote="MrGaussian"]This is one item of a question about Pontrjagin dual.   The Pontrjagin dual of an abelian group G is G*=Hom(G, Q/Z) (no matter whether G is finite or not).  I was able to prove that if G is finite then G* is isomorphic to G.  My  question is to prove that if G is finite abelian group then G/H is isomorphic to some subgroup of G.  I think the idea is to consider (G/H)* and to try to prove that (G/H)* is isomorphic to Hom(K,Z) for some K subgroup of G and because Hom(K,Z) is isomorphic to K it follows that (G/H) is isomorphic to K because G/H is finite and then (G/H )* is isomorphic to G/H.  Thank you.[/quote]

This would seem pretty easy then, since

Hom(G/H, Q/Z) is a subgroup of Hom(G,Q/Z), consisting of all maps that vanish on H,
so (G/H)* is then isomorphic to a subgroup of G*, and you should be done from there.

[quote="Nukular"]

This would seem pretty easy then, since

Hom(G/H, Q/Z) is a subgroup of Hom(G,Q/Z), consisting of all maps that vanish on H,
so (G/H)* is then isomorphic to a subgroup of G*, and you should be done from there.[/quote]

Thank you Nukular


[quote="kitsune"]Hello,

I have the following code in my LaTeX document:

\newtheorem{thm}{Theorem}

and I want to put a box around everything between \begin{thm} and \end{thm}. How can I do this?[/quote]

[code]$\boxed{BoxGame}$[/code]

$ \boxed{BoxGame}$

Thanks, but is there a way to incorporate that into the definition of the 'thm' theorem environment? I don't want to have to type that every time.

Use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=ntheorem]ntheorem package[/url]. You'll find a framed theorem example on page 19 of the [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/ntheorem/ntheorem.pdf]documentation[/url].


Let $ a,b \in \mathbb R$ s.t.  $ a^3\minus{}3ab^2\equal{}29$ and $ b^3\minus{}3a^2b\equal{}34$. Compute $ a^2\plus{}b^2$.

$ (a^3 \minus{} 3 a b^2)^2 \plus{} (b^3 \minus{} 3 a^2 b)^2\equal{}(a^2 \plus{} b^2)^3$...now did I think of this identity?

[hide]Let the equations be $ A$ and $ B$. This is equivalent to the system $ A \plus{} Bi$ and $ A \minus{} Bi$. How can you factor these equations?[/hide]

hello, from the first equation we have
$ b\equal{}\pm\frac{1}{3}\sqrt{3}\frac{\sqrt{a(a^3\minus{}29)}}{a}$, inserting this in the
second one we get
$ \frac{1}{9}\sqrt{3}\frac{(a(a^3\minus{}29))^{\frac{3}{2}}}{a^3}\minus{}a\sqrt{3}\sqrt{a(a^3\minus{}29)}\equal{}29$
or
$ \minus{}\frac{1}{9}\sqrt{3}\frac{(a(a^3\minus{}29))^{\frac{3}{2}}}{a^3}\plus{}a\sqrt{3}\sqrt{a(a^3\minus{}29)}\equal{}29$
Now you can solve these two equations with respect to $ a$.
Sonnhard.


probability
counting
distinguishability
expected value
binomial coefficients

An urn contains n+m balls, of which n are red and m are black. They are withdrawn from the urn, one at a time and without replacement. Let X be the number of red balls removed before the first black ball is chosen. Find the expected value E(X).

[hide="case by case"]
$ X \equal{} 0$: probability of happening is $ \frac{m}{m\plus{}n}$.
$ X \equal{} 1$: probability of happening is $ \frac{nm}{(m\plus{}n)(m\plus{}n\minus{}1)}$.
$ X \equal{} 2$: probability of happening is $ \frac{n(n\minus{}1)m}{(m\plus{}n)(m\plus{}n\minus{}1)(m\plus{}n\minus{}2)}$.

In general, $ X \equal{} k$ has probability $ \frac{m\binom{n}{k}}{(k\plus{}1)\binom{m\plus{}n}{k\plus{}1}}$ of happening for $ k \equal{} 0, 1,\ldots, n$. Unfortunately, I thought there was going to be an easy way to sum this up but nothing obvious comes to mind. [/hide]

We can number the red balls from 1 to n

Let i = 1, 2, ...., n, and

Xi = 1 , if red ball i is taken before any black ball is chosen
Xi = 0, otherwise

So,

E(Xi) = P(Xi = 1)

How to compute P(Xi = 1) ?

In fact, Paladin, that is a perfectly summable expression, if you apply a little creativity to it.  Some suggestions for how to make it summable:
[hide]First, write all binomial coefficients in terms of factorials.  Much will cancel.  Then, pull all the things not involving $ k$ outside of your summation.  If you divide and multiply by the appropriate thing (independent of $ k$), you'll get a nice binomial coefficient sum (some version of the Hockey Stick Identity) and thus the simple-looking result.[/hide]
A hint to a nicer approach:
[hide]Consider the following equivalent question: there are $ m\plus{}1$ distinguishable boxes among which you are to distribute $ n$ balls randomly.  What is the expected number of balls to end up in the first box?[/hide]
allrighty's method works as well, keeping in mind that the value we want is $ E(X) \equal{} E\left(\sum_{i \equal{} 1}^{n}X_{i}\right) \equal{} n E(X_{i}) \equal{} n P[X_{i}\equal{} 1]$.  In order to calculate this, [hide]realize that it is equivalent to the question, "I have $ m\plus{}1$ balls ordered randomly.  What is the probability that the (unique) ball labeled 'red' is in the first position?"[/hide]


let n be an integer \geq 210.prove or disprove: there are always at least 5 positive composite integers less than and relatively prime  to n.

Very nice problem! :D

[hide]Assume that $ n$ has $ k$ distinct prime divisors. Let $ p_i$ be the $ i$-th prime number for all $ i\in\mathbb{N}$. So $ n\ge p_1p_2\ldots p_k$. If $ n\equal{}p_1p_2\ldots p_k$, we shall prove that $ p_{k\plus{}1}^2,p_{k\plus{}1}p_{k\plus{}2},p_{k\plus{}2}^2,p_{k\plus{}1}p_{k\plus{}3},p_{k\plus{}1}^3$ are less than $ n$ (they are the five composite numbers which are relatively prime to $ n$). After we prove this, the case $ n>p_1p_2\ldots p_k$ follows easily.

We can prove that by using Bertrand's Postulate, i.e., $ p_{i\plus{}1}\minus{}p_i<p_i$ or $ p_{i\plus{}1}<2p_i$. I will write the complete proof if someone asks.[/hide]

Hmmm...

$ n$ could have any of the factors $ p_{k\plus{}1}, p_{k\plus{}2}, p_{k\plus{}3}$.

Yes, that is included in the last case, which is easier.


Gauss
number theory unsolved
number theory

check if the congruence $ x^2\equal{}12mod2989$ has a solution.
Baltic way 1999

We have $ 2989\equal{}7^2*61$, the congruence has a solution if and only if $ (12/7)\equal{}(12/61)\equal{}1$ But $ (12/7)\equal{}(5/7)\equal{}(7/7)\equal{}(2/5)\equal{}\minus{}1$ So the congruence has no solutions