[quote="2006"]$\ a,b,c \geq0$ and $\ abc=1$ prove that
$\sqrt[3]{\frac{a}{b+26}}+\sqrt[3]{\frac{b}{c+26}}+\sqrt[3]{\frac{c}{a+26}}\geq1$
        :D[/quote]
The hint see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=104676

[quote="2006"]$\ a,b,c \geq0$ and $\ abc=1$ prove that
$\sqrt[3]{\frac{a}{b+26}}+\sqrt[3]{\frac{b}{c+26}}+\sqrt[3]{\frac{c}{a+26}}\geq1$
        :D[/quote]
I am sorry. See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111873


Let $(O)$ be circumcircle of quadrangle $ACBD$ .$AB$ cuts $CD$ at $M$.Suppose that $N$ is midpoint of $BD$.$MN$ cuts $AC$ at $K$.Prove that:
$\frac{AK}{CK}=(\frac{AM}{CM})^{2}$

Help me,please!
Note that we can prove:$MA.MB=MC.MD=R^{2}-OM^{2}$

$\frac{AK}{CK}: \frac{AM}{CM}= \frac{AK}{AM}: \frac{CK}{CM}\\ = \frac{sin \angle AMK}{sin \angle AKM}: \frac{sin \angle CMK}{sin \angle CKM}= \frac{sin \angle BMN}{sin \angle DMN}: \frac{sin \angle AKM}{sin \angle CKM}\\ =\frac{sin \angle BMN}{sin \angle DMN}= \frac{sin \angle BMN}{sin \angle BNM}: \frac{sin \angle DMN}{sin \angle DNM}= \frac{BN}{BM}: \frac{DN}{DM}= \frac{DM}{BM}= \frac{AM}{CM}\\ \Rightarrow \frac{AK}{CK}= (\frac{AM}{CM})^{2}$

Hi !
Where is the necessity that $N$ is the midpoint of $BD$ ? :blush:


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G\neq \{0\}$ a closed subgroup of $ (\mathbb{R},+)$. Prove that $ G=\mathbb{R}$ or $ G\approx \mathbb{Z}$

[quote="Rust"]Let $ a=inf(x>0,x\in G)$.  If $ a=0$ then closed subgroup is R, else $ G=aZ$.[/quote]

Could you be a little more explicit, please? I was thinking quite a lot on this problem:) 10x.

If a>0, obviosly all elements are $ an, n\in Z$ (else we get easy contradition). If a=0 in all interval $ (x-\epsilon,x+\epsilon)$ we have element from G, because G is closed $ x\in G \ \forall x\in R$.


after a long and ugly solution I came to this .... lim a_n =0

and how did you do that??
what technics did you use??


Let $ f(n)$ be the number of times you have to hit the $ \sqrt {\ }$ key on a calculator to get a number less than $ 2$ starting from $ n$. For instance, $ f(2) \equal{} 1$, $ f(5) \equal{} 2$. For how many $ 1 < m < 2008$ is $ f(m)$ odd?

[hide="edit: ok im back"]
the highest integer for which it takes n presses is $ 2^{n \plus{} 1}$
so you just alternatingly subract and add powers of 2

so it's $ 2^{9\plus{}1}\minus{}2^{8\plus{}1}\plus{}2^{7\plus{}1}\minus{}2^{6\plus{}1}\plus{}2^{5\plus{}1}\minus{}2^{4\plus{}1}\plus{}2^{3\plus{}1}\minus{}2^{2\plus{}1}\plus{}2^{1\plus{}1}\minus{}2^{0\plus{}1}$
$ \equal{}1024\minus{}512\plus{}256\minus{}128\plus{}64\minus{}32\plus{}16\minus{}8\plus{}4\minus{}2$
$ \equal{}682$

[/hide]

[hide="Solution"]
If you have a number between $ 2^{2n}$ and $ 2^{4n}$, its squareroot will fall between $ 2^{n}$ and $ 2^{2n}$

We can use this to partition our restricted numberline into intervals of the form $ [2^{2^k}, 2^{2^{k \plus{} 1}} )$, namely:
$ [1,2), \; [2,4), \; [4,16), \; [16,256), \; [256, 2008]$

If we have a number in any category, if we press squareroot, we will move into the category to the left. Thus to get an odd number of key-presses, we must start in either the second or fourth intervals, counting from the left:
- Interval 2 contains integers 2 and 3, and thus has two integer elements.
- Interval 4 contains integers 16 through 255, and thus 255-15 = 240 integer elements

This gives a total of [b]242[/b] integral starting points.
[/hide]

[quote="alkjash"][quote="CircleSquared"][hide="edit: ok im back"]
the highest integer for which it takes n presses is $ 2^{n \plus{} 1}$
so you just alternatingly subract and add powers of 2

so it's $ 2^{9 \plus{} 1} \minus{} 2^{8 \plus{} 1} \plus{} 2^{7 \plus{} 1} \minus{} 2^{6 \plus{} 1} \plus{} 2^{5 \plus{} 1} \minus{} 2^{4 \plus{} 1} \plus{} 2^{3 \plus{} 1} \minus{} 2^{2 \plus{} 1} \plus{} 2^{1 \plus{} 1} \minus{} 2^{0 \plus{} 1}$
$ \equal{} 1024 \minus{} 512 \plus{} 256 \minus{} 128 \plus{} 64 \minus{} 32 \plus{} 16 \minus{} 8 \plus{} 4 \minus{} 2$
$ \equal{} 682$

[/hide][/quote]

Urgg...
I made the exact same mistake.[/quote]

that doesn't sound optimistic to me
well someone could probably tell me what i did wrong, for starters

probably a subtraction error
or a lot of them

[quote="The Zuton Force"]Nope.

You divided the number by 2 instead of taking the squareroot, thus reducing the exponent by one.
However, by taking a squareroot, you are dividing the [i]exponent[/i] by 2.

Your terms should be based on numbers of the form $ 2^{2^n}$.

In particular, the answer is $ 2^{2^3} \minus{} 2^{2^2} \plus{} 2^{2^1} \minus{} 2^{2^0}$[/quote]

ohhhhh.  thanks for pointing that out  :)

[quote="Brut3Forc3"]You know, the problem never explicitly says that $ m$ has to be an integer... :P[/quote]

That's assuming this is the original wording...

That is in fact the original wording, the word "integer" is not used. However, the function is defined as $ f(n)$, and $ n$ is generally understood to be an integer variable, implying that $ f(n)$ takes integer arguments, so I doubt anyone would get confused or contest the question  :P

The use of the phrase "how many" usually implies a discrete set, which the reals do not make.

The question is poorly worded.  That happens.  It's simultaneously completely unambiguous :)

I think the easiest way to solve this question is:
when m=2-3 f(m)=1
when m=4-15 f(m)=2
when m=16-255 f(m)=3
when m=256-2007 f(m)=4
so the odd value of f(m) is 242


let $ x,y,z$ positives real numbers such that $ xyz\equal{}1$
$ \sqrt{\frac{x^2y\plus{}x^2z}{(y\plus{}z\plus{}1)^2}}\plus{}\sqrt{\frac{y^2z\plus{}y^2x}{(z\plus{}x\plus{}1)^2}}\plus{}\sqrt{\frac{z^2x\plus{}z^2y}{(x\plus{}y\plus{}1)^2}}
 \geq \sqrt{2}$

[quote="Axelshmmy"]let $ x,y,z$ positives real numbers such that $ xyz \equal{} 1$
$ \sqrt {\frac {x^2y \plus{} x^2z}{(y \plus{} z \plus{} 1)^2}} \plus{} \sqrt {\frac {y^2z \plus{} y^2x}{(z \plus{} x \plus{} 1)^2}} \plus{} \sqrt {\frac {z^2x \plus{} z^2y}{(x \plus{} y \plus{} 1)^2}} \geq \sqrt {2}$[/quote]

By Cauchy Schwarz we get
\[ \left( {\sum {\frac {{x\sqrt {y \plus{} z} }}{{y \plus{} z \plus{} 1}}} } \right)^2 \left( {\sum {\frac {{x\left( {y \plus{} z \plus{} 1} \right)^2 }}{{y \plus{} z}}} } \right) \ge \left( {x \plus{} y \plus{} z} \right)^3
\]
So it's enough to show that
\[ \left( {x \plus{} y \plus{} z} \right)^3 \ge 2\left( {\sum {\frac {{x\left( {y \plus{} z \plus{} 1} \right)^2 }}{{y \plus{} z}}} } \right)
\]

\[ \Leftrightarrow \left( {x \plus{} y \plus{} z} \right)^3 \ge 4\sum {xy} \plus{} 4\sum x \plus{} 2\sum {\frac {x}{{y \plus{} z}}}
\]

\[ \Leftrightarrow p^3 \ge 4q \plus{} 4p \plus{} \frac {{2\left( {p^3 \minus{} 2pq \plus{} 3} \right)}}{{pq \minus{} 1}}
\]
Setting $ f\left( q \right) \equal{} p^3 \minus{} 4\left( {p \plus{} q} \right) \minus{} \frac {{2\left( {p^3 \minus{} 2pq \plus{} 3} \right)}}{{pq \minus{} 1}}$ then we have
\[ f\left( q \right) \ge \max \left\{ {f\left( 3 \right),f\left( {\frac {{p^2 }}{3}} \right)} \right\}
\]

\[ f\left( 3 \right) \equal{} \frac {{3p^4 \minus{} 3p^3 \minus{} 12p^2 \minus{} 20p \plus{} 6}}{{pq \minus{} 1}}
\]

\[ \equal{} \frac {{\left( {p \minus{} 3} \right)\left( {3p^3 \plus{} 6p^2 \plus{} 6p \minus{} 2} \right)}}{{pq \minus{} 1}} \ge 0
\]

\[ f\left( {\frac {{p^2 }}{3}} \right) \equal{} \frac {{p^3 \left( {\frac {{p^3 }}{3} \minus{} 1} \right) \minus{} 4\left( {p \plus{} \frac {{p^2 }}{3}} \right)\left( {\frac {{p^3 }}{3} \minus{} 1} \right) \minus{} 2\left( {p^3 \minus{} \frac {{2p^3 }}{3} \plus{} 3} \right)}}{{pq \minus{} 1}}
\]

\[ \equal{} \frac {{3p^6 \minus{} 4p^5 \minus{} 12p^4 \minus{} 15p^3 \plus{} 12p^2 \plus{} 36p \minus{} 54}}{{3\left( {pq \minus{} 1} \right)}}
\]

\[ \equal{} \frac {{\left( {p \minus{} 3} \right)\left( {3p^5 \plus{} 5p^4 \plus{} 3p^3 \minus{} 6p^2 \minus{} 6p \plus{} 18} \right)}}{{3\left( {pq \minus{} 1} \right)}} \ge 0
\]

[quote="Axelshmmy"]let $ x,y,z$ positives real numbers such that $ xyz = 1$
$ \sqrt {\frac {x^2y + x^2z}{(y + z + 1)^2}} + \sqrt {\frac {y^2z + y^2x}{(z + x + 1)^2}} + \sqrt {\frac {z^2x + z^2y}{(x + y + 1)^2}} \geq \sqrt {2}$[/quote]

$ \iff$   $ \sum(\frac {a}{\sqrt {b + c} + \frac {1}{\sqrt {b + c}}}\geq \sqrt {2})$

by caushy shwartz inequality :

${ \sum(\frac {a}{\sqrt {b + c} + \frac {1}{\sqrt {b + c}}})(\sum a\sqrt {b + c} + \sum\frac {a}{\sqrt {b + c}})\geq (\sum a})^2$

we have :

$ \sum\frac {a}{\sqrt {b + c}} = \frac {\sum a\sqrt {a + c}\sqrt {a + b}}{\sqrt {\prod (a + b)}}\leq \frac {\frac {1}{2}\sum a(a + b + a + c)}{2\sqrt {2}} = \frac {\sum a^2 + \sum ab}{2\sqrt {2}}$

and :

$ \sum a\sqrt {b + c} \leq (a + b + c)\sqrt {\frac {2(ab + ac + bc)}{a + b + c)}} = \sqrt {2(ab + ac + bc)(a + b + c)}$

so it's inought to prove that :

$ (\sum a)^2 \geq \sqrt {2}(\frac {\sum a^2 + \sum ab}{2\sqrt {2}} + \sqrt {2(ab + ac + bc)(a + b + c)})$

$ \iff$  $ ((\sum a)^2 + \sum ab)^2 \geq 16(ab + ac + bc)(a + b + c)$

by schur ;  $ 16(ab + ac + bc)(a + b + c) \leq 4((a + b + c)^3 + 9)$

and by AM-GM ; $ \sum ab \geq 3$

the inequality became :  $ (x^2 + 3)^2\geq 4(x^3 + 9)$ with $ x = a + b + c \geq 3$

$ \iff$  $ (x - 3)(x^2(x - 1) + 3x + 9)\geq 0$

wich is true  :)


Sum the series
\[ \sum_{k\equal{}0}^{\infty}\frac{(4k)!}{16^k \sqrt{2} (2k)! (2k\plus{}1)!}x^k\]
and for what $ x\in \mathbb{R}$ we can do it?

Let $ a_k$ be the sequence of the coefficients of the function (call it $ f$).
The following holds:
$ a_{k\plus{}1}(4k\plus{}4)(4k\plus{}6)\equal{}a_k(4k\plus{}1)(4k\plus{}3)$ (1)

Some lemmas (the 2nd, 3rd, and 5th follow from the first):
If $ f(x)\equal{}\sum a_kx^k$, then:
(i) $ xf'(x)\equal{}\sum ka_kx^k$
(ii) $ x(xf'(x))'\equal{}xf'(x)\plus{}x^2f''(x)\equal{}\sum k^2a_kx^k$
(iii) $ \frac{f(x)\minus{}a_0}{x}\equal{} \sum a_{k\plus{}1}x^k$
(iv) $ x \left( \frac{f(x)\minus{}a_0}{x} \right) '\equal{} \sum ka_{k\plus{}1}x^k$
(v) $ x\left( x \left( \frac{f(x)\minus{}a_0}{x} \right)' \right)'\equal{} \sum k^2a_{k\plus{}1}x^k$

(1) gives us:
$ 24\sum a_{k\plus{}1}x^k\plus{}40\sum ka_{k\plus{}1}x^k\plus{}16\sum k^2a_{k\plus{}1}x^k\equal{}$
$ 3\sum a_{k}x^k\plus{}16\sum ka_{k}x^k\plus{}16\sum k^2a_{k}x^k$

Assume that $ a_0\equal{}1$ ($ a_0\equal{}\frac{1}{\sqrt{2}}$ but we can multiply the solution by it later), and get:
$ 24\sum a_{k\plus{}1}x^k\equal{}24(\frac{f\minus{}1}{x})$
$ 40\sum ka_{k\plus{}1}x^k \equal{} 40(f'\minus{}\frac{f\minus{}1}{x})$
$ 16\sum k^2a_{k\plus{}1}x^k\equal{}16(xf''\minus{}f'\plus{}\frac{f\minus{}1}{x})$

$ 3\sum a_{k}x^k\equal{}3f$
$ 16\sum ka_{k}x^k \equal{} 16xf'$
$ 16\sum k^2a_{k}x^k\equal{}16(xf'\plus{}x^2f'')$

And we end up with the equation $ 24(\frac{f\minus{}1}{x})\plus{}40(f'\minus{}\frac{f\minus{}1}{x})\plus{}16(xf''\minus{}f'\plus{}\frac{f\minus{}1}{x})\equal{}3f\plus{}16xf'\plus{}16(xf'\plus{}x^2f'')$
or: $ f'(24\minus{}32x)\plus{}f''(16x\minus{}16x^2)\minus{}3f\equal{}0$

Now I am feeling stupid because it hasn't got me anywhere....

OK, it is much simpler -
Let $ c_n$ be the Catalan sequence. Let $ g(x) \equal{} \sum c_n x^n$.
Notice that $ \sqrt {2}f(16x^2) \equal{} \sum c_{2n}x^{2n} \equal{} \frac {g(x) \plus{} g( \minus{} x)}{2}$.
It is well known that $ g(x) \equal{} \frac {1 \minus{} \sqrt {1 \minus{} 4x}}{2x}$, so:
$ f(16x^2) \equal{} \frac {\sqrt {1 \plus{} 4x} \minus{} \sqrt {1 \minus{} 4x}}{4\sqrt {2}x}$
$ f(x^2) \equal{} \frac {\sqrt {1 \plus{} x} \minus{} \sqrt {1 \minus{} x}}{\sqrt {2}x}$
$ f(x) \equal{} \frac {\sqrt {1 \plus{} \sqrt {x}} \minus{} \sqrt {1 \minus{} \sqrt {x}}}{\sqrt {2x}}$ for $ x > 0$.
Not sure how to handle negative $ x$'s.
Anyway, the series converges for $ |x| < 1$ (because it converges absolutely for 0<x<1)

I thought about using the generating function of the Catalan Numbers when I tried to calculate the radius of convergence and I found that it is related to the radius of convergence of that generating function...

Answer is $ \sqrt{\frac{1\minus{}\sqrt{1\minus{}x}}{x}}$

For $ 1 > x > 0$ it's the same as mine  :wink:
What was your method? (and if you don't mind, where do you take these problems from?)

As I mentioned, I am not sure how I am supposed to extract $ f(x)$ for $ x < 0$, knowing $ f(x^2)$... Probably missing something trivial. It might be possible to put $ x \equal{} i\sqrt {t}, t \ge 0$ or something, but then $ \sqrt{1\pm\sqrt{t}i}$ has 2 possible values, no?

What happens when $ |x| \equal{} 1$?

hello, i have found $ \sum_{k\equal{}0}^{\infty}\frac{(4k)!}{16^k \sqrt{2} (2k)! (2k\plus{}1)!}x^k\equal{}\frac{1}{\sqrt{1\plus{}\sqrt{1\minus{}x}}}$.
Sonnhard.


[hide="Solution"]
First, if $4 | n$, then $a^2+b^2 \not \equiv -1 \pmod{n}$, since $a^2+b^2 \not \equiv -1 \pmod{4}$.
Now assume $4 \nmid n$
First we consider the case when n is odd.
By Dirichlet's theorem there exist a prime (infinitely many, in fact) $p \equiv  2n-1 \pmod{4n}$ since $\gcd(2n-1,4n)=1$. Since n is odd $p \equiv 1 \pmod{4}$, so by Fermat's theorem on the sum of two squares p can be represented as $a^2+b^2$, and so $p+1=a^2+b^2+1$. However, $n \mid p+1$, so the claim is proved. If n is even and equivalent to 2 mod 4 take $p \equiv n-1 \pmod{4n}$.
[/hide]

[quote=Peter]Let $n$ be a positive integer. Prove that the following two statements are equivalent. [list][*] $n$ is not divisible by $4$ [*] There exist $a, b \in \mathbb{Z}$ such that $a^{2}+b^{2}+1$ is divisible by $n$. [/list][/quote]
One direction is easy, if $4 |n$ and there exist $a,b \in \mathbb{Z}$ such that $n|a^2+b^2+1$, then $a^2+b^2+1 \equiv 0 \pmod{4}$, which easily gives a contradiction.

For the second direction, assume $4 \nmid n$. We need to show the existence of integers $a,b$ such that $a^2+b^2+1 \equiv 0 \pmod{n}$.
First assume that $n$ is odd. Consider the system ($p$ is a prime)
$$\left\{\begin{array}{l}p \equiv 1 \pmod{4}\\p \equiv -1 \pmod{n}\end{array}\right.$$
Since $(4,n)=1$, hence the system has a solution. By the first equation, we further get, by Fermat's two square theorem, that there exist $a,b \in \mathbb{Z}$ such that $p=a^2+b^2$. Hence, the second equation gives us $n|p+1=a^2+b^2+1$, as desired. Note that this method yields that $p$ is odd, and so $a^2+b^2+1$ is even. Thus, $2n|a^2+b^2+1$ is also true, and since we can do this for all $n$, hence we have also covered the case of even integers as well. This completes the proof. $\square$

For $2\implies 1$, simply assume that $n\mid a^2+b^2+1$ for some $a,b$. If $4\mid n$, we have a contradiction, since $a^2+b^2\equiv -1\pmod{4}$ is not solvable.

For the other direction, we will proceed as follows. Let $p_1,p_2,\dots,p_k$ be prime divisors of $n$, with multiplicities $t_1,\dots,t_k$. If $p_1=2$, then $t_1=1$, and we can simply let $a\equiv 1\pmod{2}$ and $b\equiv 0\pmod{2}$. Hence, it suffices to consider the odd prime divisors.

We now claim that, for every prime $p$, there exists $a,b$ such that, $p\mid a^2+b^2+1$. To prove this, consider the sets
$$
S_1=\{a^2:a=0,1,\dots,p-1\}\quad\text{and}\quad S_2=\{-b^2-1:b=0,1,\dots,p-1\}
$$
in $\mathbb{Z}_p$. Clearly, $|S_1|=|S_2|=\frac{p+1}{2}$, and therefore, since their union has a cardinality at most $p$, we deduce that $|S_1\cap S_2|\geq 1$, finishing the claim.

Now, to conclude the argument, we claim that for every prime $p$, and every $n$, there are numbers $a,b$ such that, $p^n\mid a^2+b^2+1$. We will prove this fact by induction on $n$. Base case is clear. For $n=2$, let $a,b$ be such that $p\mid a^2+b^2+1$. Let $a^2+b^2+1=pi$. Consider $a'=kp+a$, $b'=\ell p +b$ for some $k,\ell$ to be determined later. Notice that,
$$
p^2\mid a'^2+b'^2+1 \iff p\mid 2ka + 2\ell b+i.
$$
Clearly, such a choice of $k,\ell$ is possible. Similar to above, in order to prove for $p^n$, let $a'=kp^{n-1}+\widetilde{a}$, and $b'=\ell p^{n-1}+\widetilde{b}$, with $p^{n-1}\mid \widetilde{a}^2+\widetilde{b}^2+1$. One can easily check that, $k$ and $\ell$ should be prescribed such that, $2k\widetilde{a}+2\ell\widetilde{b}+i$ is divisible by $p$, which can clearly be done.


Is the B harder? Or is it at about the same difficulty.  How are these two tests different?

They are supposed to be the same difficulty. Obviously it is very difficult to make 2 tests the same in difficulty so there will always be variations in scores. However, neither is intended to be more difficult than the other.


$A,B$ two non-empty measurable subset of $\mathbb{R}^{n}$, $0<t<1$ a real. Show that:
$m^{*}(tA+(1-t)B)^{\frac{1}{n}}\geq t m^{*}(A)^{\frac{1}{n}}+(1-t) m^{*}(B)^{\frac{1}{n}}$
where $m^{*}$ is the outer Lebesgue measure. ($A+B$ is not nescessary measurable)

http://www.ac.wwu.edu/~gardner/gorizia12.pdf

see page 10


limit
integration
calculus
real analysis
real analysis solved

Note that $I_n=\int_0^1(1-x^2)^ndx.$ From that, we can show that for large $n$, $I_n$ is approximately $\frac12\sqrt{\frac{\pi}n}.$

The limit should be 1.

Yes.You are right! I have no problems for you,but I will continue to post problems.

kunny

Might as well write out the justification of the last part of what Moubinoul wrote, the asymptotic claim.

$n^{1/p}\int_0^1(1-x^p)^ndx = $

(let $x=\frac{u}{n^{1/p}}, x^p=\frac{u^p}n, dx=\frac{du}{n^{1/p}}.)$

$=\int_0^{n^{1/p}}\left(1-\frac{u^p}n\right)^ndu.$

Take the limit. $\lim_{n\to\infty}\left(1-\frac{u^p}n\right)^n=e^{-u^p}$ so the limit of the integral is

$\int_0^{\infty}e^{-u^p}du.$

Now let $t=u^p,u=t^{1/p},du=\frac1pu^{1/p-1}dt.$ The integral becomes

$\int_0^{\infty}\frac1pt^{1/p-1}e^{-t}dt= \frac1p\Gamma\left(\frac1p\right)$ and we're done.

Of course this could be done more efficiently by doing just one substitution: $x=\left(\frac tn\right)^{\frac1p}.$

As for the first part of what Moubinoul wrote: we need not restrict $p$ to being an integer, and we may relax the condition $p\ge1$ to $p>0,$


I think the fact that $x,y,z$ are positive was left aside :(
Now our inequality is equivalent to the following :
$4(\sum\frac{a^4}{b^4}+\sum\frac{a^3}{b^3})\geq3\frac{(a+b)(b+c)(c+a)}{abc}$. Or when we get it to a better form :
$4(\sum a^8c^4+\sum a^7c^4b)\geq3\sum a^5b^4c^3+3\sum a^5c^4b^3+6a^4b^4c^4$ And this I think is true only from Muirheads because
$(8,4,0)\succ(5,4,3)$ and $\succ(4,4,4)$ and so is
$(7,4,1)\succ(5,3,4)$ and $\succ(4,4,4)$.
And this completes the proof.

Non-expanding proof
\[
\begin{gathered}
  4\left( {\sum {x^4 }  + \sum {x^3 } } \right) \hfill \\
   \geqslant 4\left( {\sum {x^4 }  + 3} \right) \hfill \\
   = 4\left( {\sum {x^4  + 1} } \right) \hfill \\
   \geqslant \frac{1}
{2}\left( {\sum {\left( {x + 1} \right)^4 } } \right) \hfill \\
   \geqslant \frac{3}
{2}\left( {x + 1} \right)\left( {y + 1} \right)(z + 1)\sqrt[3]{{\left( {x + 1} \right)\left( {y + 1} \right)(z + 1)}} \hfill \\
   \geqslant \frac{3}
{2}\left( {x + 1} \right)\left( {y + 1} \right)(z + 1)\left( 2 \right)\sqrt[6]{{xyz}} \hfill \\
   = 3(x + 1)(y + 1)(z + 1) \hfill \\
\end{gathered}
\]

This is why it's important to use proper notation: $\sum_\mathrm{cyc}$ and $\sum_\mathrm{sym}$. NOT JUST $\sum$.

[quote="nickolas"]Well  RaduSorici  your  sums  are not  symmetric , how can  you use  Muirheds?
Thanks[/quote]
I am through with Muirheads :D


Find, with proof, all real numbers $ x \in \lbrack 0, \frac {\pi}{2} \rbrack$, such that $ (2 \minus{} \sin 2x)\sin (x \plus{} \frac {\pi}{4}) \equal{} 1$.

Let $ a\equal{}sin(x\plus{}\frac{\pi}{4})$. If $ x\in [0,\frac{\pi}{2}]$, then $ \frac{\sqrt 2}{2}\le a\le 1$ and $ 2\minus{}sin2x\equal{}3\minus{}2a^2$.
Therefore $ 2a^3\minus{}3a\plus{}1\equal{}0$. One root is $ a\equal{}1$ or $ x\equal{}\frac{\pi}{4}\plus{}2k\pi$. Other roots are $ 2a^2\plus{}2a\minus{}1\equal{}0\to a\equal{}\frac{\minus{}1\pm \sqrt 3}{2}$. But $ \frac{\sqrt 3 \minus{}1}{2}<\frac{\sqrt 2}{2}$.

I simplified the given expression to get $ sin^3x \plus{} cos^3x$ = $ 1/\sqrt 2$

And then let $ l \equal{} sinx \plus{} cos x$
and $ m \equal{} sinx cos x$

Edit: Here is my proof.
We know $ l^2 \minus{} 2m \equal{} 1$
and $ l(1\minus{}m)$ = $ 1/\sqrt 2$
So we get $ l \equal{} 1/\sqrt 2(1\minus{}m)$
So then if write out and expand we get $ 4m^3 \minus{} 6m^2 \plus{} 1$ = $ 0$
And then solved to get $ 4m^3 \minus{} 6m^2 \plus{} 1$ = $ 0$
which has the roots $ 1/2$,$ 1 \minus{} \sqrt 3/2$,$ 1 \plus{} \sqrt 3/2$

But observe that the maximum of m which is $ sinx cos x$ is $ 1/2$

So m cannot equal $ 1 \plus{} \sqrt 3/2$

and since x is in the first quadrant both $ sinx$ and $ cosx$ are positive

So m cannot be $ 1 \minus{} \sqrt 3/2$ which is negative.

So m is $ 1/2$

We then get l is $ \sqrt 2$

Solving we get the only solutions are $ sinx \equal{} 1/ \sqrt 2$
which implies the only solution is $ x \equal{} \pi/4$

$ QED$

Edit: Dang Rust beat me to it. :lol:  :lol:

Hey Rust, how did you get $ 2\minus{}\sin(2x)\equal{}3\minus{}2a^2$??

hello, applying the addition formulas we get
$ \sqrt{2}\sin(x)+\sqrt{2}\cos(x)-\sqrt{2}\sin^2(x)\cos(x)-\sqrt{2}\sin(x)\cos^2(x)-1=0$.
Converting in to $ \tan(\frac{x}{2})$ we get
${ 2\,{\frac {\tan \left( 1/2\,x \right) \sqrt {2}}{1+ \left( \tan \left( 1/2\,x \right)  \right) ^{2}}}+{\frac { \left( 1- \left( \tan \left( 1/2\,x \right)  \right) ^{2} \right) \sqrt {2}}{1+ \left( \tan \left( 1/2\,x \right)  \right) ^{2}}}-4\,{\frac { \left( \tan \left(1/2\,x \right)  \right) ^{2}\sqrt {2} \left( 1- \left( \tan \left( 1/2
\,x \right)  \right) ^{2} \right) }{ \left( 1+ \left( \tan \left( 1/2\,x \right)  \right) ^{2} \right) ^{3}}}-2\,{\frac { \left( 1- \left(\tan \left( 1/2\,x \right)  \right) ^{2} \right) ^{2}\sqrt {2}\tan \left( 1/2\,x \right) }{ \left( 1+ \left( \tan \left( 1/2\,x \right) \right) ^{2} \right) ^{3}}}-1=0}$
Substituting $ \tan(\frac{x}{2})=t$ factoring and simplifying we have
$ (t^4-2t^3+2t^3\sqrt{2}-2t-2t\sqrt{2}-1)(-t-1+\sqrt{2})^2=0$
Only $ t=\sqrt{2}-1$ fulfilles our equation and it gives $ x=\frac{\pi}{4}$.
Sonnhard.


Hi all

 I have the following pattern I need to try and make since of and I really need help soooo bad!

First of all given any period for n I want to be able to find P n+1 (I know the variables x and y)

Pn * x – P0*y ^ n - P0 * y^n+1=Pn+1

Where P0 is the first in the Pn sequence.

Does this make since? There is a definite pattern but I just don't seem able to crack it!

Many thanks in advance!

Ummm... this shouldn't be in Round Table. Could some mod please move it?


AMC
USA(J)MO
USAMO
number theory solved
number theory

Does there exist a multiple of 5^100 which contains no zero in its decimal representation?

USAMO 2003/1 :

For all positive integers n there is a multiple of 5n with precisely n digits, consisting only of odd digits. The result follows.


[hide]everything cancels out except for $(\frac12)(2006)$, so the value is $1003$... i think [/hide]

Your answer is correct you can post your solution

[hide]After simplifying a few parentheses we can see a pattern:(3/2)(4/3)(5/4)...(2005/2004)(2006/2005). The numerator and denomintor will cross multiply to equal one for every two fractions, so we are left with 2006/2=[b]1003[/b].[/hide]

This may be a bit off topic, but with fractions, it is a bit nicer to the eye to use vertically-adjusting parenthesis instead of normal ones. You can do that by typing \left( \frac{x}{y}\right). It should come up like $\left( \frac{x}{y}\right)$. Looks quite a bit nicer than $(\frac{x}{y})$ :P.

So, the problem nice-ified (in a hopeless attempt to appear on-topic :P):

What is the value of the expression:
$\left( 1+\frac{1}{2}\right)*\left( 1+\frac{1}{3}\right)*\left( 1+\frac{1}{4}\right)...\left( 1+\frac{1}{2004}\right)*\left( 1+\frac{1}{2005}\right)$

See? How nice.

-BLt

[quote="riddler"][quote="the broken light"]
So, the problem nice-ified (in a hopeless attempt to appear on-topic ):
[/quote]
Most problems are [i]nice-ified[/i] what are you trying to say?[/quote]

He means it looks better in $\LaTeX$ now.

Oh, don't get me wrong, it was a very nice problem to begin with. I enjoyed solving it. I was simply referring to the $\LaTeX$. :P

-BLt

By telescoping we know that \[ \prod_{k=n_1}^{n_2} \left(\left(1+\frac{1}{k}\right) \equiv \frac{k+1}{k}\right) = \frac{n_2+1}{n_1} \]and that's quite elementary as solution. (Btw, by setting $n_1 = 2$, $n_2=2005$, we get the answer)


A person was born on August 30. When does he celebrate his half-birthday?

doesn't the year matter....leap years.....i guess March 1st

what if its a leap year ,I guess then it would make a diff wouldnt it,

Then April 15th?

that way, we dont need to worry about leap year

April(4th) = 1/2 of August (8th month)
15th = 1/2 30th...

just tried to get around with the leap year..


A chess table $ 10*10$ has every square colored with one of $ 15$ different colours and every colour is use to at least $ 6$ squares. Prove that there exist $ 3$ squares colored with different colours, such that this $ 3$ squares lie on the same row or column.

[i]Suppose it's not true.[/i] We take one column and in that column must be a color, say red, that covers at least 5 squares. So red appears in at least 5 rows. In each of these 5 rows there must be at most one another color and that gives at most 6 colors in 5 rows (50 squares). But then the remaining 9 colors cover other 50 squares, which is impossible because each color covers at least 6 squares.


$m$ men and their wives are standing in a line such that no husband is standing next to his wife.  Find a formula for the number of arrangements possible for $m>3$.

You can do something recursive or with a $\prod$ symbol if you want.

[hide="Hint"]Start with $k$ husbands and their wives, and then think of how you can add another husband and wife.[/hide]

I *might* be on to something using PIE..

[hide]
I first find all possible permutations of m, which is (2m)!. 2m because for every m husband, there is also m wife.

Then I pair one husband and one wife into one unit, so the total number of people is 2m-1. I find the total permutation of that, (2m-1)!, multiply it by 2 (since the paired unit could be either husband-wife or wife-husband), and multiply that by m (since there are m pairs of husband-wife, and we want to count for every of those pair).

Then I pair up two husbands with their corresponding wifes, find the permutation of (2m-2)!, multiply that by 4 (you multiply it by 2 for each pair, and since theres 2 pairs its 2x2=4), and multiply it by m-choose-2 (to count every possible dual-pairs).

Repeat the process until you pair up all m husbands, and add and subtract the result according to PIE.

I would like to express this using summation notation, only I dont know Latex that well so it would get really ugly if I just type it out in plain text..
[/hide]

If its wrong, please dont be too harsh.

[hide="Another hint"] Start with the husbands, and consider how wives can be placed in between husbands. [/hide]

Just as a comment, I realized that my solution at first was wrong; the closed form isn't too bad.  If nobody gets it in 24 hours I'll post my solution.

[hide="Solution"]We proceed recursively.  When $m=2$, there are 8 arrangements.  If we let X and x represent a husband and wife, then we have:
ABab
AbaB
aBAb
abAB
BAba
BabA
bABa
baBA

If we added another husband and wife, then they could be placed in the following positions, where X represents a person and _ a valid position:

_X_X_X_X_

So there are $(2m-1)(2m-2)$ ways to position another husband and wife among $m$ husbands and their wives, so if $a_{m}$ represents the solution for $m$ husbands and wives, then $a_{m+1}=a_{m}(2m-1)(2m-2)$ for $m\ge3$.

So the answer would be:
$8\prod_{i=3}^{m}(2m-1)(2m-2)$
$8((4)(5)(6)(7)\ldots(2m-1))$
$8\cdot\frac{(2m-1)!}{3!}$
$\boxed{\frac{4(2m-1)!}{3}}$[/hide]

And to show that you can't always trust your TI-89 to simplify everything, plugging that into mine gets:
[hide]$\frac{m!\,4^{m}(m-0.5)!}{2m(1.5)!}$[/hide]

[hide="A solution"]
There are $m!$ ways to position the husbands.
The first wife has $m+1$ places she can go; $2$ of which are next to her husband.
The second has $m+2$ places, $2$ next to her husband.
Therefore the number of ways is $m!(m-1)(m)(m-2)(m-3)\cdots(2m-1) = m!\frac{(2m-1)!}{(m-2)!}= \boxed{\frac{m!(2m-1)!}{(m-2)!}}$
[/hide]
Is this correct?


ceiling function
logarithms
number theory unsolved
number theory

I write number $ 2^{2005}$ and $ 5^{2005}$ one before other in their decimal notations, and i form a new number with them. How many digits are in that new number?

The formula?

[hide="Formula"]

A number $ n$ has $ \left\lceil \log{n} \right\rceil$[/hide]

Unless I misinterpreted the question.

what is the demostration of that formula  :D  :?:


inequalities algebra polynomial function logarithms limit integration

In case you didn't notice, these kept getting sent to Intermediate.  You could save the moderators the work and just post them there yourself.

[quote="Spoon"]3.

Directly from Cauchy:

[tex]\displaystyle{(a_1+a_2+\ldots+a_n)\Big(\frac{1}{a_1} + \frac{1}{a_2}  + \ldots + \frac{1}{a_n}\Big) \geq \Big(\frac{a_1}{a_1} + \frac{a_2}{a_2} + \ldots + \frac{a_n}{a_n}\Big)^2}[/tex]

[tex]\displaystyle{(a_1+a_2+\ldots+a_n)\Big(\frac{1}{a_1} + \frac{1}{a_2} + \ldots + \frac{1}{a_n}\Big) \geq \Big(\sum_{i=1}^{n} 1\Big)^2 }[/tex]
[tex]\displaystyle{(a_1+a_2+\ldots+a_n)\Big(\frac{1}{a_1} + \frac{1}{a_2} + \ldots + \frac{1}{a_n}\Big) \geq n^2 }[/tex][/quote]
-edit-
Can anyone find a way without cauchy Swartz?

The proof you are quoting is with Cauchy-Schwarz, known as "Cauchy" for short.

[quote="Scrambled"]Can anyone find a way without cauchy Swartz?[/quote]

It is assumed that all the [tex]a_1,\ldots,a_n[/tex] are positive. Otherwise, the inequality is not true. Applying AM-GM  to [tex]a_1,\ldots,a_n[/tex] and [tex]1/a_1,\ldots,1/a_n[/tex] and then multiplying together.

Here's another way for #3:

[tex]\mbox{For each } \frac{a_i}{a_j} \mbox{ and } \frac{a_j}{a_1} \mbox{ in the distribution of the left side, }[/tex]
[tex] \frac{a_i}{a_j}+\frac{a_j}{a_i} \ge 2 \mbox{ because of the inequality } x^2+y^2 \ge 2xy[/tex]

Thus, for every term [tex]\frac{a_i}{a_j} \le 1[/tex], there exists a corresponding term [tex]\frac{a_j}{a_i}[/tex] s.t. their sum is greater than or equal to 2.  Therefore, the average of all the terms is greater than or equal to 1, and there are a total of n*n = n:^2: terms.

[quote="Scrambled"]1. Prove 1+1/2+1/3+1/4...+1/1,000,000<20
2. Is the sum of 1+1/3+1/6+1/10+1/15... finite?
3. (prove a1+a2 ...+an)(1/a1+1/a2...+1/an) :ge: n^2[/quote]

I can prove (3) using Chebyshev quite easily, but I dont' think that's what you want.

Wolog, let a1 >= ...ai... >=an so that 1/a1 <= ...1/ai... <=1/an

Then (a1+.....+an)/n*(1/a1+.....1/an)/n  :ge: (a1/a1 + .... an/an)/n

But (a1/a1 + .... an/an) = n, so:
(a1+.....+an)/n*(1/a1+.....1/an)/n  :ge: 1, and the result follows immediately.

Alternately:  Use AM-GM.

Then:
(a1 + .... + an) >= n*prod(ai)
(1/a1 + .... + 1/an) >= n*prod(1/ai)

Multiplying these together gives the desired inequality.

Or, directly from AM-HM, the sum divided by n is greater than or equal to n divided by the sum of the reciprocals; clearing denominators gives exactly what you wanted.

This also brings up the related problem:

Prove b1/a1 + b2/a2 +.....+ bn/an  :ge: n

where {bi | i=0,.....n} is some permutation of {a1, ..... ,an}

[quote="Hexaditidom"]How would you do #1? I don't think theres a formula for the harmonic sequence is there?

Its not aritmetic or geometric, and i tried finding a polynomial by writing out rows and subtracting every two terms, etc. That didnt work either.

Another one I've been wondering about is factorials. Does that have a formula? If so, can you use it to find the answer to:

as n -> :inf: , what is (1+2+3+4+5+6...)/(1*2*3*4*5*6...)?[/quote]

That limit is 0, because it is

lim n->  :inf: n(n+1)/2/n! = (n+1)/[2*(n-1)!] = 1/(2(n-2)!) = 1/ :inf: = 0

For the first question, see if you can find a transcendental function which is an upper bound for the sum of the harmonic series.  Then calculate that upper bound.

#1...

[tex]\displaystyle{1+\frac{1}{2}+\frac{1}{3}+\ldots +\frac{1}{n} < \ln n + 1}[/tex]

so we just need to show that [tex]\ln(1,000,000) < 19[/tex]...

[tex]\ln(10^6) < 19[/tex]
[tex]\displaystyle{\ln(10) < \frac{19}{6}}[/tex]
[tex]\displaystyle{\ln(10) < \ln(e^3) < \frac{19}{6}}[/tex] QED

[quote="Spoon"][tex]\displaystyle{1+\frac{1}{2}+\frac{1}{3}+\ldots +\frac{1}{n} < \ln n + 1}[/tex][/quote]

Why?

are you saying I need to prove that, or are you really asking?

Well, since you're asking, I'm not sure off the top of my head how to prove it.  I'm sure it's true and I imagine that I could prove it if I tried, but my point was just that it would need justification.

All right, I'll go off the assumption that

[tex]\displaystyle{\lim_{n\rightarrow \infty} \left( \sum_{k=1}^n \frac{1}{k} - \int_1^n \frac{1}{x}\; dx\right) < 1}[/tex], and this holds for all n > 2.

Hmmm... I just realized that is the same as just stating what I said before...

Ok.. how about this, the right side approximation of [tex]\displaystyle{\int_1^\infty \frac{1}{x}\; dx}[/tex] will be less than the exact value, because it's monotonically deacreasing... therefore the right side sum with width of 1 would be [tex]\displaystyle{\sum_{k=2}^n \frac{1}{k}}[/tex]... adding 1 to both sides of this inequality would prove the original conjecture.

[quote="scm007"]Yeah you can just say that the harmonic series can be bounded by ln x

integral of 1/n = ln x

But since it starts at 1, we need to add 1.[/quote]

That doesn't prove anything though...


geometry
circumcircle
incenter
trigonometry
geometry proposed

Let $ ABC$ be a triangle with $ A\equal{}60^{\circ}$ and the circumcircle $ w$ . Consider a point $ M$ which belongs to $ w$ so

that the line $ BC$ separates $ A$ , $ M$ . Prove that $ MA\equal{}MB\plus{}MC\ \Longleftrightarrow$ $ B\equal{}C$ or $ m(\widehat {MAB})\equal{}\frac B2$ .

Yes, it's easy,  just lengthen BM and discuss .

[b]Remark：[/b]
$ M$ is a Feuerbach point .
$ I$ is the incenter, $ N$ is the midpoint of $ BC$, ray $ IN$ meet the circumcircle at $ K$, then $ KA \equal{} KB \plus{} KC$ .

[u]It is nicely[/u], [b]Lym[/b] ! Please, maybe you or another user can present a simple proof of this easy "slicing" problem ...

Lengthen $ BM$ to $ D$, let $ DM\equal{}CM$, then △$ CDM$ is an equilateral triangle, and let $ BE\equal{}AC$, $ E$ is on $ BC$ .
Then we found △$ ACM$≌△$ BED$, $ DE\equal{}CM\equal{}DC$, So there are two kind of possibility：
$ (1): C\equal{}E$. for this situation, we get $ AC\equal{}BC$, △$ ABC$ is an equilateral triangle .
$ (2): D$ is the circumcenter of △$ CEM$. for this situation, we get ∠$ BCM\equal{}1/2$∠$ BDE$, scilicet ∠$ BAM\equal{}1/2$∠$ ABC$ .
Done！

[quote="Virgil Nicula"]Let $ ABC$ be a triangle with $ A \equal{} 60^{\circ}$ and the circumcircle $ w$ . Consider a point $ M$ which belongs to $ w$ so

that the line $ BC$ separates $ A$ , $ M$ . Prove that $ MA \equal{} MB \plus{} MC\ \Longleftrightarrow$ $ B \equal{} C$ or $ m(\widehat {MAB}) \equal{} \frac B2$ .[/quote]

[b][u]Proof[/u].[/b] $ m(\widehat{MAB})\equal{}x\implies\left\|\begin{array}{c}
m(\widehat{MCA})\equal{}C\plus{}x\\\\
m(\widehat{MCB})\equal{}x\\\\
m(\widehat{MBC})\equal{}60^{\circ}\minus{}x\end{array}\right\|$ . Therefore, $ MA\equal{}MB\plus{}MC\Longleftrightarrow$ $ \sin \widehat {MCA}\equal{}\sin\widehat {MCB}\plus{}\sin\widehat {MBC}\Longleftrightarrow$

$ \sin (C\plus{}x)\equal{}\sin x\plus{}\sin\left(60^{\circ}\minus{}x\right)\Longleftrightarrow$ $ \sin \left(B\plus{}60^{\circ}\minus{}x\right)\equal{}2\sin 30^{\circ}\cos\left(30^{\circ}\minus{}x\right)\Longleftrightarrow$ $ \cos\left(30^{\circ}\plus{}x\minus{}B\right)\equal{}\cos\left(30^{\circ}\minus{}x\right)\Longleftrightarrow$

$ \left\|\begin{array}{ccc}
30^{\circ}\plus{}x\minus{}B\equal{}30^{\circ}\minus{}x & \Longleftrightarrow & x\equal{}\frac B2\\\\
\left(30^{\circ}\plus{}x\minus{}B\right)\plus{}\left(30^{\circ}\minus{}x\right)\equal{}0 & \Longleftrightarrow & B\equal{}60^{\circ}\end{array}\right\|$ $ \Longleftrightarrow$ $ B \equal{} C\ \ \vee\ \ m(\widehat {MAB}) \equal{} \frac B2$ . Generally, $ \boxed {\ \tan x\equal{}\frac {\sin A\minus{}\sin C}{\cos A\plus{}\cos C\minus{}1}\ }$ .

I would love to know and I solve this problem


K=x^3+y^3   (x and y are integer)

Which one is not equal to K?  Why?

A.)238325    B.)239148   C.)288827   D.)300888

Note that cubes are $ 0, \pm1 \pmod{7}$. So, we know that $ x^3\plus{}y^3\equiv 0, \pm1, \pm2 \pmod{7}$. I believe (A) doesn't fit that (providing I don't suck at dividing), so (A).

[quote="not_trig"]Note that cubes are $ 0, \pm1 \pmod{7}$. So, we know that $ x^3 \plus{} y^3\equiv 0, \pm1, \pm2 \pmod{7}$. I believe (A) doesn't fit that (providing I don't suck at dividing), so (A).[/quote]
Use mod 9, it's much easier to calculate and has the same residues (0,1,-1) ;)


logarithms
function
trigonometry
Euler
geometry
ARML
modular arithmetic

$ \log_{\minus{}1}i \equal{} \frac{1}{2}$. But what is $ \log i$? Please enlighten me  :blush:... i have no idea what level this is so please bear with me.

Perhaps knowledge of Euler's Formula will help.

[url]http://en.wikipedia.org/wiki/Euler%27s_formula[/url]

[quote="jsc"]$ \log_{ \minus{} 1}i \equal{} \frac {1}{2}$. But what is $ \log i$? Please enlighten me  :blush:... i have no idea what level this is so please bear with me.[/quote]
First, as it is typically defined, a logarithm function is only defined for the positive real numbers, i.e. an expression like $ \log x\quad\mbox{\text{for }}x\leq 0$ makes no sense. Second, and this is just for purposes of clarification, it is conventionally understood that the expression $ \log(\cdot)$ where no base is specified means $ \log(\cdot) \equal{}\log_{e}(\cdot) \equal{}\ln(\cdot)$. Third, it is possible to extend the definition of a logarithm to the complex numbers. The reason for this is that it is that one may express a complex number, $ x \plus{} i y$, in polar form using the following parametric equations:
\[ x \equal{} r\cos t\quad y \equal{} r\sin t\]

Therefore, $ x \plus{} i y \equal{} r\cos t \plus{} i r\sin t$.

When expressed this way, $ r$ is a positive real number, since $ r \equal{} \sqrt{x^2 \plus{} y^2}$.

Leonhard Euler (1707- 1783) discovered by writing the series expansions of $ \sin$ and $ \cos$ that the complex number $ \cos\theta \plus{} i \sin\theta \equal{} e^{i\theta}$. Given Euler's formula it is possible to write a complex number $ x \plus{}i y$ as $ x \plus{} i y \equal{} r e^{i t}$.

If a complex number is expressed using Euler's formula, one can then find it's logarithm; in as much as, $ \log(r e^{i t}) \equal{} \log r \plus{} it \log e \equal{}\log r \plus{} i t$.

Having said all of that, given that we want to find, $ \log_{\minus{}1} i$ we may write,
\[ i \equal{} (\minus{}1)^{\log_{\minus{}1} i} \\ \\ \Rightarrow \log i \equal{} \log_{\minus{}1} i \cdot \log(\minus{}1) \\ \\ \Rightarrow \log_{\minus{}1} i \equal{}\frac{\log i}{\log(\minus{}1)}\]

Now the question becomes what is $ \log i$ ? and what is $ \log(\minus{}1)$ ?

Expressing $ i$ and $ \minus{}1$ using Euler's formula:
\[ i \equal{} 1\cdot e^{i\frac{\pi}{2}}\quad \minus{}1 \equal{} 1\cdot e^{i\pi} \\ \\ \log i \equal{} \log 1 \plus{} i\frac{\pi}{2}\quad \log(\minus{}1) \equal{} \log 1 \plus{}i\pi \\ \\ \Rightarrow \log i \equal{} i\frac{\pi}{2} \quad \log(\minus{}1) \equal{} i \pi\]

\[ \boxed{\therefore \log_{\minus{}1} i \equal{} \frac{i{\pi\over 2}}{i\pi} \equal{} {1\over 2}}\]

[quote="Dr. No"]Second, and this is just for purposes of clarification, it is conventionally understood that the expression $ \log(\cdot)$ where no base is specified means $ \log(\cdot) \equal{} \log_{e}(\cdot) \equal{} \ln(\cdot)$.[/quote]

Um..... I don't think so. I think it means $ \log_{10}{\cdot}$

My solution is a bit dodgy, I'm sure there's a mistake somewhere:

let $ y \equal{} \log i$. Then,

$ 10^y \equal{} i \equal{} e ^{\ln 10 y}$

Remember that $ e ^{i \frac {\pi}{2}} \equal{} i$, so we have

$ \ln 10 y \equal{} \frac {i \pi}{2}$,

$ y \equal{} \frac {i \pi}{2\ln 10}$

However, this creates a string of solutions...

DOH

the i's aren't coming up nicely  :o

Umm... you might just not want to use backslash-i. Just plain i would be fine.

This:
$ \i$

looks more like a 1 or an l.

This:
$ i$
looks like $ i$.

[quote="iniquitus"][quote="Dr. No"]Second, and this is just for purposes of clarification, it is conventionally understood that the expression $ \log(\cdot)$ where no base is specified means $ \log(\cdot) \equal{} \log_{e}(\cdot) \equal{} \ln(\cdot)$.[/quote]

Um..... I don't think so. I think it means $ \log_{10}{\cdot}$
[/quote]

It depends on the context. Usually it is $ e$ but it is often 10 as well.

In order to understand the complex logarithm you need to understand complex exponentiation.  Since $ e^{2\pi i n} \equal{} 1$ for all integers $ n$, it follows that the complex logarithm is infinitely-valued as $ e^z \equal{} w$ if and only if $ e^{z \plus{} 2\pi i n} \equal{} w$ for all integers $ n$.

In mathematics above the high school level, $ \log$ has a contextual meaning -- in most areas of mathematics (and in particular in complex analysis, the field to which this discussion most properly belongs) it means $ \log_e$, i.e. $ \ln$, although computer scientists typically use $ \log \equal{} \log_2$.

There have been topics on this subject in the past -- try a search.

[quote="timwu"]Is this correct?[/quote]  Yes.  Here are some links to related discussions:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=612265332&t=172149
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=612265332&t=149758
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=612265332&t=39969

(In the first link, you'll have to ignore the irrelevant bits.)

[quote="iniquitus"][quote="timwu"]So, here, it seems $ a^{i}$ has multiple (in fact, infinitely many) values depending on the choice of $ k$.[/quote]

Ok so how many values does $ (e^{\pi})^{i}$ have?[/quote]
It is not generally true that $ (a^b)^c \equal{} a^{bc}$.  The fundamental issue here is that the [url=http://en.wikipedia.org/wiki/Complex_logarithm#Log.28z.29_as_a_multi-valued_function]complex logarithm[/url] is, in fact, multivalued (which screws with the proof of this property in general).

[quote="Yongyi781"]Umm... $ e^{\frac {\pi}2}$ does NOT equal $ i$![/quote]

\[ e^{i \frac{\pi}{2}} \equal{} \cos \frac{\pi}{2} \plus{} i\sin \frac{\pi}{2} \equal{} 0 \plus{} i 1 \equal{} i\]

[quote="Poincare"]I am really ignorant in logs so can someone explain to me what "e" is ?

thanks,
Dhaivat Pandya(uh huh, you read right that's my name :D )[/quote]

$ e \equal{} 2.71828182846\ldots$ is a transcendental number, like $ \pi$; it is the base of the natural logarithm $ \ln$, which is itself the essential form of all logarithmic functions -- this is why it is called the natural logarithm. The number $ e$ is also the value of the exponential function, $ \exp(x)$ at $ x \equal{} 1$. The exponential function occurs frequently in calculus, because it is the function, the only function, that is equal to its own rate of change.
\[ \exp(x) \equal{}\sum_{n \equal{} 0}^{\infty}\frac{x^n}{n!}\\ \\ \exp(a) \equal{} e^a\]

[quote="Dr. No"][quote="Yongyi781"]Umm... $ e^{\frac {\pi}2}$ does NOT equal $ i$![/quote]
\[ e^{i \frac {\pi}{2}} \equal{} \cos \frac {\pi}{2} \plus{} i\sin \frac {\pi}{2} \equal{} 0 \plus{} i 1 \equal{} i
\]
[/quote]

Note that I wrote $ e^{\frac {\pi}2} \neq i$, because he forgot to add an $ i$ in the exponent.

[quote="iniquitus"]Oh ok sorry Dr No

I think I got confused cuz the "log" button on the TI84
means "log10"...

And thx for the explanation t0rajir0u[/quote]
No problem. Writing $ \log$ for $ \log_{10}$ is quite common, especially on calculator buttons, because other than $ \ln$, one is apt to use $ \log_{10}$ most often (particularly in engineering).

I once had a TI-92, it has been misplaced in multiple moves now, it had no $ \log_{10}$ key, only an $ \ln$ key. In order to use it to calculate "common logarithms," aka $ \log_{10}$, one had to know that $ \log_{10}(x) \equal{}\frac{\ln x}{\ln 10}$.

[quote="timwu"]I have a related question about complex exponentiation (I demonstrated the following to several people at ARML, including [b]iniquitus[/b] but no one agreed with me).

Question: What's $ a^{i}$, where $ a\in \mathbb{R}$?

My Solution: First rewrite $ a = |a|\cdot e^{i\theta}$, where $ \theta \equiv 0, \pi \pmod {2\pi}$. Then,
\begin{align*} a^{i} & = |a|^{i}\cdot e^{ - \theta} \\
& = e^{i\ln a}\cdot e^{ - \theta} \\
& = e^{ - \theta} \cdot \text{cis}(\ln a) \end{align*}
If $ a > 0$,
$ a^{i} = e^{2k\pi}\cdot \text{cis}(\ln a)$
If $ a < 0$,
$ a^{i} = e^{(2k + 1)\pi}\cdot \text{cis}(\ln a)$
where $ k\in \mathbb{Z}$.

So, here, it seems $ a^{i}$ has multiple (in fact, infinitely many) values depending on the choice of $ k$. Its plot in complex plane are discrete points on ray $ \arg(z) = \ln a$. Is this correct? If not, what's wrong with it?[/quote]

First, for any complex number $ z$, $ \arg(z)$ is usually restricted such that $ |\arg(z)|\leq \pi$ [i]a priori[/i] in order to avoid the kind multi-valued result you showed.

\[ a^i = (e^{\ln a})^i = e^{-\arg(a) + i\ln |a|} =e^{-\arg(a)}\left(\cos(\ln |a|) + i \sin(\ln |a|)\right)\]

Since $ a\in \mathbb{R}$,
\[ a^i = \begin{cases}  e^{-\pi}\{\cos( \ln |a|) +i\sin (\ln |a|)\}\quad\mbox{for}\, a<0 \\ 0 \quad\mbox{for}\, a = 0 \\ \cos(\ln |a|) + i\sin(\ln |a|)\quad \mbox{for}\, a > 0\end{cases}\]


Hello..I was wondering how I can make latex write the simbol "arc" over a certain group of letters. Fie example , I need to write $\widehat{AB}$, but that thing over $AB$ should be round. Does anyone know how to do this?

You have the choice between \stackrel{\frown}{BC} which gives $\stackrel{\frown}{BC}$ or you might like to install the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=arcs]arcs package[/url] so you can use \overarc{BC} which gives

Why doesn't the \overarc command work on the forums?

Because the arcs package isn't installed.


I noticed that the post numbers in threads came back recently (after about 2 years of absence), which is great. Unfortunately, they do not show in the Topic Review window. Is it possible to have them there as well? (because they are most useful when you are posting a reply and want to refer to one of the previous posts).

[quote="Valentin Vornicu"]Don't use post numbers to refer to previous post. It's merely an accommodation for people wanting to help posting problems in the resources section (faster post id link).[/quote]I just wrote a greasemonkey script that displays the post id in the bottom right hand corner (in search results too).

Would it be possible for posts to have a permanent ID (I mean that's actually displayed along with the post) so that people can do what fedja is suggesting? It really would help if you want to respond to something that's not the post above you, but don't want to vastly inflate your message size by quoting something lengthy.

That's why those are there. You click on the #5 and you get the post link, which is the link you should use. Do not use the number as posts can be arranged in different ways depending on the users prefs (author, post time, title, etc.) and some might dissapear in time (recycled, splitted, etc.)

I have no idea what greasemonkey is.

Greasemonkey lets you modify pages locally through a scripting language - so that what is displayed is not exactly what was served to the user.  It's a Firefox extension.  Not sure if it's available for any other browser.

OK but people are lazy! If I want to refer to worthawholebean's post, I want to say "what worthawholebean was talking about in post #3 was..." rather than having to go get the link and use url tags and everything to say "what worthawholebean was saying [url=http://www.artofproblemsolving.com/Forum/posting.php?p=1087750&mode=topicreview&postorder=DESC#1087750]above[/url] was...". It's also silly to use a link when the post you're referring to is on the same page.

edit: I should clarify: my suggestion is that the post number (e.g. 1087550 for worthawholebean's) or some other number which is smaller but still unique for each post (there aren't really 1.1 million posts, are there?!) be displayed in the post so that you can refer to it without ambiguity.

First of all the link you have in the "above" link, above, is wrong. It points to a bad page.
Secondly, it is not wrong.

Furthermore, you, as I said, are not anticipating all the cases (ie posts split INTO topic, other order of posts, etc.) which makes the numbering futile. Think about the numbering as a visual effect as opposed to just having a # there.

[quote="Valentin Vornicu"]First of all the link you have in the "above" link, above, is wrong. It points to a bad page.
Secondly, it is not wrong. [/quote]
I copied the link that was presented on the Topic Review page for that post. Probably, that little page should give the link to that post in that thread, rather than some weird thing (or at least both should be available). I think this is more what fedja was asking about, actually.

[quote="Valentin Vornicu"]Furthermore, you, as I said, are not anticipating all the cases (ie posts split INTO topic, other order of posts, etc.) which makes the numbering futile. Think about the numbering as a visual effect as opposed to just having a # there.[/quote]
I still don't understand the problem. I'm not requiring that the permanent ID numbers be in increasing order. If a post is moved, it ought to retain the same ID. (If this is not the case for the "post ID" that goes with posts now, then you could just have it added as text to the bottom of all messages. Then it would be retained on moves, and it would be searchable, so even if a post is split out of a thread, references to it in that thread will still be usable.)

I do not understand your problem. You have the post id (together with the link) by just clicking on the #. The format is .php?p=ID#ID just get the ID if you want to do that, but I strongly recommend you post a link because as I said posts can be moved, on a different page (users may have different pagination displaying 5 posts on a page for example, so while you say in your post "the above post ID 323221232" for some user that might be on a different page, below, or even in a different topic at one point).

Giving the full link (permalink btw) to the post is very important.

OK, I understand. But it should still be possible to obtain this link from the Topic Review window, so that if you are writing a reply, you don't have to open a new tab/window in order to find that link. As it is, you can only obtain the weird link which I mistakenly used in my post above; if I am going to have to include a link, I don't want to also be expected to mouse over that link to see the post ID number, and then type out on my own http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1087811#1087811 in order to get a link to your post. :P

Ok, I will add that feature. Remember the numbering will again be faulty (as the last post will be #1 and etc.)


email percent ooooooooooooold articles ooooooooooooooold ooooooold Old

This new thread is created for discussions post results.
Let us use it 8 AM 25th May on-ward, mainly for exchange of information about ranking and counseling.
All dear users of 'Post IIT-JEE 2009 Discussion!' thread are requested to migrate tho this thread.

Looks like results will be available http://www.jee.iitb.ac.in/result.htm.

This is the first URL which is working today and looks fast as well. There is database connect error now. IITB people will connect it to database at 8 am.

For JEE 2009 results, IITB site is migrated to nginx (pronounced as "engine X") which is a lightweight, high performance web server. Check Wiki for details. IITB site should be really fast. Rest of the sites are mostly on Apache.

The only thing I care about right now is the results.. :!:

I think there is some movement in bombay site check it out wagawaga

What are the authorities doing? Cant they finish all the tasks the prev day and just click a button at 8 am the next day. zzzz

Has anyone seen their results yet...
45 mins late already and I can't get mine...I thought someone said that the results were out 20 mins earlier last year..

All the Best..

Actually last year also it came late, like 30 mins.

At Chennai, apart from the toppers mentioned in the press note...

V Prashant (490/500 in CBSE Board as well)   AIR 5
A V Gunaa (avg on AoPS)                           AIR 16

Results have been stuck up on the notice boards at IIT Madras
Dunno abt other centres

Check this out..

http://www.iitg.ernet.in/jee/result.htm

my rank is not worth telling..729 :o  :(

classical singers come to iit  :huh:
i am a casual singer. not really besuraa, but never learned anything. can i join it too  :rotfl:

no really not joking. my mother has learned singing (classical) and sings good. so does my sis (i rarely praise her rotflmao) :)
i feel had i taken interest i cud hv done much better  :oops:
do they [b]teach[/b] singing (classical) at music club of iit-m  :maybe:

I have tried so much,but in vain
Can anyone sing figure/faint like bennnington? :rotfl:

@VP :
That really scares me...Please give some examples af rare sports that nobody else will think of applying and I may have a chance at NSO. :D

@Scary: Chester rarely sings and faint is all about the screaming not singing. :)
But I still like them..

Yah,whatever
I meant screaming,but all that comes out of me is a roar(seldom)/screech(usal)/squeak(often) :(
And my rap sucks,so becoming MS is impossible for me.

@scary
u cud try getting into cob or slipknot as the lead singer..if u can screech and scream..
kick as*..man u have a bright future unlike some of us :)
all the best

Now I get it!
You guys want me to improve my slang voccabulary like this? :mad:
Everytime I visit AoPS,I have to visit dictionary.com as well. :mad: Plz dont take advantage of a tubelight like me :furious:
[hide]never knew screeching was in demand as well :P
guess math olympiad wont be as profitable as getting into any rockband :rotfl: [/hide]

Hey guys see this-
http://jee.iitd.ac.in/jee2009.htm

Leaving for madras tonight. will join hostel tomorrow. gl hf..meet me in my room. we are ON man. :DD:D

hello guys.
hows life in iitm
life here is excellent

Enjoying hostel life... I met Conjurer here as well..
Waiting for the selections of NSO..

Great campus :coolspeak:

I met preetham at SAC
MM and conjurer pls meet me at gurunath sometime....i cant come to your hostels.....i am too scared to do that  :P

Howz life at IITM?Are you enjoying?
[hide]I meant is there ragging? :rotfl: [/hide]

Anybody with trouble [hide="sitting"]Coke bottles???  :rotfl: [/hide]?

Hi!
Do you have to study extremely high maths for IIT??

iit jee maths is an advanced version of 11th and 12th cbse maths.


geometry
geometric transformation
homothety
geometry unsolved

Dear Mathlinkers,
let ABC a triangle,
CB'B"A, AC'C"B two externally squares erected on ABC,
J, K the centers of CB'B"A, AC'C"B
Y, Z the midpoints of CA, AB
and P the midpoint of B"C'
Prove that the lines AP, JZ and KY are concurrent.
Sincerely
Jean-Louis

Dear Lym,
please, can you explain your answer?
Sincerely
Jean-Louis

I mean your problem can be solved with [b]M.vuibuit theorem[/b].

Dear Lym,
I don't know what is the "Vuibuit theorem". Tell me...
Sincerely
Jean-Louis

Dear jamey，see this picture：
[b]M.vuibuit theorem[/b] is $ AD$⊥$ BC$ .

Dear Lym,
now I understand... for me it is a Vecten's result perhaps Vuibuit in your cultur...
Your idea rejoins the one of Luis...
Sincerely
Jean-Louis

Dear Mathlinkers,
an article concerning the Vecten's figure and its developpement can be seeing on my site
http://perso.orange.fr/jl.ayme , la figure de Vecten  vol. 5 p. 51
Sincerely
Jean-Louis

Lemma: $PA \perp BC$.

Proof: Let the perpendiculars from $A, B'', C''$ to $BC$ be $H, B_1, C_1$ respectively. Since $CA=AB''$, $AH=HB_1$ and similarly $AH=HC_1$. Hence $PH \parallel B''B_1 \perp BC$ but $AH\perp BC$, thus $PAH$ is a straight line.

Let $B''B$ meet $C''C$ at $Q$ and $KY$ meet $JZ$ at $X$. By homethety, $AXQ$ is also a straight line and $AX=XQ$. Hence it suffices to show $AQ\perp BC$.

Let $AH$ meet $BB'$ and $CC'$ at $Q_1, Q_2$ respectively. Let perpendiculars from $B', C'$ to $BC$ at $B_2, C_2$ respectively. Then $BC_2=AH=CB_2$ so $CC_2=BB_2$. $HQ_1=B'B_2*\frac{BH}{BB_2}=\frac{BH\cdot HC}{BB_2}=\frac{CH\cdot HB}{CC_2}=HQ_2$. Therefore $Q_1\equiv Q_2\equiv Q$. QED


inequalities
LaTeX
geometry proposed
geometry

Let $C_1\left(O_1,\ r_1\right)$, $C_2\left(O_2,\ r_2\right)$, $C_3\left(O_3,\ r_3\right)$ be three pairwisely non-intersecting circles, such that $O_1$, $O_2$, $O_3$ are not collinear. Prove that

$\frac{\sum r_1(r_1r_2^2+r_1r_3^2+r_2^2r_3+r_3^2r_2-r_1r_2r_3) O_2O_3\ ^2}{\left(\sum r_1r_2\right)^2}\geq\frac{\left(\sum O_1O_2\right)^2}{9}$.

Matei Vlad (Own)

[color=red][Moderator edit: You don't need to post this as an attachment, you can use LaTeX on the forum.][/color]

[quote="mateivld"]Sorry,I made a mistake!In the right part the coefficient is 1/27r1r2r3[/quote]

Thanks, I just corrected this in your post. However, the inequality still cannot be true: Let $r_1$ tend to 0; then, the right hand side will go to $\infty$ as well, while the left hand side will remain bounded... Or did I misunderstand your "external" condition?

  darij


factorial
calculus
integration
function
floor function

integrate this

 $\int^{8}_{0}\frac{8!}{n!(8-n)!}dn$

xplain how you do it plz

This should be deleted!
You may not post calculus in this forum!

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=140454[/url]

While it's up though, do you want us to use the Gamma function (interpolated factorial) or some form of a discrete factorial (like $\lfloor n \rfloor !$)?

If you want the continuous definition, then you can approximate it with a trapezoidal sum and get that it is $\approx \sum-1/2 = 2^{8}-1/2 = 255.5$, which is barely over the actual answer, $255.429986...$. I don't know how to evaluate the exact answer.

If you want a discrete definition of some sort, then just find the values for the integers from $0$ to $8$ and add...


Let $ G$ be the group of order $ p^{s}$, where $ p$ is a prime number and $ s$ is a natural number. Let $ H$ be the proper subgroup of $ G$. Show, that $ H$ is the proper subgroup of its normalizer.

[hide]
Here is my solution, but I am not sure its correct.

We have $ |H| \equal{} p^{l}, 0\leq l < s$, $ N(H) \equal{} \left\{x\in G| xH \equal{} Hx \right\}$. We know, that $ H\leq N(H)\leq G$. All we have to show is $ H\neq N(H)$.

Now if there exists $ x\in Z(G)$ and $ x\notin H$, then $ x\in N(H)$ and $ H\neq N(H)$.

Otherwise $ Z(G)\leq H$, where $ |Z(G)| \equal{} p^{i}, 0 < i\leq l$. Then $ G/Z(G) \equal{} \left\{Z(G),a_1 Z(G), a_2 Z(G), \ldots, a_{p^{l \minus{} i} \minus{} 1}Z(G),\ldots,a_{p^{s \minus{} i} \minus{} 1}Z(G) \right\}$, $ H/Z(G) \equal{} \left\{Z(G),a_1 Z(G), a_2 Z(G), \ldots, a_{p^{l \minus{} i} \minus{} 1}Z(G)\right\}.$

So we have to find $ x\notin H$, which commutes with $ a_i$, $ \forall i\in\left\{1,2,\ldots, p^{l \minus{} i} \minus{} 1 \right\}$.

According to Sylow theorems there exists $ K \leq G/Z(G)$, such that
\[ H/Z(G)\triangleleft K \equal{} \left\{Z(G),a_1 Z(G), a_2 Z(G), \ldots, a_{p^{l \minus{} i} \minus{} 1}Z(G), b_{p^{l \minus{} i}}Z(G), \ldots, b_{p^{l \minus{} i \plus{} 1} \minus{} 1}Z(G)\right\}\]
Then $ (b_{j}Z(G)) (a_i Z(G)) \equal{} (a_i Z(G))(b_j Z(G))$, $ \forall i$. Hence also $ b_{j}a_i \equal{} a_i b_j$. So $ b_j \in N(H),$ but $ b_j \notin H.$[/hide]

A more general theorem says that $ H \subsetneq N_G(H)$ whenever $ H$ is a proper subgroup of a nilpotent group $ G$.
Of course $ p$-groups are an example of nilpotent groups.

[b]Lemma:[/b] If $ H$ is a $ p$-subgroup of the finite group $ G$, then $ [N(H): H]\equiv[G: H] (\mod{p})$

[i][b]Proof:[/b][/i] We define $ S=G/H$ and introduce the action of $ H$ on $ S$ as:
\[ h(gH)=(hg)H\]
Obviously $ 1(gH)=gH$ and $ a(b(gH))=(ab)(gH)$.
Then we have:
\begin{eqnarray*}
S_0&=&\left\{s\in S| hs=s, \forall h\in H \right\}=\\
&=&\left\{gH\in S| h(gH)=gH, \forall h\in H \right\}=\\
&=&\left\{gH\in S| (hg)H=gH, \forall h\in H \right\}=\\
&=&\left\{gH\in S| (g^{-1}hg)H=H, \forall h\in H \right\}=\\
&=&\left\{gH\in S| (g^{-1}hg)\in H, \forall h\in H \right\}=\\
&=^{1}&\left\{gH\in S| (g^{-1}Hg)=H \right\}=\\
&=&\left\{gH\in S| g\in N(H) \right\}=\\
&=& N(H)/H
\end{eqnarray*}

$ (1):$ We have used finiteness of $ H$.

But we know, that $ [G: H]=|G/H|= |S|\equiv |S_0| = |N(H)/H|=[N(H): H] (\mod{p})$, since $ H$ is $ p$-group.

[b]For the problem:[/b]

Since $ [G: H]=p^{k}$, where $ k>0 \Rightarrow [N: H]\equiv 0 (\mod{p})\Rightarrow [N(H): H]\neq 1 \Rightarrow H\neq N(H).$


function
induction
triangle inequality
IMO
IMO 2009
IMO Shortlist

Determine all functions $ f$ from the set of positive integers to the set of positive integers such that, for all positive integers $ a$ and $ b$, there exists a non-degenerate triangle with sides of lengths
\[ a, f(b) \text{ and } f(b \plus{} f(a) \minus{} 1).\]
(A triangle is non-degenerate if its vertices are not collinear.)

[i]Proposed by Bruno Le Floch, France[/i]

Remark: If a triangle with have sidelengths 1,a,b with a,b positive integers, then it's forced by triangle inequality that a=b.

1. Put a = 1, then remark tells us that $ f(b) \equal{} f(b \plus{} f(1) \minus{} 1)$
2. [b]Claim[/b] $ f(1) \equal{} 1$.
Otherwise $ f(1) \minus{} 1 > 0$, which means that $ f$ repeats itself every $ f(1) \minus{} 1$ numbers. This means that $ f$ can take finitely many values, so if we take $ a$ sufficiently large, $ a$, $ f(b)$, $ f(b \plus{} f(a) \minus{} 1)$ cannot be a triangle.
3. Put b = 1, then $ a, 1, f(f(a))$ is a triangle. This implies that $ f(f(a)) \equal{} a$ by remark.
4. [b]Claim[/b] $ f(n) \equal{} (n \minus{} 1)f(2) \minus{} (n \minus{} 2)$ by induction.
By 3, $ f$ is bijective, so we now know that $ a,b, f(f(a) \plus{} f(b) \minus{} 1)$ is a possible triangle. This means that $ f(f(a) \plus{} f(b) \minus{} 1) < a \plus{} b$.
Take a = b = 2, then $ f(2f(2) \minus{} 1) < 4$, i.e. it can either be 1, 2 or 3.
1 is not possible for that would mean $ 2f(2) \minus{} 1 \equal{} 1$, i.e. f(2) = 1, contradicting to bijectivity of $ f$.
2 is not possible for that would mean $ 2f(2) \minus{} 1 \equal{} f(2)$, i.e. f(2) = 1, again contradiction.
So it must be 3, i.e. $ 2f(2) \minus{} 1 \equal{} f(3)$, proving the hypothesis for $ n \equal{} 3$.

The induction step involves exactly the same argument, by taking $ a \equal{} 2, b \equal{} n$, and argue that $ f(f(2) \plus{} f(n) \minus{} 1) \equal{} n \plus{} 1$.

5. The formula for $ f$ tells us that $ f$ is strictly increasing. Suppose that $ f(i) \equal{} i$ for $ i \equal{} 1, 2, ..., k \minus{} 1$. If $ f(k) > k$, then as $ f$ is increasing nothing can map back to $ k$, contradicting to the bijectivity of $ f$. So $ f(k) \equal{} k$. By induction $ f(n) \equal{} n$ for all $ n$.

How do you come to repetition when f(1)>1?

$ f(1)>1$, then $ f$ is a peroidic funciton, and let $ k$ be the period, then $ a\plus{}nk, f(b), f(b\plus{}f(a)\minus{}1)$ are triangle's length. A contradiction when let $ a,b$ constant.

I got essentially the same solution as Soarer, but instead of constructing a recursive formula - I simply proved that the function is strictly increasing.

The problem is clearly better than the second problem - a bit harder, but still [b]classic[/b] and straightforward, nothing extraordinary.

Well, Erken, I think proving the function is strictly increasing is easier actually without using the recursive formula. You can give a proof by contradiction asuming the function to be decreasing at some point. Anyway, giving the recursion appears better. other than that, had the same solution as you both. Infact, I kind of guessed the result in the beginning first and saw. That perhaps helped.
Any other solution guys?

The 5th step of Soarer's solution can be done by setting $ n\equal{}f(2)$, which gives $ (f(2)\minus{}1)^2\equal{}1$.

All in all, this problem doesn't require a lot of ingenuity; the solution isn't as obvious as in #2, but still can be reached using only standard techniques. Both second problems were relatively easy this year, so the silver cutoff might jump up a little...

I have the same solution.But to complete the proof I proved $ f(f(n))\equal{}n$ so that $ f(n)\geq n$ imply
$ f(n)\equal{}n$ for all n.

[quote="khashi70"]Hi ... I don't agree with u bertram ... both the problems $ 4$ and $ 5$ weren't easy enough to silver cut off jump for that reason . I spoke with some IMO participant and they were also agree with me ... I think the silver cut off is gonna be 21 or 22 because only problems $ 1$ and $ 2$ were too easy .[/quote]

You should express your opinions about results and cut off-s in respective topics.

Let a=1, b=1, then 1, f(1), f(f(1)) is non-degenerate, so f(1) = f(f(1)) = k, say.

Let a=1, b=k, then 1, k, f(2k-1) is non-degenerate so f(2k-1) = k.

Let a=1, b=2k-1, then 1, k, f(3k-2) is non-degenerate, so f(3k-2) = k.

Let a=1, b=3k-2, then 1, 3, f(4k-3) is non-degenerate, so f(4k-3) = k.

Let a=4k-3, b=1 then 4k-3, k, f(2k-1) is non-degenerate, which means f(2k-1) > 3k-3.
But f(2k-1) = k, then k > 3k-3, so k = 1, i.e. f(1) = 1.

I have the same solution as in the first post and don't think it has very different solutions.

I got my solution a bit harder way, as I missed the periodicity idea:

Let $ g(x) \equal{} f(x) \minus{} 1$, then the 3-uple of numbers turns into $ a, g(b) \plus{} 1, g(b \plus{} g(a)) \plus{} 1$.

The inequality $ a \plus{} f(b) > f(b \plus{} f(a) \minus{} 1)$ transforms into $ a \plus{} g(b) > g(b \plus{} g(a))$. It is then a straightforward induction to obtain $ na \plus{} g(a) > g(a \plus{} ng(a))$ (*) for all positive integers $ a, n$.

I claim now that $ g(a) \leq a$. Assume the converse, and fix $ x$ s.t. $ g(x) > x$. Now exploit another inequality, $ g(b) \plus{} 1 \plus{} g(b \plus{} g(a)) \plus{} 1 > a$, or $ g(g(a) \plus{} b) > a \minus{} g(b) \minus{} 2$ (**). To make use of this, observe that we can write $ g(a) \plus{} b$ as $ g(x)k \plus{} x$ with $ b \in \{1, 2, ... g(x)\}$ and $ k \equal{} \frac {g(a) \plus{} b \minus{} x}{g(x)} \in \mathbb{N} \cup \{0\}$. Now we can combine both (*) and (**):

$ kx \plus{} g(x) > g(kg(x) \plus{} x) \equal{} g(g(a) \plus{} b) > a \plus{} g(b) \minus{} 2$

and plugging in $ k \equal{} \frac {g(a) \plus{} b \minus{} x}{g(x)}$ we get

$ xg(a) > g(x)(a \plus{} g(b) \minus{} 2 \minus{} f(x)) \minus{} b \plus{} x \equal{} g(x)a \plus{} (a term that is bounded)$,

thus, dividing both sides by $ x$ gives that for $ a$ large enough, $ g(a) > ca$ for some constant $ c > 1$. This, however, immediately contradicts (*) for $ a \equal{} x$ and $ n$ large enough: $ nx \plus{} g(x) > g(x \plus{} ng(x)) > c(x \plus{} ng(x)) > cx \plus{} cg(x)n$, which cannot hold for $ n$ large. This shows that $ g(a) \leq a$ (***).

Thus, $ f(a) \leq a \plus{} 1$. In particular, this implies that $ f(1)$ is either $ 2$ or $ 1$, which we deal with case by case:

In the first case, playing with small values of $ a, b$ shows it is not possible: $ f(1) \equal{} 2$, thus for $ a \equal{} b \equal{} 1$ we have the three sides of the triangle to be $ 1, 2, f(2)$, so $ f(2)$ also equals $ 2$. Plug now $ a \equal{} 1, b \equal{} 2$ to get the triple $ 1, 2, f(3)$ must comprise sides of a triangle, so $ f(3) \equal{} 2$, and inductively, $ f(n > 1) \equal{} 2$, which cannot be (f is clearly unbounded).

In the second case, $ f(1) \equal{} 1$, so making $ b \equal{} 1$ gives $ 1, a, f(f(a))$ must make sides of the triangle. That leads to
$ a \minus{} 1 < f(f(a)) < a \plus{} 1$, so $ f(f(a)) \equal{} a$. Put now $ f(a)$ instead of $ a$ in the very initial number triple to transform it into $ f(a), f(b), f(a \plus{} b \minus{} 1)$.(****) Now assume there's an $ x$ such that $ f(x) > x$, pick the minimal such $ x$. Then, by (***), $ f(x) \equal{} x \plus{} 1$. If $ x > 2$, then let $ a \equal{} 2, b \equal{} x \minus{} 1$ to turn (****) into $ f(2), f(x \minus{} 1), f(x)$. But now, $ f(2) \plus{} f(x \minus{} 1) \leq 2 \plus{} x \minus{} 1 \equal{} x \plus{} 1 \equal{} f(x)$, a contradiction. If $ x \equal{} 2$, put $ a \equal{} b \equal{} 3$, then $ 2f(3) > f(5)$, and since $ f(3) \equal{} f(f(2)) \equal{} 2$, we must have that $ f(5) \equal{} c \leq 3$. But $ 5 \equal{} f(f(5)) \equal{} f(c) \leq c \plus{} 1 \leq 4$, a contradiction. This shows that, in fact, $ f(a) \leq a$ for all $ a$, and, combined with $ f(f(a)) \equal{} a$, implies $ f(a) \equal{} a$.

I hope it's ok, please let me know if you find any flaws

I hope this is correct:
After showing that $ f(1) \equal{} 1$ and $ f$ is injective, one can proceed as follows: let $ k \equal{} f(2)$ and fix $ b$. Then $ |f(b \plus{} k \minus{} 1) \minus{} (b)| \equal{} 1$. If $ f(b \plus{} k \minus{} 1) \equal{} f(b) \minus{} 1$, then $ f(b \plus{} t(k \minus{} 1))$ has to be $ f(b) \minus{} t$, for all $ t$, for otherwise $ f(b \plus{} (m \minus{} 2)(k \minus{} 1)) \equal{} f(b \plus{} m(k \minus{} 1))$ for some $ m$, but $ f$ is injective. This is impossible. Thus $ f(b \plus{} t(k \minus{} 1)) \equal{} f(b) \plus{} t$. Let $ A \equal{} \{b \plus{} t(k \minus{} 1)|t\ge0\}$. If $ k > 2$, then $ f(A)$ contains all but finitely many positive integers, while $ \mathbb N\setminus A$ is infinite. So $ k \equal{} 2$, and plugging $ b \equal{} 2$ we conclude $ f(n) \equal{} n$.

I proved that $ f(1) \equal{} 1$ since otherwise $ f$ would be bounded above and that can't happen.
Then I proved that $ x \equal{} f(f(x))$.
Then I proved that $ f(2) \equal{} 2$ by proving that if $ f(2) \equal{}k>2$, then there exist two distinct integers, $ a\equal{}2$ and $ b\equal{}(k\minus{}1)k\minus{}(k\minus{}2)$ such that $ f(a)\equal{}f(b)\equal{}k$, contradicting the bijection.
I then used induction to show that since $ f(2) \plus{} f(x \minus{} 1) > f(x)$, we must have $ x \plus{} 1 > f(x) > x \minus{} 1$, so $ f(x) \equal{} x$.

f(1) = 1, and f(f(x))=x, as everyone else has. The second implies that f is injective.

Induct strongly; assume f(1)=1, f(2)=2, ..., f(x-1)=x-1. Assume $ f(x)\neq x$, by injectivity, f(x)=k>x. We can then get $ f(k \plus{} b \minus{} 1) \minus{} f(b)\le x \minus{} 1$ for all $ b$. We then find f(k)=x, f(k+1)=x+1, ..., f(x+k-2)=2x-2, then we can get f(2k-1)=2x-1, and keep going. Using the fact that k>x, we can check that these groups of x-1 consectuve values are disjoint. We generate, for all y>k, f(y) such that f(y)=z for all z>x. But we then get some y such that f(y)=k, but we have f(x)=k, contradicting injectivity.

Thabonk

Let $P(a,b)$ denote the assertion. $P(1,b)$ gives that $f(b)=f(b+f(1)-1).$ Suppose $f(1)>2$ then let $c=f(1)-1.$ $f$ has period $c,$ so we can choose $a$ to be larger than the sum of $f(x)+f(y)$ for all $x,y$ so that $a,f(x),f(y)$ cannot be in a triangle. Thus, $f(1)=1.$

Now, $P(a,1)$ gives $a=f(f(a)).$ Thus, $f$ is bijective. We claim that $f(2)=2.$ Let $c$ be such that $f(c)=2.$ Then, $P(a,c)$ gives $|a,f(c+f(a)-1)|\le 1.$ If $f(f(a)+c-1)=a+d$ then taking $f$ on both sides gives \[f(a)=f(a+d)+c-1.\]We have possibilities: $d=-1,0,1.$ If $d=0$ then $f(1)=2$ which is contradiction. Now, if $d=1$ then $f(a)-f(a+1)=c-1$ so $f(a)$ is nonincreasing. Thus, $f(x)=1$ for all $x,$ which doesn't work. Now, if $d=-1$ then \[f(a)-f(a-1)=c-1\]which shows that $f(a)=c+(a-2)(c-1)$ and \[f(f(a))=c+(c+(a-2)(c-1)-2)(c-1)\]which shows that $c=2.$ Thus, $f(2)=2.$

Then, $P(f(a),b)$ gives $f(a),f(b),f(a+b-1).$ Let $a=2$ to get $f(2),f(b),f(b+1)$ forms a triangle. Thus, $|f(b+1)-f(b)|\le 1.$ In particular, $f(b+1)\le f(b)+1.$ Thus, $f(b+n)\le f(b)+n$ so $f(1+n)\le f(1)+n$ which implies $f(j)\le j$ for all $j.$ Thus, $f(j)=j$ because $f(f(j))=j.$ This is a valid function.

We claim that $f(x)=x$ is the only solution, which works.

Let $P(a,b)$ be the assertion. I first claim $f(1)=1$. If $f(1)>1$, then $P(1,b)$ says that $f$ is periodic. Then taking $P(10^{10^{10^{10}}} \cdot \max \{ f \}, 1)$, we reach a contradiction. Now take $P(a,1)$ we get that $f(f(a))=a$. Taking $P(a,f(b))$, we get that $a,b,f(f(a)+f(b)-1)$ form a triangle. Let $Q(a,b)$ be the assertion of this. Let $f(2)=k>1$. We will prove this by strong induction $f(x)=(x-1)k-(x-2)$.

Base case ($x=2$): This is obvious by how we defined $k$.

Inductive Step: Assume $f(x)=(x-1)k-(x-2)$ for all $x \le n$. Then $f((x-1)k-(x-2))=x$ for all $x \le n$. Taking $Q(2,n)$, we get that $2,n,f(nk-(n-1))$. Since this isn't equal to $(x-1)k-(x-2)$ for any $x \le n$, by injectivity $f(nk-(n-1))>n$. By the triangle inequality, we get $f(nk-(n-1))=n+1$, finishing the induction.

To finish, note that $2=f(f(2)=f(k)=k^2-2k+2$, and this means $k=2$. Thus $\boxed{f(x)=x}$.

Claim: $f(1)=1$. $a=b=1$ gives $f(1)=f(f(1)).$ Letting $a=1$, we have $f(b)=f(b+f(1)-1),$ since if one side is 1, the other two sides must be equal. This shows that $f$ is periodic with period $f(1)-1$ unless $f(1)=1$. However, if $f$ is periodic, this means that $f(b)+f(b+f(a)-1)$ would have a finite maximum value, and setting $a$ to this value would guarantee that it isn't a triangle, so $f(1)=1.$\\

Now, by plugging in $b=1$, we see that $a,1,f(f(a))$ forms a triangle, so $f(f(a))=a$, so $f$ is an involution and therefore bijective.\\

Let $f(2)=n$ (so $f(n)=2$) and plug in $a=2$. We have $2, f(b), f(b+n-1)$ is a triangle, and by bijectivity, we have either $f(b+n-1)=f(b)+1$ or $f(b+n-1)=f(b)-1$.

Note that this means increasing the argument of $f$ by $n-1$ either increases or decreases the output by 1. Consider the sequence $f(1)$, $f(1+1(n-1))$, $f(1+2(n-1))$, and so on. If at any point moving forwards in the sequence decreases the value by 1, it will either make it nonpositive or repeat one of the previous terms, contradiction. Therefore, $f(1+(n-2)(n-1))=n-1$ and $f(1+n(n-1))=n+1$. Then, we have that either $f(n+1)=f(2+n-1)=n+1$ or $n-1$.

Case 1: $f(n+1)=n+1$. Then, by bijectivity, $1+n(n-1)=n+1$, so $n=0$ or $n=2$ and we drop $n=0$.

Case 2: $f(n+1)=n-1$. Then, by bjiectivity, $1+(n-2)(n-1)=n+1$, so $n=2\pm\sqrt{2}$, which cannot happen.

Therefore, $n=2$, and $f(b+1)=f(b)+1$ or $f(b+1)=f(b)-1$. From $f(b)>0$, $f(1)=1$, and bijectivity, we have $f(x)=x$, which works.

Here is my solution: [url]https://calimath.org/pdf/IMO2009-5.pdf[/url]
And I uploaded the solution with motivation to: [url]https://youtu.be/lKGbpsbWpCQ[/url]

The answer is $f(x)=x$ is the only solution.
Let the assertion be $P(a,b)$.

Claim: $f(1)=1$
Proof: $P(1,b)$ gives that $f(b+1)=f(b+f(1))$ for $b\geq 0$. This gives $f(1)=f(f(1))$ so we know that $f^n(1)=1$ for $n \geq 1$. Let $f(1)=c$. Now we know that $f(b+1)=f(b+c)$ so $f$ is periodic with period $c-1$. Now $P(a,1)$ gives $f(f(a))+c>a$. However, we can continually increase $a$ if $c>1$ by $c-1$, which increases the RHS without increasing the LHS. This forces $c=1$.

Now $P(a,1)$ gives $f(f(a))=a$ so $f$ is injective and surjective. This forces $f(2) \geq 2$.

Claim: $f(2)=2$.
Proof: Let $f(2)=d$. Then, $P(2,b)$ gives $|f(b+1)-f(b+d)|=1$ for all $b \geq 0$. FTSOC, assume $d>2$. Then, we have at least two infinitely long chains of numbers in arithmetic sequence such that their positive difference after applying $f$ onto them is $1$. If, in a chain, the difference goes down from going to the previous to the next term, it must go up again, contradicting injectivity, so it can only increase by $1$, so in reality $f(b+d)-f(b+1)=1$. However, since both chains are of infinite length, one chain must start at $1$, and wherever the other starts at, they will overlap, contradicting injectivity again. So $d=2$.

We can now proceed with induction to show that $f(n) =n$. We have the bases cases for $1,2$ proven above. Inductive step: \[ P(f(a),b) \implies f(a)+f(b)>f(a+b-1)\] so if we let $a=n-1$ and $b=2$, then we have $n+2>f(n+1) \implies n+1 \geq f(n+1)$. Since we have $f$ is surjective and all $1 \dots n$ are taken up, we have $f(n+1)=n+1$ and we are done.

We claim that $f(x) = x$ is the only solution

Let $P(a, b)$ denote the assertion

$P(1, 1)$ implies $f(1) = f(f(1))$.

[b][color=#f00]Claim:[/color][/b] $f(1) = 1$

[i]Proof.[/i] Assume FTSOC, $f(1) = c > 1$.

$P(1, x)$ implies $f(x+c-1) = f(x)$. But now, fixing $b$ and picking large enough $a \equiv 1 \pmod{c-1}$ we get a contradiction.

[b][color=#f00]Claim:[/color][/b] $f(f(x)) = x$ implying that $f$ is a bijection

[i]Proof.[/i] $P(a, 1)$

[b][color=#f00]Claim:[/color][/b] $f$ is linear

[i]Proof.[/i] $P(2, b)$ implies $f(b+f(2)-1)$ belongs to ${f(b)-1, f(b), f(b)+1}$. However, by the fact that $f$ is injective, $b+f(2)-1 \neq b$. But, it cannot be $f(b)-1$ either since by injectivity of $f$, and if we keep on adding $f(2)-1$ to the argument, we will get negative values. So, $f(b+f(2)-1) = f(b)+1$. However notice here that $f(b+k(f(2)-1))$ takes all values greater than or equal to $f(b)$. So, $f(2) = 2$ by injectivity and thus $f(b+1) = f(b)+1$. But now we are done

1,b: f(b)=f(b+f(1)-1) => f is bounded => contraction.
a,1: f(f(a))=a. Assuming f is not the identity, let a be a positive integer such that f(a)-1=:c>a-1
f(b+Kc)<=f(b+(K-1)c)+(a-1)<=...<=f(b)+K(a-1)<=
K(a-1)+const(a). So f(x)<x for large enough x, which is clearly a contradiction. Hence, the identity function is the only one that works.

[quote=Spectator][hide]Groupsolved with [b]brainfertilzer[/b] and [b]plang2008[/b]

We claim that $f(x) = x$ is the only solution. We see that this works once plugged in.
Let the given assertion be $P(a,b)$.

[color=#00f][b]Claim 1: $f(1) = 1$[/b][/color]
Assume for the sake of contradiction that $f(1)\neq 1$. Then,
\[P(1,b)\colon f(b) = f(b+f(1)-1)\] because of the triangle inequality.
Then, $f$ is periodic with period $f(1)-1$. Note, with $n$ arbitrarily large $P(1+n(f(1)-1),b)$ shows that it doesn't satisfy the triangle inequality, contradiction.

Note
\[P(x,1)\colon f(f(x)) = x\]
showing that $f$ is surjective.
Then,
\[P(f(a),b)\colon f(a), f(b), f(a+b-1)\]
Let this assertion be $Q(a,b)$.

We induct. Our base case is $f(1) = 1$. We also have $f(2) = 2$ by "I have to do homework due tomorrow and it's 10:30 pm somebody help me" theorem. Assume that the claim is true for all $1,2\cdots, 2x-2$. Then,
\[Q(x,x)\colon f(2x-1)\leq f(x)+f(x)\leq 2x \leq 2x-1\]
and
\[Q(x,x+1)\colon f(2x)\leq f(x+1)+f(x)\leq 2x+1 \leq 2x\]
Because we have no other spaces for $2x$ and $2x-1$ and $f$ is surjective, we must have $f(2x) = 2x$ and $f(2x-1) = 2x-1$, thus concluding the induction and the proof.

[hide=remarks]
i have a lab report due tomorrow somebody help me fdhafjdshfkdlsfhdlkshfjdklashfdlsajfhdlksah
also we bs-ed a lot of this proof oopsies
[/hide][/hide][/quote]
Legend says he still hasn't found time to rigorously do this

We claim the answer is f(x)=x only
By triangle inequality, for lengths m,n,1 we have m=n.
For a=1, it is easy to see that f(b)=f(b+f(1)-1). If f(1)-1 is not 0, then f(b) is periodic with period f(1)-1 not infinite, meaning f can only take on f(1)-1 values, contradiction. Then f(1)=1. Taking b=1, a=f(f(a)), so f is an involution and therefore a bijection.

We know a, f(b), f(b+f(a)-1), or a, f(f(b))=b, f(f(b)+f(a)-1) (substitute f(b) instead of b) form a triangle. a=b=2, we have f(2f(2)-1)<4. LHS cannot be 1 since that would imply 2f(2)-1=1, or f(2)=1=f(1). LHS is not 2 since that implies f(2)=3/2, not an integer. So we have f(2)=2. At a=2, f(b)-1$\le$f(b+f(2)-1)$\le$f(b)+1 by triangle inequality. Now we casework:

f(b+1)=f(b+f(2)-1)=f(b)-1, contradiction since this means f is strictly decreasing.
f(b+1)=f(b+f(2)-1)=f(b), contradiction since f is constant.
Finally, we have f(b+1)=f(b+f(2)-1)=f(b)+1, and inductively this holds so f ranges over all positive integers, and this is the only solution.

[quote name="huashiliao2020" url="/community/p27856146"]
[quote=Spectator][hide]Groupsolved with [b]brainfertilzer[/b] and [b]plang2008[/b]

We claim that $f(x) = x$ is the only solution. We see that this works once plugged in.
Let the given assertion be $P(a,b)$.

[color=#00f][b]Claim 1: $f(1) = 1$[/b][/color]
Assume for the sake of contradiction that $f(1)\neq 1$. Then,
\[P(1,b)\colon f(b) = f(b+f(1)-1)\] because of the triangle inequality.
Then, $f$ is periodic with period $f(1)-1$. Note, with $n$ arbitrarily large $P(1+n(f(1)-1),b)$ shows that it doesn't satisfy the triangle inequality, contradiction.

Note
\[P(x,1)\colon f(f(x)) = x\]
showing that $f$ is surjective.
Then,
\[P(f(a),b)\colon f(a), f(b), f(a+b-1)\]
Let this assertion be $Q(a,b)$.

We induct. Our base case is $f(1) = 1$. We also have $f(2) = 2$ by "I have to do homework due tomorrow and it's 10:30 pm somebody help me" theorem. Assume that the claim is true for all $1,2\cdots, 2x-2$. Then,
\[Q(x,x)\colon f(2x-1)\leq f(x)+f(x)\leq 2x \leq 2x-1\]
and
\[Q(x,x+1)\colon f(2x)\leq f(x+1)+f(x)\leq 2x+1 \leq 2x\]
Because we have no other spaces for $2x$ and $2x-1$ and $f$ is surjective, we must have $f(2x) = 2x$ and $f(2x-1) = 2x-1$, thus concluding the induction and the proof.

[hide=remarks]
i have a lab report due tomorrow somebody help me fdhafjdshfkdlsfhdlkshfjdklashfdlsajfhdlksah
also we bs-ed a lot of this proof oopsies
[/hide][/hide][/quote]
Legend says he still hasn't found time to rigorously do this
[/quote]

the real rigorous solve was the friends we made along the way

The only solution is $f(x) = x$ for all $x \in \mathbb{Z}_{>0}$. It is easy to confirm that this solution works.

Denote the given assertion as $P(a, b)$. $P(1, b)$ states that $1, f(b), f(b + f(1) - 1)$ are the sides of a non-degenerate triangle. Hence, $f(b) = f(b + f(1) - 1)$ for all $b \in \mathbb{Z}_{>0}$.

[b][color=#9a00ff]Case 1:[/color][/b] $f(1) > 1$. Note that $f$ is periodic with period $f(1) - 1$, and thus has a maximum $M$. However, the assertion $P(2M, 1)$ states that $2M \geq f(1) + f(f(1)) > 2M$. Contradiction.

[b][color=#9a00ff]Case 2:[/color][/b] $f(1) = 1$. $P(a, 1)$ states that $a, 1, f(f(a))$ are the sides of a non-degenerate triangle. Thus, $a = f(f(a))$ and so $f$ is a bijection. Let $g(a) = f(f(a) + f(2) - 1)$ for all $a \geq 1$. Furthermore, $P(a, f(2))$ states that $a, 2, g(a)$ are the sides of a non-degenerate triangle. By the Triangle Inequality,
\[g(a) \in \{a-1, a+1\}\]
(since $f(a) + f(2) - 1 > f(a)$ and $f$ is bijective, $g(a) \neq a$). Note that by injectivity, all $g(a)$ are distinct. Since $f(2) + f(2) - 1 \neq 1$, $g(2) \neq 1$, and so $g(2) = 3$. Since $f(3) + f(2) - 1 \neq f(2)$, $g(3) \neq 2$, and so $g(3) = 4$.

[color=#9a00ff]Claim: [/color] For all $a \geq 2$, $g(a) = a + 1$.

[color=#9a00ff]Proof: [/color] Proceed by induction. The claim is true for $a = 2, 3$. Suppose the claim is true for all $a \leq k - 1$, where $k \geq 4$. Then since $g(k) \neq g(k - 2) = k - 1$ and $g(k) \in \{k - 1, k + 1\}$, $g(k) = k + 1$, completing our induction.

Note that for all $a \geq 2$, $f(a) + f(2) - 1 \geq 2 + 2 - 1 = 3$. Hence, for all $a \geq 2$, $a + 1 = g(a) \neq f(2)$. Since $f$ is injective, $f(2) = 2$. Simple induction gives $f(x) = x$ for all $x \in \mathbb{Z}_{>0}$.

We plug $a = 1$. This gives that $f(b) = f(b+f(1) - 1)$. FTSOC assume that $f(1) \neq 1$. This means that $f(b) = f(b+c)$ for some positive constant $c$. This means that in our original equation we can increase $a$ to an arbitrarily large number and still have the same values for the other two sides. Contradiction.

We can plug in $b = 1$ to see that $f(f(a)) = a$. This means that $f$ is both surjective and injective.

We prove by strong induction that $f(x) = x$. Our base case is trivial.

Now we try to prove that $f(k) = k$ and we know that $f(x) = x$ for all $x <k$
We look at $P(2, k)$ where  and see that $f(k + f(2) - 1) \in {f(k) - 1, f(k), f(k) + 1}$. If $f(k + f(2) - 1) = f(k) - 1$ then we can keep adding $f(2) - 1$ inside the function and make it negative which is impossible. If we have $f(k+f(2) - 1) = f(k)$ then by injectivity we have $f(2) = 1$ which is impossible because $f(1) = 1$. So, we must have $f(k+f(2) - 1) = f(k) + 1$. We can see that $f(k+x(f(2) - 1))$ takes on all values of $f$ such that $f(x) \geq f(k) + 1$ by injectivity. Thus we must have $f(k) = k$ $\blacksquare$

Very cool problem! Let $P(a,b)$ denote the assertion.

Initially, we prove $f$ is unbounded. If $f$ was bounded, then take a sufficiently large $a$, and obviously the statement is not true.

We claim $f(1)=1$. Consider $P(1,b)$, then obviously $f(b) = f(b + f(1)-1)$.
If indeed $f(1)>1$, then the sequence would be periodic every $f(1)-1$ terms, implying that the image of $f$ is $f(1), f(2), \dots, f(f(1)-1)$. But this implies $f$ is bounded, contradiction.

Then taking $P(a,1)$, we have $f(f(a))=a$, so $f$ is an involution and therefore bijective.

We prove that $f((k-1)f(2) - (k-2)) = k$ for $k\geq 2$ through induction. The base case, $k=2$, is obvious.
Now assume this is true for $k=2, 3, \dots, m$. Consider $P(2,(m-1)f(2) - (m-2))$. We have $2, m, f(m f(2) - (m-1))$ are the sides of a triangle. Since $f$ is bijective, in fact, we must have $f(mf(2)-(m-1))=m+1$. This proves the claim.

But since $f((k-1)f(2) - (k-2))=k$, the set $S = \{(k-1)f(2) - (k-2) \mid k\geq 2, k \in \mathbb N \}$ must contain all the positive integers at least 2, which implies $f(2)=2$. It is easy to see this finishes.

Denote the assertion as $P(a, b)$.

First,
\[ P(1, b) \implies f(b) = f(b+f(1) - 1) \]
Now, if $f(1) \neq 1$ then $f$ is periodic. If we make $a$ very large then we find that this case fails. Thus, $f(1) = 1$. Then,
\[ P(a, 1) \implies f(f(a)) = a \]
This means that $f$ is both injective and surjective. Plug
\[ P(2, b) \implies f(b) - 1 \leq f(b + f(2) - 1) \leq f(b) + 1 \]
Note that in particular since $f(b+f(2) - 1) \neq f(b)$ by injectivity, we have that either $f(b+f(2)-1) = f(b)+1$ or $f(b+f(2) - 1) = f(b) - 1$.

Now, suppose that $f(b+f(2)-1) = f(b) - 1$ for some $b$. Denote $m = f(2) - 1$. Then $f(b+2m) = f(b+m) - 1 = f(b) - 2$. By induction, then
\[ f(b+(k+1)m) = f(b+km) - 1 = f(b) - k - 1 \]
for all positive integers $k$. But then we eventually get a nonnegative output, contradiction.

So we must have $f(b+f(2) - 1) = f(b) + 1$. If $m = f(2) - 1$, then by induction,
\[ f(b + (k+1)m) = f(b+km) + 1 = f(b) + k + 1 \]
If $f(2) - 1 > 1$ then take some $c$ such that $c \not \equiv 1 \pmod{m}$. Since $f(c) \neq f(f(2)) = 2$, then there exists nonnegative integers $k$ and $j$ such that
\[ f(c+km) = f(c) + k = 2+j = f(1+jm) \]
Then this contradicts injectivity.

Thus, $f(2) - 1 = 1$, and so we have $f(b+1) = f(b) + 1$. This implies that $f(n) = n$, which obviously works. $\blacksquare$

The only solution is $f(n)=n$ for all $n \in \mathbb{N}$. It's easy to check that this function satisfies the problem condition.

Now, let $P(a,b)$ be the assertation of there is a triangle whose sides are $a,f(b)$, and $f(b+f(a)-1)$. This would later implies $$a+f(b)\ge f(b+f(a)-1)+1, a+f(b+f(a)-1)\ge f(b)+1,\text{and } f(b)+f(b+f(a)-1)\ge a+1.$$

[b][color=#f00]Claim 1. $f(1)=1.$
Proof.[/color][/b]
Assume to the contrary, we have $f(1)=c\ge 2$. Observe that $P(a,1)$ gives $f(b)\ge f(b+c-1)$ for all $b \in \mathbb{N}$. This implies the chain of inequality: $$f(b)\ge f(b+c-1)\ge f(b+2c-2)\ge f(b+3c-3)\ge \cdots $$
It means that for every $i\in \{1,2,3,...,c-1 \}$, there is $M_i\in \mathbb{N}$ such that the sequence $f(i+M_i(c-1)), f(i+(M_i+1)(c-1)), f(i+(M_i+2)(c-1)), \cdots$ is a constant sequence. We now let every terms of that sequence be $N_i$.
Choose $r=\max(M_1,M_2,...,M_{c-1})+2$. Notice that the sequence $f(1+(c-1)r),f(2+(c-1)r),f(3+(c-1)r), \cdots$ is periodic in which the period is at most $c-1$.
Now, choose $a=2max(N_1,N_2,...,N_{c-1})$. Notice $P(a,1+(c-1)r)$ implies $$f(1+(c-1)r)+f(1+(c-1)r+f(a)-1)\ge a+1.$$
However, $f(1+(c-1)r)+f(1+(c-1)r+f(a)-1)\le a$ by the definition of $a$. This is a contradiction. $\blacksquare$

Now, we have $f(1)=1$. Notice that $P(a,1)$ tells us $$f(f(a))+1\ge a+1 \text{and } a+1\ge f(f(a))+1. $$
This implies $f(f(a))=a$ for every $a\in \mathbb{N}$, which gives us the bijectivity of $f$.
Let $k\ge 2$ be a natural number such that $f(k)=2$. We also have $k=f(2)$.

[b][color=#f00]Claim 2. $f(1+(k-1)l)=l+1$ for all $l\in \mathbb{N}_0$
Proof.[/color][/b] We will use induction. For $l=0$ and $l=1$, it is clear. Assume that we now have $f(1+(k-1)d)=d+1$ for every $d=0,1,2,...,l$. For $d=l+1$, by the bijectivity of $f$, we have $f(1+(k-1)(l+1))\ge l+1$. However, from $P(f(a),k)$, we have $f(a)+1\ge f(a+k-1)$ for all $a \in \mathbb{N}$. Substituting $a=1+(k-1)l$, we get $l+2\ge f(1+(k-1)(l+1))$. This forces $f(1+(k-1)(l+1))=l+2$. This completes the induction. $\blacksquare$

We now have $f(1+(k-1)l)=l+1$ for all $l\in \mathbb{N}_0$. This implies $f(l)=1+(k-1)(l-1)$ for every $l \in \mathbb{N}$. Now, the bijectivity of $f$ forces $k=2$ (If not, then there is no $q\in \mathbb{N}$ such that $f(q)=2$). This brings us to $f(l)=l$ for every $l\in \mathbb{N}$.

To sum up, the only function that satisfies the problem condition is $f(n)=n$ for every $n \in \mathbb{N}$. It is easy to check that this solution indeed satisfies the property given.


For $A,B,C,D>0$ Prove:

$(A^3 + B^3 + C^3)(B^3 + C^3 + D^3)(C^3 + D^3 + A^3)(D^3 + A^3 + B^3) \geq \left[(AB+CD)(AC+BD)(AD+BC)\right]^2$

I have a solution, but it is very specific and I would like to see a different one. I don't know if this is too hard?

Are you sure this is what you intended? It's not equal when A=B=C=D, so it's probably strict as well...


Hi there,

I've found a problem in a contest that I don't know how to approach.
It'd be great if you can give me some tips to know how to get started

[code]
There're 400 cards, each one has its number written in the back. We know the 400 numbers are different.
I have to order the cards from highest to lowest, following this procedure: Choose three cards and ask your assistant to inform you which one is the highest and which one the lowest (he can't tell what the numbers are)
Determine how many times this procedure should be used to order the cards.[/code]

Thanks in advance!

It is possible to arrange 400 cards in 133 comparisons [i]if[/i] you chose sets of 3 cards overlapping by 1 [i]and[/i] you chose them the correct sequential order.

I don't think so (but don't completely understand).
It seems to me as when you need only $1$ to order $4$ cards... (but even $2$ doesn't suffice)

[quote="ZetaX"]It seems to me as when you need only $1$ to order $4$ cards... (but even $2$ doesn't suffice)[/quote]

Let there four cards with numbers 1 through 4.  If you chose 1,2,3 or 1,2,4 or 1,3,4 or 2,3,4 you would know the order of only those three cards.
You might need one or two more comparsions, depending on which cards you chose to compare the forth remaining card.

So, three comparsions is the maximum (legitmately) needed in that case.

I ran a program on my computer that was able to choose 21235 triplets without ever including the same pair of cards twice in a triplet.   It was able to finish comparing every possible pair after 8120 more comparsions of triplets.

So, the necessary number of comparsions is at most 29355.


group theory
abstract algebra
superior algebra
superior algebra solved

Prove that every finit group with an odd number of elements cannot  be written as a reunion of three subgroups ,each one different form {e} or the whole group.

You do realize that if n is prime yourquestion is outrageously wrong

If you insist.
Take Z4 ( the additive group of residues mod 4) and see that you can't write it as a reunion of three proper subgroups.

Are you sure your problem is correct?

Here's what I think:

I've seen this be4. The prb is this: Prove that any group with an odd number of elements CAN'T be written as the reunion of 3 proper subgroups.

Proof:

We assume it can be done.
Let the subgroups be A, B, C. Let's assume that the intersection of B and C is NOT included in A. We take an element x from B \cap C-A and an element y from A-(B \cup C). We can see that xy can't be in A, in B or in C, so A \cup B \cup C =/= G, contradiction. This means that A \cap B \cap C = H =/=  \emptyset and, for example, B and C have nothing else in common except for H (same for A,B and A,C). Now take an element x from A-H and multiply it with every element from B-H. We will get |B-H| different elements, all from C-H, meaning that |C-H| >= |B-H| iff |C|>=|B|, but since this can be said for every pair of subgroups from {A, B, C}, it means that |A|=|B|=|C|. Since A \cap B \cap C=/= \emptyset , it means that |G| > |A| > |G|/3. We know that |A| divides |G|, so |A|=|G|/2, meaning that 2 | |G|, but this is false, since G has an odd number of elements. I think it's done.

Yes, this was the problem.I corrected it.I apologise for posting a wrong problem.

Actually, i saw this problem in that cool maths magazine you have in Galati  :D .

This problems looks familiar to me too. But I don't think I saw it in that cool Galati math magasine.
It was in a Constanta problem book.


trigonometry
calculus
integration
function
real analysis
real analysis unsolved

$ \sum \frac{\cos \sqrt{n}}{n}$
converge or .. ?

i have a feeling that it converges, but showing it is an all different matter, obviously. :(

presume dirichlet should be of help here, i.e. given $ a_n,\;b_n\,\in\mathbb R,\;\;\;\text{where}\;\;\;n\in\mathbb N$,
if $ a_n\to 0$ as $ n\to\infty$ and if $ S_{partial}\equal{}\sum_{n\equal{}1}^m\;b_k$ is bounded, then $ \sum_{n\equal{}1}^\infty\;(a_n\cdot b_n)$ converges.

let $ a_n\equal{}\frac{1}{n}$. clearly, $ a_n\to 0$ as $ n\to\infty$.
now we look at $ S_{partial}\equal{}\sum_{n\equal{}1}^m\;\cos\,\sqrt n$.

showing that the sequence $ S_{partial}$ has bounded partial sums will now require some thinking.... hmmm.... not easy! :|

this is main moment . I don't agree that this sum $ \sum \cos \sqrt {n}$ is bounded. it's enough to show that $ \sum \cos \sqrt {n} \equal{} O(n^{\delta})$ for $ \delta < 1$
or $ |\sum e^{2\pi i \sqrt {n} x }|$ is $ O(n^{\delta})$ but here doesn't work some trick which used when we considered $ \sum \frac {\sin n^{2}}{n}$ there was $ k^{2} \minus{} m^{2} \equal{} (k \minus{} m)(k \plus{} m) ...k \minus{} m \equal{} h...$ but here I can't do nothing $ \sum e^{...\sqrt {k} \minus{} \sqrt {m}...}$
or generalization for like this sum will be next :
how to find asymptotic of this
$ |\sum e^{i f(n) x}|$ where $ x$ is liovillian.  :huh:
here is another problem :
$ \sum \frac{\cos 2^{n}}{n}$ I suppose that here we can use something if we consider $ \sum e ^{...(2^{k}\minus{}2^{m})...}$

compare it with the integral of $ \frac{\cos( \sqrt{x})}{x}$. This fonction doesn't oscillate too much, so it will work  :)

$ f(x)\equal{}\frac{\cos(\sqrt{x})}{x}$

we have

(1) $ \int_{1}^{N\plus{}1}f\equal{}\sum_{n\equal{}1}^{N}f(n) \plus{} \sum_{n\equal{}1}^{N}a(n)$

$ a(n)\equal{}\int_{n}^{n\plus{}1}(n\plus{}1\minus{}x)f'(x)dx$

but
$ |a(n)|\leq \int_{n}^{n\plus{}1} |f'| \leq \frac{1}{2n^{1.5}}\plus{}\frac{1}{n^2}$

then

(2) $ \sum a(n)$ is converge absolutly

(1) and ( 2)   gives $ \sum f(n)$ is the same nature of $ \int_{1}^{\plus{}\infty}f$

the last integral converge (change variable u^2=x )

yep, if a fonction does not oscillate too much i.e. $ \int |f'| < \infty$, then $ \sum f(n)$ and $ \int f(x)dx$ have the same behavior..


vector
trigonometry
geometry
trig identities
Law of Sines

a plane leaves Versailles traveling 200mph S 45 degrees E. If a wind blows directly out of the southwest 100mph, find the resultant speed in mph.

Simple vector addition? Draw the diagram and you get a triangle with an angle and two sides given. Then just bash it out with Law of Sines twice...

Are you familiar with vectors?
The plane velocity is one vector, the wind velocity is a second vector. The vector sum gives you the net velocity.

I got 100sqrt(5)mph relative to the ground
How do you make a sqare root sign?

[quote="Icosahedroid"]How do you make a square root sign?[/quote]
[code]$\sqrt{x}$[/code] should output $ \sqrt{x}$.
So, in your case, $ 100\sqrt{5}$. Furthermore, I get the same answer.

The wind is blowing directly perpendicular to the plane (the plane is traveling in direction 315, the wind is blowing in direction 45, where 0 is East). So, the result is $ \sqrt {200^2 \plus{} 100^2} \equal{} 100 \sqrt {2^2 \plus{} 1^2} \equal{} 100 \sqrt {5}$.


help me with this integrals, I will buy you a beeer
1  [tex]\int_{-\infty}^{+\infty}e^{tx}e^{-e^{-x}}e^{-x}dx [/tex]
2 [tex]\int_{-\infty}^{+\infty}e^{tx}\frac{e^{-x}}{(1+e^{-x})^2}dx [/tex]
3 [tex]\int_{-\infty}^{+\infty}e^{tx}\frac{1-cos(ax)}{\pi ax^2}dx [/tex]

Integrals are calculus, this definitely should not be in MOEMS forum.
PS, this is middle school, people aren't supposed to be drinking beer  :lol:

Whoa!!!  Whoa!!!  Whoa!! this is for 4th-8th graders, not 12th graders


Solve the equation in $\mathbb{Z}$:
$x^y+y^z+z^x=2(x+y+z)$

It is easy to show that $min(x,y,z)\ge -1$. If one of them negative or greate 2, then we have not solution. It give x=y=z=2.


Consider cyclic quadrilateral $ ABCD$ with $ AB \equal{} 6, BC\equal{}7, CD \equal{} 8,$ and $ AD \equal{} 9$. Find $ (AC)^2$.

Label $AB=a,BC=b,CD=c,DA=d,BD=m,AC=n.$ Using the 1st and 2nd Ptolemy's theorem we get

$\frac{m^2}{n^2}=\frac{(ab+cd)^2}{(ad+bc)^2} \ , \ m^2n^2=(ac+bd)^2 \Longrightarrow n^2=\frac{(ac+bd)(ad+bc)}{ab+cd}=\frac{2035}{19}$

Let $ \alpha$ be the angle at B. Then the angle at C is $ 180\minus{}\alpha$.

By applying the cosine law twice,

$ AC^2\equal{}6^2\plus{}7^2\minus{}2*6*7cos \alpha$
$ AC^2\equal{}8^2\plus{}9^2\plus{}2*8*9cos \alpha$.

Equate, solve for $ \alpha$, and plug it back in.


A geometric sequence of positive integers is formed for which the first term is 3 and the fourth term is 192. What is the third term of the sequence?

If the first term of the sequence is $ a$, then the fourth term of the sequence is $ ar^3$, for some real numbers $ a$ and $ r$. Now, we know $ a\equal{}3$, so

$ 3(r^3)\equal{}192 \implies r^3\equal{}64 \implies r\equal{}4$.

Thus, the third term is $ 3(4)^2\equal{}\boxed{48}$.


What is the volume of the largest right rectangular prism that can be inscribed in the ellipsoid

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}= 1$

?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=134167

accordind tho brasiluca we need to use jacobian of a change basis of a ellipsoid to sphere

Didn't think of using AM-GM for this...

[hide]
Stretching the system will not effect optimization.

Inscribe a cube inside a unit sphere - it has volume $\left( 2 \frac{1}{\sqrt{3}}\right)^{3}= \frac{8}{3\sqrt{3}}$

Now take this and stretch it in the x, y, and z dimensions by factors of a, b, and c respectively to turn the unit sphere into the ellipsoid in the problem.

$V=\frac{8abc}{3\sqrt{3}}$
[/hide]


Prove that for all positive reals the following inequation holds:

\[ \sqrt{a^4+c^4} + \sqrt{a^4+d^4} + \sqrt{b^4+c^4} + \sqrt{b^4+d^4} < 4 + a^4 + b^4 + c^4 + d^4 \]

Any other solutions before posting mine?

It can be observed that in mffmartinelli's proof, a stronger inequality can be proved. :D (if we improve $x^2-16x+128>0$ to $x^2-16x+64\geq 0$)

$\sqrt{a^4+c^4}+\sqrt{a^4+d^4}+\sqrt{b^4+c^4}+\sqrt{b^4+d^4}\leq 2+a^4+b^4+c^4+d^4$

By AM-GM inequality, $\sqrt{a^4+c^4}\leq\frac{1}{2}[1+(a^4+c^4)]$.
Summing up similar inequalities gives the desired result. :)
Equality holds iff $a=b=c=d=\sqrt[4]{\frac{1}{2}}$