function
vector
inequalities unsolved
inequalities

the numbers $ x_{i} \equal{} 1 or 0 or \minus{} 1$.

Let $ S_{n}$ as sum of all numbers in the form of $ x_{i}x_{j} (1\le i < j \le{n})$.

Find the minimum of $ S_{n}$ when n=207.

[quote="bjh7790"]
Let $ S_{n}$ as sum of all numbers in the form of $ x_{i}x_{j}$.
[/quote]

Your problem description is not too precise, since it [u]might[/u] say $ S_n\equal{}\left[\sum_{i\equal{}1}^nx_i\right]^2$ with $ x_i\in\{0,\pm 1\}$ from which the answer is obvious. If you meant something different, pls fill in the gaps of your problem spec.

Okay, then $ 2S_n(\bar{x})\equal{}(\sum_{i\equal{}1}^nx_i)^2\minus{}\sum_{i\equal{}1}^nx_i^2\geq \minus{}n$ holds. Let $ m_n$ denote the minimum of the function $ 2S_n(\bar{x})$. [b]We claim $ m_{2k}\equal{}m_{2k\plus{}1}\equal{}\minus{}2k$. In particular, the minimum of $ S_{207}$ is $ \frac{m_{207}}{2}\equal{}\minus{}103$[/b].

For the case $ n\equal{}2k$ we set $ x_i\equal{}(\minus{}1)^i$, then $ \minus{}2k\leq m_{2k}\leq 2S_{2k}(\bar{x})\equal{}\minus{}2k$, which proves the first part.

For the case $ n\equal{}2k\plus{}1$ we set $ x_i\equal{}(\minus{}1)^i$ again, then $ \minus{}2k\minus{}1\leq m_{2k\plus{}1}\leq 2S_{2k\plus{}1}(\bar{x})\equal{}\minus{}2k$. If $ m_{2k\plus{}1}\equal{}\minus{}2k\minus{}1$ holds, we can determine a vector $ \bar{y}$ with $ 2S_{2k\plus{}1}(\bar{y})\equal{}\minus{}2k\minus{}1$, which implies $ \sum_{i\equal{}1}^{2k\plus{}1}y_i\equal{}0$ and $ \sum_{i\equal{}1}^{2k\plus{}1}y_i^2\equal{}\minus{}2k\minus{}1$. The second equation gives $ y_i\in\{\pm 1\}$, but then we have the contradiction $ 0\equiv\sum_{i\equal{}1}^{2k\plus{}1}y_i\equiv\sum_{i\equal{}1}^{2k\plus{}1}1\equiv 1\bmod{2}$. Therefore, $ m_{2k\plus{}1}\equal{}\minus{}2k$.


conics
hyperbola
analytic geometry
ellipse
trigonometry
trig identities
Law of Cosines

An explosion is heard by two law enforcement offivers who are 1000 meters apart. One officer heard the explosion 1.5 seconds after the other officer. If the speed of sound in air is approximately 340 meters per second, write an equation for the possible locations of the explosion, relative to the two enforcement officers.

Are you sure that's not a hyperbola?

Edit: changed

The point is on a hyperbola, because the definition of hyperbola is the set points the difference of whose distances from two distince fixed points (officers) is a positive constant (1.5 * 340).

A hyperbola with a horizontal transverse takes the general form
$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2}=1$
and the focal distance is c,. In this case, the focal distance is half the distance between two officers, 500.
Also, $(h, k)$ is the center of the hyperbola. If we set it to (0, 0), the officers will be located at (-500, 0), (500, 0).

To find a,  note that the difference between the distances is always $2a$ and it's 1.5*340 is this case. So now we solve for the hyperbola:
$a = \frac{1.5 \cdot 340}{2} = 255$
$c^2 = a^2 + b^2$
Substitute 500 for c, we get $b= \sqrt{184975}$.

So the answer is any point located on the either branch of hyperbola
$\frac{x^2}{65025} - \frac{y^2}{184975} = 1$, if the officers are located at the (-500, 0) and (500, 0) in a rectangular coordinate system.

[quote="nat mc"][quote="BornForMath"]
$\frac{x^2}{65025} + \frac{y^2}{184975} = 1$, if the officers are located at the (-500, 0) and (500, 0) in a rectangular coordinate system.[/quote]
Sory, I meant hyperbola.   BTW: Thats an ellipse  :roll:   :P[/quote]

Oops, that was a typo like you did.  :D

It doesn't seem to me that nat mc is answering the right question.  One man hears the explosion 1.5 seconds after the other.  So, if the distance from the one man (say, man A) is dA=v*tA, where v=velocity of sound=340m/s, then the distance from the other man (say B) is dB=v*(tA+1.5)=dA+v*1.5=dA+510.  Placing man A at the origin (0,0) and the man B at (1000,0), the explosion occurs at some point (x,y).  We therefore have a triangle of sides of length dA, dB, and 1000.  We can now use the law of cosines to find a relation between dA, dB and (x,y).  Note first that x=dA*cos(theta) and y=dA*sin(theta).  Moreover, from the law of cosines

dB^2=(1000^2)+dA^2 - 2*(1000)*dA*cos(theta)
(dA+510)^2 = 10^6 + dA^2 - 2000*dA*cos(theta)
Solving for dA:
dA=(739,900)/(2000*cos(theta)+1020).

Consequently,

x=dA*cos(theta)=(739,900*cos(theta))/(2000*cos(theta)+1020)
y=dA*sin(theta)=(739,900*sin(theta))/(2000*cos(theta)+1020)

Since dA must be greater than 0 it is required that 2000*cos(theta)+1020>0.  If you plot this equation you see that it is negative between about 120 degrees and 240 degrees.  This region is therefore not possible.  You can plot the above expressions for x and y vs. theta (excluding the range 120 to 240 degrees) to get the possible locations for the explosion.

Does this seem correct?

Thomas

[quote="4everwise"]I'm not sure. I applied the distance formula in the same manner as Nat MC, except I placed one man at (500,0) and the other at (-500,0).

And it is a hyperbola.[/quote]
Yea, thats exactly what I did.  :)


[quote="Virgil Nicula"]I note the circumcircle $w=C(O,R)$ of the triangle $ABC$ for which $8R^2=a^2+4m^2_a$.
Show that $b=c$ and $OA=OG$.[/quote]

$8R^2 = a^2+4m^2_a$

$4R^2 = \frac{a^2}{2}+2m^2_a$

$(2R)^2 = b^2+c^2 \ \ \ (1)$

So, the triangle $ABC$ is right $(\widehat{A}=90^0)$

But it is not necessary that $b=c$.

Indeed, for any $b,c$ satisfying $(1)$ we have

$m_a=R,a=2R\Rightarrow$

$a^2+4m^2_a = (2R)^2+4R^2 = 8R^2$

what am I doing wrong?  :huh:


Prove or find the counterexample:

$ x^6 \plus{} y^6 \plus{} 4x^3y^3 >\equal{} 3y^4x^2 \plus{} 3x^4y^2$

$ x, y$ are positive reals.

This is equivalent to $ (x\minus{}y)^2(x^4\plus{}2x^3y\plus{}2xy^3\plus{}y^4)\ge 0$

And apply that inequality we have nice inequality, which true for positive real numbers x,y,z:
$ 2\sum x^6\plus{}4\sum x^3y^3\ge3\sum xy(x^2\plus{}y^2)$ :P
But it can prove easy by AM_GM inequality.  :oops:


Six thousand points are marked on a circle, and they are colored using 10 colors in such a way that within every group of 100 consecutive points all the colors are used. Determine the least positive integer $ k$ with the following property: In every coloring satisfying the condition above, it is possible to find a group of $ k$ consecutive points in which all the colors are used.

Dear mathlinkers,
at the site
http://www.oei.es/oim/ibero2009.pdf
you can find the solutions (in spanish) of the problems proposed past september in the contest

[quote="gb2124"]If I'm not wrong,there must be some problem which is similar as this.
This appears in another competition.
(It only changes the number of it)
But I forget where.[/quote]

That's correct, This problem can be generalized.

I obtain that the answer is 89, and my answer was correct! :D

The answer is 89.

Firstly, we will make some assumptions and conventions. A $ L$-arc is a set of $ L$ consecutive points. The orientation is clockwise.

Suppose that, for some coloring, no 89-arc has all 10 colors. So we will prove these facts:

1 - For each 11-arc there is one color that doesn't appear among the 89 following points;
2 - Each 12-arc has just 2 colors.

If 1) is not valid for some 11-arc, then the 89 next points has all 10 colors (all 100-arc has all colors), contradiction!

For 2): split any 100-arc in eight 11-arcs ($ A_1,A_2, \ldots,A_8$) and a 12-arc $ A_9$. Using 1), we know that for the arcs $ A_i, i\equal{}1,2,3,4,5,6,7,8$ there is a color $ i$ that doesn't appear at $ A_{i\plus{}1},\ldots,A_9$.  Clearly all colors $ 1,2,\ldots,8$ are distinct, so there is just 2 colors at $ A_9$.
This argument works for any 12-arc. So, that there is at most 2 color at any 12-arc. Now we need to prove that the "at most" is, in fact, "exactly".

If we get a monochromatic 12-arc, just see the 88-arc following it. The 100-arc obtained "glue-ing" them has 10 colors, so the last point of the 12-arc and all 88-arc form a 89-arc with all 10 colors, contradiction!

Let it $ X_0$ a point with color 0, this color not appear at the 89-arc following it. The 11-arc following $ X_0$ form the arc $ A_1$. The arc $ X_0\plus{}A_1$ is a 12-arc, with 2 colors, namely 0 (the color of $ X_0$) and 1. By the way, $ X_0$ is the only point at this arc with the color 0. So the 11-arc $ A_1$ has color 1.
If $ X_1$ is the latest point of $ A_1$, the 89-arc following doesn´t have the 1 color at it. Applying the same reasoning above, we can see a beautiful pattern: monochromatic 11-arcs alternating this colors. But hey!, 6000 is not a multiple of 11. It is impossible!

So, we always obtain a 89-arc with all 10 colors.

The problem now is to exhibit a colouring such that a 88-arc does not have the property. I will do it later(sloth...)


Is it okay to use the blank carbon copy or carbon copy option to email multiple people the same email?

Bunch of more specific questions:
[hide]
Is it rude to do use bcc or cc?
What is the best way to email multiple people the same email?
Does it cheapen the email?
Is it like spamming?
Do you dislike it when you see that an email sent to you that was sent to twenty or so other people?
[/hide]

Check out this website:

http://www.putergeek.com/email/

Maybe that'll help you...  Personally, the chain emails about "If you don't pass this on you'll be raped and murdered in your bathroom by your neighbor" or even "If you don't pass this on you're not my friend" really tick me off.  But it you're needing to get information out to a lot of people, sending the same email to multiple people is okay for me - just make sure that people know that they're not the only one getting it.

The "B" in "Bcc" stands for "blind," not "blank."

I write a message and send it to person A. I cc to person B, and I bcc to person C. A knows I sent it to B and B sees that the primary recipient was A. However, neither A nor B knows I sent it to C. I probably have a reason for that - which might include that A is a difficult person with whom I'm trying to deal in an official capacity, C is involved in the case in some way, and I don't want A to contact C directly. Probably I'm protecting C in some way. (You can accomplish this without any formal "bcc" option in the email program just by forwarding the already-sent message to C.)

Of course, ANYTHING you write in an email to anyone can and likely will be forwarded to someone else. Remember that before you say something that you might regret saying.

I never had an occasion to use bcc:, but I used to send multiple emails (updates on a math conference) via a script that sent them out one by one. That did not work very well: some of my correspondents thought they were the only recipient of the message, and I got some "Why me? :maybe:" responses.

But if your email says "Happy New Year to my best friend in the entire world!!", a script would work better than cc:

As the president of several student groups, I use CC and BCC quite frequently. These can be very useful when you are corresponding with a particular individual and someone else needs to know what's going on. For example, one of my groups invites professors to lead discussions with students over dinner. We always bcc the officer list on our invitations to the professors; it's nicer if the prof sees him/herself as the only recipient of the email, but the other officers in the club do need to see which professors have been invited and for which dates.

I don't think it is rude at all. If I receive a message and I was cc'ed, I think of it as "this message is to someone else, but it contains information that I also need to know."


Add three equations, we get:
$(x+y+z)^{2}=(x+y+z)$, therefore $x+y+z =0$ or $x+y+z=1$
We don't find exactly value of $a,b,c$

Multiply the second equation by y and the third equation by z, then subtract. We get:
$(y-z)(y^{2}+yz+z^{2})=0$(*)
The second factor can only be 0 if y=z=0, then (from first equation): $x^{2}=x$, that is, x=0 or x=1. This case gives 2 solutions:(0,0,0) and (1,0,0).
Now suppose not(y=z=0). Then,from (*), we have y=z. The system of equations now becomes:
$x^{2}+2y^{2}=x\\ y^{2}+2xy=y$
We can divide the last equation by y, because y=0 would imply y=z=0. From this: $y=1-2x$, substitute in first equation and simplify. result:$9x^{2}-9x+2=0$. Two roots: 1/3 and 2/3. This gives rise to the other solutions (2/3,-1/3,-1/3) and (1/3,1/3,1/3).

Hien's first step is right, but not sufficient. He replaces one equation by the sum of the three, but forgot to take the other two into account...


[hide="a classic"]In essence, what we are searching for is the number of factors of five that 100! has. The reason is because for every zero at the end of a number the number has one more power of ten, which is made up of 2 and 5. Since 5 is less common we need to count those. 100/5=20. But numbers like 25 have 2 factors of five. So, we need to count those twice. 100/25=4. 20+4=24.[/hide]

[hide="easy"]
It ends in $[\frac{100}{5}]+[\frac{100}{25}]=20+4=\boxed{24}$.
[/hide]
Btw [x] means the integer part of x.

Could someone generalize it?  :)

[quote="moogra"]all correct. that was in a national problem a few years ago[/quote]
The same problem for 1998! was in Finnish olympiad and in my college mathematics competition.

[quote="bpms"][quote="daermon"]really?  I would have expected nationals to be harder.  I haven't ever seen a nationals pset though :lol:[/quote]
Um excuse me but it could have been one of the earlier problems on the nationals. I strongly doubt it was in the last ten.[/quote]

I don't think this was in a Nationals Problem, I think the actual number was like 2002! or something and it was a Target #2.

[quote="nutz_for2.718281828"]Here's the formula for the number of zeros in n!.

[hide="formula:"]$[\frac{n}{5}]+[\frac{n}{25}]+[\frac{n}{125}]+.....$[/hide]

Sorry, I can't write the greatest or least integer funtions. :([/quote]

\lfloor for $\lfloor$

\rfloor for $\rfloor$

\lceil for $\lceil$

\rceil for $\rceil$

National Problems aren't neccessarily like insanely hard. The first 5 or so sprint problems are usually like pretty easy. For example, take 2005 National Test, probably the hardest ever written. Here was the first question from the sprint round:

Find the average of the measures of the interior angles in a triangle.

Also, the sprint round had more questions in the 80s, so the problems were somewhat easier. Like #15 would be some direct fraction computation.

[quote="bpms"][quote="daermon"]really?  I would have expected nationals to be harder.  I haven't ever seen a nationals pset though :lol:[/quote]
Um excuse me but it could have been one of the earlier problems on the nationals. I strongly doubt it was in the last ten.[/quote]

1992? something liek that. ive seen it

well if that was on nationals i sure want to go...it is one of those things I probably won't ever forget.

[quote="frost13"]To end the debate, in 2002 at the National Competition in Chicago, the 2nd Target Question was :

"In how many zeros does the decimal representation of the number 2002! end?"

I remember it clearly since one of my kids asked how to do that on the bus on the way to the competition and then it was on the test and he got all 8 targets.  Funny how that can work out.

I have seen other questions that are similar that go :

"If 3^x is a factor of 500!, then what is the greatest possible value of x?"

Really the same question from a different point of view.[/quote]

I am not familiarised to this MC stuff but these questions are simple (i think it was one similar in a Romanian Olympiad for 6th graders.)

we simply do this for $n!$
$[\frac{n}{a^{1}}]+[\frac{n}{a^{2}}]+..........[\frac{n}{a^{x}}]$  and $a^{x}$ is the  highest number we can get that is lower or equal than n.

so using it for the 3^x thing we will have

[hide]$[\frac{500}{3}]+[\frac{500}{9}]+[\frac{500}{27}]+[\frac{500}{27}]+[\frac{500}{81}]+[\frac{500}{243}]=247$ You might want to check my calculations because i did them in a minute (i am really in a hurry  :D  )[/hide]


1.$ a,b,c$are nonnegetive numbers such that $ abc \equal{} 1$,prove that
\[ \frac {1}{\sqrt {a^2 \plus{} \frac {15}{c}}} \plus{} \frac {1}{\sqrt {b^2 \plus{} \frac {15}{a}}} \plus{} \frac {1}{\sqrt {c^2 \plus{} \frac {15}{b}}}\geq\frac {3}{4}\]
2.For $ x,y,z,w\geq 0$,show that
\[ 27(x^2y \plus{} y^2z \plus{} z^2w\plus{} w^2x) \plus{} 37(xyz \plus{} yzw \plus{} zwx \plus{} wxy)\leq 4(x \plus{} y \plus{} z \plus{} w)^3\]

Is it really hard? :maybe: .Can someone post a solution or give me a hint .Thank you very much.

[quote="arqady"][quote="Tourish"]1.$ a,b,c$are nonnegetive numbers such that $ abc \equal{} 1$,prove that
\[ \frac {1}{\sqrt {a^2 \plus{} \frac {15}{c}}} \plus{} \frac {1}{\sqrt {b^2 \plus{} \frac {15}{a}}} \plus{} \frac {1}{\sqrt {c^2 \plus{} \frac {15}{b}}}\geq\frac {3}{4}\]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=72297[/quote]
 :) Many thanks.But,still not have perfect solutions  :(
How about the second one ?


function
calculus
derivative
linear algebra
matrix

Let $ f,g\in\mathcal{F}(R,R)$ be the functions defined by $ f(t)\equal{}e^{rt}$ and $ g(t)\equal{}e^{st}$, where $ r\neq s$.  Prove that $ f$ and $ g$ are linearly independent in $ \mathcal{F}(R,R)$.

If $ ae^{rt} \plus{} be^{st} \equal{} 0$, then $ a \plus{} b \equal{} 0$ and $ ar \plus{} bs \equal{} 0$ (evaluate $ ae^{rt} \plus{} be^{st}$ and its derivative at 0), from which it's not hard to show $ a \equal{} b \equal{} 0$.

It can be generalized to any number of exponentials, but this requires more work (i.e. enough differentiation and Lagrange Interpolation is one way to solve that one).

scorpius' technique hints at the generalization, which relies on the fact that the [url=http://en.wikipedia.org/wiki/Wronskian]Wronskian[/url] is a [url=http://mathworld.wolfram.com/VandermondeMatrix.html]Vandermonde matrix[/url].

that's a cute way of doing this, but you can just note that one of them blows up faster than the other at infinity (and this easily generalizes to any number of functions)

We haven't touched the Vandermonde matrix yet and I have no clue what Wronskian is.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=178405&search_id=716888684]a thread where the generalization was discussed[/url].


In the circle below, the length of diameter $ CF$ is 12m. $ DE \bot CF$, and $ DE = 12$m. How many meters are in the length of $ \widehat{DF}$? Express your answer in terms of $ \pi$.
[asy]import olympiad; size(100); import geometry_dev; import graph; defaultpen(linewidth(0.8));
draw(Circle(origin,1));
draw((0,1)--(0,-1)); draw((1,0)--(-1,0));
label("$C$",(-1,0),W); label("$D$",(0,1),N); label("$E$",(0,-1),S); label("$F$",(1,0),E);[/asy]

Since the diameters are perpendicular, each sector is 1/4 of the circle, so the arcs are 1/4 of the circumference. The diameters have length 12, so the circumference is $ 12\pi$, implying that $ \widehat{DF} \equal{} \frac {1}{4} \cdot 12\pi \equal{} \boxed{3\pi}$.

EDIT: Bleh, thanks, Coptie.


Can anyone give an example of a sequence of strictly increasing $ \sigma$-algebras but their union is not a $ \sigma$-algebra?

Thanks.

Just take a sequence of finite partitions of an interval such that each partition is a refinement of the previous one and consider the (finite) $ \sigma$-algebras generated by those partitions. Actually, it seems harder to give an example of a strictly increasing sequence of $ \sigma$-algebras whose union [b]is[/b] a $ \sigma$-algebra again. ;)

Why is the union not a sigma algebra? :roll:

Let's consider a special case of fedja's example. Let $ x_0, x_1, x_2, \cdots$ be a strictly decreasing sequence whose limit is zero and $ \Sigma_n$ be the $ \sigma$-algebra generated by $ \{ [0, x_n], (x_n, x_{n \minus{} 1}], \cdots, (x_1, x_0] \}$. Then $ \textstyle C \equal{} \bigcup_n \Sigma_n$ is not a $ \sigma$-algebra, since every $ I \in C$ has positive length but $ \textstyle [0, x_n] \backslash \bigcup_{n \equal{} 1}^{\infty} (x_n, x_{n \minus{} 1}] \equal{} \{ 0 \}$ does not.

[quote="sos440"] $ \textstyle [0, x_n] \backslash \bigcup_{n \equal{} 1}^{\infty} (x_n, x_{n \minus{} 1}] \equal{} \{ 0 \}$.[/quote]

I think it should be $ \textstyle [0, x_0] \backslash \bigcup_{n \equal{} 1}^{\infty} (x_n, x_{n \minus{} 1}] \equal{} \{ 0 \}$. Isn't it?

Thanks all for ur reply.


If $ a$ is a rational number, and $ \cos\pi a$ is also rational, then prove $ \cos\pi a\in\{0,1,\pm \frac12\}$.

Should we revive this old chestnut? Tons of proofs allover... (by the way, value $ \minus{}1$ missing from your set).

I gave a proof [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=247054]here[/url] and MellowMelon gave another proof [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=246555]here[/url].

Thanks, I forgot $ \minus{}1$, thanks for answers.


algebra
polynomial
Vieta
Diophantine equation
IMO
number theory
IMO 1981

Determine the maximum value of $m^2+n^2$, where $m$ and $n$ are integers in the range $1,2,\ldots,1981$ satisfying $(n^2-mn-m^2)^2=1$.

Experimenting with small values suggests that the solutions of $n^2 - mn - m^2 = 1$ or $-1$ are successive Fibonacci numbers. So suppose $n > m$ is a solution. This suggests trying $m+n$, $n$:
$(m+n)^2 - (m+n)n - n^2 = m^2 + mn - n^2 = -(n^2 - mn - m^2) = 1$ or $-1$.
So if $n > m$ is a solution, then $m+n$, $n$ is another solution.
Running this forward from $2$,$1$ gives $3$,$24;$5$,$3$;$8$,$5$;$13$,8$;$21$,$13$;$34$,$21$;$55$,$34$;$89$,$55$;$144$,$89$;$233$,$144$;$377$,$233$;$610$,$377$;$987$,$610$;$1597$,$987$;$2584$,$1597$.

But how do we know that there are no other solutions? The trick is to run the recurrence the other way.
For suppose $n > m$ is a solution, then try $m$, $n-m$: $m^2 - m(n-m) - (n-m)^2 = m^2 + mn - n^2$
$= -(n^2 - mn - m^2) = 1$ or $-1$, so that also satisfies the equation.
Also if $m > 1$, then $m > n-m$ (for if not, then $n \ge 2m$, so $n(n - m) \geq  2m^2$,
so $n^2 - nm - m^2 \ge m^2 > 1$).
So given a solution $n > m$ with $m > 1$, we have a smaller solution $m > n-m$.
This process must eventually terminate, so it must finish at a solution $n$, $1$ with $n > 1$.
But the only such solution is $2$, $1$.
Hence the starting solution must have been in the forward sequence from $2$, $1$.

Hence the solution to the problem stated is $1597^2 + 987^2$

[hide="Solution"]

Assume $(n,m)$ works.  Then, also $(m-n, m)$ also works by Vieta Jumping.  Also, by simple testing, $(n+m, n)$ works, and then by Vieta jumping $(-m, n)$ works.  We are going to use the map $(n,m)\to (n+m, n), (m-n, m), (-m,n)$.  By Vieta Jumping, these are the only solutions of the form $(a,n)$ and $(m,b)$ and other solutions are produced by other forms.  Assume by induction that all numbers up to $(F_k, F_{k-1})$ are of the form $(F_k, F_{k-1})$, then using this process $(n+m, n)$ or $(m-n, m)$ works, however, these are both of this form (the base case is noting that $(1,1)$ works). Therefore, all numbers are Fibonacci numbers, therefore we get $F_{17}^2+F_{16}^2$.    [/hide]

The result in this IMO problem is due to J. Wasteels, from the year 1902:

J. Wasteels, Quelques Proprietes des Nombres de Fibonacci, Mathesis 3 (1902), pp 60-62

-----
Dominique Cassini proved in 1680 that two consecutive Fibonacci numbers $x$ and $y$ always satisfy the equation
\[ x^2-xy-y^2~=~\pm1. \]
The converse statement (that all solutions to this equation are consecutive Fibonacci numbers) is less obvious, and has been established much later by Wasteels in 1902.
This is also exercise 6.44 in the book "Concrete Mathematics" by Ronald Graham, Donald Knuth, and Oren Patashnik.
-----
The proof technique used by Wasteels is the technique that children in mathematical olympiad circles call "Vieta jumping".

Clearly, we have $n^2-mn-m^2= \pm 1$ so by the quadratic formula we have

$n=\dfrac{m \pm \sqrt{5m^2 \pm 4}}{2}$

It is well known (well, maybe not WELL known) that $5x^2 \pm 4$ is a square number iff $x$ is a Fibonacci number. So, $m$ must be a Fibonacci number. Clearly if $m= 1597$, $n$ will be too large and if $m>1587$ then $m$ is too large. So, considering $m=987$ we get $n=1597$ giving the answer to be $987^3+1597^3=5034507976$.

[b]Remark:[/b] The original problem statement asks for $m^3+n^3$, not $m^2+n^2$.

[quote=ZETA_in_olympiad][quote=blackbluecar]
[b]Remark:[/b] The original problem statement asks for $m^3+n^3$, not $m^2+n^2$.[/quote]

I'm pretty sure this is not the case. Where did you get it from?[/quote]

Download the official problem statements from the imo website. It says $m^3+n^3$ there.

The $k^2-5m^2 = \pm 4$ Pell-like equation has quasi-Fibonacci like solutions, which motivates the following claim:

[b][color=#f00]Claim.[/b][/color] If $(m, n)$ is a solution, then so is $(n, m+n)$.

[i]Proof.[/i] Algebra. $\blacksquare$

This means that all solutions can be generated from the minimal solution $(1, 1)$; in particular, the pair $(1597, 987)$ maximizes the desired value.

Let $\phi$ be the golden ratio.
We work in the ring $\mathbb Z\left[ \frac{1 + \sqrt{5}}{2} \right]$. Indeed note that this is a ring as it trivially closed under addition, and it is closed under multiplication since $\phi^2 = \phi+ 1$. By defining the norm
\[ \| a + b\phi\| = (a + b\phi)\left(a - \frac{b}{\phi} \right) = a^2 + ab - b^2 \text{\tiny (hey, where've I seen this before?)\normalsize}, \]
one can note that the units of this ring generate all the solutions to a slightly different equation. Even more, the ring has a fundamental unit, namely $\phi$.

For the sake of proving this I'll prove it cause its kinda neat.
$\textbf{Claim.}$ Let $a$, $b$ be integers such that $\frac{1}{a + b\phi} \in \mathbb Z \left[ \phi \right]$. Then $a + b\phi = \pm \phi^N$, for some integer $N$.
$\emph{Proof.}$ Suppose not. By negating the value we can consider the set of units with real value more than $1$. Clearly $\phi$ is in this set. It is sufficient to show that there does not exist an element with real value between $1$ and $\phi$ in this set, as otherwise we can just keep multiplying by $\frac{1}{\phi}$ till we get there. Suppose $p + q\phi$ was such an element.

We have
\[ 1 < p + q\phi < \phi. \]
On the other hand, $p - \frac{q}{\phi} = p + q - q\phi \in (\phi - 1, 1)$ because $p + q\phi$ is a unit. Multiplying by $\phi$, we have
\[ 1 < p + q\phi < \phi, \]
and
\[ 1 < p\phi - q < \phi.\]
Now if $p \ge 2$, then $q \ge \phi$, but on the other hand, $q \le 1 - \frac{2}{\phi}$.  These are incompatible.
If $p \le 0$, then $1 < -q$, but on the other than $\phi - 1 < q$. These too are incompatible. It remains to consider $p=1$. But then, we find that $0 < q < 1$, clearly absurd. As such there can be no unit between $1$ and $\phi$. $\square$

We now classify all solutions to the equation we have (but not yet the given one.) We have $(a,b) = (0,1)$ corresponding to the fundamental solution, and it's moreover easy to see that $\phi^n = F_{n - 1} + F_{n}\phi$, where $F_n$ is defined as the Fibonacci sequence for all integers $n$ with $F_1 = F_2 = 1$. Therefore, $(\pm F_{n-1}, \pm F_{n})$ are all the solutions to the original equation, up to a choice of one sign.

But then note that going the other way, the Fibonacci sequence goes
\[ 1, 1, 0, 1, -1, 2, -3, 5, -8, \dots\]
and in particular,
\[ F_{-n} = (-1)^{n + 1}F_{n}. \]
then, by flipping signs repeatedly, we obtain a slightly different solution set; namely,
\[ (n,m) = (\pm F_n, \pm F_{n - 1}). \]
At last, we compute the answer to be $\boxed{1597^2 + 987^2}$. We are done.

Let $(n, m)$ be a solution of $(n^2 - mn - m^2)^2 = 1$. Then obviously we have $n > m$, except the case $(n, m) = (1, 1)$. Now consider the following claim:

[b][color = red]Claim: [/color][/b]If $(n, m)$ is a solution, then $(n+m, n)$ is a solution.

[i]Proof. [/i] Note that $((n+m)^2 - n(n+m) - n^2)^2 = (m^2 + mn - n^2)^2 = (n^2 - mn - m^2)^2 = 1$, so $(n+m, m)$ is a solution. $\square$

Now if $(n, m)$ is a solution, then draw arrow $(n, m)$ to $(n+m, n)$. Then it's not hard to see that the above graph contains all solution of $(n^2 - mn - m^2)^2 = 1$ and if $m=n$, then obviously we have $(m, n) = (1, 1)$. So if $(n, m)$ is a solution, then $(n, m) = (F_k, F_{k-1})$ for some $k \ge 1$. Hence the maximum value of $m^2 + n^2 = 1597^2 + 987^2 = 3524578$. $\blacksquare$


Find all x such that:

$ cos3x\plus{}\sqrt{2\minus{}cos^23x}\equal{}2(1\plus{}sin^22x)$

hello, i have found $ x\equal{}2k\pi$ with $ k$ $ \in \mathbb{Z}$.
Sonnhard.


Given a,b,c natural numbers, such that $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$ and $(a,b,c)=1$. Prove that $a^2+b^2$ is a fourth degree of a natural number.

Let  (a,b)=d, (b,c)=m and  (a,c)=n . Then (d,m)=(m,n)=(d,n)=1 .
(Because of (a,b,c)=1 )

  Let  a=d.n.x , b=d.m.y and  c=m.n.z

We know that 1/a^2 + 1/b^2 = 1/c^2 .
So a^2 + b^2 = (a.b/c)^2    . Put a=d.n.x and b=d.m.y and c=m.n.z . We have that  (n.x)^2 + (m.y)^2 = (d.x.y/z)^2 .

  Then d.x.y/z is an integer.So z divides d.x.y . And we know    (z,d.x)=(z,d.y)=1 (Because (b,c)=m and (a,c)=n)

That's why z=1 . Then (n.x)^2 + (m.y)^2 = (d.x.y)^2 . Then x divides m.y  and y divides n.z . (But we know (a,b)=d and this gives (n.x,m.y)=1 ) So x=y=1 . Finally n^2 + m^2 = d^2 .
a^2 + b^2 = (d.n)^2 + (d.m)^2 =d^2 .(m^2 + n^2) = d^4 . Proof completed.

[quote="shyong"]Not (x,y,z)=1 but (x,y)=(x,z)=(y,z)=1  :)

But thats the same as (x,y,z)=1 actually :roll:[/quote]
No. $(x,y,z)=1$ is definitely NOT equivalent to $(x,y)=(y,z)=(z,x)=1$.

What I mean is (x,y)=(x,z),(y,z)=1 implies (x,y,z)=1  :P


1. A point Q is chosen in a parallelogram ABCD and is connected to all vertices such that $ \angle{AQB}\plus{}\angle{DQC}\equal{}180^\circ$, also $ \angle{QAD}\equal{}50^\circ$. Find $ \angle{QCD}$

[img]http://www.part.lt/img/1ce1483bb8965f7eeb2495174f9d0d30187.JPG[/img]

2. Solve an equation system:

ab+cd=34
ac-bd=19

If a,b,c,d - positive integers.

3. Calculate:

$ \frac{(1^4\plus{}\frac{1}{4})\cdot(3^4\plus{}\frac{1}{4})\cdot(5^4\plus{}\frac{1}{4})\cdot...\cdot(19^4\plus{}\frac{1}{4})}{(2^4\plus{}\frac{1}{4})\cdot(4^4\plus{}\frac{1}{4})\cdot(6^4\plus{}\frac{1}{4})\cdot...\cdot(20^4\plus{}\frac{1}{4})}$

 :)

[hide="Solution to 1"]

Translate $ AQD$ so that $ AD$ coincides with $ BC$. Call the new translated $ Q$, $ Q'$. $ BQ'CQ$ is a cyclic quadrilateral. Therefore $ \angle Q'BC \equal{} 50 \equal{} \angle Q'QC$. Since $ Q'Q$ is parallel to $ CD$, it must be the case that $ \angle Q'QC \equal{} \angle QCD \equal{} 50$.

[/hide]

[hide="Hint for 3"]
Use the Sophie Germaine identity: $ x^4 \plus{} 4y^4 \equal{} (x^2 \plus{} 2xy \plus{} 2y^2)(x^2 \minus{} 2xy \plus{} 2y^2)$ with $ y \equal{} \frac{1}{2}$

[hide="Answer"]$ \frac{1}{841}$[/hide]
[/hide]

[quote="JavaMan"][hide="Solution to 1"]

Translate $ AQD$ so that $ AD$ coincides with $ BC$. Call the new translated $ Q$, $ Q'$. $ BQ'CQ$ is a cyclic quadrilateral. Therefore $ \angle Q'BC \equal{} 50 \equal{} \angle Q'QC$. Since $ Q'Q$ is parallel to $ CD$, it must be the case that $ \angle Q'QC \equal{} \angle QCD \equal{} 50 degrees$.

[/hide]

[hide="Hint for 3"]
Use the Sophie Germaine identity: $ x^4 \plus{} 4y^4 \equal{} (x^2 \plus{} 2xy \plus{} 2y^2)(x^2 \minus{} 2xy \plus{} 2y^2)$ with $ y \equal{} \frac {1}{2}$

[hide="Answer"]$ \frac {1}{841}$[/hide]
[/hide][/quote]

I think the answer for Question 3 is 1/841...


ratio
real analysis
real analysis unsolved

Find an example of an infinite series with positive terms such that d'Alembert's ratio test does not exist.

For example, $ \sum_{k \equal{} 1}^{\infty} \frac {2 \plus{} ( \minus{} 1)^k}{k^2}$.  (I assume you mean "convergent series...")


function
linear algebra
matrix
search
linear algebra unsolved

let {$ a_0,a_1,...,a_n$} be an arithmetics sequence.($ a_1\equal{}a_0\plus{}d$) .find detA (as function of $ a_0$ and d)
A$ \equal{}\begin{pmatrix}a_0&a_1&...&a_n\\a_1&a_0&...&a_{n\minus{}1}\\...\\a_n&a_{n\minus{}1}&...&a_0\end{pmatrix}$


integration
trigonometry
algebra
polynomial
Putnam
calculus
real analysis

Prove that the system : $\int_0^xP(x)\sin x dx=\int-0^xP(x)\cos x dx=0$, where $P$ is anon-constant polinomial, have finitely many solutions.

Please proofread the question- that doesn't make any sense. The limits of the integral and the dummy variable can't have the same name.


I always think in my class: Wow it would be nice if I could describe this question with a picture.
Is there a way I can do this or is not able to be done?

You can do this though.
Type up your Asymptote code up on AoPS forums, click Preview and right click the image.
Save it to your computer and then upload it to a site like [url=http://photobucket.com]Photobucket[/url].
Then, in the classroom, put in
[code]][img id=http://photobucket.com/link/to/the/image/[/code]
:)

If you do that why go through all the unnecessary steps of saving and then uploading the image to Photobucket?

I didn't know if the images were temporarily stored on the server.
I guess they aren't?


I hope I put in the right place.
I stumbled upon a nice problem yesterday which goes like this :
   "Let $a_n=\{\sqrt{n}\}$ where $\{x\}$ stand for the fractional part of $x$. Prove that there exists a subsequence $k_n$ such that $a_{k_n}\to\frac{\sqrt{3}}{2}$."

  Question : is $\{\sqrt{n}\}$ dense in $[0,1)$?

try to prove that if $a_n$ is a sequence of real numbers that tends to infinity and that satisfies $lim_{n\to  \infty}a_{n+1}-a_n=0$ then $\{a_n\}$ is dense in $[0;1]$. It is often easier to solve general problem than particular cases  :)

[quote="alekk"]$\displaystyle \lim_{n\to \infty} a_{n+1}-a_n=0$ then $\{a_n\}$ is dense in $[0;1]$[/quote]

(1) How do you prove this?

(2) Our sequence does not seem to have this property...

You don't think $a_n=\sqrt n$ has the properties $a_n\to\infty$ and $a_{n+1}-a_n\to 0$?

I'm sorry. I thought that the fractional part of $\sqrt{n}$ should have these properties. For some reason I thought that $\{ a_n \}$ refers to the sequence $a_n$. Sorry again for the dumb thing I wrote.

Try proving that if $a_n\to 0$, $a_1+a_2+...+a_n\to\infty$ then $\{a_n\}$ is dense in $[0,1]$. ($\{ 1+{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{n}\}}$ is dense in $[0,1]$). Of course it is equvalet to what alekk said but I think it`s easier.


What do you think the cutoff score of the AMC12 B is for this year

100

It's quite flawed IMO to use top 5% for an individual test because you can bet that only the hardcore ppl would take both A and B and I believe most schools either offer only A or offer both A & B. Thus, the top 5% of the B is better than the top 5% of people who took the A.

that's not necessarily true, maybe some people messed up on A, and try to do better on B but did worse since it's harder.

Hm.. well, if I'm a typical B-taker, then I'm pretty sure my theory is true.

I had 105 on 12A after attempting 19 and missing 4 (including #1)
I then got 108.5 on 12B after attempting 20 and missing 4 AGAIN (arg! at least I got #1 this time:) )

After the lousy performance on the A, I studied quite a bit for the B. In fact, I took the B as a visitor at another HS. (I'm fairly certain most ppl taking the B also study pretty hard after the A round)

my school took only B. I was under the impression a lot of schools took only B due to clerical errors.

I was so close on passing the 12A, I got a 97. I did worse on B.
I just have to hope that I will get a lot better next year. Good thing is that I still have four years.

The cutoff will be a 100. It wasn't that much harder than any other 12.

I'm just glad I'm a freshman so my AMC score doesn't really matter for USAMO.

[quote="churchilljrhigh"]is it posisble to take two tests in one day. the local junior high offered it in the morning and the senior high offred in the after lunch.[/quote]

Ya, but that would be cheating because they have many problems in common.

I think it will be 100, but I think there will be very few perfect scores. It seemed to me that the majority of the the questions were very easy, but the hardest questions were very hard (for AMC 12), so I am guessing that scores will be high in general, but taper significantly at the top.


let $ a,b,c$ be positive real numbers such that: $ min(a,b,c)\ge \frac{1}{4}max(a,b,c)$. Prove that:
$ (ab \plus{} bc \plus{} ca)[\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}]\ge \frac {9}{4} \plus{} \frac {1}{16}[(\frac {a \minus{} b}{a \plus{} b})^2 \plus{} (\frac {b \minus{} c}{b \plus{} c})^2 \plus{} (\frac {c \minus{} a}{c \plus{} a})^2]$

Sharp inequality :mad: :lol:

WLOG, assume that $ c \ge b \ge a \ge \frac {1}{4}c > 0$.
Put $ b \plus{} c \equal{} 2x,c \plus{} a \equal{} 2y,a \plus{} b \equal{} 2z$ then $ a \equal{} y \plus{} z \minus{} x, b \equal{} z \plus{} x \minus{} y, z \equal{} x \plus{} y \minus{} z$, by the assumption, we conlude that $ x,y,z$ are three length sides of a triangle.

Moreover, $ c \ge b \ge a \ge \frac {1}{4}c > 0 \Rightarrow 4(y \plus{} z \minus{} x) \ge x \plus{} y \minus{} z$ or $ 3y \plus{} 5z \ge 5x$.

Our inequality is equivalent to:

$ (2xy \plus{} 2yz \plus{} 2zx \minus{} x^2 \minus{} y^2 \minus{} z^2)\left(\frac {1}{4x^2} \plus{} \frac {1}{4y^2} \plus{} \frac {1}{4z^2}\right) \ge \frac {9}{4} \plus{} \frac {1}{16}\sum_{cyc}^{a,b,c}\frac {(x \minus{} y)^2}{z^2}$.
$ \Leftrightarrow \sum_{cyc}^{a,b,c} (x \minus{} y)^2\left(\frac {2}{xy} \minus{} \frac {5}{4z^2}\right) \ge 0$

Let $ S_x \equal{} \frac {2}{yz} \minus{} \frac {5}{4x^2}$ and similary for $ S_y,S_z$, we easily have that $ S_x \ge 0$ and $ S_y \ge 0$. Now we prove:

$ y^2S_y \plus{} z^2S_z \ge 0 \Leftrightarrow 2\left(\frac {y^2}{zx} \plus{} \frac {z^2}{xy}\right)\ge \frac {5}{2} \Leftrightarrow 4(y^3 \plus{} z^3) \ge 5xyz$

But we have $ (y \minus{} z)(4y^2 \plus{} yz \minus{} 4z^2) \ge 0 \Rightarrow 4(y^3 \plus{} z^3) \ge (3y \plus{} 5z)yz \ge xyz$, so the last inequality is right.

Thus:
$ S_x(y \minus{} z)^2 \plus{} S_y(z \minus{} x)^2 \plus{} S_z(x \minus{} y)^2 \ge S_y(z \minus{} x)^2 \plus{} S_z(x \minus{} y)^2 \ge$

$ \ge S_y.\frac {y^2}{z^2}(x \minus{} y)^2 \plus{} S_z(x \minus{} y)^2 \equal{} \frac {(x \minus{} y)^2(y^2S_y \plus{} z^2S_z)}{z^2} \ge 0$.

So the inequality is proved.

Equality holds iff $ a \equal{} b \equal{} c$ or $ a \equal{} b \equal{} 4c$ or any of its permutations.


Just wondering- what type of graphing calculator does everyone have?

I have a regular TI89 and love it. I want an 89 titanium tho...

i used to have an 83 silver. i like the 89 better tho. the drop down menus are really nice.

I have an 89 from Nationals, but I never really understood it...

I won an 89 as a door prize at Mu Alpha Theta Nat'l Convention. I would have never bought one otherwise...

I got one when I won a south east asian olympiad (only because vietnam didn't go!).

I forgot which model it is. I've rarely used it and now that I'm a conscript, I never use it. *sigh*

Is it useful? I wonder if I'll need it for college. I wish I'd joined these forums a long time ago. It might have helped my preparations for the IMO. Might have won that tiebreaker for the last spot back in '01. *kick myself*

[quote="cardwacko9999"]I have an 84+ SE, it saps up so much batteries... (maybe it's just the fact that I play on it 2 hrs daily)[/quote]

I fixed my friend's 89 titanium and he owes me like 60 bucks.

I lost my 84 SE though. Someone jacked it.

[quote="cardwacko9999"]someone owns a 81? ... its so outdated (is someone lying)[/quote]

Other people have 72's. Who would buy a 92 anyway?

[quote="cardwacko9999"]someone owns a 81? ... its so outdated (is someone lying)[/quote]

It's my dad's old calculator that he nevers uses, and he just gave it to me.  I really don't have a need for any type of graphing calculator anways.

I bought a TI-83+ SE last spring for this current school year.  When they came out with the TI-84+ I was kinda bummed out, but then I looked at the stats and the TI-84+ SE is almost exactly the same.  The only difference in the internals of the calculator is that the 84+ SE has a clock.  No difference in RAM, ROM or processor.  Kinda funny in my opinion that the only real difference is styling.

I know that the thread is about graphing calculators, but as far as taking relatively un-advanced standardized tests like the SAT I or PSAT, nothing can beat the TI-30X IIS.

I got my TI-83 Silver Edition Plus from the 2004 MathCounts competition. What's great is that I have a chance of getting two more :D !

oh yeah guess what. i don't have one at all! :)

Terrific, congratulations, Treething, but be wary that a lot of people on these forums have missed said opportunities...

my brother has one, dunno which but if i did have one i probably wouldnt use it.

gues what mine is sooooooo high tech, nobody has cept me!

[quote="Gyan"]Tokenadult: [quote]HP 48GX[/quote]
Is that a bamboo slide rule model or an  abacus. :D
(I still have my Aristro-Studiio Slide-Rule!)
(Kids got a few  extra  TI-92's from mathcounts  which they left fo me - I also have TI- SR50 (Vintage 1971 (Model)), TI-57, TI-67, HP-55 (  along with a  TI-30, TI-86 ... but nothing as impressive as my slide rule...  Except may be  my Chambers 7 figure logtables  given to me by my father)[/quote]

I use an abacus in homework once in a while!!!   :D

Kind of hard to use it with trig, though.

oh yeah mines a TI-93![hide]there's no such thing[/hide]

[quote="yif man12"]Had an 89. Lost it or got it stolen.

Have an 83 silver. Provided I don't smash it to bits tonight. (which I am thinking about doing)[/quote]

You could always give it to me...

Yeah right... Just help me get mine back and i'll give you $50...
It gors in quarter-lifes (like half-lives) so now it's $12 and $4 tomorrow.

Help, please. I can't memorize with it lost. I psycologically depend on my TI-84 SE.


So we are pretty much P\/\/N50R1NG everyone in sports.

Red Sox: yay we win
Patriots: 10-0
Celtics: 8-0
Revolution: MLS Cup (tomorrow at 12)
BC Football: Blah without those two ridiculous losses we would probably be #1 seeing we were #2 and OSU lost

okay the Bruins are not part of this, but whatever.

And there's like, a 99% chance the patriots will win tommorow

And the Celtics did get a lucky break in yesterday's game, but they still won.

Hmm actually it seems that the Bruins have a winning record. But they're still in fourth in their division and 9th in the conference.

The Revolution just lost to the Dynamo...again, they blew a 1-0 lead after halftime.

BC beat Miami FL, and VA Tech > UVA so it will be BC vs. VA Tech in the ACC Title game.

Seriously, if we hadn't lost to FSU and UMD then the championship game would be like

(2) LSU def. (1) BC, 56-3

Celtics now 10-1 or 9-1, and patriots are 10-0, 11-0 if you count the bye week as a win

meh, only dolphins fans count bye weeks as wins.  :wink:

the celtics just won on a ridiculously lucky shot, so they're 11-1 I believe.

Isn't this a math fourm?
 :maybe:
Well, I don't watch sports...
All I know is that the Red Sox made in to the World Series.
Right?  :maybe:

Who even watches TV when you can do AoPS?

sarcasm
DUKES FOOTBALL TEAM PWNS ALL!

But really, why is MA pwning in sports all of a sudden?

[quote="ZhangPeijin"][quote="CatalystOfNostalgia"]Yeah seriously who watches ESPN for stuff other than Mathcounts?

Celtics 11-2
Patriots 11-0
Bruins (I don't care)
BC vs. VA Tech on Saturday for ACC Title?[/quote]

Are you mocking me?  :|
And FYI, I don't have US channels on my TV. It's all chinese stuffs.[/quote]

Nah, I was just kidding. Frankly, I should spend more time doing homework, playing violin, and playing piano and less time watching TV (or doing homework or playing violin while watching TV). Then there's the math.

i no, and frankly don't care. my dads happy bout it though

the celtics are 25-3
but yes more importantly the pats are 16-0
actually that is quite insignificant, but oh well

wait did bc win? i didn't watch. i shall check now...

ok bc won, CHAMPS SPORTS FTW

The Patriots are losing to the Giants.  No doubt 'bout it.

pft

I am trained in professional sports analysis. The Patriots will win. 17-10.

Patriots- 18-0, going to be super bowl champs.

wouldn't call it unlikely, we just need to win one more game, right? against the Packers

No way.  The Giants have a better line, a better running back (Jacobs over Maroney any day), and a better rush attack.  In addition, while with a banged-up secondary, they have actually gotten many early starters back, so, the G-men are winning, but not by 12.5 points.

I think the Pats might choke. I wonder why you say that the Giants have a better line, I mean, there's a reason why Brady gets 10 seconds in the pocket in every play. Your center is back, and that's obviously a huge boost, but the O-Line for the Giants is not better.

The Pats defense is inconsistent, to say the least. Also, Moss hasn't been doing very much in the playoffs so far. It seems that everyone: Brady, Moss, the defense has cooled down since the first...10 games? I mean look at all these close wins to Philadelphia, Baltimore, and of course, the Giants.

I would still favor New England, but definitely not at 12.5 points, maybe two.

CHOKE 17-14

meh i'm going to sleep, screw house

That's what I am talking about.  Tyree & Eli, that was friggin' amazing.

Dude, that catch on 3rd and...11? was complete luck. wtf. he caught the ball against his helmet and somehow got all the way back down. wtf.

I blame it on Bellichick's red sweatshirt

To add with that, how does Eli Manning escape a sack when he had three people grabbing his shirt?

ugh... that was painful. We had 3 mistakes in the Giants last drive, the missed INT, the outrageous play, and Ellis Hobbs vs. Plaxico Burreus at the endzone. Still, that drive probably wouldn't happen if Steve Smith actually held on to the ball early in the game on that other INT, so IMO, the last year AFCCG was much worse than this game.


If $ n$ is a positive integer of the form $ p_1p_2p_3p_4$ where $ p_1,p_2,p_3,p_4$ are distinct primes and the 16 divisors are $ 1 \equal{} d_1 < d_2 < d_3 < ....................... < d_{16}\equal{}n$ ,prove that $ d_{9} \minus{} d_{8} \neq 22$ given that $ n < 1995$

Without loss of generality, let $ {p_1<p_2<p_3<p_4}$ then $ {d_9}\equal{}p_2p_3$ and $ {d_8}\equal{}p_1p_4$.
Next, since $ {n<1995}$, the $ {p_is}$ must be  made up of two one digit and two two digit primes .
This is my starting point once i am true i'll post my remaining solution

21 7 15, just because $ p_1 < p_2 < p_3 < p_4,$ doesn't necessarily mean that $ p_2p_3 > p_1p_4.$ For instance, we could have $ p_ 1\equal{} 2, p_2 \equal{} 3, p_3 \equal{} 5, p_4 \equal{} 11.$ Then, $ p_2p_3 \equal{} 15 < 22 \equal{} p_1p_4.$

[hide="Solution"]First of all, without loss of generality, assume $ p_1<p_2<p_3<p_4$.

For sake of contradiction, let us assume $ d_9\minus{}d_8\equal{}22$, which implies we have $ p_ap_b\minus{}p_cp_d\equal{}22$ where $ a, b, c, d$ is a permutation of $ 1, 2, 3, 4$.

Note that by the above assumption, we cannot have $ p_1\equal{}2$ because then, some other $ p$ will have to be even.

Since $ n<1995\equal{}p_1p_2p_3p_4$, the largest value that $ p_4$ could take is $ 17$, when $ p_1\equal{}3, p_2\equal{}5$ and $ p_3\equal{}7$.

Now all we have to do is try out all quadruples of $ p_1, p_2, p_3, p_4\in {3, 5, 7, 11, 13, 17}$ and confirm that the value of $ p_ap_b\minus{}p_cp_d$ cannot be 22, leading a contradiction. $ \blacksquare$[/hide]

[quote="JBL"][quote="FantasyLover"]which implies we have $ p_ap_b \minus{} p_cp_d \equal{} 22$[/quote]  No: if the four primes are $ p_1 < p_2 < p_3 < p_4$ then any of $ p_4, p_1p_4, p_2p_3$ and $ p_1p_2p_3$ can be part of this middle pair (though not necessarily any pair of these can form the middle pair in any order).  For example, $ n \equal{} 2\cdot 3 \cdot 5 \cdot 31$ has middle pair $ 2\cdot 3 \cdot 5 < 31$.

Doesn't seem to be any way to do this without a bunch of casework.[/quote]
Wow, apparently I thought each of $ d_8$ and $ d_9$ are product of two primes for some reason... :|

I think I minimize the casework here, but it's not the most elegant argument.
[hide]From FantasyLover, the possible values of $ p_1,p_2,p_3,p_4$ are $ \{3,5,7,11,13,17\}$.
Note that $ p_1\equal{}3$ and $ p_2\equal{}5$ since excluding either of these from the product would force $ n$ to be too big.
Since $ d_8$ and $ d_9$ are the middle factors, $ d_8d_9\equal{}n$.
Consider $ n$ such that $ n\equiv 1\pmod 4$. Then $ d_8\equiv d_9^{\minus{}1}\equiv d_9\pmod 4$ (odd numbers mod 4 are their own inverses). So $ d_9\minus{}d_8\equiv 0\pmod 4$ and in particular, not $ 22$.
We only need to take care of $ n\equiv 3\pmod 4$. With the earlier restriction, the two cases are $ n\equal{}3\cdot 5\cdot 7\cdot 11$ and $ n\equal{}3\cdot 5\cdot 13\cdot 17$, neither of which work.[/hide]

@bictor
I think you have continued the solution from Fanatasylover ,but fantasylover has assumed that $ d_{9}$ and $ d_{8}$ are product of two primes ,which neednt be true ,
Your solution might work only if you prove $ p_{1} \equal{} 2$ gives you some other contradiction
But clearly $ p_{1}$ cannot be $ 2$ because even if $ d_{9} \equal{}p_{a}p_{b}p_{c}$ and $ d_{8} \equal{} p_{d}$ ,there will be a contradiction because by our assumption $ d_{9}$ and $ d_{8}$ should be of same parity ,but only one of the two can be even . So $ n$ is not even .
So bictor's solution is correct . :rotfl:


A small corporation borrowed $ \$$900,000: some at 7%, some at 8%, and some at 11%. How much was borrowed at 8% if the annual interest was $ \$$72,250 and the amount borrowed at 8% was $ \$$50,000 more than the amount borrowed at 11%?

[hide="Partial Solution"]
I am too lazy to post the actual answer, but I can give you the equations.

$ x \plus{} y \plus{} z \equal{} 900000\\
0.07x \plus{} 0.08y \plus{} 0.11z \equal{} 72250\\
0.08y \equal{} 0.11z \plus{} 50000 \implies 0.08y \minus{} 0.11z \equal{} 50000$

This system can probably be best solved used matrices.
[/hide]


Let $a,b,c$ be distinct positive integers. Prove that

$\frac{a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b-6abc}{a^{2}+b^{2}+c^{2}-ab-bc-ca}\geq \frac{16abc}{(a+b+c)^{2}}$


If $3x+y=6$ and $6x+y=9$ then what is $51x+13y \displaystyle{?}$

[color=darkred]
[hide="alternative solution"]
Instead of just solving the equations...

3x + y = 6
6x + y = 9

4(6x + y = 9)
9(3x + y = 6)

24x + 4y = 36
27x + 9y = 54
51x + 13y = 90

While it's just as easy, possibly easier, to solve for x and y and plug the values in, this approach may be better for certain problems.
[/hide]
[/color]

you could have done it this way..

[hide]
( 6x + y = 9 ) - ( 3x + y = 6 )

=

3x = 3

x = 1

y = 3

then, if u just plug in the numbers, 51 * 1 + 13 * 3 = 90

basically the same stuff as Treething,
[/hide]
[color=darkred]
Hidden.[/color]

Whenever there is 2 variable and 2 equations, it is able to solve. Same it goes with 3 equations and 3 variables. So try solving this...

When:
4a+3b=8c
5+4b=10d
2+9c=6a
7d-6a=3
What is a*b*c*d?

3x+y=6
6x+y=9
y=6-3x
6x+6-3x=9
add the like terms!
3x+6=9
solve the equation
x=1
so u y=3

[quote="daermon"]That was super easy :D [hide="Answer"]x=1 y=3, so answer is [color=red]90[/color][/hide][/quote]

super easy???? ajajaja lol

[quote="lotrgreengrapes7926"][quote="daermon"]That was super easy :D [hide="Answer"]x=1 y=3, so answer is [color=red]90[/color][/hide][/quote]
If you think everything's super easy, you should go to some harder forums.  :D[/quote]

Yes.......... you're right

I just did
[hide]$-6x+9=6-3x\rightarrow x=1$, after which
$y=6-3(1)\rightarrow y=3 \Rightarrow 51(1)+13(3)=\boxed{90}$[/hide] ;)

;) Hmm.. I did it in my head and got.

[hide]90 :)[/hide]

There has been increase in these types of problems. and I love it! LOL :rotfl:

There's another way to solve linear equations-Cramer's rule (although most of the time it makes things more complicated :) ). I'm too lazy to find out how to type determinants in tex, so just google it.
Edit:Here's a nice site[url=http://www.purplemath.com/modules/cramers.htm]www.purplemath.com/modules/cramers.htm[/url]

[hide]3x+y=6
6x+y=9

you can use the multiplication-addition algorithm--

multiply first equation by -1...
-3x-y=-6

add new first equation to second equation to get 3x=3 (this is where the algorithm comes in place--you cancel out one of the variables)

simplify, x=1

plug that into one of the equations--
3(1)=y=6
y=3

plug both into "51x+13y=?"
51(1)+13(3)=[b]90[/b]

[/hide][/b][/hide]

Another way of solving systems of equations (especially better for larger ones) is to use matrix properties to apply the inverse of the coefficient matrix multiplied by the answer matrix. This should give you the answer in a matrix that is $x$ by 1 form, where $x$ is the number of unknown variables. It's cool, no need for determinants.  :D

[hide]
$x=1$ by substitution, so $y=3$
Plugging them in, you get 90[/hide]

Cramer's rule and matrices are methods for solving such equations, but remember that those methods are not quite middle school level.

[color=darkred][size=75][This topic better not end up like the other one. ][/size][/color]


Are you sure there is no condition for $ a,b,c,d$?

For the inequality, $ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} a) \geq {16abcd}$ but for the rest part I suppose the ineq gets inverted...I'm not sure. :maybe:
I mean $ (1 \plus{} \sqrt [4]{abcd})^{4} \leq (1 \plus{} a)(1 \plus{} b)(1 \plus{} c)(1 \plus{} d)$....maybe.

Thank you for such a nice problem, Yulia! :)

[quote="Yulia"]For positive $ a, b, c, d$ prove that
$ (a + b)(b + c)(c + d)(d + a)(1 + \sqrt [4]{abcd})^{4}\geq16abcd(1 + a)(1 + b)(1 + c)(1 + d)$[/quote]
$ (a + b)(b + c)(c + d)(d + a)(1 + \sqrt [4]{abcd})^{4}\geq16abcd(1 + a)(1 + b)(1 + c)(1 + d)\Leftrightarrow$
$ \Leftrightarrow(a + b)(b + c)(c + d)(d + a) - 16abcd +$
$ + 4\sqrt [4]{abcd} \left((a + b)(b + c)(c + d)(d + a) - 4\sqrt [4]{a^3b^3c^3d^3}(a + b + c + d)\right) +$
$ + 2\sqrt {abcd}\left(3(a + b)(b + c)(c + d)(d + a) - 8\sqrt {abcd}(ab + ac + ad + bc + bd + cd)\right) +$
$ + 4\sqrt [4]{a^3b^3c^3d^3}((a + b)(b + c)(c + d)(d + a) -$
$ - 4\sqrt [4]{abcd}(abc + abd + acd + bcd))\geq0,$ which obvious
because $ (a + b)(b + c)(c + d)(d + a) - 16abcd\geq0$ is true by AM-GM;
$ (a + b)(b + c)(c + d)(d + a) - 4\sqrt [4]{a^3b^3c^3d^3}(a + b + c + d)\geq0$ is true since,
$ (a + b)(b + c)(c + d)(d + a) \geq(abc + abd + acd + bcd)(a + b + c + d)\Leftrightarrow$
$ \Leftrightarrow(ac - bd)^2\geq0$
and $ abc + abd + acd + bcd\geq4\sqrt [4]{a^3b^3c^3d^3}$ is true by AM-GM;
$ (a + b)(b + c)(c + d)(d + a) \geq 4\sqrt [4]{abcd}(abc + abd + acd + bcd)$ is true because $ a + b + c + d\geq4\sqrt [4]{abcd}$ is true by AM-GM;
$ 3(a + b)(b + c)(c + d)(d + a) \geq8\sqrt {abcd}(ab + ac + ad + bc + bd + cd)$ follows from
three inequalities:
$ (a + b)(b + c)(c + d)(d + a) \geq(abc + abd + acd + bcd)(a + b + c + d)$ ;
by Maclaren we obtain:
$ \frac {a + b + c + d}{4}\geq\sqrt {\frac {ab + ac + bc + ad + bd + cd}{6}}$ and
$ \frac {\frac {1}{a} + \frac {1}{b} + \frac {1}{c} + \frac {1}{d}}{4}\geq\sqrt {\frac {\frac {1}{ab} + \frac {1}{ac} + \frac {1}{ad} + \frac {1}{bc} + \frac {1}{bd} + \frac {1}{cd}}{6},}$ which equivalent to
$ abc + abd + acd + bcd\geq\sqrt {\frac {8}{3}(ab + ac + bc + ad + bd + cd)abcd}.$ :)

[quote="Yulia"]For positive $ a, b, c, d$ prove that
$ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} a)(1 \plus{} \sqrt [4]{abcd})^{4}\geq16abcd(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)(1 \plus{} d)$[/quote]
I think it can be solved by mixing variables,With $ f(a,b,c,d)\equal{}LHS\minus{}RHS$,and we will prove $ f(a,b,c,d)\geq f(a,b,\sqrt {cd},\sqrt {cd})$
It is only my ideal,sorry if it is wrong

2arqady: I like your proof!  :thumbup:
Here is the author's solution (Oleksii Klurman):
Let’s rewrite our inequality in the form
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} a)}{(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)(1 \plus{} d)}\geq \frac {16abcd}{(1 \plus{} \sqrt [4]{abcd})^4}$
We will use the following obvious lemma
$ \frac {x \plus{} y}{(1 \plus{} x)(1 \plus{} y)}\geq \frac {2\sqrt {xy}}{(1 \plus{} \sqrt {xy})^2}$
By lemma and Cauchy-Schwarz
$ \frac {a \plus{} b}{(1 \plus{} a)(1 \plus{} b)} \frac {c \plus{} d}{(1 \plus{} c)(1 \plus{} d)}(b \plus{} c)(a \plus{} d)\geq\frac {4\sqrt {abcd}(\sqrt {ab} \plus{} \sqrt {cd})^2}{(1 \plus{} \sqrt {ab})^2(1 \plus{} \sqrt {cd})^2}\geq\frac {16abcd}{(1 \plus{} \sqrt [4]{abcd})^4}$
Last one also by lemma for $ x \equal{} \sqrt {ab}, y \equal{} \sqrt {cd}$

Thank you Yulia for this beautiful proof! :lol:
Maybe $ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} e)(e \plus{} a)\left(1 \plus{} \sqrt [5]{abcde}\right)^5\geq$
$ \geq32abcde(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)(1 \plus{} d)(1 \plus{} e)$ is true too?

[quote="arqady"]Maybe $ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} e)(e \plus{} a)\left(1 \plus{} \sqrt [5]{abcde}\right)^5\geq32(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)(1 \plus{} d)(1 \plus{} e)$ is true too?[/quote]
Won't it be
$ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} e)(e \plus{} a)\left(1 \plus{} \sqrt [5]{abcde}\right)^5\geq\\
32abcde(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)(1 \plus{} d)(1 \plus{} e)??$
& can't it be solved using the $ FANTABULOUS$ technique u used to solve Yulia's problem?   :)

[quote="Sunkern_sunflora"]
Won't it be
$ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} e)(e \plus{} a)\left(1 \plus{} \sqrt [5]{abcde}\right)^5\geq \\
32abcde(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)(1 \plus{} d)(1 \plus{} e)??$[/quote]
Thank you, I have fixed it.
[quote="Sunkern_sunflora"]
& can't it be solved using the $ FANTABULOUS$ technique u used to solve Yulia's problem?   :)[/quote]
I am not succeeding. :(

[quote="Sunkern_sunflora"][quote="arqady"]Maybe $ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} e)(e \plus{} a)\left(1 \plus{} \sqrt [5]{abcde}\right)^5\geq32(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)(1 \plus{} d)(1 \plus{} e)$ is true too?[/quote]
Won't it be
$ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} e)(e \plus{} a)\left(1 \plus{} \sqrt [5]{abcde}\right)^5\geq \\
32abcde(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)(1 \plus{} d)(1 \plus{} e)??$
& can't it be solved using the $ FANTABULOUS$ technique u used to solve Yulia's problem?   :)[/quote]

What's $ FANTABULOUS$ technique ? :maybe:

[quote="arqady"]
Maybe $ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} e)(e \plus{} a)\left(1 \plus{} \sqrt [5]{abcde}\right)^5\geq$
$ \geq32abcde(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)(1 \plus{} d)(1 \plus{} e)$ is true too?[/quote]
Yes, as I know it is proved for n=5 (but I don't know the proof) and it is unsolved for general case.

[quote="shfdfzhjj"]What's $ FANTABULOUS$ technique ? :maybe: [/quote]
Umm.... nothing special. It's also mathematical!
$ FANTASTIC\plus{}FABULOUS\equal{}FANTABOLOUS$!! :wink:

[b]To Yulia:[/b]
You wrote that the problem is unsolved for general cases. May I know why?

Does the ineq holds for $ n\equal{}3$ as well?
I mean :

$ (a\plus{}b)(b\plus{}c)(c\plus{}a)\left(1\plus{} {\sqrt[3]{abc}}\right)^3 \geq {8abc(1\plus{}a)(1\plus{}b)(1\plus{}c)}$......is it true for positive $ a,b,c,d$??

[quote="Sunkern_sunflora"]

$ (a \plus{} b)(b \plus{} c)(c \plus{} a)\left(1 \plus{} {\sqrt [3]{abc}}\right)^3 \geq {8abc(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}$......is it true for positive $ a,b,c,d$??[/quote]
It's true. Try to prove it. :wink:

O.K., I'll try. But can u tell me arqady, why it is true upto $ n\equal{}5$ and not after that. I am totally confused! :huh:

Anyway, hopefully it follows from Yulia's general technique, doesn't it?
 :)

Inequality of such type is true for five variables. And if it is true for n variables then it is true for less than n variables. But I could not prove it for six variables  :) .

[quote="arqady"]Maybe
$ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} e)(e \plus{} a)\left(1 \plus{} \sqrt [5]{abcde}\right)^5$

$ \geq32abcde(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)(1 \plus{} d)(1 \plus{} e)$

is true too?[/quote]Let $ a \equal{} b \equal{} c \equal{} x^2,d \equal{} e \equal{} \frac {1}{x^3}$, then

$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} e)(e \plus{} a)\left(1 \plus{} \sqrt [5]{abcde}\right)^5}{abcde(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)(1 \plus{} d)(1 \plus{} e)} \equal{} \frac {256x\left(1 \plus{} x^5\right)^2}{\left(1 \plus{} x^2\right)^3\left(1 \plus{} x^3\right)^2} < 32$

holds for $ 0 < x < \frac {1}{8}\vee x > 8$.

Thank you very much, Ji Chen!
But your counterexample does not refute the following inequality.
Let $ a,$ $ b,$ $ c,$ $ d$ and $ e$ are non-negative numbers. Prove that:
\[ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} e)(e \plus{} a)}{32}\geq\frac {a \plus{} b \plus{} c \plus{} d \plus{} e}{5}\sqrt [5]{a^4b^4c^4d^4e^4}
\]
:(

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=218419


calculus
integration
function
inequalities
quadratics
special factorizations
real analysis

I need help with the following problem. I have seen the solution for this problem, but I still don't understand. Could someone give me a solution with
a clear explanation? Thanks!!!

$f(x)$ and $g(x)$ are both continuous over $a\leq x \leq b$, Using the fact that for any real t the inequality $\int^{a}_{b}\left\{tf(x)+g(x)\right\}^{2}dx\geq 0 $  is true, prove the following inequality
$
\left\{\int^{a}_{b}f(x)g(x)dx\right\}^{2}\leq \int^{a}_{b}\left\{f(x)\right\}^{2}dx \int^{a}_{b}\left\{g(x)\right\}^{2}dx
$

This is the Cauchy-Buniakovsky-Schwarz inequality, for integrals, which makes it one of the most famous of all inequalities. The hint is steering you towards the proof that works in any abstract inner product space; there are several other possible proofs.

To follow the hint: First assume WLOG that $f$ is not identically zero. (If it is, the inequality is trivially true.) Expand out $\int (tf+g)^2$ to get a quadratic expression in $t$. Complete the square in this expression, then set $t$ equal the value that minimizes the expression. Since it is (as the integral of a square) $\ge 0$ for all $t$, you can use that to get the inequality you need.

That's for real-valued functions. You have to be a little more careful with the idea of "completing the square" for complex valued functions, but you can still make a version of this proof work.

For a second proof: Assume to start with that $\int f^2 = \int g^2 = 1$. Prove the inequality in that special case by using the inequality $fg \le \frac12f^2+\frac12g^2$. Then prove the general case by considering $F = f/\sqrt{\int f^2}$ and $G = g/\sqrt{\int g^2}$.

Dr. Merryfield,
I didn't know this inequality actually has a name. At least now I can look it up online.
I am going to try it again following your directions. I might have to ask you more questions.
Thank you for your help! :)

hmm
don't you know Cauchy Ineq: \[(\sum a_i b_i)^2 \leq (\sum a_i^2) (\sum b_i^2) \]?

No. Not until Dr. Merryfield told me the name of the inequality. Kumon math didn't mention it.
After reading his msg, I went to Wolfram website and found some info about it. It is interesting.


geometry
circumcircle
geometric transformation
homothety
geometry proposed

I and Amir2 used to combined some facts about two transformations ,Here is the result :
[color=blue]Problem.[/color]Let $ABCD$ be a quadrilateral such that $AD$ meets $BC$ in $E$ and $AB$ meets $CD$ in $F$.
Let $M$ on $AB$ and $N$ on $CD$ be points such that \[\frac{AM}{MB}=\frac{DN}{NC}=\lambda_{1}\] and $R$ on $AD$ and $S$ on $BC$ such that \[\frac{AR}{RD}=\frac{BS}{SC}=\lambda_{2}\] Denote by $\mathcal{P}$ the intersection of the circumcircles of triangles $ERS$ and $FMN$ .(One which is not in the quadrilateral).

$1)$Prove that as $\lambda_{1},\lambda_{2}$ varies $\mathcal{P}$ remians constant.
$2)$If $ABCD$ is inscribed in a circle with center $O$ then $O\mathcal{P}$ is perpencidular to $EF$.

[quote="lomos_lupin"]... $\frac{AM}{MB}=\frac{DN}{\boxed{CD}}=\lambda_{1}$ ...[/quote] I think that $\frac{AM}{MB}=\frac{DN}{NC}\ .$ Very nice problem ! I"ll approach.

yes,very Nice but easy problem.

just notice that $\mathcal P$ is spiral homothecy center that bring $AB$ to $CD$ and $AD$ to $BC$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that if a group $G$ has square-free order and $p$ is the largest prime divisor of $|G|$, then it has a normal $p$-Sylow subgroup.

It is obvious if $|G|=p$ . Let  $|G|=pp_1\cdots p_k$,where $p_i$ all are different prime. Since $n_p\mid |G|$ then $n_p=1,p$ or $p_i$ for some $i$. Since $n_p\equiv 1\mod p$ we cannot have $n_p=p$. We cannot have $n_p=p_i$ too, since this implies that $p\mid p_i-1$ and then $p\leq p_i-1$ (contradicts the fact that $p>p_i$). So $n_p=1$ and the $p$-sylow subgroup is normal.

$n_p$ divides $p_1\ldots p_k$; it doesn't need to be a prime, as you appear to believe. It could be, for instance, $p_1p_2$, or $p_2p_4p_6$, and so on.


i need help in regards to solving number theory equations, i thought i understood it but apparently i don't, can you give me some advice
for example so you can see how i'm thinking in regards to number theory
2^1000 divided by 13 leaves a remainder of what?
i started off by saying that  2^5= 6 mod 13 and that 2^ (5k)= 6^k mod 13 and that 2 ^5(200) = 6^200 mod 13 which is clearly wrong. Can you maybe give me some pointers please?

How about starting off with $ 2^6\equiv \minus{}1 \mod 13$?

[hide]
So the question asks us to find $ x$ given that

$ 2^{1000} \equiv x \mod 13$

By Fermat's little theorem,

$ 2^{12} \equiv 1 \mod 13$
$ \Rightarrow 2^{996} \equiv 1 \mod 13$
$ \Rightarrow 2^4 \equiv x \mod 13$
$ \Rightarrow x \equiv 3 \mod 13$

So the remainder is $ \boxed{3}$
[/hide]

To understand the solution posted by RisingStar, you need to know Fermat's Little Theorem, so I'll explain it for you.

Fermats Little Theorem:

$ a^p\equiv a\pmod{p}$ when $ a$ is an integer and $ p$ is prime.

A variant of this is:

$ a^{p\minus{}1}\equiv 1\pmod{p}$ when $ a$ is coprime to the prime integer $ p$.

I hope this helps. If you want a proof for FLT I can show you.

thanks, that actually does help a lot  :lol:


Which is greater, $\log_2 3$ or $\log_3 5$? (No calculators, of course :) )

[quote="cezar lupu"]You know that $log_{n} (n+1)\geq log_{n+1} (n+2) $ ;)[/quote]
Well, even if we use it, we obtain $\log_2 3\ge \log_3 4$, but that isn't enough because $\log_3 5\ge \log_3 4$...

[quote="K81o7"]I say $\log_2{3}$.  You first of all know that $\sqrt{2}<\frac{3}{2}$.  Thus, $log_2{3} > \frac{3}{2}$.  We then know that $\sqrt{3} > \frac{5}{3}$.  Thus, $\log_3{5} < \frac{3}{2}$.  This gives us $\log_2{3} > \frac{3}{2} > \log_3{5}$ and the result follows from the transitive property.[/quote]
Yeah, that's correct. It's probably the easiest to compare two logs with $\frac{3}{2}$.

look at this,
$log_2{3}$
$=\frac{log3}{log2}$
so,
$=log\frac{3}{2}$
taking 2nd case,
$=log\frac{2}{3}$
so naturally $log_2{3}$ is bigger
correct?

[quote="riddler"][quote="frt"][quote="riddler"]look at this,
$log_2{3}$
$=\frac{log3}{log2}$
so,
$=log\frac{3}{2}$
[/quote]
Why? $\frac{\log 3}{\log 2}\ne \log \frac{3}{2}$ :?[/quote]
sorry but Im really new to logs I didnt know. :blush:[/quote]

for log we have these properties

$\log a +\log b =\log ab $
$\log a - \log b =\log \frac ab$

[quote="riddler"][quote="K81o7"]I say $\log_2{3}$.  You first of all know that $\sqrt{2}<\frac{3}{2}$.  Thus, $log_2{3} > \frac{3}{2}$.  We then know that $\sqrt{3} > \frac{5}{3}$.  Thus, $\log_3{5} < \frac{3}{2}$.  This gives us $\log_2{3} > \frac{3}{2} > \log_3{5}$ and the result follows from the transitive property.[/quote]
how did you get $log_2{3}>3/2$? from $\sqrt2<\frac{3}{2}$?[/quote]

you know that $\sqrt2<\frac{3}{2}$  :arrow: $2\sqrt{2} <3$  :arrow: $2^{\frac 32}<3$ . Then taking $\log$ both side you will get

$\log 2^{\frac 32} <\log 3$ :arrow: $\frac {3}{2} <\frac{\log3}{\log2}=log_2{3}$


Find the maximum and minimum value of $x^2+y^2$ when :
 $(x^2-y^2+1)^2+4x^2y^2-x^2-y^2 =0$

Let $X=x^2,Y=y^2\ (X\geq 0,\ Y\geq 0),$ we can rewrite the given equation as $(X+Y)^2-(X+Y)+2(X+Y)+1=0$
Set $X+Y=K\ (K\geq 0)\ \cdots [1],$ we have $K^2-K+2(X-Y)+1=0\Longleftrightarrow X-Y=-\frac{K^2-K+1}{2}\ \cdots [2].$ Solving the system equation $[1],\ [2],$ yielding $X=-\frac{K^2-3K+1}{4},Y=\frac{K^2+K+1}{4}\ \cdots[*]$
Considering the condition for which there exists the non negative real numbers $X,Y$ satisfying $[*],$ we obtain $K^2-3K+1\leq 0\Longleftrightarrow\frac{3-\sqrt{5}}{2}\leq K\leq \frac{3+\sqrt{5}}{2},$ which satisfies $K\geq 0.$
Here the equalities holds when $(X,Y)=(0,K).$ Thus  the desired maximum value is $\frac{3+\sqrt{5}}{2},$ when $(x,y)=\left(0, \pm \sqrt{\frac{3+\sqrt{5}}{2}}\right),$ and minimum value is $\frac{3-\sqrt{5}}{2},$ when $(x,y)=\left(0, \pm \sqrt{\frac{3-\sqrt{5}}{2}}\right) .$


Points $K,N$ on sides $AB,AD$ of square $ABCD$ are such that $AK\cdot AN=2BK\cdot DN$. Lines $CK,CN$ intersect diagonal $BD$ at $L,M$. Show that $K,L,M,N,A$ are cyclic.

[quote="cancer"][color=darkred]Points $ K,\,N$ on sides $ AB,\,AD$ of square $ ABCD$ are such that $ AK\cdot AN=2BK\cdot DN.$
Lines $ CK,\,CN$ intersect diagonal $ BD$ at $ L,\,M.$
Show that $ K,\,L,\,M,\,N,\,A$ are cyclic.[/color][/quote]
[color=indigo][b]Proof.[/b] Denote $ AB=1,\,AK=a,\,AN=b$
We have: $ ab=2\cdot (1-a)(1-b)\implies ab+2-2a-2b=0\implies a^{2}b^{2}+2ab-2a^{2}b-2ab^{2}=0$ $ \Rightarrow a^{2}+b^{2}=\left(2-a-b\right)^{2}\Rightarrow KN^{2}=(BK+DN)^{2}\Rightarrow KN=BK+DN$

Denote $ E\in [AB: \,EA>BA,\,BE=DN$
Easy to show that: $ \triangle ECK=\triangle NCK$ and $ \angle NCE=90^\circ$
$ \implies \angle NCK=45^\circ\implies BCMK$ is cyclic $ \implies\angle NMK=90^\circ$
Similarly, we also have $ \angle NLK=90^\circ$

Hence $ K,\,L,\,M,\,N,\,A$ lie on the circle with diameter $ KN$ [/color]


Suppose that $\alpha$ is rational. Show that $cos \alpha \pi$ is either $0, \pm 1, \pm 1/2$ or irrational.


Suppose we have the functions $ f(x)\equal{}x^{0.5}\plus{}1$ and $ g(x)\equal{}\frac{1}{x}$.
What is the domain of $ g(f(x))$?
[hide="My Difficulties"]According to my math teacher, x must be positive because $ g(x)$ is not defined for $ x\equal{}0$ and $ f(x)$ is not defined for negative values of $ x$. When you plug in $ f(x)$, however, you find that $ x\equal{}0$ produced a real number. What is the correct answer and where can I find further clarification? If my dear teacher is incorrect, would it be best to simply agree with her? She once told the class that $ \frac{1}{3}$ was irrational and argued bitterly over it with a student in my class without bothering to check the textbook. [/hide]

If your teacher's going to argue that $ \frac {1}{3}$ is irrational, you're in trouble anyway.

The domain is $ [0,\infty)$ Your teacher seemed to give an awkward explanation.

edit: sniped got it first.

[quote="SnipedYou"]Well $ g(f(x)) \equal{} \frac {1}{\sqrt {x} \plus{} 1}$ so the domain is just positive reals and $ 0$ assuming the primary root is taken.

Since $ g(f(0)) \equal{} 1$, and as $ x$ approaches $ \infty$, $ g(f(x))$ approaches $ 0$, the the range would be $ 0 < g(f(x)) \le 1$.[/quote]Just looking at $ g(f(x))$ doesn't tell you the domain of the composition.

For example, let $ f(x)\equal{}\sqrt{x},g(x)\equal{}x^2$. The domain of f is $ [0,\infty)$ and the range of f is $ [0,\infty)$. Now the domain of g is all reals and since $ [0,\infty)\subset\mathbb{R}$, it follows that domain of $ g(f(x))$ to be $ [0,\infty)$. If you just look at $ g(f(x))\equal{}x$, it seems that the domain is all real numbers.

The underlying key is that the range of f may be the restriction in the domain of g as I showed above.

[quote="aufha"]She once told the class that $ \frac {1}{3}$ was irrational and argued bitterly over it with a student in my class without bothering to check the textbook.[/quote]

Haha, I find this funny, but sadly, not surprising.

</spam>


vector
linear algebra
linear algebra unsolved

Let ST =TS
S: V--->V    T: V:--->V
Suppose that each of  S and T may be presented by diagonal Matrices
Prove that there is a basis for V with respect to which both S &T are given by diagonal matrices simultaneousely

Suppose $v$ is an eigenvector of $S$ with eigenvalue $\lambda$. Then $STv = TSv = T(\lambda v) = \lambda Tv$ and $Tv$ is also an eigenvector of $S$ with eigenvalue $\lambda$, i.e. $T$ preserves the individual eigenspaces of $S$. So for each such eigenspace, consider the restriction of $T$ that maps the eigenspace to itself and find a basis over the eigenspace that diagonalizes $T$; let the basis for the whole space be the union of these individual bases. It continues to diagonalize $T$ and also diagonalizes $S$ as all the vectors are still eigenvectors of $S$ as well.


A graph has $ n > 2$ points. Show that we can find two points A and B such that at least $ [n/2]-1$ of the remaining points are joined to either both or neither of A and B

hint

[hide]There are n people at a party. Prove that there are two people such that, of the remaining n−2 people, there are at least [n/2]−1 of them, each of whom knows both or else knows neither of the two.[/hide]


As sxoliasoume edw ta provlimata tis Valkaniadas kathws episis kai o,tidipote girw apo auti, mia kai fetos tha diorganwthei stin xwra mas meta apo 10 xronia (Kalampaka 1997).

H episimi istoselida tis Valkaniadas anartithike prosfata kai einai h http://www.hms.gr/bmo2007/main.html

[color=red][size=134][b]KALH EPITYXIA STOUS SIMMETEXONTES KAI KALA METALLIA!!![/b][/size][/color]

Filika,
Alexandros

Βάζω τα περσινά αποτελέσματα - για σύγκριση - και εύχομαι να τα πάμε φέτος πολύ καλύτερα .

[b]Μπορούμε !  [/b]

1. Romania 165
2. Moldova 122
3. Albania 116
4. Bulgaria 107
5. Greece 106
6. Serbia and Montenegro 103
7. Cyprus 43
8. FYRO Macedonia 39

and Turkey didn't participate.

Παιδιά κι εγώ θα ευχηθώ ότι κι ο Αλέξανδρος αλλά σε σωσστά Ελληνικά όπως πρ΄΄επει να λέγεται κατα τον καθηγητή μου

ΕΠΙΤΥΧΙΑ ΣΕ ΟΛΑ ΤΑ ΠΑΙΔΙΑ (ΜΕ ΣΥΝΓΚΕΝΤΡΩΣΗ ΚΙ ΠΡΟΣΟΧΗ)

Δεν ξέρω , αλλά προσπαθώντας να καταλάβω τον περίγυρο , περιμένω 4-6 μετάλλια από την Ελληνικά  ομάδα (Όχι βέβαια όλα χρυσά , αλλά πολλά χάλκινα!) . Ίσως βγούμε και δεύτεροι  ή τρίτοι! Θα δείξει.

  [color=red]Καλή Επιτυχία σε όλη την ΟΜΑΔΑ !!! [/color]

Καλή επιτυχία στα παιδιά και στους αναπληρωματικούς και στους τζούνιορ πρόμπλεμ σόλβερς. :wink:  :D

[quote="KAZA"]Ξέρει κανεις πότε γράφουν τα παιδιά και πότε βγαίνουν τα αποτελέσματα :?:[/quote]

H imera tou diagwnismou simfwna kai me to episimo programma einai to Savvato 28 Apriliou to prwi. Tin Kiriaki ksekinaei i vathmologisi twn graptwn  kai tin idia imera to vradi, tha iparxoun ta episima apotelesmata.

Alexandros

ευχαριστούμε για την ενημέρωση

Αγαπητοί φίλοι,

Σας στέλνω τα θέματα της Βαλκανιάδας 2007 που έγινε σήμερα το πρωί στη Ρόδο.

Πρόβλημα 1

Δίνεται κυρτό τετράπλευρο $ABCD$  με $AB=BC=CD, AC\neq BD$ και έστω $E$ το σημείο τομής των διαγωνίων του. Να αποδειχτεί ότι $AE=DE$ αν και μόνο αν $\angle{BAD}+\angle{ADC}=120^{\circ}$.

Πρόβλημα 2

Να βρεθούν όλες οι συναρτήσεις $f: \mathbb{R}\to\mathbb{R}$ έτσι ώστε να ισχύει \[f\left(f(x)+y\right)=f\left(f(x)-y\right)+4f(x)y\] για κάθε ${x,y \in \mathbb{R}}$.

Πρόβλημα 3

Να βρεθούν όλοι οι θετικοί ακέραιοι $n$ τέτοιοι ώστε να υπάρχει μία μετάθεση $\sigma$ του συνόλου $\{1,2,\cdots,n\}$ για την οποία ο αριθμός \[\sqrt{\sigma(1)+\sqrt{\sigma(2)+\sqrt{\cdots+\sqrt{\sigma(n)}}}}\] να είναι ρητός.

Σημείωση: Μία μετάθεση του συνόλου $\{1,2,\cdots,n\}$ είναι μία συνάρτηση ένα προς ένα του συνόλου αυτού στον εαυτό του.

Πρόβλημα 4

Έστω $C_{1},C_{2},C_{3}$ τα σύνορα τριών κυρτών $n$-γώνων στο επίπεδο, όπου $n$ θετικός ακέραιος μεγαλύτερος του $2$, έτσι ώστε τα σύνολα $C_{1}\cap C_{2}, C_{2}\cap C_{3}, C_{3}\cap C_{1}$ να είναι πεπερασμένα. Να βρεθεί ο μέγιστος αριθμός των σημείων του συνόλου $C_{1}\cap C_{2}\cap C_{3}$.

Αλέξανδρος

Όμορφο το 3ο θέμα .   :)   Ελπιζω να έχει και όμορφη λύση.

Καλή επιτυχία στα παιδιά έστω και καθυστερημένα.   :oops:

[quote="perfect_radio"]What are the expected results for the Greek team?[/quote]

 Untill now I have not comunicated with some contestant in Rhodes . I hope in the evening we ''ll have some information.

 I wish good results to the Romanian team!

 [u]babis[/u]

[quote="cretanman"]Αγαπητοί φίλοι,

Σας στέλνω τα θέματα της Βαλκανιάδας 2007 που έγινε σήμερα το πρωί στη Ρόδο.

Πρόβλημα 1

Δίνεται κυρτό τετράπλευρο $ABCD$  με $AB=BC=CD, AC\neq BD$ και έστω $E$ το σημείο τομής των διαγωνίων του. Να αποδειχτεί ότι $AE=DE$ αν και μόνο αν $\angle{BAD}+\angle{ADC}=120^{\circ}$.

Πρόβλημα 2

Να βρεθούν όλες οι συναρτήσεις $f: \mathbb{R}\to\mathbb{R}$ έτσι ώστε να ισχύει
\[f\left(f(x)+y\right)=f\left(f(x)-y\right)+4f(x)y \]
για κάθε ${x,y \in \mathbb{R}}$.

Πρόβλημα 3

Να βρεθούν όλοι οι θετικοί ακέραιοι $n$ τέτοιοι ώστε να υπάρχει μία μετάθεση $\sigma$ του συνόλου $\{1,2,\cdots,n\}$ για την οποία ο αριθμός
\[\sqrt{\sigma(1)+\sqrt{\sigma(2)+\sqrt{\cdots+\sqrt{\sigma(n)}}}}\]
να είναι ρητός.

Σημείωση: Μία μετάθεση του συνόλου $\{1,2,\cdots,n\}$ είναι μία συνάρτηση ένα προς ένα του συνόλου αυτού στον εαυτό του.

Πρόβλημα 4

Έστω $C_{1},C_{2},C_{3}$ τα σύνορα τριών κυρτών $n$-γώνων στο επίπεδο, όπου $n$ θετικός ακέραιος μεγαλύτερος του $2$, έτσι ώστε τα σύνολα $C_{1}\cap C_{2}, C_{2}\cap C_{3}, C_{3}\cap C_{1}$ να είναι πεπερασμένα. Να βρεθεί ο μέγιστος αριθμός των σημείων του συνόλου $C_{1}\cap C_{2}\cap C_{3}$.

Αλέξανδρος[/quote]

[b] Αλέξανδρε σε ευχαριστούμε πολύ ! [/b]

 Θα περιμένουμε να μας πεις τις εντυπώσεις σου εκ των ... έσω , μια και είσαι στην επιτροπή του διαγωνισμού με  σημαντικό μάλιστα ρόλο.

   Καλή διόρθωση και καλή επιστροφή !

 Μπάμπης

H geometria poli geloia.. se misi grammi!.. Niko (Rapane) pos piges???????????????????????????????Oi alloi pos pigan?

Παρακαλώ , όσοι βρίσκετε λύσεις να τις γράφετε - χωρίς latex ενδεχομένως - για να τις συγκεντρώσουμε όλες μαζί. Δε νομίζω ότι εφράσεις όπως '' πολύ εύκολο κλπ '' βοηθάνε ή έχουν αξία. Τα έχω γράψει επανειλημμένα : κάθε άσκηση είναι εύκολη , αν λυθεί , και δύσκολη αν δε λυθεί. Εκ των υστέρων όλα είναι άνευ σημασίας. Σημασία έχει να τη λύσει αυτός που διαγωνίζεται !!! Λοιπόν , αν βρείτε ή αν ακούσετε ή αν συναντήσετε  κάποια λύση σε θέματα της  ΒΜΟ 2007 ,  να την παρουσιάσετε. Ακόμα και τυχόν προσπάθειες είναι καλό να τις στείλλετε. Ίσως κάποιος άλλος να τις  συμπληρώσει και να βρει μια νέα λύση.
 Τα θέματα θα τα βάλουμε και στο βιβλίο ! Αυτός είναι και ο λόγος που κάνω την παραπάνω έκκληση. Ευχαριστώ και

[size=150]   [color=red]Καλά Αποτελέσματα !!![/color][/size]

  Μπάμπης

Συγχαρητήρια παιδιά που ανταπεξήλθατε ακόμα και σε τόσο δύσκολα θέματα και ακόμα καλύτερα στο ΙΜΟ. :wink:  :)

kseroume apo poies xwres protathikan ta themata?

1-από Αλβανία
2-από Βουλγαρία
3-από Σερβία
4-από Τουρκία

Galeos??? I guess Galanis???!!! Ela re... katalabes poios eimai pisteuo... pos pas?? Perimeno tin lisi sou sto trito! Ego tin douleia mou sto proto tin ekana. Now it's your turn...o Stergios kai ego exoume kati lisares alla perimeno prota na do tin kinisi sou.

Οι αναπληρωματικοί πως τα πήγαν :?:

Ήμουνα ένας εξ' αυτών .
[quote="KAZA"]Οι αναπληρωματικοί πως τα πήγαν :?:[/quote]

Θα δείξει  :wink:

ITA1-Maria Colombo 10-5-9-2 tot 26 argento!
ITA2-Andrea Conti 10-4-1-0 tot 15 bronzo
ITA3-Roberto Daluiso 10-5-5-1 tot 21 bronzo
ITA4-Andrea Fogari 10-5-0-1 tot 16 bronzo
ITA5-Mattia Galeotti 10-4-1-1 tot 16 bronzo (+1 sul cutoff e non -1 ... boll :p )
ITA6-Denis Nardin 10-4-1-1 tot 16 bronzo

6medaglie ma gara strana :/
comunque sesta squadra totale battendo UK (di3 punti!!!) e Kazakhstan...

BMO_Team
Leblanc, jacksparrow, scaccomatto, edriv, mattilgale, sisifo.

Αυτό είπα κι εγώ ουσιαστικά αλλά μου είπαν ότι πρέπει να γίνει κάτι άλλο πρώτα. :wink:

[quote="kpsychas"]Για τη δεύτερη άσκηση βοηθάει να θέσετε όπου y to f(x)[/quote]

Πρέπει να ελέγξεις το σύνολο τιμών της f πρώτα  :wink:

Έχει κανείς λύσεις για τα άλυτα (και κατά την ταπεινή μου γνώμη απαράδεκτα για μαθητικό διαγωνισμό) 3, 4? Πραγματικά δεν ξέρω τι να κάνω (ειδικά στο 4ο είμαι απελπιστικά μακριά...)

Kalispera se olous,

H Ellada itan stin 7h thesi anamesa se oles tis xwres pou simmeteixan stin Valkaniada kataktwntas 4 xalkina Metallia. Stin prwti thesi vrisketai i Voulgaria, stin deuteri h Roumania kai stin Triti vrisketai i Moldavia. Tin seira apo ekei kai pera den tin thimamai kathws ksexasa na parw mazi mou ta episima apotelesmata tou diagwnismou. Pisteuw omws oti sintoma tha anartithoun stin episimi istoselida tis Valkaniadas.

Aksizei na simewsoume oti ta 2 teleutaia themata tou diagwnismou xaraktiristikan ws prototipa gia tin Valkaniada kai diskolepsan arketa tous mathites. Eidika sto 3o thema den vrethike 2os tropos lisis apo tous arxigous twn omadwn opws sinithizetai ligo prin ton diagwnismo kai kata tin diarkeia tou, ektos apo ekeinon pou ipirxe stis episimes liseis tis Valkaniadas, pragma aksioperiergo. Oloi oi tropoi pou ekanan oi mathites itan paromoioi me tin episimi lisi ektos apo ena mathiti apo tin Kipro pou eixe mia prototipi idea h opoia eline to provlima. Mporei na ipirxan merika lathakia stin lisi tou alla fainotan i idea h opoia odigouse se poli kali lisi kai autos itan kai o logos pou vathmologithike me 10. (Etixe na eimai stin diorthwsi tou 3ou thematos).

Telos tha ithela na anaferw, oti sto telos tis Valkaniadas, oi arxigoi psifisan (apo tin epomeni BMO kai epeita) na grafontai ta onomata twn kataskeuastwn twn provlimatwn sto xarti me ta themata. Episis apo tin epomeni BMO tha mporoun na proteinoun themata akomi kai oi xwres pou simmetexoun unofficially.

Episis 2 mathites eixan perfect score (40/40), o enas itan apo tin Roumania kai o allos apo tin Moldavia (o opoios exei ws twra 3 xrisa metallia stis IMO (stis 2 teleutaies eixe perfect score (42/42)!!!), kai 1 asimenio,  kai mallon twra paei gia to 4o xriso metallio).

Tha ithela na afisw ta 2 teleutaia themata na ta epeksergasteite prin dwsw tin episimi lisi pou proteinan oi xwres pou ta kataskeuasan... Einai, opws anaferame, arketa prwtotipa kai aksizoun.

[color=red][b]Sigxaritiria stin Elliniki omada gia tin poli kali parousia kai euxomaste kali epitixia stin proetoimasia tis gia tin IMO!![/b][/color]

Alexandros

[quote="ambros"]xerete pou tha ginei i epomeni BMO????[/quote]

Ksexasa na to anaferw sto proigoumeno minima. H epomeni BMO tha ginei sta skopia (FYROM).

Alexandros

Ta apotelesmata tis Valkaniadas opws episis kai ta themata kai oi episimes liseis anartithikan stin episimi selida tis Valkaniadas

http://www.hms.gr/bmo2007/main.html

Alexandros


Let a,b,c>0 such that a^2+b^2+c^2=3. Then:
   12(a^4+b^4+c^4)>=27+(2a^3+2b^3+2c^3-a-b-c)^2.
       I know it looks like hell, but I posted it since I find it very strong.

It seems that this is neither hard nor wrong. Again, your solution is much more interesting than mine.

[quote="Namdung"]ii) 4(a^3+b^3+c^3)(a+b+c) >= 4(a^2+b^2+c^2)^2 = 36
iii) (a+b+c)^2 <= 3(a^2+b^2+c^2) = 9[/quote]

Maybe I really am missing something, but I don't see how you can simply add/subtract these inequalities to give the initial inequality...

Use also i) and you will find the answer.

I really wonder what I was thinking when I wrote that post... :? It all seems clear now... //embarrassed


Payne can paint 100 square feet per hour. How many hours will it take Payne to paint both sides of a rectangular wall 125 feet long by 6 feet high?

The answer is 15
125*6= 750
They want both sides of the wall, so multiply by 2 to get 1500
Then divide by 100 to get 15.

its 15 i dont know how that happens though.

It was explained in the post above yours.


The solubility of silver oxalate is $ 2.06 \cdot 10^{ - 4}$ at pH=7.0.
(1) Calculate the solubility of silver oxalate in acidified water of pH=5.0. $ K_{a1} = 5.6 \cdot 10^{ - 2}$ and $ K_{a2} = 6.2 \cdot 10^{ - 5}$
(2) In presence of ammonia in the aqueous solution, silver forms 2 complexes $ \ce {Ag(NH_3)^ +}$ and $ \ce {Ag(NH_3)_2^ +}$. The values of the stepwise stability constants for the formation of these complexes are $ 1.59 \cdot 10^3$ and $ 6.76 \cdot 10^3$. What is the stability of silver oxalate in an aqueous solution that contains 0.02M NH3 and has pH=10.8 ?

I'm a bit rusty with these equilibrium problems but I will give it a go anyway. By the way, there might be computational errors but I believe the concept is right.

Well first we should calculate the solubility product of silver oxalate:

$ K_{sp}=3.50 \times 10^{-11}=[\ce{Ag+}]^{2}[\ce{C2O4}^{2-}]$
[hide=" Long Solution"]
(1) In this part we consider three equilibriums:

$ \ce{Ag2C2O4} \rightleftharpoons 2\ce{Ag+} + \ce{C2O4}^{2-}$
$ \ce{H2C2O4} \rightleftharpoons \ce{H+} + \ce{HC2O4}^{2-}$
$ \ce{HC2O4-} \rightleftharpoons \ce{H+} + \ce{C2O4}^{2-}$

The solubility $ S = \frac{1}{2}[\ce{Ag+}] = [\ce{C2O4}^{2-}] + [\ce{HC2O4-}] + [\ce{H2C2O4}]$. Clearly directly obtaining the concentration of silver ion is impossible so instead we calculate the second expression. Given $ pH = 5.0$ we conclude that $ [\ce{H+}] = 10^{-5}M$ and thus $ S = \frac{1}{2}[\ce{Ag+}] = [\ce{C2O4}^{2-}](1+ \frac{\ce{[H+]}}{K_{a2}} + \frac{[\ce{H+}]^{2}}{K_{a1}K_{a2}})= 1.16[\ce{C2O4}]^{2-}$.

Thus we can calculate $ S = 2.16 \times 10^{-4} M$.

(2) We now consider two other equilibriums (as well as the first three):

$ \ce{Ag+} +\ce{NH3} \rightleftharpoons \ce{Ag(NH3)+}$
$ \ce{Ag(NH3)+} +\ce{NH3} \rightleftharpoons \ce{Ag(NH3)2+}$

Like before we define solubility as ${ S = \frac{1}{2}([\ce{Ag+}] + [\ce{Ag(NH3}+}] + [\ce{Ag(NH3)2+}]) = 1.15 \times 10^{27}[\ce{C2O4}^{2-}]$. Furthermore $ S=2170[\ce{Ag+}]$ and thus $ S= 5.74 \times 10^{5}M$. [/hide]

[quote="hell_ever"]Do you really think we can dissolve 1,70,000 odd kgs of silver oxalate in 1 kg of water?  :huh:

You have assumed stability constant is dissociation constant.. :P[/quote]

Oops, didn't read the question properly, my bad. Still you solve it in the same way anyway  :P .


On a twenty-question test, each correct answer is worth 5 points, each
unanswered question is worth 1 point and each incorrect answer is worth 0
points.  Which of the following scores is $ \textbf{NOT}$ possible? \\ \\

$ \text{(A)}\ 90 \qquad \text{(B)}\ 91 \qquad \text{(C)}\ 92 \qquad \text{(D)}\ 95 \qquad \text{(E)}\ 97$ \\ \\
Select the most appropriate letter.

Looking at Choice E, note how one must miss 3 points to get a 97. It is not possible to do that since one can only miss points in increments of 4 or 5.

You can get a 90 by missing 2 questions.

You can get a 91 by missing 1 question and skipping another.

You can get a 92 by skipping 2 questions.

You can get a 95 by missing 1 question.


[quote="quangpbc"]Let x_1 > 0 and x_(n+1) >= (n + 2)x_n - \sum kn_k
( k=1 ->n-1 ) for n >= 2. Prove that for any a ∈ R the
sequence x_n eventually gets bigger than a[/list][/quote]
$ x_{1}>0$
$ \forall n>1: \ x_{n+1}+\sum_{k=1}^{n-1}kx_{k}\ge (n+2)x_{n}$
we have to prove that $ \forall a\in R,\exists n\in\mathbb{N}^{*}: \ x_{n}>a$  :?:  :?:
is that what you want to say?

I have posted a problem ,i have edited it on here :
Let $ x_{1}> 0$ and $ x_{n+1}\ge (n+2)x_{n}-\sum_{k=1}^{n-1}{kx_{k}}$ for $ n\ge2$. Prove that for any $ a\in R$ the
sequence $ x_{n}$ eventually gets bigger than $ a$.
And it means $ \forall a\in R,\exists n\in\mathbb{N}^{*}: \ x_{n}>a$ :D

The solution on here :http://www.mathlinks.ro/Forum/viewtopic.php?t=53258 .


Hi, everyone  :roll:

Prove the following:
if a, b, p be the polars of the points A, B, P for a circle whose center is O,

$ \frac {(P,a)}{(P,b)}\div\frac {(A,p)}{(B,p)} \equal{} \frac {(O,a)}{(O,b)} \equal{} \frac {(B,a)}{(A,b)}$

p.s. (A,p)=distance from A to p

Thanks  :blush:

By [url=http://mathworld.wolfram.com/SalmonsTheorem.html]Salmon's theorem[/url] we have the following relations:

$ \frac {PO}{AO} \equal{} \frac {(P,a)}{(A,p)} \  , \ \ \frac {PO}{OB} \equal{} \frac {(P,b)}{(B,b)} \  , \ \ \frac {OB}{OA} \equal{} \frac {(B,a)}{(A,b)}$

$\Longrightarrow \ \frac {(P,a)}{(A,p)} \cdot \frac {(B,p)}{(P,b)} \equal{} \frac {OB}{OA}\equal{}\frac {(B,a)}{(A,b)}$

Since the polar of a point is the inverse under $(O,r^2)$ of the circe whose diameter is the perpendicular dropped from $ O$ to its pole, we have $ r^2 \equal{} (O,a) \cdot OA \equal{} (O,b) \cdot OB$

$\Longrightarrow \ \frac {(P,a)}{(A,p)} \cdot \frac {(B,p)}{(P,b)} \equal{} \frac {(O,a)}{(O,b)} \equal{} \frac {(B,a)}{(A,b)}$


If you are Republican, or watched the Republican primaries closely, what do you think about McCain? Do you think he's the best that could have been chosen? Do you think there's something messed up in the primary process that ended it too early? Do you think someone else could've been better? Discuss.

In my opinion, McCain will do better than both Democratic candidates. However, he's very boring and dull and I don't like him. I think there were many other Republican candidates that could've done better than him, somehow they split votes and ended up with McCain winning. I think McCain deserves the credit for the win because he strategically did so--everything was probably planned and he worked hard to get to where he is now. However, I think he is screwing a lot of his decisions up, and the Republicans could've been better if there is an alternative to him in case he screws anything else up.

One of the reasons I've brought him up was because of his reactions to the recent controversies and issues with the reverends that endorsed him. What do you guys think about the reverends issues?

I was disturbed by McCain's statement that it would be okay if the U.S. were in Iraq for 100 years.

I have heard that, as a member of the Keating 5, McCain was the poster-child for government corruption in the early 1990s.  It is strange that this never seems to be mentioned in the mainstream media.

I find McCain to be a typical politician, motivated by ego rather than by ideals.  I support Ron Paul.

my family goes 4 democrats, but i am neither, so i think that McCain isnt THAT bad. i like the fact that he wants to help the environment, although the 100 years thing was scaaaaaary.... u could say that now my opinion of him has changed. now i go 4 none of the candidates, republican of democrat.

There could have been a better candidate, such as Ron Paul.  :|

new opinion on McCain: BOOO MCCAIN!!!!!!!!!!!
bleh. he's a second bush.


integration
calculus
limit
factorial
complex analysis

I was trying to prove the following using the residue theorem the other night
$\int_{0}^{2\pi}\frac{\cos(3t)}{5-4\cos(t)}dt = \frac{\pi}{12}$

After the appropriate subsituitions I got the following integral
$i \int_{C}\frac{z^{6}+1}{z^{3}(4z^{2}-10z+4)}$
where C is the contour that traces the unit circle.

Now here's is the part that bugs the geebers out of me. You can write the above integral in two different ways. Either as
$\frac{i}{2}\int_{C}\frac{z^{6}+1}{z^{3}(2z-1)(z-2)}$
or as
$\frac{i}{4}\int_{C}\frac{z^{6}+1}{z^{3}(z-\frac{1}{2})(z-2)}$
The problem is that [b]I get a different answer depending on which of the two above versions I choose!![/b] Both of the integrands are different but they have the same poles!

Is this possible? Is there something obvious that I am missing here?

(I'll show the rest of my work once I know that what I did so far makes sense or not)

edit: changed the answer from $\frac{\pi}{2}$ to $\frac{\pi}{12}$

The answer you get will not depend on which one you choose. They are equal, and the residue theorem used for any of those will always give the samy result. Possibly, you calculated the residues wrong in one case (note that they will be only half that big in the earlier than in the latter case), residues don't only depend on poles or not (nor only of what type the poles are).

[quote="EUCLID 3.14"]Now here's is the part that bugs the geebers out of me. You can write the above integral in two different ways. Either as
$\frac{i}{2}\int_{C}\frac{z^{6}+1}{z^{3}(2z-1)(z-2)}\qquad (case\#1)$

or as

$\frac{i}{4}\int_{C}\frac{z^{6}+1}{z^{3}(z-\frac{1}{2})(z-2)}\qquad (case\#2)$
[/quote]
For both cases, the only poles inside the unit circle are $z = \frac{1}{2}$ and $z=0$.

I'll only show the work for the second case because I get the right answer using the first case.

for $z = \frac{1}{2}$, $Res_{z=\frac{1}{2}}\frac{z^{6}+1}{z^{3}(z-\frac{1}{2})(z-2)}= \lim_{z \to \frac{1}{2}}\frac{z^{6}+1}{z^{3}(z-2)}= \frac{-65}{12}$

for $z = 0$, $Res_{z=0}\frac{z^{6}+1}{z^{3}(z-\frac{1}{2})(z-2)}= \lim_{z \to 0}\frac{d^{2}}{dz^{2}}\Big(\frac{z^{6}+1}{(z-\frac{1}{2})(z-2)}\Big) = \frac{21}{2}$

When you put this all together you get
$\frac{i}{4}\Big(2\pi i \Big(\frac{21}{2}-\frac{65}{12}\Big)\Big) = \frac{-61\pi}{24}$
which is incorrect.

[quote="EUCLID 3.14"]
for $z = 0$, $Res_{z=0}\frac{z^{6}+1}{z^{3}(z-\frac{1}{2})(z-2)}= \lim_{z \to 0}\frac{d^{2}}{dz^{2}}\Big(\frac{z^{6}+1}{(z-\frac{1}{2})(z-2)}\Big) = \frac{21}{2}$

When you put this all together you get
$\frac{i}{4}\Big(2\pi i \Big(\frac{21}{2}-\frac{65}{12}\Big)\Big) = \frac{-61\pi}{24}$
which is incorrect.[/quote]

You forgot the factorial component for calculating a pole of high order.  Should be:

$i/4(2\pi i)(21/4-65/12)=-\pi/2(21/4-65/12)=-\pi/2(-2/12)=\pi/12$

[quote="shawtoos"]You forgot the factorial component for calculating a pole of high order.  Should be:

$i/4(2\pi i)(21/4-65/12)=-\pi/2(21/4-65/12)=-\pi/2(-2/12)=\pi/12$[/quote]
You are absolutely right! How silly of me.  :blush:  Now I get the right answer. Thanks shawtoos.  :)

May I know what substitution did you choose?


[quote="computer_nuke"]Let $\mathcal{S}$ be the set $\{1,2,3,\ldots,10\}$. Let n be the number of sets of two non-empty disjoint subsets of $\mathcal{S}$. (Disjoint sets are defined as sets that have no common elements.) Find the remainder obtained when $n$ is divided by $1000$.[/quote]

 :huh:

[hide]Let there be three subsets, where the first two are part of the $n$ sets and the third is everything left over. Then for the first numbers in the first and second sets, there are $10\cdot 9=90$ ways for that to happen. Then there are $3^{8}=6561$ to put the rest of the numbers in...but then the big set is counted twice because the first two sets of the three sets can be switched, so it's $\frac{90\cdot 6561}{2}\mod 1000=\boxed{245}$?[/hide]


I have explored several 2d software such as KSEG, GeoGebra and C.a.R.  I am looking for a free 3d Geometry software but I can't really find anything a software that you construct with. One software I downloaded is just you are just asked to enter faces, edges, etc.

Anybody knows where to find a good 3d Geometry software?

http://www.euler3d.hu/index.php?page=home&lang=EN
It seems to be freeware, but I've never tried. (however i heard it's good)

yeah, i have that one. but you can't edit it just like in 2d software


Prove that every positive integer $n>2$, there exist integers $x$, $y$ such that:
$2^n=7x^2+y^2$.

If I'm not mistaken, the original question asks to show that such solutions exist under the assumption that $ \gcd(x, y) \equal{} 1$.  This is a classic problem whose solution can be found in Engel.  One way to do it is to explicitly write down a recurrence that generates a solution for $ n \plus{} 1$ given $ n$; you'll end up needing two of these depending on the parity of $ n$.  You can motivate the form of these recurrences by working in $ \mathbb{Z} \left[ \frac{1 \plus{} \sqrt{\minus{}7}}{2} \right]$.


2. Two 5-digit numbers are called "responsible" if they are:

\[ \text {i. In form of abcde and fghij such that fghij = 2(abcde)}\]
\[ \text {ii. all ten digits, a through j are all distinct.}\]
\[ \text {iii.} a+b+c+d+e+f+g+h+i+j = 45\]

If two "responsible" numbers are small as possible, what is the sum of the three middle digits of $\text {abcde}$ and last two digits on the $\text {fghij}$? That is, $b+c+d+i+j$.

Is there any way to do this problem without using educated guessing?

Well you could solve using the fact that any digit doubled gives something even...and you also know that the maximum value of a is 4...

[quote="ElemennopTwo implies three, so you really only have two things to work with.[/quote]

that makes it even worse.  :(

You know that 0*2 must equal 1, carrying one from the previous operation.

You could also use have 5*2 = 0, and then carry a 1 over.

I know you could find out the answer by using these rules (such as the one above), but is there another way?