algorithm
geometry
algebra
polynomial
search
floor function
induction

Prison Warden Willy is in charge of guarding a not-necessarily convex polygonal prison. He is incredibly lazy, and does not want to have to manually watch the prisoners. However, he feels for safety reasons that he needs to be able to remotely see into every nook and cranny of each and every cell.

Then he goes to Home Depot and notices that they spell special "vertex-mounted" swivel cameras that have a 360 degree field of view, blocked only when their line of sight is interrupted.

Can Warden Willy devise an algorithm that will let him place these cameras on particular vertices so as to be guaranteed two things?

1) Complete visual coverage of the prison
2) Minimum number of cameras purchased

For any given prison, do there exist multiple camera configurations which use the same (optimal) number of cameras?
Example of such a set-up:
 [geogebra]9828ce4deb89c177de5f05debccbe8d371973610[/geogebra]

This is not at all a graph theory problem.  Anyway, I will go ahead and conjecture that one needs at least as many cameras as concave angles in the prison.

I think you're right. This is more of a geometry problem, in retrospect.

If you take a star shape, like take a regular $ n$-gon and join sides $ A_i$ and $ A_{i\plus{}k}$ (indices mod $ n$) for some $ k$ not equivalent to $ \minus{}1,0,1$ mod $ n$, then make the region the interior area, a camera in the center can easily cover it, but you have as many concave angles as you want.

I would guess this problem is NP-hard or something. You can probably easily come up with a suboptimal algorithm involving some sort of triangulation and then placing a camera where it covers the most triangles, but I feel like the challenge of coming up with an algorithm that guarantees the least amount of cameras without brute force is similar to the challenge for doing it for TSP and the like.

Can we replace t0r's "at least" with "at most" just by placing a camera at each angle of measure more than 180 degrees?

Edit (Re: Ravi's post that follows): I knew this problem sounded familiar!

[quote="Ravi B"]I didn't read the problem carefully, but it sounds like the [url=http://en.wikipedia.org/wiki/Art_gallery_theorem]Art Gallery Problem[/url].[/quote]

(I'm not sure if my reduction works; I took a course in algorithms over the summer and we brushed over reductions, so this might not be correct.)

[hide]It is the Art Gallery Problem, which is NP-hard as MellowMelon claimed. If you can solve the dominating set problem, then you can solve this problem as follows: determine if a dominating set of size one (one camera) suffices for the polygon, if not, then check if a dominating set of size two, and iterate up until you obtain a valid number of cameras. This takes linear time in the number of cameras, which runs in polynomial time. (This could be potentially improved for by using binary search, although it is unnecessary.) Therefore, the Art Gallery Problem is NP-hard.[/hide]

Wow! I didn't know this sort of thing was studied so intensely.

Well, I think that I have a solution for the special-case I gave here, which IS indeed a graph-theoretical solution.
If people are interested, I can post it soon.

Can these irregular n-gons intersect themselves?

Here are some of my idea's

For any irregular, non-self-intersecting n-gon, there are atmost $ \left\lfloor \frac{n}{2} \right\rfloor$ concave angles. Hence we need at most that many cameras.

If there are only $ m$ concave angles in the n-gon, and they are all next to each other then we have to have $ m$ cameras

No intersection. I think you're on the right track.

I think you guys are liking this concave angle idea a bit too much. Both your $ \lfloor n/2 \rfloor$ claim and your claim that $ m$ consecutive concave angles means $ m$ cameras are false. Consider a right triangle, but push the hypotenuse in towards the vertex a tiny bit to make an arc. You can pick an arbitrary amount of vertices on this arc, all with concave angles. We can get arbitrarily many concave angles with 3 convex angles. Also, one camera on the vertex of the triangle will suffice to cover everything.

I suppose it's been sufficiently long. I'll post what I think is one solution:
[hide]
First, take the graph and triangulate it. Then construct a three-coloring of the vertices. This will label approximately $ \frac {n}{3}$ of the total vertices with each color. (In reality, it will be slightly more for some, is the graph does not have a nodes in a multiple of three (for example, the one given). )

Determine which of the vertex colors is the least represented on the graph, and place one camera at each of these vertices. This will always be $ \lfloor\frac{n}{3}\rfloor$, though there are sometimes multiple ways to cover the prison with this many.

Short explanation of correctness:
When you triangulate the graph, you divide it into regions in which every single camera on the vertices of that triangle can see the entire interior. Therefore, for each triangle you choose, you only need to place at MOST one camera on the vertices. It would be suboptimal to have any triangle with two or more cameras, from a heuristic point of view. By three-coloring the graph, we isolate which triangles have the vertices with the fewest camera-overlaps, and from here we can generate one solution.

Here's an image of two possible solutions:
The three-coloring has labeled the vertices 1, 2, or 3. The YELLOW vertices (choosing 1's) are one optimal covering, and the large BLUE vertices (choosing 3's) are another.

[img]19146[/img][/hide]

[quote="DiscreetFourierTransform"]First, take the graph and triangulate it. Then construct a three-coloring of the vertices.[/quote]  Why is it obvious that a 3-coloring exists?

[hide="Proof a 3 coloring exists"]You can prove easily by induction that a triangulation of an $ n$-gon always has $ n\minus{}2$ triangles. If every triangle shared at most one edge with the $ n$-gon, the $ n$-gon could only have $ n\minus{}2$ edges, which is a contradiction. So at least one triangle has to share two edges with the $ n$-gon.

Triangles are clearly $ 3$-colorable. Suppose we can color all $ k$-gons. In a $ k\plus{}1$-gon, find a vertex $ V$ such that both of the edges linking to it are contained in the same triangle (exists by the above paragraph). Temporarily ignore $ V$ and use the inductive hypothesis to color everything else. $ V$ is connected to only two other vertices. Pick the color neither one uses.[/hide]

What's the complexity class of generating first a triangulation, then a three-coloring?
It can't be THAT hard. Or is it still NP?

I'm fairly sure triangulating would be linear, right?

Graph coloring is [url=http://en.wikipedia.org/wiki/Graph_coloring]pretty hard[/url], although the specific problem of finding a $ 3$-coloring isn't quite the same as the decision problems listed in the Wikipedia article.  See, for example, [url=http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WH3-4D1V78P-3&_user=501045&_rdoc=1&_fmt=&_orig=search&_sort=d&view=c&_acct=C000022659&_version=1&_urlVersion=0&_userid=501045&md5=1597f94db1962e290407227f449534c4]this paper[/url].

$ \lfloor n/3 \rfloor$ is not necessarily the minimum; take either of the two examples I've posted in this thread. The problem is constructing the actual minimum. What you have is a suboptimal algorithm. For most NP-hard problems these are pretty easy to come up with.

Feel free to correct me anywhere you find mistake.

In given set-up minimum number of cameras would be 2 so your solution isn't correct.

My knowledge is too poor to rigorously prove it but consider this algorithm:

$ 1)$ Find all concave angles and consider placing camera on any of them.

$ 2)$ Determine "field of vision" for every possible place of camera (field of vision is polygon for which every point inside the polygon can be connected with vertex of the angle with straight line without crossing sides of the original polygon).

$ 3)$ Count the number of polygons which are not included in the "field of vision". Let those be the "other" polygons.

$ 4)$ Determine angle which "field of vision" splits original polygon in least possible "other" polygons. Place a camera there. In case that some angles produce same number of the "other" polygons, there might be multiple solutions to the problem.

$ 5)$ Treat any remaining "other" polygon same way as the original one (in other words repeat the algorithm).

I'm sorry, could you demonstrate how it's possible to do the example I posted with two cameras?

DFT: Pick the two #3 vertices on concave angles in your last diagram. That covers everything.

Flame: I am 99.9% sure that with enough thought you could construct a room for which that algorithm did not come up with the [i]optimal[/i] configuration. There are some fast and very good algorithms for TSP, for example, but none of them can guarantee the minimum.

The wikipedia article has a number of good references (and I think the reference DiscreetFourierTransform describes is actually covered therein).

I agree, in the way it was, prison can be constructed so the algorithm wouldn't produce optimal solution but I think I corrected the flaws. If you still think it's wrong, please prove or demonstrate.

[hide="algorithm"]
$ 1)$ Check if there is any trivial solution (solution with only 1 camera)

$ 2)$ Find all concave angles and consider placing camera on any of them.

$ 3)$ Determine "field of vision" for every possible place of camera (field of vision is polygon for which every point inside the polygon can be connected with vertex of the angle with straight line without crossing sides of the original polygon).

$ 4)$ Count the number of polygons which are not included in the "field of vision". Let those be the "other" polygons.

$ 5)$ If there are 2, or more "other" polygons that can be seen from a single vertex, treat it as 1 "other" polygon.

$ 6)$ Determine angle which "field of vision" splits original polygon in least possible "other" polygons. Place a camera there. In case that some angles produce same number of the "other" polygons, there might be multiple solutions to the problem.

$ 7)$ Treat any remaining "other" polygon same way as the original one (in other words repeat the algorithm).[/hide]

This is my solution (one of) for the proposed set-up:

[hide]
[img]http://i217.photobucket.com/albums/cc7/Rakia69/200902211348_609.jpg[/img][/hide]

I still have a feeling I'm missing something big as this seems to easy to be right.


function
number theory
Diophantine equation

I don't see much of a difference  between infinite descent and the well-ordering-principle. I assume it's only because of the way I first saw it/used it, but I would like to know more. Could someone explain the WOP to me?

Thanks.

The phrase "well-ordering principle" is used to mean two distinct but related concepts.  The one you are thinking of is that the natural numbers are well-ordered; in other words, any subset of the natural numbers has a minimum.

Infinite descent refers to a proof technique for showing that a (usually Diophantine) equation has no solutions by showing that if that equation has solutions, then it is possible to construct a subset of the natural numbers without a minimum, which violates the well-ordering principle.  The two proofs, for any given problem, are exactly equivalent - that is, a proof phrased using infinite descent can just as well be phrased using well-ordering.  It is usually considered a little nicer to use well-ordering.

(The well-ordering principle in a more general context states that it is possible to define a well-ordering on any set.  This statement is equivalent to the axiom of choice (the least element of a set is a choice function), and in this form is quite unintuitive; what exactly does a well-ordering of the reals look like?)

Ah, thanks, the middle paragraph was exactly what I was looking for :)


Prove the following for triangular numbers $ t_{n}$:

$ t_{1}\plus{}t_{2}\plus{} ...\plus{}t_{n} \equal{} n(n\plus{}1)(n\plus{}2)/6$.

A similar proof can be found [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=306743]here[/url].


Can someone help me with this problem?
In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio 3:4:5?

[hide="Hint"]
The $ k$th element of the $ n$th row of Pascal's Triangle is $ \binom{n}{k}$.  Thus, $ \binom{n}{k\minus{}1}: \binom{n}{k}: \binom{n}{k\plus{}1}\equal{}3: 4: 5$.  Simplify these.
[/hide]

[quote="Jongao"]On the AoPS wiki it says n=62.
I also got 62 for the top number, but doesnt that mean that it is the 63rd row?[/quote]

The top "row" is considered the 0th row.

Heres how I did it. Let entries be
$\binom{x}{m-1}$, $\binom{x}{m}$ and $\binom{x}{m+1}$

So write that out.

$\frac{(x)(x-1)...(x-(m-1)+1)}{(3)(m-1)(m-2)...1}$=
$\frac{(x)(x-1)...(x-m+1)}{(4)(m)(m-1)(m-2)...1}$=
$\frac{(x)(x-1)...(x-(m+1)+1)}{(5)(m+1)(m)(m-1)(m-2)...1}$

Now let $(x)(x-1)...(x-(m-1)+1)$ which simplifies to $(x)(x-1)...(x-m+2)$ be A.

And let $(m-1)(m-2)...1$ be B. Substitute.

$\frac{A}{(3)B}$=
$\frac{A(x-m+1)}{(4)mB}$=
$\frac{A(x-m+1)(x-m)}{(5)m(m+1)B}$.

It is much easier to pair equation 1&2 and 2&3 instead of 1&3 since we only need two equation anyway. Get.

$\frac{A}{(3)B}$=$\frac{A(x-m+1)}{(4)mB}$. Simplify. Get.
$\frac{1}{(3)}$=$\frac{(x-m+1)}{(4)m}$ As you can see A and B are nice place holders simplify tricky/uglyness. Cross-multiply.

$4m=3x-3m+3$ which is $7m=3x+3.$ Next equation!

$\frac{A(x-m+1)}{(4)mB}$=$\frac{A(x-m+1)(x-m)}{(5)m(m+1)B}$.  Simplfiy.
$\frac{1}{(4)}$=$\frac{(x-m)}{(5)(m+1)}$. Cross-multiply.

$5m+5=4x-4m$ So $9m=4x-5$ Now you have two equations.

$7m=3x+3.$ and $9m=4x-5$. Solve.
$63m=27x+27$ and $63m=28x-35$. Equal.
$27x+27=28x-35$ which you get x=62. Remember what x was in beginning. Since "$x$ choose anything" is part of "row$ x+1$", answer is 63.
(1st row is row 0, as AIME15 says).


limit
integration
real analysis
real analysis unsolved

Let $f(x)=2x(1-x), x\in\mathbb{R}$ and denote $f_n=f\circ f\circ ... \circ f$, $n$ times.
(a) Find $\lim_{n\rightarrow\infty} \int^1_0 f_n(x)dx$.
(b) Now compute $\int^1_0 f_n(x)dx$.

[quote=zhaobin]ok,
$f_n(x)=\frac 1 2 -2^{2^n-1}(x-\frac 1 2)^{2^n}$
by induction. :)[/quote]

elaborate plz!

Checking inductively that zhaobin's formula works is pretty trivial. However, I can try to explain how zhaobin may have arrived at his formula.Well, start with the obvious fact that $f_{n+1}=-2f_n^2+2f_n$. Now, we complete the square:

$f_{n+1}-\frac{1}{2}=-2f_n^2+2f_n-\frac{1}{2}= -2(f_n -\frac{1}{2})^2$
Which leaves us with:
$f_{n+1}= \frac{1}{2}-2(f_n -\frac{1}{2})^2$
However,
$f_1= \frac{1}{2}-2(x-1/2)^2$, which leads pretty quickly to zhaobin's formula.


When you have a fraction in the exponent, you're supposed to evaluate the numerator before the denominator, right?

I do not believe it matters which order you evaluate but by convention, you evaluate the numerator before the denominator


linear algebra
matrix
quadratics
vector
calculus
derivative
function

Let $A=(a_{ij})_{n{\times}n}$ be a symmetric matrix with real entries.Further let's consider the quadratic form $W(\overline{x})=<\overline{x},A\overline{x}>$,for $\overline{x}{\in}L$,where $L$ is a $n$ dimensional
linear space.It is well known that all proper values of $A$(sorry if the word is not right) are real and it has a diagonal canonical(Jordanian) form.So if all proper values of $A$ are positive then $W(\overline{x})$ will be a positively defined quadratic form.
Now I came upto an interesting question.What if $A$ is not a symmetric one.I have a conjecture which seems to be right:
[i]If all real parts of proper values of[/i] $A$ [i]are positive then $W(\overline{x})$ is a positive defined.[/i]
What do you think about it?Is it right?

Does "proper values" mean "eigenvalues"?

I'm sure that's what it means (that's sort of what it sounds like in Romanian too :)).

Anyway, I don't think it's true. Think about the matrix $A=\left(\begin{array}{cc}0&1\\-2&2\end{array}\right)$. If ${\bf v}$ is the column vector $\left(\begin{array}{c}x&y\end{array}\right)$, then you get ${\bf v}^t\cdot A\cdot {\bf v}=-xy+2y^2$. Maybe I'm missing something? :?

Indeed [b]grobber[/b] the conjecture fails even when n=2.Strange that I didn't notice that :blush: .Anyway thanks for the example.Now I must think on a new way to prove the lemma in the paper. :)

What was the lemma? Could you share it with us? :) Or maybe the location of the paper, if it can be found online?

Ok,[b]grobber[/b] I'll post the lemma later.The paper is on [i]Lyapunov[/i]'s theorem for steady equilibrium of autonom differential equations.

This conjecture is, however, true for normal matrices, i.e. matrices which satisfy $AA^t=A^tA$ (a similar matrix is a special type of normal matrix).

A matrix is normal iff it's unitarily diagonalizable (we're working in $\mathbb C$ here). Given the vector ${\bf x}=(x_1\ x_2\ \ldots\ x_n)^t$, we have ${\bf x}^*A{\bf x}={\bf\tilde x}^*U^*AU{\bf \tilde x}$, where $U$ is a unitary matrix s.t. $U^*AU$ is diagonal, and ${\bf\tilde x}$ is the unique solution to $U{\bf\tilde x}={\bf x}$.

Now, if $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$, and ${\bf\tilde x}=(\tilde x_1\ \ldots\ \tilde x_n)^t$, then ${\bf\tilde x}^*U^*AU{\bf \tilde x}=\lambda_1|\tilde x_1|^2+\ldots+\lambda_n|\tilde x_n|^2\ (*)$. We've been working with complex numbers up until now, but we had started off with a real matrix $A$, and real ${\bf x}$, so $(*)$ is real, meaning that its equal to its real part, which is $\sum\Re\lambda_i\cdot|\tilde x_i|^2$, and this is always $>0$, unless ${\bf\tilde x}=0$, which is equivalent to ${\bf x}=0$.

I hope it's Ok.

Sorry that I type the lemma so late.As I told we consider the autonom system of differential equations:
$\dot{x_{1}}=f_{1}(x_{1},x_{2},\ldots,x_{n}) \\
\dot{x_{2}}=f_{2}(x_{1},x_{2},\ldots,x_{n}) \\
\vdots \\
\dot{x_{n}}=f_{n}(x_{1},x_{2},\ldots,x_{n})$,
where derivatives are taken over time $t$ and where all functions are defined in some region $\Delta{\subset}\mathbb{R}^{n}$ and have continious derivatives up to the second order.We will say that position $\vec{a}{\in}\mathbb{R}^{n}$ is an equibrilium position for this system if $\vec{x}(t)=\vec{a}$ is a solution for this system or equivalently $\vec{f}(\vec{a})=\vec{0}$.Also let's denote by $\vec{\varphi}(t,\vec{\xi})$ the solution of that system with starting parametres $0,\vec{\xi}$.
Denote $a_{ij}=\frac{\partial{f_{i}(\vec{a})}}{\partial{x_{j}}}$ and $A=(a_{ij})_{n{\times}n}$.
The position $\vec{a}$ is called steady by Lyapunov if
a)For sufficently small $\delta>0$ the solutions $\vec{\varphi}(t,\vec{\xi})$ where $|\vec{\xi}-\vec{a}|<\delta$ are defined for all $t{\geq}0$.
b)For every $\epsilon>0$ there exists a $\sigma>0(\sigma{\leq}\delta)$ s.t. $|\vec{\varphi}(t,\vec{\xi})-\vec{a}|<\epsilon$ for all $t{\geq}0$,as soon as $|\vec{\xi}-\vec{a}|<\sigma$.
If we add up third condition
c)There exists a $\sigma>0(\sigma{\leq}\delta)$ s.t. $\lim_{t{\rightarrow}\infty}{|\vec{\varphi}(t,\vec{\xi})-\vec{a}|}=0$,as soon as $|\vec{\xi}-\vec{a}|<\sigma$.
we call the position $\vec{a}$ asymptotically steady equbrilium.
Lyapunov's theorem states that if all eigenvalues of matrix $A$ have negative real parts then
$\vec{a}$ is an asymptotically steady equibrilium.Moreover there exists an $\sigma>0$ s.t.
$|\vec{\varphi}(t,\vec{\xi})-\vec{a}|{\leq}r|\vec{\xi}-\vec{a}|e^{-{\alpha}t}$ for $t{\geq}0$,as soon as $|\vec{\xi}-\vec{a}|<\sigma$,where $r,\alpha>0$ are independent of $\vec{\xi}$.
The main idea of the proof in the paper is to find some positively defined quadratic form $W$ s.t. the distance of the solution from equbrilium in terms of that quadratic form:$w(t)=W(\vec{\varphi}(t,\vec{\xi})-\vec{a})$,have a tendence to decraese i.e. $\dot{w}(t){\leq}0$.Moreover $\dot{w}(t){\leq}-2{\alpha}w(t)$ for some $\alpha>0$.If we take standart norm we come up to the lemma that I posted.


Someone has been messing with my profile and i haven't told anyone my password!

Make sure you log out whenever you are finished on the site (and if you have a computer set to automatically log in, you are giving others the opportunity to edit your profile).


Call $ D, E$ be the midpoints of $ BP, CQ$ respectively

$ QDM$ and $ MER$ are equal triangles

So $ MQ=MR$

Thanks. Too easy. stupid of me, not to see this.

Tanvir

[quote="tanvirabs"]Inside triangle ABC, P is a point such that angle(PBA)=angle(PCA). Q and R are the feet of the perpendiculars on AB and AC respectively from P. M is the midpoint of BC. Determine whether QM=RM or not.

Tanvir[/quote][/quote]

 [b]History ![/b]

 I think that this nice problem was proposed by Great Britain at an IMO ( Short List) , but I  can not remember the year.


[quote="asintota"][size=200]$ find$ $ i^{i^{i}}$[/size][/quote]

The conventiol definition of exponents is used here ...

$ i^{i}= e^{i\log i}$ now $ \log i =\ln |i|+i \arg (i) = \frac{\pi}{2}i$

And so, $ i^{i}= e^{-\pi/2}= z$

So, $ i^{z}= e^{z\log (i)}$ but $ z\log (i) =e^{-\pi/2}\cdot \frac{\pi}{2}i$

So, $ i^{z}= i^{i^{i}}= \cos \left( \frac{\pi}{2}e^{-\pi/2}\right)+i\sin \left( \frac{\pi}{2}e^{-\pi/2}\right)$


AMC
AIME
probability
USA(J)MO
USAMO
AIME II

Another tumbleweed :maybe:
so how did everyone do on the AIME I?

took it for practice and got an 8.

did some stupid stuff, but could've messed up even more.

AIME II: 6
Answered 1-7, Missed #1

Got to work hard for next year...  No more of this missing the cutoff stuff and stress about large probability of missing the cutoff...

And... Nice job, Kongster and good luck on the USAMO!

i officially got a 1.

but i put 7 correct answers

i didn't bubble my ansewrs to 1-8  :|

[quote="Kongster"]Lol thanks, im totally going to fail due to lack of studying though :P
wait unimpossible, you just didnt bubble in the answers?[/quote]

yeah i didn't bubble ansewrs for the first line of questions (#1-8) XD

I haven't posted anything in ages.
Do I still count as a student of Virginia?
Do my roots go down deep enough for me to
hold this view hundreds of miles away?
Anyways, as the year flies past and I turn
into a depressing lower, I wish all of you guys
luck as freshmen and 8th graders, and there's
definitely something to look forward to in high school...
that is, graduation and senior slackoff.
Survive for one more month~

In Young! Gosh! You're still a student of Virginia in our hearts here at TJ for those who remember you from longfellow. I would guess your AIME score this year was double digit but that's just my speculation.

Nice poem and have fun at MOP!

i got a 7 this year.

i came out of the test pretty relieved that i didn't make more than a couple mistakes.  I thought a 7 would be safe for USAMO with a 138 amc 10.

turns out 7 was pretty average.  Looks like USAMO will evade me for at least another year.

unimpossible is in 9th grade

@unimpossible, the average for AIME is between 2 and 3. I'm sure that the USAMO index will be around 204 this year, cuzz it was like a medium hard AIME.

I really really really sucked at this year's AIME,

yeah i really didn't improve much from last year.

if you want some advice, if you want to do well in extracurricular academic stuff, don't do sports.

i got pretty pathetic at math during basketball...

What extracurricular academy stuff do you do, unimpossible?


floor function
ceiling function
inequalities
algebra proposed
algebra

Suppose $x_{1},x_{2},\dots,x_{n}\in\mathbb R$ are different numbers. Prove that there exist $f(x_{1}),f(x_{2}),\dots,f(x_{n})$ such that $f(x_{i})\in\{\lfloor x_{i}\rfloor,\lceil x_{i}\rceil\}$ and for each $S\subset \{1,2,\dots,n\}$: \[|\sum_{\sigma\in S}f(x_\sigma)-\sum_{\sigma\in S}x_\sigma|\leq \frac{n+1}4\]

I don't undestand, what is x? $f(x_{i})$ integer?
If $f(x_{i})=x_{i}$ the sum is 0.

It had a little typo. I corrected it. You see $f(x_{i})\in\mathbb Z$. You must choose $f(x_{i})$ such that the inequality holds.

WLOG, suppose $0\leq x_{1},x_{2},\dots,x_{n}\leq1$ and $x_{1}<x_{2}\dots x_{n}$. If we find $k$ that $x_{1}+\dots+x_{k}\leq\frac{n+1}{4}$ and $x_{k+1}+\dots+x_{n}\geq (n-k)-\frac{n+1}4$ then problem is easily solved.
For this find largest $k$ that $x_{1}+\dots+x_{k+1}\leq\frac{n+1}4$. So $x_{1}+x_{2}+\dots+x_{k+1}>\frac{n+1}4$. Thus $x_{k+1}>\frac{n+1}{4(k+1)}$. Then $x_{k+1}+\dots+x_{n}\geq \frac{(n+1)(n-k)}{4(k+1)}\geq (n-k)-\frac{n+1}4$. And the last inequality holds with simple AM-GM.


How much do you have to score on CMO (Canadian Mathematics Olympiads) to represent Canada at IMO.
And to say, lowest required score.
I know it's based upon a percentile but from your experiences what should be the minimum score?
Thanks

To represent at the IMO, your USAMO, APMO and CMO scores are taken into consideration

So it is not only your CMO score? I thought that it was the top six who were selected annually from the CMO. Also, how do you participate in the APMO?


calculus
derivative
limit
induction
real analysis
real analysis unsolved

Say f:R->R has n derivatives at zero, and they all have value zero.  Then is it true that f is little-oh of x^n at zero?

If we have that f is n+1 times continuously differentiable in a neighborhood of zero, then by Taylor's theorem, we could show this to be true.  However, we only assume f to have an nth derivative at the point x=0 (note that for this to be true, f must have n-1 derivatives in a neighborhood of zero).

By definition of the derivative, this is true for n=1.

You sould keep disctinct Taylor theorem and Peano theorem (in Italy we use this terminology). According to the latter
$ f(x) \minus{} P_n(x) \equal{} o((x \minus{} x_0)^n)$ as soon as $ f$ admits n derivatives at $ x \equal{} x_0.$ The proof makes use of the l'Hopital's rule $ n \minus{} 1$ times on the quantity
$ {f(x) \minus{} P_n(x)\over (x \minus{} x_0)^n}$ until you obtain $ {1\over n!}\lim_{x\to x_0}\left({f^{(n \minus{} 1)}(x) \minus{} f^{(n \minus{} 1)}(x_0)\over x \minus{} x_0} \minus{} f^{(n)}(x_0)\right) \equal{} \lim_{x\to x_0} (f^{(n)}(x_0) \minus{} f^{(n)}(x_0)) \equal{} 0$ (actually an induction to make the proof rigorous would be necessary).  By the Peano theorem you just know that the remainder is $ o((x \minus{} x_0)^n)$ and nothing more. The Taylor theorem gives you more information about the remainder but at least the $ (n \plus{} 1)$-th derivative except at $ x_0$ is necessary

Without L'Hospital, you can do the following. Let me just assume $ f(0) \equal{} f'(0) \equal{} f''(0) \equal{} 0$ for simplicity. The mean value theorem gives $ f(x) \equal{} f(x) \minus{} f(0) \equal{} f'(c_x)x \equal{} [(f'(c_x)\minus{}f'(0))/c_x]c_xx$. As $ x\rightarrow 0$, the term in brackets $ \rightarrow f''(0) \equal{} 0$, and $ |c_xx| < x^2$ , so $ f(x) \equal{} o(x^2)$.

Nice.

[quote="WWW"]Without L'Hospital, you... [/quote]


limit
logarithms
calculus
integration
calculus computations

Evaluate
$\displaystyle\lim_{n\to\infty}\left(\prod_{k=1}^n\left(1+\frac{k}{n}\right)\right)^{\frac{1}{n}}$

i have been able to simplify it to $\displaystyle\lim_{n\to\infty}\left(\frac{1}{n}\right)((n+1)(n+2)\dots(2n))^{\frac1n}$
but from here i have no clue.
could someone please point me in the right direction.
thanks.

[quote="maokid7"]
could someone please point me in the right direction.
thanks.[/quote]
Take the logarithm of the product and think of the resulting expression as of a Riemann sum.

ok so i got an answer. i just want to make sure it is correct
[hide="my solution"]
so as suggested i first take the natural log.
$f(n)=\displaystyle\lim_{n\to\infty}\left(\prod_{k=1}^n\left(1+\frac{k}{n}\right)\right)^{\frac{1}{n}}$
so
$ln(f(n))=\displaystyle\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^nln\left(1+\frac{k}{n}\right)$
so turning this into an integral we have
$ln(f(n))=\displaystyle\int_{1}^2ln(x)dx$ (if i made a mistake it is most like that it was made here  :? )
so evaluating this we find that $ln(f(n))=2ln2-1$.
to find $f(n)$ we just undo the $ln$ by raising $e$ to both sides.
so $f(n)=e^{2ln2-1}=\frac{4}{e}$.
[/hide]

You did it correctly.

$\frac1{n} = dx$

Thus the limits of integration vary are from $0$ to $1$, this makes $x = k dx$.

$\frac{k}{n} = x \rightarrow \ln (f(n)) = \int_0^1 \ln (1+x) \, dx = \int_1^2 \ln x \, dx$

$\int \ln x \, dx= x\ln x - x$

$\int_1^2 \ln x \, dx = 2\ln 2 - 2 - \ln 1 + 1 = 2\ln 2 - 1$

I just wrote in the extra steps for clarification.

Good job.


geometry
3D geometry
frustum
symmetry
LaTeX

I believe that in the line "area __ square feet and __ square feet respectively", the blanks need to be filled in in order to do this problem.

b-flat: I don't believe you can copy text that has some latex and expect the latex to show up when pasted unless you copy the code itself - same with the other threads that besttate said he couldn't answer.

20. A fuel tank is in the shape of a right frustum (truncated cone) as shown. Its parallel circular bases have area square feet and square feet respectively. When the fuel tank is half filled as shown (up to the axis of symmetry), the area of the surface of the fuel (cross section perpendicular to the bases, including the axis of symmetry) is 108 sq. ft. What is the volume, in cubic feet, of the fuel in the tank?  that is the question.

A fuel tank is in the shape of a  right frustum  (truncated cone) as shown.  Its parallel circular bases have area 36 pie  square feet and   9 pie square feet respectively. When the fuel tank is half filled as shown (up to the axis of symmetry), the area of the surface of the fuel (cross section perpendicular to the bases, including the axis of symmetry) is 108 sq. ft.  What is the volume, in cubic feet, of the fuel in the tank?

now it has been fixed.


modular arithmetic
number theory
Divisibility Theory

Determine all pairs $(n,p)$ of nonnegative integers such that [list] [*] $p$ is a prime, [*] $n<2p$, [*] $(p-1)^{n} + 1$ is divisible by $n^{p-1}$. [/list]

If $ p \equal{} 2$  then $ (p \minus{} 1)^{n} \plus{} 1 \equal{} 2$ so $ n \equal{} 2$.
If $ n \equal{} 1$ then $ (p \minus{} 1)^{n} \plus{} 1$ is divisible by $ n^{p \minus{} 1}$ for any prime $ p$. If $ n \equal{} 2$, then $ (p \minus{} 1)^{n} \plus{} 1$ must be even so $ (p \minus{} 1)^{n}$ is odd so $ p$ is even but it is prime therefore $ p \equal{} 2$.
This gives the solutions $ (1,p)$ and $ 2,2$ when $ n\leq 3$ or $ p\leq 3$.

Now look at the case when $ p,n\geq 3$. Because then $ p$ is odd, $ (p \minus{} 1)^{n} \plus{} 1$ is odd so $ n$ is odd. Let $ x$ be the greatest prime divisor of $ n$, then we have $ x|n|n^{p \minus{} 1}|((p \minus{} 1)^{n} \plus{} 1)$ so $ (p \minus{} 1)^{n}\equiv \minus{} 1\pmod x$. Now because $ x|n$ then $ x \minus{} 1$ and $ n$ are coprime so there will exist positive integers $ a,b$ such that $ an \equal{} b(x \minus{} 1) \plus{} 1$. Because $ n$ is odd, $ x$ is odd so $ b(x \minus{} 1)$ is odd so $ b(x \minus{} 1) \plus{} 1$ is even hence $ a$ is odd. But then, if $ (p \minus{} 1)^{n}\equiv \minus{} 1\pmod x$, $ (p \minus{} 1)^{an}\equiv \minus{} 1\pmod x$ therefore $ (p \minus{} 1)^{b(x \minus{} 1)}*(p \minus{} 1)\equiv \minus{} 1\pmod x$. But by Fermat's Little Theorem $ (p \minus{} 1)^{x \minus{} 1}\equiv 1\pmod x$ because $ x$ and $ p \minus{} 1$ are coprime (otherwise, it would be impossible to have $ n^{p \minus{} 1}|((p \minus{} 1)^{n} \plus{} 1)$ as $ x|n$. Then, $ (p \minus{} 1)^{b(x \minus{} 1)}*(p \minus{} 1)\equiv \minus{} 1\pmod x$ gives $ (1)^{b}*(p \minus{} 1)\equiv \minus{} 1\pmod x$ so $ p \minus{} 1\equiv \minus{} 1\pmod x$ so $ x|p$ so $ x \equal{} p$. But $ x|n$ and $ n < 2p$ so $ n \equal{} p$.
Then $ n^{p \minus{} 1}|((p \minus{} 1)^{n} \plus{} 1)$ is equivalent to $ p^{p \minus{} 1 }|((p \minus{} 1)^{p} \plus{} 1)$ but if we expand $ (p \minus{} 1)^{p} \plus{} 1$ we get:
$ p^{p \minus{} 1}| p^{p} \minus{} p^{p \minus{} 1}*\binom{p}{1} \plus{} p^{p \minus{} 2}*\binom{p}{2} \plus{} ... \plus{} p*\binom{p}{1} \minus{} 1 \plus{} 1$
or:
$ p^{p \minus{} 1}|p^{2}(p^{p \minus{} 2} \minus{} p^{p \minus{} 3}*\binom{p}{1} \plus{} ... \plus{} \binom{p}{p \minus{} 2} \plus{} 1)$
and therefore $ p^{p \minus{} 1}|p^{2}$ because the expression in brackets is not divisible by $ p$ hence the only solution possible is if $ p \minus{} 1\leq 2$ so $ p\leq 3$. But we assumed $ p\geq 3$ so $ p \equal{} 3$ so $ n \equal{} 3$.

So the solutions are $ (1,p), (2,2), (3,3)$

It is an IMO problem and it also relate with an IMO problem
Find all positive integer n such that :$\frac {2^n+1}{n^2}$ is an integer $(*)$
(put p=3 in A72)...
DR.TITU ANDREESCU wrote a comment in his book(Number theory:structures,examples and problems),the condition $n<2p$ is not necessary.
Very lucky that i found a solution of $(*)$ difference from the official solution by Lifting the exponent lemma in this link
http://reflections.awesomemath.org/2007_3.html
Work similarly,i think we can prove $A72$ without condition $n<2p$

Let $q$ be the smallest prime divisor of $n$. Hence $q|(p-1)^{n}-1$. Again from F.L.T we get $q|(p-1)^{q-1}-1$. Hence we get that $q|(p-1)^{gcd(q-1,2n)}-1$. Since q is the smallest prime divisor of $n$ so $gcd(2n,q-1)=2$. Hence $q|(p-1)^{2}-1$ i.e $q|p$ or $q|p-2$. Since $gcd((p-1)^{n}-1,p-2)=1$ so we conclude that $q|p$. and so $q=p$. Hence $p|n$. Now $n<2p$.So $n=p$. Now the problem is to find all primes such that $p^{p-1}|(p-1)^{p}-1$. Now let $p>4$. Then $p^{3}|(p-1)^{p}-1$ which forces that $p^{3}|p$. a contradiction. so $p<4$ and hence $p=2,3$.Now we check that these two values satisfies the required relations and hence $(n,p)=(2,2),(3,3)$.

[quote=rem] Now because $ x|n$ then $ x \minus{} 1$ and $ n$ are coprime[/quote]

Is this always true?? What about $n=21$, and $x=7$, for example??


Hi everybody!
the BMO is near(27 april)
Who is coming!
:)

Albanian team

1- Erion Dula
2- İlirian Aliaj
3- Ersiljo Tushaj
4- Gjergji Zaimi (Albanian Eagle)
5- Keler Marku (me)
6- Klevis Ymeri (unexpected)

common isnt anyone interested on this topic.
Arent there any students partecipating to BMO this year!

is  BMO = Balkan Mathematical Olympiad ?

My friend, the Bogy Man and I are coming

Are you partecipating for the first time this year?

The UK (who for a second year running are an invited country) team consists of:

Jos Gibbons
Tim Hennock
Jonathan Lee
Daniel Lightwing
Julia Robson
Dominic Rowland

No Mathlinkers in the group unfortunately.

The Bulgarian team is also ready:
1.Boris Stranjev
2.Anton Bikov
3.Emil Baronov
4.Tzvetelina Tzeneva
5.Deyan Simeonov
6.Dimitar Simeonov(me)

The last names of BUL5 and BUL6 are same because of coincidence ;)
Hope we have fun and present ourselves well.

The Romanian team is as follows:

(11) Adrian Zahariuc (dzeta) - bronze IMO 2004, gold IMO 2005
(12) Pachitariu Marius (ikap) - silver BMO 2004, 2005, bronze IMO 2005
(10) Dumitrescu Mihai - gold JBMO 2004
(12) Bazavan Eduard
(11) Turea Lucian - gold JBMO 2003
(12) Micnea Dragos - gold JBMO 2003, IMO 2005

[quote="Albanian Eagle"]who do you think will take the first place this year! :D[/quote]Romania of course :D


Here are some interesting questions that I am not quite certain how to do. Any help would be appreciated.

http://www.math.tamu.edu/~rojas/hw3.pdf

Of all people I didn't expect you to try to cheat on your homework :) Anyway, next time attach the PDF in the topic and/or write the problem in a separate post (which is the preferred way).

Haha, it's not really hw. I'm doing research at Texas A&M and these are suggested problems to get us to try and think about a research subject. I'm not fishing for answers here. Collaboration is definitely allowed and I posted the link so people know I'm not trying to be sneaky about it. Sorry if it came off that way!  :wink:

This is for an REU?! The U in REU stands for undergraduate, but this seems to be algebraic geometry. I wish I went there!


Let $ a,b,c\geq0$ such that $ a\plus{}b\plus{}c\equal{}5$. Prove that

$ \frac{(a\minus{}1)(b\minus{}1)}{c^2\minus{}2c\plus{}2}\plus{}\frac{(b\minus{}1)(c\minus{}1)}{a^2\minus{}2a\plus{}2}\plus{}\frac{(c\minus{}1)(a\minus{}1)}{b^2\minus{}2b\plus{}2}\leq1$

It's can be write into: Given $ a, b, c \geq\ 0$ and $ a\plus{}b\plus{}c\equal{}2$. Prove that: $ \sum\ \frac{ab}{c^2\plus{}1} \leq\ 1$
It's well-known ineq  :maybe:

[quote="nguoivn"]It's can be write into: Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 2$. Prove that: $ \sum\ \frac {ab}{c^2 \plus{} 1} \leq\ 1$
It's well-known ineq  :maybe:[/quote]

Thank you very much

[quote="nguoivn"]It's can be write into: Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 2$. Prove that: $ \sum\ \frac {ab}{c^2 \plus{} 1} \leq\ 1$
It's well-known ineq  :maybe:[/quote]

It must be $ a,b,c\ge \minus{}1$


inequalities
geometry
incenter
vector
inequalities proposed

In a triangle ABC ,G denotes the centroid and I the incentre

Prove 3(AG^2 +BG^2 +CG^2)/2 > AI^2 + BI^2 +CI^2 >=AG^2 +BG^2+CG^2

This is based on well known results so probably it cant be claimed to be original.

[quote="tranminhhoang"]For all M we have : sum(MA^2)>=sum(GA^2)[/quote]

You are right!

In general, for any point M, and any reals x,y,z, using vectors, we can prove:

$(x+y+z)(xMA^2 + yMB^2 + z MC^2) \geq yza^2 + zxb^2 + xyc^2$.

Taking x=y=z=1, gives: $MA^2 + MB^2 +  MC^2 \geq \frac{1}{3}(a^2 + b^2 + c^2) = GA^2+GB^2+GC^2$.


AMC USA(J)MO USAMO search probability geometry trigonometry

What is usually needed on USAMO to make MOP and how many people are taken on average each year? Also, if anyone knows what the grade level split is it would be greatly appreciated! :lol:

Ahhh..

This topic has been asked many many times. Why don't you go to:

http://www.unl.edu/amc

or search this forum? I keep seeing the same question over and over and if people just try to find it, I know they'll find them.

For topics, ahhhh, again. All stuffs are in there. Number theory, counting, probability, geometry, trigonometry, sequences and series, and all stuffs plus your knowledge of proof writing. That's the most important part.

 ;)


LaTeX
function
combinatorics unsolved
combinatorics

Does anybody have a nice solution for this problem ( like Double Counting ) ?

Prove that :

  ${\sum_{k=0}^n } {2k \choose k} . {2n-2k \choose n-k} = 4^n$

a small hint of LaTeX : use [code]{n \choose k}[/code] to implement the binomial coefficient ${n \choose k}$.  :) [/list]

assume that $a_0=1 ,a_1,a_2,...$ is sequence st$  \sum a_k.a_{n-k}=1$ now assume g(x) is generating function of {$a_n$} so$ g(x)^2 $ is generating function of {$b_n$} st$  \sum a_ka_{n-k}=b_n=\frac{1}{1-x} ,g(x)=\frac{1}{(1-x)^\frac{1}{2}} $we can easily see that the coefficient of $x^n$ in$ (1-x)^\frac{-1}{2} $equals$ {n\choose k}\div{4^n}$
so we have$ 4^n=\sum{2k\choose k}.{2n-2k\choose n-k}$ and then we are done also in concrete mathematics at page 186 u can find another solution.

A small correction.
Sam-n wanted to say that the coefficient of $x^n$ in $(1-x)^{-1/2}$ is $\frac{(-1)^n}{4^n}{2n \choose n}$.

P.S. Russian edition of Concrete Mathematics contains this example at page 214.


algebra
polynomial
number theory
prime numbers
superior algebra
superior algebra unsolved

For which prime numbers is the quotient ring

$\mathbb{F}_{p}[x]/(x^{2}+1)$ a field??

We r given $(x^{2}+1)$ is reducible in $\mathbb{F}_{p}$
and in the first part of this question I have
Proved that the multiplicative
group of Fp contains an element of order 4 and Deduced that p is congruent to 1 modulo 4.

I know that its a field iff its irreducible , any one knows how to apply this to the question, or wat idea i must focus on .. Thanks again

[quote="ueamath"]For which prime numbers is the quotient ring

$\mathbb{F}_{p}[x]/(x^{2}+1)$ a field??
[/quote]

We need that $x^{2}+1$ to be irreducible in $\mathbb{F}_{p}[x]$

Since this polynomial has degree 2 it is necessary and sufficient that $x^{2}+1$ have no zeros in $\mathbb{F}_{p}$

That is, $x^{2}+1 = 0$ has no solutions for all $x\in \mathbb{F}_{p}$. That means, $x^{2}\not =-1$. Thus, $-1$ is a quadradic [b]non[/b]-residue modulo prime $p$.

But it is a know fact that the Legendre symbol $(-1/p)=-1$ if and only if $p \equiv 3 \mod 4$ for odd primes. And it is trivial to see that for $p=2$ (even primes) it cannot be irreducible. Thus, $p=4k+3$ are all the primes.


floor function
search
combinatorics unsolved
combinatorics

The maximum number of pairwise incomparable subsets of a given set $X$ of $n$ elements is $\displaystyle \binom{n}{\left\lfloor \frac{n}{2} \right\rfloor}$.

You are right this is Sperner's theorem about antichains.
Use the search button with keywords 'Sperner' and/or 'antichain'...
For exemple :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=sperner&t=1162

Pierre.


1. The line which contains all points $(x,y,z)$ which are of the form $(x,y,z)=(2,-2,5)+k(1,-3,2)$ intersects the plane $2x-3y+4z=163$ at $P$ and intersects the $yz$-plane at $Q$. Find the distance $PQ$. Express your answer as a simplified radical expression.

2. The intersection of the planes $2x-y-3z=8$ and $x+2y-4z=14$ is the line $L$. Find the value $A$ so that $L$ is perpendicular to the line through $(A,2,2)$ and $(6,11,-1)$.

[hide="#1"]Points of the line are $(2+k,-2-3k,5+2k)$. Plug them into the plane equation to find $k$:

$2(2+k)+3(2+3k)+4(5+2k)=163$

$19k=133$

$k=7$

Hence $P(9,-23,19)$

The equation of $yz$-plane is $x=0$, therefore $2+k=0\iff k=-2$ gives $Q(0,4,1)$

The distance is $PQ=\sqrt{9^{2}+27^{2}+18^{2}}=9\sqrt{1+3^{2}+2^{2}}=9\sqrt{14}$
[/hide]

[hide="#2"]Rewrite the equations of the planes thus:

$2x-y=8+3z$
$x+2y=14+4z$

Multiply the first by $2$ and add it to the second to obtain

$5x=30+10z\implies x=6+2z$

Multiply the second by $-2$ and add it to the first to obtain

$-5y=-20-5z\implies y=4+z$

Therefore, the general point of the line is $(6+2z,4+z,z)$. Its vector is $\langle 2,1,1\rangle$. The vector of  the other line is $\langle 6-A, 9,-3\rangle$. The dot product of the two must be zero:

$2(6-A)+9-3=0\implies 18-2A=0\implies A=9$[/hide]


The computers in a computer room are in a network such that every computer is connected by a cable to exactly three others. The computers can exchange data directly or indirectly (via other computers). Let $ k$ be the smallest number of computers that need to be removed so that two of the remaining computers can no longer exchange data or there is only
one computer left. Let $ l$ be the smallest number of cables that need to be removed so that some two of the remaining computers can no longer exchange data. Show that $ k \equal{} l$.


Columbia Princeton college articles geometry

Hey ya'll,

I just came back from a few college tours and I've been thinking about what I want to do in college. One of the things that has been in the front of my mind is really learning a language. I'm currently taking spanish in highschool and will have completed 4 years by the time I graduate, but I'm not quite sure if that's what I want to do in college.

After hearing that Columbia offers classes that teach 40+ different languages, I was trying to figure out which language I might take when I attend college.

What I want is a language that I will be able to use quite frequently, that I will get some mileage out of. I want to master this language (probably spending a semester or two doing a 'study abroad' program in the country of the language and maybe obtaining a minor in it) and if I don't put it to any use, I'm sure that I won't recall it.

So, any suggestions on what the best language to learn nowadays for an aspiring mathematician are quite welcome. Reasoning behind the suggestion is also encouraged!!

Thanks :)

[quote="manuel"]russian.[/quote]

Why russian?

Strongly encourage you to learn Chinese, or other Far East Languages like Korean or Japanese, if you want to do well in math. Not only try to speak those languages, but to memorize the characters and to read and to write them!!! Because of this:

http://www.msnbc.msn.com/id/13560741/

The linked study is probably baloney (I say this as a parent of children who are native speakers of both Mandarin and English). Chinese is a fine language to try to learn, and Columbia would be a good place to try to learn it, because it is only a subway ride away from neighborhoods in New York City where you can hear Mandarin conversations all over the place. Princeton requires its math graduate students to show proficiency in either French, German, or Russian--presumably because a lot of important articles about mathematics have been published in those languages. See what your graduate school requirements might be, but other than that, take whatever language you feel like, or even two or three.

To me, the three most useful languages to know are: English, Chinese, and Spanish.

English because you live in the U.S., and its spoken widely around the world.

Chinese because China is developing very quickly and becoming very important in the economics of the world, plus there are a lot of mathematicians in China.

Spanish because of all the Hispanics living in the U.S.  I guess I'm fairly biased, though, living in Texas, where in many communities Spanish is the primary language used.

Edit: I agree with tokenadult on the study.

I would say russian because of the beauty of the russian style of doing mathematics and of researching.
If you are interested in applications of math into other areas or in the technical part of mathematics, well, yes chinese.
But i would say, that if you want to read a lof of creative stuff in math you should mainly learn an eastern european language.
Although, it won't be "useful" in the "real world", but it will allow you to learn a lot of beautiful stuff in math.

Yeah i always was under the assumption that German is the language used quite frequently in the sciences (including mathematics, obviously).

Many graduate schools in math require you to pass (apparently very easy) exams in two of French, German, and Russian, so you may consider one of those. Also, if you want to go to Budapest Semesters in Mathematics (which is apparently an excellent program; JBL can tell you more since he attended), you may consider learning Hungarian (which is probably a more difficult language to learn than most).


f is a real-valued function on the reals. It satisfies
f(x^3+y^3)=(x+y)(f 2 (x)+f(x)f(y)+f 2 (y)) for all x,y. Prove that f(1996x)=1996f(x) for all x.

Chinese 1996.
Solution in the 1996 - 1997 book from Andreescu and Feng. See

http://www.unl.edu/amc/a-activities/a4-for-students/mc96-97-01feb.pdf .


A yo-yo has an outer "wheel" with a diameter of 7 cm and an inner "wheel" with a diameter of 1 cm.  The inner wheel is wound with 30 cm of string and there is 40 cm of string left out.  If the yo-yo is placed on a table and the string is pulled, then how far and in what direction will the yo-yo move before the string is either stretched completely out or used up.

I still can't prove the direction the yo-yo's going to go...plus there's the problem of inertia and friction, so I'm guessing all these non-mathematical/physical variables changes the answer...


function
inequalities unsolved
inequalities

Let $a,b,c$ be three positive real numbers. Prove that:
$\sqrt{\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}}+2\sqrt{\frac{ab+bc+ca}{a^2+b^2+c^2}} \geq \sqrt{6}+2$  :)

Oh no, it is dropping to the bottom  :spider: , help me out of it.

:D First I think that you had solved it  :P

[quote="Anh Cuong"]Yes, I had a solution, use ABC  :D .[/quote]

what is ABC?

please write method ABC

                                  Alisherstudios@inbox.ru

Hey Anh Chuong can you write this method for all of us?

[quote="mathmanman"]See here : http://www.mathlinks.ro/Forum/viewtopic.php?p=440463#p440463 :)[/quote]

Thank you very much !!!  :)

[quote="Anh Cuong"]Oh, apology for my lateness. It was long time since the last time i came here.
$ABC$ is my new method in proving inequality. The solution using $ABC$ for this problem is quite short, you can download this method in http://www.mathnfriend.net  :), but of course it is Vietnamese  :P .[/quote]

Can you complete this problem?

solution by ABC method :
The function in abc is sunken so it attains minimum when b=a or b=0 . Then we have to find the minimum of
$\sqrt{2x+2+\frac{2}{x}}+2\sqrt{\frac{x^{2}+2x}{2x^{2}+1}}$

where $x=\frac{a}{c}$ . The first is root has minium  $\sqrt{6}$ which can be found easy by $x+\frac{1}{x}\geq 2$ and the second has minimum $2$ and it easy to find this by derivation .

When one variable is tend to $0$ it is easy .
QED


A balloon made up of rubber permeable to all isotopic forms of hydrogen is filled with pure deuterium ($ \ce{D2}$) and is placed inside a box filled with pure hydrogen ($ \ce{H2}$). What will happen?

Since H2 has a lower molar mass than D2, it has a faster effusion rate.  Eventually, all the D2 will be replaced by H2, assuming there is enough H2 in the box to replace the D2 in the balloon.


Let $ f(G)$ be the minimal number of paths (repetition of vertices is allowed, but repetition of edges isn't) such that together they contain all the edges in the graph, each exactly one time (meaning, the paths are disjoint of one another). The paths may be open or close.

Now, let $ G$ be a graph which has at least two vertices in any connected component of it ([u]there are no isolated vertices[/u]).

Show that $ f(G) = \max (\frac{r}{2},1)$ for a [b]connected[/b] graph, where $ r$ is the number of vertices with an odd degree.

Can you give a formula for $ f(G)$ if $ G$ is not a connected graph?

I am correcting something in the statement which may not seem clear. In the formula $ f(G)=\max (r/2,1)$, $ r$ is the number of vertices with an odd degree but is not necessarily the number of all the vertices in the graph (meaning, there could be some vertices with even degree).


Let $a_{1}=1$, $a_{2}=2$, $a_{3}$, $a_{4}$, $\cdots$ be the sequence of positive integers of the form $2^{\alpha}3^{\beta}$, where $\alpha$ and $\beta$ are nonnegative integers. Prove that every positive integer is expressible in the form \[a_{i_{1}}+a_{i_{2}}+\cdots+a_{i_{n}},\] where no summand is a multiple of any other.

We had this in the QEDMO (an olympiad organized by Darij and me in the QED association), so I just copy the solution I proposed there:

We use induction:
For $n=1=2^0 3^0$, it's true.
If $n>1$ is even, then $n=2m$ with positive integer $m$. Now $m$ is by induction hypothesis of the desired form, $m=2^{a_1} 3^{b_1} + 2^{a_2} 3^{b_2} + ... + 2^{a_k} 3^{b_k}$. But then $n=2^{a_1+1} 3^{b_1} + 2^{a_2+1} 3^{b_2} + ... + 2^{a_k+1} 3^{b_k}$ is also of the desired form.
If $n>1$ is odd, then let $n=2^0 3^p$ if $n$ is a power of $3$ and we are done then.
Otherwise $n-3^p$ is even and $m=\frac{n-3^p}{2}$ is of desired type, $m=2^{a_1} 3^{b_1} + 2^{a_2} 3^{b_2} + ... + 2^{a_k} 3^{b_k}$. This gives $n=2^{a_1+1} 3^{b_1} + 2^{a_2+1} 3^{b_2} + ... + 2^{a_k+1} 3^{b_k} + 2^0 3^p$. This fulfills all requirements, because $0< a_i+1$ and $p>b_i$ fr alle $i$, such that the new summand does neither divide one of the old ones nor is divided by one of them.


geometry
circumcircle
trigonometry
perpendicular bisector
angle bisector
geometry solved

Points P, Q lie inside angle BAC such that BP = CP and BQ = CQ. Suppose that ABP + ACQ = 180. Prove that BAP = CAQ.

Here's a hint: Draw the circumcircle of triangle ABC. Pick an arbitrary point R along minor arc BC (excluding points B and C). Then the perpendicular bisector of BC meet lines BR and CR at P and Q, respectively. Now after using sine law, all that is left is to prove:

|AP|*|BQ|=|AQ|*|BP|, where || denotes the length of the line segment.  :)

Here are two solutions:

The first one:

Let $D,E$ be the points where the internal, and, respectively, external bisector of $\angle BAC$ cuts the circumcircle of $ABC$. We have $\angle ABP+\angle ACQ=\pi=\angle ABD+\angle ACD=(\angle ABP+\angle PBD)+(\angle ACQ-\angle DCQ)$, and from ehre we deduce $\angle PBD=\angle DCQ$, which is, in turn, equal to $\angle DBQ$, so $BD$ is the angle bisector of $\angle PBQ$. In the same way we prove that $BE$ is the external angle bisector of $\angle PBQ$, so the circumcircle of $ABC$ is, in fact, an apollonain circle of the triangle $PBQ$. The conclusion follows immediately from here, since the apollonian circle of $PBQ$ corresponding to vertex $B$ is the locus of those points $A$ for which $AD$ is the bisector of $\angle PAQ$, and $AD$ is also the angle bisector of $\angle BAC$.

The second one:

We move $P$ on the perpendicular bisector $\ell$ of $BC$ and move $Q$ accordingly, so as to preserve the condition given in the hypothesis. The map $P\mapsto Q$ is an involution on $\ell$, so it induces an involution $AP\mapsto AQ$ on the pencil of lines through $A$. An involution is determined by the images of two lines, so it suffices to prove the conclusion for two particular positions of $P$, chosen so that it's easier to prove there. To this end, take $P=D$ ($D$ is defined above), and then $P=O$, the circumcenter of $ABC$.

In the first case we'll have $P=Q$, so the conclusion holds automatically, while in the second case, $Q$ will go to infinity, and  $AQ$ will be the $A$-altitude of $ABC$, and it's well-known and easy to prove (very simple angle chase) that the $A$-altitude is isogonally conjugate to $AO$.

Just wanted to tell that I have updated my note "Isogonal conjugation with respect to a triangle" (avaliable at http://www.cip.ifi.lmu.de/~grinberg/geometry2.html and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18472 ), and it contains a result (Theorem 19) that slightly extends this problem:

[color=blue][b]Theorem.[/b] Let $ABC$ be a triangle, and let $P$ and $Q$ be two points on the perpendicular bisector of the segment $BC$. We use directed angles modulo $180^{\circ}$.

[b](a)[/b] The following five assertions $\mathcal{F}_{1}$, $\mathcal{F}_{2}$, $\mathcal{F}_{3}$, $\mathcal{F}_{4}$ and $\mathcal{F}_{5}$ are pairwisely equivalent:

[b]Assertion $\mathcal{F}_{1}$:[/b] We have $\measuredangle ABP=\measuredangle ACQ$.
[b]Assertion $\mathcal{F}_{2}$:[/b] We have $\measuredangle ACP=\measuredangle ABQ$.
[b]Assertion $\mathcal{F}_{3}$:[/b] We have $\measuredangle BCP+\measuredangle BCQ=\measuredangle BAC$.
[b]Assertion $\mathcal{F}_{4}$:[/b] We have $\measuredangle PBC+\measuredangle QBC=\measuredangle BAC$.
[b]Assertion $\mathcal{F}_{5}$:[/b] The points $P$ and $Q$ are inverse to each other wrt the circumcircle of triangle $ABC$.

[b]b)[/b] If one of the five assertions $\mathcal{F}_{1}$, $\mathcal{F}_{2}$, $\mathcal{F}_{3}$, $\mathcal{F}_{4}$ and $\mathcal{F}_{5}$ holds, then the lines $AP$ and $AQ$ are isogonal to each other wrt the angle $CAB$.[/color]

Your problem is a different way to state that, if Assertion $\mathcal{F}_{1}$ holds, then the lines AP and AQ are isogonal to each other wrt the angle CAB. In fact, your conditions BP = CP and BQ = CQ mean that the points P and Q lie on the perpendicular bisector of the segment BC; your condition < ABP + < ACQ = 180° becomes Assertion $\mathcal{F}_{1}$ when it is rewritten using directed angles modulo 180°; finally, your assertion < BAP = < CAQ is equivalent to saying that the lines AP and AQ are isogonal to each other wrt the angle CAB.

  darij


It's another version of the AIME for schools that are on vacation during the week of the regular AIME.

Is it your choice as to when you want to take it? I'd like some extra time to study since its my first time.

It's your school's choice more than yours, I think. You also have to pay extra to take the alternate one.

Everyone has to take the normal date, unless you have a conflict (such as your school being closed).  The alternate AIME is not like the AMC B exams -- it isn't just open to normal registration.

An extreme method will be to get out of school due to either "a dire need of a doctor" or "religious ceremony" though that might not be accepted (well the first one; the second is ensured by the Bill of Rights I guess)

I'm pretty sure schools actually do have the option of which AIME to take, but then they have to pay more to take the alternate AIME, so it doesn't seem to make much sense to do that unless there is a school holiday. I think it's even possible to split it up so that some people at a school take the normal AIME and some people take the alternate. (My teacher wanted to order one copy of the alternate AIME just so that he would have the extra problems, but then I convinced him that it would be simple enough to get the test without doing that.)

There's also the problem that, in general, the AIMEB is harder than the AIMEA, even though the AMC committee tries to make them about the same difficulty.  If I were you, I'd take the A exam.

It is possible for schools to split, but I'm fairly certain the expectation is "only with a conflict."  I took the alternate AIME one year and thought it was pretty similar to the first AIME.  I don't think you'll get a whole lot of advantage putting it off for two weeks, anyhow.


calculus trigonometry geometry rectangle perimeter inequalities

This is a problem I have been working on but just haven't quite got. Maximize the sum: $S=\sin(a)+\sin(b)+\sin(c)$, where $a + b + c = \pi$ and a,b, and c are the angles of a triangle. All angles are in radians. P.S. the sum may not have a maximum, but a value it approaches, so, yeah.

Seems to me that Lagrange multipliers (good ole cheating with calculus) gives a value of pi/3 for each of the angles.

Ok, Thanks
Here's what I did:
$S=\displaystyle \sin a + \sin b + \sin c\\
=\displaystyle \sin a + \sin b + \sin (\pi-a-b)\\
=\displaystyle \sin a + \sin b + \sin (a+b)\\
=\displaystyle \sin a + \sin b + \sin a \cos b + \sin b \cos a\\
=\displaystyle \sin a\left(1+\cos b\right) + \sin b\left(1+\cos a\right)\\
=\displaystyle \sqrt{1-\cos^{2} a}\left(1+\cos b\right) + \sqrt{1-\cos^{2} b}\left(1+\cos a\right)\\
=\displaystyle \sqrt{1-v^{2}}\left(1+w\right) + \sqrt{1-w^{2}}\left(1+v\right)\Rightarrow\cos a = v,\cos b = w\\
=\displaystyle \sqrt{(1+v)(1+w)}\left(\sqrt{(1-v)(1+w)}+\sqrt{(1+v)(1-w)}\right)\\
=\sqrt{hg}\left(\sqrt{(2-h)(g)}+\sqrt{(h)(2-g)}\right)\Rightarrow v+1=h,w+1=g\\
=g\sqrt{(2-h)(h)}+h\sqrt{(g)(2-g)}\Rightarrow0<g,h<2$
My hunch was that the max occurs when $a=b=c$, kind of like maximum area of a rectangle given perimeter occurs when it is a square-type thing. But it was just a hunch. If I had an 89 I could've found out, but I don't. But, after reducing $S$ to that, I feebly attempted to find a max by using the general mean inequality to say:
$\displaystyle\frac{\displaystyle g\sqrt{(2-h)(h)}+\displaystyle h\sqrt{(g)(2-g)}}{2}\leq\sqrt{\displaystyle\frac{g^{2}(2-h)(h)+\displaystyle h^{2}(g)(2-g)}{2}}$

$\displaystyle\frac{S}{2}\leq\sqrt{gh(g+h-gh)}$
And that's where I've stopped. Maybe someone can take it further. But, please, no calculus. If you can't straight up find the maximum, maybe prove it is $\displaystyle\leq\frac{3\sqrt{3}}{2}$.
Thanks

i think this one is somewhat elegent:

\begin{eqnarray*}
& &\sin A + \sin B + \sin C + \sin \frac{A+B+C}{3}  \\
&=&2\sin\frac{A+B}{2}\cos\frac{A-B}{2} + 2\sin\frac{A+B+4C}{6}\cos\frac{2C-A-B}{6}\\
&\leq& 2\sin\frac{A+B}{2} + 2\sin\frac{A+B+4C}{6}\\
&=& 4\sin \frac{A+B+C}{3} \cos \frac{A+B-2C}{6}\\
&\leq& 4\sin \frac{A+B+C}{3}
\end{eqnarray*}

thus $\sin A + \sin B + \sin C \leq 3\sin \frac{A+B+C}{3}$

I am not very familiar with inequalities, so correct me if im wrong, but sin(x) is concave down on x=[0, :pi: ]
and I think by jensen's

(sin(a)+sin(b)+sin(c))/3 :le: sin((a+b+c)/3)

with equality if a=b=c

yes, you are right.
i just would like to use some more elementary method.


Ok....I doubt this is right, but I think it may be 9.  Why?  I used the pytogerom theorm and yeah....I actually got $ \sqrt{80}$ but i think rounding it off, u would get really close to 9.  Not sure.....

its pretty straighforward if you know the theorem.

*cough* cyclic *cough*

The point is... no one knows it =).  And of course, if you do know it, I'll force you to prove it xD.  :roll:

This is a good hint that you all should be aware of: Check triangle $ \triangle ABC$. If you look closely, you'll get an immediate answer.

ehehe. sorry. I should edit the problem. I originally had different side lengths and they actually worked to form triangles xD.

[quote="mathcrazed"]Quadrilateral ABCD is cyclic with $ AB \equal{} 6,BC \equal{} 6,CD \equal{} 8,AD \equal{} 7,$ and diagonal $ AC \equal{} 10$.  Find the length of $ BD$.  Well... I'm not entirely sure this is doable without trig or the happy theorem which I will refrain from proving although it's really fun xD

Edited: I think the triangles work now =)[/quote]With these sides, the quadrilateral is not cyclic :(

Why??? I'm probably missing something obvious -_-
Edit: mmm.... opposite supplementary angles ~_-...

[quote="mathcrazed"]Quadrilateral ABCD is cyclic with $ AB \equal{} 6,BC \equal{} 6,CD \equal{} 8,AD \equal{} 7,$ and diagonal $ AC \equal{} 10$.  Find the length of $ BD$.  Well... I'm not entirely sure this is doable without trig or the happy theorem which I will refrain from proving although it's really fun xD

Edited: I think the triangles work now =)[/quote]

what does cyclic mean?
yeah im stupid i know.

[quote="zolojetto"]it means that all of its verticies lie on the circumference of one circle.[/quote]
yeah, that's basically it

Ignore the problem.  If the opposite angles were supplementary, the cosines of the angles would be negatives.  However, if you apply laws of cosines, it doesn't work T.T... Sorry ~_-


"Mind Boggler 5"

If $N$, $N+92$, $2N+1$, and $3N+2$ are all 4-digit prime numbers, what is $N$?

the number is less than 3332....i think the only way to do these problems is to write a program

997 is the 168th prime, 3331 is the 470th prime:

[code]f[x_] := f[x] = Which[And[PrimeQ[Prime[x]],
      PrimeQ[Prime[x] + 92], PrimeQ[Prime[x]*2 +
        1], PrimeQ[Prime[x]*3 + 2]], x, True, f[x - 1]];

f[470][/code][b]Output:[/b] 320

So it's the 320th prime: $\boxed{2129}$

Checking, input "f[319]" returns 129, but the 129th prime has only 3 digits, so this is the unique solution.

Is there even a non-computer way to do this?  :huh:


Use quadratic formula and find roots a and b. then, it is just     (x-a)(x-b)

[hide="hint"]You can't factor it into linear factors (over the integers), so try quadratic factors.[/hide]
[hide="Big hint"]A nice little identity:
$(ax^{2}+bxy+cy^{2})(ax^{2}-bxy+cy^{2})=a^{2}x^{4}+(2ac-b^{2})x^{2}y^{2}+c^{2}y^{4}$[/hide]
[hide="Answer"]$(x^{2}+2xy+3y^{2})(x^{2}-2xy+3y^{2})$[/hide]


logarithms
number theory proposed
number theory

All of us know the density of the set of those numbers that can be written as a sum of 1,2, 4,5,... squares. What about 3?

Rust, I am sorry, but you are wrong. This is not the good answer. Not to mention that just posting answers and answers and facts that are probably trivial for you (but which are sometimes wrong) doesn't help a lot. I am still trying to understand what you wrote in that $b^{n}-1|a^{n}-1$ problem, but your arguments are absolutely incomprehensible.

If $A_{k}(N)$ is number of naturals $\{n\le N|n=x_{1}^{2}+...+x_{k}^{2}, \ x_{i}\in Z$
then
1. $A_{1}(N)=[\sqrt N ]$ density is 0,
2. $n=x_{1}^{2}+x_{2}^{2}$ if and only if all prime divisors p of n form p=3(mod 4) had even degree. It give $A_{2}(N)=O(\frac{N}{\sqrt{\ln N}})$, because \[\prod_{p\le x, p=3(mod 4)}(1-1/p)=O(\sqrt{\prod_{p\le x}(1-1/p))}=O(\frac{1}{\sqrt{\ln x}})\] , density 0,
3. $n=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ if and only if $n\not =4^{k}(8r+7)$ (proved by Lagranje). Therefore $A_{3}(N)=2N/3+O(1)$ density is 2/3,
4. For all natural number n exist integers $x_{1},x_{2},x_{3},x_{4}$, suth that $n=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}$. Therefore $A_{k}(N)=N$ if $k\ge 4$
Were is wrong?

Man, when will you understand that proofs need arguments? What is wrong is that you get $2/3$ by giving one argument, when the density is in fact $5/6$.


[hide="Solution"]The logarithmic notation doesn't tell us much, so we'll first convert everything to exponents.

$x^{24}=w$, $y^{40}=w$, and $(xyz)^{12}=w$. If we now convert everything to a power of $120$, it will be easy to isolate $z$ and $w$.

$x^{120}=w^5$, $y^{120}=w^3$, and $(xyz)^{120}=w^{10}$.

With some substitution, we get $w^5w^3z^{120}=w^{10}$ and $\log_zw=60$.[/hide]

[quote="4everwise"]Let $x$, $y$, and $z$ all exceed 1 and let $w$ be a positive number such that $\log_x w = 24$, $\log_y w = 40$, and $\log_{xyz} w = 12$.  Find $\log_z w$.[/quote]

[hide="Solution"]
Note that $\log_w x = \dfrac{1}{24}$ and $\log_w y = \dfrac{1}{40}$. Also note that $\log_{xyz} w = \dfrac{1}{\log_w xyz} = \dfrac{1}{\log_w x + \log_w y + \log_w z} = \dfrac{1}{12}$.

Therefore, $\dfrac{1}{24} + \dfrac{1}{40} + \log_w z = \dfrac{1}{12}$, meaning that $\log_w z = \dfrac{1}{60}$, so $\log_z w = \boxed{060}$.
[/hide]

[hide=My Solution]
I had a similar first approach as 4everwise: convert the statements into $x^{24}=w$, $y^{40}=w$, and $(xyz)^{12}=w$.
Now, we square the statement $(xyz)^{12}=w$, giving $(xyz)^{24}=w^2$. Now, we substitute the right side for $x^{24}y^{40}$ (which we can validate via previous equations. Once we plug this in, the equation simplifies to $z^{24}=y^{16}$. Since we know $y^{40}=w$, we can raise each side of the equation by the power $2.5$ (since $16 \cdot 2.5 = 40$). Then, we can just substitute the right side with $w$, and evaluate the exponent on the left. This results in $\log_zw=60$.
[/hide]

Well, as long as we're reviving the thread, here's my solution:

[hide]When I have tons of logarithmic bases, especially variable bases, I convert everything into a neutral base.  Natural log works nicely.  So, we have:

$\frac{\ln w}{\ln x} = 24$

$\frac{\ln w}{\ln y} = 40$

$\frac{\ln w}{\ln xyz} = 12$

We want to find $\frac{\ln w}{\ln z}$.  To make typing easier, let $W = \ln w$, etc.

Using the log-of-product-to-sum-of-log formula on the last equation, we have $W = 24X$, $W = 40Y$, $W = 12X + 12Y+ 12Z$

We want to convert the last statement into one with just $Z$ and $W$ in it.  Plugging in the first two equations into the first two terms on the RHS yields:

$W = \frac{12}{24}W + \frac{12}{40}W + 12Z$

A little bit of algebra yields $W = 60Z$, or $\frac{W}{Z} = \boxed{60}$, which is the exact quantity that we want.[/hide]

Wasn't this a #1 on some AIME a while back?

$ w^2=w.w=x^{24} y^{40}$ substitute we get

$z^{60}=w \implies \log_z{w}=60$

We have the equations:
x^24 = w
y^40 = w
x^12y^12z^12 = w

Dividing the first equation by the 3rd equation yields:
x^12 = y^12z^12

Hence:
y^24z^24 = w

y^40/y^24z^24 = y^16/z^24 = 1
y^16 = z^24
y^2 = z^3

Raise both sides to the 20th power.
w = z^60

Hence our answer is 60

[quote=bookstuffthanks][hide = Solution]We have the equations:
\begin{align*}
x^{24} &= w\\
y^{40} &= w\\
x^{12}y^{12}z^{12} &= w
\end{align*}

Dividing the first equation by the 3rd equation yields:
$x^{12} = y^{12}z^{12}$

Hence:
$y^{24}z^{24} = w$

\begin{align*}
\frac{y^{40}}{y^{24}z^{24}} &= \frac{y^{16}}{z^{24}} = 1\\
y^{16} &= z^{24}\\
y^2 &= z^3
\end{align*}

Raise both sides to the 20th power.
$w = z^{60}$

Hence our answer is $60$[/hide][/quote]

Fixed \LaTeX, also hid solution

[quote=michaelchang1]How did this thread get revived after 10 years?[/quote]

because its allowed

This is a very popular AIME problem.

Just convert the logarithms into exponential form.

"I need 5832 posts quick, so I should probably just post a ton of the solutions I have written for the sake of postfarming but still making high-quality posts." -me

[b]Solution from Hamst's 60 AIME grind:[/b]

The term that the logarithms are taking the same, but the base isn't, and to use the logarithm identities we devised, we need the bases to be equal. This tempts us to use the identity $\log_b a=\frac 1{\log_a b}$. In fact, that's put to good use.

We have $\log_w x = \frac 1{24}$, $\log_w y = \frac 1{40}$, and $\log_w xyz = \frac 1{12}$. It's immediately clear how we would use this; Since $$\log_w x + \log_w y + \log_w z = \log_w xyz,$$ $$\frac 1{24}+\frac 1{40}+\log_w z = \frac 1{12}$$ from substituting. This solves to get $\log_w z = \frac 1{60}$, so our answer is $\boxed{060}$.

The inverse logarithm identity is very applicable when the argument of multiple logarithms are constant.

I just did $log_wx = 1/24$, $log_wy = 1/40$, $log_wxyz = 1/12$. $log_wxyz - log_wx - log_wy = log_wz = 1/60$, so $log_zw=60$.

it is better to convert them into exponential form from logarithmic form as we are more used to exponents than logarithms
 :D

I'm new to logarithms so I converted all into exponential form *eye roll
$x^{24}=w,
y^{40}=w,
x^{12}y^{12}z^{12}=w$
so $y^{12}z^{12}=w^{1/2}$ so $z^{12}=y^8=w^{1/5}$ so $z^{60}=w$ so the answer is $\boxed{60}$

[quote=1983 AIME P1]Let $x$, $y$, and $z$ all exceed 1 and let $w$ be a positive number such that \[\log_x w = 24,\quad \log_y w = 40 \quad\text{and}\quad \log_{xyz} w = 12.\]  Find $\log_z w$.[/quote]

[hide=Solution]Converting to exponential form, we have $x^{24}=w$, $y^{40}=w$, $(xyz)^{12}=w$. Then, $x=w^{\frac1{24}}$, $y=w^{\frac1{40}}$, $xyz=w^{\frac1{12}}$. Multiplying the first 2 of those equations, we have $xy=w^{\frac1{15}}$, and dividing the third equation by this quantity gives $z=w^{\frac1{60}}$, implying $\log_z{w}=\boxed{60}$.[/hide]

[hide=Solution] We can convert to exponential form to get three equations, which are \begin{align*} x^{24} &= w \\ y^{40} &= w \\ (xyz)^{12} &=w .\end{align*} Now because these all have a LCM of $120$ we can make all these exponents equal that and work from their. So raising our first equation to the $5th$, our second to the $3rd$ and our last equation to the $10th$ we get three equations. Our equations are \begin{align*} x^{120}  &= w^5 \\ y^{120} &= w^3 \\ (xyz)^{120} &= w^{10}. \end{align*} Now we can divide the third equation by the first and second equations to get $$z^{120} = w^2.$$ Now taking the square root we get $$z^{60} = w.$$ Taking the logarithm we get $$\boxed{\log_{z} w = 60}.$$ $\square$

[hide=my sol]
After converting into exponents, we get the following equations:

$x^{24}=w$

$y^{40}=w$

$x^{12}*y^{12}*z^{12}=w$

After switching out the $x^{12}$ term in terms of $w$ from the first equation and doing the same thing with $y$, we are left with:
$w^{1/2}*w^{3/10}*z^{12}=w$

Because,
$1-\frac{1}{2}-\frac{3}{10}=\frac{1}{5}$

So,
 $z^{12}=w^{\frac{1}{5}}$

Which means $\log_z w=\boxed{060}$

[/hide]

[hide=mysol]
I did the same thing a1001 did

The given equations rewrite as $x^{24}=w$, $y^{40}=w$, and $x^{12}y^{12}z^{12}=w$, respectively. Raising the third of these equations to the $10$th power, we have that $x^{120}y^{120}z^{120}=w^{10}$, and substituting $x^{24}=w\implies x^{120}=w^5$ and $y^{40}=w\implies y^{120}=w^3$ yields $w^8z^{120}=w^{10}$ or $z^{120}=w^2$. Hence, $z^{60}=w\implies \log_zw=\boxed{60}$, as requested.

We have that $x^{24}=y^{40}=w$, and $x^{12} \cdot y^{12} \cdot z^{12}=w$, so $z^{60}=w$, by some boring computations, which I do not want to write. So we have $\boxed{60}.$

[hide]With logarithms, it is much easier to work with logs of the same base than logs of the same argument based on the properties we have. With this in mind, we take the reciprocals of all three equations to get $w$ as the base. We now have $\[\log_w x = \frac{1}{24},\quad \log_w y = \frac{1}{40} \quad\text{and}\quad \log_{w} xyz = \frac{1}{12}.\]$ Now that we have multiple logs of the same base, we can do operations like adding the first two equations, which gives us $\log_{w} xy = \frac{1}{15}$. Finally, we can subtract this from the last equation to get $\log_{w} z = \frac{1}{60}$, so our answer is $60$. Note that we didn't have to do any hard application of logs after the initial realization to take the reciprocal.[/hide].


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a finite group of order $ n$, $ k|n$ s.t. $ \left(k,\frac{n}{k}\right) \equal{} 1$.
(a) Suppose there exists a normal subgroup $ N$ of $ G$ of order $ k$. Prove that $ N$ is the only subgroup of order $ k$.
(b) Prove or disprove: all subgroups of order $ k$ are conjugate.

(a) suppose $ U$ is a subgroup of order $ k$. we want to show $ U \equal{} N$; it's sufficient to prove $ U \subseteq N$. if $ a \in U$, then $ a^k \equal{} 1$. now take integers $ u,v$ such that $ uk \plus{} v(n/k)\equal{}1$. in the group $ G/N$ of order $ n/k$ we compute $ a \equal{} (a^k)^u (a^v)^{n/k} \equiv 1$ mod $ N$, so $ a \in N$.

You might like to read [url=http://en.wikipedia.org/wiki/Hall_subgroup]here[/url] for (b). There is much to say.
Btw $ PSL(2,7)$ is the simple group of order $ 168$.

(a) can be generalized to the following: let $ G$ be a group, $ N,H$ subgroups, s.t. $ N$ is finite, $ H$ has finite index and $ (|N|,[G: H]) \equal{} 1$. Suppose further that either $ N$ or $ H$ is normal. Prove that $ N < H$.
Thanks for the link, olorin, I didn't know about that. Problem (a) just led me to this question.

This has been discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=215646]here[/url].
And we only need $ NH\le G$ ($ \Leftrightarrow NH\equal{}HN$) instead of either group being normal.

Thanks - I have exactly the same solution, but I didn't want to assume that $ HN$ is a subgroup because then it's easier to get the right idea.


MIT
college
summer program
Mathcamp
AMC
USA(J)MO
USAMO
\/closed

Will there be some sort of Programming Math Jam or class in the future? since programming also requires a lot of creativity and problem solving?

It would be even better to have a programming class :)

Although this is AoPS...

but it is more related to math then some of the other Math Jams

well, it is art of problem solving, not art of math problem solving so...

But, didn't they have USACO math jam once? I think they did...

[quote="PowerOfPi"]What's USACO? Also, I think it would be possible, as long as the connection to problem solving is there and there has to be someone that volunteers to host it. I think that last part and just the time it consumes are the reasons there aren't many math jams like these. There have been non-AoPS math jams, like linguistics, MIT, AMC 10/12, and Mathcamp math jams.[/quote]

USA Computing Olympiad.

Basically, usaco problems are not programming problems. (Though you do need decent programming skills code them. Most of the difficulty comes from finding the algorithms. (Unless you happen to be bad at programming... being good at programming helps a lot.)

And yes, there has been a usaco math jam. There wasn't exactly a lot of programming in the math jam though.

Also: There is a high correlation between people good at usamo and people good at usaco:

IOI USA team this year:

Brian Hamrick
Neal Wu
Travis Hance
Wenyu Cao

All of these are either moppers or formerly moppers, and Hamrick and Wenyu are usamo winners.

Well, the level of programmers, from "what's programming" to "I created a web 2.0 browser" is too variable for this to come true.


[quote="#H34N1"]What is the least number of persons required if the probability that two or more of them have the same birthday exceeds $ \frac 12$[/quote]

[hide]
$ \frac{366}{366}\cdot\frac{365}{366}$ .... and on and on

[/hide]

[hide="hint"] Count the complement [/hide]
[hide="complete solution"] This problem is easy if you count the complement. For nobody to share a birthday, there are 365 possibilities for the first person (forgetting about leap years), then 364 for the next, then 363, and so on. Multiplying these together yields $ \displaystyle \frac{365!}{(365\minus{}n)!}$, where $ n$ is the number of people. The total possible combinations of birthdays is $ 365^n$, so we need to find when $ \displaystyle \frac{365!}{365^n(365\minus{}n)!}$ first is less than 1/2, which by trial and error, occurs at $ \boxed{n\equal{}23}$[/hide]


trigonometry
inequalities proposed
inequalities

$Prove:$ $\frac{cosx}{sin^{2}x(cosx-sinx)}>8$ $for$ $x\in (0,\frac{\pi}{4})$


The 39th. Rudolf Ortvay international theoretical physics competition for university students (phd. students included) starts tommorow.
The competition is strictly INDIVIDUAL. NO groups alowed.
The problems are not textbook style, but lots of thinking required ones.

The previous problems and further info can be seen [url=http://ortvay.mafihe.hu/main.html]here[/url].

The new problems will also appear on this website.

nice questions Thaakisfox are u writing it?

well yes I will try to solve as many as I can. :D
What about you? :)

i will try to solve but i am in my last year of high school :(


In my book it says this:  $({A\cap}{B}\cap{X})\cup({A}\cap{B}\cap{C}\cap{X}\cap{Y})\cup({A}\cap{X}\cap\bar{A})$ can be reduced to this: $({A\cap}{B}\cap{X})$ can some one show me the steps to do this (really how to get the $Y$ and $C$ out)?

Thank You

$A\cap B\cap C\cap X\cap Y$ is contained in $A\cap B\cap X$, and $A\cap X\cap \bar{A}$ is contained in $A\cap \bar{A}$ which is clearly the empty set. The union of these two with $A\cap B\cap X$ must then be $A\cap B\cap X$.


For fixed $ \lambda \in (0,1)$ and $ n,N \in \mathbb{N}$ consider the polynomial
$ P(\lambda) \equal{} \sum_{i\equal{}0}^n a_i \lambda^i$
where $ a_i \in \mathbb{Z}$ and $ \minus{}N \leq a_i \leq N$ $ \forall$ $ i \equal{}0,1,...,n$

Given that $ P(\lambda) > 0$ :
(i) Prove that if $ \lambda$ is a rational number $ \lambda \equal{} \frac{p}{q}$ then $ P(\lambda) \geq \frac{1}{q^n}$
(ii) Find a positive lower bound on $ P(\lambda)$ for irrational $ \lambda$.

Its really easy for rational $ \lambda \equal{} \frac{p}{q}$ , since $ P(\lambda)q^n$ is a positive integer it must be atleast $ 1$.

Which gives $ P(\lambda)\geq\frac{1}{q^{n}}$

But this result cannot be trivially extended to irrationals.

This problem will be solved if one can find a rational approximation $ \frac{p}{q}$ for $ \lambda$ such that

$ |\lambda \minus{} \frac{p}{q}| < \frac{C}{q^n}$

where $ C \equal{} \frac{1}{2n^2N}$ is a constant.

I am not sure if one can always do this. Need Help.

This depends on exactly what $ \lambda$ is. For any real $ n>2$, for almost every $ \lambda$, there is some $ C$ so that $ \left|\lambda\minus{}\frac pq\right|>\frac{C}{q^n}$ for all integers $ p,q$. At $ n\equal{}3,C\equal{}\frac1{18}$, the set of values in $ [0,1]$ with even one approximation better than that has measure less than $ \frac1{10}$.

So no; you can't find those approximations most of the time. Even for $ n\equal{}2$, you can only get within $ \frac{1}{kq^2}$ if there is an entry of at least $ k$ in the continued fraction.


group theory
abstract algebra
number theory unsolved
number theory

Hi,

How can we show that for all integers n and m we have

n!m!(n+m)! divide (2n)!(2m)!

for example, if n=m we have   n!^2  (2n)!  divide (2n)!^2, it's clear because (2n)!/(n!)^2  is a binomial number

thanks

$ [2x] \plus{} [2y] \ge [x] \plus{} [y] \plus{} [x \plus{} y]$ where $ [x]$ is the integer part of $ x$.

I know this proof. And I thank you very much.

I'm looking for a proof that enables to see (2n)!(2m)! / (n!m!(m+n)!)  as a cardinal of a finite set.

For example, (2n)!/ (n!^2) is exactly the cardinal of sets of cadinal n in a big set of 2n elements.

Or maybe, for another proof, does there exists a subgroup of S_(2n) X S_(2m) which is isomorph to S_n x S_m x S_(n+m), where S_n is the symmetric group ?

Thanks

$ \binom{2n \plus{} 2m}{m,n,m \plus{} n}$ is a multinomial, which is clearly integer.

If I recall correctly, $ \binom{2n \plus{} 2m}{m, n, m \plus{} n}$ is not the same thing as $ \frac{(2n)! (2m)!}{m! n! (m \plus{} n)!}$.


complex analysis
complex analysis unsolved

For which sets of points on the boundary of a circle can one find a power series that converges exactly on those boundary points (and in the interior)?

I posted it some time ago and Fedja told me it is still an open problem.

Is it known for finite and cofinite sets?

Assume the disk is the unit disk. Then the Maclauren series of $\sum_{k=1}^{N}\log(1-z/z_{k})$, with distinct $z_{k}$ on the unit circle $\mathbb{T}$, converges precisely on $\mathbb T\setminus\{z_{k}\colon k=1,\dots,N\}$. So the answer is yes for cofinite sets.

An attempt at generalization: $f(z)=\sum_{k=1}^{\infty}c_{k}\log(1-z/z_{k})$, where $z_{k}\in\mathbb T$ and $\sum_{k}|c_{k}|<\infty$. Now what?  :huh:


We have rescheduled the AIME Math Jam for [b]Monday, March 24[/b] at [b]7 PM Eastern / 4 PM Pacific[/b]. During the Math Jam we will discuss the solutions to all 15 problems on the AIME.

The [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=220]transcript of the Math Jam[/url] is now available.


For every integer $ a>1$ an infinite list of integers is constructed $ L(a)$, as follows:

[list] $ a$ is the first number in the list $ L(a)$.[/list]
[list] Given a number $ b$ in $ L(a)$, the next number in the list is $ b\plus{}c$, where $ c$ is the largest integer that divides $ b$ and is smaller than $ b$.[/list]
Find all the integers $ a>1$ such that $ 2002$ is in the list $ L(a)$.

[hide]If $ b$ is a number in the sequence, and $ p$ is the smallest prime dividing $ b$, then the next number is $ b\plus{}\frac{b}{p}\equal{}b\cdot\frac{p\plus{}1}{p}$. From this, we immediately see that the list is increasing, so we any other $ a$ must be less than 2002.

Say that we had an even number in the sequence. Let this number be $ 2^i \cdot m$, where $ m$ is odd. Clearly, for the next $ i$ steps in the list, 2 will be the smallest prime, so we will multiply by $ \frac{3}{2}$ each time. After these $ i$ steps, we will be left with $ 3^i \cdot m$. Then, 3 will be the smallest prime, so the next number will be $ 2^2\cdot3^{i\minus{}1} \cdot m$. Then, the next numbers will be $ 2\cdot 3^i \cdot m, 3^{i\plus{}1}\cdot m$, and from here, a pattern of the smallest prime being 3,2,2 will repeat forever. Each number in this sequence is divisible by $ 4$ or $ 3$, and since $ 2002\equal{}2\cdot7\cdot11\cdot13$ is not divisible by $ 4$ nor $ 3$, it follows that no even number can precede 2002.

If the first number is even, then it must be $ 2002$, giving us one valid value of $ a$.

Otherwise, if $ a$ is odd, then the next number in the list must be even, and thus must be 2002. In this case, we must have $ 2002\equal{}a\cdot\frac{p\plus{}1}{p}$, or $ 2002p\equal{}a(p\plus{}1)$. Since $ a$ is odd and $ \gcd(p,p\plus{}1)\equal{}1$, we see that $ p\plus{}1$ must be an even factor of 2002. Testing out all 8 even factors of 2002, we see that the only such factors that are 1 greater than a prime are $ 2\cdot7\equal{}14$ and $ 2\cdot7\cdot13\equal{}182$. The first case corresponds to $ a\equal{}11\cdot13^2\equal{}1859$, and the second corresponds to $ a\equal{}11\cdot 181 \equal{} 1991$. But in both these cases, $ 11$ is the smallest prime, and the next value should be $ \frac{12a}{11}$, which will not give a value of 2002.

Thus, our only valid value of $ a$ is $ 2002$.[/hide]

Let $a_i$ be the number on the $i-th$ position on $L(a)$, so $a=a_0$. Do $a_{t}=2012$ and $a_{t-1}=bp$ with $b$ the largest proper divisor of $a_{t-1}$ and $p$ prime so is obvious $b>p$. From this we have;

$2002=b(p+1)$, the divisors of $2002$ are $1,2,7,11,13,14,22,26,77,91,143,154,182,286,1001,2002$ the posible pairs of integers $(b,p+1)$ that satisfy $b>p$ are $(77,26), (91,22), (143,14), (154,13), (182,11), (286,7), (1001,2), (2002,1)$ and the only one that satisfy that $p$ is prime is $(143,14)$, so this leaves us with $bp=1859$ whose largest proper divisor is $169$ which is a contradiction, what means that there can´t be any number before $2002$ on $L(a)$, so $a=a_{0}=2002$


A polynomial $ F(x)$ of degree $ n$ satisfies $ F(k) \equal{} F_k, k \equal{} 0, 1, 2, ... n$, where $ F_k$ is the Fibonacci sequence with $ F_0 \equal{} 0, F_1 \equal{} 1$.  Compute $ F(n\plus{}1)$.

[hide="Big hint"] See the technique described [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=211560]here[/url].  [/hide]

It is easily verified that $ \Delta^k(p(x))_{x\equal{}0}\equal{}F_{\minus{}k}$

Then since $ p(x)\equal{}\sum_{k\equal{}0}^n\frac{ x_{(k)} \Delta^k(p(x))_{x\equal{}0}}{k!}$,

The problem can be solved by the method that I gave in your other thread:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212474

(we have to write $ a_k\equal{}F_{\minus{}k}$ as an linear recurrence with increasing indices but thats easy enough...

[hide="more generally"] one can establish
\[ f_n(x) = \begin{cases} \sum_{k = 1}^{n}{( - 1)^{k + 1}F_k\frac {\Gamma(x + 1)}{\Gamma(x + 1 - k)k!} & x\notin \mathbb{Z}\cap \lbrace x<n \rbrace  \\
F_x & x\in \mathbb{N}\cap \lbrace 0\leq x\leq n \rbrace \end{cases}
}\]
where $ f_n(x)$ agrees everywhere except on $ x\in \mathbb{Z}^{ - }$with the $ n$th degree polynomial satisfying $ F(j) = F_j$ for $ j = 0,1,2...,n$.  Letting $ x = n + 1$ gives the desired result.

Though in general this requires some analysis, it can be established on $ \mathbb{N}$ just by induction.[/hide]

$ x_{(k)}\equal{}(x)(x\minus{}1)...(x\minus{}k\plus{}1)$ which is equivalent to your use of the gamma function.

I like the form that I used because it resembles the Maclaurin series forumula.

By employing the Lagrange Interpolation Formula, we have the equalities $ F(n\plus{}1)\equal{}\sum_{k\equal{}1}^{n\plus{}1}\binom{n\plus{}1}{k}(\minus{}1)^{n\minus{}k}F_k\equal{}F_{n\plus{}1}\plus{}(\minus{}1)^{n\plus{}1}F_{n\plus{}1}$, all of which can easily be obtained.  Of course, this is correct if and only if I have not made any computational mistakes.


algebra
polynomial
quadratics
Gauss
trigonometry
complex numbers

Let $ \alpha$ be the $ 7^{th}$ root of 1 which $ \alpha$ isn't $ 1$
find the two roots of the equation $ z^2\plus{}z\plus{}2\equal{}0$ in a polynomial form of $ \alpha$ with the lowest degree

$ z^2 \plus{} z \plus{} 2 \equal{} 0$ is a quadratic equation and can be solved by entirely elementary means, so a constant-degree polynomial can be found.  I don't follow.

I presume that he is saying: write the roots of $ z^2\plus{}z\plus{}2\equal{}0$ as a polynomial in $ \alpha$ with integer coefficients with minimal degree.

Ah.

[hide="Solution"] The roots of $ z^2 \plus{} z \plus{} 2$ are $ \beta \equal{} \frac { \minus{} 1 \plus{} \sqrt { \minus{} 7}}{2}$ and its conjugate, and it is well-known that $ \mathbb{Z}[\beta] \subset \mathbb{Z}[\alpha]$.  This is a specialization of the [url=http://en.wikipedia.org/wiki/Kronecker-Weber_theorem]Kronecker-Weber theorem[/url].  In the discussion that follows $ \alpha \equal{} e^{ \frac{2 \pi i}{7} }$.

It turns out that $ \boxed{\beta \equal{} \alpha \plus{} \alpha^2 \plus{} \alpha^4 }$ (verify that the conjugate doesn't work because the imaginary part is positive).  Since $ \bar{\alpha} \equal{} \alpha^6$, it follows that $ \bar{\beta} \equal{} \alpha^6 \plus{} \alpha^5 \plus{} \alpha^3$.  Then $ \beta \plus{} \bar{\beta} \equal{} \minus{} 1$ and

$ \beta \bar{\beta} \equal{} 3 \plus{} \alpha \plus{} \alpha^2 \plus{} \alpha^3 \plus{} \alpha^4 \plus{} \alpha^5 \plus{} \alpha^6 \equal{} 2$

so $ (z \minus{} \beta)(z \minus{} \bar{\beta}) \equal{} z^2 \plus{} z \plus{} 2$ as desired.  This approach generalizes; see [url=http://en.wikipedia.org/wiki/Quadratic_Gauss_sum]quadratic Gauss sum[/url]. [/hide]

Ahh very cool! This is actually related to a problem that I was thinking about a while ago:

given $ p$th roots of unity, can take two sets that have a product and sum that are integers. And that theorem gives a partial answer (of course we are looking for nontrivial subsets)!

We construct a quadratic that has a discriminant $ b^2\minus{}4ac\equal{}\minus{}p$ (note $ p\equiv 3\mod 4$ for this construction). We may choose $ a\equal{}1$, $ b\equal{}1$, and $ c$ is an easily found integer. Then the roots of this quadratic are expressible in terms of the $ p$th roots of unity, but they are also conjugates. Furthermore, these roots have to have more that one $ p$th root of unity because otherwise they'd be roots of a quadratic which is nonsense, so this is a nontrivial example).

I believe this problem came from one of those ugly trig sums where cos 2kpi/7, etc summed up is a nested square root. I'm very surprised to see such a connection here!


Consider the fibonacci sequence defined by a_1 =a_2=1, a_(n+1)=a_(n-1)+a_n. Prove that for any n, there is a fibonacci number ending with n zeros.

well, since every positive integer divides infinitely many Fibonacci numbers, $ 10^{n}\mid F_{k}$ for infinitely many $ k\in \mathbb N$.

EDIT: For the proof of the fact "every positive integer divides infinitely many Fibonacci numbers", you can take a look at [url=http://www.mathlinks.ro/Forum/weblog_entry.php?t=156030]here[/url]. Also, my signature has something to do with this. :wink:


We have the set 1,2,...,49. We choose six numbers from these (a1,..a6)
Then we begin to extract numbers.What is the average number of extractions such that all the six chosen numbers are extracted?

Very nice problem. We need such a kind of problems. The answer is 27.5

The problem is equivalent to find average number of elements of all subset of E={1,2, ..., 49} which contain {a1, a2, ..., a6}. This average number is equal:
      A = Sum{k=0,43}(6+k)C(k,43)/Sum{k=0,43}C(k,43)
Note that C(k,43) = C(43-k,43) we have
      Sum{k=0,43}(6+k)C(k,43) = (1/2)Sum{k=0,43}(6+k+6+43-k)C(k,43) = (55/2)Sum{k=0,43}C(k,43)
So A = 55/2 = 27.5


Ν.δ.ο. για εκεινα τα $ a,b,c>0$  που οι παρανομαστες της παρακατω ανισωσης ειναι θετικοι ισχυει

$ \frac{(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2}\leq\frac{1}{ac\plus{}3ab\minus{}b^2}\plus{}\frac{1}{ab\plus{}3cb\minus{}c^2}\plus{}\frac{1}{bc\plus{}3ac\minus{}a^2}$

Η ανισότητα γράφεται ισοδύναμα

$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ac \plus{} 3ab \minus{} b^2} \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} 3cb \minus{} c^2} \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{bc \plus{} 3ac \minus{} a^2}\geq (a \plus{} b \plus{} c)^2$ στην οποία λόγω ομοιογένειας μπορούμε να υποθέσουμε ότι $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.

Έτσι η προς απόδειξη ανισότητα γίνεται

$ \frac {1}{ac \plus{} 3ab \minus{} b^2} \plus{} \frac {1}{ab \plus{} 3cb \minus{} c^2} \plus{} \frac {1}{bc \plus{} 3ac \minus{} a^2}\geq 1 \plus{} 2(ab \plus{} bc \plus{} ac)$

Όμως λόγω της ανισότητας Andreescu έχουμε ότι

$ LHS \geq \frac {(1 \plus{} 1 \plus{} 1)^2}{4(ab \plus{} bc \plus{} ac) \minus{} 1} \equal{} \frac {9}{4(ab \plus{} bc \plus{} ac) \minus{} 1} \stackrel{(\star)}{\geq} \frac {9}{4\cdot 1 \minus{} 1} \equal{} 3 \stackrel{(\star)}{\geq} 1 \plus{} 2(ab \plus{} bc \plus{} ac)$,

$ (\star)$ Διότι ισχύει $ ab \plus{} bc \plus{} ac \leq a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.

Ελπίζω να μη μου έχει ξεφύγει κάτι.

Αλέξανδρος

Ο.Κ. Αλεξανδρε
Προφανως ενοεις μια ισοδυναμη αυτης ειναι ομογενης..
Η δικια μου λυση ειναι περιγραφικα ελλατωση του Β μελους με καταλληλη αυξηση των παρανομαστων και μετα με Αndreescu τα ιδια...

ρε παιδια ποια ισοδυναμη ανισοτητα ειναι ομογενης στην λυση του αλεξανδρου; :|

η ανισοτητα γραφεται
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ac \plus{} 3ab \minus{} b^2} \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} 3cb \minus{} c^2} \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{bc \plus{} 3ac \minus{} a^2}\geq (a \plus{} b \plus{} c)^2$
απο την ΑΜ-ΓΜ παιρνουμε
$ 3ab\leq a^2\plus{}b^2\plus{}ab$etc
αρα το LHS ειναι(ANDREESCU and $ 3(a^2\plus{}b^2\plus{}c^2)\geq (a\plus{}b\plus{}c)^2$
$ LHS\geq \sum \frac {a^2 \plus{} b^2 \plus{} c^2}{ac \plus{}a^2 \plus{}ab}\geq \frac {(3(a^2\plus{}b^2\plus{}c^2))^2}{(a\plus{}b\plus{}c)^2}\geq (a\plus{}b\plus{}c)^2$
QED :D

Ευχαριστω για το ''παιδι" ειχα χρονια να το ακουσω
Εχουμε την ιδια λυση
Για την ομογενεια για να ειμαι ειλικρινης υπεθεσα οτι θα υπαρχει καποια ισοδυναμη γιατι εκεινη που αναφερεται ο Αlex εκ πρωτης οψεως .....
Θα μας πει ο Αλεξανδρος ...

Έχετε απόλυτο δίκιο. Η κατάσταση δε σώζεται λόγω του δεξιού (στη δική μου "λύση") μέλους που κάνει την ανισότητα να μην είναι ομογενής! Ήταν κεκτημένη ταχύτητα! Ζητώ συγνώμη που σας ταλαιπώρησα. Η λύση που έγραψε ο Athinaios είναι πάρα πολύ καλή!

Φιλικά,
Αλέξανδρος