[quote="Devendra Patel"]At what timjes apart from 12:00 do the three hands of the clock exactly overlap?[/quote]It depends a lot on the movement of the hands of the clock that you assume. If you assume that the three hands have a discrete movement (that, for example, the hand of the ours move in an instant from 1 to 2 when it is 2:00, and it does not stop in no intermedial point), then it would be 1:05:05, 2:10:10,..., 11:55:55. If you asssume a continuous movement of the three hands of the clock, then, there is no other moment when the three hands. I find kind of boring to prove all the other 6 cases (some of which are absourd) but you can try.

;) it is basicaly not possible because each time the second hand moves the minute hand moves too,and hence forth

Moderator Edit: Double post erased!

[quote="Devendra Patel"]At what timjes apart from 12:00 do the three hands of the clock exactly overlap[/quote]

You can follow this link to the proof
[url]http://www.mathforum.com/library/drmath/view/56819.html[/url]


The problem is to break a square of nXn dimension into minimum number of smaller squares. Every nXn where n is even can be broken into 4 (n/2)X(n/2) squares.
3X3 can be broken into 1 2X2 and 5 1X1.
For 7X7 minimum number of squares required are 9. Try it :-)

What will be the minimum number of squares for pXp, where p is prime?

For 7x7, the minimum is achieved by 1 4x4, 2 3x3, 3 2x2s, 3 1x1s
For pxp
I had an idea, but there are too many possibilities...cant figure it out...must think more

No one interested in solving this? :(  :(

I thought I would get an answer to this here.

Min. number of squares for pXp, p is prime?

[quote="seamusoboyle"]I think it would be easier to prove it for odd numbers, as opposed to primes[/quote]

21 = 3*7, we can do in same number of squares as we need for 3X3. Hence we are interested in smallest prime that divides that odd number.

[quote="seamusoboyle"]But that wont be the best configuration, for example if you did yours your way, i could just group a 9 by 9 square as one. Aslo, this still may not be the best configuration...[/quote]

Interesting!

Refute with an example that the configuration may not be the best for oddXodd.

My claim is that we need to search for smallest prime in that odd number and we are done.

Okay, it is not hard to show that for an odd x odd square, we cannot do better than 6 pieces.  (Pf: That 4 is insufficient is clear.  Assume you can do 5.  One needs to be in each corner.  Consider the largest of these and draw a picture and it's immediately clear.)  So, the only counter-examples would require squares whose side-lengths were not divisible by 3.  I do not, off hand, have one available, and it's rather difficult to test in any scientific manner, but I'll see what I can do.  (It is an interesting question!)


Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ (a^2\minus{}bc)\sqrt{a^2\plus{}4bc}\plus{}(b^2\minus{}ca)\sqrt{b^2\plus{}4ca}\plus{}(c^2\minus{}ab)\sqrt{c^2\plus{}4ab} \ge 0\]
VasC has a nice proof by Chebyshev in his book. Now I have found a proof by AM-GM and Cauchy Schwarz for it. :)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ (a^2 \minus{} bc)\sqrt {a^2 \plus{} 4bc} \plus{} (b^2 \minus{} ca)\sqrt {b^2 \plus{} 4ca} \plus{} (c^2 \minus{} ab)\sqrt {c^2 \plus{} 4ab} \ge 0
\]
[/quote]
Here is my proof:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157776

Can someone post the solution using Chebyshev Inequality about was talking can_hang2007 about in post 1?


One gear turns $ 33\frac{1}{3}$ times in a minute.  Another gear turns
45 times in a minute.  Initially, a mark on each gear is pointing
due north.  After how many seconds will the two gears next
have both their marks pointing due north?

You can probably use mods but the way I would do it is look for the LCM.  Say the LCM is $ n$.  Then after $ n$ turns of the first gear, it is back where it started, and after $ n$ turns of the second gear, it is back where it started.

I'm just basically carrying out the solution outlined in the post above. By taking the inverse of the two numbers, you get the amount of time it takes for the mark on each gear to face north again.

The values are 3/100 and 1/45 minutes, or 9/5 and 4/3 seconds. The LCM of these two numbers can be found by getting rid of the fractions by multiplying by 15, and then dividing by 15 later. 15*9/5 = 27 and 15*4/3 is 20. The LCM of these numbers is 540. 540/15 is 36. So the answer is that in 36 seconds, the marks on both gears will face north again.

[hide="formal solution and answer"]One gear turns $33\frac{1}{3}=100/3$ times in 60 seconds, so it turns 5/9 times in one second, or 5 times in 9 seconds. The other gear turns 45 times in 60 seconds, so it turns 3/4 times in one second, or 3 times in 4 seconds. To find out after how many seconds the two gears next have both their marks pointing due north, we have to find the least common multiple of $4=2^2$ and $9=3^2$, which is $2^2\cdot3^2=36$. Therefore, the two gears next have both their marks pointing due north after $\boxed{36}$ seconds. (One gear turns exactly $5\times4=20$ times, and the other gear turns exactly $3\times9=27$ times.)[/hide]


Find a closed formula for

1) $ D_n\equal{}1\minus{}2\plus{}3\minus{}4\plus{}...\plus{}(\minus{}1)^{n\minus{}1}n$

2) $ T_n\equal{}1^2\minus{}2^2\plus{}3^2\minus{}4^2\plus{}...\plus{}(\minus{}1)^{n\minus{}1}n^2$

hello, $ D_n\equal{}\frac{1}{4}\left((\minus{}1)^{n\plus{}1}(2n\plus{}1)\plus{}1\right)$,
$ T_n\equal{}\frac{1}{2}(\minus{}1)^{n\plus{}1}(n\plus{}1)n$.
Sonnhard.


it's the smallest common divisor of their differences..

[quote="13375P34K43V312"]When the three integers 618, 343, 277 are divided by a positive integer d where d>1, the remainders are the same. What is the smallest possible value of d?[/quote]

hey which problem set?


Prove that if $x,y,z \in R$ then:
$ \sum (x+y-z)^2 \geq \sum xy$.

here is what i found:
we replace $x+y-z=a$, $-x+y+z=b$, $x-y+z=c$
we have then: $ x= \frac{a+c}{2} $  and the rest
so we have:
$ \sum a^2 \geq \sum \frac{(a+c)(a+b)}{4}$
and after some simple computations we get to:
$ \sum (a-b)^2 \geq 0$.

that's all.
do u find smth different?
cheers! :D :D

Here is a different solution.By Cauchy,we have:
3.( \sum (x+y-z) 2 ) \geq (x+y+z) 2  \geq 3.(xy+yz+xz).
So we have what we need. :)


function
algebra
functional equation
algebra solved

Find all f(x) such that for all reals x, y,

f(x - f(y)) = f(x) + xf(y) + f(f(y)).

Let us assume that there is y0 such that f(y0)!=0. Otherwise f=0 is obviously a solution.

f(x-f(y))=f(x)+xf(y)+f(f(y)) ===> f(f(x)-f(y))=f(f(x))+f(x)f(y)+f(f(y)) (1)
Taking x=y in (1) we get
f(0) = 2f(f(x)) + f^2(x) ===> by using this in (1) we obtain :
f(f(x)-f(y)) = (f(0)-f^2(x))/2 + (f(0)-f^2(y))/2 + f(x) f(y)  ===>
f(f(x)-f(y)) = f(0) - (f(x)-f(y))^2/2 (2)
So f(z) = f(0) - z^2/2 , for all z in {f(x)-f(y) | y,z in R}
f(x-f(y0))-f(x) = xf(y0) + f(f(y0)) (3)
The function g(x)=xf(y0) + f(f(y0)) takes R to R bijectively ===> for every t in R there is an x in R such that f(x-f(y0))-f(x)=t. (1)+(2)+(3) ===> f(x)=c-x^2/2 for all x in R.

Getting back to the initial relationship we have:
c - (x-f(y))^2/2 = c-x^2/2 + xf(y) + c-f^2(y)/2 ===>
c - x^2/2 - f^2(y)/2 + xf(y) = c - x^2/2 + xf(y) + c-f^2/2 ===>
2c = c ===> c = 0

So, the solutions of the functional equation are :
(I) f(x) = 0, for all x in R
(II) f(x) = -x^2/2 , for all x in R

Hmmm, should it be a little bit similar to IMO 1999 problem 6?

this is the problem harazi is talking about!

Problem 6

Determine all functions f: R -> R such that f(x - f(y) ) = f( f(y) ) + x f(y) + f(x) - 1 for all x, y in R. [R is the reals.]

Solution

Solution communicated by Ong Shien Jin

Let c = f(0) and A be the image f(R). If a is in A, then it is straightforward to find f(a): putting a = f(y) and x = a, we get f(a - a) = f(a) + a2 + f(a) - 1, so f(a) = (1 + c)/2 - a2/2 (*).

The next step is to show that A - A = R. Note first that c cannot be zero, for if it were, then putting y = 0, we get: f(x - c) = f(c) + xc + f(x) - 1 (**) and hence f(0) = f(c) = 1. Contradiction. But (**) also shows that f(x - c) - f(x) = xc + (f(c) - 1). Here x is free to vary over R, so xc + (f(c) - 1) can take any value in R.

Thus given any x in R, we may find a, b in A such that x = a - b. Hence f(x) = f(a - b) = f(b) + ab + f(a) - 1. So, using (*):   f(x) = c - b2/2 + ab - a2/2 = c - x2/2.

In particular, this is true for x in A. Comparing with (*) we deduce that c = 1. So for all x in R we must have   f(x) = 1 - x2/2. Finally, it is easy to check that this satisfies the original relation and hence is the unique solution.

from kalva site

so it is simmilar to it
:)

is the imo problem considered a difficult one? i have little experience with functional equations, but this problem seems difficult and its solution a creative one.

I don't know other people opinions but to me this was one of the most difficult functional equations ever seen.

Jeje, talking about that, I feel so proud of myself cuz that problem was the first problem 6 of an IMO I could solve. I got nice solution, very similar to the one in kalva, but much more understandable :D:D:D:D:D


In the triangle ABC we have a^2>=4bc.Prove that there exists M on BC such thata MB*MC=b*c.

Again, unsolved.

Funny, this "discriminant" of a triangle ;)


The diagonal matrix can be found by placing the eigenvalues on the diagonal of a identity matrx.  How do you know what order to place them?

You don't "know what order" to put the eigenvectors. Each possible order is just as valid as any other possible order and the diagonalization is not unique.

Suppose $A$ is diagonalizable, meaning that we can find a basis of $F^{n}$ consisting of eigenvectors. (Here, $F$ is $\mathbb{R}$ or $\mathbb{C},$ or whatever other field this is a matrix over.) Stack these eigenvectors (column vectors) together in some order to get an invertible matrix $Q$. Then $Q^{-1}AQ=D,$ where $D$ is diagonal. The diagonal elements of $D$ are the eigenvalues, in the same order as their corresponding eigenvectors.

Now suppose that you put those eigenvectors together into a matrix in some different order. The result is the matrix $QP,$ where $P$ is a permutation matrix. (A permutation matrix is a matrix with one 1 in each column and one 1 in each row, with all other elements being zero. Permutation matrices are a very special case of orthogonal matrices, meaning that $P^{T}P=I,$ or the transpose is the inverse.)

So now, $(QP)^{-1}=P^{T}Q^{-1}$ and

$P^{T}Q^{-1}AQP=P^{T}DP.$

Multiplying on the right by $P$ permutes the columns of a matrix; multiplying on the left by $P^{T}$ does the same permutation of the rows. The result is that $P^{T}DP=D',$ where $D'$ is a diagonal matrix with the same diagonal elements as $D$ but in a different order.


find all $f$ from positive reals into themselves such that:

$x^2(f(x)+f(y))=(x+y)f(f(x)y)$ for all $x,y>0$

$x^2(f(x)+f(y))=(x+y)f(yf(x))$
Let $g(x)=xf(x)$ we have:
$x^2(\frac{g(x)}{x}+\frac{g(y)}{y})=(x+y)\frac{g(yf(x))}{yf(x)}$
$\leftrightarrow xg(y)+yg(x)=(x+y)g(y\frac{g(x)}{x})$
let $x=y$ we have:
$g(g(x))=1$
Let $y->xg(y)$ we have:
$xg(x)g(y)+xg(xg(y))=(x+xg(y))g(g(x)g(y))$

$\leftrightarrow g(x)g(y)+g(xg(y))=(1+g(y))g(g(x)g(y))$

Let $y-> f(y)$ we have:
$g(x)+g(x)=2 \rightarrow g(x)=1$

Then
$f(x)=\frac{1}{x}$

Excuse me [b]hunghktn[/b], but I don't understand one thing, you wrote that: $x^{2}(\frac{g(x)}{x}+\frac{g(y)}{y})=(x+y)\frac{g(yf(x))}{yf(x)}\Leftrightarrow
xg(y)+yg(x)=(x+y)g(y\frac{g(x)}{x})$ but I think that we have $xg(y)+yg(x)=\frac{(x+y)}{g(x)}g(y\frac{g(x)}{x})$ and after putting x=y we obtain $g(g(x))=g^{2}(x)$. What am I missing?

I think I have  a nice solution for this one. Denote $g(x)=xf(x)$. We have $xg(y)+yg(x)=\frac{x+y}{g(x)}g(\frac{yg(x)}{x})$ (*). Now put $y=\frac{x^{2}}{g(x)}\Rightarrow g(\frac{x^{2}}{g(x)})+x=1+\frac{x}{g(x)}$.Put $x=1$ then $f(t)=t$, where $t=\frac{1}{g(1)}$.Putting $x=y$ in (*) we obtain $g(g(x))=g^{2}(x)$, $x=t\Rightarrow t=t^{2}\Rightarrow t=1$.Put $x=1$ in (*)$\Rightarrow g(y)+y=(y+1)g(y)$ thus $g(y)=1$ for all y, so $f(y)=\frac{1}{y}$ for all $y$.QED

Sorry,it's a stupiad mistake.
Your solution is very nice,Armo.


function
real analysis
real analysis unsolved

show that a real valued funtion bounded and nondecreasing is uniformly continuous

Say that $ \inf_{x}f(x) \equal{} m$ and $ \sup_{x}f(x) \equal{} M$.  Fix $ \epsilon>0$ and pick $ N>0$ large enough that $ |f(\minus{}N)\minus{}m|<\epsilon/2$ and $ |f(N)\minus{}M|<\epsilon/2$.  Note that $ f$ restricted to $ [\minus{}N,N]$ (a compact set) is uniformly continuous.  If $ x,y\in [\minus{}N,N]$, there exists $ 2N>\delta>0$such that $ |x\minus{}y|<\delta$ implies $ |f(x)\minus{}f(y)|<\epsilon/2$.

If $ x,y\in (\minus{}\infty,\minus{}N)$ or $ x,y\in (N,\infty)$, then $ |f(x)\minus{}f(y)|<\epsilon/2$ (regardless of $ |x\minus{}y|$)

If $ x\in (\minus{}\infty,\minus{}N)$ and $ y\in [\minus{}N,N]$ with $ |x\minus{}y|<\delta$, then $ |f(x)\minus{}f(y)|\leq |f(x)\minus{}f(\minus{}N)|\plus{}|f(\minus{}N)\minus{}y|$.  Since $ |\minus{}N\minus{}y|<|x\minus{}y|<\delta$, $ |f(\minus{}N)\minus{}y|<\epsilon/2$, and since $ |f(x)\minus{}f(\minus{}N)|<|m\minus{}f(\minus{}N)|<\epsilon/2$ (as $ f$ is nondecreasing), we get that $ |f(x)\minus{}f(y)|<\epsilon$.  The other case is analogous.

It's only uniformly continuous if it's continuous. You need that hypothesis too, or the Heaviside function $ f(x)\equal{}\begin{cases}1&x\ge 0\\0&x\le 0\end{cases}$ is a counterexample.


calculus
integration
Diophantine equation
number theory unsolved
number theory

Solve this equation in Z
$ 1^2\plus{}2^2\plus{}....\plus{}n^2\equal{}m^2$

And next time, start a new thread for your problems (instead of posting them into other peoples threads).

Hi,
by using the other expression of sum(k²),I found that the solution is{(1,1),(-1,1),(1,-1),(-1,-1)}

There is exactly one other solution $ (24, 70)$ which is responsible for at least some of the properties of the [url=http://en.wikipedia.org/wiki/Leech_lattice]Leech lattice[/url].

n.(n+1).(2n+1)=6m^2  2n+1 equil 1(mod2)
(n,n+1)=(2n+1,n+1)=(2n+1,n)=1
i)n=3s^2 n+1=2r^2 2n+1=t^2  2r^2-3s^2=1 -r^2 equil 1(mod3) contradiction!
ii)n=2s^2 n+1=3r^2 2n+1=t^2  4s^2+1=t^2 (t-2s)(t+2s)=1  t=1 s=0 n=0 contradiction!
iii)n=s^2 n+1=2r^2 2n+1=3t^2  2s^2+1=3t^2 it is a pell equation! 3t^2-2s^2=1 if there is a solution ,then there is enternity!
(1,1) is a solution  (9,11) is a solution infinite solution
iv)n=3s^2 n+1=r^2 2n+1=3t^2  2r^2 -3t^2=1 it is a pell equation too.
v)n=6s^2 n+1=r^2 2n+1=t^2  t^2-12s^2=1 it is a pell equation too.
vi)n=s^2 n+1=6r^2 2n+1=t^2 12r^-t^2=1 it is a pell euation ,too
 infinite solution!
CÜNEYD ÖZTÜRK

[url=http://en.wikipedia.org/wiki/Siegel%27s_theorem_on_integral_points]Siegel's theorem[/url] guarantees that there are only finitely many integer points on this curve, so there's clearly something wrong with your analysis.

For example, case iii) is not one Pell's equation but two:  $ s^2 \plus{} 1 \equal{} 2r^2$ and $ (2s)^2 \plus{} 1 \equal{} 3t^2$, and the intersection of the solutions to two Pell's equations is not guaranteed to be infinite.  $ (9, 11)$ is also not a solution:  $ 9 \cdot 10 \cdot 19 \neq 6 \cdot 11^2$.

excuse me 12s^2-t^2=1 isn't pell equation

It can be solved using identical methods; what's your definition of a Pell's equation?  That doesn't invalidate my point; not every solution to $ s^2 \plus{} 1 \equal{} 2r^2$ is a solution to the original equation.

2s^2+1=3t^2 is a pell equation so (11,9) is a solution 242+1=243
pell equation :
ax^2-by^2=1 a,b e Z  b isn't perfect square

That's even broader than my definition of a Pell's equation (which is $ a \equal{} 1$ or $ b \equal{} 1$) and it includes the case that you just said isn't a Pell's equation, so I'm not sure what you're trying to say.

And you'll notice that $ (11, 9)$ doesn't correspond to a value of $ n$ in the original equation, so again, I'm not sure what you're trying to say.

$ excuse$  $ me$  $ pell$ $ equation$ $ is$ $ x^2 \minus{} ay^2 \equal{} 1$ $ a$ $ isn't$ $ perfect$ $ square$


A 3-foot high tree was planted, and grows by an equal number of feet each year. At the end of the seventh year, it is $ \frac{1}{9}$ taller than at the end of the sixth year. How many feet tall will it be at the end of the 13th year?

[quote="GameBot"]A 3-foot high tree was planted, and grows by an equal number of feet each year. At the end of the seventh year, it is $ \frac {1}{9}$ taller than at the end of the sixth year. How many feet tall will it be at the end of the 13th year?[/quote]

Say it grows x feet a year. Thus,

$ 3\plus{}5x\equal{}\frac{9}{10}(3\plus{}6x)$
$ 3\plus{}5x\equal{}\frac{27}{10}\plus{}\frac{27}{5}x$
$ \frac{3}{10}\equal{}\frac{2}{5}x$
$ x\equal{}\frac{3}{4}$

$ 3\plus{}12x\equal{}3\plus{}9\equal{}12$.

Note that is asked for the [b]end[/b] of the thirteenth year. So we could set up an equation somewhat similar to bogtro's:
$3+7d=\frac{10}{9}*(3+6a)$
$3+7d=\frac{10}{3}+\frac{20d}{3}$
$9+21d=10+20d$
$d=1$
So we plug d into this equation: $3+13d=3+13=\boxed{16}$


function
number theory unsolved
number theory

Find all function f:N->N such that  $ f(n)=f(f(n-1))+f(f(n+1))$ for n>1

First let $ m=min\{f(n)|n\geq 1\}$.Now there is $ a$ such that $ f(a)=m$.Now if $ a>1$ we have that
$ f(a)=f(f(a-1))+f(f(a+1))$ and we get that $ f(f(a+1))<m$ wich is a contradiction.So the only $ a$ is $ 1$.Now we define $ m_{1}=min\{f(n)|n> 1\}$.Note that there is $ b>1$ such that $ f(b)=m_{1}$ so
$ f(b)=f(f(b-1))+f(f(b+1))$.Now we have that $ f(f(b+1))<m_{1}$ and if $ f(b+1)>1$ we have a contradiction with how we chose $ m_{1}$.So it remains that $ f(b+1)=1$ but we have $ f(b+1)=f(f(b))+f(f(b+2))=1$ again a contradiction.So no such function exists.

Another way: since we need $ f(n\plus{}1) > f(f(n))$, for any $ n\in\mathbb{N}\equal{}\{1, 2,\ldots\}$, and the only function $ g:\mathbb{N}\mapsto\mathbb{N}$ such that $ g(n\plus{}1) > g(g(n))$, for each $ n \equal{} 1, 2,\ldots$, is the identity $ \mathbb{N}\mapsto\mathbb{N}: n\to n$, it's plain the actual problem has no solution.


analytic geometry
graphing lines
slope
MATHCOUNTS

This question is quite confusing, you cannot really talk about the slope of two points.

I think the slope should be indefined :maybe: ,since slope is defined by the change in the direction of Y over the change in the direction of X.

I hope this can help you! :P

It's undefined. Consider substituting in the coordinites to the slope formula. You have to divide by 0, which is undefined.

[quote="tinadoty"][quote="funcia"][quote="tinadoty"]Hey, confusing question at a competition i had (not mathcounts, it's my county competition).

It said list the slope of the following coordinates.
(-1,0) (-1, 7)

Me & My team put No Slope and got it wrong?
Should we have wrote undefined slope or, 0/7.

Thx[/quote]

plot it that would help the slope is actually 0
because it doesn't "slope" or whatever
i had a question similar to this
in a final exam once[/quote]

No all you do is 7-0 = 7 and -1 - -1 which is 0. So its 7/0 and when the zero is on the bottom its undefined / no slope. When its on top its 0 slope. So yeah 0/7 is 0 slope but 7/0 is undefined. But they messed up on a few questions too.[/quote]
no i am absolutely sure it is 0
true 0/7 is undef. 7/0 is also but it has no slope thus slope=0

Think about it this way. If you were to grapht he line, it would be vertical. Slope means the amonut of units that the line rises per unit it runs to the right. Well it rises indefinitely, and never runs to the right, so the slope is undefined.

[quote="perfect628"]So 7/0 is 0 is what you're saying...[/quote]
no 7/0 is undef. 0/7 is 0 sorry about my REALLY BAD WAY of explaining myself

it has no slope
the slope formula does not work for this case

[quote="funcia"][/quote][quote="tinadoty"][quote="funcia"][quote="tinadoty"]Hey, confusing question at a competition i had (not mathcounts, it's my county competition).

It said list the slope of the following coordinates.
(-1,0) (-1, 7)

Me & My team put No Slope and got it wrong?
Should we have wrote undefined slope or, 0/7.

Thx[/quote]

plot it that would help the slope is actually 0
because it doesn't "slope" or whatever
i had a question similar to this
in a final exam once[/quote]

No all you do is 7-0 = 7 and -1 - -1 which is 0. So its 7/0 and when the zero is on the bottom its undefined / no slope. When its on top its 0 slope. So yeah 0/7 is 0 slope but 7/0 is undefined. But they messed up on a few questions too.[/quote][quote="funcia"]
no i am absolutely sure it is 0
true 0/7 is undef. 7/0 is also but it has no slope thus slope=0[/quote]

7/0 is undefined while 0/7 is 0 because 7*0=0......

it is? wow! now i don't have to sue the author of my old algebra book for teaching me lies that ruined my soul!!!

hi ,

i think this line in perpendicular to x axis

what u say?  :?:

[quote="humblemastermind"][quote="#H34N1"][quote="tinadoty"]Hey, confusing question at a competition i had (not mathcounts, it's my county competition).

It said list the slope of the following coordinates.
(-1,0) (-1, 7)

Me & My team put No Slope and got it wrong?
Should we have wrote undefined slope or, 0/7.

Thx[/quote]

[hide]
The slope would be no slope. Only vertical lines have no slope.[/hide][/quote]

That's not quite true. Their slope is undefined. It's not really correct to say that they have "no slope".[/quote]

Vertical lines have undefined slopes. Oh wait. I typoed (again!)

What I meant is that horizontal lines have no slope aka a slope of 0

[quote="tinadoty"][quote="funcia"][quote="tinadoty"]Hey, confusing question at a competition i had (not mathcounts, it's my county competition).

It said list the slope of the following coordinates.
(-1,0) (-1, 7)

Me & My team put No Slope and got it wrong?
Should we have wrote undefined slope or, 0/7.

Thx[/quote]

plot it that would help the slope is actually 0
because it doesn't "slope" or whatever
i had a question similar to this
in a final exam once[/quote]

No all you do is 7-0 = 7 and -1 - -1 which is 0. So its 7/0 and when the zero is on the bottom its undefined / no slope. When its on top its 0 slope. So yeah 0/7 is 0 slope but 7/0 is undefined. But they messed up on a few questions too.[/quote]

I've had teachers who say a slope of 7/0 is undefined and some who prefer no slope.

the competition coordinators probably prefer undefined

a slope of zero is zero.


Given 4 numbers from 0 to 10, using only add, subtraction, multilplication and division, what is the criterion to get 24?

e.g.  given 1, 2, 3, 4, we get 1*2*3*4=24, but if given 1,1,1,1, it's impossible to get 24.

Well an easy way for the multiplication/division part is to use its prime factorization.
$ 24 \equal{} 2^3\cdot 3$. So you can play with that. ie $ 2*2*2*3,2*2*6,8*3,2*4*3$
May also use division: $ \frac {2^3\cdot2^3\cdot3}{2^3}$, that is $ \frac {8*8*3}{8}$
Addition and multiplication, that I think is just a lot of trial and error.

Original poster,

you posted as if there WERE a source that
claimed that there WERE a way to make
24 using those constraints.

I never heard that there ever existed that
claim.

the criteria seem to be pretty complex
i.e. maybe the only way is to list them all and test

i dont think there is a explicit condition

A solution is given [url=http://reijnhoudt.nl/24game/index.html]here[/url].  Because the search space is finite, it is possible to simply list the valid and invalid quadruples.  However, there does not appear to be a particularly succinct way to describe them.

The only quadruples with four distinct numbers that do not have a valid solution are $ (1, 6, 7, 8)$ and $ (3, 4, 6, 7)$, which I suppose is a nice bit of trivia you can use to confound your friends.

But there are many more which are not 4 distinct integers.

I meant that we expect that as more repeated values are added it becomes harder to produce $ 24$ because our options are severely limited.  In the extreme case, given four of the same number $ a, a, a, a$ our only options are numbers like $ 4a, 2a^2, 2a \plus{} a^2, 2a \plus{} 1$, and so forth.  So in some sense these are the most surprising quadruples without solutions (although $ 7$ is always tough to work with).


[quote="color"]Determine a branch of $z^{i}$ that is defined for $z=i$. Also detrime its value for $z=i$.[/quote]
I am not sure what you mean by "branch" my books does a bad job explaining that.

But for $z\not = 0$ we have,
$z^{i}= e^{i\ln z}$ by definition.
Now the branching of $\ln z$ occurs on $(-\infty,0)$
So thats your branch.

To find $i^{i}$ do the same thing $i^{i}= e^{i\ln i}$.
Now by definition,
$\ln i = \ln |i|+i \arg (i) = \ln (1)+i (\pi/2) = i(\pi/2)$
So,
$i^{i}= e^{i\cdot i(\pi/2)}= e^{-\pi/2}$


if $p$ is a prime and $a$ a real number prove the below sequence has infinetly multiple of $p$:
[$a$],[$2a$],[$3a$],$......$

it is easy :
it is enough to show that for many $k$
:{$k${$a$}}$\le\frac{1}{p}$
then choose $n=pk$


Proove that:

$ \sum _{n\equal{}1}^{\infty } \frac{(\minus{}1)^{n\plus{}1}}{\Gamma (n)}\equal{}\frac{1}{e}$

Since $ \Gamma \left( {n \plus{} 1} \right) \equal{} n!$ then it is enough to show that
\[ \sum\limits_{n \equal{} 0}^{ \plus{} \infty } {\frac{{\left( { \minus{} 1} \right)^n }}
{{n!}}}  \equal{} \frac{1}
{e}.
\]
This comes from the fact that
\[ e^x  \equal{} \sum\limits_{n \equal{} 0}^{ \plus{} \infty } {\frac{{x^n }}
{{n!}}} .
\]


[color=blue]Creo que existía un teorema que afirmaba que para todo [b][i]a[/i][/b], [b][i]b[/i][/b], y [b][i]c[/i][/b] reales, se cumple que el cuadrado de la sumatoria de los términos es mayor o igual que la sumatoria de los cuadrados.

¿Es eso cierto? Voy a revisar en mis apuntes. De todas formas, es mejor si alguien responde con antelación, pues, de ser esto cierto, la solución es inmediata.[/color]

Bueno, una forma de resolverlo, pero un poco larga, es desarrollar completamente ambos lados, y finalmente aplicar Muirhead.
Como una observacion, creo k la igualdad se da si y solo si $a=b=c=1$

[quote="conejita"]Bueno, una forma de resolverlo, pero un poco larga, es desarrollar completamente ambos lados, y finalmente aplicar Muirhead.[/quote]
[color=blue]Perdona mi ignorancia. ¿Qué es Muirhead?[/color]

[quote="aev5peru"]sean $a,b,c,$ pertenecientes a los reales positivos
demostrar que
$((a+b-c)(a-b+c)(-a+b+c))^{2}\geq (a^{2}+b^{2}-c^{2})(a^{2}-b^{2}+c^{2})(-a^{2}+b^{2}+c^{2})$[/quote]

Esta es una forma de acortar un poco la expansion.
Notemos que si $a,b,c$ no son las longitudes de los lados de un triangulo tenemos que:
$a+b<c \implies a^{2}+b^{2}\le (a+b)^{2}< c^{2}$
Como el LHS es mayor que $0$ basta probar la desigualdad para $a,b,c$ las longitudes de los lados de un triangulo.
Entonces existen $x,y,z \in \mathbb{R}^{+}$ que cumplen que:
$a=x+y$
$b=y+z$
$c=z+x$
Reemplazando obtenemos que el problema es equivalente a:
$(2y\cdot 2z \cdot 2x)^{2}\ge \prod_{cyc}2(y^{2}+xy+yz-zx) \iff$
$8x^{2}y^{2}z^{2}\ge \prod_{cyc}(y^{2}+xy+yz-zx) \iff$
$\sum_{sym}x^{2}y^{2}z^{2}+\sum_{sym}x^{3}y^{3}+\sum_{sym}x^{4}y^{2}\ge 2\sum_{sym}x^{3}y^{2}z+\sum_{sym}x^{4}yz$
Y Muirheads no funciona ahi.. entonces no creo que al expandir el original funcione solo con muirheads. La otra desigualdad que conozco para estas cosas simetricas es Schur que tampoco funciona. Si la desigualdad es cierta entonces la igualdad se da si y solo si $a=b=c$
Alguien ayude a terminar.

Aqui hay una explicacion de Muirheads:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=15896[/url]

[quote="R.G.A.M."]Creo que existía un teorema que afirmaba que para todo a, b, y c reales, se cumple que el cuadrado de la sumatoria de los términos es mayor o igual que la sumatoria de los cuadrados.[/quote]

Esto es falso.
Si tomamos $a=b=1$ y $c=-2$
Tenemos $a+b+c =0$ y $a^{2}+b^{2}+c^{2}= 1+1+4= 6$
Lo que si es cierto es que si:
$a,b,c$ son reales no negativos entonces:
$a^{2}+b^{2}+c^{2}= (a+b+c)^{2}-2(ab+bc+ca)\le (a+b+c)^{2}$

Como a lo sumo uno de los factores de la derecha es no positivo y el termino de la izquierda es no negativo, basta suponer que dichos factores son positivos.
(a + b - c)²(a- b + c)² = (a² - (b - c)²)²
                                   = (a^4 - (b² - c²)²) + 2(b - c )²(-a² + b² + c²)
                                   >= (a² + b² - c²)(a² - b² + c²)
Analogamente:
(a + b - c)²(-a+ b + c)² >= (a² + b² - c²)(-a² + b² + c²)
(a - b + c)²(-a+ b + c)² >= (a² - b² + c²)(-a² + b² + c²)
y del producto obtenemos la desigualdad pedida.


Suppose $ a, b, c$ are three lenghts of sides of a triangle.Prove that
$ \frac {a}{\sqrt {a^2 \plus{} 3bc}} \plus{} \frac {b}{\sqrt {b^2 \plus{} 3ca}} \plus{} \frac {c}{\sqrt {c^2 \plus{} 3ab}} \ge \frac {3}{2}$

[quote="carlebee"]Suppose $ a, b, c$ are three lenghts of sides of a triangle.Prove that
$ \frac {a}{\sqrt {a^2 \plus{} 3bc}} \plus{} \frac {b}{\sqrt {b^2 \plus{} 3ca}} \plus{} \frac {a}{\sqrt {c^2 \plus{} 3ab}} \ge \frac {3}{2}$[/quote]
I think it can be $ \frac {a}{\sqrt {a^2 \plus{} 3bc}} \plus{} \frac {b}{\sqrt {b^2 \plus{} 3ca}} \plus{} \frac {c}{\sqrt {c^2 \plus{} 3ab}} \ge \frac {3}{2}$
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=163238


Hello,

Please I think there is a relation between Sine and $ \sqrt{3}$ because I can easily figure this in your solutions but I don't what is it? Could you tell me?

Thanks

Well, $ \sin\frac{\pi}{3}\equal{}\sqrt{3}.$ Not really sure what you're asking though.

[quote="RoFlLoLcOpT"]Well, $ \sin\frac {\pi}{3} \equal{} \sqrt {3}.$ Not really sure what you're asking though.[/quote]

$ \sin\frac {\pi}{3} \equal{} \frac{\sqrt {3}}{2}$ and not $ \sqrt{3}$

There's no relation between $ \sqrt{3}$ and sine as the max value of sine is $ 1$.

OK if we have a $ \triangle ABC$ then can $ \sin \left(\frac {A \plus{} B \plus{} C}{3}\right) \equal{} \frac {\sqrt {3}}{2}$

Yes we can have this because this gives $ A\plus{}B\plus{}C\equal{}180^\circ$


Let $ a$ and $ b$ be two integers. Suppose $ x^2\minus{}x\minus{}1$ is a factor of the polynomial $ ax^5\plus{}bx^4\plus{}1$. Find the value of $ a$.

$ Q(x)\equal{}ax^5 \plus{} bx^4 \plus{} 1$
$ P(x)(ax^3)\equal{}(x^2 \minus{} x \minus{} 1)(ax^3)\equal{}ax^5\minus{}ax^4\minus{}ax^3$
$ Q(x)\minus{}P(x)(ax^3)\equal{}(b\plus{}a)x^4\plus{}ax^3\plus{}1$
$ P(x) (b\plus{}a)x^2\equal{}(x^2 \minus{} x \minus{} 1)(b\plus{}a)x^2\equal{}(b\plus{}a)x^4\minus{}(b\plus{}a)x^3\minus{}(b\plus{}a)x^2$
$ Q(x)\minus{}P(x)(ax^3\plus{}(b\plus{}a)x^2)\equal{}(b\plus{}2a)x^3\plus{}(b\plus{}a)x^2\plus{}1$
$ P(x) (b\plus{}2a)x\equal{}(x^2 \minus{} x \minus{} 1)(b\plus{}2a)x\equal{}(b\plus{}2a)x^3\minus{}(b\plus{}2a)x^2\minus{}(b\plus{}2a)x$
$ Q(x)\minus{}P(x)(ax^3\plus{}(b\plus{}a)x^2\plus{}(b\plus{}2a)x)\equal{}(2b\plus{}3a)x^2\plus{}(b\plus{}2a)x\plus{}1$
$ P(x) (2b\plus{}3a)\equal{}(x^2 \minus{} x \minus{} 1)(2b\plus{}3a)\equal{}(2b\plus{}3a)x^2\minus{}(2b\plus{}3a)x\minus{}(2b\plus{}3a)$
$ Q(x)\minus{}P(x)(ax^3\plus{}(b\plus{}a)x^2\plus{}(b\plus{}2a)x\plus{}(2b\plus{}3a))\equal{}(3b\plus{}5a)x\plus{}(2b\plus{}3a\plus{}1)$

Then:
$ 3b\plus{}5a\equal{}0$, impies that     $ 6b\plus{}10a\equal{}0$
$ 2b\plus{}3a\plus{}1\equal{}0$, impies that $ 6b\plus{}9a\plus{}3\equal{}0$
$ a\equal{}3$ $ b\equal{}\minus{}5$

$ 3x^5 \minus{}5x^4 \plus{} 1\equal{}(x^2 \minus{} x \minus{} 1)(3x^3\minus{}2x^2\plus{}x\minus{}1)$

Is there any solution without using long division?

$ ax^5\plus{}bx^4\plus{}1$
$ \equal{}(ax^3\plus{}(a\plus{}b)x^2\plus{}(2a\plus{}b)x\plus{}3a\plus{}2b)(x^2\minus{}x\minus{}1) \plus{} (5a\plus{}3b)x\plus{}3a\plus{}2b\plus{}1$
So,
$ \left\{\begin{array}{ll}
5a\plus{}3b \equal{} 0 \\
3a\plus{}2b\plus{}1 \equal{} 0
\end{array}\right.
\iff
\left\{\begin{array}{ll}
a \equal{} 3 \\
b \equal{} \minus{}5
\end{array}\right.$

Thus, $ a \equal{} 3$.

$ \phi\equal{}\frac{1\plus{}\sqrt{5}}{2}$

$ P(\phi)\equal{}0 \Rightarrow Q(\phi)\equal{}0$

using that $ \phi^{n\plus{}1}\equal{}\phi^{n}\plus{}\phi^{n\minus{}1}$

$ \phi^4\equal{}\phi^3\plus{}\phi^2\equal{}2\phi^2\plus{}\phi\equal{}3\phi\plus{}2$
$ \phi^5\equal{}5\phi\plus{}3$

$ (5a\plus{}3b) \phi \plus{}3a\plus{}2b\plus{}1\equal{}0$
$ (10a\plus{}6b) (1\plus{}\sqrt{5})\plus{}6a\plus{}4b\plus{}2\equal{}0$
$ (10a\plus{}6b) \sqrt{5}\plus{}(16a\plus{}10b\plus{}2)\equal{}0$
then $ 10a\plus{}6b\equal{}0$ and $ 16a\plus{}10b\plus{}2\equal{}0$
$ a\equal{}3$ $ b\equal{}\minus{}5$

It is not difficult to generalize this approach for $ ax^k\plus{}bx^{k\minus{}1}\plus{}1$

mszew's approach is readily generalizable by proving that $ \phi^k \equal{} F_k \phi \plus{} F_{k\minus{}1}$ (either inductively or directly).


The subset $ S$ of $ \{1,2,...,2006 \}$ is such that if $ x,y \in S$ and $ x>y$, then $ x\minus{}y$ is not in $ S$. What is the maximum number of elements $ S$ can have?

[hide="solution"]
A subset consists of  $ 1003$ odd elements $ \{1,3,5,...,1003\}$ satisfies the condition because $ x \minus{} y$ is always even.
hence it suffices to prove a subset of $ 1004$ elements is impossible.

Suppose there exists such a subset,call it $ S_0 \equal{} \{a_1,a_2,...,a_{1004}\}$ ($ a_1$ is the minimal element.)
then at least one of distinct $ 1003$ numbers $ a_2 \minus{} a_1,a_3 \minus{} a_1,...,a_{1004} \minus{} a_1$ is an element of $ S_0$ because $ S_0$ has $ 1004$ elements,which is contradiction

therefore the maximum is $ 1003$.[/hide]

[quote="dgreenb801"]That subset only has 502 elements.[/quote]

oops,my stupid mistakes :blush:

that should be $ \{1,3,5,...,2005\}$


If $ a,b,c$ are nonnegative real numbers, no two of which are zero, then

$ \frac {a(4b\plus{}4c\minus{}a)}{b^2\plus{}c^2}\plus{}\frac {b(4c\plus{}4a\minus{}b)}{c^2\plus{}a^2}\plus{}\frac {c(4a\plus{}4b\minus{}c)}{a^2\plus{}b^2} \le \frac {21}{2}$.

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, no two of which are zero, then

$ \frac {a(4b \plus{} 4c \minus{} a)}{b^2 \plus{} c^2} \plus{} \frac {b(4c \plus{} 4a \minus{} b)}{c^2 \plus{} a^2} \plus{} \frac {c(4a \plus{} 4b \minus{} c)}{a^2 \plus{} b^2} \le \frac {21}{2}$.[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ $ abc \equal{} w^3$ and $ u^2 \equal{} tv^2.$ Hence, $ t\geq1$ and
$ \sum_{cyc}\frac {a(4b \plus{} 4c \minus{} a)}{b^2 \plus{} c^2}\leq \frac {21}{2}\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(a^6 \minus{} 8a^5b \plus{} 23a^4b^2 \minus{} 8a^3b^3 \minus{} 16a^3b^2c \plus{} 8a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow w^6 \minus{} 2(39u^2 \minus{} 22v^2)uw^3 \plus{} 162u^6 \minus{} 540u^4v^2 \plus{} 657u^2v^4 \minus{} 246v^6\geq0\Leftrightarrow$
$ \Leftrightarrow w^3\leq39u^3 \minus{} 22uv^2 \minus{} \sqrt {1359u^6 \minus{} 1176u^4v^2 \minus{} 173u^2v^4 \plus{} 246v^6}.$
But $ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0$ gives $ w^3\leq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}.$
Thus, it remains to prove that $ 3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\leq$
$ \leq39u^3 \minus{} 22uv^2 \minus{} \sqrt {1359u^6 \minus{} 1176u^4v^2 \minus{} 173u^2v^4 \plus{} 246v^6}.$
But $ 3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\leq$
$ \leq39u^3 \minus{} 22uv^2 \minus{} \sqrt {1359u^6 \minus{} 1176u^4v^2 \minus{} 173u^2v^4 \plus{} 246v^6}\Leftrightarrow$
$ t\left(41t \minus{} 25 \minus{} 2\sqrt {\frac {(t \minus{} 1)^3}{t}}\right)^2\geq1359t^3 \minus{} 1176t^2 \minus{} 173t \plus{} 246\Leftrightarrow$
$ \Leftrightarrow(t \minus{} 1)(163t^2 \minus{} 280t \plus{} 125)\geq2(41t \minus{} 25)\sqrt {t(t \minus{} 1)^3}\Leftrightarrow$
$ \Leftrightarrow(21t \minus{} 25)^2(45t^2 \minus{} 66t \plus{} 25)\geq0.$ Done! :)

Arqady: Your proof is so hard to follow. It is just amazing! :first:
Do you use a calculater or do you do it by hand?
When I try to multiply out, or write $ a,b,c$ as $ u,v,w$ instead it often takes very long time, and I frequently make mistakes.. Do you have any tips or tricks concerning the topic?

(Btw. you made a minor mistake. It should be $ w^3 \le 3uv^2 \minus{} 2u^{\color{red}3} \plus{} 2\sqrt{(u^2\minus{}v^2)^3}$)

[quote="Mathias_DK"]
Do you use a calculater or do you do it by hand?[/quote]
I do it by hand with calculator.
[quote="Mathias_DK"]
Do you have any tips or tricks concerning the topic?[/quote]
It's easy to check that:
$ \sum_{cyc}a^6\equal{}729u^6\minus{}1458u^4v^2\plus{}729u^2v^4\minus{}54v^6\plus{}(162u^3\minus{}108uv^2)w^3\plus{}3w^6,$
$ \sum_{sym}a^5b\equal{}243u^4v^2\minus{}324u^2v^4\plus{}54v^6\minus{}(27u^3\minus{}63uv^2)w^3\minus{}3w^6,$
$ \sum_{sym}a^4b^2\equal{}81u^2v^4\minus{}54v^6\minus{}(54u^3\minus{}36uv^2)w^3\minus{}3w^6,$
$ \sum_{cyc}a^4bc\equal{}(27u^3\minus{}27uv^2)w^3\plus{}3w^3,$
$ \sum_{cyc}a^3b^3\equal{}27v^6\minus{}27uv^2w^3\plus{}3w^6,$
$ \sum_{sym}a^3b^2c\equal{}9uv^2w^3\minus{}3w^6.$
Hence, $ (a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2\equal{}\sum_{sym}(a^4b^2\minus{}a^4bc\minus{}a^3b^3\plus{}2a^3b^2c\minus{}a^2b^2c^2)\equal{}$
$ \equal{}27(\minus{}w^6\plus{}2(3uv^2\minus{}2u^3)w^3\plus{}3u^2v^4\minus{}4v^6)\geq0.$ It gives:
\[ 3uv^2\minus{}2u^3\minus{}2\sqrt{(u^2\minus{}v^2)^3}\leq w^3\leq3uv^2\minus{}2u^3\plus{}2\sqrt{(u^2\minus{}v^2)^3}\]
It's ugly, but very useful.
[quote="Mathias_DK"]
(Btw. you made a minor mistake. It should be $ w^3 \le 3uv^2 \minus{} 2u^{\color{red}3} \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}$)[/quote]
I have fixed it. Thank you!

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, no two of which are zero, then

$ \frac {a(4b \plus{} 4c \minus{} a)}{b^2 \plus{} c^2} \plus{} \frac {b(4c \plus{} 4a \minus{} b)}{c^2 \plus{} a^2} \plus{} \frac {c(4a \plus{} 4b \minus{} c)}{a^2 \plus{} b^2} \le \frac {21}{2}$.[/quote]
The inequality still holds for $ a,b,c$ are real numbers, but it is just a trivial corollary of the case nonnegative real numbers as
\[ \frac{a(4b\plus{}4c\minus{}a)}{b^2\plus{}c^2} \equal{}\frac{4ab\plus{}4ac\minus{}a^2}{b^2\plus{}c^2} \le \frac{4|ab|\plus{}4|ac|\minus{}a^2}{b^2\plus{}c^2}.\]
Here is my proof for the case nonnegative real numbers (by SOS):
As
\[ \sum \frac{a(b\plus{}c)}{b^2\plus{}c^2} \minus{}3 \equal{}\sum \frac{b(a\minus{}b)\minus{}c(c\minus{}a)}{b^2\plus{}c^2} \equal{}\sum (a\minus{}b)\left( \frac{b}{b^2\plus{}c^2} \minus{}\frac{a}{a^2\plus{}c^2}\right) \equal{}\sum \frac{(a\minus{}b)^2(ab\minus{}c^2)}{(a^2\plus{}c^2)(b^2\plus{}c^2)}\]
and
\[ \sum \frac{a^2}{b^2\plus{}c^2} \minus{}\frac{3}{2} \equal{}\frac{1}{2} \sum \frac{(a^2\minus{}b^2)^2}{(a^2\plus{}c^2)(b^2\plus{}c^2)}\]
We can rewrite our inequality as
\[ \sum (a\minus{}b)^2(a^2\plus{}b^2)(a^2\plus{}b^2\plus{}8c^2\minus{}6ab) \ge 0\]
Assume that $ a \ge b \ge c$, then it is easy to verify that
\[ (b\minus{}c)^2(b^2\plus{}c^2\plus{}8a^2\minus{}6bc) \ge 0\]
and we can reduce the inequality into
\[ (a\minus{}c)^2(a^2\plus{}c^2)(a^2\plus{}c^2\plus{}8b^2\minus{}6ca) \plus{}(a\minus{}b)^2(a^2\plus{}b^2)(a^2\plus{}b^2\plus{}8c^2\minus{}6ab) \ge 0\]
which is true as
\[ a\minus{}c \ge a\minus{}b \ge 0, \quad (a\minus{}c)(a^2\plus{}c^2) \ge (a\minus{}b)(a^2\plus{}b^2) \ge 0\]
and
\[ (a^2\plus{}c^2\plus{}8b^2\minus{}6ca)\plus{}(a^2\plus{}b^2\plus{}8c^2\minus{}6ab)\equal{}(a\minus{}3b)^2\plus{}(a\minus{}3c)^2 \ge 0.\]
This completes our proof. :)
Equality holds when $ a\equal{}b\equal{}c$ or $ a\equal{}3b\equal{}3c$ and its cyclic permutations. :)

Nice proof, Can_hung2007.
A related inequality is the following:

If $ a,b,c$ are the side-lenghts of an acute triangle, then

$ \frac {a(2b\plus{}2c\minus{}3a)}{b^2\plus{}c^2\minus{}a^2}\plus{}\frac {b(2c\plus{}2a\minus{}3b)}{c^2\plus{}a^2\minus{}b^2}\plus{}\frac {c(2a\plus{}2b\minus{}3c)}{a^2\plus{}b^2\minus{}c^2} \le 3$.

[quote="Vasc"]Nice proof, Can_hung2007.
A related inequality is the following:

If $ a,b,c$ are the side-lenghts of an acute triangle, then

$ \frac {a(2b \plus{} 2c \minus{} 3a)}{b^2 \plus{} c^2 \minus{} a^2} \plus{} \frac {b(2c \plus{} 2a \minus{} 3b)}{c^2 \plus{} a^2 \minus{} b^2} \plus{} \frac {c(2a \plus{} 2b \minus{} 3c)}{a^2 \plus{} b^2 \minus{} c^2} \le 3$.[/quote]
By AM-GM, we have
\[ \frac{a(2b\plus{}2c\minus{}3a)}{b^2\plus{}c^2\minus{}a^2} \equal{}\frac{2ab\plus{}2ac\minus{}3a^2}{b^2\plus{}c^2\minus{}a^2} \le \frac{(a^2\plus{}b^2)\plus{}(a^2\plus{}c^2)\minus{}3a^2}{b^2\plus{}c^2\minus{}a^2} \equal{}1\]
and applying for the similar terms, we get the result. :)

The following is true

\[ \frac {a(b\plus{}c\plus{}2a)}{b^2\plus{}c^2}\plus{}\frac {b(c\plus{}a\plus{}2b)}{c^2\plus{}a^2}\plus{}\frac {c(a\plus{}b\plus{}2c)}{a^2\plus{}b^2} \ge 6\]

My proposed inequalities are the simplest particular cases of the following inequality

$ \sum\frac {4(1\minus{}k)a(b\plus{}c)\plus{}(\minus{}k^2\plus{}10k\minus{}1)a^2}{ka^2\plus{}b^2\plus{}c^2} \le \frac {3(\minus{}k^2\plus{}2k\plus{}7)}{k\plus{}2}$,

which holds for $ 0\le k<1$, but also for $ \minus{}1\le k<0$ if the denominators are positive. Equality occurs for $ a\equal{}b\equal{}c$, and again for $ \frac a{w}\equal{}b\equal{}c$ etc., where $ w\equal{}\frac {k\plus{}3}{k\plus{}1}$.

From this  it follows the greatest and the smallest values of $ k$ such that

$ \sum\frac a{ka^2\plus{}b^2\plus{}c^2}\le \frac 9{(k\plus{}2)(a\plus{}b\plus{}c)}$;

they are $ k\equal{}7\plus{}2\sqrt{11}$  and $ k\equal{}7\minus{}2\sqrt{11}$.

[quote="encyclopedia"]The following is true
\[ \frac {a(b \plus{} c \plus{} 2a)}{b^2 \plus{} c^2} \plus{} \frac {b(c \plus{} a \plus{} 2b)}{c^2 \plus{} a^2} \plus{} \frac {c(a \plus{} b \plus{} 2c)}{a^2 \plus{} b^2} \ge 6
\]
[/quote]

This inequality is a particular case of the inequality

$ \frac {(1 \minus{} k)a(b \plus{} c) \plus{} 2(2k \plus{} 1)a^2}{ka^2 \plus{} b^2 \plus{} c^2} \ge 6$,

which holds for $ 0\le k < 1$, with equality for $ a \equal{} b \equal{} c$, and again for $ a \equal{} 0$ and $ b \equal{} c$ etc.

Moreover, it is a particular case of the inequality

$ \frac {(2k \plus{} 1)a(b \plus{} c) \plus{} 2(1 \minus{} k)a^2}{b^2 \plus{} kbc \plus{} c^2} \ge 6$,

which holds for $ \frac {1\minus{}\sqrt{13}}{2}\le k \le 1$, with equality for $ a \equal{} b \equal{} c$, and again for $ a \equal{} 0$ and $ b \equal{} c$ etc.

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, no two of which are zero, then

$ \frac {a(4b \plus{} 4c \minus{} a)}{b^2 \plus{} c^2} \plus{} \frac {b(4c \plus{} 4a \minus{} b)}{c^2 \plus{} a^2} \plus{} \frac {c(4a \plus{} 4b \minus{} c)}{a^2 \plus{} b^2} \le \frac {21}{2}$.[/quote]

[quote="encyclopedia"]The following is true
\[ \frac {a(b \plus{} c \plus{} 2a)}{b^2 \plus{} c^2} \plus{} \frac {b(c \plus{} a \plus{} 2b)}{c^2 \plus{} a^2} \plus{} \frac {c(a \plus{} b \plus{} 2c)}{a^2 \plus{} b^2} \ge 6
\]
[/quote]

I see it as the particular of the inequality
\[ \frac {a(k(b \plus{} c) \minus{} a)}{b^2 \plus{} c^2} \plus{} \frac {b(k(c \plus{} a) \minus{} b)}{c^2 \plus{} a^2} \plus{} \frac {c(k(a \plus{} b) \minus{} c)}{a^2 \plus{} b^2} \le \frac {3}{2}(2k \minus{} 1)
\]
which holds for $ k\ge \minus{} \frac {1}{2}$ when $ k \equal{} \minus{} \frac {1}{2}$ we get the above, $ k \equal{} 4$ we get yours.

[quote="encyclopedia"] I see it as the particular of the inequality
\[ \frac {a(k(b \plus{} c) \minus{} a)}{b^2 \plus{} c^2} \plus{} \frac {b(k(c \plus{} a) \minus{} b)}{c^2 \plus{} a^2} \plus{} \frac {c(k(a \plus{} b) \minus{} c)}{a^2 \plus{} b^2} \le \frac {3}{2}(2k \minus{} 1)
\]
which holds for $ k\ge \minus{} \frac {1}{2}$ when $ k \equal{} \minus{} \frac {1}{2}$ we get the above, $ k \equal{} 4$ we get yours.[/quote]
Indeed, this inequality holds only for $ \minus{}\frac 1{2}\le k \le 4$.

And the inequality

\[ \frac {a(4(b \plus{} c) \plus{} ka)}{b^2 \plus{} c^2} \plus{} \frac {b(4(c \plus{} a) \plus{} kb)}{c^2 \plus{} a^2} \plus{} \frac {c(4(a \plus{} b) \plus{} kc)}{a^2 \plus{} b^2} \le \frac {3}{2}(k \plus{} 8)
\]

Holds for $ k\le \minus{}1$

Is it the best constant for ?

\[ \frac {a(p(b \plus{} c) \plus{} qa)}{b^2 \plus{} c^2} \plus{} \frac {b(p(c \plus{} a) \plus{} qb)}{c^2 \plus{} a^2} \plus{} \frac {c(p(a \plus{} b) \plus{} qc)}{a^2 \plus{} b^2} \le \frac {3}{2}(2p \plus{} q)
\]

The following problem is also interesting.

Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Find the minimum and the maximum value of $ k$ such that

$ \sum\frac {ka(b\plus{}c)\minus{}a^2}{b^2\minus{}bc\plus{}c^2} \le 3(2k\minus{}1)$.

[hide]$ \frac 1{4}\le k\le \frac {5\plus{}2\sqrt 5}{2}$[/hide]

[quote="Vasc"]The following problem is also interesting.

Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Find the minimum and the maximum value of $ k$ such that

$ \sum\frac {ka(b \plus{} c) \minus{} a^2}{b^2 \minus{} bc \plus{} c^2} \le 3(2k \minus{} 1)$.

[hide]$ \frac 1{4}\le k\le \frac {5 \plus{} 2\sqrt 5}{2}$[/hide][/quote]

And the inequality

\[ \sum\frac {a(b \plus{} c) \plus{} ka^2}{b^2 \plus{} kbc \plus{} c^2} \le 3\]

holds for all $ k\le 1\minus{}\sqrt{5\minus{}2\sqrt{3}}$


How many zeros does the number $ 2009!$ end with?

[hide="Solution"]There are $ 401$ appearances of $ 5^1$, $ 80$ appearances of $ 5^2$, $ 16$ appearances of $ 5^3$, and $ 3$ appearances of $ 5^4$. $ 401\plus{}80\plus{}16\plus{}3\equal{}\boxed{500}$.[/hide]

[hide="Nifty formula"]
If $ p$ is a prime and $ p^k|n!$, then
\[ k\leq\sum_{i \equal{} 0}^{\infty}\left\lfloor\frac n{p^i}\right\rfloor.\]
[/hide]

[hide="Solution"]
The highest power of $ 5$ that divides $ 2009!$ is $ 401 \plus{} 80 \plus{} 16 \plus{} 3 \equal{} \boxed{500}$.

This is the answer because there are clearly more $ 2$s than $ 5$s.
[/hide]

Edit: Bleh, beaten

Mr. Rusczyk had an Alcumus video about this, and the easy way to do it is simple division.  Divide 2009/5, = 401.8  Take the integer part, 401.  401/5= 80 (taking the integer part.) 80/5=16. 16/5=3 3/5=0.

So, $ 401\plus{}80\plus{}16\plus{}3(\plus{}0)\equal{}\fbox{500}.$

Easy as that :).


MATHCOUNTS
inequalities
algebra
polynomial
absolute value

In the MATHCOUNTS forums, one method of solving inequalities such as $ |2x|\leq |x\minus{}3|$ was to square both sides. Although this method does provide the correct answer for this (and many similar inequalities), I was wondering if there was a proof that this method works or doesn't work.

Example:

$ |2x| \leq |x\minus{}3|$
$ 4x^2 \leq x^2 \minus{} 6x \plus{} 9$
$ 3x^2 \plus{} 6x \minus{} 9 \leq 0$
$ 3(x^2 \plus{} 2x \minus{} 3) \leq 0$
$ 3(x\plus{}3)(x\minus{}1) \leq 0$

Now, for the product to be less than or equal to zero, one of the terms must be zero or one must be positive while the other is negative. If $ x$ is less than $ \minus{}3$, then both terms will be negative. So we want $ x \geq \minus{}3$ (because when $ x\equal{}\minus{}3$ one of the terms is $ 0$). Similarily, if $ x$ is greater than $ 1$, then both terms will be positive. That means $ x \leq 1$. Therefore, the restrictions on $ x$ are $ \minus{}3 \leq x \leq 1$.

Since there is no limit to what an absolute value sign can contain, we'll have to use the standard notation for a polynomial, ($ a_nx^n \plus{} a_{n\minus{}1}x^{n\minus{}1} \plus{} \cdots \plus{} a_2x^2 \plus{} a_1x \plus{} a_0$), and squaring, combining like terms, and also finding the solution through the usual way.

This is often very tedious though, which reminds me similarly of [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=225423]this problem[/url].


number theory
relatively prime
algebra
polynomial
Rational Root Theorem

Find a rational number m such that ecuation $ 2*x^3\minus{}x^2\plus{}mx\minus{}1\equal{}0$ has a rational root

For every rational x there is a rational m that meets the given requirement...

by rational root theorem the only possible rational roots are $ \pm 1$ and $ \pm 1/2$.  From there it is casework: plug in each possibility for $ x$, and solve for $ m$.

ACTUALLY, what I just wrote isn't quite right.  rational root theorem assume the coefficients are integers.  so if $ m\equal{}r/s$ (r,s relatively prime integers) then we would multiply the equation through by $ s$ and then $ m$ would be $ \pm$ a factor of $ s$ over a factor of $ 2s$.  But if we constrain ourselves to integer $ m$'s then rational root theorem will give $ 4$ answers.


Does anyone have a copy of last year's F=Ma exam?

You could've done a simple google search, but anyway, here you go.

I didn't actually download the file, but based on the name, it looks like the 2008 one rather than the 2009 one.  I could not find the 2009 one either until I used the WebAssign which has all the past exams for a couple of years and even a practice one.


for participation in the amc/aime/usamo etc. do you have to be a
us citizen?
permanent resident?
or just legal resident?

This is the first question in their FAQ:
[url]http://www.unl.edu/amc/f-miscellaneous/faq.shtml[/url]


geometry
algebra
polynomial
AMC 10
arithmetic sequence
AMC

Since the AMCs are coming up, I have made this thread.  :) There are a lot of others going on. The problems can be AMC 10 or 12 level.

$Problem$ $1:$ How many 3 letter words with the middle letter a vowel are there?

[hide="Problem 1"]
$26\cdot 5\cdot 26=\boxed{3380}$[/hide]

[size=134][b]Problem 2.[/b][/size]

Find the units digit in the decimal expansion of

\[(15+\sqrt{220})^{19}+(15+\sqrt{220})^{82}\]

:|

[hide="Problem 2 outline"]
Let $x=a+\sqrt{b}$ and $y=a-\sqrt{b}$.

Then, $x^{n}+y^{n}=2\left(a^{n}+\dbinom{n}{2}a^{n-2}b+\dbinom{n}{4}a^{n-4}b^{2}+\cdots\right)$.

$5|a$, $5|b\implies 10|(x^{n}+y^{n})$.

$x^{19}+x^{82}=10k-\left(y^{19}+y^{82}\right)$.

Since $\left(y^{19}+y^{82}\right)<1$, the units digit will be $\boxed{9}$.[/hide]

[size=134][b]Problem 3.[/b][/size]

If $x$, $y$, and $z$ are positive numbers satisfying
\[x+\frac{1}{y}=4 \]

\[y+\frac{1}{z}=1 \]

\[z+\frac{1}{x}=\frac{7}{3}\]
then what is the value of $xyz$?

[hide="Problem 3"]
after isolating all three equations with x,y,and z on each side, you can try to attack one of the three variables

pour moi, I chose x and, after extreme substitutions, I got

$x=3/2$

now after another round of extreme substitutions, you get

$z=5/3$

$y=2/5$

they all cancel out nicely when you multiply them.....

so it's $1$

I'm pretty sure I took the hard way of solving it :oops:
but if there is much more elegant way of solving it, please post post and post!!!

[/hide]

[b][size=150]PROBLEM 4[/size][/b]

The student lockers at Olympic High are numbered consecutively beginning with locker number 1. The plastic digits used to number the lockers cost two cents apiece. Thus, it costs two cents to label locker number 9 and four cents to label locker number 10. If it costs 137.94 dollars to label all the lockers, how many lockers are there at the school?

I stole this problem from my school math club's forum :D

[hide="Elegant solution to Problem 3"]
Adding the equations we get,

$x+\frac{1}{y}+y+\frac{1}{z}+z+\frac{1}{x}=\frac{22}{3}$.

Multiplying the equations we get,

$\left(x+\frac{1}{y}\right)\left(y+\frac{1}{z}\right)\left(z+\frac{1}{x}\right)=\frac{28}{3}$.

Expanding,

$x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+xyz+\frac{1}{xyz}=\frac{28}{3}$.

$xyz+\frac{1}{xyz}=2\implies (xyz-1)^{2}=0\implies \boxed{xyz=1}$.
[/hide]
[hide="Problem 4"]
$\frac{137.94}{0.02}=6897$, so 6897 digits must have been used.

$1\rightarrow 9\implies$ 9 digits.
$10\rightarrow 99\implies$ 180 digits.
$100\rightarrow 999\implies$ 2700 digits.
$1000\rightarrow 2001\implies$ 4008 digits.

So there are $\boxed{2001}$ lockers. [/hide]

[size=134][b]Problem 5.[/b][/size]

If $8$, $x+9$, and $3x+7$ are 3 consecutive terms (but not necessarily in the order given) of an arithmetic sequence, find the sum of all distinct possibilities for the value of $x$.

[hide="Solution to Problem 5"]

Their are 6 possible orders for the terms:
8, x+9, 3x+7
8, 3x+7, x+9
x+9, 8, 3x+7
x+9, 3x+7, 8
3x+7, x+9, 8
3x+7, 8, x+9

For each of these, the second term minus the first term must equal the third term minus the second term. Using this method shows that $x=3,\frac{3}{5},0$. Therefore, their sum is $\frac{18}{5}$.

[/hide]

Someone else can post the next problem.

For solution five, I did it the same way, but I'm wondering if there is a less "messy" solution.

[quote="JavaMan"][quote="rnwang2"]:|

[hide="Problem 2 outline"]
Let $x=a+\sqrt{b}$ and $y=a-\sqrt{b}$.

Then, $x^{n}+y^{n}=2\left(a^{n}+\dbinom{n}{2}a^{n-2}b+\dbinom{n}{4}a^{n-4}b^{2}+\cdots\right)$.

$5|a$, $5|b\implies 10|(x^{n}+y^{n})$.

$x^{19}+y^{82}=10k-\left(y^{19}+y^{82}\right)$.

Since $\left(y^{19}+y^{82}\right)<1$, the units digit will be $\boxed{9}$.[/hide]
[/quote]

I don't quite get what you did there, could you explain a little more? Why is $y = a-\sqrt{b}$?[/quote]
It is defined that way so that when you add $x^{n}$ and $y^{n}$ there will not be any radicals (Binomial Theorem).

[quote="JavaMan"][quote="rnwang2"]:|

[hide="Problem 2 outline"]
Let $x=a+\sqrt{b}$ and $y=a-\sqrt{b}$.

Then, $x^{n}+y^{n}=2\left(a^{n}+\dbinom{n}{2}a^{n-2}b+\dbinom{n}{4}a^{n-4}b^{2}+\cdots\right)$.

$5|a$, $5|b\implies 10|(x^{n}+y^{n})$.

$x^{19}+y^{82}=10k-\left(y^{19}+y^{82}\right)$.

Since $\left(y^{19}+y^{82}\right)<1$, the units digit will be $\boxed{9}$.[/hide]
[/quote]

I don't quite get what you did there, could you explain a little more? Why is $y = a-\sqrt{b}$?[/quote]

We are just letting $\sqrt{b}=-120$, so everything conveniently cancels.  This problem seemed un-AMC-ish, but I can't talk after my Mock AMC  :wink:

Sorry, just ignore me, I overlooked something....

[quote="JavaMan"][quote="rnwang2"]
$x^{19}+y^{82}=10k-\left(y^{19}+y^{82}\right)$.
[/quote]

Did you mean $x^{19}+x^{82}=10k-\left(y^{19}+y^{82}\right)$? (Which was another source of my concern).[/quote]
Ah! Yes, I did.  Sorry. :wallbash:

Anyway...

[size=134][b]Problem 6.[/b][/size]

Let $M$ and $N$ be points on the sides $AB$ and $BC$ of the square $ABCD$. Let $P=AN\cap DM$, $Q=AN\cap CM$, and $R=CM\cap DN$. Given $[AMP]=20$ and $[BMQN]=30$, and $[CNR]=40$, what is the area of $DPQR$?

Forget about it.

It is pretty easy to see that

$[DPQR]=[AMP]+[BMQN]+[CNR]$.

Someone post another problem.  :)

Umm, due to a serious flaw in rnwang's solutions, number 2 is still unsolved
 :wink:
[quote="JavaMan"][quote="rnwang2"]

[size=134][b]Problem 2.[/b][/size]

Find the units digit in the decimal expansion of
\[(15+\sqrt{220})^{19}+(15+\sqrt{220})^{82}\]
[/quote]

You guys are confusing me :P

If $\sqrt{b}=-120$ , then what's $a$?

Is x/y supposed to be $15+\sqrt{220}$?[/quote]

[quote="rnwang2"][quote="JavaMan"]Still though, can someone define $a$ and $\sqrt{b}$ in terms of numbers?  :what?:[/quote]
Uhh...

$a=15$, $b=220$.[/quote]

Unless there was a typo in the problem statement, the problem was done to solve instead: \[(15+\sqrt{220})^{19}+(15-\sqrt{220})^{82}\]
notice the significant difference... in fact, I believe its a typo because when I made a program to calculate it on my calculator I got at number like $...16.65194725...$
but I could be mistaken

The solution calculates the right thing. Are you sure your calculator program isn't busted?

[quote="mysmartmouth"]USC invited each South Carolina high school to send up to 39 students to watch a football game. A section which has 199 seats in each row is reserved for those students. What is the least number of rows needed to guarantee that if 2006 students show up, then all students from the same high school can be seated in the same row?

A) 11
B) 12
C) 13
D) 14
E) 15[/quote]
[hide]
If we're assuming that each high school that shows up consists of around 39 students, we can fit 5 high schools comfortably, with 4 seats extra, in the same row.
$5*39=195$
There are $2006+39=2045$ total students.
$2045/195=10.xxx$  Because we cant have a fraction of a row, we round up to the next integer, which is 11 so the answer is A?[/hide]

According to my computer, the last 2 digits are 52. And 2 makes more sense for a #2 anyway...

[quote="chess64"]Well the answer turns out to be the same... besides, notice that you solved the wrong problem, as me@home pointed out above.[/quote]
If you follow the reasoning, you see that

$x^{19}+x^{82}=10k-\left(y^{19}+y^{82}\right)$.

$x^{19}+x^{82}$ is the desired expression in the original problem.

[quote="chess64"][quote="paladin8"][quote="chess64"]The answer to that problem is 0.[/quote]

I think rnwang2's answer and solution are fine... note that my calculator verifies that

$(15+\sqrt{220})^{n}$ for $n = 1, 2, 3$

have units digit $9$, so yeah.[/quote]

$\left\lfloor(15+\sqrt{220})^{19}\right\rfloor=10449002663787619784174075904$

$\left\lfloor(15+\sqrt{220})^{82}\right\rfloor=84028818356889573885330541997707872070\\181977404881649234386648064111867552216374\\82815685456035656151268654283095116087296$[/quote]

RAndom, but coudn't the remaining digits not included in the floor be important in the exta digit for the one's.

$15^{19}+(19)(15^{18})(\sqrt{220})+(\binom{19}{2})(15^{17})(\sqrt{220}^{2})$

everything after and including the third term will have units digit zero

the first term obviously has units digit 5, and the second will have units digit 4 because its slightly less than what you would get if you assumed $\sqrt{220}$ (5).  So when n is large the units digit will be 9, so the answer should be 8.

[quote="chess64"][quote="paladin8"][quote="chess64"]The answer to that problem is 0.[/quote]

I think rnwang2's answer and solution are fine... note that my calculator verifies that

$(15+\sqrt{220})^{n}$ for $n = 1, 2, 3$

have units digit $9$, so yeah.[/quote]

$\left\lfloor(15+\sqrt{220})^{19}\right\rfloor=10449002663787619784174075904$

$\left\lfloor(15+\sqrt{220})^{82}\right\rfloor=84028818356889573885330541997707872070\\181977404881649234386648064111867552216374\\82815685456035656151268654283095116087296$[/quote]

What precision is this? On an online calculator I used, it showed up like I expected, ending in $\ldots 9.999 \ldots$.

Has any resolve been reached? So far I've seen like 4 different answers (mine, the popular 9, 8, and even 2)

Does anybody happen to have the answer sheet or anything? The reason for my skepticism was that I was so sure I had seen the problem before but as more of a conjugate style, like I wrote, with a $-$ in the second term...

This was an example in the AMC 12 prob series, perhaps someone taking that could clarify.

[quote="paladin8"][/quote][quote="chess64"][quote="paladin8"][quote="chess64"]The answer to that problem is 0.[/quote]

I think rnwang2's answer and solution are fine... note that my calculator verifies that

$(15+\sqrt{220})^{n}$ for $n = 1, 2, 3$

have units digit $9$, so yeah.[/quote]

$\left\lfloor(15+\sqrt{220})^{19}\right\rfloor=10449002663787619784174075904$

$\left\lfloor(15+\sqrt{220})^{82}\right\rfloor=84028818356889573885330541997707872070\\181977404881649234386648064111867552216374\\82815685456035656151268654283095116087296$[/quote][quote="paladin8"]

What precision is this? On an online calculator I used, it showed up like I expected, ending in $\ldots 9.999 \ldots$.[/quote]

OK, maybe it is 9; my 0 probably came from rounding error.

So for problem 2, the idea given at the start can be made to work as follows.
Consider ( 15+$\sqrt{220}$ )^19 + (15 - $\sqrt{220}$ )^19. Expanding we get
2*(15^19 + ${19}\choose{2}$ 15^17*220+...)
which clearly has units digit 0.
We can do the same thing for ( 15+$\sqrt{220}$ )^82, and we see that the units digit of
( 15+$\sqrt{220}$ )^82 + (15 - $\sqrt{220}$ )^82
is also 0.
Now note that (15 - $\sqrt{220}$ ) is less than 1. (Its actually roughly 1/6th because it is $\frac{5}{15+\sqrt{220}}$ which is roughly $\frac{5}{15+15}$).
So the end of ( 15+$\sqrt{220}$ )^19 looks like
... 9.999999...9999xyz
and the end of ( 15+$\sqrt{220}$ )^82 looks like
... 9.9999...9999wabc (with very many more 9s before the 1st non 9 digit)
So we now have to do this addition.
9.99999xyz...
9.99999999...
(the number of 9s is for illustration only. There are atleast 9 9s though)
Note that the carry during the addition can at most be 1 (worst case is all 9s after the decimal point). So the carry from the addition of digit x and 9 is 0 or 1. But whatever this carry the addition from the last few 9s before the decimal point results in a carry of 1, and so 1 is carried over into the units digit.
So the units digit is 9+9+1 (mod 10) which is 9.

I hope I haven't made a mistake, but given the number of math geniuses on this site I'm sure I'll be found out reasonably quickly!
Also, in general it is a good idea to be very careful when using computers on floating point values, it is too easy to lose precision even if you're using something like Mathematica but particularly when you're writing your own C program to do the numbers.

The answer above is basically wang's answer, just with a bit more explanation. To show the units digit is 9, its more elegant to go his way.
Consider x^19+y^19 + x^82+y^82. You see that the units digit of each of (x^19+y^19) and (x^82+y^82) is 0. So
x^19+y^19 + x^82+y^82= 10k
so,
x^19+x^82=10*k-(y^19+y^82) and the result follows as y^19+y^82 <1.

Me@home was wrong, he did indeed solve the correct problem.

In fact, that's the beauty of it...he used the conjugates of the numbers involved to end up churning out the answer. What he essentially argued is that because the number given is equal to an integer multiple of 10 minus a very small positive number, its units digit must equal 9. This is because the rational terms of the number given clearly add up to a multiple of 10, and by adding the conjugate, we cancel the irrational terms out, thus giving a multiple of 10. Really nice solution, I didn't even think of that!  :lol:

Looks like we need a new problem now. :laugh:

[b][size=134]Problem 8.[/size][/b]

Let a sequence $\{u_{n}\}$ be defined by $u_{1}=7$ and $u_{n+1}-u_{n}=n+(n-2)^{2}$ for $n=1,2,3,4,\ldots$ For this sequence, $u_{n}$ can be expressed as a completed simplified polynomial such that the coefficient of $n^{2}$ is -2. Find the coefficient of $n$ in this polynomial.

just use the stacks method (degrees of differences) and set up a system


$n \geq 5$ football teams participate in a round-robin tournament.  For every game played, the winner receives 3 points, the loser receives 0 points, and in the event of a draw, both teams receive 1 point.  The third-from-bottom team has fewer points than all the teams ranked before it, and more points than the last 2 teams; it won more games than all the teams before it, but fewer games than the 2 teams behind it.  Find the smallest possible $n$.

A Round robin tournament means that every team plays every other team exactly once.  In particular, in an $ n$ team tournament each team will only play $ n\minus{}1$ games.

For $ n\equal{}13$:

Posit       Win Draw Loss Points
1/10.           2       9        1       15
11.              3        4        5       14
12.             4       0        8       12
13.             4       0        8       12

Cross table too much work to write down.

Dear Mathlinkers,My solution in mind is really nasty.Could some one give a brief and clean solution.For the answer I think is $13$.


Problem:

We have two horizontal metal bars fixed parallel on a table. They are perfect conductors, so we can assume their resistance is 0. A third metal bar is lying perpendicularly on them, and is free to move. The three bars form an image that's similar to the letter H. The section of the third bar is $A$, its resistivity is $k$, its mass is $m$. The friction constant between the third bar and the other bars is $\mu$. Now we link a generator (voltage=$V$) to the two fixed bars. We want to make the third bar move, but it must keep on being perpendicular to the fixed bars, therefore we apply a magnetic field $B$.

Find the minimum magnetic field $B_{min}$ necessary to make the third bar move. (Warning: you also have to find its orientation)

I apologize for my horrible english; if you have doubts about the text, please ask me.

If the length of the third bar is $\ell$, its resistance is $R = \frac{k\ell}{A}$; thus, the current through the whole system is $I = \frac{V}{R_{\mathrm{total}}} = \frac{V}{R} = \frac{VA}{k\ell}$. Now, we know that the force induced by a magnetic field will be $I \overline{\ell} \times \overline{B}$, so $F_{\mathrm{magnetic}} = \frac{VA}{k\ell} \ell B \sin \theta$; if $B$ acts at an angle $\psi$ from the vertical, we see that the induced force will act at an angle $\psi$ from the horizontal. Now, the forces acting on the third bar in the vertical direction are its weight $mg$ down, the vertical component $F_{\mathrm{magnetic}} \sin \psi$ of the magnetic force, and the normal force from each of the two bars $2N$, so $N = \frac{mg - F_{\mathrm{magnetic}} \sin \psi}{2}$ and thus the static frictional force from each bar should be $f \le \mu_{\mathrm{s}} N$, making the total frictional force $F_{\mathrm{frictional}} = 2f \le \mu_{\mathrm{s}} (mg - F_{\mathrm{magnetic}} \sin \psi)$. In order for the bar to move, the horizontal component of the magnetic force must overcome this static frictional force, so if we let $\phi = \sin^{-1} \frac{1}{\sqrt{\mu_{\mathrm{s}}^2 + 1}}$, we note that $F_{\mathrm{magnetic}} \cos \psi \ge \mu_{\mathrm{s}} (mg - F_{\mathrm{magnetic}} \sin \psi)$ $\Leftrightarrow$ $mg \mu_{\mathrm{s}} \le F_{\mathrm{magnetic}} (\cos \psi + \mu_{\mathrm{s}} \sin \psi)$ $=$ $\frac{VAB \sin \theta}{k} \sqrt{\mu_{\mathrm{s}}^2 + 1} \sin(\psi + \phi) \le \frac{VAB \sqrt{\mu_{\mathrm{s}}^2 + 1}}{k}$ $\Leftrightarrow$ $B \ge \frac{kmg\mu_{\mathrm{s}}}{VA\sqrt{\mu_{\mathrm{s}}^2 + 1}}$; this is attainable if $\theta = \frac{\pi}{2}$ and $\psi = \frac{\pi}{2} - \phi$, i.e. if the magnetic field is directed at an angle of $\phi$ from the horizontal but perpendicular to the third bar.

Perfect. Using the derivative of $B= f(\alpha)$, I found a minimum for $\alpha = tan^{-1} \mu$ that's the same.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Hi, its me again, and i've got another problem

B=
[ t   2t  -t   ]
[ t   1   -2  ]
[ 0  0   t+6]

a) for which values of t is matrix b invertible
b) for which values of t does the matrix B have rank 2

Many thanks for the help

For a) : $B~ \text{is inversible} \Leftrightarrow \det{B} \neq 0$

and for b/, there are at most 3 interesting cases. So, just compute the characteristic polynomial ..


function
probability
integration
probability and stats

Let $ \varphi$ characteristic function of some probability distribution. Prove that \[ \xi (t)\equal{}\int_0^{\infty}\varphi(ut)e^{\minus{}u}du\] is also a characteristic function of some probability distribution.

Two approaches:
1. Use Bochner theorem;
2. Suppose that for every $ u \ f(u,t)$ is a characterestic function(as a function of $ t$) of some distribution $ F_u(x)$, then look at the c.f. of the distribution $ \int F_u(x)dG(u)$, where $ G$ is an arbitrary distribution function.


An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is 2, find the area of the hexagon.

[hide]I used an example to figure the ratio between the area of an equilateral triangle and a regular hexagon. Given a triangle and hexagon with perimeters of $3$, the areas are $\frac{\sqrt{3}}{4}$ and $\frac{3\sqrt{3}}{8}$ respectively. Then the ratio of the area of the hexagon to the triangle is $\frac{3}{2}$. Then since the area of the given equilateral triangle is $2$, the area of the hexagon must be $3$.[/hide]

[hide]
$\frac{s^2\sqrt{3}}{4}=2$
$s^2\sqrt{3}=8$
$s^2=\frac{8\sqrt{3}}{3}$
$s=\sqrt{\frac{8\sqrt{3}}{3}}=\frac{\sqrt{8\sqrt{3}}}{\sqrt{3}}=\frac{2\sqrt{6\sqrt{3}}}{3}$
Since there are three sides to a triangle, the perimeter is $2\sqrt{6\sqrt{3}}$.
A hexagon has six sides and since the hexagon is regular each side is a side to an equelateral triangle: $6*\frac{s^2\sqrt{3}}{4}$ and $s=\frac{2\sqrt{6\sqrt{3}}}{6}$ (from before)
$6*\frac{(\frac{2\sqrt{6\sqrt{3}}}{6})^2\sqrt{3}}{4}=\frac{\frac{2\sqrt{3}}{3}*3\sqrt{3}}{2}=\sqrt{3}\sqrt{3}=3$

[/hide]

[quote="Tecnogram888"][quote="amirhtlusa"]in general, the area of a equilateral triangle with side $2a$ is four times the area of a equilateral triangle with side $a$, also the hexagon is made of 6 equilateral triangles with side $a$, there fore the area is: $6*0.5=3$[/quote]
i dont understand how u got the 6*0.5=3.[/quote]
the hexagon is made of $6$ equilateral triangles with side length of $a$. the area of each triangle is $0.5$ so the area of hexagon is $3$


[quote="Rust"]$a_{i}=\sum_{j=1}^{i}j(i-j)=\frac{(i+1)i(i-1)}{6}$[quote]

Can you explain this,I cant see it right away...

Btw. thanks for a quick answer.


guys plz post in here the dates you are gonna write your bitsat lets see what day is so crowded up with mathlinks citizens!!!!

lemme start first May16th morning!!!! :D

:) me- the 14th of may in the evening :)

this has already been discussed before, see there :)


Intr-un cerc se duc doua coarde care se intalnesc, $\ AB,CD$.Pe segmentul $\ AB$ s-a luat punctul $\ M$ astfel incat $\ AM=AC$ iar pe segmentul $\ CD$, punctul $\ N$ astfel incat $\ DN=DB$.Sa se demonstreze ca, daca punctele $\ M,N$ nu coincid, atunci $\ MN$ este paralela cu $\ AD$.

Esential este sa arati ca <SMN=<SAD unde S este AB intersectat cu CD
Se observa cu usurinta ca  triunghiul ACM asemenea cu DBN deoarece sunt isoscele si <CAM=<BDN
ACBD inscriptibil de unde <ACD=<ABD dar <ACM=<NBD de unde ne rezulta ca <MCN=MBN deci MNBC inscriptibil
Avem <BCN=<SMN dar <BCN=<SAD de unde concluzia problemei

In sfarsit postezi si tu pe sectiunea romaneasca, Bianca.Solutia e corecta.In mare masura asa am solutionat-o si eu. :)


algebra
polynomial
function
calculus
calculus computations

prove the following:
A polynomial is any function p: R ---------> R of the form p(x) = anx^n + ... + a2x^2 + a0 for some n in the natural numbers including zero where an is in the real numberes for each n.  (the letter or number next to "a" is sub)
Prove that polynomials are continous.

Show that the function $ x$ and the constant function $ a_i$ are continuous.
Then proceed by showing that the sum and the product of continuous functions is continuous.


Does there exist a function $f: R^{+}\to R^{+}$ such that $f(x+y)\geq y(f(x))^2\forall x,y\in R^{+}$?

Does $f(x)=e^{e^x}$ work?

(I double-checked it, but I'm still not sure that it works.)

Okay, here's the proof:
The function will satisfy the requirements if $\frac{f(x+y)}{y*f(x)^2}$ be at least 1 for all $x>0$ and $y>0$.

If we let $f(x)=e^{e^x}$, then $\frac{f(x+y)}{y*f(x)^2}=\frac{e^{e^{x+y}}}{y*e^{e^x}}=\frac{e^{e^xe^y-2e^x}}{y}=e^{e^x(e^y-2)-ln(y)}$

----If you accept that the above is at least 1, then you don't need to read any more.----

The above is at least 1 if $e^x(e^y-2)-ln(y)$ is nonnegative.

For the defined $x$ and $y$, the minimum value of $e^x(e^y-2)-ln(y)$ occurs when $x=0.$So, our solution for $f$ is valid if $e^y-2-ln(y)$ is nonegative.

The line tangent to $e^X$ at $X=1$ is $Y=e*(x-1)+e=e*x$.  Since $e^X$ is concave up (since $e^X$ is always postive), the tangent line is below $e^X$.
The line tangent to $2+ln(X)$ at $X=1$ is $Y=1*(X-1)+2=X+1$.  Since $ln(X)$ is concave up (since $-1/X^2$ is always negative), the tangent line is above $2+ln(X)$.
Since these two tangents meet only at $\frac{1}{1-e}$ which is negative, $e^X$ is greater than $2+ln(X)$ for all positive $X$.

Therefore, $e^y-2-ln(y)$ is nonegative for all postive y, proving our proposed solution correct.


function
complex numbers
algebra unsolved
algebra

Is there exist continous function f: R -> R, f is not constant such that
f(x) + f(2x) + f(3x) = 0
for all x from R.

<i>Source: Newman seminar </i>

I dont have the sln.

Namdung

I have seen this movie, there exist Namdung

f(x)=Real part of |x|^z where z \in C s.t. 1+2^z+3^z=0

I don't  have mapple to get the value of z

A nicer one will be f:R->R, C s.t.
for any x \in R
f(x)+f(2x)+...+f(2004x)=0

What is the Newman seminar ?

No, there is a problem book from D.J.Newman called 'A problem seminar' (Springer), and Namdung's problem may be found in it.

Moubi, you may find this book at the library (chevaleret).

Pierre.

Ah, I should have guessed it could be a name. Would be sloppy to give the name of a seminar without any further description.

[quote="Moubinool"]f(x)=Real part of |x|^z where z \in C s.t. 1+2^z+3^z=0[/quote]
Incredible!

[quote]A nicer one will be f:R->R, C� s.t.
for any x \in R
f(x)+f(2x)+...+f(2004x)=0[/quote]
What is the difference?


induction
number theory proposed
number theory

Let $ k>1$ be an integer. Prove that there exists infinitely many natural numbers such as $ n$ such that: \[ n|1^n\plus{}2^n\plus{}\dots\plus{}k^n\]

A lemma :
If $ p$ is an odd prime and $ p|x\plus{}y$ then we have
\[ p^k|x^{p^k}\plus{}y^{p^{k}}
\]
It is an well know result .We can prove it by induction on k .
Now consider our problem . We consider two cases :
[b]Case 1 [/b] $ k\equiv 0(\mod 2)$
Call p is a prime divisor of $ k\plus{}1$ . From condition k is an even number so $ p$ is an odd prime .
From lemma we have :
\[ p^a|x^{p^a}\plus{}(k\minus{}x)^{p^a},\forall x\equal{}1,..,\frac{k\minus{}1}{2}
\]
Therefore  :
$ p^a|\sum_{x\equal{}1}^k x^{p^a}$
It means that $ n\equal{}p^a,a>0$ satisfy condition .
[b] Case 2[/b] $ k\equiv 1 (\mod 2)$
Similar the case 1 we consider an odd prime of $ k\plus{}1$ then we have
\[ p^a|x^{p^a}\plus{}(k\minus{}1\minus{}x)^{p^a}
\]
And
\[ p^a|k^{p^a}
\]
Therefore :
$ p^a|\sum_{x\equal{}1}^k x^{p^a}$
It means that $ p^a,a>0$ satisfy condition .
Problem claim .

the  same as he said but
we can use Hensel lemma:

Hensel lemma:if we denote $ \parallel{}x\parallel{}(p)$ the largest power of p that divide $ x$
and if p is odd prime number such that$ p|a\minus{}b$
we have

$ \parallel{}a^n\minus{}b^n\parallel{}(p)\equal{}\parallel{}a\minus{}b\parallel{}(p)\plus{}\parallel{}n\parallel{}(p)$
so if we put $ n\equal{}p^t$where $ p|k\plus{}1$
we reach to soloution

[quote="TTsphn"]A lemma :
If $ p$ is an odd prime and $ p|x \plus{} y$ then we have
\[ p^k|x^{p^k} \plus{} y^{p^{k}}
\]
It is an well know result .We can prove it by induction on k .
Now consider our problem . We consider two cases :
[b]Case 1 [/b] $ k\equiv 0(\mod 2)$
Call p is a prime divisor of $ k \plus{} 1$ . From condition k is an even number so $ p$ is an odd prime .
From lemma we have :
\[ p^a|x^{p^a} \plus{} (k \minus{} x)^{p^a},\forall x \equal{} 1,..,\frac {k \minus{} 1}{2}
\]
Therefore  :
$ p^a|\sum_{x \equal{} 1}^k x^{p^a}$
It means that $ n \equal{} p^a,a > 0$ satisfy condition .
[b] Case 2[/b] $ k\equiv 1 (\mod 2)$
Similar the case 1 we consider an odd prime of $ k \plus{} 1$ then we have
\[ p^a|x^{p^a} \plus{} (k \minus{} 1 \minus{} x)^{p^a}
\]
And
\[ p^a|k^{p^a}
\]
Therefore :
$ p^a|\sum_{x \equal{} 1}^k x^{p^a}$
It means that $ p^a,a > 0$ satisfy condition .
Problem claim .[/quote]
Nice,TTsphn,But is it useful with
$ n^2|1^n\plus{}2^n\plus{}..k^n$
I had a solution for it :lol:

[quote="sumita"]the  same as he said but
we can use Hensel lemma:

Hensel lemma:if we denote $ \parallel{}x\parallel{}(p)$ the largest power of p that divide $ x$
and if p is odd prime number such that$ p|a \minus{} b$
we have

$ \parallel{}a^n \minus{} b^n\parallel{}(p) \equal{} \parallel{}a \minus{} b\parallel{}(p) \plus{} \parallel{}n\parallel{}(p)$
so if we put $ n \equal{} p^t$where $ p|k \plus{} 1$
we reach to soloution[/quote]

Dear Sumita , I think if $ k\plus{}1\equal{}2^{m}$ ur solution is wrong .

I prove this problem with Primitive root
if $ p$ is odd prime number such that $ p|1\plus{}2\plus{}\cdots\plus{}k$ then u can prove : $ n\equal{}p^{a}$ for each $ a$ is a good number .

Allnames,can you post your solutions with your own problem?
I'm really interested in it.


function vector integration limit algebra polynomial geometry

let $a_1,..,a_n$ be $n$ real numbers that are independant over $\mathbb{Q}$.
Is it true that $(\{ka_1\},\{k{}_a{}_2\},..,\{ka_n\})$ is dense in $[0;1]^n$

It's called Kronecker's Theorem (it's better known in its one-dimensional version), and the answer depends on what you mean by $k$ :). If $k$ is allowed to be any real, yes. This is one version of the theorem. The other, equivalent version gives the positive answer when $k$ is an integer, but the necessary condition is that the $a_i$ [i]and[/i] $1$ be independent over $\mathbb Q$.

You can find at least three different proofs in Hardy and Wright's "Introduction to the Theory of Numbers".

Edit:

Indeed, Weyl's approach is surprisingly accessible :).

Yes, as long as "independent over the rationals" includes the condition that none of the $a_i$ is itself rational.

Suppose we have the condition that for integers $m_1,m_2,\dots,m_n,$ if $m_1a_1+m_2a_2+\cdots+m_na_n\in\mathbb{Z},$ then $m_1=m_2=\cdots=m_n=0.$ (That's what I'm re-interpreting "independent over the rationals" to mean.)

Now consider the function $f(x_1,x_2,\dots,x_n)=e^{2\pi i(m_1x_1+m_2x_2+\cdots m_nx_n)}$ for $m\in\mathbb{Z}^n.$ There are two cases. If $m$ is the zero vector, then $f$ is identically equal to 1 and $\int_{[0,1]^n}f=1.$

Otherwise, if $m$ is not the zero vector,

$\int_{[0,1]^n}f=0.$

Look at the limit $\lim_{N\to\infty}\frac1N\sum_{k=1}^nf(ka_1,ka_2,\dots,ka_n).$

If $m$ is the zero vector, this is 1. Otherwise, we're looking at the limit of

$\frac1N\sum_{k=1}^Ne^{2\pi ik(m_1a_1+m_2a_2+\cdots+m_na_n)} =\frac{e^{2\pi im\cdot a}-e^{2\pi i (N+1)m\cdot a}}{N(1-e^{2\pi i(m_1a_1+m_2a_2+ \cdots+m_na_n))}}$

We are assuming (see our initial supposition) that the denominator is nonzero. It follows that this expression tends to zero as $N\to\infty.$

Now let $f$ be a finite linear combination of these functions - that is, a multidimensional trigonometric polynomial. For a finite set $E\in\mathbb{Z}^n,$ we define $f: [0,1]^n\mapsto\mathbb{C}$ by

$f(x)=\sum_{m\in E}c(m)e^{2\pi im\cdot x}.$ Then,

$\lim_{N\to\infty}\frac1N\sum_{k=1}^N f(ka)=c(0)=\int_{[0,1]^n}f.$

Note that I'm emplying the shorthand of writing $ka$ when I mean the sequence alekk was asking about: $(\{ka_1\},\{ka_2\},\dots,\{ka_n\}).$

By a version of the Weierstrass theorem, these trigonometric polynomials are dense in $C([0,1]^n).$ We could either prove this directly, or we can quote Stone-Weierstrass.

Taking a uniformly convergent sequence of trigonometric polynomials and proceeding to the limit, we get for every continuous function $f$ the result:

(*) $\lim_{N\to\infty}\frac1N\sum_{k=1}^N f(ka)=\int_{[0,1]^n}f.$

If the sequence $(ka)$ were not dense in the cube, we could find a continuous function which was positive in a neigborhood not visited by $(ka)$ and zero elsewhere. This function would have a positive integral, but the left side of (*) would be zero. So we can't have that.

(Note that we've actually proved more than mere density: we've proved uniform distribution.)


superior algebra
superior algebra unsolved

learning groups, how do I find all the one dimensional representations of say Z5?

for every positive natural number $ n$ and a group $ G$, the homomorphisms $ \mathbb{Z}/n \to G$ correspond to elements $ g \in G$ with $ g^n\equal{}1$. apply this to $ n\equal{}5$ and $ G\equal{}\mathbb{C}^*$.

You only need the generators since Z5 is cyclic. Thus you're looking for representations T where T(x)^5=1

1)  T(x)=1
2) T(x) = e^(2pi*i/5)
3) T(x)= e^(4pi*i/5)
4) T(x)=e^(8pi*i/5)
5)T(x)= e^(2pi*i)

I think that does it? (those are all five one dimensional representations).


I'm looking for a degree symbol in LaTeX, but I couldn't find one in the guide.

Look at the following: $90^{\circ}$.

[code]$90^{\circ}$[/code]

  Darij


Prove that for $ a_1\geq a_2\geq...\geq a_n>0$ and $ n\in N, n\geq2$, the following inequality holds:

${ \frac{a_1}{a_2}(a_1-a_2)+\frac{a_2}{a_3}(a_2-a_3)+\cdots+\frac{a_{n-1}}{a_n}(a_{n-1}-a_n)}+\frac{a_n}{a_1}(a_n-a_1)\geq\frac{4(a_1-a_n)^2}{a_1+a_2+\cdots+a_n}$

[quote="marin.bancos"]Prove that for $ a_1\geq a_2\geq...\geq a_n > 0$ and $ n\in N, n\geq2$, the following inequality holds:

${ \frac {a_1}{a_2}(a_1 - a_2) + \frac {a_2}{a_3}(a_2 - a_3) + \cdots + \frac {a_{n - 1}}{a_n}(a_{n - 1} - a_n)} + \frac {a_n}{a_1}(a_n - a_1)\geq\frac {4(a_1 - a_n)^2}{a_1 + a_2 + \cdots + a_n}$[/quote]

The inequality is equivalent to

$ \sum_{k=1}^n(\frac{a_{i}^2}{a_{i+1}}-2a_{i}+a_{i+1}) \ge \frac {4(a_1 - a_n)^2}{a_1 + a_2 + \cdots + a_n}$

Or ,

$ \sum_{k=1}^n\frac{(a_{i+1}-a_i)^2}{a_i} \ge \frac {4(a_1 - a_n)^2}{a_1 + a_2 + \cdots + a_n}$

By Using The Cauchy-Shwarz inequality we have:

$ \sum_{k=1}^n\frac{(a_{i+1}-a_i)^2}{a_i}(\sum_{k=1}^na_i) \ge (\sum_{cyc}|a_{i+1}-a_i|)^2=(2a_1-2a_n)^2$

and the result follows  :P

This is the solution to my problem. Well done, [b]Abdek[/b]! Thanks!


whats the best way to practice for USAMO if i can't solve a single USAMO problem?
to give you a little bit about my background, i usually get 9's on AIMEs (except for the WOOT practice AIMES.)

Do Canada or old USAMOs - those are easier. Honestly, a lot of it is just building up the mental endurance to spend an hour plus on a single problem. You should spend at least half an hour on a hard problem and give up only if you have made zero progress.

http://www.mathlinks.ro/Forum/viewtopic.php?t=238886
(some discussion here)


algebra function domain calculus probability ratio geometry

1.) If $(1-x+x^2)^n =a_0+a_1x+a_2x^2 +.....+a_{2n}x^{2n}$ then $a_0+a_2+a_4+...........+a_{2n}$ equals.

2.) What is the domain of the function: $y = \sqrt {{{x - 1} \over {x - 2\{ x\} }}}$ where { } denotes the fractional part function. (sorry... i don't know if this requires calculus)

3.) First 2n natural numbers are written down on 2n cards.  One card is selected at random.  If the probability of drawing a number 'i' is proportional to 'i', then the probability of drawing an odd number in one draw is

4.) The circle $x^2+y^2+4x-6y+9=0$ undergoes the following transformation:
$3f(x,y)-f(x+1,y)-f(x,y+1) = 0 $
Then the ratio of the areas of the new circle is to original circle is

3) [hide]The total number of possible outcomes is equal to 2n(2n+1)/2, the sum of all the integers from 1 to 2n. 2n of these will result in the number 2n, 2n-1 of these outcomes will result in 2n-1, etc. The number of 'success' outcomes is equal to n(2n-1+1)/2, the sum of all odd integers from 1 to 2n. When you simplify and divide the successes by the possible outcomes, you get n/(2n+1). [/hide]
That was really horribly stated, but I hope you got my meaning...

2
[hide]
Since there's only a square root, the domain is all real numbers unless the value under the root is negative.
For $\frac{x-1}{x-2\{x\}} < 0$, either the numerator or denominator must be < 0.

I don't know any formula for ${x}$.  I know TI calculators do, and as such, this function could be graphed.

Anyway, there are two possible ways for it to be negative

$x - 1 < 0$
and
$x - 2\{x\} > 0$

or

$x - 1 >= 0$
and
$x - 2\{x\} < 0$

Solving the first one, we know $x < 1$, let's look at the second statement of the first possibility.
For any positive number (0 < x < 1), x IS {x}, and x - 2x < 0, so no positive numbers work here.
Now we need to find any negative numbers that will work for x.
Let's keep in mind that the highest {x} can be is .999..., so the highest 2{x} can be is, well, very close to 2.
Although, since we're only working with negative numbers, -2 < {x} < 0.
Since x - 2{x} > 0, x can't be less than -2.
For any x that -1 < x < 0, x > 0 because, as before, x = {x}, and x - 2x > 0 is always true when x is negative.
Now we need to look at when -2 < x < -1.  As x decreases from -1, x - 2{x} decreases from -1 to 0.
As x approaches -2 from the right, 2{x} also approaches -2, getting ever closer to 0, which is the value when x = -2.
So the non-working (I know there's a word...) values for the domain here are:
$-1 < x < 0$

Solving for the first part of the second possibility, we see x >= 1.
However, plugging in 1 for x in the second equation, we get 0 < 0, which is not true.
As x gets larger, we find the same problem as we did before.
We can prove that x > 2{x} for all x >= 1 (which I probably should've done before).
Say 1 < x < 2.  For 2{x} to even break one, {x} must be > 1/2.
Yet we can cleary see that 1.5 > 1.  As x rises from 1.5 to 2, so does 2{x}, but at twice the rate.
This is similar to Zeno's Paradox of Hercules and the Tortoise.
As in the paradox, the values get closer and closer together, and when the reach the end, they are at the same spot.
Except in our case, {x} resets back to 0, instead of 1, so x and 2{x} are never equal when x >= 1 or x <= -1.
Therefore the second possibility for the fraction to be negative can never happen, and we have our answer

$Domain: (-\inf, -1] , (0, \inf)$

[/hide]

4 is real easy, didn't see it first time...
[hide]
$f(x,y)$ is the function $x^2+y^2+4x-6y+9=0$ and we need to find the area for the equation $3f(x,y)-f(x+1,y)-f(x,y+1) = 0$.

First let's find the area of the first one.  Completing the square, we get
$x^2+4x+4+y^2-6y+9 = -9+4+9$

$(x+2)^2+(y-3)^2 = 4$

So the area is $\pi r^2 = 4\pi$

On the second one, we merely multiply the equation by 3, then subtract the equation when (x+1) is substituted for x and again when (y+1) is substituted for y.  This is not something I really feel like typing in.  Once you write those down and simplify the resulting equation, you'll get
$(x-1)^2+(y-4)^2 = 2$

So the area this time is $2\pi$
And thus the ratio of the first area to the second area is $\frac{4\pi}{2\pi} = 2$
[/hide]

[hide="Hint for 1"]What happens when x=1  and  -1 ( is substituted in the equation? )  [/hide]

Please post solutions along with your answers.

We use {x}=x-[x] and thus we have $\frac{x-1}{2[x]-x}$
if x<0 then the raport is positive  because $[x] \leq -1$ and $2[x]-x \leq x<0$
if x=0 the raport does not exist
if $0<x \leq1$ then [x]= 0 and the raport is positive or 0
if 1<x<2 then [x]=1 and also the raport is positive
if $x \geq 2$ we have $ 2[x]>2x-2\Rightarrow 2[x]-x>x-2 \geq 0 $and the raport is positive.

hope is good because otherwise :mad:

[quote="sweetonect"]i think that the solution for problem 2 is R-{0}[/quote]

Yes this was the answer given
does this mean all R except 0?

And sorry but can someone please explain what 3 is asking for... I'm not sure if I understand the wording of the problem.

Number 4 was easy-I just wanted to make sure answer key was wrong. It said 1:1

is there an easier solution for number 4??

Well, I don't know how the answer key got 1:1 for the 4th one.  I got the same answer as before when I did it again.  And I think it might be wrong on the second one also... try plugging in any number $-1 < x < 0$.  Of course, I'm assuming the fractional part of a negative number is negative... and checking my calculator, it thinks the same thing.  Not that it's necessarily right, but I wouldn't be too happy if it wasn't.
Anyway, if you plug one of the said numbers, the top part is always negative, and the bottom part is always positive, resulting in a negative number, and square root of a negative number is non-real.
Plus, why would the answer key say "all real numbers minus 0"?  Wouldn't it put it in the proper notation?  Or probably you just said it that way, I'm dumb.

no the answer key had "R-{0}"
i just wanted to make sure that that was what it was saying.

And 4 [i]IS[/i] wrong in the answer key
Answer is [b]1:2[/b]

And New Problem:
From the set of first 100 natural numbers, x is chosen at random.
The probability that  $\frac{x^{2}-60 x+800}{x-30}<0$ is

And can someone please explain what number 3 (the first probability question) is asking? (Maybe rephrase it in some way)

For #3, the probability of drawing a card k is equal to $\frac{k}{n(2n+1)}$.  So you sum the odd numbers' probabilities and get

$\displaystyle \frac{1}{n(2n+1)} \sum_{k=1}^n (2k-1)$

$= \frac{n^2}{n(2n+1)}$

$= \frac{n}{2n+1}$

[quote="qweretyq"]
And New Problem:
From the set of first 100 natural numbers, x is chosen at random.
The probability that  $\frac{x^{2}-60 x+800}{x-30}<0$ is [b]0.3[/b]
[/quote]

$\frac{x^{2}-60 x+800}{x-30}<0$

[b]Case [/b]$1 \le x \le 29$

$x^2 - 60x + 800 > x - 30$

$x^2 - 61x + 830 > 0$

The quadratic formula yields

$x < \frac{61 - \sqrt{401}}{2}$ and $x > \frac{61 + \sqrt{401}}{2}$

Note that $\sqrt{401}$ is only slightly larger than 20.  So we may say

$x \le 20$ and $x \ge 41$

Clearly, the only possible x are those s.t. $1 \le x \le 20$, and there are 20 such x.

[b]Case [/b]$31 \le x \le 100$

$x^2 - 60x + 800 < x - 30$

$x^2 - 61x + 830 < 0$

The quadratic formula yields

$\frac{61 - \sqrt{401}}{2} < x < \frac{61 + \sqrt{401}}{2}$

$21 \le x \le 40$

The only x that will work are those s.t. $31 \le x \le 40$.  There are 10 such x.

That gives a total of 30 possible x.  Therefore, the probability is $\frac{30}{100} = 0.3$.

Probability (the person), plugging 40 in for x yields 0, which is not < 0, so I think the answer is $\frac{29}{100}$

You're right, I forgot to take into account the zeroes of the numerator.  In fact, neither 20 nor 40 work, so the answer should be instead $\frac{28}{100} = \frac{7}{25}$.

edit-

ok nvm i didnt see the second page


we have 2 objects and a spring moving on a plane with friction(i will describe later the movement).
we are given $\mu,m_1,m_2,k$

first , the object with mass $m_2$ has a speed $v_0$ and it moves in the direction of $m_1$ with friction
of the coef $\mu$.
after this,at a certain moment,the spring pushes it back in the direction it came from.
the problem asks how much should $v_0$ be so that object $m_1$ is moved because of the object $m_2$
wich actually carries it away at the right.

Shouldn't there be $k$ in the problem you given??

Well, I've got a very ugly solution that is probably not right:

First let's glue the $M_1$ to the ground. Let's denote Y the maximal displacement of the $M_2$ to the left and Z the maximal displacement it gains to the right side (away from $M_1$). Clearly, the $M_1$ would move iff

$kZ > \mu M_1 g$

We can calculate Z from the energy conservation law:

$\frac{1}{2}M_2v^2 = \frac{1}{2}kZ^2 + \mu M_2g(2Y+Z)$

Similarly for Y:

$\frac{1}{2}M_2v^2 = \frac{1}{2}kY^2 + \mu M_2gY$

That's a quadratic equation, by solving for Y we get

$Y = \frac{\sqrt{\mu^2M_2^2g^2 + M_2v_0^2K}-\mu M_2 g}{K}$

From the first equation for Z we get similarly

$Z = \frac{\sqrt{\mu^2M_2^2g^2 + M_2v_0^2K - 4\mu M_2 g K Y}-\mu M_2 g}{K}$

substitute for Y:

$Z = \frac{\sqrt{\mu^2M_2^2g^2 + M_2v_0^2K - 4\mu M_2 g (\sqrt{\mu^2M_2^2g^2 + M_2v_0^2K}-\mu M_2 g)}-\mu M_2 g}{K}$

Back to the original condition:

$\sqrt{\mu^2M_2^2g^2 + M_2v_0^2K - 4\mu M_2 g (\sqrt{\mu^2M_2^2g^2 + M_2v_0^2K}-\mu M_2 g)}-\mu M_2 g > \mu M_1 g$
$\mu^2M_2^2g^2 + M_2v_0^2K - 4\mu M_2 g (\sqrt{\mu^2M_2^2g^2 + M_2v_0^2K}-\mu M_2 g) > \mu^2 g^2 (M_1 + M_2)^2$
$M_2v_0^2K - 4\mu M_2 g (\sqrt{\mu^2M_2^2g^2 + M_2v_0^2K}-\mu M_2 g) > \mu^2 g^2 ((M_1 + M_2)^2 - M_2^2)$

Substitute $X = (M_2v_0^2K)/(\mu^2g^2)$ for a while:

$X - 4M_2(\sqrt{M_2^2 + X} - M_2) > ((M_1 + M_2)^2 - M_2^2)$
$4M_2\sqrt{M_2^2 + X} < X + (5M_2^2 - (M_1 + M_2)^2)$
$16M_2^4 + 16M_2^2X < X^2 + 2X(5M_2^2 - (M_1 + M_2)^2) + (5M_2^2 - (M_1 + M_2)^2)^2$
$X^2 - X(2(M_1 + M_2)^2 + 6M_2^2) + (4M_2^2 + 2M_1M_2 - M_1^2)^2 - 16M_2^4 > 0$

Which as a quadratic equation for X yields (for positive X):

$X > 3M_2^2 + (M_1+M_2)^2 + 4M_2\sqrt{(M_1+M_2)^2 - 2M_2^2}$

Therefore

$v_0 >\sqrt{\frac{\mu^2g^2}{M_2K}(3M_2^2 + (M_1+M_2)^2 + 4M_2\sqrt{(M_1+M_2)^2 - 2M_2^2})}$

... which doesn't sound too believable for me ;P

Does anyone have a better/nicer solution? Or can you find a mistake in mine?

If we know $k$, then we can have four equations:
(1) $2fx+fy+\frac12 ky^2= \frac12 m_2v_0^2$
(2) $ky=\mu m_1 g$
(3) $\frac12 m_2v_0^2=fx+\frac12 kx^2$
(4) $f=\mu m_2 g$

I have no idea if I can solve this system of equations...

wouldn't you need some relation between m1 and m2? Let m2 be infinitely bigger than m1, then no v0 will solve your problem. There has to be some restriction.

Then just give infinity $v_0$.
I don't think that we should think too much about the problem itself......

[quote="spx2"]how can you solve it if you dont think bout it ?[/quote]
We we talk about the movements, we often assume that it will happen. If you wanna do physics, you will have to obey the rules. If not, why are you learning physics.

There are all kinds of ugly considerations, like static friction.  I don't like this problem lol.

spx, do you have the solution? If so, could you provide some incite.

This problem does not involve too much physic. It's actually more like an equation-solving problem.


modular arithmetic
number theory proposed
number theory

[quote="moldovan"]If for an integer $ n$ the tens digit of $ n^2$ is $ 7$, what is the units digit of $ n^2$?[/quote]

Let the unit digit $ \equal{}x$, then $ n^2 \equiv \overline {7x} \equiv 0,1 \pmod{4}.......(1) , \equiv 0,1,\minus{}1 \pmod{5} ........(2)$

Only $ x\equal{}6$ satisfies both $ (1)$ and $ (2)$, so the units digit of $ n^2\equal{}6$


Among $ n$ apparently equal coins, less than a half are fake differing from the regular coins by weight. Prove that it is possible to find at least one regular coin in at most $ n \minus{} 1$ measuring with a balance without scale.

[hide="Open question"] Knowing the number of real coins, $ k > \frac n2$, what is the minimum number of weights we need to find all real/fake coins.[/hide]

I think we need to be told that the bad coins all weigh the same, right? if that's the case...

Take an arbitrary coin and weigh it against every other coin (for a total of $ n\minus{}1$ weighings). we say a coin "matches" another if it weighs the same as it, otherwise we say it "mismatches" it.

If the chosen coin is a bad coin, it will necessarily match less than half of the other coins, since less than half of the coins are bad.

If the chosen coin is a good coin, it will necessarily match at least half of the other coins. (equality occurs when there are $ k\plus{}1$ good coins, $ k$ bad coins.)

based on these observations, we get a good coin as follows. if the coin we choose matches a minority of the other coins, we know it must be bad, so any of the coins (which are the majority) that it mismatched must be a good coin. on the other hand, if the chosen coin matches at least half of the others, it must be good, so again we've found a good coin.

[quote="pluristiq"]
I think we need to be told that the bad coins all weigh the same, right?
[/quote]

No, we do not need this information.

We start with $ n$ groups, each consisting of a single coin.
These groups will change over time, but we will make sure that at every moment in time all
coins in a group have the same weight.

We go through a number of steps.
In every step, we take two groups of equal cardinality and compare them against each other.
If they weigh the same, we merge them into a single group.
If they have different weights, we discard both groups (at least one of the groups must
consist of fake coins; hence we discard at least as many fake as good coins, and among
the remaining coins still less than half are fake).

The prodedure terminates, when all groups have distinct cardinalities.
Since we only merge groups of equal cardinality, the possible group sizes are powers of 2.
When all groups have distinct cardinalities, the largest one contains more than half of all coins.
These coins all have equal weight and must all be good.

Since after every weighing at least one group disappears, there are at most $ n\minus{}1$ steps.


a,b,c,x,y,z > 0 prove that :
[img]http://www.gnux.be/latex/data/e1640c58a7494eb6608ec130968edccd.png[/img]

What are you doing here ? You have already posted the same problem at least two times

http://www.mathlinks.ro/viewtopic.php?t=181385
http://www.mathlinks.ro/viewtopic.php?t=181386


This is a problem with an elementary solution.

[b]Problem.[/b] Prove that for the n-th prime number $p_{n}$, the following relation holds:
\[p_{n}<2^{2^{n}}.\]

How elementry? Like theres always a prime between $n$ and $2n$ elementry?

It only requires mathematical knowledge of very basic Number Theory. It is only in the Pre-Olympiad forum since it requires some creativity.

Yea can you cite that fact without proving it?

Of course you can.  It is a somewhat trivial fact.

Well, I mean the proof is definetely not trivial, but intuitively I guess you can consider it trivial.  Well anyway this problem is just induction $p_{1}<2^{2^{1}}$, $2<4$.  Now we can assume $p_{n}< 2^{2^{n}}$ and by Chebyshev's theorem(or w/e you want to call it) $p_{n}<p_{n+1}<2p_{n}<2^{2^{n}+1}$.  Now we just need to prove that $2^{n}+1<2^{2^{n+1}}$ which is easy because rearranging that gives $2^{n}-1 >0$which is clearly true for all $n>0$

There's another proof that does not use "Chebyshev's Theorem."  In fact, it does not use any theorems that are not very basic.  I won't post anything in this thread until I give people some time.

Using Betrand's Postulate (or Chebyshev's Theorem) gives you the vastly stronger $p_{n}\leq 2^{n}$.  In order to get boxedexe's result, you only need [hide="the key idea"]to modify Euclid's argument for the infinitude of the primes.[/hide]

Now, I wonder if you can derive $p_{n}\leq 2^{n}$ [i]without[/i] using Bertrand's Postulate.  (Note that this result is weaker, i.e. it does not directly imply Bertrand's Postulate, so a "not hard" proof is something one could reasonably hope to find.)

Your key idea is what I used.  Indeed, $p_{n}\leq 2^{n}$ is a generalization, but without Bertrand's postulate, the idea becomes tricky since induction doesn't work as well.  We get $p_{n+1}< 2^{\frac{n(n+1)}{2}}+1$.

[quote="JBL"][hide="the key idea"]to modify Euclid's argument for the infinitude of the primes.[/hide][/quote]

[hide="I see"] In other words, we have $p_{1}p_{2}... p_{k}< 2^{1+2+...+2^{k}}< 2^{2^{k+1}}$. [/hide]


function
real analysis
topology
real analysis unsolved

If $ f$ is a real continuous function defined on a closed set $ E \subset X$, for some metric space $ X$, prove there exists a real continuous function $ g$ defined on the entire metric space $ X$, so that $ g(x) \equal{} f(x)$ for all $ x \in E$.

Is this from Principles of Mathematical Analysis? I can't seem to find it. I'm assuming it is a problem in Ch 4.

As a side note, this is actually a special case of a much more general theorem called Tietze's Extension Theorem. The techniques for that proof are beyond what would be expected in Rudin, so they can't be mimicked like I originally thought.

Just thought I'd mention that to get this back to the top of the list since I've been seeing this all day and wondering how to do it myself.

Edit: Actually, I got it.

$ \mathbb{R}\setminus E$ consists of $ \bigcup I_n$ where $ I_n$ are open intervals. For every $ I_n\equal{}(a,b)$ let $ g(x)\equal{}\frac{(x\minus{}b)}{(a\minus{}b)}f(a)\plus{}\frac{(x\minus{}a)}{(b\minus{}a)}f(b)$ so that $ g(b)\equal{}f(b)$ and $ g(a)\equal{}f(a)$ and it is a straight line in between, i.e. continuous.

So the extension is $ h(x)\equal{}\begin{cases} f(x) \ \text{if} \ x\in E\\
g(x) \ \text{if} \ x\in\mathbb{R}\setminus E \\ \end{cases}$

you only proved the case of $ X\subset\mathbb{R}$, which is trivial (the only condition on $ X$ is that it is a metric space).


function
real analysis
real analysis unsolved

Does there exsist function f which is inverse of Rieman's function? For  the additional information , Rieman function is continuous in irrational set(R/Q)  and discontinuous in Q.

By definition, $ f(x)=0$ for all $ x\in\mathbb R\setminus\mathbb Q$. Do you want the inverse of that?