Okay hurt=-2 and heal=+1

Only sixteen countries;

U.S. - 21
U.K. - 20
Ireland - 20
China - 20
Japan - 20
India - 20
France - 20
Canada - 20
Belgium - 20
Mexico - 20
Brazil - 20
Russia - 20
Australia - 20 (country/continent)
Finland - 20
Spain - 20
Poland - 18

Don't got anything against Poland, just random choice of subtraction.

U.S. - 21
U.K. - 20
Ireland - 20
China - 20
Japan - 20
India - 21
France - 20
Canada - 20
Belgium - 20
Mexico - 20
Brazil - 20
Russia - 18
Australia - 20 (country/continent)
Finland - 20
Spain - 20
Poland - 18

U.S. - 21
U.K. - 20
Ireland - 21
China - 20
Japan - 20
India - 21
France - 20
Canada - 20
Belgium - 20
Mexico - 20
Brazil - 20
Russia - 18
Australia - 20 (country/continent)
Finland - 20
Spain - 18
Poland - 18

omg, how many of these competitions like these are we gonna have. i

#3U.S. - 19
U.K. - 20
Ireland - 21
China - 20
Japan - 20
India - 21
France - 20
Canada - 20
Belgium - 20
Mexico - 20
Brazil - 20
Russia - 18
Australia - 20
Finland - 20
Spain - 18
Poland - 19

U.S. - 17
U.K. - 20
Ireland - 21
China - 20
Japan - 20
India - 21
France - 20
Canada - 20
Belgium - 20
Mexico - 20
Brazil - 20
Russia - 19
Australia - 20
Finland - 20
Spain - 18
Poland - 19

[color=red]U.S. - 15[/color]
U.K. - 20
Ireland - 21
[color=green]China - 21[/color]
Japan - 20
India - 21
France - 20
Canada - 20
Belgium - 20
Mexico - 20
Brazil - 20
Russia - 19
Australia - 20
Finland - 20
Spain - 18
Poland - 19

meh. even tho i live in it, dont like US much...
i wish i could add one to japan also...

[quote="jjx1"]U.S. - 15
U.K. - 20
Ireland - 21
China - 22
Japan - 20
India - 19
France - 20
Canada - 20
Belgium - 20
Mexico - 20
Brazil - 20
Russia - 19
Australia - 20
Finland - 20
Spain - 18
Poland - 19

HOW COUDL PEOPLE BE TAKING POINTS AWAY FROM THE U.S!!!!!!!!!!!!![/quote]

Because we don't like U.S. education.

U.S. - 13
U.K. - 20
Ireland - 21
China - 22
Japan - 20
India - 19
France - 20
Canada - 20
Belgium - 20
Mexico - 20
Brazil - 20
Russia - 20
Australia - 20
Finland - 20
Spain - 18
Poland - 19

U.S. - 13
U.K. - 21
Ireland - 21
China - 22
Japan - 20
India - 19
France - 20
Canada - 20
Belgium - 20
Mexico - 20
Brazil - 18
Russia - 20
Australia - 20
Finland - 20
Spain - 18
Poland - 19

I'm not doing anything to USA because I feel nice today.  :D

U.S. - 12
U.K. - 21
Ireland - 21
China - 22
Japan - 20
India - 19
France - 20
Canada - 20
Belgium - 20
Mexico - 20
Brazil - 18
Russia - 20
Australia - 20
Finland - 20
Spain - 18
Poland - 19

-2 U.S. + 1 U.S. (Because I also feel nice today.)

[size=0][anti-patriotism]And the U.S. is a nation of consumers, the republicans and the democrats screwed us over basically(two parties trying to get their own elected, not necessarily for the good of the country), and negligence for [i]real[/i] healthcare*cough*FDA*cough*.[/anti-patriotism][/size]
-2 US +1 Russia
U. S. - 10
U.K. - 21
Ireland - 21
China - 22
Japan - 20
India - 19
France - 20
Canada - 20
Belgium - 20
Mexico - 20
Brazil - 18
Russia - 21
Australia - 20
Finland - 20
Spain - 18
Poland - 19

Just because Russia is awesome.

U. S. - 11
U.K. - 21
Ireland - 21
China - 22
Japan - 20
India - 19
France - 20
Canada - 20
Belgium - 20
Mexico - 20
Brazil - 18
Russia - 21
Australia - 20
Finland - 18
Spain - 18
Poland - 19

don't take your anger out on U.S. if U.S. randomly disappeared, we'd all drown.
Get rid of finland. Nobody really cares about it right now.

#13U. S. - 11
U.K. - 21
Ireland - 21
China - 22
Japan - 20
India - 19
France - 20
Canada - 20
Belgium - 20
Mexico - 20
Brazil - 18
Russia - 22
Australia - 20
Finland - 16
Spain - 18
Poland - 19

Russia.

Sure, let's kil  finland and U.s.

. S. - 11
U.K. - 21
Ireland - 21
China - 23
Japan - 20
India - 17
France - 20
Canada - 20
Belgium - 20
Mexico - 20
Brazil - 18
Russia - 22
Australia - 20
Finland - 16
Spain - 18
Poland - 19

South Africa - 21
Germany - 20
Argentina - 18
Ireland - 18
Brazil - 20
Australia - 20
Canada - 21
Russia - 18
Mexico - 16
Spain - 20
France - 20
Poland - 20
China - 13
United Kingdom - 20
Finland - 20
Japan - 20
U.S. - 28
Belgium - 20
Singapore - 16
Indonesia - 20
Italy - 20
Norway - 20
Greece - 19

South Africa - 21
Germany - 20
Argentina - 18
Ireland - 18
Brazil - 20
Australia - 20
Canada - 21
Russia - 18
Mexico - 16
Spain - 20
France - 20
Poland - 20
China - 11
United Kingdom - 20
Finland - 20
Japan - 20
U.S. - 29
Belgium - 20
Singapore - 16
Indonesia - 20
Italy - 20
Norway - 20
Greece - 19

South Africa - 21
Germany - 20
Argentina - 18
Ireland - 18
Brazil - 20
Australia - 20
Canada - 21
Russia - 18
Mexico - 16
Spain - 20
France - 20
Poland - 20
China - 11
United Kingdom - 20
Finland - 20
Japan - 20
U.S. - 30
Belgium - 20
Singapore - 14
Indonesia - 20
Italy - 20
Norway - 20
Greece - 19

South Africa - 21
Germany - 20
Argentina - 18
Ireland - 18
Brazil - 20
Australia - 20
Canada - 21
Russia - 18
Mexico - 16
Spain - 20
France - 20
Poland - 20
China - 9
United Kingdom - 20
Finland - 20
Japan - 20
U.S. - 31
Belgium - 20
Singapore - 14
Indonesia - 20
Italy - 20
Norway - 20
Greece - 19

South Africa - 21
Germany - 20
Argentina - 18
Ireland - 18
Brazil - 20
Australia - 20
Canada - 21
Russia - 18
Mexico - 16
Spain - 20
France - 21
Poland - 20
China - 9
United Kingdom - 20
Finland - 20
Japan - 20
U.S. - 31
Belgium - 20
Singapore - 14
Indonesia - 20
Italy - 20
Norway - 20
Greece - 19

South Africa - 21
Germany - 20
Argentina - 18
Ireland - 18
Brazil - 20
Australia - 20
Canada - 21
Russia - 18
Mexico - 16
Spain - 20
France - 21
Poland - 20
China - 10
United Kingdom - 20
Finland - 20
Japan - 18
U.S. - 31
Belgium - 20
Singapore - 14
Indonesia - 20
Italy - 20
Norway - 20
Greece - 19

South Africa - 21
Germany - 20
Argentina - 18
Ireland - 18
Brazil - 20
Australia - 20
Canada - 21
Russia - 18
Mexico - 16
Spain - 20
France - 21
Poland - 20
China - 8
United Kingdom - 20
Finland - 20
Japan - 18
U.S. - 33
Belgium - 20
Singapore - 14
Indonesia - 20
Italy - 20
Norway - 20
Greece - 19

South Africa - 21
Germany - 20
Argentina - 18
Ireland - 18
Brazil - 20
Australia - 20
Canada - 21
Russia - 18
Mexico - 16
Spain - 20
France - 21
Poland - 20
China - 9
United Kingdom - 20
Finland - 20
Japan - 16
U.S. - 33
Belgium - 20
Singapore - 14
Indonesia - 20
Italy - 20
Norway - 20
Greece - 19

South Africa - 21
Germany - 20
Argentina - 18
Ireland - 18
Brazil - 20
Australia - 20
Canada - 21
Russia - 18
Mexico - 16
Spain - 20
France - 21
Poland - 20
China - 9
United Kingdom - 20
Finland - 20
Japan - 16
U.S. - 34
Belgium - 20
Singapore - 12
Indonesia - 20
Italy - 20
Norway - 20
Greece - 19

South Africa - 21
Germany - 20
Argentina - 18
Ireland - 18
Brazil - 20
Australia - 20
Canada - 21
Russia - 18
Mexico - 16
Spain - 20
France - 21
Poland - 20
China - 10
United Kingdom - 20
Finland - 20
Japan - 14
U.S. - 34
Belgium - 20
Singapore - 12
Indonesia - 20
Italy - 20
Norway - 20
Greece - 19

South Africa - 21
Germany - 20
Argentina - 18
Ireland - 18
Brazil - 20
Australia - 20
Canada - 21
Russia - 18
Mexico - 16
Spain - 20
France - 21
Poland - 20
China - 8
United Kingdom - 20
Finland - 20
Japan - 14
U.S. - 35
Belgium - 20
Singapore - 12
Indonesia - 20
Italy - 20
Norway - 20
Greece - 19

South Africa - 21
Germany - 20
Argentina - 18
Ireland - 18
Brazil - 20
Australia - 20
Canada - 21
Russia - 18
Mexico - 16
Spain - 20
France - 21
Poland - 20
China - 8
United Kingdom - 20
Finland - 20
Japan - 14
U.S. - 36
Belgium - 20
Singapore - 10
Indonesia - 20
Italy - 20
Norway - 20
Greece - 19

South Africa - 21
Germany - 20
Argentina - 18
Ireland - 18
Brazil - 20
Australia - 20
Canada - 21
Russia - 18
Mexico - 17
Spain - 20
France - 21
Poland - 20
China - 8
United Kingdom - 20
Finland - 20
Japan - 14
U.S. - 34
Belgium - 20
Singapore - 10
Indonesia - 20
Italy - 20
Norway - 20
Greece - 19

South Africa - 19
Germany - 20
Argentina - 18
Ireland - 18
Brazil - 20
Australia - 20
Canada - 21
Russia - 18
Mexico - 17
Spain - 20
France - 21
Poland - 20
China - 8
United Kingdom - 21
Finland - 20
Japan - 14
U.S. - 34
Belgium - 20
Singapore - 10
Indonesia - 20
Italy - 20
Norway - 20
Greece - 19

South Africa - 19
Germany - 20
Argentina - 18
Ireland - 18
Brazil - 20
Australia - 20
Canada - 21
Russia - 18
Mexico - 17
Spain - 20
France - 21
Poland - 20
China - 8
United Kingdom - 21
Finland - 20
Japan - 15
U.S. - 34
Belgium - 20
Singapore - 8
Indonesia - 20
Italy - 20
Norway - 20
Greece - 19


calculus
derivative
trigonometry
function
absolute value
calculus computations

I was just wondering how to solve problems like these:

Positions of 2 particles on a line are Y1=sin(t) and Y2=Sin(T+Pi/3)

At what times b/t 0 and 2pi do these particles meet?

What is the fartest apart these particles ever get?

When in the interval 0 to 2pi is the distance between the particles changing the fastest?

I just need to know a general way to do these problems, I know that I have to find the derivative but what do I do after that? Thanks.

Once you find the derivative, you set it equal to 0 and find the roots.  These roots represent the critical points.  You then need to test these points to determine if they are maxima or minima.

For this problem, you can also use the identity
\[ \sin \alpha \minus{} \sin \beta \equal{} 2\cos \frac {\alpha \plus{} \beta }{2}\sin \frac {\alpha \minus{} \beta}{2}
\]
So $ y_2 \minus{} y_1$ is equal to
[hide]
\[ 2\cos \left( T \plus{} \frac {\pi}{6}\right) \sin \frac {\pi}{6} \equal{} \cos \left( T \plus{} \frac {\pi}{6}\right)
\]
[/hide]
and its absolute value gives the distance.

Both the above function and its derivative (the rate of change of distance) are given as a single trig function, so you can find their zeros, maxima, and minima, and when they occur.


geometry
perimeter
geometric transformation

I had a math test in school today. I think I did pretty good but there was (at least) two tasks I didn't manage to solve. A class mate of mine solved one of them and explained how he did for me. But I don't think anyone in my class solved the other one. I would appreciate if you could show me the solution.  :)

"In a right-angled triangle the hypotenuse is twice as long as the shortest cathetus. The longer cathetu's length is the average value of the other sides. The triangle's perimeter is 90 cm. What is the area of the triangle?" (Translated from Swedish, ask if you don't understand my translation.  :P)

Anyway... I got the equation
$ x \plus{} 2x \plus{} \frac {(2x \plus{} x)}{2} \equal{} 90$
$ 4.5x \equal{} 90$
$ x \equal{} 20$
And thus the sides are 20 cm, 30 cm and 40 cm.

But that isn't the right answer since $ 20^2 \plus{} 30^2 \not \equal{} 40^2$.

How do you solve this problem without brute force?

I don't think there is a solution to this

Here's what I did so far and I can't find a solution or even if I did anything wrong.

[hide]Since this is a right triangle with the short leg half as long as the hypotenuse then the other leg can be written as $ x\sqrt{3}$

So for the perimeter we have the equation

$ x \plus{} 2x \plus{} x\sqrt{3} \equal{} 90$

and for the third side we have

$ x\sqrt{3} \equal{} \frac{x \plus{} 2x}{2}$[/hide]

Watch your translation too, triangles don't have circumferences, circles do, triangles have perimeters.

[quote="SnipedYou"][hide]Since this is a right triangle with the short leg half as long as the hypotenuse then the other leg can be written as $ x\sqrt {3}$

So for the perimeter we have the equation

$ x \plus{} 2x \plus{} x\sqrt {3} \equal{} 90$

and for the third side we have

$ x\sqrt {3} \equal{} \frac {x \plus{} 2x}{2}$[/hide][/quote]

[hide]Hmm... I don't understand how you get $ x\sqrt{3}$...[/hide]

[quote="SnipedYou"]Watch your translation too, triangles don't have circumferences, circles do, triangles have perimeters.[/quote]Aha. My dictionary fooled me. :blush:

[url=http://en.wikipedia.org/wiki/Cathetus]Cathetus[/url]

In short terms, a cathetus is a leg of a triangle.

Was the question misread?
[quote="Adam4444"]"In a right-angled triangle the hypotenuse is twice as long as the shortest cathetus. The longer cathetu's length is the average value of the other sides. The triangle's perimeter is 90 cm. What is the area of the triangle?"[/quote]
From the given, there are [b]no[/b] possible solutions that satisfy the given since they contradict each other.

There are multiple ways to prove this:

[b]Quick and Direct Way[/b]: $ a\sqrt {3} \equal{} a\frac {3}{2} \implies \sqrt {3} \equal{} \frac {3}{2}$. Which is absurd.

[b]Another Way[/b] with the Pythagorean Theorem.
$ \frac {a \plus{} 2a}{2} \equal{} \frac {3a}{2} \implies a^2 \plus{} \left(\frac {3a}{2}\right)^2 \equal{} (2a)^2$
$ \frac {13a^2}{4} \equal{} 4a^2 \implies \frac {13}{4} \equal{} 4$, which is clearly absurd ($ a\not \equal{} 0$ because this triangle has a perimeter).

I talked to my teacher today. I did well on the test. But as you say, there is no solution to this problem. He just thought it was funny to test how critical we think. He got me. :blush:


calculus
derivative
limit
calculus computations

hi all,
how can I find the derivative of $ f(x)\equal{}1/x$ for $ x\not\equal{}0$ [u]using only [/u]the definition of derivative...?

Easy.  The quotient you desire is:
$ \frac {\frac {1}{x \plus{} h} \minus{} \frac {1}{x}}{h} \equal{} \frac{\frac{x}{x(x \plus{} h)}\minus{}\frac{x\plus{}h}{x(x\plus{}h)}}{\frac{h(x(x\plus{}h))}{x(x\plus{}h)}}\equal{} \frac {x \minus{} x \minus{} h}{h(x(x \plus{} h))} \equal{} \minus{} \frac {1}{x^{2} \plus{} xh}$
Take the limit as $ h$ goes to $ \infty$ and you get $ \minus{} \frac {1}{x^{2}}$

[quote="grn_trtle"]Don't you mean $ \lim_{h\to0}$?[/quote]typo imho....

Yes, typo.  You know what I mean. :oops:


Let $n$ black and $n$ white objects be placed on the circumference of the circle. Let's call any set of $m$ consecutive objects an $m$-chain.
a) Prove that for any natural $k\le n$ there is a chain of $2k$ consecutive pieces on the circle of which exactly $k$ are black.
b) Prove that there are at least $2$ such chains that are disjoint if $k\le \sqrt{2n+2}-2$

Here is the solution for (a):
We will first prove that there exists a chain of $2k$ objects that has at least $k$ black objects and a chain that has at most $k$ black objects.
Let's prove that there exists a chain of $2k$ objects that has at least $k$ black objects. We will assume the opposite. Than every one of $2n$ different $2k$-chains has less than $k$ black objects. Adding up all the objects in all the chains, we will count less than $2kn$ black objects. However, each object will be counted $2k$ times so we will count $2k*n$ black objects in total. This is a contradiction, so there exists a chain of $2k$ objects that has at least $k$ black objects. By analogy, there exists a chain of $2k$ objects that has at most $k$ black objects.
We will number the $2n$ objects around the circle in clockwise direction from $1$ to $2n$. Then, take the $2k$-chain  has the first object (counting in clockwise direction) being object $1$. If we take the $2k$ chain that has first object $2$, then the number of black objects in that chain will be equal to the number of objects in $2k$-chain that has first object $1$ minus $0$, if the object $1$ is white or $1$ if object $1$ is black, and plus $0$ if object $2k+1$ is white or plus $1$ if object $2k+1$ is black. Therefore the difference in the number of black objects for two $2k$-chains that share $2k-1$ objects, is at most $1$. Therefore if a given $2k$-chain starts with object $p$, and it has $x$ black objects, and another $2k$-chain starts with object $q$, and it has $y$ black objects( $y\ge x$), then among the sequences that start with objects $p+1,p+2,...,q$ there will exist at least on sequence $a_{i}, 0 \le i\le y-x$ that has $x+i$ black objects.
Now let's look at the $2k$-chain that has $v$ black objects, the most black objects among all $2k$-chains and the $2k$-chian that has $w$ black objects, the least black objects among all $2k$-chains. Because $w\le j\le v$ and there will exist a seuqnce that has the number of black objects equal to any number from $v$ to $w$, there will exist a sequence that has $k$ black objects.


function
Euler
number theory unsolved
number theory

Find the last five digits of the number in the form

$    9^{9^{9^{9^9}}} $

for 1001   9's

This problem can be found in Shklarsky, Chentzov and Yagloms book "The USSR Olympiad Problem Book" on page 15. The solution is kinda long so I won't post it here, but now you at least know where to find it.

thanks for the info.
I don't have the book, anybody care to post it?

I'll give you three digits- the method easily extends to more, with annoying computation. It also doesn't matter whether you take 1001, 2005, or 10 9's; the last few digits stabilize very quickly.

Let $a_0=1$, $a_1=9$, $a_{n+1}=9^{a_n}$.
We want to find $a_N$ mod 1000 for some large $N$. In order to do this, it suffices to find $a_N$ mod 8 and mod 125. It is clear that $a_n\equiv1\mod 8$ for all $n$. To find $a_N$ mod 125, we need to find $a_{N-1}$ mod $\phi(125)=100$. Again, factor this 100 as $4\cdot25.$ $a_{N-1}\equiv1\mod4$. To find $a_{N-1}$ mod 25, we need $a_{N-2}$ mod $\phi(25)=20$. Factor 20 as $4\cdot5.$ $a_{N-2}\equiv1\mod4$. To find $a_{N-2}$ mod 5, we need $a_{N-3}$ mod $\phi(5)=4$. $a_{N-3}\equiv1\mod4$.
Now, we build back up.
$a_{N-2}\equiv1\mod4$ and $a_{N-2}\equiv9^1\equiv4\mod5$, so $a_{N-2}\equiv9\mod20$.
$a_{N-1}\equiv1\mod4$ and $a_{N-1}\equiv9^9\equiv14\mod25$, so $a_{N-1}\equiv89\mod100$.
$a_N\equiv1\mod8$ and $a_N\equiv9^{89}\equiv39\mod125$, so $a_N\equiv289\mod1000$.

For five digits, the power of 2 branch would take two steps to clear up, and the computation would be harder, but the principle is the same.

[quote="jmerry"]

To find $a_N$ mod 125, we need to find $a_{N-1}$ mod $\phi(125)=100$. Again, factor this 100 as $4\cdot25.$ $a_{N-1}\equiv1\mod4$. To find $a_{N-1}$ mod 25, we need $a_{N-2}$ mod $\phi(25)=20$. Factor 20 as $4\cdot5.$ $a_{N-2}\equiv1\mod4$. To find $a_{N-2}$ mod 5, we need $a_{N-3}$ mod $\phi(5)=4$. $a_{N-3}\equiv1\mod4$.
[/quote]

I don't understand what you are doing with the phi function, I've not seen it used like this.
Can someone explain this please?

[quote="jmerry"]
Now, we build back up.
$a_{N-2}\equiv1\mod4$ and $a_{N-2}\equiv9^1\equiv4\mod5$, so $a_{N-2}\equiv9\mod20$.
[/quote]

how do you get $a_{N-2}\equiv9^1\equiv4\mod5$ from $a_{N-3}\equiv1\mod4$, which was at the end of the first section.

Any help welcome, thanks.

This is simply repeated use of the identity $a^{\phi(n)}\equiv 1\mod n$ when $a$ and $n$ are relatively prime. With this identity, all you need to calculate $a^b$ mod $n$ is the value of $a$ mod $n$ and the value of $b$ mod $\phi(n)$.

I was working with some related problems for a local contest, but rejected them as being too hard. Finding the last three digits of $7^{7^{7^\dots}}$ is particularly nice.

one last thing, how do you know

9^89 = 39 mod 125       I  cant calculate it

[quote="jmerry"] With this identity, all you need to calculate $a^b$ mod $n$ is the value of $a$ mod $n$ and the value of $b$ mod $\phi(n)$.

[/quote]

This is getting to me. I just can't get my head around it. Is there a proof. How do you get to $b$ mod $\phi(n)$
from what you have

Look for the theorem of Euler-Fermat, I think there will be something here and on the web

nice,jmerry.
and the answer is none of $1001's$ business.right?

Right- the answer is essentially independent of $N$, as long as $N>5$.

There are shortcuts for computing $9^{89}$ mod 125: it's $9^{-11}\equiv14^{11}$, and that can be reasonably done by hand.

The computations for 5 digits are not reasonable by hand, but they're easy to set up on a computer. The answer comes out to $45289.$ As for why you wouldn't do it by hand- that involves computing $9^{289}$ mod 625, and then $9^{289}$ mod 3125. Finally, you get the chinese remainder problem of finding what is 1539 mod 3125 and 9 mod 32.

If you want six digits, that's $745289.$ I won't go into the details there.


email
calculus
integration
limit
logarithms
geometric series
calculus computations

#1,3,4 are all routine comparisons- limit comparison is well-suited to them.
#2 is different; the idea you mention is inconclusive.
For #5, split it up- do you recognize the pieces? There aren't very many series you know how to sum exactly.

For series number 2 you can use the Integral Test or the Cauchy Condensation Test (with this last one its very simple to decide).
For series number 4,

$\sum_{n=1}^{\infty}\frac{n+5}{\sqrt{n^{7}+n^{2}}}$,

let's compare with the convergent series $\sum_{n=1}^{\infty}\frac{1}{n^{5/2}}$:

$\lim_{n\to\infty}\frac{\frac{n+5}{\sqrt{n^{7}+n^{2}}}}{\frac{1}{n^{5/2}}}= \lim_{n\to\infty}\frac{n^{5/2}[n(1+5/n)]}{\sqrt{n^{7}(1+1/n^{5})}}= \lim_{n\to\infty}\frac{n^{7/2}(1+5/n)}{n^{7/2}\sqrt{1+1/n^{5}}}$
$= \lim_{n\to\infty}\frac{1+5/n}{\sqrt{1+1/n^{5}}}= 1$.

As this limit is a nonzero real number, then both series have the same nature.

The limit comparison test is the great "de-clutterer." We use it to ignore irrelevant clutter. Take that sum Carcul just showed. What do we do with $\frac{n+5}{\sqrt{n^{7}+n^{2}}}$ ?

Within a product or quotient, we look at each factor separately, and as what it looks like for large $n.$ Since $n$ is eventually much larger than $5,$ then $n+5$ might as well be $n.$ And since $n^{7}$ is eventually much larger than $n^{2},$ then $n^{7}+n^{2}$ might as well be $n^{7}.$ That means the denominator looks like $\sqrt{n^{7}}=n^{7/2}.$ Then $n$ divided by $n^{7/2}$ is $\frac1{n^{5/2}}.$

So that's what we'll compare to: $\sum\frac{1}{n^{5/2}},$ which is a convergent $p$-series.

This whole thought process can all be done mentally, and should, with a little practice, become very quick. Carcul almost certainly thought this way - why else would he have chosen $n^{-5/2}$ to compare to? In fact, the entire argument that Carcul presented is an afterthought - a formal justification of something he already knew was true.

[quote="dukebdx12"]I have a test tomorrow and got some quick questions on some problems. I don't really need them worked out I just need help starting them. Thanks

Determine whether the series is convergent or divergent.

1)seriesSum((5-2 sqrt(x))/x^3, n, 1, infinity)
* Do I use p-series here and do I do limit comparison test?

2)seriesSum(1/(n ln(n)^2), n, 2, infinity)
* same question as above with p-series.

3)seriesSum((n+2)/(n+1)^3, n, 3, infinity)
*same as above also.

4)seriesSum((n+5)/root(n^7+n^2, 3), n, 1, infinity)
* p-series and how would I go about determining?...what can I use?

5)seriesSum(1/e^n+1/(n (n+1)), n, 1, infinity)
* I don't know where to start on this one. And find sum.

If need help tonight and am going to class now. If someone knows how to help me on these questions in more detail you can reach me at aim: xxcrew10 or my email: dukebdx12@yahoo.com Thanks[/quote]

I'm sorry that I'm responding after the test, but it is good practice for me (and I have a test this weekend :D )

[hide="1."]\[\sum_{n=1}^{+\infty}\frac{5-\sqrt{n}}{n^{3}}\]

We know that $\sum_{n=1}^{+\infty}\frac{1}{n^{3}}$ converges. Thus, $\sum_{n=1}^{+\infty}\frac{5}{n^{3}}$

must also converge. Now we may apply the comparison test:
$\sum_{n=1}^{+\infty}\frac{5}{n^{3}}< \sum_{n=1}^{+\infty}\frac{5-\sqrt{n}}{n^{3}}$

, and we conclude that this series is $\textrm{CONVERGENT}$.[/hide]

[hide="2."]\[\sum_{n=2}^{+\infty}\frac{1}{n(\ln n)^{2}}\]
Just use the integral test:
$\sum_{n=2}^{+\infty}\frac{1}{n(\ln n)^{2}}\textrm{ converges}$
$\Leftrightarrow \int_{2}^{+\infty}\frac{1}{x (\ln x)^{2}}dx\textrm{ converges}$

The integral simplifies to:$-\frac{1}{\ln(x)}+C$, so it does $\textrm{CONVERGE}$.[/hide]

[hide="3."]Notice that for all but finitely many $n$,
\[\frac{n+2}{(n+1)^{3}}< \frac{n+n}{(n+n)^{3}}= \frac{1}{4n^{2}}\]

On the right is a convergent $p$-series, so by the comparison test, it is $\textrm{CONVERGENT}$[/hide]

[hide="4."]As Kent said, compare to $\sum \frac{1}{n^{5/2}}$, a convergent $p$-series. \[\textrm{CONVERGENT}\][/hide]

[hide="5."]This sum can be separated into two different sums:
$\sum \frac{1}{e^{n}}$
$\sum \frac{1}{n(n+1)}$

The first is a convergent geometric series, and the second one can be compared with $\sum \frac{1}{n^{2}}$, a convergent $p$-series. So, the sum has to converge. $\textrm{CONVERGENT}$[/hide]

In your solution for 1, it should be

$\sum_{n=1}^{\infty}\frac{5-\sqrt{n}}{n^{3}}< \sum_{n=1}^{\infty}\frac{5}{n^{3}}$

and so we can conclude that the left series must converge.

Uh, wait a minute ...

Comparison tests are for positive term series; this can be extended to comparison tests being used (with absolute values) in comparisons to determine absolute convergence.

But $\frac{5-\sqrt{n}}{n^{3}}$ is not positive, and $\left|\frac{5-\sqrt{n}}{n^{3}}\right|<\frac{5}{n^{3}}$ is false.

The correct comparison and statement:

$\sum_{n=1}^{\infty}\frac{5-\sqrt{n}}{n^{3}}$ converges absolutely by limit comparison to $\sum_{n=1}^{\infty}\frac{1}{n^{5/2}}$

Yes Professor Merryfield, you are obviously right. Thanks for the correction.


Let $a;b;c;d$ be real numbers satisfying $0<a, b, c, d<1$. prove that:
$(1-a)(1-b)(1-c)(1-d)>1-a-b-c-d$

Well it simplifies to

$ab+ac+ad+bc+bd+cd+abcd > abc+abd+acd+bcd$

which is clearly true if $0 < a, b, c, d < 1$ (each third degree term is less than a corresponding second degree term).

[quote="Protonios"]And a difficult one I constructed. I think it is IMO level inequality.

In any acute angle triangle ABC prove that the following inequality holds:

3abc < sqrt{R*[a/cosA + b/cosB + c/cosC][a^4+b^4+c^4]}

where R is the circumradius.

Explain why equality can't hold.[/quote]
Why a 3 on the left hand side instead of the stronger $\sqrt{6\sqrt{3}}$? (There's equality in this one.)
[hide="hint"]
We can AM-GM all over the place. Then it just remains to prove upper bounds for $\sin A\sin B\sin C$ (easy) and $\cos A\cos B\cos C$ (not as easy, but doable).
[/hide]


I counted $ \boxed{13}$; did anyone find more?

[quote="luiseduardo"]How many triangles do you find in this picture ? Why ?

[img]http://img24.imageshack.us/img24/4226/calso.jpg[/img][/quote]

Label the top line vertices left to right A, B

Label the middle line vertices left to right C, D, E, F

Label the bottom line vertices left to right G, H, I

Triangles:  ACD, CDH, CGH, ACH, AGH, BEF, EFH, FHI, BFH, BIH, DEH, ABH,
CEH, DFH, CFH

15 triangles total

[quote="luiseduardo"]I counted 13, but I think the correct answer is 15 triangles, but I don't understand how to find it.[/quote]Follow Arrange your tan's solution with the labels and you should be able to find that there are 15 triangles.

Yes, that's correct. Thanks now I understand :)


geometry
AMC
AIME
percent
AMC 10
AMC 12
AMC 10 A

I have heard rumours in my school and in general around the area that the AMC scoring will change... to mods: if this causes issues, please delete this post, but I'm just wondering as to whether this is true or not. I heard that for every correct answer to an AMC (if they change the rules) is worth 5 points and every one unanswered is now worth 2.5 points. Again, this may just be a nasty rumour that gets spread around, and I do not know that this is true for sure.  :(

That is exactly what AMC Director said.

This makes people have to answer atleast 14 questions instead of 11.

The scoring on the 2007 AMC 10 A, 2007 AMC 10 B, 2007 AMC 12 A, 2007 AMC 12 B will be 6 points for a correct answer, 1.5 points for a blank answer, and 0 points for an incorrect answer.

The background reasons for the change are detailed in the Message from t he AMC 12 Chair on pages 14-15 of the 2006 High School Summary of Results and Awards (just recently finished and mailed to School Contest Managers):
" ...beginning in 2002...students could qualify for the AIME by solving as few as 11 problems. Analysis of student performance since then reveals that the average number of blank responses by AIME qualifiers increased from 8 to 11.  This trend, suggesting that many bright students are ignoring problems of even modertate difficulty, runs counter to our goals.  For that reason, beginning in 2007, the reward for a blank response will decrease from 2.5 to 1.5 points."

The decision for the change was made in April 2005, announced in the High Summary of Awards and Results, again announced this year in the 2006 High School Summary of Awards and Results.  I have also announced it in this Forum.  The point scoring system is described on the front cover of each 2007 contest pamphlet.

Steve Dunbar
AMC Director

Thank You

Now this means I have to get a perfect on AMC 10 to improve my score.

That's a brilliant idea, Mr. Dunbar.

Although that makes my chances of qualifying for the AIME significantly lower, since I usually leave 6-7 answers blank on the AMC 12s.

so now we need to answer a few more problems to qualify on amc 12? thats not cool  :D

With the new scoring, it makes it impossible to get a 99.5.

Also, isn't the expected value of each question now zero?  Thus, if  you can eliminate one choice, it's to your benefit to guess.

[quote="mistapotta"]Also, isn't the expected value of each question now zero?  Thus, if  you can eliminate one choice, it's to your benefit to guess.[/quote]
It's still "guessing negative". That is, you get 1.5 points for leaving a problem blank (37.5 for leaving them all blank) and an expected 1.2 for guessing blindly (an expected 30 for guessing on all of them).

Don't guess (you'll guess wrong), but do be a little more aggressive, especially in answering questions that where you think you know the answer but maybe haven't proved it with full rigor to your own satisfaction.

I think it's a good change. I was one of the coaches who was aware of the consequences of the former scoring rule, and advised the students on my team to work conservatively when taking the AMC 10 and AMC 12 with a goal to qualifying for the AIME. My strongest student would typically do a few more problems than the minimum necessary for qualifying for the AIME, getting all of those right, and leaving all other problems blank. (He sometimes skipped ahead to later problems that were easier--for him--than lower-numbered problems, but he definitely disregarded some problems entirely.) I'm all for the rules encouraging all test-takers to attempt somewhat more problems.

My top students are also encouraged, as a point of personal development, to improve their scores year-on-year, and to attempt to gain a state championship score on one of the tests in the set {AMC 10A, AMC 12A, AMC 10B, AMC 12B}, as one of my students has done more than once. I am still waiting for one of the students on my team to ace an AMC test, which is a reasonable goal for a few of them now.

I agree with tokenadult: it's a good change. In fact, if you read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24818&start=20]this old topic[/url] you'll see that I advocated exactly this change. For anyone seeking to understand this, I suggest you read that old topic, paying particular attention to AMCDirector's posts.

When I said "be more aggressive" I really meant "go ahead and answer it, and move on to another problem."

Yeah I guess they did give a bit too much credit for questions left blank.

Will AIME qualifying scores stay the same - 120 for AMC 10 and 100 for AMC 12 or top 1% and top 5% respectively? (If so, it will be more difficult to qualify this year. Do you expect fewer AIME participants and therefore AIME participation will be more valuable?)

Will it be hard to compare 2007 results with the previous results and make any sense out of it?

[quote="Xantos C. Guin"][quote="randomdragoon"]They should make it so that you start with 37.5 points, get +4.5 for a right answer, 0 points for a blank, and -1.5 points for an incorrect answer. This way, the "guessing expected value" is clearly negative (while it looks 0 on the old [new] system if you don't think hard enough) :P[/quote]

I think that they should change it so you start with 150 points, and lose 4.5 for a blank, and lose 6 for a wrong answer.  :wink:[/quote]

Same thing, but with more pessimism  :|

[quote="Go Around the Tree"]lol then everyone would receive a perfect score if they wanted[/quote]

 :spider:

(the above picture is pretty confused looking  :P )

how could everyone receive a perfect score then?

[quote="Phelpedo"][quote="randomdragoon"]They should make it so that you start with 37.5 points, get +4.5 for a right answer, 0 points for a blank, and -1.5 points for an incorrect answer. This way, the "guessing expected value" is clearly negative (while it looks 0 on the old [new] system if you don't think hard enough) :P[/quote]

Does it really require much thought to realize that $5(1.5) > 6$?[/quote]
You might think that there are 4 incorrect responses and 4(1.5)=6. I've seen some people reason this way..

(like one of my grandparent's friends bargained something down from $\$$2 each to 4 for $\$$10...)

I shall score a 130+ on this year's AMC 10!

lol  :rotfl:

[quote="Boy Soprano II"][quote="randomdragoon"]You might think that there are 4 incorrect responses and 4(1.5)=6. I've seen some people reason this way..[/quote]
Umm, well, I think most people capable of qualifying for the AIME would be quite capable of working all these excruciatingly difficult details of expected value, and anyhow, unless I'm mistaken, these subtle minutiae of adding 0 to your score are not exactly the raison d'être of the AMC.[/quote]
Well, considering that you used to be able to qualify for the AIME by answering only 11 questions...

i believe the change to 1.5 was almost to cut down a little?
specifically cut down on the people who were settling for like, 12 answered and 13 blank on the AMC 12

[quote="jmerry"]I wouldn't be surprised to see the qualification line dip below 100 this year- the AMC can't precisely calibrate the difficulty of their tests, and the new scoring system will make comparisons to old tests less relevant. The 100 line is simply the guarantee- they won't raise the bar any higher if the test is too easy, but they will lower it if the test is too hard.[/quote]

The AIME qualification cutoff is supposed to be the top one percent or the above 100, whichever is easier. Last year nearly 10% of people who took either the AMC10 or 12 (about 26,000) qualified for the AIME, pretty much twice as many as any other year. So I doubt if they'll need to lower the bar.

No, it's top 5% for AMC12.  AMC10 is top 1% or anything over 120, although until recently, it was just the top 1%.

With the new scoring, I'd be very surprised if the AMC10 cutoff isn't lowered.  The new "minimal strategy" for the 10 requires getting 19 questions right, and that's if you leave at least 4 of the remaining 6 blank.  Otherwise you need 20 correct answers, which is not that easy.

With regards to AMC 12, will the problems get minutely easier or remain the same difficulty?

[quote="randomdragoon"]Well, considering that you used to be able to qualify for the AIME by answering only 11 questions...[/quote]
Okay, perhaps I should change my "would" to a "should"? :wink:

I agree with Phelpedo on the difficulty.

The one thing you can do, get more people to participate, lowering the 1%/5% bar.

i think changing massive amounts of people's minds to participate in AMC is quite  harder than studying problems :D

I think one of the reasons why this year's AMC 8 was so easy was because they raised the bar for AMC 12.....(in return giving easy problems so you'd still average about the same......so I'm predicting that this year's AMC12 will be easy).. hopefully...

[quote="Xantos C. Guin"][quote="randomdragoon"]They should make it so that you start with 37.5 points, get +4.5 for a right answer, 0 points for a blank, and -1.5 points for an incorrect answer. This way, the "guessing expected value" is clearly negative (while it looks 0 on the old [new] system if you don't think hard enough) :P[/quote]

I think that they should change it so you start with 150 points, and lose 4.5 for a blank, and lose 6 for a wrong answer.  :wink:[/quote]

Sounds good. Haha...

Really, though, I think it's a great change. I'll do a bit worse, but that's alright. Now I'm absolutely encouraged to divide my time wisely enough to attempt all of the problems.


Thanks to tetrahedr0n who has just sent me the TEXed copy of the test, I am now ready to allow people to take it early.

The official date of the Mock AMC I will be:

Saturday, January 29, 2005 at 4:00 PM Eastern Time.

If this date and time is not good for you, you may send me a message and I will allow you to take it early so long as I get your results before 5:15 PM.

This AMC is roughly of a Mock AMC-10 difficulty level... maybe slightly harder but it is by no means anywhere near the level of difficulty of AMC12.

Also, can anyone think of a better name than Mock AMC I? As I stated in the other thread, I really don't like I because it looks too much like 1.

[quote="Lucky707"]Also, can anyone think of a better name than Mock AMC I? As I stated in the other thread, I really don't like I because it looks too much like 1.[/quote]

Mock AMC 9?  Mock AMC IX?

Sent it to me tomorrow if possible

like to do it before school start...

I would also like it tomorrow, please

Thanks!

I will send it to you now and you can take it tomorrow whenever you'd like.

What will we call our mock AMC's after we run out of letters?

Greek letters. Mock AMC $\alpha, \beta, \gamma, \phi, \theta ...$

Or we could just repeat letters, like AA, BB, CC...

Please send it to me on the 24th.  Thanks.

-interesting_move

Hi
On saturday i will be gone for most of the day and all of sunday i will be gone so i won't have time to take it that weekend

could you send it to me when you see this message? thanks

i'm pmed you the same thing, sorry

could you send me the test when you get this message? thanks :D

Just found out I have to go to a last-minute-practice math club meeting at the same time.  If it's not too much trouble, Lucky707, please send me the test as soon as you can.

Thanks,

mathfanatic

Can send it to me asap.. As I am in SIngapore.. My time dun match ur time..

Can I take part as well? PM it, if possible.

Hello lucky

I won't be able to do it at 4pm tomorrow so the only time I can possibly work on it is either now around 10pm or tomorrow after 6pm.

Either way, can you still send me the questions?  I'd like to have a challenge even if I don't get my score back.

Thanks. :)


calculus
integration
limit
logarithms
trigonometry
function
calculus computations

I'm having trouble solving the following integral:
\[ \int_{ - \infty}^\infty\cfrac{1}{e^x + e^{ - x}}\: dx
\]
My attempt:
\begin{align*} \int_{ - \infty}^\infty\cfrac{1}{e^x + e^{ - x}}\: dx & = \int_{ - \frac {\infty}{i}}^\frac {\infty}{i}\cfrac{i}{e^{iu} + e^{ - iu}}\: du \\
& = i\int_{ - \frac {\infty}{i}}^\frac {\infty}{i}\sec{u}\: du \\
& = i\left(\lim\limits_{a\to - \frac {\infty}{i}}\int_a^0\sec{u}\: du + \lim\limits_{b\to\frac {\infty}{i}}\int_0^b\sec{u}\: du\right) \\
& = i\Bigg[\bigg(\ln\left|\sec{0} + \tan{0}\right| - \ln\left|\cfrac{1 + \sin{( - \frac {\infty}{i})}}{\cos{( - \frac {\infty}{i})}}\right|\bigg) \\
& \quad + \bigg(\ln\left|\cfrac{1 + \sin\frac {\infty}{i}}{\cos\frac {\infty}{i}}\right| - \ln\left|\sec{0} + \tan{0}\right|\bigg)\Bigg]\qquad\qquad\text{ This step may be a big mistake : (} \\
& = i\left(\ln\left|\cfrac{1 + \sin\frac {\infty}{i}}{\cos\frac {\infty}{i}}\right| - \ln\left|\cfrac{1 - \sin{( \frac {\infty}{i})}}{\cos{( \frac {\infty}{i})}}\right|\right) \\
& = i\ln\left|\cfrac{1 + \sin\frac {\infty}{i}}{1 - \sin\frac {\infty}{i}}\right| \end{align*}
Dunno where to go from there.. Is it even right to work with trig functions with $ \infty$ in them? Please help. :)

$ \int \frac {1}{e^x \plus{} e^{ \minus{} x}}\ dx \equal{} \int \frac {1}{1 \plus{} t^2}\ dt\ (Let\ t \equal{} e^x)$

First notice that this function is even

$ \therefore\quad\int_{-\infty}^{\infty}\frac{dx}{e^x+e^{-x}}$

$ =2\int_0^{\infty}\frac{dx}{e^x+e^{-x}}$

$ =2\int_0^{\infty}\frac{dx}{e^x+e^{-x}}\cdot\underbrace{\frac{e^x}{e^x}}_{\text{multiplicative identity}}$

$ =2\int_0^{\infty}\frac{e^x}{e^{2x}+1}\,dxdx$

$ \underbrace{=}_{u=e^x}2\int_1^{\infty}\frac{du}{1+u^2}$

$ =2\arctan(u)\bigg|_1^{\infty}$

$ =2\bigg[\arctan(\infty)-\arctan(1)\bigg]$

$ =2\bigg[\frac{\pi}{2}-\frac{\pi}{4}\bigg]$

$ =\frac{\pi}{2}$

[quote="undefined117"]Okay that is [i]awesome[/i] :lol:

So:
\begin{align*} \int_{ - \infty}^\infty\cfrac{1}{e^x + e^{ - x}}\: dx & = \int_{ - \infty}^\infty\cfrac{1}{1 + t^2}\: dt \\
& = \int_{ - \frac {\pi}{2}}^\frac {\pi}{2}\cfrac{\sec^2\theta}{1 + \tan^2\theta}\: d\theta \\
& = \bigg[\theta\bigg]_{ - \frac {\pi}{2}}^\frac {\pi}{2} \\
& = \boxed{\pi} \end{align*}
ehh I did something wrong.. my calculator says the answer is $ \cfrac{\pi}{2}$..

btw you're up pretty late :D[/quote]

Your error is in your limits.  Note that you let $ e^x = \tan \theta$, and so $ \lim_{x \rightarrow \infty} \arctan e^x = \pi/2$ and $ \lim_{x \rightarrow -\infty} \arctan e^x = 0.$

whoops, yeah I realized that mistake after I got in bed. :blush: I should probably stop going crazy with the complex form and look for easier alternatives like this. XD That makes sense. Thanks all!


1,2 Dimethyl cyclohexan-1,2-diol + $ H^{\plus{}}$

Isnt that pinnacol pinnacalone rearrangement  :D
[hide]
you get 2,2 dimethyl cyclohexanone  if initial compd is cis
you get 1 acetyl 1 mwthyl cyclopentane if initial compd is trans  :)
This is bcos the migrating grp enters trans to the departing grp
[/hide]

dont blabber -48 cry baby ranga who watches chutti tv (yes i heard you singing the song that comes i that dreadful channel today). I even know the rn mechanism want it?  :D  :D

And btw after reading this plz do not cry  :rotfl:


I was wondering if we should have a Games forum replacemants thread like in the Mafia forum. That way all empty slots can be found in the same place.


Find the values of $a,b$ such that $\lim_{x\rightarrow 0}\frac{1}{x^{3}}\left(\frac{1}{\sqrt{1+x}}-\frac{1+ax}{1+bx}\right)$ exists, then find the limit value.

That's essentially correct, but if $8b-8a=4$ and $3=4b$, then $b=3/4$, $a=1/4$, and the limit is $9/32$.

Hm, there must be a mistake in the earlier post, because the limit is actually $-1/32$.  I just followed hydro's work without doing any of my own :)

ok, to complete my solution I should add one term in the Taylor's series of $(1+x)^{-\frac{1}{2}}$. So $o(x^{2})=-\frac{5}{16}x^{3}+o(x^{3})$

If we substitute it in my earlier post, we get $\lim_{x \rightarrow 0}\frac{(8b-4-8a)x+(3-4b)x^{2}+(3b-\frac{5}{2})x^{3}+8o(x^{3})+8bxo(x^{3})+o(x^{4})}{8x^{3}+bx^{4}}$

So if $8b-4-8a=0$ and $b=\frac{3}{4}$, the limit is given by $\frac{(3b-\frac{5}{2})x^{3}}{8x^{3}}=-\frac{1}{32}$

Maybe Kunny wants something without Power Seres ?  :D
[hide]
it is equivalent to

$\lim_{x\to 0}\frac{1}{x^{3}}\left(\frac{x(-a^{2}x^{2}+(b^{2}-a^{2}-2a)x+2b-2a-1}{\sqrt{1+x}(1+bx)((1+bx)+(1+ax)(\sqrt{1+x})}\right)$

So if limit exist , then we need to solve

$b^{2}-a^{2}-2a=0$ and $2b-2a-1=0$

 :)
[/hide]


Find a function $f: \mathbb N \to \mathbb N$ such that \[f(f(x))=1993\cdot x^{1945}\] for all $x \in \mathbb N$.

[quote="rahulakaneo"]Find a function f(x) such that f(f(x))=1993*(x^1945) for all x in N(The Set Of Natural Numbers) ?[/quote]

This is a classical problem with classical solution :

[hide="My solution"]
Let $ u(x)\equal{}1993x^{1945}$.
Let $ A\equal{}u(\mathbb N)$
Let $ B\equal{}\mathbb N\backslash A$

$ u(x)$ is injective and $ u(x)>x$ $ \forall x>0$
So, for any $ x>0$, there is a unique couple of integer $ n(x)\geq 0$ and $ b(x)>0$ such that $ b(x)\in B$ and $ x\equal{}u^{[n(x)]}(b(x))$ where $ u^{[m]}$ is the composition $ u\circ u\circ u ... \circ u$ $ m$ times (with $ u^{[0]}(x)\equal{}x$ and $ u^{[1]}(x)\equal{}u(x)$)

Cardinal($ B$) is infinite and so we can split $ B$ in two infinite equipotent sets $ B_1$ and $ B_2$ :
$ B_1\cup B_2\equal{}B$
$ B_1\cap B_2\equal{}\emptyset$
$ \exists$ a bijection  $ h(x): B_1\to B_2$

Then we can easily define $ f(x)$ as :

If $ b(x)\in B_1$ : $ f(x)\equal{}u^{[n(x)]}(h(b(x)))$

If $ b(x)\in B_2$ : $ f(x)\equal{}u^{[n(x)\plus{}1]}(h^{[\minus{}1]}(b(x)))$
[/hide]


[quote="sri340"]Suppose all the three points (–2, 10), (1, –8) and (4, 10) lie on the graph of  y =   a^x2 + bx + c .
What is abc ?[/quote]

$ 10\equal{}4a\minus{}2b\plus{}c$      (i)
$ \minus{}8\equal{}a\plus{}b\plus{}c$         (ii)
$ 10\equal{}16a\plus{}4b\plus{}c$    (iii)

from (i)(ii) and (iii) we get $ a\equal{}2,b\equal{}\minus{}4,c\equal{}\minus{}6$

$ \frac{\minus{}2\plus{}4}{2}\equal{}1$.
So vertex is $ (1, \minus{}8)$.
Therefore $ y\equal{}a(x\minus{}1)^2\minus{}8$.
Now $ 10 \equal{} a(4\minus{}1)^2\minus{}8 \iff a\equal{}2$ by $ (4, 10)$.
Thus $ y \equal{} 2(x\minus{}1)^2\minus{}8 \equal{} 2x^2\minus{}4x\minus{}6$.

$ \boxed{(a, b, c)\equal{}(2, \minus{}4, \minus{}6)}$


Καλησπέρα. Με προβληματίζουν κάποιες ασκήσεις στις οποίες δεν έχω καμία ιδέα τι να κάνω και θα εκτιμούσα την οποιαδήποτε βοήθεια. :)

-----------------------------------------

Έστω $ \gamma$ απλή, κλειστή και τμηματικά λεία καμπύλη στο $ \mathbb{C}$ με θετική φορά διαγραφής. Υποθέτουμε ότι η συνάρτηση f είναι αναλυτική πάνω και στο εσωτερικό της $ \gamma$ και ότι τα $ z_1.z_2,\ldots,z_n$ στο εσωτερικό της $ \gamma$ είναι διάφορα μεταξύ τους. Αν:

$ Q(z)\equal{}(z\minus{}z_1)(z\minus{}z_2)\cdots(z\minus{}z_n)$

να αποδειχθεί ότι το

$ P(z)\equal{}\frac{1}{2\pi i}\oint_{\gamma}\frac{f(w)}{Q(w)}\frac{Q(w)\minus{}Q(z)}{w\minus{}z}dw$

είναι το μοναδικό πολυώνυμο n-1 βαθμού που έχει τις ίδιες τιμές με την f στα σημεία $ z_1.z_2,\ldots,z_n$, δηλαδή $ P(z_k)\equal{}f(z_k),\: k\equal{}1,2,\ldots,z_n$

[quote="modulo"]Καλησπέρα. Με προβληματίζουν κάποιες ασκήσεις στις οποίες δεν έχω καμία ιδέα τι να κάνω και θα εκτιμούσα την οποιαδήποτε βοήθεια. :)

-----------------------------------------

Έστω $ \gamma$ απλή, κλειστή και τμηματικά λεία καμπύλη στο $ \mathbb{C}$ με θετική φορά διαγραφής. Υποθέτουμε ότι η συνάρτηση f είναι αναλυτική πάνω και στο εσωτερικό της $ \gamma$ και ότι τα $ z_1.z_2,\ldots,z_n$ στο εσωτερικό της $ \gamma$ είναι διάφορα μεταξύ τους. Αν:

$ Q(z) \equal{} (z \minus{} z_1)(z \minus{} z_2)\cdots(z \minus{} z_n)$

να αποδειχθεί ότι το

$ P(z) \equal{} \frac {1}{2\pi i}\oint_{\gamma}\frac {f(w)}{Q(w)}\frac {Q(w) \minus{} Q(z)}{w \minus{} z}dw$

είναι το μοναδικό πολυώνυμο n-1 βαθμού που έχει τις ίδιες τιμές με την f στα σημεία $ z_1.z_2,\ldots,z_n$, δηλαδή $ P(z_k) \equal{} f(z_k),\: k \equal{} 1,2,\ldots,z_n$[/quote]

Ok, πρέπει λοιπόν να κάνεις τρία πράγματα:
1. Να δείξεις ότι η $ P$ είναι πολυωνυμο βαθμού $ n \minus{} 1$
2. Να δείξεις ότι $ P(z_k) \equal{} f(z_k)$
3. Nα δείξεις ότι είναι το μοναδικό τέτοιο πολυώνυμο.

Τα 1-2 είναι απλή εφαρμογή λογισμού υπολοίπων και του ολοκληρωτικού τύπου του Cauchy (πες μου να γράψω την πλήρη λύση αν κολλήσεις παρολ'αυτά). Όσο για τη μοναδικότητα του 3... αν ένα πολυώνυμο βαθμού $ n \minus{} 1$ έχει $ n$ ρίζες... :D

Εύκολο μεν... αλλά πολύ όμορφο και σημαντικό αποτέλεσμα!

Cheerio,

Durandal 1707


A square with side $ n$ is divided into $ n^2$ unit squares. What is the largest $ n$ for which one can mark $ n$ unit squares such that every rectangle, having sides at gridlines and area at least $ n$, contains at least one marked square in its interior?

I must be misreading the problem; it seems to be trivial - 1!

It's definitely not - please read the problem carefully.

[hide="really?"]The set of rectangles in question includes unit squares, so all unit squares must be filled...[/hide]


logarithms
complex numbers
topology
complex analysis
complex analysis unsolved

Sketch an image under $ w \equal{} \text{Log} (z)$ of the half disk                $ |z| < 1, \text{Re} (z) > 0$.

To this point, I do not have much experience with mappings under the principal value mapping of the logarithm.  I know that

$ \text{Log} (z) : \equal{} \ln | z | \plus{} i \cdot \text{Arg} (z)$.

However, I do not see what this image would look like.

So $ |z|<1$ implies that $ Log|z|$ is in $ \minus{}\infty$ to $ 0$ (but not including zero)

Now if you have the part of the disk where the real part is positive, that means that the angle is between $ \minus{}\frac{\pi}{2}$ and $ \frac{\pi}{2}$

So the answer is that the graph is the strip $ x\in (\minus{}\infty,0)$ and $ y\in (\minus{}\pi/2,\pi/2)$

(note this is an open set)


Euler
modular arithmetic
number theory
greatest common divisor
calculus
derivative
function

Whew, it's been a long time since I made a post around here. Anyways, I am wondering whether anyone happens to know the proof that...

If [tex]F_n[/tex] is the nth Fermat number, then any factor of [tex]F_n[/tex] must be of the form [tex]k2^{n+2}+1.[/tex]

Or, better yet, does anyone know this proof for a generalized Fermat number of the form [tex]a^{2^n}[/tex]? If so, please let me know.

Thanks.

this is quite easy:

let $F_n=2^{2^n}+1$ and suppose $p|F_n$ where $p$ is a prime. the cases $n\leq 1$ are easily checked by hand. so suppose $n\geq 2$.
then we know that the order of 2 in $\mathbb{Z}/p\mathbb{Z}$ is exactly $2^{n+1}$. therefore $2^{n+1}|p-1$. in particular, $p\equiv 1\mod 8$, therefore 2 is a square $\mod p$. therefore the order of $\sqrt{2}\mod p$ is exactly $2^{n+2}$, so $2^{n+2}|p-1$, so we're done.

As a little FYI, in this context, 'order' means the smallest $t$ such that $a^t\equiv 1\bmod p$

Saying the samethig but  it may be clear enough for those whe just understand Fermat't  Little theorem.

If p   is prime, and divides (2^(2^n) + 1)

(We assume all primes here are >2,  etc..)

Since p is prime, FLT gives 2^(p-1) == 1 (mod p)    (eq 1)

if h is order of 2, that is  h is smallest  positive number such that 2^h = 1 then  h must divide (p-1).
We will proe that h=2^(n+1)

proof:  since p divides F(n) we also have 2^(2^n) == -1  (mod p)
or (2^(2^(n+1)) == +1  (mod p)
so  h divides (2^(n+1)) ... ( but does not divide 2^n)

So  2^(n+1) is  "order of 2" (smallest such number where you will get  2^somenumber_greater_than_zero ==1)

(Why? only factors of 2^(n+1) are 2 (or powers of 2 ) so if any power of 2  was +1 earlier, 2^(2^n) would be +1 too, but we kno w that 2^( 2^n ) is  -1) ) hence  2^(n+1) is order of 2.

So we have h = 2^(n+1) and since h divides (p-1)  , 2^(n+1) divides  (p-1)   or p=k*2^(n+1)+1.

This was known long ago,  Further factor of 2 comes because: ...

We can go a little  more, b   by obserrving  that (for n>2) p is 1 mod 8
And for   odd p which is 1 mod 8,  2 is a  square (This is a little advanced, but a standard result of number theory)

which tells up  2^((p-1)/2) is  == +1 (mod p)

(The general result is  2^((p-1)/2) is either +1 or -1 depending on whether  p is +-1 mod 8 or +-3 mod 8)
so you can now say that     2^(n+1) is a factor of  ((p-1)/2)))

QED.

PS.. That is why  Euler tested only the  numbers of the form  (64k+1) for  F_5, actually he  need have tried only 128k+1 type factors to reach 641 as a factor of (2^32 + 1)

Hope this helps.

The similar result will also hold  greneralized (a^(2^n)+1) type  numbers,  Factors would be of the type k*(2^(n+1)+1
The extra factor of 2 may not  always come...

I appreciate everyone's detailed responses. Thanks for the help. I just have one question:

[quote]if h is order of 2, that is h is smallest positive number such that 2^h = 1 then h must divide (p-1). [/quote]

Why must h divide p-1?

edit: Nevermind about my question, I figured it out. The key to the proof of the lemma involves writing p-1 as qd+r.

The order of any element mod $p$ must always divide $p-1.$ More generally, if $a$ is relatively prime to $n$, the order of $a$ mod $n$ must divide $\varphi(n).$

You need to first prove this lemma: if $a^j\equiv 1\pmod{n}$ and $a^k\equiv 1\pmod{n}$ and $m=\mathrm{gcd}(j,k),$ then $a^m\equiv 1\pmod{n}.$

Then by Fermat and/or Euler, the order of $a$ must be the gcd of the order itself and $\varphi(n)$ (or $p-1),$ so it must divide $\varphi(n)$ (or $p-1).$

To add to  Kent's post above,  This method, sometimes provide a powerful  well known method to prove, if a number is prime.

Assuming  gcd(a,p)=1 and
Suppose we know that order of    a  is (p-1)

This means p is prime.

If (p-1) is easily factorable , then it  it becomes easy to check if  (p-1) is   the order or not. For example,  for every factor   q of  (p-1) if we can show that a^((p-1)/q) is not 1  (and of course, a^(p-1) is 1) that means p is prime.

Hence the famous test,   F_n if prime if and only if  (3)^((F_n -1)/2 )  == -1 (mod F_n) .
(I  wonder, how come  Fermat did not check that for F_5, as it would have been easy to check  this by hand) and he certainly should have known the above.)

[quote="Gyan"](I  wonder, how come  Fermat did not check that for F_5, as it would have been easy to check  this by hand) and he certainly should have known the above.)[/quote]

A historian could possibly resolve this issue, if the primary sources are detailed enough. Most likely, Fermat's attention shifted elsewhere, and he didn't return to that problem with the best tools he had after he first announced his conjecture.

A careful consideration of dates is why we can be quite sure that Fermat [b]didn't[/b] really have a proof for his conjecture (later proved by Andrew Wiles) that became known as Fermat's Last Theorem. He published later a proof of a weaker special case, which is good indirect evidence that by then he knew he didn't have a proof of the general case.

That's not necessarily true. Apparently Fermat liked to challenge other mathematicians (especially the English ones) to prove things that he proved. So it is possible that he published the proof of a special case and left the rest for others to figure out on their own. Back then, people didn't always publish results immediately (or at all) because there weren't such strange rules about becoming tenured professors. In fact, Fermat wasn't a professional mathematician: he was a lawyer.

[quote="ComplexZeta"]That's not necessarily true. Apparently Fermat liked to challenge other mathematicians (especially the English ones) to prove things that he proved. . . . In fact, Fermat wasn't a professional mathematician: he was a lawyer.[/quote]

That doesn't apply to the "Fermat's Last Theorem" problem statement. The statement of the full problem occurs only as a marginal note in a book Fermat read EARLY in his investigations. It was never published to anyone until after Fermat died,  when the annotated book was published by his son. Fermat was surely mistaken (well, surely to me) at the moment he wrote that note in thinking that he had a full proof of the full statement, because he did publish only the special case quite a while later, when he had more paper at hand and had had more time to think about the issue.   :D

I have known for years that Fermat was a lawyer by occupation (a judge, actually) and that is indeed why many of his conjectures didn't pan out as he supposed--he was just too busy to investigate all of them fully. He had immense insight into math--the best insight any amateur has ever had, some say--but he didn't have enough hours in the day to tie up all his loose ends.

To add some more comments here:

Tokenadult - Apart from  Fermat's Last Theorm, ("which he didn't have enough margin to write a simple proof " etc..) he is also famous for: Fermat's Numbers, and Fermat's Little Theorem (Some how, in our family it is called "Little Fermat's Theorem" :)  - a name adopted when my son was litttle when he came to know about it.. that's another story...) . And  I was commenting about the   these two aspects

Fermat's  numbers  are      $2^{2^n} + 1 $           His conjucture, that all such numbers are prime,  was proven wrong for n=5 by Euler.

Ferma't Little Theorem that   if n is prime then  $ (a^n - n )$   is divisibble by  $n$  wasn't  "his" but he certainly was the one who popularized it.

What I was wondering  in the last  post was,  assuming  Fermat  knew the  Little therorm well,  how come he din not check  F_5  by  checking  $3^{2^{32})}$ . It would have taken only 32 squaring, and  even without  a calculator it would have taken at best a few hours.

I  may   seem a little harsh here, (hind sight being 20-20 etc.. and even great mathematicians may misss simple things etc..) but  it seems to me that that would have been a very natural thing to  try.  After all, it is one of the first , practical use of the theorm, that a a person learns when he  when (s)he wants to check  if a number could be prime. (well actually  the 'second ' thing, because 'first' thing   being trial divison  which  is not practical  for large numbers - and hence finds the practical use of the theorem)

When I was   in middle school( (This was before the age of calculators and computers :))
and learned the  Fermat;s Little theorem,  one of the first  thing, I remember  I did with it was to settle a question which was being discussed  among us ( the question was raised by  some one and no  one among us - including all the teachers etc - knew the answer ) was if "11111111111" was prime or not , I remember showing that  it was not a prime  using the theorem and it was really not that   time consuming.. . and  this number is not less than F_5  that's make me wonder if Fermat really played around   with this theorem .

Interesting part about trying  with the number  3 as  witness, is one can  settle the question about primaliy of Fermat's number.  ( That is,  $F_5$  is prime if and only if  $3^{2^{31}} $  is $  -1 (mod F_5) $ ) This, of course, is known as  Pepins test, and  was not known at Fermat's time.   but if one tried and saw the number is not -1, it would have definitley proved that the number is not prime, and Fermat should have known that..

Hence I wonder...

Yes, Gyan, I understand you were writing about other theorems of Fermat's. I think the answer is still the same: Fermat considered one question at one time, and considered the theorem about the tool that can answer that question at another time, and just didn't happen to apply the tool (which, as you point out, he surely knew about) to the problem he considered on the other occasion. I suggest that the way to confirm this conjecture of mine about the historical facts would be to check the exact dates on which he was writing about each issue. Maybe the surviving sources don't make completely clear what Fermat knew when, but they might.

When I teach our History of Mathematics course, one line I like to use is this: number theory as we know it originated with Euler's encounter with Fermat's posthumously published annotated copy of Diophantus's [i]Arithmetica[/i]. As far as I know, it was Euler who first wrote out a satisfactory proof of 'Fermat's Little Theorem." The special place of "Fermat's Last Theorem" is that of all of Fermat's unsupported or poorly-supported statements about number theory, it's the only one the Euler didn't settle one way or the other.

The 17th century is a tremendously important time for the development of mathematics, but it wasn't known for particularly high standards of proof and rigor. In fact, you could argue that calculus could only surge forward by papering over issues about limits, infinitesimals, etc. with obscuring rhetoric rather than truly facing them. (Hence, Bishop Berkeley's scathing attack on "ghosts of departed quantities.")

Fermat himself was a major contributor to the development of calculus. He could draw tangent lines essentially by finding derivatives, and he recognized the principle that a function has a maximum of minimum only if its derivative is zero. (That was anachronistic language: there were no functions in the 17th century, only variables.) I've even seen an argument attributed to Fermat that (in our terms) computes the integral of $x^n$ for any $n$ (improving on Cavalieri who could to it for integer $n$ between 1 and 6 - or was that 9.)

But Fermat was handicapped by not really having the full power of modern algebraic notation at his disposal. His notation was mostly inherited from his predecessor, Viete. The first to consolidate an algebraic notation that we would recognize was Descartes - and although Descartes and Fermat were both Frenchmen whose lifespans overlapped, they had very little, if any, interaction.

[quote=Quaratinium]I mean, there is many theorems that Descartes is known for.[/quote]

??? why the revive?

Clearly he wanted you to know about the many theorems of Descartes, including analytic geometry (wait, no, that isn't a theorem, that a tool/branch), whatever "virtual velocities" are, Bell (wait, no, that's physics), "Cogito ergo sum" (wait, no, that's philosophy, and also kinda circular). Uh, something about the difference between mind and body? Optics? That soda bottle experiment middle schoolers do where they squeeze the bottle and some pocket of air goes shooting down?

Huh. I can name a bunch of stuff Descartes is known for, but not a single theorem.

[hide = Okay, to be fair]
he, like Fermat, helped popularize algebraic notation and the rudiments of calculus. Also, something about modernizing the Church's philosophies, which wasn't quite as important, because much of it went unpublished.

And there is in fact one theorem named after him: [b]Descartes' rule of signs[/b].
[/hide]

What about [url=https://en.wikipedia.org/wiki/Descartes%27_theorem]Descarte's kissing circles theorem?[/url] :D


integration
algebra
function
domain
calculus
advanced fields
advanced fields unsolved

[b]A[/b]) Let $u \in \mathcal{C}^{2} (\mathbb{R}^{2})$ satisfying $u(x,0) = u_{t}(x,0) = 0$. Consider for each fixed $(x,t) \in \mathbb{R}^{2}$, with $t \neq 0$ the region $\mathcal{D}(x,t)$ delimited by the x axis and by the lines passing trhu the points $(x,t), (x-t,0)$ and $(x,t), (x+t,0)$. Show that

$u(x,t) = \frac{1}{2}\int_{\mathcal{D}(x,t)}(u_{tt} - u_{xx})dxdt$

[b]B[/b]) Let $q \in \mathcal{C}(\mathbb{R}^{2})$ with
$\int_{\mathbb{R}^{2}} |q(x,t)|dxdt \leq 1$.

Use item (A) to show that if $v \in \mathcal{C}^{2}(\mathbb{R}^{2})$
is such that

$v_{tt} - v_{xx} + q(x,t)v = 0$ in $\mathbb{R} \times \mathbb{R}$
$v(x,0) = 0 , x \in \mathbb{R}$
$v_{t}(x,0) = 0, x \in \mathbb{R}$

then $v \equiv 0$

Thanks everybody!

Ok, I was struggling with this problem and I realised some facts I would like to share with you and hear some opinions.

For [b]A[/b] I realised that this is just the Duhamel Principle applied to solve an particular wave equation in one dimensional space.  In other words, I tried to solve
$u_{tt} - u_{xx} = w(x,t)$
$u(x,0) = 0$
$u_{t}(x,0) = 0$

Using Duhamel principle I find out that the solution $u$ to this problem is given by the formula
$u(x,t) = \int_{\mathcal{D}(x,t)} w(x,t) dxdt$

Now my problem is item [b]B[/b].

Using the preceeding result we arive at

$v(x,t) = \int_{\mathcal{D}(x,t)}-q(x,t)v(x,t)dxdt$

Now I reasoned as follows.

I) If $\int_{\mathbb{R}^{2}} |q(x,t)|dxdt \leq 1$ then we must have $\int_{\mathcal{D}(x,t)} |q(x,t)|dxdt \leq 1$, since we are restricting the domain of integration of non negative function from the whole space to an isosceles triangle.

II) Since $v \in \mathcal(C)^{2}(\mathbb{R}^{2})$ then $v$ must be bounded in every  $\mathcal{D}(x,t)$ since the triangle is a closed set.

III) Fix $x \in \mathbb{R}$ we have
$|v(x,t)| \leq \frac{1}{2} \sup_{0 < \tau \leq t}|v(x,\tau)|*M$ where $M \leq 1$. Then, let $K \doteq M\frac{1}{2} < 1$
$|v(x,t)| < \sup_{0 < \tau \leq t}|v(x,\tau)|$

Since this holds for every $< 0 \tau \leq  t$, let $\tau \to  0+$ , since $v(x,0) = 0$ for every $x \in \mathbb{R}$ and $v$ is continous then $\sup_{0 < \tau \leq t}|v(x,\tau)|$ must be 0 ..but I think there something wrong with this since $|v(x,t)|$ cannot be $< 0$...Any suggestions?

Thanks


a,b are two real numbers satisfy:
[tex]a^2 \ge b+2[/tex]
[tex]b^2 \ge a+2[/tex]
[tex]a+b=1[/tex]
show that: [tex]a^2b^2+5ab+4 \ge a^3+b^3[/tex]

Do we need all those conditions?

It's equivalent to $(ab+1)(ab+4)-(a+b)((a+b)^2-3ab)\geq 0$

After all this, we get: $[a(1-a)]^2+8a(1-a)+3=9a^2(a-1)^2+3>0$

I don't think equality is attainable.

Your last line is wrong though change x to a (little mistake).
All conditions are useful...

The way of huhuhu is true.I will write it again clearly:
We need to prove:
$a^2b^2+8ab+3 \ge 0$
then we can assume that:
$a \ge 0 \ge b$,and we solve same problem when $a,b \ge 0$
$b^2\ge 2+a$
$a-b=1$
We need proof:$a^2b^2+3 \ge 8ab$ which is equal $ab \le 4+\sqrt{13}$
But $b^2 \ge 3+b$ then $b\ge \frac{1+\sqrt{13}}{2}$
$\rightarrow a \ge \frac{3+\sqrt{13}}{2}$
It brings result.


geometry
trapezoid
ratio
geometry unsolved

A circle is inscribed in isoceles trapezoid ABCD, diagonal AC intersects the circle at K and L, with K between A and L. Find
$ \frac {AL \cdot KC}{AK \cdot LC}$

It comes out to an exact number. There is a solution on [url]http://books.google.com/books?id=ojmmAgGhfdoC&printsec=frontcover&dq=olympiad+problems+around+the+world#PPA3,M1[/url] which is messy and involves a lemma. There is a second solution that isn't shown that I'd assume would be nicer. Does anyone think they can find it?

This is a hard one!
It needs projective transformation. :ninja:


trigonometry
function
modular arithmetic
floor function

I want to see how many different solutions people can come up with to this.
For how many integer values of $x$, $0\geq x \geq 2007$ is it true that $\tan(7x)=\tan(29x)$? (The tangent functions use degrees)

I assume $\ x$ and $\ y$ have to be integers?

0, no $x$ satisfies $0 \ge x \ge 2007$
 :D
Also, is this in degrees or radians?

[hide="assuming degrees..."]
We have $7x \equiv 29x \pmod{180}$, since tan values repeat every 180 degrees.  This is the same as $22x \equiv 0 \pmod{180}$
$180 = 2^2 \cdot 3^2 \cdot 5$
$22=2\cdot 11$
$22$ shares a factor of $2$, so $x$ must make up for the other factors, which is equal to $90$.  Then $x \equiv 0 \pmod{90}$, which has $\left \lfloor \frac{2007}{90} \right \rfloor$ values, plus 1 for $x=0$, or $23$ values.
[/hide]

That's right.

I guess I should have been more clear. I'll edit.


Determine all $m, n \in \mathbb N$ such that $mn - 1$ divide $m^2 + n^2$.

This problem has appeared many many times on this forum. These are just some of the links:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=ab-1&t=26232[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=ab-1&t=22810[/url]


The BMO is taking place these days (5-10 may 2005) at Iassy, Romania... here are the problems:
First the [url=http://www.mathlinks.ro/Forum/download.php?id=2115]PDF Version[/url] of the problems :)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35934]Problem 1[/url]

1. Let $ABC$ be a triangle whose inscribed circle touches $AB$ and $AC$ at $D$ and $E$, respectively. Let $X$   and $Y$ be the points of intersection of the bisectors of the angles $\angle ACB$ and $\angle ABC$ with the line $DE$ and let $Z$ be the midpoint of $BC$. Prove that the triangle $XYZ$ is equilateral if and only if $A$ is $60^o$.

    [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35935]Problem 2[/url]
    2. Find all primes $p$ such that $p^2-p+1$ is a perfect cube.

    [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35937]Problem 3[/url]

    3. Let $a$, $b$, $c$ be positive real numbers. Prove the inequality:

   $\frac{a^2}b+\frac{b^2}c+\frac{c^2}a\geq a+b+c+\frac{4(a-b)^2}{a+b+c}$

    When does the equality hold?

    [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35936]Problem 4[/url]

    4.  Let $n\geq 2$ be  integer. Let $S$ be a subset of $\{1,2,...,n\}$ such that $S$ neither contains two elements
    one of which divides the other nor contains two elements which are co-prime. What is the maximal number of elements of such a set $S$?

Indeed, Congratulation on this result [b]Dzeta[/b] !!!
Any other results available ?

Great, Dzeta!!!!  ;)  ;)  :D . Hey, but we should congratulate all members of romanian team, they had huge scores.  :first:

I'm not really sure about the full results, because I've been in Paris (France) until today :D, but Andreis told me that he got 38 points (3rd place and gold medal :first:), Sasha got also 39 points and gold medal.

Kappa got 32 points and he is 7th. I am not sure if he got a gold or a silver :? ... Iura got 29 points, which means silver. I also know that Octavian Ganea (2nd team) got 23 points and Ungureanu Andrei got 14 points.

The Romanian team crushed all the other teams :first: :P :)

Thanks very much for the quick responce, guys! Congrats to [b]Dzeta[/b]!

Could you please post or link to the results of the other competitors/teams if they are available.

Many thanks!
Zhivko

Yes,I also wanna know if possible what our boys did there.Congrats to the Romanian team.It's their second victory in a row

[quote="ikap"]Sasha ain't got 39, but 33. It's Michnea Dragos, who got the 39.

I got 26 :(

Gold medals are: the 3 romanians mentioned up, Dust, Sasha and someone from Greece, all with 33 points. Kappa will take the first silver :(

Bul4 took 29, i don't know his name, and there was a 27, a 26 among the bulgarians, and the rest of them around 20-23 points I think, except Bul1 who took 17. They all missed problem 3, i can't explain why.

Hope this information will help.[/quote]

Hmmm, thanks very much ikap, anyway congrats to you and your teammates, you made very good results.
I wonder why our team made such results, but sometimes everything can happen and they are not probably not used to non-symmetric inequalities(so do I :-().
Anyway, 10x again

[quote="ikap"]Sasha ain't got 39, but 33. It's Michnea Dragos, who got the 39.[/quote]Sorry, I did not understand over the phone :) Congrats to Micnea then! (he previously went to 3! editions of the JBMO and got 2 silvers and 1 gold!).

So we have 3 Romanians, 2 Moldovians and 1 Greek getting gold medals! Who would have thought that there is going to be no Bulgarian? :)

I've solved the 1st problem from BMO  :D

Wow,that's confusing...Anyone knows the pointsq needed for silver?
 Anyway,congrats to the romanian and moldovian team.You seem to overcome losing some of your 12-grade stars-good for you,now,hopefully,there will be some more IMO top 5 contenders from Balkans.
 Some team scores?I wonder if moldovians were able to beat us...

[quote="Wolfy"] Aome team scores?I wonder if moldovians were able to beat us...[/quote]Looking at the medals, they did. But at the general points I think not, as I understood that one of the Moldovian team members scored 4 points :(

Can anyone list all the countries participating the BMO?
BTW, is this only the first tour? Or it consist of just 4 problems?

Wow! I didn't know there were so many bulgarians in the forum... Shall we make a bulgarian community then? What do you think, folks?

Does anyone care to link to the place where the general ranking should appear?

Regards,
Zhivko

[quote="n0vakovic"][quote="M@re"]does anybody know wich country proposed problems (all of them)[/quote]

3rd problem was proposed by Serbia & Montenegro, author is Dusan Djukic. He also made G5 and G8 from ISL 2004.

to M@re: taj sam[/quote]
 A pa cuo sam ja za tebe  :)  :D  :lol: , zezam se (mislim ne zezam se stvarno sam cuo)

So, Serbia gave the hardest problem.....  AGAIN
What about others, is 1st problem from romania, it seems to me like it is.

How did the moldavian team?
does someone know the complete results (at least partaial)?
congrats to the romanian team!

Congratulations to Sasha and Dust for gold medals! And Iura for silver!
How diid the other moldavian contestants solve?

Well,the full results are already available [url=http://www.rms.unibuc.ro/balkan_results.html]here,at the official site[/url]

Congratulations to the moldavian team!!  All of them!!!!
now, good luck at the imo!!!

Is there any rule about the number of medals at the BMO? I see that there are very few persons that didn't get a medal.. it's not that I don't agree with the idea..   :D  .. but it's a bit strange

(this is my first post  :blush: I'm a little nervous)

Yup there is a rule... It's customary that each country gets at least a medal ;) ... joking :D
   The exact numbers are decided during a special meeting after coordination, but the numbers are roughly (correct me if i'm wrong): 2/3 of the contestants get a medal and the gold:silver:bronze ratio is about 1:2:3...

This year, the official results were: 6 gold 14 silver 17 bronze 20 no medal

[quote="hardsoul"][quote="w13"][quote="hardsoul"]Hello.Congratulations for Romanian team.
Does anyone know the results of the Kazakhstan team.[/quote] :) Зачем они тебе?Дж.Б.[/quote]

Зачем Румынии результаты Молдовы? ;)[/quote]
Они участвовали в этой олимпиаде,а ты-нет.

congrats to everyone , especialy to dzeta!!!!  :D

[quote="toumaf"]Sorry to revive a dead topic, but since I'm training for IMO 2006 and I still find IMO shortlists very difficult, I would like to know if someone had some BMO shortlist (if it exists).[/quote]It exists but it's not public.

Even for problems proposed a few years ago?


trigonometry
geometry
analytic geometry
geometry proposed

In $\triangle ABC$, $AD$ is the bisector of $\angle A$, and $E$, $F$ are the feet of the perpendiculars from $D$ to $AC$ and $AB$, respectively. $H$ is the intersection of $BE$ and $CF$, and $G$, $I$ are the feet of the perpendiculars from $D$ to $BE$ and $CF$, respectively. Prove that both $AFEH$ and $AEIH$ are cyclic quadrilaterals.

It can be shown(eg by Ceva trigonometric form) easily that $H$ lies on the altitude $AA'$( $A'\in{BC}$) from $A$ to $BC$.
Then since $A'HGD$ and $AFDE$ are cyclic  quadrilaterals :
$BH\cdot{BG}=BA'\cdot{BD}=BF\cdot{BA}$ hence $AFHG$ is cyclic.
In a similar way:
$CI\cdot{CH}=CE\cdot{CA}=CD\cdot{CA'}$ and hence $AEIH$ is cyclic, too.

[quote="Sailor"]$H$ lies on the altitude $AA'$( $A'\in{BC}$) from $A$ to $BC$.[/quote]

That is the key to this problem. Many contestants failed to notice this and thus did not get this problem right. It can be proved by trigonometry, pure geometry, or even using Cartesian coordinates!


1.  Find all triples $(x,y,z)$ of positive rationals such that $x + y + z$, $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$ and $xyz$ are all integers.

2. Let be given two parallel lines $k$ and $l$, and a circle not intersecting $k$. Two tangents from a variable point $A \in k$ to the circle intersect the line $l$ at $B$ and $C$. Let $m$ be the line through $A$ and the midpoint of $BC$. Prove that all the lines $m$ (as $A$ varies) have a common point.

3.  $k$ is a fixed positive integer. Let $a_n$ be the number of maps $f$ from the subsets of $\{1, 2, ... , n\}$ to $\{1, 2, ... , k\}$ such that for all subsets $A, B$ of $\{1, 2, ... , n\}$ we have $f(A \cap B) = \min (f(A), f(B))$. Find $\lim_{n \to \infty} \sqrt[n]{a_n}$.

4.  $m, n$ are relatively prime. We have three jugs which contain $m$, $n$ and $m+n$ liters. Initially the largest jug is full of water. Show that for any $k$ in $\{1, 2, ... , m+n\}$ we can get exactly $k$ liters into one of the jugs.

5.  A parallelepiped has vertices $A_1, A_2, ... , A_8$ and center $O$. Show that:
\[ 4 \sum_{i=1}^8 OA_i ^2 \leq \left(\sum_{i=1}^8 OA_i \right) ^2  \]

6.  The distinct reals $x_1, x_2, ... , x_n$ ,($n > 3$) satisfy $\sum_{i=1}^n x_i = 0$, $\sum_{i=1}^n x_i ^2 = 1$. Show that four of the numbers $a, b, c, d$ must satisfy:
\[ a + b + c + nabc \leq \sum_{i=1}^n x_i ^3 \leq a + b + d + nabd  \].

1. http://www.mathlinks.ro/Forum/viewtopic.php?p=342751#p342751
2. http://www.mathlinks.ro/Forum/viewtopic.php?p=342752#p342752
3. http://www.mathlinks.ro/Forum/viewtopic.php?p=342753#p342753
4. http://www.mathlinks.ro/Forum/viewtopic.php?p=342754#p342754
5. http://www.mathlinks.ro/Forum/viewtopic.php?p=342756#p342756
6. http://www.mathlinks.ro/Forum/viewtopic.php?p=342758#p342758


algebra
polynomial
number theory unsolved
number theory

Let p be a prime .Dertermine the maximum deg of the polynomial T(x) with integer coefficient in{0,1,...,p-1} and deg<p sastisfying the condition if T(n)-T(m) divisible by p then n-m divisible by p

I think this has been posted before, and the answer is $p-2$.

Yes i think so but you can explain for you.


floor function
combinatorics proposed
combinatorics

Let $ n$ be a positive integer. Develop a general formula for positive divisors of $ n^2$ less than $ n$ that donot divide $ n$.

My answer:
[hide]Let $ n\equal{}\prod_{i\equal{}1}^r p_i^{x_i}$ where $ p_i$ is prime.

General formula is $ \sum_{i} \lfloor \frac{n}{p_i^{x_i\plus{}1}} \rfloor$ - $ \sum_{i,j} \lfloor \frac{n}{p_i^{x_i\plus{}1}p_j^{x_j\plus{}1} } \rfloor$ + ...[/hide]

I don't think that's right, but even if it is it's not as simple as possible.  (For one thing, it involves $ n$ and the $ p_i$, but we can write a formula without them.)

[hide]Let $ n \equal{} p_1^{a_1} \cdot \cdots \cdot p_k^{a_k}$.  Then $ n^2 \equal{} p_1^{2a_1} \cdot \cdots \cdot p_k^{2a_k}$.  Note that aside from the factorization $ n^2 \equal{} n \cdot n$, every factorization of $ n^2$ into two parts includes exactly one part less than $ n$.  Thus, the number of factors of $ n^2$ less than $ n$ is exactly half the number of factors of $ n^2$ excluding $ n$, and this number is $ \frac {1}{2}\left( \minus{} 1 \plus{} \prod_{i \equal{} 1}^k (2a_i \plus{} 1)\right)$.  However, we have to exclude those factors that divide $ n$.  There are exactly $ \minus{} 1 \plus{} \prod_{i \equal{} 1}^k (a_i \plus{} 1)$ of these.  The difference is $ \frac {1}{2}\left(1 \plus{} \prod_{i \equal{} 1}^k (2a_i \plus{} 1) \minus{} 2\prod_{i \equal{} 1}^k (a_i \plus{} 1)\right)$, the number we seek.[/hide]

I agree that your solution is right. I shall write my solution( with a rectification of what I wrote on my previous post), lets see where I am wrong.

Let $ n\equal{} p_1^{a_1}p_2^{a_2} \dots p_k^{a_k}$
$ \Rightarrow n^2\equal{} \prod_{i\equal{}1}^k p_i^{2a_i}$

$ \lfloor \frac{n}{p_i^{a_i \plus{}1}} \rfloor$ - $ \lfloor \frac{n}{p_i^{2a_i\plus{}1}} \rfloor$ gives the natural numbers less than $ n$ that divide $ n^2$  and have the power of $ p_i$ with $ a_i\plus{}1\leq power \leq 2a_i$.

By Inclusion-exclusion principle,

$ \sum_{i} \{\lfloor \frac{n}{p_i^{a_i\plus{}1}} \rfloor \minus{} \lfloor \frac{n}{p_i^{2a_i\plus{}1}} \rfloor\}$ - $ \sum_{i,j} \{\lfloor \frac{n}{p_i^{a_i\plus{}1}p_j^{a_j\plus{}1}} \rfloor \minus{} \lfloor \frac{n}{p_i^{2a_i\plus{}1}p_j^{2a_j\plus{}1}} \rfloor\}$ + ...

[quote="srikanth"]$ \lfloor \frac {n}{p_i^{a_i \plus{} 1}} \rfloor$ - $ \lfloor \frac {n}{p_i^{2a_i \plus{} 1}} \rfloor$ gives the natural numbers less than $ n$ that divide $ n^2$  and have the power of $ p_i$ with $ a_i \plus{} 1\leq power \leq 2a_i$.[/quote]
This appears to be nonsense, and there are plenty of counter-examples: for $ n \equal{} 30$, $ p \equal{} 2$ and $ a \equal{} 1$ we have $ \left\lfloor \frac {n}{p^{a \plus{} 1}} \right\rfloor \minus{} \left\lfloor \frac {n}{p^{2a \plus{} 1}}\right \rfloor \equal{} \left\lfloor \frac {30}{4} \right\rfloor \minus{} \left\lfloor \frac {30}{8} \right\rfloor \equal{} 7 \minus{} 3 \equal{} 4$.  The factors of 900 less than 30 divisible by $ p^{a \plus{} 1} \equal{} 4$ are $ 4, 12, 20$.


Let $ H$, $ I$ and $ O$ the orthocenter, incenter and circumcenter of a triangle $ ABC$, and let $ M$ the middpoint of $ AC$.

Prove that:

a)$ \angle{BAH} \equal{}\angle{CAO}$;
b) $ \angle{HAI} \equal{}\angle{OAI}$;
c) $ \angle{ABC} \equal{}\angle{AOM}$.

[hide="Note:"]I used [url=http://www.geogebra.org/cms/index.php?lang=en.]Geogebra[/url] and looks like it's true only if $ ABC$ is acute.[/hide]


[hide="Solution-CLICK ME!"]From the inequality,

$mn<m+n$

$mn-m-n+1<1$

$(m-1)(n-1)<1$

Since $m,n>0$, we must have that one of them is $1$. WLOG, let $m=1$

Thus,

$1+27n+n^{3}=729$

A little guess and check gives us that $n=8$. Thus, $m+n=9$.[/hide]

[quote="VIPeter"]In $m$ and $n$ are positive integers satisfying $m^{3}+27mn+n^{3}=729$ and $m+n>mn$, what is the value of $m+n$?[/quote]

yay! i got the answer!

[hide="click ME; I got the answer "]$m= 8$ $n=1$ or vice-versa

You basically need to play around with the numbers. You know that one of them has to be the # 1

${m}+{n}{>}{mn}$

so if you put in numbers, like 1 + 4, it is greater than 1*4 and so on....I knew this, and i substituted numbers from 1-9....finally to get 8..however vishalarul's method is a lot quicker  i don't think it matters which one 8 or 1 is for n or m. It can be either one, because it says $n^{3}$ and $m^{3}$ [/hide]


algebra
polynomial
number theory proposed
number theory

Let $ n>1$ and $ A=\{1,2,...,n\} $. Let $ m(B)=\sum_{x \in B} {x} $ for any non-empty subset B of A. Let F the collection of those subsets B of A for which $ 2|m(B) $. Find $\sum_{B \in F}{m^2(B)} $.

Consider polynomial $f(x)=(1+x)(1+x^2)\dots(1+x^n)$. We have $\sum_{B\in F}m^2(B)=\frac{1}{2}((xf'(x))'|_{x=1}-(xf'(x))'|_{x=-1})$. But it is clear second summand is 0 for all $n\geq 5$. So we need to find first summand for $n\geq 5$ (if $n\leq 4$ we may manually find answer).
Quite easy computations gives us following formula:
\[\sum_{B\in F}m^2(B)=\frac{2^{n-5}n(n+1)(3n^2+7n+2)}{3}.\]

That was also my solution. I like this problem so much, especially because in one of our famous books even for the TST 1994 problem it is given a quite long and boring solution. This is a very short and nice solution.


[hide="Two keywords"]Convex and Jensen[/hide]

Thanks, but I prefer more elementary methods :)

Not to steal your thread or anything, but what does...

$\mathbb{R_{+}}$ and $\mathbb{Z^{+}}$ mean? What does the $+$ symbolize?

[quote="maokid7"]another question what is jensens?[/quote]

[url]http://mathworld.wolfram.com/JensensInequality.html[/url]

usually if you have a theorem question, mathworld will have it

[hide="Here is my elementary solution"]Recall AM-HM inequality, i.e. $\frac{2ab}{a+b}\le\frac{a+b}2$ for positive reals $a,b$. This should be easy for you.

Now, for each $i\in\{1,\ldots,99\}$, by AM-HM

$f(x_i)=\frac{x_i\cdot\frac1{99}}{x_i+\frac1{99}}\le\frac{x_i+\frac1{99}}4$

Putting everything together,

$\sum_{i=1}^{99}f(x_i)\le\sum_{i=1}^{99}\frac{x_i+\frac1{99}}4=\frac14\left(\left(\sum_{i=1}^{99}x_i\right)+99\cdot\frac1{99}\right)$

Since we are given that $\sum_{i=1}^{99}x_i=1$,

$\sum_{i=1}^{99}f(x_i)\le\frac24=\frac12$.[/hide]

PS: I could have written a one-liner proof without words but I chose to be wordy :D

[b] sirge [/b] , that $+$ means the set has only positive numbers, so $\mathbb{R_{+}}$ means " real plus numbers set".
[b]10000th User [/b] - thanks, this solution is really elementary, because I understand it :)


linear algebra
matrix
search
linear algebra unsolved

Let $ A,\ B \in \mathcal{M}_{n}\left(\mathbb{C}\right)$ be such that $ AB \equal{} BA$. Prove that $ \det \left(A\overline{A} \plus{} B \overline{B}\right) \geq 0$, where $ \overline{X}$ is the matrix formed by conjugating the elements of $ X$.

pohoatza,
I recommend you have a look at http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2048574659&t=197496

I knew that topic, thanks. Still, is the one from this topic true?

I think the answer is yes.
Try to strat with $ A\in \mathrm{GL}_n(\mathbb{C})$ and write $ A\overline{A} \plus{} B \overline{B} \equal{} A(I_n\plus{}C\overline{C})\overline{A}$ and then use the link.

Question : $ B\equal{}AC$ ,$ \overline{B}\equal{}(\overline{A})(\overline{C})$ and as you : $ (\overline{A})(\overline{C}) \equal{}(\overline{C})(\overline{A})$ ?

[quote="pohoatza"]Let $ A,\ B \in \mathcal{M}_{n}\left(\mathbb{C}\right)$ be such that $ AB = BA$. Prove that $ \det \left(A\overline{A} + B \overline{B}\right) \geq 0$, where $ \overline{X}$ is the matrix formed by conjugating the elements of $ X$.[/quote]
 I can explain very implies , I hope that you  accept It .
 The first $ \forall X \in M_n{\mathbb{(C)}}$ then $ \det(X\overline{X} + I_n) \geq 0$
Lemma1) $ X,Y \in M_n{\mathbb{(C)}}$ then $ \overline{X^{ - 1}} = {\overline{X}}^{ - 1}$ (If $ X \in GL_n{\mathbb{(C)}})$ ),
 $ (\overline{X})({\overline{Y}) = \overline{XY}}$ , $ \det(X\overline{Y} + I_n) = \det(\overline{Y}X + I_n)$
Lemma2) Let $ P \in \mathbb{R}[X]$ if and only if ($ \forall z \in \mathbb{C},\overline{P(z)} = P(\overline{z})$)
Lemma3) $ P(x) \in \mathbb{R}[X]$ such that $ P(x) \geq 0, \forall x \in ( - \infty , \lambda_1); (\lambda_1, \lambda_2) ;.... ; (\lambda_m, + \infty)$ ,here $ \lambda_k \in \mathbb{R},\lambda_1 < \lambda_2 < .... < \lambda_m$ then $ P(x) \geq 0$ ,$ \forall x \in \mathbb{R}$ , very implies  .
Lemma1) you can prove that it and read links of [b]JC_math[/b]
 Proof lemma2) Put $ P = \sum_{k = 0}^{m}{a_kX^k}$ ,$ (a_0,a_1,...,a_m) \in \mathbb{C}^{m + 1}$
We have $ \overline{\overline{P(z)} - p(\overline{z})} = \sum_{k = 0}^{m}{(a_k - \overline{a_k})z^k}$ ,$ \forall z \in \mathbb{C}$
 Hence $ (\forall z \in \mathbb{C} ,\overline{P(z)} = p(\overline{z}))$ if only if $ \forall z \in \mathbb{C} ,\sum_{k = 0}^{m}{(a_k - \overline{a_k})z^k} = 0$ if only if $ \forall k \in \{0,1,....,m\},a_k - \overline{a_k} = 0$ if only if $ P \in \mathbb{R}[X]$
Now
case 1) If $ B \in GL_n(\mathbb{C})$ then $ \det(A\overline{A} + B\overline{B}) = \det(B)\det(\overline{B})\det(B^{ - 1}A\overline{B^{ - 1}A} + I_n) \geq 0$ ,Note that $ \overline{AB^{ - 1}} = (\overline{A})(\overline{B^{ - 1}}) = (\overline{B^{ - 1}})(\overline{A}) = \overline{B^{ - 1}A}$ if $ AB = BA$
 Case 2) If $ 0 \in Spectrum{(B)}$ , call $ \lambda_1,\lambda_2,...,\lambda_n$ are eigenvalues of $ B$ , have a some point $ \lambda_i = 0$ ,
 $ \forall \lambda \notin \{\lambda_1,...,\lambda_n\}$ we have $ B - \lambda I_n \in GL_n{(\mathbb{C})}$
  easy see that $ P(\lambda) = \det(A\overline{A} + (B - \lambda I_n)(\overline{B} - \overline{\lambda} I_n)) \geq 0$ (1)
 You can prove that $ P(\lambda) \in \mathbb{R}[X]$ , You can check it by $ P = QR$
  $ Q(\lambda) = \det(B - \lambda I_n)(\overline{B} - \overline{\lambda} I_n) \in \mathbb{R}[X]$ use  lemma 2)
 .... $ R(\lambda) \in \mathbb{R}[X]$ same use lemma 1) lemma2)
 For (2) shown take $ \lambda \in \mathbb{R} ,\notin spectrum (B)$ , use lemma3) eassy $ P(\lambda) \geq 0 , \forall \lambda \in \mathbb{R}$ Implies $ P(0) \geq 0$ ....
 --------------------
 We have 1) $ \forall A,B,C \in M_n{\mathbb{(R)}}$ such that $ AB = BA,BC = CB,CA = AC$ then $ \det(A^2 + B^2 + C^2) \geq 0$
 Here $ A_1 = A + iB,B_1 = C,A_1B_1 = B_1A_1,\overline{B_1} = B_1 = C$
  $ A,B,C,D \in M_n{\mathbb{(R)}}$ sucht that $ AB = BA ,,,....,CD = DC$ then $ \det(A^2 + B^2 + C^2 + D^2) \geq 0$
 By $ A_1 = A + iB,B_1 = C + iD,A_1B_1 = B_1A_1$ ,....,
 Question for $ A_k \in M_n{\mathbb{(C)}}, k \in \{1,2,...,m\}$,such that $ A_iA_k = A_kA_i ,\forall (k,i) \in \{1,2,..,m\}^2$ then $ \det(\sum_{k = 1}^{m}{A_k\overline{A_k}}) \geq 0$
 With Kronecker product is same  $ \det(\sum_{k = 1}^{m}{A_k\otimes{\overline{A_k}}}) \geq 0$ and If $ \forall A_k \in M_n{\mathbb{(R)}}$ then $ \overline{A_k} = A_k$ and $ \det(\sum_{k = 1}^{m}{A_k\otimes{A_k}}) \geq 0$
  :play_ball: ......

Note that in lemma2) for $ z \in \mathbb{R}$ , $ (\forall z\in\mathbb{R},\overline{P(z)} \equal{} P({z}))$ then $ P \in \mathbb{R}[X]$ need for
$ P(\lambda) \equal{} \det(A\overline{A} \plus{} (B \minus{} \lambda I_{n})(\overline{B} \minus{} \lambda I_{n}))\geq 0$ ....   :play_ball:


A 0.50 kg projectile is given an initial speed of 20 m/s at an angle of 37 degrees to the horizontal surface and lands a certain distance from its launch point. How much work is done on the projectile?


Find the sum of the reciprocals of the five roots of the quintic $ x^{5}-x^{4}+Tx^{2}-Tx=(2x-2)/T$

Another approach...
[hide]
$ x^{5}-x^{4}+Tx^{2}-(T+\frac{2}{T})x+\frac{2}{T}=0$
The polynomial with roots that are reciprocal of this is:
$ \frac{2}{T}x^{5}-(T+\frac{2}{T})x^{4}+Tx^{3}-x+1=0$
And by vieta's, the sum of the roots is $ \boxed{\frac{T^{2}+2}{2}}$[/hide]

As red_dog pointed out, $ \frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{5}}=\frac{x_{2}x_{3}x_{4}x_{5}+\ldots+x_{1}x_{2}x_{3}x_{4}}{x_{1}x_{2}x_{3}x_{4}x_{5}}$. Since the quintic is a product of its factors which are x minus each of its roots, the d and e coefficients of the quintic are $ {x_{2}x_{3}x_{4}x_{5}+\ldots+x_{1}x_{2}x_{3}x_{4}}$ and $ {x_{1}x_{2}x_{3}x_{4}x_{5}}$ respectively. The sum of the reciprocals is therefore $ \frac{d}{e}=\frac{{T^{2}}+2}{2}$


Just in case you're interested:

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=furman&t=137956


function
integration
calculus
real analysis
real analysis unsolved

Construct a function in $ L^1 (\minus{}\infty,\plus{}\infty)$ which is not in $ L^2(a,b)$ for any $ a<b$.

Something like $ |x|^{\minus{}2/3}e^{\minus{}x^2}$, translated by rationals and rescaled so that the sum converges.

Can you explain a bit further, I don't understand what you're aiming at.

Here's what mlok is saying. Let $ g(x)\equal{}|x|^{\minus{}2/3}e^{\minus{}x^2}.$ The particular thing to note about this function is that $ g\in L^1(\mathbb{R})$ but if $ 0\in[a,b],$ then $ \int_a^bg(x)^2\,dx\equal{}\infty.$ (The exact details of $ g$ can be varied; all we really need are those two properties.)

Let $ r_k$ be an enumeration of the rational numbers. (Any other countable dense subset of $ \mathbb{R}$ would do just as well.) Let

$ f(x)\equal{}\sum_{k\equal{}1}^{\infty}2^{\minus{}k}g(x\minus{}r_k).$

Then $ \int_{\mathbb{R}}f\equal{}\sum_{k\equal{}1}^{\infty}
\int_{\mathbb{R}}2^{\minus{}k}g(x\minus{}r_k)\,dx$

$ \equal{}\sum_{k\equal{}1}^{\infty}2^{\minus{}k}\int_{\mathbb{R}}g,$ which converges.

(The interchange of integral and sum is justified by the Monotone Convergence Theorem since $ g$ is nonnegative.)

But on any interval $ (a,b),$ chose some $ r_m\in(a,b).$ On that interval, $ f(x)>2^{\minus{}m}g(x\minus{}r_m)$ and

$ \int_a^b(f(x))^2\,dx\ge2^{\minus{}2m}\int_a^b(g(x\minus{}r_m))^2\,dx\equal{}\infty.$

What's interesting to me is, that every person I talked to, about this problem, gave the same example as Mlok did :huh: ( of course they didn't know I already got the answer, it was more an informal talk). What's up with that?

Thanks!

I would think there would at least have been some variety on the specific formula for $ g.$ Had I answered first, I probably would have said that $ g(x) \equal{} \frac1{\sqrt {x}}$ for $ x\in(0,1)$ and zero otherwise.

And you need a countable dense subset of $ \mathbb{R}.$ There are plenty of countable dense subsets, but the example of the rationals just comes to mind so easily.

And you need a convenient convergent series for the scaling. Say "convenient convergent series" to me and I think of $ \sum 2^{ \minus{} k}.$ Someone else might have said $ \sum\frac1{k^2}.$

I will add that the example had to be unbounded in every interval, which means it must be discontinuous everywhere. We're going to have to use some sort of limit or infinite sum to build such a thing - that's why the examples all came to you the same way.


geometry
probability
trigonometry
MATHCOUNTS
trapezoid
exterior angle

I was doing this thingamajig where it gave you the problems but not the answers.  So I was wondering what other people got, and if it matched, MOST probably it's right.  Here are the questions:

These are my answers:
1. 80
2. 8
3. 11/16
4. 4,5,6
5. no idea

2,3, and 4 are correct, 1 I am too lazy to do.

For 5, since afg=agf, they are both 80º

Then look at triangle BFD. Since m<afg is 80, m<GFD must be 100, so using the triangle sum theorem, 180-100=80 so m<B+m<D=80.

[hide="5"]Draw down the altitudes.  You could either see the 5,12 and 39 and guess that both are 5-12-13 right triangles. Which they are =D.

Or: Let one length be $ x$, then the other must be $ 13 \minus{} x$.
Thus, $ x^2 \plus{} h^2 \equal{} 25$
$ (x \minus{} 13)^2 \plus{} h^2 \equal{} 144$.
Solve from there by subtracting equations.

If you do, the height will turn out to be $ \frac {60}{13}$.
Thus, the area is $ \frac {91}{2}*\frac {60}{13} \equal{} 210$.
[/hide]

[hide="2"]The units of digit of $ 9^n$ repeats every other power.

8 is the answer[/hide]
[hide="4"]
Let the middle integer be $ x$.

Then $ x(x^2 \minus{} 1) \equal{} 24x$
$ x^2 \minus{} 1 \equal{} 24$, can divide by $ x$ because it must be positive.
$ x \equal{} \pm5$, but $ x > 0$, so $ x \equal{} 5$.

The integers are then $ 4,5,6$[/hide]

1: Exterior angle theorem.
[hide="3"]
The probability that he gets as many heads as tails is the same as getting as many tails as heads. Thus, we only need to find the probability that the same number occurs and then divide by two, but then we must add back the number of equal ways.

There are $ 6$ different ways for them to be equal. From here, you could easily figure out the rest, or use $ (16\minus{}6)/2\equal{}5$.
11 total ways gives $ \frac{11}{16}$[/hide]

[quote="mathcrazed"][hide="5"]Draw down the altitudes.  You could either see the 5,12 and 39 and guess that both are 5-12-13 right triangles. Which they are =D.

Or: Let one length be $ x$, then the other must be $ 13 \minus{} x$.
Thus, $ x^2 \plus{} h^2 \equal{} 25$
$ (x \minus{} 13)^2 \plus{} h^2 \equal{} 144$.
Solve from there by subtracting equations.

If you do, the height will turn out to be $ \frac {60}{13}$.
Thus, the area is $ \frac {91}{2}*\frac {60}{13} \equal{} 210$.
[/hide]

[hide="2"]The units of digit of $ 9^n$ repeats every other power.

8 is the answer[/hide]
[hide="4"]
Let the middle integer be $ x$.

Then $ x(x^2 \minus{} 1) \equal{} 24x$
$ x^2 \minus{} 1 \equal{} 24$, can divide by $ x$ because it must be positive.
$ x \equal{} \pm5$, but $ x > 0$, so $ x \equal{} 5$.

The integers are then $ 4,5,6$[/hide]

1: Exterior angle theorem.
[hide="3"]
The probability that he gets as many heads as tails is the same as getting as many tails as heads. Thus, we only need to find the probability that the same number occurs and then divide by two, but then we must add back the number of equal ways.

There are $ 6$ different ways for them to be equal. From here, you could easily figure out the rest, or use $ (16 \minus{} 6)/2 \equal{} 5$.
11 total ways gives $ \frac {11}{16}$[/hide][/quote]

ny subtracting equations, do u mean doing system of equatioons??

also do u mena 13-x not x-13?

[quote="micromath"]
ny subtracting equations, do u mean doing system of equatioons??
also do u mena 13-x not x-13?[/quote]
actually if u expand, you get the same thing, and sine they are opposites, there squares are equal
$ (x \minus{} 13)^2 \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} (13 \minus{} x)^2$
$ x^2 \minus{} 13x \minus{} 13x \plus{} 169 \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} 169 \minus{} 13x \minus{} 13x \plus{} x^2$
$ x^2 \minus{} 26x \plus{} 169 \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} x^2 \minus{} 26x \plus{} 169$
and by subtracting equations, one equation is multiplied by -1 and added to the other equation.

It might be. I'm sure I came across it while practicing mathcounts or amc =D

[quote="xpmath"]Heh, 5 was #25 on some old AMC 10, I think.[/quote]

this is from the Math Awareness Month competition at KU

umm no. It's #25 on 2002 AMC 10A.  Where do you think they get those problems anyway? Math awareness... They just take problems from other contest  :wink:

There is a formula to calculate the area of a trapezoid with the sides, which is:
$ A\equal{}\frac{a\plus{}c}{4(a\minus{}c)}\sqrt{(a\plus{}b\minus{}c\plus{}d)(a\minus{}b\minus{}c\plus{}d)(a\plus{}b\minus{}c\minus{}d)(\minus{}a\plus{}b\plus{}c\plus{}d)}.$ Note that a and c are the parallel sides of the trapezoid. You can then plug in the values.

[quote="uldivad9"]There is a formula to calculate the area of a trapezoid with the sides, which is:
$ A \equal{} \frac {a \plus{} c}{4(a \minus{} c)}\sqrt {(a \plus{} b \minus{} c \plus{} d)(a \minus{} b \minus{} c \plus{} d)(a \plus{} b \minus{} c \minus{} d)( \minus{} a \plus{} b \plus{} c \plus{} d)}.$ Note that a and c are the parallel sides of the trapezoid. You can then plug in the values.[/quote]

This is next to useless, your time is better spent learning to do the problem than memorizing the formula.

[quote="uldivad9"]Well, formulas can be helpful. Especially when you're unskilled like me and can't figure out 'how to do the problem'.[/quote]

its like herons formula? except for trapeziod?

could u post it because that would help me, and to me since i am also not that good, it gwould help me a lot!!

would this formula work for all quadrilaterals or just for trapezoids??

I sort of posted it already, it's a little up in an earlier post. :D
It's for trapezoids only, there are a number of different formulas for quadrilaterals.

[quote="uldivad9"]I sort of posted it already, it's a little up in an earlier post.
It's for trapezoids only, there are a number of different formulas for quadrilaterals.[/quote]

i just loked at this post a minute ago and found that out!  BTW: wikipedia says that  a is longest parrell side

:rotfl:  :rotfl:  :rotfl:  :rotfl:

Mathcrazed for the AMC 10A 2002, could u use a calculator because when u assumed the the length of the legs of the trapezoid as x u could have assigned x in two ways since there are 2 legs, either way u get the same answer but for my own curiosity i tryied it the second way and found that the calculation was slightly more tedious and without a calculator i couldnt have done it that way. Also this is slightly off topic but does your first name start with a V if not im sorry im wasting your precious time.

No, not with a V.
Yes, you could use calculators.

BTW PEOPLE:

If you use the formula similar to Heron's except for a trapezoid, you get 210 as the area.


algebra
polynomial
complex analysis
complex analysis unsolved

Let $ p(z)$ be a polynomial of degree $ N \geq 1$. Let $ L\equal{}\{ z \in \mathbb{C} \ | \ p(z)|\equal{}1  \}$.
Prove that $ \mathbb{C}\setminus L$ has at most $ N\plus{}1$ connected components.

There is one unbounded components. Each of the bounded components must contain a zero of $ p$, as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176901]discussed here[/url].


Should this be $3^{u_{1}}\times 2^{v_{1}}+3^{u_{2}}\times 2^{v_{2}}+..........3^{u_{k}}\times 2^{v_{k}}$ (ie. the u and v swapped on the second term)?

Otherwise, I don't think 5 can be expressed in this form.

[quote="Farenhajt"][quote="rjt"]Otherwise, I don't think 5 can be expressed in this form.[/quote]

$3^{1}2^{0}+3^{0}2^{1}=5$[/quote]

Yes, but isn't $u_{2}< u_{1}$?

[quote="Farenhajt"]Last time I checked, $1>0$.[/quote]
Hah, yes, but in the original post, $u_{1}$ is the first power of 3, and $u_{2}$ is the second power of [i]2[/i]. Hence my first post.

[quote="Farenhajt"]It's quite obvious that that's a typo.[/quote]
I wasn't sure. I thought that perhaps the $u$s and $v$s switched for each term. No harm in asking for clarification, right?

Of course. And in the meantime the statement has been edited.

I think it's not too hard with strong induction.


For 0<x<y<z<1 show that

x*x+y*y+z*z<xy+xz+yz+z-x

[hide]Let $ y-x=a$ and $ z-y=b$

The inequality transforms into $ a^{2}+b^{2}+ab<a+b$

Since $ 0<a<1$, $ a^{2}<a$

Since $ a+b=z-x<1$, $ b^{2}+ab<b^{2}+(1-b)b=b$

Summing gives us the desired inequality.[/hide]

An inequality I came up with while doing this problem.

Given $ 0<b<a<a+b<1$, and positive integers $ x>y$, prove that $ a^{x}-b^{x}<a^{y}-b^{y}$

It is equivalent with

0< X(Y-X) + Y(Z-Y) + (1-Z)(Z-X)