An infinite sheet of charge with uniform surface charge density pn= 10
C/m2 is located at z = 0 (x-y plane), and another infinite sheet ps2 = -10
C/m2 is located at z = 4 m, both in free space. Using Gauss's Law, determine
the electric field intensity, j? in all regions.

First find the electric field intensity $ E$ of a single infinite sheet of charge with uniform surface charge density $ \sigma$. To do so imagine a cylinder whose bases are on opposite sides of the sheet and perpendicular to the sheet. The electric field is perpendicular to the side of the cylinder so the flux through the side is zero. Thus the total electric flux through the cylinder is equal to that through both bases, $ 2E \pi r^2$ and, by Gauss' law, $ \dfrac{q_{in}}{\epsilon_0} \equal{} \dfrac{\sigma \pi r^2}{\epsilon_0}$. Equating the two expressions and solving for $ E$ we get $ E \equal{} \dfrac{\sigma}{2\epsilon_{0}}$.
By the formula the two sheets have equal magnitudes of electric field, in all regions. When z<0 or z>4 m, the electric fields point in opposite directions and cancel out, so $ j \equal{} 0$. When 0<z<4 the electric fields both point in the positive $ z$ direction so they add. In this case $ j \equal{} \dfrac{10 \frac {C}{m^2}}{2\epsilon_{0}} \plus{} \dfrac{10 \frac {C}{m^2}}{2\epsilon_{0}} \equal{} \dfrac{10 \frac {C}{m^2}}{\epsilon_{0}}$.


conics
parabola
algebra unsolved
algebra

I've been working on this for about an hour or so, and I can't get this problem. I know the answer, but no idea on how to find it.
Please help?

2x-y^2=4
x^2+y=14

What are you supposed to find? When the two lines cut? If that is it try to find y in one of the equtations and then substitute it into the other one so that you only have x'ses and then find the value of x and then you can find the value of y as well.

[quote="JCE"]I've been working on this for about an hour or so, and I can't get this problem. I know the answer, but no idea on how to find it.
Please help? $2x-y^2=4\ \ ,\ \ x^2+y=14\ .$[/quote][color=darkblue]Graphically, this system is the intersection of two parabolas: $y^2=2(x-2)$, where $x\ge 2$ and with the symmetrical axis $Ox$ ; $y=14-x^2$, where $x\in R$ and with the symmetrical axis $Oy$. Thus, [b][u]this system has at the most two solutions (intersections)[/u][/b]. Eliminate the dependent variable $y$ between this equations and obtain the equation $(x^2-14)^2=2x-4\ ,\ x\ge 2$ $\Longleftrightarrow$ $x^4-28x^2-2x+200=0\ ,\ x\ge 2$ $\Longleftrightarrow$ $(x-4)(x^3+4x^2-12x-50)=0\ ,\ x\ge 2$ $\Longleftrightarrow$ $\boxed {\ x_1=4\ (y_1=-2)\ }$ and $f(x)\equiv x^3+4x^2-12x-50=0\ \ ,\ \ x\ge 2\ .$ Observe $f\left(\frac 72\right)=-\frac 18<0$ and $f\left(\frac{15}{4}\right)=\frac{895}{64}>0\ .$ Therefore, the second solution is $x_2\in \left(\frac 72,\frac{15}{4}\right)\ .$
[b][u]Remark.[/u][/b] For the exact value of the root $x_2$, you can use the [u]Cardan's formula[/u] for the resolving the equation $f(x)=0,\ x\ge 2\ .$[/color]


[hide="easy"]
Well we suppose that "x" is the number of kids
$x-\frac{x}{3}-\frac{x}{2}=50\\ \frac{x}{6}=50\Rightarrow x=300$
[/hide]

[hide]
 Jilly Job has 1/3 the votes, Billy Bob has 1/2 the votes
=> Killy Kob has: 1-1/3-1/2= 1/6 the votes
=> 1/6 the votes = 50 votes
=> Jilly Job has 100 votes
     Billy Bob has 150 votes.
 Sorry but I don't know how to hide my answer. Please show me. Thanks a lot. :)[/hide]

[quote="Chocolate milk"]
 Jilly Job has 1/3 the votes, Billy Bob has 1/2 the votes
=> Killy Kob has: 1-1/3-1/2= 1/6 the votes
=> 1/6 the votes = 50 votes
=> Jilly Job has 100 votes
     Billy Bob has 150 votes.
 Sorry but I don't know how to hide my answer. Please show me. Thanks a lot. :)[/quote]
put hide tags around your answer.

[code][hide]hi[/hide][/code] becomes [hide]hi[/hide]

Oh sorry I forgot the question. There are 300 people voted. Is that right ?

[hide][quote="Arvind_sn"]Billy Bob is running for class president. He sees how many votes each person has. He comes up with this. Jilly Job has one third of the votes. Billy Bob has 50 percent of the votes. Killy Kob has the remaining 50 votes. How many people have voted? Explain.[/quote]

1/3x+1/2x+50=x
5/6x+50=x
1/6=50
x=300[/hide]

[hide="solution and proof."]We see that x/6=50, x=300. We can substitute:

150+100+50=300[/hide]


search
inequalities
calculus
integration
Gauss
number theory
geometric series

Show that the infinity series sum from 1 to infinity of 1/p_k, where p_k is the kth prime, is divergent.

Is this a challenge, or do you want to know how it's done? (I have seen it before, so I will only post a solution in the latter case.)

I don't have a solution myself, but it looks like an interesting series so I'd be I'd like to see how its done, if the solution isn't too lengthy or cumbersome.

In that case, I'll give a hint in spoiler:

[hide]Suppose :Sigma: (1/pk) converges. Then we can find some N so that :Sigma: (1/pk) from k=N to :inf: < 1/2.[/hide]

See what you can do from there.

that would be the definition of a limit applied to the remainder of a sum.

I gave this problem as a Challenge a long time ago. I think around the time I first signed up here. If you do a search you might be able to find it.

I don't immediately see how ComplexZeta's hint helps - so here's another hint: [hide]Try using the Prime Number Theorem to estimate the size of the kth prime.[/hide].

Using the prime number theorem is awfully hand-wavy. But anyway, here's a second hint:

[hide]Let Q be the product of all the primes not in the ``tail'' that sums to less than 1/2. Consider the sum of 1/(1+nQ).[/hide]

No, the PNT leads to a pretty nice, sound proof. All you'll need, actually, is an even weaker version of the PNT (due to Chebyshev) which says: c1*n/ln n < :pi:(n) < c2*n/ln n, where c1 and c2 are constants, and :pi:(n) is the prime counting function.. (A proof of this can be found in G.E. Andrews' book [u]Number Theory[/u]). You can use this theorem to put a lower bound on 1/pk and compare it to another series.

Solving it this way requires a pretty good understanding of at least BC-level Calculus though; because you'll end up with a comparison it to a series that's simple in form - but not immediately obvious whether it converges or not. You can just apply a standard convergence test to determine whether it does.

Okay, enough hints. Here's my proof of the assertion using the hint that I gave.

We only need the left hand inequality of Chebychev's Theorem to prove it. This makes sense, because it's the fact that the primes are too "dense" that makes the series diverge. (The constant c1 can actually be taken to equal 1 if n > 17, but we don't need that fact here).

So suppose there exists a constant c such that c*n/ln n < :pi:(n) for n sufficiently large.

Then letting n = pk,
c*pk/ln pk < k-1
pk < 1/c*(k-1)*ln pk < 1/c*k*ln k
So 1/pk > c/(k*ln k).

But :Sigma: 1/(k*ln k) can be shown to diverge by the integral test.
Therefore :Sigma: 1/pk also diverges.

G.E. Andrews gives another proof of the divergence of the series which does not rely on the PNT.

His proof goes:

Since 1/(1-1/p) = 1 + 1/p + 1/p2 + 1/p3 + ..., by the formula for geometric series,
:pi:{k=1 to s} [1/(1-1/pk)] = :pi:{k=1 to s} [(1 + 1/pk + 1/pk2 + ...)]
= :Sigma:A [1/n], where A is the set of all integers all of whose prime factors are less than ps.
> :Sigma:{n :le: ps} [1/n].

Since this last series diverges as ps -> :inf:, we must have :pi:{k:ge:1} [(1-1/pk)] = 0; from which we can deduce :Sigma:{k:ge:1} [1/pk] diverges. This last deduction is actually a theorem, but it's messy to prove. Andrews proves it in the appendix  to his book.

I like my proof using the PNT better because it's more intuitive. You can physically [i]see[/i] that it's the density of the primes making the series diverge.

Ok. Here's mine (from Apostol's [i]Analytic Number Theory[/i]):

Pick k so that  :Sigma: (1/pm, m=k+1 to :inf:)<1/2. Let Q be the product of the first k primes. Consider the sequence of numbers 1+nQ, where n is a positive integer. None of these numbers will ever be divisible by one of the first k primes. Thus all the prime factors of these numbers will be from the tail. Then for any positive integer r,

 :Sigma: (1/(1+nQ), n=1 to r) <=  :Sigma: ( :Sigma: (1/pm, m=k+1 to :inf:)t, t=1 to :inf:),

as all the terms in the LHS are also among the terms on the RHS. But the RHS <= :Sigma: (1/2t, t=1 to :inf:). Thus :Sigma: (1/(1+nQ), n=1 to r) converges. But this is ridiculous since it is easy to show that the sum of the reciprocals of any arithmetic sequence diverges.

[quote="fuzzylogic"][quote="gauss202"]Since 1/(1-1/p) = 1 + 1/p + 1/p2 + 1/p3 + ..., by the formula for geometric series,
:pi:{k=1 to s} [1/(1-1/pk)] = :pi:{k=1 to s} [(1 + 1/pk + 1/pk2 + ...)]
= :Sigma:A [1/n], where A is the set of all integers all of whose prime factors are less than ps.
[/quote]

This could be used to prove there are infinitely many primes :-)[/quote]

Assuming that the PNT didn't already assume it ;)

I always thought (although I've never proved it) that if [tex]0<a_i<1[/tex] then the series [tex]\displaystyle \sum_{i\geq 1}a_i[/tex] converges iff the product [tex]\displaystyle \prod_{i\geq 1} \frac 1{1-a_i}[/tex] is finite. If we apply this to [tex]a_i=\frac 1{p_i}[/tex] we're done because the product is, in this particular case, [tex]1+\frac 12+\frac 13+\cdots=\infty[/tex].

Do you have any idea whether that thing is true or not? It can be restated like this: If [tex]0<a_i<1[/tex], then [tex]\displaystyle \sum_{i\geq 1}a_i[/tex] converges iff [tex]\displaystyle \prod_{i\geq 1}(1+a_i+a_i^2+\cdots)[/tex] converges.

Yes, its true grobber. That's basically the same theorem that I mentioned is used by G.E. Andrews in his proof. The proof of it is provided in the appendix to his book. It could be a good exercise for you to try to prove it yourself though.

[quote="Fermatprime"]Assuming that the PNT didn't already assume it ;)[/quote]

Gauss didn't assume PNT in his second proof.

[quote="gauss202"]G.E. Andrews gives another proof of the divergence of the series which does not rely on the PNT.
[/quote]


I dont think you should post this until after tomarrow, this would be cheating if someone used a solution posted here for ARML homework.  (I am assuming you are on chicago ARML as this is a problem due tomarrow and it says your from naperville)

[quote="gotztahbeazn"]If $\tan \Big( \frac{2\pi}{3}-x \Big) = \frac{ \sin \frac{2\pi}{3}-\sin x}{\cos \frac{2\pi}{3}-\cos x}$, where $0 < x < \pi$. Compute $x$.[/quote]

WOOO I AM ON CHI-TOWN ARML TOO

man come on...use aim or something...that is "discussing"

Hmm, this is interesting. Does anyone have a solution?

(BTW shame on the OP for posting ARML HW)

[quote="Altheman"]you just might have to use some difference identities....he solved it...[/quote]
Yeah that's what I was thinking, it looks like a direct application of formulas but I'm too lazy to work it out...


[hide="EPIC hint"]British flag theorem[/hide]

eureka!

first let us set some variables...
let the height h=a+b, and the base b=c+d
where h passes point E and c^2+a^2=14^2
                                      c^2+b^2=x^2
                                     =a^2-b^2=14^2-x^2
Note that:
              a^2+d^2=4^2
              b^2+d^2=12^2
             a^2-b^2=16-144 or -128

by substitution:
                      -128=14^2-x^2
                      x^2=272
                      x=4sqrt17

Not quite.

[hide="Solution"]
By the British Flag Theorem, $ 14^2\plus{}12^2\minus{}4^2\equal{}180\plus{}144\equal{}324\equal{}x^2\implies \boxed{x\equal{}18}$.
[/hide]

Draw lines perpindicular to AD and AB that intersect at point I. We will call the new point on line AD "F" and the point on AB "G".

By substituting lines that are the same length, we get

$ AF^2 \plus{} AG^2 \equal{} 14^2$
$ AG^2 \plus{} FD^2 \equal{} 4^2$

$ (AF^2 \plus{} AG^2) \minus{} (AG^2 \plus{} FD^2) \equal{} 14^2 \minus{} 4^2$

$ AF^2 \minus{} FD^2 \equal{} 14^2 \minus{} 4^2$
$ FD^2 \plus{} BG^2 \equal{} 12^2$

$ (AF^2 \minus{} FD^2) \plus{} (FD^2 \plus{} BG^2) \equal{} (14^2 \minus{} 4^2) \plus{} 12^2$
$ AF^2 \plus{} BG^2\equal{}14^2 \minus{} 4^2 \plus{} 12^2$

$ AF^2 \plus{} BG^2  \equal{} X^2$

By substitution
$ 14^2 \minus{} 4^2 \plus{} 12^2 \equal{} X^2$
$ 196 \minus{} 16 \plus{} 144 \equal{} X^2$
$ 324 \equal{} X^2$
$ 18 \equal{} X$

whoah! im very sorry!
the last equation must be
x^2=14^2+128
x^2=196(my mistake)+128
x=18

yes it is X=18
i have solved this today at school. :D


Find a value $ k>0$ such that $ f'(k)\equal{}\minus{}9$ for $ f(x)\equal{}1/x$.

hello
$ f'(x)\equal{}\minus{}\frac{1}{x^2}$
$ k\equal{}\frac{1}{3}$
thank u


integration
calculus
logarithms
algebra
linear equation
calculus computations

can you solve :lol:

$ y\cdot y''\;\equal{}\;4\plus{}y'^2$

Make $ y' \equal{} p(y)$, and so $ y'' \equal{} pp'$: $ ypp' \equal{} 4 \plus{} p^2 \equal{} > p' \minus{} \frac {1}{y}p \equal{} \frac {4}{yp}$ which is a Bernoulli differential equation. To solve this diff. eq. make the substitution $ z \equal{} p^2$, and we get the linear equation $ z' \minus{} \frac {2}{y}z \equal{} \frac {8}{y}$, whose solution is $ z \equal{} y^2 \left(8 \int y^{ \minus{} 3}\,dy \plus{} C_1\right) \equal{}$
$ \equal{} y^2 \left(C_1 \minus{} \frac {4}{y^2}\right) \equal{} > z \equal{} C_1y^2 \minus{} 4$, and so $ p \equal{} \pm \sqrt {C_1y^2 \minus{} 4}$. This means that $ \frac {dy}{dx} \equal{} \pm \sqrt {C_1y^2 \minus{} 4} \equal{} >$
$ \equal{} > \frac {dy}{\sqrt {C_1y^2 \minus{} 4}} \equal{} \pm dx$, from which we get after integration $ \ln\left|y \plus{} \sqrt {y^2 \minus{} C_1^2}\right| \equal{} C_2 \pm \frac {2x}{C_1}$.


[quote="bodan"]$\mathrm{Let}\;\; a,b,c>0,\;\; a^2+b^2+c^2=1.\;\;\; \mathrm{Prove\; that}$ \[ a+b+c+\frac{1}{abc}\geq4\sqrt{3}.  \][/quote]

With AM-GM we have:
$a+b+c+\frac{1}{abc}=a+b+c+\frac{1}{9abc}+\frac{1}{9abc}+...+\frac{1}{9abc}\geq12\left(\frac{abc}{9^9(abc)^9}\right)^\frac{1}{12}$

So we have inequality:
$\left(\frac{1}{9^9(abc)^8}\right)^\frac{1}{12}\geq\frac{\sqrt{3}}{3}$

Taking away power $\frac{1}{12}$ and multiplying by $9^9$ we have:
$\frac{1}{abc^8}\geq3^{12}$

Next we have:
$\frac{1}{abc}\geq3\sqrt{3}$

This also can be proved with AM-GM:
$\left(\frac{1}{3}\right)^\frac{3}{2}=\left(\frac{a^2+b^2+c^2}{3}\right)^\frac{3}{2}\geq abc$ so $\frac{1}{abc}\geq\left(\frac{3}{1}\right)^\frac{3}{2}=3\sqrt{3}$ and we are done.

The thesis $a^2+b^2+c^2=1$ follows $1 \ge 3\sqrt[3]{(abc)^2} \implies abc \le \frac{1}{3\sqrt{3}}$.
Thus, we have:
\[ \frac{1}{abc} \ge \frac{3\sqrt[3]{3}}{\sqrt[9]{abc}}  \]
And:
\[ a+b+c+\frac{1}{abc} \displaystyle_\ge ^{AM-GM} 3\sqrt[3]{abc}+ \frac{3\sqrt[3]{3}}{\sqrt[9]{abc}} \newline = 3\sqrt[3]{abc}+ +\frac{\sqrt[3]{3}}{\sqrt[9]{abc}}+\frac{\sqrt[3]{3}}{\sqrt[9]{abc}}+\frac{\sqrt[3]{3}}{\sqrt[9]{abc}} \newline \displaystyle _{\ge}^{AM-GM} 4\sqrt{3}  \]
We are done.


My friend told me that in GHP (governor's honors program in Georgia), the Art of Problem Solving classes are sponsored by the people who run GHP. Is this true?

How are Governor's Schools different from state to state?  I got accepted to the TN one and they're offering Matrix Algebra for the math class.  I haven't heard anything about taking an aops class over there though.

[quote="quantum leap"]How are Governor's Schools different from state to state?[/quote]

Well, I heard for example that there are many more math classes in Georgia (GHP). (and there are different governors  :P ). By the way, which college is your GHP at? (I didn't know other states had GHP... ) :D

the TN Governor school for the sciences is being held at University of TN at Knoxville.  Dunno how much prestige that has, it's not like vanderbilt or anything

But the thing is, while the college itself might not have much prestige (mine is Valdosta State University... not really a whole lot of prestige...), the program itself has quite a bit of merit. Colleges look at it as: "this kid knows how to pass interviews, is really good at a subject (in this case, math), and actually knows how to talk!" Rather than only being good at math or any other subject for that matter, colleges not only look at that, but also look at the fact that you were able to answer questions successfully that interviewers asked and were one of the best to do so. Colleges also look at the fact that rather than being a math nerd and not being able to talk and have no friends, you are a math nerd, are able to talk well, and have friends (though the last part may not necessarily be revealed to the interviewers  :D ). So in short, while the college may not have much prestige, GHP itself does. Good luck! :)

So who here's taking what at GHP and what else does it offer?


x^4 + y^4 + z^4 = 1, x,y,z positive reals

minimize $\sum_{cyc} \frac{x^3}{1-x^8}$.

[quote="Singular"]can you explain more?[/quote]
for which step?

All of it! $\frac{f(x)}{g(x)}$ is not necessarily (and not usually!) minimised when $g(x)$ is maximised and you just ignore $f(x)$..

I believe Jensen's will work out but it is ugly.

Maybe Chebyshev will help:
assume x >= y >= z,then x(1-x^8) <= y(1-y^8) <= z(1-z^8),so the sequence is oppositely sorted.
LHS >= (x^4+y^4+z^4)(1/(x-x^9)+y/(y-y^9)+z/(z-z^9)) >= 8/3*(1/3)^(3/4)  (I didn't check the minimum!)

I don't have time to do it right now but weighted Jensen's (with weights $x^4,y^4,z^4$) might work.

Oops I made a mistake in the post.
I think that zhaobin's solution is correct,because x^4/x(1-x^8) >= x^4*(8/((8/9)^(9/8))) for all x.(I didn't check the minimum)

The proof of Zhaobin is wrong !
 He just proved $ x(1-x^8) \leq \frac{(\frac{8}{9})^{\frac{9}{8}}}{8} $ . He can't use the equality for $ x^4 $

[quote="nttu"]The proof of Zhaobin is wrong !
 He just proved $ x(1-x^8) \leq \frac{(\frac{8}{9})^{\frac{9}{8}}}{8} $ . He can't use the equality for $x^4 $[/quote]
$8x^8(1-x^8)^8 \le (\frac{8x^8+8-8x^8}{9})^9$,
so we can find out the min of   $ x(1-x^8)$
for it be hold it
$x^4=\frac{1}{3}$  then  we get  $\frac{x^4}{x(1-x^8)}\geq\frac{x^4}{\frac{(\frac{8}{9})^{\frac{9}{8}}}{8} }$   it is clear

[quote="zhaobin"]([i]now fixed[/i])[/quote]
:rotfl: :rotfl: :rotfl: zhaobin, you will need to fix the $\LaTeX$. Somehow, there is a mix-up between braces and parentheses in the order that caused this big problem.

and what can I do to cancel the big mistake?

[quote="Singular"]can you use (1-x) , (1-x) , ... , (1-x) , 8x^8  for amgm?? isnt 1-x negative?[/quote]
I think it is sooooooooooooooooooooooooooooooooooo  clear. :)

but we note that $x^4 \leq 1$, so $x \leq 1$


linear algebra
matrix
combinatorics unsolved
combinatorics

There is an expression for the number of triangles in a graph given by $ n \equal{} 1/6 tr A^3$. Does a similar expression exist for the number of simple 5-cycles in a graph? And if it exists, then how does it look? I got stuck trying to rule out all those pesky triangles...

Unless your graph is regular, I don't think you're going to be able to get a nice expression in terms of just the adjacency matrix (but I could be wrong!).  In particular, there are lots of graphs (like $ K_4$, or a graph with a triangle and every other vertex is connected to exactly one of the vertices of the triangle) where $ \frac{1}{5} \tr A^5$ counts an awful lot of stuff you aren't interested in.  Do you have a reason to suppose there is a nice such expression?

If in addition to the trace $ \operatorname{Tr}(A)\equal{}\sum_i a_{ii}$ you allow the bilinear operation $ T_2(A,B)\equal{}\sum_i a_{ii}b_{ii}$, you can get the desired answer in terms of $ \operatorname{Tr}(A^5)$, $ \operatorname{Tr}(A^3)$ and $ T_2(A^3,A^2)$ ;). This makes me wonder whether you can count $ n$-cycles using only the operations of matrix multiplication, taking trace and extracting the diagonal :?.


probability
counting
distinguishability
expected value
advanced fields
advanced fields unsolved

1. a total of $n$ balls, numbered 1 throuh $n$, are put into $n$ urns, also numbered 1 through $n$ in such a way that ball $i$ is equall y likely to go into any of the urns $1,2,...,i$. Find the expected number of urns that are empty.

2. an urn has $m$ black balls. at each stage a black ball is removed and a new ball, that is black with probability $p$ and white with probability $1-p$, is put in its place. find the expected number of stages needed until there are no black balls in the urn.

Can somebody help me with the 2 questions? for the first part, i tried to find a pattern by setting the number of balls from 1 to 3, then when $n=4$ i realised that the probability get more and more complex, and i couldn't solve them. for the second one, i don't even have the slightest idea... thanks in advance

1) The expected number of empty urns after $k$ balls are placed is $n$ times the probability that a particular urn (say, urn 1) is empty. This probability follows a geometric distribution.

2) This is the "negative binomial" distribution, which can be realized as a sum of $m$ independent geometric distributions.
How many balls until one is white?
How many after that until two are white?
How many after that until three are white?
...

[quote="jmerry"]1) The expected number of empty urns after $k$ balls are placed is $n$ times the probability that a particular urn (say, urn 1) is empty. This probability follows a geometric distribution.[/quote]Umm ... that can't be right. A rv with a geometric distribution would have to have range $\mathbb{N}$, which clearly can't be the case.

Here's a way to do this: First off, there are $\binom{2n-1}{n-1} = \binom{2n-1}{n}$ ways to place $n$ balls into $n$ urns, assuming that the balls are indistinguishable and the urns are distinguishable. Imagine $2n-1$ tokens lined up, and select $n-1$ tokens. These tokens are used to separate urns. The non-selected tokens become balls placed in these urns. E.g. for n = 5, there are 9 tokens, labelled 1, 2, ..., 9. If tokens 3, 4, 6, 8 are selected, there are two balls (1 and 2) in urn 1, 0 balls in urn 2 (since tokens 3 and 4 are next to each other), 1 ball (#5) in urn 3, 1 ball (#7) in urn 4, and 1 ball (#9) in urn 5. All possible placements are described in this way, and they are all equaly likely.

To find our how many ways there are to place $n$ balls in $n$ urns such that  $k$ urns are empty, first select k of nthe n urns ($\binom{n}{k}$ ways) and then place the n balls in the remaining $n-k$ urns such that no urn is empty. There are $\binom{n-1}{n-k}$ ways to do this: Imagine the $n$ balls lined up and place $n-1$ barriers between them. Now select $k-1$ of the barriers and remove all the others. The remaining barriers separate $k$ urns, all of which are non-empty. There are $\binom{n-1}{k-1} = \binom{n-1}{n-k}$ ways to do this.

Altogether then the probability of having $k$ empty urns is
\[ \frac{\binom{n}{k} \binom{n-1}{n-k}}{\binom{2n-1}{n}}  \]
which is a hypergeometric distribution. The expectation then is $\frac{n^2}{2n-1}$ by well known properties of the hypergeometric distribution. About half of the urns are therefore expected to be empty.

[quote="jmerry"]After $k$ balls are placed, the probability that urn 1 is empty is $\left(\frac{n-1}n\right)^k$. Multiply by $n$ and evaluate at $k=n$ to get $n\left(\frac{n-1}n\right)^n\approx \frac ne$ as the expected number of empty urns.

The geometric distribution here is the number of balls it takes to put the first in a particular urn.
Your interpretation is wrong because of the way the balls are placed; they are placed in order, which makes them distinguishable.[/quote]Right. But this is a question about empty urns, and empty urns are indistinguishable, aren't they? So the statement that the balls are distinguishable may be a red herring.

In any case, there's an error in my expected value since gives a value > 1 for n=2, which is impossible. Will have to rethink this one :)


Let f(x)=sinx/sinx.Is f(x) periodic?What is it's period?

Do you mean $f(x) = \frac{\sin{x}}{\sin{x}}$? Are you sure this is the question you want to ask?

i suppose that you could say that it is peroidic because every 180 degrees, there is a hole (division by 0)


counting
derangement
geometry
rectangle
combinatorics open
combinatorics

How many pairs $ \{p, q\}$ are there of two derangements of $ \{1, \ldots, n\}$ such that they have no common fixed point?

Equivalently, how many ways are there to fill up the first three rows of a [[Latin square]]?

The answers for $ n \equal{} 1, \ldots, 7$ are 0, 0, 1, 12, 276, 10640, 536880.

(A friend of mine has been working with long cycles whose product is a long cycle ... I suppose there are a huge array of related questions to be asked here.  I don't have any idea if there are nice answers to be expected.)

Edit: Shame on me: I didn't try enough variations in the [url=http://www.research.att.com/~njas/sequences/A000186]OEIS[/url].

This logic is flawed, but I would apprciate someone point out where it has gone wrong.

For each of the $ n!$ permutations, p, there will be $ n! \sum_{i\equal{}0}^n \frac{(\minus{}1)^i}{i!}$ dearrangements, q, of p.

However, we have counted each pair twice (once when p is the permutation and once when q is the permutation)

Therefore $ S\equal{} \frac{n!n! \sum_{i\equal{}0}^n \frac{(\minus{}1)^i}{i!}}{2}$

Unfortunately this does not give matching solutions :(

I don't follow.  $ p, q$ both have to be derangements.

JBL:  I was just about to try doubling your numbers :)

t0r: I wish I had an excuse for not thinking of doubling them earlier.  (Or, for that matter, for why I didn't click on the link to "Sequences pertaining to Latin squares and rectangles" while I was looking for it.  Oh, well.)

When I wrote "$ p$ and $ q$ don't have any common fixed points," I meant that there is no $ i$ such that $ p_i \equal{} q_i$.  (Since $ p$ and $ q$ are derangements, what I actually wrote is vacuously true.)

SimonM: you've ensured that $ q$ is deranged relative to $ p$ (i.e., it satisfies the condition in my preceding paragraph) but you haven't ensured that $ q$ is actually a derangement.

Ah, that helps now I've read it from that point. So really I am trying to count all sets of 3 pairwise dearrangements?

Yes.  (If you give up, the last link in my first post contains some information about it, including (old! -- one from 1928) references in the literature.


Find the number of polynomials of degree $ 5$ with DISTINCT coefficients from the set $ \{1, 2, 3, 4, 5, 6, 7, 8 \}$ that are divisible by $ x^{2}\minus{}x\plus{}1$.

Let our polynomial be $ P(x)\equal{}ax^5\plus{}bx^4\plus{}cx^3\plus{}dx^2\plus{}ex\plus{}f$. Polynomial division by $ x^2 \minus{} x\plus{}1$ (which is actually not so bad) yields:

$ \dfrac{ax^5\plus{}bx^4\plus{}cx^3\plus{}dx^2\plus{}ex\plus{}f}{x^2\minus{}x\plus{}1} \equal{}$

$ \equal{}ax^3\plus{}(b\plus{}a)x^2\plus{}(b\plus{}c)x\plus{}(d\plus{}c\minus{}a)\plus{}\dfrac{(d\plus{}e\minus{}a\minus{}b)x\plus{}(a\plus{}f\minus{}d\minus{}c)}{x^2\minus{}x\plus{}1}$.

Thus, for P to be divisible by $ x^2\minus{}x\plus{}1$ we need that

$ d\plus{}e\minus{}a\minus{}b\equal{}0 \Leftrightarrow d\minus{}a\equal{}b\minus{}e$, and
$ a\plus{}f\minus{}d\minus{}c\equal{}0 \Leftrightarrow d\minus{}a\equal{}f\minus{}c$.

These together are equivalent to $ d\minus{}a\equal{}f\minus{}c\equal{}b\minus{}e$. It is now simple casework to find the number of possible sets of coefficients for $ P$. If $ d\minus{}a\equal{}1$ then there are $ 10\times3!$ choices for the coefficients; if $ d\minus{}a\equal{}2$ then there are $ 5\times3!$ choices for the coefficients; if $ d\minus{}a\equal{}3$ then there are $ 4\times3!$ choices for the coefficients;  if $ d\minus{}a\equal{}4$ then there are $ 4\times3!$ choices for the coefficients; if $ d\minus{}a\equal{}5$ then there are $ 1\times3!$ choices for the coefficients; if $ d\minus{}a>5$ then there are no choices for the coefficients. Thus, the number of polynomials is

$ (10\plus{}5\plus{}4\plus{}4\plus{}1)\times 3!\equal{}144$.


geometry trigonometry calculus MATHCOUNTS probability

i was wondering what math class everyone is in. BTW, the classes might not have the same name so:

Algebra I = 9th grade math
Algebra II / Geometry = 10th grade math
Algebra III / Trigonometry = 11th grade math
Precalculus = 12th grade math
Calculus and Above = College math
Statistics = Any grade

Oh, please post who you are as well.

edit: Confuted put this here just so it's at the top. Forgive him. If you're in more than one of those classes in the poll, pick whichever one is the most challenging.

Wait, how come each choice is doubled? And how is it possible to have 12 choices? That's more then 10...or so I think? :lol:

I'm currently in 10th Honors but I'm doing the third and fourth category. (Since the curriculums are all messed up we're ahead in one subject and behind on another but usually honor classes finish everything but calculus by the end of 10th grade)

AP Calc BC and AP Statistics as a Senior in HS. I think I'll put a poll on this.

edit: Hmmm.... the poll is being VERY strange. I'll try again.

OK now it's just plain ridiculous, not only are there duplicates there's triplicates (a vocab word :) ) too...

Addendum: Now it's gone...

Indeed. It's [i]out of control[/i]. Maybe it has something to do with trying to add a poll after the thread was created. Regardless, it's somehow my fault  :roll:

Ok, it's been fixed. That was certainly strange. If you voted on the old one, vote again, your vote has been deleted. (There was only one vote).

so would geometry be a part of algebra 1 or 2?

Depends, triangles tend to be more over II while parallels and right angle and all the simple ones tend to be Algebra I. (correct me if I'm wrong)

Hey for once we actually agree on something! :idea: "You: Join me...for you have agreed with me, and that means I am your father" "Me: Nooooooooooooooooooo, not really." :lol:

:roll: ok........ what was that all about.

AP Calc BC

But Geometry really isn't a big part of either Algebra I or II.  That is to say, there is some algebra involved in the formulas for geometry, and there is some geometry involved in certain applications of algebra, but with respect to the concept of proofs, Algebra requires very little, if any, of this.  So noo, I don't agree with you entirely.  [hide]Yay, I broke the agreement-fatherage-thing that was somehow going on.[/hide]

Mmm, the fatherage thing just reminded me that ripply's my [b]dad [/b]:shock:

Who's my mommy? :lol:

By the way that last thing was an idea for the Saga. (parofy of the "Luke I am your father" scene)

I guess Geometry is in a class of it's own (or classes)

*($#@!*(!   :lol:  We'll have to change that!  Although does it really matter?  We don't find out the identity of Luke/Leia's mom in Episodes IV-VI anyway...

I voted for the last class I TOOK, back in the 1970s. That was a precalculus class that was titled "analysis" in my high school. I had that in eleventh grade, after taking geometry in tenth grade, and trying to self-study algebra 2 in the early part of eleventh grade, I think. (It's all a bunch of bad memories now.)

There's a reason why my .sig file includes the quotation that it does . . .

Don't worry, it's not what level math you're in or what grade you're in, it's how good you are at math, and I'd say you're great at it so you have nothing to worry about.

wow, lots of precalc... hmmmm... which means lots of competition for me at Mu Alpha Theta :P

I'm in America, but I can't compete in Mu Alpha Theta because my school doesn't have a MAO chapter...

[quote="WarpedKlown1335"]Precalc here.  And they have regular precalc classes and then accelerated precalc classes...and I'm in accelerated.  We're doing some calculus at the end of the year.

Jeff, consider yourself lucky I'm in Tokyo so I can't compete in Mu Alpha Theta :-).[/quote]
You mean honors? Or are they different from that...
In my school we've got Honors, Curriculum 1, Curriculum 2, and then like 2 or 3 more levels after that.

I would kick but in MAO except my school doesn't offer it either so I'll have to convince it to do MAO but I'm not sure if I can do it in time...(my VISA (no, not the credit card :) ) is running out...)

I spend most of my day with my regular 7th grade class in a really small school that can't offer anything but basic 7th grade math.  But I am able to take my science & math classes through a local high school where I'm taking what they call Alg 3-4.  Last two years I took Alg 1-2 then Geom.  It has been really fun to be the peon among the giants.

Interesting, I thought we were the only ones who have our middle-schools (two) and high school close enough to do that...I wonder how many other schools offer that kind of system.

heh, you are quite wrong... i did that last year as well, though the high school class was just as easy as any class, as i've noticed this year. i cant wait for calc cuz then the not capable people will finally be gone :wink:

Evil you, you know that without the stupid people we'll look dumb, just like here :twisted:

Here we're evaluated relatively (guess why :wink: ) so every time someone drop to Curriculum 1 our grades change slightly. So some people go as far as spreading propaganda and confuse people...harsh, but that's life here. (One thing in defense: This does [b]not[/b], I repeat [b]not[/b] apply to all people or schools, so keep that in mind.)

people at our middle school goes to the high school to take classes such as algebra II spanish II or frecnch II. (i go to Miller Middle School in San Jose, CA the high school nearby is Lynnbrook High)

The set-up for me is a little different.  I go to a really small Catholic school that simply can't offer any type of advanced classes, so to be able to stay with my age group and still take some challenging courses, the administration and teachers let my parents get me to and from a local high school for math and science, for other academic subjects I actually do independent homeschooling curriculum in my 7th grade classroom, but I participate in stuff like art, gym, computer, lunch, chorus, and stuff like that.  It's not a perfect set-up, and actually, I wish I could spend more time in the high school classes.  But this way I get to do the academic team things like Mathcounts and a Reading Classics team that I like a lot.

I'm the only student at the high school that isn't a traditional high school age kid, and I started there last year.  I'm short for my age too, so it is fun for me.  Everyone is super nice, except for the very first day when a kid in my geom. class wanted to know if I was "regulation size".  After he found out my name he realized his mom was a hygenist in my dad's dental office and never gave me any trouble for the rest of the year.

what if we're not taking math this year?

Is that possible? (I mean sure if you've taken all the way up to AP but I somehow doubt you have)

AP Calc AB!! I kind of wish my school offered AP Calc BC too or something where I can learn the whole first year of Calculus instead of just the first semester...

Oh accelerated precalc here is about the equivalent of honors precalc, or harder.

hey im in 7th grade n i take Algebra I so im not sure if ur rite about da grades... but then again im in advanced math...

Calc BC...doesn't mean much though, I only usually get like fifth or sixth at Mu Alpha Theta and the other state competitions.

I'm in algebra 2, although I can do math harder than that.My teacher last year didn't think I should take a more advanced class...

Hikari

The normal sequence for AAST is:

9th grade: Advanced/Normal Analysis I
10th grade: Advanced Analysis II
11th grade: AB, BC Calc
12th grade: Probability, Statistics, Combinatorics, Linear Algebra... (and everything else)

But, I'm taking Adv. Analysis II in ninth grade.. so I get calc next year! =)

[quote="MathFiend"]I'm the only person in Algebra 1!!! I feel very lonely...[/quote]
Are not!!!  Me too!!!!

Im in Algebra one although my teacher says I should be in Algebra 2/ gemometry. Actually Tare, there is not much gemometry in algebra 1 or 2. I know at our highschool they are three seperate classes. Im thinking about doing a couple of independent studies over the summer. I learn math very quickly and Im almost done reading the algebra 1 text book.

I just found out I was good at math. I didnt really figure this out until 6th grade so Im a starter.

Pre-Algebra though my teacher as well thinks that I should be in a hogher math class.  In my case she thinks that I should be in Algebra 1.


trigonometry calculus algebra function domain derivative limit

Find the minima and maxima of $ f(x) \equal{} \left ( \frac{\sin 10x}{\sin x } \right)^2$ in the interval $ [0,\pi ]$

[hide="most probably calculus wont help much, though"]
I differentiated $ f$ and got that maxim/minima is attained when $ 10\tan x\equal{}\tan 10x$. But I don't know how to solve this equation. Can anyone help? :maybe:
[/hide]

[b]Squeezeing of domain on $ (0,\pi)$:[/b]

Since $ \sin{x}=0$ at $ x=0$ and $ x=\pi$, $ f(x)$ is not defined for those values, hence we have to squeeze the interval on $ (0,\pi)$.

[b]Differentiation:[/b]

Now we differentiate and obtain

$ f^{\prime}(x)=2\frac{\sin{10x}}{\sin^{3}{x}}(10\cos{10x}\sin{x}-\sin{10x}\cos{x})=0 \Leftrightarrow 2\sin{10x}(10\cos{10x}\sin{x}-\sin{10x}\cos{x})=0$

Note, that we could have multiplied with $ \sin^{3}{x}$, since $ \sin{x}\neq 0$ on $ (0,\pi)$.

Now we have two cases to consider:

[b]1. case:[/b] $ \sin{10x}=0$

$ \Rightarrow 10x=k\pi,$ where $ k\in \mathbb{Z}$
$ \Rightarrow x=\frac{k\pi}{10}$

Now since $ x\in (0,\pi),$ we have $ k\in\{1,2,\ldots,9\}$.

Now we just put those $ x$ in $ f(x)$ and get $ f(x)=0$. Since $ f(x)\geq 0,$ the minimum of $ f(x)$ is $ 0$ and is obtained at least in all points $ x=\frac{k\pi}{10}$, where $ k\in\{1,2,\ldots,9\}$.

[b]2. case:[/b] $ 10\cos{10x}\sin{x}-\sin{10x}\cos{x}=0$ and $ \sin{10x}\neq 0$

We can divide with $ \sin{10x}\sin{x}\neq 0$ and obtain $ 10\tan{x}- \tan{10x}=0$.

It follows to define a new function $ g(x)=10\tan{x}-\tan{10x}$ on $ (0,\pi)\setminus \{\frac{k\pi}{10}\}$, where $ k=\{1,2,\ldots,9\}$ and study its nature.

We get ${ g^{\prime}(x)=\frac{10}{\cos{x}}(\frac{1}{\cos{x}}-1})$.

We see that $ g^{\prime}(x)>0$ on $ (0,\frac{\pi}{2})$ and $ g^{\prime}(x)<0$ on $ (\frac{\pi}{2},\pi)$. Hence $ g(x)$ strictly grows on $ (0,\frac{\pi}{2})$ and strictly falls on $ (\frac{\pi}{2},\pi)$. Since $ g(0)=0$ and $ g(\pi)=0$, $ g(x)\neq 0$ on $ (0,\pi)$.

Therefore it remains to study the behaviour of $ f(x)$ on the edges of the interval. So $ \lim_{x \to 0}\frac{\sin^{2}{10x}}{\sin^{2}{x}}$. Since this is an indeterminate form, we use L`Hospital twice and obtain $ \lim_{x \to 0}(\frac{\sin^{2}{10x}}{\sin^{2}{x}})=\lim_{x \to 0}(100\frac{\cos^{2}{10x}-\sin^{2}{10x}}{\cos^{2}{x}-\sin^{2}{x}})=100.$ The same is true also for $ \lim_{x \to \pi}f(x)=100.$

[b]Conclusion:[/b]

The global minimum of $ f(x)$ is $ 0$, being taken in $ x=\frac{k\pi}{10}$, where $ k=\{1,2,\ldots,9\}$. But there are no other local minima or local maxima. Line $ y=100$ is the vertical asymptote of $ f$.

[quote="tmrfea"]there are no other local minima or local maxima.[/quote]  There are plenty of local maxima of this function on this interval.  And $ y \equal{} 100$ is neither vertical nor an asymptote.  :maybe:

$ \left(\frac {\sin 10x}{\sin x}\right)^2 \equal{} U_9^2\left(\cos x\right)$ where $ U_9$ is the ninth Chebyshev polynomial of the second kind.  Therefore, $ f'(x) \equal{} \minus{} 2 U_9(\cos x)U'_9(\cos x)\sin x$.  The roots of $ U_9$ are simply the roots of $ \sin 10x$, which tmrfea already identified as points of minima, while the roots of $ U'_9(\cos x)$ correspond to the $ 8$ maxima on the interval to which JBL alludes.  Finding the actual values comes down to finding the roots of $ U_9'(\cos x)$, a $ 4$th degree polynomial in $ \cos ^2x$.


Hi!

Can anyone tell me the fractional representation x/y of a rational number:

32.47222222.......

If you could explain also how you achieve this?  I'm interested to know and am kind of stumped.

Thanks!

Apparently, it is not for "Advanced Fields Unsolved Problems" ;)

I will move it to the Intermediate Section.

A way to do it quick is that the number of numbers that the repetend is over is the number of 9's in the denominator. The number of number's without the repetend is the number of zero's. Subtract the number that does not have a repetend from the entire decimal and put that over the denominator.

So for 32.472222...

$472 - 47 = 425$
$\frac {425}{900} = \frac {17}{36}$

So $32.47222... = \frac {1169}{36}$

This is an infinite geometric series, of which the formula is $\frac{a}{1-r}$.  So we can write $32.47222222...$ as $32.47+(.002+.0002+.00002...)$.  So using the formula we get $\frac{.002}{1-.1}=\frac{2}{900}=\frac{1}{450}$.  Now we have to write $32.47$ as a fraction.  We can write it as $\frac{3247}{100}$, and the entire number as $\frac{29225}{900}=\frac{1169}{36}$.

[quote="nat mc"]$x=32.472222222
10x=324.722222222
10x-x=9x
\displaystyle 9x=292.25=\frac {1169} {4}
\displaystyle x=\frac {1169} {36}$[/quote]

i prefer this way, because i can use that method to also get the sums of very long fractions...hehe :D


Consider the problem:
There are two urns, both empty.There are two permissible operations: adding equal number of balls to both the urns, and halving the number of balls in either of the urns. Show that, if after an arbitrary number of operations, there are a , b, balls in the two urns respectively; a and b can take all possible values.

Figured I would dispose of an easy one.   :D

Just perform the process in reverse, i.e. start with two urns with $a$ and $b$ balls in them, you are allowed to either double the number of balls in one urn, or subtract the same number of balls from both urn.  Suppose urn $A$ has fewer balls in it.  If necessary, double the number of balls in $A$ until there are more than half as many balls in $A$ as in $B$.  If there are the same number in both, subtract them all;  otherwise, subtract balls until $B$ has twice as many balls as $A$.  Then double the number of balls in $A$, and finally subtract them all.


conics
parabola
analytic geometry
function

We have $f(x)=ax^2+x-a$ where $\left|{x}\right|\leq1$ and $a$ belongs to the set of all real numbers.
(1) If $\left|{a} \right|\leq1$ Prove that $\left|{f(x)}\right| \leq \frac54$
(2) The max of $f(x)$ is $-\frac{17}{8}$, find the value of $a$.

I can't get part a but part b doesn't seem so bad.

[url][hide]We see that f(x) is a parabola, so in order for a parabola to have a maximum value the coefficient a must be negative and this maximum value will occur at the vertex where the x-coordinate is given by the equation -b/2a.  So if the maximum occurs at y value of f(-b/2a).  So plugging -b/2a into f(x) and making a common denominator and simplifying we get (b^2-2b-16)/4a=-17/8.  Setting a=2, and b=1, we find a solution to this equation.  So a=2.[/hide][/url]

Regarding $x$ as constant and $(x^2-1)a+x$ as the function of $a$.


I think the reality show games are good. That's whay we need a Big Brother competition for people in MathCounts!!! Anyone interested?

how're u going to do this? one set of problems, first to submit in with perfect gets HOH, scores & times displayed of each person, HOH nominates 4 for eviction and 2 get evicted each week?

countdown to apprentice contest...

LOL or biggest loser

I guess I'll do it.

I don't think he's going to run to it, I think he's looking for someone to run it. I could be wrong though.

If I'm wrong, then I'll do it too.

I cant run this competition. I dont have time. Anyone want to take this over and run it?

i guess i'll volunteer, but someone tell me what type of work i have to do
-jorian

Running a competition is alot more work than you think...just a heads up

SD II was a pain in every body part...sending generic pms to 30 people every 4 days...

jhredsox, fanatic's rules are pretty good.

ok, so someone want to pm me what they think the guidelines should be (i used to always watch bb) for the math version?
ty,
-jorian :lol:

Here's a suggestion...

HOH-1 Set of Problems requiring solutions, highest score gets HOH, HOH puts up 2 people, lowest scorer automatically gets put up for eviction, previous HOH doesn't compete. Maybe the second highest score is immune from being put up for eviction?

POV-HOH, lowest scorer don't compete, maybe like 15 quick sprint level questions, winner has the power to veto one of the two people that were put up by the HOH, HOH selects one of the remaining houseguests, including HOH 2nd highest scorer for eviction. 2nd place in the POV competition is immune from getting put up this way, if he/she was already up for eviction, he/she automatically comes off, the HOH has to pick someone else from them too.

wait, is pov the challenge, or what
-jorian

pov=power of veto. Yes, it's a challenge.

When we get someone who knows what pov stands for...I'll do it, but I don't think it will work.(No offense) If you ask me, these comps that aren't all math should be adjusted.(Again, winner and evictees voted on) That's why I chost TAR-it can be all math. The Apprentice may work...

im not doing it, anyone wanna take over, i can pm u the rule i wrote up
-jorian :blush:

for ease, maybe we can work the POV into the HOH contest.

For example, 2nd highest score of the "HOH" contest wins POV, is immune, can veto a nomination.

Can you host this competition, Fanatic? :maybe:

someone please want to take over. i have a really good gameplan, and worked really hard on the first part :( -don't want to waste it!
-jorian


Let $a>0$, the function $f: (0,+\infty) \to R$ satisfies $f(a)=1$, if for any positive reals $x$ and $y$, there is \[f(x)f(y)+f \left( \frac{a}{x}\right)f \left( \frac{a}{y}\right) =2f(xy)\] then prove that $f(x)$ is a constant.

I think $a$ is fixed, so you can't take $a=1$.

You can reach $f^{2}(x)=1$ anyway:
$x=y=1$ $\to$ $f(1)=1$ (using $f(a)=1$).
$y=1$ $\to$ $f(x)=f(a/x)$ (using $f(1)=1$, $f(a)=1$).
$y=a/x$ $\to$ $f^{2}(x)=1$ (using $f(x)=f(a/x)$, $f(a)=1$).

But from here, for each separate $x$ there's $f(x)=\pm 1$.
So $f(x)=1$ if $x\in P$ and $f(x)=-1$ if $x\not\in P$.
There's still some work to do to prove that we must have $P\equiv\mathbb{R}$.

Maybe it helps that $f(xy)=f(x)f(y)$ (using $f(x)=f(a/x)$ in the functional equation).

I think this is all a which satisfies the condition and 1 is.

[quote="lordWings"]
Maybe it helps that $f(xy)=f(x)f(y)$ (using $f(x)=f(a/x)$ in the functional equation).[/quote]
Take $x=y$ then $f(x^{2})=f^{2}(x)$, we have $f(x)=f^{2}(x^{\frac{1}{2}})\geq 0$. So $f(x)=1$?? :|
Correct me if I am wrong...

Let $P(x,y)$ be the given assertion, \[P\left(x,\frac{a}{x}\right)\implies f(x)f\left(\frac{a}{x}\right)=1 \]  Hence, we have \[P(x,x)\implies f(x)^2+\frac{1}{f(x)^2}=2f(x^2)\geq 2 \implies f(x)\geq 1 \ \forall x\in \mathbb{R^+}\] by AM-GM. However, \[f(x)f\left(\frac{a}{x}\right)=1\] so, this means \[f(x)\equiv 1. \square\]


geometry
calculus
trigonometry
ratio
area of a triangle
Heron\u0027s formula

Find the max area of a triangle with tho medians of length 1 and 2.

This is from a  NMO don't remember the year.

Nice    :) !

I thought of it like this and i am not sure....
[hide]Let ABC be our triangle and G the point of interesection AM=2 BN=1 M and N are middles.
AG=4/3 BG=2/3  MG=2/3
The area of ABM=area of AMC
Area{ABG}=\frac{\frac{4}{3}* \frac{2}{3}*\sin{AGB}{2}
Can we supose that AGB=90 so sinAGB=1 it makes sense, but i am not sure.I say this because for any other angles the sin for AGB and BGM is a subunitary fraction and would make our area less than max.
If this is right the solution isn't such a big deal![/hide]

[quote="tim1234133"][hide="It is fairly..."]
...well-known that the area of a triangle is $\frac{4}{3}$ the area of the triangle formed by its medians.
So we merely wish to maximise the area of a triangle with side lengths $1,2,a$ (where $a$ is the length of the third median).

Using Heron's formula, $\Delta_{m}=\frac{1}{4}\sqrt{(a+3)(3-a)(a+1)(a-1)}$

To maximise $(a+3)(3-a)(a-1)(a+1)=-(a^{2}-9)(a^{2}-1)=-a^{4}+10a^{2}-9$ we can use calculus methods, and the maximum is at $a=\sqrt{5}$.

Putting this back into the formula for the area of the triangle of the medians gives $\Delta_{m}=1$.

Thus the maximum area of the original triangle is $\boxed{\frac{4}{3}}$
[/hide][/quote]

I didn't know about the median area ratio thing, but isn't it easier to use $A=\frac{1}{2}ab\sin C$ instead of herons

The area of a triangle with two given sides is maximised when the triangle is right.

Thus, the maximum area of a triangle with given side-lengths 1 and 2 is $\frac{1*2}{2}=1$, and the area of the triangle in question is $\frac{4}{3}\frac{1*2}{2}=\frac{4}{3}$.


calculus
integration
logarithms
trigonometry
calculus computations

1) Substitute $ x \equal{}\sin\theta$

2) Substitute $ x \equal{}\sin\theta$ and use $ \ln |\sin\theta| \equal{}\minus{}\ln 2\minus{}\sum_{k \equal{} 1}^{\infty}\frac{1}{k}\cos 2k\theta$ and $ \cos^{2}\theta \equal{}\frac{1\plus{}\cos2\theta}{2}$.

[quote="sos440"]1) Substitute $ x \equal{}\sin\theta$
[/quote]
Yeah, I was thinkin' about this change (I forgot to set the new integration limits  :mad: ), just I tried $ \theta \equal{}\arcsin x$ which is the same, the integral becomes to

$ \int_{0}^{1}{\frac{{\ln x}}{{\sqrt{1\minus{}x^{2}}}}\,dx}\equal{}\int_{0}^{\pi /2}{\ln (\sin\theta )\,d\theta }$

Now it follows that $ u \equal{}\frac{\pi}{2}\minus{}x$, isn't it?


Round 2 problems are now available on the [url=http://www.usamts.org]USAMTS website[/url].

The solution submission deadline is November 24.

Thanks! I've really been awaiting eagerly for the problems!

Now I'm hoping to download and print them out now, and begin working...So I don't fail again.

Shouldn't Sarah be the loser? She doesn't appear to be that smart, after all.

I think worthawholebean was refering to Sarah Palin and Joe Biden. Their first names are the names in Problem 1...

What? [url=http://en.wikipedia.org/wiki/Non_sequitur_(logic)]It does not follow.[/url]  :wink:
And here I was thinking math people were logical!

I think the clear message here is that neither player is using any real strategy. :P

[quote="Brut3Forc3"]I think the clear message here is that neither player is using any real strategy. :P[/quote]Correct. :)

[i]Don't ask questions about the Round 2 problems  -Admin[/i]

By the way, when is the score of Round 1 available?

In another thread:
[quote="DPatrick"]The Round 1 scores and comments will hopefully be released during the first week of November. [/quote]


algebra
polynomial
Gauss
Vieta
number theory
prime numbers
number theory proposed

Let $ a,b,c$ be integers, so that the polynomial $ x^3\plus{}ax^2\plus{}bx\plus{}c$ has three different pairwise coprime roots, which are positive integers. Also the polynomial $ ax^2\plus{}bx\plus{}c$ has a positive integer root. Prove that $ |a|$ is composite.

[i]S. Berlov[/i]

Hi  :)
Let $ x_1$ , $ x_2$,$ x_3$ the four integers roots of $ x^3 \plus{} ax^2 \plus{} bx \plus{} c$
Such that :
$ (x_1 , x_2) \equal{} (x_2 , x_3) \equal{} (x_3 , x_1) \equal{} 1$
Let $ y_1$ an integer root of $ ax^2 \plus{} bx \plus{} c$
we have
$ \delta \equal{} b^2 \minus{} 4ac$
$ y_1 \equal{} \frac { \minus{} b \plus{} / \minus{} \sqrt {\delta}}{2a}$
so we have $ m \in \mathbb{Z}$ such that $ m \equal{} \sqrt {\delta}$
=> $ y_1*2a \plus{} b \equal{} \plus{} / \minus{} m$
if $ m \neq 0$ :
$ y_2 \equal{} \frac {p}{q}$ where $ p$, $ q$ $ \in \mathbb{Z}$ and $ q\neq 1$
By viets relations :
$ |a| \equal{} (1) |(x_1 \plus{} x_2 \plus{} x_3)| \equal{} (2) |\frac {x_1*x_2*x_3}{y_1*y_2}| \equal{} |\frac {x_1x_2 \plus{} x_2x_3 \plus{} x_1x_3}{y_1 \plus{} y_2}|$
By [b](1)[/b] :
$ |p|*|a|*|y_1| \equal{} |x_1x_2x_3*q|$ => $ |p|*|a|*\frac {| \minus{} b \plus{} / \minus{} m|}{2|a|} \minus{} \equal{} |p|*| \minus{} b \plus{} / \minus{} m| \equal{} 2*|x_1|*|x_2|*|x_3|*|q|$
By gauss (because $ (p , q) \equal{} 1$)
we have
$ \minus{} b \plus{} / \minus{} m \equal{} v*q$($ v \in \mathbb{Z}$)
By [b](2)[/b] we have
$ |a|*|( \minus{} b \plus{} / \minus{} m)*q \plus{} p)| \equal{} |a|*|q^2*v \plus{} p| \equal{} |q|*|x_1x_2 \plus{} x_2x_3 \plus{} x_1x_3|$
and because $ (p , q) \equal{} 1$, we cant have $ \exists q_i$ prime numbers which divide $ q$ such that $ q_i$ $ |$ $ |q^2*v \plus{} p|$
so we have $ \forall q_i$, $ q_i$ $ |$ $ |a|$

if m = 0,
we have $ 2a$ | $ \minus{}b$ => $ 4a^2$ | $ b^2$ => $ b^2$ = $ k*4a^2$
$ k \in \mathbb{N}$
or $ b^2 \equal{} 4ac$
=> $ k*a \equal{} c$
if $ \forall p_i$  prime numbers such that $ p_i$ | $ c$,  $ a$ is not divided by $ p_i$,
=> $ k \equal{} c$
=> $ a \equal{} 1$ and $ 1$ is not prime.

Nice problem,[b]freemind[/b] :) .

[b]Solution:[/b]
Suppose that $ q_1,q_2,q_3$ are the roots of $ x^3 + ax^2 + bx + c$.
As one can see by Vieta formula:

$ 3\leq q_1 + q_2 + q_3 = - a$,

$ 3\leq q_1q_2 + q_1q_3 + q_2q_3 = b$,

$ 1\leq q_1q_2q_3 = - c$,

,therefore $ a,c < 0$ and $ b > 0$.
So let's change the notations as follows:
Let $ P(x) = x^3 - ax^2 + bx - c$ be polynomial with three diffrent pairwise coprime roots,which are positive integers and $ a,b,c\in\mathbb{N}$.And polynomial $ Q(x) = ax^2 - bx + c$ has a positive integer root.

Now suppose to the contrary that $ a = q_1 + q_2 + q_3$ is prime greater than $ 3$,so $ a$ is odd.Hence there is even number of even numbers among $ q_1,q_2$ and $ q_3$ and they are pairwise coprime,therefore,all the roots are odd integers and $ b,c$ are odd integers,too.

If $ Q$ has positive root,it means that

$ a| b + \sqrt {b^2 - 4ac}$,or $ a| b - \sqrt {b^2 - 4ac}$,in both case,$ \sqrt {b^2 - 4ac}$ must be integer.

Suppose that $ a|b + \sqrt {b^2 - 4ac}$.Second case is totally similar.

If $ Q$ has only one root,it follows that:

$ q_1q_2 + q_2q_3 + q_1q_3$ is divisible by $ q_1 + q_2 + q_3$, and $ q_1q_2 + q_2q_3 + q_1q_3 = 4\cdot q_1q_2q_3\cdot (q_1 + q_2 + q_3)$.
From where we conclude that:

$ q_1 + q_2 + q_3|4\cdot q_1q_2q_3$,which is impossible,as $ q_1 + q_2 + q_3$ is odd prime and $ q_1,q_2,q_3$ are pairwise coprime.And as a consequence $ a$ doesn't divide $ b$.

If $ Q$ has two positive integer roots,call them $ r_1,r_2$.
By [b]Vieta's[/b] formula:

$ r_1\cdot r_2 = \frac {c}{a}$,
hence $ c$ is divisible by $ a$,but as it was observed,the last relation is impossible.

If $ Q$ has one positive integer and one positive rational root,call them $ n,r$,where $ n\in\mathbb{N}$ and $ r = \frac {x}{y}$,$ x,y\in\mathbb{N}$,$ q > 1$.
By [b]Vieta's[/b] formula:

$ \frac {nx}{y} = \frac {c}{a}$,or

$ \boxed{n\cdot x\cdot (q_1 + q_2 + q_3) = c\cdot y}$.

As we know $ \gcd(q_1 + q_2 + q_3,c) = 1$,it follows that:

$ q_1 + q_2 + q_3$ divides $ y$.

On the other hand:
$ \frac {x}{y} = \frac {b - \sqrt {b^2 - 4ac}}{2a}$,where $ b - \sqrt {b^2 - 4ac}$ is even,hence $ y\leq a$,combining it with fact $ q_1 + q_2 + q_3| y$,we finally conclude that:

 $ \boxed{y = q_1 + q_2 + q_3}$,

We will use[b] Vieta's[/b] formula one more time to obtain:

$ \{\begin{array}{cccc} n\cdot x = q_1\cdot q_2\cdot q_3 \\
 \\
n\cdot a + x = q_1q_2 + q_2q_3 + q_1q_3\end{array}$

Therefore,first equation implies that $ n,x$ are odd,as $ c$ is odd.On the other hand second equation with fact that $ a$,$ b$ are odd integers,imply that one of the $ n,x$ is even.Contradiction.
$ \blacksquare$

Unexpectable solution :wink: .Let $ x_1,x_2,x_3$ be the roots of the polynomial $ x^3\plus{}ax^2\plus{}bx\plus{}c$. By the Vieta Theorem; $ \minus{}a\equal{}x_1\plus{}x_2\plus{}x_3,b\equal{}x_1x_2\plus{}x_2x_3\plus{}x_1x_3,\minus{}c\equal{}x_1x_2x_3$. Let $ y$ be the positive integer root of the polynomial $ ax^2\plus{}bx\plus{}c$. So $ (x_1\plus{}x_2\plus{}x_3)y^2\minus{}(x_1x_2\plus{}x_2x_3\plus{}x_1x_3)y\plus{}x_1x_2x_3\equal{}0$. If $ |a|$ be a prime number, must be a odd number $ (x_1\plus{}x_2\plus{}x_3\geq3)$ and because of $ x_1,x_2,x_3$ are pairwise coprime so all of $ x_1,x_2,x_3$ are odd numbers. Hence $ x_1\plus{}x_2\plus{}x_3$ , $ x_1x_2\plus{}x_2x_3\plus{}x_1x_3$ and $ x_1x_2x_3$ are odd numbers. Hence $ (x_1\plus{}x_2\plus{}x_3)y^2\minus{}(x_1x_2\plus{}x_2x_3\plus{}x_1x_3)y\plus{}x_1x_2x_3 \equiv y^2\minus{}y\plus{}1 \equiv 1 (mod 2)$ is a contradiction.

[quote="Erken"][quote="Mathias_DK"]$ x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} (x \minus{} r_1)(x \minus{} r_2)(x \minus{} r_3)$ AND $ ax^2 \plus{} bx \plus{} c \equal{} a(x \minus{} k_1)(x \minus{} k_2)$
Say that |a| is prime:
$ c \equal{} a \cdot k_1 \cdot k_2 \equal{} \minus{} r_1 \cdot r_2\cdot r_3 \equal{} > a|r_1$ OR $ a|r_2$ OR $ a|r_3$
Contradiction since $ a \equal{} r_1 \plus{} r_2 \plus{} r_3$

So we don't have to use that they are pairwise coprime.[/quote]
Your solution is incorrect,because you don't know whether $ k_1,k_2$ are both integers. :P[/quote]
Well. It is still correct:P

Asume that $ k_2$ is not an integer.
We have $ \minus{} a(k_1 \plus{} k_2) \equal{} a \cdot k_1 \plus{} a \cdot k_2 \equal{} b$ Since $ a \cdot k_2$ is an integer $ k_2 \equal{} \frac {p}{a}$ With p and a coprime. But we also have $ k_1 \cdot k_2 \equal{} \frac {p}{a} \cdot k_1 \equal{} c$ Which implies $ a|k_1$ and therefore $ k_1 \equal{} n \cdot a$

So we have $ c \equal{} a \cdot k_1 \cdot k_2 \equal{} a \cdot n \cdot p \equal{} \minus{} r_1 \cdot r_2\cdot r_3$ yielding $ a|r_1$ OR $ a|r_2$ OR $ a|r_3$
(And again we have contradiction) :)

[quote="Erken"][quote="Mathias_DK"] therefore $ k_1 \equal{} n \cdot a$

So we have $ c \equal{} a \cdot k_1 \cdot k_2 \equal{} a \cdot n \cdot p \equal{} \minus{} r_1 \cdot r_2\cdot r_3$ yielding $ a|r_1$ OR $ a|r_2$ OR $ a|r_3$
(And again we have contradiction) :)[/quote]
Again incorrect :rotfl:
Why is $ k_1\cdot k_2 \equal{} c$ true?[/quote]
LOL.. :P

$ a \cdot k_1 \cdot k_2 \equal{} c$ Hmm... Maybe I can stil fix it?:)

[quote="lapras"][quote="Erken"][quote="lapras"]
$ |a|*|( \minus{} b \plus{} / \minus{} m)*q \plus{} p)| \equal{} |a|*|q^2*v \plus{} p| \equal{} |q|*|x_1x_2 \plus{} x_2x_3 \plus{} x_1x_3|$
[/quote]
Here is mistake,as $ |a|*|( \minus{} b \plus{} / \minus{} m)*q \plus{} p)| \equal{} |q^2*v \plus{} pa|$.[/quote]
I dont understand why ? :huh:[/quote]
Because $ 2$ implies that
$ |a|\cdot \left(\frac { \minus{} b \plus{} / \minus{} m}{2a}\cdot q \plus{} p\right) \equal{} |q|\cdot|x_1x_2 \plus{} x_2x_3 \plus{} x_1x_3|$,if you substitute
$ \minus{} b \plus{} / \minus{} m \equal{} vq$,then you will obtain
$ \frac {vq^2}{2} \plus{} pa \equal{} |q|\cdot|x_1x_2 \plus{} x_2x_3 \plus{} x_1x_3|$


algebra
polynomial
Galois Theory
algebra unsolved

solve a equation :
$ \sqrt{ 5\plus{} 4\sqrt{9\minus{}2\sqrt{x}}} \equal{} 2\sqrt{13}(13\minus{}x)$

hello,i have got $ x\approx 12.54989098671480864969234$ as the solution.
Sonnhard.

[quote="bunhiacovski"]solve a equation :
$ \sqrt { 5 \plus{} 4\sqrt {9 \minus{} 2\sqrt {x}}} \equal{} 2\sqrt {13}(13 \minus{} x)$[/quote]

I suggest you learn nested radicals.

There exists only one real solution $ x\approx{12.55}$ (cf sonnhard).
$ x$ is a root of the following polynomial:
$ p\equal{}\minus{}3664076399009504*x\plus{}5950725395493025\minus{}151902517297024*x^3
\plus{}986964519794864*x^2\plus{}14610765321568*x^4\minus{}899343391232*x^5
\plus{}34595754752*x^6\minus{}760408064*x^7\plus{}7311616*x^8$.
$ p$ is an irreducible polynomial with galois group $ S_8$. This group is not solvable and we can't write $ x$ with radicals.


calculus
integration
logarithms
limit
inequalities
calculus computations

Let $ n\geq 2$ be positive integers.

(1) Prove that $ n\ln n \minus{} n\plus{}1<\sum_{k\equal{}1}^n \ln k<(n\plus{}1)\ln n \minus{} n\plus{}1$.

(2) Find $ \lim_{n\to\infty} (n!)^{\frac{1}{n\ln n}}$.

we have" the inequality  is  equivalent to:
$ n^n < e^{n \minus{} 1} n! < n^{n \plus{} 1}$
 2.
$ \lim_{ n \to \infty} n!^{n\ln n} \equal{} exp( \lim \frac {n!}{n \ln n}) \equal{} e$


modular arithmetic
algebra solved
algebra

Sequence [tex] \lbrace{a_i}\rbrace [/tex] is defined by:
[tex] a_1=1,a_2=9,a_3=9,a_4=4 [/tex] and [tex] a_n [/tex] is remainder when we divide [tex] a_{n-4}+a_{n-3}+a_{n-2}+a_{n-1} [/tex] for [tex] 10 [/tex].
Is there exist [tex] n [/tex] satisfying [tex] a_n=1,a_{n+1}=2,a_{n+2}=3,a_{n+3}=4 ? [/tex]. :)

Is it exactly like that? Look at the indices: $a_n,a_{n+1},a_{n+2},a_{n+4}$. I haven't tried to solve it so far, but the fact that we skip over $a_{n+3}$ looked suspicious to me :).

In that case, there's a simple reason why it can never happen: forget about $\pmod {10}$. Whenever $a_n$ is even $\pmod {10}$, it's $0\pmod{2}$ because $10$ is even. We just look at the sequence $\pmod{2}$. We have $a_1=a_2=a_3=1,\ a_4=0,\ a_5=a_6=a_7=a_8=1,\ a_9=0$, so we're back to the quadruple $1,1,1,0$ and the sequence is periodic $\pmod{2}$. In particular, this shows there are no four consecutive terms $1,0,1,0$, and we're done (actually, there's no isolated $1$; each $1$ must appear near at least another $1$).

A simple check of parity since (10 preserves parity) gives that the sequence must follow:
o,o,o,e,o,o,o,o,e,o,o,o,o,e,...

Therefore we cannot have 1,2,3,4 since that is o,e,o,e

Note: I had this "brainstorm" after calculating the first 564 terms of the sequence and noticed very few evens.

edit for 2 reasons:
1) o is for odd, e for even
2) dang me getting beat by 1 minute


tan^2(2x)-tan^2(x)]/[1-tan^2(2x) tan^2(x)]=tan(3x)tan(x)

 prove it using RHS ONLY

post here--u'll get much better response

http://joigroup.freezoka.com/

i am sure u'll get a better response than what u get by posting in the india forum.....this forum is dead  :(   )
[try posting in the pre-olympiad forum ]

And what stupid condition is that?? USE RHS???  :mad:
I have never seen such stupid conditions anyhwere  :rotfl:

anyway here is the soln,

RHS = $ \tan 3x . \tan x$

$ \equal{} \tan (2x \plus{} x) . \tan (2x \minus{} x)$

$ \equal{} \frac {\tan 2x \plus{} \tan x}{1 \minus{} \tan 2x. \tan x}.$   $ \frac {\tan 2x \minus{} \tan x}{1 \plus{} \tan 2x. \tan x}$

$ \equal{} \frac {\tan^2 2x \minus{} \tan^2 x}{1 \minus{} \tan^2 2x . \tan^2 x}$  = LHS
[size=200]
MOD EDIT: For Personal reasons[/size] :P  :P

my dear friends no question is stupid.it is the thinking that makes the thing ugly. i learn the importance of patience from impatient ,silence from talktative ,and peace from angry person.how could be i unthankful to such beautiful teacher.[b][i][u][quote]
regards from ur friend[/quote][/u][/i][/b]

ok sorry  :blush:

but i thin k you made that up "use rhs"  :D

and i see my post is edited!!  :rotfl: sorry par!!

[quote="ayush_2008"]sorry my friend if i heart u.
love from ur friend.
 pl visit india
regards from mun[/quote]
:rotfl: He is in India :D :D

[quote="ayush_2008"]sorry my friend if i heart u.
love from ur friend.
 pl visit india
regards from mun
[/quote]

yes.i live in india :D  :P
but i am originally from nepal :)

[quote="Da Vinci"]but i am originally from nepal :)[/quote]
You're as much from Nepal as is pardesi from Virgin Islands :D :P

[quote="madness"]
You're as much from Nepal as is pardesi from Virgin Islands :D :P[/quote]
yayayayayay :rotfl:  :rotfl:  :D


If you don't know any specific points, how do you draw circumcircles without finding any points. For example, I would like to draw a circle around the dodecagon. Thanks.

[asy]size(200);
D(polygon(12));[/asy]

If you have an dodecagon, you know the center and the radius; so you know the circumcircle.
[asy]size(200);
draw(polygon(12));
draw(unitcircle);

draw(scale(2)*unitsquare, red);
draw(shift(1,1)*scale(sqrt(2))*unitcircle, red);[/asy]

I don't quite get your code. Can you explain more a bit? Thanks.

[quote="dragon96"]I don't quite get your code.[/quote]
Click on the picture in order to get the code.

[code]draw(polygon(12));
draw(unitcircle);
[/code]
The center of [b]polygon(12)[/b] is (0,0) and the radius is 1.
[b]unitcircle[/b] is the circle centered at (0,0) with radius 1.
[code]draw(scale(2)*unitsquare, red);
draw(shift(1,1)*scale(sqrt(2))*unitcircle, red);[/code]
[b]unitsquare[/b] is the square (0,0)--(1,0)--(1,1)--(0,1)--(0,0), its center is (0.5,0.5) and its radius is $ \frac{\sqrt{2}}{2}$
[b]scale(2)*unitsquare[/b] returns the square (0,0)--(2,0)--(2,2)--(0,2)--(0,0), its center is (1,1) and its radius is $ \sqrt{2}$
[b]shift(1,1)*scale(sqrt(2))*unitcircle[/b] returns the circle of radius $ \sqrt{2}$ centered at (1,1); so it's the circumcircle of [b]scale(2)*unitsquare[/b].


Prove that there are infinite integers n such $ n^2\plus{}(n\plus{}1)^2$ is a perfect square.

[quote="danilorj"]Prove that there are infinite integers n such $ n^2 \plus{} (n \plus{} 1)^2$ is a perfect square.[/quote]

Let $ u,v$ such that $ u^2 \equal{} 2v^2 \minus{} 1$. Then :

$ (u \plus{} v)^2 \minus{} v^2 \plus{} 1 \equal{} 2(u \plus{} v)v$
$ ((u \plus{} v)^2 \minus{} v^2 \plus{} 1)^2 \equal{} 4(u \plus{} v)^2v^2$
$ (u^2 \plus{} 2uv \plus{} 1)^2 \equal{} 4(u \plus{} v)^2v^2$
$ (u^2 \plus{} 2uv \plus{} 1)^2 \equal{} 4(u \plus{} v)^2v^2$
$ (u^2 \plus{} 2uv)^2 \plus{} (u^2 \plus{} 2uv \plus{} 1)^2 \equal{} (u^2 \plus{} 2uv \plus{} 2v^2)^2$

So $ u^2 \equal{} 2v^2 \minus{} 1$ implies $ n \equal{} u(u \plus{} 2v)$ is solution of the required equation.

And it is well known that $ u^2 \equal{} 2v^2 \minus{} 1$ has infinitely many solutions :
$ (1,1)$ is solution
$ (u,v)$ solution implies $ (3u \plus{} 4v,2u \plus{} 3v)$ is solution too.

Examples :
$ (u,v) \equal{} (1,1)$ implies $ n \equal{} 1(1 \plus{} 2) \equal{} 3$ implies $ 3^2 \plus{} 4^2 \equal{} 5^2$
$ (u,v) \equal{} (3\cdot 1 \plus{} 4\cdot 1,2\cdot 1 \plus{} 3\cdot 1) \equal{} (7,5)$ implies $ n \equal{} 7(7 \plus{} 2\cdot 5) \equal{} 119$ and so $ 119^2 \plus{} 120^2 \equal{} 169^2$
$ (u,v) \equal{} (3\cdot 7 \plus{} 4\cdot 5,2\cdot 7 \plus{} 3\cdot 5) \equal{} (41,29)$ implies $ n \equal{} 41(41 \plus{} 2\cdot 29) \equal{} 4059$ and so $ 4059^2 \plus{} 4060^2 \equal{} 5741^2$
and so on ...

[quote=jwt]You showed infinitely many solutions, but you didn't find all, for example, $20^2+21^2=29^2$[/quote]
I just solved the proposed problem ... :
 [quote=danilorj]Prove that there are infinite integers n such ...[/quote]
... which is the main purpose of this forum.

[quote=jwt]You showed infinitely many solutions, but you didn't find all, for example, $20^2+21^2=29^2$[/quote]

Why do you think this? Try to find this solution!  ;)


inequalities
induction
inequalities proposed

Hi !
Recently I've invented some nice looking inequality -  It's some kind of generalization of Shapiro's inequality [Shapiro's with n/4 rather than 0,4 n on the right side] - we get Shapiro for k=1. Here it comes:

$\sum_{i=1}^{n} \frac{\prod_{m=i}^{k+i-1}x_m}{\sum_{p=i+1}^{k+i}x_p} > \frac{n}{2(k+1)}$

Inequality is cyclic: $x_{n+1} = x_1$ and so on
I've tried to prove it by induction but I get too many conditions which I would like to avoid. If any of you have some comments I'll be glad. Obviously it might not even be true,but... - and I'm not sure about right side of this inequality

Maybe one could try to prove it  :maybe:

I've found a little mistake in my inequality and I've corrected it - but anyway, does anyone try to do something with this ?


Now I'm moving on from my old avatar (Riemann);  feel free to use him now.

BTW, what is your avatar? Mine is based on my screen name.

I made my lightbulb myself, from scratch, on Microsoft Paint. I use MS Paint a lot, but this is the only one I've done that's small enough for this website.

A lighthouse from my state.

Wow, A+MATH's is very cool.  You expect to see something scary written in that font/colors, but it says "A+ Math"

My avatar is Marie Curie also known as Madame Curie, my idol.

Mine is a cartoon character that I first drew in fourth grade. This version is changed from the original. His name is Jimmy and it was made in Flash.

My avatar is a photograph of Israel M. Gelfand, one of my favorite authors of math textbooks. I will probably never change it because I refer to it in some threads here.

Take a wild guess.  Jackie Chan's my hero.  I think the photo is from <I>The Legend of Drunken Master</I>, though I'm not sure.

[quote="xqpx"]Can anyone guess what my avatar is?  :)[/quote]

It looks like a tree, but I'm not sure

I think it's a close up of some type of insect or something.

Anything I don't recognize I say is an alien, so alien.

anyone kno where a person can get a goode, free avatar?

[quote="Osiris"]Several members of this forum should be able to read my other avatar, which is an excerpt from one of the most famous Chinese poems of all time.

The background is supposed to be a tree ring / Jupiter bands motif, but it doesn't work that well.

[img]http://www.0t8.net/images/avatar.gif[/img]

The new avatar is really just a boring scenery thing.[/quote]

Do you understand the words in it?

BTW, my avatar based on my favrite game----[hide]MUonline[/hide]

Mine is an MFG photo of the glasses I wear, heh.

mine is of one of the greatest rock bands ever: rage against the machine

My avatar is a picture of what I want to look like, with blue skin and grey hair  :D

My avater is my name, using math symbols I generated using latex.

Mine is a fractal I developed myself  :)

[quote="Scrambled"]did you take a picture of yourself and invert it on paint?[/quote]

Yeah.

[quote="Danbert"]My avatar is a cool fractal I found online.  tetrahedr0n's used to be (still is?) the same as it.[/quote]And all this time I was thinking your avater was a picture of a fish or eel or something  :blush:

I use the cat I frequently doodle on the back of my homework.
Also shown here in case I've changed it since I made this post...
[img]http://team.melonisland.net/pteromys/winicon.png[/img]

On a side note, what programs do you guys use to generate your fractals?

I have a [hide]TI [hide]89[/hide] [hide]titanium[/hide][/hide]

[quote="pteromys"]On a side note, what programs do you guys use to generate your fractals?[/quote]

[url=http://www.apophysis.org]Apophysis[/url]

@mathgeek2006:  Great calculator.

[quote="Treething"]No, my avatar is.[/quote]

?????

According to my computer screen, you don't have an avatar...

my avatar is Sanzo; a hero in Saiyuki
do you know Saiyuki?? :P


trigonometry
geometry
trig identities
Law of Cosines

In the adjoining figure of tthe rectangular solid, $\angle DHG =45 ^\circ$ and $\angle FHB = 60 ^\circ$.  Find the cosine of $\angle BHD$.  (ASHME 1982)

[quote="jli"][quote="NoSoupForYou"][hide="My Solution"]Let $HF$ be $1$. Then, $BF=\sqrt{3}$, $DG=\sqrt{3}$, $GH=\sqrt{3}$, $DH=\sqrt{6}$, $BH=2$, and $BD=2$.
Using the Law of Cosines, $\cos BHD=\frac{(\sqrt{6})^2+2^2-2^2}{2\cdot 2\cdot\sqrt{6}}=\frac{6}{4\sqrt{6}}=\boxed{\frac{\sqrt{6}}{4}}$[/hide][/quote]

Great job...is there another way to do it without Law of cosines? :?[/quote]
there maybe one with geometric methods but its going to be long compared to trig method
would you like me to present a geometric solution?

Yes Palytoxin. That would be interesting.

i got it using the law of cosines.
would have solved it quicker if i didn't let $a^2=a$ and $b^2=b$ :D

well the last part we need the definition of cosine so a little trig is unavoidable, but we can do it without law of cosine
as before, we can partially use [b]NoSoupForYou's[/b] solution:
Let $HF$ be $1$. Then, $BF=\sqrt{3}$, $DG=\sqrt{3}$, $GH=\sqrt{3}$, $DH=\sqrt{6}$, $BH=2$, and $BD=2$.
since $BH=BD$ we have an isosceles triangle and so $\angle BHD=\angle BDH$.
from geometry, we know that the altitude of an isosceles triangle intersects the base at its midpoint, so let the midpoint of $DH$ be $M$. then $MH=\frac12 DH=\frac12\sqrt{6}$.
by definition of cosine, $\cos\angle BHD=\frac{MH}{BH}=\frac{\frac12\sqrt{6}}{2}=\frac{\sqrt6}{4}$
do you like this way?  :whistling:


trigonometry
real analysis
real analysis unsolved

Given the equation:

\[ \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial v}{\partial r}\right)\plus{}\frac{1}{r^2}\frac{\partial^2 v}{\partial \phi^2}\equal{}0\]

find the solutions which have the form: $ v(r,\phi)\equal{}X(r)Y(\phi)$.

What have you done until now. This is the standard separation of variable method, and it should have been described in your course, isn't it ?

$ v(r,\phi)\equal{}\frac{A_1}{r}\left(A_2\cos\phi \plus{} A_3\sin \phi\right)$ for $ A_i\in \mathbb{R}.$


can u guys plz help me with problem 17 on this test?(link below)

[url]http://www.mathedge.org/2007%20PMWC/2007%20PMWC%20tryouts%20problems.pdf[/url]

i think im being stupid, cuz i have no idea how to solve this without counting the triangles.

thank u in advance.

-shaggy

Someone correct me if I'm wrong, but I believe the only way to solve it is to count them.

Notice that many triangles can be formed by choosing 2 of 4 lines from the left vertex and 1 from the other. So thats 6*4=24 then you count the rest which are made from 2 of 4 lines from the right vertex and 1 from the other. Since we don't want to count doubles we count the rest individually. Notice that that would only happen if you don't choose any of the bottom lines so that would be 2 of 3 x 3 which is 3x3=9 so the total triangles is 24+9=33. Sorry if my english is bad. :blush:

Ok think of it like this : to make a triangle you need to have at least three vertices(ignore the spelling  :D  ). Count the number of vertices in the whole figure there are say $ x$ vertices. So the answer is $ (x)(x\minus{}3)(x\minus{}6)(x\minus{}9)...(3)(2)(1)$

Umm... I don't think so... for one, your formula comes up with the answer of 880 when there are 11 vertices... and that definitely doesn't make sense... :huh:

The formula is inaccurate. For these problems, it is almost always best to look for some clever combinatorial approach or an organized casework approach.

Correct me if I am wrong, but aren't there 9 with both of the base vertices, 9 with just the left one, 9 with just the right one, and none without either, making a total of 27?


superior algebra
superior algebra unsolved

[quote="Ronx"]Let $ A$ be a commutative ring, $ 1\ne0$. Call $ a$ a zerodivisor if $ \exists b\ne0$ such that $ ab \equal{} 0$(0 is also a zerodivisor). Let $ F$ be the set of ideals in which every element is a zerodivisor. It is easy to show that $ F$ has maximal elements. Show then that every maximal element of $ F$ is a prime ideal.
The problem for me was that if $ p_1\in F, \ p_2\in F$ does not imply $ p_1 \plus{} p_2\in F$. I appreciate any comments.[/quote]

Hello.
A classical use of Zorn's lemma shows that there exists such a maximal element (which is not necessarily the set of all zerodivisors).
Each (maximal) element of a set of ideals is an ideal.
Thus, the only fact which must be proved is that such a maximal element, say $ M$, has the following property : if $ x,y$ are elements of $ A$ such that $ xy \in M$, then $ x$ or $ y$ is in $ M$.
If $ x$ is not in $ M$, then, by maximality of $ M$, the ideal $ M \plus{} Ax$ contains an element $ x' \equal{} m_{1} \plus{} ax$, with $ m_{1} \in M$ and $ a \in A$, which is not a divisor of zero.
If $ y$ is not in $ M$, there exists similarly an element $ y' \equal{} m_{2} \plus{} by$, with $ m_{2} \in M$ and $ b \in A$, which is not a divisor of zero.
Since $ x'y' \equal{} (m_{1} \plus{} ax)(m_{2} \plus{} by)$ and $ xy \in M$, it is clear that $ x'y' \in M$ and is thus a zerodivisor.
But $ x'y'$ is the product of two non zerodivisors, thus is not a zerodivisor. (If there exists a nonnull element $ c$ such that $ x'y'c \equal{} 0$, then if $ y'c \equal{} 0$, $ y'$ is a zero divisor; else $ x'(y'c) \equal{} 0$ with $ y'c$ nonnull, thus $ x'$ is a zerodivisor.)
The contradiction proves that $ M$ is a prime ideal.

Friendly,
P.

the zero divisors may be replaced by a complement of an arbitrary multiplicative subset not containing 0. (if this is generated by one element, you get the destription of the nilradical as the intersection of all prime ideals)

[quote="Ronx"]Can we say then than the set of zerodivisors is the union of prime ideals of set F?[/quote]

Yes. Each zerodivisor $ a$ is contained in the ideal $ aA$, and each element of $ aA$ is a zerodivisor, thus $ aA$ belongs to $ F$.

Friendly,
P.


Show why $ A$ is never similar to $ A \plus{} I$

Proof:

Assume $ A$ is similar to $ A\plus{}I$. Then there exists some invertible matrix $ M$ such that $ A \equal{} M^{\minus{}1}(A\plus{}I)M$. Choose $ M \equal{} I$, and we get $ A \equal{} A \plus{} I$, clearly a contradiction.

Now I'm not sure if the definition of similar matrices is whether $ A$ and $ B$ are similar if for any invertible matrix $ M$, we have $ B \equal{} M^{\minus{}1}AM$ or just for some invertible M. My proof rests on the case that we can pick M to be any invertible matrix.

Also, the way the textbook does the proof is that if $ \lambda_{1}, \lambda_{2}, ... , \lambda_{n}$ are the eigenvalues of $ A$, then $ \lambda_{1} \plus{} 1, \lambda_{2} \plus{} 1, ... , \lambda_{n} \plus{} 1$ are the eigenvalues of $ A \plus{} I$, and that similar matrices cannot have different eigenvalues.

But how do I show that the eigenvalues of $ A \plus{} I$ are simply the eigenvalues of $ A$ each plus $ 1$?

You play a devil's advocate by assuming that $ A$ is similar to $ A \plus{} I$. Then there exists $ M$ such that... -- but it's the devil who chooses $ M$, not you. You don't get to choose $ M \equal{} I$.

Instead of eigenvalues, I would look at the traces. Using the commutativity of trace, we get $ \mathrm{tr}(M^{\minus{}1}AM)\equal{}\mathrm{tr}(MM^{\minus{}1}A)\equal{}\mathrm{tr}(A)$. On the other hand, $ \mathrm{tr}(A\plus{}I)\equal{}\mathrm{tr}(A)\plus{}n$.

If you still want to try the proof by eigenvalues, look at the characteristic equations of $ A$ and $ A \plus{} I$.

Just to mess with the original poster's mind: $ A$ is never similar to $ A+I$ [i]over a field of characteristic zero.[/i] Over a field of finite characteristic, all bets are off.

For instance, over $ \mathbb{Z}/(3\mathbb{Z}),$

$ A=\begin{bmatrix}{0}&0&0\\0&1&0\\0&0&2\end{bmatrix}$ is similar to $ A+I=\begin{bmatrix}{1}&0&0\\0&2&0\\0&0&0\end{bmatrix}.$

What happened to mlok's trace argument? Just that $ \text{tr}(I)=0.$

(I know. The original poster was probably thinking only of matrices over $ \mathbb{R}$ and this excursion was a little cruel.)

Indeed.  Over a field of finite characteristic, adding $ 1$ to each eigenvalue may simply permute them without the set of eigenvalues actually changing.  In particular, this is true of any $ p \times p$ matrix with eigenvalues $ 0, 1, 2, ... p\minus{}1$ over a field of characteristic $ p$, which is Kent Merryfield's example (with $ p \equal{} 3$).


$x,y > 0$
Find all rational $x$ and $y$, which satisfie $x^{x+y}=(x+y)^y$

[hide]x=y=2  is the only solution with integers ...
[hide][hide]Notice the fact that (1+1/k)^k is rational  if and only if  k is integer.
 This will give  x=(1+1/n)^n and y =- x+(1+1/n)^(n+1)  type solution....[/hide][/hide]
 [/hide]

Edited later:

Sorry I misread the original, and made the solution for  $ x^{(x+y)} = (x+y)^x $ which is a little harder.. for the problem  one  given above,   the solution for x  is a little simpler  $(1+n)^n $

a=y/x  (y=ax) then the equation becomes x^(ax+x)=(ax+x)^(ax) => x^(a+1)=x^a.(a+1)^a => x=(a+1)^a and y=a.(a+1)^a where a is  a positive rational :)

[quote]x=(a+1)^a and y=a.(a+1)^a where a is a positive rational[/quote]

Just be a little careful, if a is rational (say 1/2)  it does not follow x would be rational too...
[hide]See the comment above ....[/hide]

[quote="cosinerburc"][hide=""Forgot to hide""]$a=y/x$  $(y=ax)$ then the equation becomes $x^{ax+x}=(ax+x)^{ax} \Rightarrow x^{a+1}=x^a\cdot(a+1)^a \Rightarrow x=(a+1)^a$ and $y=a\cdot(a+1)^a$ where $a$ is  a positive rational :)[/hide][/quote]
Please use hide/spoiler whenever you can and I also altered your quote to look nicer. Also since your solution is already revealed, note that $a$ must be a positive integer. Why? Ask Gyan.  :P

thanks for help 3x.lich;and you are right ;i've just proved that a is an integer


Prove that:

For $ x_{i}, y \in \mathbb{N}, n \in \mathbb{Z}^{\plus{}}$, we have: $ \sum_{x_{1}\plus{}...\plus{}x_{n}\equal{}y}\frac{1}{\prod_{i\equal{}1}^{n}(x_{i}!)}\equal{}\frac{n^y}{y!}$

Multiply up by the $ y!$, and consider the following combinatorial interpretation. The R.H.S. is the number of ways to make a string of y objects, where each object in the string can be any one of n different objects.

Each term on the L.H.S. can be interpreted as the number of ways to arrange the objects in a line if there are $ x_i$ of object i. Hence, since the sum is taken over all $ x_i$ that sum to y, this is the total number of ways of placing the objects in the line.

Note, that $ x_i$ must also be allowed to be 0 as well, so it should be corrected so that $ x_i \in \mathbb{Z}^{\ge 0}$.


$\triangle ABC$ is given, with $\widehat{A}<90^0$

Construct two points $D,E$ on the sides $AB,AC$ respectively, such that $DE \parallel BC$ and $BE \perp CD$

See : [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=52603[/url]

$DE\|BC$ is equivalent to saying that $BE\cap CD$ lies on the $A$-median of $ABC$, so we can simply take $M$ to be the intersection between the $A$-median of $ABC$ and the circle with diameter $BC$ which lies inside $ABC$ (the condition $\angle A<\frac\pi 2$ ensures the existence of $M$), and then take $E=BM\cap AC,D=CM\cap AB$.


geometry
geometric transformation
reflection
Pythagorean Theorem
geometry unsolved

[img]http://img457.imageshack.us/img457/4585/dfgdpn3.png[/img]

[list=1]
[*]Semicircle of diameter $AB$
[*]$\overline{CD}\perp \overline{AB}$
[*]Semicircles of diameters $AD$ and $DB$
[*]Circles E and F are inscribed in the arbelos[/list]
Prove that circle $E$ = circle $F$

Inversion with negative power $-r^{2},$ where r is radius of the inversion circle, is equivalent to ordinary inversion with positive power $r^{2}$ followed by reflection in the inversion center. Inversion with negative power carries a circle intersecting the inversion circle in diametrically opposite points into itself.

Let (X), (Y), (Z) be the circles with diameters AC = 2x, BC = 2y, AB = 2z = 2(x + y). Invert the arbelos in a circle (D) with center D and radius r = DC, using inversion with negative power. The circle (Z) intersects the inversion circle (D) in diametrically opposite points, hence it is carried into itself. The circles (X), (Y) tangent to (Z) and passing through the inversion center D go into tangents b, a of (Z) at B, A, respectively, perpendicular to AB. The line DC passing through the inversion center D goes into itself. The circles (E), (F) with radii $r_{e},\ r_{f}$ go into circles (E'), (F') tangent to (Z) and to the parallel lines b, DC resp. a, DC. Thus radii of the inverted circles (E'), (F') are $r_{e}' = y,\ r_{f}' = x,$ respectively. The inversion center D is similarity center of (E), (E') and of (F), (F') with similarity coefficients

$-\frac{r_{e}}{y}=-\frac{r_{e}}{r_{e}'}=\frac{-r^{2}}{p(D, (E'))}$ and $-\frac{r_{f}}{x}=-\frac{r_{f}}{r_{f}'}=\frac{-r^{2}}{p(D, (F'))}$

(power of inversion divided by power of inversion center to the inverted circle), hence

$\frac{r_{e}}{r_{f}}= \frac{y}{x}\cdot \frac{p(D, (F'))}{p(D, (E'))}$

Let S', T' be tangency points of (E'), (F') with DC. Then $p(D, (E')) = DS'^{2}= YE'^{2},$  $p(D, (F')) = DT'^{2}= XF'^{2}.$ By Pythagorean theorem for the right $\triangle YZE',\ \triangle XZF',$

$YE'^{2}= ZE'^{2}-ZY^{2}= (z+r_{e}')^{2}-x^{2}= (x+2y)^{2}-x^{2}= 4y (x+y)$

$XF'^{2}= ZF'^{2}-ZX^{2}= (z+r_{f}')^{2}-y^{2}= (2x+y)^{2}-y^{2}= 4x (x+y)$

$\frac{r_{e}}{r_{f}}= \frac{y}{x}\cdot \frac{XF'^{2}}{YE'^{2}}= \frac{y}{x}\cdot \frac{4x (x+y)}{4y (x+y)}= 1$

It was posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=126728]here[/url]


Given $ \minus{}f_{D} \equal{} bv \plus{} cv^2$
$ F_{D} \equal{} bv$ for slow moving objects
$ F_{D} \equal{} cv^2$ for fast moving objects
and $ c \equal{} \frac{1}{2}DpA$
where D = drag coefficient
p = density
A = surface area of the object

1.) For slow moving objects show that $ v(t) \equal{} \frac{mg}{b}(1\minus{}e^\frac{\minus{}bt}{m})$
Derive an expression for the velocity as a function of time for a falling, fast-moving object.

2.) A force F (larger than the largest possible force of static friction) is applied to the left to an object moving to the right on a horizontal surface. Then:

A) the object must be moving at constant speed
B) F and the friction force act in opposite directions
C) the object must be slowing down
D) the object must be speeding up
E) the object must come to rest and remain at rest

3.) Block A, with mass $ m_{A}$, is initially at rest on a horizontal floor. Block B, with mass $ m_{B}$, is initially at rest on the horizontal top surface of A. The coefficient of static friction between the two blocks is $ \mu_{s}$. Block A is pulled with a horizontal force. It begins to slide out from under B if the force is greater than:
A) $ m_{A}g$
B) $ m_{B}g$
C) $ \mu_{s}m_{A}g$
D) $ \mu_{s}m_{B}g$
E) $ \mu_{s} (m_{A} \plus{} m_{B})g$

4.) A desks rests on a rough horizontal surface ($ \mu_{s} \equal{} 0.50$, $ \mu_{k} \equal{} 0.40$). A constant horizontal force, just sufficient to start the desk in motion, is then applied. The acceleration of the desk is:
A) 0
B) 0.98 m/s^2
C) 3.3 m/s^2
D) 4.5 m/s^2
E) 8.9 m/s^2

5.) When the frictionless system shown in the attached picture is accelerated by an applied force of magnitude F, the tension in the string between the blocks is
A) 1/3F
B) 2F
C) F
D) 2/3F
E) 1/2F

[hide="1"]
1. draw free-body diagram
2. get equation mg-bv=ma
3. a=$ \frac {dv}{dt}$ so we have a differential equation
4. Solve the differential equation, find the constant, then solve for v(t), and we're done.

Part 2:
do the same
mg-$ cv^2$=ma.
Use a differential equation.
Then, use Wolfram Alpha to solve that diffeq.
Not sure if this is correct, but I got
v(t)=$ \sqrt {\frac {mg}{c}}tanh({\frac {t}{m}\sqrt {mgc}})$
Then, I plug in c=$ \frac {1}{2}D\rho A$, which is pretty ugly, which is why I feel skeptical.
[/hide]
[hide="2"]
Draw a free-body diagram (I am unsure of how to post pictures on aops, so I guess I'll describe the box)
If an object is moving to the right, then the static friction must have been overcome earlier and the box experiences kinetic friction to the left.  Now, we add a force F also to the left, causing the box to decelerate even more.  If we continue to apply this force, the box will begin to have a velocity to the left, so immediately after we apply the force, the answer is C.  However, later the answer is B and D. So is this a trick question or is it just unclear about when this happens?
[/hide]
[hide="3"]
Drawing another free-body diagram gets me D.
[/hide]
[hide="4"]
Do you notice I'm growing increasingly lazy?  :P B  (will post reason if necessary)
[/hide]
[hide="5"]
A
[/hide]

1. no clue-haven't learned drag yet, maybe ill go do that now
2. I get 2 answers.  There is a net force, so there is an acceleration which makes D true and the force F and the frictional force always act in opposite directions making B true
3. D
4. B
5. A

#2 is just really ambiguous. candyhyperalert, B is not true because in this case the force F and fricitonal force are acting in the same direction. The object is moving opposite the direction of the force.
However, the problem does not specify if the object was previously moving with constant velocity, or with a force. It's not the net force that is necessarily to the left, so I conclude that there is not enough information to solve this.

2.  When you apply a force F > max{static friction}   leftward on an object moving rightward, then what'll happen is this:
         The block initially slows down and momentarily comes to rest after that it speeds up rightward since the force applied is more than the static friction force
           None of the options truly satisfies with the happening but options B, C, D are valid for some period of time.


[quote="kostas"]Ενας επιχειρειματιας θελει να διαφημισει το καταστημα του σε ολο τον γαλαξια!ζητα λοιπον απο μια εταιρια να φτιαξει μια απειρη διαφημιστικη ταμπελα.Μαλιστα αυτος για να τους βοηθησει λεει οτι θα τους δωσει τη μπογια.Ομως τους δινει μονο 4 κιλα.Αυτοι ισχυριζονται οτι δε φτανει αυτη η ποσοτητα αφου η ταμπελα ειναι απειρη, ενω αυτος οτι φτανει.Τελικα ποιος εχει δικιο?πιστευω οτι ειναι δυσκολοτερο για τους μικροτερους.[/quote]

Η ταμπέλα είναι άπειρη αλλά μόνο ένα πεπερασμένο μέρος της είναι γραμμένο το οποίο μπορεί να βαφεί από τα 4 κιλα μπογιάς.

[quote="kostas"]Οι τιμες για την μπογια αυξηθηκαν!Τωρα ο επιχειρηματιας δινει το 1/4 του κιλου!Λοιπον φτανει αυτη  ποσοτητα για να βαψει την απειρη ταμπελα?Ελατε δεν ειναι δυσκολο...πολυ![/quote]

Για το 1/4 του κιλού απλώς θα μικρύνουμε τα γράμματα  :P

Καλως ηρθες φιλε ΚΜΝ.Ο καταστηματαρχης θελει να τη βάψει ολοκληρη την ταμπελα.Κάθε σημειο της!!!Μπορει τελικά? Να θυμίσω οτι η ταμπελα ειναι απειρη!

Νομίζω  ότι θα μπορούσε να  βαφτεί  η ταμπέλα, αν  υπήρχε μια  ένα προς ένα αντιστοιχία  μεταξύ των σημείων της ταμπέλας και των σημείων του κυλίνδρου(?) του δοχείου που περιέχει την μπογιά. Στην περίπτωση αυτή, δεν έχει σημασία αν  ο όγκος της  μπογιάς είναι 4 κιλά  ή  4  δακτυλήθρες. Το πρόβλημα νομίζω είναι  τοπολογικό....

Φίλε Νικόλα η λυση ειναι σιγουρα ποιο απλη.Μπορούν και παιδια Β λυκειου.Θα με ενδιέφερε πολύ μια λυση με την απεικονηση που ειπες.Το ακουσα εχθες απο εαν επισης σπουδαιο μαθηματικο!

Άθροισμα απείρων όρων γεωμετρικής προόδου  ?? ;)

Τι εννοείς Σιλουανε? :?:

...Μην ξεχνατε οτι η σωστη λυση μιας ασκησης ειναι η πιο απλη απο ολες αυτες που εχεις σκεφτει!!! Ετσι Αρχιμήδη 6? ;)

Ετσι ακριβως Κωστα.Η ασκησή σου ειναι πολυ καλη χωρις να απαιτεί ιδιαιτερες γνώσεις.Αντε γειά τωρα σε ολους τους χρηστες του φορουμ.Θα τα ξαναπουμε.

Οποιοδήποτε ανοικτό υποσύνολο του επιπέδου με άπειρη διάμετρο και πεπερασμένο εμβαδό λύνει το πρόβλημα. Π.χ. η περιοχή μεταξύ του γραφήματος της συνάρτησης x^(-a), a και x μεγαλύτερα του κάποιου θετικού αριθμού, και του άξονα των x.

Περισσότερο ενδιαφέρον έχει "η σάλπιγκα του Αρχαγγέλου Γαβριήλ" που σχηματίζεται από την περιστροφή του κλάδου της ισοσκελούς υπερβολής y = 1/x, x>1, περί τον άξονα των x. Έχει πεπερασμένο όγκο αλλά άπειρο εμβαδόν επιφανείας! Έτσι μπορείς να τη γεμίσεις με χρώμα αλλά θέλεις άπειρη ποσότητα χρώματος να βάψεις το δοχείο!

πλασάρει...και γκόλλλλλ!!!Πολύ καλά Dulac.Ακόμα και εδώ το α δεν παιρνει οποιεσδήποτε τιμές.Το παράδοξο φαινομενικά ειναι οτι αν ισχύει για α μπορεί να να μην ισχύει για α+1.Μπορεί κανεις να βρεί μια συνάρτηση για το α?

Πάρε την συναρτηση χ^(-2) φέρε την χ=4 .Το χωρίο που ορίζεται απο την χ=4,χ^(-2).και τον αξονα των χ απο το 4 και με μεγαλύτερες τιμες του χ ειναι μια ζητούμενη ταμπέλα. Το Εμβαδον ειναι πεπερασμένο γιατί  η μαγική λεξη που ακούει στο ΟΛΟΚΛΗΡΩΜΑ δίνει lim του ολοκληρωματος της συναρτησης μας απο 4 ως τ οταν τ τείνει στο απειρο =1/4.Ενα ακομα παραδοξο είναι οτι το αλλο μερος της συναρτησης με τους αξονες(ανω χωριο) εχει απειρο εμβαδον. Το -2 μπορει να αωτικαταστησει το -2ν, ν>0.


I have thought of five numbers. If I add three of them in each possible way, I get $10, 14, 15, 16, 17, 17, 18, 21, 22$ and $24$. What are my numbers?

Because we have 10, the smallest three t,w,x :

t+w+x = 10
x+y+z=24
so, t+w+2x+y+z=34
Sum(all cases) = 174 = 6t+6w+6x+6y+6z
t+w+x+y+z=29
so, x=5.

y+z=19
t+w=5

w+y+z = 22
w=3
t=2

t+y+z=21 for sure since t= w-1

t+w+y=14
y=9
z=10
Voila!


Odell and Kershaw run for 30 minutes on a circular track.  Odell runs clockwise at 250 m/min and uses the inner lane with a radius of 50 meters.  Kershaw runs counterclockwise at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial line as Odell.  How many times after the start do they pass each other?

$ \textbf{(A) } 29 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 45 \qquad \textbf{(D) } 47 \qquad \textbf{(E) } 50$

[hide]

suppose the angle form by Odell (in radian) after time $t$ is $\theta_1$ . Angle form by Kershaw after time $t$ is $\theta_2$ . If they meet at time $t$ , then we have

$50\theta_1=\frac{250}{t}$ or $\theta_1=\frac{5}{t}$

And $60\theta_2=\frac{300}{t}$ or $\theta_2=\frac{5}{t}$

From above , $\theta_1=\theta_2$

They can meet only when $\theta_1+\theta_2=2\pi$ or $\theta_1+\theta_2=0$

So we have $\theta_1=\theta_2=\pi$ or $0$ .

Since Odell runs 7500m or $\frac{75}{\pi}$ round , which is about 23.9 rounds , hence they meet $23x2+1=47$ times (since for each round they meet twice and for the last round , they only meet once)
[/hide]

[hide]Note that both runners have the same angular speed but in opposite directions, because radius is proportional to circumerence. They will cross twice per lap. Odel runs at 250 m/min for 30 mins = 7500 m. The track length is $100 \pi$, so Odel runs $\frac{75}{pi}$, or about 23.87 laps. Thus they cross $2*23.87=47.75 \rightarrow \boxed{47}$ times.[/hide]

Hi,
What is a generalization to find how many times the runners will pass each other per lap? I know that
[quote="AMC Solution"]
Since Odell’s rate is 5/6 that of Kershaw, but Kershaw’s lap distance is 6/5 that of Odell, they each run a lap in the same time.[/quote]

but I would like to know, what happens if Odell's rate is 1/8 of that of Kershaw, etc...

TIA

[hide=Solution]
$\lfloor\frac{30}{\frac{120\pi}{\frac65\cdot250+300}}\rfloor=\boxed{\textbf{(D)} 47}$
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