Doesn't everyone love musician jokes?
Or maybe I'm just weird like that.
Anyways...post your musician jokes here

What's an oboe good for?
Firewood!

Why's a bassoon better than an oboe?
More firewood!

What's the difference between a trombonist lying dead on the road and a snake lying dead on the road?
There are skid marks before the snake!

What did they do with the drummer who was stupid enough to only bring one stick to rehearsal?
They made him the conductor!

What do you call someone who hangs around with musicians?
A drummer!

The violist happily exclaimed to the orchestra that he can play 32nd notes.  The orchestra didn't believe him so the violist played one.

How do you make a trombone sound like a french horn?
Put your hand in the bell and play all the wrong notes

Come on!
None of you know any other musician jokes?
I guess I am wierd.

[quote="cosinator"]
The violist happily exclaimed to the orchestra that he can play 32nd notes.  The orchestra didn't believe him so the violist played one.
[/quote]

Hahaha...

What's the difference between a viola and a trampoline?

[hide="Answer"]You take off your shoes when you jump on the trampoline.[/hide]

What do you call five baritones at the bottom of the ocean?
A good start.

I may be stupid but I did not get the 3rd one in the first post.

Anyway, here is one:

How did the conductor get the attention of the sleepy people?
By announcing that the orchestra was about to play the very last symphony of the night!

If you were lost in the woods, who would you trust for directions: an in-tune violist, an out of tune violist, or Santa Claus? The out of tune violist, the other two are make-believe.

Concept borrowed from http://www.ducksdeluxe.com/jokes.html

How can you tell if someone on the playground is a trombonist?

He can't swing and he doesn't know how to use the slide.

What's the difference between a bass trombone and a chainsaw?

Vibrato.

What does the conductor do with a dead violist?

He moves him back a chair.

Nobody better make fun of the cello... grrrrr.

How do you make two flutes play in tune?

You shoot one.

[quote="jka386"]what do you do when a violist comes to your house?

You ask him for the pizza.[/quote]

On the concept but off the topic, we often use a similar concept to make fun of schools.  In Michigan, two rival state universities are University of Michigan (w000t) and Michigan State University (booo).

Joke: How do you get a Michigan State graduate off your porch?

[hide="answer"] Pay him for the pizza.[/hide]

ahh, well, I learned the joke from an instrument repairman... Apparently, failed musicians often take pizza related jobs.

[quote="kyyuanmathcount"]In Michigan, two rival state universities are University of Michigan (w000t) and Michigan State University (booo).[/quote]

What do you have against MSU?!? :mad:

What's Beethoven doing right now?

DeComposing

[quote="jka386"]A cellist is fed up by all those offensive jokes directed at the cellists. So he decides to play the violin. He goes into a shop and says to the clerk, "I'd like to buy a violin."
The clerk says, "You must be a cellist."
Astonished he replies, "Why yes, I am. How did you know?"
"Because this is a fish and chips shop."[/quote]

You should replace every "cellist" in that joke by "violist"
It would make a lot more sense :lol:

Why do violists stand for long periods of time outside other people's houses?

They can't find the key and they don't know when to come in

After a concert, one of the members of the audience goes up to the principle violist and tells him his fingers were like lighning.  The violist blushes and thanks him.  The man then says, "no, what I meant is that your fingers never struck the same place twice."

These music jokes and more at [url]http://www.mit.edu/~jcb/jokes/viola.html[/url]


Explicate! Is it on a lattice grid? If so, must the squares be parallel to the lattice gridlines? If not, what restrictions are there?

[quote="mdk"]How many squares (of all possible sizes) are there in a 10 x 10 square?[/quote]

The big square 10 x 10 square is 1.

The 9x9 squares there are $ 2^{2}$

The 8x8 square there are $ 3^{2}$

...

The 1 x 1 square there are $ 10^{2}$

In total we have,
$ \sum_{n=1}^{10}n^{2}= \frac{10(21)(11)}{6}$


algebra
polynomial
function
induction
real analysis
real analysis unsolved

Assume $(P_{n}\in \mathbb{R}[X])$ converges pointwise to a function $f$ on $\mathbb{R}$, such that the set of the degrees of the $P_{n}$ is bounded. Show that $f$ is a polynomial.

It's trivial by induction if we assume that $(P'_{n}(x_{0}))$ converges for at least one real $x_{0}$. What if we don't make this assumption ?

We may assume that $P_{n}$ share the same degree by taking a subsequence.  By solving a system of linear equations about the coefficients or whatever, you may prove that a sequence of polynomials of degree $n$ converges to a polynomial as long as there exists $n$ points $x_{1}, x_{2}\ldots x_{n}$ s.t. on these points the polynomials converge.  Then the result follows.

Yeah actually the proof by induction works fien on $n$ points. I was out of my mind, thank you.


我想向大家推荐一个国内数学竞赛的网站。这个网站的主办人有很高的水平。版主也都很认真。这里有很多试题可以免费下载。有数学竞赛的讲座材料，还帮你解决数学难题。很快很及时。希望对大家有用。

http://www.aoshoo.com/bbs1/index.asp

[quote="ifai"]請問有誰可以介紹一些MathLinks外的國際性的math forums :?:
(不一定要跟競賽有關的)[/quote]

http://planetmath.org    is quite but they haven't got latex converter installed ;)

[quote="georgew_bush"]大家好！我想强调一点，aoshoo的论坛太慢了，一个网页宽带平均要40秒左右
不过数学竞赛网站好的很多呦！比如
http://www.gjmath.cn/bbs/index.asp[/quote]

同意！！！

[quote="lxd123321"][quote="lxd123321"][quote="georgew_bush"]大家好！我想强调一点，aoshoo的论坛太慢了，一个网页宽带平均要40秒左右
不过数学竞赛网站好的很多呦！比如
http://www.gjmath.cn/bbs/index.asp[/quote]

[size=150]同意！！！[/size]
[/quote][/quote]
What are you doing?

[quote="lxd123321"][quote="lxd123321"][quote="georgew_bush"]大家好！我想强调一点，aoshoo的论坛太慢了，一个网页宽带平均要40秒左右
不过数学竞赛网站好的很多呦！比如
http://www.gjmath.cn/bbs/index.asp[/quote]

[size=150]同意！！！[/size]
[/quote][/quote]
同意！

奥数之家我也去的.
不算太慢(或者说速度完全正常)
我喜欢的是里面老师的态度...
可惜我年龄太小不能讨论高中的东西

奥数网的网速很正常啊,而且也很好,题目很多

[quote="mathsfun_Tony"]奥数网的网速很正常啊,而且也很好,题目很多[/quote]
现在是正常了。你不知道一开始有多慢，绝对爽死你。

我觉得在这些网里面很难找到自己需要的，希望它们能把网站资源列到一块来，方便检索

这个网不错啊，英语不好的同学都很喜欢（比如说我）

哎, 大家喜欢去哪玩就去哪玩嘛, 各有所好, 自己喜欢就好, 别争了...

还有, 这个贴怎么界面这么奇怪? :|


linear algebra
matrix
vector
superior algebra
superior algebra solved

Find all matrices with complex coefficients that have the class of similitude bounded.

I just noticed that they must be diagonal matrices up until now, which can be seen by computing $SAS^{-1}$, where $S$ has $\alpha\ne 0$ as one of its diagonal entries, $1$ as its other diagonal entries, and $0$ everywhere else.

From that we find that any similar matrix must also be diagonal, so for any two orthogonal vectors $(x_1,x_2,\ldots,x_n)$ and $(x'_1,x'_2,\ldots,x'_n)$, the vectors $(a_1x_1,\ldots,a_nx_n)$ and $(x'_1,x'_2,\ldots,x'_n)$ are also orthogonal, where our matrix is $diag(a_1,\ldots,a_n)$. I think it's not hard to show from this that our matrices are complex multiples of $I_n$ (i.e. that all the $a_i$ are equal).

If it's not hard, then maybe you have the time to complete the proof.

Consider, for example, the vectors $(1,1,0,\ldots,0)$ and $(1,-1,0,\ldots,0)$. These are orthogonal, so $(a_1,-a_2,0,\ldots,0)$ and $(1,1,0,\ldots,0)$ are also orthogonal, which means that $a_1-a_2=0$. This works for any $i\ne j$, of course, so in this manner we show that $a_i=a_j,\ \forall i,j$.


please help on the question in the attached file.

Here is the problem
If $ abc\equal{}1$,prove that
$ \sum_{cyc} \frac{ab}{a^{5}\plus{}b^{5}\plus{}ab} \leq 1$.
My sol :
$ \sum_{cyc} \frac{ab}{a^{5}\plus{}b^{5}\plus{}ab} \leq \sum_{cyc} \frac{ab}{a^{4}b\plus{}b^{4}a\plus{}ab} \equal{}\sum_{cyc} \frac{1}{a^{3}\plus{}b^{3}\plus{}1}\equal{}\sum_{cyc} \frac{abc}{a^{3}\plus{}b^{3}\plus{}abc} \leq \sum_{cyc} \frac{abc}{a^{2}b\plus{}b^{2}a\plus{}abc}  \equal{} \sum_{cyc} \frac{c}{a\plus{}b\plus{}c} \equal{} 1$

[quote="FOURRIER"]Similar problem:

if $ a,b,c$ are positive reals, prove:

$ \sum \frac {1}{a^3 \plus{} b^3 \plus{} abc} \leq \frac {1}{abc}$[/quote]

It can also be solved using
$ a^{3}\plus{}b^{3}\geq a^{2}b\plus{}b^{2}a$. :)

[quote="stephen38"][quote="FOURRIER"]Similar problem:

if $ a,b,c$ are positive reals, prove:

$ \sum \frac {1}{a^3 \plus{} b^3 \plus{} abc} \leq \frac {1}{abc}$[/quote]

It can also be solved using
$ a^{3} \plus{} b^{3}\geq a^{2}b \plus{} b^{2}a$. :)[/quote]

Yes, and $ a^{3} \plus{} b^{3}\geq a^{2}b \plus{} b^{2}a$ is obtained by rearangement right?

I wonder if we can suppose $ a\geq b \geq c$

[quote="stephen38"]Here is the problem
If $ abc \equal{} 1$,prove that
$ \sum_{cyc} \frac {ab}{a^{5} \plus{} b^{5} \plus{} ab} \leq 1$.
[/quote]
BTW,It is an ISL1996. :wink:

[quote="FOURRIER"]Yes, and $ a^{3} \plus{} b^{3}\geq a^{2}b \plus{} b^{2}a$ is obtained by rearangement right?

I wonder if we can suppose $ a\geq b \geq c$[/quote]
$ a^3\plus{}b^3\geq a^2b\plus{}ab^2\iff (a\plus{}b)(a\minus{}b)^2\geq 0$


I'm looking at this problem and I look at the answer section in the back of the book but there is no graph...

Directions:
Write each equation in general form, if necessary, and graph each equation using the intercept method...

x - y = 7

I notice there is a negative before the y and I wouldn't know exactly how i could get this into Ax+By=C

I don't think a person can, can they?

There is a graph in the back, but there is no line..

Any answers as to why not?

Eh, the missing graph was probably a mistake or something.
Notice that x-y=7 can still be written in standard form: 1x+(-1)y=7. A=1, B=-1, and C=7. This is just simplified when you write x-y=7, as adding a negative is the same as subtracting.
Hope it helps!

Maybe it's invalid, as you just stated, and they intended to trick you................


everybody steal magixter's avatar, (which is mine at the moment.) just right-click my avatar and save it as a picture. then load as your own!!! i wonder what magixter will say when half of all Aospers has his avatar!!!  :D

Or, just go to properties and copy the url.


factorial
function
algebra
floor function

How many zeros are on the end of 2004! (factorial)

[hide]10=2*5, and there are going to be more 2s than 5s when we expand 2004!, so we just find the highest power of 5 that 2004! is divisible by. this is just floor(2004/5) + floor(2004/25) + floor(2004/125) + floor(2004/625) = 400+80+16+3=499.[/hide]

Unless I am missing something, this should be in intermediate or even just starting.

This is a type of question that appears on many competitions.

Harder related question:  Find the number of 0's on the end of 2004! [i]when it is written in base 12[/i].

For 2004!:
Count how many times 5 appears as a factor.  A pair of 5 and a 2 is necessary and sufficient for 0, and we can be sure that there will be more 2's than 5's, so the number of 5's will determine it.

Use the greatest int function on:
2004/5
2004/25
2004/125
2004/625
and all of them together to get:
400+80+16+3=499

For 2004! base 12:
Count how many times 2^2 and 3 appear, and find which is smaller.  By the same logic as before, 2^2 will be the limiter.  To make my life easier, we'll convert 2004 into base 10 to get 3460.
Same greatest integer function thing:
3460/4
3460/16
3460/64
3460/256
3460/1024
865+216+54+13+3=1151

Good thought, but the problem successfully tricked you.  Look more closely.

Shoot.  You're right.  Grrrr.
You can't just count the 2^2's, you have to count all the 2's then divide by two.

Doing that, with the same old method, we (should anyway, I worked it out by hand this time) get 3355 2's, divide by 2 and floor it to get 1677 2^2's.  Now, (again worked by hand) we get 1726 3's.  So the 2^2's still limit, but there are 1677 of them this time.


hi,
for what $ n\in\mathbb{N}$ does there exist $ a\in\mathbb{N}$ such that $ Q(a\cdot n) \equal{} n$, where $ Q(n)$ is the digital sum of $ n$.

for every n consider the number $ 10^{K}\sum_{i \equal{} 1}^{n}10^{k_{i}}$ which has digit sum n and is divisible by n.
(here $ k_{i}$ are different numbers divisible by $ \phi\left( f(n)\right)$ where $ n \equal{} 2^{a}5^{b}f(n)$)


floor function
trigonometry
AMC
AIME
quadratics
algebra
quadratic formula

$R$ is a solution to $x+\frac{1}{x}=\frac{ \sin210^{\circ}}{\sin285^{\circ}}$. Suppose that $\frac{1}{R^{2006}}+R^{2006}=A$ find $\lfloor A^{10}\rfloor$ where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$.

Just do $210 \cdot 2006$ and $285 \cdot 2006$ mod 360.

It simplifies to $\frac{\sin{60}}{\sin{30}}$, so $A$ is like $\sqrt{3}$, so $A^{10}=\boxed{243}$

Substitute $x=e^{yi}$, then see $\frac{\sin 210}{\sin 285}=2\cos 75$, and then it's obvious

^^^^ that kid's right

Could someone post a solution instead of just saying that it is obvious :blush:

[quote="Altheman"][quote="Iversonfan2005"]hm, wonder why that greatest integer thing is there....

hmmmmmmm :|

[quote="Elemennop"]
^^^^ that kid's right[/quote]

DUH  :P[/quote]

ha...i just did that to limit the guessing because if you know that x^10 is a positive integer <1000, the possibilities are pretty limited[/quote]

wait, is this like one of your mock aime problems? If so, penguinintegral can look at the solution document. It can be done through what Seva said by substituting $e^{i\theta}$

$=\frac{ \sin210^{\circ}}{\sin285^{\circ}}=\frac{-\sin 150^\circ}{-\sin 75}=\frac{2\sin 75^\circ\cos 75^\circ}{\sin75}=2 \cos75$

Then using the quadratic formula, $x=\text{cis}75^\circ$. Now we can easily evaluate $A$ using $\text{cis}$ exponent rules.

$A=\text{cis}75*2006+\text{cis}-75*2006=2\cos 330=\sqrt{3}$

It follows that $A^{10}=\boxed{243}$


$a,b,c$ are sides of a triangle. Prove that

$(a+b-c)^{a}(b+c-a)^{b}(a+c-b)^{c}\leq a^{a}b^{b}c^{c}$

[quote="PenguinIntegral"][quote="scorpius119"][hide]
Letting $b+c-a=2x$ etc. $x,y,z$ are positive reals and we need to prove
\[(x+y)^{x+y}(x+z)^{x+z}(y+z)^{y+z}\geq (2z)^{x+y}(2y)^{x+z}(2x)^{y+z}\]
AM-GM on $x+y$, $x+z$, and $y+z$ show that the left hand side is at least
\[(2z)^{z+\frac{x+y}{2}}(2y)^{y+\frac{x+z}{2}}(2x)^{x+\frac{y+z}{2}}\]
So we will be done if we can prove
\[(2z)^{2z}(2y)^{2y}(2x)^{2x}\geq (2z)^{x+y}(2y)^{x+z}(2x)^{y+z}\]
which is equivalent to
\[(2z)^{3z}(2y)^{3y}(2x)^{3x}\geq (8xyz)^{x+y+z}\]
But this is just Chebyshev after taking natural logarithm of both sides.
[/hide][/quote]
Wow that's a tough nut.

Is there anything without tangential calculus?[/quote]

You can use base 10 log, or base $2^\sqrt{2}$ log, or whatever. Showing that a logarithm to base greater than 1 is increasing isn't hard to prove algebraically (of course, a continuity argument would be needed, but without continuity or defining $e^{x}$ and $\ln x$ by calculus properties, irrational exponents have no meaning anyway).


The longest side of an isosceles right triangle has length $ \sqrt[3]{2}$ cm. Another side has length $ 2^a$. Find the length of $ a$ expressed as a common fraction.

[quote="kabi"]hello
$ 2^{1\plus{}2a}\equal{}2^{\frac{2}{3}}$
$ a\equal{}\frac{\minus{}1}{6}$
thank u[/quote]
To be more specific, we arrive to $ 2^{1\plus{}2a}=2^(\frac{2}{3})$ using the Pythagorean Theorem.

Since it is an isosceles right triangle, both legs' length is $2^a$ and the longest side must be the hypotenuse.  So:

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

$2^a+2^a=\left(2^{1/3}\right)^2$

$2(2^{2a})=2^{2/3}$

$2^{2a+1}=2^{2/3}$

$2a+1=\frac{2}{3}$

$2a=-\frac{1}{3}$

$a=\boxed{-\frac{1}{6}}$


The average amount of money spent by a person who attended
a local sporting event in 2000 was $ \$8.00$, of which 75$ \%$ was
the ticket price. In 2005, the average amount of money spent
by a person who attended a local sporting event increased
by 50$ \%$, but the ticket price did not increase. By how many
dollars did the non-ticket costs of 2000 increase to become the
non-ticket costs of 2005?

At beginning, 6 ticket, 2 non

Now 12 dollars total 6 is ticket so 6 non. 6-2=4

The ticket price is $ .75\cdot\$8\equal{}\$6$, so the non-ticket price for $ 2000$ was $ \$2$. The total price in $ 2005$ increased to $ \$12$, so the non-ticket price for $ 2005$ was $ \$6$. Our final answer is $ \$6\minus{}\$2\equal{}\$\boxed4$.


function
number theory
prime factorization

Find the sum of all the proper divisors (that is, those excluding $ 1$ and $ 2010$) of $ 2010$.

[hide="Solution"]
We know that the prime factorization of $ 2010$ is $ 2\cdot 3\cdot 5\cdot 67$.
Thus the [url=http://www.artofproblemsolving.com/Wiki/index.php/Divisor_function]sum of all its factors[/url] is,
\[ (1\plus{}2)(1\plus{}3)(1\plus{}5)(1\plus{}67)\equal{}3\cdot 4\cdot 6\cdot 68\equal{}4896\]
But this includes the proper divisors, therefore we have to subtract them to get the right answer.
\[ 4896\minus{}(1\plus{}2010)\equal{}\boxed{2885}\]
[/hide]


function
inequalities
inequalities proposed

Let $a,$ $b,$ $c$ are non-negative real numbers and $ab+ac+bc=3.$
Prove that $\frac{a+b+c}{3}\geq\sqrt[18]{\frac{a^3+b^3+c^3}{3}}.$ :D  :)

[quote="Davron"]Sir dont you wanna solve this problem for us, please. :)
Davron[/quote]
Davron, you can find the hint here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78236 ;)

More one:
$a>0,b>0,c>0,$ $ab+ac+bc=3.$ Prove that
$\frac{a+b+c}{3}\geq\sqrt[36]{\frac{a^4+b^4+c^4}{3}}.$

 [b]The next is wrong. Sorry.[/b]
By the way, $(\frac{a+b+c}{3})^{39}\geq\frac{a^4+b^4+c^4}{3}\cdot(\frac{ab+ac+bc}{3})^{17.5}$   is not true.
Likewise, $(\frac{a+b+c}{3})^{17}\geq\frac{a^3+b^3+c^3}{3}\cdot(\frac{ab+ac+bc}{3})^7$ is not true.

hello michael

why did u change ur qs and made them special cases

i hope der was nothing wrong in previous qs or if der was den plz inform

i hv already put mor than 20-30 hrs on ur 1st problem :!:  :!:

Finally i got the 1st one
i used a similar technique
its very long and quite messy

$a+b+c=p$
  $ab+bc+ca=q$
  $abc=r$

we have to prove
$(a+b+c)^{36}\geq3^{19}(a^3+b^3+c^3)^2(ab+bc+ca)^{15}$

then the ineq becomes

$p^{36}\geq 3^{19}(p^3-3pq+3r)^2q^{15}$

divide both sides by $3^{15}q^{15}$
and let
$\frac{p^2}{3q}=m$
 it can be seen $m\geq1$
now divide both sides by $p^6$

so finally we get to solve
$m^{15}\geq3^4(1-\frac{1}{m}+\frac{3r}{p^3})^2$

multiply both sides by $m^2$ and put $\frac{mr}{p^3}=n$
so we get

$m^{17}\geq3^4(m+3n-1)^2$
take sq root to get
$m^{8.5}\geq9(m+3n-1)$
ie. $m^{8.5}+9\geq9(m+3n)$

now i'll show
$3n=\frac{r}{pq}\leq\frac{1}{9m}...................................1$

by Murheads ineq we have for symmetric functions
$\sum(a^2bc)\leq\sum(a^2b^2)$
ie.$3\sum(a^2bc)\leq(ab+bc+ca)^2$
which gives $3pr\leq q^2$
ie.$\frac{r}{pq}\leq\frac{q}{3p^2}=\frac{1}{9m}$
hence we get 1

so $m^{8.5}+9\geq9m+\frac{1}{m}$

now $m<2$ otherwise this inequality is always true

observe $m^{8.5}+7.5\geq8.5m$ by AM>=GM
ie. we hv to prove
$m(3-m)\geq2$
ie. $(m-1)(m-2)\leq 0$  which is true :)  :lol:

arpit goel, that's a nice proof, not messy at all. :)
arqady, do you think that more generally the following holds:
for $n\ge0$, let $k: =k(n)=\frac13(n^3+6n^2-13n+12)$. Then

$\frac{a+b+c}{3}\geq\sqrt[k]{\frac{a^n+b^n+c^n}{3}}$ and $k$ is best possible. :?:
(I have no proof, this is just 'educated guess')

[quote="spanferkel"]
arqady, do you think that more generally the following holds:
for $n\ge0$, let $k: =k(n)=\frac13(n^3+6n^2-13n+12)$. Then

$\frac{a+b+c}{3}\geq\sqrt[k]{\frac{a^n+b^n+c^n}{3}}$ and $k$ is best possible. :?:
(I have no proof, this is just 'educated guess')[/quote]

well i dont think so

i m using quite big values though small may work
1st let $n=6$, so $k=122$

now let the fol
$b=10^{44}a$
$c=10^{45}a$

so $a^2=\frac{3}{10^{44}+10^{45}+10^{99}}$  as $ab+bc+ca=3$
put these in ur eqn and i think u will get a contradiction if i dint go wrong in calculations though was using a calculator :roll:

arpit goel, congratulations!
In fact, you have proved that
If $a,b,c$ are non-negative real numbers and $ab+ac+bc=3$ then
$\frac{a+b+c}{3}\geq\sqrt[17]{\frac{a^3+b^3+c^3}{3}}.$ ;)

I don't think he has proved it for the exponent 17.
$(m+3n-1)=\frac{p^3+3r-3pq}{3pq}$. Arqady, didn't you miss the denominator  here? Apart from that I would be happy to know that 18 is [b]not [/b]best possible. because it seems crude to me that the best exponents should be a cubic function of "n" (sorry I called that "n" as well). :? :
As to the counter-example: I think in your denominator it should read $10^{89}$ instead of $10^{99}$. But don't those things go far beyond the limitations even of our sophisticated machines due to rounding errors anyway?

[quote="spanferkel"] Apart from that I would be happy to know that 18 is [b]not [/b]best possible.[/quote]
Let $a>0,b>0,c>0$ and $ab+ac+bc=3.$ Prove that
$\frac{a+b+c}{3}\geq\sqrt[17]{\frac{a^3+b^3+c^3}{3}}.$
[b] Proof[/b] (similar to arpit goel's)
Let $a+b+c=3u,$ $ab+ac+bc=3v^2$ and $abc=w^3.$
$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+ac+bc)+3abc=27u^3-27uv^2+3w^3.$
Hence, $\frac{a^3+b^3+c^3}{3}=9u^3-9uv^2+w^3\leq9u^3-9uv^2+\frac{v^4}{u}.$
Id est, enough to prove that  $u^{17}\geq9u^3-9uv^2+\frac{v^4}{u}.$
But $u^{17}\geq9u^3-9uv^2+\frac{v^4}{u}\Leftrightarrow u^{18}\geq(9u^4-9u^2v^2+v^4)v^{14}\Leftrightarrow f(t)\geq0,$
where $f(t)=t^9-9t^2+9t-1,$ $t=\frac{u^2}{v^2}\geq1.$
$f'(t)=9(t^8-2t+1),$ $f''(t)=9(8t^7-2)>0.$
Hence, $f'(t)\geq f'(1)=0.$
Hence, $f(t)\geq f(1)=0.$ :)

I will prove the inequality :$a>0,b>0,c>0,$ $ab+ac+bc=3.$ Prove that
$\frac{a+b+c}{3}\geq\sqrt[36]{\frac{a^4+b^4+c^4}{3}}$ (*)
Solution:
As you work,put $a+b+c=3u$;$abc=3$ then
            $a^4+b^4+c^4=(9u^2-6)^2-18+12uv \leq(9u^2-6)^2-18+12=81u^4-108u^2+30$
So that:$a^4+b^4+c^4 \leq \frac{81u^4-108u^2+30}{3}=27u^4-36u^2+10$

we need to prove that $u^{36}\geq 27u^4-36u^2+10$
Let $f(x)=x^{36}-27x^4+36x^2-10$.we will prove that $f(x) \geq 0$ for all $x \geq 1$
     $f'(x)=36x^{35}-108x^3+72x=x(36x^{34}-108x^2+72)$
Apply AM_GM's inequality we have $36x^{34}+72 \geq 108\sqrt[108]{x^{34.36}} \geq 108x^2$ (because $x \geq 1$)
So that $f'(x) \geq 0$ for all $x \geq 1$
Hence $f(x) \geq f(1)=0$ for all $x\geq 1$
Therefore the problem is proved  :)  :)
After your problem I find some nice problem:
a) $a,b,c,d >0$ satisfy $ab+ac+ad+bc+bd+cd=6$.Prove that:
            $\frac{a+b+c+d}{4} \geq \sqrt[27]{\frac{a^3+b^3+c^3+d^3}{4}}$
b) $a,b,c,d >0$ satisfy $ab+ac+ad+bc+bd+cd=6$.Prove that:
            $\frac{a+b+c+d}{4} \geq \sqrt[64]{\frac{a^4+b^4+c^4+d^4}{4}}$
Cheer,Cheer :lol:  :lol:  :lol:

[quote="nhat"]
After your problem I find some nice problem:
a) $a,b,c,d >0$ satisfy $ab+ac+ad+bc+bd+cd=6$.Prove that:
            $\frac{a+b+c+d}{4} \geq \sqrt[27]{\frac{a^3+b^3+c^3+d^3}{4}}$
b) $a,b,c,d >0$ satisfy $ab+ac+ad+bc+bd+cd=6$.Prove that:
            $\frac{a+b+c+d}{4} \geq \sqrt[64]{\frac{a^4+b^4+c^4+d^4}{4}}$
[/quote]
Yes, this is true, nhat. I checked it. :lol:

Yes, it's also my proof, can_hang. Use this hint, I proved the stronger:
[quote="nguoivn"]
Given $ a, b, c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$. Prove that: $ \sqrt [5]{\frac {a \plus{} b \plus{} c}{3}} \ge \sqrt [78]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}$  :)[/quote]
But with the last ineq, how did you solve it ? I solved it by Am-Gm again  :lol:
[quote="can_hang2007"]
$ \frac {13(a \plus{} b \plus{} c)^2}{ab \plus{} bc \plus{} ca} \ge \frac {54(a^3 \plus{} b^3 \plus{} c^3)}{(a \plus{} b \plus{} c)^3} \plus{} 33.$
The last one is not hard to prove.[/quote]

The minimal $ k$ for which the inequality
\[ \left(\frac {a \plus{} b \plus{} c}{3}\right)^k\geq\frac {a^3 \plus{} b^3 \plus{} c^3}{3}\]
is true for all non-negative $ a,$ $ b$ and $ c$ such that $ ab\plus{}ac\plus{}bc\equal{}3$
is $ k\equal{}\max_{t\geq0}\frac{3(t^3\plus{}4t^2\plus{}6t\plus{}4)}{t^3\plus{}2}\equal{}15.10503624...$

[quote="Lymacsau29"]I don't prove
For a, b, c, d positive
Prove that
$ \frac {1}{\frac {1}{a} \plus{} \frac {1}{b}}$+$ \frac {1}{\frac {1}{c} \plus{} \frac {1}{d}}$>=$ \frac {1}{\frac {1}{a \plus{} b} \plus{} \frac {1}{c \plus{} d}}$[/quote]
I then prove
inequality is equivalent to
$ (ac \minus{} bd)^2 > \equal{} 0$(obviously trues)

[quote="fjwxcsl"]Suppose
\[ \frac {a + b + c}{3}\ge{\sqrt [K(n)]{\frac {a^n + b^n + c^n}{3}}(K(n)\in{N})}\]
hold  forall  $ {a,b,c\ge0},ab + bc + ca = 3$ ,then we have
\[ K(2)_{min} = 6,K(3)_{min} = 16,K(4)_{min} = 29,K(5)_{min} = 47,K(6)_{min} = 68,
.....\][/quote]
We'll prove that $K(6)_{min} = 68.$
Let $a,$ $b$ and $c$ are non-negative numbers such that $ab+ac+bc=3.$ Prove that:
\[ \frac{a+b+c}{3}\geq\sqrt[68]{\frac{a^6+b^6+c^6}{3}}\]
Let $a+b+c=3u,$ $ab+ac+bc=3v^2$ and $abc=w^3.$
Hence, $\frac{a+b+c}{3}\geq\sqrt[68]{\frac{a^6+b^6+c^6}{3}}\Leftrightarrow$
$\Leftrightarrow u^{68}\geq(243u^6-486u^4v^2+243u^2v^4-18v^6+54u^3w^3-36uv^2w^3+w^6)v^{62}\Leftrightarrow$
$\Leftrightarrow f(w^3)\leq0,$ where $f$ is a convex function. Hence, $f$ gets a maximal value when $w^3$ gets an extremal value,
which happens when two numbers from $\{a,b,c\}$ are equal or when $w^3=0$ (the last case is obvious).
Since $f(w^3)\leq0$ is homogeneous, it enough to prove him for $b=c=1.$
In this case $f(w^3)\leq0\Leftrightarrow\left(\frac{a+2}{3}\right)^{68}\geq\frac{a^6+2}{3}\left(\frac{2a+1}{3}\right)^{31}\Leftrightarrow g(a)\geq0,$
where $g(a)=68\ln(a+2)-\ln(a^6+2)-31\ln(2a+1)-36\ln3.$
We have: $g'(a)=\frac{68}{a+2}-\frac{6a^5}{a^6+2}-\frac{62}{2a+1}=$
$=\frac{2(a-1)(31a^6-12a^5-18a^4-18a^3-18a^2-18a+56)}{(a+2)(2a+1)(a^6+2)},$
which gives $a_{min_1}=1$ and  $a_{min_2}=1.14197...$
Hence, $g(a)\geq\min\{g(1),g\left(a_{min_2}\right)\}=g(1)=0.$
But  $\left(\frac{a+2}{3}\right)^{67}\geq\frac{a^6+2}{3}\left(\frac{2a+1}{3}\right)^{30.5}$ is wrong for $a=1.14.$ Done!

[quote="arqady"]Let $ a,$ $ b,$ $ c$ and $ d$ are non-negative numbers such that $ ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \equal{} 6.$
Prove that:
\[ \frac {a \plus{} b \plus{} c \plus{} d}{4}\geq\sqrt [22]{\frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3}{4}}
\]
[/quote]
I just have found a very nice proof for this inequality (by similar to the [hide="Can's way"]
 http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1625103#p1625103
[/hide]).

[quote="arqady"]Certainly!
nhat, I proved that
if $x_1,$ $x_2,...,$ $x_n$ are positive real numbers, $n\geq2,$ $n\in\mathbb N$
and $\sum_{sym}x_1x_2=\binom{n}{2}$ then
$\frac{x_1+x_2+...+x_n}{n}\geq\sqrt[4n^2]{\frac{x_1^4+x_2^4+...+x_n^4}{n}}.$ :P  :)[/quote]

Can you please tell me how did you prove this thing. Do you have something similar to your uvw method for n variables or you used something else?

Thanks in advance!

[quote="MathUniverse"] Do you have something similar to your uvw method for n variables or you used something else?
[/quote]
$uvwt$, where $\sum_{i=1}^nx_i=nu,$ $\sum_{1\leq i<j\leq n}x_ix_j=\frac{n(n-1)}{2}v^2,$... :wink:

[quote="nhat"]Aqady,I have just found the quite general inequality:
Given $n \in N \geq 3$ and $x_{1},x_{2},...,x_{n}$ are $n$ positive real satisfy $\sum x_{1}x_{2}=\binom {n}{2}$.Prove that:
              $\frac{x_{1}+x_{2}+...+x_{n}}{n} \geq \sqrt[n^2+3n-1]{\frac{x_{1}^3+x_{2}^3+...+x_{n}^3}{n}}$  :lol:  :lol:[/quote]
[quote="arqady"]Certainly!
nhat, I proved that
if $x_1,$ $x_2,...,$ $x_n$ are positive real numbers, $n\geq2,$ $n\in\mathbb N$
and $\sum_{sym}x_1x_2=\binom{n}{2}$ then
$\frac{x_1+x_2+...+x_n}{n}\geq\sqrt[4n^2]{\frac{x_1^4+x_2^4+...+x_n^4}{n}}.$ :P  :)[/quote]

I have just found that these nice results can be proved by Cauchy-Schwarz. I will post it later (after the English version of my CS book is out... Maybe next year...).

The following inequality is also true.
Let $ a,$ $ b,$ $ c$ and $ d$ are non-negative numbers such that $ ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \equal{} 6.$
Prove that:
\[ \frac {a \plus{} b \plus{} c \plus{} d}{4}\geq\sqrt [42]{\frac {a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4}{4}}
\]
By the way, $ \frac {a \plus{} b \plus{} c \plus{} d}{4}\geq\sqrt [40]{\frac {a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4}{4}}$  is wrong already.

[quote="arqady"]Certainly!
nhat, I proved that
if $x_1,$ $x_2,...,$ $x_n$ are positive real numbers, $n\geq2,$ $n\in\mathbb N$
and $\sum_{sym}x_1x_2=\binom{n}{2}$ then
$\frac{x_1+x_2+...+x_n}{n}\geq\sqrt[4n^2]{\frac{x_1^4+x_2^4+...+x_n^4}{n}}.$ :P  :)[/quote]
if $n\geq 4$,then my result will be much better than yours:
\[\frac{x_1+x_2+...+x_n}{n}\geq\sqrt[n^2+4n+11]{\frac{x_1^4+x_2^4+...+x_n^4}{n}}\]
And I think it is impossible to find the exact and obvious expression for it.For each given $n,k$,however,we can get the precise result...

[quote="arqady"]I have proved that for all non-negatives $x_1$, $x_2$,..., $x_n$ such that $\sum_{1\leq i<j\leq n}x_ix_j=\frac{n(n-1)}{2}$
and for all $n\in\mathbb N$ such that $n=4$ or $n\geq7$ the following inequality is true:
\[\frac{x_1+x_2+...+x_n}{n}\geq\sqrt[n^2+4n+10]{\frac{x_1^4+x_2^4+...+x_n^4}{n}}\]
Thank you Tourish for your interest![/quote]
if $n\geq 8$,I guess $n^2+38$ is enough.... :lol:

[quote=arqady]Let $ a,$ $ b,$ $ c$ and $ d$ are non-negative numbers such that $ ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \equal{} 6.$
Prove that:
\[ \frac {a \plus{} b \plus{} c \plus{} d}{4}\geq\sqrt [22]{\frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3}{4}}
\]
By the way, $ \frac {a \plus{} b \plus{} c \plus{} d}{4}\geq\sqrt [20]{\frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3}{4}}$ with the same conditions is wrong already.[/quote]
Let $4u=a+b+c+d$, $6v^2=ab+bc+ca+d(a+b+c)$ and $4w^3=abc+bcd+cda+dab$, then
$u^{44}\geq (16u^3-18uv^2+3w^3)^2v^{38}$, by Rolle's theorem there is exists $x,y,z$ such that $3u=x+y+z$, $3v^2=xy+yz+zx$ and $w^3=xyz$. Since it's decreasing function of $w^3$, enough to check when it's reach maximum. WLOG $y=z=1$
$(\frac{x+2}{3})^{44}\geq(\frac{16x^3-12x^2+3x+20}{27})^2(\frac{2x+1}{3})^{19}$
or
$(x+2)^{44}\geq 3^{19}(16x^3-12x^2+3x+20)^2(2x+1)^{19}$, so we need to prove $g(x)\geq0$, where $g(x)=44\ln(x+2)-2\ln(16x^3-12x^3+3x+20)-19\ln(2x+1)-19\ln3$
$g'(x)=\frac{2(x-1)(304x^3-444x^2-159x+326)}{(x+2)(16x^3-12x^3+3x+20)(2x+1)}$
$g(x)\geq g(1)=0$


[quote="cincodemayo5590"]Find x: $x^{2^{x^{...}}}=20x$  (if it is possible...)[/quote]

i dont think so because if $x > 1$ then it will $x \to  infinity$ and if $x < 1$ then $x \to  0$, and if $x = 1$ then 1 doesnt = 20

[quote="Altheman"]is the pattern x,2,x,2...

then it is just $20x=\log_2\log_x(20x)$

but i am not sure that has solutions[/quote]

oh i thought it was $x{}^2{}^x{}^x{}^x$ my bad lol

[quote="lasagnaman"]My calculator shows the right side of Altheman's equation (with which I heartily agree) to have domain x > 1, and for that value x ~= 5, and is monotonic decreasing. At that x-value, 20x = 20 and is monotonic increasing, so there doesn't appear to be a solution. Is there anyway to show this mathematically?[/quote]

i guess you could show that the difference between the two sides of the equation is constantly increasing from x=1 and never hits 0... does that make sense?

Yes, I believe it does have a real solution, but it may not be explicitly solvable.

Since $x^{2^{x^{2^{...}}}} = 20x$ (and assuming the tower of exponents does not go to infinity), you can replace the top part of the tower with 20x.

So $x^{2^{20x}} = 20x$.

Taking the $\log_{20}$ of both sides gives $2^{20x}\log_{20}{x} = \log_{20}{20x}$.

Or $2^{20x} = \frac{1}{\log_{20}{x}} + 1$

By looking at the graphs of both sides of this equation you can see that they intersect just a liiiiittle above x = 1. Whether the tower of exponents goes off to infinity for that value of x will determine if it is a true solution or an extraneous solution. I have my doubts as to whether that value of x can be solved for exlicitly.

x=0 is a trivial solution, because 2 to any power can't equal 0, and 0 to any nonzero power is 0

By replacement, $x^{2^{20x}}=20x$. x = 1 is too low (1 vs 20), but 2 raised to a power over 20 is a gigantic number. This suggests that there will be another solution slightly over 1.

This solution is about 1.00000285686.

There are no other solutions.

[quote="cincodemayo5590"]lol i changed a problem that altheman had told me a bit... guess it kinda died  :P

(and alex, on the history... sHaMe!  :D )[/quote]

 :D


Each of four students hands in a homework paper. Later the teacher hands back the graded papers randomly, one to each of the students. In how many ways can the papers be handed back such that every student receives someone else's paper? The order in which the students receive their papers is irrelevant.

There are a total of 24 ways the homework can be handed back.
There is 1 way the homework can be given back correctly to all 4 students.
There are 4C3=4 ways that exactly 3 students get theirs back.
There are 4C2=6 ways that exactly 2 students get theirs back.
There are 4 ways for exactly 1 person to get their paper back.
24-1-4-6-4=9 ways for the teacher to get the papers all mixed up.


inequalities
geometry
circumcircle
LaTeX
inequalities unsolved

Prove that in an acute triangle with sidelenghts $a,b,c$ and circumcircle radius $R$:
$\frac{R}{a}\sqrt{\frac{R-\frac{a}{2}}{R+\frac{a}{2}}}+\frac{R}{b}\sqrt{\frac{R-\frac{b}{2}}{R+\frac{b}{2}}}+\frac{R}{c}\sqrt{\frac{R-\frac{c}{2}}{R+\frac{c}{2}}}\ge 2\sqrt{3}+3$, when do we have equality?

[quote="vardanch"]Let the triangle's angles be x,y,z
$R/a*sqrt((R-a/2)/(R+a/2))=2R/a*sqrt((1-(a/2R))/(1+(a/2R)))/2=1/2sin(x)*sqrt((1-sin(x))/(1+sin(x)))=(1-sin(x))/(2sin(x)cos(x))=(1-sin(x))/sin(2x)$
$\Rightarrow$ our inequality <=>
$(1-sin(x))/sin(2x)+(1-sin(y))/sin(2y)+(1-sin(y))/sin(2y)\geq 2sqrt(3)+3$
<=>
$sum(1/sin(2x))-sum(1/2cos(x))\geq 2sqrt(3)+3$

when we take x=y=z=pi/3 then we have that our inequality=2sqrt(3)-3, I think this inequality is wrong.

$sum(R/a*sqrt((R-a/2)/(R+a/2)))\geq 2sqrt{3}-3$ is true
This inequality we can prove by YENSEN
$sum(1/sin(2x))\geq 2sqrt{3}$
$sum(1-cos(2x))\leq 3$ $\Rightarrow$ $sum(1/sin(2x))-sum(1/2cos(x))\geq 2sqrt(3)-3$ Q-E-D
The inequality
$sum(R/a*sqrt((R+a/2)/(R-a/2)))\geq 2sqrt{3}+3$ is true too.[/quote]
I think you have some problems with $Latex$ :huh: .


algebra
polynomial
ratio
factorial
geometric series
absolute value
calculus

hi, i need help finding out whether the following series absolutley converges, converges but not absolutly, or if they diverge.

sigma n=1 to infinity 1.) (((-1)^n)(4^(n-1)))/(((4^(n+1))(n^(1/3)))
                                 2.) ((n+2)!)/((4^n)(n!))
                                 3.) ((-1)^n)/((4^n)(n!))
                                 4.) ((-1)^n)*(n!/4^n)
                                 5.) (n^7)/(3^n)

any help again would be great, thanks again.

Several of these call for algebraic simplifications before deciding what to do. Again, size is the main issue, and "de-cluttering" is the right mode of though.

#1: Note that $\frac{4^{n-1}}{4^{n+1}}=\frac1{16}.$ In size, this is simply a constant times $\frac1{n^{1/3}}.$ That's a "$p$-series" with $p<1,$ so the terms go to zero too slowly for absolute convergence. But they do at least go to zero, and we have an alternating sign so by the alternating series test we have conditional (and very slow) convergence.

#2. Note that $\frac{(n+2)!}{n!}=(n+2)(n+1).$ Our terms are $\frac{\text{polynomial}}{4^n}.$ We have a favorable comparison to a geometric series with a ratio slightly above $\frac14,$ and hence absolute convergence.

#3. That factorial in the denominator ovewhelms everything. Absolute convergence.

#4. Write the absolute value of the terms as $\frac{4^n}{n!}.$ It's still true that the factorial overwhelms the geometric numerator, although you can use the ratio test if it will make you feel better. Absolute convergence.

#5. Like #2. $\frac{\text{polynomial}}{3^n}.$ Favorably comparable to a geometric series. Converges absolutely.

thanks for the help, great explanation. thanks again.


Find all possible integer solutions of a,b and c satisfying the Diophantine equation:  $2^a+3^b=5^c$

I assume that a>=3.
then modulus 4: 3^b = 1 so b is even.
Modulus 8: 1= 5^c so c is even.
so let b=2x and c=2y
then: 2^a = (5^x - 3^y) (5^x + 3^y) = XY
so since gcd(X,Y)=2, we have that
X=2^1
Y=2^s
So 5^y = 3^y + 2.
and 5^y = 2^s - 3^y
so 2^s -2=2(3^y) or that 2^t =3^y +1.
so (y,t)=(0,1),(1,2) or t>=3 and so mod 4: y is odd. so modulus 8: 0 = 4. contradiction.
hence (y,s)=(0,2),(1,3)
Finally if a<=2,
3^b + 4 = 5^c
so mod 4:b is even
mod 3: c is even.
we can proceed as follows: c=2t and 4=2^2 and as before...
if a=1, then 3^b +2= 5^c
and one has to solve this to end this..

Just to finish off we need to solve
2 + 3^b =5^c
assume c>=2
mod 25: b=20t +13
now b (mod 5) =3 so 3^b = 5 mod 11.
hence 5^c =7 mod 11 which is impossible: 5^c= {1,5,3,4,9}
-Ali


Denote: $I_{\omega}, I_{\alpha}, I_{\alpha'}$ be inversions with respect to appropriate circles. $O$ - center of $\omega$, $A$ - center of $\alpha$, $A'=I_{\omega}(A)$, $P_{\infty}$ - point at infinity

Prove that : $I_{\alpha'}(O)=A'$
I cannot understand the reasoning, could you explain it to me :?:

The proof is following:

$\alpha$ and the two inverse (in $\alpha$) points $A$ and $P_{\infty}$ invert (in  $\omega$) into $\alpha'$ and the two inverse (in $\alpha'$) points $A'$ and $O$. Thus $A'$ (the inverse of $A$ in $\omega$) is the inverse (in $\alpha'$) of $O$.

my english is not good enought but i didn't understood where is the pole of the inversion...? :rotfl:

I think everything is clear,  the pole of inversion is a centre of a circle, but the centres are named, so what's the problem :?:

Excuse me, could you give the number of the page where you found the exercise? I didn't find it in the book :blush:

It's on page 116

In this proof the following fact is used, but not mentioned in the book (don't know why).
We are given two circles: $\alpha=(O_{1},r_{1}) , \beta=(O_{2},r_{2})$ and $O_{2}\not\in\alpha$. Pair of points $(P,P')$ is and inversive pair with respect to $\alpha$.
We perform the inversion w.r.t. $\beta \quad I_{\beta}(\alpha)=\gamma \quad I_{\beta}(P)=(X) \quad I_{\beta}(P')=(X')$
Then $X$ and $X'$ is an inverse pair w.r.t. $\gamma$

Recall the book's [b]inversive definition of inversion[/b]:

[i]Any point on $\omega$ is its own inverse; the inverse of any other point $P$ is the second intersection of any two circles throught $P$, orthogonal to $\omega$.[/i]

So consider a circle $\alpha$, and a point $P$ not in $\alpha$. We know we can determine $P'$, the inverse of $P$ in $\alpha$, by tracing through $P$ two circles $\beta_{1}$ and $\beta_{2}$, orthogonal to $\alpha$. But when we invert on $\omega$, $\alpha$ goes to $\alpha'$, and $\beta_{1}$, $\beta_{2}$ go to circles $\beta_{1}'$, $\beta_{2}'$. But since inversion preserves angles, $\beta_{1}'$ and $\beta_{2}'$ are still orthogonal to $\alpha'$; hence their points of intersection are inverse points in $\alpha'$ by the inversive definition of inversion.

Now, if we take the two inverse points in $\alpha$ as $A$ and $P_{\infty}$, inversion in $\omega$ send them to $A'$ and $O$, respectively. By our previous result we get that $A'$ and $O$ are inverse points in $\alpha'$, as we wanted to show. :wink:

Thanks a lot  :lol:  I've been trying to prove it for few days, but unfortunately I didn't focus on another definition of inversion


Hello, this topic is 2 discuss INO's

Since I gave INPhO and INChO i'll tell u sumthing.

Bansal topper is getting 75/80 in INPho and and 75/100 in INChO (my marks r not worth telling).Post ur marks  as well as that of people u kno. :lol:

is he getting 75 now ?

ie wrt to performance card  :D


Two integers are randomly selected from the set of integers greater than or equal -9 and less than or equal to 9 . The two numbers need not be different. What is the probability that the sum of the two integers is less than their product? Give your answer as a common fraction in lowest terms.

[hide]
162/361[/hide]

[hide="a start"]361 possible selections
$ x\plus{}y<xy$
$ x<xy\minus{}y$
$ x<y(x\minus{}1)$
$ y>\frac{x} {x\minus{}1}$[/hide]

[hide]
The total number of possible selections of integers is $ 19\times19\equal{}361$
If both numbers are negative, then the sum is obviously less than the product.
The number of possible ways to select two negative integers is $ 9\times9\equal{}81$
If one integer is negative and one is positive, then the product will always be less than the sum, so we don't have to worry about that.
If one or both numbers is 0, then the product will always be 0, so whenever one number is 0 and the other is negative, the sum will be less than the product. There are $ 9\times2\equal{}18$ possibilities for that.
With both numbers positive:
None of the numbers can be 1 if we want the sum to be less than the product.
If one of the numbers is 2, then the other number can be 3-9, so there are 7 possibilities for that.
For any other number greater than 2, any number from 2-9 will work, so there are 8 possibilities for each number and $ 7\times8\equal{}56$ possibilities in total.
Therefore, the probability is
$ \frac{81\plus{}18\plus{}7\plus{}56}{361}\equal{}\boxed{\frac{162}{361}}$
[/hide]
Hope that helps!

[quote="andersonw"][hide]
The total number of possible selections of integers is $ 19\times19 \equal{} 361$
If both numbers are negative, then the sum is obviously less than the product.
The number of possible ways to select two negative integers is $ 9\times9 \equal{} 81$
If one integer is negative and one is positive, then the product will always be less than the sum, so we don't have to worry about that.
If one or both numbers is 0, then the product will always be 0, so whenever one number is 0 and the other is negative, the sum will be less than the product. There are $ 9\times2 \equal{} 18$ possibilities for that.
With both numbers positive:
None of the numbers can be 1 if we want the sum to be less than the product.
If one of the numbers is 2, then the other number can be 3-9, so there are 7 possibilities for that.
For any other number greater than 2, any number from 2-9 will work, so there are 8 possibilities for each number and $ 7\times8 \equal{} 56$ possibilities in total.
Therefore, the probability is
$ \frac {81 \plus{} 18 \plus{} 7 \plus{} 56}{361} \equal{} \boxed{\frac {162}{361}}$
[/hide]
Hope that helps![/quote]

So, does order matter in this problem?  In the beginning, I think that it did (according to the above solution), but later on,

[quote="andersonw"]If one of the numbers is 2, then the other number can be 3-9, so there are 7 possibilities for that.[/quote]

I don't think that order was accounted for.

Could someone please help me?

Thanks.

[quote="andersonw"]Order does count, (i.e. 7,8 is different from 8,7.)
For example, the pair 2,7 works. And later on, I count the pair 7,2 here
[quote]For any other number greater than 2, any number from 2-9 will work[/quote][/quote]

I got it now.  Thanks a lot!


Let number sequence $ (u_n)$ so that: $ \begin {cases} u_1 \equal{} 3 \\
u_n  \equal{} 2u_{n \minus{} 1}\plus{}5n \end {cases}$

Find formula of $ u_n$ by $ n$?

[quote="PeterP"]For example:
Substitute $ u_n \equal{} 2^nv_n$,
then $ 2^nv_n \equal{} 2^nv_{n \minus{} 1} \plus{} 5n$,
or $ v_n \equal{} v_{n \minus{} 1} \plus{} \frac {5n}{2^n}$

$ v_n \minus{} v_1 \equal{} v_n \minus{} \frac32 \equal{} \sum_{k \equal{} 2}^n (v_k \minus{} v_{k \minus{} 1})$ etc.[/quote]

Thank PeterP very much. Can you post full solution? And can you tell me:

1) $ u_n \equal{} 2^nv_n$?

2) $ v_n \minus{} v_1 \equal{} v_n \minus{} \frac32 \equal{} \sum_{k \equal{} 2}^n (v_k \minus{} v_{k \minus{} 1})$ ?


trigonometry
geometry
trig identities
Law of Sines

The sides of a triangle are consecutive integers n, n+1, and n+2, and the largest angle is twice the smallest angle. Find n.

Let the smallest angle be $\theta$ then the largest is $2\theta$. By law of sines we have $\frac{n}{\sin{\theta}}= \frac{n+2}{\sin{2\theta}}$. So then $\frac{\sin{2\theta}}{\sin{\theta}}=1+\frac2n$.  Using $\sin{2\theta}=2\sin{\theta}\cos{\theta}$ this becomes $2\cos{\theta}=1+\frac2n$, thus $\cos{\theta}=\frac12+\frac1n$. Then maybe we can use cos formula, solve for cos there and set that equal to the above. Sounds pretty ugly though...

Yeah, seeing it done out it isn't that bad. I didn't have pen/paper with me so I didn't actually do it out to see, just sounded ugly.


function
calculus
derivative
induction
integration
real analysis
real analysis solved

Suppose that $f''(x)$ is the continuous function at $(-\infty,\infty)$ such that  $x_1<x_2\longrightarrow f''(x_1)<f''(x_2)$ and $f(0)=f'(0)=0$.

Compare $\frac{f(\sqrt{2004})}{2004}$ and $\frac{f(\sqrt{2005})}{2005}$.

you have mentioned 3 functions here.
f(x), f'(x), and f"(x)
edit your question pls

Hello,paganinio!

$f'(x)$ is the derivative of $f(x)$,similarlly,$f''(x)$ is that of $f'(x)$.

kunny

The conditions implies that $f,f'$  are increasing functions on $[0,+\infty[$

I think the following generalization holds: if $f:[0,\infty)\to\mathbb R$ is of class $C^n$, $f^{(n)}$ is increasing and $f(0)=f'(0)=\ldots=f^{(n-1)}(0)=0$, then $\frac f{x^n}$ is increasing.

It holds for $n=1,2$, and I think the induction step shouldn't be significantly different from the proof for $n=2$, knowing that it holds for $n=1$, but I'm not even sure the whole thing is true :).

How nice! I didn't know this..  :blush: Thanks! I used Induction. It's not that tedious, but this is much nicer.


inequalities
calculus
integration
geometry
geometric transformation
trigonometry
inequalities unsolved

Sa se arate ca pentru orice x>0 avem
[i]Mod Edit - Translation, Show for any x>0, we have[/i]

\[\displaystyle\int_{x}^{x+1}\sin( y^n) dy \leq\frac{2+(x+1)^n-x^n-nx^{n-1}}{n x^{n-1}},n\in N^*\]

gah what happened to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26438[/url]? :?

Okay, I suppose I will move it... altough the question should be where, romanian section, calculus tutorials, or college.

please put  this problem in calculus.thanks


advanced fields
advanced fields unsolved

I don't know where this kind of questions should post on. Please move it to a suitable forum if it is necessary.

What is the meaning of "well-founded"? The following is the original exercise and I can't guess its meaning.
The property of being well-founded is an abstract property which applies also to some but not all graphs which are different from the universe of all sets. Here some examples of graphs. Which of them are well-founded? The answers should be proven.
1. ...;
2. the set {0,1,2,3} with the edges (0,1), (1,0), (2,3);
3. the set [b]N[/b] of the natural numbers with every edge being of the form (n,n+1);
... ...

Another question is that how to prove there is exactly one empty set? It seems obvious, but how do we explain it?

In those cases, you need to prove or disprove the existence of an infinite descending chain (it's not the general definition, but in these cases it's equivalent).
In problem 2 there is an infinite descending chain: 0, 1, 0, 1, ..., therefore the graph is not well founded.
In problem 3 there are no infinite descending chains, so it's well founded.

Proving the unicity of the empty set is trivial with the Zermelo-Fraenkel axioms. If that's what you're looking for, just go read the axioms (wikipedia) and try proving it yourself.


inequalities
function
inequalities unsolved

Let $u_i = \frac {x_i} {x_{i+1}}$ where $x_{n+1} = x_1$.
Thus $u_1 \cdots u_n = 1$.
Now let $t_i = ln(u_i)$. Thus $t_1 + ... + t_n = 0$.

Moreover $ \frac {x_i} { \sqrt {x_ix_{i+1} + x_{i+1}^2 } } = \frac {e^{t_i}} {\sqrt {e^{t_i} + 1} }$.

Now, it suffices to use Jensen's inequality to the convex function $f(t) = \frac {e^t} { \sqrt {e^t + 1} }$ to get the desired inequality.

An additional remark : This inequality is due to a member of ML :) , namely Kantor. Now, guess who he is  ;)

Pierre.

O, so Kantor is Mohammed Aassila (I hope I wrote his name correctly), the man who proposes some of the craziest (this means hard and beautiful) problems in Crux. Anyway, I think that this inequality was first proposed by Gheorghe Szollossi in Gazeta Matematica, but I don't know the year. I think it's 2001.


In any  triangle  ABC  show that
$ (a^2+b^2+c^2)^2\leq4m_bm_ca^2+4m_cm_ab^2+4m_am_bc^2  $
Thanks

I think change the triangle ABC to its median triangle should help since it can be transformed to a simpler inequality. That means we  'll replace $a, b, c$ by $m_a, m_b,m_c$ and $m_a, m_b, m_c$ by $ka, kb,kc$ where k is a constant (I have no paper now , so please find k yourself :D ).

Using treegoner idea we obtain a median triangle
with sides Ma, Mb, Mc and k =  and our inequality becomes :

(Ma^2+ Mb^2+Mc^2)^2  =< 4*(3/4)^2*(bc*Ma^2+ac*Mb^2+ab*Mc^2);
 and we know Ma^2+ Mb^2+Mc^2 = ()*(a^2+b^2+c^2);

So we need to prove :    (a^2+b^2+c^2)^2 =<
     =< 2bc(b^2+c^2)–a^2*bc+2ac(a^2+c^2)–b^2*ac+2ba(b^2+a^2)–c^2*ba
or    3abc (a+b+c) =< (a^2+b^2+c^2)(2bc+2ac+2ab – (a^2+b^2+c^2))

Using  bc+ac+ab = p^2+r(r+4R);  a^2+b^2+c^2 = 2*(p^2 - r(r+4R));

 3*8Rr*(p^2) =< 8*(p^2 - r(r+4R))*(4R+r)*r  =>
 3*R*p^2 =< (p^2 - r(r+4r))*(4R+r);             =>
 or simplifying (4R+r)^2*r =<p^2*(R+r);

Here we use Blundon inequality:
      p^2 => 2*(R^2)+10Rr - (r^2) -2(R-2r)*\|(R(R-2r).   (\| = Sqrt)
As a result :
     [(4R+r)^2]*r =< (R+r)[2*(R^2)+10Rr - (r^2) -2(R-2r)*\|(R(R-2r) ] =>
     2(R-2r)*\|R(R-2r)*(R+r) =< (R-2r)[2(R^2) +(r^2)]
Hence  we need to prove :
 2\|R(R-2r)*(R+r) =< (2R^2 +r^2) or 4R(R-2r)*[(R+r)^2] =< [2(R^2) +(r^2)]^2
 which yields to 16(R^2)*(r^2)+r^4+(8R^3)r >= 0

So inequality is true and equality when R=2r – for equilateral triangle.
I think that’s all. Can anyone check if it has no mistakes.


Given that $0 \le p \le 1$, find the smallest value of $K$ (in terms of $p$) such that

$\frac{a+b}{a+b+pc}+\frac{b+c}{b+c+pa}+\frac{c+a}{c+a+pb}\le K$

for all $a, b, c > 0$.

[quote="N.T.TUAN"]$K\geq\max (\frac{6}{2+p},1+\frac{2}{1+p},2)$. Am I wrong?  :maybe:[/quote]
Id est, $K\geq\frac{6}{2+p}.$ :D
And easy to prove ( Cauchy ) that $K_{min}=\frac{6}{2+p}.$ :wink:

I used following lemma: Let $f(x,y,z)=a\sum_{\text{sym}}x^{3}+b\sum_{\text{sym}}x^{2}y+cxyz$, then two following assertions are equivalent
1)$f(x,y,z)\geq 0$ for all $x>0,y>0,z>0$.
2)$f(1,1,1)\geq 0,f(1,1,0)\geq 0,f(1,0,0)\geq 0$.

The answer is $\frac{6}{p+2}$, but I can't get the right solution. because I got the answer by using Jensen's theorem and Nesbit's inequality ironically. :lol:
I do know this approach doesn't make sense.


A piece of paper is folded in half.  A second fold is made at an angle $ \phi$ ($ 0^\circ < \phi < 90^\circ$) to the first, and a cut is made as shown below.
[img]12881[/img]
When the piece of paper is unfolded, the resulting hole is a polygon.  Let $ O$ be one of its vertices. Suppose that all the other vertices of the hole lie on a circle centered at $ O$, and also that $ \angle XOY \equal{} 144^\circ$, where $ X$ and $ Y$ are the the vertices of the hole adjacent to $ O$. Find the value(s) of $ \phi$ (in degrees).

[quote="CircleSquared"][quote="worthawholebean"]([b]5[/b]) A piece of paper is folded in half.  A second fold is made at an angle $ \phi$ ($ 0^\circ < \phi < 90^\circ$) to the first, and a cut is made as shown below.
[img]http://www.artofproblemsolving.com/Forum/album.php?mode=attach&id=12881[/img]
When the piece of paper is unfolded, the resulting hole is a polygon.  Let $ O$ be one of its vertices. Suppose that all the other vertices of the hole lie on a circle centered at $ O$, and also that $ \angle XOY \equal{} 144^\circ$, where $ X$ and $ Y$ are the the vertices of the hole adjacent to $ O$. Find the value(s) of $ \phi$ (in degrees).[/quote]

[hide]
is theta 18 by any chance?
if it is i might post a solution (because i don't want to in case it's wrong.  it's really complicated.)

[/hide][/quote]

[hide]I got 81, is that right? If you actually try folding it (cause you're noob like me) then the figure is a kite, and I went from there (and drawing diagrams).[/hide]

There seems to be an issue with linking to images from the galleries.

I get 81 for the convex kite, and 54 for the concave kite.

Does anyone have a proper solution for this?

I actually did all the folding too. I think the answer they want is 81.
What I was wondering is if the concave case actually works- 54 degrees works out numerically, but is it possible to draw a circle through the other three vertices? It didn't really work when I did it...

A piece of paper is folded in half.  A second fold is made at an angle $ \phi$ ($ 0^\circ < \phi < 90^\circ$) to the first, and a cut is made as shown below.
[img]https://cdn.artofproblemsolving.com/attachments/c/b/5f2a2f269484cb7f8c3473f7489f870b5413cd.png[/img]
When the piece of paper is unfolded, the resulting hole is a polygon.  Let $ O$ be one of its vertices. Suppose that all the other vertices of the hole lie on a circle centered at $ O$, and also that $ \angle XOY \equal{} 144^\circ$, where $ X$ and $ Y$ are the the vertices of the hole adjacent to $ O$. Find the value(s) of $ \phi$ (in degrees).

im pretty sure parmenides51 is just adding it to a collection

how did he find it in the first place lol

This problem is HMMT 2008 , Geometry Round, p5. I just added the figure.
I am collecting all HMMT Geometry Rounds  among others [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url] ([url=https://artofproblemsolving.com/community/c2746635_geometry_from_usa_contests]here[/url]) and asked this to be added in related post collection [url=https://artofproblemsolving.com/community/c40244h1064480p30059908]here[/url].


January 1, 2007 was a Monday. In 2007, how many Fridays are
there?

Since only the first (and in leap years, second as well) go into the $ 53^\text{rd}$ week of an year, the are as many Fridays as there are full weeks: $ \fbox{52}$.


logarithms
inequalities proposed
inequalities

\[ \left\{\begin{array}{l}x>3,y>2\\x\plus{}y\plus{}z\equal{}6\end{array}.  \right.\]Prove that:\[ \ln(\frac{xyz}{6})^{2}\leq (3\minus{}x)(2\minus{}y)(z\minus{}1).\]
:ddr: :ddr: :ddr: :ddr:


group theory
abstract algebra
geometry
topology
calculus
integration
set theory

When reading advanced mathematical texts, and it's said "it's easy to show that ..." what do you do about this?

My approach is to meticulously write down a proof of the fact, and to think about what my proof means. However, this approach is slow as ; it can take forever to read a text written by an author that seems to have smoked pot while writing and hence left out most of the text. Should one learn to accept to not understand everything, or should you get more literature or what?

To whine a little, I am particularly upset right now for two reasons:
1) Over the past two days I've spent something like 15 hours on reading four pages about free groups, where a sample of the things I did as easy exercises were: Decipher an abstract definition of what a free group is without any aid of explanation, prove that the rank of a free group is well-defined, prove that a free group modulo it's commutator subgroup is free abelian and to show that a group on more than two free generators has subgroups of all countable ranks. Is it really normal for it to take this long to read in a book that is supposed to be read by students?

2) This semester (6th semester) I intended to take four courses: groups and rings, hardy spaces, riemann geometry and web programming (not as unserious as it sounds). However I quickly realized that I could not do these without killing myself. Riemann geometry in particular was all about proving all but the most difficult theorems yourself, and I simply could not fit this in a schedule without working 8 hours a day 7 days a week.

I guess the questions end up as:

1) What is a good way to deal with statements left to the reader)
2) Am I too meticulous about filling in details relative to my ability?

I don't usually do those things at all, and while this allows me to read relatively quickly, I think it seriously impairs my ability to understand the mathematical papers and books that I read. Maybe you do it too much, I don't know. But I don't think you should stop doing it completely.

If I'm serious about what I'm reading, I'll usually think about anything like that when I encounter it until I think I can prove it, but I don't actually carefully write out the proof (and often it turns out that my thoughts were wrong). If you read something like that and you are actually pretty sure you understand how to show it, then maybe you shouldn't take the time to write out the details. But if you are not too sure of how to do it, then working it out in detail probably is worthwhile because it forces you to understand what has happened before you continue.

Presumably after you do this for some time in the same book, you should get better at understanding the topic and the sort of ideas that the author doesn't like to elucidate, and you should get faster at this and you should be able to skip more things yourself. If this isn't happening, you should probably ask yourself why. (Are you not understanding how to apply the same ideas in various situations? Are you forgetting stuff too quickly? Maybe the book is just getting a lot harder. There ought to be reasons though.)

I usually fill in details on a per chapter basis instead of as they come up. Read a chapter/section. If you don't understand or you have to fill in details, just keep going. Often a concept I don't understand or proof I didn't fill in will clarify itself by the end of the section. The stuff that is still fuzzy I then fill in.

To me (this may not be the best approach for everyone) it seems that filling in the details wastes time if it will just clarify itself by seeing it in examples and proofs later on in the chapter. In your case, it seems like so many details were left out that this may not have been a good approach (was the Massey's Algebraic Topology book?  :lol: jk).

I always get VERY angry, I actually feel betrayed by the author. :mad:

Usually the "exercises" I struggle with are usually "almost obvious", but they get hard when you really want to get all the details.

I had an argument with some people recently about how authors should deal with those situations. One of my friends said that an author should never say "it is clear that" (or something to that effect) without giving any kind of hint as to what to use. So, it's acceptable to say something like "Using Lemma 2.3 and Corollary 2.5, it is clear that ...", but if the thing is just a direct application of those two things, the author should not say the result is clear without admitting what the reader should look at. Otherwise the reader will end up trying to prove it from Theorem 2.1 directly or wonder why Corollary 2.4 won't work or whatever, thus wasting lots of time. On the other hand, I personally think that it's those activities, that active engagement with the reading, that forces the reader to really understand the material. (This does make it hard to skim math text though.)

For example, I'm reading some notes by Bernard Host for my junior independent work this semester. In the notes, he takes some measure space $ X$ and inductively does the product of the space with itself and constructs a new measure on the product space. Then he defines a seminorm which involves integration against this measure. But the measure is somewhat obscure. So he makes all of these statements about properties of the measures and seminorms, and I can't understand any of them. They are all trivial to prove, but I have spent the past 3 days trying to figure them out. Some of them are simply 1 liners where you put two definitions together or whatever. But since I didn't understand what these things were, I couldn't see these proofs. Now that I have proved them all, I have some intuition for what the measures and seminorms are. The cost is that it took me 3 days to read 2 pages of some lecture notes that are just definitions preparing for the actual math later on, but I think it is worthwhile that I have done this and excusable that he makes me.

I suppose you must make a judgment when you are reading about how thoroughly you want to understand the material. Since this is for my research project, and it is actually my main interest in mathematics right now, I am not going to let any little thing slip by. But when I read some paper about nilmanifolds which is just to get an idea of what they are, I am not going to worry about these group theoretic lemmas that I don't care about whatsoever. So that paper I just skimmed, and I understood almost nothing of the proofs, but it only took a couple of hours and I got some vague ideas about nilmanifolds out of it.

Another important point is that the author already understands everything, and it's very hard to guess what will be clear to the reader and what will not be clear when everything is clear to you. So you cannot entirely blame the author for this; maybe the author is actually very conscious of the fact that readers will find some things difficult, but unintentionally gives excessive detail about some things and too little about others because he cannot easily judge how hard will be the ideas and proofs for a reader that is unfamiliar with the subject.

Things which I don't find interesting I take the author on trust.
My only deep interests for the time being in maths are in axiomatic set theory and logic, with a side interest in cryptography, lambda calculus and computability.
In my labourings through elementary set theory, I've come to one very important realization - if I am to learn the subject, or any subject for that matter, I must go very very slow - you can only be an expert in a small area.
You can only master one subject in a semester I think, and relaxing the definition of mastery you can extend that upto two unrelated subjects, but not more.
That, I'm afraid, is life. :(


search
geometry
geometric transformation
reflection
3D geometry
prism

How long do you need to solve all the puzzles by YOURSELF?
Please change the extension to exe.

how can this be a game? It's hardly 40 kb? :huh:

[quote="peter"]This file doesn't open.  Leastways, not with Ultimate Zip.[/quote]

All you do is change the extension from .rar to .exe and run it. Fun game, but pretty difficult after only the first few levels...

Yeah, I'm on levels 11 and 12 (woo, now 2 at a time!) and they're tricky.. for level 10 I had to start over and work backwards. That's fun.

level 11 is the first one that really required a trick. You know, this game is pure algebra :!:  :!:  :!: I love it! :)

Indeed a very nice game :D Maybe someone can port this to j2me or symbian? :)

:( I got stuck on #9.  Maybe I should try it again.

it won't let me download. it keeps logging me out and saying that i'm unauthorized to view or download it. i keep logging back in and it still doesn't work  :(

[quote="WindSlicer"]I don't get how to get it to work (i.e. open), heh.[/quote]Just rename the file from .rar in .exe and run it. Make sure you make a special directory for it because it creates an extra file (a database containing all the levels you solved and your solutions).

[quote="WindSlicer"]How do you rename it? It doesn't have .rar when you click "rename" but it's in the file name like if I rename it to "chromatron.exe" it is really "chromatron.exe.rar" I dunno, I'm no good with anything that takes more than a few clicks[/quote]Ah it seems that your windows is not set up to show the exensions of the most common files. You have to disable this option. Open a file explorer window. Go to Tools -> Folder Options -> Second Tab, scroll down a bit, and click off the "Hide extension of the common used files" option. Then you will be able to rename the file.

Thanks a mil Valentin!

On lv 10... soooo close yet so far, heh.

This is sad... I'm in the 30's but I still can't get 22 :(.

20, 22, 27.  i really should quit while i still can ;)

[quote="NoSoupForYou"]How do you do #9?[/quote]

[hide="hint"]dont go for the target in the second column.  you can come back for it later ;)[/hide]

I am now completely and utterly stuck on 11 and 12. Sad.
EDIT: Just got 11.

w00ts 27 28 29.  but the quantum entangler's got me confused :huh:

I finished the game. Now I'll start the search for sols with less pieces. :)

Now I'm stuck on 12 and 16. Why can't I get 12? It's starting to seem impossible! Still no hints please...
[quote="jli"]help on 13 and 15 please[/quote]
[hide="13"]There isn't really a trick here. You will use 3 benders and one splitter on blue, and the rest on red.[/hide]
[hide="15"]Put the blue filter between the leftmost two--I forget what they're called--things that only let through light in a certain direction. It doesn't matter that the white pinwheel isn't lit anymore. You can light it by making separate blue, red, and green rays come together later.[/hide]

[quote="LynnelleYe"]Now I'm stuck on 12 and 16. Why can't I get 12? It's starting to seem impossible! Still no hints please...[/quote]
If you change your mind, there are already hints that I gave out on 12 and 16 in this topic. :D

[quote="LynnelleYe"]I forget what they're called--things that only let through light in a certain direction.[/quote]
There are called conduits. :D

Sorry for this double post. But this post is huge.
These are the hints for the levels where I personally felt challenged.

#9
[hide="first hint"] Note that the conduits on this level only let through the light from northwest to southeast(and vice versa). So getting the highest one is hard thing to do.[/hide]
[hide="second hint"]But the bottom ones are easy to get. First try to get the two top ones.[/hide]
[hide="third hint"] The leftmost pinwheel is reached from right side[/hide]

#11
[hide="first hint"] The red and blue beam [b]BOTH[/b] needs to go in diagonally.[/hide]
[hide="second hint"]Therefore, [b]BOTH[/b] beams need to use the bender.[/hide]
[hide="third hint"]In my solution, green beam only hits bender and nothing else.[/hide]

#12
[hide="first hint"] Top blue pinwheel do not get hit by the blue beam directly.[/hide]
[hide="second hint"]Both red and blue beams hits [b]ALL[/b] splitters.[/hide]
[hide="third hint"]Think about how to get the top red. It might help.[/hide]

#16
[hide="first hint"]To get white ones, zigzag![/hide]
[hide="second hint"]Blue can be reached with one bender.[/hide]
[hide="third hint"]Last bender is used twice, like #11[/hide]

#18
[hide="first hint"]Forget about white. Where should the prism be? Where should the light come from?[/hide]
[hide="second hint"]Once you figure out the prism and the direction of the light, the last white pinwheel is easy to deduce.[/hide]
[hide="third hint"]Go straight to hit the bottom white, make some turns to get the top white.[/hide]

#19
[hide="first hint"]A prism can separate magenta beam into blue and red beams.[/hide]
[hide="second hint"]You can get two magenta beams with a splitter.[/hide]

#21
[hide="first hint"]We need to get two reds and two teals from two white beams, and we have a prism.[/hide]
[hide="second hint"]
[color=white]_[/color]RGB
+[color=white]_[/color]BGR
______
[color=white]_[/color]RTTR
[/hide]
[hide="third hint"]There is a perfect position for the prism. Find it.[/hide]

#22
[hide="first hint"]We need to do reverse of #21, make 2 whites with two reds and two teals.[/hide]
[hide="second hint"]
How do we make two teals? #19 may refresh your mind.
[/hide]
[hide="third hint"]A bender is used twice by a teal and a red.[/hide]

#27
[hide="first hint"]It is easy to see where the quantum tangler and four reflectors go.[/hide]
[hide="second hint"]
Can't get one pinwheel? See what happens if one of the beams from quantum tangler is reflected back.
[/hide]
[hide="third hint"]
Red beam becomes,
RBG[b]B[/b]
RGB[b]G[/b]
[/hide]

#28
[hide="first hint"]If you still don't get how the quantum tangler works, you can easily experiment with it in this level.[/hide]
[hide="second hint"]If you know how it works, but not getting this level, this word might help, [b]TIME[/b].[/hide]
[hide="third hint"]
If you are trying to get,
RGB
BGR
this is impossible.
Try to get
[color=white]_[/color]RGB
BGR
[/hide]

#30
[hide="first hint"]You don't need bender.[/hide]
[hide="second hint"]Can you get the white pinwheel with white beam and the other beams with any beam?[/hide]
[hide="third hint"]
The doppler goes into very weird spot, [color=white]above the white pinwheel[/color].
[/hide]

#32
[hide="first hint"]It's getting harder!!! First note that three of the pinwheels are collinear diagonally. You can get them easily with a bender, a quantum tangler, and a doppler.[/hide]
[hide="second hint"]Forget about the top blue, get others.[/hide]
[hide="third hint"]A doppler and a prism should be left by now. And you got everything except the top blue. Make magenta somewhere using the doppler, and use prism to get the last blue.[/hide]

#35
[hide="first hint"]The white beam go through the quantum tangler. One beam will go through the pinwheels and the other beam will go through a prism.[/hide]
[hide="second hint"]Beams that are separated by prism need to have a lot of space, so that we can doppler them to get colours we want for the other side[/hide]
[hide="third hint"]Only one doppler is used to get the first pinwheel. Also for the second pinwheel.[/hide]

I am stopping here since I don't have #39.

I gave up and got #12 with the help of your first hint and now I'm feeling stupid for not realizing the contents of your hint myself. Now on 16 and 18...

I'm stuck on #10 and i do the hints and i can't win!


Maxwell's equations relating the electric field E and the magnetic field H as they vary with time in a region containing no charge and no current can be stated as follows:

div $ \textbf{E}$ = 0
curl $ \textbf{E}$ = $ \minus{}\frac{1}{c}\frac{\partial \textbf{H}}{\partial t}$
div $ \textbf{H}$ = 0
curl $ \textbf{H}$ = $ \frac{1}{c}\frac{\partial \textbf{E}}{\partial t}$

where $ c$ is the speed of light. Show that $ \bigtriangledown \times (\bigtriangledown \times \textbf{E}) \equal{} \minus{}\frac{1}{c}\frac{\partial^{2} \textbf{E}}{\partial t^{2}}$

I know that $ \bigtriangledown \times (\bigtriangledown \times \textbf{E})$ can be simplified down into curl $ \frac{\partial \textbf{H}}{\partial t}$, but how can I continue from there?

Another possible way is to calculate $ \minus{}\bigtriangledown^{2}\textbf{E}$ which is also the same as $ \bigtriangledown \times (\bigtriangledown \times \textbf{E})$, but I couldn't proceed. How could I go on from here?

[quote="1/(ln x)"]Show that $ \bigtriangledown \times (\bigtriangledown \times \textbf{E}) \equal{} \minus{} \frac {1}{c}\frac {\partial^{2} \textbf{E}}{\partial t^{2}}$[/quote]

you should start with the cross product  identity

$ \nabla \times (\mathbf{a} \times \mathbf{b}) \equal{} \mathbf{a} (\nabla \cdot \mathbf{b}) \minus{} \mathbf{b} (\nabla \cdot \mathbf{a}) \plus{} (\mathbf{b} \cdot \nabla) \mathbf{a} \minus{} (\mathbf{a} \cdot \nabla) \mathbf{b}$

in your case, $ \mathbf{a}$ is the vector differential operator [b]"del"[/b] ( i.e. $ \nabla$ ) and $ \mathbf{b}$ is  the electric field ( i.e. $ \textbf{E}$ )

you should eventually get

$ \nabla \times \nabla \times \mathbf{b} \equal{} \nabla(\nabla \cdot \mathbf{b}) \minus{} \nabla^{2}\mathbf{b}$

which is often called [b]curl of the curl[/b]  :D

in the absence of charges and currents,

$ \nabla^2 \mathbf{E} \;\equal{}\; \mu_0 \epsilon_0\quad \frac{\partial^2 \mathbf{E}}{\partial t^2} \qquad (SI)$

or

$ \nabla^2 \mathbf{E} \;\equal{}\; \sqrt{\mu_0 \epsilon_0}\quad \frac{\partial^2 \mathbf{E}}{\partial t^2} \qquad (CGS)$

[quote="misan"]
you should eventually get

$ \nabla \times \nabla \times \mathbf{b} \equal{} \nabla(\nabla \cdot \mathbf{b}) \minus{} \nabla^{2}\mathbf{b}$

which is often called [b]curl of the curl[/b]  :D

[/quote]

First we know that $ \rm div \mathbf{b} \equal{} 0$, which is given.

So simplifying that I get $ \nabla \times \nabla \times \mathbf{b} \equal{} \rm grad (\rm div \mathbf{b}) \minus{} \nabla^{2}\mathbf{b} \equal{} \minus{} \nabla^{2}\mathbf{b}$.

Now I am stuck, how can I continue?


algebra
polynomial
function
trigonometry
real analysis
real analysis unsolved

I know a nice solution to this one but still want to share it with mathlinkers.

Let $ f$ be a continuous periodic function such that for every $ x\in\mathbb R$, the partial sums of the series $ \sum_{k\equal{}1}^\infty f(kx)$ are bounded (but the bound may depend on $ x$). Prove that $ f$ is a trigonometric polynomial.

I haven't gotten anywhere yet, but being a trig polynomial can't be the only condition. We'd better have $ f(0)\equal{}0.$ Else just let $ x$ equal the period.

If you want the full description, it must be a finite sum of sines but to show that once you already know that $ f$ is a trigonometric polynomial is fairly easy and not half as interesting as the first step :).

this one reminds me your solution of problem which was posted by me :)
http://www.mathlinks.ro/viewtopic.php?t=147733
maybe in similar way  need to suppose that  $ f$ is not trigonometric polinomial,. Then as in your observation 2 we can show that
$ \sum_{j\ne 0}\left|\frac {a_{j}}{e^{2\pi ij\,x} \minus{} 1}\right|^{2}\le 4M_{1}^{2}$
now to get contradiction we need to  use the fact that infinitely much   $ a_{j}$ is not equal to zero.
But I don't  have any good information about $ a_{j}$, I just know that $ f$ is continuous function..
if $ f$ is very good function then $ a_{j}$ tends to zero very good so serie will be of course bounded and so I can't get contradiction.
So tehrefore we need to choose in serie such x wich are very well aproximated via rational numbers. For example The numbers with irrationality measure $ \mu \equal{}\infty$  I think then we can get contradiction with boundness of serie   $ \sum_{j\ne 0}\left|\frac {a_{j}}{e^{2\pi ij\,x} \minus{} 1}\right|^{2}\le 4M_{1}^{2}$

[quote="fedja"]I know a nice solution to this one ...

... [/quote]
fedja     please show us your solution  :blush:

You almost had it yourself. The statement to prove is the following. If $ f$ is 1-periodic and the partial sums of the series $ \sum_k f(kx)$ are bounded for all [b]irrational[/b] $ x$, then $ f$ is a trigonometric polynomial" (this also makes the statement an "if and only if" one).

[b]Claim 1:[/b] If $ x$ is irrational and the sums of the above series are bounded by $ M$, then for all $ y\in [0,1]$ the sums of the series $ \sum_k f(y\plus{}kx)$ are bounded by $ 2M$.

Indeed, we can just approximate $ y$ by $ \ell x$ modulo $ 1$ with an integer $ \ell$ with arbitrary precision and use continuity of $ f$.

[b]Claim 2[/b] If $ m\in\mathbb Z$, then the partial sums of the series $ \widehat f(m)\sum_k e^{2\pi i kmx}$ are bounded by $ 2M$.

Indeed, just integrate the result of Claim 1 against $ e^{\minus{}2\pi i my}$ over $ [0,1]$.

[b]Claim 3[/b] If $ m_s$ is any infinite sequence of pairwise different integers and $ \delta_s$ is any sequence of positive numbers, then there exists an irrational $ x$ and a subsequence $ s_p$ such that the distance from $ m_{s_p}x$ to the nearest integer is less than $ \delta_{s_p}$ for all $ p$.

This is just induction. Put $ s_1\equal{}1$ and choose any closed interval $ I_1$ of positive length such that $ m_1 x$ is an integer with precision $ \delta_1$ for all $ x\in I_1$. Now choose $ s_2$ so that $ m_{s_2}|I_1|>2$. Then $ m_{s_2}x$ runs over an interval of length at least $ 2$ when $ x$ runs over $ I_1$. Hence, we can find 2 disjoint options for $ I_2\subset I_1$ such that $ m_{s_2}x$ is an integer with precision $ \delta_{s_2}$ for all $ x\in I_2$, and so on. We get a sequence of nested intervals and their common point will satisfy the main condition of the claim. Since we have two disjoint choices at each step, we can make sure that this point is irrational (say, by enumerating the rationals and choosing $ I_n$ that does not contain $ q_n$ at each step).

[b]Claim 4:[/b]. If infinitely many $ \widehat f(m)\ne 0$, then the conclusion of Claim 2 cannot hold.

Indeed, just choose $ \delta_m$ in such way that $ |\widehat f(m)|\delta_m^{\minus{}1}\to\infty$ along the indices of non-zero Fourier coefficients of $ f$ and choose $ x$ from Claim 3.

Ok. I just was thinking that you have different solution.
Really good problem (since this one includes such parts as fourier series, irrationality measure ... )


So...um...yeah...guys...

Toronto Maple Leafs

2-0 after 2 periods vs. Atlanta.

Sounds like a win right?

ATL wins 5-2 after 3rd.

Right at the ACC.

I've got tissues if anyone needs any.

Just incase you guys were in danger of forgetting.

I haven't seen a more significant collapse this season.

I mean, the Montreal game...at least we continued fighting after playing poorly. But yesterday, it was just a pathetic effort. Maurice was right to say we lost it in the third...we took what? 9 penalties during that period?

And even more bad news? We're slipping in the standings with Ottawa now tied with us with one game in hand...and Bruins poised to do the same. We better get our act together for Thursday; learn how to hang onto leads and play 60 minutes instead of 20.

[quote="SM4RT"]The leafs started out strong but now just suck. They cant beat anyone as of late it doesnt even matter what they are up by, they end up blowing it. I hope they make the playoffs this year.[/quote]

Ahem.

It's not that they suck.

It's that they CHOKE.

CHOKE CHOKE CHOKE.

Well a team that consistently chokes kinda sux..lol unless if they pull it together

CANUCKS!!! GO CANUCKS!!! canucks? YEA CANUCKS!!!!! we suck this year though... :(


let x,y,z>0 and $ \sum x^2\plus{}2xyz\equal{}1$
prove that:
$ 3xyz \leq 2\sum x^2y^2$

Let $ x\equal{} \sqrt[]{\frac{ab}{(a\plus{}c)(b\plus{}c)}}$ ...
This ineq becomes:
$ \sum\ \frac{a}{b\plus{}c} \geq\ \frac{3}{2}$
(obviously trues)  :wink:


Compute (without simple brute force) $\sum_{k=1}^{89}\sin^{2}k$ ?

[hide]
Pair up outside terms, and use the identities $\sin(\theta) = \cos\left(\frac{\pi}{2}-\theta\right)$ and $\sin^{2}\theta+\cos^{2}\theta = 1$ to get
\[\sum_{k=1}^{89}\sin^{2}k = 44(1)+\sin^{2}45 = 44.5 \]
[/hide]

[hide=Attempt at a solution]
How about using
$\sin^{2}\theta = \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^{2}$
Then we have a finite geomtetric series
$\sum_{x=1}^{89}\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^{2}$,
which is easily summable using the formula

$\frac{a(1-r^{n})}{1-r}$

I'm assuming you mean for k to be in degrees?

This problem showed up in a local contest (WoCoMaL) this year, although the index varied from 1 to 90 instead of 1 to 89.

Sadly, it was on an open-calculator round, and many just put their TI-89's to work while working out the other 2 problems in the round by hand.

I was staring at that wocomal problem for a minute before I realized that the writers used $5^{0}$ to mean 5 degrees, not 1.

A similar problem:

Compute $\prod_{k=1}^{89}\tan k^\circ$.


geometry
calculus
integration
function
projective geometry

Heya, I'm doing the IB Diploma programme, and have decided to do an Extended Essay in Mathematics (HL). Which means, I need to have between 3500-4000 written words, and loads of maths.
I would like it to be interesting, and possible to understand for other students who don't do HL Mathematics...I've read a couple Extended Essays in maths...and few of them have been interesting.
So I need some help in defining a topic/research question. :) I have an easy one, that I know I will be able to do, if I can't find anything interesting... It's just to do it, and I can get it to be 3500words I think... Estimating pi/proving pi/deciding whether polygon inscribed or circumscribed a circle gives me pi fastest...

Seriously though, I would like to do this well :) And I guessed this would be a smart place to ask for help. Seems like there are a lot of amazing math people here ;)

Some interesting ideas:

Projective Geometry (there's a lot of historical content to be found here, and some very cool results)

Evaluating Trigonometric Integrals using the Residue Theorem (possibly too much analysis required)

The Riemann-Stieltjes Integral (this probably is not a great idea because of the analysis content, although the ability to invert sums and integrals is very, very cool)

Generating Functions (ahem, everything you need here can be found in Generatingfunctionology)

Constructible Regular Polygons (again a lot of historical content)

I like the first topic the most.  I've given kids EE topics before, so unfortunately a lot of my favorite topics are gone already.


[quote="cindy28"]im just a novice here and i dont know were i can write some solutions to answer the questions of someone. im hopping someone could teach me :(

thanks!!! :D[/quote]
Well, you just click the Topicreply button and then enter your solutions...put the dollar signs if necessary.

To learn how to write solutions in LaTeX form, I suggest you visit  [url=http://www.artofproblemsolving.com/Forum/index.php?f=220&sid=b66327321367169884e217e4e448ddb2]here[/url], it teaches you how to use it.

Make sure to hide your answers though if you're posting an answer to a challenge that someone has given. If you're not sure whether you've hidden something or not, click the "preview" button and you can see if you have.

Have fun! :)

still didnt get it!!!! :wacko:
i dont understand a word!!! :(

Um, I think that she understands topic reply, or she wouldn't reply. I think she wants to know how exactly to write a concise solution. For that, you might want to read some of the articles on the site, and nothing beats starting simple.

Explore the website... :?

Click resources, because you need resources, and then click w.e you need.

Also, you should try reading some of the solutions to problems posted on this site--it'll let you get a feel for it.

[quote="cindy28"]but where can i find it?[/quote]
i wonder how you reply a topic.

[quote="cindy28"]i know how to reply but i dont know where i can write the solutions.

someone has told about using the Latex :lol:  i think i understand it better. but some are not. thier so....so....difficult to understand.[/quote]
Where did you wrote this reply?

[quote="cindy28"]in the post reply of course!

ok lets get it straight!

What i need to know now is how to use the latex.[/quote]

Also, look at the link I provided in my previous post

thanks for the link that you posted me i hope it'll help :)

[quote="xxreddevilzxx"][quote="shobber"][quote="cindy28"]in the post reply of course!

ok lets get it straight!

What i need to know now is how to use the latex.[/quote]
On the top of the page, you can find $Latex$ help, click it.
And don't forget to put dollar signs when using $Latex$[/quote]

you mean $\LaTeX$?[/quote]
Yep......
I always wonder how to type $\LaTeX$......

hope the Latex helpS :?

whats with the use of the dollar sign and how do you use it like this? $

The dollar signs are used for "math mode."  You put one at the beginning of the stuff you want in LaTeX and one at the end.  For example,  you would type in:
$x^2+y^2=0$

to get:
$x^2+y^2=0$

Hope that helps.

(sorry about the double post--I had to turn off the BBCode in the last post)

[quote="AoPS website"]BBCode is a special implementation of HTML.  BBCode itself is similar in style to HTML: tags are enclosed in square braces [ and ] rather than < and > and it offers greater control over what and how something is displayed.[/quote]
 :)

A quick way to learn new Latex is copying other people's posts that have interesting Latex into Notepad or something and examining how they did it :)

[quote="cindy28"]ohh so without the bbcode latex wont help.[/quote]
Yes, if you have your BBCode turned off, the LaTeX won't "work." You can turn on the BBCode by going to the top of the page to click on "Options."  Then, on the left, under "Profile: Options", click on "Posting a Message" and click "Yes" to the "Always Allow BBCode."  Then click "Submit."

[quote="cindy28"]so wats d BIG difference about the $Latex$ and the normal type?[/quote]
It makes your math expressions more readable and neater.  $\displaystyle \frac{x^2}{a}=7$ looks nicer than (x^2)/a=7.  Of course, you don't have to use it.

Latex looks loads better O.o

Suppose you have $\frac{1}{2+\frac{3}{4+\frac{5}{6}}}$

Imagine how that would look without Latex!

Maybe that wasn't a good example lol. It looks easier in non-Latex   :blush:

$f(x)=\{\begin{array}{cc}1,&\mbox{ if }
x\geq 2\\-1, & \mbox{ if } x<2\end{array}$

So much easier to understand, thanks to $LaTeX$

(Thanks to whoever wrote the .pdf file for that one)

but isn't it harder to write $Latex$ cause you still have to write some dashes?

Specially dollar signs.

It might take a little more time to write

but for the most part, it is easier to read for others...

cant there be another window where you can write it better?

cauze it would be better if there would be another window


[quote="auhongan_au"]Let positive real numbers a,b,c prove that (a+b)^2(b+c)^2(c+a)^2>=(a+b+2c)(b+c+2a)(c+a+2b)[/quote]
If $a=b=c=1/2$ then...

[quote="chien than"]I think this problem may be :
Prove that $ (a\plus{}b)^{2}(b\plus{}c)^{2}(c\plus{}a)^{2}\geq abc(a\plus{}b\plus{}2c)(b\plus{}c\plus{}2a)(c\plus{}a\plus{}2b)$
where $ a;b;c$ be positive real numbers[/quote]
If so, it's equivalent to $ \sum_{cyc}(a^{4}b^{2}\plus{}a^{4}c^{2}\plus{}2a^{3}b^{3}\minus{}a^{3}b^{2}c\minus{}a^{3}c^{2}b\minus{}2a^{2}b^{2}c^{2})\geq0.$


[quote="warut_suk"]For $a,b,c\in R$, prove that
$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq \dfrac{3}{2}$[/quote]

Hi warut_suk!

Nice ineq!

Using Minkowski's ineq one obtains

$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq \sqrt{(a+b+c)^2+(3-a-b-c)^2}=\sqrt{x^2+(3-x)^2}$
where $x=a+b+c$.

Now we are to prove that
$\sqrt{x^2+(3-x)^2}\geq\dfrac{3}{2}$ or $x^2+(3-x)^2\geq \dfrac{9}{4}$ or $2x^2-6x+9\geq\dfrac{9}{4}$ or $8x^2-24x+27\geq 0$
which is obviously true since its discriminant is $D=24^2-4\cdot 8 \cdot 27<0$

I think there is the following stronger ineq

$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq \dfrac{3\sqrt{2}}{2}$

Which can be proven by the same way!
 :)

Hi,

Even easier using the concavity of $x\to \sqrt {x}$:

$\sqrt {a^{2}+(1-b)^{2}} \geq \dfrac {\sqrt {2}}{2}(1+a-b)$, and summing gives the lower bound $\dfrac {3 \sqrt {2}}{2}$.


Let $ (X,d_1), (Y,d_2)$ are metric space. $ f: (X,d_1)\rightarrow
(Y,d_2)$ continous, open. Prove: $ X$ local compact $ \Rightarrow$
$ f(X)$ local compact

let $ f(x)\in f(X)$. since $ X$ is locally compact, there exists a neighborhood $ U$ of $ x$ which is compact. we'll show that $ f(U)$ is a compact neighborhood of $ f(x)$. let $ \{V_i\}_{ i\in I}$ be an open cover of $ f(U)$. Clearly, $ \{f^{\minus{}1}(V_i)\}_{i\in I}$ is an open cover of $ U$, and there's a finite subcover, $ \{f^{\minus{}1}(V_i)\}_{i\in F}$, of $ U$ since $ U$ is compact. then $ f(U)\subset f(\bigcup_{i\in F}f^{\minus{}1}(V_i))\equal{}\bigcup_{i\in F}f(f^{\minus{}1}(V_i))\equal{}\bigcup_{i\in F}V_i$, so we have a finite subcover of $ f(U)$, QED.

This only proves that $ U$ is compact, not that it is a neighborhood of $ f(x)$.

you mean $ f(U)$. and it is a neighborhood because $ f$ is open.

[quote="limsup"][quote="pluristiq"] then $ f(U)\subset f(\bigcup_{i\in F}f^{ \minus{} 1}(V_i)) \equal{} \bigcup_{i\in F}f(f^{ \minus{} 1}(V_i)) \equal{} \bigcup_{i\in F}V_i$,[/quote]

The last equality should be an inequality  $ \subseteq$ isn't it?! :)[/quote]

no. $ f(f^{\minus{}1}(Y_i))\equal{}Y_i$. in other words, the image of the pre-image of a set equals the set. there's no way the set can get smaller, and actually it stays the same. on the other hand, when you take the preimage of the image, you usually get a bigger set than what you started with.

No. If you think carefully you will understand that your argument should be exactly reversed.
For example, let $ f: \mathbb{R}\to \mathbb{R}$ be given by $ f(x) \equal{} x^2$ and let $ Y$ be say $ [ \minus{} 4,4]$. Find $ f(f^{ \minus{} 1}(Y))$

wow, that was silly of me! you're obviously right. thanks for the correction! :)


ABC is an isosceles triangle with AB=AC.A circle through B touches the side AC at D and intersects AB at P.If D is the mid point of AC, then prove that AB=4AP.

This is based on Power of a Point.

[hide]
Since $ \text{AD}$ is tangent to your circle and $ \text{AB}$ crosses through, then

$ \text{AD}^2\equal{}\text{AB} \times \text{AP}$

Then, use your given information ($ \text{AD}\equal{}\text{AC}/2\equal{}\text{AB}/2$) to work this equation into terms of $ \text{AB}$ and $ \text{AP}$.
[/hide]


[quote="nikos"]Soup              18 ευρώ
Salad             23  ευρώ
Roast Beef      41 ευρώ
Ice Cream      36  ευρώ
Coffee            27 ευρώ

How much would oysters cost ? (y is not a vowel) .[/quote]

diladi???........... :maybe:  :maybe:  :huh:

Δηλαδή πόσο κοστίζει .
Είχε η ερώτηση και πιθανές απαντήσεις :

α) 27 ευρώ     β) 30 ευρώ      γ) 33       δ) 36        ε) 39

Aπλά λύνουμε ένα σύστημα  :wink:

Δηλαδή Σιλουανέ?

re niko etsi pu to egrayes to provlima nomiza oti to ka8e grama itan diaforetiki metavliti kai skeftomuna pos 8a apaliyo merikes metavlites pu emfanizonte mono mia fora opos p.x.to d apo to salad...
anyway.simfono=5,fonien=4,ara to c=33ine to sosto...
arxika dokimasa tin antistixia a=1,b=2..klp..klp ala den voi8use +poli
meta evala z=1 klp..klpkaime mia praksi..mu isxian ola ektos apo to mpifteki,kai leo ti ginete edo re gi ayto estila ti enois..
pantos 8a itan pio diskolo an den anefere oti to y ine simfono gt den 8a skefti kanis +poli eykola oti ine diaxorismos simfono fonienton...

 :D

δεν καταλαβα τι εννοεις Ραλλη   :blush:

kosta,stin arxi skeftika na antistixiso to gramma a=1 to b=2 klp...z=26kai na do me pia praksi metaksi ayton ton ari8mon (p.x gia to salad s=19,l=12,a=1,d=4)8a proekipte to zitumeno poso..p.x (19+1+12+1+4)/4=17,75 pu me strogilopiisi dini ta 18 eyro...
meta skeftika mipos(kai edo kolisa perisotero/...)to ka8e grma paristani diaforetiki metavliti (ala oxi kataanagki opos tis times tis alfabitas..)diladi
oti s+o+u+p=18 (1)
s+a+l+a+d=23 (2)
.......................

kai vasi ayton na emfanisume to o+y+s+t+e+r+s...
ala iparxun metavlites opos to d sto salad pu iparxun mono mia fora i ales opos to y pu den iparxun kan stiis arxikes..
opote meta afu egrafe kai oti to y den ine fonien,e, leo tote 8a ine ola ta simfona mia sigekrimeni timi kai ola ta fonienta mia ali timi...opote
apo tin supa +tin salata pernis oti simfono =5 kai fonien =4 pou blepis aplos oti epali8eyun kai to mpifteki,to pagoto kai ton kafe kai ara vgeni eykola meta oti ta stridia ine 33...euro..

popoooo......mu anikse i oreksi............ :lol:

[quote]pantos 8a itan pio diskolo an den anefere oti to y ine simfono gt den 8a skefti kanis +poli eykola oti ine diaxorismos simfono fonienton... [/quote]

Έχεις δίκιο .Απλά την έβαλα όπως την βρήκα .

Congrats RAKAR! Πολύ καλή σκέψη και αντιμετώπιση του προβλήματος!

Απλώς εκτός από το ότι σου ανοίγει η όρεξη με αυτόν τον τρόπο, σου ανοίγει ταυτοχρόνως και η... τσέπη!

Σκεπτόμενος ότι είμαστε Έλληνες και στις ταβέρνες κανείς δεν φέρνει απόδειξη, ότι το Φ.Π.Α. είναι 18%, θα έχουμε:

$33-33\cdot\frac{18}{100}= 33-5,94 = 27,06$

Στις επιλογές που έδωσε ως προς απάντηση ο Νίκος (χρόνια του πολλά όπως και σε όλους τους Νίκους) βρίσκεται και το 27. Επομένως, η σωστή απάντηση είναι το α), όταν αναφερόμαστε στα ελληνικά πάντα πλαίσια!  :P  :P

J/K  :rotfl:

Ευχαριστώ mostel .

Εύχομαι και εγώ χρόνια πολλά σε όλους Νικόλαους του forum
και επί τη ευκαιρία εύχομαι σε όλα τα παιδιά καλή επιτυχία
στους διαγωνισμούς της ΕΜΕ που αρχίζουν από το Σάββατο .


Imagine a smooth ball the size of earth with a steel band strapped tight around the middle. If you added a yard to the length of the steel band, would it then be loose enough all around to slip a your finger under it?, your fist?  Your little brother’s head?

By using $ p\equal{}2\pi r$ one can easily get $ \frac{1}{2\pi}$ yards
The answers are yes, yes, no(not sure about this one).

Quite interesting  :)

Original poster,

you are making an assumption that the new
length is suspended the same height uniformly
above the sphere, BUT you did not state that.

Without that assumption, the gap between
the sphere's surface and the underside of
the steel band can be significantly larger
on one side of the sphere.

I know, incredibly enough, adding 36 inches to the length of the steel band would lift it 5 ¾ inches above the surface of the ball all the way around. Whether your little brother’s head will fit under that is something you will just have to experiment with. You can also try it on a basketball whose circumference is 29.5 inches. The expanded band will be 65.5 inches, the diameter will be 1/ π  (20.8 inches) and the radius will be 10.4 inches. Or 5 3/4 inches more than the standard basketball.