quadratics
number theory solved
number theory

Let $n$ be a positive integer. Prove that if $3n+1$ and $4n+1$ be perfect squares, then $n$ is divisible by $56.$

It's easy to check that 8|n. So 7|n remains.
Probably this can be done by considering quadratic residues,
though I don't see immediately a short solution. It comes down to prove that the cases
(1) x = 0, y = 2 (mod 7)
(2) x = 2, y = 0 (mod 7)
cannot occur. Probably I'm overlooking something obvious again :?

Here?s what I did: 3n+1 is a perfect square so it must be (3k+1)^2 or (3k-1)^2.

Let?s take the case 3n+1=(3k+1)^2. Then n=3k^2+2k so 4n+1= 12k^2+8k+1= 16k^2+8k+1-4k^2= (4k+1)^2-(2k)^2=a^2 because 4n+1 is a perfect square. Then we get a^2+(2k)^2=(4k+1)^2. Since (2k, 4k+1)=1 we get that (a, 2k)=1 so from Pythagoras? thm we find a=2pq and 2k=p^2-q^2 or a=p^2-q^2 and 2k=2pq.

In both cases we have 4k+1=p^2+q^2. In the first case 2p^2=6k+1 which is odd, so false, so we are in the second case: a=p^2-q^2, 2k=2pq and 4k+1=p^2+q^2. This means that p^2+q^2-1=4pq, so p^4+q^4+1-2p^2-2q^2=14p^2q^2=0 (mod 7), so (p^2-1)^2+(q^2-1)^2=1 (mod 7). Let p^2-1=s and q^2-1=t. Then we have s^2+t^2=1 (mod 7). This can only happen if s=0 (7) and t=1 or ?1 (7) (or the other way around) because a perfect square can only be 0, 1, 2, or 4 (7) and the only way to get 1 (7) by adding two of thee numbers is 0+1. Let?s assume s=0 and t=1 or ?1 (7). Then we get 4n+1=a^2=(p^2-q^2)^2=(s-t)^2=1 (7). This means that 4n+1= 1 (7), so n= 0 (7).

For the case 3n+1=(3k-1)^2 we do the same calculations and the only difference is that we get p^2+q^2+1=4pq instead of p^2+q^2-1=4pq. But the following thing was proved on this forum (twice I think): if pq|p^2+q^2+1 then (p^2+q^2+1)/pq=3, so it can?t be 4, so we have a contradiction in this case.

Here I proved that 7|n. Like Arne said, for 8|n it?s enough to see what happens if n=1, 2, .., 7 (8). In those cases at least one of the numbers 3n+1 and 4n+1 will not be a perfect square, since squares can only be 0, 4, 1 (8).


[quote="nobody1"]Calculate:
\[\lim_{x\to+\infty}\left[x\left(\sqrt{x^{2}+2x}-\sqrt{x^{2}+x}+x\right)\right] \]
[/quote]
I think You mean
[quote="nobody1"]Calculate:
\[\lim_{x\to+\infty}\left[x\left(\sqrt{x^{2}+2x}-\sqrt{x^{2}+x}\right)\right] \]
[/quote]
Answer is $\frac{1}{2}$

Extremal, your solution for your revised problem is wrong. Look at the x term outside the parentheses. This thing isn't even bounded. Well, both sequences are not bounded.

I mean this
[/quote]
I think You mean
[quote="nobody1"]Calculate:
\[\lim_{x\to+\infty}\left[\left(\sqrt{x^{2}+2x}-\sqrt{x^{2}+x}\right)\right] \]
[/quote]
first case is obviusly limit equal $+oo$


Let $ a,b,c,d$ be positive real numbers, such that $ a\plus{}b\plus{}c\plus{}d\equal{}4$.
Prove that the following inequality holds

$ \frac{1}{3a^4\plus{}1}\plus{}\frac{1}{3b^4\plus{}1}\plus{}\frac{1}{3c^4\plus{}1}\plus{}\frac{1}{3d^4\plus{}1}\geq 1$

[quote="marin.bancos"]Let $ a,b,c,d$ be positive real numbers, such that $ a \plus{} b \plus{} c \plus{} d \equal{} 4$.
Prove that the following inequality holds

$ \frac {1}{3a^4 \plus{} 1} \plus{} \frac {1}{3b^4 \plus{} 1} \plus{} \frac {1}{3c^4 \plus{} 1} \plus{} \frac {1}{3d^4 \plus{} 1}\geq 1$[/quote]

We can easily prove Acording to Am-Gm inequality : $ \frac {1}{3a^4 \plus{} 1} \ge 1 \minus{} \frac {3}{4}a$ ...

and the conclusion follows  :P


[img]http://webwork2.math.utah.edu/webwork_system_html/tmp/equations/28/6ea204870c5ad3d2686aec05d6662f1.png[/img]

X_1=______

X_2=_____

X_3=____

X_4=____

Apparently x has 4 answers but i dont know how to get them. I understand how to do these problems when there is just x^2 but not theses. help please

[hide]
This problem can be expressed as $(x^{2})^{2}-10x^{2}+9$. Now it is like a quadratic equation. It can be factored into $(x^{2}-1)(x^{2}-9)$ Thus $x = 1,-1, 3,-3$[/hide]


linear algebra
matrix
algebra
polynomial
linear algebra solved

$A\in M_4(Q)$ such that $det(\sqrt{2}A+(1+i)I_4)=0$

Prove that $det(A^2-I_4)=4$

I have already solved this one on the romanian forum.

I don't understand all romanian word
could you post here in english

It is very simple: it is clear that $ (-1-i)/( sqrt 2)$ is a proper value for the matrix and since the characteristic polynomial has rational coefficients, it follows that it must be 0 in the points $ (-1+i)/(sqrt 2), (1-i)/(sqrt2), (1+i)/sqrt (2)$ Thus these are all proper values of A. A simple calcul finishes the solution.


the web site for the contest is down (http://www.octmtournament.org/) and I just realized that the contest is this Saturday but I have absolutely no clue what time it is (help please) and I don't know where I might get some past contest problems (thank you so much this forum rocks)

Funny, I can access the site just fine.

[url]http://www.octmtournament.org/testsandresults.cfm[/url]


Prove that we cannot find consecutive factorials with first digits 1, 2, ... , 9.

hint

[hide]
We can write any integer y in scientific notation as y = n x 10^k, where 1 <= n < 10 is a real number and k is a positive integer[/hide]

[quote="mdk"]hint

[hide]
We can write any integer y in scientific notation as y = x x 10k, where 1 <= x < 10 is a real number and k is a positive integer[/hide][/quote]

$ x\cdot10k$ isn't scientific notation, unless you mean $ x\cdot10^{k}$...please use LaTeX.


inequalities
triangle inequality
inequalities proposed

Prove the inequality

$ \sqrt {a^2 \plus{} (1 \minus{} b)^2} \plus{} \sqrt { b^2 (1 \minus{} c)^2} \plus{} \sqrt {c^2 \plus{} (1 \minus{} a)^2} \ge \frac { 3 \sqrt {2}}{2}$

holds for arbitrary real numbers $ a,b,c$

Power Mean;$ \sqrt{\frac{a^2\plus{}(1\minus{}b)^2}{2}} \geq \frac{|a|\plus{}|1\minus{}b|}{2}$
\[ \therefore \sqrt {a^2 \plus{} (1 \minus{} b)^2} \plus{} \sqrt { b^2 (1 \minus{} c)^2} \plus{} \sqrt {c^2 \plus{} (1 \minus{} a)^2} \ge \frac { 3 \sqrt {2}}{2}\] by triangle inequality


quadratics
number theory unsolved
number theory

If $ (6\sqrt{6}\plus{}14)^{2n\plus{}1}\equal{}N$ and $ F$ be the fractional part of $ N$,show that $ NF\equal{}20^{2n\plus{}1}$

[quote="RAGHAV GROVER"]If $ (6\sqrt {6} \plus{} 14)^{2n \plus{} 1} \equal{} N$ and $ F$ be the fractional part of $ N$,show that $ NF \equal{} 20^{2n \plus{} 1}$[/quote]

Let the sequence $ a_n$ defined as :

$ a_0 \equal{} 2$
$ a_1 \equal{} 28$
$ a_{n \plus{} 2} \equal{} 28a_{n \plus{} 1} \plus{} 20a_n$   (*** [u][b]edited [/b][/u]: $ 20\to 20a_n$. Thanks, Albanian eagle ! ***)

$ a_n$ is always an integer and, since $ 6\sqrt 6 \plus{} 14$ and $ 14 \minus{} 6\sqrt 6$ are the two roots of the quadratic $ X^2 \minus{} 28X \minus{} 20 \equal{} 0$, the general value is $ a_n \equal{} (6\sqrt 6 \plus{} 14)^n \plus{} (14 \minus{} 6\sqrt 6)^n$

So $ (6\sqrt 6 \plus{} 14)^{2n \plus{} 1} \equal{} a_{2n \plus{} 1} \plus{} (6\sqrt 6 \minus{} 14)^{2n \plus{} 1 }$

And, since $ a_{2n \plus{} 1}$ is an integer and $ (6\sqrt 6 \minus{} 14)^{2n \plus{} 1 }\in(0,1)$, we get that $ \{(6\sqrt 6 \plus{} 14)^{2n \plus{} 1}\} \equal{} (6\sqrt 6 \minus{} 14)^{2n \plus{} 1 }$

and so $ (6\sqrt 6 \plus{} 14)^{2n \plus{} 1}\{(6\sqrt 6 \plus{} 14)^{2n \plus{} 1}\} \equal{} (6\sqrt 6 \plus{} 14)^{2n \plus{} 1}(6\sqrt 6 \minus{} 14)^{2n \plus{} 1 }$ $ \equal{} 20^{2n \plus{} 1}$

Q.E.D.

Typo: You mean $ a_{n\plus{}2}\equal{}28a_{n\plus{}1}\plus{}20a_n$ right?

[quote="Albanian Eagle"]Typo: You mean $ a_{n \plus{} 2} \equal{} 28a_{n \plus{} 1} \plus{} 20a_n$ right?[/quote]

ooops, you're right.
Edited, thanks


function
geometry
algebra solved
algebra

Functions f and g are defined on the whole real line and are mutually inverse: g(f(x))=x, f(g(y))=y for all x, y. It is known that f can be written as a sum of periodic and linear functions: f(x)=kx+h(x) for some number k and a periodic function h(x). Show that g can also be written as a sum of periodic and linear functions. (A functions h(x) is called periodic if there exists a non-zero number d such that h(x+d)=h(x) for any x.)

This problem was solved before (by me :) ). Author of topic was Moubinool.

[b]CORRECTED VERSION OF THE POST![/b]

A related problem is problem 1 of the Bundeswettbewerb Mathematik 2003, 2nd round (actually this was the easiest problem of this 2nd round; I had solved it even before I did the geometry problem):

[i]Let $f:\mathbb{R}\to\mathbb{R}$ be a function whose plot has at least two centers of symmetry.
Prove that the function f can be written as the sum of a linear and a periodic function.[/i]

Once you have solved this problem, the Tournament of Towns problem becomes clear to you.

  Darij

???

BTW, can you find my post about this problem? ;)

[quote="darij grinberg"]A related problem is problem 1 of the Bundeswettbewerb Mathematik 2003, 2nd round (actually this was the easiest problem of this 2nd round; I had solved it even before I did the geometry problem):

[i]Let $f:\mathbb{R}\to\mathbb{R}$ be a function whose plot has at least one center of symmetry.
Prove that the function f can be written as the sum of a linear and a periodic function.[/i]

Of course, this problem makes the idea of the Tournament of Towns problem clear.
  Darij[/quote]
But it is impossible to write $x^3$ as sum of linear and periodic function.

Sorry for writing nonsense. I've now corrected my post above.

  Darij

[quote="darij grinberg"][b]CORRECTED VERSION OF THE POST![/b]

A related problem is problem 1 of the Bundeswettbewerb Mathematik 2003, 2nd round (actually this was the easiest problem of this 2nd round; I had solved it even before I did the geometry problem):

[i]Let $f:\mathbb{R}\to\mathbb{R}$ be a function whose plot has at least two centers of symmetry.
Prove that the function f can be written as the sum of a linear and a periodic function.[/i][/quote]
It is just obvious.


geometry
perimeter
circumcircle
trigonometry
geometry proposed

Three points $P,\ Q,\ R$ move on the perimeter of the regular octagon $A_{1}A_{2}\cdots A_{8}$ of side-length $1.$

(1) Find the maximum value of the area of $\triangle{PQR}.$
(2) If $Q$ coincides with the vertex $A_{1}$ of the regular octagon and $\angle{PQR}=90^\circ,$ then find the maximum value of the area of $\triangle{PQR}.$

1) Assume that the maximum area requires vertex $ P$ not to be on a vertex of the octagon.  Labeling wlog the vertices of the octagon so that $ P$ is on the segment $ A_1A_2$, it is trivial that $ QR$ is parallel to $ A_1A_2$, since otherwise the distance from either $ A_1$ or $ A_2$ to side $ QR$ would be larger than from any other point on segment $ A_1A_2$.  Therefore, the maximum area can also be achieved with $ P$ on a vertex of the octagon.  Proceed similarly with $ Q$ and $ R$ to find that the maximum area can be achieved with $ P,Q,R$ all vertices of the octagon, which we will assume henceforth.

Assume that $ PQR$ is contained within a quadrilateral $ A_1A_2A_3A_4$.  Then, its area is at most the area of $ A_1A_2A_4$.  But the distance from $ A_2$ to $ A_1A_4$ is smaller than the distance from $ A_7$ to $ A_1A_4$, and $ PQR$ needs to extend at both sides of any diameter $ A_iA_{4 \plus{} i}$, or have one side equal to such a diameter.

Call $ r$ the radius of the circumcircle of the octagon, then theorem of the cosine yields
\[ 1 \equal{} 2r^2 \minus{} 2r^2\cos\frac {\pi}{4} \equal{} r^2\left(2 \minus{} \sqrt {2}\right).
\]
Call now $ S_2,S_3,S_4$ the areas of triangle $ A_1A_2A_3$, quadrilateral $ A_1A_2A_3A_4$ and pentagon $ A_1A_2A_3A_4A_5$.  It is trivial to find that
\[ S_2 \equal{} \frac {\sin\frac {\pi}{4}}{2} \equal{} \frac {\sqrt {2}}{4},
\]

\[ S_3 \equal{} 3\frac {r^2\sin\frac {\pi}{4}}{2} \minus{} \frac {r^2\sin\frac {3\pi}{4}}{2} \equal{} r^2\sin\frac {\pi}{4} \equal{} \frac {\sqrt {2} \plus{} 1}{2},
\]

\[ S_4 \equal{} 4\frac {r^2\sin\frac {\pi}{4}}{2} \equal{} \sqrt {2}r^2 \equal{} \sqrt {2} \plus{} 1.
\]
The area of $ PQR$ is then the area of the octagon minus $ S_i \plus{} S_j \plus{} S_j$, where $ i \plus{} j \plus{} k \equal{} 8$, none of $ i,j,k$ is larger than 4, and $ S_1$ is by definition the area of segment $ A_1A_2$, ie 0.  We need therefore to minimize, under these constraints, $ S_i \plus{} S_j \plus{} S_j$.  Since $ S_4 \equal{} 2S_3$, it is obvious that $ i \equal{} j \equal{} 3$, $ k \equal{} 2$ will yield a better result than any combination of $ i \equal{} 4$, $ j \plus{} k \equal{} 4$.  Or, all of $ i,j,k$ are at most 3, and their sum is 8, so two of them are 3, and the maximum area is the the total area of the octagon $ 2S_4$ minus the area $ 2S_3 \plus{} S_2$, ie, the maximum area is $ S_4 \minus{} S_2 \equal{} 1 \plus{} \frac {3\sqrt {2}}{4}$.

2) Since $ \angle PQR \equal{} \frac {\pi}{2}$, it is obvious that the area of $ PQR$ is $ \frac {PQ\cdot QR}{2}$, while $ PQ^2 \plus{} QR^2$ is at most the square of the diameter of the circumcircle of the octagon.  Using the AM-QM, $ PQ\cdot QR\leq 2r^2$, with equality iff $ PQ \equal{} QR$ and simultaneously $ PR$ is a diameter of the circumcircle of the octagon.  This is achieved when $ P \equal{} A_3$, $ R \equal{} A_7$, for a maximum area of $ S_4 \minus{} 2S_2 \equal{} 1 \plus{} \frac {\sqrt {2}}{2}$, and we are done.


Ahmed and Beth have respectively $p$ and $q$ marbles, with $p>q$. Starting with Ahmed, each in turn gives to the other as many marbles as the other already possesses. It is found that after $2n$ such transfers, Ahmed has $q$ marbles and Beth has $p$
marbles.
Find
$\frac{p}{q}$
in terms of $n$.

I consider stages in the process. obviously, the sum of the no of marbles is an invariant( I haven't though of its possible use as yet)

The first stage :- A has p-q marbles and B has 2q marbles.
2nd stage :- A 2p-2q , B 3q-p
3rd stage - A 3p-5q , B 6q-2p

and so on

2Kth stage :-A  $a_{2k} p -\ (\ a_{2k+1} -1)\ q $ , B  $ (\ a_{2k+1} q -( a_{2k} -1)\ p $

2K+1 th stage :-A  $a_{2k+1} p -\ (\ a_{2k+2} -1)\ q $ , B  $ (\ a_{2k+2} q -( a_{2k+1} -1)\ p $

where $ a_{2k} =\ 2a_{2k-1} $ and $ a_{2k+1} =\ 2a_{2k} -1$

The first few terms of $a_{n}$ are 1,2,3,6,11,22,43,86,171,342,...

I am right now trying to figure out $a_{2k}$ and $a_{2k+1}$ .But as you can clearly see, once figured out , the problem reduces to simple manipulation to get required answer.

Hope I'll be able to figure out the series soon. Post if anyone else can with the recurrence relation I have given . i do not know (have no experience) how to solve recureence relations for even and odd terms. Probably that's the toughest part of the problem :)

My each stage corresponds to each transfer.

Thus I need to find $ a_{2n} \ and\ \ a_{2n+1} $ to solve the problem.

well, that is the major problem, this is an olympiad problem. How long will it take to find the series! :wallbash_red:  :wallbash:  :wallbash_red:  :wallbash:
[u][b]Rushil[/b][/u], I tried to split the series into $2$, one with a increasing value and one with a decreasing value,
for eg.  series is... $5,2,6,1,7,0$
so splitting it into $2$ series like $5,6,7$ and $2,1,0$ can you do that?

Ahh.. I got it!!! Yes, it made use of splitting the series but not in the way you said. I considered the even and odd terms seperately.

[u]For the EVEN terms:-[/u]

The even terms are 2, 6, 22, 86, 342, 1366, ..... You will find that the difference between consecutive terms is
4, 16, 64,256,.... i.e. powers of 4. Now such a series can be easily summed do as to find the Nth term by the Method of Differences(a standard method. If you don't know , ican tell you later).
[u]For the ODD TERMS:-[/u]
Similarly, the difference between consecutive odd terms is 2 ,8, 32, 128,512,... .e. odd powers of 2. Again, by the method of differences , we can find Nth terms. (I am not including details sice they are a bit tedious - I am sure you can figure them out)

In this way, $ a_{2k} = \frac{( 4^k +\ 2 )}{3} $ and $ a_{2k+1} = \frac{(2 \times  4^k +\ 1)}{3} $ ( I may have got the computation wrong but the method is basically this)

Thus using that $ a_{2n}p -\ (a_{2n+1} - 1)q =\ q $ (as I have proved before)

we get $ \frac{p}{q} =\ \frac{(2 \times 4^n +1 )}{( 4^n +2)} $ again there might be a calculation error for n might have to be replaced by n-1 etc. So in the end, the problem didn't turn out to be that much  :wallbash_red:

Hope this is correct. [b]Which olympiad is this question from??[/b]
[b]INDIA ROCKS!!!![/b]

I doubt this should appear in getting started  :P  . Anyway i have an idea about looking the question backward . Instead of starting to find the sequence from beginning to end , maybe we can start from the back . Well , I cant figure out how to do that but maybe you guys out there would know :D .So what do you think ?

[quote="Rushil"]I really don't understand what point you are trying to make. Can you clarify?[/quote]

What I mean is , maybe we can just start with Ahmed ($q$ marbles) and Beth ($p$ marbles) then working backward to Ahmed ($p$ marbles) and Beth($q$ marbles)

By the way riddler , which year is this BMO ?

[quote="shyong"]By the way riddler , which year is this BMO ?[/quote]
2001 :D

[quote="riddler"][b][u]Rushil[/u][/b] how did you guess the values of the sequence??? It is not given in the question! :huh:
can you explain how you got the answer properly, what you are taking into consideration.... etc. It is from the BMO.[/quote]

I didn't guess these values, I calculate values for some of the starting terms. Till n=7 and till the limit of my patience, I calculated some terms and looked for patterns. Once I got a possible conjecture , I established it via the Mathematical Induction and using the recurrence relations I had derived above.I believe this is a standard procedure in doing such problems - see some values, form a conjecture and prove it.

[b]Is my answer correct??!![/b]

Do you want me to post the entire solution or which part. It could take me some time!!!1

sorry  :rotfl:  :rotfl:  :oops:  :rotfl:  :rotfl:  but I dont have the solutions.

For an introduction to method of differences see this site

[url=http://www.madras.fife.sch.uk/maths/spiral2.html]www.madras.fife.sch.uk/maths/spiral2.html[/url]

Another one is [url=http://mathworld.wolfram.com/FiniteDifference.html]Finite differnece[/url]

[quote="Rushil"]can you tell me the site you got these questions from[/quote]
look here for the questions, but as I said I dont have the solutions.
[url]http://www.bmoc.maths.org/home/bmo2-2001.pdf[/url]

Rushil, learn to edit your posts :!:  You double posted, then double posted, then quadruple posted!!!


Find all triples (a,b,c) of positive integers such that for every positive integer n
a^n + b^n = c^(n+1)

$ a\equal{}x(x^n\plus{}y^n),b\equal{}y(x^n\plus{}y^n),c\equal{}x^n\plus{}y^n$ is solution for any $ x,y\in Z$.

$ a\equal{}b\equal{}c\equal{}2$ is solution.

for $ n \equal{} 1 ,a \plus{} b \equal{} c^2$ , for $ n \equal{} 2 : a^2 \plus{} b^2 \equal{} c^3$ and for $ n \equal{} 3$ we have $ a^3 \plus{} b^3 \equal{} c^4$ so $ c^2(c^3 \minus{} ab) \equal{} c^4$
and $ ab \equal{} \frac {1}{2}(c^4 \minus{} c^3)$ so $ c \equal{} {0,1,2}$ if  $ c \equal{} 0$ then $ a \equal{} b \equal{} 0,$ if $ c \equal{} 1$ then $ {a,b} \equal{} {0,1}$ if $ c \equal{} 2$ we have $ a \equal{} b \equal{} 2$ .

Not necessary.!
Use :
$ 0<a<1$ then $ \lim_{n\to\infty}a^n\equal{}0$
$ a>1$ then $ \lim_{n\to\infty}a^n\equal{}\plus{}\infty$

Let $ A_n\equal{}a^n\plus{}b^n$, then $ A_{n\plus{}2}\equal{}(a\plus{}b)A_{n\plus{}1}\minus{}abA_n$. Therefore $ C^{n\plus{}3}\minus{}(a\plus{}b)c^{n\plus{}2}\plus{}abc^{n\plus{}1}\equal{}0$ or $ (c\minus{}a)(c\minus{}b)\equal{}0$.
If c=b, then $ b^n\equal{}a^n(a\minus{}1)\to a\equal{}b\to a\equal{}b\equal{}c\equal{}2$.

[quote="Rust"]Let $ A_n \equal{} a^n \plus{} b^n$, then $ A_{n \plus{} 2} \equal{} (a \plus{} b)A_{n \plus{} 1} \minus{} abA_n$. Therefore $ C^{n \plus{} 3} \minus{} (a \plus{} b)c^{n \plus{} 2} \plus{} abc^{n \plus{} 1} \equal{} 0$ or $ (c \minus{} a)(c \minus{} b) \equal{} 0$.
If c=b, then $ b^n \equal{} a^n(a \minus{} 1)\to a \equal{} b\to a \equal{} b \equal{} c \equal{} 2$.[/quote]
Oh yes I know that . My solution and your is choose for $ a,b,c\in R$ .
But it can be solve only condition for $ n\equal{}1,2,3$

For a=b=c=2, we get l.h.s=2^n+2^n=2*2^n=2^(n+1)=R.H.S
Hence it is satisfied for all real n  :lol:


Find all continuous functions $f: \mathbb{R}\to\mathbb{R}$ such that \[f(2x-y)=2f(x)-f(y)\;\forall x,y \in\mathbb{R}.\]

A general problem: $f(\alpha x+\beta y)=\alpha f(x)+\beta f(y)$

[quote="pohoatza"]denote $g(x)=f(x)-f(0)$, therefore $2g(x)=g(2x-y)+g(y)$  $(1)$.
now by taking $y=0$ in $(1)$, we have $2g(x)=g(2x)$, and taking $x=0$, we have $g(x)=-g(-x)$.

therefore we can write $(1)$ as $g(2x)=g(2x-y)+g(y)$, and let $t=2x-y$, therefore $g(t+y)=g(t)+g(y)$, but $g$, is continous, therefore $g(x)=ax$, where $a=g(1)$, therefore $f(x)=ax+b$, where $b=f(0)$[/quote]
$f(0)=0$.

it really doesn't matter, but i think you are wrong.

:blush: ah sorry i did forget the qestion .


limit
trigonometry
logarithms
calculus
calculus computations

Suppose that $ \lim_{x\to \infty}{a_{n+1}- a_n = 0}$, and $ a_n$ is bounded. Must $ a_n$ converge?

No.

Let $ a_n \equal{} \sin\left(\sqrt {n}\right).$

Or, if you take your metric space to be $ M\equal{}(0,1]$, let $ a_{n}\equal{}1/n$.

This is random but how do you get the millennium prize names next to your name?

[quote="ejr322"]Ok, thanks. I figured the counterexample would be something like $ a_n \equal{} sin(f(n))$.[/quote]
You just have to make sure that $ f(x)\to\infty$ while $ f'(x)\to 0.$ I used $ \sqrt {x},$ but I could also have used $ \ln x.$ There are many choices.

JRav: I'm assuming the context here was $ \mathbb{R}$ (or $ \mathbb{C}.$) Otherwise, the word "bounded" doesn't make a lot of sense.


Dear all...

Can any one suggest good books for pre-calculus, AP calculas AB and BC

plz send the titles along with the name of the publishers and also the cost of the book if you know it (approximately is also ok :P )

I'm also interested in gaining knowledge from the best pre-calc books...

I also have to learn precalc over the summer, assuming I meet the requirements to take the test to test out of precalc.

The calculus book I used to self-teach myself is: [u]Calculus and Analytic Geometry[/u] by Thomas and Finney. It is great! And Pre-cal umm who cares its hardly needed. :rotfl:


Let $ x,y,d \in N$ such that:$ (\frac{x}{y}\plus{}\frac{y}{x}\plus{}\frac{x\plus{}y}{dxy})$ $ \in N$
Prove that $ d \le x\plus{}y$

[hide]the given expression equals $ \frac {dx^2 \plus{} dy^2 \plus{} x \plus{} y}{dxy}$, so $ d$ divides the numerator, since it divides the denominator. then $ d$ must divide $ x \plus{} y$ and thus $ d\leq x \plus{} y$[/hide]


Yep, guess who. :P

[hide="Hint"]I wonder who wrote the first comment on the video.[/hide]

Blah no guesses here
Lol he's craaazy
The end is worth millions: that using derivatives he proved that top and bottom are extremes

HMM I WONDER WHO THAT MIGHT BE

Oh, and nikhil: a torus is a donut.  Or a bagel.

[quote="math154"]I know right? It helps cover up his derivative bashing error. :rotfl:  :rotfl:  :rotfl:[/quote]

What all errors were clearly intentional. Also where was this error.

Well obviously the top and the bottom are not the extremes, so there MUST have been an error somewhere.

Why do I have to say everything once in Gmail and then again elsewhere...

[hide]The equation of a torus is $ f(x,y)^2 \plus{} (R \minus{} \sqrt {x^2 \plus{} y^2})^2 \equal{} r^2$

$ \implies\frac\partial{\partial x}f(x,y)^2 \plus{} \frac\partial{\partial x}R^2 \minus{} 2R\frac\partial{\partial x}(x^2 \plus{} y^2)^{\frac12} \plus{} \frac\partial{\partial x}(x^2 \plus{} y^2) \equal{} \frac\partial{\partial x}r^2$

$ \implies2f(x,y)\frac {\partial f(x,y)}{\partial x} \plus{} 0 \minus{} 2R\left(\frac12(x^2 \plus{} y^2)^{ \minus{} \frac12}\frac\partial{\partial x}(x^2 \plus{} y^2)\right) \plus{} 2x \equal{} 0$

$ \implies2f(x,y)\frac {\partial f(x,y)}{\partial x} \equal{} 2R\left(\frac12(x^2 \plus{} y^2)^{ \minus{} \frac12}(2x)\right) \minus{} 2x \equal{} 2x(R(x^2 \plus{} y^2)^{ \minus{} \frac12} \minus{} 1)$

$ \implies\frac {\partial f(x,y)}{\partial x} \equal{} \frac {x(R(x^2 \plus{} y^2)^{ \minus{} \frac12} \minus{} 1)}{f(x,y)} \equal{} \frac {x(R(x^2 \plus{} y^2)^{ \minus{} \frac12} \minus{} 1)}{\pm\sqrt {r^2 \minus{} (R \minus{} \sqrt {x^2 \plus{} y^2})^2}}$

$ \equal{} \boxed{\pm\frac {x(R \minus{} \sqrt {x^2 \plus{} y^2})}{\sqrt {(x^2 \plus{} y^2)(r^2 \minus{} (R \minus{} \sqrt {x^2 \plus{} y^2})^2)}}}.$

Yay![/hide]

If you had gotten something completely wack as your extrema the video would get more views.

[quote="Nlakkamraju"]Is he seriously an 9th grader?
What is a torus? I still don't get it?
What is up with that huge set of equations?
I don't know what more to say.[/quote]

I HAVE TO SAY YOUR AVATAR IS BEASTLY.

[quote="pianoforte"]OH GOD SO HOTTTTTTT!!!! <333[/quote]

OMG IKR? WE SHOULD TOTALLY LIKE HAVE FAN CLUB MEETINGS <333

edit: this will probably be rated as spam but I also have to say I love the fact that mewto and piano's comments are the only ones not rated as spam =P

[quote="Truffles"]

[quote="pianoforte"]OH GOD SO HOTTTTTTT!!!! <333[/quote]

OMG IKR? WE SHOULD TOTALLY LIKE HAVE FAN CLUB MEETINGS <333

[/quote]

QFT.

That was the awesomest wrap-up I have ever seen.

Sup guys when you are done watching my calculus mastery you can see 2 famous Missouri aops users take down a troll right [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&t=360898&start=0]here.[/url]

Also as a mod I probably should not be endorsing that, so let that be a lesson of what **SHOULD NOT** happen in the Missouri forum.

1. Only 1 Missourian AoPS user is successful.
2. I believe the proper term for the target is "attempted troll," commonly referred to as "troll-bait."
3. Yes, clearly that should [color=red]not[/color] happen in the Missouri forum.

Mewt[color=#FF0000]w[/color]o55555 is [color=#FF0000]beast and awesome[/color].

[quote="ksun48"]Mewt[color=#FF0000]w[/color]o55555 is nonexistent.[/quote]
QFT.

[quote="Mewto55555"]Actually I think the above might be good video material, if we had people say the posts instead of making them.

I could play me.[/quote]

Dude I'm sure Arrange your Tan will take part if you ask him.

I'm too lazy to click... :) Whatsit about?


Let $ A \equal{} 11^{10} \minus{} 1$. Prove $ A \vdots 600$.

[quote="stevenmeow"]well $ 11^2 \equal{} 1 \plus{} 120$
and $ 11^{10} \equal{} 120^5 \plus{} 5 \times 120^4 \plus{} 10 \times 120^3 \plus{} 10 \times 120^2 \plus{} 5 \times 120 \plus{} 1$
so $ 11^{10} \minus{} 1 \equal{} 120^5 \plus{} 5 \times 120^4 \plus{} 10 \times 120^3 \plus{} 10 \times 120^2 \plus{} 5 \times 120$
and every term is divisible by 600[/quote]

Thanks. You are good! :lol:

$ x^{10} \minus{} 1 \equal{} (x \minus{} 1) (x \plus{} 1) (x^4 \minus{} x^3 \plus{} x^2 \minus{} x \plus{} 1)(x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1)$

[u]For $ x \equal{} 11$:[/u]

$ \blacktriangleright \; \; (x \minus{} 1)(x \plus{} 1) \equal{} 120$

$ \blacktriangleright \; \; x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1 \equiv 1 \plus{} 1 \plus{} 1 \plus{} 1 \plus{} 1 \equiv 0 \pmod 5$

Therefore, collectively it is divisible by $ 5 \cdot 120 \equal{} 600$


calculus
integration
trigonometry
calculus computations

Find the positive value of $ k$ for which $ \int_0^{\frac {\pi}{2}} |\cos x \minus{} kx|\ dx$ is minimized.

For some $ a\in\left(0,\frac {\pi}{2}\right):$
$ \int_0^{\frac {\pi}{2}} |\cos x \minus{} kx|\ dx \equal{} \int_0^{a} (\cos x \minus{} kx) dx \plus{} \int_a^{\frac {\pi}{2}} (kx \minus{} \cos x) dx \equal{} \sin a \minus{} k\frac {a^2}{2} \plus{} k\frac {\pi^2}{8} \minus{} 1 \minus{} k\frac {a^2}{2} \plus{} \sin a \equal{} 2\sin a \minus{} k a^2 \plus{} k\frac {\pi^2}{8} \minus{} 1$

We know that $ \cos a \equal{} k a$, so $ k \equal{} \frac {\cos a}{a}:$

$ f(a) \equal{} 2\sin a \minus{} a\cos a \plus{} \frac {\pi^2}{8}\cdot\frac {\cos a}{a} \minus{} 1$

Now it should be minimized... :huh:

$ f'(a) \equal{} \cos a \plus{} a\sin a \minus{} \frac {\pi^2}{8a^2}(\cos a \plus{} a\sin a) \equal{} (\cos a \plus{} a\sin a)(1 \minus{} \frac {\pi^2}{8a^2})$  :w00t:

This gives $ a \equal{} \frac {\pi}{2\sqrt {2}}$. Our $ k$ will be $ k \equal{} \frac {2\sqrt {2}}{\pi}\cos\left(\frac {\pi}{2\sqrt {2}}\right)$.


Claudia wants to use 8 indistinguishable red beads and 32 indistinguishable blue beads to make a necklace such that there are at least 2 blue beads between any 2 red beads. In how many ways can she do this?

[hide="I hope this is right"]We start out by inserting the required beads.  First insert the red beads

R R R R R R R R

and then stick two blue beads between each red bead.

RBB RBB RBB RBB RBB RBB RBB RBB

This takes care of all the red beads and $16$ of the blue beads.  Now all we have to do is see how many ways there are to distribute the rest of the blue beads among the eight "spaces" we have created.

Each of the $16$ remaining blue beads can go in either one of the eight spaces, so there are $8^{16}$ ways to do this.  However, since the necklace is circular we have to divide this number of ways by $8$ (since RBBBRBB is really the same as RBBRBBB).  So I think the answer is $8^{15}$.

[/hide]

Good start - don't forget to account for cases with rotational symmetry!


function
analytic geometry
real analysis
real analysis unsolved

I really don't manage to make some progress on this one:
  Find the continuous functions $ f:[0,1]\to[0,1]$ such that for all $x\in [0,1]$, the set of numbers $ x, f(x), f(f(x)),...$ is finite.

If there is an $x$ with period $3$, the iteration is "chaotic" and there are periodic points of every order.   So I guess that the answer is: modulo a change of coordinates (conjugation by continuous increasing function), the only solutions are $f(x) = x$ and $f(x) = 1 - x$.   This is, of course, very far from being a proof.

Edit: the iteration is also chaotic if there is an $x$ with odd period, and the idea of the guess above is that this would imply existence of an aperiodic point (so that there is no solution of this type).  This leaves the possibility of all points having purely even period $2^k$.


Show that, if $\sqrt{2}$ is written in decimal notation, there is at least one nonzero digit at the interval of 1,000,000-th and 3,000,000-th digits.

[quote="Rust"]By theory continuosly fractions may be proved one of digits $c_n,c_{n+1},\dots,c_{2n}$ is not equal 0.[/quote]

Can you give some hint about it?

[quote=MNJ2357]If $$\{10^{1000000} \sqrt{2}\}<10^{-2999999}$$ where $\{x\}$ is the decimal part of $x$, There exist a natural number $m$ such that
$$m<10^{1000000}\sqrt{2}<m+10^{-2999999} \Longrightarrow m^2<2 \cdot 10^{2000000}<m^2+1$$ Contradiction!![/quote]

Actually all the digits at the interval of 1,000,000-th and 3,000,000-th digits are zeros iff $$\{10^{1000000} \sqrt{2}\}<10^{-1999999}.$$


Corners are sliced off a unit cube so that the six faces each become regular octagons. What is the total volume of the removed tetrahedra?

$ \textbf{(A)}\ \frac {5\sqrt {2} \minus{} 7}{3}\qquad \textbf{(B)}\ \frac {10 \minus{} 7\sqrt {2}}{3}\qquad \textbf{(C)}\ \frac {3 \minus{} 2\sqrt {2}}{3}\qquad \textbf{(D)}\ \frac {8\sqrt {2} \minus{} 11}{3}\qquad \textbf{(E)}\ \frac {6 \minus{} 4\sqrt {2}}{3}$

[hide]
Assume each side of the octagon has a length of s. Then,
s+2*s*sqrt(2)/2 = 1
By simplification,
s*(1+sqrt(2)) = 1
By solving for s, we get,
s = sqrt(2) - 1
Since each tetrahedron has a height of s*sqrt(2)/2 and a base consisting of a right triangle whose lengths are the same, we need to solve for s*sqrt(2)/2. Doing this, we get,
s*sqrt(2)/2 = (2-sqrt(2))/2
Finally, we must solve for the volume of each tetrahedron. Since the general formula is 1/3*base*height and the sides in these specific tetrahedra are at right angles to each other, we get
V = 1/3*(2-sqrt(2))/2*1/2*((2-sqrt(2))/2)^2
   = 1/6*((2-sqrt(2))/2)^3
   = (10-7*sqrt(2))/24
Since there are 8 such tetrahedra (one for each corner), we multiply this value and get
V = [b](10-7*sqrt(2))/3[/b], or [b]B[/b]
[/hide]

[hide]Look at a side, if the octagon has side x then the corners have length 1-2x so $x\sqrt{2}=1-2x\rightarrow x=\frac{1}{\sqrt{2}+2}$
$1-2x=\frac{2\sqrt{2}+3}{2\sqrt{2}+4}$
The total volume is then $\frac{8(\frac{2\sqrt{2}+3}{2\sqrt{2}+4})^{3}}{2}=\frac{10-7\sqrt{2}}{3}$
(since 1 tetrahedron is half a square of side length $\frac{2\sqrt{2}+3}{2\sqrt{2}+4}$)[/hide]

[quote="Totally Zealous"]
The total volume is then $ \frac {8(\frac {2\sqrt {2} \plus{} 3}{2\sqrt {2} \plus{} 4})^{3}}{2} \equal{} \frac {10 \minus{} 7\sqrt {2}}{3}$
(since 1 tetrahedron is half a square of side length $ \frac {2\sqrt {2} \plus{} 3}{2\sqrt {2} \plus{} 4}$)[/quote]
This is wrong. The volume of each tetrahedron is not half that of a cube with the same side length. You have to do what echatterjee did, which is a bit of annoying work.


AMC
USA(J)MO
USAMO
geometry
articles
AIME
pigeonhole principle

I am a Junior in High School and recently took the AMC and AIME.  I got a 111.5 on the AMC and a 3 on the AIME.   I did not prepare for either of them at all. I was wondering if it is possible for me to make USAMO next year.  I know it will probably be a lot of work but I am up for it.  Also, is it harder for Seniors to make USAMO rather than Juniors.   Just want to know if it is possible.  Thanks for the help.

no, seniors are taken to the same standards as juniors in terms of USAMO qualification

of course its possible for you to make USAMO!

i was about the same as you last year(my junior year),
this year i got 150/10

so yeah, everything is possible :)

yeah, i think amir spent quite a lot of time on aops during the summer (i tried to do a few of his weekly contest and marathon problems) heh

also, have you looked into any summer camps?

awesome math is a specific one that is geared largely towards honing problem solving ability, along with other interesting areas of math

if you have any other questions about resources, strategies, be sure to read the articles section of aops, it has many good recommendations about books to read, and other topics

amirhtlusa - I was wondering if you could tell me what you did to raise your score.  Also congrats on making USAMO.  Nice Job!  Anything you could tell me about what helped you raise your score would be great.  Thanks!

Wait, but I can prove that it's false by contradiction! Suppose those were indeed 3 sure-fire steps to making USAMO. Then consider an arbitrary set of 4000 high schoolers who are physically able to perform those steps. We then must have 4000 USAMO qualifiers over 4 years, so by the pigeonhole principle, at least one year has at least 1000 qualifiers. But there are only ~375 spots! CONTRADICTORY CONTRADICTION!

hmm, I don't think the surefire steps are wrong; rather, the supposition that there exists 4000 such highschoolers are wrong. As I understand it, there really aren't that many who are committed enough to fully work through both AoPSes and practice relentlessly on old tests. In the future, of course, if AoPS or other means encourages more serious competitors, the sure-fire steps may not apply, but for the moment, they look pretty good.

[quote="MithsApprentice"]Three Sure-fire Steps for Making USAMO

1) Order AoPS 1 and 2.
2) Go through both books.
3) Practice every old AIME and AMC you can lay hands on.
[/quote]

Hey, that's what I did this year!

[quote="MithsApprentice"](barring some accident on test day or myriad careless errors).[/quote]

Oh, that explains it. :roll:

barce 11,I'm in the same boat as you.  gotta starting preparing for next year i guess...

The thing is a senior can't make mosp unless he's a usamo winner.  the maa's trying to train the "young blood" for next yr

[quote="Xantos C. Guin"][quote="MithsApprentice"]Three Sure-fire Steps for Making USAMO

1) Order AoPS 1 and 2.
2) Go through both books.
3) Practice every old AIME and AMC you can lay hands on.

After you're done, try taking this year's tests under normal testing conditions again. If you don't have an index that would qualify, we need to talk. With this type of qualifying index I'm confident that as long as you're willing to dedicate enough time to learning and practicing math, you'll make it next year (barring some accident on test day or myriad careless errors).[/quote]

Thanks, I should do that instead of being a lazy penguin.
I took the AMC10B and the AIMEI and got 146.5/7. Either I will barely make the floor or miss it by 1.[/quote]

or 2 :|


I am so upset My last two post is both be rated at "below average"!!!!
I don't think this kind of post should be rated！And this two post makes my rating frome 4.7 to 3.9.
If no one can deal with this problem ,I will not post on this section.
I think only some person can rate.
EDIT: And I found that maybe only one or two person who rated all the post here.
Becuase when I rate Tiks's post (which has been rated at spam) at Excellent(Just a check) ,Then it comes to "average" .

Yeah I know that's true. Like nearly all the posts in this forum are rated below average, many even rated as spam even though it is not spam at all...
But I usually don't really look at the ratings because who knows who is rating these posts, and I think in this topic it is quite hard to post a long and well-structured post.

Guys chill out. The post rating is dynamic. For now it is only on [b]testing[/b] purposes. In some time I will remove irrelevant forums from rating stats (for example this one), and all the users who abused it will be deprived of rating possibilities ;)

Anyway I'll be in Vietnam as well, so looking forward to meeting some of you guys there. Don't ask me about my role there, because at this moment, it might just be tourist :)

Thanks.And I think you should post this sentence"Anyway I'll be in Vietnam as well, so looking forward to meeting some of you guys there. Don't ask me about my role there, because at this moment, it might just be tourist "again at the orginal topic,orthewise somebody won't know you'll be there. :)

Well the ratings don't really do anything at the moment. All they do is tell you which posts are of good quality so when you're sifting through 10 pages of spammy and excelent posts, you will save time.

However, if there is an option of reorganizing posts by their rating numbers (ie view the very good ones first and ignore the last 3 pages of spam) then it could be useful.

[quote="Treething"]...

I can't believe anybody cares about rating this much. Contrary to popular belief, $\text{rating}^{2}$ does not equal mansausage length (in centimeters).[/quote]
You are right.Maybe I care the rating so much :D

[quote="7h3.D3m0n.117"]However, if there is an option of reorganizing posts by their rating numbers (ie view the very good ones first and ignore the last 3 pages of spam) then it could be useful.[/quote]There will be one very soon. For now we are gathering data to see how people generally vote, establish boundaries, and see who abuses the system.


Been doing these sheets of problems by UKMT

[url]http://www.mathcomp.leeds.ac.uk/Senior%20Mentoring%20QP.htm[/url]

Probably pretty easy for some of you guys but I can't do:

November Q 5 (c)
and December Q3 (can do second part but not first!)
and December Q4 (c)

Any help or solutions would be much appreciated! :lol:

Thanks

Are you allowed outside help on these mentoring sheets?

Yeah, just doing them for my own interest

Do you actually have a mentor? I am both an Advanced mentee and an Intermediate mentor. If you were to post in the British section of the forum, then there are likely to be more people likely to be answer your question. I shall try to give you some advice on these questions during the next 24 hours!
Regards

Thanks inquisitor,

No, I don't have an actual mentor, like I said I came across these sheets and am doing them without any feedback etc. That is why any help would be very useful.

for the question about two squares summing to 2003 - try writing the numbers a and b as 4m,4m+1,4m+2,4m+3 and then squaring them then you should see that any square is of the form 4n or 4n+1. (Hence two cant sum to number of the form 4k+3)

Yeah already did that, (looked at the equation mod 8, basically same thing) It was the part before that I am stuck on:

Find integers a, b and c such that

a*a + b*b + c*c = 2003

Thanks tho!

oops, sorry. For the inequality with w,x, y and z note that 1+x is greater than or equal to 2sqrt(x). Apply this inequality to each of the bracketed terms to get the desired result.

Thanks inquisitor, but do you mind elabourating on how you found those numbers.

I just used a calculator, and tried a few big values, then tried to fill in the other two!
Regards

Ah ok cool, I was thinking maybe there was an elegant approach etc. How about December Q4 (c) any ideas on this one, really bugging me!

Thanks


calculus
integration
parameterization
trigonometry
calculus computations

find
$ \int \ e^{acosx} cosbx \,dx$
a,b-parameter

integrate by parts...

This is a bit more complex than I first thought
$ \int e^{a \cos (x)} \cos (bx) dx \equal{} \frac{1}{b} e^{a \cos (x)} \sin (bx) \plus{} \int \frac{a}{b} \sin(x) \sin (bx) e^{a \cos(x)} dx$
$ \equal{} \frac{1}{b}e^{a \cos x} \sin (bx) \plus{} \int \frac{a}{2b} (\cos((b\minus{}1)x)\minus{}\cos((b\plus{}1)x))e^{a \cos (x)} dx$

we can rearrange to find that
$ \int \cos((b\plus{}1)x)e^{a \cos (x)} dx$
$ \equal{} \frac{2}{a}e^{a \cos x} \sin (bx) \plus{} \int e^{a \cos x}\cos((b\minus{}1)x) dx \minus{}  \frac{2b}{a}\int e^{a \cos(x)} \cos(bx) dx$

so if we can find the value of the integral for 2 values of b that differ by $ 1$, we can find the value for any $ b$ that differs from both those values by an integer.  It may even be possible to find a general formula by induction.

I'm stuck here.  If we try $ b\equal{}1$, we are integrating $ \int e^{a \cos x} \cos x dx$.  If we attempt the substitution $ u\equal{}\cos x$ we have $ \int ue^{au} dx$, but $ \minus{}\sin x dx \equal{} du$ and thus we have $ \int \minus{}\frac{ue^{au}}{\sqrt{1\minus{}u^2}} du$... I don't think this is solvable elementarily.


geometry
trapezoid
ratio
congruent triangles

In a trapezoid, the lengths of the bases are $ 4$ and $ 16$, and the lower base angles are $ 30^{o}$ and $ 60^{o}$. What is the distance between the midpoints of the two bases?

Easy!
Basically, we are trying to find the height of the trapezoid.  We drop altitudes from the endpoints of the shorter base to the longer base and set them equal to h.  We have split the base into segments of lengths hxsqrt 3, 4, and h/sqrt 3.  Solving for h, we get that h=3 sqrt 3. :D

i dont think your trying to find the height. i think its a lot more complex than that

[hide="what im pretty sure is the solution"]we have a trapezoid $ ABDE$. extend both legs until they intersect. that should get you a $ 30, 60, 90$ triangle $ ABC$. let the midpoints of the shorter and longer legs be Q and P respectively. we figure out the other sides are $ 8$ and $ 8\sqrt3$. draw a perpendicular down from the right angle and we have another $ 30, 60, 90 BOC$. we find $ CO$ is $ 4\sqrt3$ and $ BO$ and $ OA$ are 4 and 12 respectively. we also know $ O$ is the midpoint of $ BP$. therefore triangle $ BOC$ and $ COP$ are congruent and $ PQ$ lies on the hypotenuse of triangle $ COP$. let M be the intersection of $ DE$ and $ CO$. since triangle $ BOC$ and $ COP$ are congruent, $ M$ must be the midpoint of $ QD$. now we have, yet again, ANOTHER $ 30, 60, 90$ triangle $ CMQ$. we know $ QM=1$ so $ CQ=2$. $ PQ=8-CQ$ therefore $ PQ=6$.

answer $ 6$[/hide]

I like your setup, abacadaea, but there's an easier way.

You extended the legs to make a 30 60 90 triangle, which was nice. However, instead of going into congruent triangles and such, you can see that the ratio of the sides of the smaller triangle at top and the large triangle are 1:4. Also, the line that was drawn to connect the midpoints of the sides of the trapizoid will meet the large triangle at its right angle. This line thus becomes a median for that triangle, and we can immediately see that the length of it is 8 (one half of the hypotenuse). The rest is trivial, using the fact that the ratio is 1:4, and we can see that the answer is 6.

[hide]
You can simply draw the diagram, extend the perpendicular line through the top midpoint to the other base, and connect the midpoints to have a right triangle with legs $ 3$ and $ 3\sqrt{3}$, which I think is simpler than extending the sides.
[/hide]

Your right, that may be a more orthodox approach, and easier to see. I was sort of feeding off of abacadaea's post, and i didnt have paper or pencil with me...

I guess lots of people would also see the 30 and 60 and immediately think of the 30 60 90 triangle as well.

one part of the problem im not understanding-do we know that the trapezoid is isosceles? does knowing this affect the answer?

Actually, we know for sure that its NOT isoceles, because it is given that the base angles are 30 and 60 degrees.

yes, if i had been isosceles, Math Geek's answer would be right, but it's not, as the angles are different

oops...my mistake..i misread the question. i thought it said both base angles are 30 or something like that..i made a careless and false assumption..my bad :blush:


Bonnie makes the frame of a cube out of 12 pieces of wire that are each six inches long. Meanwhile Roark uses 1-inch-long pieces of wire to make a collection of unit cube frames that are not connected to each other. The total volume of Roark's cubes is the same as the volume of Bonnie's cube. What is the ratio of the total length of Bonnie's wire to the total length of Roark's wire? Express your answer as a common fraction.
[asy]size(50);
draw((0,0)--(4,0)--(4,4)--(0,4)--cycle);
draw((3,1)--(7,1)--(7,5)--(3,5)--cycle);
draw((0,0)--(3,1));
draw((4,0)--(7,1));
draw((4,4)--(7,5));
draw((0,4)--(3,5));[/asy]

Bonnie volume: $ 6^3\equal{}216$. Her wire length: $ 6\cdot 12\equal{}72$
Roark volume: $ 216\equal{}216\cdot 1^3$, so he makes 216 cubes, and for every he takes 12 1 inch parts, so $ 216\cdot 12$
Frac: $ \frac{72}{216\cdot 12}\equal{}\frac 1{36}$


I'm looking for an article

[b]Cobeli, Cristian; Zaharescu, Alexandru An exponential congruence with solutions in primitive roots.
Rev. Roumaine Math. Pures Appl. 44 (1999), no. 1, 15--22.[/b]

I'll be very glad if someone could tell me where I can find it or advise where I should look for it.
It's available  on http://www.ams.org, but I have no access for it, so if someone can get it from there and give me, I would be very glad too.

Thank you very much.

I don't know how old you are, but I recommend you go to a college library.  A large research university should have what you are looking for.

[quote="rassvet"]To my great disappointment, this number of the journal isn't available neither at my university(MSU), nor at the main library of my country(Russia)
 :lol:[/quote]

You might have online access via JSTOR or a similar archive, look around for options.


The sum of the positive integers $ a_{1}, ...,a_{r}$ is $ a_{1}+...+a_{r}=2007$ . What is the largest possible product $ a_{1}...a_{r}$ that can be formed under this condition?

[hide]We want to mazimize r in order to get the largest possible product. However the integer 1 will not increase the product, so we want to use as many 2's as possible and one 3. So, $ a_{1}=a_{2}=...=a_{1002}=2$ and $ a_{1003}=3$. The largest possible product is therefore $ 2^{1002}*3$[/hide]

[quote="danthemanvt"]We want to mazimize r in order to get the largest possible product. However the integer 1 will not increase the product, so we want to use as many 2's as possible and one 3. So, $ a_{1}=a_{2}=...=a_{1002}=2$ and $ a_{1003}=3$. The largest possible product is therefore $ 2^{1002}*3$[/quote]
$ 3^{2}>2^{3}$ and $ 3>2^{1.5}$, danthemanvt, thus your analysis and your answer is wrong

[quote="chickendude"][hide]
I'm sorry, but I can't think of any calculator-less way to go about this

if 2007 was replaced with a perfect square, then the optimum product would have $ a_{1}$ to $ a_{r}$ all be the square root
Since it's not a perfect square, we do the best we can

$ \sqrt{2007}\approx 44.800$

Make the numbers in a $ 4: 1$ ratio between 45 and 44

36(45) + 9(44) = 2016 is too big
27(45) + 18(44) = 2007

so your product would be

$ 45^{27}\cdot 44^{18}$
[/hide][/quote]
Chickendude, I dont think that answer is correct.  First, you do not need a caculator to do this.  You just have to make the observation that its a function that increases for 1-3 but decreases after 3, so you want to use as many 3s as possible.  Thus its $ 3(669)=2007$ so its $ 3^{669}$ as the product

so whats the right answer?  I get a huge number

Ordinary boy is correct, the answer is $ 3^{669}$. :)

I thought of a problem like this some time ago, and proved that using $ 3$'s gives the largest product.

[quote="archimedes1"][quote="i_like_pie"] I ... proved that using $ 3$'s gives the largest product.[/quote]
You mean like $ 2^{3}<3^{2}$ and $ 2+2+2>3+3$?[/quote]

2+2+2=3+3, not greater

[quote="i_like_pie"][quote="archimedes1"][quote="i_like_pie"] I ... proved that using $ 3$'s gives the largest product.[/quote]
You mean like $ 2^{3}<3^{2}$ and $ 2+2+2>3+3$?[/quote]
I'm pretty sure $ 2+2+2>3+3$ was supposed to be something else...

I mean the largest possible value of $ x^{\frac{n}{x}}$, for any $ n$, is when $ x=3$.[/quote]

No.
$ x=e$
produces a bigger result, but since $ 3$ is the closest integer to $ e$, that works out nicely.

EDIT:
For those of you curious, that can be proved with calculus.
[hide="Warning! Calculus content ahead"]Let $ f(x) = x^{n/x}$, $ n$ is constant and $ x\neq 0$. Then:
$ \ln f(x) = n/x\cdot \ln x$

Differentiate both sides
$ \frac{1}{f(x)}\cdot f'(x) = n\cdot \left(\frac{1-\ln x}{x^{2}}\right)$
$ f'(x) = n\cdot \left(\frac{1-\ln x}{x^{2}}\right)f(x)$
$ f'(x) = n\cdot \left( \frac{1-\ln x}{x^{2}}\right)x^{n/x}$

Set this equal to $ 0$ (to find the max/min), and since $ x^{n/x}\neq 0$ ($ x\neq 0$), we must have the numerator of that fraction equal to $ 0$:
$ 1-\ln x = 0$
$ x=e$

This clearly produces the maximum.[/hide]

[quote="vishalarul"][/quote][quote="i_like_pie"][quote="archimedes1"][quote="i_like_pie"] I ... proved that using $ 3$'s gives the largest product.[/quote]
You mean like $ 2^{3}<3^{2}$ and $ 2+2+2>3+3$?[/quote]
I'm pretty sure $ 2+2+2>3+3$ was supposed to be something else...

I mean the largest possible value of $ x^{\frac{n}{x}}$, for any $ n$, is when $ x=3$.[/quote][quote="vishalarul"]

No.
$ x=e$
produces a bigger result, but since $ 3$ is the closest integer to $ e$, that works out nicely.

EDIT:
For those of you curious, that can be proved with calculus.
[hide="Warning! Calculus content ahead"]Let $ f(x) = x^{n/x}$, $ n$ is constant and $ x\neq 0$. Then:
$ \ln f(x) = n/x\cdot \ln x$

Differentiate both sides
$ \frac{1}{f(x)}\cdot f'(x) = n\cdot \left(\frac{1-\ln x}{x^{2}}\right)$
$ f'(x) = n\cdot \left(\frac{1-\ln x}{x^{2}}\right)f(x)$
$ f'(x) = n\cdot \left( \frac{1-\ln x}{x^{2}}\right)x^{n/x}$

Set this equal to $ 0$ (to find the max/min), and since $ x^{n/x}\neq 0$ ($ x\neq 0$), we must have the numerator of that fraction equal to $ 0$:
$ 1-\ln x = 0$
$ x=e$

This clearly produces the maximum.[/hide][/quote]
Notice that the problem says positive integers, not irrational numbers...

[quote="i_like_pie"][/quote][quote="vishalarul"][/quote][quote="i_like_pie"][quote="archimedes1"][quote="i_like_pie"] I ... proved that using $ 3$'s gives the largest product.[/quote]
You mean like $ 2^{3}<3^{2}$ and $ 2+2+2>3+3$?[/quote]
I'm pretty sure $ 2+2+2>3+3$ was supposed to be something else...

I mean the largest possible value of $ x^{\frac{n}{x}}$, for any $ n$, is when $ x=3$.[/quote][quote="vishalarul"]

No.
$ x=e$
produces a bigger result, but since $ 3$ is the closest integer to $ e$, that works out nicely.

EDIT:
For those of you curious, that can be proved with calculus.
[hide="Warning! Calculus content ahead"]Let $ f(x) = x^{n/x}$, $ n$ is constant and $ x\neq 0$. Then:
$ \ln f(x) = n/x\cdot \ln x$

Differentiate both sides
$ \frac{1}{f(x)}\cdot f'(x) = n\cdot \left(\frac{1-\ln x}{x^{2}}\right)$
$ f'(x) = n\cdot \left(\frac{1-\ln x}{x^{2}}\right)f(x)$
$ f'(x) = n\cdot \left( \frac{1-\ln x}{x^{2}}\right)x^{n/x}$

Set this equal to $ 0$ (to find the max/min), and since $ x^{n/x}\neq 0$ ($ x\neq 0$), we must have the numerator of that fraction equal to $ 0$:
$ 1-\ln x = 0$
$ x=e$

This clearly produces the maximum.[/hide][/quote][quote="i_like_pie"]
Notice that the problem says positive integers, not irrational numbers...[/quote]

Yes, but you didn't specify that $ x$ was an integer. You gave the impression that $ x^{n/x}$ was maximized for $ x=3$, whether $ x$ was rational or not.
But for this problem, this type of solution is okay.
How did you exactly "prove" that using $ 3$s gives the largest product? Did you use a non-calculus method?

[quote="ordinary boy"]$ 2k=a$, so the product is $ 2^{\frac{a}{2}}$
$ 3k=a$, so the product is $ 3^{\frac{a}{3}}$
$ 4k=a$ so the product is $ 4^{\frac{a}{4}}$, which is $ 2^{\frac{a}{2}}$.

Notice this function is stictly increasing then decreasing and theres only 1 top bump, so its when its 3, as

$ 2^{\frac{a}{2}}<3^{\frac{a}{3}}$
and
$ 3^{\frac{a}{3}}>4^{\frac{a}{4}}$

To prove this raise both sides to the $ \frac{6}{a}$ power and you are left with $ 3^{2}>2^{3}$[/quote]
I already used an "elementary" way of proving it.

Although I forgot to mention that in this case $ a$ is $ 0 \mod 3$, as if $ a$ were something like $ 1 \mod 3$, you dont use maximum number of 3s.

For example if the problem had been $ a1+a2+...=4$, cleary $ 3*1$ is not greater than $ 2^{2}$
if its $ 2 \mod 3$ then you want to use maximum number of 3s than one 2.

So basically, we proved(up there) that you want to use any many 3s as possible for $ 0 \mod 3$, but this can help us prove the other two cases.

For $ 1 \mod 3$ is $ 4 \mod3$ which is $ 0 \mod3+2 \mod3+2 \mod3$  since for $ 0 \mod 3$, you use as many cases as possible for the other 2, you use two 2s for for $ 1 \mod 3$ to maximize, its [b]all possible 3s and two 2s[/b]

For $ 2 \mod3$ its $ 0 \mod3+2 \mod3$, since $ 0 \mod 3$ to maximize is all 3s we have for $ 2 \mod 3$ to maximize [b]all possible 3s and one 2s[/b]

For $ 0 \mod 3$, its obviosuly [b]all possible 3s[/b]

you would use 2s instead of 4s,5s,6s...for the same reason as i proved for $ 3$.

This isnt a complete proof but the understanding is easier this way

[quote="vishalarul"]How did you exactly "prove" that using $ 3$s gives the largest product? Did you use a non-calculus method?[/quote]
I did something similar to ordinary boy, but slightly different. This post is too small to contain my solution. :P


geometry
3D geometry
sphere
conics
ellipse
Pythagorean Theorem
power of a point

The distance between the centers of two spheres of radii 6 inside a cylinder of radius 6 is equal to 13.  A plane tangent to both spheres intersects the cylinder in an ellipse.  What is the sum of the major and minor axes of this ellipse?

[hide]Consider a cross-section of the cylinder that cuts through the centers of the two spheres.  As in the below diagram (left side), let A and B the the centers of the spheres, and let D be the midpoint of AB.  The desired plane appears in this cross-section as FG; it is tangent to sphere (circle in the cross-section) A at C.

[img]http://www.artofproblemsolving.com/Forum/files/ellipse.jpg[/img]

AB = 13, so $AD = \frac{13}{2}$.  Since AC is a radius, AC = 6.  Finally, AC is perpendicular to FG, since FG is tangent at C.  Thus ACD is a right triangle, so from the Pythagorean theorem (or recognizing as 5-12-13), $CD = \frac52$.

Now let EF = x (diagram, right side).  From Power of a Point, FC = EF = x as well.  Draw the perpendicular from F to AD, and let the point of intersection be H.  So $FH = AE = 6, HD = AD - AH = 6-x,$ and $DF = DC + CF = \frac52 + x$.  So from the Pythagorean theorem again, we have
\[\left(\frac52+x\right)^2 = 6^2 + (6-x)^2 \]
\[\left(\frac52+x\right)^2 - (6-x)^2 = 6^2\]
\[\left(\frac52+x+6-x\right)\left(5/2+x-6+x\right)  = 6^2\]
\[\left(\frac{17}{2}\right)\left(2x-\frac72\right) = 6^2\]
\[2x-\frac72 = \frac{72}{17}\]
\[2x = \frac{263}{34}\]
\[x = \frac{263}{68}\]

So now $DF = \frac52 + x = \frac{433}{68}$ and $GF = 2DF = \frac{433}{34}$.  This is the major axis of the ellipse; the minor axis is simply the diameter of the sphere, or 12.  $12+\frac{433}{34} = \frac{841}{34}$.[/hide]

mathfanatic,

There are few students on this site whom I think would benefit from this suggestion more than you -- it's time to learn TeX.  Trust me, it won't take nearly as long as you think to learn enough to write 90% of your posts and everything you write will be easier to read.  Also, you know you're going to learn it sooner or later.  Holiday break seems like a reasonable time...

I do know TeX; I just try to avoid using it to save your bandwidth ;)

Don't worry about the bandwidth.  A TeXed up post is going to be read far more often and is therefore more valuable to the community.  Bandwidth is cheap.  We can handle it.

That certainly looks nicer, though the next thing I'll start badgering you about is spacing and centering  ;)


I would like to study C language program.What book can yu advise me.

"The C Programming Language"  by By Brian W. Kernighan and Dennis M. Ritchie.

Also read this nice article, "The Tao of Programming" at http://www.canonical.org/~kragen/tao-of-programming.html

If you are a beginner in programming, "C how to program" by Deitel & Deitel would be very good.

what book would you recommend for intermidiate programmers

If someone is already a programmer and just wants to learn about C, what he said is good: "The C programming language". It has everything about C.

Do you know any other programming languages ? I'll tell you which books you need depending on that...


geometry
rectangle
AMC
AIME
algebra
polynomial
geometric transformation

The sides of rectangle $ ABCD$ have lengths 10 and 11.  An equilateral triangle is drawn so that no point of the triangle lies outside $ ABCD.$  The maximum possible area of such a triangle can be written in the form $ p\sqrt {q} \minus{} r,$ where $ p, q,$ and $ r$ are positive integers, and $ q$ is not divisible by the square of any prime number.  Find $ p \plus{} q \plus{} r.$

I think you are suppose to do something with the imaginary plane but I dont know exactly how to.

If you're looking for a solution, there's one on the wiki [[1997 AIME Problems/Problem 15]] (I highly doubt so, since you mentioned the complex plane).

[hide="Motivation"]
It's not too difficult to determine the position of the equilateral triangle, since the rectangle looks almost like a square.
[asy]pair A=(0,0),B=(11,0),C=(11,10),D=(0,10);
draw(A--B--C--D--cycle);
pair E=(11,20-11*3^.5), F=E*expi(pi/3);
draw(A--E--F--cycle);
label("\(A\)",A,SW);
label("\(B\)",B,SE);
label("\(C\)",C,NE);
label("\(D\)",D,NW);
label("\(E\)",E,NE);
label("\(F\)",F,N);[/asy]
In particular, the equilateral triangle is determined by $ CE, CF$, but if we try to solve for the lengths in terms of these two variables, we end up with a fourth degree polynomial (kalva proceeds by factoring it into irrational roots, but that would be pretty difficult to do while taking an AIME :P).

So the method of attack is incorrect; but as the wiki solution suggests, a $ 60^{\circ}$ rotation about $ A$ also seems to determine the equilateral triangle. Rotations on the complex plane are pretty easy with $ \text{cis}$ and such, so we try drawing the rectangle on the complex plane with $ A$ at the origin. Its probably unnecessary to use the complex plane; I would guess that vectors could do the same, but I don't know anything about vectors so I won't say anymore.
[/hide]

The following solution utilizes the diagram hidden in azjps's "Motivation"

[hide="Solution with Complex Numbers"]Let $ A$ be the origin. Then, wlog $ E \equal{} (11,a)$ and $ F \equal{} (b,10)$

Thus, $ b \plus{} 10i \equal{} \left(\frac {1}{2} \plus{} \frac{\sqrt{3}}{2} i \right)(11 \plus{} ai) \equal{} \frac {11 \minus{} a \sqrt {3}}{2} \plus{} \frac {11 \sqrt {3} \plus{} a}{2} \cdot i$

By equating imaginary components, $ \frac {11 \sqrt {3} \plus{} a}{2} \equal{} 10 \implies a \equal{} 20 \minus{} 11 \sqrt {3}$

Thus, $ AE^2 \equal{} a^2 \plus{} 11^2 \equal{} 884 \minus{} 440 \sqrt {3}$

The area of an equilateral triangle in terms of the side length is $ \frac {\sqrt {3}}{4} s^2$, so here we have:
$ \frac {\sqrt {3}}{4} \left(884 \minus{} 440 \sqrt {3} \right) \equal{} 221 \sqrt {3} \minus{} 330$

Hence $ (p, q, r) \equal{} (221 ,3 ,330) \implies \boxed{p \plus{} q \plus{} r \equal{} 554}$[/hide]

[hide="keeping it real"]good thing someone else drew the diagram cause i don't know how. using azjps's diagram, let $ A$ be the origin, $ AB = 11$, $ BC = 10$, and $ \angle EAB = \theta$. Then E is at $ (11, 11\tan\theta)$. Drawing a line through E parallel to the x-axis, we get the trigonometric equation $ \sin(60 - \theta) = \frac {10 - 11\tan\theta}s$, where $ s$ is the side length of the triangle. So $ s = \frac {10 - 11\tan\theta}{\sin(60 - \theta)} = \frac {10 - 11\tan\theta}{\frac {\sqrt 3}2\cos\theta - \frac 12\sin\theta}$. Aslo notice that $ s = \frac {11}{\cos\theta}$.
So now we can do some math...
\begin{eqnarray*} \frac {10 - 11\tan\theta}{\frac {\sqrt 3}2\cos\theta - \frac 12\sin\theta} & = & \frac {11}{\cos\theta} \\
10\cos\theta - 11\sin\theta & = & \frac {11\sqrt 3}2\cos\theta - \frac {11}2\sin\theta \\
20\cos\theta - 22\sin\theta & = & 11\sqrt 3\cos\theta - 11\sin\theta \\
\cos\theta(20 - 11\sqrt 3) & = & 11\sin\theta \\
\tan\theta & = & \frac {20 - 11\sqrt 3}{11}.\end{eqnarray*}
From which we get $ \cos\theta = \frac {11}{\sqrt {(20 - 11\sqrt 3)^2 + 11^2}}$. Recalling that $ s = \frac {11}{\cos\theta}$, we get $ s = \sqrt {(20 - 11\sqrt 3)^2 + 11^2}$, so $ s^2 = (20 - 11\sqrt 3)^2 + 11^2 = 400 - 440\sqrt 3 + 363 + 121 = 884 - 440\sqrt 3$. The area of the triangle is $ \frac {s^2\sqrt 3}4 = \sqrt 3(221 - 110\sqrt 3) = 221\sqrt 3 - 330$. $ p + q + r = 221 + 3 + 330 = \boxed{554}$.
[/hide]


probability
calculus
derivative
integration
function
probability and stats

I'm working my way through some problems before an exam tomorrow and I could use some help.

The problem i'm trying to solve is this:

$ \frac {\partial F}{\partial t} \plus{} rx\frac {\partial F}{\partial x} \plus{} \frac {1}{2} \sigma^2 x \frac {\partial^2 F}{\partial x^2} \minus{} rF \equal{} 0$
$ F(T,x) \equal{} log(x)$

So i apply Feynman-Kac and get:

$ F(t,x) \equal{} e^{ \minus{} r(T \minus{} t)} E(log(x))$

$ dX_s \equal{} rx ds \plus{} \sigma x dW_s$
$ X_t \equal{} x$

Then $ X \sim N[x \plus{} rx(T \minus{} t), \sigma^2 x (T \minus{} t)]$

So basically my question is, if $ X \sim N[\mu,\sigma^2]$, how is $ log(x)$ distributed?

Also, how would I go about doing things if $ F(T,x) \equal{} (log(x))^2$...then I'd need to know how $ (log(x))^2$ is distributed...

oh! No, you have to consider the diffusion
\[ dX_s \equal{} rX_s \, ds \plus{} \sigma X_s dW_s\]
\[ X_t\equal{}x\]
instead. And you should know that
\[ X_s \equal{} X_t \exp( (r\minus{}\frac{\sigma^2}{2})(s\minus{}t) \plus{} \sigma W_{s\minus{}t})\]
so that $ \log(X_T) \equal{} \log(x) \plus{} (r\minus{}\frac{\sigma^2}{2})(T\minus{}t) \plus{} \sigma W_{T\minus{}t}$, wich shows that $ \log(X_T)$ is distributed according to a normal distribution with mean $ \log(x) \plus{} (r\minus{}\frac{\sigma^2}{2})(T\minus{}t)$ and variance $ \sigma^2 (T\minus{}t)$.

Does this help a little bit ?

:bye:
[quote]So basically my question is, if X \sim N[\mu,\sigma^2], how is log(x) distributed? [/quote]
Very easy  :moose:
$ \mathcal{P}(\log(X)<x)\equal{}\mathcal{P}(X<\exp(x))$ this probability you can calculate using the fact that X is normally distributed.You can also calculate PDF by taking a derivative from the integral (as a function of an upper bound).


[quote="zhaobin"]$a,b>0$,prove:
$\frac{1}{1+a^2+a^4+a^6}+\frac{1}{1+b^2+b^4+b^6} \geq \frac{1}{1+a^3b^3}$[/quote]

Hi, Zhao Bin.

I find a way to solve this problem. But it's very difficult and complicated. Do you have any easy way? Do you know whether the problem was solved? Please tell me. Thanks a lot!

Long time ago, I used this inequality to prove that $abcd=1$ implies

$\frac{1}{1+a+a^2+a^3}+\frac{1}{1+b+b^2+b^3}+\frac{1}{1+c+c^2+c^3}+\frac{1}{1+d+d^2+d^3} \ge 1$.

sorry.i have no solution,althougt i spend a lot of time.and i don't know whether it have been solved.sorry again.

Yeah, it is at least partially a consequence of one of the [i]incredibly beautiful [/i]ineqs mentioned in http://www.mathlinks.ro/Forum/viewtopic.php?p=226013#p226013. In part (c) of the proposition put $n=2, k=3, x_1=a^2,x_2=b^2$ then we have $r=ab$. The problem is just that $r\ge1$ is requested. (I wonder by the way if that condition is really necessary for the proposition to hold.)
So at least for $ab\ge1$ this yields
$\frac{1}{1+a^2+a^4+a^6}+\frac{1}{1+b^2+b^4+b^6} \geq \frac{2}{1+r+r^2+r^3}.$ Now just observe that $\frac{2}{1+r+r^2+r^3}\geq\frac{1}{1+r^3}$ as this is equivalent to $(r+1)(r-1)^2\geq0$ and so we are done for this case.

If $ab<1$, applying the same proposition with $x_1=a^{-2},x_2=b^{-2}$ we obtain $\frac{(\frac ab)^3}{1+a^2+a^4+a^6}+\frac{(\frac ba)^3}{1+b^2+b^4+b^6} \geq \frac{1}{1+a^3b^3}$, which is better than nothing... Any ideas?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=18857]http://www.mathlinks.ro/Forum/viewtopic.php?t=18857[/url]
some days ago I find the solution in the link.
but sorry I haven't wroten anything here :)


analytic geometry
function
calculus
derivative

Find the shortest distance between (11,60) and 12x+35y+25=0.

As a follow up...

2. Find the shortest distance between $ (11,60)$ and $ y\equal{}x^2\plus{}3$.

hello, converting your equation in the Hessian normal form we get
$ \frac {12x \plus{} 35y \plus{} 25}{ \minus{} 37} \equal{} 0$ and plugging the coordinates of the given point in this equation we get
the searched distance as
$ d \equal{} 61$.
Sonnhard.

hello, the distance is given by $ d(x,y) \equal{} \sqrt{(x \minus{} 11)^2 \plus{} (y \minus{} 60)^2}$
plugging $ y \equal{} x^2 \plus{} 3$ int this term we get the following function
$ f(x) \equal{} \sqrt {(x \minus{} 11)^2 \plus{} (x^2 \minus{} 57)^2}$.
Now you must discuss this function.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, the distance is given by $ d(x,y) \equal{} \sqrt {(x \minus{} 11)^2 \plus{} (y \minus{} 60)^2}$
plugging $ y \equal{} x^2 \plus{} 3$ int this term we get the following function
$ f(x) \equal{} \sqrt {(x \minus{} 11)^2 \plus{} (x^2 \minus{} 57)^2}$.
Now you must discuss this function.
Sonnhard.[/quote]

hello , how would you discuss a function? I understood all till that part, and how you reach your solution

hello, i mean to differentiate the given function to find the extremal points, the first derivative is given by
$ f^'(x)=\frac{1}{2}\frac{-226x-22+4x^3}{\sqrt{(x-11)^2+(x^2-57)^2}}$
Now you must solve the equation $ f'(x)=0$ for $ x$.
Sonnhard.

For the first one you could use the distance of a point to line formula which is given by $ \frac{|Ax_1\plus{}By_1\plus{}C|}{\sqrt{A^{2}\plus{}B^{2}}}$ where the line is $ Ax\plus{}By\plus{}C\equal{}0$ and the point $ (x_1,y_1)$. Plugging everything in we get
$ \frac{132\plus{}2100\plus{}25}{\sqrt{1225\plus{}144}}\equal{}\frac{1357}{37}\equal{}61$.


pigeonhole principle
linear algebra
group theory
linear algebra unsolved

Does anyone know if there is an easy way to do this problem?

Let the sequence $ a_k$ be defined by $ a_k \equal{} a_{k \minus{} 1}*a_{k \minus{} 2}$ where each member of a is an element of the set of n by n matrices that contain exactly one 1 in each row and column and have the rest filled with 0s.

Find the maximum possible period of $ a_k$ in terms of n.

Find a formula for $ a_k$ in terms of $ a_1, a_2$.  What do you know about the periods of permutation matrices?

I tried to find a formula, but I couldn't. As far as I can tell, it doesn't simplify, although I'm no expert on Linear Algebra.

This has very little to do with linear algebra [i]per se[/i]. We can write permutations as matrices, or we can write them in other notations; either way, we have the same thing, the symmetric group $ S_n.$ The maximum period of any single element (which may be what t0rajir0u was talking about) isn't particularly relevant, since we are not talking about multiples of any single permutation, and $ S_n$ is not abelian. In fact, I think that non-commutativity will defeat the attempt to write a simple formula for $ a_k.$

I can give a crude upper bound. The state of the system at any point is the two matrices $ (a_{k\minus{}1},a_k)$ as that suffices to determine the future of the sequence. Since the cardinality of $ S_n$ is $ n!,$ there are $ (n!)^2$ such pairs of permutations. On the other hand, if both matrices were the identity, then the entire future and past of the sequence would be the identity. For a non-constant sequence, that can't occur. Each step turns the pair $ (a_{k\minus{}1},a_k)$ into the pair $ (a_k,a_{k\plus{}1}).$ If we ever return to a state we've been in before, then the sequence must repeat - and since the recursion can be read backwards as well as forwards, then the sequence must be strictly periodic. By the pigeonhole principle, this must happen.

This argument gives a longest possible period of $ (n!)^2\minus{}1.$ I would be open to arguments showing that it must be smaller than that. (If $ n\equal{}2,$ there is a case - essentially the only nontrivial case - which achieves the maximum period of $ 3.$ But beyond that I don't know.)

Wouldn't it be
$ (n!\minus{}1)^2$
because if at any point, either of the terms is the identity, then it should immediately start repeating?

No. The sequence could go $ I,A,A,A^2,\dots.$ That's not an immediate repeat. In fact, that's what happens for $ n\equal{}2,$ where if $ A$ is the only permutation other than the identity, the sequence goes $ I,A,A,I,A,A,\dots$ and has period $ 3\equal{}(2!)^2\minus{}1.$ It takes two consecutive identities to force an immediate repeat.

On the other hand, if one identity appears, then all of the permutations in the sequence commute and it the estimate of maximum length would be quite a bit shorter, in general - we'd get the period of the Fibonacci numbers mod the order of the element $ A.$


We have $ 6$ possibilities for the first spot, and after this, we have $ 5$ possibilities for the remaining spots since there can be no two equal, consecutive digits.  Our answer is then $ 6 \bullet 5^{4}\equal{}3750$.

[quote="kurt.math"]Hi,

Zorgo is a planet where there is only 6 numerals in the number system, {1,2,3,4,5,6}.
If their can be no consecutive numbers, than how many numbers with 5 digits can be made?
(For example, 55456 is not allowed, but 14634 is)[/quote]

14634 isn't one...  :P

[quote="J.Mayers"]We have $ 6$ possibilities for the first spot, and after this, we have $ 5$ possibilities for the remaining spots since there can be no two equal, consecutive digits.  Our answer is then $ 6 \bullet 5^{4} \equal{} 3750$.[/quote]

Are you sure this is right?

Let's say you pick the number $ 2$ for the first digit, this means that the second digit can only be one of the following: $ 2,4,5,6$ - 4 possibilities (this means that the second digit doesn't have to have 5 possible digits). Also if you look at any digit in the middle you also have to take in count the previous digit. For example if the first digit is 2 and the second digit is 4  then the third digit can only be $ 1,2,4,6$ - 4 possibilities.

The confusion is in what is the meaning of "there can be no [b]consecutive[/b] numbers"?

You would (probably) think that "no consecutive numbers" means "no two adjacent digits differ by one", as it is used when you say "no consecutive integers". This is tonypr's interpretation of the problem.

But from his example ("14634 is [allowed]"), you can [i]infer[/i] that he means for "no consecutive numbers" to mean "no two adjacent digits are the same". This was J. Mayer's interpretation of the problem.

J. Mayer's answer is correct for the latter interpretation. The first interpretation is a more difficult (but still interesting) problem.

Thankyou guys, sorry for the confusion  :blush:

I also thought that it would be $ 6 \times 5^{4}$, I just wasn't sure.  :)

the soltution is the following:
observe that we can have at most $ 3$ different digits in the number.we shall divide the problem in cases: $ 1,2,3$ different digits.
if we have $ 1$ digit, there are $ 6$ possible numbers. if we have $ 2$ different digits, there are the fpllowing popssible combinstions of doubles:{1,3},{1,4},{1,5},{1,6} and so on.we see that we´ll have a total  of $ 300$ posiible numbers.if we have $ 3$ different digits , we´ll have a total of $ 2400$ different possible numbers.so we´ll have a total of $ 2706$ possible numbers.

i´m not quite sure if this is the final answer, because i  did not check if i made any little calculation mistakes. :|

[quote="v235711"]...
observe that we can have at most $ 3$ different digits in the number...[/quote]

An obvious rebuttal of this is to look at the example given.


hi all,

what about the following pb: find all integer n so that there exist integers (x,y) :
3/n=1/x+1/y

bye.

exactly those integers n are expressible which have a divisor which is not 1 mod 3. one quite easily sees that writing the equation as
$y(3x-n)=nx$,
so there is a y if and only if $3x-n$ divides $nx$. but $gcd(3x-n,x)=gcd(n,x)$ is a divisor of n, thus if n is expressible, then 3x-n divides n^2, thus equals a divisor of n^2. if all divisors of n are 1 mod 3, this is not possible, since 3x-n is 2 mod 3. thus in order to be expressible, one prime divisor of n must be 3 or 2 mod 3, call it p. if p is expressible, then n is, too(just multiply x,y by n/p). if p=3, then choose x=y=2. if p=2 mod 3, then choose $x=\frac{p+1}{3}$, $y=p\frac{p+1}{3}$.


The circumference inscribed on the triangle $ABC$ is tangent to the sides $BC$, $CA$ and $AB$ on the points $D$, $E$ and $F$, respectively. $AD$ intersect the circumference on the point $Q$. Show that the line $EQ$ meet the segment $AF$ at its midpoint if and only if $AC=BC$.

Let $M$ be the intersection of $EQ$ and $AB$. We have \begin{eqnarray*} & & AM = MF \\ & \iff & AM^2 = MF^2 \\ & \iff & AM^2 = MQ \cdot ME \\ & \iff & \Delta AMQ \sim \Delta EMA \\ & \iff & \angle AEM = \angle QAM \\ & \iff & \angle EDA = \angle DAB \\ & \iff & DE \parallel AB \\ & \iff & \angle EDC = \angle ABC \\ & \iff & 90^{\circ} - \frac{1}{2} \angle ACB = \angle ABC \\ & \iff & \angle ACB + 2 \angle ABC = 180^{\circ} \\ & \iff & \angle ABC = \angle BAC \\ & \iff & AC = BC. \blacksquare \end{eqnarray*}

[hide="Harmonic Solution"]
Clearly $QEDF$ is harmonic, so take pencil $E(QEDF)$ and intersect it with $AB$. We must have that $(AFQ\infty)=(AFQ ED\cap AB)$, so $ED$ is parallel to $AB$. As $EDC$ is isosceles, the result follows. $\Box$[/hide]

$$AM \cong FM$$ $$\Leftrightarrow AM^2 = MQ \cdot ME$$ $$\Leftrightarrow \triangle MQA \sim MAE$$ $$\Leftrightarrow \angle MAQ \cong \angle AEM$$ $$\Leftrightarrow \angle FAD \cong \angle AEQ$$ $$\Leftrightarrow \angle BFD - \angle QDF = \angle AEF - \angle QEF$$ $$\Leftrightarrow \alpha + \gamma - \angle QDF = \beta + \gamma  - \angle QDF$$ $$\Leftrightarrow 2\alpha = 2\beta$$ $$\Leftrightarrow \angle BAC \cong \angle CBA$$ $$\Leftrightarrow BC \cong AC$$ $$\square$$

$\rightarrow$
Let $M$ be the midpoint of AF
$(A,L;Q,D) \overset{E}{=}(A,F;M,DE \cap AF) \implies DE \parallel AF \implies AC=BC$.
$\leftarrow$
$AC=BC \implies ED \parallel AF$
Note that: $(A,L;Q,D)=-1 \overset{E}{=}(A,F;M,ED \cap AF=\infty_{AF}) \implies AM=MF$.
and we finished :).


geometry
quadrilateral
area
maximization
IMO Shortlist
IMO Longlist

Find whether among all quadrilaterals, whose interiors lie inside a semi-circle of radius $r$, there exist one (or more) with maximum area. If so, determine their shape and area.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

The existence of a maximum is warranted by a compactness argument.
Trivially, a quadrilateral achieving maximum area must be convex and have all its vertices on the boundary of the semi-circle.
We may assume two vertices lie on the diameter of the semi-circle, otherwise consider the side with least distance from the center of the circle of radius $ r$ and prolong it to a full chord; then the quadrilateral is also inscribed in the semi-circle with diameter this chord $ \leq 2r$.

Now symmetrize with respect with the diameter. The result is a polygon with $ 6$ vertices, of double the area, contained in the circle of radius $ r$.
By a celebrated result of Steiner, the largest area of an $ n$-gon contained in a circle is realized by the regular $ n$-gon, therefore in our case the maximum area is given by a half regular hexagon.


There are three sequences $ \{A_{n}\},\{B_{n}\},\{C_{n}\}$ such that $ A_{1}\equal{} a > 0, B_{1}\equal{} b < 0, C_{1}\equal{} c > 0, b^{2}< 4ac$

$ A_{n}\equal{} A_{n\minus{}1}C_{n\minus{}1}$
$ B_{n}\equal{} 4A_{n\minus{}1}C_{n\minus{}1}\plus{}B_{n\minus{}1}(A_{n\minus{}1}\plus{}C_{n\minus{}1})$
$ C_{n}\equal{} (A_{n\minus{}1}\plus{}B_{n\minus{}1}\plus{}C_{n\minus{}1})^{2}.$

Prove that there is a  positive integer $ t$ such that $ B_{t}\geq 0$.

Hi dear Yangyao. It is not as hard as you have said about. Now I'll show you that, my doc.
Assume D_n is the negative form of B_n such that D_n = -B_n. Then D_1 = d = -b > 0. We know that in we have a. [code] $ \ $ $[/code] = b^2 - 4ac < 0 and this discriminant is of the equation ax^2 + bx + c =0.
                                             Having some operations out of the equation will be probably the solution of our problem
                                             i.e., it will show the wanted something

here's my solution:
Assume that there isn't any number that B_t>0
Therfore we have: B_2<0
So we have:4A_1C_1+B_1(A_1+C_1)<0
That's equivalent to: b<-(a+c)-4ac<0
Then we square both sides and get:b^2>(a+c+4ac)^2
Then we easily infer:4ac>(a+c+4ac)^2(contradiction)  :)


Let a,b,c>0 such that a+b+c=1.
Prove that $ (1\minus{}a)(1\minus{}b)(1\minus{}c)\geq 8abc$

We claim that $ (a \plus{} b)(b \plus{} c)(c \plus{} a) \ge 8abc$ for all $ a, b, c > 0$ from which the conclusion follows.  But this is just $ \prod_{cyc} (a \plus{} b) \ge \prod_{cyc} 2 \sqrt {ab}$.

Kouichi Nakagawa is just factoring.

[quote="facis"]Kouichi Nakagawa, how do you go from your second step to your result?[/quote]

When you expand $ (a\plus{}b)(b\plus{}c)(c\plus{}a)\minus{}8abc$ and arrange it, you will split $ 6abc$ into $ 2abc,\ 2abc,\ 2abc$,then you will get $ a(b^2\minus{}2bc\plus{}c^2)$ etc.

t0rajir0u, then doesn't the result follow from AM-GM? Because your result looks like the product of AM's is greater than or equal to the product of GM's.

[quote=natural123]Let a,b,c>0 such that a+b+c=1.
Prove that $$ (1\minus{}a)(1\minus{}b)(1\minus{}c)\geq 8abc$$[/quote]
[url]https://artofproblemsolving.com/community/c4h514306p2890445[/url]
Let $x$, $y$, $z$ be positive numbers for which\[\frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z} = 1\]
Prove that $xyz \geq 8$.
[url]https://artofproblemsolving.com/community/c6h1416508p7976991[/url]
Given $a,b,c >0$ such that $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} =2$ Prove that : $$abc \leq \frac{1}{8} $$
[url]https://artofproblemsolving.com/community/c4h1583571p9785950[/url]
Let $a,b\geq 0$ and $a+b>0.$Prove that$$a(\frac{1}{a^2+b^2}+\frac{b^2}{2})+a+b\geq 3\sqrt{ab}$$
Let $a,b\geq 0$ and $a^2+b^2=2.$Prove that$$2a+b\geq \sqrt{ab}+2 \sqrt{a}$$

$$(1-a)(1-b)(1-c)\geq 8abc \Leftrightarrow$$
$$\frac{-abc+(ab+bc+ac)-(a+b+c)+1}{abc}\geq 8 \Leftrightarrow$$
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 9 \Leftrightarrow$$
$$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9$$

Which is true by Cauchy-Schwarz

[quote=Kouichi Nakagawa]$ (1\minus{}a)(1\minus{}b)(1\minus{}c) \ge 8abc$
$ \iff (b\plus{}c)(c\plus{}a)(a\plus{}b) \ge 8abc$
$ \iff a(b\minus{}c)^2 \plus{} b(c\minus{}a)^2 \plus{} c(a\minus{}b)^2 \ge 0$[/quote]
Just use Am-Gm for b and c, c and a, a and b and we have what we need


are there any mathcounts questions in the contests araea? there used to be, but somehow it just disappeared. :(

Some good websites are

http://goodlymaths.0catch.com

http://mathcounts.saab.org/mc.cgi

http://www.freewebs.com/tzpublicsite/index.htm

[quote="anirudh"]i know websties about mc, but what i mean is what happened to th4e mathcounts problems in the contest button (resources.php)[/quote]

there were never any, contest button is for olympiad resources

I think that's mostly a copyright issue, though I'm not entirely sure.  Stuff has been changing lots recently.

We do not have permission to list MATHCOUNTS problems in our resources section -- the problems are copyrighted.


inequalities
LaTeX
function
inequalities unsolved

Every factor (a+1/b) is nonegative. The GM-HM-inequality gives:
$ \sqrt[3]{(a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a})} >\equal{} \frac {3}{1/(a\plus{}1/b)\plus{}1/(b\plus{}1/c)\plus{}1/(b\plus{}1/a)}$

It is however only necessary to know that equality holds if $ (a \plus{} \frac {1}{b})\equal{}(b \plus{} \frac {1}{c})\equal{}(c \plus{} \frac {1}{a})$, which is satisfied if a=b=c. Cubing the inequality gives the desired inequality:
$ (a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a}) >\equal{} {(1/3\plus{}1/(1/3))}^{3}$

having equality condition doesnt mean that we should have the inequality that you use :S
or does it :S

It does!, The GM expression assumes its least value under the equality condition.
Besides, the GM-HM is a valid inequaliy, since every factor is non-negative.

[quote="URBAN1990"]Every factor (a+1/b) is nonegative. The GM-HM-inequality gives:
$ \sqrt [3]{(a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a})} > \equal{} \frac {3}{1/(a \plus{} 1/b) \plus{} 1/(b \plus{} 1/c) \plus{} 1/(b \plus{} 1/a)}$

It is however only necessary to know that equality holds if $ (a \plus{} \frac {1}{b}) \equal{} (b \plus{} \frac {1}{c}) \equal{} (c \plus{} \frac {1}{a})$, which is satisfied if a=b=c. Cubing the inequality gives the desired inequality:
$ (a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a}) > \equal{} {(1/3 \plus{} 1/(1/3))}^{3}$[/quote]
The solution is not valid! You are assuming $ a\equal{}b\equal{}c$, in which the ineq is obvious.

mathias_dk expressed my feelings well :D

[quote="mehmetcantu"]if a,b,c are positive real numbers with a+b+c=1

Prove this inequality
$ (a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a}) > \equal{} (10/3)^3$
(dont know is it possible to learn solutions of  mathematics and youth magazine 2007)[/quote]
Here are listed all the mistakes i made solving this one :) (And the solution of course)

There is equality if $ a \equal{} b \equal{} c$.
Then $ a : \frac {1}{b} \equal{} \frac {1}{3} : 3 \equal{} 1 : 9$. And the same with the others.

This motivates me to use cauchy schwartz at first:
$ (a \plus{} \frac {1}{b})(1 \plus{} 9) \ge (\sqrt {a} \plus{} \frac {3}{\sqrt {b}})^2$.
Do the same with others and get:
$ (a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a}) \ge \frac {(\sqrt {a} \plus{} \frac {3}{\sqrt {b}})^2(\sqrt {b} \plus{} \frac {3}{\sqrt {c}})^2(\sqrt {c} \plus{} \frac {3}{\sqrt {a}})^2}{10^3}$
So we just have to prove:
$ (\sqrt {a} \plus{} \frac {3}{\sqrt {b}})^2(\sqrt {b} \plus{} \frac {3}{\sqrt {c}})^2(\sqrt {c} \plus{} \frac {3}{\sqrt {a}})^2 \ge \frac {10^6}{3^3}$.

That isn't easier afaik, so another approach. Homogenizing:
$ a \plus{} \frac {1}{b} \equal{} \frac {ba \plus{} (a \plus{} b \plus{} c)^2}{b(a \plus{} b \plus{} c)}$.
So $ (a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a}) \equal{} \frac {(ab \plus{} (a \plus{} b \plus{} c)^2)(bc \plus{} (a \plus{} b \plus{} c)^2)(ca \plus{} (a \plus{} b \plus{} c)^2)}{abc(a \plus{} b \plus{} c)^3}$

Then hölder gives us: $ (ab \plus{} (a \plus{} b \plus{} c)^2)(bc \plus{} (a \plus{} b \plus{} c)^2)(ca \plus{} (a \plus{} b \plus{} c)^2) \ge (\sqrt [3]{abc}^2 \plus{} (a \plus{} b \plus{} c)^2)^3$

So it is left to prove:
$ \frac {(\sqrt [3]{abc}^2 \plus{} (a \plus{} b \plus{} c)^2)^3}{abc(a \plus{} b \plus{} c)^3} \ge \frac {10^3}{3^3}$.
Or written without fractions:
$ 3^3(\sqrt [3]{abc}^2 \plus{} (a \plus{} b \plus{} c)^2)^3 \ge 10^3abc(a \plus{} b \plus{} c)^3$.

When $ a \equal{} b \equal{} c$ we have $ \sqrt [3]{abc}^2 : (a \plus{} b \plus{} c)^2 \equal{} 1 : 9$. So using AM-GM gives us:
$ \sqrt [3]{abc}^2 \plus{} 9(\frac {a \plus{} b \plus{} c}{3})^2 \ge 10 \sqrt [10]{(abc)^{2/3}(\frac {a \plus{} b \plus{} c}{3})^{18}}$

Thus we need only to prove:
$ (10 \sqrt [10]{(abc)^{2/3}(\frac {a \plus{} b \plus{} c}{3})^{18}})^3 \ge 10^3abc(\frac {a \plus{} b \plus{} c}{3})^3$.

This is equivalent to:

$ (abc)^{2/10}(\frac {a \plus{} b \plus{} c}{3})^{27/5} \ge abc(\frac {a \plus{} b \plus{} c}{3})^3$ $ \iff$
$ (\frac {a \plus{} b \plus{} c}{3})^{24/10} \ge abc^{8/10}$
Which is obvious by AM-GM. Equality iff $ a \equal{} b \equal{} c$. QED

[quote="URBAN1990"][quote="Mathias_DK"][quote="URBAN1990"]Every factor (a+1/b) is nonegative. The GM-HM-inequality gives:
$ \sqrt [3]{(a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a})} > \equal{} \frac {3}{1/(a \plus{} 1/b) \plus{} 1/(b \plus{} 1/c) \plus{} 1/(b \plus{} 1/a)}$

It is however only necessary to know that equality holds if $ (a \plus{} \frac {1}{b}) \equal{} (b \plus{} \frac {1}{c}) \equal{} (c \plus{} \frac {1}{a})$, which is satisfied if a=b=c. Cubing the inequality gives the desired inequality:
$ (a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a}) > \equal{} {(1/3 \plus{} 1/(1/3))}^{3}$[/quote]
The solution is not valid! You are assuming $ a \equal{} b \equal{} c$, in which the ineq is obvious.[/quote]

I am not assuming that a=b=c. I only assume that the equality between the geometric and harmonic mean is then $ (a \plus{} \frac {1}{b}) \equal{} (b \plus{} \frac {1}{c}) \equal{} (c \plus{} \frac {1}{a})$ and that the maximum of the harmonic mean is the minimum of the geometric mean. Isnt this correct?
a=b=c is a solution for the above equation, and does therefore, like anu other solution to the equation give the minimum of the Geometric mean.
What is wrong with my solution?[/quote]
It's harder to explain why something is wrong than why something is right, but I'll give it a shot.

You have proved that:
$ (a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a}) \ge (\frac {3}{1/(a\plus{}1/b)\plus{}1/(b\plus{}1/c)\plus{}1/(b\plus{}1/a)})^3$
But if $ b \to 0$ then $ \frac {3}{1/(a\plus{}1/b)\plus{}1/(b\plus{}1/c)\plus{}1/(b\plus{}1/a)} \to 0$, so you have only proved:
$ (a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a}) \ge 0$.

nice solution mathias i made your mistakes too :)
but im sorry i couldnt learn holder inequality well :P
i've tried to do it by muirhead inequality after homogenizing:S

[quote="mehmetcantu"]if a,b,c are positive real numbers with a+b+c=1

Prove this inequality
$ (a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a}) > \equal{} (10/3)^3$
(dont know is it possible to learn solutions of  mathematics and youth magazine 2007)[/quote]
$ T\equal{}(a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a})\equal{}\frac{(ab\plus{}1)(bc\plus{}1)(ac\plus{}1)}{abc}\equal{}
\frac{a^2b^2c^2\plus{}1\plus{}abc(a\plus{}b\plus{}c)\plus{}ab\plus{}bc\plus{}ac}{abc}$
but
1) $ a^2b^2c^2\plus{}\frac{1}{27^2}\ge \frac{2}{27}abc$
2) $ abc(a\plus{}b\plus{}c)\equal{}abc$
3) $ ab\plus{}bc\plus{}ac\equal{}(ab\plus{}bc\plus{}ac)(a\plus{}b\plus{}c)\ge 9abc$ (AM-GM)
4) $ \frac{27^2\minus{}1}{27^2}\equal{}\frac{27^2\minus{}1}{27^2}(a\plus{}b\plus{}c)^3\ge (27\minus{}\frac{1}{27})abc$
adding this 4 ineqs we get
$ T\ge \frac{(27^2\plus{}27*9\plus{}27\plus{}1)abc}{abc}\equal{}\frac{10^3}{3^3}$ and we are done :D

[quote="mehmetcantu"]If a,b,c are positive real numbers satisfying  a+b+c=1, then prove the inequality:
$ \left(a + \frac {1}{b}\right)\left(b + \frac {1}{c}\right)\left(c + \frac {1}{a}\right) > = \left(\frac {10}{3}\right)^3$[/quote]

I did it with the weighted AM-GM inequality:
$ \prod_{cyc} \left(a + \frac {1}{b} \right) = \prod_{cyc}\left(a + 9\cdot {\frac {1}{9b}\right) \geq 10^3 \cdot \prod_{cyc} \sqrt [10]{\frac {a}{(9b)^9}} = 10^3 \sqrt [10]{\frac {1}{9^{27}(abc)^8}} = \left(\frac {10}{3} \right)^3\sqrt [10]{\frac {1}{9^{12}(abc)^8}}}$
So it's left to prove that
$ \frac {1}{9^{12}(abc)^8} \geq 1$ or equivalently $ \frac {1}{27} \geq abc$ and this inequality follows immediately from the given condtition, because $ \frac {1}{3} = \frac {a + b + c}{3} \geq \sqrt [3]{abc}$.

thanks for the shortest solution felixd.

I think the inequality is not correct.. but i'm most likely wrong.

$ (a\plus{} 1/b)(b \plus{} 1/c)(c \plus{} 1/a) >\equal{}(10/3)^3$  <=>
$ (ab \plus{} a/c \plus{} 1 \plus{} 1/bc)(c \plus{} 1/a) >\equal{} (10/3)^3$ <=>
$ (abc \plus{} ab/a \plus{} ac/c \plus{} a/ac \plus{} c \plus{} 1/a \plus{} c/bc \plus{} 1/abc) >\equal{} (10/3)^3$ <=>
$ (abc \plus{} 1/abc) \plus{} (a \plus{} 1/a) \plus{} (b \plus{} 1/b) \plus{} (c \plus{} 1/c) >\equal{} (10/3)^3$ <=>

We know that if $ x>0$ then $ x \plus{} 1/x >\equal{} 2$ so here we know $ a,b,c>0 \equal{}> abc>0$
So:
$ 2 \plus{} 2 \plus{} 2 \plus{} 2 >\equal{} (10/3)^3$ <=>
$ 8>\equal{} 1000/27$ <=>
$ 8>\equal{} 37,....$ contradiction.

[quote="Papalakis"]I think the inequality is not correct.. but i'm most likely wrong.

$ (a \plus{} 1/b)(b \plus{} 1/c)(c \plus{} 1/a) > \equal{} (10/3)^3$  <=>
$ (ab \plus{} a/c \plus{} 1 \plus{} 1/bc)(c \plus{} 1/a) > \equal{} (10/3)^3$ <=>
$ (abc \plus{} ab/a \plus{} ac/c \plus{} a/ac \plus{} c \plus{} 1/a \plus{} c/bc \plus{} 1/abc) > \equal{} (10/3)^3$ <=>
$ (abc \plus{} 1/abc) \plus{} (a \plus{} 1/a) \plus{} (b \plus{} 1/b) \plus{} (c \plus{} 1/c) > \equal{} (10/3)^3$ <=>

We know that if $ x > 0$ then $ x \plus{} 1/x > \equal{} 2$ so here we know $ a,b,c > 0 \equal{} > abc > 0$
So:
$ 2 \plus{} 2 \plus{} 2 \plus{} 2 > \equal{} (10/3)^3$ <=>
$ 8 > \equal{} 1000/27$ <=>
$ 8 > \equal{} 37,....$ contradiction.[/quote]
I think you had mistake, when you opened brackets

for dima he didnt make mistake :) its true
if a+1/a = 2 then a must be equal to 1    and in your work a,b,c are all equal to 1 which contradict with a+b+c=1 condition.

mehmetcantu i dont understand what you mean. I didnt say that $ a \plus{} 1/a \equal{} 2$ in the problem.. i used the fact that every number plus itself reversed always surpases or equals 2.
To be more precise, i said that $ a \plus{} 1/a >\equal{} 2$ similarily for $ b,c, abc$. Now in the problem, i have to add those four numbers with their reversed and show that the sum is greater or equal to $ (10/3)^3$ But since those four numbers are greater or equal to 2 when their reversed are added to them, all i have to show is that $ 2 \plus{} 2 \plus{} 2 \plus{} >\equal{} (10/3)^3$ because $ (a \plus{} 1/a) \plus{} (b \plus{} 1/b) \plus{} (c \plus{} 1/c) \plus{} (abc \plus{} 1/abc) >\equal{} 2 \plus{} 2 \plus{} 2 \plus{}2 >\equal{} (10/3)^3$..

i never said that $ a,b,c \equal{} 1$

[quote="URBAN1990"]Every factor (a+1/b) is nonegative. The GM-HM-inequality gives:
$ \sqrt [3]{(a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a})} > \equal{} \frac {3}{1/(a \plus{} 1/b) \plus{} 1/(b \plus{} 1/c) \plus{} 1/(b \plus{} 1/a)}$

It is however only necessary to know that equality holds if $ (a \plus{} \frac {1}{b}) \equal{} (b \plus{} \frac {1}{c}) \equal{} (c \plus{} \frac {1}{a})$, which is satisfied if a=b=c. Cubing the inequality gives the desired inequality:
$ (a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a}) > \equal{} {(1/3 \plus{} 1/(1/3))}^{3}$[/quote]

This may be wishful thinking but could this proof be saved if we show at the minimum a=b=c ?

we have at the minimum  (1)     $ (a \plus{} \frac {1}{b}) \equal{} (b \plus{} \frac {1}{c}) \equal{} (c \plus{} \frac {1}{a})$

assume without loss of generality $ b > c$ thus from (1) we get $ \frac {1}{c} < \frac {1}{a} \Leftrightarrow a < c$ so $ b > a$ again from (1) $ \frac {1}{c} < \frac {1}{b} \Leftrightarrow b < c$ which contradicts our assumption. From the fact this argument applies whichever variables we choose we get a=b=c at the minimum.

[quote="mehmetcantu"]if a,b,c are positive real numbers with a+b+c=1

Prove this inequality
$ (a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a}) > \equal{} (10/3)^3$
(dont know is it possible to learn solutions of  mathematics and youth magazine 2007)[/quote]

$ (a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{a})$
$ \equal{}abc\plus{}\frac{1}{abc}\plus{}10\ \cdots [*]$ with $ 0<abc\leq \frac{1}{27}$ by A.M-G.M.

Since $ y\equal{}x\plus{}\frac{1}{x}$ is decreasing function for $ 0<x\leq \frac{1}{27}$,

$ \therefore [*]\geq \frac{1}{27}\plus{}27\plus{}10\equal{}\frac{1000}{27}$. Q.E.D.