inequalities
three variable inequality
imo 2000
IMO
IMO Shortlist
Hi

Let $ a, b, c$ be positive real numbers so that $ abc \equal{} 1$. Prove that
\[ \left( a \minus{} 1 \plus{} \frac 1b \right) \left( b \minus{} 1 \plus{} \frac 1c \right) \left( c \minus{} 1 \plus{} \frac 1a \right) \leq 1.
\]

There's no need to quote my post for giving a solution Peter ;)

I'm sorry, it's an old habit. :(

Maybe you can configure the resource page so that it ignores quotes? (just like it ignores attachements)

From problem every one can take as
$abc\geq (a+b-c)(a-b+c)(-a+b+c)$ so we have to prove this
WLOG we can assume $a\geq b\geq c$ then $a+c>b$ and $a+b>c$
1) $b+c>a$ an others then it ia sides of triangle so take
$a=x+y, b=y+z, c=z+x$ so from it follows. (we will look next step)
2) $a\geq b+c$ . Since $a,b,c$ are positive so
$abc>0\geq (a+b-c)(a-b+c)(-a+b+c)$
Abdurashid

[hide="note that..."]$p+q=2c>0, \ q+r=2a>0, \ r+p=2b>0$

So for all positive we apply AM-GM, for one negative we have $LHS<0<RHS$[/hide]

[quote="José"][quote="Peter VDD"] \[8pqr \le (p+q)(q+r)(r+p)\] which is obvious from the AM-GM since at most one of $p,q,r$ can be negaitve..[/quote]

I don't understand this step. Please, can anyone explain me?[/quote]
The numbers $(p+q),(q+r),(r+p)$ are positive by definition.

Now obviously at most one of the numbers $p,q,r$ can be negative. If one of the numbers is negative, then it is trivial since the left side would be negative and the right side positive.

Hence we can assume that they are are all nonnegative and then it is an application of AmGm.

I feel ignorant. How do I pass from AM-GM inequality to the inequality that Peter gave? I understood everything in his proof but this (the passage)

[quote="Peter VDD"]you can expand it and apply AM-GM with 8 variables
or you can apply $p+q\ge2\sqrt{pq}$ and multiply each factor[/quote]
Nice options... :rotfl:

I think this is easier______CHN  student

oh wow thats nice also
hehe i was stuck with that inequality for a while...

In fact, it's not entirely correct. Why would $x+y-z\ge0$ for example? If not, you can't appy AM-GM. There is a solution in that sense, but not as trivial as the one posted there.

Let $a=\frac{x}{y}$, $b=\frac{y}{z}$, and $c=\frac{z}{x}$ for some real $x$, $y$, $z$. Expanding the inequality we have currently with substitutions, we are now left with proving that
\[(x+y-z)(x-y+z)(-x+y+z)\leq xyz.\]
However, by AM-GM, we have that
\[\sqrt{(x-y+z)(x+y-z)}\leq x,\]
so applying this cyclically and multiplying, we have that
\[(x+y-z)(x-y+z)(-x+y+z)\leq xyz,\]
finishing the problem.

To eliminate the condition, let $a,b,c=\frac{x}{y},\frac{y}{z},\frac{z}{x},$ respectively. It then suffices to show:
\[\prod_{\text{cyc}}(x-z+y)\leq xyz.\]
By AM-GM:
\[x\geq \sqrt{(x-z+y)(z-y+x)},\]
so our result is true, as desired.

Substitute $a=x/y$, $b=y/z$, and $c=z/x$. Then, it suffices to prove

\[(x+y-z)(x-y+z)(-x+y+z) \le xyz\]

which is a direct occurrence of Schur's when $r=1$. $\square$

From $abc=1$, we substitute $(a,b,c) = \left(\frac yz, \frac zx, \frac xy\right)$. Then we require

\[\left(\frac yz - 1 + \frac yz\right) \left(\frac zx - 1 + \frac yz\right) \left(\frac xy - 1 + \frac zy\right) \leq 1\]
\[(x+y-z)(-x+y+z)(x-y+z) \leq xyz,\]

which is a direct consequence of the degree 3 form of Schur's. $\blacksquare$

The original problem is a special case of the following generalization where
$$\lambda =1$$.

[b]Generalization 1[/b]
Let $a,b,c,\lambda$ be positive reals ($\lambda \leq 1$) such that $abc=1$. Then prove that

$$\left(a-\lambda +\dfrac{\lambda}{b}\right)\left(b-\lambda +\dfrac{\lambda}{c}\right)\left(c-\lambda +\dfrac{\lambda}{a}\right)\leq \lambda \sqrt{\lambda
\prod{\left(\dfrac{a+1-\lambda}{c}+\left(\sqrt{\lambda}-1\right)^2\right)}}$$

Because $abc = 1$, we can make the substitution $a = x/y$, $b = y/z$, $c = z/x$, where $x, y, z$ are positive reals. The inequality becomes$$1\ge\left(\frac xy + \frac zy - 1\right)\left(\frac yz + \frac xz - 1\right)\left(\frac zx + \frac yx - 1\right) = \frac{(x-y+z)(x+y-z)(-x+y+z)}{xyz}.$$Multiplying both sides by $xyz$ and expanding/regrouping terms, we can obtain the following inequality:$$x^3+y^3+z^3 + 3xyz \ge x^2y + xz^2 + yx^2 + yz^2 + zx^2 + zy^2.$$However, this is obviously true by expanding Schur's with $t = 1$. $\blacksquare$

Letting $a=\tfrac xy$, $b=\tfrac yz$, and $c=\tfrac zx$ transforms the desired inequality into proving
\[(-x+y+z)(x-y+z)(x+y-z)\le xyz\]
which can be rearranged as
\[\sum_{cyc}x(x^2+yz)\ge \sum_{cyc}x^2(y+z).\]
This is the $r=1$ case of Schur. $\mathbb{Q.E.D.}$

First, substitute $a=x/y$, $b=y/z$, and $c=z/x$ so that the inequality becomes $(x+y-z)(y+z-x)(z+x-y)\le xyz$. Now if WLOG $z$ is the max, then each factor on the LHS must be positive except possibly for $x+y-z$, but if this is nonpositive then the LHS is nonpositive and the RHS is positive, so the inequality is clear in that case; hence, assume that each factor in the LHS is positve, and substitute $p=\frac{x+y-z}2$, $q=\frac{y+z-x}{2}$, and $r=\frac{z+x-y}{2}$. Then the inequality becomes $8pqr\le (p+q)(q+r)(r+p)=\sum_{text{sym}}p^2q+2pqr$, so we want $6pqr\le \sum_{text{sym}}p^2q$ which is true by Muirhead.

since $abc=1$, we make the substitution $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$ for $x,y,z \ge 0$
we rewrite the given inequality in terms of $x ,y,z $:
$$1\ge\left(\frac xy + \frac zy - 1\right)\left(\frac yz + \frac xz - 1\right)\left(\frac zx + \frac yx - 1\right)$$
$$\Longrightarrow xyz\ge (x-y+z)(x+y-z)(-x+y+z)$$
which is true by the theorem of A.Padoa

Let $a=\frac{x}{y}$, $b=\frac{y}{z}$, and $c=\frac{z}{x}$. Then, we want to prove that $(\frac{x-y+z}{y})(\frac{y-z+x}{z})(\frac{z-x+y}{x})\le 1$ or $(x-y+z)(y-z+x)(z-x+y)\le xyz$. By AM-GM, $\sqrt{(x-y+z)(y-z+x)}\le x$, proving the result.

We expand the given and cancel out (using $abc= 1$) to get that we need \[\sum_{\text{cyc}} \frac{1}{a} + a - \frac{a}{b} \leq 3\] We let $a = \frac{x}{y}$, $b = \frac{y}{z}$, and $c = \frac{z}{x}$, and we scale this so that $xyz = 1$. We can see that our given becomes $ \sum_{\text{cyc}} x^3 - x^2y+y^2x + xyz \geq 0$ which is just shur. $\blacksquare$

[quote=Valentin Vornicu]Let $ a, b, c$ be positive real numbers so that $ abc =1$. Prove that
\[ \left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leq 1\][/quote]
Let $ a, b, c$ be positive real numbers so that $ abc =1$. Prove that
$$\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leq \frac{1}{a+2}+ \frac{1}{b+2}+\frac{1}{c+2}\leq 1$$
(Ji chen)

[quote=Peter]Since $abc = 1$, there exist positive real numbers $x$, $y$ and $z$ such that $a = x/y$, $b = y/z$ and $c = z/x$. The inequality then reduces to

\[ \left( \frac{x-y+z}y \right) \left( \frac{y-z+x}z \right) \left( \frac{z-x+y}x \right) \leq 1.  \]

Substitute $p= b + c - a$, $q = c + a - b$ and $r = a + b - c$. The inequality then reduces to \[ 8pqr \le (p+q)(q+r)(r+p)  \] which is obvious from the AM-GM since at most one of $p,q,r$ can be negaitve..[/quote]

Wonderful proof

Let $ a, b, c$ be positive real numbers so that $ abc = 1$. Prove that
$$ \left( a^2-1+\frac {1}{b} \right) \left( b^2-1+\frac {1}{c }\right) \left(c^2-1+\frac{1}{a} \right)  \leq 1$$


logarithms
vector
trigonometry
complex numbers
complex analysis

1) Compute $ Log[(1\plus{}i)^5]$, where Log is the principal branch of the logarithm.

2) Find all values of $ (1\plus{}i)^{2\minus{}i}$.

it's enough to find arg.
what is definition of complex logarithm ?
$ Log z \equal{} \log |z| \plus{} i arg z$
there is many way to find argument . Usualy I use geometric interpretation of vector $ 1 \plus{} i$ on complex plane.
in this case it's good to use such representation of complex numbers as :
$ 1 \plus{} i \equal{} \sqrt {2} (\cos \pi /4 \plus{} i\sin \pi/4 )$ and use Formula Muavra . I mean :
$ (\cos x \plus{} i\sin x)^{n} \equal{} \cos nx \plus{} i\sin nx$
in 2-nd case rewrite it as $ e^{(1\plus{}i)Log(2\minus{}i)}$ and find argument of $ (1\plus{}i)Log(2\minus{}i)$

The principal branch of the complex logarithm, Log(z), is

$ Log(z) \equal{} \ln|z| \plus{} i Arg(z)$,

where $ \minus{} \pi < Arg(z) \leq \pi$.

For the second, it should be $ exp[(2\minus{}i) \log(1\plus{}i)]$.

[quote="Carcul"]

For the second, it should be $ exp[(2 \minus{} i) \log(1 \plus{} i)]$.[/quote]
yes, sure  :lol: .
I don't know Why I wrote $ exp[(1\plus{}i) \log(2 \minus{} i)]$ maybe I was reading $ (2\minus{}i)^{1\plus{}i}$

hello,for the second i have
$ (1+i)^{2-i}=2\exp(\frac{\pi}{4})(\sin(\frac{\ln(2)}{2})+i\cos(\frac{\ln(2)}{2})$.
Sonnhard.

That's right, except the sign before i should be negative.

times any integer power of $ \exp (2 \pi)$ ?

hello, the result of the first question should be
$ \log\left((1 \plus{} i)^5\right) \equal{} \minus{} \frac {3\pi\cdot i}{4} \plus{} \frac {5}{2}\log(2)$
Sonnhard.


induction
LaTeX
calculus
calculus computations

I'm sort of a bit confused this part of the induction in this case.

Basically the question is this.

A summation from 1 to n,  n^2 = Big-Theta of n^3.

The base P(1) is done which shows true in this case.
I then do the P(n) -> P(n + 1) and rewrite it so it's this
A summation from 1 to (n + 1),  (n + 1) ^2 = Big-Theta of (n + 1) ^3.

How do I continue from there to finish the proof?

Thanks

First of all, it is not clear what you are trying to prove (what is Big-Theta, for example? And what is your sum: $\sum_{i=1}^n i^2$ or $\sum_{i=1}^n n^2$?). Please, write the statements so that they could be understood (otherwise it is very hard to help you) and use $\LaTeX$ if you know it (if you don't, just write clearly in some unambiguous way). Secondly, whatever you meant, I doubt it qualifies as an open question; it rather looks like a standard calculus exercise so I'm moving it to Calculus Computations and Tutorials. Just clarify what you mean and somebody will answer you there.   :)

Of course, it's well known (and easily proved as one of those exercises when you're just learning induction in the first place) that

$\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6=\frac{n^3}3+\frac{n^2}2+\frac{n}6$

And it's easy to show that $\sum_{i=1}^nn^2=n^3.$

Both formulas are asymptotically comparable to a constant times $n^3.$


With the AMC 12 B cutoff being 100, do you think there is still a possibility of having a "curve" on the AMC 12 B? I personally thought the B was harder to score higher on.

[quote]It seemed a tiny bit harder than usual, so I'm guessing the cutoff will be 99 or 99.5[/quote]

The Cutoff was announced today: it was 100..

[quote="da-kid"][quote="Art of Owna"]What do you mean curve-like for USAMO index?[/quote]

Yes. I thought it was a harder test to get a higher score on. Since the majority of the USAMO qualifiers (or even hopefuls) go for the 130+. Do you think the MAA will curve the test even the playing field for the AMC 12 B takers?[/quote]

I doubt it. I don't think the difficulties differed enough for them to care.

[quote="randomdragoon"]The USAMO qualifiers for the A test and the USAMO qualifiers for the B test are proportional to the people who took the A and the B I thought...

What I mean is if 3x more people take the A test than the B, they will take USAMO qualifiers who took the A test verus the B test in a 3:1 ratio.

I might be wrong though.[/quote]

It's more likely that the number of USAMO qualifiers is proportional to the number of people who take the AIME I and the AIME II.

[quote="superwizard188"]my sister who took it said it was easier than the 12A. She actually got a higher score... o-0

I took the 10B.[/quote]

It was. on the 12a, I got 108. on the 12b, I got 111.


Let be given two positive integers $ m$, $ n$ with $ m < 2001$, $ n < 2002$. Let distinct real numbers be written in the cells of a $ 2001 \times 2002$ board (with $ 2001$ rows and $ 2002$ columns). A cell of the board is called [i]bad[/i] if the corresponding number is smaller than at least $ m$ numbers in the same column and at least $ n$ numbers in the same row. Let $ s$ denote the total number of [i]bad[/i] cells. Find the least possible value of $ s$.


A base two numeral consists of 15 digits all of which are ones. This number when tripled and written in base two, contains how many digits?

go to the next post. sorry. admin please delete.

hope this is right...
[hide="solution"]

So given the sequence $2^{n}+2^{n-1}...+2^{0}$
We use the sum for geometric sequences $\frac{a(1-r^{n})}{1-r}$
We have the first term, a, as $2^{n}$, the ratio, r, as $2^{-1}$, and the number of terms, n, as n+1.

$\frac{2^{n}(1-(2^{-1})^{n+1})}{1-2^{-1}}=\frac{2^{n}(1-2^{-n-1})}{2^{-1}}=\frac{2^{n}-2^{-1}}{2^{-1}}=2^{n+1}-2^{0}$

So...

$2^{n}+2^{n-1}...+2^{0}=2^{n+1}-1$
---------------------------------------------------------

Generalization:

Lets call the total number of digits $x$, which is also equal to $n+1$.

[b]Given a string of $x$ 1's in base two, when x is greater than or equal to 3, that is tripled and put into base 2 again, the number of digits in the new string is $x+2$.[/b]

Explanation:

So the sum of the original sequence is $2^{n+1}-1$, as shown above.

$(2^{n+1}-1)\cdot3=3\cdot2^{n+1}-3$.

If the power of two, n, is greater than 1, the -3 becomes irrelavent, since it will not demote the number down to a lower power of two.

So we have $3\cdot2^{n+1}$, instead of $2^{n}$.
Since we have an extra power of 2, this adds a new digit in the string for a $2^{n+1}$.

Then we multiply by three.  This can also be expressed as multiply by $2$ and then $\frac{3}{2}$.

So since we are multiply by $2$, this adds yet another power of 2, as $2^{n+2}$.  The $\frac{3}{2}$ is smaller than 2, and thus does not add a power of two.

So we started from $2^{n}$ to $2^{0}$, and now we have $2^{n+2}$ to $2^{0}$.  The total digits in the first will be $n-0+1$, or $n+1$, and the total digits in the second is $n+2-0+1$, or $n+3$.
$n+3-(n+1)=n+2$

Therefore, we added two extra places in our total number of digits.  So a string of $x$ digits, is now a string of $x+2$ digits.

So $15+2=\boxed{17}$[/hide]


A paint brush is swept along both diagonals of a square to produce the symmetric painted area, as shown. Half the area of the square is painted. What is the ratio of the side length of the square to the brush width?
[asy]unitsize(15mm);
defaultpen(linewidth(.8pt));

path P=(-sqrt(2)/2,1)--(0,1-sqrt(2)/2)--(sqrt(2)/2,1);
path Pc=(-sqrt(2)/2,1)--(0,1-sqrt(2)/2)--(sqrt(2)/2,1)--cycle;
path S=(-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle;

fill(S,gray);
for(int i=0;i<4;++i)
{
fill(rotate(90*i)*Pc,white);
draw(rotate(90*i)*P);
}
draw(S);[/asy]$ \textbf{(A)}\ 2\sqrt {2} \plus{} 1 \qquad \textbf{(B)}\ 3\sqrt {2}\qquad \textbf{(C)}\ 2\sqrt {2} \plus{} 2 \qquad \textbf{(D)}\ 3\sqrt {2} \plus{} 1 \qquad \textbf{(E)}\ 3\sqrt {2} \plus{} 2$

DPatrick gave a good solution in the [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=161]Math Jam[/url][/hide]

[hide]maybe you can divide it into 8 pieces that'd be a lot easier[/hide]

I didn't get that question on the REAL amc that was last week.

Brute forced this one.

Does anyone have a clear solution of this problem?

Finding the side length of the paintbrush swipe was pretty invigorating.

Assume WLOG that the square has side length $2$. Let $x$ be the length of each edge that is not painted (hypotenuse of white triangles). The area of each white triangle is $\frac{x^{2}}{4}$, so the total is $x^{2}$. Since this has to be half of $4$,

$x^{2}= 2 \Rightarrow x = \sqrt{2}$.

Then the brush width is just the hypotenuse of the little gray triangles in the corner, which have side length

$\frac{1}{2}\cdot (2-\sqrt{2}) = 1-\frac{1}{\sqrt{2}}$ so hypotenuse has length $\sqrt{2}-1$.

The ratio is then

$\frac{2}{\sqrt{2}-1}= 2+2\sqrt{2}$.

Simple enough?

[quote="King_James23"]Guys can you explain to me very slowly[/quote]Read paladin8's post or the MATH JAM link


trigonometry
vector
linear algebra
matrix

1. Simplify $ sin(a \plus{} b \plus{} c) \plus{} 4sin(\frac {b \plus{} c}{2})(\frac {c \plus{} a}{2})(\frac {a \plus{} b}{2})$

2. If $ tan(a)$ and $ tan(b)$ are the roots of $ x^2 \plus{} px \plus{} q \equal{} 0$, what is $ sin(a \plus{} b)sin(a \plus{} b) \plus{} psin(a \plus{} b)cos(a \plus{} b) \plus{} cos(a \plus{} b)cos(a \plus{} b)$?

3. Vector v is defined by $ < \minus{} 2,5,1 >$ and vector w is definted by $ < 4,0,3 >$. Find the cosine of the measure of the angle between the vectors.

EDIT: reddog you're right, it's actually $ sin(a \plus{} b \plus{} c) \plus{} 4sin(\frac {b \plus{} c}{2})sin(\frac {c \plus{} a}{2})sin(\frac {a \plus{} b}{2})$

Another typo: In number 2, the question is asking for $ sin(a \plus{} b)sin(a \plus{} b) \plus{} psin(a \plus{} b)cos(a \plus{} b) \plus{} qcos(a \plus{} b)cos(a \plus{} b)$ It is the product of q and $ cos^2(a\plus{}b)$

[hide="3"]$ \vec{v}\equal{}\minus{}2i\plus{}5j\plus{}k$

$ \vec{w}\equal{}4i\plus{}3k$

$ \vec{v}.\vec{w}\equal{}|\vec{v}| |\vec{w}| cos\theta$

where $ \theta$ is the angle between $ \vec{v}$ and $ \vec{w}$

$ (\minus{}2i\plus{}5j\plus{}k).(4i\plus{}3k)\equal{} \left(\sqrt{4\plus{}25\plus{}1}\right) \left(\sqrt{16\plus{}9}\right)cos\theta$

$ (\minus{}5)\equal{}\sqrt{30} (5) cos\theta$

$ \theta\equal{}arc cos\left(\minus{}\frac{1}{\sqrt{30}}\right)$[/hide]

[hide="#3"]$ \cos{\theta} \equal{} \frac{\vec{v} \cdot \vec{w}}{\parallel{}v\parallel{} \parallel{}w\parallel{}}$

$ \vec{v} \equal{} \begin{pmatrix}
\minus{}2 \\
5 \\
1 \\
\end{pmatrix}$

and

$ \vec{w} \equal{} \begin{pmatrix}
4 \\
0 \\
3 \\
\end{pmatrix}$

so

$ \cos{\theta} \equal{} \frac{\begin{pmatrix}
\minus{}8 \\
0 \\
3 \\
\end{pmatrix}}{5\sqrt{30}} \equal{} \frac{\minus{}5}{5\sqrt{30}} \equal{} \boxed{\minus{}\frac{1}{\sqrt{30}}}$[/hide]

@picture74: They're asking for the cosine of that angle, not the angle itself.

[quote="modularmarc101"]
@picture74: They're asking for the cosine of that angle, not the angle itself.[/quote]

Sorry,didn't read the question properly  :oops:

[hide="1"]
Probably it's $ 4\sin\frac{b+c}{2}\sin\frac{c+a}{2}\sin\frac{a+b}{2}$

$ 4\sin\frac{b+c}{2}\sin\frac{c+a}{2}\sin\frac{a+b}{2}=2\sin\frac{b+c}{2}\left(\cos\frac{b-c}{2}-\cos\frac{2a+b+c}{2}\right)=$

$ =\sin a+\sin b+\sin c-\sin(a+b+c)$

Then ${ \sin(a+b+c)+4\sin\frac{b+c}{2}\sin\frac{c+a}{2}\sin\frac{a+b}{2}=\sin a+\sin b+\sin c}$
[/hide]

[hide="2"]
We have $ \tan a\plus{}\tan b\equal{}\minus{}p, \ \tan a\tan b\equal{}q$

$ \sin^2(a\plus{}b)\plus{}p\sin(a\plus{}b)\cos(a\plus{}b)\plus{}\cos^2(a\plus{}b)\equal{}1\plus{}\frac{p}{2}\sin 2(a\plus{}b)\equal{}$

$ \equal{}1\plus{}\frac{p}{2}\cdot\frac{2\tan(a\plus{}b)}{1\plus{}\tan^2(a\plus{}b)}$

$ \tan(a\plus{}b)\equal{}\frac{\tan a\plus{}\tan b}{1\minus{}\tan a\tan b}\equal{}\frac{p}{q\minus{}1}$

Now replace $ \tan(a\plus{}b)$
[/hide]


salam bacheha!
Drooz dashtam roo 1 msale fek mikardam ke 1ho in masale be zehnam resid ke aya Bnahayat adade avale p vojood dare ke p+1 / 2 ham adaD aval beshe?!!!  khodam ziad roosh fek nakardam,nemidoonam..shayad kheili tablo bashe!:D  vali khob..age kasi ghablan shenide ino,lotfan biad bege! age ham nashinidid va hallesh karDd,bazam khaheshan biain beGn!:D
mamnoon!

masale moadele ine ke aya binahayat adade avale P vojud darad ke 2p-1 ham aval bashad?ke ye masaleye hal nashodast!

rahe man ine ke fekr mikonam doroste
farz mikonim ke[color=blue] q[/color] & [color=blue]p[/color] aval bashand:

[color=blue]q=(p+1)/2[/color] pas [color=blue]2q=p+1[/color] [color=blue]pas p=-1(mod q)[/color]

pas [color=blue]p^(q+1)=1(mod q) [/color]  &    [color=blue]p^(q-1)=1(mod q)[/color]

hal farz mikonim ke martabeye [color=blue]p[/color] be hang [color=blue]q [/color]barabar [color=blue]d [/color]bashad pas

[color=blue]q-1=0(mod d)[/color]     &     [color=blue]q+1=0(mod d)  [/color]   yani  [color=blue](p-1)/2=0(mod d)[/color]       &       [color=blue](p+3)/2=0(mod d)[/color]

pas natije migirim ke[color=blue] d[/color], [color=red]yek [/color]ra ad mikonad ke tanaghoz ast

albate dar in beyn 3 va 5 va 13 fekr konam estesns hastand(ba in vojood shayad in rah dorost bashe  :roll: )

(age jaee joob zadam behem begid :oops: )

[quote="m bahador"]

pas natije migirim ke[color=blue] d[/color], [color=red]yek [/color]ra ad mikonad ke tanaghoz ast

[/quote]
che juri natije gerefti d yek ra ad mikone?man fek mikonam natije mishe ke d 2 ra ad mikonad!  albate az aval ham vazeh bud ke d=2!va hich tanaghozi ham nadarim...!

are gereftam, tooye oonja ba vojood in ke jam kardam ,p haro ham hazf kardam ke hamin eshtebaham bood :blush:

ensafan bebakhshid eshtebahe mofti bood

dar zemn vaghean man vase inke sabet konam d=2  ajab khar kari kardam(aslan havasam be oon hamneheshti aval nabood)

kholase jamian bebakhshid :oops:


I posted a similar post to this before.

I gave this question (in which I randomly made up) to stevenmeow, Kevin Y. Chen, and a few other people at da Texas Junior Mathworks Summer Camp...

But now it comes back...with a more challenging version!

Original Question:

A digital clock is being sold at a general store in which when the hour number and the minute number have a common factor, they immediately divide themselves by their GCF. For example, when it is 3:12, the clock immediately becomes 1:04. AM and PM don't show up on the clock. Note: Once the 3:12 becomes 1:04, the clock continues to 1:05, not 3:13.

How many non-existing times are there? (Examples: 3:15, 4:16, 7:49. Existing time examples: 2:13, 3:14, 1:54)

DA NEW QUESTION:

Toadoncart walks into the store and buys the clock for four gold toads (the cashier was extremely happy). He takes it home and studies it carefully, and then decides to write all the times down (if he writes down 3:15, he would write 1:05 instead). What is the probability that he would get the time 1:04? (For example, 1:04 would have a greater probability than 1:59 because 1:04 would count for 1:04, 2:08, 3:12, 4:16, and so on)

I would think of some more questions for the odd clock, but this is it for now....

I guess i didn't explain this well...

Toadoncart walks into the store and buys the clock for four gold toads (the cashier was extremely happy). He takes it home and studies it carefully, and then decides to write all the times down (if he writes down 3:15, he would write 1:05 instead) on seperate sheets of folded-up paper. What is the probability that he would get the time 1:04 if he places all of the paper into a basket and randomly picks one out? (For example, 1:04 would have a greater probability than 1:59 because 1:04 would count for 1:04, 2:08, 3:12, 4:16, and so on)


linear algebra
matrix
Ring Theory
IMC
college contests

Let $A$ be an $n$x$n$ matrix with integer entries and $b_{1},b_{2},...,b_{k}$ be integers satisfying $detA=b_{1}\cdot b_{2}\cdot ...\cdot b_{k}$. Prove that there exist $n$x$n$-matrices $B_{1},B_{2},...,B_{k}$ with integers entries such that $A=B_{1}\cdot B_{2}\cdot ...\cdot B_{k}$ and $detB_{i}=b_{i}$ for all $i=1,...,k$.

Of course,for diagonal matrices the problem is easy.Now if $A$is an $n*n$ matrix,according to $Smith$ $Normal$ $Form$:

If A is an $n*n$matrix defined over $PID$,then there exist invertible $n*n$ matrices $P$,$Q$ such that $PAQ=D$ where $D$is a diagonal matrix.And in sequence $D_{(}11),D_{(}22),...,D_{(}nn)$ every member devides the next one.
 Now based on this form,( $Z$ is PID),problem reduces to diagonal matrices and we are done!!! :)

Did you notice the entries have to be integer? ;)

i have a long solution for this but i am lazy to write it here :blush:

To peter.VDD
MY post is clear and concise.Note that I am working with smith form on $Z$(as a PID).If you are confused,you would better review proof of smith form.

All proofs of [url=http://mathworld.wolfram.com/SmithNormalForm.html]Smith Normal Form[/url] I have are valid only over [url=http://mathworld.wolfram.com/Field.html]fields[/url], which $\mathbb Z$ clearly isn't... but that may of course be due to my lack of good proofs. Please show me your proof then. :)

The result is mentioned [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=structure&t=101625]here[/url] by eugene & harazi. But there's no proof there.

A proof of existence of Smith Normal form over PID can be found in "Basic Abstract Algebaa", by Bhattacharya, Jain and Nagpaul

Note that it suffices to prove the result for $k=2$: if $det A=b_1...b_k$ write $det A=(b_1...b_{k-1})b_k$ then we could find integer matrices $B_{k-1}'$ and $B_k$ such that $A=B_{k-1}'B_k$ and $det B_{k-1}'=b_1...b_{k-1}$ and $det B_k=b_k$, now repeat.

Now the case $k=2$.

Multiplying a matrix $A$ by the elementary matrices representing row-addition operations does not alter the value of the determinant of $A$. That is, if $A'=E*A$ is a row reduced form of $A$ with $E$ a product of the aforementioned elementary matrices , then $det E=1$. Furthermore $E$ has integer entries. Hence it suffices to prove the problem for upper triangular matrices.

Suppose $A$ is an upper triangular with integer entries and $det A=bc$ for some integers $b,c$. Let's show that there exist upper triangular integer matrices $B$, $C$ such that $A=BC$ and $det B=b$, $det C=c$. We will induct on $n$.

First the case $n=2$. The idea is simple. Suppose

\[A=\begin{pmatrix}
m & x  \\
0& n \\
\end{pmatrix}\]

Since $A=BC$, we are looking for $B$ and $C$ of the form:

$$B=\begin{pmatrix}
t_1 & y  \\
0& t_2 \\
\end{pmatrix} ,  C=\begin{pmatrix}
\frac{m}{t_1} & z  \\
0& \frac{n}{t_2} \\
\end{pmatrix}$$

for some integers $t_1,t_2,y,z$ such that $t_1|m$ and $t_2|n$. We have $det A=mn=bc$ and thus we need $t_1t_2=b$. Finally $y$ and $z$ must satisfy $x=t_1z+\frac{n}{t_2}y$. For a given $x$, we can guarantee a solution $(y,z)$ exists only if $\gcd(t_1,\frac{n}{t_2})=1$.

We just need the following:
-----------
[b]Lemma:[/b] If $b|mn$ then we can find $t_1|m$ and  $t_2|n$ such that $b=t_1t_2$ and  $\gcd(t_1,\frac{n}{t_2})=1$.
Proof: First note that we may assume WLOG that $mn$ and $b$ have the same set of prime factors. Next it suffices to prove the lemma in the case where $b$,$m$, and $n$ are all powers of the same prime. In this case either $b|n$, in which case we choose $t_1=\frac{b}{n},t_2=n$, or $n|b$, in which case we choose $t_1=1,t_2=b$.
-------------
The case $n=2$ is done. The inductive step is very easy. Suppose $A$ is an $n$ by $n$ matrix and let $A'$ be the lower-right $(n-1)$ by $(n-1)$ submatrix of $A$. Denote $m=A_{1,1}$ and $n=det A'$. Thus in block form we have

\[A=\begin{pmatrix}
m & X  \\
0& A' \\
\end{pmatrix}\]

Since $A$ is upper triangular, we have $det A=A_1,1detA'=mn=bc$. By the Lemma, we can find $b_1,c_1,b_2,c_2$ such that $m=b_2c_2$ and $n=b_1c_1$ and $gcd(b_2,c_1)=1$. By the inductive hypothesis we may find $n-1$ by $n-1$ matrices $B'$ and $C'$ such that $A'=B'C'$ and $det B'=b_1$, $det C'=b_2$. Let's take $B$ and $C$ of the form (expressed below in block form):

$$B=\begin{pmatrix}
b_2 & Y  \\
0& B' \\
\end{pmatrix} ,  C=\begin{pmatrix}
c_2 & Z  \\
0& C' \\
\end{pmatrix}$$

We just need to two row vectors (of size $n-1$) $Y$ and $Z$ such that $b_2Z+YC'=X$ (*). It's not hard: write $Y=[y_1,...,y_{n-1}]$ and $Z=[z_1,...,z_{n-1}]$ and $X=[x_1,...,x_{n-1}]$.

Matching the first coordinate we must have $b_2z_1+y_1C'_{1,1}=x_1$. There exists an integer solution $(y_1,z_1)$ because $gcd(b_2,C'_{1,1})|gcd(b_2,det(C'))=gcd(b_2,c_1)=1$. Matching the second coordinate we must have $b_2z_2+y_1C'{1,2}+y_2C'_{2,2}=x_2$, which becomes $b_2z_2+y_2C'_{2,2}=x_2-y_1C'{1,2}$, which admits a solution $(y_2,z_2)$ for the same reason as above.

Continuing in this manner, we determine $Y$ and $Z$ satisfying the relation (*). The key is that $b_2$ is relatively prime to $c_1=det C'$ and thus to all the diagonal entries of $C'$.


group theory
abstract algebra
advanced fields
advanced fields unsolved

Given a topological group G and a subgroup H, how do I show that the quotient map $G \rightarrow G/H$ is an open map?

Is there any other condition on $H?$ In particular is $H$ a closed subgroup?

The canonical projection $p: G \rightarrow G/H$ is open $\Leftrightarrow p^{-1}(p(U))$ is open for any $U$ open in $G$.
Note $\phi_{h}: G \rightarrow G$, $\phi_{h}(g)=gh^{-1}$ is continuous, for any $h \in H$. We have $\cup_{h \in H} {\phi_{h}^{-1}(U)}$ open in $G$. But also $\cup_{h \in H} {\phi_{h}^{-1}(U)}=\{x; \exists u \in U, \exists h \in H, x = uh\}=p^{-1}(p(U))$, which is then open.


i have submitted this problem quite a few times only to get a tle on the 17th test case...now i'm plain frustrated

here is the problem statement

You are given a sequence A of n integer numbers (1<=Ai<=n). A subsequence of A has the form Au, Au+1 ... , Av (1<=u<=v<=n). We are interested in subsequences that are permutations of 1, 2, .., k (k is the length of the subsequence).

Your task is to find the longest subsequence of this type.
Input

    * Line 1: n (1<=n<=100000)
    * Line 2: n numbers A1, A2, ... ,An (1<=Ai<=n)

Output

A single integer that is the length of the longest permutation
Example

Input:
5
4 1 3 1 2

Output:
3

now according to my algo,,,i start from every index,,,check if the current running sum is of the form n(n+1)/2 where n is the length.. and all number are distinct..if there are duplicates i break;

here is the code

[code]#include<iostream>
#include<vector>
#include<map>
using namespace std;
#define REP(i,n) for(int i=0;i<(n);i++)
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
int a[100010],ldist[100010];
int c[100001];
long long sums[100001];
int main()
{
  int n,j,kk;
  long long sum,len,maxi=0;
  scanf("%d",&n);
  REP(i,n)
	{
		scanf("%d",&a[i]);
		if(a[i]==1)kk=i;
	}
  memset(c,0,sizeof(c));
   sums[0]=a[0];
  REP(i,n)sums[i]=sums[i-1]+a[i];
  REP(i,n)
    {
      int t=a[i];
      if(c[t]!=0)
	{
	  ldist[i]=i-c[t];
	  c[t]=i;
	}
      else
	{
	  c[t]=i;
	  ldist[i]=-1;
	}
    }

  REP(i,n-maxi)
    {
      len=0;
      j=i;
if(i>kk)break;
	bool flag=false;
      while(1)
	{
          if(j>=n)break;
	    if(j-ldist[j]>=i && j-ldist[j]<=j)break;
	    else
	    {
		if(a[j]==1)flag=true;
	      len++;
	      if(flag)
		{
			if(sums[j]-sums[i-1]==(len*(len+1))/2)maxi=max(maxi,len);
		}
	      j++;
	    }
	}
    }
  printf("%lld\n",maxi);
  return 0;
}
[/code]

now i want a faster algo
please help

I just used a reasonably brute-force algorithm, and it was accepted.
All you need to do is keep track of which numbers you've seen, and where the start of the sequence you are currently looking at is (which starts at 0). I also had a variable saying what the smallest unseen number was (starting at 1).

As you read the next number:
If we have already seen it, increment the 'start' variable, marking each of those numbers as unseen, until we pass the first occurrence of the number we just read. That way, we never have duplicates. Update the smallest unseen number as you go.
If we have seen it, mark it as seen. If its the smallest unseen number, increment that smallest unseen number until we actually reach something that is unseen. The answer will be the biggest value that variable ever gets to (minus 1).

Woah, that probably doesn't make much sense.
For the example they gave: 4 1 3 1 2
start = 0, minunseen = 1
read 4: haven't seen this yet, mark it as seen
read 1: haven't seen this yet, mark it as seen. increment minunseen to 2.
read 3: haven't seen this yet, mark it as seen.
read 1: have already seen this. Move the start forward to the '3', marking the 4 as unseen.
read 2: haven't seen this yet, mark it as seen. increment minunseen to 3; seen that, increment minunseen to 4.
Answer = 4-1 = 3.


[color=cyan]1. If   x + yz = -2,  x2 + y2 + z2 = 3, and  y + z = 2, compute both possible values for the complex number x.

2. 4a + b = 1/(2a) + 2/b.  If a and b are positive, compute the value of ab.

3. If  r = 20/(r + 5) and  s = -5 + 20/s, where r and s are two distinct real numbers, compute r + s.   (I may have posted this one before.)

4. A Martian, 3 Plutians, and 6 Earthlings are sitting at a round table.  If the Martian will not sit next to any Plutian, how many different seating arrangements are possible?[/color]

Quote:3. If r = 20/(r + 5) and s = -5 + 20/s, where r and s are two distinct real numbers, compute r + s. (I may have posted this one before.)

[hide]r^2 + 5r - 20 = 0

s^2 + 5s - 20 = 0

Sum = -5[/hide]

Quote:2. 4a + b = 1/(2a) + 2/b. If a and b are positive, compute the value of ab.

[hide]4a+b=(4a+b)/(2ab)

1=1/(2ab)

ab=1/2[/hide]

For #4 :

We must have

E - A - E and the remaining places are not important

If |E| is the number of Earthlings, ...

My guess is :  |E| * (|E|-1) * (|M| + |E|-2)! / ((|M|! * (|E|-2)!)

In our case it is 1050.

It is very easy to get wrong on combinatorics  :D ! So is it correct ?

[quote="JBL"]1. If   x + yz = -2,  x2 + y2 + z2 = 3, and  y + z = 2, compute both possible values for the complex number x.

[/quote]

Assuming x2 means x^2 etc., then:
(y + z)^2 = 4
y^2 + 2yz + z^2 = 4
y^2 + z^2 = 4 -2yz

but x^2 + y^2 + z^2 = 3, so y^2 + z^2 = 3 - x^2

from the above 2 lines
4 - 2yz = 3 - x^2, or x^2 - 2yz +1 = 0
but since x + yz = -2, yz = -2 - x
and therefore x^2 - 2(-2 - x) + 1 = 0
x^2 + 2x + 5 = 0
x = -1 [u]+[/u] 2*[b]i[/b]

JBL wrote:1. If   x + yz = -2,  x2 + y2 + z2 = 3, and  y + z = 2, compute both possible values for the complex number x.

2. 4a + b = 1/(2a) + 2/b.  If a and b are positive, compute the value of ab.

3. If  r = 20/(r + 5) and  s = -5 + 20/s, where r and s are two distinct real numbers, compute r + s.   (I may have posted this one before.)

4. A Martian, 3 Plutians, and 6 Earthlings are sitting at a round table.  If the Martian will not sit next to any Plutian, how many different seating arrangements are possible?

[hide]#1. squaring the last expression gives y:^2:+z:^2:=4-2yz. Substitute this into the second equation, and let yz=a. Now, we have the two equations x+a=-2 and x:^2:+4-2a=3 => x:^2:-2a=-1. In the first one, we take the a to the other side to get an expression for x:^2: in terms of a (x=-2-a => x:^2:=a:^2:+4a+4). We then substitute this into the second equation to get a:^2:+4a+4-2a=-1 => a:^2:+2a+1=(a+1):^2:=-4 => a=-1:pm:2i. Substituting back in for x gives x=-1:pm:2i.

#2. 4a+b=(4a+b)/2ab => either 4a+b is 0, in which case either a or b has to be nonpositive, or 2ab=1 => ab=.5

#3. r:^2:+5r=20 and s:^2:=-5s+20.  They are the two roots to the equation x:^2:+5x-20 =0. Using the formula for the sum of the roots gives their sum to be -5.

#4. Put the Martian down and define that to be the head of the table. Surrounding him must be two earthlings (fools) (ok sorry i'll keep my preferences out of it). Anyways, we have 3 plutians and 4 earthlings left who can be arranged in any which way. this will be 7!/3!4! or um 35 I think. If the members of each race are distinguishable, then you multiply that 35 by 1!3!6! which will give you some big number[/hide]

[quote="JBL"]4. A Martian, 3 Plutians, and 6 Earthlings are sitting at a round table.  If the Martian will not sit next to any Plutian, how many different seating arrangements are possible?[/quote]

I get a different answer than Belenos, but I agree that combinatorics are tricky!

There are [size=75]10[/size]P[size=75]1[/size] ways to place the Martian; once he-she-it is placed there are [size=75]6[/size]P[size=75]2[/size] ways to place two of the earthlings, one on each side of the Martian.  The remaining 7 organisms can be placed without restriction, but the above will count each particular arrangement 10 times so the product of the above terms must be divided by 10:
[size=75]10[/size]P[size=75]1[/size]*[size=75]6[/size]P[size=75]2[/size]*[size=75]7[/size]P[size=75]7[/size]/10
(10)(15)(5040)/10
(15)(5040)
75600

One last thought: I thought it was "Plutonian" :)

[color=cyan]When you write the question, you can call it whatever you want :).  Good job, all.
Naismith, I believe your value 15 for 6P2 is incorrect -- you meant 30, so your answer is half what it should be, also the same thing that MysticTerminator would have gotten had he multiplied his numbers together.  Belemos, I'm not quite sure where you went wrong, but if you could explain what you did I might be able to give you some ideas.[/color]


3. a) Show that a power of 3 can never be a perfect number.

b) More generally, if p is an odd prime, show that a power pk can never be a perfect number.

c) Show that a number of the form 3i5j can never be a perfect number.

d) More generally, if p and q are distinct odd primes, show that the product piqj can never be a perfect number.

I have looked at this problem for over an hour and I have no idea where to begin does anyone have any suggestions?

[quote="t0rajir0u"]Explicitly sum up the divisors of $p^{i}q^{j}$.  They're given by

$\sigma(p^{i}q^{j}) = (1+p+p^{2}+...+p^{i})(1+q+q^{2}+...+q^{j})$

(Can you see why?)  A perfect number is a positive integer $n$ such that

$\sigma(n) = 2n$[/quote]

Actually that's the point where I'm stuck I was doing the 3 and 5 part, and I used the p^k+1 formula that simplifies primes and I get (3^i+1)-1 * (5^j+1)-1 all divided by 8, but I can't think of what to do next.

c) We have $\sigma(3^{i}5^{j})=2\cdot 3^{i}5^{j}$ iff $\frac{3^{i+1}-1}{3-1}\cdot \frac{5^{j+1}-1}{5-1}=2\cdot 3^{i}\cdot 5^{j}$ iff $(3^{i+1}-1)(5^{j+1}-1)=2^{4}\cdot 3^{i}\cdot 5^{j}$ then $5^{j}\mid 3^{i+1}-1...(I)$ and $3^{i}\mid 5^{j+1}-1...(II)$. From $(I)$ and $(II)$ is easy to prove that
$\frac{3^{i+1}-1}{2\cdot 5^{j}}\cdot \frac{5^{j+1}-1}{4\cdot 3^{i}}=2$  where each factor in LHS is an integer, so, each factor is 1 or 2. Consider separately this cases($1\cdot2$ or $2\cdot1$) , and solve the systems of equations, you will get a contradiction.

MangoMan do you know where does this problems come?

$Tipe$

Any odd perfect number has at least three distinct prime divisors, because if $n=p^{a}q^{b}$ then
$\frac{\sigma (n)}n = (1+\frac{1}{p}+\cdots+\frac 1{p^{a}})(1+\frac{1}{q}+\cdots+\frac 1{q^{b}})$
$\leq (1+\frac{1}{3}+\cdots+\frac 1{3^{a}})(1+\frac{1}{5}+\cdots+\frac 1{5^{b}})$
$< (\Sigma_{k=0}^{+\infty}\frac 1{3^{k}})(\Sigma_{k=0}^{+\infty}\frac 1{5^{k}}) = \frac{15}8 < 2.$

As far as I know, the current bound for an odd perfect number (if any) is at least $8$ distinct prime divisors.

Pierre.

An odd perfect number has to be a perfect square right...not sure where that gets us generally


I was curious as how to solve this kind of problem... a sort of linear and geometric sum-ish thingamajiggy

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86052[/url]

note that this problem is less messy when the sum approaches infinity

If you want to use calculus, there's a very easy way to find the sum using derivatives.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87042]hint[/url]
This will help.

Alterman, the given topic is interesting, but for this question, there exists a finite-sum function. I'll look at it. May post too.


Given a hexagon $ ABCDEF$ such that

$ AB \equal{} BC,CD \equal{} DE,EF \equal{} FA$, and $ \angle A \equal{} \angle C \equal{} \angle E$

Prove that lines $ AD,BE,CF$are concurrent.

Join the vertices to make the triangles $ ACE$, note $ A, C$ and $ E$ its angles, and $ a, C$ and $ e$ its side-lengths. See that the triangles $ ABC, CDE$ and $ EFA$ are isosceles and note $ \alpha, \beta$ and $ \gamma$ their respective base angles. From the given double equality we get the following three equalities:
$ A+\alpha=E+\beta; E+\gamma=C+ \alpha$ and $ C+\beta=A+\gamma \ ( \ 1 \ )$.
If $ M = AD\cap CE$, then ${ \frac{CM}{ME} = \frac{S_{ACD}}{S_{AED}}}=\frac{AC\cdot CD\cdot sin \ ( C+\beta )}{AE \cdot DE \cdot sin \ ( E+ \beta )} \ ( \ 2 \ )$; similarly we find other two equalities, their product being $ 1$, hence the conclusion.

Best regards,
sunken rock

Let $ O$ be the circumcenter of $ \triangle ACE$. It's easy to see $ BO,FO,DO$ to be interior bisector of $ \angle ABC,\angle CDE,\angle AFE$ respectively. According $ \triangle AOE \equal{} \triangle FOE,\triangle EOD \equal{} \triangle COD,\triangle AOB \equal{} \triangle COB$ and $ \angle A \equal{} \angle C \equal{} \angle E$ We'll get $ AO,CO,EO$ to be interior bisector of $ \angle A,\angle C,\angle E$, It's implies that $ ABCDEF$ is a tangency hexagon.Applying Brianchon's theorem we have done :)
 Sorry for my bad english  :)  :)


A geometric sequence is given. Find the common ratio and write out the first four terms.

(1) {(-5)^n}

(2) {(2^n)/(3^(n - 1)}

Please do not post homework problems. At least show some work.


[hide="Solution"]Denote $a=6^{1989}$. We'll solve the problem using $\mod{4}$ and $\mod{25}$.

It's obvious that $a\equiv 0\pmod{4}$. As for $\mod{25}$, we can note that $6^5\equiv 1\pmod{25}$, which can be proven through reducing the binomial expansion of $(5+1)^5$ modulo $25$. Hence $a=6^{397\cdot 5+4}\equiv 6^4\pmod{25}$, and $(5+1)^4=5^4+4\cdot 5^3+6\cdot 5^2+4\cdot 5+1$, which means that $6^4\equiv 21\pmod{25}.$

From $a\equiv 0\pmod{4}$ and $a\equiv 21\pmod{25}$ we can make two equations:

$a=4k\land a=25l+21$ where $k,l\in\mathbb{Z}\quad (*)$

From there we derive the diofantine equation $4k-25l=21$ from which $k=6l+5+\frac{l+1}{4}$. It means that $\frac{l+1}{4}$ is an integer, hence $l=4m-1,m\in\mathbb{Z}$. Substituting $l$ in $(*)$ we get $a=100m-4$, which means that $a\equiv -4\equiv 96\pmod{100}.$

Therefore, the last two digits of $6^{1989}$ are $96$.[/hide]

Farenhajt: $a^{\phi(n)} \equiv 1 \pmod {n} \forall a \perp n$ by Euler, which you can use to find that $6^{20} \equiv 1 \pmod {25}$ and get the same result. ;)

[quote="silouan"]Find the two last digits of  $6^{1989}$.  If you want solve it with the method of mod 100. ;)

BTW, please don't post only the answer but the hole job[/quote]

since $2^{1989}$ is a factor, it is $0 \mod 4$, then looking mod 25, and by euler's generalization, $6^{20}\equiv 1 \mod 25$, say it is $k$, then we have $6k\equiv 1 \mod 25\implies k \equiv 21 \mod 25$ so combine those with chinese remainder theorem to get $k \equiv 96 \mod 100$

[quote="chess64"]$a^{\phi(n)} \equiv 1 \pmod {n} \forall a \perp n$[/quote]

What does $a\perp n$ mean?

relatively prime

oops, sorry to repeat, I didn't see the previous post.


Let x,y,z are positive numbers such that $ xy\plus{}yz\plus{}zx\equal{}1$. Prove that
$ 8x^2y^2z^2\ge (1\minus{}x^2)(1\minus{}y^2)(1\minus{}z^2)$

[quote="thegod277"]Let x,y,z are positive numbers such that $ xy \plus{} yz \plus{} zx \equal{} 1$. Prove that
$ 8x^2y^2z^2\ge (1 \minus{} x^2)(1 \minus{} y^2)(1 \minus{} z^2)$[/quote]
It's equivalent to $ \sum_{sym}(x^4y^2 \minus{} x^3y^3 \plus{} x^4yz \minus{} 2x^3y^2z \plus{} x^2y^2z^2)\geq0,$
which is obvious by Muirhead and Schur.
I did not see it before.

[quote="thegod277"]Let x,y,z are positive numbers such that $ xy \plus{} yz \plus{} zx \equal{} 1$. Prove that
$ 8x^2y^2z^2\ge (1 \minus{} x^2)(1 \minus{} y^2)(1 \minus{} z^2)$[/quote]

This one is easy. :lol:

It looks very much similar to me for sure. Possibly it appeared in one of the regional olympiads. I can post that problem here if necessary, though it was rather easy, even easier than this one.
 :)

[b]To arqady:[/b]
I have seen you to do simultaneous use of such muirhead and schur in some inequalities. Can u please teach me?

Sunkern_sunflora, about Schur see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1166

about Muirhead see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15896


I got a score of 8 on AIME and i am in 8th grade, but still didn't get selected in USAMO,  how could this happen?

it means you got less than a 132 on the 10 or 12 (You have to get a 212 index meaning your amc score plus 10*AIME score)

YOU HAD TO HAVE GOTTEN AT LEAST 132 ON AMC. :(

[quote="inyoung"]I feel the same way here. I'm in seventh grade, got an 8, and missed it by 2 stupid errors on the 10. What I don't get is that the 10 and 12 were counted equally[/quote]
I don't want to start an argument, but two stupid errors on the AMC 10 would give you the score I got, a 138. Unless you didn't answer other questions? (By the way, I didn't qualify, nowhere close. I got a 1.  :P )
Have fun!


since i am really damn and lazy person, i can't solve a simply quesiton.
i am  always %%%%  with mathematics. i am not  sure if i am really stupid guy or what. :D
may be i don't have the talent for %%% math. i want some links of basic maths on powers , exponents and Sqaure Root.

so here is the problem
7x-18/2 - 3x = 2x-4/6 - 1/8 (4x-16)+3

I am not sure if my post is against the rules of this forum not.
so i would be sorry if it is.

[hide]7x-18/2 - 3x = 2x-4/6 - 1/8 (4x-16)+3 in a nicer format (not really): $7x-\frac{18}2-3x=2x-\frac46-\frac18(4x-16)+3 \Rightarrow 7x-9-3x=2x-\frac23-\frac{x}2+2+3 \Rightarrow 4x-9=\frac{3x}2+\frac{13}3 \Rightarrow \frac{5x}2=\frac{40}3 \Rightarrow x=\frac{16}3$[/hide]

I hope I didn't make any careless errors.  :D

thanks alot for you reply.

can you please tell me how to write like that
cause i have totolly different                question
the question is like this
(7x-18)/2 - 3x = (2x-4)/6 - 1/8 (4x-16)+3

sorry for inconvience
 :(

Just click on the link on the left panel (with logout and site guide etc.) that says LaTeX.

Btw, your new equation will be $\frac{7x-18}{2}-3x=\frac{2x-4}{6}-\frac{1}{8}(4x-16)+3$. Use the distributive property to solve like what I did in the last quesiton. If I didn't make any careless errors  :D , you should get $x=20$.

Here is the answer to your question.

7x-9-3x=2x-2/3-1/8(4x-16)+3
x=16/3

P.S. I would show you the work, but that takes too long and I have a life.

[quote="José"]I got X=20.[/quote]

How? It doesn't work when I plug it in the original equation.

I think you're looking at the wrong equation. :P

Jose was probably talking about...uh...the equation Robinhe stated.

Sorry i dont know how to use latex :( [/quote]

[quote="most-wanted"]since i am really - and lazy person, i can't solve a simply quesiton.
i am  always %%%%  with mathematics. i am not  sure if i am really stupid guy or what. :D
may be i don't have the talent for %%% math. i want some links of basic maths on powers , exponents and Sqaure Root.

so here is the problem
7x-18/2 - 3x = 2x-4/6 - 1/8 (4x-16)+3

I am not sure if my post is against the rules of this forum not.
so i would be sorry if it is.[/quote]

[hide]
7x-9=2x-2/3-1/2x+2+3 (simplify)
5.5x=13 1/3
33x=80
x=80/33

numbers too large so not sure


linear algebra
matrix
linear algebra unsolved

Find the connected components of the set of invertible real symetric matrices of order $n$. The same for complex ones. What about cyclic matrices (real and complex)?

for the real case, $M \in S_n(\mathbb{R})$.
There exists $U \in SO_n(\mathbb{R})$ such that $U^tMU=D$ is diagonal. Because $SO$ is connected, one can suppose that $M=Diag(e_1, e_2,..,e_n)$ where $e_i \in \{-1; +1\}$.

Because the set of symetrique invertible matrices $C_s$ with a given signature $s$ is open in $M_n(\mathbb{R})$, to conclude one just have to see that $C_s$ is connected. This is easy because in $Diag(-1,+1)$ is connected by a path to $Diag(+1,-1)$ in $GL_2{\mathbb{R}} \cap S_2(\mathbb{R})$. Hence I think the connected component of $GL_n{\mathbb{R}} \cap S_n(\mathbb{R})$ are the set of matrices with a given signature.

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=cyclic&t=55872 for the complex cyclic matrix case.


After I did a problem, I checked out what videos I had. Later, I checked again, and two disappeared! What happened? (I have only recently became addicted to Alcumus, so I'm kinda a newbie at this) :)
Thanks!

Videos will occasionally disappear based on how Alcumus thinks of your needs; if your subject is passed, the video could be moved further down in the page into the "Passed" or later to the "Mastered" section.

Also if you do really horrible on problems Alcumus will stop quizzing you for a while on that subject and remove that subject from your videos.

[quote="maybach"]Will there be no more new videos?[/quote]
Not until (maybe) next year.


graph theory
combinatorics unsolved
combinatorics

A communications network consisting of some terminals is called a [i]$3$-connector[/i] if among any three terminals, some two of them can directly communicate with each other. A communications network contains a [i]windmill[/i] with $n$ blades if there exist $n$ pairs of terminals $\{x_{1},y_{1}\},\{x_{2},y_{2}\},\ldots,\{x_{n},y_{n}\}$ such that each $x_{i}$ can directly communicate with the corresponding $y_{i}$ and there is a [i]hub[/i] terminal that can directly communicate with each of the $2n$ terminals $x_{1}, y_{1},\ldots,x_{n}, y_{n}$ . Determine the minimum value of $f (n)$, in terms of $n$, such that a $3$ -connector with $f (n)$ terminals always contains a windmill with $n$ blades.

Let the communication network be a graph.
Let $ n\equal{}1$. Because a cycle of length $ 5$ is a 3-connector, but doesn't contain a triangle, $ f(n)>5$. We show that $ f(1)\equal{}6$. Take a graph with $ 6$ vertices. Take a vertex $ A$. It is clear that there either exist at least $ 3$ neighbours of $ A$, or at least $ 3$ nodes not being joined to $ A$. In the first case, some two of these neighbours have to be joined by an edge, forming a triangle together with $ A$. In the second case, the set comprising $ A$ and any two of them can have at most one edge - the one between the two of them. So all three non-neighbours of $ A$ then form a triangle.

For $ n>1$ we show that $ f(n)\equal{}4n\plus{}1$. Firstly we show that $ f(n)>4n$. Let $ K_{m}$ denote the complete graph with $ m$ vertices. Then a graph consisting of two disjoint not connected $ K_{2n}$'s obviously is a 3-connector and since it hasn't a connected subgraph with more than $ 2n$ vertices, while a windmill is a connected graph and has $ 2n\plus{}1$ vertices; we have $ f(n)>4n$.
Now take a $ 3$-connector with $ 4n\plus{}1$ vertices. Take a vertex $ A$. Let $ V_{A}$ the set of its neighbours and $ NV_{A}$ the set of nodes not connected to $ A$.
Take $ X,Y\in NV_{A}$. Because neither $ AX$ nor $ AY$ exists, $ XY$ must exist. $ X,Y$ were arbitrarily chosen, thus $ NV_{A}$ is actually a complete graph. If $ |NV_{A}|\ge2n\plus{}1$, then $ NV_{A}$ obviously contains a windmill with $ n$ blades and we are done. Suppose $ |NV_{A}|\le2n$. Then $ |V_{A}|\ge2n$. Take nodes $ V_{2n\minus{}2}, V_{2n\minus{}1},V_{2n}$. Some two of them are connected by edge. By an eventual rearrangement of the indices, we can suppose that there is an edge between $ V_{2n\minus{}1}$ and $ V_{2n}$. We are left with $ 2n\minus{}2$ neighbours of $ A$. Repeat the process. If $ |V_{A}|>2n$ then after 'creating' $ n\minus{}1$ edges we'll be left with some 3 vertices $ V_{0}, V_{1},V_{2}$. Some two of them are connected by an edge, say $ V_{1}$ and $ V_{2}$. We obtained a windmill with central node $ A$ and blades $ \{V_{1},V_{2}\},$...$ ,\{V_{2n\minus{}1},V_{2n}\}$. Done. We're left with the case $ |V_{A}|\equal{}2n$ and thus $ |NV_{A}|\equal{}2n$. As in the previous case we build $ n\minus{}1$ edges: $ V_{2n\minus{}1}V_{2n}$, ... ,$ V_{3}V_{4}$. We are left with $ V_{1}$ and $ V_{2}$. If they're connected then we have a windmill. Suppose they're not. Remember that $ NV_{A}$ is a complete graph with $ 2n$ vertices. Now take the $ 2n$ sets $ X_{i}\equal{}\{V_{1},V_{2},NV_{i}\}$. Clearly one (or both) of the edges $ V_{1}NV_{i}$ and $ V_{2}NV_{i}$ must exist. Thus, WLOG, $ V_{1}$ has $ n\ge2$ neighbours in the set $ NV_{A}$. Let two of them be $ NV_{1}$ and $ NV_{2}$. Then we have a windmill centered at $ NV_{1}$ with blades: $ \{V_{1},NV_{2}\}$ and other $ n\minus{}1$: $ \{NV_{3},NV_{4}\},\ldots\{NV_{2n\minus{}1},NV_{2n}\}$.
It's also clear why the case $ n\equal{}1$ has been treated separately.

edit: this bad

Consider the complement of the graph. It is triangle free and hence can be split into a bipartite graph with no edges in group $A$ and $B$ for some sets $A$ and $B$ whose union is the entire graph. Now, we claim the answer is $4n+1$.

In a graph with $4n+1$ vertices, either $A$ or $B$ has size at least $2n+1$, and then the entire set $A$ or entire set $B$ can be taken as a windmill.

In a graph with $k \leq 4n$ vertices, take $A$ and $B$ as complete graphs with vertices split as evenly as possible.

We claim that the minima are $6$ and $4x+1$ when $x=1$ and $x\ge 2$, respectively.
\\$~$\\
Let's take care of the case $x=1$ first. If there are $5$ terminals, then a simple $C_5$ would be a $3$-connector and would have no windmill with $1$ blade, otherwise known as a triangle. For $k\le 5$ we can delete terminals, and the network stays $3$-connector and clearly no triangle arise. Now, assume there are at least $6$ terminals. Then, $R(3,3)=6$ along with every three points having at least one edge implies that there is a triangle.
\\$~$\\
Now, suppose $x>1$ and there were $4n$ terminals. Then, simply make two $K_{2n}$s. As before, for less than $4n$ terminals just delete terminals from the original $4n$ terminal construction. If there were $4n+1$ terminals, pick any terminal $T$. There must either be $2n+1$ terminals such that none can communicate with $T$, or $2n$ such that all can communicate with $T.$
\\$~$\\
Suppose it is that $2n+1$ terminals cannot communicate with $T$ then the $2n+1$ terminals form a $K_{2n+1}$ which one can derive an $n$-windmill from.
\\$~$\\
Suppose it is that at least $2n$ terminals can communicate with $T.$ Then, if there exists a matching of size $n$ among those terminals, we are done. Consider any three terminals, two of them must be able to communicate. Disregard these and apply the same with the remaining until we have created $n-1$ pairs of distinct terminals, where the two terminals in each pair can communicate.
\\$~$\\
If we are left with at least $3$ terminals, we are done, so suppose we are left with $2$ terminals, say, $A,B$. Note that the terminals that cannot communicate with $T$ forms a $K_{2n}$, and for each terminal which cannot communicate with $T$, it must be able to communicate with $A$ or $B.$ By pigeonhole, one of $A$ or $B$ has at least two neighbors in the terminals that cannot communicate with $T.$ From here we derive an $n$-windmill so we're done.


limit
topology
real analysis
real analysis unsolved

Let $K\subset \mathbb{R}$ be compact. Prove that the following two statements are equivalent
to each other.
(a) For each point $x$ of $K$ we can assign an uncountable set $F_{x}\subset \mathbb{R}$ such that
$dist(F_{x}, F_{y}) \geq |x-y|$
holds for all $x, y \in K$;
(b) $K$ is of measure zero.

One way is fairly straightforward. Note that each uncountable set $F$ has an uncountable intersection with one of the intervals $[m,m+1]$, $m\in\mathbb Z$. Let $K_{m}$ be the set of points $x\in K$ such that $F_{x}\cap [m,m+1]$ is uncountable. It suffices to show that each $K_{m}$ has measure $0$. For each point $x\in K_{m}$ and each $\delta>0$ consider the intersection of the $\delta$-neighborhood of $F_{x}$ with $[m,m+1]$, which we'll denote by $G(x,\delta)$. It is a nice (relatively) open set in $[m,m+1]$ and $\lim_{\delta\to 0+}\frac{|G(x,\delta)|}\delta=\infty$. Now choose big $M>0$ and consider all possible intervals $(x-\delta,x+\delta)$ with $x\in K_{m}$, $\frac{|G(x,\delta)|}\delta>M$. They form a covering of $K_{m}$. Choose a subcollection of pairwise disjoint intervals $(x_{n}-\delta_{n},x_{n}+\delta_{n})$ such that the union of $5$ times larger intervals contains $K_{m}$ (Vitali lemma). Then the sets $G(x_{n},\delta_{n})$ are pairwise disjoint and contained in $[m,m+1]$, so we have
\[|K_{m}|\le 5\sum_{n}\delta_{n}\le\frac{5}{M}\sum_{n}|G(x_{n},\delta_{n})\le \frac{5}{M}\to 0\text{ as }M\to+\infty\]
and we are done.

I don't know if it was so straightforward, but it is very nice  :lol:

Here is the sketch of the construction for the other direction. Let $I$ be an interval such that $|I|>2\text{\,diam\,}K$. We'll try to construct $F_{x}$\subset $I$ for all $x\in K$. Let $J_{1},\dots, J_{m}$ be disjoint closed intervals such that $\cup J_\ell\supset K$ and $|J_{1}|+\dots+|J_{m}|$ is very small. For each $\ell=1,\dots,m$, choose a pair of subintervals $I_\ell'$ and $I_\ell''$ of the interval $I$ such that $|I_\ell'|=|I_\ell''|=2|J_\ell|$, $\text{dist\,}(I_\ell',I_\ell'')=|J_\ell|$ and $\text{dist\,}(I_\ell, I_{m})\ge \text{dist\,}(J_\ell,J_{m})+|J_\ell|+|J_{m}|$ where $I_\ell$  stands for any of $I_\ell'$ and $I_\ell''$ (if the total length of $J_\ell$ is small enough, such choice is possible). Note now that if we choose $F_{x}$ to be a subset of $I_\ell'\cup I_\ell''$ for $x\in J_\ell$, then, for sure, we'll have $\text{dist\,}(F_{x}, F_{y})\ge |x-y|$ for $x,y$ from different $J_\ell$. Also, if $x,y$ are in the same $J_\ell$, we can make the choice of $F_{x}$ and $F_{y}$ in $I_\ell'$ and $I_\ell''$ independently. Thus, it remains to choose $F'_{x}\subset I_\ell'$ (and $F_{x}''\subset I_\ell''$) for $x\in J_\ell\cap K$. But now we got exactly the same problem as the one we started with, only for a smaller set. So, we can repeat our construction infinitely many times. In the end we'll get for each $x\in K$ a Cantor type set $F_{x}$ with the desired property. It would be nice to draw a couple of pictures here but I hope that the argument is understandable even without them...


[quote="hwenterprise"]I have no clue if my title has anything to do with my problem:

Sorry I don't know LaTex...

a/b=1/2
b/c=3/4
c/d=6/7

What is, under the above circumstances:

a/d+b/d+c/d?[/quote]

[hide]
I just plugged in the minimum whole number that each variable can be. I don't know how to "formally" do it...

7c = 6d
3c = 4b
2a = b

a = 9
b = 18
c = 24
d = 28

so...

9/28 + 18/28 + 24/28 = 51/28
[/hide]

[quote="socrates"]Multiplying all equalities we have a/d=9/28.
similarly we have b/d=9/14 etc and we add. I hope this is not wrong! :)[/quote]

Yes, you are correct, since we know what c/d is.

[quote="hwenterprise"]I have no clue if my title has anything to do with my problem:

Sorry I don't know LaTex...

a/b=1/2
b/c=3/4
c/d=6/7

What is, under the above circumstances:

a/d+b/d+c/d?[/quote]

2a=b
4b=3c
7c=6a

4b=8a
b=2a
a=42
b=84
c=112
d=392/3

Ans. 51/28

so,  :rotfl:

[quote="hwenterprise"]I have no clue if my title has anything to do with my problem:

Sorry I don't know LaTex...

a/b=1/2
b/c=3/4
c/d=6/7

What is, under the above circumstances:

a/d+b/d+c/d?[/quote]

Latex for dummies (dont take this offensivly(sp?))

To get the fraction $\frac{2}{3}$, you type \frac{numerator}{denomentator} inside of dolllar signs.
[/b]


I'll post the PDF document of the solution packet of COPSD very soon.

Till then, you're free to discuss the problems here. Only 2 people qualified for it this time. :)


ARML MATHCOUNTS AMC USA(J)MO USAMO blogs AIME

This game is based on the sig "This has been a public shaming of ___. Thank you". Credits go to those who began that sig.
Very simple. Shame the person above you with one sentence. And try not to be overly rude. It's only meant to be fun.

Since I start, I'll shame a person of my choice.

pythag011 has scored lower than me on the Mandelbrot Round 3. This has been a public shaming of pythag011. Thank you (and yes, say ty and the end).

Don't know anything about TPAM, so, look at my sig. :)

1 does noe equal 2. This has been a public shaming of 1=2. Thank You.

I have more posts than erabel.
This has been a public shaming of erabel.  Thank you.

I have more whole words in my sn than lingomaniac88.  This has been a public shaming of lingomaniac88. ty

I have more moving smilies in my sig than Math Geek.

This has been a public shaming of Math Geek. Thank you.

I scored higher in State MC than lotsofmath.

This has been a public shaming of lotsofmath. Thank you.

pacman2812 have 4 numbers, and distracted523, 3. This has been a public shaming of distracted523. ty

pacman2812 was the last poster. This has been a public shaming of pacman2812. Thank you.

Zephy posted at exactly 8:00 :!:  :!:  :!: . this has been a public shaming of zephy.

pythag011 feels he (she?) is too good to post zephyredx's full name.
This has been a public shaming of pythag011.  Thank you.

lingomaniac88 can't even determine the gender of pythag011. This has been a public shaming of lingomaniac88. Thank you.

Brut3Forc3 can't find the "e" key on his computer.  This has been a public shaming of Brut3Forc3.  Thank you.

cf249 failed to make his goal (to not make stupid mistakes) for the week. This has been a public shaming of cf249. Thank you.

ProtestanT missed a perfect score on the AMC 10 by one question.
This has been a public shaming of ProtestanT.  Thank you.

You don't have an avatar either, you have just a picture.  thbapsotpam ty.

My picture at least makes more sense than ernie's. THBAPSO ernie, TY.

My avatar outlines your personality.  thbapsotpam, ty.

ernie does not know what my personality is . THBAPSO ernie, TY.

evilhamster's personality is clearly revealed as random.  thbapsotpam, ty.

I didn't lose, so ernie's sig is worthless. THBAPSO ernie. Thank you.

Bugi posts too much on Beat the Mod that is a game he isn't even in. THBAPSO Bugi. TY.

Wickedestjr. is located on only one point. TY

Please do not bump 2-year threads. Mods, please lock/delete.

3333 thinks those 2 states are adjacent rather than far away.

[quote="3333"]andrewjjiang forgot that this was fun factory. TY[/quote]
Even if it is FF, you shouldn't bump threads. Create a new one with [REVIVAL] next to it.

It is revived...can't we just continue it??? I must say...it looks fun... :!:

3333 doesn't know how to properly publicly shame someone. THBAPSO 3333, thank you very much.
That being said, revivals from this far back are unnecessary.

You guys should know forum rules. Seriously.

Oh well, thanks to about-to-become future posters on this revival thread for actually reading AIME15's post (even that's rare these days), and thanks to AIME15 for making the post, I guess.


The opposite angular points of a square are [size=150](2,0)[/size]and [size=150](5,1).[/size]Find the remaining points. :lol:

[hide]The 2 points connected is the diagonal of the square. The midpoint is $(3.5,0.5)$. The length is $\sqrt{10}$. The slope is $\displaystyle\frac{1}{3}$.

The other diagonal must be the same length, go through the same midpoint, and be perpendicular to the given segment. The equation of the line of the second diagonal is $y=-3x+11$. The other 2 points are $(3,2)$ and $(4,-1)$.[/hide]


Mr. Ropper is paid time-and-a-half for hours worked in excess of 40 hours and double-time for hours worked on Sunday.  If Mr. Ropper had gross weekly wages of 342 dollars for working 50 hours, 4 of which were on Sunday, what is his regular hourly rate?

[hide]
Let the regular wage be $r$
Then $r ( x+1.5y+2z ) = 342$
$x = \text{Regulars Hours}= 40$
$y = \text{Hours Over 40}= 50-40-4 = 6$
$z= \text{Hours on Sundays}= 4$
Substituting $r = \frac{342}{57}= 6$
[/hide]

[quote="sharkman"]Is there another way to answer this question by creating a chart or table?[/quote]
It's trivially easy to make pointless charts and tables to solve problems, just draw a graph and a line and label a couple points that have the answer as coordinates and stuff.

bpms,

Math is better appreciated by people who are not math majors when concepts are explained using easy methods.

Your view?

I think charts and tables help students to grasp what's going on in word problems.

My view is that eventually these concepts get worked out in your own head. 36 posts ago I didn't get my responses either!

Convert his over-time hours into regular hours by multiplying by the appropriate modifier. So, 4(2) and 6(1.5) That make sense?


inequalities
logarithms
function
inequalities proposed

Remember |a|+|b|+|c| >= |a+b+c|. For an arbitrary real x,
\[
\begin{array}{l}
 \frac{1}{2}\left( {|a_i  - x + a_j  - x| + |a_j  - x + a_k  - x| + |a_k  - x + a_i  - x|} \right) \\
  \ge |a_i  - x + a_j  - x + a_k  - x| \\
 \end{array}
\]
Sum them up we have
\[
\begin{array}{l}
 \frac{{|a_1  - x + a_2  - x + a_3  - x|}}{3} + \frac{{|a_2  - x + a_3  - x + a_4  - x|}}{3} + ... + \frac{{|a_n  - x + a_1  - x + a_2  - x|}}{3} \\
  \le \frac{{|a_1  - x + a_2  - x|}}{2} + ... + \frac{{|a_n  - x + a_1  - x|}}{2} \\
 \end{array}
\]
By symmetric majorization criterion, we have
\[
\left( {\frac{{a_1  + a_2  + a_3 }}{3},\frac{{a_2  + a_3  + a_4 }}{3},...\frac{{a_n  + a_1  + a_2 }}{3}} \right) \prec \left( {\frac{{a_1  + a_2 }}{2},\frac{{a_2  + a_3 }}{2},...\frac{{a_n  + a_1 }}{2}} \right)
\]
Then, consider \[
f(x) = \ln x
\]
by majorization inequality we get the desired result.

I think there is a mistake in your proof.

At first, when you was summing them up, where did you find $\frac{1}{3}$?

But this is not all.

For example, for $i=1$, $j=2$ and $k=3$ we have $|\frac{a_1+a_2}{2}-x|+|\frac{a_2+a_3}{2}-x|+|\frac{a_3+a_1}{2}-x| \geq |a_1+a_2+a_3-3x|$. Ok, there is no $\frac{1}{3}$. But there is also no $\frac{a_3+a_1}{2}$ in the original inequality. What are you going to do with it during summing? It's a cyclic inequality. There are $\frac{a_1+a_2}{2}$, $\frac{a_2+a_3}{2}$, $\frac{a_3+a_4}{2}$, $\frac{a_4+a_5}{2}$, ... . But there is no $\frac{a_3+a_1}{2}$!

[quote="harazi"]Isn't this the famous inequality with wrong official solution?[/quote]

Yes, exactly. The article you mentioned was

?. Храбров, [i]Вокруг монголь?кого неравен?тва[/i], in: Задачи Санкт-Петербург?кой олимпиады по математике 2002, p. 146-167.

This was the article from which I learned majorization theory; however, I have never read the proof of the Mongolian inequality in that article, since it looks not really attractive to me (I generally don't like cyclic inequalities with restraints).

  darij

[quote="harazi"]Isn't this the famous inequality with wrong official solution? The famous problem that was open for a few years? I doubt this will work so easy. I remember to have seen a solution using Karamata inequality in a russian article, but I forgot the idea. Does anyone have that article? I think it is from Leningrad Olympiad preparation.[/quote]
Yes, it is that inequality. The article is here (in Russian):
ftp://ftp.mccme.ru/users/vyalyi/matpros/i8149162.ps.zip (Zipped ps file)
ftp://ftp.mccme.ru/users/vyalyi/matpros/i8149162.pdf.zip (Zipped pdf file)

There are two solutions in this article. One of them uses majorization theory. Note that the idea of using this theory was proposed only one year later than the first proof had been obtained. And we see that siuhochung had an idea of using this theory in 30 minutes! Siuhochung, good luck in solving this really hard problem!

P.S. This inequality is not from 'Leningrad Olympiad preparation'. It is indeed from Mongolian olympiad! And V. Adyaasuren is true author of this problem.

Ah, i don't know this inequality is so famous.... I shouldn't have tried this ...  :blush:
Maybe I try it later.... over 99% i will get nothing.  :D

May I know when is this problem proposed?

I conjectured in this forum that for $n=5$ the inequality holds for any positive numbers $a_i$. But this
fact is  already known.

[quote="Vasc"]I recently conjectured in this forum that for $n=5$ the inequality holds for any positive numbers $a_i$.[/quote]

Yes, this is correct. A proof can be found in §5.2 of the paper cited above.

  darij

[quote="darij grinberg"]

Yes, exactly. The article you mentioned was

?. Храбров, [i]Вокруг монголь?кого неравен?тва[/i], in: Задачи Санкт-Петербург?кой олимпиады по математике 2002, p. 146-167.

  darij[/quote]

I remember the article. The solution took several pages. And it can be downloaded at the link indicated by Remike.

Yes, actually if I remember correctly the Mongolian olympiad problem was proposed for arbitrary $n$. And in recent years several special or similar cases have been posted on the 239MO.  :)

[quote="Vasc"]I recently conjectured in this forum that for $n=5$ the inequality holds for any positive numbers $a_i$.[/quote]
Yes, it is proved in that article. The proof uses Jensen's inequality for the function $f(x)=\ln\left(\frac{1}{x}-1\right)$ and numbers $\frac{S-a_k}{2S}$, where $S=a_1+a_2+a_3+a_4+a_5$.

[quote="manlio"]This articles looks very nice, but this language is really impossible to understand so I

must only be satisfied with looking at formulas.

Has anyone, in the next future, the time and patience to translate it from russian?

I would appreciate very much the reading of this very interesting article.

But this is only a my opinion  ;)[/quote]
Ok, [b]manlio[/b]. I'll translate it in few days , if I have time, but I can't translate all article, I'll translate exactly the solution of this nice inequality. :)

OK thank you very much

If I were younger, I would like to learn your language ; I like a lot your types, they are so artistical  ;)

Can you, please, tell me something about your language?

I am very interested to know something about it. :)

Thank you.

[quote="Myth"]Hmm... I think, Armo's mother tongue is armenian, not russian ;)
Am I right?[/quote]
Yes, [b]Myth[/b], you are right (as always) , but in Armenia russian is very popular , almost 90% know russian here.
To [b]manlio[/b]: So it is better to ask [b]Myth[/b] or other russian guys in mathlinks to tell you something about their language, :)

[b]A Mongolian Inequality[/b]
Translated by Myth from Hrabrov's article.

Let $a_1,\ldots,a_n$ be positive real numbers such that $a_1 \geq a_2 \geq \ldots \geq a_n$. Prove the inequality \[\frac{a_1+a_2}{2} \cdot \frac{a_2+a_3}{2} \cdot \ldots \cdot \frac{a_n+a_1}{2} \leq \frac{a_1+a_2+a_3}{3} \cdot \frac{a_2+a_3+a_4}{3} \cdot \ldots \cdot \frac{a_n+a_1+a_2}{3}.\]

[b]Proof.[/b]
Let $x_i=\dfrac{a_i+a_{i+1}}{2}$ and $y_i=\dfrac{a_i+a_{i+1}+a_{i+2}}{3}$.

It is easy to see that
\[x_1\geq y_1\geq x_2\geq y_2\geq ...\geq x_{n-2}\geq y_{n-2}\geq x_{n-1},\]
\[x_n\geq x_{n-1}\mbox{ and }y_n\geq y_{n-1}\geq y_{n-2}\geq x_n.\]
Moreover,
\[x_1+x_2+...+x_n=a_1+a_2+...+a_n=y_1+y_2+...+y_n.\eqno (1)\]
Let $X^*=(x_1^*,...,x_n^*)$ and $Y^*=(y_1^*,...,y_n^*)$ be nonincreasing permutations of $X=(x_1,...,x_n)$ and $Y=(y_1,...,y_n)$ respectively. We will check that $X^* \succ Y^*$. Let's compare sums $\sum_{i=1}^k x_i^*$ and $\sum_{i=1}^k y_i^*$.

First of all, we will find exact expression for the first sum. Indeed, since numbers $x_1$, $x_2$, ..., $x_{n-1}$ go in nonincreasing order, then it is necessary that $k$ largest numbers $x_1^*$, ..., $x_{k}^*$ contain numbers $x_1$, ..., $x_{k-1}$, and a remaining number is a maximum of $x_k$ and $x_n$. It follows that
\[\sum_{i=1}^k x_i^*=\sum_{i=1}^{k} x_i+(x_n-x_k)_+.\eqno (2)\]
Analogously, since numbers $y_1$, ..., $y_{n-2}$ go in nonincreasing order, then it is necessary that $k$ largest numbers $y_1^*$, ..., $y_{k}^*$ contain $y_1$, ..., $y_{k-2}$, and remaining two numbers are two maximums of $y_{k-1}$, $y_k$, $y_{n-1}$ and $y_n$. It follows that
\[\sum_{i=1}^k y_i^*=\sum_{i=1}^k y_i+(y_n-y_k)_++(y_{n-1}-y_{k-1})_+.\eqno (3)\]

[b]Lemma 1.[/b] If $k\leq n-2$ then
\[\sum_{i=1}^k(x_i-y_i)=\frac{1}{6}(a_1+2a_2-a_{k+1}-2a_{k+2}).\]
[b]Proof.[/b] Direct check. $\blacksquare$

[b]Lemma 2.[/b] $X^*\succ Y^*$.

[b]Proof.[/b] We are to show that
\[\sum_{i=1}^k x_i^*\geq \sum_{i=1}^k y_i^*\eqno (4)\]
for all $k\leq n-1$ and
\[\sum_{i=1}^n x_i^*=\sum_{i=1}^n y_i^*.\eqno (5)\]
But last equality is obvious due to (1). Moreover, inequality (4) for $k=n-1$ is true, since it can be obtained from (5) by mean of substraction of correct inequality $x_{n}\leq y_{n}$.

Rewrite (4) using (2) and (3):
\[\sum_{i=1}^{k} x_i+(x_n-x_k)_+ \geq \sum_{i=1}^k y_i+(y_n-y_k)_++(y_{n-1}-y_{k-1})_+\]
for $k\leq n-2$. Applying Lemma 1 here we obtain the following equivalent inequality:
\[a_1+2a_2-a_{k+1}-2a_{k+2}+3(a_1+a_n-a_k-a_{k+1})_
\geq 2(a_n+a_1+a_2-a_k-a_{k+1}-a_{k+2})_+
+2(a_{n-1}+a_n+a_1-a_{k-1}-a_k-a_{k+1})_+.\]
But it is the sum of the following inequlities, which can be easily derived using $(\ )_+$ operation's properties:
\[2(a_2-a_{k+2})+2(a_1+a_n-a_k-a_{k+1})_+\geq 2(a_n+a_1+a_2-a_k-a_{k+1}-a_{k+2})_+,\]
\[a_1-a_{k+1}=(a_1-a_{k+1})+(a_{n-1}+a_n-a_{k-1}-a_{k})_+
\geq (a_{n-1})+a_n+a_1-a_{k-1}-a_k-a_{k+1})_+,\]
\[(a_n+a_1-a_k-a_{k+1})_+=(a_n+a_1-a_k-a_{k+1})_++(a_{n-1}-a_{k-1})_+
\geq (a_{n-1}+a_n+a_1-a_{k-1}-a_k-a_{k+1})_+.\]
$\blacksquare$

Initial inequality is $x_1x_2\dots x_n\leq y_1y_2\dots y_n$, or, if we take logarithms,
\[\ln x_1+\ln x_2+\dots+\ln x_n \leq \ln y_1+\ln y_2+\dots+\ln y_n.\]
We know that function $\ln t$ is concave and we have proved $X^*\succ Y^*$, so from Karamata's inequalities we conclude the result.

No, you should prove it for arbitrary convex function, then, indeed, $X^*\succ Y^*$. To take only $x^2$ is not sufficient for that.

[quote="darij grinberg"]?. Храбров, [i]Вокруг монголь?кого неравен?тва[/i], in: Задачи Санкт-Петербург?кой олимпиады по математике 2002, p. 146-167.[/quote]

I just saw that Alexander Khrabrov (?лек?андр Игоревич Храбров), the author of this article, joined MathLinks! He has a very interesting website: http://www.math.spbu.ru/user/aih/ . In particular, you can find there the [url=http://www.math.spbu.ru/user/aih/papers/mongol.zip]unabridged version of the article I cited above[/url] (the abridged version is the one Remike posted a link to). Note that the article has a few mistakes:

- In the "Замечание" on page 3, the phrase "что чи?ла из набора X более близки друг к другу по величине, чем чи?ла из набора Y" should be replaced by "что чи?ла из набора Y более близки друг к другу по величине, чем чи?ла из набора X".
- On page 4, the first of the three formulas coming directly after the "Определение 3" is incorrect: I am speaking about the formula $\left(\alpha\right)_++\beta\geq \left(\alpha+\beta\right)_+$. One possible way to correct this formula is to replace it by $\left|\left(\alpha\right)_++\beta\right|\geq \left(\alpha+\beta\right)_+$, but I don't know whether this is what the author intended to say.
- At the beginning of the solution of Пример 4, the formula $a=a_1-a_2+a_3-a_4+...-a_{2n-1}$ should be replaced by $a=a_1-a_2+a_3-a_4+...+a_{2n-1}$. Maybe a similar correction should be done for Упражнение 5, I don't know.
- The solution of Пример 4 is incomplete. The order of the numbers $a_2$, $a_4$, ..., $a_{2n-4}$, $a_{2n-2}$, a may be different from what the author thinks. However, it is not difficult to repair the solution.

With all the mistakes (and maybe there are some more), the paper remains a wonderful introduction into majorization theory!

  Darij

Sorry may I know why $x_n\geq x_{n-1}\mbox{ and }y_n\geq y_{n-1}\geq y_{n-2}\geq x_n$ is true?

[quote="baysa"]This inequality can be also proved by using Chebyshev inequality.[/quote]

Please write a little bit details.


algebra
binomial theorem
combinatorics unsolved
combinatorics

Suppose you have 2n books. There are n identical copies of book 1 and 1 copy of the other n books. For example: You have 10 books = 2(5), there are 5 copies of book 1 and 1 copy of book 2, 3, 4, 5 & 6. Find a formula (in terms of n) for the number of ways to choose any n amount of items.

in case one needs a sentence that justifies the answer Pierre has given,just take any subset of the books of single copies.the missing many books will come from copies of book 1.

ok 2(or more) sentences  :)
let the books be numbered $1,1,1,..,1,2,3,...,n+1$ with $n$ copies of $1$, take any subset $S$ of $2,3,...,n+1$.If $|S|=s$ then take $n-s$ copies of $1's$.this makes a choice of $n$ books.Conversely for any choice of $n$ books,take away all the copies of $1's$.what remains will be a subset of $2,3,..n+1$.so the number of choices of $n$ books simply corresponds to the distinct subsets of $2,3,..n+1$.and there are $2^n$ of them.

Soccer:

If it helps, we're using a consequence of the binomial theorem which states that:

$2^n=\binom{n}{0}+\binom{n}{1}+\ldots+\binom{n}{n-1}+\binom{n}{n}$

Does this make it more clear?


abstract algebra
group theory
superior algebra
superior algebra unsolved

Why is it that if G is a finite cyclic group acting on a module A, then the cohomology groups $ H^{n}(G,A)$ has period 2 for n≥1, that is $ H^{n}(G,A) \equal{} H^{n\plus{}2}(G,A)$ ?

It is assumed common knowledge when defining the Herbrand quotient  h(G,A).

See chapter 2, proposition 3.4 (page 78) of http://www.jmilne.org/math/CourseNotes/CFT400.pdf .


G is the central point of $ \bigtriangleup ABC$. AG, BG, CG meets BC,CA, AB at D, E, F respectively, and meets with the circumstribed circle of $ \bigtriangleup ABC$ at A', B', C'.

Find the minimum of $ \frac{A'D}{DA} \plus{} \frac{B'E}{EB} \plus{} \frac{C'F}{FC}$.

[quote="bjh7790"]G is the central point of $ \bigtriangleup ABC$. AG, BG, CG meets BC,CA, AB at D, E, F respectively, and meets with the circumstribed circle of $ \bigtriangleup ABC$ at A', B', C'.

Find the minimum of $ \frac {A'D}{DA} \plus{} \frac {B'E}{EB} \plus{} \frac {C'F}{FC}$.[/quote]
$ G\minus{}$ centroid of $ \triangle ABC$ ; $ AD\equal{}m_a$ , $ BE\equal{}m_b$ , $ CF \equal{} m_c$;
$ AD*DA'\equal{}\frac{a^2}{4}$ and $ \frac {A'D}{DA}\equal{} \frac{a^2}{4m_a^2}$    . . .
So : $ \frac {A'D}{DA} \plus{} \frac {B'E}{EB} \plus{} \frac {C'F}{FC}\equal{}$ $ \frac{a^2}{4m_a^2}\plus{}\frac{b^2}{4m_b^2}\plus{}\frac{c^2}{4m_c^2}\equal{} \frac{a^2}{2b^2\plus{}2c^2\minus{}a^2}\plus{}\frac{b^2}{2c^2\plus{}2a^2\minus{}b^2}\plus{}\frac{c^2}{2a^2\plus{}2b^2\minus{}c^2} \ge 1$


An antique dealer buys a chair for 20 dollars and sells it for 85 dollars.  The dealer later buys back the same chair for 110 dollars and sells it for 135 dollars.  How many dollars total profit did the dealer make on this chair?

[hide]
20 dollars and sells it for 85 dollars. The dealer later buys back the same chair for 110 dollars and sells it for 135 dollars

-20+85-110+135=220-130=90[/hide]

[hide]

-20 + 85 = 65 [profit first time]
-110 +135 = 25 [profit second time]
65+25 = 90 [total profit]

[/hide]


[quote="R.G.A.M."][color=darkblue]   Buenas a todos. Supongo que algunos pocos me conocen, como Daniel Chen, a quien agradezco por haberme invitado al foro.

   Soy Rafael Arias, olímpico matemático nacido en Paraguay. Para mi entrenamiento, me han recomendado que investigue sobre todas las propiedades especiales que puedan tener los puntos concíclicos. Conozco varias de ellas, como las que se cumplen en un cuadrilátero cíclico, mas supongo que ha de haber alguna propiedad especial cuando se generaliza.

   Ya de antemano, muchísimas gracias.

   Un pedido más: ¿Alguien me puede ayudar a manejarme por este servicio. Nuevamente, muchas gracias.[/color][/i][/b][/quote]

Laspropiedades mas interesantes de puntos concliclios se obtienen de diferentes formas.

Una generalizacion del teorema de Ptolomeo es el teorema de Furhman el cual muestra una condicion necesaria y suficiente para que 6 puntos sean conciclicos.

Otraspropiedades fuertes pueden ser obtenidas apartir de la inversion aprovechando que puntos conciclicos se transforman en puntos colineales donde hay muchas relaciones conocidas.
si quieres puedo enviarte algunos items en puntos conciclicos.

[quote="Tony2006"]podrias postear el teorema de furhman??[/quote]

[b]Teorema. [/b](Fuhrmann) Dado el hexagono $ABCDEF$, es  cíclico si y solo si $CF\cdot AD\cdot BE=AB\cdot DE\cdot CF+BC\cdot EF\cdot AD+CD\cdot FA\cdot BE+AB\cdot EF\cdot CD+DE\cdot BC\cdot FA$.

[b]Demostración. [/b]Sean $AB=a$, $BC=b$, $CD=c$, $DE=a'$, $EF=b'$,
$FA=c'$, $CF=e$, $AD=f$, $BE=g$, $BD=x$, $AC=y$, $AE=z$ y $EC=w$.
Aplicando el teorema de Ptolomeo en $\square AFEC: ez=b'y+c'w$...(I), en
$\square ADCB: xy=ac+bf$...(II), en $\square BEDC: xw=ba'+cg$...(III), en $\square AEDB: fg=aa'+xz$...(IV).
Multiplicando (IV) por $e$ obtenemos que $aa'e+exz=efg$,
multiplicando (I) por $x$ y reemplazando tenemos
$aa'e+b'xy+c'wx=efg$. Finalmente, reemplazando las ecuaciones (II) y
(III) obtenemos que $aa'e+b'(ac+bf)+c'(cg+ba')=efg$ de ahi que
$efg=aa'e+bb'f+cc'g+ab'c+a'bc'$. $\blacksquare$


logarithms
inequalities unsolved
inequalities

Solve for all real $ x$, $ y > 0$:

$ 2xy\ln{\frac{x}{y}} < x^2 \minus{} y^2$.

Set t=x/y>0 and get the equivalent condition \[ f(t): \equal{}\ln(t)\minus{}\frac{1}{2}\left(t\minus{}\frac{1}{t}\right)<0\,.\] f' is strictly negative such that f is strictly decreasing on the set of positive reals. Since f(1)=0, f(t)<0 iff t>1, i.e. x>y.


Find explicitly $ X,Y,Z\in\mathcal{M}_{2}(\mathbb{C})$ s.t. $ \{XY\minus{}Z^{2}\equal{}I_{2},YZ\minus{}X^{2}\equal{}I_{2},ZX\minus{}Y^{2}\equal{}I_{2},XY\not\equal{}{YX},XZ\not\equal{}{ZX},YZ\not\equal{}{ZY}\}.$

Cf. also :
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142221

Hint:
take $ X\equal{}\begin{pmatrix}0&0\\0&i\end{pmatrix}$. ($ i^2\equal{}\minus{}1$).


integration
algebra
linear equation
calculus
calculus computations

solve the equation

$ t^2x'\plus{}2tx\minus{}4x^3\equal{}0$ for $ t>0$

Rewriting your ode we get a Bernoulli differential equation: $ x' \plus{} \frac{2}{t}x \equal{} \frac{4x^3}{t^2}$. Now make the substitution $ z(t) \equal{} x^{\minus{}2}$, and we obtain the linear equation $ z' \plus{} \frac{2}{t}z \equal{} \frac{4}{t^2}$, whose solution is

$ z \equal{} \frac{1}{t^2}\left(\int \frac{4}{t^2} \cdot t^2\,dt \plus{} C\right) \equal{} \frac{1}{t^2}(4t\plus{}C)$,

and so the solution is $ \frac{1}{x^2} \equal{} \frac{4t\plus{}C}{t^2} \equal{}> x^2 \equal{} \frac{t^2}{4t\plus{}C}$.


Each morning of her five-day workweek, Jane bought either a $ 50$-cent muffin or a $ 75$-cent bagel.  Her total cost for the week was a whole number of dollars.  How many bagels did she buy?

$ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

$ 0.75\times{2}\plus{}0.50\times{3}\equal{}3.00$. Therefore, she bought 2 bagels (B).

I died on this problem. I thought that the muffins cost 75 cents. They taste better, after all!

I died even more on this problem.  I thought that bagels cost either 50 or 75 cents.  So when I put down 5, I was wondering if something was wrong, but I didn't bother to check my answer.  I'm an idiot...  And if it weren't for this problem, my score would be higher than my 10A score....

This was the easiest problem.

What I did was the same as tinytim's.  I knew that she couldn't have an odd number of bagels, since then the muffins can't make the total a whole number.  Four bagels + 1 muffin isn't whole either, so the answer was 2.

[hide="my solution"]Let us assume that she bought $ x$ muffins. Therefore it cost $ \frac{x}{2}\plus{}\frac{15\minus{}3x}{4}\equal{}\frac{15\minus{}x}{4}$ dollars. Therefore $ 15\minus{}x$ must be a multiple of 4 and less than or equal to 6. Therefore $ x\equal{}3$, and she brought $ 5\minus{}3\equal{}\boxed{2}$ bagels.[/hide]

On the test I just did guess-and-check.

Yeah, I did guess and check on the test.  But here I have another solution.

[hide="another solution"]The number of bagels must be even, so the number of muffins is odd.  Then the total of these end in .50.  Thus the number of bagels is 2 mod 4, so it is 2.[/hide]


Assuming that the probability of the birth of a male or a female is equally likely, what is the probability that a family with three children has exactly two daughters?

Since the probability is equally likely, we would have $ \frac{1}{2}^3$. However, there are $ \binom{3}{2}$ ways to "choose" two children to be girls. Thus, we multiply this by 1/8 to get $ \frac{3}{8}$.


algebra
function
domain
linear algebra
linear algebra theorems

Please, can anyone send me file of the book $ \text{[Flatland: A Romance of Many Dimensions]}$ by Edwin A. Abbott if you have? Or help me finding the book?

Thanks.

Because this is book is public domain, You could find it on the Project Gutenberg site:[url=http://www.gutenberg.org/etext/97]here[/url] or [url=http://www.gutenberg.org/etext/201]here[/url]. HTML version - [url]http://www.geom.uiuc.edu/~banchoff/Flatland/[/url]

Thanks, I'm interested in reading it as well, off the recommendation of one of my math teachers.


Given that $ a\plus{}b\plus{}c\plus{}d\le\frac{16(\sqrt[4]{abcd})}{a\plus{}b\plus{}c\plus{}d}$, prove that $ a\plus{}b\plus{}c\plus{}d<0$.

I think this goes here, since it's not nearly Olympiad level.  :maybe:

[quote="Temperal"]Given that $ a\plus{}b\plus{}c\plus{}d\le\frac{16(\sqrt [4]{abcd})}{a\plus{}b\plus{}c\plus{}d}$, prove that $ a\plus{}b\plus{}c\plus{}d < 0$.

I think this goes here, since it's not nearly Olympiad level.  :maybe:[/quote]

Excuse my impatient tone of voice, but I just spent around 15 minutes first thinking "man. this'll be easy. AM-GM trivial.", then "uhh...", and then "god I'm dumb."

But um. That clearly isn't true, like especially if $ a \equal{} b \equal{} c \equal{} d \equal{} 1$.

Also, let $ a \equal{}\minus{}b$, with $ |a|$ arbitrarily (and by arbitrarily I mean "incredibly") large. Now let $ c \equal{}\minus{}1$ and $ d > 1$.

Maybe you'd be better off restricting $ a,b,c,d$ to positive values and restating the question to "prove that $ a\plus{}b\plus{}c\plus{}d\leq4$."

[quote="Temperal"]Given that $ a\plus{}b\plus{}c\plus{}d\le\frac{16(\sqrt [4]{abcd})}{a\plus{}b\plus{}c\plus{}d}$, prove that $ a\plus{}b\plus{}c\plus{}d < 0$.

I think this goes here, since it's not nearly Olympiad level.  :maybe:[/quote]

Perhaps he means $ a\plus{}b\plus{}c\plus{}d <\frac{16(\sqrt [4]{abcd})}{a\plus{}b\plus{}c\plus{}d}$... :maybe:

Do you mean integer solutions? Otherwise, how is this a number theory problem...?

[quote="xpmath"]He posted this on another website too; I think he meant: Given that $ a\plus{}b\plus{}c\plus{}d\le\frac{4(\sqrt [4]{abcd})}{a\plus{}b\plus{}c\plus{}d}$, prove that $ a\plus{}b\plus{}c\plus{}d < 0$.[/quote]

Yes, that's what I meant. Sorry for any inconveniences.

[quote="pianoforte"]i don't think that's true for a=b=c=d=1[/quote]

Er...? Now that I've updated the equation, $ a \equal{} b \equal{} c \equal{} d \equal{} 1$ does not satisfy the equation.

I couldn't edit my post, though....

You've got a problem that the inequality isn't homogeneous. All terms on the RHS are degree 0, while all terms on the LHS are degree 1. That means that given any set of positive $ \{a,b,c,d\}$, you can multiply each one by a constant that's arbitrarily close to 0, and eventually the condition will be satisfied. But since $ a,b,c,d$ are positive, clearly the conclusion isn't going to follow.

Yeah the problem is totally ridiculous. If we write it as

$ (a\plus{}b\plus{}c\plus{}d)^{2}\le 4\sqrt [4]{abcd}\Rightarrow a\plus{}b\plus{}c\plus{}d < 0$

we immediately see that by switching the signs of each of $ a, b, c, d$ preserves the given condition but switches the signs of $ a\plus{}b\plus{}c\plus{}d$. Here's a salvage that is lame and weak: given reals $ a, b, c, d$ such that $ abcd\ge 0$, show that

$ a\plus{}b\plus{}c\plus{}d < 4\sqrt [4]{abcd}\Rightarrow\min\{a,b,c,d\}< 0$.

EDIT: Yeah, my bad, see comment below.

For the salvage, $ a \equal{} b \equal{} c \equal{} d$ is a counterexample. If you make the condition a strict inequality then I think it is true.

Proof: Suppose not. AM-GM. Contradiction. Whee.

That's what I had in mind.... but I guess this is a pretty horrible problem... sorry.


Define f(x)=asinx + bxcosx - 2x, where a and b are constants. Given that f(2)=10, compute f(-2).

Solution:

[hide]Substituting,

10 = asin(2) + 2bcos(2) - 4

f(-2) = asin(-2) - 2bcos(-2) + 4

Noting that cos(x) = cos(-x) and sin(x) = -sin(-x) and adding equations,

10 + f(-2) = a(sin(2) + sin(-2))

10 + f(-2) = 0

f(-2) = -10[/hide]

...or [hide]just note that all three terms are odd functions[/hide]

well basically, but I like one-liners better ;).


group theory
abstract algebra
superior algebra
superior algebra theorems

Apologies in advance for posting such an elementary question that all of you will be able to answer with a wave of your Magic Theorem-Proving Stick.

So there's this one homomorphism theorem that goes "if $H$ is a subgroup and $N$ a normal subgroup of a group $G$, then $HN$ is a subgroup of $G$, $H \cap N$ is a normal subgroup of $G$, and $H/(H \cap N)$ is isomorphic to $(HN)/N$.  My textbook left it as an exercise to the reader, so apparently the proof is trivial :maybe:   How do you prove it?  I've managed to prove the first two parts but not the third.  Also, how can you interpret the third statement semi-intuitively?  I just can't wrap my mind around it.

Thanks  :lol:

That isomorphism needs to be canonical, and it's going to come from the quotient relations.
For all $a\in H$, map $a(H\cap N)$ (in $H/(H\cap N)$) to $aN$ (in $HN/cap N$).
Note that $a(H\cap N)$ is $aN\cap H$.

This is the isomorphism: verify the necessary properties.


Find all n so that 19n+1 and 95n+1 are both perfect squares

Hint: Pells

Let $95n+1=u^{2},19n+1=v^{2},\ u,v\in\mathbb{Z}$. Note that $u^{2}-5v^{2}=-4$.  Solve this and you're done.

Masoud Zargar


Find all a,b,c are possible integer numbers such as:$ a^2\minus{}b,b^2\minus{}c,c^2\minus{}a$are perfect quare numbers

If $ a^2 \minus{} b,b^2 \minus{} c,c^2 \minus{} a$ is a perfect square then $ 4a^2 \minus{} 4b,4b^2 \minus{} 4c,4c^2 \minus{} 4a$ is a perfect square .
Let $ a \equal{} \max\{a,b,c\}$ then we have :
If $ a > b$ then :
$ (2a \minus{} 1)^2<4a^2 \minus{} 4b < (2a)^2$
Imply that $ 4a^2 \minus{} 4b$ can not be a perfect square.
So $ a \equal{} b$
Because $ a^2 \minus{} a$ is a perfect square then $ a \equal{} 1$
So $ a \equal{} b \equal{} c \equal{} 1$
This equation has solution $ a \equal{} b \equal{} c \equal{} 1$


Merged with [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265273[/url] and locked.


Find all non-negative rational solutions of this equation:

$ a^b\cdot b^a \equal{} b^b$

[hide="logs"]
Take the base a log of both sides.
$ ab\log_a{b}\equal{}b\log_a{b}$
$ a\equal{}1$
Finish with a substitution

[/hide]

Uhh, I don't see what's wrong with it, but that's probably just me.
[hide="because"]
$ \log_a{a^b} \cdot \log_a{b^a}$
$ \equal{}b\log_a{a} \cdot a\log_a{b}$
$ \equal{}ba\log_a{b}$
[/hide]

[hide="It is just you :)"]$ \log(a^bb^a)\neq\log a^b\cdot\log b^a$[/hide]

:rotfl: Okay, you win.  Still wondering how I managed to miss that two days after learning it in class.

Give it some time to sink in ;)

(And another detail - your final conclusion that $ a\equal{}1$ would imply you have an illegitimate log base.)

Thanks Farenhajt
With your hint, b=ak, I got this solution

$ \\a^b\cdot b^a \equal{} b^b\rightarrow\\ a^{ak}\cdot (ak)^a \equal{} (ak)^{ak}\rightarrow\\ a^k\cdot ak \equal{} (ak)^k\rightarrow\\ a \equal{} k^{k\minus{}1}\rightarrow\\ b \equal{} k^k$

and it seems right

You just need to discuss when those are rational (confer the original text).


If true in general, this would be very nice: Let $a$ and $b$ be nonnegative reals and let $k \ge -1$ be an integer. Then

$(1+k)(a-b)^2 + 8ab \ge 2(a+b)\sqrt[k]{2^{k-1}a^k+2^{k-1}b^k}$

With equality iff $a=b$ excepting where $k=\pm 1$, which are identities. I lack a general proof but have shown the statement for $k=-1$ (identity), $k = 0$ (where we interpret the zero-radical as the suitably weighted geometric mean $2\sqrt{ab}$), and $k = 1$ (identity), $2, 3, 4, 5, 6, 7$.

...

Edit - In an effort to deduce how this behaves for $k$ real, based on graphs from Mathematica, it appears that for the intervals $(-\infty,-1]$, $[-1,1]$, $[1,1.5]$, and $[1.6,\infty)$ the inequality signs are $\le$, $\ge$, $\le$, and $\ge$ respectively. For $k \in (1.5,1.6)$ the behavior is seriously complicated.

(It is, in my opinion, entirely uninspiring in terms of leading to a general solution.) With the help of Mathematica, observe that

$(8(a - b)^2 + 8ab)^7 - (2(a + b))^7(64a^7 + 64b^7) = $

$8192(a-b)^4(255a^{10}-779a^9b+2501a^8b^2-4049a^7b^3$ $+ 6608a^6b^4 - 6608a^5b^5  + 6608a^4b^6 - 4049a^3b^7 + 2501a^2b^8-779ab^9+255b^{10}) = $

$8192(a - b)^4(255(a - b)^{10}) + 57344(a - b)^4(253ab(a - b)^8 + 742a^2b^2(a - b)^6 + 1161a^3b^3(a - b)^4) + 917504a^4b^4(a - b)^4(61a^2 - 100ab + 61b^2) \ge 0$

A fact about the inequality that is perhaps more inspiring is that upon dividing through by $b^2$, we obtain $(1+k)\left(\frac{a}{b}-1\right)^2+\frac{8a}{b} \ge 2\left(\frac{a}{b}+1\right)\sqrt[k]{2^{k-1}\frac{a^k}{b^k}+2^{k-1}}$. Since the original inequality is symmetric, we have that this inequality holds iff it holds upon exchanging $a$ and $b$. If $x = \frac{a}{b}$, and $f(x) = LHS - RHS$, this causes $f'(1)=0$. Moreover, the second-degree convexity of both sides was made equal by equating $f''(1)=0$ and choosing the appropriate weight for $(a-b)^2$.

can you solve it without the help of mathematica? Expanding by hand for such a larger power is quite unreasonable i think. If it's possible to solve it by hand for the case k=7, that solution might give us some insights on the general case.

I think because the inequality is so sharp, some expansion/factoring will be necessary. Here is the best idea I have so far: because of the way the inequality is parameterized (see my earlier post regarding the fact $f(1) = f'(1) = f''(1) = f'''(1) = 0$), we know that $(a-b)^4$ will factor the expansion for all $k$. We remains to see is that the left over factor is positive. Hence, all we need is to show that this factor is in the span of

$\sum_{i=0}^{n}c_ia^ib^i(a-b)^{2n-2i}$

where $c_0,\dots, c_n \in \mathbb{R}^{+}_0$. Actually, we could apply these very identities recursively to combine terms and allow some negative $c_i$'s - a daunting prospect. :surrender: However, it is true that terms corresponding to $c_i$ for low values of $i$ are much stronger than straight AM-GM.  With this in mind, if we apply AM-GM weighted by a factor that we know is close to 0 for near equality cases, we increase the sharpness of AM-GM. As an example, observe that

$a^6 - 9a^4b^2 + 16a^3b^3 - 9a^2b^4 + b^6 = (a-b)^6 + 6ab(a-b)^4 \ge 0$. :yoda:

Straight AM-GM would give $a^6 + a^3b^3 + a^3b^3 \ge 3 a^4b^2$ or $a^6 - 3a^4b^2 + 4a^3b^3 - 3a^2b^4 + b^6 \ge 0$, which is too weak. A little supporting algebra, however, gives

$a^6 - 9a^4b^2 + 16a^3b^3 - 9a^2b^4 + b^6 \ge 0$
$\iff a^6 - 2a^3b^3 + b^6 \ge 9a^4b^2-18a^3b^3 + 9a^2b^4$
$\iff (a^3-b^3)^2 = (a-b)^2(a^2+ab+b^2)^2 \ge (a-b)^2(3ab)^2 = 9a^2b^2(a-b)^2$

as desired. :coolspeak:

I am trying to approach the problem in this manner. (Okay, so I also wanted to see the new emoticons.)

If the $k$th power mean of two variables is a concave function of $k$ for large enough $k$, then I can show it. In particular, I am only lacking only a proof of the following:

[b]Lemma.[/b] For any positive integer $k$ and real number $a$ with $0 \le a \le 1$,

$ 2\left(\frac{a^{k+1}+1}{2}\right)^{\frac{1}{k+1}} \ge \left(\frac{a^{k}+1}{2}\right)^{\frac{1}{k}} + \left(\frac{a^{k+2}+1}{2}\right)^{\frac{1}{k+2}} $

Just an update, this is no longer a conjecture. It is true as conjectured in post 1 for the sets $(-\infty,-1],\{-1,0,1\},[1,3/2],[2,\infty)$. The factor $1+k$ is optimal because it leads to a $LHS - RHS$ difference of $c_1(a-b)^4$ near equality for all real $k$ except $k=3/2$, where it leads to $c_2(a-b)^6$. The inequality is sharp globally for $|k|$ small. (By inspection, as $|k|$ grows large, the magnitude of the left approaches infinity while the right is bounded.)

I have completed a full proof and submitted it to American Mathematical Monthly. Either they will publish it or I will post it on my website.

The note was not accepted. Therefore, I have posted it at http://web.mit.edu/~tmildorf/www/ASharpBound.pdf Enjoy!


Let x1, x2, . . . , xn be positive numbers such that S = x1+x2+· · ·+xn =
1
x1
+· · ·+
1
xn.
Prove that
n Xi=1
1
n − 1 + xi
n Xi=1
1
1 + S − xi.

Bomb

Well it is the 5th mathlink contest 3rd round problem 3. we r discussing there. so u may look at there! ;)


vector
rotation
linear algebra
calculus
calculus computations

please be rigorous proving this problem:

suppose there are two vectors, $ \vec x$ and $ \vec y$ in $ \mathbb{R}^n$, n>2, with the Euclidean metric, and $ d(\vec x, \vec y) \equal{} d$. Let $ 2r > d$ and prove that there exist infinity many vectors $ \vec z$ such that
$ d(\vec x, \vec z) \equal{} d(\vec y, \vec z)\equal{}r$

geometrically, the solution seems obvious, but how does one prove the statement rigorously?

Can you describe what exactly are the symbols $ d(\vec x,\vec y)$ and $ r$?

$ d(\vec x, \vec y)$ is the Euclidean metric,

$ d(\vec x, \vec y) \equal{} \sum_{i \equal{} 1}^n \left(x_i \minus{} y_i\right)^2$

I apologize for the double meaning $ d$ has... I should've been a bit clearer.
$ d$ is also the distance between $ \vec x$ and $ \vec y$

$ r$ is a real number $ >d/2$

[quote="jmerry"]The problem with $ r$ is that you haven't actually used it. Is it $ d(x,z)$?

One way to do this is to explicitly find the circle of points satisfying this condition. Once you find one point, you can rotate it around the line between $ x$ and $ y$; there are infinitely many angles we could rotate by.

That's the $ \mathbb{R}^3$ version- more dimensions means even more points.[/quote]

yes :blush: ... I forgot the r.

I know that the solution seems very obvious geometrically, but how can one prove it completely rigorously? I mean there has got to be a better way than finding the solutions explicitly... afterall, this is suppose to be an analysis problem.

just a side note... the problem is from Rudin's book for analysis. no linear algebra should be required to solve this... i'm still looking for a solution.