calculus
integration
limit
function
real analysis
calculus computations

Let $ f$ be defined on $ [1, \infty]$ and suppose $ I_{n}= \int_{1}^{n}f$ exists for all $ n \in \mathbb N$.  Decide whether the statement is true or false with either a (short) proof or counterexample:

1.  If $ f$ monotonic decreasing and if $ \lim_{n \rightarrow \infty}I_{n}$ exists, then $ \int_{1}^{\infty}f$ converges.

2.  If $ \lim_{x \rightarrow \infty}f(x) = 0$ and $ \lim_{n}I_{n}= I$, then $ \int_{1}^{\infty}f$ converges to $ I$.

3.  If $ I_{n}$ converges, then so does $ \int_{1}^{\infty}f$.

4.  Suppose that $ f'$ exists and is bounded on $ [1, \infty]$, and that $ \lim_{n}I_{n}= I$.  Then $ \int_{1}^{\infty}f$ converges to $ I$.

5.  If $ \int_{1}^{\infty}f$ converges, then $ \lim_{x \rightarrow \infty}f(x) = 0$.

#3 is false. Consider the following counterexample:

Let $ f(x)=\sin(2\pi x).$ Then $ I_{n}=\int_{1}^{n}\sin(2\pi x)\,dx=0$ for all $ n\in \mathbb{N}.$

However, $ \int_{0}^{\infty}\sin(2\pi x)\,dx$ does not converge.

This shows what the issue is for #1, #2, or #4, but the extra conditions in those problems rule out this exact example. I'll let others decide whether they are true.

[hide="Hint."]Of the five statements, at least two are false.[/hide]

[quote="arp"]
5.  If $ \int_{1}^{\infty}f$ converges, then $ \lim_{x \rightarrow \infty}f(x) = 0$.[/quote]
This [b]can[/b] be made true with enough initial conditions on the hypothesis.
But consider a silly counterexample,

$ f(x) = \{ \begin{array}{c}1 , \ x\in \mathbb{N}\\ 0 , \ x\not \in \mathbb{N}\end{array}$

This integral is clearly zero because one isolated point doest change the integral. But $ \lim_{x\to \infty}f(x)$ doesnt even exist.

--------
For #1 consider $ f(x) = 1$.

sylow_theory's example is not continuous. Very well, then, let us require continuity.

Define $ T(x)=\max(1-|x|,0).$ This is the piecewise-linear continuous function with vertices at $ (-1,0),(0,1),$ and $ (1,0),$ taking on the value zero outside of $ [-1,1].$ ("T" is for "tent".)

Let $ f(x)=\sum_{n=1}^{\infty}nT(n^{3}(x-n))$

Once $ n$ is large enough for the tents not to overlap, this a piecewise linear continuous function with vertices at $ (n-\frac1{n^{3}},0),\,(n,n)$ and $ (n+\frac1{n^{3}},0).$

$ \int_{0}^{\infty}f(x)\,dx=\sum_{n=1}^{\infty}\frac1{n^{2}}$ converges.

But not only does the function fail to tend to zero, it is unbounded as $ x\to\infty.$

The fact that this example is not smooth is unimportant; by replacing $ T$ by a $ C^{\infty}$ "bump" function and repeating the idea, we could create a $ C^{\infty}$ example with the same flaws.

---

[quote="sylow_theory"]For #1 consider $ f(x) = 1.$[/quote]
But then the sequence $ I_{n}$ would diverge, so that's not a counterexample - at least if we read the problem as requiring $ \lim I_{n}$ to exists as a finite number.

It seems like interest has been lost in this question, so I'll try to answer two more and see if someone wants to do the last.

For 1, sylow, kent is right that we are to assume $ \lim I_{n}= I$ exists as a finite number.  In fact, this makes the statement true.  First, note that $ f \geq 0$ everywhere because of monotocity, for if $ f(x_{0}) = l < 0$ for some $ x_{0}$, then $ f(x) \leq l$ for all $ x \geq x_{0}$ and so $ \lim I_{n}$ could not exist.  There is some $ N$ s.t. for $ n \geq N$,
 $ |I_{n}-I| < \epsilon$ for every $ \epsilon > 0$.  Let $ I(b) = \int_{1}^{b}f$, $ b \in \mathbb R$.  For any $ b \geq N$, there is an $ n \geq N$ such that $ n \leq b < n+1$ ($ n = [b]$).  Then $ I_{n}\leq I(b) \leq I_{n+1}$ since $ f$ is nonnegative everywhere, so we get $ I-\epsilon < I_{n}\leq I(b) \leq I_{n+1}< I+\epsilon$, which means $ \int_{1}^{\infty}f$ converges to $ I$.

I'll cheat off Kent to answer 4: False. If $ f(x) = \sin(2\pi x)$, then $ f'$ is bounded and and $ \lim I_{n}= 0$, but $ \int_{1}^{\infty}f$ doesn't even exist.


combinatorics solved
combinatorics
graph theory

Here's a question I found in a paper download from this site a long time ago (I don't know if you can find the paper here now):

Prove that in a group of 2n people there are 2 people which have an even number of common acquaintances, where, as usual, if A knows B then B knows A.

It looks nice and easy, but I've been thinking abt it for quite some time and I don't even know if my soln is correct. Anyway, it inspired me to think of this:

1. Is it true that in every graph there is a vertex for which the number of chains of length 2 starting from that vertex is even? I have to mention that I don't know the answer for this one but it seems to be 'yes'. Could ne1 help me with it? It's closely related to this:

2. Is it true that in every graph there is a vertex linked to an even number of vertices of even degree? (the answer for 1 is yes iff the answer for 2 is yes, if you think abt it for a while). Any ideas?

The initial problem appeared in the Tounament of towns in 1995 (with n = 25, but it doesn't matter).
Since the solution is a little boring to write here, you may find it (in french) at :
http://boumbo.salle-s.org/xavier/maths/stmalo/

Choose the paper on graph theory. This is exercise #4.

[Moderator edit: Or see http://www.mathlinks.ro/Forum/viewtopic.php?t=68109 ]

Pierre.

Thanx a lot, but, like I said, I think I have a soln for that one (it's also a bit boring so i didn't exactly check everything out but...). I think the other problems I wrote there are quite interesting so maybe some1 here has a cute or interesting or useful idea abt them?

Hmmm...did I miss something or the answer for your pb 1 is trivially 'no' as shown by the following graph :
A-B-C-D

Pierre.

Thanks a lot! It's like taking a load off my chest. Even if the answer is no, at least I know that! Thx again!


induction
number theory solved
number theory

Prove that s_n = \sum^{2n-1}_{k=1} 2^(k-1) * Cn(4n-2,2k) is a perfect square for all n = 1,2,3,...

the sum clearly equals 1/4((1+ \sqrt 2)^(4n-2) + (1- \sqrt 2)^(4n-2) -2)
define an=2an-1 +an-2 with a1=6 and a2=198
it remains to show that a4n-2 -2 is perfect square.
we show by induction that a4n-2 = a2n-1^2 +2 ,a4n=a2n^2 -2
a4n-1=1/2(a2n^2-a2n-1^2)-2 and that a4n+1=1/2(3a*2n^2+a2n-1^2)+6
proof: lets assume that the hypothesises are true for all n<=4k.
a4k+1= 2a4k+a4k-1 = 2a2k^2-4 + 1/2(a2k^2-a2k-1^2)-2
=1/2(3a2k^2+a2k-1^2)+6
similarly you can do it for a4k+2, a4k+3, a4k+4.

a type mistake .
it should be a1=2, a2=6
the rest is the same.


I like the ones that aren't to fake like things that you know will never happen. The type of movie that makes you afraid of the dark for like 2 weeks lol...

[quote="1=2"]Re: Thriller/Horror movies
What do you think is the best thriller/horror movie composed of?
I'm just wondering.[/quote]
Well, Have you watched any horror/thriller movies lately? (I mean, do you watch horror or thriller movies?) If you have watched some good horror movies, you wouldn't have asked a question like this.

(Watch some horror movies, say like:The Exorcist, Evil Dead, The Ring etc; They're good horror movies)

By far the best horror movie is The exorcist 1. I mean , I couldn't sleep well for three entire weeks  :P

In second place I would place H2O , but just the first one,  because the following H2O suck  :D

In third place I would place Scream , The ring , The profecy 1 and Poltergeist 1


geometry
incenter
circumcircle
power of a point
radical axis

[color=cyan]So, here they be.  The rules are simple:  you can draw a circle centered at a given point through a given point.  You can draw a circle centered at a given point with a radius a given length.  And you can draw a straight line through two given points.  That's it.  You can use previous problems to do later problems.

1.  Given 2 points, construct the midpoint of the segment joining them.

2.  Given a point and a line, construct the unique perpendicular to the line passing through the point.

3. Construct the incenter of a given triangle.

4. Construct the circumcenter of a given triangle.

5.  Construct the tangent to a given circle from a point not inside that circle.  (Note that this has two cases.)

6. Given a unit length, construct a length of :sqrt:n where n is any positive integer.

7. Construct a regular hexagon of given side length.

8. Construct a regular pentagon of given side length at least 2 different ways.

9. Construct the radical axis of two circles.  (The radical axis is the set of points whose powers with respect to the two circles are equal.  That is, given 2 circles O1 and O2 with radii r1 and r2 respectively, the radical axis consists of all points P such that PO1:^2: - r1:^2: = PO2:^2: - r2:^2:.)

10. Divide a segment into n equal pieces for any positive integer n.

11. Construct the common tangents of 2 circles in all 3 cases for which tangents exist (0, 1, and 2 intersection points).

Look in the Advanced Forum for three more difficult problems on construction, plus whatever anyone else has added.[/color]

Another one I like is: given a line segment, divide it into n equal parts for any positive integer n.

[color=cyan]Oops, as I was writing this I was going to put that one in near the bottom but then I forgot.  So why not make that number 10?[/color]

Hm..that's some pretty interesting constructions.  I'm taking drafting this year, and it'll probably pretty useful for it!

-interesting_move


Given the quadrilateral $ABCD$ with $\angle DAB=150^{o}$, $\angle DAC+\angle ABD=120^{o}$ and $\angle DBC-\angle ABD=60^{o}$. Evaluate $\angle BDC$.

Answer 45 :D
Putting $\angle{BAC}= t$ then $\angle{CAD}=150-t$, $\angle{ABD}=t-30$, $\angle{DAC}=t+60$, \angle{ADB}=60-x, \angle{ACB}=150-3t.

If $\angle{BDC}=x$ then
$\frac{BC}{CD}=\frac{BC}{AC}. \frac{AC}{CD}$ and after few sine laws we get
(1): $\frac{sin x}{sin(60-t+x)}=\frac{sin t. sin(t+60)}{sin (2t+30). sin (t+30)}$

We can check that if $x=45$ then (1) is true for every $t$.

From here $\frac{sin x}{sin(60-t+x)}=\frac{sin 45}{sin(60-t+45)}$ and therefore $x= 45$.


trigonometry
limit
real analysis
real analysis unsolved

Find a real number $ r\ne1$ for which $ \lim_{n\to \infty}\sin(r^n)$ converges to a number unequal to 0,
(or show that such a number does not exist).

no  :(

(The message was too small. So I made the message longer before submitting).

Maybe this is idea (only idea not the full solution, I really don't wanr to spend time to solving this problem  :lol:  )
I will speak roughly by the approximation language).
suppose there  exist limit, then we can write
$ r^{n} \equal{} 2\pi k_{n} \plus{} b \plus{} \eta_{n}(\pi \minus{} 2b) \plus{} \varepsilon_{n}$
where $ k_{n} \in \mathbb{Z}$,  $ \eta_{n}$  takes value from the set $ \{0,1\}$  or  $ \eta_{n} \in \{0,1\}$  and  $ \varepsilon_{n} \to 0$ as $ n \to \infty$
I also suppose that $ \eta_{n} \equal{} 0$ for all $ n$ (I believe that  such steps as $ r > 1$, $ \eta_{n} \equal{} 0$ is true and one can explain it.  :lol: )
also suppose that $ k_{0} \equal{} 0$ then we have :
$ r^{n \plus{} 1} \equal{} 2\pi k_{n \plus{} 1} \plus{} b \plus{} \varepsilon_{n \plus{} 1}$ thus we have :
$ 2\pi k_{n \plus{} 1} \plus{} b \plus{} \varepsilon_{n \plus{} 1} \equal{} r(2\pi k_{n} \plus{} b \plus{} \varepsilon)$ hence :
for sufficiently large $ n$ we have :
$ k_{n \plus{} 1} \approx \frac {b(r \minus{} 1)}{2 \pi} \plus{} rk_{n}$ or
$ k_{n \plus{} 1} \approx \frac {b(r^{n \plus{} 1} \minus{} 1)}{2\pi}$
on the other hand we have : $ r^{n \plus{} 1} \equal{} 2\pi k_{n \plus{} 1} \plus{} b \plus{} \varepsilon_{n \plus{} 1}$ hence,   for the sufficiently large $ n$ we have
$ 2\pi k_{n \plus{} 1}\approx 2\pi k_{n \plus{} 1}b \plus{} b^{2}$, but it's impossible , since   such $ b$ does not exist.


If I want to teach myself Computer Science to the point of being able to do well on an AP Computer Science [i]course[/i] that's online, what should I learn?

It's Java based from what I hear, so is learning Java sufficient?

Two great books:

Concrete Mathemathics by Knuth,Graham,Patashnik - contains exercises + solutions to all of them
Introductio to Algorithms by Cormen, Leiserson, Rivest, Stein - a lot of useful stuff.

Another great resource:

ocw.mit.edu

And judges( in case you'd like to practice):

train.usaco.org - worth doing
acm.uva.es
spoj.pl
....


pigeonhole principle
geometry
analytic geometry
AMC
USA(J)MO
USAMO

1) Suppose a triangle can be placed inside a unit square such that the center of the square is not inside the triangle. Show that one side of the triangle has length less than 1.

2) A region M in the plane has area > 1. Prove that there are two points (a, b) and (c, d) in M such that a - c and b - d are both integers.

3) In a party, n girls and n boys are paired. One observes that in each pair, the difference in height is < 10 cm. Show taht the difference in height for the kth tallest boy and the kth tallest girl is also < 10 cm for k = 1, 2, ..., n.

4) Prove that from any sequence of 2003 reals, one can choose a consecutive block whose sum differs from an integer by at most 1/1000.

5) In a contest with 8 problems, 8 students took part. Each problem was solved by 5 students. (a) Prove that there are two students who between them solved all eight problems. (b) Is this true if 5 is replaced by 4 ?

6) Prove that among any 6 integers, there will be a pair whose sum/difference is divisible by 9.

(6)[hide]more generally, if we are given n numbers, there exist 2 whose sum/diff. is divisible by 2n-3. We construct (n-1) boxes, labeled 0, 1&(2n-4), 2&(2n-5), 3&(2n-6), ..., (n-2)&(n-1). We put each of the n numbers in the box that contains its residue mod (2n-3). Now, by pigeonhole, as we have n numbers and (n-1) boxes, two must be in the same box. Either their sum or difference is 0, mod 2n-3. (here n=6)[/hide]

(4) [hide]let the sequence of reals be given by ak, 1:le:k:le:2003. define si as {:Sigma:(k=1 to i) ak}, where {x} represents the fractional part of x, or x-[x]. now, divide the real number line from 0 to 1 into 1000 parts. If one of the si is in the first or last block, from 0 to 1/1000, we're done. If not, there exist two of the si in another block. Taking the difference yields a consecutive block of that sequence, whose sum is within 1/1000 of an integer.[/hide]

(2) [hide]let us proceed by contradiction. superimpose (is that a real word?) this region on a cartesian plane tiled by horizontal and vertical points through the lattice line, subdividing the plane into 1 by 1 unit squares. now, if (x,y) is the ordered pair corresponding to a point in the plane, define its SPECIAL coordinates (creative, aren't i?) as ({x},{y}), where {a}=fractional part of a=a-[a]. then, assuming that there are no two points of m satisfying the given condition, there cannot be any two points in M having the same pair of SPECIAL coordinates. So take any 1 by 1 square in the plane, and note that M can be mapped into this square without any overlap by the preceding observation. however, this would mean the max area of M is 1, a contradiction.[/hide]

I'm afraid I don't understand number 4 at all.

Mystic, I'm not entirely convinced of your number 2 -- I can design plenty of injective maps from regions of area greater than 1 to a 1x1 square -- you have to prove that just because two points aren't mapped to the same place means that the initial area is less than 1.

Okay, so lets do number 5.

[hide]If any student solved all 8 problems, we're done.  If any student solved 7 problems, some other student must have solved the 8th, so we take those 2 students and we're done.  If any student solved 6 questions, consider the other 2 students.  Each was solved by 5 people, thus they must both have been solved by at least 1 of the same people.  Take that person and our six-question guy, and we're done.

Thus, the only scenario we are left with is that in which each student solved exactly 5 questions.  Now then, returning to this much-neglected problem:

Take any of the students, and WLOG he has solved questions 1, 2, 3, 4 and 5.  We wish to show that some other student solved questions 6, 7, and 8.  Now, between them, questions 6, 7, and 8 have been solved a total of 15 times by the 7 other students.  By the pigeonhole principle, one of the students must have accounted for 3 of these solutions.  Since he can't solved the same question 3 times (I would have had a perfect score on USAMO last year if I could have done that ), he must have solved each of 6, 7 and 8.  Thus, we are all done.

I highly suspect the same is not true for 4.  I'll get on it now.[/hide]

Arne wrote:An extra question:

The bound 1/1000 for problem 4 can obviously be sharpened.
What is the sharpest possible bound ?

[hide]If the bound is of the form 1/k, with k an integer, we construct k boxes, and define si as before. now, we have that if any of the si are in the first or last box, we're automatically done. now, if not, we want at least 2 of the si in one of the remaining boxes so that we can take their difference. So i think k is 2004, and the lowest bound is 1/2004. however, i don't know why the bound should be of the form 1/k. sort of non-rigourously, we can see that if k is not an integer, we will have [k] boxes and then a mini-box in the middle, that will not sharpen the pigeonholing argument. but that's not too terribly rigorous.[/hide]

ugh, JBL, you're right on my #2. if i said [hide]construct the mapping from M to the unit square with vertices at (0,0), (1,1), (0,1), and (1,0), such that any point in M is mapped into its SPECIAL coordinates, and then note that this mapping must be injective by the above reasoning, and then note that this preserves area because basically splitting M based on the grid lines and then reassemble it in one square won't change area[/hide] would that fix it?

Yeah, I think that will do it.  The way I would phrase it is to say that it preserves area over the open lattice squares, and since the other parts (all the grid lines) don't have any area, it does in fact preserve area.  You might have to mess that around a little bit when you talk about the boundaries, but I don't think that's a really major point.

For 4), a hint :

Consider the sequence 1/2004, 1/2004, ..., 1/2004 (2003 terms).
This sequence has k = 1/2004.

"between them" implies when the solutions from two students are put together, all 8 problems are solved.

In your particular example, students 1 and 8 (among several other pairs) between them solved all 8 questions.


Solve
\[ \left\lfloor {\frac{{x \minus{} p}}
{p}} \right\rfloor  \equal{} \left\lfloor {\frac{{ \minus{} x \minus{} 1}}
{p}} \right\rfloor
\]

for real $ x$ and nonzero integer $ p$

I think $ x$ belongs to $ [0 , \frac {p\minus{}1}{p}]$


A salesman buys a coat at $ \$64$ less than $ 12.5 \%$. He then sells the coat at a gain of $ 25\%$ of his cost after allowing a $ 20\%$ discount on the marked price. What is the marked price, in dollars, of the coat?

i know... the first sentence doesn't make sense... maybe it's saying he bought it at a 12.5% discount, and the final price was 64? 0.o sometimes the questions and solutions for FTW are a bit shaky, like for some reason, the answers are all like $ DFRAC(7)(11)$. 0.o it would be greatly appreciated if all of those bugs could be worked out, especially in the Alpha Test version. -.-

The first sentence actually means "The salesman buys a coat at \$64 [i]minus[/i] 12.5%".

12.5% = $ \frac{1}{8}$, so $ 64 \cdot \frac{7}{8} \equal{} \$56$
He gains 25%, so $ \$56 \cdot 25\% \equal{} \$14$ , $ \$56  \plus{} 14 \equal{} 70$.
To find the marked price, we must reverse the discount, so we divide $ \$70$ by $ \frac{80}{100}$. The answer and marked price is $ \$87.5$.


A surveyor is mapping a triangular plot of land. He measures two ofthe sides and the angle formed by these two sides and finds that thelengths are 400 yards and 200 yards and the included angle is 50°.What is the measure of the third side of the plot of land, to the nearest yard? What is the area of this plot of land, to the nearest square yard?

Can you set up the equation using the law of cosines?
I would be able to answer the question if I had the right equation.


calculus
inequalities unsolved
inequalities

Find the maximum value of the expression: $ (x^3\plus{}1)(y^3\plus{}1)$ when $ x\plus{}y\equal{}1$. Without using calculus.
I was trying cauchy-schwarz, but no success.

I solved, the answer is $ 2$.
What do you mean by a solution without calculus? :?:
Edit: don't read, I thought that $ x,y\geq 0$.

[quote="danilorj"]Find the maximum value of the expression: $ (x^3 \plus{} 1)(y^3 \plus{} 1)$ when $ x \plus{} y \equal{} 1$. Without using calculus.
[/quote]
$ (x^3 \plus{} 1)(y^3 \plus{} 1)\leq4\Leftrightarrow3 \minus{} (x^3 \plus{} y^3) \minus{} x^3y^3\geq0\Leftrightarrow$
$ \Leftrightarrow3(x \plus{} y)^6 \minus{} (x \plus{} y)^3(x^3 \plus{} y^3) \minus{} x^3y^3\geq0\Leftrightarrow$
$ \Leftrightarrow(x^2 \plus{} 3xy \plus{} y^2)^2(2x^2 \plus{} 3xy \plus{} 2y^2)\geq0,$ which true.
The equality holds when $ x^2 \plus{} 3xy \plus{} y^2 \equal{} 0$ id est, when $ x \equal{} \frac {1 \plus{} \sqrt5}{2},$ $ y \equal{} \frac {1 \minus{} \sqrt5}{2}$ or
$ y \equal{} \frac {1 \plus{} \sqrt5}{2},$ $ x \equal{} \frac {1 \minus{} \sqrt5}{2}.$
Thus, the answer is $ 4.$

[quote="danilorj"]But how do you assure that 4 is the maximum?[/quote]

aquardy has proved that
$ (x^3 \plus{} 1)(y^3 \plus{} 1) \le 4$
So that the maximum is $ \le 4$

But he also show example when the  equality holds.
Hence 4 is the answer.

i have   solution for  this problem:
 Let $ xy\equal{}t \leq \frac{1}{4}$
and $ P\equal{} 1\plus{}x^3y^3\plus{}(x\plus{}y)^3\minus{}3xy(x\plus{}y)\equal{}t^3\minus{}3t\plus{}2.$
We have:$ f'(t)\equal{}3t^2\minus{}3\equal{}0 \leftrightarrow t\equal{}\minus{}1 ,t\equal{}1$
Thus,$ MaxP\equal{}f(\minus{}1)\equal{}4$
Last, we solve equation $ t^2\minus{}t\minus{}1\equal{}0$, to find values$ x$ and $ y$

Let $x,y $ be reals such that $ x + 2y = 1$. Prove that
$$ (x^3 + 1)(y^3 + 1)\leq \frac{11}{2}+3\sqrt 2$$


AMC
USA(J)MO
USAMO
MATHCOUNTS
AMC 10
\/closed

In the middle School Tournaments, there have been some tourneys that involved  USAMO and IMO problems, and I was wondering if there could be a forum such that they are separated?  (ie. High School Tournaments)

Well, not sure exactly which forum you are referring to. But the mods of the respective forum should rather execute their moderator rights more effectively by moving all USAMO and IMO problems to the olympiad section, or other suitable fora possibly. If you see such cases you may also want to pm the mods in charge.

007math wants to create a USAMO/IMO Tournaments forum.

But how well would that work? A tournament of IMO problems? I think maybe IMO problems should just be presented individually.

gah...i mean an overall High School Tournaments Forum (like middle school  :lol: )

[quote="007math"]gah...[/quote]

oh... sorry if i frustrated you by overgeneralizing your suggestion. :)

Subsection of MATHCOUNTS - Middle School Tournaments
There could be a section of High School for tournaments, as tournaments in MST are very boring with only MATHCOUNTS problems, and if someone tries harder problems, mods rush in and warn that those tournaments shouldn't be there.

Well we could start some tournaments in the high school forums, but how much would a separate tournaments forum be used?

Reasons against:

Newbies will promptly post all their AMC problems in this forum, since "it's for High School Competitions, and the AMC is a High School Competition", thus making it an interminable moderator headache for whoever is unlucky enough to mod that forum

There really aren't that many fullblown AoPS competitions in high school, whereas there are quite a few active mock Mathcounts competitions and they clutter up the main Mathcounts forum. Thus, a high school forum wouldn't be helpful.

[quote="solafidefarms"] thus making it an interminable moderator headache for whoever is unlucky enough to mod that forum
[/quote]

I would mod it :D
And also, every tournament with AMC problems, or similar could easily be locked, or set to doom.

[quote="solafidefarms"]
whereas there are quite a few active mock Mathcounts competitions and they clutter up the main Mathcounts forum. [/quote]

Most users in MST are probably bored of easy MC problems (only newbies are the problem, but mostly the same people join every tournament). 007math pointed out that in his first post (MC problems are just too boring and easy to just post them in tournaments)

Well, I wouldn't have a problem modding it.

A lot of people are past MC level, and would like a tournament with harder problems.

Heh, sometimes modding is more annoying than it looks.

Perhaps instead tournaments with more difficult problems could just be posted in Middle School Tournaments.

Making harder problems that aren't computationally intensive or instantly solved by one application of something is fairly difficult. No offense to anyone, but the problems in many mock mathcount tests are quite computationally intensive. However, not all Mathcounts problems are boring. They might not be as difficult, but there are several with quite elegant solutions.

I'm not sure about adding a high school tournaments forum...there are already some mock tests, although, to be fair, there aren't too many. I guess the main thing that I'm worrying about is that a lot of the tests that will come up are going to have half-baked problems. I agree with solafidefarms.

But hey, if you think that this is a good idea, go for it.

[quote="AIME15"]We could create a whole new forum altogether, split into Middle School Tournaments and High School Tournaments.[/quote]

This is kinda a good idea.

I'm not sure that making a whole new forum for HST is the right way to go. I think that we should keep our Middle school tournament forum and the next time someone makes a mock test, they will just specify how hard the problems are going to be so that people who can't do USAMO problems don't sign up to some mock MATHCOUNTS test with USAMO level problems.

xpmath, again I say that less tournaments are probably caused by lack of MC level problems.
mathemagician1729, OK, but maybe renaming the forum MST to M&HST (ok, it's not good name)

[quote="vallon22"]couldn't you just make a mock test in the high school forum and lable it "mock test" like everyone used to do in mathcounts forum before the new subforum?

or is there a rule in the high school forum forbidding this?[/quote]

i think this is a good idea. mock tests have helped me a tremendous amount for mathcounts, so im sure a mock amc 10 or 12 wouldnt hurt. or maybe, a subforum entitled High School Tournaments.

You could probably post Mock AMC's in the AMC forum.

My idea is just to move middle school tournements and put it into a forum called Tournements, which would also include High school tournements.


geometry
circumcircle
ratio
trigonometry
conics
parallelogram
geometry unsolved

There is a circle with its circumcenter as O and its radius is 1. There is a point A outside the circle. AM,AN is the tengents to the circle O with M N as the tangent points respectively. There is a point L in bad arc MN (not equal to M,N). A line passing point A parallel to line MN intesects with ML,NL at point P,Q respectively. if PQ=2, cos(angle POQ)=1/3, find the length of AM.

i have no idea to use the given condition cos(angle POQ)=1/3. i dun know how to translate it.

Assuming that "bad" arc $ MN$ should be "back" arc $ MN$ (more distant arc from $ A.$ Let $ AL$ cut the circle $ (O)$ again at $ K$ on the front arc $ MN$ (closer to $ A$) and let $ X \equiv MN \cap AL$. Using directed line segments:

$ \overline{AP} \cdot \overline{AQ} \equal{} \frac{\overline{AL}^2}{\overline{XL}^2} \cdot \overline{XM} \cdot \overline{XN} \equal{} \frac{\overline{AL}^2}{\overline{XL}^2} \cdot \overline{XL} \cdot \overline{XK} \equal{} \frac{\overline{AL}^2}{\overline{XL}} \cdot \overline{XK} \equal{}$

$ \equal{} \frac{\overline{AL}}{\overline{AK}} \cdot \frac{\overline{XK}}{\overline{XL}} \cdot (\overline{AL} \cdot \overline{AK}) \equal{} \minus{}\overline{AM}^2$

because the cross ratio $ \frac{\overline{AL}}{\overline{AK}} \cdot \frac{\overline{XK}}{\overline{XL}} \equal{} \minus{}1$ is harmonic (easily seen by projecting the circle $ (O)$ to a circle and the line $ PQ$ to infinity). Giving $ \cos \widehat{QOP}$ is obfuscation, we need $ \tan \widehat{QOP} \equal{} ... \equal{} 2 \sqrt{2}.$

$ \tan \widehat{QOP} \equal{} \frac{\tan \widehat{QOA} \plus{} \tan \widehat{AOP}}{1 \minus{} \tan \widehat{QOA} \cdot \tan \widehat{AOP}} \equal{} \frac{\frac{\overline{QA}}{\overline{OA}} \plus{} \frac{\overline{AP}}{\overline{OA}}}{1 \minus{} \frac{\overline{QA}}{\overline{OA}} \cdot \frac{\overline{AP}}{\overline{OA}}} \equal{}$

$ \equal{} \overline {OA} \cdot \frac{\overline{QA} \plus{} \overline{AP}}{\overline{OA}^2 \minus{} \overline{QA} \cdot \overline{AP}} \equal{} \frac{\overline{OA} \cdot \overline{QP}}{\overline{OA}^2 \minus{} \overline{AM}^2} \equal{} \frac{\overline{OA} \cdot \overline{QP}}{\overline{OM}^2} \equal{} 2 \sqrt{2}$

Substituting $ \overline{OM}^2 \equal{} 1,$ $ \overline {QP} \equal{} 2$ yields $ \overline{OA} \equal{} \sqrt{2}$ and $ \overline{AM}^2 \equal{} \overline{OA}^2 \minus{} \overline{OM}^2 \equal{} 1.$ Done. The problem does not ask for this, but from $ \overline{QA} \cdot \overline{AP} \equal{} \overline {AM}^2 \equal{} 1$ and $ \overline{QA} \plus{} \overline{AP} \equal{} \overline {QP} \equal{} 2,$ we get $ \overline {QA} \equal{} \overline {AP} \equal{} 1,$ points $ P, Q$ are symmetrical with respect to the line $ OA$ and $ L \in OA.$

[quote="yetti"]
because the cross ratio $ \frac {\overline{AL}}{\overline{AK}} \cdot \frac {\overline{XK}}{\overline{XL}} \equal{} \minus{} 1$ is harmonic (easily seen by projecting the circle $ (O)$ to a circle and the line $ PQ$ to infinity).[/quote]

i dun understand the reason for the harmonic division..(i am still a beginner of projective geom), can yo explain a little bit more?

Suppose you have 2 planes $ \pi, \pi',$ not parallel, and a point $ S$ not on either of them. Arbitrary line through $ S$ cuts $ \pi, \pi'$ at $ P, P'.$ $ P'$ is image of $ P$ in the central projection of $ \pi$ to $ \pi'$ from $ S.$ But lines through $ S$ parallel to $ \pi'$ do not cut it, even though they cut $ \pi.$ Collection of all lines parallel to $ \pi'$ forms a plane $ \sigma \parallel \pi'$ and the plane $ \sigma$ cuts the original plane $ \pi$ in a line $ l.$ The image $ l'$ of $ l \in \pi$ does not show in $ \pi'.$ We say that the line $ l \in \pi$ was projected to the line at infinity $ l' \in \pi'.$ Each line through $ S$ parallel to $ \pi'$ defines a direction in $ \sigma$ or in $ \pi'$ and collection of all these lines is a collection of all directions in $ \sigma$ or in $ \pi'.$ Therefore, point at infinity is the same thing as direction in a plane and line at infinity is collection of all directions in a plane. For example, 2 parallels have a common point at infinity - they have the same direction. Any 2 lines intersecting on the line $ l$ projected to infinity become parallel, because their common point is at infinity.

Generally, central projection takes a circle to a conic section and it preserves tangencies. I used a well known theorem (with long 3D proof): Given a circle and a line not intersecting this circle, a suitable central projection exists, which takes the given circle to another circle and the given line to infinity. In your problem, the circle $ (O)$ can be projected to another circle $ (O')$ and any line through $ A$ not intersecting the circle $ (O),$ for example the line $ PQ,$ to infinity. Since the image $ A'$ of $ A$ is at infinity, images $ A'M', A'N'$ of the circle tangents $ AM, AN$ and image $ A'K'L'$ of the circle secant $ AKL$ become parallel, $ M'N' \perp K'L'$ becomes a diameter of the projected circle $ (O').$ The cross ratio of 4 points is preserved in a central projection,

$ \frac {\overline{AL}}{\overline{AK}} \cdot \frac {\overline{XK}}{\overline{XL}} \equal{} \frac {\overline{A'L'}}{\overline{A'K'}} \cdot \frac {\overline{X'K'}}{\overline{X'L'}} \equal{} \frac {\overline{X'K'}}{\overline{X'L'}} \equal{} \minus{} 1$

$ \overline{A'L'}, \overline{A'K'}$ are infinite segments with the same direction, they cancel out, because their difference is insignificant compared to infinity. $ X'$ becomes the midpoint of $ K'L',$ $ \overline{X'K'}, \overline{X'L'}$ are oppositely directed segments of equal length, that's why $ \minus{} 1.$
__________________________________________

You could also use the following theorem: 2 diagonals of a complete quadrilateral cut the 3rd one harmonically. Let $ AB, BC, CD, DA$ be 4 lines forming a complete quadrilateral. $ X \equiv AB \cap CD,$ $ Y \equiv BC \cap DA,$ $ Z \equiv AC \cap BD.$ Lines $ AC, BD, XY$ are the 3 diagonals. How do you prove the theorem ? Project the line $ XY$ to infinity, the image $ A'B'C'D'$ of $ ABCD$ becomes a parallelogram, and use invariance of the cross ratio in a central projection.

In your problem, you also have to know that $ MN$ is a polar of the point $ A$ with respect to the circle $ (O).$ Let $ UV$ be another circle secant from $ A.$ Then $ B \equiv KV \cap LU,$ $ C \equiv KU \cap LV$ lie on the polar $ MN.$ Let $ KV, LU, KU, LV$ are the 4 lines forming a complete quadrilateral, then $ KL, UV, BC$ are the 3 diagonals. The 2 diagonals $ UV, BC \equiv MN$ cut the 3rd diagonal $ KL$ harmonically at $ A, X,$ so that

$ \frac {\overline{AL}}{\overline{AK}} \cdot \frac {\overline{XK}}{\overline{XL}} \equal{} \minus{} 1.$

The 1st method (projecting the circle $ (O)$ to a circle and the line $ PAQ$ to infinity is much quicker and easier to imagine.


Okay firstly please be honest. And don't post srini does not sleep at all and rakesh sleeps only for 2.5 hours every year. Post proper stuff here.

I used to be a typical late night sleeper. Sleeping after 2:30 and sometimes going on till 3:30. And then most of my morning was spoiled. I used to get up at 8:30 mull around for another hour. go back to sleep and then start productive work only at 1:00. However I wanted to shift my biological clock because this was not really helping me in comp exams and I was feeling sleepy and I started sleeping at 12:00 itself with disastrous results. I still feel sleepy in the morning. :D:D

So what time do you goto sleep? And how long do you sleep?

Used to be from 11:00 to 6:30 earlier abt a month before :D
But bcos of the influence of my friends n shre in particular, i went to late night and am not able to sleep till 12:00 am :( I try to but still, dont have the inclination!
Nowadays its been from 12:30 am to 8:30 am

PS: Elastiboysai had written he slept from 3:50 to 9:30 in another post :)
And ofc for the sake of saying it i shall say:
srini does not sleep at all and rakesh sleeps only for 2.5 hours every year  :P

ha well, my sleeping patern is shockingly similar to shreyas's :D
especially in the last few months, i've been sleeping at around 2- 2.30 and wake up at 1030- 11. Effectively, i start studying properly only at around 1 :(

and yesterday, i promised myself that from now on, i'd sleep at 12 itself to "tune my biological clock" :D
but it was a DISASTER. i just couldn't get proper sleep and for at least an hour or 2, i was just lying down and yawning. as a result i got up only at 930 today :(
*yawns*

Baby, why did you post the same thing so many times? :D

My sleeping pattern is probably the most erratic but still.. I usually have dinner and go to sleep at around 9 and then get up at 11-11.30, then study/do something that people would tend to mistake for studying till 2.30, watch Dawson's Creek from 2.30 -3 and then go back to sleep and then get up in the morning at 7 - 7.30...

Green tea is a very potent stimulant :D

The mod can edit that :D Vetti mod is always here to the rescue :D
That was a consequence of the stupid server here which was swaying between various links all by itself without me touching anything :D

And incase u dont believe that it wasnt my fault, i couldnt have posted all that as consecutive posts wont be allowed for less that 10 mins :P

hey to be frank i go to sleep at around 9 and wake up at 7 am  :P and rohit dont tell i dont sleep and all for sleep is the most important aspect of my life da and one small correction rakesh sleeps for only 1:00:00:02 hrs and please dont exxagerate it to 2.5 hrs and all!!! :D

[quote="madness"]Dei srini, just bcos shre said u can write nething u feel is right, doesnt mean u can lie :D
[quote="lankies"]
You lie like a brute :P
[/quote][/quote]
hey it shows u are offline but how did you post now???

dei its saying the truth da :D
im offline :P :P

[quote="madness"]dei its saying the truth da :D
im offline [/quote]
hey is there any facilities for you to change your active state!!! :?:

[quote="Albert Einstien"][quote="madness"]dei its saying the truth da
im offline [/quote]
hey is there any facilities for you to change your active state!!! [/quote]

well there is pathetic fallacy :P :D
you cant change the active state of a compd atleast :P

i don't any compromise with my sleep--i sleep at 11.30 and would have woken up at 7.30 if it weren't 4 school
i've to wake at 6:15 but i do take a nice nap at school :D  :D

Now that u all r talkin abt it,
i used 2 sleep frm 11.30 to 6.30-7.00 normally :wink:
for the past 1.5 months my sleep pattern has drastically changed.
Somthin like 3.30 ,4 to 9.40 10.
Funnily  I too tried to sleep early yest- 11.40, it worked bt i woke up only at 9. :rotfl:

I'm scared I'll sleep off during JEE paper I. :( My full momentum is usually gained only after 3. :(

I am scared abt paper 2 :oops:
I really slept off during FT5 paper 2 (thats bcoz i had eaten 2 much n was ready 2 burst) :rotfl:

oh my god u all study till 3 and all ah ???how is it possible ??? u all kattans!!!!!!!!!!! :o  :o  :o

Look who is talking. Thats all I have to say

[hide="TO SRINI AND ROHIT:"]
I DON'T SLEEP AT ALL DA :P  :P
[/hide]

Anyway the truth is I sleep around 12:00,12:30 and ofc get up only at 8:30 in the morning. It has been my routine for about 2 months now. :rotfl:

I generally sleep at around 12- 1230 and get up between 6:30 to 7:00 .

very few times i have tried 11 - 4:45 .

Both have equal productivity

Shifted to my old self. Sleeping only after 2.  :rotfl:  :rotfl:

u r damaging ur pineal gland

i think soumya is practising to prevent jetlags :rotfl:  :rotfl:


I was looking through the USAMTS webiste, and I noticed it said that one must send both the pdf and the latex file if using LaTeX. I didn't do this when I submitted it, does that mean my work will get disqualified?

No, it won't be disqualified.  If we don't have the TeX code, then we probably won't publish your solution.  But your paper will still be graded.


Show that there are an infinity of natural numbers wich cannot be written as a^3+b^3 with a,b naturals but can be written in the same form with a,b rationals(positive).

However any natural number gives residue 0,1,8 mod9
the number 9k+3 can't be written as a^3+b^3 with a,b naturals.
Now let a=9k^3-1,b=9k^3+1,then n=(a^3+b^3)/9k^3 is natural and gives residue 3 mod9,that we had to prove.


A fair, 6-sided die is thrown and a card is drawn from a 52-card, standard deck. Find the probability that a multiple of 3 appears on the die and a face card (Jack, Queen, King, or Ace) is drawn.

[quote="i_like_pie"][hide]There are two multiples of 3 on a die, so the probability is $\frac12.$

There are 12 face cards, so the probability is $\frac{12}{52},$ or $\frac{3}{13}.$

$\frac12\cdot\frac3{13}=\boxed{\frac3{26}}$[/hide][/quote]

wait... wouldn't the probability of the die thing be $\frac13.$, and the face card one be $\frac{16}{52},$ or $\frac{4}{13}.$
so then it would be $\frac13\cdot\frac4{13}=\boxed{\frac4{39}}$
am I wrong???...

[quote="Necrowarrio0"]xxxxx, It would not be 4/39 because you plus the two probabilities together, not times.[/quote]
You multiply, and xxxxx is correct. The correct answer is ${4\over39}.$

Oh is it multiply? I thought it was plus.....  :|

The above was edited.

AH SORRY!!!! IT IS MULTIPLY....................  :(

In probablity "and" usually tells you to use multiplication
while "or" usually tells you to use addition

Aren't face cards just jack, queen, and king?  :P

I guess it doesn't matter because the question tells you that it includes ace as a face card


abstract algebra
superior algebra
superior algebra solved

for $S= \{m/n \in Q| n Is Odd  \}$, under usual operations on real numbers. I must determine if this is a ring.  Is it sufficient to show that $S$ satisfies the ring axioms? ie: $(m_1/n_1) + (m_2/n_2) = (m_2/n_2) + (m_1/n_1)$ so $S$ is abelian under addition, then show Closure, Associativity, and distributivity

[quote="zargon"] I must determine if this is a ring.  Is it sufficient to show that $S$ satisfies the ring axioms?[/quote]

Well if an object satisfies all the properties in the definitition of a class, then it belongs to the class. Basic logic.

The least obvious thing to check is closure.
Take $a/b,c/d\in S$. Then $b,d$ are odd numbers. $a/b+c/d=(ad+bc)/bd$. This may not be a reduced form for $a/b+c/d$, but since $bd$ is odd, no matter by what you divide it, you still get an odd number, so $a/b+c/d\in S$.


[quote="frt"][hide]
$1995=3\cdot 5\cdot 7\cdot 19$, so $1995^{10}$ is a multiple of $9$, which means the sum of all the digits must be divisible by $9$.

Adding all the digits, we get $154\equiv 1\pmod{9}$, so $N=8$.
(I hope I added them correctly... :roll: )[/hide][/quote]

I was so intimidated by the number that I didn't think to use that simple method.

(and you added correctly according to my super [2 dollar] calculator)

[quote="4everwise"][quote="frt"]
$1995=3\cdot 5\cdot 7\cdot 19$, so $1995^{10}$ is a multiple of $9$.[/quote]You mean it's a multiple of 3. ;)[/quote]

Well, since 1995 has one multiple of 3, then $1995^{10}$ has a factor of $3^{10}$ and $3^2=9$.

Right. I'm losing it today, lol.  :blush:  :rotfl:


$ P$ and $ Q$ are each a distinct member of the set $ \{6, \minus{}\frac 12, \minus{}3, \frac 13\}$. What is the least possible value of the quotient $ P\div Q$?

To minimize first we try to make $ P\div Q$ negative.

If $ P$ is positive and $ Q$ is negative, then the minimum possible value of $ P\div Q$ is $ \frac {6}{ \minus{} \frac {1}{2}} \equal{} \minus{} 12$.

If $ P$ is negative and $ Q$ is positive, then the minimum is $ \frac { \minus{} 3}{\frac {1}{3}} \equal{} \minus{} 9$.  $ \minus{} 12 < \minus{} 9$, so $ \boxed{ \minus{} 12}$


inequalities
function
algebra solved
algebra

you can consider wlog that

a+b+c+d=1 (there is a post here that explains why this is a "WLOG" consideration)
then you use Jensen for  the convex function f : (0,+oo)-->R, f(x)=1/x with wieghts a,b,c,d.

cheers!

I think Cauchy also works. (All sums are cyclic.)
Let L denote the left - hand side.
Put A = a + b + c + d and B = ab + ac + ad + bc + bd + cd.
Then we get L *  \suma(b + 2c + 3d) \geq (a + b + c + d).
Now \suma(b + 2c + 3d) = 4B.
So we have L \geq (a + b + c + d)/(4B).
So we want to prove that (a + b + c + d)/(4B) \geq 2/3.
This is equivalent to 3A + 6B \geq 8B or 3A \geq 2B.
Now we have
1) A = a + b + c + d \geq ab + bc + cd + da
2) A = a + c + b + d \geq ac + bc + bd + da
3) A = b + a + c + d \geq ab + ac + cd + bd
and summing these gives indeed 3A \geq 2B.

PS. Nice solution Lagrangia !

This is from a shortlist isn't it? I think I've seen it before.

Yeah, I guess "IS3" means IMO 1993 Shortlist.

Here is my solution:
we have :

a/(b+2c+3d)+b/(c+2d+3a)+c/(d+2a+3b)+ d/(a+2b+3c)= a 2 /(ab+2ac+3ad) +b 2 /(bc+2bd+3ab)+ c 2 /(cd+2ac+3bc)+ d 2 /(da+2bd+3cd)  \geq (a+b+c+d) 2 /4(ab+bc+cd+da+ac+bd)
So now we have:
(a+b+c+d) 2 /4(ab+bc+cd+da+ac+bd)  \geq 2/3
3* \sum a 2 + 6* \sum ab  \geq  8* \sum ab
3* \sum a 2  \geq 2 *\sum ab
(a-b) 2 + (b-c) 2 + (c-d) 2 + (d-a) 2 + (a-c) 2 + (b-d) 2   \geq 0


algebra
polynomial
function
superior algebra
superior algebra unsolved

Let $ p$ be a prime. Here $ a_p$ is the number of irreducible polynomails of degree $ n$ in $ \mathbb F_p[X]$.

$ a_p(n) \equal{} \frac{1}{n} \sum_{d|n}\mu(d)p^{n/d}.$
here $ \mu$ is the mobius function.

How can I show that $ a_p(n)$ is positive for all $ n$ and also when $ p\geq 2$ is not a prime number.

It's a multiplicative function, and you can easily check that it's not zero when $ n$ is a prime power.
See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=30473]here[/url].

Well, you can just construct an irreducible polynomial in every field.
[quote="peteryellow"]Let $ p$ be a prime. How can I show that is positive for all and also when $ p$ is not a prime number.[/quote]
Don't use same letters in different meanings. Often denotes $ \mathbb{F}_p[X]$ the ring of polynomials over a field. In this case $ p$ is a power of prime. There are different rings with $ p$ elements if $ p$ is composite.

but can somebody give a hint of how to  show it.

One way is to show the existence of the field $ \mathbb{F}_{p^n}$ as the splitting field of $ x^{p^n}\minus{}x$. Then since the multiplicative group of a field is cyclic, there must be some generator $ a$, so $ a$ is a primitive element, and thus the minimal polynomial of $ a$ over $ \mathbb{F}_p$ must be an irreducible polynomial of degree $ n$.

Alternatively, you can show directly that that formula gives a positive number. I clicked jmerry's link, but I'm slightly confused why you can do that, since $ f(k) \equal{} p^k$ is not a multiplicative function.

[quote="peteryellow"]but my question is different. I want to show that this is positive? how can I do that?[/quote]
If you already know that it counts something, then it cannot be negative; so only 0 must be excluded,

[quote="calc rulz"]I clicked jmerry's link, but I'm slightly confused why you can do that, since  is not a multiplicative function.[/quote]
You're right, and I made a mistake the first time around. Two quick fixes:
(1): $ \sum_{d|n}\mu(d)q^{n/d}\ge q^n\minus{}q^{n\minus{}1}\minus{}q^{n\minus{}2}\minus{}\cdots\minus{}1\equal{}q^n\minus{}\frac{q^n\minus{}1}{q\minus{}1}\ge 1$
(2): If $ D$ is the largest squarefree divisor, the sum is equal to $ \mu(d)q^{n/D}$ mod $ q^{n/D\plus{}1}$, and is not zero. Since it counts something, it's positive.

how do you get that:

$ \sum_{d|n}\mu(d)q^{n/d}\ge q^n\minus{}q^{n\minus{}1}\minus{}q^{n\minus{}2}\minus{}\cdots\minus{}1$

$ \mu(k)\ge \minus{}1$ for all $ k$, and subtracting the terms that don't match to divisors makes it smaller than not subtracting them. In fact, we never have equality, and I don't need that last $ \minus{}1$ term. It comes closest for $ n\equal{}2$, with a value of $ q^2\minus{}q$ in that case.

can this result be related in any way to the number of irreducible polynomials with coefficients in the set $ {0,1,2,...,9}$ ?

[quote="t0rajir0u"]No, but it can be related to something close.  $ a_p(n)$, for arbitrary $ p$, also counts the number of Lyndon words on $ p$ letters of length $ n$, which when $ p$ is prime can be put in bijection with irreducible polynomials; see the definition and proof [url=http://qchu.wordpress.com/2009/11/03/the-cyclotomic-identity-and-lyndon-words/]here[/url].

To prove that $ a_p(n)$ is always positive when $ p \ge 2$, it suffices to show that Lyndon words of every length exist.  But this is clear; if $ a < b$ are two letters in the alphabet, then $ aaa...b$ is always a Lyndon word.  (This is actually not enough to prove that irreducible polynomials of every degree exist over $ \mathbb{F}_p$; the bijection between the two requires the existence of finite extensions of every degree.)[/quote]

yes but can I compute the number of irreducibles with coeffs in 0..9 using $ a_p(n)$ ?

No; at best, you can take $ p$ to be a prime power.  The problem is that these are the only values of $ p$ for which there exists a finite field of cardinality $ p$.  There are two other interpretations of "irreducible polynomial with coefficients in $ \{ 0, ... p\minus{}1 \}$" when $ p$ is not a prime power: irreducible in $ \mathbb{Z}[x]$ or irreducible in $ \mathbb{Z}/p\mathbb{Z}$, and neither of them are counted by this expression.


Ellen baked $ 2$ dozen cupcakes of which half contained chocolate, two-thirds contained raisins, one-fourth contained chocolate chips, and one-sixth contained nuts.  What is the largest possible number of cupcakes that had none of these ingredients?

Since $ \frac23$ have raisins, $ \frac23\times{24}\equal{}16$ have raisins. Each of the other ingredients can be mixed in with any of these cookies, so the maximum is $ 24\minus{}16\equal{}\boxed{8}$.

This is one of those "trick" problems that love to show up on math contests. Before you plunge right in and start drawing a four-circle Venn diagram, you need to stop, take a step back, take a deep breath, and then re-read the problem.

I approached this problem in the same way as gaussintraining.

In the end, we have 16 cupcakes with raisins and maybe or maybe not something else, and therefore 8 cupcakes with none of the ingredients. We don't have anywhere near enough information to fill in a four-circle Venn diagram. For one, it's entirely possible that all 16 cupcakes have raisins paired with at least one other ingredient, and it's also possible that 4 of those cupcakes are raisin only. Don't even try to wander into that quagmire; you'll never be seen again! ;).

Tim

[hide]I had a quick way, the largest venn diagram is 16, to make outside portiion the largest, we can put all other venn diagrams inside the largest one, like 4,6,12 all goes inside the 16, that way, the outside will be the largest which has none of these 4 ingredients at all, so 24-16=[b]8![/b][/hide]

[quote=no_name]Was it really necessary to revive a four-year old thread with a solution?[/quote]

Yes, it is okay to revive alcumus threads if you are actually posting something relevant to the thread.

I think he meant that the answer was 8 with an exclamation mark behind it because he was excited or something.

:( could someone explain this. I don't get it to well.

[hide=Solution]Because we are trying to find the maximum number of plain cupcakes Ellen could have made, we can assume that she will try and mix fillings whenever possible. Because the largest fraction of cupcakes ($2/3$) contain raisins, if she tries to mix fillings, all of the cupcakes that contain chocolate will contain raisins, too. This also applies to all of the other fillings, so we can ignore those and just focus on the raisins. $2/3$ of the cupcakes contain raisins, so $1/3$ don't. $12 * 1/3= \boxed{4}$ [/hide]

[quote=HWGelber]Let me know if that solution is too wordy.[/quote]

(You're probably not going to see this, but..)

Your solution is not wordy, but you said $1$ dozen cupcakes instead of $2$. The answer should be $8$. (But you were on the right track  :) )

[quote name="jcl-12" url="/community/p15744495"]
[quote=elevate]Tim is right, bust I really think that this shouldn't be a level 19 problem...it even gives 100+ points.[/quote]

it is a level 23 problem
[/quote]

hi, there's no point of responding when they probably haven't been on the site for 8-9 years

also alcumus was probably different then with different levels and stuff

[quote=arguewithplato]Should I quit math if I got this one wrong after spending an hour on it? Seems like I'm not cut out for this.[/quote]
You can't avoid maths. It will crop up wherever you go; when you go to the shops or become a scientists, maths will crop up [b]everywhere[/b] you go. It's in your genetics (literally).

tldr: no.

Hi,
I believe there’s another way to do it, but it doesn’t have anything to do with counting and probability.
I might be wrong.

Let the total cupcakes be the area of a square with side a, which is a^2
So the ones with other ingredients in it be squares or rectangles.
(You could draw a square to start then cut that square out to get pieces that represent others, like the chocolate is half of that square, to get the nuts first cut the square a half then cut that half into three pieces. But note that all of this is to get the visual pieces of each category not taking pieces out of the square a^2 piece)
Then laying all the pieces out
(a x a) - all the cupcakes,
(a x a/2) - chocolate,
(a x (2/3)a) - raisins,
(a/2 x a/2) -  chocolate chips,
(a/3 x a/2) - nuts,

The question ask what is the largest possible number of cupcakes that have none of these ingredients is the same as how to crunch all the pieces together so that when fit it in the big piece, the a^2 square, it left the largest area. To get that we crunch all the smaller pieces into that biggest piece in all those smaller piece (a x (2/3)a).
So the largest possible space left is a^2 - (2/3)a^2 = (1/3)a^2
And a^2 = 24 so the space is 24/3 = 8


Let $x,y,z$ be positive reals such that $4xyz = x+y+z+1$.  Prove that

\[(x+1)(y+1)(z+1) \geq 8.\]

Note that $x+y+z\geq 3(xyz)^{1/3}$. Let define $a:=(xyz)^{1/3}$, then by assumption we have
\[
(a-1)(4a^2+4a+1)=4a^3-3a-1=4xyz-3(xyz)^{1/3}-1\geq 0
\]
It follows that $a^n\geq a\geq 1$ for every $n$. Now
\[
(x+1)(y+1)(z+1)=xyz+(xy+xz+yz)+4xyz\geq 5xyz+3(xyz)^{2/3}=5a^3+3a^2\geq 8.
\]

[quote="ThAzN1"]Cool.  My (contrived) solution uses the substitution $a=\frac{1}{2x+1}$, etc so our condition (after some algebra) rewrites as $a+b+c=1$.  Our inequality becomes
\[\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) \geq 64.\]
However, this follows easily by homogenizing and using AM-GM.[/quote]
huygens and amgm works as well.

[quote="ThAzN1"]Cool.  My (contrived) solution uses the substitution $a=\frac{1}{2x+1}$, etc so our condition (after some algebra) rewrites as $a+b+c=1$.  Our inequality becomes
\[\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) \geq 64.\]
However, this follows easily by homogenizing and using AM-GM.[/quote]

Hi ThazN1, this problem seem closely related to yours, check if your method also working for this problem

http://www.mathlinks.ro/Forum/topic-41928.html

Let me know it it is working. Thanks


It is given that $x$ and $y$ are positive integers and $3x^{2}+x=4y^{4}+y$. Show that: $x-y$, $3x+3y+1$ and $4x+4y+1$ are squares of integers.

3[(x^2)-(y^2)]+(x+y)=y^2
(x-y)(3(x-y)+6y+1)=y^2
if d=(x-y,3(x-y)+6y+1) if d>1, an p is prime number.
p devides d and p devides (x-y) and p devides y^2 so p devides y.
so p devides 3(x-y)+6y and we know p devides d so p devides 3(x-y)+6y+1 so p=1 but it is impossible so d=1 .
we know [3(x-y)+6y+1)].[(x-y)]=a^2 so (x-y)=m^2.
other questions is same .


Princeton
college
search
articles
real analysis
Duke
geometric series

UP Math prof proves Princeton man wrong...

http://www.manilatimes.net/national/2005/may/05/yehey/top_stories/20050505top4.html

I'm not too sure what to believe... I know the website looks bogus, but if you search for it online, there are some other pages with the same story.

What do you guys think?

*i* coulda told you that!

-peter ruse
-kamaldeep gandhi
-ricky sharma
-rathanak chuon

Listen ... false axioms?  Gimme a break.  This guy sounds just like Doron Shamdi (perhaps that is misspelled, but you can find his posts in Other Problem Solving Topics yourself) who has decided that mathematics is false.  I mean, fine, you don't like the axioms, but that doesn't mean that your own axioms are [i]truer[/i], just that they are an alternative choice.  This is a fairly classic example of a crank and you shouldn't worry about it.

If it would comfort you that modern mathematics will remain intact despite his antics, I could go through one of his articles (such as the editorial in the Manila Times found [url=http://www.manilatimes.net/national/2004/jun/18/yehey/opinion/20040618opi7.html]here[/url]) and explain why the various things he says are either nonsense or at least don't demonstrate the kind of thing he's trying to show.  Or you could just take my word for it :)

lol.  Everyone go to the link that JOel sent you.  There are about 10000000 ways to prove 1 = 0.9999999...

you could use the standard let x = 0.999999, then 10x = 9.9999999, etc.

or you could use a1=0.9, a2=0.09, etc. and use sum of an infinite geometric series and more

Wow, thank you Joel.  Thou hast saved us all.

[quote="MithsApprentice"]Dunno...that letter by Wiles sure sounds fake...[/quote]

Yes it did....it sounded very "unprofessional"

eww... this guy needs some common sense!

a) that guy is a quack (he's a fairly regular poster on sci.math)

b) that newspaper should be completely ashamed of itself, but certain ethnic groups tend to take excessive pride in the 'accomplishments' of members of said ethnic group.  A publication with higher standards, however, would have laughed that article away.

haha he says that average of two numbers must lie right in between them and then states that when you average 0.99999.... and 1 you get 0.99999..... which is a contradiction since it doesn't lie in between

yet he just proved exactly what he tried to disprove: that 0.99.... =1 since two same numbers averaged together will give you that number itself

[quote]The most advanced computer today can only deal with numbers having no more that 11 digits.[/quote]

Somehow, I doubt this...

But anyway, yeah, his "proof" of the nonsense of, well, mathematics is crazy. This guy needs to go to an asylum... for math. Why can't he just accept the beauty and logic of it all like the rest of us? :D

Actually, blah^3, he was the [i]editor[/i] of the math and science portion of that paper, if I read correctly.  Which explains in part why he managed to get so much space from them for his silliness.

Well, I guess every paper has its employees that they would like swept under the carpet (although it seems that they don't even know the magnitude of the absurdities there)

The thing that I find really ironic is that our Edgar Escultura got his Ph.D the same year, and from the same class, as Walter Rudin, the famed author of those difficult-to-the-point-of-unreadable analysis texts that every undergrad runs into sooner or later.

[quote="blahblahblah"]The thing that I find really ironic is that our Edgar Escultura got his Ph.D the same year, and from the same class, as Walter Rudin, the famed author of those difficult-to-the-point-of-unreadable analysis texts that every undergrad runs into sooner or later.[/quote]

I'm not sure what you mean, but Walter Rudin got his PhD from Duke in 1949. He was a professor at UW-Madison in the 70s, but I don't think he was Escultura's advisor. Escultura's advisor appears to have been Laurence Young - who was a very famous and well respected mathematician. I can imagine he would roll over in his grave if he could see what his former student is doing today.

Yeah, somehow the list of Ph.D candidates and their advisors blurred together in my head :| .  I know this because I did a google search for escultura's name and Rudin's name jumped out at me on the same webpage, and I must have just mentally associated them as getting their Ph.D's at the same time.

I have to admit that the complete article seems to be fake. Remember that sometimes this articles are written for jurnalists that seem not to understand what happened, even when it seems not to be the case. I remember that once I read an article about a girl, 22 in Sweden that had proven the Riemann Hypothesis, and at the end it was that she had earned a price for some research in something related (as many prices that are given to young mathematicians) and the jurnalist had listened about the Clay institute price the same day and mixed the information. It was not too long ago, three years or less. Any way, I would not trust too much such article!

these quacks are quack quacks.  (yes indeed!)  how?  they don't know the first thing about good quacking or *where* to start quacking.  there's nothing wrong with BS'ing if you do it skillfully, in which case it's not really so much BS anymore, but stuff actually worthy of attention.  i mean, there's Godel, who allows us to BS on certain levels.  but these superficial ones!  it's gotta be more polyphonic than that, oh hotdammit!

-petee outta the blue

The damage has already been done:

[url=http://houseonahill.net/index.php/blog/wiles-escultura-and-pierre-de-fermats-equation/]houseonahill.net/index.php/blog/wiles-escultura-and-pierre-de-fermats-equation/[/url]

[quote]No wonder I couldn’t understand algebra.[/quote]

Edit:  The first time I read Wiles' letter, I didn't get that he was being sarcastic.  Woops.  I wouldn't have guessed him for it.  That was harsh: "Also I’d like to have the address of the guy who let you get a PhD 30 years ago. I’d like to discuss few things with him. . . "

So then I was right not to have guessed him for it.

i highly doubt that the letter is from Wiles, it is totaly unprofessional..., it just doesnt make sense
on the other hand, i would be very happy to see one of the counter examples, that they have proposed... ;)

Well...if Wiles's proof is true, and as ameteurs, we certainly have no right to assert that it's faulty, then there are no real counterexamples.


find all [b]x,y[/b] in integer number:
 [b]xy[/b] divisible by [b]x+y[/b].

How do you think he can answer you if it is an unsolved problem for him? :D  :D  :D

Please post complete solution, otherwise this one will remain here for years before going into the solved section.

Pierre.

Let $d=(x,y)$, $x=du$, $y=dv$, then $u+v\mid duv$, but $(u,u+v)=(v,u+v)=1$, so $u+v\mid d$, i.e. $d=ab$, $u+v=b$. Denote $c=u$.
We see $x=abc$, $y=ab(b-c)$.

That's all.


Is a person who scored a 132 on AMC 10 and 7 on AIME (me) this year good enough to go to this camp? I just want a rough idea...

i think grade might matter..
anyway you shouldnt have a problem getting in
they accept way worse people
(except if youre a senior or something)
last year i only got slightly higher and i got in without testing

So AwesomeMath isn't like ridiculously high level? IF I do get in, I'm not sure if I can learn with the rest of the people there.

There are classes of varying difficulties. You can always find things more difficult if you ask. I think that the easier classes should be manageable for students who did well at mathcounts.


linear algebra
matrix
combinatorics unsolved
combinatorics

There is a n*n matrix filled with 0 and 1.We can do the following operation:
If there is more than m (m<=n) 1 in a row (or a column), then we can change all the zeros in this row (or column) into 1.
Prove that:
If there is at most (m^2-1)  1s in the matrix at the very beginning then we can never change the matrix into all 1s.
(Here m^2-1 is the best number)

Just a first observation: $ m^2\minus{}1$ is not the"best number" (also, I think you mean if there are at least $ m$ ones in a row or column we can change it into all ones). For a simple example, take $ n\equal{}10$ and $ m\equal{}2$. Clearly $ m^2\minus{}1 \equal{} 3$ ones are not enough, while $ m^2 \equal{} 4$ are enough if we position them in a $ 2 \times 2$ subsquare. But if we take the diagonal of the square to be all ones ($ 10$ in all), we can still not make even one move. We may need to rephrase this as $ m^2\minus{}1$ is the "best" in the sense that, for more, [b]there exists[/b] a configuration that allows the square to be filled with ones, but that is trivial, since a $ m \times m$ subsquare of ones does the job.

Dear mavropnevma
What you say is right.Here the best means that there exists a condition such that it can be turned into all 1s
For all number less than m^2-1,one can never turn it into all 1s.
Did you find the solution to this?
I think I got it. :D

This is one of those delightful cases when one needs to prove a stronger statement in order the get the required result as a particular case, due to the easier, unrestricted manipulation allowed by the general case. This general context therefore will be

[color=blue]Given integers $ h \geq c \geq 1$, $ w \geq r \geq 1$, a rectangular $ h \times w$ array, containing $ 0 \leq N \leq rc$ entries equal to $ 1$, and the rule that anytime a row contains at least $ r$ ones it may be filled with ones, while anytime a column contains at least $ c$ ones it may be filed with ones, then the only case that allows the full array to end up with all entries equal to $ 1$ is $ N \equal{} rc$ (for some special configurations).[/color]

Denote by $ x$ the number of those rows containing each at least $ r$ ones, and by $ R$ the set of these entries, while we denote by $ y$ the number of those columns containing each at least $ c$ ones, and by $ C$ the set of these entries. Then $ N \geq | R \cup C | \equal{} | R | \plus{} | C | \minus{} | R \cap C | \geq xr \plus{} yc \minus{} xy$. It is easily seen that the problem now reduces to filling with ones a $ (h' \equal{} h\minus{}x) \times (w' \equal{} w\minus{}y)$ array, containing $ 0 \leq N' \equal{} N \minus{} | R \cup C | \leq rc \minus{} (xr \plus{} yc \minus{} xy) \equal{} (r\minus{}y)(c\minus{}x) \equal{} r'c'$ entries equal to $ 1$. When $ x \equal{} y \equal{} 0$ there is status-quo, therefore no move can be made, and the array remains unfilled, while otherwise the problem reduces in steps to the base case of $ N\equal{}0$, which is trivial.

The particular case is for $ h\equal{}w\equal{}n$ and $ r\equal{}c\equal{}m$; when $ 0\leq N \leq m^2\minus{}1$ the array cannot therefore be filled up, while when $ N\equal{} m^2$ the array may be possibly filled up, for example when the entries equal to $ 1$ make up a $ m \times m$ subarray (but in other cases also, although not for all configurations).

Dear mavropnevma
That's exactly the way how I did that.
It's really interesting,isn't it?


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

I know that there are several proofs for $\sum_{n\in\mathbb{N}}n^{-2}=\pi^{2}/6$.  I was trying to prove something, but I'm not sure if I've made an error here:

Let $f\left(x\right)=x^{2}-x+1/6$ for $0\leq x\leq 1$ and require that $f\left(x\right)=f\left(x-n\right)$ for $n\leq x\leq n+1$ where $n\in\mathbb{Z}$.  Part integration shows us that

\[\int_{0}^{1}f\left(x\right)e^{-2\pi inx}\,dx=\begin{cases}1/2\pi^{2}n^{2}& n\neq 0 \\ 0 & n=0\end{cases}.\]
The next step is to show that

\[f\left(x\right)=\frac{1}{\pi^{2}}\sum_{n\in\mathbb{N}}\frac{\cos\left(2\pi nx\right)}{n^{2}},\]
which clearly implies that $\sum_{n\in\mathbb{N}}n^{-2}=\pi^{2}/6$.  I know that since $f$ is a continuous function, we can say that

\[f\left(x\right)=\sum_{n\in\mathbb{Z}}a_{n}e^{-2\pi inx},\]
where $a_{n}=1/2\pi^{2}n^{2}$.  However, what is the next step?

[quote="amcavoy"]I know that there are several proofs for $\sum_{n\in\mathbb{N}}n^{-2}=\pi^{2}/6$.  I was trying to prove something, but I'm not sure if I've made an error here:
[/quote]

What about this?

Define $f(x)=x^{2}$ on $[-\pi , \pi]$.
Define $f(x+2\pi n)=x^{2}$ for all $n\in \mathbb{Z}$.

This function satisfies the "Dirichlet conditions".

Expand it as a Half-Range Cosine Series.

Play around with it until you bring it to the desired form.

Oh yeah:
\[f\left(x\right)=\sum_{n\in\mathbb{Z}}a_{n}e^{2\pi inx}=\frac{1}{\pi^{2}}\sum_{n\in\mathbb{Z}}\frac{e^{2\pi inx}}{2n^{2}}=\frac{1}{\pi^{2}}\sum_{n\in\mathbb{N}}\frac{\cos\left(2\pi nx\right)}{n^{2}}, \]
so $f\left(0\right)=1/6=\left(1/\pi^{2}\right)\sum_{n\in\mathbb{Z}}n^{-2}$.


function
real analysis
real analysis unsolved

If $ f(x) \in C(\minus{}\infty,\infty)$

can we prove that there exists a interval $ (a,b) \subset C(\minus{}\infty,\infty)$ ,such that

$ f(x)$ is monotony as $ x \in (a,b)$

 :idea:  :roll:

If $ \ f(x)$ happens to be an open map, we can say that the function is monotonic.

No, we can not, because this is not true (just take continuos function which is nowhere differentiable) :wink:


Let $ a,b,c\geq 0$ and $ ab\plus{}bc\plus{}ca\equal{}1$.prove that $ \sum \frac{ab\plus{}1}{a\plus{}b}  \geq 3$

[quote="euler_vn"]Let $ a,b,c\geq 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$.prove that $ \sum \frac {ab \plus{} 1}{a \plus{} b} \geq 3$[/quote]

Very nice  :P
$ < \equal{} > p^2 \plus{} pr \plus{} 2 \plus{} 3r \ge 3p$
With:$ p \equal{} a \plus{} b \plus{} c;q \equal{} ab \plus{} bc \plus{} ac \equal{} 1;r \equal{} abc$
$ \plus{} ) p \ge 2$
$ \equal{} > p^2 \plus{} pr \plus{} 2 \plus{} 3r \equal{} (p \minus{} 1)(p \minus{} 2) \plus{} pr \plus{} 3r \plus{} 3p \ge 3p$
$ \plus{} ) p \le 2$
$ \equal{} > r \ge \frac {p(4 \minus{} p^2)}{9}$ by shur :)
$ \equal{} > p^2 \plus{} pr \plus{} 2 \plus{} 3r \ge p^2 \plus{} \frac {p^2(4 \minus{} p^2)}{9} \plus{} 2 \plus{} \frac {3p(4 \minus{} p^2)}{9} \ge 3p$
We have prove that:
$ p^2 \plus{} \frac {p^2(4 \minus{} p^2)}{9} \plus{} 2 \plus{} \frac {3p(4 \minus{} p^2)}{9} \ge 3p$
$ < \equal{} > (p \minus{} 2)(3p \minus{} p^3 \minus{} 5p^2 \minus{} 9) \ge 0$-It's true by $ p \le 2$
  :P

[quote="caubetoanhoc94"][quote="euler_vn"]Let $ a,b,c\geq 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$.prove that $ \sum \frac {ab \plus{} 1}{a \plus{} b} \geq 3$[/quote]

Very nice  :P
$ < \equal{} > p^2 \plus{} pr \plus{} 2 \plus{} 3r \ge 3p$
With:$ p \equal{} a \plus{} b \plus{} c;q \equal{} ab \plus{} bc \plus{} ac \equal{} 1;r \equal{} abc$
$ \plus{} ) p \ge 2$
$ \equal{} > p^2 \plus{} pr \plus{} 2 \plus{} 3r \equal{} (p \minus{} 1)(p \minus{} 2) \plus{} pr \plus{} 3r \plus{} 3p \ge 3p$
$ \plus{} ) p \le 2$
$ \equal{} > r \ge \frac {p(4 \minus{} p^2)}{9}$
$ \equal{} > p^2 \plus{} pr \plus{} 2 \plus{} 3r \ge p^2 \plus{} \frac {p^2(4 \minus{} p^2)}{9} \plus{} 2 \plus{} \frac {3p(4 \minus{} p^2)}{9} \ge 3p$
We have prove that:
$ p^2 \plus{} \frac {p^2(4 \minus{} p^2)}{9} \plus{} 2 \plus{} \frac {3p(4 \minus{} p^2)}{9} \ge 3p$
$ < \equal{} > (p \minus{} 2)(3p \minus{} p^3 \minus{} 5p^2 \minus{} 9) \ge 0$-It's true by $ p \le 2$
  :P[/quote]
why $ p\leq 2$  :?:

[quote="peine"][quote="caubetoanhoc94"][quote="euler_vn"]Let $ a,b,c\geq 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$.prove that $ \sum \frac {ab \plus{} 1}{a \plus{} b} \geq 3$[/quote]

Very nice  :P
$ < \equal{} > p^2 \plus{} pr \plus{} 2 \plus{} 3r \ge 3p$
With:$ p \equal{} a \plus{} b \plus{} c;q \equal{} ab \plus{} bc \plus{} ac \equal{} 1;r \equal{} abc$
$ \plus{} ) p \ge 2$
$ \equal{} > p^2 \plus{} pr \plus{} 2 \plus{} 3r \equal{} (p \minus{} 1)(p \minus{} 2) \plus{} pr \plus{} 3r \plus{} 3p \ge 3p$
$ \plus{} ) p \le 2$
$ \equal{} > r \ge \frac {p(4 \minus{} p^2)}{9}$
$ \equal{} > p^2 \plus{} pr \plus{} 2 \plus{} 3r \ge p^2 \plus{} \frac {p^2(4 \minus{} p^2)}{9} \plus{} 2 \plus{} \frac {3p(4 \minus{} p^2)}{9} \ge 3p$
We have prove that:
$ p^2 \plus{} \frac {p^2(4 \minus{} p^2)}{9} \plus{} 2 \plus{} \frac {3p(4 \minus{} p^2)}{9} \ge 3p$
$ < \equal{} > (p \minus{} 2)(3p \minus{} p^3 \minus{} 5p^2 \minus{} 9) \ge 0$-It's true by $ p \le 2$
why $ p\leq 2$  :?:
  :P[/quote][/quote]

I Consider two cases $ p \le 2;p \ge 2$ :lol:

[quote="caubetoanhoc94"][/quote][quote="peine"][quote="caubetoanhoc94"][quote="euler_vn"]Let $ a,b,c\geq 0$ and $ ab + bc + ca = 1$.prove that $ \sum \frac {ab + 1}{a + b} \geq 3$[/quote]

Very nice  :P
$ < = > p^2 + pr + 2 + 3r \ge 3p$
With:$ p = a + b + c;q = ab + bc + ac = 1;r = abc$
$ + ) p \ge 2$
$ = > p^2 + pr + 2 + 3r = (p - 1)(p - 2) + pr + 3r + 3p \ge 3p$
$ + ) p \le 2$
$ = > r \ge \frac {p(4 - p^2)}{9}$
$ = > p^2 + pr + 2 + 3r \ge p^2 + \frac {p^2(4 - p^2)}{9} + 2 + \frac {3p(4 - p^2)}{9} \ge 3p$
We have prove that:
$ p^2 + \frac {p^2(4 - p^2)}{9} + 2 + \frac {3p(4 - p^2)}{9} \ge 3p$
$ < = > (p - 2)(3p - p^3 - 5p^2 - 9) \ge 0$-It's true by $ p \le 2$
why $ p\leq 2$  :?:
  :P[/quote][/quote][quote="caubetoanhoc94"]

I Consider two cases $ p \le 2;p \ge 2$ :lol:[/quote]
okay, I didn't see that you considered two case, now I understand, thank's.

[quote="peine"][/quote][quote="caubetoanhoc94"][/quote][quote="peine"][quote="caubetoanhoc94"][quote="euler_vn"]Let $ a,b,c\geq 0$ and $ ab + bc + ca = 1$.prove that $ \sum \frac {ab + 1}{a + b} \geq 3$[/quote]

Very nice  :P
$ < = > p^2 + pr + 2 + 3r \ge 3p$
With:$ p = a + b + c;q = ab + bc + ac = 1;r = abc$
$ + ) p \ge 2$
$ = > p^2 + pr + 2 + 3r = (p - 1)(p - 2) + pr + 3r + 3p \ge 3p$
$ + ) p \le 2$
$ = > r \ge \frac {p(4 - p^2)}{9}$
$ = > p^2 + pr + 2 + 3r \ge p^2 + \frac {p^2(4 - p^2)}{9} + 2 + \frac {3p(4 - p^2)}{9} \ge 3p$
We have prove that:
$ p^2 + \frac {p^2(4 - p^2)}{9} + 2 + \frac {3p(4 - p^2)}{9} \ge 3p$
$ < = > (p - 2)(3p - p^3 - 5p^2 - 9) \ge 0$-It's true by $ p \le 2$
why $ p\leq 2$  :?:
  :P[/quote][/quote][quote="caubetoanhoc94"]

I Consider two cases $ p \le 2;p \ge 2$ :lol:[/quote][quote="peine"]
okay, I didn't see that you considered two case, now I understand, thank's.[/quote]

Okay.
P/s:Hỏi mãi mới bik xét 2 TH tiếng anh là thế :(


modular arithmetic
number theory proposed
number theory

Let $p$ be an odd prime. Assume $\sigma$ is a permutation of $\{1,2,\ldots,p\}$. Show that we can find $i\ne j\in\{1,2,\ldots,p\}$ s.t. $i\sigma(i)\equiv j\sigma(j)\pmod p$.

P.S. People will probably find this a joke :)

Oh, really  :lol:  :lol:  :lol:
Suppose the contrary. It is clear $\sigma(p)=p$. Then $\sigma(1),2\sigma(2),...,(p-1)\sigma(p-1)$ is a permutation of $1,2,...,p-1$. Multyplying all number $i\sigma(i)$ we obtain $(p-1)!^2\equiv (p-1)!\pmod{p}$. Contradiction with Wilson'n theorem.

First assume $\sigma (p) =0$, because otherwise its trivial.
Now assume that all $ i\cdot \sigma (i), i=1,...,p-1$ are different.
Then you get $ \prod  i\cdot \sigma (i) \equiv -1 \not\equiv 1 \equiv \prod i \cdot \prod \sigma (i) \mod p$

nice joke  :lol:


blogs MATHCOUNTS function algebra polynomial AMC AIME

I am interested in collecting a small but fun list of topics for write about in my blog.  In particular, I would like to hear about topics that middle school math teamers typically say "wow!" about upon first discover.  What would students here suggest?

Maybe this is a little advanced, but a lot of people in my class were really suprised when they found out that some infinities are bigger than others and the teacher proved it.

[hide="Also Interesting"]There was a problem in my class: \[x^{x^{x^{\cdots}}}=2\]

Who's answer is: \[\sqrt{2}\]

That people found interesting.

However, they found this:

What is x if \[x^{x^{x^{\cdots}}}=y\]

even more interesting because it's answer is \[^y\sqrt{y}\] because then \[\sqrt{2}^{\sqrt{2}^{\cdots}}=2=4\] which obviously isn't true because \[2\ne 4\]

What happened after that if that most people claimed that math as we know it was false, and as such, we should not learn math in school. :D

And then the teacher explained how infinity can't be treated like a normal number, so we hadn't "disproved" math.[/hide]

What I'm getting at is that proofs of things that aren't true are always interesting.

I like RC-7th 's Idea....

Infinity can be tough, but I think it would be fun to learn it...

While a perfectly good subject to blog about, I think that kind of self-similarity and infinity discussion is more advanced that what I'm asking about.

Picture that MATHCOUNTS student who can score around half the points at the Chapter competition.  That's the audience I'm interested in.

Well, according to Treething, you have to solve infinity problems in MATHCOUNTS...

I looked at some MATHCOUNTS practice problems and they definitley had infinity problems..

like I expected..

and in most schools, you don't get to learn about infinities..

I would love to learn about infinities, i think it's hard to get around the fact that it's really going on FOREVER... most middle schools don't teach much about it because it's kinda hard except for like MATHCOUNT people... so I think it would be great. But I don't think you should start off with problems like the ones RC-7th mentioned, they are a little to hard to start off with

[quote="LuCy4EvA"]I think people would love stuff like how pi goes on forever, how to prove anything to the 0th power is 1, and how 0.999999 = 1. We've always like to ask these questions in math class, but the teachers usually don't know. I was amazed when I found out how to prove x^0 = 1... when I told my classmates, they thought that was really cool[/quote]

$0^0\ne 1$

The problem I posted was the only one the teacher had given to us that related to infinity.

Game theory is pretty fun. There was a question from mathcounts that went like this : Two people are playing a game with 40 counters. On each turn, a person may take 1,2,3,4, or 5 counters. The person who takes the last counter wins. How many counters should the person who goes first take to ensure that they win?

okay first of all

0^0

means nothing to the nothing power

do you understand the concept X^0 = 1 ?

but it only happens when X is like numbers not including 0

0 means nothing..

there won't be nothing to power over...

but when you set a group like this, {0}

you always have to count 0 as one of the elements...

god!! this is hard to tell..

wait, I get ur explanation, but is it 1 or undefined?

$0^0$ is undefined.

For most numbers you can product chains of exponents using multiplication and go backwards in these chains using division:
\[ \frac{2^3}{2} = \frac{8}{2} = 4 = 2^2. \]
\[ \frac{2^2}{2} = \frac{4}{2} = 2 = 2^1. \]
\[ \frac{2^1}{2} = \frac{2}{2} = 1 = 2^0. \]
\[ \frac{2^0}{2} = \frac{1}{2} = \frac{1}{2} = 2^{-1}. \]
\[ \frac{2^{-1}}{2} = \frac{\frac{1}{2}}{2} = \frac{1}{4} = 2^{-2}. \]

Division by 0 is undefined, so this reversal of the multiplication chain is not possible.

[quote="kidnappedturtle"]I remember one of the things that first wowed me was combinatorics, especially counting arguments and (simple) generating functions. Like seeing combinatoric identities proved in three different ways: 1) just using normal algebra, 2) block-walking on Pascal's triangle, and 3) counting arguments. Finding out that those "How many ways can you make 83 cents using pennies, nickels, and dimes" questions could actually be solved through a real method (!) was also neat. Not sure if that stuff is too advanced, though.[/quote]

This are topics I would call "Getting Started" topics.  A little beyond the basics.  Generating functions I would call Intermediate -- even at the simple level -- because they involve using polynomials together with counting arguments.  Students need to know two subjects fairly well and be able to go between then easily -- the kinds of skills that are required for AIME problems.  Those are certainly topics I might blog about however.  Thanks for the suggestions.

Things that blew my hair back:

1.) Proof by contradiction.
2.) Euclid's proof for infinitely many primes.
3.) Fundamental theorem of arithmetic.( Every integer factors as a product of primes)
4.) Parity arguments.(The checkerboard tiling one comes to mind)

Things that blow my hair back:
1) hair dryer
2) riding in convertible
3) roller coaster

Seriously (and on topic)
I went to a middle school that didn't go very far into much of anything so I was ecstatic when we learned graphing. While the class was struggling with lines, I graphed my first *gasp* parabola! It amazes me that 4 years ago, I couldn't find slope. :what?:

But since we're talking about MathCounts, I really liked counting and geometry problems. I had a teacher who showed us how to derive the area formula for a circle, essentially doing some integral calculus for 7th graders. :wacko: He wasn't using anything we didn't understand, though. He just introduced the idea of making the pieces of the circle small enough and sticking them back together so that it would look enough like a rectangle to use base x height.
So I guess I'm really talking about limits and approximations, and proofs to some extent.

I did a bunch of parity argument proofs with my middle school students, and the ones who I imagine could get about 1/4 or more of the points on a MATHCOUNTS Chapter competition really seemed to enjoy them.  My favorite is one about whether or not it is possible to create a magic square with the first n2 primes.

There are also things in the MATHCOUNTS handbook that could use expanding upon.  Kids are often given formulas from there without understanding why they work, like the formula for the sum of all the factors of a number.

My favorite math proof of all time was the proof that there is a countable infinity of rational numbers  (Cantor's diagonal argument).  I was much older than middle school when I first encountered it, but I think that it could be presented in a way that the MATHCOUNTS kids could understand.

When I was in middle school myself, I was really hung up on 00.  I would have loved to see a good explanation of why it is neither 0 nor 1, both of which seemed like perfectly reasonable values.  Similarly, 0!=1 intrigues my middle schoolers, and telling them that it just is that way because it makes all the formulas work out properly isn't very satisfactory to them :)

By the way, I should mention that many mathematicians, including myself, define $0^0$ to be 1.  My arguments, and counterarguments by Simon and others, appeared in a topic on a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11999]Limit Problem[/url].

Basically, some mathematicians say it is undefined, and some set it to 1.  It's a matter of what is the most convenient definition.

Many kids are exposed to Magic Squares in elementary school.  But Middle School is when you can begin talking about nested patterns and recursion and algorithms.  Attached is a 12x12 Magic Square as an example.  Ask the students to find smaller magic squares hidden in this big magic square.  Either show the kids the pattern I used to make this square or challenge them to find the pattern for themselves.  In either case, ask [b]why[/b] does it work?  Invite them to create their own pattern that will create a 9x9, 12x12, or 15x15 square.  There are [b]many[/b] different patterns / algorithms to make these larger squares.
$\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline 113   & 1     & 81    & 127   & 15    & 95    & 126   & 14    & 94    & 116   & 4     & 84 \\
\hline 33    & 65    & 97    & 47    & 79    & 111   & 46    & 78    & 110   & 36    & 68    & 100 \\
\hline 49    & 129   & 17    & 63    & 143   & 31    & 62    & 142   & 30    & 52    & 132   & 20 \\
\hline 124   & 12    & 92    & 118   & 6     & 86    & 119   & 7     & 87    & 121   & 9     & 89 \\
\hline 44    & 76    & 108   & 38    & 70    & 102   & 39    & 71    & 103   & 41    & 73    & 105 \\
\hline 60    & 140   & 28    & 54    & 134   & 22    & 55    & 135   & 23    & 57    & 137   & 25 \\
\hline 120   & 8     & 88    & 122   & 10    & 90    & 123   & 11    & 91    & 117   & 5     & 85 \\
\hline 40    & 72    & 104   & 42    & 74    & 106   & 43    & 75    & 107   & 37    & 69    & 101 \\
\hline 56    & 136   & 24    & 58    & 138   & 26    & 59    & 139   & 27    & 53    & 133   & 21 \\
\hline 125   & 13    & 93    & 115   & 3     & 83    & 114   & 2     & 82    & 128   & 16    & 96 \\
\hline 45    & 77    & 109   & 35    & 67    & 99    & 34    & 66    & 98    & 48    & 80    & 112 \\
\hline 61    & 141   & 29    & 51    & 131   & 19    & 50    & 130   & 18    & 64    & 144   & 32 \\
\hline
\end{tabular}$

what's the magic square?

I don't really get it....

Is it like a matrix or something? :?

All the rows, columns and diagonals add up to the same number.

to see the different solutions of the same problem was always aweful for me.particularly different solution of the same problem using different field of maths.intutive and easy solution using elemantry language r also helpful.

[quote="A+MATH"]Proofs of irrationality are some of my favroites.[/quote]

Well, whether or not that is middle-school level depends.  The proof that $\sqrt{2}$ is irrational is very clever and uses very simple math.  Many middle schoolers could understand that.

A proof that $\pi$ is irrational would go into one of the advanced/college boards...

Things that surprised and intrigued me in my middle school years:

1) Figuring out how to calculate the number of shortest paths from one corner to another on a grid (already mentioned)

2) Finding that the sum of factors of a number like 120 is (1+5)(1+3)(1+2+4+8)

3) Finding the proof that 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6

4) Proof of the Pythagorean theorem

5) A non-calculus proof of the volume of sphere formula (given the area of circle formula)

6) Discovering the concept of instantaneous slope

In general, I usually found it most fascinating when I read about or discovered a neat way to calculate something that would normally be tedious or impossible.

I also think it would be great if you could, instead of just showing a proof or idea, walk viewers through to make them discover it as much by themselves as possible.  Of course you're probably already doing that.

[quote="jimli"]I think that you should host online 24 games.  You know that game where you make 24 with four numbers 1-10.  That would be lots of fun.[/quote]

Hosting games is somewhere down our list of priorities.

My FAVORITE is the decimal expansion of $\frac{1}{n}$, where n is a prime number. Some of them are pure awesome. For example:
\[\frac{1}{7}=0.142857. . . . . . . .\]
\[\frac{2}{7}=0.285714. . . . . . . .\]
\[\frac{3}{7}=0.428571. . . . . . . .\]
\[. . . . . . . . . . \]
Other primes such as 13 and 17 have similar characteristics.  :)

lots and lots of factoring things

$(x+y)^2 = x^2 + 2xy + y^$

$(x+y)(x-y) = x^2 - y^2$

those sorts of things plus the fact that the coefficients of $(x+y)^n$ is the $n^{{th}}$ row of pascals triangle. I thought that that was really cool. (note that the top row, 1, is row #0.

Pascal's triangle and Fibonacci numbers always caught my attention.  There are some very basic but cool things you might be able to do with the golden ratio.

Clever factorization is always cool.

Basic number theory or rearranging of numbers to make multiplication easier (that was pretty cool, you already did that).

Different properties of numbers, perfect numbers, or the dividing by 11 rule, etc.

Perhaps go over some cool things with palindromes.

I still don't really get magic squares much  :P

[quote="LuCy4EvA"]is 0^0 undefined?[/quote]

0^0=0/0 -->undefined.

The most fun thing to do for me was play with a graphing calculator for the first time, after some instructions.


geometry
algebra
polynomial
inequalities
trigonometry
vector
analytic geometry

Math Zoom Year-round courses in Southern California for Fall 2008 will start in September.  The following are class schedules.  For further information, visit [url]http://www.mathzoom.org/mathzoom_yearround.html[/url].

[b]- Math Challenge II[/b]
In Math Challenge II, students learn and practice in areas such as algebra and geometry on the high school level, as well as advanced number theory and combinatorics.  Topics include polynomials, inequalities, special algebraic techniques, trigonometry, triangles and polygons, collinearity and concurrency, vectors and coordinates, numbers and divisibility,  modular arithmetic, residue classes, advanced counting strategies, binomial coefficients, pigeonhole principle, sequence and series, and various other topics and problem solving techniques involved in math contests such as the American Mathematics Competition (AMC) 10, 12 and ARML, and also the beginning AIME level.
[u]MC II schedules:[/u]
  [b]Arcadia[/b]: Saturdays starting 9/13, 9am-11am, 41 W Santa Clara St, Arcadia, CA 91007
  [b]Irvine[/b]: Fridays starting 9/5, 7:30pm-9:30pm, 4255 E Campus Drive, Suite 290, Irvine, CA 92612

[b]- Math Challenge III[/b]
This course is for experienced students who can solve at least a few problems in the AIME contest, and up to the USAMO level. In this course, the students learn more in-depth math concepts and problem-solving strategies that involves more rigorous mathematical writing and a broad range of topics. The topics includes everything covered in MC II with more depth, and additional topics such as quadratic residue, diophantine equations, recurrence, proving inequalities, mathematical induction, functional equations, generating functions, inversion, projective geometry, and rigorous proof-writing practices. Through effective guidance and direction by our experienced teachers, students develop strong problem solving skills that make them perform well in AIME and USAMO.
[u]MC III schedule:[/u]
[b]Arcadia[/b]: Saturdays starting 9/20, 2pm-4pm, 411 E Huntington Dr, Suite 309, Arcadia, CA 91006

[b]- Math Challenge I[/b] (grades 6-8)
Students learn the Math skills at a deeper level with topics in beginning algebra, fundamental geometry, basic number theory concepts and counting strategies, as well as logic and probabilities. The students not only learn practical skills of challenging problem solving that are supplemental to their school curricula, but also develop skills in creative thinking, logical reasoning, oral and written presentation, and team work. This course helps 6th to 8th graders (some advanced 5th graders may attend upon approval) to participate in the American Mathematics Competition (AMC) 8, MathCounts and Math Olympiads for Elementary and Middle Schools MOEMS. Through effective guidance and direction by our experienced teachers, students develop strong problem solving skills that make them perform well in the MathCounts, AMC 8, and MOEMS Contests.
[u]MC I schedules:[/u]
[b]Arcadia:[/b] Saturdays starting 9/13, 11:15am-1:15pm, 41 W Santa Clara St, Arcadia, CA 91007
[b]Diamond Bar:[/b] Saturdays starting 9/13, 6:30pm-8:30pm, 315 S. Diamond Bar Blvd., #C, Diamond Bar, CA 91765

[b]- Young Math Olympians[/b] (grades 3-5)
This course is focused on nurturing the interests of young kids from 3rd to 5th grades in challenging Math problem solving. With the effective guidance of our experienced teachers, kids explore the fun of solving interesting and meaningful Math related problems, and develop their logical and critical thinking skills. Some basic concepts covered in the Math Olympiads for Elementary School (MOEMS) are introduced.
[u]YMO schedules:[/u]
[b]Arcadia:[/b] Saturdays starting 9/13, 9:30am-10:30am, 41 W Santa Clara St, Arcadia, CA 91007


Em có 1 bài toán sau :
cho 2 số N và k, điều kiện N > k và N không chia hết cho k.

tức là N chia k dư Q ( hay N = Q mod k )

Lúc đầu N số đang ở trạng thái A, mỗi lần ta đổi được k số sang trạng thái B. ngược lại ta cũng thể đổi từ B về A, nhưng mỗi lần bắt buộc phải đổi k người. Tức là nếu đổi i người bên B về A thì tương đương với đổi k - i bên A về B. hix ! Em nói lung tung quá nhỉ !
CMR : nếu k chẵn và Q lẻ thì không bao h đổi hết N số sang trạng thái B được !
Em thử rất nhiều test và cảm thấy nó đúng, nhưng chưa tìm ra 1 cách CM nào tổng quát. Các anh giúp em với nhé !

Ví dụ : Nếu A = 333 và k = 50 thì không có cách.
           Nếu A = 15 và k = 10 thì không có cách...

Nếu A = 1000 và k = 35 thì có cách : đầu tiên ta đổi 28 lần được 980 số sang trạng thái B. Còn dư lại 20 số. Lần kế tiếp ta lấy 10 số bên B và 25 số bên A đổi lại. Ta được 25 số + 10 số về trạng thái A. Lần cuối đổi lại 35 số về B xong.
Nếu em ghi có chỗ nào không rõ ràng thì mấy anh cứ nói ạ ! Em sẽ sữa lại. Cám ơn máy anh rất nhiều !

Các bài toán trong box Việt Nam sẽ được post trong box này http://www.mathlinks.ro/Forum/index.php?f=284 . Mà theo anh không nên thảo luận về Toán trong box Việt Nam, cái này sang ddth tốt hơn, em nên post bằng tiếng Anh ra bên ngoài.

Da em hiểu rồi ! Do lần đầu em chưa biết nên post nhầm chỗ ! Srry anh nhé  :oops:


function
number theory unsolved
number theory

What is the number of partitions of a positive integer in $ k$ integers, such that there aren't two equal numbers in the partition.

Example. n=7, k=3
I need only 1+2+4.

If we ignore the condition(to be not equal each other), it would be a well-known open problem. However, your question may have a solution. Try to use "generating functions".

I forgot how to manipulate the generating functions regarding the partitions :blush:. I think it's very difficult to find a closed form for the partitions I need.

Your question is equivalent to the following question "What is the coefficient of  [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/a/2/9/a297bb91c9703af1975402dded3ab9b7e6431dde.gif[/img] in [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/6/f/c/6fc57c3597157b964300706a9f5848b93f22e8c8.gif[/img] ?"

Moreover, you can generalize the problem allowing to use i [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/1/b/a/1ba9b59bdee92f38c1698c784b67ba70f803331d.gif[/img] times. Then the question becomes "What is the coefficient of [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/5/3/8/5381feaab83ed3656546a8cdfac424159d680155.gif[/img] in [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/9/8/d/98d7dd9bab313baf7c0edde3b6e61c4bbbd35ff1.gif[/img] ?"

I've found on the internet a closed form. It's $ p_{3}(6j\plus{}k)\equal{}3j^{2}\minus{}(3\minus{}k)j$, for $ k\in [1,5]$ and $ p_{3}(6j\plus{}6)\equal{}3j^{2}\plus{}3j\plus{}1$. Nice

[quote="matehan"]Your question is equivalent to the following question "What is the coefficient of $ x^{n}$ in $ (1\plus{}x)(1\plus{}x^{2})\ldots(1\plus{}x^{n})$ ?"[/quote]

This is wrong.He asked for the number of partitions in exactly $ k$ integer.

His question is equivalent,for example,to find the coefficient of $ x^{n}y^{k}$ in

$ (xy\plus{}1)(x^{2}y\plus{}1)\ldots (x^{n}y\plus{}1)$

Yes, you are right. I didn't see that condition.   :blush:


Let $a,b,c,d$ be non-negative numbers sucht that $a^2+b^2+c^2+d^2=1$.
Prove that
$a^3+b^3+c^3+d^3+abc+bcd+cda+dab \geq 1$.

[b]Remark1[/b]. I am interested in a solution without squaring and expanding.

[b]Remark2[/b]. If $x_1^2+x_2^2+...+x_n^2=1$ and $p=\frac 6{(n-2)(\sqrt{n}+1)}$, then
$\sum x_1^3+p\sum x_1x_2x_3 \leq 1$.

i think we can use AC method just like in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90766[/url] what nhat did i.e let

$f(a,b,c,d)=a^3+b^3+c^3+d^3+abc+abd+acd+bcd$ then

$f((a+b)/2,(a+b)/2,c,d)-f(a,b,c,d)=\frac{1}{4}(a-b)^2(c+d-3(a+b))$
$f(0,a+b,c,d)-f(a,b,c,d)=ab(3(a+b)-(c+d))$

So $f(a,b,c,d)\leq Max\{f((a+b)/2,(a+b)/2,c,d),f(0,a+b,c,d)\}$

So by AC method ,

$f(a,b,c,d)\leq Max\{f(0,0,0,1),f(0,0,1/\sqrt{2},1/\sqrt{2}),f(0,1/\sqrt{3},1/\sqrt{3},1/\sqrt{3}),f(1/2,1/2,1/2,1/2)\}=1$

equality case when $a=b=c=0,d=1$(and its permutation) or $a=b=c=d=1/2$

Maybe Vasc want another solution , i guess ?

It is wrong, because AC-Theorem is valid if $a+b+c+d=constant$.
We must have $a^2+b^2+c^2+d^2=(a+b)^2/4+(a+b)^2/4+b^2+c^2$ and $a^2+b^2+c^2+d^2=0^2+(a+b)^2+b^2+c^2$ .
See:
http://www.mathlinks.ro/Forum/viewtopic.php?p=532271#p532271


See http://www.mathlinks.ro/Forum/viewtopic.php?t=38369 and http://www.mathlinks.ro/Forum/viewtopic.php?t=43403 for an even stronger inequality which is still unsolved.

  Darij

[quote="arqady"]I hope, this is unknown:
$a>0,b>0,c>0.$ Prove that $\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\geq\sqrt[4]{27(a^4+b^4+c^4)}.$[/quote]
I've found a proof of that one.  :)
 edit: :( No, sorry. It seems I have not.


is there any trick to do this problem:

what is the sum of the digits in (111111111) squared?

All numbers of this form (edit:less than or equal to 9 digits) squared gives us a pattern of 123... up to the number of digits and then down to one, e.g. $1111^{2}=1234321$.  Thus. $111111111=12345678987654321$.  The sum of the digits is $\frac{9(10)}{2}+\frac{8(9)}{2}=81$.

Masoud Zargar

Smart method. What if there are more than 9 1's?

Thank You Robin

That is a good thing to remember.

I like that trick.

111^2=12321


[quote="N.N.Trung"]Let $ f: [0;1] \rightarrow \mathbb{R}$ be a function such that:
$ (i)$ $ f(1) \equal{} 1$,
$ (ii)$ $ f(x) \ge 0$ for all $ x \in [0,1]$,
$ (iii)$ if $ x,y$ and $ x \plus{} y$ all lie in $ [0,1]$, then $ f(x \plus{} y) \ge f(x) \plus{} f(y)$.
Prove that $ f(x) \le 2x$ for all $ x \in [0;1]$.[/quote]

Let $ a\in[0,1]$ such that $ f(a)>2a$
If $ a\leq\frac 12$, and using $ f(x \plus{} y) \ge f(x) \plus{} f(y)$, we get $ f(2a)\geq 2f(a) > 4a\equal{}2(2a)$. Using this process as much as necessary, we get $ b\in(\frac 12,1]$ such that $ f(b)>2b>1$
Then $ f(1)\equal{}f(b\plus{}(1\minus{}b))\ge f(b)\plus{}f(1\minus{}b)>1\plus{}0$ (remember $ f(x) \ge 0$ for all $ x \in [0,1]$) which is a contradiction with $ f(1)\equal{}1$

So no such $ a$ exists and $ f(x)\leq 2x$ $ \forall x\in[0,1]$


The absolute value of every number in the sequence $\{a_n\}$ is smaller than 2005, and \[a_{n+6}=a_{n+4}+a_{n+2}-a_n.\] holds for all positive integers n. Prove that $\{a_n\}$ is periodic.

Incredibly, this was probably the most difficult problem of our independent study problems in the 1st TST (excluding the final exam).

so just define two sequences $b, c$ the even and odd terms so that $b_{n+3} = b_{n+2} + b_{n+1} - b_n$, similar one for $c$. Then characteristic equation is $r^3 - r^2 - r + 1 = 0 \Rightarrow (r-1)(r^2-1) = 0 \Rightarrow r = -1, 1, 1$ so $b_n = \alpha + n\beta + \gamma (-1)^n$; if $\beta \ne 0$, then for $n$ arbitrarily large, the modulus will be unbounded, which cannot happen, so $\beta = 0$ and $b_n = \alpha + \gamma (-1)^n$ which has period 2; a similar analysis takes place for the $c_n$ so $a_n$ has period $4$.


Problem 2 (Group II). The glass of a fluorescent lamp is composed mainly by silica (SiO2), alumina (Al2O3), calcium oxide (CaO), magnesium oxide (MgO), and sodium oxide (Na2O).

2.1 Regarding those compounds, select from the following statements the correct one:

A - Alumina is the responsible for the formation of the net structure of glass.
B - The ion Na+ is one of the responsibles for the breaking of Si-O bonds in the silica structure.
C - Alumina is added in order to reduce the melting point of glass, facilitating its production.
D - The less the quantity of CaO and Na2O, the smaller will be melting point of glass.

2.2 Consider that the general formula of the glass used in those lamps production is $ SiO_{2}(Na_{2}O)_{m}(CaO)_{n}$. Supposing now that, in the above formula, m = 0.08 and n = 0.06, compute the minimum mass of sodium sulphate (Na2SO4) necessary for the production of 1.0 kg of that glass.

$ Na_{2}SO_{4}(s)+heat \longrightarrow Na_{2}O(s)+SO_{3}(g)$.

Let $ M$ be the molar mass of glass..$ r$ is molar mass of $ Na_{2}O$ .$ k$ molar mass of $ Na_{2}SO_{4}$
in 1 mole of glass i.e in $ \frac{M}{1000}$ Kg we have 0.08 moles $ Na_{2}O$ hence in 1kg we have
have $ \frac{80m}{M}$ g of $ Na_{2}O$
in $ k$ g of $ Na_{2}SO_{4}$ we have $ m$ g of $ Na_{2}O$
so minimum mass of $ Na_{2}SO_{4}$ required is $ \frac{80k}{M}$

Your answer is correct. How about question 2.1?