trigonometry
function
geometry
geometric transformation
reflection
absolute value

I am graphing trigonometric functions in math class and I would like to know how you can graph a trig equation in the formula

y = A sin B (X - C) + D where sin can be replaced with cos, tan, sec, csc, cot.

And how would one derive a formula from looking at a trig graph? I know these are basic questions related to transformations but I would really appreciate some help

We actually just started this topic in class the other day.  I"ll summarize my notes for you, and give a couple examples from them.  A note, we only covered sin and cos with the general equation you gave, so I can't speak for the effectiveness of what follows with the other functions.

The equation you gave:
y = a sin b(x + c) + d

Describes a sinusoidal function with the following transformations:

-a vertical stretch by a factor of a, assuming a > 1
-a horizontal compression by a factor of 1/b, assuming b > 1
-a shift right/left c units
-a shift up/down d units

The shift right/left is also called a phase shift.

Sinusoidal graphs have three basic traits...
-amplitude
-period
-phase shift

amplitude - one half the vertical distance between the maximum and minimum values of a sinusoidal graph.  In the parent sin graph, y = sin x, it's 1.  Amplitude relates to the vertical stretch or compression of a graph.

Vertical stretch when |a|>1
Vertical compression when |a|<1
by a factor of a in both cases.

Examples: graph y = sin x
                       y = 4sin x
         and         y = (1/4)sin x
On your calculator.  Observe how the functions with the transformations are different from the parent graph, y = sin x.

period - the shortest horizontal interval in which the graph repeats.  Basically, the interval in which there is one 'unit' of the graph's pattern.  Example: The period of y = sin x is 2(pi).

Period = 2(pi)/|b|, period is equal to 2 pi divided by the absolute value of b.  If you found the period of a graph you were looking at, you would be able to use that general equation to find b.

b produces a horizontal stretch when |b| < 1
b produces a horizontal compression when |b| > 1
both by a factor of 1/|b|.

A side note: Reflections
A negative value for a reflects over the x-axis, a negative value for b reflects over the y-axis.

Phase shift - the horizontal distance a graph is translated left/right from its original position, that of the parent graph

That's basically the same as transformations you should have learned when doing other functions, like rational functions.
Examples:

y = sin (x - 3) would produce a shift right 3 units.
y = sin (x + 3) would produce a shift left 3 units.
Graph them for yourself and check it out.  Make sure to put the parent graph on there too.

A side note about transformations: They're always applied in the order of operations.  So any transformation involving "multiplication" will come before one involving "addition/subtraction".  For example, you'd do a horizontal compression before you did a phase shift.

All reflections must also be made before their corresponding shift.  For example:

y = - sin(x) - z.
Perform the reflection before the shift down z units.

y = sin-(x + z)
Perform the reflection before the shift left z units.

So, that's my notes.  It's basically doing a lot of examples and knowing what the transformations look like.  Read all the examples in your book, and if it has answers in the back, read and look at the graphs of all those too.

You can also practice by making up equations yourself, predicting what the transformations will look like (e.g. "This function will have a period of x ... and amplitude of y....) and seeing if you're right.   Really look at the graph, and explain to yourself how you know what the transformation is.  For example, if you had a graph of a sine function with a max of 10 and a minimum of 4, you could use a = (max - min)/2 = (10 - 4)/2 = 3, to come up with a =3.   If there were no other transformations on the graph, the final equation would be y = 3sin(x)

Hope that helps!


uh...I don't know if this question suitable to be posted here, because I encounter this problem at work. If not, please remind me, thanks

There are two square waves, [u]wave 1[/u] and [u]wave 2[/u]
wave 1 has active period : a1, and deactive period d1
wave 2 has active period : a2, and deactive period d2
a1, a2, d1, d2 are natural numbers

If wave 1 start at time t0, find a start time for wave 2 such that wave 1 and wave 2 has maximum intersection.

Definitely not HSB.

Looks like sound? waves, or something.

Where should I move this?

The waves are look like that
[img]http://www.csie.nctu.edu.tw/~yangyc/9117091/rect_wave.JPG[/img]
a1, d1, a2, d2 are natural numbers
only t2 can be adjusted to let the intersection of a1 and a2 is maximum


There is an x*x square.  Integers are written in each square so that in every two squares with a common side, the neighbouring number is one higher or one lower or the same.
Prove that at least one number is written in the x*x square at least x times.

Is it pidgeon hole or induction or something compeltely different?

but seems like it can't be proven by that easily.
I tried total number of ways to fill in the squares with the limited range of numbers, but the probability is too big to generalise.  I think I am missing something or a proof to limit the number of times each number can be used, which differs for all numbers according to case.
Any ideas or hints?

[hide]

Just notice that we can sort the rows from the lowest number to the highest one, from left to right. Same with columns from up to down:

1 2 3 2
2 3 4 3
3 4 5 4
2 3 4 5

becomes

1 2 2 3
2 3 3 4
2 3 4 5
3 4 4 5

Now the first number in the first row is 1, let k be the last number in the first column. Because the last row is ordered, the last number of the last row >= k. Let the last number of the first row be p. Then if p <= k then we have k appear at least once on every column, so at least n times in the whole board.

If OTOH p > k then we have p appear on every row at least once, so at least n times in the whole board.

[/hide]


In the Pariliament of Sikinia, each member has at most three enemies. Prove that the house can be separated into two houses, so that each member has at most one enemy in his own house.

Separate them into two houses arbitrarily. If a person has two enemies in the same house, he should switch. By doing this again and again, it can be done.


[quote="minsoens"]ehh hmm.. here is my instinctive way to do it:
$AD \cong ME$ because $\text{Given}$
$AD \parallel ME$ because $\text{Given}$
$\angle DAM = \angle EMB$ because $\text{Alternate Interior Angles}$ and $\text{Vertical Angle Theorem}$ (I don't know what else to use here)
$AM = MB$ because $\text{Definition of Midpoint}$
$\triangle DAM \cong \triangle EMB$ because $\text{SAS}$
$\angle DAM = \angle EMB$ because $\text{CPCTC}$
$DM \parallel EB$ because $\text{Alternate Interior Angles}$ and $\text{Vertical Angle Theorem}$ (again, this should be sorta intuitive because the lines are parallel and another line crosses them.)[/quote]

oh good ol' cpctc...this brings me back...i have not written those theorem names in years...

[quote="learnmath2"]given AD is congruent to ME --------- AD is parallel to ME........... M is the midpoint of AB

prove;;;;;; MD is parallel to BE

[url]http://www.geocities.com/hazelmilano23/untitled.bmp[/url][/quote]
Um, is this your homework?


induction
logarithms
limit
real analysis
real analysis unsolved

Prove, that the sequence $ \{x_n\}$, where $ x_{n\plus{}1}\equal{}x_n\plus{}\frac{x_{n}^2}{n^2}$ and $ 0<x_1<1$, is limited.

Let $ x_1 \equal{} a\in(0,1)$. By induction we conclude that $ x_n\le an$ for all $ n$. Now let $ \beta \equal{} \limsup\frac {x_n}{n}\le a < 1$. Then for every $ \gamma > \beta$, we have $ x_{n \plus{} 1}\le x_n \plus{} \gamma^2$ for all sufficiently big $ n$. Hence, $ \beta\le\gamma^2$ for every $ \gamma > \beta$. This is possible only if $ \beta \equal{} 0$. Thus, $ x_n \equal{} o(n)$. Now rewrite the recurrence relation as $ x_{n \plus{} 1} \equal{} x_n\left(1 \plus{} \frac {x_n}{n^2}\right)\le x_n e^{x_n/n^2}$. We immediately get $ x_n\le \exp\left\{\sum_{k \equal{} 1}^{n \minus{} 1}\frac {x_k}{k^2}\right\}$. Since $ x_n \equal{} o(n)$, the sum in the exponent is $ o(\log n)$ yielding $ x_n \equal{} o(n^{1/3})$. Now return to the recurrence relation as it was originally written and observe that the series $ \sum_n \frac {x_n}{n^2}$ is dominated by the series $ \sum_n n^{ \minus{} 4/3}$. The rest should be clear.

A follow-up question: Let $ f(a) \equal{} \lim x_n$ when $ x_1 \equal{} a$. We know that $ f(a)$ is well-defined for $ a < 1$. How fast does $ f(a)$ tend to infinity as $ a\to 1 \minus{}$? ;)

Why if the sum in the exponent is $ o(\log n)$ then $ x_{n} \equal{} o(n^{1/3})$? :)

Using computer I tried to find $ f(1 \minus{} 10^{ \minus{} x})$ for positive integer $ x$. I got:
$ f(1 \minus{} 10^{ \minus{} 1})\approx 15.84$
$ f(1 \minus{} 10^{ \minus{} 2})\approx 193.61$
$ f(1 \minus{} 10^{ \minus{} 3})\approx 1991.27$
$ f(1 \minus{} 10^{ \minus{} 4})\approx 19988.62$
$ f(1 \minus{} 10^{ \minus{} 5})\approx 199946.73$
$ f(1 \minus{} 10^{ \minus{} 6})\approx 1995992.46$
We can see that $ f(1 \minus{} 10^{ \minus{} x})\approx 2\cdot 10^{x}$. Or using the substitution $ a \equal{} 1 \minus{} 10^{ \minus{} x}$ we get: $ f(a)\approx \frac {2}{1 \minus{} a}$. I know that this is not exact result, but maybe not bad.

[quote="Yuriy Solovyov"]Why if the sum in the exponent is $ o(\log n)$ then $ x_{n} \equal{} o(n^{1/3})$? :)
[/quote]
$ e^{o(\log n)}\le e^{\frac 14\log n}\equal{}n^{1/4}\equal{}o(n^{1/3})$ for large $ n$. Of course, there is nothing sacred about $ \frac 13$ here; any number less than $ 1/2$ will work.

We can write: $ \frac {1}{x_n} \minus{} \frac {1}{x_{n \plus{} 1}} \equal{} \frac {x_{n \plus{} 1} \minus{} x_n}{x_n x_{n \plus{} 1}} \equal{} \frac {x_{n}^2}{n^2 x_n x_{n \plus{} 1}} \equal{} \frac {x_{n}}{n^2 x_{n \plus{} 1}} \equal{} \frac {1}{n^2 \plus{} x_n} < \frac {1}{n^2}$.
Or: $ \frac {1}{a} \minus{} \frac {1}{f(a)} < \sum_{n \equal{} 1}^{\infty}{\frac {1}{n^2}} \equal{} \zeta (2)$ or $ f(a) < \frac {a}{1 \minus{} a\cdot\zeta (2)}$, but this works only if $ a < \frac {1}{\zeta (2)}$ :( But we alsow can write: $ \frac {1}{a} \minus{} \frac {1}{f(a)} < \frac {1}{1^2 \plus{} a} \plus{} \zeta (2) \minus{} 1$. This gives $ a < \frac { \minus{} 1 \plus{} \sqrt {1 \plus{} \frac {4}{\zeta (2) \minus{} 1}}}{2}$ and this is stronger result then in first case. But this is general does not work...

In the other case: $ f(a) \equal{} \frac {a}{1 \minus{} a\cdot\sum_{n \equal{} 1}^{\infty}{\frac {1}{n^2 \plus{} x_{n}}}}$ and I'll correct my previous version of $ f(a)$: maybe it should be $ f(a)\approx\frac {2a}{1 \minus{} a}$ and this gives $ \sum_{n \equal{} 1}^{\infty}{\frac {1}{n^2 \plus{} x_{n}}}\approx\frac {1 \plus{} a}{2a}\to 1$ when $ a\to 1 \minus{}$.

Alsow when $ a\to 1 \minus{}$ then $ x_{n}\approx n$ and $ \sum_{n \equal{} 1}^{\infty}{\frac {1}{n^2 \plus{} x_{n}}}\approx\sum_{n \equal{} 1}^{\infty}{\frac {1}{n^2 \plus{} n}} \equal{} 1$. I'll think about approximation of the $ \sum_{n \equal{} 1}^{\infty}{\frac {1}{n^2 \plus{} x_{n}}}$.

Maybe this substitution helps: let $ x_{n}\equal{}n^2\cdot y_{n}$, then: $ (n\plus{}1)^2 \cdot y_{n\plus{}1}\equal{}n^2\cdot y_{n}\plus{}n^2\cdot y_{n}^2\equal{}n^2\cdot y_{n}(1\plus{}y_{n})$.


i was wondering who else was planning to bring their violin, viola etc to camp.

i've never brought a musical instrument to summer camp before because i always thought it was too cumbersome.   now i'm debating whether or not to bring my er hu or violin.....

anyone else have any thoughts?

Haha yes I just might bring my cello. :) I probably won't be that bored though. And I dunno what Continental will think about me dragging that thing alongside my carry-on.

But you should definitely bring a violin Susah. Practicing is a different story. Lots of people will be working on their piano skills though.

(Oh and btw, why arent you on AIM?)

There's always a talent show at the end of camp, so definitely bring your instrument if you think you'd be interested in playing!

I'm bringing my violin! I don't know how much I'm going to practice though... :?


geometry
perimeter
symmetry
inequalities
inequalities proposed

Let $ p $ be the perimeter of polygon $ A_1A_2...A_n $ . $ S $ is the are of polygon $ A_1A_2...A_n $ . Prove that : $ S \leq \frac{p^2}{4n.cot \frac{(n-2)\pi}{2n}} $

This follows from the fact that for n-gon with fixed perimeter, the regular one has the largest area.

Please post a solution for your idea , Wong .
  The folowing ineq is harder :For every polygon :  $  p^2 \geq 4\pi S $

[quote="nttu"]Please post a solution for your idea , Wong .
[/quote]
Well, I think this is a well-known fact. The idea of the most common proof is to join the segment $A_iA_{i+2}$ and maximise the area of $\triangle A_iA_{i+1}A_{i+2}$ when $A_{i+1}$ varies for fixed perimeter.

Harder???
From isoperimetric theorem, for closed curve with fixed perimeter, circle has the largest area. This follows since a circle with perimeter p has area $\frac{p^2}{4\pi}$.
Actually are you asking for a proof for isoperimetric theorem? For figure with only straight lines, it is pretty easy.
For figure "really" formed by curve, there is no simple elementary proof-the proof I've read before involved Calculus.

[quote="Diarmuid"]Actually, I'm pretty sure I have seen such an elementary proof - it didn't use anything much more advanced than symmetry considerations, and the fact that the anlge in a semi-circle is a right angle.  Maybe I've forgotten another piece...[/quote]
I've read that proof before-however, it has a very very small flaw.
Refer to the paragraph right after the proof of lemma 2 for more details in this [url=http://www.cut-the-knot.org/do_you_know/isoperimetric.shtml]link[/url].

Yeah, I think that's the one I had seen.

Nikolaos gave an elementary proof for this problem  ;)  I'm sure that both of you haven't seen it before

Nickolas, isn't this Blaschke's inequality?  ;)
I have a proof I think. :P


Sorry, the first inequality in question dosen't always hold true.
(To be honest, I was fairly certain that the first inequality was false when I first looked at it.)

Consider the following case:
a1=1, b1=0, c1=0
a2=1, b2=0, c2=0
x1=1, y1=1, c1=1
x2=2, y2=0, c2=0

It satisfies:
a1>=b1>=c1>=0  -->  1>=0>=0>=0
a2>=b2>=c2>=0  -->  1>=0>=0>=0
x1>=y1>=z1>=0  -->  1>=1>=1>=0
x2>=y2>=z2>=0  -->  2>=0>=0>=0

a1+a2+a3>=a2+b2+c2  -->  1+0+0>=1+0+0  -->  1>=1
x1+y2+z3>=x2+y2+z2   -->  1+1+1>=2+0+0  -->  3>=2

Yet,
(a1)(x1)+(b1)(y1)+(c1)(z1)>=(a2)(x2)+(b2)(y2)+(c2)(z2)
  (1)(1)  + (0)(1)  +  (0)(1)  >= (1)(2)  +  (0)(0)  +  (0)(0)
      1     +     0     +     0      >=     2     +      0     +     0
                                         1>=2
which is false.

I suspect (but I'm not betting money) that the other inequalities have similar fallacies.

(And yet, this seemed to be correct when I first looked at it...)

ok,thanks.draperp
I will think about it again :(

Sorry, it seemed like a great inequality to prove.  I didn't think it was false until awhile later.
I wonder if there's additional contraints that could be put in place to make it true...I haven't found any good ones yet...


[quote="Utkirstudios"]let a,b,c are real numbers then prove that:
\[ 7(a^{4}+b^{4}+c^{4})+10(a^{3}b+b^{3}c+c^{3}a) \ge 0 \]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112423


Let P and Q  be the two centers of two circles p and q. Define under which conditions is possible and construct (by euclidean tools only) the parallel line r to PQ s.t. the segment intersection of r, p and q be twice the distance PQ.

[quote="grobber"]What exactly do you mean by "segment intersection of r,p,q"? r cuts p,q in $4$ points, so which segment should we consider? Or maybe we can consider any of them?[/quote]

I mean the points which are in p^q^r, where p and q are the disks.

I hope is it clear now. If not I have to rephrase the claim.

It is clear $p\cap q\neq\varnothing$. Let $A$ and $B$ be points of intersection of $p$ and $q$. Then desired line $r$ is a parallel line to $PQ$ through $A$.  :)

I'm not so sure about that.. There are lots and lots of cases when the parallel through $A$ to $PQ$ won't even intersect $p\cap q$.

I think sprmnt21 means: $r\cap (p\cup q)$.
But you are right: there are a lot of cases. My solution assume $P$ outside of $q$ and $Q$ outside of $p$.

[quote="Myth"]I think sprmnt21 means: $r\cap (p\cup q)$.
[/quote]

No! I mean $r\cap (p\cap q)$.

[quote="sprmnt21"][quote="Myth"]In this case I think there is one possibility:
$P\in q$, $Q\in p$, $PQ\leq \frac{1}{3}(R_p+R_q)$.

Am I right?[/quote]

May be  :) .

I could answer more precisely if you explain HOW you get this result.[/quote]
It is just condition of existance of such chord $PQ$.

[quote="Myth"]

...

It is just condition of existance of such chord $PQ$.[/quote]

Ok! I see. But my question was: way if $PQ > \frac{1}{3}(R_p+R_q)$ there is not such a chord that solve the problem?

Note that greatest common chord of $p$ and $q$ (which parallel to $PQ$) is $PQ\cap (p\cap q)$.
If one of following conditions is false:
$P\in q$, $Q\in p$, $PQ\leq \frac{1}{3}(R_p+R_q)$,
then greatest common chord of $p$ and $q$ (which parallel to $PQ$) is less than $2PQ$.

Ok, now I can say: "you are rigth"!  :) .

furtermore, I can add that in this condition there is a really good hint for the solution of the problem.

If AB is such a chord s.t. AB=2PQ and M and N are the points of intersection between c(P) and c(Q) let G=MN^AB and let R and S be the (other) intersections of AB with c(P) and c(Q) respectively.
Prove that RA/AG = GB/BS.


Hello!
Do you know problems of CWMO2006?
If you know, please post them in English!
 I'll be very happy :lol:
How about rezults?

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=118085[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=118087[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=118086[/url]
......

You may find them in Olympiad Section.

Hong Kong had got the champion YEAH!

I've just found a PDF document. Unluckily it's in Chinese. Solutions were also posted.

我去参加了
题目挺简单的

[quote="modeler"][quote="ychjae"][quote="lovejrz"]我去参加了
题目挺简单的[/quote]
這樣你拿了多少分？
小弟是87/120 :P[/quote]
金牛別吵![/quote]

好像你也是個金牌得主吧...

[quote="ychjae"][/quote][quote="modeler"][quote="ychjae"][quote="lovejrz"]我去参加了
题目挺简单的[/quote]
這樣你拿了多少分？
小弟是87/120 :P[/quote]
金牛別吵![/quote][quote="ychjae"]

好像你也是個金牌得主吧...[/quote]

我也希望是這樣，可事與願違，差點銀也沒有。 :(

[quote="modeler"][quote="ychjae"]
好像你也是個金牌得主吧...[/quote]
我也希望是這樣，可事與願違，差點銀也沒有。 :([/quote]
我說的是中國東南地區數學奧林匹克 :lol:


AMC USA(J)MO USAMO inequalities function calculus

When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Please use LaTeX for posting solutions. Thanks.

Note, that we can assume $a+b+c=1$. Then $a,b,c\in (0,1)$ and our inequality appears this way:
$\frac{(a+1)^2}{2a^2+(1-a)^2}+\frac{(b+1)^2}{2b^2+(1-b)^2}+\frac{(c+1)^2}{2c^2+(1-c)^2}\leqslant 8$.
Let's put $f(x)=\frac{(x+1)^2}{2x^2+(1-x)^2}$ for $x\in (0,1)$. This function has the following properties:

1) Its derivative: $f'(x)=-\frac{4(x+1)(2x-1)}{(3x^2-2x+1)^2}$. We see that for $x\in (0,\frac{1}{2})$ function $f$ is increasing; for $x\in (\frac{1}{2},1)$ it is decreasing. Its maximum is reached for $x=\frac{1}{2}$ and equals $f_{max}=f(\frac{1}{2})=3$,

2) $f(\frac{1}{3})=\frac{8}{3}$, $f(\frac{1}{4})=2\frac{3}{11}<\frac{7}{3}$ and $f(\frac{1}{5})=2$,

3) By calculating the second derivative one may check that $f$ is convex in $(\frac{1}{5},\frac{1}{2})$.

Now let's see what follows from above. If at least one of the numbers $a,b,c$ does not exceed $\frac{1}{5}$, our ineq. obviously holds.
Assume $a,b,c>\frac{1}{5}$. If we have e.g. $a\geqslant \frac{1}{2}$, then $b,c<\frac{1}{3}$ and more - one of $b,c$, for example $b$ must obey $b\leqslant \frac{1}{4}$ - and then of course the ineq. holds.
Finely, if all three numbers are s.t. $a,b,c\in (\frac{1}{5},\frac{1}{2})$ - then use Jensen's ineq. to prove the given ineq.

I'd be happy to see some more interesting solution...

See also

http://www.mathlinks.ro/Forum/viewtopic.php?t=46
http://www.mathlinks.ro/Forum/viewtopic.php?t=28678
http://www.mathlinks.ro/Forum/viewtopic.php?t=43278
http://www.mathlinks.ro/Forum/viewtopic.php?t=48989

  darij

[quote="MithsApprentice"]Let $a, b, c$ be positive real numbers. Prove that
\[ \frac{(2a+b+c)^2}{2a^2+(b+c)^2} + \frac{(2b+c+a)^2}{2b^2+(c+a)^2} + \frac{(2c+a+b)^2}{2c^2+(a+b)^2} \le 8.  \][/quote]
$\frac{(2a+b+c)^2}{2a^2+(b+c)^2}\leq\frac{4}{3}\cdot\frac{4a+b+c}{a+b+c}\Leftrightarrow(2a-b-c)^2(5a+b+c)\geq0.$

[quote="dieuhuynh"]thank you [i]arqady[/i]  your solution very nice[/quote]
I am glad, that it was pleasant to you. :lol:

[quote="paladin8"][hide]
WLOG, assume $a+b+c = 3$.

Then the LHS becomes $\sum \frac{(a+3)^2}{2a^2+(3-a)^2} = \sum \frac{a^2+6a+9}{3a^2-6a+9} = \sum \left(\frac{1}{3}+\frac{8a+6}{3a^2-6a+9}\right)$.

Notice $3a^2-6a+9 = 3(a-1)^2+6 \ge 6$, so $\frac{8a+6}{3a^2-6a+9} \le \frac{8a+6}{6}$.

So $\sum \frac{(a+3)^2}{2a^2+(3-a)^2} \le \sum \left(\frac{1}{3}+\frac{8a+6}{6}\right) = 1+\frac{8(a+b+c)+18}{6} = 8$ as desired.[/hide][/quote]
Is it true if we similar to above as$a+b+c=3$or$1$ If it is true than I can do $a+b+c=k$ where $k$ is element of Real number
 Abdurashid

Yes, for any real number k we can set it to a+b+c (this follows from the fact that we can rescale the inequality). In other words, see what happens when you plug in ka, kb, and kc for a,b,c....

can you set a value to $abc$ too?, not that i'm saying you would in this problem.

if you could set up the same scaling type of argument on that term

[quote=quantumbyte]Can someone tell me why we are aloud to let $a+b+c=3$.[/quote]

It's because LHS is homogenous, the pair $(a,b,c)$ produces the same value as $(ak,bk,ck)$. Hence we can set $k = \frac{3}{a+b+c}$ to get $a+b+c=3$

[quote=eibc]The inequality is homogenous, so WLOG let $a + b + c = 3$. Then we wish to prove that $\sum_{\text{cyc}} \frac{(a + 3)^2}{2a^2 + (3 - a)^2} \le 8$.

Using the tangent line trick, we can prove the inequality
$$\frac{(a + 3)^2}{2a^2 + (3 - a)^2} \le \frac{4a + 4}{3} \iff 3(a - 1)^2(4a + 3) \ge 0,$$
which is clearly true. Summing cyclically over $a$, $b$, and $c$ gives the desired result.[/quote]

Yea ,is true ,moreover this posts:#15,#26,#41,#42,#45-#51,#54,#57,#58,#59,#62-#65

sorry, this is just a sketch because I wrote this in a few minutes in a private message to myself. $\sum_{cyc}\frac{(3+a)^2}{2a^2+(3-a)^2}\le 8$
$\frac{(3+a)^2}{2a^2+(3-a)^2}\le h(1)+(x-1)h'(1)=8/3+4(x-1)/3=4\frac{x+1}{3}$
f(x)=(3+a)^2, g(x)=3a^2-6a+9, h(x)=f(x)/g(x); h'(1)=4/3

Summing cyclically, LHS\leq4/3(6)=8, we can prove this is actually true by factoring

3-variety inequalities is no more important but yep

[quote=MithsApprentice]Let $ a$, $ b$, $ c$ be positive real numbers. Prove that
\[ \dfrac{(2a \plus{} b \plus{} c)^2}{2a^2 \plus{} (b \plus{} c)^2} \plus{} \dfrac{(2b \plus{} c \plus{} a)^2}{2b^2 \plus{} (c \plus{} a)^2} \plus{} \dfrac{(2c \plus{} a \plus{} b)^2}{2c^2 \plus{} (a \plus{} b)^2} \le 8.
\][/quote]
Let $a,\,b,\,c$ are real numbers such that $a+b+c \ne 0$ and $ab+bc+ca \geqslant 0.$ [url=https://artofproblemsolving.com/community/c6h2368407p19337233]Prove that[/url]
$$\frac{(2a+b+c)^2}{2a^2+(b+c)^2}+\frac{(2b+c+a)^2}{2b^2+(c+a)^2}+\frac{(2c+a+b)^2}{2c^2+(a+b)^2} \geqslant 4.$$
Let $ a$, $ b$, $ c$ be positive real numbers. [url=https://artofproblemsolving.com/community/c6h3097917p28009101]Prove that[/url]
$$ \dfrac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \dfrac{(2b + c + a)^2}{2b^2 + (c + a)^2} + \dfrac{(2c + a + b)^2}{2c^2 + (a + b)^2} \ge 12-\dfrac{(a+b)(b+c)(c+a)}{2abc}$$
[url=https://artofproblemsolving.com/community/c6h419560p2369204]h[/url] [url=https://artofproblemsolving.com/community/c6h2500980p21109224]h[/url] [url=https://artofproblemsolving.com/community/c6h3095901p28009246]h[/url] [url=https://artofproblemsolving.com/community/c6h2381045p19876507]h[/url]

Have made a generalization for this one :)

[b]Generalization 1[/b]
Let $a,b,c$ be positive reels. Then prove that

$$\sum_{cyc}{\dfrac{\left((k+1)a+kb+kc\right)^2}{(k+1)a^2+kb^2+kc^2}}\leq 6k+\dfrac{3}{2k+1}+1$$

As the fraction’s terms have the same powers, we can set $a+b+c=3$, so that the interval we’re working with is $(0, 3)$. It suffices to prove
\[\sum_{cyc}\frac{(3+a)^2}{2a^2+(3-a)^2}\le 8.\]
Note that $f(x)=\tfrac{(3+x)^2}{2x^2+(3-x)^2}$ is not concave (as in, 2nd derivative isn’t obviously negative), which calls for the tangent line trick (showing that $f(x)\le f(1) + f'(1)(x-1)$). Using some calculus and simplification, this turns into
\[\frac{x^2+6x+9}{x^2-2x+3}\le 4x+4\Longleftrightarrow 0\le (4x+3)(x-1)^2\]
which is clearly true. Since $3f(1)=8$, this proves the result. $\square$

Wait this is very natural.

$a+b+c=3$ since there's an absurd amount of symmetry.

\[\frac{a^2+6a+9}{2a^2+(3-a)^2}\]
and apply Tangent line trick to kill problem.

Let $a+b+c=3$
This becomes $\sum_{cyc} \frac{(a+3)^2}{2a^2+(3-a)^2}$
Let $f(a)=\frac{(a+3)^2}{2a^2+(3-a)^2}$
Then, $\frac{4x+4}{3} \ge f(x)$ because $\frac{4x+4}{3}-f(x) \ge 0$ because take the derivative to get that $0$ and $1$ are the local maximum/minimums, but at $0$ and $1$, the difference is positive, so the difference is always positive.
So $f(a)+f(b)+f(c) \le \frac{4(a+b+c)}{3}+3*\frac{4}{3} = 4+4 = 8$

[quote=MithsApprentice]Let $ a$, $ b$, $ c$ be positive real numbers. Prove that
\[ \dfrac{(2a \plus{} b \plus{} c)^2}{2a^2 \plus{} (b \plus{} c)^2} \plus{} \dfrac{(2b \plus{} c \plus{} a)^2}{2b^2 \plus{} (c \plus{} a)^2} \plus{} \dfrac{(2c \plus{} a \plus{} b)^2}{2c^2 \plus{} (a \plus{} b)^2} \le 8.
\][/quote]
Note that:
$$3-\frac{(2a+b+c)^2}{2a^2+(b+c)^2}=\frac{2(a-b-c)^2}{2a^2+(b+c)^2}$$
So we just need to prove:
$$\sum \frac{(b+c-a)^2}{2a^2+(b+c)^2} \ge \frac{1}{2}$$
But:
$$\sum \frac{(b+c-a)^2}{2a^2+(b+c)^2} \ge \frac{\sum (b+c-a)^2}{2(a^2+b^2+c^2)}$$
and:
$$\sum (b+c-a)^2 \ge \sum a^2$$
Equality only holds iff $a=b=c>0$


How many positive integers ( single and multidigit ) between $1$ and $123456$ have all different digits ( ie no digit repeated .... such as $559$ where $5$ is repeated ) ?

For example both $1$ and $123456$ have no digits repeated. $0$ is a digit.

?

As a related problem:

How many positive integers are there from 1 to 123456789 (inclusive) such that each integer has digits which are in ascending order?

[hide]
For each combination of digits, there is only one way to arange them in ascending order. The total number of such numbers is then: $\displaystyle{9\choose 2} +\displaystyle{9\choose 3} + \displaystyle{9\choose 4} + \displaystyle{9\choose 5} + \displaystyle{9\choose 6} + \displaystyle{9\choose 7} + \displaystyle{9\choose 8} + \displaystyle{9\choose 9} = 2^9 - 9 - 1 = 502$
[/hide]

EDIT: Oops, didn't see the solution being posted as I typed up mine. And I didn't include single digit numbers.

[quote="chesspro"][quote="4everwise"][quote="chesspro"]As a related problem:

How many positive integers are there from 1 to 123456789 such that each integer has digits which are in ascending order?[/quote]That's what I was thinking. You still have to mix around the numbers a couple of times though to cover all the numbers, right?

[hide="Answer to Extension"]$2^9$, each digit can be there or not be there[/hide][/quote]

Minus one, you can't have nothing there.  ;)[/quote]I consider all integers from 0 to 9 ascending.  :D

[quote="4everwise"][/quote][quote="chesspro"][quote="4everwise"][quote="chesspro"]As a related problem:

How many positive integers are there from 1 to 123456789 such that each integer has digits which are in ascending order?[/quote]That's what I was thinking. You still have to mix around the numbers a couple of times though to cover all the numbers, right?

[hide="Answer to Extension"]$2^9$, each digit can be there or not be there[/hide][/quote]

Minus one, you can't have nothing there.  ;)[/quote][quote="4everwise"]I consider all integers from 0 to 9 ascending.  :D[/quote]

The problem asks for positive integers, you can't count 0.  :P

[quote="xxxyyyy"][quote="Adunakhor"][quote="beta9"]9 one-digit numbers
9x9=81 two-digit numbers
9x9x8=648 three-digit
9x9x8x7= 4536 four-digit
9x9x8x7x6= 27216 five-digit
9x9x8x7x6x5 = 136080 six-digit

total = 168570[/quote]

The question is only asking for possible numbers in the range $1-123456$. The number of 6 digit numbers in the range that have all distinct digits must start with either $10$ or $12$. With the second case, there is only one number that works: $123456$. With the first case, the number is $8\times\times6\times5=1680$.[/quote]

But a six digit number starting with 12 can be 120345 ???

so I think the answer you have given is incorrect?[/quote]

Oops, you're right :P . For numbers starting with:

$120 \Longrightarrow 7\times 6\times 5$ numbers
$1230 \Longrightarrow 6\times 5$ numbers
$12340 \Longrightarrow 5$ numbers
$12345 \Longrightarrow 2$ numbers, $0$ or $6$

[quote="Adunakhor"][/quote][quote="xxxyyyy"][quote="Adunakhor"][quote="beta9"]9 one-digit numbers
9x9=81 two-digit numbers
9x9x8=648 three-digit
9x9x8x7= 4536 four-digit
9x9x8x7x6= 27216 five-digit
9x9x8x7x6x5 = 136080 six-digit

total = 168570[/quote]

The question is only asking for possible numbers in the range $1-123456$. The number of 6 digit numbers in the range that have all distinct digits must start with either $10$ or $12$. With the second case, there is only one number that works: $123456$. With the first case, the number is $8\times\times6\times5=1680$.[/quote]

But a six digit number starting with 12 can be 120345 ???

so I think the answer you have given is incorrect?[/quote][quote="Adunakhor"]

Oops, you're right :P . For numbers starting with:

$120 \Longrightarrow 7\times 6\times 5$ numbers
$1230 \Longrightarrow 6\times 5$ numbers
$12340 \Longrightarrow 5$ numbers
$12345 \Longrightarrow 2$ numbers, $0$ or $6$[/quote]

so the final answer would be ... ?

[quote="mathjay616"][hide]The final answer is 33599

First include all 5 (or less)-digit numbers, of which there 10*9*8*7*6 - 1, since 0 is not included

Then add all the 6 digit numbers, they are all of the type 10---- or 12----, so there are 2(8*7*6*5) more

10!/5! + 2*8!/4!-1 = [b]33599[/b][/hide][/quote]

You are including 6 digit numbers greater than 123456 in your counting. Also, $10\times 9\times 8 \times 5\times 6 -1$ will not include 3, 2, or 1 digit numbers because those will require more than $1$ zero. So zero itself would not be counted either... To answer xxxyyyy's question, the final answer would just be the sum of all different subcases, which is $\boxed{34417}$ (hopefully didn't miscount  :P )

[quote="DanK"]I got $\boxed{34418}$...  Did you forget to include 123456 maybe?[/quote]

I included 2 cases for numbers that began with 12345 - 123450 and 123456.

So 247 numbers that begin with 12, 1680 that begin with 10, and 32490 numbers with 4 or less digits, all distinct. Adding them all up I get $34417$. Unless I missed a different number?  :?

[quote="Adunakhor"]

You are including 6 digit numbers greater than 123456 in your counting. Also, $10\times 9\times 8 \times 5\times 6 -1$ will not include 3, 2, or 1 digit numbers because those will require more than $1$ zero. So zero itself would not be counted either... To answer xxxyyyy's question, the final answer would just be the sum of all different subcases, which is $\boxed{34417}$ (hopefully didn't miscount  :P )[/quote]

Thanks, now I see my mistakes  :oops:


Source: Romania 1999
7.1 Determine the side lengths of a right trianlge if they are intgers and the product of the leg lengths is equal to three times the perimeter.

[hide="Solution"]Use $a$, $b$, and $c$.  Clearly, $c=\sqrt{a^{2}+b^{2}}$.
Set up the equation $ab=3a+3b+3\sqrt{a^{2}+b^{2}}$.  Solve for the radical and we get $3\sqrt{a^{2}+b^{2}}=ab-3a-3b$.  Square both sides to get $a^{2}b^{2}-6a^{2}b-6ab^{2}+9a^{2}+18ab+9b^{2}=9a^{2}+9b^{2}$, or $a^{2}b^{2}-6a^{2}b-6ab^{2}+18ab=0$.
Factor to get $ab(ab-6a-6b+18)=0$.  Clearly, $a$ and $b$ must be positive, so $ab\neq0$.  Divide by it to get $ab-6a-6b+18=0$.  Add 18 to both sides and factor to get $(a-6)(b-6)=18$.
Since $a,b\in\mathbb{Z}$, $a-6$ and $b-6$ are integers as well.  We have the following possibilities for $a-6$ and $b-6$:
$\pm1\text{ and }\pm18$
$\pm2\text{ and }\pm9$
$\pm3\text{ and }\pm6$
No negative solution works, because if we have -6, -9, or -18, either $a$ or $b$ will not be a positive integer.
Trying out the three that remain, we get $\boxed{7,24,25}$, $\boxed{8,15,17}$, and $\boxed{9,12,15}$.  And guess what: THEY ALL WORK![/hide]

[hide="my way"]
in other words we have:
$2\triangle = 6s$
$\triangle = 3s$
$rs = 3s$
$r = 3$.

Now draw the diagram, since $r = 3$, and tangent lines from the same point are congruent, we deduce that $a+b-6 = c = \sqrt{a^{2}+b^{2}}$.  Squaring and using Simon's favorite trick we get $(a-6)(b-6) = 18$ and get the sides lengths:
$(9, 12, 15)$
$(7, 24, 25)$
$(8, 15, 17)$
[/hide]


algebra
polynomial
function
superior algebra
superior algebra unsolved

A truly magnificient problem: suppose that the field $K$ is not algebraically closed. prove that one can find $n$ and a homogenous polynomial of degree $n$ and in $n$ variables such that the equation $f(x_1,...,x_n)=0$ has only a trivial solution.

Is there anything wrong with this? :?

Let $x^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0\in K[x],\ n\ge 2$ be an irreducible polynomial over $K$, and let $\alpha_1,\ldots,\alpha_n$ be its roots in an algebraic closure $\overline K$ of $K$ (they may not be distinct if $K$ is not a perfect field, but that doesn't matter). Now take $f(x_1,\ldots,x_n)$ to be $\prod_{i=1}^n(x_1+x_2\alpha_i+\ldots+x_n\alpha_i^{n-1})$.

Clearly, the coefficient of every monomial in the $x_j$'s is a symmetric function of the roots $\alpha_i$, and hence an element of $K$. The only solution to $f=0$ is the trivial one because for each $i,\ (1,\alpha_i,\ldots,\alpha_i^{n-1})$ for a linearly independent system over $K$.

Yes, it's ok, only they wanted an absolutely elementary proof, with no algebraic closure.  :)

Well, any splitting field suffices... what exactly did they allow/want¿


superior algebra
superior algebra unsolved

As a result of the discussion on the Burnside problem, I decided to look into the maximal length of of words with no $ k$-th order repetition.

This approach fails in the most spectacular way possible, for over an alphabet of just three letters, there exist infinite words with no contiguous block repeated twice in a row.

I was wondering if anybody had insight into how to prove this.  There seem to be a lot of more advanced results along these lines, (for example, that the number of squarefree words of given length grows exponentially) but this seems like it should be such an elementary construction...


trigonometry
geometry
geometric transformation
reflection
quadratics
inequalities

In triangle ABC   cotA + cotB + cotC <2
Prove the greatest angle <90 degrees, and cot(least angle) < 4/3

[quote]In $\Delta ABC: \cot{A} + \cot{B} + \cot{C} < 2$
Prove that the greatest angle is smaller than $\frac{\pi}{2}$[/quote]

We say that $\angle A \geq \angle B \geq \angle C$.

[b][u]If $\angle A = \frac{\pi}{2}$:[/u][/b]

$\cot{A}+\cot{B}+\cot{C} = 0 + \cot{B} + \cot\displaystyle\left(\frac{\pi}{2}-B\right) = \cot{B}+\tan{B}$

Since $B, C \in [0, \frac{\pi}{2}[$, $\cot{B}, \cot{C} > 0$, we are allowed to apply $AM-GM$:
\[ \frac{\cot{B}+\tan{B}}{2} \geq \sqrt{\cot{B}.\tan{B}}  \]
\[ \cot{A}+\cot{B}+\cot{C}=\cot{B}+\tan{B} \geq 2  \]
Contradiction

[b][u]If $\angle A > \frac{\pi}{2}$:[/u][/b]

[u]Lemma 1: [/u] If two sharp angles (meaning contained in $]0,\frac{\pi}{2}[$) sum up to $y$, then \[ 2\cot{\frac{y}{2}} \leq \cot{z} + \cot{(y-z)}  \]
Proof: Look at picture 1 of a unit circle. Two angles sum up to $2 \angle FAD$

$\angle FAC + \angle FAB = 2 \angle FAD$

We will prove that $EB+EC > 2 ED$.

It is trivial that $AB < AC$, and because $AD$ is the angle bissector of $\angle BAC$, it is true that $BD < CD$.

\[ \cot{FAB} + \cot{FAC} = EB + EC  \]
\[ = (ED - BD) + (ED + CD)  \]
\[ = 2 ED + (CD - BD) > 2 ED = 2\cot{FAD}  \]

So we have proven Lemma 1.

Now, We draw again a unit circle (picture 2)

We reflect point A with respect to the y-axis. We call the reflection A'
Now, we see that $\cot{BCA} = -\cot{BCA'}$

$\angle BCD = \frac{\pi - \angle BCA}{2}$, so we notice that $2 \angle BCD = \angle BCA'$.

Lemma 2: We will show that for angles in the interval $]0,\frac{\pi}{4}[$, the following is true:
\[ \cot{2a} + 2 < 2\cot{a}  \]
We use the t-formula's $(t=\tan{a})$:
\[ \iff \frac{1-t^2}{2t} + 2 < \frac{2}{t}  \]
Since we are working in the interval $]0,\frac{\pi}{4}[$, we know that $t > 0$, so we can multiply:
\[ \iff -t^2+4t-3<0  \]
This quadratic is always negative, except when $t \in [1,3]$
But we knew that $t \in ]0,1[$, so it is true that $\cot{2a} + 2 < 2\cot{a}$

$\frac{\pi}{2}> \angle BCA' = 2 \angle BCD > 0$
So $\angle BCD \in ]0,\frac{\pi}{4}[$ and we can use Lemma 2.

So finally we get our result by using lemma 1 and 2: \[ \cot{A}+\cot{B} + \cot{C} \geq \cot{A} + 2\cot\displaystyle\left(\frac{\pi - \angle BCA}{2}\right)  \]
\[ > \cot{A} - \cot{(\pi-A)} + 2 = 2  \]

Contradiction, so it must be true that if \[ \Delta ABC: \cot{A} + \cot{B} + \cot{C} < 2  \]
it is true that $\angle A < \frac{\pi}{2}$, QED and we are done!

Wasn't that a nightmare?  :D

i know the problem wasn't intended to be solved this way, but I thought it would be fun to do so...
Other, (more conventional, elegant or short) solutions?  :)

[quote="Jan"][u]Lemma 1: [/u] If two angles sum up to $y$, then \[ 2\cot{\frac{y}{2}} \leq \cot{z} + \cot{(y-z)}  \]
Proof:[/quote]No, I think that's true for acute angles only. Reason for this is that $\cot$ is convex only in $]0,\frac\pi2[$. It doesn't flaw your proof though, as you only use it on acute angles.

[quote]No, I think that's true for acute angles only. Reason for this is that $\cot$ is convex only in $]0,\frac{\pi}{2}[$. It doesn't flaw your proof though, as you only use it on acute angles.[/quote]
Yes, of course you're right (cfr. my picture: We can only say that $AB < AC$ if they are on the same side of the y-axis. Note that if they are on the left side, they wouldn't be angles of a triangle anymore.)
Thanks for correcting me!

I'll edit it, and I will change 'goniometric circle', appearantly it's not frequently used in English  :D

As for the 'sequel':

[quote]$\Delta ABC: \cot{A} + \cot{B} + \cot{C} < 2$
Prove that $\cot{C} < \frac{4}{3}$[/quote]

To prove this, let's take a look at the case $\cot{C} \geq \frac{4}{3}$:

Lemma 1 (see first part of the question) tells us that:
\[ \cot{A} + \cot{B} + \cot{C} \geq 2 \cot{\frac{\pi - C}{2} + \cot{C} (1) }\]

By the use of the t-formula's $(t = \tan{\frac{C}{2}} )$:
In this case, $\cot{C} = \frac{1-t^2}{2t} \geq \frac{4}{3}$
So we get that $-3 \leq t \leq \frac{1}{3}$.

Now, we will know prove that $2 \cot{\frac{\pi - C}{2} + \cot{C} \geq 2}$:

\[ 2 \cot{\frac{\pi - C}{2} + \cot{C} = 2 \tan{\frac{C}{2}} + \cot{C} }\]
Now by the t-formula's $(t = \tan{\frac{C}{2}} )$:
\[ 2t + \frac{1-t^2}{2t} \geq 2 \]
Since $t > 0$:
\[ \iff 4t^2 + 1 - t^2 \geq 4t \iff 3t^2-4t+1 \geq 0 \]
\[ \iff t \leq \frac{1}{3} \vee t \geq 1 \]

We already know that $-3 \leq t \leq \frac{1}{3}$, so it is true that $2 \cot{\frac{\pi - C}{2} + \cot{C} \geq 2}$  $(2)$.

Combine $(1)$ and $(2)$ and get: \[ \cot{A} + \cot{B} + \cot{C} \geq 2 \]

Contradiction, so it must be true that $\cot{C} < \frac{4}{3}$!

Jan, maybe you should put such long posts between [hide] tags, that makes the pages much easier to scroll. ;)

An extension: [well, at least in some sense an extention]
In any triangle ABC we have $\cot(A)+\cot(B)+\cot(C)\ge\sqrt{3}$.

:D I know this is not the proof you had in mind:

[hide]Again, say $\angle A \geq \angle B \geq \angle C$

In the first proof, we already showed that if $\angle A \geq \frac{\pi}{2}$, then \[ \cot(A)+\cot(B)+\cot(C)\ge 2 > \sqrt{3}  \]

So it's enough to show that in an acute angled triangle, it is true that $\cot(A)+\cot(B)+\cot(C)\ge\sqrt{3}$

By lemma 1, we know that $\cot{A}+\cot{B}+\cot{C} \ge 2 \cot{\frac{\pi - C}{2}} + \cot C$.
We will show that $2 \cot{\frac{\pi - C}{2}} + \cot C \ge 3 \cot{\frac{\pi}{3}}$ (Lemma 3) :

Look at picture 3 of a unit circle:

$\angle ABC = \frac{\pi}{3}$

$2 \angle CBD = \angle EBC$

So we see that $3 \angle ABC = 2 \angle ABD + \angle ABE = \pi$.

We try to show that $2 \cot{\frac{\pi - C}{2}} + \cot C \ge 3 \cot{\frac{\pi}{3}}$
In other words: \[ 2 FD + FE \ge 3 FC  \]
Which is equivalent to \[ 2 CD \le CE  \]

Use the sine law in triangles $BEC$ and $BCD$:

\[ \frac{CD}{\sin{CBD}} = \frac{BD}{\sin{\frac{\pi}{3}}}  \]

and \[ \frac{CE}{\sin{2 CBD}} = \frac{BD}{\sin{\frac{2 \pi}{3}}}  \]
So get \[ CD = \frac{2 BD.\sin{CBD}}{\sqrt{3}}  \]
and \[ CE = \frac{2 BD.\sin{2 CBD}}{\sqrt{3}} = \frac{4 BD.\sin{CBD}\cos{CBD}}{\sqrt{3}}  \]

So \[ CE = 2 CD.\cos{CBD}  \]
And finally \[ CE \geq 2 CD  \]
We proved Lemma 3!

\[ \cot{A}+\cot{B}+\cot{C} \ge 2 \cot{\frac{\pi - C}{2}} + \cot C \ge 3 \cot{\frac{\pi}{3}} = \sqrt{3}  \]

QED! [/hide]
Comments welcome

[quote="Jan"]So it's enough to show that in an acute angled triangle, it is true that $\cot(A)+\cot(B)+\cot(C)\ge\sqrt{3}$[/quote]For the acute-angled case, we have by Jensen (on $f(x)=\cot x$) that $\cot(A)+\cot(B)+\cot(C)\ge3\cot\left(\frac{A+B+C}{3}\right)=\sqrt{3}$ ;)


Support
Stanford
college
Ross Mathematics Program
ratio

Polls just closed in SC.  Post preliminary results from your state here, and let's see if the Republicans can maintain control, or if the Democrats will take over!

In SC, the big race is for Governor:
Tommy Moore is a Democrat, but a lot of Republicans (including myself) support him over Mark Sanford, incumbent Republican, who was named one of the 3 worst governors in America by TIME Magazine.

Preliminary Results:

SC is a REPUBLICAN state.  Yet Democrats are leading every race in SC save 2, but the governor's race is going back and forth, it's about 50-50 right now.

How are you a Republican if you aren't old enough to vote yet?

Can you explain how that's a big race? It's been a republican stronghold this entire election season as the incumbent has mantained a lead of over 20% in the polls.

Anyway, polls don't close until 8 here so not results yet.
But my state has the governor, a senator, and five seats up for reelection.

The governor is a shoo in republican, no big deal there.

The senate race is closely contested between an Independent and a Democrat. While the Independent seems to have a shoo-in win, depending the source, polls have been fluctuating a lot so there's not quite consensus yet.

Out of the five house seats up for reelection, two of them have Democrat incumbents and three of them have Republican incumbents.
For both of the Democrat incumbent races, the Democrat has a lead of over 70%.
All three of the Republican incumbent house seats, are rated by the NY times as tossups and considering there are only 17 toss up seats, we have a very close election here for all 3 of those districts. Only two states have more than 1 tossup state and that is us, and Florida and we have about 1/5th the number of House seats as Florida.

What's happenning in Va is indeed very interesting. On one end, you have a racist, George Allen, and the other hand, a mysogynist. The race is too close to call. Who do you think will win?

Being very economically conservative, I definitely am pulling for George Allen in that race, but it looks like he won't win as of now.

Umm.. no it isn't.

Allen can still request a recount because of the margins in Virginia law and the counties still have to do a recanvassing.

Also, the reason Lieberman lost the primary is because he's too conservative on a lot of social issues so counting Lieberman as a Democrat is misleading because he could easily vote with the Republicans.

[quote="Ignite168"]How are you a Republican if you aren't old enough to vote yet?[/quote]

In the United States, unlike other countries, there are no qualifications or requirements to join a political party.  Simply put:  If you claim you are a member, then you are a member.

Also, I believe mdk's use of the word "almost" was meant to address your point.

[quote="mysmartmouth"]Considering that Lieberman is a Democrat listed as an Independent, Democrats pretty much already controlled the Senate.

(Lieberman lost the Democratic primary, so he ran as an independent, and somehow won the overall election.)[/quote]

Well Liebermann is probably the least-democratic democrat ever, he supports the war like alot. That's why Bush went up there to support him, cause he knew the republican candidate had no chance of winning.

Anyway MA results for governor:

Deval Patrick (D) 55%
Kerry Healey (R) 36%
Christy Mihos (I) 7%
Grace Ross (Rainbow Party) 2%

2nd black governor!

And... the senate

Ted Kennedy (6th reelection) (D) 70%
Some random dude with a 5-letter last name (R) 30%

[quote="13375P34K43V312"][quote="mysmartmouth"]Considering that Lieberman is a Democrat listed as an Independent, Democrats pretty much already controlled the Senate.

(Lieberman lost the Democratic primary, so he ran as an independent, and somehow won the overall election.)[/quote]

Well Liebermann is probably the least-democratic democrat ever, he supports the war like alot. That's why Bush went up there to support him, cause he knew the republican candidate had no chance of winning.[/quote]

Please take comedy to the Round Table, thanks.

On another note, RUMSFIELD RESIGNED! WOOHOO! He kept making shitty decisions that Bush got blamed for.

I know abbreviation is considered sacred in the USA, but is it possible to write the names of the states in full at least once? :(

Anyway, I would just like to say that you guys have a BAD system in the USA, I'd dare to call it undemocratic.

As far as I understand it, your House of Representatives consists of people representing one congressional district within their own state.  There is exactly one member per district.
So in theory, it's possible to have a completely democratic house of representatives even though 49% of the entire USA population voted democratic.

In France they have a somewhat similar, where the National Assembly has 577 members, each representing a different constituency...but there's a two round system : when no candidate obtained more than 50% of the votes, the two best scoring candidates face each other in a second round.

The system can be considered just as bad : the "Presidential Majority", a union of right wing parties obtained about 43% of all votes, yet they have 399 seats of all 577 seats! :ninja:   That gives them an absolute majority in the National Assembly even though they don't even represent half of the population.

I think one candidate per constituency is good, either you split the country (or state in the USA case) into several electoral districts (yeah I know, some USA states only have a representative)... or just one.

The system in the Netherlands works about that way (Belgium itself is way too complicated with its double federal structure and seven parliaments) :
all parliaments presents their lists for the Second Chamber (comparable with your House of Representatives). Each party gets the number of seats corresponding to its own ratio of votes (approximately).  Within the party there is a system to calculate who should take those seats.

The government has to consists of a group of parties whose sum of the number of seats exceeds 75.

[quote="fredbel6"]As far as I understand it, your House of Representatives consists of people representing one congressional district within their own state.  There is exactly one member per district.
So in theory, it's possible to have a completely democratic house of representatives even though 49% of the entire USA population voted democratic.
[/quote]

The US is NOT a democracy. We are a republic. So yes, our elections are democratic within their own respective boundaries but the nation is governed by individuals who represent certain portions of the populace.

Your theory is correct and a rather large problem. We call it gerrymandering when congressional districts are manipulated to favor one party over another. In fact, the Republicans cut their losses this time around because they were able to gerrymander so many districts during their reign over Congress. We'll probably be seeing a reversal of that if the Democrats keep Congress for a while.

I'm sure many people (including Republicans like myself) wanted the Democrats to take a house. And thank goodness they did. Over the past dozen years Republicans have all but forgotten their Contract with America. They've begun to neglect the pragmatic and moral issues that make parties successful and abuse the trust of the American people. We've seen that with campaign finance scandals, porkbarrels, pedophiles...the list goes on. This isn't what the Republican party represents, and it certainly isn't what they campaign under. Nor it is what they will win by.

[quote="MithsApprentice"][quote="fredbel6"]As far as I understand it, your House of Representatives consists of people representing one congressional district within their own state.  There is exactly one member per district.
So in theory, it's possible to have a completely democratic house of representatives even though 49% of the entire USA population voted democratic.
[/quote]

The US is NOT a democracy. We are a republic. So yes, our elections are democratic within their own respective boundaries but the nation is governed by individuals who represent certain portions of the populace.[/quote]
I was distracted , I meant : a completely republican house with 49% of the population voting for the democrats.

Ignite, you were right. :wink:

[quote]Your theory is correct and a rather large problem. We call it gerrymandering when congressional districts are manipulated to favor one party over another. In fact, the Republicans cut their losses this time around because they were able to gerrymander so many districts during their reign over Congress. We'll probably be seeing a reversal of that if the Democrats keep Congress for a while.[/quote]
Who gets to decide these districts?  The state governor?

The Turkish system is pretty bad too, they have hundreds of parliament members but the seats are only given to parties scoring at least 10% of ALL votes, so there are only two parties with seats. :huh:  (very convenient if you wanna keep minorities out)

[quote]I'm sure many people (including Republicans like myself) wanted the Democrats to take a house. And thank goodness they did. Over the past dozen years Republicans have all but forgotten their Contract with America. They've begun to neglect the pragmatic and moral issues that make parties successful and abuse the trust of the American people. We've seen that with campaign finance scandals, porkbarrels, pedophiles...the list goes on. This isn't what the Republican party represents, and it certainly isn't what they campaign under. Nor it is what they will win by.[/quote]
Well I can say that in many European countries, the Republican government (the Bush administration) is making a pretty poor or shall we even say... catastrophic impression. :|

[quote="fredbel6"]Who gets to decide these districts?[/quote]
It's usually the state legislatures. The process is normally done once every ten years (the same frequency as the official census), and there are civil rights laws regulating it. These laws have prevented racial gerrymandering, but not districts designed to be "safe" for one party.
The last time California districts were adjusted, both parties cooperated to make nearly every seat safe for the party that held it at the time. This was extremely effective.

[quote="Xevarion"][quote="Ignite168"][quote]As far as I understand it, your House of Representatives consists of people representing one congressional district within their own state. There is exactly one member per district.
So in theory, it's possible to have a completely democratic house of representatives even though 49% of the entire USA population voted democratic. [/quote]

Please explain a case where that could happen because for a completely Democratic House, we would need $>50\%$[/quote]
Why shouldn't it be possible with less than 50%? Remember we have other parties. Every Democrat could win with less than 50% because he was still more than the Republican and the remaining votes went to independents/third parties. Suppose, for example, that a conservative independent ran in every district and drew votes from the Republican candidates.

So this is even worse than what fredbel was imagining.[/quote]

But that could happen in every place couldn't it? Also, many states have a rule that if no canindate has a majority, there is a runoff between the top two canindates. In fact, someone could win with under 1% if there are over a hundred parties that evenly drew votes, but even if this happened, the Democrats' actions would still be serving the MOST people over anybody else because they received a plurarilty of the vote.

By Arrow's theorem, every democratic election scheme has similar potential failures.  And the vast majority of the time, those flaws end up not mattering because people aren't distributed so that 51% of the population in every single district leans the same way.
(In fact, one criticism of the American electoral system with respect to the House of Representatives is that it is supposed to be the branch most immediately responsive to the will of the people, but the power of incumbency seems to have overwhelmed this fact in recent years, so that a representative running for re-election hardly ever loses.  One suggested correction for this is to intentionally make districts which break down at close to 50%-50%, so they will be more competitive.  The obvious downside to this, of course, is that almost 50% of voters in such a district will end up with a representative they don't want, while in uncompetitve districts like my own, more than 80% of voters agree enough to vote for their representative.)

How does the Belgian election system work?

[quote="JBL"]
How does the Belgian election system work?[/quote]
You really wanna know? :|  :o Sure?

Well there are 150 federal parliament members, elected every four years unless the government falls. Belgium is divided into 11 electoral districts, corresponding to the ten provinces...but there is a complicated electoral district around the bilingual capital Brussels.  Normally each electoral district has a fixed number of seats, and people cannot votre for candidates in others, and The d'Hondt method is used to distribute the seats within each district. I do think there is a 5% threshold.
But Brussels is part of a weird electoral district Brussels-Halle-Vilvoorde, the rest of which is in completely Flemish territory.  I don't even completely understand the situation, but as I get it, its 22 seats are split according to language, and the Flemish votes are combined with those of the rest of Flemish Brabant and the Francophone votes with those of Walloon Brabant.
This electoral district is intensely disliked by Flemish politicians as it crosses linguistic borders and gives francophones rights in Flanders that Flemings in Wallonia don't have.  It's considered unconstitutional by a court of arbitration but the Frenchspeaking politicians refuse to solve the matter.  It's been postponed until the federal elections of 2007.  I wonder whether or not they will do it.  If they don't, it's possible that it's the beginning of the end of this nation.

And that's just the federal level.  Belgium has invented some sort of double federalism : territorial and "people" autonomy.
Each five years the three regions (Flanders, Brussels and Wallonia) elect their own parliaments.  In Brussels, 89 seats are to be distributed, In Flanders 118 and in Wallonia 75.
However, those who voted for a Dutchspeakingparty in Brussels, can also fill in their choice for the Flemish Council.  If they didn't vote for a Dutchspeaking party, that means they didn't declare themselves to be a Fleming and so they can't have a say. (So one chooses what he is everytime he votes in Brussels) Based on those votes for the Flemish Council, six Flemish politicians in Brussels, together with the 118 members of the Flemish Region, form the Flemish Council.
Those six can only participate in votes about "us", the Flemings (our public television, our education, culture,...), they can't vote on territorial matters in Flanders.
The Flemish Parliament is in Brussels, quite close to the Federal Parliament.  When the 118 members of the Flemish Region vote there, they are deciding on other territory.  The absurdity of making a city capital of a region it's not part of is typically Belgian.

In Flanders they use electoral districts as well ,and this time neatly corresponding with the five Flemish provinces and there is a fixed number of seats for each province.  Again d'Hondt is used to distribute the seats, and I think that there is a 5% threshold here as well.  Personally I think that is silly.  I will care whether or not the candidate I vote for is Dutchspeaking or not, but I don't care what province within Flanders he is from.

At the same time, those Frenchspeaking members of the Brussels Parliament choose 19 members, who join up with the 75 members of the Walloon Parliament to form the French Community.
The Walloon Parliament itself does not meet in Brussels, but in Namur, in Wallonia itself.

The people of the small Germanspeaking community (consequence of the First World War, only 70,000 of them) only have people autonomy, no territorial autonomy, and they elect a parliament too.

I told you Belgium is more confusing! :ninja:

Many people say it's not a federation, or a confederation either, but a "contra-federation" instead, and I think they are right.


Im struggling to solve this problem.anyone please help.

Paint costing $19/L is to mixed with 5L of a $3/L thinner to make a paint that can be sold at $14/L. How much paint will be produced?

Let x = be number of paint

I cant seem to figure out an equation for it.

Let $ N$ be the number of liters of paint initially mixed in.

So, Value=Price/Quantity.
$ 14\equal{}\frac{3\cdot5\plus{}19N}{N\plus{}5}$

$ 14N\plus{}70\equal{}15\plus{}19N$

$ 55\equal{}5N$

$ N\equal{}11$

So, we initially use $ 11$ liters of paint to produce $ 16$ liters of mixed paint.

Is that what you were asking?  :)

where did you get 3.5 from?

think the question was asking how much paint i need to be mixed with the thinner to get $ 14/L.
think should be like this:
Cost = equation 1
litre = equation 2
Cost at$14/L = 14(equation 2)

So we have the equation

equation 1 = equation 2
and solve it for X

Don't LaTeXify your text.

Because when we find the price to add, the price of the thinner is 3 dollars per liter, and you get 5 liters of it. Thus, 3 times 5.

Also, we are NOT trying to find cost=liter number. (If that's the case, The cost would be 1 dollar per liter, because the numbers cancel out). We want to find a set cost for one liter, so we divide cost by the number of dollars by the number of liters, to get the number of dollars for ONE liter.

think the question was asking how much paint i need to be mixed with the thinner to get $14/L.
think should be like this:
Cost = equation 1
litre = equation 2
Cost at $14/L = 14(equation 2)

So we have the equation

equation 1 = equation 2
and solve it for X

No, Textangle is right. Total price equals total cost divided by total amount. That's the equation he used.


Given $ \sqrt[6]{\sqrt[8]{\sqrt[4]{7^3}}}\equal{}7^x$, find $ x$. Express your answer as a common fraction.

You can change the radical signs to fractional exponents:

$ \sqrt[6]{\sqrt[8]{\sqrt[4]{7^{3}}}}\equal{}(((7^3)^\frac14)^\frac18)^\frac16\equal{}7^\frac{3}{4\cdot8\cdot6}\equal{}7^\frac{1}{64}$

So we know:

$ x\equal{}\frac1{64}$


Does anyone know how to do this question? Please help.

[quote="Sandie18"]Suppose $A \in R^{n \times m}$ has singular values $\sigma_1 \geq..\geq \sigma_r > 0$ and $0 < r \leq min(n,m)$

Find the non-zero eigenvalues of $A^TA$[/quote]Take a look at the proof for the singular value decomposition of $A$. It uses the matrix $A^TA$ at a crucial point. That should give you the answer.


How am I supposed to change the time given on the forum?   :huh:

To change the time, go to Profile--->Preferences--->Internationalisation

You can change the time zone there to whatever you wish.

Thanks, I've been wondering how to do that as well.

Thanks, a mod can now lock this, I guess.


polynomial
algebra
search
superior algebra
superior algebra unsolved

Is $ x^3 \plus{} 17x \plus{} 36$ an Eisenstein translate?

A monic polynomial can only be a Eisenstein translate (by the way, I never heard this term before and needed to take som search; please define it next time) of those primes that divides it discriminant. Thus you have to check this finite number of primes.

PS: A Eisenstein translate is a polynomial $ f(x)$ such that there exists some $ b$ with the property that $ f(x\plus{}b)$ can be shown to be irreducible by Eisensteins criterion.

Moreover, in the case $ f(x)\equal{}x^3\plus{}Ax\plus{}B$, one can check that $ f$ is an Eisenstein translate at $ p$ iff:
$ p\neq 3$ and $ p|A$, $ p\parallel{}B$, or
$ p\equal{}3$ and $ p|A$, $ B\not\equiv 0,\pm1\mod 9$.

Thanks for the hints. Sorry about the notation, I've never experienced before that I used a term not known


Show that for all positive integers $m$ and $n$, \[\gcd(m, n) = m+n-mn+2\sum^{m-1}_{k=0}\left \lfloor \frac{kn}{m}\right \rfloor.\]

[quote="Peter"]Show that for all positive integers $ m$ and $ n$,
\[ \gcd(m, n) \equal{} m \plus{} n \minus{} mn \plus{} 2\sum^{m \minus{} 1}_{k \equal{} 0}\left \lfloor \frac {kn}{m}\right \rfloor.
\]
[/quote]
This problem must be :
$ d \equal{} \gcd(m,n)$ so
$ [\frac {kn}{m}] \equal{} [\frac {ka}{b}]$
But $ \{ka\}_{k \equal{} 0}^{b \minus{} 1}$ is an complete reside mod $ b$ then
$ \sum_{k \equal{} 0}^{b \minus{} 1}\{\frac {ka}{b}\} \equal{} \frac {b(b \minus{} 1)}{2b}$
So $ \sum_{k \equal{} 0}^{b \minus{} 1}[\frac {ka}{b}] \equal{} \frac {(a \minus{} 1)(b \minus{} 1)}{2}$
$ \Rightarrow \sum_{k \equal{} 1}^{m \minus{} 1}[\frac{ka}{b}] \equal{} d(a \minus{} 1)(b \minus{} 1) \equal{} d \plus{} mn \minus{} m \minus{} n$
So the equality is true.

[quote="TTsphn"]So $ \sum_{k \equal{} 0}^{b \minus{} 1}[\frac {ka}{b}] \equal{} \frac {(a \minus{} 1)(b \minus{} 1)}{2}$
$ \Rightarrow \sum_{k \equal{} 1}^{m \minus{} 1}[\frac {ka}{b}] \equal{} d(a \minus{} 1)(b \minus{} 1) \equal{} d \plus{} mn \minus{} m \minus{} n$
So the equality is true.[/quote]
Exactly $ 2\sum_{k\equal{}1}^{m\minus{}1}[\frac{kn}{m}]\equal{}n(m\minus{}1)\minus{}d(b\minus{}1)\equal{}d\plus{}nm\minus{}n\minus{}m.$

Please see http://www.mathlinks.ro/Forum/viewtopic.php?t=59747 .


How does qualification for people 10th grade and under work?

Last year, i believe it was something like everyone that gets a certain AIME score has a chance to qualify.  However, since there were too many people, they added their AMC score + AIME score.

When they do this, to narrow down the people, do they only use the AMC + AIME scoring format on the people who were exactly on the cutoff AIME score?

Meaning, if the cutoff was 6, is the AMC + AIME thing only used for people that got a 6 for AIME?  Or is it used for everyone that got 6 or above on AIME?

I have this question because if it were the second option, that would discourage everyone from taking the AMC 12 because its harder to get higher scores on  AMC 12.  I was wondering this so to see if I should take the AMC 12 or not.

For example, if the cutoff score was 186 and the AIME score was 6...

That means that someone that got a 126 on AMC 10 + 6 on AIME is considered to get in and not a person with 105 on AMC 12 + 8 on AIME when clearly the second is more impressive.

I'm not completely sure, but I think that from the group of people that took the AMC 12, they will take the 160 who got the highest AIME scores, and the lowest score among those 160 people is the floor.  Then from the group of people who took the AMC 10, they take those who got an AIME score higher than the floor.

Read the Selection Process, Section II, page 4 of the 2007 USAMO Teacher's Manual,
at
http://www.unl.edu/amc/d-publication/d1-pubarchive/2006-7pub/07-USAMO-TM/TM-USAMO,2007.pdf
The process will be the same again this year.

Steve Dunbar
MAA Director, American Mathematics Competitions


AwesomeMath
summer program
geometry
FTW
MathZoom
AMC
AIME

Hey, I think that it would be a good idea for former students to give advice to incoming students. I have put in a start from a list that I posted elsewhere in this forum.

1.	Be proactive about asking questions or telling somebody that you don't understand. I am going to look into getting a system together to combat this. Asking questions isn’t always easy, but it is important to think about.

2.	I thought that the food was generally good. I ordered out with the kids in my suite occasionally to spice things up. You can also order food (or other general items like toothpaste) from Walmart.

3.	We are responsive to feedback. We do our best to make a good environment but we still want to hear what you all think so we can continue to improve. A good person to talk to would be your counselor (we have meetings where we can bring these things up). Also I’ll make sure to be around. The camp coordinator or whatever they call it would be good too.

4.	Go to sleep so you can stay awake in class. Kids will always resist the counselors…it just seems stupid to me.

5.	Try some activities (in something like volleyball, you can just join in and there is no pressure, and it is easier to talk to people).

6.	Bring things to do like games.

7.	Get a basic idea of the layout of the parts of campus that we will be using (i.e. the classrooms, dorms, eating areas, campus store, pool, music room, etc). It is nicer to know where you are going.

P.S. If you post advice, please type it nicely. I'll just delete posts if they are not written with reasonable sentences, etc.

Over a long period of time, the sleep issue becomes much bigger.  That is, when the counselors tell you it's past 10:30, it means "go to sleep now or you'll get less than 8 hours and/or miss breakfast".  Perhaps you can get away with it at first, but treat it like school.  You have your day off on the weekend, but you have to shift back into a natural rhythm.
i.e.  Losing an hour each night is apprx *21.  Remember that.  It's fun to party the last night and you don't want to have to miss out.

Bring money for food, and preferably a list of numbers to order from (the internet is useful before hand).  Share and split costs, and don't waste it all week 1.  Stockpile food in the day for nighttime snacks.  There's no internet, except for cases like online classes, so get numbers (and [i]map[b]S[/b][/i]) ahead of time.

Bring some sort of way to tell time.  Otherwise you may end up running to class after lunch.  It's a large campus and there's an assumption that you'll be responsible for your own whereabouts, since most people there are in high school.

[quote]7. Get a basic idea of the layout of the parts of campus that we will be using (i.e. the classrooms, dorms, eating areas, campus store, pool, music room, etc). It is nicer to know where you are going. [/quote]

It's also nice to not miss breakfast the second day there.  Try to stay in groups.  Room / find friends you can meet up with in the mornings.  Bring a cell phone just in case, and a registry of friends at camp you can call.

Often brought up: As of the summer of 2008 at the UT Dallas campus, there are two practice pianos (which are usually always available) in very poor condition (keys that won't play, sticky keys, etc.) in the music building in the underground floor, and your usual nice grand piano in the concert hall, which may be used when there aren't college kids there (more than half the time)... although I always felt bad playing on the grand piano, what with my bad piano-playing skills and...

[quote="Altheman"]
P.S. If you post advice, please type it nicely. I'll just delete posts if they are not written with reasonable sentences, etc.[/quote]
So, with proper capitalism and punctuation?

The campus isn't big, so finding your way around shouldn't be all that hard. Travel in groups at first if you need too.
At the beginning, most people exchange cell numbers and what not, which can turn out to be pretty useful.

[quote="alexhhmun"] There's no internet, except for cases like online classes
[/quote]
Really?? Well, for a lot of people, we still got internet even though no one needed to take classes.

[quote="serialk11r"] By practice piano, do you mean upright junk built in korea or 1900? Or is there something slightly better, like junk grand?
[/quote]
upright junk ftw! ><
well, put it this way, it's playable at least.
I'm not sure how good the grand piano was in the actual concert hall, but it looked really nice. so i'm assuming it's pretty decent.

About sleep, I felt like spending time with the people there was really important, so if you lose like a couple hours, it's not all that bad. Because you can catch up at sleep at home, but hanging out with everyone is limited to only those 3 weeks..... I feel like I'd value spending time/hanging out for an extra hour with my friends rather than get an extra hour of sleep or something I could do at home. But that's only my opinion.
And obviously, if you're so tired that you're passing out in class/dropping dead while studying/something like that [aka my roommate], obviously get some sleep. But if you're just a little tired, grab some coffee and enjoy the all the time you can spend awake at camp.

Grand piano is fine... Sometimes these random college kids come in and listen if you happen to be playing Saint-Saens ~_- or a Chopin Ballade... Some guy was like "are you a student here?" and I mutely shook my head. it's pretty funny.

Never use any piano except the grand piano. not worth it.

Advice:
1. Bring a cell phone. Definitely if you want to order pizza at night. lol.

2. Make sure to order lots of water from Walgreens... and OJ or any other drink you want. It gets really hot there and sometimes it's kinda annoying to not have anything decent to drink.

3. Bring lots of money if you intend to order food =D. Or just bring it so you can get the amazing milkshakes at the Pub.

4. Ermm. Don't play Mao until 11:30 every night if you're on the 2nd floor since the counselor below will come up and start screaming at you.

5. ... Never let 4 texas kids pull you into the pull if you have your cell phone in your pocket... it leads to lost glasses, a broken cell, and really really wet clothes.

6. uhh... Pay attention in class and actually try or you might not gain much ~_-... and the point of awesome math is math right... kinda.

7. Party it up =P

Still can't decide on AwesomeMath or MathZoom....

anyone help?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200592

I've never been to MathZoom, so I can't really make a comparsion.
But i can tell you that AMSP is really amazing.
1) Classes are really intense & interesting. Your professors will probably be well-known & have a deep interest & knowledge in math. Your TA's will as well.
2) The people there are really really cool. Like your roomies & friends you meet there becomes second family xD
3) You learn a lot, if you go there willing and with a open mind [and try not to goof off too much in class ]

A lot of people have gone to AMSP, and said it's been a really amazing experience [or the best 3 weeks of their lives].
xD
you should definitely try it out.

my advices would be...

1. BRING SUNSCREEN [especially if youre planning to go to the pool during daytime...]

2. BRING DEODRANT PLEASE. PERSONAL HYGIENE!

3. BRING MONEY YES YES MONEY. for food, junk food, drinks, whatever you need.

4. definitely bring a cellphone unless you want to become a lost child.

5. bring a camera :] to capture all your memories :)

6. don't stay up past midnight on weekdays. weekends, do as you wish, i suppose.

7. bring prep books, books, magazines, summer homework, etc to keep you occupied when youre not doing math..and you get bored.

8. participate in evening social activities!

9. bring a backpack!!!!!

10. drink lots of water. keep yourself hydrated. dont walk in wet grass. you'll slip and fall and whine.

11. um try not to set your dorm's AC to 50 degrees all day long. and do not definitely keep the AC at 50 degrees ALL NIGHT LONG. it's freezing [bbbbrrrrrrr]

12. bring pillows, blankets, sheets. they're not provided.

13. don't lose your sanity from having too much fun at awesomemath :]

(14) enjoy ur time there. its quite stress-free =]]

AwesomeMath is pretty cheap for summer camp actually. Have you seen EPGY? It's more expensive by a lot :O

[quote="ReseesPieces"]i was gonna go but its obnoxiusly expensive...is there anyway i could get price reduced or somethin?or try work at camp while im there???[/quote]

i think its worth its price & serialk11r is right.
if you had applied early, you would have gotten like a 10% discount i think.
but i think that's all the discounts there are. unless you apply for financial aid.

But always always always, 0% is bad toed.  :rotfl:

How I can write solutions to mathematical reflection problems?

I'm going to refer you to this [url=http://www.artofproblemsolving.com/Forum/viewforum.php?f=150]forum[/url], where you will get more answers.

Don't just study all day, go have fun and socialize too.
Make sure you have at least one other person with you at all times, all the campuses are big and it's easy to get lost.
Make sure you go to office hours and ask questions if you don't understand!
Don't go off campus. Once there were these 2 people who were craving beef jerky and ended up getting lost and spending 10 hours in town before coming back.
Get you counselor's phone number. If you ever get lost on a field trip, it will come in handy.
Don't feel bad if you get a bad score on a test. This is much harder than school.
Make sure you don't eat only french fries for the 3 weeks there *cough cough french fries cough cough*

Protip: Get milk at least 5 times a week.

Hmm also if you're going to cornell, make sure to take LOTS of fortune cookies.
Also, make sure to buy me mountain dew.
Also, make sure you buy me tons of deodorant spray (:P).

[quote="nsun48"]Also, make sure to buy me mountain dew.[/quote]

Interesting that by reading one line of the post, we can tell whether every (or none of the) suggestion is a good idea. :)

^ :rotfl: Never get near someone who wants you to buy lots of MD...

A few questions that may or may not have already been answered :

1.) Can you choose your roommate if you're going with a friend?
2.) What are RAs and TAs?
3.) How do you use your laptops in the dorms if the website says there is no internet in the dorms?

Yes
Residential assistants / teaching assistants
There's Internet; I don't know why they claim that they don't.

Sorry for reviving. I am thinking of applying to Cornell AMSP through regular application. But I have a couple questions: Do you get homework from the classes in AMSP(to work on in the evenings)? Also, what are the forums? Finally, What level in Ideamath summer camp corresponds to the level 2 courses in AMSP? Thanks!

There are daily problem sets that you can work on everyday. The forums are like optional lectures about various topics that you can attend around once a week.


Does anyone know if there is a formula to determine how many squares are there in an x-by-y grid (or even in an x-by-x grid, for that matter) when squares with non-integer sides are counted. For small numbers, it is not difficult to do case study, but how would you do it if for example, with an 11-by-11 grid?

for 11, it would be $1^{2}+2^{2}+3^{2}...+11^{2}=506.$

for rectangles...well...casework isn't that bad.

hmmm, now that i think of it, it might be:

xy+(x-1)(y-1)+(x-2)(y-2)...until either (x-n) or (y-n) is equal to 1.

indianamath, your method only counts the squares with integer sides and sides parrallel to the edges of the grid.
I don't think there is a formula.

For an x by y grid, squares:

The formula is $(x)(y)+(x-1)(y-1)+(x-2)(y-2)\cdots(x-x)(y-x)$.

For a x by y grid, rectangles:

The formula is $(1+2+3+\cdots+x)\cdot(1+2+3+...+y)$, or $\frac{(x)(y)(x+1)(y+1)}{4}$.

...I think.

EDIT: You're right, alkjash, this only applies to squares with sides parallel to the sides and with integer lengths.  Oh right, for the 1st one, I'm assuming x is the lower length side.

EDIT again:  As the answer to the 11 x 11 grid question, there are 506 squares and 4356 rectangles.

uh rectangles is much simpler, just pick two of the $m+1$ and two of the $n+1$ to be our edges...

${m+1\choose2}{n+1\choose2}$

which is equivalent to your thing

Well, for the rectangles, I would use constructive counting:

Firstly, there are $x \cdot y$ places for the first point.

Then, for the second point, which is on the same column as the first, there are $y-1$ choices.

Then, for the third point, which is in the same row as the first, there are $x-1$ choices.

The last point has only one choice.

And then you divide by 4 for overcounting.

Edit: The answer is therefore $\frac{xy(y-1)(x-1)}{4}$
Again, this only counts the rectangles with integer side lengths and sides parrellel to the edges of the grid.

Umm... Everybody wrote about squares with integer side lengths parallel to the gridlines. But what about squares like, for example, $\sqrt{2}$ by $\sqrt{2}$?

welll, how do you have a root2 by root2 grid? if you're just asking about how many squares an drectangles in a root2 square, the answer would be infinity.

[quote="diophantient"][quote="erdogankerem123"][quote="CircleSquared"]I think he means the units of the grid is $\sqrt{2}$.

In other words, each unit square in the grid has length not 1, but $\sqrt{2}$

Well, I'll say this:
The formula holds for any unit length, as long as it is actually a square grid and there are no missing segments :jump:[/quote]
i think he means the squares not parallel to the sides of the grid, where you get a $\sqrt{2}$ by $\sqrt{2}$ square with the diagonal of the unit squares of length 1. [quote="13375P34K43V312"]There is a closed form formula for the number of squares, of any size, in a n by n grid. But it's ugly (fourth-degree), and not worth memorizing.h[/quote]
What's a fourth-degree formula. Where can you find it?[/quote]
4th degree means of the form
$\sum^{5}_{k=0}a_{k}x^{k}$[/quote]
Isn't it
$\sum^{4}_{k=0}a_{k}x^{k}$

[quote="erdogankerem123"][quote="CircleSquared"]I think he means the units of the grid is $\sqrt{2}$.

In other words, each unit square in the grid has length not 1, but $\sqrt{2}$

Well, I'll say this:
The formula holds for any unit length, as long as it is actually a square grid and there are no missing segments :jump:[/quote]
i think he means the squares not parallel to the sides of the grid, where you get a $\sqrt{2}$ by $\sqrt{2}$ square with the diagonal of the unit squares of length 1. [quote="13375P34K43V312"]There is a closed form formula for the number of squares, of any size, in a n by n grid. But it's ugly (fourth-degree), and not worth memorizing.h[/quote]
What's a fourth-degree formula. Where can you find it?[/quote]

I think he means what [b]is[/b] the fourth degree formula, and kohjhsd should be right with his representation of a fourth degree formula.

[quote="ZzZzZzZzZzZz"][hide="Etc"][quote="erdogankerem123"][quote="CircleSquared"]I think he means the units of the grid is $\sqrt{2}$.

In other words, each unit square in the grid has length not 1, but $\sqrt{2}$

Well, I'll say this:
The formula holds for any unit length, as long as it is actually a square grid and there are no missing segments :jump:[/quote]
i think he means the squares not parallel to the sides of the grid, where you get a $\sqrt{2}$ by $\sqrt{2}$ square with the diagonal of the unit squares of length 1. [quote="13375P34K43V312"]There is a closed form formula for the number of squares, of any size, in a n by n grid. But it's ugly (fourth-degree), and not worth memorizing.h[/quote]
What's a fourth-degree formula. Where can you find it?[/quote][/hide]

I think he means what [b]is[/b] the fourth degree formula, and kohjhsd should be right with his representation of a fourth degree formula.[/quote]

Oh. I think this was already posted before in this thread.

[quote="Pancho Bigotes"]If we consider a x-by-x grid we can use the formula for the sum of the squares of the first $x$ numbers:

$\frac{x\cdot(x+1)\cdot(2x+1)}{6}$[/quote]


geometry
perimeter
linear algebra
matrix
circumcircle
inequalities
incenter

feizhenpeng had posted the problems of China TST 2005 quizzes in:
http://www.mathlinks.ro/Forum/viewtopic.php?t=30704
http://www.mathlinks.ro/Forum/viewtopic.php?t=32012
http://www.mathlinks.ro/Forum/viewtopic.php?t=32916

But they are in Chinese Version. What a pity!

I have translated them into English, following are the English version.

I had also translated the China TST quizzes of 2003 and 2004 in
http://www.mathlinks.ro/Forum/viewtopic.php?t=26393
http://www.mathlinks.ro/Forum/viewtopic.php?t=5283

China TST 2005, Test 1, 8:20-12:20, March 16, 2005

Problem 1:  http://www.mathlinks.ro/Forum/viewtopic.php?t=98961
Point $P$ lies inside triangle $ABC$. Let the projections of $P$ onto sides $BC$,$CA$,$AB$ be $D$, $E$, $F$ respectively. Let the projections from $A$ to the lines $BP$ and $CP$ be $M$ and $N$ respectively. Prove that $ME$, $NF$ and $BC$ are concurrent.

Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?t=30719
Let $a,b,c > 0$ and $ab+bc+ca=\frac{1}{3}$, prove that:
\[\frac{1}{a^{2}-bc+1}+\frac{1}{b^{2}-ca+1}+\frac{1}{c^{2}-ab+1}\leq 3 \]
Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?t=98963
Let $a_{1},a_{2}\cdots a_{n}$ and $x_{1}, x_{2}\cdots x_{n}$ be integers and $r \geq 2$ be an integer. It is known that $\sum_{j=0}^{n}a_{j}x_{j}^{k}=0$ for $k=1,2, \cdots r$. Prove that $\sum_{j=0}^{n}a_{j}x_{j}^{m}\equiv 0 (\mod m)$ for all $m \in \{ r+1, r+2, \cdots, 2r+1 \}$

China TST 2005, Test 2, 8:20-12:20, March 17, 2005

Problem 4: http://www.mathlinks.ro/Forum/viewtopic.php?t=98965
Let $k$ be a positive integer. Prove that one can partition the set $\{ 0,1,2,3, \cdots ,2^{k+1}-1 \}$ into two disdinct subsets $\{ x_1,x_2, \cdots, x_{2k} \}$ and $\{ y_1, y_2, \cdots, y_{2k} \}$ such that $\sum_{i=1}^{2^k} x_i^m =\sum_{i=1}^{2^k} y_i^m$ for all $m \in \{ 1,2, \cdots, k \}$.

Problem 5: http://www.mathlinks.ro/Forum/viewtopic.php?t=98966
Cyclic quadrilateral $ABCD$ has positive integer side lengths $AB$, $BC$, $CA$, $AD$. It is known that $AD=2005$, $\angle{ABC}=\angle{ADC} = 90^o$, and $\max \{ AB,BC,CD \} < 2005$. Determine the maximum and minimum possible values for the perimeter of $ABCD$.

Problem 6: http://www.mathlinks.ro/Forum/viewtopic.php?t=98967
Let $a,b,c,d >0$ and $abcd=1$. Prove that:
\[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \geq 1 \]

China TST 2005, Test 3, 8:20-12:20, March 21, 2005

Problem 7: http://www.mathlinks.ro/Forum/viewtopic.php?t=98968
Triangle $ABC$ is inscribed in circle $\omega$. Circle $\gamma$ is tangent to $AB$ and $AC$ at points $P$ and $Q$ respectively. Also circle $\gamma$ is tangent to circle $\omega$ at point $S$. Let the intesection of $AS$ and $PQ$ be $T$. Prove that $\angle{BTP}=\angle{CTQ}$.

Problem 8: http://www.mathlinks.ro/Forum/viewtopic.php?t=98969
Determine whether $\sqrt{1001^2+1}+\sqrt{1002^2+1}+ \cdots + \sqrt{2000^2+1}$ be a rational number or not?

Problem 9: http://www.mathlinks.ro/Forum/viewtopic.php?t=98970
We call a matrix $\textsl{binary matrix}$ if all its entries equal to $0$ or $1$. A binary matrix is $\textsl{Good}$ if it simultaneously satisfies the following two conditions:
(1) All the entries above the main diagonal (from left to right), not including the main diagonal, are equal.

(2) All the entries below the main diagonal (from left to right), not including the main diagonal, are equal.

Given positive integer $m$, prove that there exists a positive integer $M$, such that for any positive integer $n>M$ and a given $n \times n$ binary matrix $A_n$, we can select integers $1 \leq i_1 <i_2< \cdots < i_{n-m} \leq n$ and delete the $i_i$-th, $i_2$-th,$\cdots$, $i_{n-m}$-th rows and $i_i$-th, $i_2$-th,$\cdots$, $i_{n-m}$-th columns of $A_n$, then the resulting binary matrix $B_m$ is $\textsl{Good}$.

China TST 2005, Test 4, 8:20-12:20, March 22, 2005

Problem 10: http://www.mathlinks.ro/Forum/viewtopic.php?t=98971
Prove that for any $n$ ($n \geq 2$) pairwise disdinct fractions in the inteval $(0,1)$, the sum of their denominators is no less than $\frac{1}{3} n^{\frac{3}{2}}$.

Problem 11: http://www.mathlinks.ro/Forum/viewtopic.php?t=98972
Let $\omega$ be the circumcircle of acute triangle $ABC$. Two tangents of $\omega$ from $B$ and $C$ intersect at $P$, $AP$ and $BC$ intersect at $D$. Point $E$, $F$ are on $AC$ and $AB$ such that $DE \parallel BA$ and $DF \parallel CA$.
(1) Prove that $F,B,C,E$ are concyclic.

(2) Denote $A_1$ the centre of the circle passing through $F,B,C,E$. $B_1$, $C_1$ are difined similarly. Prove that $AA_1$, $BB_1$, $CC_1$ are concurrent.

Problem 12: http://www.mathlinks.ro/Forum/viewtopic.php?t=98973
Let $n$ be a positive integer, set $S_n = \{ (a_1,a_2,\cdots,a_{2^n}) \mid a_i=0 \ \text{or} \ 1, 1 \leq i \leq 2^n\}$. For any two elements $a=(a_1,a_2,\cdots,a_{2^n})$ and $b=(b_1,b_2,\cdots,b_{2^n})$ of $S_n$, define
\[ d(a,b)= \sum_{i=1}^{2^n} |a_i - b_i|  \]
We call $A \subseteq S_n$ a $\textsl{Good Subset}$ if $d(a,b) \geq 2^{n-1}$ holds for any two distinct elements $a$ and $b$ of $A$. How many elements can the $\textsl{Good Subset}$ of $S_n$ at most have?

China TST 2005, Test 5, 8:20-12:20, March 25, 2005

Problem 13: http://www.mathlinks.ro/Forum/viewtopic.php?t=98975
Find all positive integers $m$ and $n$ such that the inequality:
\[ [ (m+n) \alpha ] + [ (m+n) \beta ] \geq [ m \alpha ] + [n \beta] + [ n(\alpha+\beta)] \]
is true for any real numbers $\alpha$ and $\beta$. Here $[x]$ denote the largest integer no larger than real number $x$.

Problem 14: http://www.mathlinks.ro/Forum/viewtopic.php?t=98976
Given prime number $p$. $a_1,a_2 \cdots a_k$ ($k \geq 3$) are integers not divible by $p$ and have different residuals when divided by $p$. Let
\[ S_n= \{ n \mid 1 \leq n \leq p-1, (na_1)_p < \cdots < (na_k)_p \} \]
Here $(b)_p$ denotes the residual when integer $b$ is divided by $p$. Prove that $|S|< \frac{2p}{k+1}$.

Problem 15: http://www.mathlinks.ro/Forum/viewtopic.php?t=98977
Find the least positive integer $n$ ($n\geq 3$), such that among any $n$ points (no three are collinear) in the plane, there exist three points which are the vertices of a non-isoscele triangle.

China TST 2005, Test 6, 8:20-12:20, March 26, 2005

Problem 16: http://www.mathlinks.ro/Forum/viewtopic.php?t=98978
Find all positive integer triples $(a,m,n)$ such that $a > 1$, $m<n$, and the prime factor(s) of $a^m-1$ are the same as the prime factor(s) of $a^n-1$.

Problem 17: http://www.mathlinks.ro/Forum/viewtopic.php?t=98979
In acute angled triangle $ABC$, $BC=a$,$CA=b$,$AB=c$, and $a>b>c$. $I,O,H$ are the incentre, circumcentre and orthocentre of $\triangle{ABC}$ respectively. Point $D \in BC$, $E \in CA$ and $AE=BD$, $CD+CE=AB$. Let the intersectionf of $BE$ and $AD$ be $K$. Prove that $KH \parallel IO$ and $KH = 2IO$.

Problem 18: http://www.mathlinks.ro/Forum/viewtopic.php?t=98980
$n$ is a positive integer, $F_n=2^{2^{n}}+1$. Prove that for $n \geq 3$, there exists a prime factor of $F_n$ which is larger than $2^{n+2}(n+1)$.


The angles of quadrilateral $ ABCD$ satisfy $ \angle A \equal{} 2 \angle B \equal{} 3 \angle C \equal{} 4 \angle D$.  What is the degree measure of $ \angle A$, rounded to the nearest whole number?

$ \textbf{(A)}\ 125 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 153 \qquad \textbf{(D)}\ 173 \qquad \textbf{(E)}\ 180$

[hide]
$A=2B=3C=4D$

Different ways of solving this but in this case, we will look for $D$ first.

The sum of the four angles of the quadrilateral is $360$ degrees.

$A=4D$
$B=\frac{4}{2}D$
$C=\frac{4}{3}D$
$D=D$

$D+4D+2D+\frac{4}{3}D=360$
$7D+\frac{4}{3}D=360$
$D=43.2$
$A=4D$
$D=43.2*4$
$D=172.8$ or $173$ degrees.
[/hide]

[hide="Solution"]Express them in terms of $a$. You then get $a+1/2 a+1/3 a+1/4 a = 25/12 a = 360$. Next, $25a = 4320$. Rounding, we get $a = 172$[/hide]


function factorial calculus integration quadratics limit algebra

A hopefully interesting power series I came up with after fiddling around:

For $ x > \minus{}\frac{1}{4}$, find the function $ f(x)$ to which the following power series converges:

$ \sum_{n \equal{} 0}^{\infty} \frac{(\minus{}1)^n}{n\plus{}1} \binom{2n}{n} x^n$

Then evaluate

$ \int_{\frac{1}{4}}^{\frac{3}{4}} f(x) \ dx$

this probably isn't what you want but...

$ \sum_{k\equal{}0}^{\infty}(\minus{}1)^k\frac{\Gamma(2k\plus{}1)(3^{k\plus{}1}\minus{}1)}{(2^k(k\plus{}1)\Gamma(k\plus{}1))^2}$

That series only converges in the reals for $ x \in (-\frac{1}{4}, \frac{1}{4})$.  (The function it converges to can be extended nicely to $ (-\frac{1}{4}, \infty)$.)

This is well-known to (enumerative & algebraic) combinatorialists, and in fact it's appeared on the forum a few times before.  Here's the first part of your question:

[hide]The numbers $ \frac {1}{n + 1}\binom{2n}{n}$ are the Catalan numbers -- they satisfy the recursion $ C_{n + 1} = \sum_{k = 0}^n C_k C_{n - k}$ with $ C_0 = 1$.  If we let $ C(x) = \sum_{n \geq 0} C_n x^n$, it follows immediately that $ C(x) = 1 + xC(x)^2$.  This is a simple quadratic equation in $ C(x)$ that we can solve to get $ C(x) = \frac {1 \pm \sqrt {1 - 4x}}{2x}$.  Since $ \lim_{x \to 0}C(x) = 1$, we only have one choice for the sign; $ C(x) = \frac {1 - \sqrt {1 - 4x}}{2x}$.  The function you're interested in is $ f(x) = C( - x) = \frac {1 - \sqrt {1 + 4x}}{2x}$.[/hide]

[quote="0714446459923"]this probably isn't what you want but...

$ \sum_{k \equal{} 0}^{\infty}( \minus{} 1)^k\frac {\Gamma(2k \plus{} 1)(3^{k \plus{} 1} \minus{} 1)}{(2^k(k \plus{} 1)\Gamma(k \plus{} 1))^2}$[/quote]

How can you justify that?

I see a possible answer, but no solution.
I'm interested to see how you got here.

term by term integration

Then I just replaced all the factorials with gammas


Μέχρι σήμερα 103 χώρες δήλωσαν ενδιαφέρον, 91 χώρες αποδέχτηκαν συμμετοχή και μόνο 68 έχουν δηλώσει Αρχηγό ή Υπαρχηγό ή Παρατηρητές. Η προθεσμία ήταν 30 Απριλίου. Μεταξύ των αργοπορούντων συγκαταλέγονται Κίνα, Κούβα, Αλβανία, Δανία, Ολλανδία, ΠΓΔΜ, Μολδαβία, Μεξικό, Μογγολία, Ιράν, Καζακστάν, Πολωνία, Σερβία, Ουκρανία, Κορέες και το Βιετ Ναμ (που υποτίθεται ότι θα διοργανώσει την επόμενη ΔΜΟ!).
Η Ελλάδα έχει δηλώσει μόνο Αρχηγό και έναν Παρατηρητή με τον Αρχηγό. Δεν αναφέρονται ονόματα.
Οι πληροφορίες αυτές βρίσκονται στον ιστότοπο της ΔΜΟ.

Update

Μέχρι σήμερα 103 χώρες δήλωσαν ενδιαφέρον, 91 χώρες αποδέχτηκαν συμμετοχή και μόνο 74 έχουν δηλώσει Αρχηγό ή Υπαρχηγό ή Παρατηρητές. Η προθεσμία ήταν 30 Απριλίου. Μεταξύ των αργοπορούντων συγκαταλέγονται Κίνα, Κούβα, Αλβανία, Δανία, Γεωργία, ΠΓΔΜ, Μολδαβία, Μεξικό, Μογγολία, Ιράν, Καζακστάν, Φιλιππίνες, Τρινιδάδ και Τομπήγκο, Ουκρανία, Κορέες και το Βιετ Ναμ (που υποτίθεται ότι θα διοργανώσει την επόμενη ΔΜΟ!).
Η Ελλάδα έχει δηλώσει μόνο Αρχηγό και έναν Παρατηρητή με τον Αρχηγό. Δεν αναφέρονται ονόματα.
Μέχρι στιγμής έχουν εγγραφεί 63 μαθητές, 70 αρχηγοί, 67 υπαρχηγοί, 29 παρατηρητές με αρχηγό, 18 παρατηρητές με υπαρχηγό, και 6 παρατηρητές με μαθητές.
Οι πληροφορίες αυτές βρίσκονται στον ιστότοπο της ΔΜΟ.

Update (18/05/2006)

Μέχρι σήμερα 103 χώρες δήλωσαν ενδιαφέρον, 96 χώρες αποδέχτηκαν συμμετοχή (δεν δήλωσαν ακόμα Αλγερία, Μπαχρέϊν, Χιλή, [b]Κούβα[/b], Β. Κορέα, Νικαράγουα, και [b]Φιλιππίνες[/b]) και μόνο 87 έχουν δηλώσει Αρχηγό ή Υπαρχηγό ή Παρατηρητές. Η προθεσμία ήταν 30 Απριλίου. Εκτός των ανωτέρω αργοπορούντες είναι Βολιβία, Μπρουνάϊ, [b]Κίνα[/b], Γουατεμάλα, Κυργυζιστάν, Μακάου, [b]Ουκρανία[/b], Ενωμένα Αραβικά Εμιράτα και το [b]Βιετ Ναμ[/b] (που υποτίθεται ότι θα διοργανώσει την επόμενη ΔΜΟ!).
Η Ελλάδα έχει δηλώσει Αρχηγό, Υπαρχηγό, έναν Παρατηρητή με τον Αρχηγό και έναν Παρατηρητή με τον Υπαρχηγό και 2 μαθητές. Δεν αναφέρονται ονόματα.
Μέχρι στιγμής έχουν εγγραφεί 251 μαθητές, 86 αρχηγοί, 84 υπαρχηγοί, 32 παρατηρητές με αρχηγό, 24 παρατηρητές με υπαρχηγό, και 9 παρατηρητές διάφοροι.
Οι πληροφορίες αυτές βρίσκονται στον ιστότοπο της ΔΜΟ.

Ποιοί συμπληρώνουν την εξάδα ?

1. Κουφογιάννης Φραγκίσκος

2. Ράπανος Νικόλαος

3. Μπουντάς Νικόλαος

4. Ψύχας Κων/νος

5. ??????????????

6. ??????????????

Ο Κουφογιάννης όχι πάει ΜΙΤ νομίζω .Άρα τρεις θέσεις περιμένουν :D

Πιθανότατη Ομάδα για την Διεθνή ΜΟ:

ΡΑΠΑΝΟΣ
ΜΠΟΥΝΤΑΣ
ΨΥΧΑΣ
ΚΑΡΑΜΙΧΑΛΗΣ
ΣΤΑΥΡΟΠΟΥΛΟΥ
ΡΩΜΑΝΙΔΗΣ

ΕΥΧΟΜΑΙ ΚΑΛΗ ΕΠΙΤΥΧΙΑ ΣΤΗΝ ΟΜΑΔΑ!

[quote="RAKAR"]ΒΓΗΚΕ Η ΟΜΑΔΑ ΠΟΥ ΘΑ ΠΑΕΙ ΣΤΗΝ ΙΜΟ(ΟΧΙ ΣΤΟ SITE THS EME)

          1)ΡΑΠΑΝΟΣ ΝΙΚΟΣ
          2)ΜΠΟΥΝΤΑΣ ΝΙΚΟΣ
          3)ΨΥΧΑΣ ΚΩΝ/ΝΟΣ
          4)ΡΩΜΑΝΙΔΗΣ ΝΙΚΟΣ
          5)ΚΑΡΑΜΙΧΑΛΗΣ ΡΑΛΛΗΣ
          6)ΣΤΑΥΡΟΠΟΥΛΟΥ ΕΛΕΝΗ

ΔΕΝ ΓΝΩΡΙΖΩ ΤΟΥΣ ΑΝΑΠΛΗΡΩΜΑΤΙΚΟΥΣ.....

ΡΑΛΛΗΣ :D[/quote]

Καλή επιτυχία !!  :)

Συγχαρητήρια στα παιδιά και Καλή επιτυχία !!

Συγχαρητηρια σε ολους!ειναι μεγάλη τιμή μην το ξεχνάτε!

Καλή επιτυχία στην ΙΜΟ ομάδα μας με τις ευχές όλων μας να τους συνοδεύουν να τους σκίσουν γερά παιδιά με τσαμπουκά
Γειά σας

Από τις 103 χώρες που δήλωσαν ενδιαφέρον, 4 δεν συμπλήρωσαν καν το έντυπο πληροφοριών (Αλγερία, Χιλή, Β. Κορέα, και Νικαράγουα), από τις υπόλοιπες 99, 3 χώρες δεν αποδέχτηκαν συμμετοχή ακόμη (Μπαχρέϊν, [b] Κούβα[/b],  και [b]Φιλιππίνες[/b]) και από τις υπόλοιπες 96, 4 δεν έχουν δηλώσει Αρχηγό ή Υπαρχηγό ή Παρατηρητές (Βολιβία, Μπρουνάϊ, [b]Ουκρανία[/b], και Ενωμένα Αραβικά Εμιράτα). Η προθεσμία ήταν 30 Απριλίου.
Η Ελλάδα έχει δηλώσει Αρχηγό, Υπαρχηγό, έναν Παρατηρητή με τον Αρχηγό και έναν Παρατηρητή με τον Υπαρχηγό και 5 μαθητές (δεν έχει εγγραφεί ο Καραμιχάλης ακόμα). Αρχηγός είναι ο Θ. Μπόλης (τέως Πρόεδρος της Επιτροπής Διαγωνισμών της ΕΜΕ) και Υπαρχηγός ο Γ. Τασσόπουλος (μέλος ιδίας). Παρατηρητές είναι οι σύζυγοι αυτών που πληρώνουν 2000 και 1650 ευρώ αντίστοιχα.
Μέχρι στιγμής έχουν εγγραφεί 412 μαθητές, 91 αρχηγοί, 89 υπαρχηγοί, 33 παρατηρητές με αρχηγό, 26 παρατηρητές με υπαρχηγό, και 14 παρατηρητές διάφοροι. Από τις 92 χώρες που δήλωσαν Αρχηγό ή Υπαρχηγό, 16 δεν έχουν δηλώσει ούτε έναν μαθητή ακόμα. Η προθεσμία ήταν 31 Μάη.
[b]Το Υπουργείο Εθνικής Παιδείας και Θρησκευμάτων δεν έχει εκδώσει την Υπουργική Απόφαση για τη Βαλκανική Ολυμπιάδα ακόμα και τα παιδιά ήταν στην Κύπρο χωρίς καμμία κάλυψη. Τα εισιτήρια δεν έχουν πληρωθεί ούτε έχει εγκριθεί η πίστωση για τα εισιτήρια της ΔΜΟ![/b]

Συγνώμη ρε παιδιά με όλο  το συμπάθειο κι χώρις παρεςξήγηση αύτο το περι παρατηρητών συζύγων  δεν είναι ψιλοξεφτίλα? νομίζω ότι πρέπει να γίνουν κάποιο άλλοι παρατηρητές ακόμη κι για λόγους ευνόητους αν θέλετε
Χαρετώ ελπίζω να μην θιχτούν οι ΚΚ Μπόλης Κι Τασσόπουλος

Υπάρχει μια παρεξήγηση εδώ. Παλιά τα συγγενικά πρόσωπα του Αρχηγού και του Υπαρχηγού λεγόταν "Accompanying persons" και οι παρατηρητές ήταν Μαθηματικοί. Όλοι οι εκτός οκτάδας πληρώνουν και πληρώνουν ακριβά (κάπου 180 ευρώ την ημέρα). Οποιαδήποτε χώρα μπορεί να στείλει όσους παρατηρητές θέλει, αρκεί να μπορεί να πληρώνει. Για τις συζύγους ή τέκνα του Αρχηγού ή του Υπαρχηγού πληρώνει ο Αρχηγός ή ο Υπαρχηγός και όχι η διοργανώτρια οντότητα!

Δείτε την ομάδα μας .
http://imo2006.dmfa.si/participants/GRC.html

ΡΕ  ΝΙΚΟ ΤΙ ΕΙΝΑΙ ΑΥΤΑ ΡΕ ? ΔΕΝ ΕΙΧΕΣ ΛΕΦΤΑ ΓΙΑ ΦΩΤΟ ΡΕ?  :rotfl:  :rotfl: ΑΝΤΕ ΑΠΟ ΕΚΕΙ

κ.Άρχηγέ ο ΜΠΟΥΝΤΑΣ είναι ΝΙΚΟΛΑΟΣ και όχι ΚΩΝΣΤΑΝΤΙΝΟΣ !

Από τις 103 χώρες που δήλωσαν ενδιαφέρον, 4 δεν συμπλήρωσαν καν το έντυπο πληροφοριών (Αλγερία, Χιλή, Β. Κορέα, και Νικαράγουα), από τις υπόλοιπες 99, 3 χώρες δεν αποδέχτηκαν συμμετοχή ακόμη (Μπαχρέϊν, Κούβα, και Φιλιππίνες) και από τις υπόλοιπες 96, 3 δεν έχουν δηλώσει Αρχηγό ή Υπαρχηγό ή Παρατηρητές (Μπρουνάϊ, Ουκρανία, και Ενωμένα Αραβικά Εμιράτα). Η προθεσμία ήταν 30 Απριλίου.
Η Ελλάδα έχει δηλώσει Αρχηγό, Υπαρχηγό, έναν Παρατηρητή με τον Αρχηγό και έναν Παρατηρητή με τον Υπαρχηγό και 5 μαθητές (δεν έχει εγγραφεί ο Καραμιχάλης ακόμα). Αρχηγός είναι ο Θ. Μπόλης (τέως Πρόεδρος της Επιτροπής Διαγωνισμών της ΕΜΕ) και Υπαρχηγός ο Γ. Τασσόπουλος (μέλος ιδίας). Παρατηρητές είναι οι σύζυγοι αυτών που πληρώνουν 2000 και 1650 ευρώ αντίστοιχα.
Μέχρι στιγμής έχουν εγγραφεί 443 μαθητές, 92 αρχηγοί, 89 υπαρχηγοί, 33 παρατηρητές με αρχηγό, 28 παρατηρητές με υπαρχηγό, και 15 παρατηρητές διάφοροι. Από τις 92 χώρες που δήλωσαν Αρχηγό ή Υπαρχηγό, 13 δεν έχουν δηλώσει ούτε έναν μαθητή ακόμα. Η προθεσμία ήταν 31 Μάη. Στον ιστότοπο της ΔΜΟ μπορείτε να δείτε όλες τις αποστολές και τις φωτογραφίες τους πηγαίνοντας στο Registration, πατώντας status και μετά πατώντας το όνομα της χώρας μετά τη σημαία της.
[b]Το Υπουργείο Εθνικής Παιδείας και Θρησκευμάτων δεν έχει εκδώσει την Υπουργική Απόφαση για τη Βαλκανική Ολυμπιάδα ακόμα και τα παιδιά ήταν στην Κύπρο χωρίς καμμία κάλυψη. Τα εισιτήρια δεν έχουν πληρωθεί ούτε έχει εγκριθεί η πίστωση για τα εισιτήρια της ΔΜΟ![/b]

[quote="DuLac"]
[b]Το Υπουργείο Εθνικής Παιδείας και Θρησκευμάτων δεν έχει εκδώσει την Υπουργική Απόφαση για τη Βαλκανική Ολυμπιάδα ακόμα και τα παιδιά ήταν στην Κύπρο χωρίς καμμία κάλυψη. Τα εισιτήρια δεν έχουν πληρωθεί ούτε έχει εγκριθεί η πίστωση για τα εισιτήρια της ΔΜΟ![/b][/quote]

 :huh:  :|  Γνώριμη κατάσταση.

Θα ήθελα με τη σειρά μου να ευχηθώ καλή επιτυχία στη νέα ομάδα για την ΔΜΟ!!!
Λυπάμαι που δε θα βρίσκομαι μαζί τους (μια ατυχία είναι αρκετή σ'αυτό το επίπεδο...),αλλά έτσι τουλάχιστον δίνεται η δυνατότητα και σε άλλα παιδιά να συμμετέχουν σε ένα διεθνή διαγωνισμό,που είναι και το πιο δίκαιο.
Καλή επιτυχία λοιπόν και πολλά μετάλλια!!!  :winner_first:

Στέλιος... :icecream:

http://imo2006.dmfa.si/participants/GRC.html

Καθότι μπορείτε να δείτε τα πράγματα πλέον είναι φυσιολογικά,,,,,,,,,,,, :D

Από ότι είδα κι εγώ όλα βαίνουν καλώς οι εικόνες είναι πιο καλές από τις προηγούμενες καλή επιτυχία κι πάλι

Λοιπόν από ότι είδα τα Τρίκαλα ήθελαν δεν ήθελαν θα εκπροσωπηθούν στην ΙΜΟ  :D
Αφού ένας γνωστός μου ο κ .Κοντοκώστας θα είναι Deputy Leader .


integration
trigonometry
real analysis
real analysis unsolved