Given $ a, b, c > 0$. Prove that: $ \sqrt[4]{\frac{a^3\plus{}b^3\plus{}c^3}{abc}} \ge 2\sqrt[4]{3}\left(\frac{a}{a\plus{}2b\plus{}3c}\plus{}\frac{b}{b\plus{}2c\plus{}3a}\plus{}\frac{c}{c\plus{}2a\plus{}3b}\right)$
:)

[hide]

hint : use the trivial (strict) ineq :
$ \frac{ (a\plus{}b\plus{}c)^9}{2} \geq  3abc( (a\plus{}b\plus{}c)^2\plus{}3(a^2\plus{}b^2\plus{}c^2))((a\plus{}b\plus{}c)^4\plus{}9(a^2\plus{}b^2\plus{}c^2)^2)$ :wink: [/hide]

Can you post your full proof, anas? By Am-Gm, I can prove the strongers easily  :wink:

Yes, it's also the way I created it, 2424  :lol:

well it's easy to prove that :  $ RHS \leq   \frac{ 3(a^2\plus{}b^2\plus{}c^2)}{ 2(a\plus{}b\plus{}c)^2}$  and now it's  enough to prove that :
 $ \frac{a^3\plus{}b^3\plus{}c^3}{3abc} \geq  ( \frac{3(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2})^4$ that is :

$ ( \sum{a^2}\minus{}\sum{ab})( M) \geq 0$  when $ M$ is the expression that I mentionned in the hint   :P

Thank you, anas. And will your way work for the stronger:  :)
$ \sqrt[4]{\frac{a^{3}\plus{}b^{3}\plus{}c^{3}}{abc}\plus{}1}\ge 2\sqrt[4]{4}\left(\frac{a}{a\plus{}2b\plus{}3c}\plus{}\frac{b}{b\plus{}2c\plus{}3a}\plus{}\frac{c}{c\plus{}2a\plus{}3b}\right)$

[quote="2424"][quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \sqrt [4]{\frac {a^3 \plus{} b^3 \plus{} c^3}{abc}} \ge 2\sqrt [4]{3}\left(\frac {a}{a \plus{} 2b \plus{} 3c} \plus{} \frac {b}{b \plus{} 2c \plus{} 3a} \plus{} \frac {c}{c \plus{} 2a \plus{} 3b}\right)$
:)[/quote]
By CS,we have
$ LHS^4 \leq 48(a^2 \plus{} b^2 \plus{} c^2)^2( \frac {1}{(a \plus{} 2b \plus{} 3c)^2} \plus{} \frac {1}{(b \plus{} 2c \plus{} 3a)^2} \plus{} \frac {1}{(c \plus{} 2a \plus{} 3b)^2})^2$
and we have
 $ \frac {1}{(a \plus{} 2b \plus{} 3c)^2} \plus{} \frac {1}{(b \plus{} 2c \plus{} 3a)^2} \plus{} \frac {1}{(c \plus{} 2a \plus{} 3b)^2} \leq \frac {1}{4(ab \plus{} bc \plus{} ca)}$
(it's not hard by AM-GM)
and now,we only need prove
 $ \frac {a^3 \plus{} b^3 \plus{} c^3}{abc} \geq 3( \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca})^2$
$ < \equal{} > (a^3 \plus{} b^3 \plus{} c^3)(ab \plus{} bc \plus{} ca)^2 \geq 3(a^2 \plus{} b^2 \plus{} c^2)^2abc$
it's true because
$ (a^3 \plus{} b^3 \plus{} c^3)(ab \plus{} bc \plus{} ca)^2 \geq 3(a^3 \plus{} b^3 \plus{} c^3)(a \plus{} b \plus{} c)abc \geq 3(a^2 \plus{} b^2 \plus{} c^2)^2abc$
So, we are done! :)[/quote]
$ \frac{1}{(a+2b+3c)^{2}}+\frac{1}{(b+2c+3a)^{2}}+\frac{1}{(c+2a+3b)^{2}}\leq\frac{1}{4(ab+bc+ca)} $
The inequality is wrong with a=b=k,c=0

[quote="vthiep1994"]
$ \frac{1}{(a+2b+3c)^{2}}+\frac{1}{(b+2c+3a)^{2}}+\frac{1}{(c+2a+3b)^{2}}\leq\frac{1}{4(ab+bc+ca)} $
The inequality is wrong with a=b=k,c=0[/quote]
Are you sure?

from sos ,i think it is true for three real numbers $a$,$b$,$c $with $ab+bc+ca>0$?

[quote=arqady][quote="2424"]
and we have
 $ \frac {1}{(a \plus{} 2b \plus{} 3c)^2} \plus{} \frac {1}{(b \plus{} 2c \plus{} 3a)^2} \plus{} \frac {1}{(c \plus{} 2a \plus{} 3b)^2} \leq \frac {1}{4(ab \plus{} bc \plus{} ca)}$
[/quote]
The following inequality is also true.
For non-negatives $a$, $b$ and $c$ such that $ab+ac+bc\neq0$ prove that:
\[\frac {1}{(a \plus{} 2b \plus{} 4c)^2} \plus{} \frac {1}{(b \plus{} 2c \plus{} 4a)^2} \plus{} \frac {1}{(c \plus{} 2a \plus{} 4b)^2} \leq \frac {9}{49(ab \plus{} bc \plus{} ca)}\]
It seems that I've seen this inequality before. :maybe:[/quote]
 Let $a$, $b$ and $c$ be positive numbers. [url=https://artofproblemsolving.com/community/c6h1537579p9284628]Prove that:[/url]
$$\frac{1}{(4a+4b+c)^3}+\frac{1}{(4a+4c+b)^3}+\frac{1}{(4b+4c+a)^3}\leq\frac{1}{243abc}.$$


$1$.ask a person to think of a three digit number (say $abc$)

$2$. below it write the number with the first two digits interchanged ($bac$)

$3$. next write down the second number with the last two digits exchanged ($bca$)

$4$.below it write the third number with the first two digits interchanged ($cba$)

$5$.below it write the fourth number with the last two digits interchanged ($cab$)

$6$.below it write the fifth number with the first two digits interchanged ($acb$)

Basically this permutes the digits. if the original number had two digits identical, then you have $2$ sets of three numbers. if all the digits were identical you have the same number repeated $6$ times.

$7$ now erase the first number(the number thought of) and add up the rest. suppose the total is $T$.

find out an easy way to determine the number thought of, given the number $T$

I believe there's an error in the example for step 6...
shouldn't it be acb?

[quote="robinhe"]Yeah probably.  He said all of the permutations of a number $abc$.

[hide]Let $abc=100a+10b+c$.  If we add up $abc+acb+...$ we get $100a+10b+c+100a+10c+b+100b+10c+a+...=222(a+b+c)$.

So $T+abc$ is divisible by 222.

So I think from there you just plug in numbers for $a+b+c$ (the whole thing, not plugging in numbers for each of a, b, and c) that when multiplied by 222 is greater than T.

Basically guess-and-check after the initial calculating parts.[/hide][/quote]

that was a typo error, Thanks for notifying.. I have already corrected it.

[quote="calculuslover800"][quote="robinhe"]

[hide]Let $abc=100a+10b+c$.  If we add up $abc+acb+...$ we get $100a+10b+c+100a+10c+b+100b+10c+a+...=222(a+b+c)$.

So $T+abc$ is divisible by 222.

So I think from there you just plug in numbers for $a+b+c$ (the whole thing, not plugging in numbers for each of a, b, and c) that when multiplied by 222 is greater than T.

Basically guess-and-check after the initial calculating parts.[/hide][/quote]

[hide="expanding on that,"]
Let $x = 100a + 10b + c$. If we find the remainder when $T$ is divided by 222, then we know that the remainder when $x$ is divided by 222 is going to be 222 - ($T$'s remainder).  For example, if $T=3798$ then $T \equiv 24 \mod{222}$ and $x \equiv 198\mod{222}$.  In this case it leaves 4 possiblilites for $x$: 198, 420, 642, 854.  Now, divide every possible value $T+x$ by 222.  If the result is the sum of the digits of the $x$ you used, then that possibility works.  In this case,
$(3798+198)/222 = 18$
$(3798+420)/222 = 19$ (Since 420 is 222 more than 198, all you do is add 1)
$(3798 + 642)/222 = 20$
$(3798 + 854)/222 = 21$

Since 1 + 9 + 8 = 18, and none of the other sums work, 198 is the only possible solution.

[/hide][/quote]

This was good... and fast.


http://www.spoj.pl/problems/PALSEC/
First I tried bruteforce, but that didn't really work. I checked the polish site, but the solutions were only in polish, so I couldn't understand a thing. Can someone help?

My first thought is something like dp[a][b] = number of ways to get palindromes only using words between indices a and b. Then we can find the relation between dp[a][b] and dp[a+1][b-1] depending on what $x_{a}$, $y_{a}$, $x_{b}$, and $y_{b}$ are.

EDIT: Looks like that is a bit too simple since words can be of different lengths...


I think problem is not difficult:

$\sum\limits_{cyc} \frac{a_1^n}{p_1a_2 + p_2a_3 +...+p_ma_1} \geq \frac{\sum\limits_{i=1}^{n} a_i^{n-1}}{\sum\limits_{i=1}^{m}p_i}$

[quote="Nameless"]$n\geq 1$?
$a_i>0$?
$n=m$? :huh:[/quote]

$n\geq 1$ -> OK
$a_i>0$ -> OK
$n=m$ -> OK :roll:

[quote="Nameless"]I think you misunderstand($a_i$ co lon hon 0 khong ? n co lon hon 1 khong? Va tu gia thiet ta co the suy ra m=n)[/quote]

Yes :) All everything are true, (tớ cũng không hiểu ý của cậu là gì ??)
I think you proved following : $\sum\frac{a^2}{b+c} \geq \frac{a+b+c}{2}$
(bài này tổng quát hơn)

Where are they?[b] zhaobin, Hungkhtn, Namless...[/b]


geometry
geometric transformation
reflection
rotation
abstract algebra
group theory
superior algebra

{[b]Background[/b]}
Let S be the plane \[ S={(x,y)}\] where x and y are real
and consider \[ f,g \in A(S)\] defined by f(x,y) = (-x,y) and g(x,y) = (-y,x);  f is the reflection about the y axis and g is the rotation through 90 degrees.  We then define
	\[ G={f^{i}g^{j}|i=0,1; j=0,1,2,3}\]

	for n>2 and let h be the rotation in the plan about the origin through an angle of \[ 2\pi/n\]

	We then define  	\[ G={f^{k}h^{j}|k=0,1; j=0,1,2,....,n-1}\]

	One can verify that \[ f^{2} = h^{n}= \]identity  mapping
	and  \[ fh=h^{-1}f\]

[b]Problem[/b]
For n=3, find a formula that expresses
	\[ {f^{i}h^{j} \ast\[{f^{s}h^{t}=f^{a}h^{b}}}\]

	Show that G is a non abelian group of order 6

I have worked the problem to get to the point
	\[ h^{t}f = fh^{-t}\]  and

	\[ f^{i}h^{j}\ast fh^{t}= f^{i}fh^{-j+t}= f^{i+1}h^{t-j}\]

	My question is with the next step how do they get to
	\[ f^{i}h^{j}f^{0}h^{t}= f^{i}h^{j}f^{s}h^{t}= f^{a}h^{b}\]

	and the next step of
	a=i+j and
	\[ b = t + (-1)^{s} j\].  Thus G is closed under the product of mappings. Since
	\[ e=f^{2}h^{3}\] , e is in G. .....???

Remember that $ f^i$ is either $ f$ or the identity.

You've already verified that you can pull all the f's to the left.  Hence a is either 0 or 1 (in particular, a=i+s(mod2)).

Now think about what happens when you switch the order of $ h^j f^s$.  If s is even, $ f^s \equal{} 1$ and j stays as is.  Otherwise, the sign of j is reversed.  We can capture this by writing
$ h^j f^s$ = $ f^s h^{(\minus{}1)^s j}$

Verifying that G is a group is easy now, since we can write down all the elements in the form $ f^i h^j$.

To show it has order six, write down the elements:
$ G\equal{} \left\{ f^i g^j | i \equal{} 0,1 , j \equal{} 0, 1, 2 \right\}$
which gives six distinct elements.  Showing that it's nonabelian is also simple ($ fg \neq gf$).

Thanks so much for clarifying, and the quick response.  I am into the next section, and with your input this makes much more sense.   I really appreciate it.


calculus
integration
logarithms
derivative
function
calculus computations

27. $\frac{d}{dx}\left( \int^{x^{3}}_{0}\ln{(t^{2}+1)}dt \right)$

$\text{(A)}\ \frac{2x^{3}}{x^{6}+1}$
$\text{(B)}\ \frac{3x^{2}}{x^{6}+1}$
$\text{(C)}\ \ln(x^{6}+1)$
$\text{(D)}\ 2x^{3}\ln(x^{6}+1)$
$\text{(E)}\ 3x^{2}\ln(x^{6}+1)$

I'm not good with integral and derivative at the same time as this. All I know for this problem is that $d(x \ln x-x) = \ln x$.

Look up "Fundamental Theorem of Calculus" in your textbook - it should be in the index.

Let $G(u)=\int_{0}^{u}\ln(1+t^{2})\,dt$

The FTC tells you what $G'(u)$ is. (Remember that that's a function of $u.$

You're actually asked for the derivative of $G(x^{3}).$ So you must use the chain rule.

Note: you do [i]not[/i] differentiate the function inside the integral. The answer should have a logarithm in it.

If $f\in\mathrm C$ and $\{u,v\}\subset \mathrm C'$, then

$\boxed{G(x)=\int_{u(x)}^{v(x)}f(t)\ dt\implies G'(x)=f(v(x))\cdot v'(x)-f(u(x))\cdot u'(x)}$.

sometimes this is called the second fundamental theorem of calculus

it basically says that the derivative of the integral of a function is that function once again

$\frac{d}{dx}\int_{C}^{x}{f(t)dt}= f(x)$

C is any constant
this is then generalized with the chain rule

$\frac{d}{dx}\int_{C}^{u}{f(t)dt}= f(u) * u'$

and then generalized even more by making both bounds into functions

$\frac{d}{dx}\int_{v}^{u}{f(t)dt}= f(u) * u'-f(v) * v'$

Hmm. This theorem sounds really familiar but I don't think I really remember what it was.  :roll:

So I looked it up and book clarified it little bit. But I still want to know if this is right.

$\frac{d}{dx}\left( \int^{x}_{0}\ln(t^{2}+1) dt \right) = \ln(x^{2}+1)$

But when I change $x$ to some other expressions.. say $x^{5}+4x^{4},$ then my answer will be $\ln(x^{2}+1) \cdot (5x^{4}+16x^{3})$. Correct?

You have to replace all of the $x$ values.

$\ln\left((x^{5}+4x^{4})^{2}+1\right)\cdot(5x^{4}+16x^{3}).$

Oh yes. That's what I meant. I just forgot to substitute that. Thank you.

[quote="Silverfalcon"]Hmm. This theorem sounds really familiar but I don't think I really remember what it was.  :roll: [/quote]

It's Leibniz integral rule. See [url=http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign]this page[/url] for the more general case. You can easily work out the proof considering $F = F(x,b(x),a(x))$ and noting that $\frac{1}{\delta b(x)}\int_{b(x)}^{b(x)+\delta b(x)}f(x,t) dt = f(x,b(x))$.


6^17 + 17^6 is divisble by 3 or 7?

6^1 = 0(mod 3)
(6^1)^17=0(mod 3) so 6^17 is divisible by 3
how do u do this?

[hide]$6^{17}=36^{8}\cdot6$
You can divide each of this numbers by 7:
$36^{8}(mod 7)\equiv 1^{8}(mod 7)$, So you multiply that one by the other six we had alone. $6^{17}\equiv 6 (mod 7)$

$17^{6}=289^{3}$
Now you divide $\frac{289}{7}$ to get a remainder of $2$.
$2^{3}\equiv 1 (mod 7)$

Add this $1$ to our other $6$ to get $7$. This shows that seven evenly divides $6^{17}+17^{6}$[/hide]

$6^{17}+17^{6}\text{, divisible by 3 or 7?}$

Solution:
[hide][u]For $3$, note that:[/u]
$6 \equiv 0 \text{mod }3 \text{ and }17 \equiv 2 \text{mod }3$

Thus:
$6^{17}\equiv 0^{17}\text{mod }3 \text{ and }17^{6}\equiv 2^{6}\text{mod }3$
$6^{17}\equiv 0 \text{mod }3 \text{ and }17^{6}\equiv 64 \text{mod }3$
$[6^{17}+17^{6}] \equiv [0+64] \text{mod }3$
$[6^{17}+17^{6}] \equiv 64 \text{mod }3$

$64 \equiv 1 \text{mod }3$

$\rightarrow [6^{17}+17^{6}] \equiv 1 \text{mod }3$

Thus it is not divisible by $3$.

[u]For $7$, note that:[/u]
$6 \equiv (-1) \text{mod }7 \text{ and }17 \equiv 3 \text{mod }7$

Thus:
$6^{17}\equiv (-1)^{17}\text{mod }7 \text{ and }17^{6}\equiv 3^{6}\text{mod }7$
$6^{17}\equiv-1 \text{mod }7 \text{ and }17^{6}\equiv 729 \text{mod }7$
$[6^{17}+17^{6}] \equiv [-1+729] \text{mod }7$
$[6^{17}+17^{6}] \equiv [728] \text{mod }7$

Note that $728=700+28$ and:
$700 \equiv 0 \text{mod }7 \text{ and }28 \equiv 0 \text{mod }7$ thus
$728 \equiv 0 \text{mod }7$

$\rightarrow [6^{17}+17^{6}] \equiv 0 \text{mod }7$

Thus our number is divisible by $\boxed{7}$.[/hide]

[hide]Note that you can save work because $17^{6}\equiv 1 \pmod{7}$ by Fermat's Little Theorem[/hide]

[u]Follow-up question[/u]
What is the prime factorization of $6^{17}+17^{6}$?

So we know its divisible by 7.

We also know that $17^{2}=289 \equiv-1 \pmod{5}\implies N \equiv 1^{17}+(-1)^{3}\equiv 0 \pmod{5}$

So far, we've got that its divisible by $5$ and $7$. Ok, so now what?

[hide="Answer"]Actually, it turns out that $N=16926683582305=(5)(7)(305717)(1581919)$, using a computer.[/hide]

If they have the same parity its divisible by $2$... lots of little cases. You can't really generalize this that well; ex. if you let $b=1$, you get $b+1$, and you can't really tell how many distinct prime factors that has... what exactly do you want?

[quote="miyomiyo"]Hmmmmm.....

In general, how many distinct prime factors does $a^{b}+b^{a}$ have?[/quote]

Erm, take $a=p-1$ and $b=1$. I don't think it generalizes.  :)

I'm not asking for a general formula if it sounded like that.  I'm merely asking for a way of finding the prime factorization.  I don't know the answer or I wouldn't be asking.


Can experts in number theory post the proof of "every Mersenne prime corresponds to exactly one perfect number".

Thank you very much! :P

see http://mathworld.wolfram.com/PerfectNumber.html

or http://en.wikipedia.org/wiki/Perfect_number#Even_perfect_numbers

I would like to have the proof of "every Mersenne prime corresponds to exactly one perfect number", but I do not think they are in wikipedia or mathworld, can someone please posts the proof here? :)

Thank you very much! :P

[quote="K81o7"]Here's a proof.  Consider the monotonically increasing sequence of Mersenne primes, and the monotonically increasing sequence of perfect numbers.  Now pair them up, one by one!  :)
Unless you want a proof that any perfect number is of the form $2^{p-1}(2^{p}-1)$ where $2^{p}-1$ is a Mersenne prime?[/quote]

K81o7, can you please post the proof that any perfect number is of the form $2^{p-1}(2^{p}-1)$, this is what I am looking for. :)

Thank you very much! :P

[quote="ZetaX"]There is no such proof known, it's only proven that any [b]even[/b] perfect number is of that form.[/quote]

Mathematicians believe that all the perfect numbers are even (though it has not been proven), so I was actually thinking of the proof any [b]even[/b] perfect number is of the form $2^{p-1}(2^{p}-1)$. :maybe:
Can you please post the proof here?

Thank you very much! :P

You see section 16.8 page 239 in book follow
[quote="N.T.TUAN"]Here is a book good and It has got a section on perfect number
http://modular.math.washington.edu/scans/papers/hardy/[/quote]


The probability that a family in a town has at least 2 children is 1/2. If we take 30 families from this town , which is the probability that between them there are 5 with at least 2 children?

Let families be represented as coins, and "having atleast 2 children" be heads, and having less than 2 children as tails.

I am interpreting the question as "the probability -there exists- atleast.."

Then we are looking for the probability there are atleast 5 heads among 30 coins.  Its 1 - (exactly 4 tails + exactly 3 tails + exactly 2 tails + exactly 1 tails + exactly 0 tails)

or $1 - \frac{\sum_{j=0}^4 \binom{30}{j}}{2^{30}}$

Otherwise, if it exactly 5 heads, then the probability is $\binom{30}{5} / 2^{30}$

ps. this is not preolympiad

Thank you very much.I was sure that this is not preolympiad although I did not know the solution , but early in the morning and before going to school I could not find the wright place(I did not see any sub forum for Probability in the college section).Now ,seeing the solution , I think it is  simply high school(in Greece Probability and Bernoulli are sections in school books).
The problem was given to some friend by a stundent and he gave me it late ,  last night. I looked at the problem for a while.Finally, I thought it was Bernoulli( my friend ,  too) and gave the same solution as you( 5 means exactly 5 and I should write it, sorry).But  I was   not sure for the solution because  I was not sure for the negation of sentense  <<  at least 2 >> and its probability.

Any way , now I have your solution and I' ll give it to my friend.
Thanks and I wish you the best.
 - Babis


Let $f: Z$ x $Z\to R$ be a function. Let for f be true that:
1) for every two integers $m,n$:
$nmf(n,m)=(n+m)(mf(n-2,m)-nf(n,m-2))$
2) the set of pairs $(m,n)$ for which $f(n,m)\neq0$ is finite
Describe all such function f.

Could someone please try to solve it?
I've been tryin' to solve for the past two days, and I got nothing.

i think there may be some mistake in the question....
because this equation is hold for all m,n then this must hold for
n=0;and m = real number
but this equation is not holding for above set.......

put n = o in the given equation
The equation becomes
$0 = m^{2}*f(-2,m)$
now it is given that $f(n,m)\neq0$
hence this equation is not hold for all m......

From $0=m^{2}*f(-2,m)$ you'd get $f(-2,m)=0$ for all $m\neq0$, which is not a contradiction. Condition 2 only says $f(m,n)\neq0$ is finite, the set (n,m),f(n,m)=0 is obviously infinite, so...


[quote="calc rulz"][quote="ScottGarrabrant"]On the first question I proved that if s=p-1 then it works.  I am hoping for 1 point.

On the second question, I got the right answer, but my explenation was probably not the best, is that a guaranteed 7?

On the third, I got the equation, of f(x)=integer*(4x-odd square), but not much more, I also described the factors of this function.

On the fourth, I got the right answer, but I did not prove 3 or 5

On the fifth, I basically said that the frog would have to jump over the point 2^i, and that would put it in the exact same possition mod whatever distance jumped, which means it would still have to eventually get to a point that is a multiple of the distance it jumped, and it would be just as easy to get to point 2^i from the place the frog was before it jumped over 2^i as it was for the frog to get from the place it landed to another multiple of the distance it jumped((2^i)K)

I did not answer the sixth

I would greatly appreciate anyones oppinion on how many points I can expect for each problem, also if anyone else has any other specific answers that they gave, and would like an oppinion about their score, feel free to ask here. Thanks.[/quote]

1: Umm, it doesn't work if $s=p-1$...

2: 5-6, 7 if you're really lucky

3: 1-2

4. 2

5. Umm....I have no clue, I didn't solve this one and haven't managed to figure it out after USAMO...

6. 0[/quote]

a 2 for number 4? I say 5-6, after all, he got the right answer; he just didn't prove that 3 or 5 didn't work.

[quote="Phelpedo"]
a 2 for number 4? I say 5-6, after all, he got the right answer; he just didn't prove that 3 or 5 didn't work.[/quote]

If all that was missed were 3&5, I agree with Phelpedo - I'll be surprised if that's not a 5 or a 6.

What do you think a pretty nice proof that all composites can be obtained if all a are integers would merit? 1 or 2 points? :roll:

I'm not sure what the composites only would merit - I would guess 1 or 2, but I don't really know what their rubric will be.

I don't think you'll get any points for #1.

For #4, proving composites should be worth a point, I think. It's definitely the easiest part of the proof.

On #3, I got the right answer, but my proof was faulty (I used a theorem that turns out to be an open problem and proceeded from there to get the same answer). What do you think I'll get for that?

I would guess that's basically equivalent to just writing down the right answer alone. Do they give points for this?

It depends... if you did the "mod a huge prime" part, you should get maybe 2-3 points. Now I wonder what they give for proving $(4x - k^2) | f(x)$...


algebra
polynomial
induction
algebra solved

Consider the sequence of polynomials :
P1 = x-1,P2 = x2 - x -1, Pn+2 = xPn+1 - Pn.
For all n find all roots of Pn.

i still havent completed the solution but made plenty of progress .so i am posting all i cud find
Define some new polynomials g_0(x)=g_1(x)=1 and suppose tht they satisfy the same recurrence relation. then one can prove using induction that (for n>=3)  p_n(x)=g_n(x)-g_(n-1)(x).since we interested in finding the roots of p_n(x)we are searching for all x such that g_n(x)=g_(n-1)(x).
Now intuition tells me that the roots of p_n(x) lie in the segment  [-2,2].put x=2cos A.and u get using induction that    sin nA = sin A * g_(n-1)(2cos A). also I 've found a explicit form for g_n(x) for l x l > 2.i am too tired to complete the problem.i hope sum1 picks up from here and completes this .time for me to sleep.(and as always(most of the ppl on the Forum  say this :D , i hope this is correct)
Cheers

Hi galois,

i tried your trick and it worked perfectly.
I found that the only roots of Pn are:
R_k=2*cos( Pi*(2*k-1)/(2*n+1)) for 1<=k<=n

regards.

thanx alekk for completing the proof .i was just too lazy to do that :D
best regards


The convex polygon $A_{1}A_{2}\ldots A_{2006}$ has opposite sides parallel $(A_{1}A_{2}||A_{1004}A_{1005}, \ldots)$.
Prove that the diagonals $A_{1}A_{1004}, A_{2}A_{1005}, \ldots A_{1003}A_{2006}$ are concurrent if and only if opposite sides are equal.

The "if" part is obvious: if the opposite sides are congruent, every pair of consecutive diagonals cut each other in their common midpoint.

Conversely, assume that all diagonals are concurrent in $P$. Then, given consecutive vertices $A_i,A_{i+1}$ (the indices are taken modulo $2006$), we have $\frac{A_iP}{PA_{i+1003}}=\frac{A_{i+1}P}{PA_{i+1004}}=t$ because $A_iA_{i+1}A_{i+1003}A_{i+1004}$ is a trapezoid. If we denote $\frac{A_iP}{PA_{i+1003}}$ by $t_i$, we thus have $t_i=t,\ \forall i$, but $t_{i+1003}=\frac 1t$, so $t=1$, and we're done: the two diagonals considered above cut each other in their common midpoint $P$, so $A_iA_{i+1}=A_{i+1003}A_{i+1004}$.

[hide="sol"]I'll use grobber's solution :)

W.T.S. $ A_{1}A_{1004}, A_{2}A_{1005},\ldots A_{1003}A_{2006}$ are concurrent $ \leftrightarrow |A_1 A_2 |\equal{}|A_ {1004} A_{1005} |, |A_2 A_3 |\equal{}|A_{1005} A_{1006}| ,\cdots,|A_{1003} A_{1004} |\equal{} |A_{2006}A_1 |$.

$ \rightarrow$ ) (grobber's sol)

[quote="grobber"]The "if" part is obvious: if the opposite sides are congruent, every pair of consecutive diagonals cut each other in their common midpoint.

Conversely, assume that all diagonals are concurrent in $ P$. Then, given consecutive vertices $ A_i,A_{i \plus{} 1}$ (the indices are taken modulo $ 2006$), we have $ \frac {A_iP}{PA_{i \plus{} 1003}} \equal{} \frac {A_{i \plus{} 1}P}{PA_{i \plus{} 1004}} \equal{} t$ because $ A_iA_{i \plus{} 1}A_{i \plus{} 1003}A_{i \plus{} 1004}$ is a trapezoid. If we denote $ \frac {A_iP}{PA_{i \plus{} 1003}}$ by $ t_i$, we thus have $ t_i \equal{} t,\ \forall i$, but $ t_{i \plus{} 1003} \equal{} \frac 1t$, so $ t \equal{} 1$, and we're done: the two diagonals considered above cut each other in their common midpoint $ P$, so $ A_iA_{i \plus{} 1} \equal{} A_{i \plus{} 1003}A_{i \plus{} 1004}$.[/quote]

$ \leftarrow$ )
If $ |A_{1}A_{2}|\equal{}|A_{1004}A_{1005}|, |A_{2}A_{3}|\equal{}|A_{1005}A_{1006}| ,\cdots,|A_{1003}A_{1004}|\equal{} |A_{2006}A_{1}|$,

quadrangle $ A_1 A_2 A_{1004}A_{1005}$ becomes a parallelogram. Then, if we denote X as an intersection of $ A_1 A_{1004}$ and $ A_2 A_{1005}$, X is a midpoint of $ A_1 A_{1004}, A_2 A_{1005}$. Similarly, we can show that an intersection of $ A_2 A_{1005}$ and $ A_3 A_{1006}$ is a midpoint of $ A_2 A_{1005}$, and this point is also X. Now we can repeat this, and finally $ A_{1}A_{1004}, A_{2}A_{1005},\ldots A_{1003}A_{2006}$ are concurrent at X.

QED.[/hide]


find the remainder when $ \frac{L-1}{49}$ where $ L={1,2,3,4,5,6\cdots49}$

What do you mean by find the remainder?It is L-1 isn't it?

find the remainder when $ L\minus{}1$ is divided by $ 49$. when $ L\equal{}1,2,3,4\cdots49$

[quote="binomial_4eva"]find the remainder when $ L - 1$ is divided by $ 49$. when $ L = 1,2,3,4\cdots49$[/quote]

[quote="binomial_4eva"]find the remainder when $ \frac {L - 1}{49}$ where $ L = {1,2,3,4,5,6\cdots49}$[/quote]

Ah, thanks for the clarifaction  :wink:  [/sarcasm].

And there is no remainder because any given value for $ L$ will result in a fraction.

Why shouldn't there be a remainder?L-1=49*0+l-1.So, can't it be l-1.

The remainder in a fraction can be found by writing the fraction as a mixed number. The numerator is the remainder, the denominator is the divisor, and the integer part is the quotient.

So, yes, for all of the cases for L given here, it would have to be L-1, because no L-1 is greater than 49.


If I am in the calculus forum and I want to search, I click search and then I have to click on the dropdown and scroll all the way down to find calculus - even though I was just in the calculus forum. Some code should be implemented so that when you click search from some forum it's already selected in the scroll down box.

Especially because searching for past questions is something so often done. If the list of forums wasn't so darn long I wouldn't complain but I don't think it would hurt to make this improvement.

Why don't you just search in the general forum?

Because you will get approximately 100 times as much results  :wink:

You'll get more results but it'll be far more likely to find what you're looking for, even if it means you have to sift through several pages.

Thats completely wrong.
Thats like: if you are looking for a needle in a haystack, you have better chances without a magnet because only then you may see interesting stems of hay.

There are plenty of terms that give thousends of results on full page, but less than ten on some forums.

The point is it'd be best if the search function was simplified (as suggested) :)


probability
Pythagorean Theorem
geometry
power of a point

1. poker chips come in three colors, red, white and blue. Compute the number of different combinations there are of ten poker chips.
2. Circles O and P have radii 16 and 9 respectively, and are tangent externally at point A. Point B is chosen on circle O such that AB=8 and BC is drawn tangent to circle P at point C. Compute the length of BC.
3. A palindrome is a number that reads the same backwards and fowards, such as 23432. 0<n<100,000 and n is an integer. If n is randomly chosen and p/q is the probability in simplest form that n is a palindrome, compute p/q.

so yea guys...a little help please? thanks  :lol:

[hide="1."]1 is just $ 3^{10}$, isn't it?[/hide]

[hide="A tentative 3."]I think there are 99,998 numbers between 0 and 100000, is this true?

Anyway, if the number is a five digit number $ abcde$, then the only important digits are $ a,b,c$, since $ d,e$ are just copies of $ a,b$. So, assuming $ a$ isn't 0, there are $ 9\cdot10^2$ possibilites. For a four digit palindrome $ abcd$, only $ a,b$ matter. Assuming $ a\not\equal{}0$, there are $ 9\cdot10$ possible 4 digit palindromes. For a three digit palindrome $ abc$, $ a,b$ matter. There are once again $ 9\cdot10$ possibilites. For 2 digit palindromes, there are 9 possibilites. Now, my question. Is a one digit number a palindrome? If so, there are 9 1 digit palindromes...so the probability is:

$ \frac{9\cdot10^2\plus{}9\cdot10\plus{}9\cdot10\plus{}9\plus{}9}{99998}\Leftrightarrow$
$ \frac{549}{49999}$

Is this right? I wouldn't honestly believe so. I've learned to distrust my answers.[/hide]

[quote="Icy"]I think there are 99,998 numbers between 0 and 100000[/quote]

There are $ 99999$ such numbers.  These are the numbers $ 1, 2, 3, ... 99999$.

Haha, oops. I forgot to think about that. But from your lack of other corrections, I'm guessing the rest of my method was correct? If so, I'm pretty satisfied with my almost right answer, as dumb as that sounds :D

The correct answer then, is

$ \frac{122}{11111}$?

[hide="2"]
Center of O and P is $ O$ and $ P$ Respectively.
I believe Triangle $ OPC$ has a right angle at Vertex $ C$ (Tangent)
Thus we use Pythagoras $ {OC}^2\equal{}{25}^2 \minus{}9^2 \equal{} 544$
Since $ OC\minus{}AB\equal{}BC$ $ BC\equal{} \sqrt{544} \minus{}8$
[/hide]

[quote]1 is just $ 3^{10}$, isn't it?[/quote]

I think that he means the [b]combination[/b], so it's just:

$ C(3 \plus{} 10 \minus{} 1,10) \equal{} 12*11 \equal{} 132$ different combinations

[quote]I believe Triangle OPC has a right angle at Vertex C (Tangent) [/quote]
Why do you believe that? It doesn't say that O, B, C are collinear.

Hint: Connect BP, since BC is tangent to circle P, angle C is a right angle. Calculate BP first, then use Pythagorean Theorem to solve for BC.

[quote]$ 122\over 11111$?[/quote]
You maybe type a number wrong, I guess? My answer is $ 222\over 11111$

I dunno, I'm getting $ \frac{1098}{99999}$, which simplifies to $ \frac{122}{11111}$

[hide="my (almost certainly wrong) 3"]
$ \frac {1}{100} \equal{} \frac {1}{10}\frac {1}{10}$, the probability that $ d \equal{} b$ and $ e \equal{} a$, is the probability that a 5 digit number will be a palindrome.  That makes 900 5-digiters.
Similar reasoning for all the other digit sets gives:
4-digit: 90
3-digit: 90
2-digit: 9
1-digit: 9

Totals up to ... 1098 palindromes.
So the probability that a number will be a palindrome is $ \frac {1098}{99999} \equal{} \frac {122}{11111}$.

[/hide]

Sooo I agree with Icy.

[hide="time to try numero dos"]
Draw perpendicular BD to AO.  The length of BD is $ 4\sqrt{3}$.  Why?  Draw median BE of triangle BAO and find out.  Now draw triangle BDP.  We can find the hypotenuse of this triangle because we know DP.  Why?  The same reason we know the length of BD.  The hypotenuse turns out to be $ \sqrt{217}$, I think.  Then, draw triangle BCP.  What do you know, it's a right triangle with hypotenuse BP!  Since we know CP=9, we find that $ BC\equal{}\sqrt{136}$?

By the way, triangle BEA (which you should have drawn as I mentioned it twice) is equilateral.  Makes things so simple.

[/hide]

[quote="timwu"]I think that he means the combination, so it's just:

$ C(3 \plus{} 10 \minus{} 1,10) \equal{} 12*11 \equal{} 132$ different combinations [/quote]

At risk of sounding obtuse, why is it $ C(3 \plus{} 10 \minus{} 1,10)$? By the way, $ \dbinom{12}{10}\equal{}66$.

Sorry, I fogort to divide it by $ 2!$...

IMO, this question is equivalent to: If $ a \plus{} b \plus{} c \equal{} 10$, how many non-negative integer solutions are there?

There is a formula(and it's easy to derive) for this kind of questions: $ E(m,r) \equal{} C(m \plus{} r \minus{} 1,r)$, where $ m$ means the number of variables, and $ r$ means their sum.

To @Icy and @CircleSquared:

5-digit palindromes are also acceptable, aren't they? so I add another $ 9*10^2$ on the numerator, which makes $ 222$.

@tim:
If you look at my solution (Not sure about Icy's), I included the 5-digit palindromes.  Maybe you included them twice on accident?

[hide="I figured out numero uno"]
Balls and urns with 10 balls, 2 dividers.
$ \binom{12}{2}\equal{}66$

Balls and urns is useful, I suggest you look it up in the AoPS Wiki if you don't already know it.
[/hide]

Um...I think you're counting 5-digit palindromes twice or something...the first $ 9\cdot10^2$ is for five-digit, then $ 9\cdot10$ for 4-digit, $ 9\cdot10$ for 3-digit, $ 9$ for two digit and $ 9$ for one digit...this sums to $ 1098$.

I'm pretty sure I included 5 digit palindromes as well :D

EDIT:
Oops, didn't notice CircleSquared posted before me, saying the exact same thing :P

Oh...okay. Damn balls-and-urns.  :P I should've seen that. But my excuse is that it was very late at night when I posted that.

Oops...I really did calculate them twice, sorry...it's exactly $ 122$ :oops:

For no. 2:

Draw a line $ BH$ perpendicular to $ OA$. Let $ AH \equal{} x$, then $ OH \equal{} 16 \minus{} x$.
By Pythagorean Theorem, $ OB^2 \minus{} OH^2 \equal{} BA^2 \minus{} HA^2 \equal{} BH^2$.
$ 16^2 \minus{} (16 \minus{} x)^2 \equal{} 8^2 \minus{} x^2$,
$ x \equal{} 2, BH^2 \equal{} 60$.
Similarly, $ BH^2 \plus{} PH^2 \equal{} BP^2 \equal{} BC^2 \plus{} PC^2, 60 \plus{} 11^2 \equal{} BC^2 \plus{} 9^2, BC \equal{} 10$.

...So all we three get different answers?....

Geometer Sketchpad likes timwu's answer. Here are what I think are the mistakes in the other solutions.

Triangle $ OPC$ does not have a right angle at $ C$ by tangents. That's actually triangle $ BPC$ with this property.

The distance of $ B$ to line $ OP$ isn't $ 4\sqrt{3}$. If you draw the median from $ B$ in triangle $ OAB$, you don't get an equilateral triangle. Just an isosceles one.

:wallbash: Okay I agree with tim.

@Mellow:
How would you use GSP to do this?

I used it to compute $ BP^2$, after which the answer follows by either Power of a Point or the Pythagorean Theorem (equivalent methods). Here are the steps I took.
1. Make a line with points $ O,P$. Put point $ A$ in between $ O$ and $ P$.
2. Tell GSP to measure the distances $ OA$ and $ PA$. Adjust $ P$ and $ O$ until the distances are $ 16$ and $ 9$ respectively (as close as possible, exactness is difficult with GSP).
3. Select $ O$, then $ A$, and construct a circle. Select $ P$, then $ A$, and construct a circle.
4. Put a point on the circle around $ O$ called $ B$, and have GSP measure $ BA$. Adjust $ B$ until that length is $ 8$.
5. Have GSP compute $ BP^2$, then $ \sqrt(BP^2 \minus{} 81)$.

I ended up getting a final answer of $ 9.95$, due to roundoff errors from setting the points beforehand.


rotation
geometry
congruent triangles
combinatorics unsolved
combinatorics

I need to write a paper on Kirkman's Schoolgirl Problem...

[b]Fifteen schoolgirls walk each day in five groups of three. Arrange the girls' walk for a week so that in that time, each pair of girls walks together in a group just once. [/b]

the problem itself is quite interesting...
and I go to the web and do some research,
but in this page,
[url]http://www.delphiforfun.org/Programs/kirkman.htm[/url]

there is a paragrph,

[i]By the way, the number of ways to select the 35 groups of girls that define a solution from the 455 possible groups is about 10^27 - approximately the diameter of the universe in inches! [/i]

where is the 10^27 coming from? From the wording, the number should be 455 choose 35, which is equals to 10^52... then I checked with my calculator, 10^27 = 455 choose 15....
but then...why???

[quote="math beginner"][i]By the way, the number of ways to select the 35 groups of girls that define a solution from the 455 possible groups is about 10^27 - approximately the diameter of the universe in inches! [/i]

where is the 10^27 coming from? From the wording, the number should be 455 choose 35, which is equals to 10^52... then I checked with my calculator, 10^27 = 455 choose 15....
but then...why???[/quote]
They mean that the number of solution to your problem is $10^27$, and, of course, [i]arbitrary[/i] choice of 35 groups from 455 doesn't give solution.

ok...now this page show me a way to solve the problem
[url]http://home.wlu.edu/~mcraea/Finite_Geometry/Applications/Prob31SchoolGirl/problem31.html#Anchor-Kirkman's-40966[/url]

Does it mean that if I find another set of 5 non-congruent triangles in my STS system as my generator that is different than the one shown, and rotate it 7 times, then I'll have another unique solution?


AMC
AMC 12
AIME
geometry
probability
calculus
trigonometry

I'm a little confused as to how i should study for the AMC12.  I'm taking it in February and as far as i can see, from my current knowledge, i can probably get 3 problems correct -.-

Can anyone help me in choosing books to prepare me for the AMC12 so i can qualify for the AIME by February?

Thanks!

Well, I'd recommend any of the following:  Introduction to : Algebra, Geometry, Probability and Combinations. Also, "Art of Problem Solving: Volume 1" is a must.

Once you get the basics down, do as many tests possible. You can look them up in AoPS Wiki (on the side bar) by searching AMC 12 and look under that page.

What grade are you in? Have you thought about the AMC 10?

Working through volume 1 and parts of volume 2 should be sufficient for qualifying for AIME. By that, I mean read thoroughly, understand everything, and do the problems--you might not solve all of them (I solved about half my first time through), but you should at least attempt them. The introductory/intermediate subject books are good for going more thorough into certain weak subjects, but I don't feel they are necessary--AoPS plus a lot of AMC12/AIME problems should be good enough. You want to build up a knowledge base and experience solving the problems.

[quote="cheeseyicecream"]For the AMC 12, the Intermediate series may also help. I do not own any of them, truthfully, but I've heard nothing but great things about them. If you want to do well on the AIME, they are a must.[/quote]

I'm pretty sure the Intro helps with most, you don't need much Intermediate but you should if you have time because of AIME.

Well, what's your level in normal math?

Note that between 18/150 and 120/150 (qualifying score) is a fairly large gap, so you would need to study pretty hard.

They cover a lot of material, but don't go in detail very much.

The main difference between the AMC12 (and competition math in general) and Calc BC (and curriculum math in general) is [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php]problem solving[/url]. The curriculum focuses a lot on memorization and formulas... Competitive math focuses more on problem solving. Also, some topics are not even taught in curriculum math, like [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_DiscreteMath.php]counting, probability, and number theory[/url].

I don't know what to recommend, but I can say that you need to study things that isn't taught at school, like what was above.

Hmm, thanks powerofpi.

I understand what you're saying and i completely agree.  But do you suggest i study ALL of the books available on this website, as in, all the intermediate books PLUS the AoPS books?  Doesn't that seem a bit overkill?

I don't think you would have time for it. Reading Volumes 1 and 2 would need some reading. The whole intermediate series would be too long in my opinion.

What grade are you in? Can you participate next year? If so, then doing the Intermediate series would be good.

[quote="DevilsAdvocate"]well, I'm currently in calculus BC (calc 1 and 2 together), but the questions on the AMC are just completely different from anything I've seen; and of course i was exaggerating with the 2 questions.  I can probably answer anywhere between 6-10.

I don't mind studying, i just need to know what materials will maximize my score.  I'm thinking of ordering volumes I + II of AoPS and just going through those until February, does that sound good?[/quote]
I hear that this one http://www.amazon.com/First-Steps-Math-Olympians-Competitions/dp/088385824X is really good. You would probably only need that one.

I recommend it, as I also suggested it.
[quote="gaussintraining"]First Steps for Math Olympians by J. Douglas Faires is a good book for preparing for the AMC, especially if you're fairly new to problem solving.[/quote]
You can buy it through Amazon or this site.

[quote="DevilsAdvocate"]powerofpi, I'm a junior and i can participate next year but if i can get a potentially high score this year, why not?

I guess you're right about finishing all those books, I've decided to only buy Volume 2 after some reading, now i just need one more book!

Can anyone comment on the book ihatepie suggested?[/quote]

It's a decent book, but it's not really that useful past amc. (and maybe basic aime)

Books aren't the only way to study, doing problems is often more helpful (though books are also great).

Basically do a lot of problems.

[quote="DevilsAdvocate"]Ok, I'm sorry, I'm just slightly confused.  Would skipping AoPS: Volume I have a negative impact on my AMC12 score?  If i were to just jump to Volume II and do all the problems for these next 3 months, would i be well off for the test, to pass into AIME range?[/quote]

If you can do most of the problems in Volume I, it should be pretty easy for you to make aime. So If you think Volume I will be pretty easy for you, you probably can already pass the amc12.

[quote="DevilsAdvocate"]Ok, I'm sorry, I'm just slightly confused.  Would skipping AoPS: Volume I have a negative impact on my AMC12 score?  If i were to just jump to Volume II and do all the problems for these next 3 months, would i be well off for the test, to pass into AIME range?[/quote]

In my opinion that would be a mistake.  The problems in AOPS V1 are pretty difficult.  Seeing as you can only get 3 questions correct, I don't think you would be able to solve all the problems in AOPS V1.  I'm not trying to put you down, but it would be a good idea to go through the problems.

[quote="DevilsAdvocate"]of course i was exaggerating with the 2 questions.  I can probably answer anywhere between 6-10.
[/quote]
He was kidding about only getting 3 right, but I still agree, if someone could solve all of the problems from Volume I, they would consistently score above 36-60 on the AIME.

I think each AIME problem is worth 10 points towards your index thingy for qualification for the USAMO

[quote="DevilsAdvocate"]AIME15USAMO, you said if someone could answer all the Volume I questions they would consistently score 30-60 on the AIME?  I thought the AIME was out of 15?  Did you mean 3-6?[/quote]
No, he meant to say AMC 12. Notice that 6-10 questions right corresponds to 36-60 points (although he wasn't counting questions left blank.

Hi, I'm not DevilsAdvocate, but I'm in somewhat the same position; do you think that the AoPS books get into enough depth at the different subjects? From past AMC 12, AIME, and ARML problems, I see that I have an affinity for number theory, probability, and combinatorics, but I am just awful at geometry. Will this cure or at least greatly help that?
I was wondering because I don't want to buy more textbooks than I absolutely have to.

Yes, except for the Intermediate has no NT and no Geometry, but I've been told (by ruscyk) that Intro to Geometry will cover most non-trig geo on AIME. Also, for AIME, you should look to solve by analytical geometry. :)

Get Art of Problem Solving Volume 1, it is well worth the investment

It was designed with helping students to do well.
If you want to study online tests, look only at the first 15 if you are having trouble. These tests were designed to provide challenge in the last 5-10 questions, so when you are just starting out don't worry too much about these last questions.

Good luck!

I've never heard of the PSS. Can someone tell me what PSS stands for?

PSS is a book, people also call it Engel because that was the author of the book. PSS stands for Problem Solving Strategies, it's a highly advanced book mainly used for preparing for olympiad-level math.

Thanks, AIME15USAMO. I never would've guessed the title of the book. A lot of it seems to be online on google books.

Sure. PM sent.

Edit: For anyone else who wants part of Engel's PSS on Google Books for free, follow this link:

[url]http://books.google.com/books?id=wj3_FxnRz5kC&dq=Problem+Solving+Strategies+engel&printsec=frontcover&source=bn&hl=en&ei=HCwIS4ndIYaWtgeZ7vHACg&sa=X&oi=book_result&ct=result&resnum=4&ved=0CBYQ6AEwAw#v=onepage&q=&f=false[/url]

that is a very helpful way to study, ive used it very frequently and i've noticed that it helps significantly  :)


Is there a palindromic power of $5$ other than $5$?

[quote="AstroPhys"][hide="Answer"]The answer is no.[/hide][/quote]Because...?

I dont know is there?  there problaby isnt a proof.

Well you could show that the first two digits can't be 12.

Unfortunately, there are powers of 5 that start with 52.  For example, $5^{74}$.

More generally, it's a well-known result that there are powers of 5 that start with any sequence of digits.  For example, there is a power of 5 that starts with 123456789.  One way to prove it is with the pigeonhole principle.

Generic palindrome --> \[\sum a_{i}(10^{n-i}+10^{i})\]
Now we just have to show that repeatedly dividing $1.0000\cdots$ by $2$ and ignoring the decimal point doesn't yields a palindrome.
First, there must be some digit at least $5$ in the palindrome (okay, this leads nowhere).
Fundamentally, what happens when a palindrome is doubled? If we interpret the palindrome written in base ten as written in base two, then clearly the new number is still a palindrome. Thus we see that $1000000000\cdots_{10}$ is a palindrome base $2^{n}\times 10$ for some $n$. This is as far as I get

Like I said, there problaby isnt a proof.

I have a feeling like a little inegenuity will get proof of this problem, possibly invariants.
But then again, I am reminded of that open problem of taking a number and adding its flip repeatedly to yield a palindrome...


function
trigonometry
integration
real analysis
real analysis unsolved

$ a_{n} \geq 0$ and $ \sum_{n \equal{} 1}^{\infty} a_n < \infty$. Prove that function:
\[ f(x)\equal{}\sum_{n \equal{} 1}^{\infty} a_{n}|\sin(nx)|
\]
satisfies Lipschitz condition if and only if $ \sum_{n \equal{} 1}^{\infty} na_{n} < \infty$

[hide="one part"]Let $ x,y \in \mathbb{R}$.
$ |f(x)\minus{}f(y)|\equal{}\left| \sum_{n\equal{}1}^{\plus{}\infty}a_n|\sin(nx)|\minus{}\sum_{n\equal{}1}^{\plus{}\infty}a_n|\sin (ny)| \right|$ $ \equal{}\left|\sum_{n\equal{}1}^{\plus{}\infty}a_n(|\sin (nx)|\minus{}|\sin (ny)|) \right| \leq \sum_{n\equal{}1}^{\plus{}\infty}a_n| |\sin (nx)|\minus{}|\sin (ny)| |$ $ \leq \sum_{n\equal{}1}^{\plus{}\infty}a_n|\sin (nx)\minus{}\sin (ny)|\equal{}$ $ \equal{}\sum_{n\equal{}1}^{\plus{}\infty}a_n \left| \int_y^x \minus{}n \cos(nt)dt \right| \leq
 \sum_{n\equal{}1}^{\plus{}\infty}na_n \left| \int_y^x |\cos(nt)|dt \right|\equal{}\sum_{n\equal{}1}^{\plus{}\infty}na_n |x\minus{}y|$, and so if $ \sum_{n\equal{}1}^{\plus{}\infty}na_n < \plus{}\infty$, $ f$ satisfies Lipschitz's condition. Note that the passage$ \left| \sum_{n\equal{}1}^{\plus{}\infty}a_n|\sin(nx)|\minus{}\sum_{n\equal{}1}^{\plus{}\infty}a_n|\sin (ny)| \right|\equal{}\left|\sum_{n\equal{}1}^{\plus{}\infty}a_n(|\sin (nx)|\minus{}|\sin (ny)|) \right|$ is possible because $ \sum_{n\equal{}1}^{\plus{}\infty}a_n < \plus{}\infty \Longrightarrow \sum_{n\equal{}1}^{\plus{}\infty}a_n|\sin(nx)|$ converge uniformly on $ \mathbb{R}$ by Weierstrass' criterion[/hide]


inequalities
logarithms
search
inequalities unsolved
109

Well, problem 2 follows from Iran 1996:

$\sum \frac{(a+b+c)^2}{b^2+bc+c^2}\geq \sum \frac{4(ab+bc+ca)}{(b+c)^2} \geq 9$.

The first inequality follows from $(a+b+c)^2(b+c)^2 - 4(ab+bc+ca)(b^2+bc+c^2) = (b^2+c^2-ab-ac)^2\geq 0$.

first ineq is equivalent to
$\sum a(a+(\sqrt{b}-\sqrt{c})^2)(\sqrt{b}-\sqrt{c})^2 \ge 0$

third ineq has a stronger version:
$\prod (a+b)^2 \ge \frac{4}{3}abc(a+b+c)^3$, this is exactly popoviciu's inequality for $f(x) = \ln x$

and, finally, the fourth inequality is trivial in the way written there, but if you really meant

$(ab+bc+ca)^2(a+b+c)\geq 9abc(a^2+b^2+c^2)$,

this inequality is wrong. Just take $a=1, b=1-x, c=1+x$.

[quote="Nameless"][b]4[/b]. Let $ a,b,c > 0$. Prove

$ (ab \plus{} ac \plus{} bc)^{2}(a \plus{} b \plus{} c) > abc(a^{2} \plus{} b^{2} \plus{} c^{2}).$[/quote][quote="Peter Scholze"]the fourth inequality is trivial in the way written there, but if you really meant[/quote]$ (a \plus{} b \plus{} c)(bc \plus{} ca \plus{} ab)^2 > 4abc(a^2 \plus{} b^2 \plus{} c^2),$

this inequality is also true, and the constant $ 4$ is the best possible.

By the way, the second inequality see also here : http://www.mathlinks.ro/viewtopic.php?t=13280

$ \sum \frac{4(ab\plus{}bc\plus{}ca)}{(b\plus{}c)^2} \geq 9 ?$

It'a Iran 1996 inequality. Use "Search" to find the proof.

[quote="Nameless"][b]2[/b]. Let $ a,b,c > 0$. Prove

$ \frac {1}{b^{2} \plus{} bc \plus{} c^{2}} \plus{} \frac {1}{c^{2} \plus{} ca \plus{} a^{2}} \plus{} \frac {1}{a^{2} \plus{} ab \plus{} b^ {2}}\ge\frac {9}{(a \plus{} b \plus{} c)^2}.$

I solved this inequality but my solution is very long.[/quote]Without using [url=http://www.mathlinks.ro/viewtopic.php?t=32470]Iran 96 inequality[/url],

$ \frac {1}{b^2 \plus{} bc \plus{} c^2} \minus{} \frac {28a^2 \minus{} 5b^2 \minus{} 5c^2 \plus{} 20bc \plus{} 8ca \plus{} 8ab}{2(a \plus{} b \plus{} c)^4}$

$ \equal{} \frac {(b \minus{} c)^2\left[2(b \plus{} c \minus{} a)^2 \plus{} b^2 \minus{} bc \plus{} c^2\right] \plus{} (a \minus{} b)^2\left[(a \plus{} 2b \minus{} c)^2 \plus{} 8ab\right] \plus{} (a \minus{} c)^2\left[(a \plus{} 2c \minus{} b) ^2 \plus{} 8ac\right]}{2(a \plus{} b \plus{} c)^4\left(b^2 \plus{} bc \plus{} c^2\right)}$

$ \geq 0\Longrightarrow$

$ \sum{\frac {1}{b^2 \plus{} bc \plus{} c^2}}\geq\sum{\frac {28a^2 \minus{} 5b^2 \minus{} 5c^2 \plus{} 20bc \plus{} 8ca \plus{} 8ab}{2(a \plus{} b \plus{} c)^4}} \equal{} \frac {9}{(a \plus{} b \plus{} c)^2}.$

By the way,

$ \sum{\frac {1}{b^2 \plus{} bc \plus{} c^2}}\geq\sum{\frac {23}{3\left(8b^2 \plus{} 7bc \plus{} 8c^2\right)}}$

$ \geq \sum{\frac {3\left[62a^2 \plus{} 14a(b \plus{} c) \minus{} 7b^2 \plus{} 38bc \minus{} 7c^2\right]}{2(a \plus{} b \plus{} c)^2\left[8\left(a^2 \plus{} b^2 \plus{} c^2\right) \plus{} 11(bc \plus{} ca \plus{} ab)\right]}} \equal{} \frac {9}{(a \plus{} b \plus{} c)^2}.$

See also here : http://www.mathlinks.ro/viewtopic.php?t=28706


conics parabola symmetry geometry rectangle ratio circumcircle

Let there be the parabola ABC , and tangents ADE, CFE and DBF.
Prove that CF:FE=ED:DA=FB:BD

Please help me  :blush: I will be happy for your help

Perpendicular $ o$ to the parabola directrix through the focus $ O$ is the parabola axis. Parabola tangents $ t_a, t_c$ at $ A, C$ intersect at $ E.$ Parallels $ p_a, p_c$ to $ o$ through $ A, C$ cut the parabola tangents $ t_c, t_a$ at $ X, Z$ and $ AXCZ$ is a parallogram. This is because $ C$ can be projected by a parallel projection to $ C'$ in the direction of $ o,$ so that the $ \triangle A'E'C'$ becomes isosceles ($ A' \equiv A$ can stay in place). The original parabola $ \mathcal P$ goes to a parabola $ \mathcal P'$ with axis $ o' \parallel o$ and tangents $ t_a \equiv AEZ, t_c \equiv CEX$ of $ \mathcal P$ go to tangents $ t_a' \equiv A'E'Z',$ $ t_c' \equiv C'E'X'$ of $ \mathcal P'.$ By symmetry, $ A'X'C'Z'$ is a rectangle, its diagonals $ A'Z', C'X'$ cut each other in half at $ E'$ $ \Longrightarrow$ $ AZ, CX$ also cut each other in half at $ E,$ $ \overline{EX} \equal{} \minus{} \overline{EA},$ $ \overline{EY} \equal{} \minus{} \overline {EC}.$ This (trivial) result holds for any pair of parabola tangents.

Since parallel projection preserves ratios of segments on a line, it is sufficient to prove the proposition for the projected figure. In other words, we skip primes in the notation and assume that the parabola tangents $ t_a \equiv ADE, t_c \equiv CEF$ are symmetrical with respect to the parabola axis $ o.$ Let $ V$ be the parabola vertex and $ t_v$ parabola tangent at $ V.$ Let the tanget $ t_b$ at $ B$ cut $ o$ at $ K$ and $ t_v$ at $ L.$ Let the tanget $ t_b$ cut parallels $ p_a, p_c \parallel o$ through $ A, C$ at $ M, N.$ By the above lemma applied to parabola tangent pairs $ (t_b, t_a),$ $ (t_b, t_c),$ $ (t_b, t_v),$ respectively, we have $ \overline {DB} \equal{} \minus{} \overline{DM},$ $ \overline {FN} \equal{} \minus{} \overline{FB},$ $ \overline {LK} \equal{} \minus{} \overline{LB}.$ According to your previous problem, $ DEFO$ is cyclic: Feet of perpendiculars from $ O$ to $ t_a, t_b, t_c$ are on $ t_v,$ collinear, $ t_v$ is Simson line of the $ \triangle DEF$ with the pole $ O,$ which lies on it circumcircle. By symmetry, $ E \in o$ and $ o \equiv EO$ bisects the angle $ \angle DEF$ $ \Longrightarrow$ $ O$ is midpoint of the circumcircle arc $ FD$ opposite to $ E$ $ \Longrightarrow$ $ OL \perp t_b \equiv FD$ bisects the circumcircle chord $ FD$ $ \Longrightarrow$ $ \overline{LF} \equal{} \minus{} \overline {LD}.$ Combined with $ \overline {LK} \equal{} \minus{} \overline{LB},$ we get $ \overline {KF} \equal{} \minus{} \overline{BD},$ $ \overline {BF} \equal{} \minus{} \overline{KD}.$ In conclusion,

$ \frac {\overline{ED}}{\overline{DA}} \equal{} \frac {\overline{KD}}{\overline{DM}} \equal{} \frac {\overline{FB}}{\overline{BD}} \equal{} \frac {\overline{NF}}{\overline{FK}} \equal{} \frac {\overline{CF}}{\overline{FE}}$

at first paragraph "C can be projected by a parallel projection to C' in the direction of o, so that △A'E'C' becomes isosceles"
is very hard for me to imagine the figure.

A ---> A' , E ---> E' , C ---> C'  What are the points A' , E' , C' on ?
Curiosity excites me  :ewpu:

Parabola has one point at infinity. This point at infinity is the same thing as direction of the parabola axis $ o$ (not the axis itself). If direction of the parallel projection is the same, the projected parabola has the same point at infinity, the same axis direction. Define the parallel projection as follows:

1. Direction of the parallel projection is parallel to $ o.$
2. All points on the line $ p_a \parallel o$ through the tangency point $ A$ stay in place.
3. Tangency point $ C$ goes into $ C'$ in the direction of $ o,$ such that $ AC' \perp o.$

To construct the projected figure (including the projected parabola):
1. Construct $ C'$ according to the definition. Parabola tangent $ t_c$ at $ C$ cuts $ p_a$ at $ X,$ which stays in place. $ t_c' \equiv C'X$ is the projection image of $ t_c.$
2. Parallels to $ o$ through $ E, F$ cut $ t_c' \equiv C'X$ at the images $ E', F'$ of $ E, F.$ $ A \in p_a$ stays in place and $ t_a' \equiv AE'$ is the image of $ t_a.$
3. Parabola tangent $ t_b$ at $ B$ cuts $ p_a$ at $ M,$ which stays in place. $ t_b' \equiv F'M$ is the projection image of $ t_b.$ Check: $ t_a', t_b'$ intersect at $ D'$ and $ DD'$ must be parallel to $ o.$ Parallel to $ o$ through $ B$ cuts $ t_b'$ at the image $ B'$ of $ B.$
4. You could similarly find the projection image of the parabola vertex tangent, only to find that it is a general tangent, not the vertex tangent of the projected parabola, but it is not necessary. Direction of the axis and 3 tangents with tangency points is more than enough to construct the projected parabola.
5. Reflect lines $ p_a, p_b, p_c \parallel o$ through $ A \equiv A', B', C'$ in the projected tangents $ t_a', t_b', t_c'.$ (Two are sufficient, say the reflections of $ p_a, p_c$ in $ t_a', t_c'.$) These intersect at the focus $ O'$ of the projected parabola. Note that $ O'$ is not projection image of the original focus $ O.$ Check: $ AO', C'O'$ must be equal and $ E'O'$ must be parallel to $ o.$
6. Conststruct circles $ (A'), (B'), (C')$ with radii $ A'O', B'O', C'O'.$ (Again, two are sufficient, say the congruent circles $ (A'), (C').$) Their common external tangent is the directrix $ d'$ of the projected parabola. Again, note that $ d'$ is not projection image of the directrix $ d$ of the original parabola.

Let $ a, b, c, d$ be 4 fixed tangents of a circle $ (O)$ at $ A, B, C, D$ and let $ p$ be a variable tangent of this circle at $ P.$ Let $ p$ cut $ a, b, c, d$ at $ X, Y, Z, T.$ Then $ OX \perp PA, OY \perp PB, OZ \perp PC, OT \perp PD.$ It follows that the line pencils $ O(X, Y, Z, T) \sim P(A, B, C, D)$ are similar. But angles of $ P(A, B, C, D)$ are independent of the variable point $ P \in (O)$ $ \Longrightarrow$ the cross ratio $ G(X, Y, Z, T) \equal{} k \equal{} \text{const}.$

Suppose the tangent $ d$ is projected to infinity, taking the circle $ (O)$ to a parabola $ \mathcal P.$ Cross ratios are preserved $ \Longrightarrow$ the ratio $ \frac {\overline{XZ}}{\overline{YZ}} \equal{} k \equal{} \text{const}$ is constant for any $ P \in \mathcal P.$  Tangent to any curve cuts itself at the tangency point. Moving the arbitrary parabola point $ P$ (together with its tangent $ p$), consecutively, to coincide with the tangency points $ A, B, C$ of $ a, b, c$ then yields the desired result.


Σας παρακαλω παιδια οποιαδηποτε διευκρινηση θα βοηθουσε.

[quote="therock003"]Σορρύ παιδιά,αλλά θα ήθελα καποιες διευκρινισεις σε 2 απλά πραγματάκια.

Α)Τι καλείται Σαγματικό σημείο?Σε περσινά θέματα δόθηκε μια πολυμεταβλητή συνάρτηση και ζητήθηκε να βρεθούν τα σαγματικά σημεία.

Β)Επίσης υπήρχε ένα θέμα όπου δινόταν μια διαφορική εξίσωση (γραμμική αν θυμάμαι καλά) και ζητόταν να βρεθεί η οικογένεια καμπυλών.Τι νοείται με το "οικογένεια καμπυλών"?[/quote]

Vasika se ti diagonismous ta vrikes afta sta opoia mporei na simetixes kai na mi gnorizes tis duo erwtiseis pou diatiposes???

Sagmatiko simio mias sinartisis pollwn metavlitwn einai to antistixo tou stasimou simiou mias metavlitis , diladi otan ena simio den einai oute topiko megisto oute topiko elaxisto paroti einai krisimo simio....

Oikogenia kampilwn legete mia sinartisi tis morfis F(x,c)=0 (c einai mia afthereti stathera) ..Gia tin akrivia onomazete monoparametriki oikogenia kampilwn afou exei mia mono parametro...An exoume F(x,y,c)=0 tin leme diparametriki oikogenia kampilwn kok.

[quote="therock003"]Δεν συμμετεχω σε διαγωνισμους,αυτα ειναι θεματα εξετασεων σε διαγωνισμα της σχολης μου,απλως δεν εχω μπροστα μου το φυλαδιο αυτην την στιγμη για να διατυπωσω επακριβως τα ερωτηματα.

Πως μπορεις να αποδειξεις δηλαδη πως ενα σημειο είναι σαγματικό όταν σου δινεται μια συναρτηση πολλων μεταβλητών?

Όσο για το δευτερο θα δώσω την εξίσωση όταν βρω το φυλλάδιο.[/quote]

Den xriazete na vreis to thema , i lisi tis diaforikis eksisosis pou tha vreis , einai mia oikogenia kampilwn otan den sou dinete kapoia arxiki sinithiki wste na prosdiorisis tin stathera...

Oso anafora ta sagmatika simia , pistevw pws ksereis tin diadikasia na dikseis an ena simio einai topiko elaxisto megisto...Des oti einai krisimo simio (na midenizete i paragogos 'h oles oi merikes paragogoi na einai 0) meta des an einai megisto 'h elaxisto kai an den einai oute to ena oute to allo , tote einai sagmatiko...Opws tha ekanes kai se sinartisi 1 metavlitis

[quote="SorrowMan"]
Den xriazete na vreis to thema , i lisi tis diaforikis eksisosis pou tha vreis , einai mia oikogenia kampilwn otan den sou dinete kapoia arxiki sinithiki wste na prosdiorisis tin stathera...[/quote]

Mou dinei mia arxiki timi,kai mporo na ipologiso kai to c,ara i lisi poia tha einai?


Jane spent the following amounts on her school lunches last Monday through Thursday:
$ \$3.29$ - Mon., $ \$1.79$ - Tues., $ \$2.65$ - Wed., and $ \$3.29$ - Thurs. After she bought her
lunch on Friday, she calculated that her average lunch cost for that week was $ \$2.80$
per day. How much did her lunch cost on Friday?

First, add up all the prices.

$ 3.29 \plus{} 1.79 \plus{} 2.65 \plus{} 3.29 \equal{} 11.02$

For five days, we are given that the average price is $ 2.80$. Our answer is then $ 2.80 \times 5 \equal{} 14 \minus{} 11.02 \equal{} \boxed{2.98}$.


calculus
integration
limit
logarithms
symmetry
real analysis
real analysis unsolved

Hello.  Can someone suggest to me how to prove convergence (in the principle sense) for the Logarithmic integral:

$Li(x)=\mathop\lim_{\epsilon\to 0}\left[\int_{0}^{1-\epsilon}\frac{du}{Log[u]}+\int_{1+\epsilon}^{x}\frac{du}{Log[u]}\right]$

Where I'm using Log to represent natural log.

Note that $\frac1{\log t}-\frac1{t-1}$ has a removable singularity at $t=1$. The principal value integral of $\frac1{1-t}$ converges by symmetry, and the difference doesn't introduce any problems.

Near zero, the improper integral converges by comparison to $u^{-c}$, with $c$ between zero and 1.

Thanks jmerry.  I'm not quite following your explanation.  Here's my understanding of your suggestion.

$f[x]=\frac{1}{Log[x]}-\frac{1}{t-1}$

is continuous if f(1) is defined to be 1/2.  Thus the integral:

$\int_{0}^{x}\left[\frac{1}{Log[u]}-\frac{1}{u-1}\right]du$

converges.

Thus:

$\text{pr.v. }\int_{0}^{x}\left[\frac{1}{Log[u]}-\frac{1}{u-1}\right]du$

exists.

Which implies:

$pr.v. \left[\int_{0}^{x}\frac{1}{Log[u]}du-\int_{0}^{x}\frac{1}{u-1}du\right]$

exists.

Now, the second integral exists by symmetry and thus the first must exists also.  Not sure this is sufficient though.


[quote="DParrish"]Is there any way we can have a daily/weekly math problem emailed to us, or anything like that?[/quote]

That's an idea we discussed in the office a long time ago and the answer is that it might be part of technology that we develop in the future, but it's not coming any time soon.

[quote="DParrish"]Is there any way we can have a daily/weekly math problem emailed to us, or anything like that?[/quote]There is an experimental feature called PotD (problm of the day). You can check it out on the portal page ( http://www.mathlinks.ro/Forum/portal.html ) or http://www.mathlinks.ro/Forum/resources.php?potd=1&week=-1


geometry
circumcircle
parallelogram
geometric transformation
reflection
geometry proposed

The line S is tangent to
the circumcircle of acute triangle ABC
at B. Let K be the projection of the
orthocenter of triangle ABC onto line S
(i.e. K is the foot of perpendicular from
the orthocenter of triangle ABC to S).
Let L be the midpoint of side AC.
Show that triangle BKL is isosceles.
(Source: 2000 Saint Petersburg City
Math Olympiad)
enjoy this one

We have to show that L', the proj of L on S, is the midpt of KB. it's easy to see that this is equivalent to the parallel to OB through L passing through the midpt of OH.

To prove this last thing We take Hb to be the midpt of HB. It's well-known that BHb=OL, and since BHb || OL, it means that BHbLO is a parallelogram, so HbL || OB. Let X=HbL \cap OH. It follows rather easily that OXL and HXHb are congruent, so X is the midpt of OH, which is what we wanted.

Other solution: call $M$ and $N$ the projections of $A$ and $C$ to $S$. As we know that the projection of $L$ will be the midpoint of $MN$, we have to show that $BM=KN$.
Take now $P$ the projection of $O$ – the circumcenter of $\odot ABC$ onto $AB$ ( i.e. midpoint of $AB$) and $Q$ the reflection of $O$ w.r.t. $P$, so $AQBO$ is a rombus. But $CH = 2 OP = OQ$ and $CH\parallel OQ$ ( both perpendicular onto $AB$) and now we get that the projections of two equal and parallel segments onto a straight line are equal, hence $MB=NK$, done.

Best regards,
sunken rock


$ a,b,c$are real positive numbers
proof
$ \frac{ab}{c(c\plus{}a)}\plus{}\frac{bc}{a(a\plus{}b)}\plus{}\frac{ca}{b(b\plus{}c)}
\geq
\frac{a}{c\plus{}a}\plus{}\frac{b}{a\plus{}b}\plus{}\frac{c}{b\plus{}c}$

O~
I have another answer
$ LHS \equal{} \sum_{cyc}\frac {a^2b^2}{abc(c \plus{} a)} \plus{} \sum_{cyc}\frac {c^2b^2}{a^2b^2 \plus{} abc^2} \geq 3$
by Cauchy
$ LHS \geq \frac {4(ab \plus{} bc \plus{} ca)^2}{3\sum_{cyc}abc^2 \plus{} \sum_{cyc}a^2b^2 } \geq 3$


inequalities
Ring Theory
algebra
polynomial
inequalities proposed

[quote="hungkhtn"]This follows immediately my following proposition in my book

[b]Proposition (Hungkhtn).[/b]
Consider the following symmetric non-homogeneous expression
\[F(a,b,c)=m\sum_{cyc}a^{3}+n\sum_{cyc}a^{2}(b+c)+pabc+q\sum_{cyc}a^{2}+r\sum_{cyc}ab+s\sum_{cyc}a+t. \]
For $t=a+b+c$, prove that the inequality $F(a,b,c) \ge 0$ holds if and only if
\[F\left(\frac{t}{3},\frac{t}{3},\frac{t}{3}\right) \ge 0 \ ; \ F\left(\frac{t}{2},\frac{t}{2},0\right) \ge 0 \ ; \ F(t,0,0) \ge 0 \ ; \]
[/quote]
It's just an application of Hojoo Lee's theorem with some normalization! Don't expect that!
For $t=a+b+c$
\[F(a,b,c)=m\sum_{cyc}a^{3}+n\sum_{cyc}a^{2}(b+c)+pabc+\frac{1}{t}(a+b+c)\left(q\sum_{cyc}a^{2}+r\sum_{cyc}ab\right)+\frac{1}{t^{2}}(a+b+c)^{2}s\sum_{cyc}a+\frac{1}{t^{2}}(a+b+c)^{3}. \]
It's easy to transform $F(a,b,c)$ to symmetric form!

... interesting theorems... but I don't think the application to prove the above inequality is interesting.  I am waiting a simple solution...  :roll: (I don't like use any theorem, and I want see an elementary proof  =  non-theorem :D)
For myself, classical methods is more interesting than new methods  :P And I hope everybody felt as same as me, too
sorry ... my english is very bad :P So  you maybe misunderstand my idea  :blush:
Now, let kill it by another method  :wink:

Because PID proved by classical method so...
@zazai-hoang: [hide]Dùng vô chiêu mà thắng hữu chiêu[/hide]

Colud you excaltly formulate Hojoo Lee's theorem? And a prove would be welcome.  :lol:

Yes, The soul's way to prove my theorem is really unexpected. I don't imagine it before. In my book, still the following general theorem of PID is shown

[b]Proposition. (Hungkhtn)[/b]
Let $P(a,b,c,d)$ be a symmetric homogeneous polinomial of degree $3$.
The inequality $P \ge 0$ holds for all non-negative variables $a,b,c,d$ if and only if
\[P(1,1,1,1) \ge 0 \ ;\ P(1,1,1,0) \ge 0\ ;\ P(1,1,0,0) \ge 0 \ ; \ P(1,0,0,0) \ge 0 \ ;\]

By this theorem, using the Soul' way, we can conclude that all symmetric (homogeneous+non-homogeneous) inequalities of degree $3$ can be proved.

This looks very usefull. :) Hungkhtn, can you post a proof of your proposition?

[quote="hungkhtn"]Yes, The soul's way to prove my theorem is really unexpected. I don't imagine it before. In my book, still the following general theorem of PID is shown

[b]Proposition. (Hungkhtn)[/b]
Let $P(a,b,c,d)$ be a symmetric homogeneous polinomial of degree $3$.
The inequality $P \ge 0$ holds for all non-negative variables $a,b,c,d$ if and only if
\[P(1,1,1,1) \ge 0 \ ;\ P(1,1,1,0) \ge 0\ ;\ P(1,1,0,0) \ge 0 \ ; \ P(1,0,0,0) \ge 0 \ ; \]
By this theorem, using the Soul' way, we can conclude that all symmetric (homogeneous+non-homogeneous) inequalities of degree $3$ can be proved.[/quote]
This is also an application of PID
Let $f(x,y,z)=A\sum_{cyc}x^{3}+B\sum_{sym}x^{2}y+Cxyz$
Because $P(1,1,1,0)\ge 0$, $P(1,1,0,0)\ge 0$, $P(1,0,0,0)\ge 0$ so $f(x,y,z)\ge 0$
But $P(1,1,1,0)=P(1,0,1,1)=...$ so $P(x,y,z,t)=f(x,y,z)+f(y,z,t)+f(z,t,x)+f(t,x,y)\ge 0$. QED

[b]Notice that the general statement can be solved by this way.[/b]

You may right in proving my theorem of 3 variable (although it is not clear), but in this proof, you are wrong  :P . 4 variables can not be solved by this way :P

[quote="dule_00"]And now a stupid question... What is PID theorem?  :([/quote]
Give a symmetric polynomial with 3-th degree $P(x,y,z)$. The $P(x,y,z)\ge 0$ if and only if
$P(1,1,1)\ge 0$,$P(1,1,0)\ge 0$,$P(1,0,0)\ge 0$

Thank you. What about equality case? Is it true that it can be obtained only at points $(1,1,1), (1,1,0), (1,0,0)$?  And just for the confirmation, for PID   polynomial deos not need to homogeneous?

It must be homogeneous in this case!
If it's not homogeneous, see Hungkhtn's second post and my first post in this topics

Hungthn could you post solutions to these two propositions ?Thank you

[quote="The sould of rock"]Anybody can check my solution? I can't find the error. Help me please![/quote]

I think it is perfect :D

Thanks for rely on me but my solution is false because $f(1,1,1) \neq P(1,1,1,0)$

see also here,
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=76508&hilit=8+abc&start=20

[quote="zaizai-hoang"][color=blue]okie, with hungkhtn's theorem, we can conclude that all symmetric (homogeneous+non-homogeneous) inequalities of degree 3 can be proved. Great!
Then I will post the first solution (one of all 3 proof I have :wink: )
We have:
$12(a^{2}+b^{2}+c^{2})+6abc+48-30(a+b+c) \\ \ge 12(a^{2}+b^{2}+c^{2})+9\sqrt[3]{(abc)^{2}}+45-5[(a+b+c)^{2}+9] \\ \ge 7(a^{2}+b^{2}+c^{2})+\frac{27abc}{a+b+c}-10(ab+bc+ca) \\ \ge 2[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}] \ge 0$
The second solution, I used to mixing variable method
The third solution, we can put $a=x+1,b=y+1,c=z+1$ then by expanding expression we are done :wink:
To the soul of rock: okie, Can you post yours? (note that: non-PID :P) Okie?  :wink: [/color][/quote]
____________
Very nice solution! Congratulations!

The following inequality is stronger:
$ a,b,c\ge 0\Rightarrow abc+a^2+b^2+c^2+5\ge 3(a+b+c) $
This inequality appeared Mathlink forum, do not remember the exact link.
I will briefly sketch the proof:
$ 2LHS=(2abc+a^2+b^2+c^2+1)+(a^2+b^2+c^2+9)\ge 2ab+2bc+2ca+(a^2+b^2+c^2+9= $
$ =(a+b+c)^2+9\ge 6(a+b+c) $
We applied a well-known result, namely:
$ a,b,c\ge 0\Rightarrow a^2+b^2+c^2+2abc+1\ge 2(ab+bc+ca) $
(see V.Cirtoaje, Algebraic Inequalities, pb.5, pag,5 )
With $ (a+b+c)^2+9\ge 6(a+b+c)\Leftrightarrow (a+b+c-3)^2\ge 0 $
the proof is ended!
We:
$ abc+a^2+b^2+c^2+5\ge 3(a+b+c)\Leftrightarrow abc+2(a^2+b^2+c^2)+8\ge 3(a+b+c)+a^2+b^2+c^2+3 $
But:
$ 3(a+b+c)+a^2+b^2+c^2+3\ge 5(a+b+c)\Leftrightarrow a^2+b^2+c^2+3\ge 2(a+b+c)\Leftrightarrow $
$ \Leftrightarrow (a-1)^2+(b-1)^2+(c-1)^2\ge 0 $
__________
Sandu Marin

[quote="zaizai-hoang"]$\bigstar$ Let $a,b,c$ are non-negative real numbers. Prove that:
$abc+2(a^{2}+b^{2}+c^{2})+8\ge 5(a+b+c)$
I have three solution for above inequality. And what do you think ? :roll:[/quote]

$\bigstar$ Let $a,b,c\geq 0$ . Prove that best ineq:

\[8abc+9(a^2+b^2+c^2)+64+7(ab+bc+ca) \geq  40(a+b+c)\]

[quote="xzlbq"]
$\bigstar$ Let $a,b,c\geq 0$ . Prove that best ineq:

\[8abc+9(a^2+b^2+c^2)+64+7(ab+bc+ca) \geq  40(a+b+c)\][/quote]
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality is equivalent to $f(v^2)\geq0$, where $f$ is a linear function.
Hence, by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=278791]uvw[/url] it remains to prove our inequality for $b=c$, which gives
$9a^2+2(4b^2+7b-20)a+25b^2-80b+64\geq0$.
If $4b^2+7b-20\geq0$ then our inequality is true.
But for $4b^2+7b-20<0$ we need to prove that $(4b^2+7b-20)^2-9(25b^2-80b+64)\leq0$, which is
$(b-1)^2(2b^2+11b-22)\leq0$, which is true for $4b^2+7b-20<0$ and $b\geq0$.

[quote="hungkhtn"]This follows immediately my following proposition in my book

[b]Proposition (Hungkhtn).[/b]
Consider the following symmetric non-homogeneous expression
\[F(a,b,c)=m\sum_{cyc}a^{3}+n\sum_{cyc}a^{2}(b+c)+pabc+q\sum_{cyc}a^{2}+r\sum_{cyc}ab+s\sum_{cyc}a+t.\]
For $t=a+b+c$, prove that the inequality $F(a,b,c) \ge 0$ holds if and only if
\[F\left(\frac{t}{3},\frac{t}{3},\frac{t}{3}\right) \ge 0 \ ; \ F\left(\frac{t}{2},\frac{t}{2},0\right) \ge 0 \ ; \ F(t,0,0) \ge 0 \ ;\]

By this theorem, all symmetric inequalities of degree $3$ (3variable) is solved.

To solve Namdung's one, we have

1, If $a=b=c=t$ then
\[t^{3}+6t^{2}-15t+8 \ge 0,\]
which is true,
2, If $a=b=t,c=0$ then
\[4t^{2}+8 \ge 10,t\] which is true, too.
3, If $a=t,b=c=0$ then
\[2t^{2}+8 \ge 5t\]
which is true, too.

 :lol:[/quote]
_________
Your end result is very powerful! You can post a short demo for this result?


Show that $\sigma (n) -d(m)$ is even for all positive integers $m$ and $n$ where $m$ is the largest odd divisor of $n$.

[quote="Peter"]Show that $ \sigma (n) \minus{} d(m)$ is even for all positive integers $ m$ and $ n$ where $ m$ is the largest odd divisor of $ n$.[/quote]
$ n \equal{} 2^{k}m$
$ \sigma (2^km) \equal{} \sigma (2^k)\sigma (m)\equiv \sigma (m) (\mod 2)$
Let $ m \equal{} \prod_{i \equal{} 1}^{r} p_i^{a_i}$  where $ p\equiv 1 (\mod 2)$
We has $ \sigma (m) \equal{} \prod _{i \equal{} 1}^r\frac {p_i^{a_i \plus{} 1} \minus{} 1}{p \minus{} 1}\equiv (a_1 \plus{} 1)...(a_r \plus{} 1) \equal{} \tau(m)$
So $ \sigma(n)\equiv \tau(m)(\mod 2)$
This problem was be prove.

So the problem boils down to the trivial $1+t+t^2+\cdots +t^s \equiv s+1 \pmod{2}$ for odd $t$.
Then $\sigma(n) =\sigma(2^km) = \sigma(2^k)\sigma(m) \equiv \sigma(m) \equiv \tau(m) \pmod{2}$,
since $\sigma(m) = \prod_{p \textrm{ prime, } p\mid m} \sum_{i=0}^{v_p(m)} p^{i} \equiv $ $\prod_{p \textrm{ prime, } p\mid m} (v_p(m) + 1) =$ $ \tau(m) \pmod{2}$.


when $ \mathcal{A}\equal{} \emptyset$, why $ \bigcap_{A\in \mathcal{A}} A\equal{} X$ ?
thanks

As the text says, it's convenient to assume it's like this. Don't bother with the argumentation, which pertains to fundamentals of logic and set theory (there is some discussion in Bourbaki).

In a similar way, for numbers in a set $ X$, if $ X$ is empty it is convenient to take $ \sum_{x \in X} x : \equal{} 0$ and $ \prod_{x \in X} x : \equal{} 1$.

[color=green]Moderator says: never quote the previous post in full, you are answering it by default[/color]

Thanks very much!


Hi. I'm new on this forum, so I don't know if this topic has been presented before or what.

Well, I'm a pretty decent violinist, but can't decide between music or math. I don't want to go to a music conservatory because I'd lose out on other subjects, and I couldn't see myself doing that. My ideal school would be MIT but with high-level music at least somewere near, or access to a really good teacher. I hear things like since MIT has all these music/math/science geniuses that they end up having a good orchestra even though they don't mean to.
Also, the ideal school would have to have a flexible schedule (which probably won't happen) because I have an active music agenda with orchestras and gigs and stuff.
I'm just looking for peoples opinions if they already are in college.

Music performance programs demand an enormous amount of time from their students, and that drives away most people who have merely casual interests or want to do something else like math. There's very little correlation between serious math programs and serious music programs, because students don't do both. I spent two years as a violist in the university orchestra at Berkeley- there were many math/computer science people in the viola section, and those were entirely disjoint from the music specialists.
If you're serious about the music, you'll probably have to do some research to find a school with music and math programs that are good enough for what you want. If not, you can get a pretty good orchestra at almost any large enough school.

As for schedule flexibility: It's generally easy to miss a few scattered days. Colleges don't force you to attend classes, and you'll generally get reasonable accommodations if you inform professors in advance.

[quote="JBL"]You might look at Oberlin, which is a small liberal arts school attached to an excellent music conservatory.  The math there is not comparable to MIT, but they are definitely qualify as one of the strong liberal arts schools in the country (I believe they have among the highest rates of students going on to Ph.D.s), and I know they have some very competent math students and faculty members.[/quote]
What if one wishes for a math focus with some music, rather than a music focus with some math? (Which I take Oberlin is, from your description.)

Yes, Oberlin certainly doesn't have a "math focus" (and it needn't have a music focus, if you're only enrolled in the college and not the conservatory).  Really, I only know anything about it because my brother and a friend or two go there, and I know very little about music programs anywhere else, MIT included.  I do think that jmerry is certainly correct when he says, [quote]If not, you can get a pretty good orchestra at almost any large enough school.[/quote]Harvard, for instance, has excellent math and some very good student orchestras, but there isn't a major in music performance.

Carnegie Mellon U. in Pittsburgh has a fantastic school of fine arts, including a  very good music program. CMU is strange in that it excels in both the arts and the sciences. I am currently in the math program here, and I know quite a few people who are comp sci and engineering majors as well as fine musicians. The music school here offers plenty of instruction and ensembles for those who are not actually enrolled.

Bottom line: check out CMU.

Does anybody hear anything about Case Western? I know their sciences are way up there. Plus, Case is close to CIM (Cleveland Institute of Music). Studying under David Cerone would be awesome! :lol:

[quote]Carnegie Mellon U. in Pittsburgh has a fantastic school of fine arts, including a very good music program. ~ curl[/quote]
Yeah, they do in general, but there's really no teachers that catch my eye there. I've heard great things about their chamber ensembles, though.

Well, college I guess is really what you make of it. It sorta matters where you go, but ultimately, it's how much you work you put in is how far you're going to get.
I'm coming to realize that if you want a top-notch education in music/math, then you have to go to conservatory early, finish early, then go to a university. :rotfl:  :roll: What's scary is I know people who've done it.  :|

If you can get in, how about those dual-enrollment programs (I think that's what they're called) like:

Juilliard/Columbia
Harvard/NEC
John Hopkins/Peabody

Those are just the ones I've heard in passing; there are probably other places that offer similar programs.

Music theory or music performance?

edit: this was in response to manuel's comment that he is at MIT and majors in math and music.  That post was deleted because [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135229]something weird[/url] happened to it.

[quote="curl"]Carnegie Mellon U. in Pittsburgh has a fantastic school of fine arts, including a  very good music program. CMU is strange in that it excels in both the arts and the sciences. [/quote]I went to CMU.  I was a math major.  My roommate was a music major. :)

My major warning was that both I and my roommate seemed pretty busy with schoolwork, and we had (after freshman year) absolutely no overlapping classes.  I can't imagine doing both majors at the same time.  I suppose if you took the bare minimum necessary for each degree, you could manage it.  I was in the CS track in the math department, plus I was in the Math Studies program (which is a more theory-based curriculum), so I had a heavier workload than the "typical" math major.  My roommate was also pretty busy with practicing, taking theory classes, and doing concerts.  Plus the admissions process is totally different for the College of Science and the College of Fine Arts.

Recently MIT is very glad to admit students with a strong interest in music.


Hi,

I'm totally new at this and don't know where to start. I have a C++ project and used Doxygen to convert to to .tex format, intermediate format for pdf. I downloaded MikTex and WinEdt, but don't know how to turn my .tex into the pdf files :blush:

Could anyone help?

I tried opening one of the .tex using WinEdt, and then
Accessories -> PDF -> PDFTex,

But the console that appears says

!Undefined Control Sequence
1.1 \section
{MyProject Class List}
?

And now I'm kinda stuck.

Does anyone know how I'm meant to convert these .tex to pdf ?

Thanks in advance for any help.

$\LaTeX$ is a formatting language. Much like C++, you have a set of instructions you need to follow to obtain the desired output. Try looking at the $\LaTeX$ guide we have here on the site: http://www.mathlinks.ro/LaTeX/AoPS_L_About.php and/or the simple introduction I've made in this forum (the very first topic).

Thanks for the response.  I'll give it a read, problem is I'm still writing the thesis and may not have time to work it out as deadline is soon  :?

I may have to settle for printed html output from Doxygen.

I've tried a few things from the command line and in WinEdt but hey all give th above error.

Doxygen says in the help file

PDF
    Generated from the LaTeX output by running make pdf in the output directory.

However being stupid I don't know what "running make pdf" means.  I have the output directory full of .tex files though.

I found a site with a Doxygen config file on it and in it is said

# If the PDF_HYPERLINKS tag is set to YES, the LaTeX that is generated
# is prepared for conversion to pdf (using ps2pdf). The pdf file will
# contain links (just like the HTML output) instead of page references
# This makes the output suitable for online browsing using a pdf viewer.

PDF_HYPERLINKS         = YES

# If the USE_PDFLATEX tag is set to YES, pdflatex will be used instead of
# plain latex in the generated Makefile. Set this option to YES to get a
# higher quality PDF documentation.

USE_PDFLATEX           = YES

Does ps2pdf use .tex files that seem to be generated? I thought it only worked on .ps files...

I'm confused and buried in C++ and a Word doc, not good.

I'll include some info out of one of the .tex files incase it seems familiar.

Thanks again for any further information.

[code]
\section{Reasoning Class Reference}
\label{class_reasoning}\index{Reasoning@{Reasoning}}
{\tt \#include $<$Reasoning.h$>$}

\subsection*{Public Member Functions}
\begin{CompactItemize}
\item
{\bf Reasoning} ()
\item
...more of this here...
\item
vector$<$ string $>$ {\bf description\-Insertion\-Sort} (vector$<$ int $>$ name\-Ordering, vector$<$ string $>$ veng\-Desc)
\end{CompactItemize}

\subsection{Detailed Description}
This class is meant to simulate the reasoning ability of our Artifical Animal.

Definition at line 22 of file Reasoning.h.

\subsection{Constructor \& Destructor Documentation}
\index{Reasoning@{Reasoning}!Reasoning@{Reasoning}}[/code]

There are several paths to get from Latex to a PDF.

1) [b]Latex[/b] [i]Filename.tex[/i]  --> [i]Filename.dvi[/i]
Then, [b]dvi2pdf[/b] [i]Filename.dvi[/i] --> [i]Filename.pdf[/i]

2) [b]Latex[/b] [i]Filename.tex[/i] --> [i]Filename.dvi[/i]
Then, [b]DVIPS[/b] [i]Filename.dvi[/i] --> [i]Filename.ps[/i]
Then, [b]ps2pdf[/b] [i]Filename.ps[/i] --> [i]Filename.pdf[/i]

3) [b]Pdflatex[/b] [i]Filename.tex[/i] --> [i]Filename.pdf[/i]

I would recommend option 3.

Also, I have run into complaints about undefined control sequences when switching between document classes.  Make sure that the document class you are using supports the command that the compiler is complaining about.  Also, verify what version of Tex that the program you are using outputs.  I'm assuming here that it outputs Latex.


Let $ X$ and $ Y$ be topological spaces, and let $ f: X\to Y$. Show that $ f$ is continuous iff $ f(int E) \subset int(f(E))$ for every $ E \subset X$, iff $ int(f^{\minus{}1}(E))\subset f^{\minus{}1}(int E)$ for every $ E\subset Y$.

Note that $ int E$ is the closure of $ E$.

:?:

Usually $ int E$ means the interior of $ E$ and $ cl(E)$ or $ \overline{E}$ means closure of $ E$.

sorry about that, i will revise the problem

Let $ X$ and $ Y$ be topological spaces, and let $ f: X\to Y$. Show that $ f$ is continuous iff $ f(\overline{E}) \subset \overline{f(E)}$ for every $ E \subset X$, iff $ \overline{f^{ \minus{} 1}(E)}\subset f^{ \minus{} 1}(\overline{E})$ for every $ E\subset Y$.


Suppose a point dipole is located at the centre of a grounded spherical conducting shell. Find the potential and electric field at points inside the shell. (Hint: Use zonal harmonics that are regular at the origin to satisfy the boundary conditions on the shell.)


conics
parabola
integration
calculus
geometry theorems
geometry

please teaching me the formula for calculating the length of the given parabola!
thanks

If $ f: [a,b]\to \mathbb{R}$ is continuous and differentiable on $ ]a,b[$ then the length of the graph of $ f$ between $ a$ and $ b$ is $ \int_a^b\sqrt {1 \plus{} (f'(x))^2}dx$. So you have to compute the integral $ \int_a^b\sqrt {1 \plus{} (cx\plus{}d)^2}dx$.


i love coming to the site because i love watching the animated proofs as well as other parts of the site

i was wondering if there was a way to choose which one i saw or look at them at once
sometimes they go by to fast for me to notice what they are

Right [url=http://www.artofproblemsolving.com/Gallery/AoPS_G_Gallery.php]here[/url], all in one place.


group theory
abstract algebra
superior algebra
superior algebra unsolved

(i) Let $ H$ be a subgroup of a finite group $ G$. Let $ P$ be a Sylow subgroup of $ G$. Prove that there exists a conjugate $ P'$ of $ P$ in $ G$ s.t. $ P'\cap H$ is a $ p$-Sylow subgroup of $ H$.

(ii) Assume $ H$ normal and $ (G: H)$ prime to $ p$. Prove that $ H$ contains every $ p$-Sylow subgroup of $ G$.

Both things are trivial:
Let $ Q$ be a p-sylowgroup of $ H$. Then there is a conjugate $ P'$ of $ P$, s.t. $ Q\subseteq\ P'\Rightarrow\ Q\equal{}P'\cap H$.

If p doesn't divide |G:H|, then every p-sylowgroup of H is a p-sylowgroup of G. Since all the p-sylows are conjugated in G and H is normal, all p-sylows lie in H.


inequalities
logarithms
real analysis
real analysis unsolved

Prove following two inequalities:

1) $ \forall y>x>0:$  $ \left(\frac{y\minus{}x}{\ln y \minus{} \ln x}\right)^{y\minus{}x}\leq\frac{(y/e)^y}{(x/e)^x}\leq\left(\frac{x\plus{}y}{2}\right)^{y\minus{}x}$

2) $ \forall y>x>0:$  $ (xy)^{y\minus{}x}\leq\left(\frac{(y/e)^y}{(x/e)^x}\right)^2\leq\left(\frac{y^2\plus{}xy\plus{}x^2}{3}\right)^{y\minus{}x}$

I believe these are genuinely mine, but if someone has happened to see them elsewhere, I would be thankful if he or she could point me to his/her source :)

Good luck and have fun!


Take any three digit number, say $521$.
write the digits in the opposite direction to get $125$.
Then subtract the smaller number from the larger one: $521-125=396$
again write the digits in the opposite direction to get $693$
add the two numbers $396+693=\boxed{1089}$.
independent of the initial choice, in the end you'll always get $1089$, as long as the result after the subtraction is a three digit number. can you give a mathematical explanation of this fact?

ps. there are many other interesting things about this number.

[hide]

Take a number $ABC$

This number is
$100A+10B+C$
flip it backwards you get
$100C+10B+A$
subtract em and you get

$|100A+C-(100C+A)| = |99A-99C| = |99(A-C)|$

let $Q = |A-C|$

we want to take the sum of $99Q$ and $99Q$ backwards

in digits, 99Q is in the form
$(Q-1)100+90+(10-Q) = 99Q$
flip it backward and you get
$(10-Q)100+90+(Q-1) = 1089-99Q$

add em together and you get $1089$ regardless of Q
[/hide]


Let $ t(A)$ denote the sum of elements of a nonempty set $ A$ of integers, and define $ t(\oslash ) \equal{} 0$. Find a set $ X$ of positive integers such that for every integer $ k$ there is a unique ordered pair of disjoint and finite subsets $ (A_k,B_k)$ of $ X$ such that $ t(A_k) \minus{} t(B_k) \equal{} k$.

The set $ X \equal{} \{ 3^n \ \mid \ n \equal{} 0,1,2,\ldots \}$ should work, since any integer, written in basis $ 3$ (thus using digits $ 0$, $ 1$ and $ 2$), may be re-written using "digits' $ 0$, $ 1$ and $ \minus{}1$, and that uniquely.

It seems like that set should be unique, but at first glance I have no idea how to prove such a thing (if it's actually true) -- anyone have any ideas?

I think that on the same line, the set $ X \equal{} \{2, 3, 9, \ldots, 3^n, \ldots \}$ also qualifies as an answer; in general the sets $ X \equal{} \{ \alpha_n3^n \mid n \equal{} 0,1,2,\ldots \}$, with $ \alpha_n \in \{ 1, 2 \}$ but infinitely many of them equal to $ 1$. Again, the ternary writing of an integer can be uniquely turned into one using as "digits" $ 0$ and $ \pm \alpha_n$.

Okay, very nice, but unfortunately this also doesn't capture everything because $ \{3,4, 9, 27, 81, \ldots\}$ seems to work, as does $ \{6, 8, 9, 27, 81, 243, \ldots\}$.  It does certainly give the feeling that we must eventually have powers of three, but I have no feeling at all about whether or not these should really be characterizable.


There are $ 6$ blue marbles, $ 5$ red marbles, $ 4$ green  marbles, and $ 3$ white marbles in a bag. If I am blindfolded, how many marbles must I choose such that I am assured of having at least three marbles of the same color?

Worst case scenario:  You choose $ 2$ blue, $ 2$ red, $ 2$ green, $ 2$ white (in any order) and then another one (which makes $ 3$ for a certain color guaranteed).  Therefore, our answer is $ 2\plus{}2\plus{}2\plus{}2\plus{}1\equal{}\boxed{9}$


The sequence of palindromes is written in increasing order 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, . . . .
Thus 11 occupies the tenth position, 22 the eleventh, etc.
Which position is occupied by 1003001?

There are 9 1-digit palindromes, 9 2-digit palindromes, 90 3-digit palindromes, 90 4-digit palindromes, 900 5-digit palindromes, and 900 6-digit palidromes. Of the 9000 7-digit palindromes, 1003001 is the 4th.

9 + 9 + 90 + 90 + 900 + 900 + 4 = [b]2002[/b]


polynomial
geometry
3D geometry
number theory unsolved
number theory

Does anybody know whether there is rational point other than $ (1,1)$ on the curve $ x^3\plus{}y^3\equal{}2$?
Thanks a lot.

hmm... i don't know if my proof is correct but here it goes:

Let $ x\equal{}\frac{a}{c}$, $ y\equal{}\frac{b}{d}$. Then the equation becomes $ \left(\frac{a}{c}\right)^{3}\plus{}\left(\frac{b}{d}\right)^{3}\equal{}2 \Rightarrow a^{3}\plus{}b^{3}\equal{}2(cd)^{3}$ for $ a, b, c, d \in N$. We will denote $ cd\equal{}z$. Then the problem becomes finding all whole solutions to $ a^{3}\plus{}b^{3}\equal{}2z^{3}$. Since the equation is homogenous, we can assume $ \gcd(a,b,z)\equal{}1$ (since we can divide out all common factors). We now have 2 cases:

Case 1: $ \gcd(a,b)\equal{}2$. This implies that if $ a\equal{}2a'$ and $ b\equal{}2b'$, $ 8a'^{3}\plus{}8b'^{3}\equal{}2z^{3}$, $ 4a'^{3}\plus{}4b'^{3}\equal{}z^{3}$, contradiction.

Case 2: $ \gcd(a,b)\equal{}\gcd(b,z)\equal{}\gcd(c,a)\equal{}1$. In this case we factorize the expression into Eisenstein numbers in the following way: let $ \omega$, $ \omega^{2}$ represent the roots of unity of $ 3$ which are not $ 1$. Then the expression is factorized as \[ (a\plus{}b)(a\plus{}b\omega)(a\plus{}b\omega^{2})\equal{}2z^{3}\]. Let $ d\equal{}\gcd(a\plus{}b\omega,a\plus{}b\omega^{2})$. We know $ d|(a\plus{}b\omega)\plus{}(a\plus{}b\omega^{2})\Rightarrow d|2a\minus{}b$. Also, $ d|(a\plus{}b\omega)\minus{}(a\plus{}b\omega^{2})\Rightarrow d|b\omega(1\minus{}\omega)$. We can discard the $ \omega$ factor since it is a unit. Since $ 1\minus{}\omega$ is an Eisenstein prime, it cannot be broken up further. Therefore, we have the following cases: $ d\equal{}b\Rightarrow b|2a\minus{}b \Rightarrow b|a$, which is a contradiction since $ \gcd(a,b)\equal{}1$.

The second case is $ d\equal{}1\minus{}\omega$. We know $ 1\minus{}\omega$ is associated only with the prime $ 3$. However, $ (a\plus{}b\omega)(a\plus{}b\omega^{2})$ holds at least twice the number $ 1\minus{}\omega$ but never its conjugate, $ 1\minus{}\omega^{2}$ (because otherwise $ a\plus{}b\omega$ or $ 1\plus{}b\omega^{2}$ would be divisible by $ 1\minus{}\omega^{2}$, implying $ a$ and $ b$ are both divisible by $ 3$, which is a contradiction. But this would imply $ a^{2}\minus{}ab\plus{}b^{2}$ is not an integer, which is another contradiction).

Therefore $ \gcd(a\plus{}b\omega, a\plus{}b\omega^{2})\equal{}1$. In a similar manner you can end up proving all three factors of the original factorization must be relatively prime.

We now return to the original factorization. you notice, since $ x\plus{}y$ is obviously even, $ \frac{x\plus{}y}{2}$, $ a\plus{}b\omega$, $ a\plus{}b\omega^{2}$ must all be perfect cubes which, through algebraic manipulation, you find is never true. We need to note we never took into account the only case where $ a\equal{}b$ is permissible, which is $ 1$. We can easily verify this case is a solution.

Again, I went through this proof pretty quickly, so I might have forgotten something


Find the sum of all four digit positive integer palindromes.

there are two cases
1)palindromes of the form aaaa,for this ther are 9possibles
2)abba,for this ther are 9*10 possibles living a total of 99
 i will post the sum later

Actually, I think there are 81 palindromes of form ABBA (A can be any of 9 digits, B can be any of 9 digits).
The 9 digits for A are 1,2,3,...,9.
The 9 digits for B are 0,1,2,...,8.
So the total number of palindromes is 90.

As for the sum... I'll get back to you  :lol:

Let's sum the first and last digits, then the middle digits.

We can easily see that we sum 1001 ten times, 2002 ten times, 3003 ten times, and so on. Therefore the sum of the end digits is

$ 1001 * 10 \plus{} 2002 * 10 \plus{} 3003 * 10 \plus{} ... \plus{} 9009 * 10 \equal{}$
$ 10010 (1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} 6 \plus{} 7 \plus{} 8 \plus{} 9) \equal{}$
$ 10010 * 45$

For the middle digits, we sum 0110 nine times, 0220 nine times, and so on. The sum of those is

$ 110 * 9 \plus{} 220 * 9 \plus{} ... \plus{} 990 * 9 \equal{}$
$ 990 (1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} 6 \plus{} 7 \plus{} 8 \plus{} 9) \equal{}$
$ 990 * 45$

The end sum is $ 45 (10010 \plus{} 990) \equal{} 45 * 11000 \equal{} \boxed{495000}$, which agrees with aleph0's sum (although I don't know how he got it exactly)

[quote="ernie"]Actually, there is no restriction saying that $ a \neq b$, so there are $ 90$ palindromes, as 21 7 15 said before.[/quote]

Actually, 21 7 15 said there were 99 total. I said that there are 90 total, but 81 of form ABBA and 9 of form AAAA.
Anyway, it's not important.

Another solution:

The generic form of our palindrome is $ ABBA\equal{}A\plus{}10B\plus{}100B\plus{}1000A\equal{}1001A\plus{}110B$.
Suppose we fix $ B$ and sum all the palindromes from $ A\equal{}1$ to $ A\equal{}9$. Then we have:

\[ \sum_{k\equal{}1}^{9}(1001k\plus{}110B)\]

But $ B$ can range from $ 0$ to $ 9$, so we have:

\[ \sum_{j\equal{}0}^{9}\sum_{k\equal{}1}^{9}(1001k\plus{}110j)\\
\equal{}\sum_{j\equal{}0}^{9}\left(9(110j)\plus{}1001\sum_{k\equal{}1}^{9}k\right)\\
\equal{}\frac{10\cdot 1001\cdot 9\cdot 10}{2}\plus{}990\sum_{j\equal{}0}^{9}j\\
\equal{}\frac{10010\cdot 90}{2}\plus{}\frac{990\cdot 9\cdot 10}{2}\\
\equal{}\boxed{495000}
\]

But ffao's solution is simpler.


function
integration
analytic geometry
probability and stats

Let $ (X,Y)$ be a random variable uniformly distributed on polygonal having vertices (0,0), (1,0), (1,1).
Find $ E(XY)$.

Length of sides are: 1, 1, $ \sqrt{2}$
Corresponding probabilities are $ 1/(2\plus{}\sqrt{2})$, $ 1/(2\plus{}\sqrt{2})$, $ \sqrt{2}/(2\plus{}\sqrt{2})$

Conditional expectations are $ 0$, $ E(U)\equal{}1/2$, $ E(U^{2})\equal{}1/3$, where U is uniform [0;1]

So i got $ \frac{1}{(2\plus{}\sqrt{2})}0\plus{}\frac{1}{(2\plus{}\sqrt{2})}\frac{1}{2}\plus{}\frac{\sqrt{2}}{(2\plus{}\sqrt{2})}\frac{1}{3}$

suppose that $ Z: (\Omega, P) \to F$ is a random variable, and that you know $ P_Z$ the law of $ Z$. Then for any good function (positive, or bounded, or ..) $ f: F \to \mathbb{R}$:
$ E[f(Z)] \equal{} \int_{F} f(z) P_Z(dz)$

Here you can take $ Z\equal{}(X,Y): (\Omega, P) \to \mathbb{R}^2$ and $ f(x,y) \equal{} xy$. Can you see how to conclude ?

[b]Roah[/b] is right, because [b]alekk[/b] cited the wrong density.

For a slightly more formal repetition of [b]roah[/b]'s argument:

Define the events $ A=\{X=Y\}\,,\,B=\{Y=0\}\,,\,C=\{X=1\}$. They describe the three edges and trivially intersect in the common corner, which is a zero set. Because of the uniformness these three events have likelihood proportional to their [i]lengths[/i], i.e. $ \frac{\sqrt{2}}{2+\sqrt{2}},\frac{1}{2+\sqrt{2}},\frac{1}{2+\sqrt{2}}$ respectively. Moreover, the conditional distribution on each event is uniform as well. Then

$ \begin{array}{rl}\mathbb{E}[\,XY\,]= & \frac{1}{2+\sqrt{2}}\cdot\{\,\sqrt{2}\cdot\mathbb{E}[\,XY\,|\,A\,]+\mathbb{E}[\,XY\,|\,B\,]+\mathbb{E}[\,XY\,|\,C\,]\,\}\\[0.25cm]
 = & \frac{1}{2+\sqrt{2}}\cdot\{\,\sqrt{2}\cdot\mathbb{E}[\,U(0,1)^2\,]+0+\mathbb{E}[\,U(0,1)\,]\,\}\\[0.25cm]
 = & \frac{1}{2+\sqrt{2}}\cdot\{\,\frac{\sqrt{2}}{3}+\frac{1}{2}\,\}\end{array}$

The distribution of $ X$ is
$ F_X(x) \equal{}$
$ 0$ if $ x \le 0$
$ \frac {x}{2}$ if $ 0 < x < 1$
$ 0$ if $ x \ge 1$.
The distribution of $ Y$ is
$ F_Y(y) \equal{}$
$ 0$ if $ y \le 0$
$ \frac {1}{2} \plus{} \frac {y}{2}$ if $ 0 < y < 1$
$ 0$ if $ y \ge 1$.
Hence $ X,Y$ have not uniform distribution.
So
$ E(XY) \equal{} E_X(E(XY|X)) \equal{} \frac {1}{2}E(XY|X \equal{} 1) \equal{} \frac {1}{2}\int_0^1ydy \equal{} \frac {1}{4}$.

[b]Spasty[/b], you got your distributions wrong, e.g., for $ X$ you quote: $ 0\leq x<1: F_X(x)\equal{}\frac{x}{2}$, this would mean that the set $ \{X\equal{}1\}$, which is one of your (shorter) edges of you triangle has likelihood $ \frac{1}{2}$, a contradiction. Actually, I just see you claim $ F_X(x)\equal{}0$ for $ x\geq 1$, which does not make sense, since a distribution is non-decreasing.

P.S. As [b]roah[/b] and I have elaborated the key is to identify uniform random variables upon appropriate conditioning. There was no statement about uniform coordinate RV's $ X,Y$.

Ok.
The distribution of $ X$ is
$ F_X(x) \equal{}$
$ 0$ if $ x \le 0$
$ \frac {x}{2}$ if $ 0 < x < 1$
$ 1$ if $ x \ge 1$.
The distribution of $ Y$ is
$ F_Y(y) \equal{}$
$ 0$ if $ y \le 0$
$ \frac {1}{2} \plus{} \frac {y}{2}$ if $ 0 < y < 1$
$ 1$ if $ y \ge 1$.
Hence $ X,Y$ have not uniform distribution.
So
$ E(XY) \equal{} E_X(E(XY|X)) \equal{} \frac {1}{2}E(XY|X \equal{} 1) \equal{} \frac {1}{2}\int_0^1ydy \equal{} \frac {1}{4}$.

$ X,Y$ have distributions of mixed type (discrete and continuous).

it is still not correct as previously stated: you wrongly assign a likelihood of $ \frac{1}{2}\equal{}F_X(1)\minus{}F_X(1^{\minus{}})$ to the "edge" $ \{X\equal{}1\}$, which is one of the (shorter) sides of your triangle.

I think that "uniformly distributed on polygonal having vertices (0,0), (1,0), (1,1)" means that we must consider the segment with vertices (0,0) (1,0) and the segment with vertices (1,0), (1,1).

If we consider three sides, I'm agee with you.


Can someone tell me something more about the tournament of the towns ?

Arne, I hope that these two websites may prove useful:

http://www.math.toronto.edu/oz/turgor/about1.php
http://www.amt.canberra.edu.au/imtot.html