sum of 3 #s is 61. the second #is 5 times the first and the third is 4 more than the second .find the #s

Let the first number be $ \alpha$.

$ \alpha \plus{} 5\alpha \plus{} 4 \plus{} 5\alpha \equal{} 61$

$ 11\alpha \plus{} 4 \equal{} 61$

$ 11\alpha \equal{} 57$

$ \alpha \equal{} \frac{57}{11}$

The numbers are $ \frac{57}{11}$, $ \frac{285}{11}$, and $ \frac{329}{11}$.

@fantasy  -> y are u posting when isabella has already posted the solution!

They were typing at the same time, and izzy beat fantasy.

Or you can check out the time stamps.

They're in the lower right of each post.

thanks now i realize
 by the way both their time has clashed at 4:15  :P  :P  :P
so who's the winner


How well do u have to do in USAMO to qualify for MOP???

Depends on how hard the test is and how well people do. Last year, IIRC, it was 7+ (least sure about this number) for freshmen and younger, 21+ for sophomores and juniors, and 28+ for seniors (i.e., top 12).


[quote=chien than]Prove that:
$\sqrt{x+(y-z)^{2}}+\sqrt{y+(z-x)^{2}}+\sqrt{z+(x-y)^{2}}\geq \sqrt{3}$
where $x;y;z$ are positive real numbers satisfying $x+y+z=1$[/quote]
For  $x,y,z\ge0 $ and  $ x+y+z=1 $ . Prove that$$ \sqrt{x+(y-z)^{2}}+\sqrt{y+(z-x)^{2}}+\sqrt{z+(x-y)^{2}}\le\sqrt{10}$$

[quote=chien than]Prove that:
$\sqrt{x+(y-z)^{2}}+\sqrt{y+(z-x)^{2}}+\sqrt{z+(x-y)^{2}}\geq \sqrt{3}$
where $x;y;z$ are positive real numbers satisfying $x+y+z=1$[/quote]
2011，I prove
For  $x,y,z\ge0 $ and  $ x+y+z=1 $ . Prove that$$ \sqrt{\lambda x+(y-z)^2}+\sqrt{\lambda y+(z-x)^2}+\sqrt{\lambda z+(x-y)^2}\ge\sqrt{3\lambda}.(0<\lambda\leq 1)$$
In particular
For  $x,y,z\ge0 $ and  $ x+y+z=1 $ . Prove that$$ \sqrt{x+2(y-z)^{2}}+\sqrt{y+2(z-x)^{2}}+\sqrt{z+2(x-y)^{2}}\ge\sqrt{3}$$

[quote=sqing]For  $x,y,z\ge0 $ and  $ x+y+z=1 $ . Prove that$$ \sqrt{x+(y-z)^{2}}+\sqrt{y+(z-x)^{2}}+\sqrt{z+(x-y)^{2}}\le\sqrt{10}$$[/quote]
2007，I prove
For  $x,y,z\ge0 $ and  $ x+y+z=1 $ . Prove that$$ \sqrt{\lambda x+(y-z)^2}+\sqrt{\lambda y+(z-x)^2}+\sqrt{\lambda z+(x-y)^2}\le\sqrt{2(\lambda+4)}.(0<\lambda\leq 5)$$

[quote=sqing][quote=arqady][quote=sqing]
For  $x,y,z\ge0 $ and  $ x+y+z=1 $ . Prove that$$ \sqrt{x+(y-z)^{2}}+\sqrt{y+(z-x)^{2}}+\sqrt{z+(x-y)^{2}}\le\sqrt{10}$$[/quote]
$\sum_{cyc}\sqrt{x+(y-z)^2}\leq3$ is also true (with the same conditions).[/quote]
For  $x,y,z\ge0 $ and  $ x+y+z=1 $ . Prove that$$ \sqrt{x+(y+z)^{2}}+\sqrt{y+(z+x)^{2}}+\sqrt{z+(x+y)^{2}}\le 3$$[/quote]

x+(y+z)^{2}《=1

[quote=arqady][quote="chien than"]Prove that:
$\sqrt{x+(y-z)^{2}}+\sqrt{y+(z-x)^{2}}+\sqrt{z+(x-y)^{2}}\geq \sqrt{3}$
where $x;y;z$ are positive real numbers satisfying $x+y+z=1$[/quote]
$\sqrt{x+(y-z)^{2}}+\sqrt{y+(z-x)^{2}}+\sqrt{z+(x-y)^{2}}\geq \sqrt{3}\Leftrightarrow$
$\Leftrightarrow2\cdot\sum_{cyc}\sqrt{(x+(y-z)^{2})(y+(x-z)^{2})}+\sum_{cyc}(x-y)^{2}\geq2\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\sqrt{(x+(y-z)^{2})(y+(x-z)^{2})}\geq3(xy+xz+yz).$
But $\sum_{cyc}\sqrt{(x+(y-z)^{2})(y+(x-z)^{2})}\geq\sum_{cyc}\left(\sqrt{xy}+|(y-z)(x-z)|\right)\geq$
$\geq\sum_{cyc}\sqrt{xy}+|\sum_{cyc}(x-z)(y-z)|=\sum_{cyc}(\sqrt{xy}+x^{2}-xy).$
Thus, remain to prove that for all non-negative $a,$ $b$ and $c$ holds:
$(ab+ac+bc)(a^{2}+b^{2}+c^{2})+a^{4}+b^{4}+c^{4}\geq4(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}),$ which triviality.[/quote]

Dear Arqady,
Your last ineq holds by means of Schur's and AM-GM , right?

[quote=chien than]Prove that:
$\sqrt{x+(y-z)^{2}}+\sqrt{y+(z-x)^{2}}+\sqrt{z+(x-y)^{2}}\geq \sqrt{3}$
where $x;y;z$ are positive real numbers satisfying $x+y+z=1$[/quote]

[img]http://s16.sinaimg.cn/middle/006ptkjAzy76bAJBybl2f&690[/img]


rotation
geometry
geometric transformation
reflection
combinatorics unsolved
combinatorics

- A figure its said to be an "$ (a,b)$ell" if it has $ a \plus{} b \minus{} 1$ unit squares and it has $ a$ vertical squares and $ b$ horizontal squares in this way:

x x x x x x ....( $ b$ horizontal squares)
x
x
x
x
.
.
.
($ a$ vertical squares)
x
x

Let $ B$ be a $ m \times n$ board with $ mn$ unit squares such that $ B$ can be covered with only "$ (a,b)$ells" (this means, only "ells" of the $ (a,b)$ type) without overlapping, and the "$ (a,b)$ells" are completely inside the board. Show that $ a\equal{}2$ or $ b\equal{}2$.

can you rotate / mirror the ells?

if yes then you can use $ (2,b)$ or $ (a,2)$ ells...

[hide="Very Unrigorous"]Consider a tiling of a mxn board with wrong types of ells (assume there exists a tiling). Consider any of the ells in the tiling. Let's ssay it's oriented Like:

XXXXXX
X1
X
X

Where the X's are part of the ell. Now, Consider the ell containing the square with the 1. It either has the same orienation or it doesn't. If it does,

111111
12
1
1

Consider the 2 instead. (All the ones belong to the same ell.) So we can continue this until we get a ell with different orientation.

Then we have something like:

11111
1222
1  32
     2
     2

In the example above, consider the ell containing the square with the 3. It obviously can't fit in.
[/hide]

why i can't place the ell in the spot with a 3 like this:

11111
1222
1 32
2
2

to

11111
1222
132
23
23333333

11111
1222
1432
2 3
2 3333333

Then consider the square with the 4, Now it doesn't work.

ok... but, are you assuming that this arrangement

11111
1222
1432
2 3
2 33333

is in a corner? because you said that we continue in putting ones until changing direction of the "ell", so maybe we can set:

x11111
x1222
x1432
x243
x2433333
xx4
444

and, how does this generalize the (a,b) ell? I don't think I'm understanding you well.

Note: as a matter of fact, the problem statement is wrong. if a or b = 1, we can obviously tile the arrangement. so it rephrases like:

Show that $ a\leq 2$ or $ b\leq 2$.


Hi , i don't know how to solve this problem :

At the square side 6 cm, is registred a triangle ABC.
AQ = BR = 1/2QR, AP = CS = 1/2PS and PQ //RS// BC.
Thus, we can state that the area of trapeziu PQRS is : ??

[img]http://img12.imageshack.us/img12/5607/triangulon.th.jpg[/img]

I wait.Thank you.

[hide="Hint"]$ \triangle AQP : \triangle ARS : \triangle ABC \equal{} QP^2 : RS^2 : BC^2$[/hide]

[quote="Cafofo"]I must be stupid ... this is not helped :(  :rotfl:[/quote]
What do you want more? Kouichi Nakagawa's hint is enough to calculate.

Really friend, i don't know how to continuos with this hint ( in fact, I do not understand what exactly this meant hint [like where has this proportion?] ) ; sorry if this stupid quests, but i really want to learn that.
I've never calculate some area just using proportions of a triangle into a square.

$ Q1.$ Find the area of $ \triangle ABC$.
$ Q2.$ Find the length of $ RS$, $ QP$.
$ Q3.$ Find the area of $ \triangle ARS$, $ \triangle AQP$.
$ Q4.$ Find the area of trapezium $ PQRS$.

In fact i couldn't find the lenght of RS and QP, i don't know where im missing...
but i solve this by this way:

$ \triangle AQP$ ~ $ \triangle ABC$ , you get the proportion = 1 / 4  ;
The same with $ \triangle ARS$ ~ $ \triangle ABC$ , proportion = 3/4 ;

$ Area$ $ \triangle ABC$ = 18, just do $ \triangle ARS$ - $ \triangle AQP$ = 9 cm²

Thanks for the hep


Why does $ \sum_{n\equal{}0}^{\infty} \frac{z^n}{n!} \equal{} \sum_{n\equal{}0}^{\infty} \frac{nz^{n\minus{}1}}{n!}$?

Because you can differentiate inside the sum, but that requires justification.

Besides $ \sum\limits_{n\equal{}0}^{\infty }{\frac{n{{z}^{n\minus{}1}}}{n!}}\equal{}\sum\limits_{n\equal{}1}^{\infty }{\frac{{{z}^{n\minus{}1}}}{(n\minus{}1)!}}\equal{}\sum\limits_{n\equal{}0}^{\infty }{\frac{{{z}^{n}}}{n!}}.$


[quote="quykhtn-qa1"]Let a,b,c be none-gative numbers such that:ab+bc+ca=1.Prove that:
2a+3b+4c ≥ √23[/quote]

$ a\equal{}\frac{1\minus{}bc}{b\plus{}c}$----> $ 3b^{2}\plus{}b(5c\minus{}\sqrt{23})\plus{}2\minus{}\sqrt{23}c\plus{}4c^{2}\geq 0$

$ D\equal{}(5c\minus{}\sqrt{23})^{2}\minus{}4*3*(2\minus{}\sqrt{23}c\plus{}4c^{2})\equal{}\minus{}(\sqrt{23}c\minus{}1)^{2}\leq 0$

[quote="zaya_yc"][quote="quykhtn-qa1"]Let a,b,c be none-gative numbers such that:ab+bc+ca=1.Prove that:
2a+3b+4c ≥ √23[/quote]

$ a \equal{} \frac {1 \minus{} bc}{b \plus{} c}$----> $ 3b^{2} \plus{} b(5c \minus{} \sqrt {23}) \plus{} 2 \minus{} \sqrt {23}c \plus{} 4c^{2}\geq 0$

$ D \equal{} (5c \minus{} \sqrt {23})^{2} \minus{} 4*3*(2 \minus{} \sqrt {23}c \plus{} 4c^{2}) \equal{} \minus{} (\sqrt {23}c \minus{} 1)^{2}\leq 0$[/quote]
VERY NICE SOLUTION :)


calculus
function
inequalities unsolved
inequalities

Using no calculus find maximum value of $ f(x)= \frac{1}{4+x}+\frac{1}{4+\frac{1}{x}}$ for $ x\in R^{+}$

Since this function is always equal to 1...

Sorry! I made a mistake. I corrected it.

The Maximum Value is: 0.4  and occur for x=1,
 Try: (X+1/X )=t  Then use  for All positive values of x , t>=2  .

Not sure whether this is "no calculus" or not, but for any $ x>0$ it is true that $ f(x) = \frac{1}{4+x}+\frac{1}{4+\frac{1}{x}}= \frac{2}{5}-\frac{3(x-1)^{2}}{5(4+x)(4x+1)}$.  This identity proves that for any $ x>0$ it is true that $ f(x) \leq \frac{2}{5}$, and $ f(x) = \frac{2}{5}$ if and only if $ x=1$.


Find all function $ f : \mathbb{R} \ \to \  \mathbb{R}$ such that :
\[ f(x^2 \plus{} f(y) \minus{} y) \ \equal{} \ (f(x))^2 \minus{} f(y), \quad \forall  x; y  \in \mathbb{R}.\]

[quote="nguyenvuthanhha"][i]Find all function $ f : \mathbb{R} \ \to \ \mathbb{R}$ such that :

   $ f(x^2 \plus{} f(y) \minus{} y) \ \equal{} \ (f(x))^2 \minus{} f(y) \ \ \forall \ \ x; y \in \mathbb{R}$[/i][/quote]

Let $ P(x,y)$ be the assertion $ f(x^2\plus{}f(y)\minus{}y)\equal{}f(x)^2\minus{}f(y)$

Comparing $ P(x,y)$ and $ P(\minus{}x,y)$ implies $ f(x)^2\equal{}f(\minus{}x)^2$
Suppose now $ \exists u$ such that $ f(\minus{}u)\equal{}\minus{}f(u)$
$ P(0,u)$ $ \implies$ $ f(f(u)\minus{}u)\equal{}f(0)^2\minus{}f(u)$
$ P(0,\minus{}u)$ $ \implies$ $ f(\minus{}f(u)\plus{}u)\equal{}f(0)^2\plus{}f(u)$
But, since either $ f(\minus{}f(u)\plus{}u)\equal{}f(f(u)\minus{}u)$, either $ f(\minus{}f(u)\plus{}u)\equal{}\minus{}f(f(u)\minus{}u)$, we get that either $ f(u)\equal{}0$, either $ f(0)\equal{}0$ and so two cases :

[u][b]Case 1[/b] : [/u]$ f(0)\neq 0$ and so $ f(\minus{}u)\equal{}\minus{}f(u)$ $ \implies$ $ f(u)\equal{}0$
So $ f(\minus{}x)\equal{}f(x)$ $ \forall x$
Suppose now that $ \exists u$ such that $ f(u)<0$
$ P(\sqrt{\minus{}f(u)},u)$ $ \implies$ $ f(\minus{}u)\equal{}f(\sqrt{\minus{}f(u)})^2\minus{}f(u)$ $ \implies$ $ 2f(u)\equal{}f(\sqrt{\minus{}f(u)})^2\geq 0$, hence contradiction.
So $ f(x)\geq 0$ $ \forall x$
So $ f(x)^2\geq f(y)$ $ \forall x,y$
So $ f(x)\in[0,M]$ is bounded.

Comparing $ P(x,y)$ and $ P(x,\minus{}y)$, we get that $ f(x^2\plus{}f(y)\minus{}y)\equal{}f(x^2\plus{}f(y)\plus{}y)$ $ \forall x,y$
So $ f(x)\equal{}f(x\plus{}2y)$ $ \forall x\geq M$, $ \forall y>0$ and so $ f(x)\equal{}c$ $ \forall x\geq M$
Now, for any $ y$, it is possible to choose $ x$ such that $ x^2\plus{}f(y)\minus{}y>M$ and $ x>M$.
Then $ P(x,y)$ $ \implies$ $ c\equal{}c^2\minus{}f(y)$ $ \forall y$

And so we get the solution $ f(x)\equal{}c$ $ \forall x$ with $ c\equal{}c^2\minus{}c$
And so $ \boxed {f(x)\equal{}2\forall x}$ (we dont consider the solution $ f(x)\equal{}0$ here since in the case 1 we have $ f(0)\neq 0$

[u][b]Case 2[/b] :[/u] $ f(0)\equal{}0$
Then $ P(x,0)$ $ \implies$ $ f(x^2)\equal{}f(x)^2$ and so $ f(x)\geq 0$ $ \forall x\geq 0$
Then, suppose $ f(u)>0$, then $ P(0,u)$ $ \implies$ $ f(f(u)\minus{}u)\equal{}\minus{}f(u)< 0$ and so $ f(u)\minus{}u<0$ and so $ u>f(u)>0$
So, $ f(\minus{}u)\equal{}\minus{}f(u)$ $ \forall u$. Else, we'd get a $ v<0$ such that $ f(v)>0$

Then $ P(x,x^2)$ $ \implies$ $ f(f(x^2))\equal{}f(x)^2\minus{}f(x^2)\equal{}0$ And so $ f(f(x))\equal{}0$ $ \forall x\geq 0$ and so (since odd) $ f(f(x))\equal{}0$ $ \forall x$
Let then $ y>0$, and so $ f(y)\geq 0$
Then $ P(\sqrt{\frac{f(y)}{2}},f(y))$ $ \implies$ $ f(\minus{}\frac{f(y)}{2})\equal{}f(\frac{f(y)}{2})^2\geq 0$ and so $ f(y)\leq 0$ and so $ f(y)\equal{}0$

And so the only solution in this case n^° 2 : $ \boxed{f(x)\equal{}0\forall x}$

[quote="Dumel"][quote="pco"]
[u][b]Case 1[/b] : [/u]$ f(0)\neq 0$ and so $ f( \minus{} u) \equal{} \minus{} f(u)$ $ \implies$ $ f(u) \equal{} 0$
So $ f( \minus{} x) \equal{} f(x)$ $ \forall x$[/quote]Why?  :huh:[/quote]

In the four first lines, we shown :

That $ f(x)^2\equal{}f(\minus{}x)^2$ and so that $ \forall x$ either $ f(\minus{}x)\equal{}f(x)$, either $ f(\minus{}x)\equal{}\minus{}f(x)$
Then, we said that if $ f(\minus{}u)\equal{}\minus{}f(u)$ for some $ u$, then either $ f(0)\equal{}0$, either $ f(u)\equal{}0$
Then, in case 1, we condidered $ f(0)\neq 0$. So, if $ f(\minus{}u)\equal{}\minus{}f(u)$, then $ f(u)\equal{}0$
So, $ \forall x$ either $ f(\minus{}x)\equal{}f(x)$, either $ f(\minus{}x)\equal{}\minus{}f(x)$ but then $ f(x)\equal{}0$ and so $ f(\minus{}x)\equal{}\minus{}f(x)\equal{}0\equal{}f(x)$

So, in case 1, we have $ f(\minus{}x)\equal{}f(x)$ $ \forall x$

[quote="EastyMoryan"]Nevermind this ><[/quote]

I dont understand what this mean, sorry

He probably posted something, then realized that it was a mistake or something, then edited the post rather than deleting it.


Let $ a,b,c,d,e$ be positive reals such that $ abcde \equal{} 1$.
Prove that
$ 81\prod (a^{3} \plus{} 2) \ge (7 \plus{} 2\sum ab)^{3}$

This is not my own problem...

But $ a\equal{}b\equal{}c\equal{}d\equal{}e\equal{}1$ then $ LHS<RHS$?

[quote="tdl"]But $ a \equal{} b \equal{} c \equal{} d \equal{} e \equal{} 1$ then $ LHS < RHS$?[/quote]
Sorry, It is my stupid mistake.... I edited.


I am preparing for a physics olympiad competition.Does anyone here know of any forums that might help me?

Thanks

Have you seen [url=http://www.physicsforums.com]http://www.physicsforums.com[/url]?

i suggest that we should have a physics forum here??
There are so many trig and calculus stuff in physics problems...

[quote="Psycho"][quote="shobber"][quote="kentliu"]haha.. man.. havent done physics in so long.. failed the olympiad test so bad.. if i took it may last  year tho.. would have aced it like nothing.. i have done all the problems before.. just dont remember how to do them..[/quote]

well, i am curious to know what you take in a physics olympiad?? There is going to be a contest about mechanics in Beijing, so will you introduce something about your contest so that we can share our expierences and knowledges?[/quote]
we have a physics olympiade in flanders as well... got through the first round, but i didn't feel like going to the second one  :blush:  i knew i was no match against the other competitors (i only have 1 hour of physics a week, and most others 3...). if you want some questions, i can translate some of the flemish problems for you. i only don't know where to post them  :?  or should i pm some?[/quote]

You can't be serious about that? The 2nd prize last year went to a someone in third class, they don't even have physics yet in school!!! Why do people always think you need school to learn things? Like if teachers have any use.. :D

well i didn't feel like it either  :?
besides, i already had promised i'ld see my girlfriend that day  :P

that has all the questions to previous tests..

http://www.compadre.org/psrc/evals/olympiad.cfm

and i'll tell u a few questions from this times test..
i remember one of the was about a spring that has been damped.. and u have to calculate the total distance it travels before it stops as well as amplitude at certain time.
i also remeber on Multiple choice there was questions about the thread, similar to these that i posted a while ago
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14942
there was another free response about a space ship traveling past a planet, and whether or not it hits, what speed must it be to hit it, etc.. its mainly about gravitational potential..

i used the text book "physics for scientists and engineers" by serway and this one other guy.. 4th edition i believe, and the cover was black with this circle rolling across making this wavy line... tats a pretty good book.. if u could do the red questions in the excercises, then you should be more than okay.

[quote="Psycho"][quote="Peter VDD"]didn't? Is it past yet? Last year it was only after easter holidays?[/quote]
the second round was last wednesday... 9 pupils from my school could go, but only 2 actually went  :D  Physics aren't as popular as maths here  ;)  (and oh yeah, it's probably the first time they work with two rounds and a final in the Flanders Physics Olympiade. last year, i guess it was only one round and then immediatly to Brussels or where-ever the final was  :? )[/quote]

Why physics are not popular??
In my school, most students take physics contest course rather than math, and my school have lots of students win in the physics contest each year while only few win in the math...

i dunno  :?  lack of interest probably...
besides, maths contest are 'easier' in that way that you just can go to them, and see how good you are. of course you have to know some formulas and such, but since we have 8 hours of maths a week, you just KNOW all those formulas. unlike physics... two years ago, we saw gases and such, last year electricity, and this year speed, etc. All three years have only few in common. So when you go to a physics contest, you have to refresh all those formulas from past years if you want to acheive something...

yeah, but still... if you don't know how to use them properly  :?  what's the point then?
i didn't study anything in advance and i got 80/100 in the first round... (and the limit was 80%  :P ) i got most of my points because a lot of things were almost pure maths. the second round is WAY more difficult...

for the vietnamese comunity, a friend of mine is administrator of a good physics forum:
http://www.vatly.org/

[quote="ln(dx/dy)"]I think there should be a physics forum on this site  ;)[/quote]We are planning to realease several non-math, but math-related forums, such as computer science forum and physics forums on the site. However we are going to hold them off a little bit :)


Hej! Kakšno se vam je zdelo izbirno za na državno?  :read:

Mislim da bo ta tema bolj zanimiva ko bodo znani rezultati  :lol:

Zdaj je postala tema zanima  :lol:  ...  men so se zdele tok težke da lahko o državnem samo sanjam  :(

Sej če so bile naloge težke zate, so bile verjetno tudi za druge... :showoff:
Tako da lahko tudi skromen izkupiček točk pelje na državno tekmovanje... :pilot:

bi morala biti pa omejitev res zelo skromna  :?

aja nino, če ti je uspelo poiskat rešitve za 3m - 9 + 1*2*...*m = n^3, bi bil vesel razlage, ker uradna rešitev vsebuje po mojem mnenju malo "sumljiv" sklep  :?:


graph theory
combinatorics
Extremal combinatorics
IMO Shortlist

Once again, it looks like it's time to count the same thing twice.  This time we will count ordered triples of distinct people $ (A,B,C)$ such that $ B$ greeted both $ A$ and $ C$.  Let $ m$ be the number of people who exchanged greetings with any two given people.

[b]Method 1:[/b] There are $ 12k$ choices for $ A$.  Once $ A$ is chosen there are $ 3k\plus{}6$ choices for $ B$, and once $ B$ is chosen there are $ 3k\plus{}5$ choices for $ C$.  This gives $ 12k(3k\plus{}5)(3k\plus{}6)$.

[b]Method 2:[/b] There are $ 12k$ choices for $ A$.  There are then $ 12k\minus{}1$ choices for $ C$.  Once $ A$ and $ C$ are chosen, there are $ m$ choices for $ B$, for a total of $ 12k(12k\minus{}1)m$.

Solving, $ m\equal{}\frac{(3k\plus{}5)(3k\plus{}6)}{12k\minus{}1}\equal{}\frac{1}{4}(35\plus{}11\plus{}\frac{3k\plus{}131}{12k\minus{}1})\equal{}\frac{1}{4}(3k\plus{}11\plus{}\frac{1}{4}(1\plus{}\frac{525}{12k\minus{}1}))$.

Now $ 525\equal{}3*5^2*7$, and it can easily be checked that $ 35$ is the only factor of $ 525$ that is one less than a multiple of $ 12$.  Therefore $ k\equal{}3$, and $ 36$ people are at the meeting.

To show that this is actually possible, consider each person as being labeled by an ordered pair $ (x,y)$, where $ 1 \leq x,y \leq 6$.  Person $ (x,y)$ shakes hands with person $ (w,z)$ if either $ x\equal{}w$, $ y\equal{}z$, or $ x\plus{}y$ is congruent to $ w\plus{}z$ modulo $ 6$.

I will repeat my post on a topic with same statement [url]http://www.mathlinks.ro/viewtopic.php?t=301854[/url]

The problem describes a graph $ G \equal{} (V,E)$ with $ |V| \equal{} n \equal{} 12k$ vertices and $ d \equal{} 3k \plus{} 6$, the degree of each vertex (i.e. the number of edges incident in that vertex). Such a graph is called $ d$-regular. Moreover, it is given that for any two vertices, the number of other vertices joined to both those two is the same, be that constant $ \lambda$. Usually we denote the degree of a vertex $ x$ by $ \deg x$, and the set of its neighbors (joined to him by edges) by $ N(x)$, so we are given that $ \deg x \equal{} d$ for all $ x \in V$ (i.e. $ d(G) \equal{} d$), and $ | N(x) \cap N(y) | \equal{} \lambda$ for all distinct $ x,y \in V$.

In fact what it is described here is a symmetric balanced incomplete blocks design (BIBD), with parameters $ (n, d, \lambda)$. But it is immediate, by a little double-counting, that the relation $ \lambda(n \minus{} 1) \equal{} d(d \minus{} 1)$ is necessary; in our case $ \lambda(12k \minus{} 1) \equal{} (3k \plus{} 6)(3k \plus{} 5)$. It is relatively easy to prove that this may happen only for $ k \equal{} 3$ (therefore $ \lambda \equal{} 6$).
The question now is if a $ (36, 15,6)$ symmetric BIB actually exists (the fact that the numbers check the relation above is not always sufficient). The good news is that these numbers verify the condition of the Bruck-Ryser-Chowla theorem ($ 36$ is even, and $ 15 \minus{} 6 \equal{} 9$ is a perfect square), but that is also just necessary, not sufficient.

I cite from the link is [url]http://crosbi.znanstvenici.hr/prikazi-rad?chset=ASCII&lang=EN&rad=56134[/url]
Up to isomorphism there are $ 4$ symmetric $ (36, 15, 6)$ designs with automorphisms of order $ 7$, and $ 38$ symmetric $ (36, 15, 6)$ designs with automorphisms of order $ 5$. For those designs full automorphism groups are determined. Also, all symmetric $ (36, 15, 6)$ designs having automorphisms of order $ 3$ acting with $ 9$ and $ 6$ fixed points, or cyclic automorphism groups of order $ 4$ acting standardly are constructed and orders of their full automorphism groups are determined.

See also, best, [url]http://www.uow.edu.au/~jennie/WEB/WEB69-93/max/005_1969.pdf[/url]

I cite from IMO-Compendium (SL-95-19, p. 596):

For each two people let $ n$ be the number of people exchanging greetings
with both of them. To determine $ n$ in terms of $ k$, we shall count in two
ways the number of triples $ (A,B,C)$ of people such that $ A$ exchanged
greetings with both $ B$ and $ C$, but $ B$ and $ C$ mutually did not.
There are $ 12k$ possibilities for $ A$, and for each $ A$ there are $ (3k \plus{} 6)$ possibilities
for $ B$. Since there are $ n$ people who exchanged greetings with both
$ A$ and $ B$, there are $ 3k \plus{} 5 \minus{} n$ who did so with $ A$ but not with $ B$. Thus
the number of triples $ (A,B,C)$ is $ 12k(3k \plus{} 6)(3k \plus{} 5 \minus{} n)$. On the other
hand, there are $ 12k$ possible choices of $ B$, and $ 12k\minus{}1\minus{}(3k\plus{}6) \equal{} 9k\minus{}7$
possible choices of $ C$; for every $ B$, $ C$, $ A$ can be chosen in $ n$ ways, so the
number of considered triples equals $ 12kn(9k \minus{} 7)$.
Hence $ (3k \plus{} 6)(3k \plus{}5 \minus{} n) \equal{} n(9k \minus{} 7)$, i.e., $ n \equal{}\frac{3(k\plus{}2)(3k\plus{}5)} {12k\minus{}1}$.
This gives us that $ \frac{4n} {3} \equal{} \frac{12k^2\plus{}44k\plus{}40} {12k\minus{}1} \equal{} k \plus{} 4 \minus{} \frac{3k\minus{}44} {12k\minus{}1}$ is an integer too. It is directly
verified that only $ k \equal{} 3$ gives an integer value for $ n$, namely $ n \equal{} 6$.

Man, I really liked this problem... such a simple idea, but still took me quite a while to find it- anyways here's a sketch of my proof.
First let's construct a graph with vertices $a_1,a_2,a_3,...,a_{12k}$ such that there exists an edge between $2$ vertices iff they greet each other- hence there are $3k+6$ edges emanating from each vertex.

The key observation here comes really early in this problem, which is defining a sort of "point tally" which represents the sum of the cardinalities of all intersections that we get for each 2-element subset of the vertices of our graph- think of it like the number of vertices belonging to the neighbourhood of any 2-element set such that both elements of the set map to it, and sum this over all such 2-element sets. The key insight here is noticing that if there is an edge connecting $a_1$ to some $n$ elements, then $\binom{n}{2}$ points must be added to our points tally, as we will include $a_1$ as one of the vertices in the "intersection" of $\binom{n}{2}$ 2-element subsets ([hide=Note]I know it seems trivial, but it was really cool when I figured this out [/hide]). But, we already know the graph is $(3k+6)$-regular, and as there are $12k$ vertices, there must be $(12k) \cdot \binom{3k+6}{2} =\frac{12k(3k+6)(3k-5)}2$ points in total!

This almost solves the problem for us, all we have left is figuring out how to exploit this information about the point tally and perhaps a little bashing at the end...
To overcome the first dilemna, first let's define the magnitude of the intersection of a two-element set $=t$ (using the fact that the intersection of a $2$ element of a set always has the same magnitude).
Thus, the total number of points must just simply be $t \cdot (\text{number of 2-element subsets}) =t \cdot \binom{12k}{2}$ (looking back, this was the motivation for defining our point tally the way we did) $=> \frac{12k(12k-1)}{2} \cdot t = \frac{12k(3k+6)(3k-5)}2 => t \cdot (12k-1) = 3(k+2)(3k+5)$. Time to gcd bash a little, perhaps?
As $12k-1 \mid (3k+6)(3k+5) => 12k-1 \mid (k+2)(3k+5) =>12k-1 \mid 4(k+2)(3k+5) =>12k-1 \mid 12k^2+44k+40 => 12k-1 \mid 45k+40$.
Thus $45k+40 =m(12k-1) => k(12m-45)=m+40$- obviously, for $m \geq 9$, $k$ can't be an integer, thus we only need to check for $m$ taking values from $1$ to $8$, and we get our integral solution when $m=5$ $=>k=3$.

Hence there are $12 \cdot 3 = \fbox{36}$ people at the meeting.

[hide=Solution][quote=ISL 1995 C5]At a meeting of $ 12k$ people, each person exchanges greetings with exactly $ 3k\plus{}6$ others. For any two people, the number who exchange greetings with both is the same. How many people are at the meeting?[/quote]
[b][color=#000]Solution:[/color][/b] In graph-theoretic terminologies, For a graph $G$, $\mid V(G) \mid$ $=$ $12k$, $d(v_i)$ $=$ $3k+6$ and $\mid N(v_i)$ $\cap$ $N(v_j)$ $\mid$ $=$ $\ell$ is constant. (Where, $\mid N(x) \mid$ denotes the neighbourhood of vertex $x$). From the given condition, we have: \begin{align*} \ell \binom{12k}{2} = 12k \binom{3k+6}{2} \Longleftarrow 12k-1 \mid 9k^2+33k+30 \implies 12k-1 \mid 3k+131\end{align*}
Quick check reveals that, $k=3$ works! Hence, the total no. of people are $36$ $\qquad \blacksquare$[/hide]

Such a good exercise for double counting :-D
[b]Solution:[/b] We will count the quantity $S=$number of triplets $(A,B,C)$ of peoples in the meeting. such that $C$ exchanges greetings with both $A$ and $B$
Let: For each pair $A,B$ in the meeting, There are $m$ peoples greeting with both of them.
[b]First,[/b] Select pairs $A,B$ first then, choose $C$. There are$S=\binom{12k}{2}m$ ways
[b]Second,[/b] Select $C$ first, Then Choose pairs $A,B$ with both of them greeting with $C$. There are $S=12k\binom{3k+6}{2}$ ways
Therefore, $$\binom{12k}{2}m=12k\binom{3k+6}{2}$$
After some calculation and divisible theory, We get $12k-1|175$
Therefore, $k=3$.
[b]The Construction for $k=3$[/b](to see that this actually work)
Consider the following $6\times 6$ matrix: $$
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 3 & 4 & 5 & 6 & 1 \\
3 & 4 & 5 & 6 & 1 & 2 \\
4 & 5 & 6 & 1 & 2 & 3 \\
5 & 6 & 1 & 2 & 3 & 4 \\
6 & 1 & 2 & 3 & 4 & 5
\end{pmatrix}  $$
Each indice represents peoples, For the people labeled $i$ greeting with the peoples in the same row/column and the peoples with the same labeled
So, there are $12\times3=36$ peoples in this meeting.

Quite simple and cute.
Let the set $F=(1,2,... 12k)$,which denotes the people in the meeting.
Let set $A_{i}$ denote the people with whom person $i$ shared greetings with.
It is known $|A_{i}|=3k+6$ and $|A_{i} \cap A_{j}|=s$,for every $(i,j)$
Now counting the number of triples $(A_{i},A_{j},l)$ where the first two sets share an element $l$.
We get that $ s \binom{12k}{2} = 12k \binom{3k+6}{2}$,which ultimately results in finding $k$ such that
$12k-1|3k^2+11k+10$ which after some division algorithm,results in finding $k$ such that
$12k-1|9k+43$ which can be computed manually to give $k=3$ and hence the answer is $36$.


Let a and b be integers such that

$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{580}+\frac{1}{581}= \frac{a}{b}$

If a and b have no common prime factors, what is the largest power of 3 dividing b?

A) $3^{0}$
B) $3^{1}$
C) $3^{3}$
D) $3^{4}$
E) $3^{5}$

let's see...

[hide]
$1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{581}=\frac{\frac{581!}{1}+\frac{581!}{2}+\frac{581!}{3}+\dots+\frac{581!}{581}}{581!}$

We need to find the term in the numerator with the greatest $x$ such that $3^{x}$ is a factor of the term. There is a two-way tie between $\frac{581!}{3^{5}}$ and $\frac{581!}{(2)3^{5}}$.

This means we can divide each term in the numerator by $3^{5}$ and still have all integers in the numerator.

Thus when the numerator and denominator of the big fraction are simplified into relative primes, the denominator will have a factor of $3^{5}$. Thus the answer is $3^{5}$.

[/hide]

That answer is incorrect, but I don't know why.

Anyone do it a different way?

Hmm, let's try this:

[hide]
I think I found the error in the above reasoning. Here it goes:

$1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{581}=\frac{\frac{581!}{1}+\frac{581!}{2}+\frac{581!}{3}+\dots+\frac{581!}{581}}{581!}$

We need to find the term in the numerator with the greatest $x$ such that $3^{x}$ is a factor of the term. There is a two-way tie between $\frac{581!}{3^{5}}$ and $\frac{581!}{(2)3^{5}}$.

This means we can divide each term in the numerator by however many 3s there are in 581! other than 5 and still have all integers.

The question asks for the largest POWER OF 3, and in that case $3^{k}$ is less than or equal to 5. Thus $3^{1}$ is correct.

I think.

[/hide]

[hide]
We see that the largest power of 3 less than or equal to 581 is 243. But $\frac{1}{243}+\frac{1}{486}=\frac{1}{162}$ with denominator only divisible by 81. So now we consider all terms with denominator divisible by 81, which is all up to 81*7=567. This gives us
\[\frac{1}{81}\left( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right) \]
Now we use 1/2+1/3+1/6=1, and get
\[\frac{1}{81}\left( 2+\frac{1}{4}+\frac{1}{5}+\frac{1}{7}\right) \]
and this turns out to be equal
\[\frac{1}{81}\cdot \frac{363}{140}=\frac{121}{140\cdot 27}\]
Okay, this is getting annoying. The power of 3 is at most 27 now! This time, direct computation would be painful in figuring out whether 3 divides the numerator of
\[\sum_{n=1}^{21}\frac{1}{n}\]
But we can apply a trick here. The terms with multiples of 3 add up to 121/140, as shown above, and can pair the other terms: 1 with 1/20, 1/2 with 1/19, 1/4 with 1/17, and so on up to 1/10 with 1/11. We get a fraction with numerator divisible by 3 added to 121/140, so this fraction does [b]not[/b] have numerator divisible by 3. When we add in the other terms and put them under a common denominator, all their numerators will be divisible by 27, so the huge sum has no factor of 3 in the numerator with a denominator containing a factor of 27, and we're done! This makes the answer $\boxed{C}$.
[/hide]


geometry
circumcircle
ratio
geometry unsolved

Let ABC be a traingle and P a point inside it such that  angle  PBC=PCA<PAB

the line PB cuts the circumcircle of ABC at B and E ,and the line CE cuts the circumcircle

of APE at E and F . Show that [APEF]/[ABP] does not depend on the choice of P

Triangle ABP is similar to ACF
and [ACF]=[APEF] due to AE//CQ
thus the ratio is AC^2/AB^2


I am seeking problems which are solved with Bernoulli inequality! Can you help me to find them? And I want to find solution on that inequality. Proof that:
\[(1+x)^n< 1+nx,\quad x> -1,\quad 0<n<1\].
I want that inequality to be solved by the Bernoulli's inequality!

[color=blue]Problem:[/color]

$\forall n \in \mathbb{N},n \ge 1 \Rightarrow \sum\limits_{k = 1}^n {\left( {1 + k} \right)^{k - 1} }  \ge \frac{{n\left( {n^2  + 2} \right)}}{3}$

$\forall n \in \mathbb{N},n \ge 1 \Rightarrow \prod\limits_{k = 0}^n {\left( {1 + \left( \begin{array}{l}
 n \\
 k \\
 \end{array} \right)} \right) \ge 1 + 2^n } $

If you want some problem:
Prove that $a^b+b^a >1$ for all $a,b \in \mathbb{R}^+$

ZetaX, can you give me the solution for youe inequality

I need a solution for the inequality
\[n^{n+1}>(n+1)^n,\quad n>2\]
Can you help me?

Here's a proof using Bernoulli's inequality:
\begin{eqnarray*}
\left( \frac{n}{n+1} \right)^{n+1}
  & = & \left( 1 - \frac{1}{n+1} \right)^{n+1} \\
  & = & \left[
              \left( 1 - \frac{1}{n+1} \right)^{(n+1)/2} \right]^2 \\
  & > & \left( 1 - \frac{1}{n+1} \cdot \frac{n+1}{2} \right)^2 \\
  & = & \left( \frac{1}{2} \right)^2 \\
  & = & \frac{1}{4} \\
  & \ge & \frac{1}{n+1} \, .
\end{eqnarray*}

proof of $a^b+b^a>1$ when $a,b \in \mathbb{R}^+$:
You can assume that $a,b<1$ because otherwise its trivial.
Then you get:

${{{{{{a^b+b^a=\frac{a}{a^{1-b}}+\frac{b}{b^{1-a}}=\frac{a}{(1+(a-1))^{1-b}}+\frac{b}{(1+(b-1))^{1-a}}>\frac{a}{1+(a-1)(1-b)}}+\frac{b}{1+(b-1)(1-a)}}=\frac{a}{a+b-ab}}+\frac{b}{a+b-ab}}=\frac{a+b}{a+b-ab}}=1+\frac{ab}{a+b-ab}}$.

Now $1>b \implies a>ab \implies a+b-ab>0 \implies a^b+b^a > 1$.

(Here I used the 'second' bernoulli inequality $(1+x)^n< 1+nx,x> -1,0<n<1$ which you stated in your post:
When $x>-1$ then also $nx>-1$ and $\frac{1}{n}>1$, so you get:
$(1+nx)^\frac{1}{n} > 1+x \implies  1+nx > (1+x)^n$ )

[quote="Ravi B"]I think billzhao may have given a proof on this site.  Induction on n works fine.[/quote]

Yes, I misinterpreted the condition at the end, it's a bit misleading.


Alright, so, we see that 24=2^3*3. This implies that, for a positive integer, n, to have its square be a multiple of 24, n must be greater than or equal to 2^2*3=12. Thus, we find the number of multiples of 12 less than 10^3, and we get 12.

you have everything right except the end. there are 83 multiples of 12 less than 10^3


$x+0=0$ and $x_{1}=1$ and $x_{n+1}=\frac{x_{n}+nx_{n-1}}{n+1}$ for $n>1$. Find $\lim_{n\to\infty}x_{n}$

[hide="so far i have"]
$\frac12<x_{n}<1$ for all $n>2$ and then i showed that $\lim_{n\to\infty}|x_{n+1}-x_{n}|=0$
so the limit does exist. What I am having trouble with is actually evaluating it.
[/hide]

[hide="Hint"]Consider the difference $x_{n+1}-x_{n}$.[/hide]

[quote="tjhance"][quote="The Zuton Force"][hide="Solution"]
So, your  sequence limit is $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}= \text{ln}2$
[/hide][/quote]

How do you know that the limit of that is $\text{ln}2$?[/quote]

Calculus, probably.


topology
real analysis
real analysis solved

Let  $A_n$ a closed set in [0,1] for every n  in $N$ .Show that is not posible to have   $U$($A_n$)=[0.1].

What's $U$? At first I thought it was supposed to mean reunion, but if, for example,  $A_n=[0,1],\ \forall n$, then it doesn't hold :?.

I think Adyo is referring to the following problem:
let An be a COUNTABLE family of DISTINCT closed sets of [0;1].
Prove that [0;1] is not equal to the union of the An.
(i think there is one solution with Baire theorem but i don't remember ..)

Distinct? Maybe you mean disjoint? I have no idea whether it's true or not, but the problem as you have stated it is, again, wrong:

[$0,1$]$=\{1\}\cup [0,\frac 12]\cup[\frac 12,\frac 12+\frac 14]\cup[\frac 12+\frac 14,\frac 12+\frac 14+\frac 18]\cup\ldots$.

The correct statement is: [0,1] cannot be partitioned into a numerable family of non-vide closed sets. As far as I remember, long time ago I asked about such a decomposition into a non-numerable family of compact sets and grobber showed that the answer is positive. I am also interested about this nice problem, since I really cannot find any solution using Baire theorem.

Oral examination Ecole Normale Suprieure MP* (45 rue d'Ulm 75005 Paris, 5 minutes walking near Louis Le Grand high school),
also Ecole Polytechnique MP* (at Palaiseau south of Paris, take RER B from Luxembourg station stop at Lozre, and climb 300 steps to get at l'X)

Suppose $[0,1]=\cup_{n\geq 0}C_n$ where $C_n$ are non empty closed sets.

The idea is to construct a family of intervals $I_n$ and a sequence of integer
$0=k_0<k_1<...<k_n<...$
such that
i) $I_n \subset \overline{I_n} \subset I_{n-1}$
ii) $I_n$ does not meet $C_0,C_1,...,C_{k_n}$
iii) $\overline{I_n}$ meet $C_{k_n}$

And consider the intersection $\overline {I_n}$ to get a contradiction.

---------------------------------------------------------------------------------
Exercices de mathmatiques, oraux X-ENS, analyse 1
by
Francinou (math teacher MPSI at Henri IV high school)
Gianella (math teacher MPSI at Louis Le Grand high school)
--------------------------------------------------------------------------------

Generalisation:

$E$ a metric and complete space connex and locally connex, then $E$ cannot
partition into a reunion of numerable non empty closed sets.

Suppose contrary: $[0,1]\cup\limits_{n}C_n$.
Let $D_n$ be the set of inner points of $C_n$, $E_n=C_n\setminus D_n$.
We have $E=\cup\limits_{n}E_n=[0,1]\setminus \cup\limits_{n}D_n$ is nonempty closed set, hence $(E,|x|)$ is a complete space.
We will show that $E_n$ is non-dense in $E$ (then we apply Baire theorem to obtain contradiction). Suppose $x\in E_n$, $\epsilon>0$. Consider ball in $E$ with radius $\epsilon$: $B=(x-\epsilon,x+\epsilon)\cap ((0,1)\setminus C_n)$. Since $x$ is a frontier point of $C_n$ we conclude $B\neq\varnothing$ and $B$ is an open set. Suppose $y\in B$. There are two cases
Case 1. $y\in (E\setminus E_n)$ then, obviously, we are done, i.e. we can find ball inside $B$ with center $y$ s.t. it doesn't intersect with $E_n$ (we are using $B$ is an open set);
Case 2. $y\not\in (E\setminus E_n)$, then $y$ is an inner point of $C_m$ $\Rightarrow$ $C_m$ has frontier point $z$ between $y$ and $x$ (exercise!) $\Rightarrow$ $z\in E_m$ and $z\in B$, so we are done (see Case 1).

Sorry to bother you like this, but could you please explain it a bit more?

I began in exactly the same way, but this is what I wanted to do (how I wanted to obtain contradiction): we have $E=[0,1]\setminus \bigcup D_n=\bigcup E_n$. If we show that the set $E$ has non-void interior, then we can apply Baire's theorem to get a contradiction because all $E_n$ have void interior. In order to show that $S$ has non-void interior I had to show that $\bigcup D_n$ is non-dense in [$0,1$]. Is this even true?

Maybe you apply some Baire theorem I don't know about? I'm talking about the one stating that the union of countably many closed sets with void interiors has void interior.

Baire Category Theorem.
Complete space is a second category set, i.e. it cannot be obtained as union of countable many non-dense subsets.

I proved that $E$ is complete space and $E_n$ is [b]non-dense in[b] $E.


This is the play my friends and I wrote for our Language Arts assignment. Read it over and tell me what you think. I have to warn you, it'll probably take a while to read.

Edit: some of the dialogue was clumsy and out of charector, so I changed it. I also added a prologue.

It's pretty good. How long did this take you?

It took be the better part of this month(June). We had three people working on it, so yeah. This actually wasn't the version we turned in. After it was due, I made quite a few changes since I got sorta obsessed. Any suggestions?


In a circle, chords AB and CD intersect at R. If AR:RB=1:4 and CR:RD=4:9, compute the ratio AB:CD.

[hide="Umm"]
By power of a point, $(AR)(RB)=(CR)(RD)$.  Letting $CD=13$, $CR=4$ and $RD=9$.  So $(AR)(RB)=36$.  Since $AR: RB=1: 4$, $AR=1y$ and $RB=4y$.  Substituting, $4y^2=36$, so $y=3$.  Thus, $AB=3(1+4)=15$.

So $AB: CD=15: 13$?
[/hide]

[hide]I used the property that $AR \cdot RB=CR \cdot RD$.  So $4x^2$ something is 36 where x is just a dummy variable. $x=3$, so AR=3 and RB=12.  AB=15 and CD=13, so the ratio is $\frac{15}{13}$?[/hide]

I know, bad explanation.  :(


For the first inequality

$ \frac {1}{2} \plus{} \frac {\sum a^2}{\sum ab} \geq \sum \frac {a}{b \plus{} c}$

it can be rewritten as

$ \sum ab \plus{} 2\sum a^2 \geq 2\sum \frac {a}{b \plus{} c}\sum ab$

or

$ \sum ab \geq 2abc\sum \frac {1}{a \plus{} b}$

or

$ \sum \frac {1}{a} \geq 2\sum \frac {1}{a \plus{} b}$

or

$ \sum(\frac {1}{a} \plus{} \frac {1}{b}) \geq 4\sum\frac {1}{a \plus{} b}$

which is true as

$ (a \plus{} b)^2 \geq 4ab$

For the second inequality

$ \sum \frac{a}{b\plus{}c} \plus{} \frac{1}{2}\frac{\sum ab}{\sum a^2} \geq 2$

By Cauchy

$ \sum \frac{a}{b\plus{}c} \geq \frac {(\sum a)^2}{2\sum ab}$

so it suffices to prove

$ \frac{\sum a^2}{\sum ab} \plus{} \frac{\sum ab}{\sum a^2} \geq 2$

which is true by AM-GM

$ \frac{1}{2}\plus{}\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\geq \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{a}{a\plus{}b}$

$ \Longleftrightarrow \sum{bc(c\plus{}b)(a\minus{}b)(a\minus{}c)}\plus{}2abc\sum{(a\minus{}b)(a\minus{}c)}\geq 0$

obvious true .

$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{a}{a\plus{}b}\geq 2\minus{}\frac{1}{2}\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}$

$ \Longleftrightarrow 2\sum{a^3(a\minus{}b)(a\minus{}c)}\plus{}\sum{bc(c\plus{}b)(a\minus{}b)(a\minus{}c)}\geq 0$

obvious true .


I'm trying to convert PNG to EPS by safe, free program (those are image files, BTW).

I know the topic of KSEG is described before and recently Rushil posted a thread about documentation of KSEG drawing to EPS but the program in the site only has "Free Trial."

I know many people use KSEG in this site and I wonder how you use it to put on LaTeX. Thanks for your help.

You can include .png files directly in your .pdf output (using pdflatex). This is the simplest way to go around. Or start learning metapost :)

What's PDFLaTeX?

Is it regular LaTeX you install from AoPS or something different? I don't know much of these stuffs..  :oops:

[quote="Valentin Vornicu"]You don't install $\LaTeX$ from AoPS! :) You install a bunch of executables and packages (macros) at one time called Miktex. After that you usually also install a nice editor (Texnic center for example) so that you can nicely edit files.

But the compilator can be either latex.exe (installed by Miktex) or pdftex.exe or pdflatex.exe. The first one produces .dvi files (which can be afterthat transformed into .ps, .pdf or whatever by means of other programs) while the last 2 produce directly .pdf files.[/quote]

Haha.. I know you don't install from AoPS but links are basically there.

I think I have Miketex because when I use TeXnicCenter, I can convert files from LaTeX to PDF,PS, and DVI. So does that mean I can simply put PNG drwaing into my LaTeX?

Nver mind. I got it.

The easiest way is simply print the seg. drawing from KSEG into PDF using virtual printer and do the same thing for general PDF.

I guess you can try to do something after converting to PNG but that's too complicated for me.  :D

Yes. Have the .png file in the same directory as the .tex file. Use pdflatex or pdftex to compile (so create directly a .pdf)

Use \usepackage{graphicx} in the preamble and where you want the image put

\includegraphics[width=3in]{image.png}

And that's it. You can play around with the width and height arguments ;)

If you have Adobe Photoshop, things are mcuh easier! Open the .png file and save it as a .eps file  - this option is there in all versions!! If you dont have Photoshop, you can try similar picture editing software. They are bound to have this conversion option. You can also try the Advanced Batch converter with great conversion capabilities! It's a atrial version - get it at http://www.batchconverter.com  -- but its definitely cheaper than Photoshop!  :lol:


This is similar to that game Mastermind.

[quote="1=2"]That was the game I was playing with my little brother when I found out that he could have ended the game there and then and didn't.[/quote]

Isn't your little brother like 5???

@ernie: yes... what's wrong with that???

lol...

i bet i couldnt have cracked the code

lol ur bro's smarter than me :roll:

that doesnt fit the third clue-- i agree with lotsofmath

[quote="1=2"]Crack the color code given the following information:

[hide="black blue yellow blue"]Two of the colors are correct and they are in the wrong position, the other two pegs are wrong.[/hide]
[hide="black white red yellow"]All of them are the right color, and exactly one of them is in the right position.[/hide]
[hide="green yellow yellow yellow"]One of the pegs is the right color and in the wrong position, the rest are wrong.[/hide]
[hide="black blue red green"]One of the colors is in the right position, one of the colors is in the wrong position, the rest are wrong.[/hide][/quote]

Everyone's just givin answers....heres a SOLUTION

Second clue tells us which COLORS are right.  so we know that black and yellow arent in 1st and 3rd spots respectively becuz of first clue.  there is no green from second clue, so yellow is in the second or 4th spot.  black cant be first, there is no blue, so red must be 3rd spot.  easily solved from there.

lotsofmath is rite.

No, it doesn't--it breaks the third clue because the yellow is in the right spot.

Yea it has to be yellow black red white.

Yellow has to be first because of the third clue. Then go from there...


$ \frac{1}{z}-\frac{4z}{2x-1}=\frac{1}{x}-\frac{4x}{2y-1}=\frac{1}{y}-\frac{4y}{2z-1}$
$ x+y+z=0\Longrightarrow$ $ x+y=-z$,$ y+z=-x$,$ x+z=-y$.
Solve for $ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.
(This was my post from the Intermediate Topics forum but I just posted it here because I was afraid maybe the mods would think I was being snobby to say "Maybe you should move this to Pre-Olympiad")


geometry
rectangle
geometric transformation
rotation
trigonometry

[quote="MSkumar"]I have to rotate a rectangle in some angle and after rotation I have to calculate howmuch width and height that recangle going to occupy. How to calculate height and width after rotation?[/quote]

After a rotation, I don't think the length and width change.

Isn't distance preserved in rotations?

[quote="vishalarul"]Isn't distance preserved in rotations?[/quote]

Yes, distance is preserved. The only thing that changes is the position.

I don't think he means the height and width of the rectangle being rotated (which obviously remains the same), but rather the height of the space it occupied if represented by an upright rectangle.

[whoa my wording is weird]

For example: If you rotated a square with height (and length) $r$ such that its diagonal is directly aligned vertically (and its other diagonal directly aligned horizontally), rotating it 45 degrees, than it would occupy a space with height (and length) of $r\sqrt{2}$. The square still retains its length, but it occupies a different rectangular area...

Draw it out - it will look like this

[img]http://motivate.maths.org/conferences/conf66/Images/non-symmetry.gif[/img]

[quote="Anaxerzia"]I don't think he means the height and width of the rectangle being rotated (which obviously remains the same), but rather the height of the space it occupied if represented by an upright rectangle.

[whoa my wording is weird]

For example: [b]If you rotated a square with height (and length) $r$ such that its diagonal is directly aligned vertically (and its other diagonal directly aligned horizontally), rotating it 45 degrees, than it would occupy a space with height (and length) of $r\sqrt{2}$. The square still retains its length, but it occupies a different rectangular area...[/b]

Draw it out - it will look like this

[img]http://motivate.maths.org/conferences/conf66/Images/non-symmetry.gif[/img][/quote]

I don't quite understand you. You are saying that a square with a side length of $r$, would have a side length of $r\sqrt{2}$ under a rotation of $45^{\circ}$? Distance is preserved in rotations.

How can the square retain in length and have a different area?  :huh:

Like this: the blue rectangle has a certain length and width, but the black rectangle can vary according to the angle (but I don't know where the angle is in the problem).

In other words, you want to calculate the height and width of the [b]bounding rectangle.[/b]

[hide="Solution"] Say that we rotated the blue rectangle with height $h$ and width $w$ through some angle $\theta$ counterclockwise.  It is not hard to see from the diagram that the width of the black rectangle is then given by

$W = w \cos \theta+h \sin \theta$

And that the height of the black rectangle is given by

$H = w \sin \theta+h \cos \theta$ [/hide]

Right kohjhsd!

Which makes the [hide="solution:"]

If the dimensions of the rectangle are $l$ and $w$, then through some trigonometry, we see that each side of the surrounding rectangle is composed of two parts that are each of a right triangle, with dimensions $L$ and $W$. In this diagram, I considered $\theta$ to be the smaller angle shown, as if the rotation was that of a unit circle: counter-clockwise, like in most applications of this (i.e. drawing programs). Thus the first part of $L$ (called here $L_{1}$) can be found by:

$\sin\theta = \frac{L_{1}}{w}$

and then...

$w\sin\theta ={L_{1}}$.

For the other part of $L$ ($L_{2}$), we use -

$\cos\theta = \frac{L_{2}}{l}$

and then...

$l\cos\theta = L_{2}$.

So $L = l\cos\theta+w\sin\theta$.

Using a similar method, we find that $W = w\cos\theta+l\sin\theta$.

[/hide]


Let pn be the nth prime number (p1 = 2). Let A be a set of positive integers, none of which has a prime factor greater than pn. Let f(n) be the smallest positive integer such that any such set A with f(n) elements contains four numbers whose product is a perfect fourth power. Find f(n) in terms of n.

Why doesn't anyone try this ? I thought it was nice :s

:( I think this "selfmade" problem deserves better :(

Dunno if it's right:

In the "worst" case (i.e. when we f(n) is the greatest) we will have 3 times each prime factor less than p_n (each as a number on its own). Then, we will have to add just one prime factor to become a perfect fourth power, as needed.
Therefore, the minimal value for f(n) for which we are sure that we'll have a perfect fourth power, is 3n+1.

Kurt, your solution fails for $f(1)$; only powers of 2 are allowed, and the set $S = \{ 2^1, 2^2, 2^3, 2^4 \}$ with $3n+1 = 4$ elements has no four elements that multiply to a fourth power.


Show that for all $a$, $b$, $c \ge 0$,
\[\sqrt[3]{\frac{(a + b)(a + c)(b + c)}{8}} \ge \frac{\sqrt{ab} + \sqrt{ac} + \sqrt{bc}}{3}.\]

I don't think this one is complicated.
 You should prove this one which is much more simple and easier to prove
     $LHS \geq \sqrt{\frac{ab+bc+ca}{3}} \geq RHS$

Question 2 of MathLinks Contest Edition 3 Round 1: http://www.mathlinks.ro/Forum/files/3rd_edition_-_round_1_solutions.pdf


Let A, B be square matrices(n x n)
Prove that:  A.B = B.A <=> det(A.A + B.B) >=0
(I'm so sorry because I don't know to use Latex)

[quote="antueminh"]Let A, B be square matrices(n x n)
Prove that:  A.B = B.A <=> det(A.A + B.B) >=0[/quote]
Wrong! You should be able to give a counterexample yourself! Following these questions:

a)Where do the entries of matrices come from?ie are they real or complex?
b)Should the statement be one-sided or two-sided?

Now that you have the correct form of the question, solve it using these;

c)Commuting matrices are simultaneously triangularizable.
d)Another method is to complexify the problem.

Thanks The Captain but I am not wrong. You can give some examples and the theory is true.
A and B are square matrices (n x n) with real elements.

^
Try and find a matrix that does not commute with  $ A\equal{}\begin{pmatrix}0 & 1 \\  0 & 0\end{pmatrix}$. If you do you will get a counterexample.

Your statement should be one-sided("$ \Rightarrow$").


number theory
Diophantine equation
number theory proposed

Hello! :)
(As a Chinese student,please forget about my poor English :D )
Could somebody help me solve the equation 9x^4-2y^2=1 in Z(as elememtary as possible)

I'm not sure what you mean by "elementary" but you can try x = 1, y = 2.

i couldn't find a complete solution but i'll try to help you a bit :)

$ 9x^{4}-2y^{2}=1$

if $ 9x^{4}=t^{2}$

then we have

$ t^{2}-2y^{2}=1$

which is the Pell's equation but i don't know exactly how to solve it...i hope this will help you  :)

[quote="riddler"]Yeah from there it's easy
[hide]
$ (t-\sqrt2y)(t+\sqrt2y)=1$
then follow the procedure to generate values of t and y
[/hide][/quote]

could you please tell me the procedure cause i'd like to learn how to solve it now  :lol:

Well you find the 1st pair of values..  So for are $ t^{2}-2y=1$ they are 3 and 2.
Then,
$ (t-\sqrt{2}y)=\frac{1}{(t+\sqrt{2}y)}$
Square both sides
and put in the values of 3 and 2,
To get,
$ 17-12\sqrt{2}=....$
So, 17 and 12 are your new values of t and y
You can do this infinitely to find infinite values.

You can also multiply both sides by $ (t-\sqrt{2}y)$ and keep finding new values...
Hope you understand...

thank you very much for your explanation  :lol:

[quote="alex90"]i couldn't find a complete solution but i'll try to help you a bit

$ 9x^{4}-2y^{2}=1$

if $ 9x^{4}=t^{2}$

then we have

$ t^{2}-2y^{2}=1$

which is the Pell's equation but i don't know exactly how to solve it...i hope this will help you[/quote]

[quote="riddler"]Yeah from there it's easy
[hide]
$ (t-\sqrt2y)(t+\sqrt2y)=1$
then follow the procedure to generate values of t and y
[/hide][/quote]

Now, I don't see why it is easy.  You have solved another equation (the Pell equation).  You have integer solutions in $ t$ and $ y$, but this does not necessarily give integer solutions for $ x$ and $ y$.  For example, when $ t=17$ then $ x=\sqrt{17/3}$.   Or, is there something else you were thinking? :)

[quote="Rust"]Only one solution $ x=1,y=2$ in $ N$.
It may be prove from $ x^{2}=\frac{(1+\sqrt 2)^{4k+2}+(1-\sqrt 2 )^{4k+2}}{6}$.[/quote]
Thanks, Rust.  How does one do that?  Don't you mean the solutions are given by
$ x^{2}=\frac{(3+2\sqrt 2)^{n}+(3-2\sqrt 2 )^{n}}{6}$

$ (9,2)=1$ , $ 1|1$ and as a result, this diophantine equation has solutions.

One obvious solution is $ (x_{o},y_{o})=(1,2)$

The other solutions are given by: $ (x,y)=(1+2t, 2-9t)$

But $ t\in\mathbf{Z}$, so:

$ 9(1+2t)^{4}-2(2-9t)^{2}> 9\cdot-2\cdot 2 = 1$ and we are done!

Sorry if i have missed something.

[quote="mostel"]$ (9,2)=1$ , $ 1|1$ and as a result, this diophantine equation has solutions.

One obvious solution is $ (x_{o},y_{o})=(1,2)$

The other solutions are given by: $ (x,y)=(1+2t, 2-9t)$

[/quote]
This works only for linear diophantine equations. And how do you get from $ x^{2}=\frac{(1+\sqrt 2)^{4k+2}+(1-\sqrt 2 )^{4k+2}}{6}$ that there are no other solutions?


geometry
power of a point
radical axis
geometry proposed

Let $w_k=C_k(O_k,r_k),\ k\in \overline {1,2}$ two circles $(\ O_1\ne O_2\ )$, a fixed point $A\in w_1$and a mobile point $M\in w_2$.
Denote: the tangent $t$ in the point $M$ at the circle $w_2$; the second intersection $N\in AM\cap w_1$;
the point $P\in w_1,\ NP\parallel t$; the second intersection $R\in MP\cap w_1$.
Ascertain the geometrical locus of the point $L\in AR\cap t$.

[b]
[hide="case-1"][/b]
$1)$ Suppose that $R$ is between $P$ and $M$.

Let $X = LR \cap NP$
From the parallel lines $PX,LM$ and the segment $PM$ we know that
$\widehat{LMP}=\widehat{MPX} \Rightarrow$
$\widehat{LMR}=\widehat{RPX}$

Now, if $N$ is between $P$ and $X$, then $\widehat{RPX}=\widehat{RAM}\Rightarrow \widehat{LMR}=\widehat{LAM}$
[b](see fig 1-1)[/b]

if $P$ is between $N$ and $X$ [b](see fig 1-2)[/b]
then $\widehat{RPX}=\widehat{NAR}$ and $\widehat{NAR}=\widehat{LAM}$
hence $\widehat{LMR}=\widehat{LAM}$[/hide]

[b]
[hide="case-2"][/b]
$2)$ Now suppose that $P$ is between $R$ and $M$.

From the parallel lines $PX,LM$ and the segment $PM$ we know that
$\widehat{LMP}=\widehat{RPX}$
or else $\widehat{LMR}=\widehat{RPX}$

if $N$ is between $X$ and $P$ [b](fig 2-1)[/b], then $\widehat{RPX}=\widehat{LAM}$
$\Rightarrow \widehat{LMR}=\widehat{LAM}$

if $P$ is between $X$ and $N$ [b](fig 2-2)[/b], then $\widehat{RPX}= \widehat{RAN}=\widehat{LAM}$
$\Rightarrow \widehat{LMR}=\widehat{LAM}$[/hide]

[b][color=blue]Conclusion[/color][/b]

At any case we have  $\widehat{LMR}=\widehat{LAM}$

which means that the triangles $LMR,LAM$ are similar (since they have a common angle at $L$)$\Rightarrow$

$\frac{LR}{LM} = \frac{LM}{LA} \Rightarrow LM^2 = LA \cdot LR$

$LM^2$ is the power of $L$ w.r.t. $w_2$
$LA \cdot LR$ is the power of L w.r.t. $w_1$

$\Rightarrow L$ is on [color=blue]the radical axis[/color] of $w_1,w_2$.

There is one case more [case-$(3)$] when $M$ is between $R$ and $P$ (so $M$ is in the interior of $w_1$)

But I'll leave it to someone else.

[quote="Virgil Nicula"]Very nice (solutions, cases and graphics), Pontios ! Thank you ![/quote]my pleasure!  :)

This problem is very nice, but it had too many cases to check.

Anyway, here is the third case (actually, I didn't solve it before, because I was bored and tired  :D ).

[b][hide="case-3"][/b]$3)$ $M$ is between $P$ and $R$.

From the parallel lines $PX,LM$ and the segment $PM$ we know that $\widehat{LMR}=\widehat{XPR}$

if $N$ is between $X$ and $P$ [b](fig 3-1)[/b], then $\widehat{XPR}=\widehat{NAR}=\widehat{LAM}\Rightarrow \widehat{LMR}=\widehat{LAM}$

if $P$ is between $X,N$ [b](fig 3-2)[/b], then $\widehat{RAN} = \widehat{RPN}\Rightarrow$

$\pi - \widehat{RAN} = \pi - \widehat{RPN}\Rightarrow$

$\widehat{LAM} = \widehat{XPR}\Rightarrow$

$\widehat{LAM}=\widehat{LMR}$[/hide]

I erred the address. The solution of this problem is the my last reply
from the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=72866


algebra
polynomial
complex numbers
algebra unsolved

1. Let $ f(x) \equal{} x^n \plus{} 5x^{n\minus{}1} \plus{} 3$ where $ n > 1$ is an integer. Prove that $ f(x)$ cannot be expressed as the product of two polynomials, each of which has all its coefficients integers and degree at least 1.

2. Let $ p(x)$ be a polynomial with integer coefficients satisfying $ p(0) \equal{} p(1) \equal{} 1999$. Show that $ p(x)$ has no integer zeros.

3. Let $ P(x)$ be a polynomial with real coefficients. Show that there exists a nonzero polynomial $ Q(x)$ with real coefficients such that $ P(x)Q(x)$ has terms that are all of a degree divisible by $ 10^9$.

1) Eisenstein's criterion does it.

2) if there were a root $ r$, then $ r\equal{}r\minus{}0|P(r)\minus{}P(0)\equal{}\minus{}1999$ and $ r\minus{}1|P(r)\minus{}P(1)\equal{}\minus{}1999$, which is impossible because there is no $ r$ such that both $ r$ and $ r\minus{}1$ divide 1999.

What prime are you applying Eisenstein's with respect to?  (A much nicer criterion for this problem is Perron's.)

the prime 3. i was referring to extended eisenstein ...

we can in fact replace $ 10^9$ with an arbitrary natural number $ k$.

let the complex roots of $ P(x)$ be $ z_1,\overline{z}_1 ..., z_s, \overline{z}_s$ and the real roots be $ r_1, ..., r_t$. then $ (x^k \minus{} z_1^k)(x^k \minus{} \overline{z}_1^k)\cdots (x^k \minus{} z_s^k)(x^k \minus{} \overline{z}_s^k)(x^k \minus{} r_1^k)\cdots (x^k \minus{} r_t^k)$ is a polynmial with real coefficients which is a multiple of $ P(x)$ and all of whose terms have degree divisible by $ k$.

Sorry I'm such a noob, how did you get the x^k and why did you use complex numbers?

i introduced complex numbers because $ P(x)$ may have complex roots.  i used $ x^k$ in order to make every term have degree divisible by $ k$.


linear algebra
matrix
induction
algebra
function
domain
Gauss

Suppose that $ a$ and $ b$ are elements of a ring R for which $ (a,b)\equal{}R.$
Show that $ (a^{m},b^{n})\equal{}R$ for any positive integers $ m$ and $ n$.

I need help with this, Im stuck. Thx

Seems surprising to me, without constraints on $ R$.

Here's an easy proof when $ R$ is commutative with $ 1$: write $ 1\equal{}ax\plus{}by\equal{}(ax\plus{}by)^{n\plus{}m}$.

If $ (a,b)$ means the ideal generated by $ a,b$,
then take $ R\equal{}\mathcal{M}_2(\mathbb{F}_2)$ (which is a simple ring) and $ a\equal{}b\equal{}\left(\begin{matrix}0&1\\0&0\end{matrix}\right)$.
Then $ (a,b)\equal{}R$ and $ (a^2,b^2)\equal{}0$.

OK, so $ R$ is commutative.  Does it need a unity?

[quote="LydianRain"]Does it need a unity?[/quote]
No it doesn't.
If $ R\equal{}aR\plus{}bR\subseteq R^2$, i.e. $ R\equal{}R^2$,
then $ R\equal{}R^n$ for all $ n\in\mathbb{N}$ by induction.
(if $ R^n\equal{}R$, then also $ R^{n\plus{}1}\equal{}R^n\cdot R\equal{}R\cdot R\equal{}R^2\equal{}R$)
This gives $ R\equal{}R^{m\plus{}n}\equal{}(aR\plus{}bR)^{m\plus{}n}\subseteq \sum_{k\equal{}0}^{m\plus{}n}a^kb^{m\plus{}n\minus{}k}R\subseteq a^mR\plus{}b^nR$
for all $ m,n\in\mathbb{N}$.

[quote="LydianRain"]Seems surprising to me, without constraints on $ R$.

Here's an easy proof when $ R$ is commutative with $ 1$: write $ 1 \equal{} ax \plus{} by \equal{} (ax \plus{} by)^{n \plus{} m}$.[/quote]

Why does this solve the Problem?
 I dont understand, do I have to use the binomial formula? and then some kind of induction?

I dont see from that how to put $ 1$ as a combination of $ a^{n}$ and $ b^{m}$.

Just consider expanding that expression out (by the binomial theorem) -- every term contains either a factor of $ a^n$ or of $ b^m$.

Yes, any term has $ a^{s}b^{t}$ and since $ s \plus{} t \equal{} n \plus{} m$, then its impossible to have $ s < n$
and $ t < m$,  so I can always factorize either $ a^{n}$ or $ b^{m}$.

Thankyou  :P

See also my following sci.math post on the Freshman's Dream, which
mentions connections to Prufer domains, CRT,  Gauss' Lemma, etc.

$ (A\plus{}B)^n \equal{} A^n \plus{} B^n$ for GCDs/Ideals, if $ A\plus{}B$ cancellative, sci.math, 19 Nov 2008
http://google.com/groups?selm=y8zej18x733.fsf%40nestle.csail.mit.edu
http://google.com/group/sci.math/msg/ba5ac69dddfc4a67


geometry
trapezoid
geometric transformation
reflection
projective geometry

Let $ ABC$ be a triangle. Denote $ D\in BC\ ,\ AD\perp BC$ . For $ M\in AD$ define $ \|\begin {array}{c} X\in AB\cap CM \\
 \\
Y\in AC\cap BM\end{array}$ .

Prove that the quadrilateral $ BXYC$ is cyclically $ \Longleftrightarrow$ the point $ M$ is the orthocenter of $ \triangle ABC$ .

[hide="Short and sweet"]Notice $ XY$ passes through the harmonic conjugate of $ D$ wrt $ B, C$ (Ceva/Menelaus).  However, if $ BXYC$ is to be cyclic, then $ XY$ must be antiparallel to $ BC$.  There is only one line through the harmonic conjugate of $ D$ which is antiparallel to $ BC$.[/hide]

[quote="K81o7"] There is only one line through the harmonic conjugate of $ D$ which is antiparallel to $ BC$.[/quote]

Thats not true. Additionally, the problem is incorrect for the same reason.

K81o7, can you please explain what you mean by "harmonic conjugate" "wrt", and "antiparallel"?

This problem correct when $ \triangle ABC$ is scalene.
[hide="Solution"]
Assume that there are points $ X'$ on $ AB$ and $ Y'$ on $ AC$ such that $ BCY'X'$ is cyclic and $ CX'$, $ AD$, and $ BY'$ are concurrent. It follows that $ \angle BX'Y' \equal{} 180 \minus{} \angle ACB \equal{} \angle BXY$, so $ X'Y'\parallel XY$ is cyclic.  By Ceva's Theorem on concurrent cevians $ CX$, $ BY$, and $ AD$ at $ M$, we have that
\[ \frac {BD}{CD}\cdot \frac {CY}{AY}\cdot \frac {AX}{XB} \equal{} 1\implies \frac {BD}{CD} \equal{} \frac {AY}{AX}\cdot \frac {XB}{CY}
\]
Since $ \triangle AX'Y'\sim \triangle AXY$, we have that $ \frac {AX}{AY} \equal{} \frac {AX'}{AY'}$, so $ \frac {BD}{CD} \equal{} \frac {AY'}{AX'}\cdot \frac {XB}{CY}$.  By Ceva's Theorem on concurrent cevians $ CX'$, $ BY'$, and $ AD$, we have that
\[ \frac {AX'}{BX'}\cdot \frac {BD}{DC}\cdot \frac {CY'}{AY'} \equal{} 1\implies \frac {AY'}{AX'}\cdot \frac {BX}{CY} \equal{} \frac {BD}{DC} \equal{} \frac {AY'}{AX'}\cdot \frac {BX'}{CY'}
\]
It follows that
\[ \frac {BX}{CY} \equal{} \frac {BX'}{CY'}\implies \frac {BX}{XX'} \equal{} \frac {CY}{YY'}
\]
$ X'Y'\parallel XY$ gives that $ \frac {XX'}{AX} \equal{} \frac {YY'}{AY}$, so
\[ \frac {BX}{XX'}\cdot \frac {XX'}{AX} \equal{} \frac {YY'}{AY}\cdot \frac {CY}{YY'}\implies \frac {BX}{AX} \equal{} \frac {CY}{AY}
\]
Since $ \angle XAY \equal{} \angle BAC$, we conclude that $ \triangle XAY\sim \triangle BAC\implies XY\parallel BC$.  Yet, this will imply that $ BCYX$ is a cyclic trapezoid, so $ XB \equal{} CY\implies AB \equal{} BC$, which is a contradiction since $ \triangle ABC$ is scalene.  Thus, there is only one point $ M$ such that the condition holds and the orthocenter clearly works for $ M$ because that will give that $ \angle BXC \equal{} 90 \equal{} \angle BYC$, so $ M \equal{} H$, where $ H$ is the orthocenter of $ \triangle ABC$. [/hide]

To [b]facis[/b]: "wrt" means "with respect to."  "Antiparallel" is defined as follows.  If quadrilateral $ ABCD$ is cyclic, then $ AB$ and $ CD$ are defined to be antiparallel.  In the given question, $ XY$ and $ BC$ are antiparallel.  I don't know what a harmonic conjugate is though...  :(

Sorry, you are right ! O.K., $ ABC$ is $ A$ - nonisosceles triangle, i.e. $ AB\ne AC$ .

[quote="Altheman"]Right...the problem fails when AB=AC. Lines are anti parallel, not segments. You need to reference an angle for anti parallel to be meaningful.[/quote]

Yes...but "antiparallel" in this case means "parallel to the reflection of $ BC$ over the bisector of $ \angle{BAC}$".

I guess I also assumed Virgil Nicula was talking about a scalene triangle...otherwise it obviously breaks down as the harmonic conjugate is at infinity and the antiparallel line is, in fact, parallel.  I didn't even consider the degenerate case.

Harmonic conjugate: Let $ A, B, C,$ and $ D$ be points on a line.  Then we say $ C$ and $ D$ are harmonic conjugates with respect to $ A$ and $ B$ if and only if $ \frac{AC}{CB}\equal{}\minus{}\frac{AD}{DB}$ (with a negative sign because we use directed lengths).  Note that if we are given $ A, B,$ and $ C$, then there exists a unique point $ D$ which is the harmonic conjugate of $ C$ with respect to $ A$ and $ B$.  Also notice that if $ C$ and $ D$ are harmonic conjugates with respect to $ A$ and $ B$, then $ A$ and $ B$ are harmonic conjugates with respect to $ C$ and $ D$.  It also turns out that if we construct the circle with diameter $ AB$, then $ C$ and $ D$ are inverses with respect to this circle.

So notice that if $ AD$, $ CX$, and $ BY$ concur, then you can extend $ XY$ to a point $ D'$ on $ BC$, and if you now apply Ceva's Theorem to $ M$ in $ \triangle{ABC}$ and Menelaus to line $ XY$ in $ \triangle{ABC}$, you get that $ D$ and $ D'$ are harmonic conjugates with respect to $ AB$.

Harmonic conjugates are just a special, nice case of the cross ratio.  Projective geometry stuff. :D


limit
logarithms
number theory
number theory unsolved

Prove that for each $m \in \mathbb{N}$ there exists integer $a$ such that none of numbers $1^2+a$, $2^2+a$, ... , $m^2+a$ is divisible by perfect square greater than $1$.

Let's take different primes p[1], p[2], ... , p[m].
If we find  [b]a[/b] , such that a=-i*i(mod p[i]*p[i]) for each i=1, 2, ... , m, we'll solve problem.
But such  [b]a [/b] exists by Chines theorem about rests(excuse me, if I'm wrong in writing of name for this theorem).

Excuse me. I'm wrong in reading problem's condition :(

I think the problem is there in the book "Elementary Theory of Numbers" by W. Sierpinski. I am not sure though, and I don't have the book with me now so cannot check.

Well the problem is indeed in the Sierpinski's book (though in "200 problems in elementary number theory") but it is stated without solution as a note to some other problem

Here is a "solution".  It's not completely rigorous but I think it could be made so.

Let $M$ be the number of primes $\leq \sqrt{2m}$ and let $p_1$, $p_2$, ... $p_k$ be the primes between $\sqrt{2m}$ and $\sqrt{n}$.  If $n$ is an integer much larger than $m$, we estimate the number of integers less than or equal to $n$ with the given property at
\[ \frac{n}{2^M}(1-\frac{m}{p_1^2})(1-\frac{m}{p_2^2})\cdots(1-\frac{m}{p_k^2})  \]
because the proportion of all integers congruent to one of $\{-1^2,-2^2,\ldots, -m^2\}$ mod $p^2$, where $p$ is some prime, is about $\frac{1}{2}$ for $m\geq \frac{1}{2}p^2$, and about $\frac{m}{p^2}$ for larger $p$.

Clearly $\frac{1}{2^M}$ is finite, independent of $n$.
\[ \lim_{n\to\infty}(1-\frac{m}{p_1^2})(1-\frac{m}{p_2^2})\cdots(1-\frac{m}{p_k^2})  \]
is also finite and nonzero, because the negative of its logarithm is of the same order of magnitude as $\sum^{\infty}\frac{1}{p_i^2}$ which is finite.
This means that as $n\to\infty$ the number of numbers less than or equal to $n$ with the given property tends to a constant fraction of $n$, and hence for sufficiently large $n$ there are numbers less than or equal to $n$ with the given property.

Edit:  typos in two of the formulas fixed.

It's good example of an application of the very useful asymptotic method.


function
limit
logarithms
calculus
calculus computations

let $ f$ be incresing fuction defined from $ R^\plus{}$ into $ R^\plus{}$ such that $ \lim_{x \to \infty }f(2x)\minus{}f(x)\equal{}0$  ,prove that :
$ \lim_{x \to \infty } \frac{f(x)}{ln(x)} \equal{}0$.

$ f(x)\approx f(2x)$
hence
$ f(x)\approx f(2^{n}x)$ for all $ n\in N$
Hence
for big $ x>>x_{0}$
$ \frac{f(x)}{\ln x} \approx \frac{f(x_{0})}{n\ln2 \plus{} \ln x_{0}}$ and let $ n\minus{}>\plus{}oo$
but I note that this is only idea and can be replaced to full solution after adding some  words.


calculus
integration
number theory proposed
number theory

Prove that for $ x,y,z,t \in \mathbb{Z}$
the equation $ x^3 \plus{}y^3\plus{}z^3\plus{}t^3 \equal{} 2008$ has infinitely many integral solutions.

Take for examlpe $ x\equal{}10\minus{}60a^3$, $ y\equal{}10\plus{}60a^3$, $ z\equal{}\minus{}60a^2$, $ t\equal{}2$ for any $ a \in \mathbb{Z}$.

[quote="FelixD"]Take for examlpe $ x \equal{} 10 \minus{} 60a^3$, $ y \equal{} 10 \plus{} 60a^3$, $ z \equal{} \minus{} 60a^2$, $ t \equal{} 2$ for any $ a \in \mathbb{Z}$.[/quote]

Quite nice, congrats ! :)

Extremely good!!1 Tried to think why you taught of that but do not seem to be getting. Can you say the motivation behind your proof?

Of course I can. :)
My first thought was to reduce the equation to one variable (maybe it's possible, if so, I'd be happy^^ and indeed it is). If we take $ x\equal{}a\minus{}b$ and $ y\equal{}a\plus{}b$, then $ x^3\plus{}y^3\equal{}2a^3\plus{}3ab^2$. The equation gets $ 2a^3\plus{}3ab^2\plus{}z^3\plus{}t^3\equal{}2008$. Hence if $ 2a^3\plus{}t^3\equal{}2008$, then we must have $ 3ab^2\equal{}(\minus{}z)^3$, where $ a$ is constant. But $ t\equal{}2$ and $ a\equal{}10$ is a solution of $ 2a^3\plus{}t^3\equal{}2008$. Then we can take $ b\equal{}3ak^3$ and we get infinitely many solutions for $ k \in \mathbb{Z}$ (in my example above I used $ a$ as the variable...^^)

So the first thought was a success...  :P


group theory
superior algebra
superior algebra unsolved

This seems interesting...
 Its a well known fact that the automorphism of the Symetric group, Sn, are conjugating by someone.
  ¿What about Hom(Sn,Sm)?

 I think the following is true, but failed to prove it, if the symmetric group acts transitivily on a set of m elements, then m=2 or m=n.
  Anyone knows something? Any ideas?

[quote]if the symmetric group acts transitivily on a set of m elements, then m=2 or m=n. [/quote]

$S_4$ acts transitively on $\big\{ \{\{1,2\},\{3,4\}\}, \{\{1,3\},\{2,4\}\}, \{\{1,4\},\{2,3\}\}\big\}$.

Yes. small n gives counterexamples. Anyway, this morning i realize that it wasn´t true, take for instance Sn acting on Sn by left multiplication.
  On the other hand, ¿does anybody knows any general result about this?

If $m<n$ and $n>4$, $m=2$ is the only possibility. On the other hand, for all $n$ and $k$, $S_n$ acts transitively on $\binom{n}{k}$ elements, and those aren't the only examples.

[quote="Charlydif"] Its a well known fact that the automorphism of the Symetric group, Sn, are conjugating by someone. [/quote]

Aut(S_n) is inner only for $n > 6$, a famous and subtle fact.

[quote]What about Hom(Sn,Sm)?[/quote]

The answer is known as the "Burnside ring of $S_n$".  This is the analogue of the representation ring, but for permutation representations rather than linear representations, and is discussed in books on permutation groups.


I am taking the intro number theory class, and I wanted to compile a PDF from my TeXnic Center.  Problem is, I typed a really long list of numbers and the end result said "0 Errors 0 Warnings 2 Bad Boxes 3 pages", and the bad boxes were:

"Overfull \hbox (55.55638 pt too wide) in paragraph at lines 81--88
\OTI/cmr/m/n/10 19..."

and

"Overfull \hbox (35.6678 pt too wide) in paragraph at lines 81--88

\OTI/cmr/m/n/10..."

Have any ideas?

You'd have to show us lines 81-88 before we can diagnose the precise problem.

Overfull \hbox means that the text is really a bit too long for the line and TeX is having difficulty as it can't hyphenate lists of numbers, and it is leaving it up to you to sort out line-breaks. However, if the resulting pdf looks OK then I wouldn't worry about it. It isn't even given the status of a warning, so is safe to ignore.

If you want to discover what's behind the message then have a look at [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=overfull](La)TeX makes overfull lines[/url] in the very useful [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?introduction=yes]TeX FAQ[/url].


This problem is easy but I think it was solved without a lot messy algebra (which I tried and harshly not worked) so it was nice logarithrm problem.

Wording has been changed.

x,y,z, and w are real numbers. If x,y,z lies in interval of $(1, \infty)$ and w is in interval of $(0, \infty)$, and that:

$\log_x  w = 24\\
\log_y w = 40\\
\log_{xyz} w = 12$

Find the value of $\log_z w$.

P.S. All those intervals and stuffs don't mean anything much. They're actually simple things. Just made it fancy. :P

For an AIME, this is pretty easy... (assuming I did this right..)

[hide]Converting the logs into exponential form, we get $ w = x^{24} = y^{40} $ and $ (xyz)^{12} = w $
Substituting the first equation into the second gives $z^{12} = w^\frac 15$
$w = z^{60}$ back into logs, $\log_z w = 60$[/hide]

[hide="Answer"]60[/hide]That is right answer.

Here is how I learned to do it from Kalva. I wish I got this on my own but oh well :( .

[hide="Sol"]

\[x^{12} y^{12} z^{12} = w\\
x^{120} y^{120} z^{120} = w^{10}\\
w^5 w^3 z^{120} = w^{10}\\
z^{120} = w^2\\
z^{60}  = w\\
\log_z w = 60\]

$\fbox {60}$ is final answer.[/hide]

[hide]Theres also a slightly easier way of doing it (using logx[w]=24 Therefore  logw[x]=1/24 and so on)[/hide]

[hide="Kinda Messy Solution"]

If you rewrite the final equation as ln w/(ln x + ln y + ln z) = 12

12 = ln w/(2ln x) so 2ln x = ln x + ln y + ln z

Substituting ln y = (ln w)/40 you get:

ln x = (ln w)/40 + ln z        (ln x)/(ln w) = 1/40 + (ln z)/(ln w)

1/24 - 1/40 = 1/60 = (ln z)/(ln w)

so log[size=75]z[/size]w = 60

[/hide]


The chief cook at Burgers-R-Us can cook eight hamburgers in 5 minutes. At the same rate, how many hours would it take her to cook 672 hamburgers?

It would take $ \frac{672}{8}\times5\equal{}420$ minutes to cook the hamburgers. This is equivalent to $ \boxed{7}$ hours.


algebra
polynomial
trigonometry
inequalities
algebra unsolved

1. Find all polynomials $ P(x) \in R[x]$ such that $ P(x^2\minus{}2x)\equal{}\left( P(x)\minus{}2 \right)^2$
2. $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)\plus{}2xy$.

Maybe all are easy but I don't know any way to solve these problems. Can anyone give me a general way or something for this type of problems? Thanks.

The second problem is $ f(x)\equal{}x^2\plus{}g(x)$, where $ g$ is any solution to $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$.

[quote="Erken"]Are there any comments for the second problem?[/quote]

Sorry. $ f$ is polynomial with real coefficients.

Keep x=y=0 => f(0)=0
Keep y=-x => f(x)+f(-x)=2x^2
So, f(x)=x^2+cx, c is real constant

1) P(x)=x+1 satisfies the first equation

$ P(x^2 \minus{} 2x) \equal{} \left( P(x) \minus{} 2 \right)^2$
This problem is very complex, but I have already solved a similar one [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=751225#751225]here[/url].
First replacing $ x$ with $ x\plus{}1$ gives $ P(x^2\minus{}1)\equal{}\left( P(x\plus{}1) \minus{} 2 \right)^2$,
and then $ Q(x): \equal{}P(x\plus{}1)\minus{}2\in\mathbb{C}[x]$ satisfies $ Q(x^2\minus{}2)\equal{}Q(x)^2\minus{}2$, which I already solved.

This gives $ Q(x)\equiv \minus{} 1$, $ Q(x)\equiv 2$ or $ Q(x) \equal{} 2T_{n}({x\over 2})$ for some $ n\in\mathbb{N}$
and with $ P(x)\equal{}Q(x\minus{}1)\plus{}2$ we get the solutions
$ P(x)\equiv 1$, $ P(x)\equiv 4$ and $ P(x) \equal{} 2T_{n}({x\minus{}1\over 2})\plus{}2, n\in\mathbb{N}$.

Here $ T_{n}(x) \equal{} \cos (n\cdot \arccos x)$ denotes the [url=http://en.wikipedia.org/wiki/Chebyshev_polynomial]Chebyshev Polynomial[/url] of degree $ n$.


probability
counting
distinguishability
combinatorics proposed
combinatorics

Five marbles are distributed at a random among seven urns. What is the expected number of urns with exactly one marble?

[hide="Too Easy?"]If I am not mistaken, by linearity of expectations, this is $ 7$ times the probability that an arbitrary urn has one marble, which follows by balls-and-urns (consider the bijection between arrangements of $ 6$ dividers and $ 5$ marbles and the total number of "distributions" and the bijection between arrangements of $ 5$ dividers and $ 4$ marbles and the number of "distributions" with one marble in the first urn.)

\[ 7\cdot \frac{\binom{5\plus{}4}{4}}{\binom{6\plus{}5}{4}}\equal{}\frac{21}{11}\]

But this seems too easy.[/hide]

When I think of five marbles distributed at random among seven urns, I think of $ 7^5$ arrangements, not $ \binom{11}{6}$.

Well, of course the answer depends on whether the marbles are distinguishable or not. I guess it's somewhat ambiguous for this problem, although "marbles" sounded generic for me. Perhaps the original poster can weigh in?


Let x,y,z in R such that xy+yz+zx=-1.Find min
 P=x^2+5y^2+8z^2

Let $x,y,z$ in $R$ such that $xy+yz+zx=-1$.Find min value of $P={x^2}+5{y^2}+8{z^2}$


geometry
geometric transformation
homothety
parallelogram
trapezoid
angle bisector
geometry proposed

The incircle of triangle $ABC$ tangents its sides in points $A_1,B_1,C_1$ respectively. $A$-outcircle tangents its sides in points $A_2,B_2,C_2$ respectively. Lets $A_3$ be a middle of arc $C_1B_1$, without point $A_1$. Let $C_3$, $B_3$ be middles of arcs $A_2B_2$(without point $C_2$) and $A_2C_2$(without point $B_2$) respectively. $AB_3$ intersects $BC$ in a point $B_4$. $A_3B_4$ intersects $AB$ in point $B_5$. Similarly we construct point $C_5$. Prove that lines $BC$, $B_3C_3$, $B_5C_5$ are concurrent.

Have you labelled the points correctly, Andrey?  According to my diagram, the result isn't true...

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=677[/img]

By the way, this is the first geometry question I've ever seen that had to use a subscript 5. :D

[quote="Valiowk"]Have you labelled the points correctly, Andrey?  According to my diagram, the result isn't true...
[/quote]

I guess it's correct now :blush:

First, we tackle a simpler (almost trivial) problem: Let the lines $ A_3B_3,\ A_3C_3$ meet the lines AB, AC at P, Q, respectively. Then the lines $ B_3C_3,\ BC, PQ$ are concurrent.

Let $ (A_3),\ (B_3),\ (C_3)$ be 3 circles centered at $ A_3, B_3, C_3$ and touching the line pairs $ (AB, AC),\ (AB, BC),\ (AC, BC)$. The point $ R \equiv BC \cap B_3C_3$ is the external homothety center of the circles $ (B_3),\ (C_3)$. The points P, Q are the external homothety centers of the circle pairs  $ [(A_3), (B_3)],\ [(A_3), (C_3)]$, respectively, so that the points P, Q, R are collinear, which means that the lines $ BC,\ B_3C_3,\ PQ$ are concurrent.

In order to prove the problem proposition, we have to exhibit a different circle $ (A_0)$ touching the lines AB, AC, i.e., centered on the internal bisector AI of the angle $ \angle A$, the external homothety centers of which with the circles $ (B_3),\ (C_3)$ are the points $ B_5, C_5$, respectively. In other words, we have to show that the lines $ B_3B_5,\ C_3C_5$ meet at a point $ A_0 \in AI$ on the angle bisector $ AI \equiv AI_a.$ This is a projective problem:

[b]Lemma:[/b] Let $ AI_a,\ BI_a,\ CI_a$ be 3 arbitrary concurrent cevians of a triangle $ \triangle ABC$. Let $ A_3 \in AI_a,\ B_3 \in BI_a,\ C_3 \in CI_a$ be arbitrary points on these cevians different from $ A, B, C, I_a$. Let the lines $ AB_3, AC_3$ meet the line BC at $ B_4, C_4$ and the lines $ A_3B_4,\ A_3C_4$ meet the lines $ AB,\ AC$ at $ B_5, C_5$, respectively. Then the lines $ B_3B_5,\ C_3C_5$ meet at a point $ A_0 \in AI_a$.

Project the line $ AI_a$ to infinity by a central projection. Any 2 lines intersecting on the line $ AI_a$ are projected into 2 parallel lines. Denote $ L \equiv AC \cap BE_a,\ M \equiv AB \cap CE_a$ the feet of the cevians $ BE_a,\ CE_a$, so that $ BM \equiv AB,\ CL \equiv AC$. After the projection, we get a parallelogram $ B'L'C'M'$ (because $ A', I_a'$ are at infinity), intersected by 2 parallel lines, both parallel to $ B'M' \parallel CL'$ (because $ A'$ is at infinity), through some arbitrary points $ B_3' \in B'L',\ C_3' \in C'M'$. These 2 parallel lines meet the parallelogram diagonal $ B'C'$ at $ B_4', C_4'$ and some other 2 parallel lines through $ B_4', C_4'$ (because $ A_3'$ is at infinity) meet $ B'M',\ C'L'$ at $ B_5', C_5'$.

Quadrilaterals (trapezoids) $ B'B_3'B_4'B_5' \sim C'C_3'C_4'C_5'$ with the parallel sides $ B'B_3' \equiv B'L' \parallel C'M' \equiv C'C_3',$ $ B_3'B_4' \parallel C_3'C_4',$ $ B_5'B_4' \parallel C_5'C_4',$ $ B'B_5' \equiv B'M' \parallel C'L' \equiv C'C_5',$ and common corresponding diagonal $ B_4'B' \equiv B'C' \equiv C_4'C'$ are centrally similar. This means that their remaining corresponding diagonal lines $ B_3'B_5' \parallel C_3'C_5'$ are also parallel. Thus these 2 lines intersect on the line at infinity, which is the image of the line $ AI_a,$ so that the original lines $ B_3B_5,\ C_3C_5$ intersect on this line at a point $ A_0 \in AI_a,$ which is what we were supposed to prove.


A car travels $ d$ miles in $ \frac{3}{5}$ hours. What is the speed of the car in miles per hour? Express your answer as a common fraction in terms of $ d$.

i was doing this prob and it came up blank1!!!!
i hope i put thios in the right forum


geometry
trigonometry
perpendicular bisector

Among triangles with one side of a given length L and with given area S, determine all of those for which the product of the lengths of the three altitudes is maximum.

Nice question.
[hide]It seems intuitively clear that the maximum occurs when the vertex not on the side of length L lies on its perpendicular bisector, or in other words when the triangle is isosceles.  We prove it as follows:
Note that one altitude has a fixed length of $\frac{2S}{L}$.  The other altitudes satisfy $h_a\cdot a = 2S, h_b\cdot b = 2S$, so $h_a\cdot h_b\cdot h_c = \frac{8S^3}{abL} = \frac{4S^2\sin{\theta}}{L}$, where $\theta$ is the angle between sides a and b.  Thus, we want $\theta$ to equal ninty degrees, or, barring that, to be as large as possible.  To finish off, we just need to consider the set of circles passing through the endpoints of the side of length L to see that if we cannot have a right triangle ($2S > L^2$), the maximum occurs when the vertex is on the perpendicular bisector.  Otherwise, it occurs when the triangle is right.[/hide]


[quote="chien than"]Let  $ABCD$ be a square. Draw $Ax,Ay$ are 2 ray inside odf square. Draw $BF, DH\perp Ax, BE, DK\perp Ay. CQ\perp BF, CI\perp BE. DN\perp IC, IM\perp CQ$.
$HK \cap EF, MN=\{L,X\}. IQ\cap EF, MN=\{G,Y\}$
Prove that: $LGYX$ là hình vuông.[/quote]

I don't speak Vietnamise, my friend!

Let  $ABCD$ be a square. Draw $Ax,Ay$ are 2 ray inside odf square. Draw $BF, DH\perp Ax, BE, DK\perp Ay. CQ\perp BF, CI\perp BE. DN\perp IC, IM\perp CQ$.
$HK \cap EF, MN=\{L,X\}. IQ\cap EF, MN=\{G,Y\}$
Prove that: $LGYX$ be a square
Help me,please!

Oh! Chien than, this is my problem :mad: ? I won't post the solution. This problem is very hard! :D . I think you should study harder! :roll:

Cant anyone send a diagram?I am a bit confused about the problem statement.:-s

[quote="chien than"]Let  $ABCD$ be a square. Draw $Ax,Ay$ are 2 ray inside odf square. Draw $BF, DH\perp Ax, BE, DK\perp Ay. CQ\perp BF, CI\perp BE. DN\perp IC, IM\perp CQ$.
$HK \cap EF, MN=\{L,X\}. IQ\cap EF, MN=\{G,Y\}$
Prove that: $LGYX$ be a square[/quote]

 My figure works only if $DM \perp CQ$ and not if $IM\perp CQ$ , as you write. Please clear this point.

thank u for ur nice diagram. :coolspeak: ..ya thats why i thought that thr r some problems... :ninja:


function
integration
calculus
real analysis
real analysis unsolved

$f: [0;1] \to \mathbb{R}$ a lipschitz function. Prove that there exists a measurable function $g$ such that $f(x)=f(0)+\int_0^x g(t)dt$.

It's a consequence of the well-known theorem which states that an absolutely continuous function is differentiable a.e. and equal to the integral of its derivative. Clearly, a Lipschitz function is absolutely continuous.


Twice the supplement of an angle equals 8 times the complement of hte angle.  Find the angle.

hey thats really easy because we're learning that in math class.

Let x be the angle. then 2(180-x)=8(90-x) and 180-x=4(90-x) so 180-x=360-4x. Then 3x=180 so x=60.

The angle measures 60 degrees.

[hide="Solution"]
$ 2(180 \minus{} x) \equal{} 8(90 \minus{} x)$

$ 360 \minus{} 2x \equal{} 720 \minus{} 8x$

$ 6x \equal{} 360$

$ x \equal{} 60$
[/hide]

edit:someone beat me to it...that's what i get for leaving this window open while I eat dinner  :oops:


My first is in little, but not in stand.
My second is in finger, but not in brand.
My third is in big, but not in bite.
My fourth in home, but not in mite.
My fifth in think, but not in hide.
My whole what you need, at night, inside.

[quote="BOGTRO"]Is it light?[/quote]

True.

[hide]light[/hide]
is the intended answer to the puzzle.


function
limit
search
real analysis
real analysis unsolved

A function $ f(x)$ is continuous for all $ x>0$ , and $ \forall a>0$ , the sequences $ \{f(n\cdot a)\}$ all converge to $ 0$

$ \mathbf{Prove : }\;\lim_{x\to+\infty}f(x)=0$

[hide]
I'm not sure how rigorous this is, but here I go

$ \lim_{a\to0^{+}}f(n\cdot a) = f(x)$
as a goes to zero, the points in the sequence become closer and closer together.
As you take the limit as the density approaches infinity, you are left with a continuous function, f(x).

Since $ a$ remains greater than 0 as we take the limit, the original condition also holds, so it follows trivially that since $ \lim_{n\to\infty}f(n \cdot a)=0$
then $ \lim_{x\to\infty}f(x)=0$
[/hide]

It's not that easy; see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1925654751&t=137954]this topic.[/url]


If $ f(x) \equal{} x^4 \plus{} x^2 \plus{} 5x$, evaluate $ f(5) \minus{} f(\minus{}5)$.

hello, we get
$ f(5)\minus{}f(\minus{}5)\equal{}675\minus{}625\equal{}50$.
Sonnhard.

OR,

Notice that the $ x^4 \plus{} x^2$ will cancel out, so it's actually just

$ 5(5) \minus{} 5(\minus{}5) \equal{} \boxed{50}$


Mai demult Cezar Lupu zicea intr-un "thread"(ca sa vbesc mathlinkeste :D ) "Bucurati-va de asta",referindu-se la o problema de la Inter cu multe solutzii misto..Eu acum va zic "Bucurati-va si de asta":
a,b,c reale positive cu $a^{2}+b^{2}+c^{2}=3$,sa se demonstreze inegalitatea: \[4(a^{3}+b^{3}+c^{3})\geq3(a+b+c+abc)\] Problema asta are o groaza de solutzii frumoase..nu ma sfiiti sa le postatzi :)
Am pus problema si la "Pre-Olimpiad",da` m-am gandit ca sunt romani care frecventeaza mai putzin zonele alea..
Bafta la solutzii :)

iese misto cu chebisev $3\sum{a^{3}}\geq (\sum{a})(\sum{a^{2}})$, dar $\sum{a^{2}}=3$, si dupaia curge cu $AM-GM$.


Determine the next term in the sequence:

$1, 1, 4, 27, 3125, \ ?$

[hide]
nevermind, perfect628 is right  :blush: [/hide]


why is the reactivity order of alcohols the way it is

SN2 reactions in which the hydroxil group was previously converted in a better leaving group (a tosylate for example), or SN1 reactions promoted by acid?

Then carbocation stability should dictate alcohol reactivity.

why do acid catalyzed have have to be unimolecular

Because for a $ S_{N}2$ reaction to be of pratical use it cannot be runned with acid catalysis: under these conditions the nucleophile would be converted into the conjugate acid (the best nucleophiles are also reasonably good bases), which would greatly decrease its nucleophilicity. On the other hand, $ S_{N}1$ reactions of alcohols are acid catalysed, because these reactions don't require a good nucleophile but a good leaving group (such as an water molecule).

what if i carry oput the raetion in aprotic polar solvent?

The reaction must still be acid catalysed.

ok can't it be both acid catalyzed and in aprotic polar solvent at low temp.

The solvent makes no difference, because the incoming nucleophile is still protonated (and probably also the solvent) as the reaction continues to be acid catalysed. Low temperature makes every possible reaction even slower.


[hide]
1.Sin 30 =1/2
2.Sin 49=.75(to nearest tenth)
3.Add the second and last to get 4x+5z=-26. Add the first and the second to get x+5z=-18. Use system of equations on those two to find x and z. Z=-12/7. X=-33/7. Then plug those in back to any of the first three to find y. Y=10/7.[/hide]

[quote="jenny221"]hey everyone!

1.) Without using a calculator solve sin 30°

2.) Use a calculator to find the function value of sin 49°

3) Solve the following system:
   X + y + z = -5
   X – y + 4z = -13
   3x + y + z = -13[/quote]
Is this your homework?
To ckck the nearest tenth means to one decimal place, just so you know...


S24

How many functions $ f(x) = ax^{2}+bx+c$ are there with the property that, for all $ x$, $ f(x) \times f(-x) = f(x^{2})$?

Please note that the "AMC" forum on our site is for the [b]American[/b] Mathematics Competitions, and is labeled as such.  As we are a US-based site, the majority of our members are in the US, and so we have a special forum for the US AMC/AIME/USAMO contests.

I have moved all of these (Australian) AMC problems to the appropriate high school forums.

[quote="MellowMelon"]$ f(x) \cdot f(-x) = f(x^{2}) %Error. "Rightleftarrow" is a bad command.
$
$ (ax^{2}+bx+c)(ax^{2}-bx+c) = ax^{4}+bx^{2}+c$
$ a^{2}x^{4}+(2ac-b^{2})x^{2}+c^{2}= ax^{4}+bx^{2}+c$
$ a^{2}= a, c^{2}= c, 2ac = b^{2}+b$
This means $ a$ and $ c$ can only be $ 0$ and $ 1$. If either is 0, then $ b$ can only be $ 0$ or $ -1$. If $ a = c = 1$, then $ b$ can be $ -2$ or $ 1$.

So replace the $ x^{2}-1$ in the above post with $ x^{2}+x$ and I agree with it.[/quote]
I don't understand how
$ (ax^{2}+bx+c)(ax^{2}-bx+c) = ax^{4}+bx^{2}+c$
$ a^{2}x^{4}+(2ac-b^{2})x^{2}+c^{2}= ax^{4}+bx^{2}+c$
implies
$ a^{2}= a, c^{2}= c, 2ac = b^{2}+b$ , because I thought it would mean that
$ a^{2}x^{4}+(2ac-b^{2})x^{2}+c^{2}= 0$, which doesn't necessarily mean that $ a^{2}= a, c^{2}= c, 2ac = b^{2}+b$, unless $ ax^{4}+bx^{2}+c$ = 0

Well, the coefficients of the $ x^{4}$ term, $ x^{2}$ term, and constant term have to be equal to their partners in order for the two expressions to be equal for all values of x...

And plus, you've got it wrong. $ a^{2}x^{4}+(2ac-b^{2})x^{2}+c^{2}= ax^{4}+bx^{2}+c$ doesn't mean that $ a^{2}x^{4}+(2ac-b^{2})x^{2}+c^{2}= 0$.

It means that $ (a^{2}-a)\cdot x^{4}+(2ac-b^{2}-b)\cdot x^{2}+(c^{2}-c)\cdot x^{0}=0$.

[quote]Please note that the "AMC" forum on our site is for the American Mathematics Competitions, and is labeled as such. As we are a US-based site, the majority of our members are in the US, and so we have a special forum for the US AMC/AIME/USAMO contests.

I have moved all of these (Australian) AMC problems to the appropriate high school forums.[/quote]

Apologies..

Anyway i got 8 as well


Welcome to My Tournament.  My tournament is a simple tournament, I will post 10 questions, and depending on the number of people I have playing, some number of people will be eliminated (Lowest scores).  This will go until two people remain.  The questions will not be very hard, most of them will be chapter/state level.  If there are ties, a tiebreaker question will be used.  When the first round is posted, I will also state the number of people playing and how many people will be eliminated.  To sign up, please go to the sign up link at the end of this post.  I fyou have any questions, just post on this forum.

NOTE: Reminder PM's will not be sent out, but if anyone would like to do that for me, I can send that person the things to be sent and to which people.

[url=http://www.emailmeform.com/fid.php?formid=125675]Sign Up![/url]

SIGN UPS END ON AUGUST 12 (TUESDAY)!!!!!!!

[quote="1=2"]Will calculators be allowed???[/quote]

On the second thought, there will be no calculators allowed, just to make complex computations (multiplication and division mostly) hard.  (I dont know what i said, but no calculators)

This is the list of people who have signed up, it will be updated:

1. vallon22
2. AIME15
3. Urc
4. Poincare
5. Shentang
6. bachukas
7. isabella2296
8. 1=2
9. 5849206328x

We need more people!!!  (Atleast 20)!!!!!

Todays the last day, but because of the lack of players, I will extend the date to this Friday!!!!

Xpmath has joined.

[quote="ernie"]I think I'll rather sign up here. :P[/quote]

Okay....and looking at the lack of players, I'll also allow you to PM me or post that your joining on this topic.

List of players who have joined:

1. vallon22
2. AIME15
3. Urc
4. Poincare
5. Shentang
6. bachukas
7. isabella2296
8. 1=2
9. 5849206328x
10. xpmath
11. tinytim
12. ernie
13. FantasyLover
14. suma_milli
15. abacadaea

Please tell me if I have missed anyone.

[quote="herefishyfishy1"]What kind of a name is "My Tournament?"[/quote]

I couldn't think of a nice name, but if you have one, I can rename it.

Do you want to participate too?

[quote="ernie"]When will it start?[/quote]

Tomorrow (Friday), but if at least one more person doesn't join, I'm gonna have to extend the sign up date.  But I think 14 is enough.  I'll send PM reminders.  You'll be able to download the PDF test either today at night or tomorrow around 10-12 central time.

Here is Round 1 in PDF.  Four people will be eliminated in this round, the rest will advance to Round 2.  There are 10 questions, and no calculators are allowed.  There is no time limit, but don't spend hours trying to get all of them right.  This round has fairly easy problems, so I'm expecting ties.  There will be tie breaker problems posted here later.  Please PM the answers to me.  This is due on August 20, Wednesday.  I will post the scores here.  If you haven't signed up, but you would like to, just OM me the answers and I'll include you in the tournament.  So here is the test in PDF.

Okay, I'm extending deadline to Saturday because school just started and its kinda hectic.  So most likely scores will be posted on Friday once i get settled into 8th grade.

gah
there's too much stuff going on
srry, but i'm gonna have to drop out

I signed up!
Do we have to write solutions( :mad: ), or just answers( :lol: )

Is it OK if we are in middle school and taking WOOT? Because if not, abacadaea and I are out.


algorithm
combinatorics unsolved
combinatorics

the question is

"prove that any map can be colored with six colors in such a way that no two neighboring countries have the same color."

also, if anyone can breifly show the proof of the theorem itself, it will be greatly appreciated.

Thank you for your interests.

http://research.microsoft.com/~gonthier/4colproof.pdf

Not sure if this is of any use to you but I thought I'd try.

Look at ex.#17 here :
http://www.animath.fr/cours/graphes.pdf

Pierre.

Five color theorem of maps was an old result due to Heawood. Therefore first you can color any map with five colors and then change any region color to a new sixth color. But if you implied that four coloring of the maps is difficult by the coloring algorithms developed by Appel and Haken (1976) and Robertson et.al. (1996) then I would suggest spiral chain coloring algorithm.

see http://arxiv.org/abs/math/0507127 and http://arxiv.org/abs/0710.2066

see also http://www.flickr.com/photos/49058045@N00/  for many graphical illustrations. :)