How many triples of non-empty subsets from the set {1,2,...,2008} such that any pairs of the triple don't have common element?

Consider a table with three columns labelled $ A, B, C$ (where $ (A, B, C)$ is the triple of sets), and 2008 columns labelled with the numbers from 1 to 2008. Each cell of the table is filled in with either a 0 or a 1. A 1 in a cell means that the corresponding element (the row number) is in the corresponding set (the column label), while a 0 means it's not. each possible triple of sets corresponds to such a table, and each such table corresponds to a triple, under the conditions that no row have 2 1's in it and no column have all 0's in it.

(1) the number of ways to fill in the table with NO restrictions on the columns (but with the proper restrictions on the rows) is $ 4^n$ (because each row may be either $ 001$, $ 010$, $ 100$, or $ 000$).

(2) the number of ways to fill in the table so that EXACTLY 3 columns are all 0's is 1.

(3) the number of ways to fill in the table so that EXACTLY 2 columns are all 0's is $ 3(2^{2008}\minus{}1) \equal{} 3\cdot 2^{2008}\minus{}3$.

(4) the number of ways to fill in the table so that EXACTLY 1 column is all 0's is $ 3[3^2008 \minus{} 1 \minus{} 2(2^{2008}\minus{}1)] \equal{} 3^{2009}\minus{}6\cdot 2^{2008}\plus{}3$

So the answer is $ 4^{2008} \minus{} 1 \minus{} (3\cdot 2^{2008} \minus{} 3) \minus{} (3^{2009} \minus{} 6\cdot 2^{2008} \plus{}3) \equal{} 4^{2008}\minus{}3^{2009}\plus{}3\cdot 2^{2008}\minus{}1$

[b]Generalization:[/b]

For $ 1\le m\le n$ and $ |X| \equal{} n$, Find the number of all n-tuples $ (X_1, X_2, ...,X_m)$  such that $ X_i\subset X, |X_i|\neq 0$ $ X_i\cap X_j \equal{} \emptyset$ where $ i\neq j$ and $ i, j, \equal{} 1,2,...,m$.

You can kill this problems with Principle of Inclusion-Exclusion!

Notice that $ (m \plus{} 1)^n$ is the number of all n-tuples $ (X_1, X_2, ...,X_m)$ such that $ X_i\subset X$, $ X_i\cap X_j \equal{} \emptyset$, $ i\neq j$ and $ i, j, \equal{} 1,2,...,m$

Anybody likes solving another prolems, go here:

http://www.mathlinks.ro/weblog_entry.php?t=254299


Find the limit of this sequence of numbers:
            \[ u_n=\sum_{i=1}^n \frac1{i!}\]

This one is famous and basic : $e-1$  :D  :D  :D  :D


A board whose length is 84 inches is cut into three pieces in the
ratio $ 1: 2: 3$. What is the number of inches in the length of the
shortest piece?

The ratio of the smallest piece to the total is 1/(1+2+3)=1/6.
84*1/6=14.

Here's how I solve these problems.

$ x \plus{}2x \plus{} 3x \equal{} 84$

$ 6x \equal{} 84$

$ x \equal{} 14$

So the smallest piece is $ 1(14) \equal{} 14$.

A little shortcut:  We can say there are 6 equal sized pieces.  Divide 84/6 to get the length of one piece.

Note that these are all essentially the same.


Find all positive primes $p$ for which there exist integers $m,n$ satisfying:
[b]1.[/b] $p=m^2+n^2$
[b]2.[/b] $m^3+n^3-4$ is divisible by $p$

$m^2+n^2|m^3+n^3-4 \Longrightarrow m^2+n^2|(m+n)(mn-m^2-n^2)+4 \Longrightarrow m^2+n^2|(m+n)mn+4 \Longrightarrow m^2+n^2|m^2n+n^2m+4\Longrightarrow k(m^2+n^2)=m^2n+n^2m+4$

Suppose $n>m\geq2 ,k\geq2$
If $k>n$ then
$(n+1)(m^2+n^2)\leq k(m^2+n^2)=m^2n+n^2m+4 \Longrightarrow nm^2+n^3+m^2+n^2\leq m^2n+n^2m+4 \Longrightarrow n^3+m^2+n^2\leq n^2m+4\leq n^3+m^2$

If $k=n$ then:
$n(m^2+n^2)=m^2n+n^2m+4 \Longrightarrow n^2(n-m)=4 \Longrightarrow 9\leq 4$

Hence $k<n$:

$k(m^2+n^2)=m^2n+n^2m+4 \Longrightarrow m^2(n-k)+m(n^2)+(4-kn^2)$

Hence $(n^2)^2-4(n-k)(4-kn^2)$ sould be a perfect square.
$(n^2)^2-4(n-k)(4-kn^2)= n^4+4kn^3+4k^2n^2-16n+16k=A$

But $(n^2+2kn-1)^2<A<(n^2+2kn)^2$

We can easily check  if $k=1$ or $m=1$

Nice problem  ;)


trigonometry
Ring Theory
Mandelbrot Competition
National Assessment

What non-team math competitions for high school students are out there? I know of AMC/AIME/USAMO and USAMTS...

Also, if anyone has any information on non-team Chemistry competitions or Foreign Language competitions, I'd be really grateful, just post or message me -- I'm a homeschooled student, the local schools are pretty clueless, and Google hasn't been all that helpful... Thanks!

Mandelbrot competition (has a team element, but it's only for 4 people and I know someone who did it individually)
http://www.mandelbrot.org

Math League
http://www.mathleague.com
Specific stuff for homeschoolers:
http://www.mathleague.com/homeschool.htm

Atlantic-Pacific Mathematics Competition (probably not worth taking if you're very into this stuff -- relatively low level)
Don't know the website, if one exists

US Chemistry Olympiad, National French Exam, National Latin Exam, dont know if there are contests like these for spanish and others.
French thing ---> [url]http://www.frenchteachers.org/concours/[/url]

My company, National Assessment & Testing, offers several non-team competitions.  One of them has already occurred for this year, but our Mu Alpha Theta Contest and Ciphering Time Trials will be occuring soon...  The Mu Alpha Theta Contest is free, but is only for schools with Mu Alpha Theta (a mathematics honor society) chapters... I'm not sure if your homeschool is a member, but it can be.  Anyhow, more information on all of these is on our website, so I won't clutter up the board with it.

There's the USNCO (US National Chemistry Olympiad).  Apparently there's a written part and a laboratory practical part, which can get ugly...but anyway, on to information:

Information page at [url]http://www.chemistry.org/portal/a/c/s/1/acsdisplay.html?DOC=education%5Cstudent%5Ccorules.html[/url]
-->Practice Exams at [url]http://www.chemistry.org/portal/a/c/s/1/acsdisplay.html?DOC=education\student\co_lsexam.html[/url] (Local) and http://www.acs.org/portal/Chemistry?PID=acsdisplay.html&DOC=education\student\co_ntexam.html (National).

An unofficial page with some practice exams, more information, etc. at [url]http://nhb.topcities.com/Chemclub/database/olimpiads/Ncho/USA/ChemTeamIndex2.html[/url]

And (oh, wow, I don't have these saved to my hard drive - quick!  quick!) a treasure trove of old USNCO questions at
[url]http://dbhs.wvusd.k12.ca.us/NChO/NChO-Menu.html[/url]

And as for spanish exams, there is the National Spanish Exam which I've mentioned before...oh, here we go: [url]http://www.2nse.org/[/url].  But it looks like you need a sponsor (i.e. a school, an AATSP member, or something) and it costs $65.00 so you'd want to do it with some school.  Maybe if you check around at the local high schools.


abstract algebra
algebra
polynomial
group theory
superior algebra
superior algebra solved

Let $\displaystyle \mathcal K$ be a finite field. Prove that the following statements are equivalent:

(a) $\displaystyle 1+1=0$;

(b) for all $\displaystyle f \in \mathcal K \left[ X \right]$ with $\displaystyle \textrm{deg} \, f \geq 1$, $\displaystyle f \left( X^2 \right)$ is reducible.

$a\to b:$ $f(X^2)=f(X)^2$.

$b\to a:$ Let $G=K^*$ be the multiplicative group of the field $K$. Then because $K$ is finite, $G$ is cyclic (you can find this on the forum). If $char K\neq 2$, then $|K|$ is odd and the order of $G$ is even (being one less than $|K|$). Since $G$ is cyclic of even order $g\to g^2$ cannot be an automorphism of $G$ (for example because $G$ must have a nontrivial element of order 2 by Cauchy and then the kernel of the group homomorphism $g\to g^2$ is not trivial) hence it is not "even" a bijection, so there exists $g\in G$ which is not a sqare. But then for $f=X-g$, $f(X^2)=X^2-g$ is irreducible because it has no roots since $g$ is not a square.

$f(x)=ax+b\Rightarrow f(x^2)=ax^2+b$ but $(f(x))^2=a^2x^2+b^2$ which are not equal.

Yes, in order to have $f(X^2)=f^2(X)$ you would need the coefficients of $f$ to be roots of $x^2-x=0$, i.e. they should be $0$ or $1$, but the idea is there; it just needs to be slightly modified :).

Let $f(x)=a_nx^n+\ldots+a_1x+a_0$, so $f(x^2)=a_nx^{2n}+\ldots+a_1x^2+a_0$. We look for a polynomial $g(x)=b_nx^n+\ldots+b_1x+b_0$ such that $f(x^2)=g^2(x)\ (*)$. In a finite field of size $2^t$, all elements are squares. Just choose $b_j$ such that $b_j^2=a_j$, and it's easy to see that we do indeed get $(*)$.

yes, of course :P ... I was saying that $f(X^2)\neq (f(X))^2$... I had few moments in which I thought that I can't se the obvious :))

More generally:

Let $\mathbb{F}_{p^r}$ be a finite field with $p^r$ elements and let $f$ be a polynomial over $\mathbb{F}_{p^r}$.
Then $f\left(X^p\right)$ is reducible.

In fact it's not hard to see that $f\left(X^p\right)$ is the $p$-th power of a polynomial (use the Frobenius automorphism).


How many zeros are at the end of $ (100!)(200!)(300!)$ when multiplied out?

We simplify want to find out the number of powers of 10 in these factorials. There are obviously more 5 than 2s, so we count the number of power of 5s.

There are 100/5=20, 20/5=4, or 24 powers of 10 in 100!
There are 200/5=40, 40/5=8, 8/5=1.6, or 49 powers of 10 in 200!
There are 300/5=60, 60/5=12, 12/5=2.4, or 64 powers of 10 in 300!

Thus there are $ 24\plus{}49\plus{}64\equal{}\boxed{137}$ zeros at the end of $ (100!)(200!)(300!)$.

Thanks I have seen that problem all three years I was in Mathcounts, and I didn't know how to do it.

In the question, there are !'s after every number. What do those mean? I think I have seen them before, but I never knew what they were.
(100!)(200!)(300!)

Factorial

$ 1!\equal{}1$
$ 2!\equal{}1\times2$
$ 3!\equal{}1\times2\times3$

and so forth.

[quote="LLin1"]Also how do you get stars under your name, like Poincare Conjecture, or something like that?[/quote]

The stars under one's name represent the number of posts he/she has posted. The more posts, the more stars.


Determine the permutation [tex](x_1,x_2,...,x_4_0)[/tex] of numbers 1,2,...,40 such that the expression [tex]x_1 + x_2 / x_3 - x_4 * x_5 + x_6 / x_7 - x_8 * x_9 + ... + x_3_8 / x_3_9 - x_4_0[/tex] has the maximal possible value. (the operators +,/,-,* alternate regularly.)
It can be generalized for [tex]4n[/tex] (where [tex]n[/tex] is an arbitrary natural)?

But I must confess, that I have a lot of problems with the simpler task... I think that more than one permutation give the same maximum. I have one suggestion, but I can't prove that it belongs to the best of all.
It's based on:
let (a,b,c,d,e,f) be a permution of (1,2,3,4,5,6)
minimum of ab + cd + ef occurs for 6*1 + 2*5 + 3*4
maximum of a/b + c/d + e/f occrus for 6/1 + 5/2 + 4/3
But the generalization...

Thanks for help.

I should be more concrete in my suggestion:
I think that the maximum looks like:
40+(39/20)+(38/21)+...(30/29)-(1*18)-(2*17)-...-(10*9)-19 and the generalization is (according to my opinion) similar... It's easy to prove that numbers 40,39,...,30 must be in numerators of fractions for reach the maximum, but I don't know how to prove that the "position" of other numbers as optimal as possible for reach the maximum...

Thanks for each help that I really need.


The area of a circular plate is $ 200\pi$ $ cm^2$. What is the number of centimeters in the radius of the plate? Express your answer in simplest radical form.

The area of a circle with radius $ r$ is $ \pi r^2$. Thus we take the square root of 200, leaving us with $ \boxed{10\sqrt {2}}$.

EDIT: If you don't know how to simplify square roots, there's an entry on it in my blog. The link's in my signature. The entry is titled either "Square Roots - "Now That's Radical!" Part 1" or "Square Roots - "Now That's Radical!" Part 2".


Think about ur post battered. Was it really that needed?
(and yes he is correct, we all know that :) )

[quote="small_unicorn"]Simplify
$ a) \frac {1}{1 \plus{} \frac {1}{ 2 \plus{} \frac {1}{3} }}$

$ b) \frac {1}{2 \plus{} \frac {1}{ 2 \plus{} \frac {1}{2} }}$

$ b) \frac {1}{3 \plus{} \frac {1}{ 3 \plus{} \frac {1}{3} }}$[/quote]

What "c" problem are you guys talking about? :P

cgyao15, please don't respond to spam...
I'm assuming you already know, pink, but in case you didn't, there are two "b"'s so by c, i think they mean the second b.

ummmmmm..... of course I'm being sarcastic.....  :|

These problems have been answered multiple times, there is no need for further discussion.


limit
number theory
least common multiple
number theory unsolved

In China TST 2007 there is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=247844]such a problem[/url]:

Consider the largest subset $ X$ of $ \{1,2,\ldots,n\}$ such that for arbitrary two elements from $ X$ their lowest common multiple doesn't exceed $ n$. donote $ A_n\equal{}|X|$. Prove that

$ A_n \le 1.9\sqrt{n}\plus{}5$.

Now I wanna ask, what is the exact value of

$ a\equal{}\lim_{n\rightarrow \infty}\frac{A_n}{\sqrt{n}}$

(About the lower bound I know that $ a \ge \frac{3\sqrt{2}}{4}\equal{}1.060\cdots$. I think $ a$ is very closed to $ 1$.)

For arbitrary a and b, does that mean for any two elements of $ X$ ?
If so, I can only think of {1,2} because for any $ n$ and $ n \minus{} 1$, the $ lcm(n,n \minus{} 1) \geq n$
And well... $ A_n$ is obvious after.

[quote="shalex"]I don't understand what Euclid's Turtle meant.  :blush:[/quote]

Which part of what I said ? Well the question was find a subset of the natural numbers (starting from 1) up to n where the $ lcm(a,b) \leq n$
and $ a,b\in \mathbb{N}$

I am just trying to figure out when the $ lcm$ would exceed $ n$. Then I just though, except for $ 1$ and $ 2$, $ lcm(n,n\minus{}1)> n$. I am just assuming by arbitrary two elements, it means for everyone pairwise elements.

Except for $ 1$ and $ 2$, $ lcm(n,n\minus{}1) > n$ This follows from $ gcd(n,n\minus{}1) \equal{} 1$ so $ n\minus{}1 \nmid n$.

If anything is unclear with what I am trying to say, say what you don't understand in your post so I don't have to ask/assume :P


calculus
integration
trigonometry
logarithms
calculus computations

Find the following indefinite integrals.

(1) $ \int \frac{\sin x}{3\plus{}\sin ^ 2 x}\ dx$.

(2) $ \int e^{2x\plus{}e^x}\ dx$.

[hide]

$ \int \frac{\sin x}{3 \plus{} \sin^2 x} \, dx \equal{} \minus{}\int \frac{\minus{}\sin x \, dx}{3 \plus{} (1 \minus{} \cos^2 x)}$

$ \equal{} \minus{}\int \frac{du}{4 \minus{} u^2} \equal{} \frac{1}{4} \int \left(\frac{1}{u\minus{}2} \minus{} \frac{1}{u\plus{}2}\right) du$

$ \equal{} \frac{1}{4} \log\left|\frac{u \minus{} 2}{u \plus{} 2}\right| \plus{} C \equal{} \frac{1}{4} \log\left|\frac{\cos x \minus{} 2}{\cos x \plus{} 2}\right| \plus{} C$

[/hide]

[hide]
$ \int e^{2x\plus{}e^x} \, dx \equal{} \int e^x \left( e^x e^{e^x} \right) \, dx \equal{} \int u e^u \, du \equal{} (u\minus{}1) e^u \plus{} C \equal{} (e^x \minus{} 1)e^{e^x} \plus{} C$
[/hide]

Good! :)  You can get rid the sign of the absolute value for the first one.

Kunny, just a question: in Japan students learn to compute primitives like that in 12th grade?

Very interesting. What other topics do they learn?


Find all functions $f: Q*\rightarrow Q*$ such that
$f(x+1)=f(x)+1$
$f(x^{2})=[f(x)-1]^{2}+1$
 (:what?:)
[i]Louis Latin and Vicky[/i]

no solution

f(0)= 0 and f(n) = n

but the second condition then fails...

:huh:  :wink:

How did you conclude $f(0)=0$?  :?:

If $f(0)=C$, the second equation gives $C=(C-1)^{2}+1\iff C=1\lor C=2$, and then the first gives $f(n)=n+C$ for integer $n$. Obviously, the second equation is satisfied if $C=1$, hence $f(n)=n+1$ works for integer $n$, but how to prove that's the only solution on $\mathbb{Q}$?

EDIT: I think I got it.

[hide="Conclusion"]Put $f(x)=g(x)+x\implies g(x+1)+x+1=g(x)+x+1\implies g(x+1)=g(x)$. Hence $g(x)$ is periodic with period $1$. The second condition becomes

$g(x^{2})+x^{2}=(g(x)+x-1)^{2}+1$

$g(x^{2})+x^{2}=g^{2}(x)+x^{2}+1+2xg(x)-2g(x)-2x+1$

$g(x^{2})=g^{2}(x)+2xg(x)-2g(x)-2x+2$

$g(x^{2})=g^{2}(x)+2(x-1)(g(x)-1)\qquad(*)$

Putting $x+n$ instead of $x$, where $n$ is a non-zero integer, we get

$g(x^{2}+2nx+n^{2})=g^{2}(x+n)+2(x+n-1)(g(x+n)-1)$

Because of the periodicity, we have $g(x^{2}+2nx+n^{2})=g(x^{2}+2nx)$ and $g(x+n)=g(x)$, hence the last equation becomes

$g(x^{2}+2nx)=g^{2}(x)+2(x+n-1)(g(x)-1)\qquad(**)$

Subtracting $(*)$ from $(**)$ we get

$g(x^{2}+2nx)-g(x^{2})=2n(g(x)-1)$

for every rational $x$ and integer $n$.

Putting $x={m\over n}$ where $m$ is an integer, we get

$g\left({m^{2}\over n^{2}}+2m\right)-g\left({m^{2}\over n^{2}}\right)=2n\left(g\left({m\over n}\right)-1\right)$

However, the LHS is zero, because of the periodicity, hence the RHS must be zero too, which gives $g\left({m\over n}\right)=1$ for any rational ${m\over n}$. Hence it's proven $f(x)=x+1$ for any rational $x$, and that's the unique solution.[/hide]


Given three coordinates in $ xyz$ space, how may one determine the equation of the plane which passes through the three coordinates?

[hide="Without vectors"]
A plane is represented as:
$ Ax+By+Cz = D$
One of the coefficients must be non-zero, say $ A$. We can divide through by $ A$, and get the equation into the form:
$ x+Uy+Vz = K$
You have 3 $ (x,y,z)$ points, and 3 unknowns, then it's just solving a linear equation.
[/hide]

[hide="Vectors"]
Say you have the points $ P, Q, R$ (which are given).
Find the normal to the plane (which is normal to $ \vec{PQ}$ and $ \vec{PR}$)
$ \vec{n}= \vec{PQ}\times \vec{PR}$

Assume a point on the plane is $ (x,y,z)$. Then, $ \vec{n}\cdot (\vec{P}-(x,y,z)) = 0$ which gives the cartesian equation of the plane.
[/hide]

[hide="for aufha"]
You know how a line is in the form $ Ax+By = C$

this is a relation in two variables. It is graphed in two dimensions as a line.

The standard form of a plane is
$ Ax+By+Cz = D$

It is graphed in three dimensions and will give a plane
[/hide]


[quote="Lawrence Wu"][quote="leoxnlin"]If you make a 100 on the test (note: you may take it multiple times) you get a smiley face. If you make a 101 on the test, you get a super smiley face.

1. What is the meaning of life?
a) 42
c) 6*7
q) the number of holes in a connect four board
z) 24
b) stupidity

2. What is the answer to this question?
a) The answer cannot be determined.
b) All of the above.
c) None of the above.
d) All of the above.
e) All of the above, except d.
f) a and c only
g) all of the above except e

3. Which is larger, pi, e or the golden ratio, when the digits are all reversed?
a) pi
b) a.
c) e.
d) b and c.
e) pie.
f) The answer cannot be determined because the meaning of choice c is ambiguous.

4. What is the largest possible score on this test?
a) 18.2
b) 99.9999999999999999999
c) 4 + 3i

5. What is the meaning of death?
a) 42
c) 6*7
q) the number of squares in tic-tac-toe
z) 24
b) intelligence

6. Why is leoxnlin doing this?
a) leoxnlin is stupid.
b) a, obviously
c) b, obviously
d) leoxnlin is really smart.
e) leoxnlin is procrastinating
f) all of the above except d[/quote]
GUESSAGE
1.b
2.c
3.f
4.c
5.b
6.f[/quote]

There is no (b) for 1 lol

Note: the answers may change as time passes. They mostly stay the same though.

I'll guess!

1.b
2.b
3.b
4.b
5.b
6.b

 :lol:

[quote="leoxnlin"]Sorry, distracted you only got a 16.66666666666666666666666666666666.[/quote]

You need to be a little more familiar with $ \text{\LaTeX}$, so that you could post $ 16.\overline{6}\%$ instead of a truncation. :roll:

acazbf

I like how everyone is saying f for the last one :P

Fine, distracted.

academics rule, your score is $ 50\%$.
Hamster1800, your score is $ 16.\overline{6}\%$
cf249, your score is $ 33.\overline{3}\%$

1) b
2) c
3) f
4) [color=#ff0000]a[/color]
5) [color=red]q[/color]
6) [color=#aaaa00]d[/color] (dissapointing with myself)

yellow=edited
red=re-edited

There is only $ 1$ b in here... so,

1. b
2. g
3. d
4. a
5. b
6. f.

Well, if I can submit more then one test:

1. a
2. a
3. a
4. a
5. a
6. a

=D :)  :)  :)

pacman: 2
Yongyi: $ \sqrt{e^{\pi i}}$
hadasah: 2
1=2: 0
BOGTRO: unfortunately, I have to be fair to Yongyi. You have posted more than one set of answers, and therefore I can't answer any of them.

I still have the highest score? amazing.
i'll try again. knowing leoxnlin, its probably the self-contradictory answers
bfecbf

[quote="Lawrence Wu"]knowing leoxnlin, its probably the self-contradictory answers[/quote]

You know me?!?!????????????!!!!!!!!!!!!!!!!!!!!!  :o

[quote]
The answer may not always be the most ridiculous, or funny, or stupid.
[/quote]
or self contradictory. But may be. I wish they all were. But I don't want to change the answers. Unless other people want me to.

kimmystar: 1
Lawrence Wu: 3  :|

@leoxnlin: No, I read your signature. :)
i still am the only one to have gotten a 4 so far. but now i must get 0.
bdeaqf
yay.

No. I'm sure that the first is b.

uhmmm...
b
a
e
{101}(if not valid, b)
q
e

[quote="lotsofmath"]i change to...
bdeaqf[/quote]
uh dude? you copied my answers
oh well i intended to fail anyway.


If this problem had been posted before, or it's too easy for this section, please either delete it or move it

let $P(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_0$ with integer coefficients.  If $q$ is a prime such that it divides $a_{n-1}, a_{n-2}, \dots, a_{0}$ but does not divide $a_n$, and $q^2$ does not divide $a_0$, prove that the polynomial is irreducible over rationals, i.e. it cannot be written as the product of two non-constant polynomials with rational coefficients.

For some of you this "may" be a hint:

[hide]
It was a practice problem in the "Extreme Principle" section
[/hide]

It's Eisenstein's Irreducibility Criterion.  Basically, you suppose for a contradiction that you can do it with integer coeff. polynomials U and V. (multiply by denominator)  Wlog, the constant term of U has factor q, and V doesn't.  Then the $x^1$ term of U must have a factor q, $x^2$,.. etc.  But the leading term of U can't have a factor q.  Contradiction.


Prove that \[\sum_{cyc}\frac{2a^{3}b^{3}}{a+b}\geq 3,\forall a,b,c >0 , abc =1\]

We can see that
\[\sum_{cyc}{\frac{a^{3}b^{3}}{a+b}=\sum_{cyc}{\frac{a^{2}b^{2}}{ac+bc}}}\]
So by Cauchy-Schwarz and AM-GM:
\[\sum_{cyc}{\frac{a^{2}b^{2}}{ac+bc}\geq \frac{ab+bc+ac}{2}\geq \frac{3}{2}}\]

I don't understand how this one can be OWN inequality

It is almost same as IMO 1995. problem 2.


If a, b,c are real positive integers prove that:
$ 27(a+b)^2(b+c)^2(c+a)^2\geq 64abc(a+b+c)^3. $

Brute force : Expand, it may be rewritten as
$27 \sum a^4b^2 + 27 \sum a^3b^3 \geq 19 \sum a^2b^2c^2 + 30 \sum a^3b^2c + 5 \sum a^4bc$, where the sums are symmetrics

Since $(4,2,0)$ majorizes $(4,1,1)$, and since both $(4,2,0)$ and $(3,3,0)$ majorize each of $(2,2,2),(3,2,1)$ you only have to use Muirhead'inequality to conclude.

Pierre.

This is Popoviciu inequality. An elegant solution:
   We use the fact that $ (a+b)(b+c)(c+a)\geq\frac{8}{9}\cdot (a+b+c)(ab+bc+ca) $. Then it is enough to prove that $ (ab+bc+ca)^2\geq 3abc(a+b+c)$ which is trivial.


inequalities
induction
inequalities unsolved

Prove that
$ a_1^7 \plus{} a_2^7 \plus{} ... \plus{} a_n^7 \plus{} a_1^5 \plus{} a_2^5 \plus{} ... \plus{} a_n^5 \geq 2\left(a_1^3 \plus{} a_2^3 \plus{} ... \plus{} a_n^3\right)^2$,
if $ a_1$, $ a_2$, ..., $ a_n$ are $ n$ distinct natural numbers.

[i]Alternative formulation:[/i] If $ a_1,a_2,...,a_n$ is a sequence of distinct natural numbers, then prove that
$ \sum_{k\equal{}1}^n {a_k}^7 \plus{} \sum_{k\equal{}1}^n {a_k}^5 \geq 2 \left(\sum_{k\equal{}1}^n {a_k}^3\right)^2$.

I remember we had this one on the IMO training in Oberwolfach, but of course I forgot the solution and had to find it again.

[color=blue][b]Problem.[/b] Prove that, if $a_1$, $a_2$, ..., $a_n$ are n distinct natural numbers, then

$a_1^7+a_2^7+...+a_n^7+a_1^5+a_2^5+...+a_n^5 \geq 2\left(a_1^3+a_2^3+...+a_n^3\right)^2$.[/color]

[i]Solution.[/i] We will prove the inequality by induction over n.

For n = 1, the inequality is $a_1^7+a_1^5\geq 2\left(a_1^3\right)^2$, what is trivial from AM-GM: $a_1^7+a_1^5\geq 2\sqrt{a_1^7\cdot a_1^5}=2a_1^6=2\left(a_1^3\right)^2$.

Now assume that the inequality holds for some natural n, so we have $a_1^7+a_2^7+...+a_n^7+a_1^5+a_2^5+...+a_n^5 \geq 2\left(a_1^3+a_2^3+...+a_n^3\right)^2$ for any distinct natural numbers $a_1$, $a_2$, ..., $a_n$. We will show that it also holds for n + 1; in other words, we will prove

$a_1^7+a_2^7+...+a_n^7+a_{n+1}^7+a_1^5+a_2^5+...+a_n^5+a_{n+1}^5 \geq 2\left(a_1^3+a_2^3+...+a_n^3+a_{n+1}^3\right)^2$

for any distinct natural numbers $a_1$, $a_2$, ..., $a_n$, $a_{n+1}$. In fact, we can WLOG assume that $a_{n+1}$ is the greatest among these distinct numbers $a_1$, $a_2$, ..., $a_n$, $a_{n+1}$, so we have, particularly,

$1^3+2^3+...+\left(a_{n+1}-1\right)^3\geq a_1^3+a_2^3+...+a_n^3$.

Using the formula $1^3+2^3+...+\left(a_{n+1}-1\right)^3=\frac{\left(a_{n+1}-1\right)^2a_{n+1}^2}{4}$, this becomes $\frac{\left(a_{n+1}-1\right)^2a_{n+1}^2}{4}\geq a_1^3+a_2^3+...+a_n^3$, so that $\left(a_{n+1}-1\right)^2a_{n+1}^2\geq 4\left(a_1^3+a_2^3+...+a_n^3\right)$. Hence,

$a_1^7+a_2^7+...+a_n^7+a_{n+1}^7+a_1^5+a_2^5+...+a_n^5+a_{n+1}^5$
$=\left(a_1^7+a_2^7+...+a_n^7+a_1^5+a_2^5+...+a_n^5\right)+a_{n+1}^7+a_{n+1}^5$
$\geq 2\left(a_1^3+a_2^3+...+a_n^3\right)^2+\underbrace{a_{n+1}^7+a_{n+1}^5}_{=2a_{n+1}^6+\left(a_{n+1}-1\right)^2a_{n+1}^5}$      (since $a_1^7+a_2^7+...+a_n^7+a_1^5+a_2^5+...+a_n^5\geq 2\left(a_1^3+a_2^3+...+a_n^3\right)^2$ by the induction assumption)
$=2\left(a_1^3+a_2^3+...+a_n^3\right)^2+2a_{n+1}^6+\left(a_{n+1}-1\right)^2a_{n+1}^5$
$=2\left(a_1^3+a_2^3+...+a_n^3\right)^2+2\left(a_{n+1}^3\right)^2+\underbrace{\left(a_{n+1}-1\right)^2a_{n+1}^2}_{\geq 4\left(a_1^3+a_2^3+...+a_n^3\right)}\cdot a_{n+1}^3$
$\geq 2\left(a_1^3+a_2^3+...+a_n^3\right)^2+2\left(a_{n+1}^3\right)^2+4\left(a_1^3+a_2^3+...+a_n^3\right)\cdot a_{n+1}^3$
$=2\left(\left(a_1^3+a_2^3+...+a_n^3\right)+a_{n+1}^3\right)^2=2\left(a_1^3+a_2^3+...+a_n^3+a_{n+1}^3\right)^2$,

and the induction step is complete. So the problem is solved.

  Darij

[quote="ehsan2004"]I think it's a Romani MO in 1995  ;)[/quote]

As I remember it also appears in Moscow olympiad before 1980´th.


It seems like I'm spending a lot of time on the USAMTS problems.  Usually I'll devote half a day (4-5 hours) to solving all of the problems from a round.  Then, I'll spend another 3 or 4 hours writing out complete, thorough, neat solutions.   Does anyone else spend this much time?

Yeah, I do.  For the first two rounds, I spent over 10, each.  The third round, I found to be especially easy, and I did it in two-three days, for about 5 hours or so.  The fourth round took me the longest to complete.  I spent about 10 hours on it in just two days... and I worked on it for the whole month.

I really spend a lot of time working on them.  I'm glad I'm not the only one :)

Last yr when i did USAMTS, i basically spent like a morning...usually saturday...and that was it...what i didnt finish just didnt get finished...lol


$\triangle{ABC}$ has $AB = AC$. $P$ lies on $AC$, and $Q$ lies on $AB$. We have $AP = PQ = QB = BC$. Find $\angle{ACB}/\angle{APQ}$.

WOW, BornForMath came back after like a year...wow

hints? i really want to know how you're supposed to solve this

[quote="pkothari13"]likewise. I drew a huge picture and labeled every possible angle I could construct and/or find to no avail  :([/quote]

this problem is just an angle chase...you are missing something...(let $\angle ABC=x$, then you can find every other angle in terms of $x$

There are  a few versions of this  problem and , unless you have seen it before,   you may find this a little tricky .. there are quite a few fun ways to solve these   but  I have also seen people getting lost in "angle chase" ..

The answer, hidden below,  may give you a hint to find the solution..
[hide]$\angle{ACB}= 80^{o}$ and
$\angle{APQ}$  =[hide] $140^{o}$
[/hide][/hide]
I want to see if  there are any fun ways to get this, I may put one of my favorite solution, it there is an interest.

[quote="Gyan"] there are any fun ways to get this, I may put one of my favorite solution, it there is an interest.[/quote]

Yes, definitely!  :)  When I went "angle chasing," all of my equations ended up equalling each other, so I am curious to see what I overlooked.

[hide="Okay Here it is"]Let  us call  $\angle{BAC}= \alpha$
Let point $R$  be on $AC$ (Between $P$ and $C$ ) such that $PQ=QR$

Easy to see  $\angle{RBC}= \alpha$  (Or $\angle{BRC}=\angle{BCR}$)
[hide="Because"]$\angle{QAP}=\angle{AQP}=\alpha$
$\angle{QPR}= \angle{QRP}= 2\alpha$
$\angle{BQR}= 180^{o}-\alpha-(180^{o}-4\alpha) = 3\alpha$
$\angle{QBR}= (180^{o}-3\alpha) /2$
$\angle{RBC}= (180^{o}-\alpha)/2-(180^{o}-3\alpha)/2 = \alpha$ [/hide]
So $BR=BC=(BQ=QR)(=QP=PA) \implies \triangle{BRQ}$ is equilateral $\implies \angle{QBR}= \angle{BQR}=\angle{QRB}= 60^{o}\implies \alpha = 20^{o}$
Now you know everything you need to know :)[/hide]


function
logarithms
geometric sequence
real analysis
real analysis solved

Find $f:[0,1]\rightarrow [0,1]$ bijective such that
$\forall x \in [0,1]$ , $\sqrt{f^{-1}(x)f(x)}=x$

Suppose we want to do it with a power function: $f(x)=x^r$ for some $r>0.$

Then we are lead to $r+\frac1r=2$ and there's only one $r$ that satisfies that, namely $r=1.$ This particular approach gives us only the identity function.

I haven't yet proved that this answer is unique, but I suspect it is unique.

The problem is saying that $f^{-1}(x)$, $x$, $f(x)$ forms a geometric sequence for every x.  By iterating in both directions, we get a bi-infinite geometric sequence.  The only way such a sequence can stay between 0 and 1 is to be a constant sequence.  Thus $f(x) = x$.

I think there was a similar problem a day or two ago, except it involved arithmetic mean instead of geometric mean.  In fact, you can probably show the two problems are equivalent by looking at a transformation like $g(x) = \ln f(e^x)$.


A non-convex quadrangle ABCD is given. If angle(BAD) = 60 , AB=4=AD and BC=3=DC, then what is the lenght of AC?

Choices:
1.) (2*sqrt3)-sqrt5
2.) (2*sqrt3)+sqrt5
3.) (2*sqrt3)-5
4.) (2*sqrt3)+5

[/list]

[hide="Solution"]
ABD is an equilateral triangle by SAS, so BD is 4. In BCD, the altitude from C is $ \sqrt{5}$ by the Pythagorean Theorem. In BAD, the altitude from A is $ 2\sqrt{3}$, again by the Pythagorean Theorem. Thus the answer is $ 1) 2\sqrt{3}\minus{}\sqrt{5}$.[/hide]


AMC
AIME
geometry
MATHCOUNTS
inequalities
geometric transformation
rotation

This year I qualified for the AIME for the first time, and I had a ton of fun practicing for it, so I decided to look into doing more math outside of what I did in math class.  Mainly I've been doing that by hanging around here, but even though I've had a lot more fun and have learned a ton of things, it hasn't really helped me on the AMC-12's I've taken to practice and guage where I am and such, and one of my goals is to do well on the AMC and AIME next year (like everyone else...).  So I was wondering what type of content I should study to do better on them.  Also, what resources would you recommend to work on this stuff?  I'm looking into buying the AoPS books, but since I haven't really done much math outside the classroom until now, I don't really know much about things like this...

Yeah, I think AoPS books are good points to start.
For AIME, do a lot of past AIMEs and most of them u can get in the resource page.

Is it just the experience that gets you better when you do past AIMEs?

[quote="Xantos C. Guin"]Not really.

If you don't know how to bubble in answers then experience might help.
Otherwise, it is more knowledge based.

Someone at my school has taken the AIME 3 times and never scored above a 5.
I took it once and got a 7.

I think AoPS vol2 would be a good place to start. So would past AIME's and maybe AMC 12's.[/quote]

When I say experience, I mean taking mutliple past tests and going through the solutions. And what do you mean when you say "knowledge based?"

Well, remembering a solution isn't exactly the best idea, mate. You need to be able to know why solutions work. Learn why each problem works, as each problem features a unique trick or property. Learn it, and you should be well off. Now, the question is what are the properties, and how do you use them? If anyone has any useful techniques, please post them here.  :)

Naturally, doing past AIMEs is a good way to get practice. To raise AIME score particularly, I would recommend AIME Problem Series. As well, intermediate (or introductory if you need them!) courses are good and/or AoPS textbooks. Going through one or both of the AoPS textbooks is invaluable to your problem solving ability.

Thanks, I think I have a pretty good idea what to do now, although feel free to post any other ideas that might help.  Now I just need to get those AoPS books...

There are already like 686592434152 threads about this...

[quote="Xantos C. Guin"]Not really.

If you don't know how to bubble in answers then experience might help.
Otherwise, it is more knowledge based.

Someone at my school has taken the AIME 3 times and never scored above a 5.
I took it once and got a 7.

I think AoPS vol2 would be a good place to start. So would past AIME's and maybe AMC 12's.[/quote]

So it can be a good idea to start AoPS 2 without having done AoPS 1? (That's what I may do.)

But I'm guessing that you already know most of the material in both AoPs books then...

[quote="Xantos C. Guin"]That's possible, I haven't seen the 1st page of AoPS vol 1 and I got a 7.[/quote]

Same, I have never seen either of the AoPS books and got an 8 this year.

I seem to know most of the material on the TOC of AoPS 1 so I'm guessing it wouldn't be a good idea to get, I think I'll get AoPS 2 first instead.

[quote="Xantos C. Guin"]Atleast vol 1.

Volume 1 is MATHCOUNTS, easy AMC, 1-5ish AIME problems.

If you can get 5+ on AIME, skip vol 1 and start with vol 2.[/quote]

Volume 1 contains some fundamental stuff that you can get by with on AIME but don't want to skip (modular arithmetic, inequalities, rotations, etc.). If you have time, you should read it.

You guys are underestimating AoPS 1. If you really understand, very well, everything in AoPS 1, you can get at least an 8.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48774

[quote="Karth"][quote="chess64"]If you're weak in geometry, I would strongly recommend going through the relevant chapters in AoPS 1. In most other cases, I think it would be okay to skip to AoPS 2, if you can already score at least 7 or 8.[/quote]

7 or 8 ONLY WITH AOPS 1??  :ninja:[/quote]

I've only read AoPS 1, and I'm pretty sure I don't know a lot of the stuff in AoPS 2, but I still got a 7.

Of course, this year's AIME was pretty easy.

I'm pretty sure you can get a 7 or 8 just with AoPs 1, but wouldn't that be on only the easiest AIMEs?  :huh:

No, on the easiest AIMEs you can get 10, if you read what Mr. Rusczyk wrote...

[quote="chess64"]You guys are underestimating AoPS 1. If you really understand, very well, everything in AoPS 1, you can get at least an 8.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48774]$\textsc{Click HERE}$[/url][/quote]

Hmm... what does he mean when saying "deep mastery?" I pretty much know the stuff and know how to use it... is that good enough?

You also have to know how to apply it.

And sometimes more than one concept per problem.

Those kind are my favorites.

A problem that can be solved by "deep mastery" of AoPS 1 has a solution that does not involve any complex math that is not taught in the book.


Find the smallest positive number $n$ such that if $10^{n}=M\times N$, where $M$ and $N$ are positive integers, then at least one of $M$ and $N$ must contain the digit $0$.

[quote="roadnottaken"]The smallest possible product $M*N$ based on the given information is $1*10=10$.  Therefore $n=1$.  Too easy. :sleep2: The shear number of topics you have posted in such a short period of time forces me to the conclusion that you are spamming. :spam:  :spam:  :spam:  :spam:[/quote]

It could be 2*5, so therefore $n \neq 1$.

YOu shouldn't be saying that this is easy if you got it wrong.

The answer is $n=8$, since you will have to have a 0 in the decimal expansion.

$10^{8}=5^{8}*2^{8}=390625*256$.

So the answer is 8.

[quote="roadnottaken"]The smallest possible product $M*N$ based on the given information is $1*10=10$.  Therefore $n=1$.  Too easy. :sleep2: The shear number of topics you have posted in such a short period of time forces me to the conclusion that you are spamming. [/quote]

Yes. The post above mine is by me right now.

Furthermore, each post has a legitimate use, and is not considered spamming.

Maybe look up [b]forum spamming[/b] on google.

Spam is the posting of advertisements, abusive, or unneeded messages on Internet Forums by automated spambots.

I have not posted any of those, but rather legitimate posts.

Okay. To get to the main point, the correct answer is 8. :roll:

[quote="roadnottaken"]I misread the question :oops: .  When will I learn not to try to solve problems late at night?  I appologize; probably the term "spam" was too harsh, but I really do think that Anirudh's posting habits are excessive.[/quote]

That, my friend, is true.

He definitely didn't need to post that many questions.

[quote="anirudh"]Maybe look up forum spamming on google.

Spam is the posting of advertisements, abusive, or unneeded messages on Internet Forums by automated [b]spambots.[/b]

I have not posted any of those, but rather legitimate posts.
[/quote]
The definitions of spamming on usual forums and this particular forum(AoPS) are different(esp that bolded part). In particular, posts that are made for the sole purpose of increasing your post count is DEFINITELY spam.
And instead of posting like 12379515 different topics, you could have combined the aime problems into one post, practice problem into one post, and so on.
Posting that many problems in such a short time is definitely spam.

[quote="mathgeniuse^ln(x)"]The thing is that I've heard many people saying to have only one problem per post.[/quote]
yea but if you're going to post like um 30 problems in less than an hour, then eh

True, True.

But still the problems are good.  the dpamming is not.


A point $ O$ inside an equilateral triangle $ PQR$ and the perpendiculars $ OL$, $ OM$ and $ ON$ are drawn to the sides $ PQ$, $ QR$, and $ RP$ respectively.

The ratios of lengths of the perpendiculars $ OL: OM: ON$ is $ 1: 2: 3$.

If $ \frac{\text{area of }LONP}{\text{area of }\Delta PQR} \equal{} \frac{a}{b}$, where $ a$ and $ b$ are integers with no common factors, what is the value of $ a \plus{} b$?

My answer is 47.

Let OL, OM and ON be 1 unit, 2 units and 3 units respectively. Extend the line OL vertically, draw a vertical line which passes through P, and draw a horizontal line passes through N. Let the intersection of the line OL and the horizontal line passing through N be A and let the intersection of the vertical line passing through P and the horizontal line passing through N be B. As a result, we obtained a rectangle with length AB=PL and breadth BP=AL. One can easily figure out that angle BPN= 30 degree and angle AON=60 degree. From here, we can make use of the length ON and OL to find the length of AN, BN and BP using simple trigonometry. Then, deduct the area of triangle BNP and ANO from the area of the rectangle to find the area of the shaded area.

The lenght of the equilateral triangle PQR can be easily found by finding the sum of triangle POQ, ROQ and ROP in terms of length of the equi. triangle PQR and equate it to the area of triangle PQR (find it using sine rule, also in terms of its length).

I have a Ptolemy solution!
Let OL=1, OM=2, ON=3, OP=x, OR=y, OQ=z. We have
$ sin(NPL) \equal{} sin(sin^{ \minus{} 1}(\frac {1}{x}) \plus{} sin^{ \minus{} 1}(\frac {3}{x})) \equal{} \frac {\sqrt3}{2}$ which simplifies to
$ PN \plus{} 3PL \equal{} x^2 \cdot \frac {\sqrt {3}}{2}$
But by Ptolemy on ONPL, we have
$ PN \plus{} 3PL \equal{} x \cdot LN$
So $ LN \equal{} \frac {x\sqrt {3}}{2}$
But $ LN \equal{} \sqrt {13}$ by the law of cosines.
So $ x \equal{} \frac {2\sqrt {39}}{3}$
Similarly, $ z \equal{} \frac {2\sqrt {21}}{3}, y \equal{} 4$
I'm too lazy too finish.


prove that
 \sum tg 2 (A/2)+8sin(A/2)sin(B/2)sin(C/2) \geq 2.
good luck. :D  :D

This one is Jack Garfulkel's inequality. Are you sure you create it, arosisi? If yes, I really admire you. :)
   My solution is in Inequality-Unsolved Problem-Special problem.
    What about your solution? :D

sorry i don't know it's Jack Gafulkel's ineq. I just found it when doing my exercise.
from this inequality
 \sum (a 2 )/ \(sum (ab))+8abc/(prod(a+b)) \geq 2 (*)
by letting ab+bc+ca =1 then
a=tg(A/2)
b=tg(B/2)
c=tg(C/2)
i have
\sum tg  2  (A/2)+8sin(A/2)sin(B/2)sin(C/2)   \geq 2.
(*) can be proved easily
(*) <=> \sum (a 2 ))/ \(sum (ab))-1
+8abc/(prod(a+b))-1 \geq 0
<=>  \sum (a-b) 2 (a 2 b+a 2 c+b 2 c+b 2 a-c 2 b-c 2 a) \geq 0. :D  :D


algebra
polynomial
sum of roots
complex numbers

[quote="modularmarc101"]Prove that it is impossible to form a cubic equation  with real coefficients and with roots $ 2, 3,$and$ 2 \minus{} 3i$.[/quote]

The equation: $ a_3 x^3\plus{}a_2 x^2\plus{}a_1 x\plus{}a_0\equal{}0$
$ \frac{a_2}{a_3}\equal{}\minus{}$ sum of roots$ \equal{}\minus{}(2\plus{}3\plus{}(2\minus{}3i))\equal{}\minus{}7\plus{}3i$ which is not real.

As $ a_3$ is real leads to $ a_2$ is not real,  $ a_2,a_3$ cannot be both real, done.

By the [url=http://en.wikipedia.org/wiki/Complex_conjugate_root_theorem]complex conjugate root theorem[/url], if $ 2\minus{}3i$ is a root of a polynomial with real coefficients, then so must $ 2\plus{}3i$.

Since $ 2\plus{}3i$ is not a root of the polynomial, there are no polynomials with real coefficients and roots $ 2, 3,$ and $ 2\minus{}3i$. Q.E.D.


Given that $a , b , c $ and $ d $ are real numbers such that $\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \frac{d}{a}$
Find the largest possible value of -
$\frac{ab - 3bc + ca }{a^2 - b^2 + c^2}$

Could someone give a detailed explanation , please ??
 :)

[b]I'm a noob, disregard and read my post below.[/b]

Ya, i think treething is right...
[hide] If you consider at first just $\frac{a}{b} = \frac{b}{c}$ you can cross multiply and simply until you arrive at the conclusion that $ac-b^2=0$ Then if you do the same with $\frac{c}{d}=\frac{d}{a}$ you arrive at $ac-d^2=0$. Then you can conclude that $b^2=d^2$, then $b=d$ from this, if you consider $\frac{b}{c}=\frac{c}{d}$, you can see that $bd-c^2=0$ and since $b=d$, you can say that $b^2-c^2=0$ Then you can see that $b=c=d$ Then using the previously acquired $ac-b^2=0$, you can substitue and obtain $ab-b^2=0$, then you can conclude that $a=b$. Now, you can see that inorder for the equality $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}$ to be true you can see that the following must be true: $a=b=c=d$. Then the largest value is $-1$[/hide]

Mass cross multiplying means a^2 = b^2 = c^2 = d^2.

So wait. The only requirements are that a = c = +/- b = +/- d.

That is, a = -b = c = -d works, as well as a = b = c = d.

b = d = -a
a = c

-a^2 + 3a^2 + a^2 / a^2 - a^2 + a^2

3a^2 / a^2 = [b]3, not -1.[/b]

Winnity.


ceiling function
probability
probability and stats

A professor teaching a Mathematics course gives a multiple choice quiz that has 10 questions,each with four possible answers.What is the minimum number of students that must be in the professor's class in order to guarantee that at least 3 paper sheets are identical?
(Assume that no answers are left blank)

I know that some how the Pigeon Hole principle will be used in solving the question.But, I don't know where to start tackling the question.

Can anyone give some hints?

Well, you have to count the number of possible paper sheets that can be made and after to use the general Pigeon Hole Principle: "If there's $ A$ pigeons and $ B$ pigeon holes, there's a pigeon hole which contains at least $ \left\lceil \frac{A}{B} \right\rceil$ pigeons, and then find the best boundary.

Note that this isn't exactly a probability question, although some probability questions could be posed from this setup.

There are $ 4^{10} \equal{} 2^{20}$ possible ways to answer all of the questions. To guarantee at least one instance of three identical papers, we must have one more than twice that many students - that is, $ 2^{21} \plus{} 1,$ which is slightly more than 2 million.

Here's one possible probability question based on this: If all students were totally clueless about the content of the test and chose answers based on some random mechanism, how many students would there need to be for the probability of an instance of three identical papers to exceed one half? (Hint: this answer will be much smaller than 2 million.)


A writer with an audience of millions insisted that he was never to be interrupted while writing. After the day when he actually was interrupted, he never wrote again.
How is this possible?

[hide="Uhhhhhh..."]He was in a room full of assassins and he was shot.[/hide]

[quote="Inspired By Nature"][quote="7h3.D3m0n.117"][quote="E^(pi*i)=-1"][hide]
He is an pilot who writes in the air with exhaust clouds.  When he was interrupted he lost control of the plane, it crashed, and he died.
[/hide][/quote]
lol nice one.....how was he interrupted thou?[/quote]

Maybe that can be up to reader to decide. :lol:[/quote]

  :lol:  Who knows, maybe he got a call on his cell phone?  :rotfl:


Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \frac {(4\sqrt2 \minus{} 3)(ab \plus{} ac \plus{} bc)}{a^2 \plus{} b^2 \plus{} c^2}\geq4\sqrt2
\]
There is a proof by one line. :wink:

I'm really curious how you can solve this in exactly one line (assuming the line is not 2000 characters long).

Who knows the ineq might solve it in one line but who doesn't cannot :lol: .

It's equivalent to $ \sum a(b^2 \minus{} ab \minus{} \sqrt 2 bc \plus{} \sqrt 2 ac)^2 \ge 0$ wich is true...

Lol, in this way any inequality is somehow equivalent with $ a^2\geq 0$ and they are all qed. This is not a one line solution in my opinion.

I also agree that it's not a one line solution :mad: .

Vasc  was the first:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176214
 :)


modular arithmetic
function
number theory proposed
number theory

Let $r>1$ be an integer, and let $p:= 4^r - 2^r + 1$. If $p$ is prime, show that $p \equiv 1 \pmod{6r}$.

Suppose there is a prime $q>3$ which divides $r$. We have $4^r-2^r+1=(2^r-\omega)(2^r-\bar\omega)=(2^{qt}-z^q)(2^{qt}-\bar z^q)$, where $\omega$ is a root of $x^2-x+1$ and $z$ is either $\omega$ or $\bar\omega$, depending on the residue of $q$ modulo $6$. In both cases, we find that $(2^t-z)(2^t-\bar z)=4^t-2^t+1|p$, and this is a contradiction to the primality of $p$. This shows that $r$ is of the form $2^a3^b$.

We now want $6r|2^r(2^r-1)\iff 2^{a+1}|2^{2^a3^b}\ (*),\ 3^{b+1}|2^{2^a3^b}-1\ (**)$. It's easy to see that if $a=0$ (i.e. if $r$ is odd), then $3|p$, so $p=3\Rightarrow r=1$. Otherwise, we must have $a\ge 1$. This means that $a+1\le 2^a\le 2^a3^b$, which proves $(*)$, and also $\varphi(3^{b+1})=2\cdot 3^b|2^a3^b$, which proves $(**)$.

At first this problem
[hide="looked obviously false"]because from "$2^r$ is a primitive $6$-th root of unity" we cannot conclude "$2$ is a primitive $6r$-th root"[/hide]

but then it was clear that it is
[hide="true for a simple beautiful reason"]
The function $f(r) = 4^r - 2^r + 1$ is increasing, so it is impossible for $2$ to be a primitive $6k$-th root of unity with $k | r$ a proper divisor, because $f(k) < p$.  (We already know $2^{2r}$ and $2^{3r}$ are not equal to 1, so this is all that one needs to check.)[/hide]

As an intermediate exercise one can prove a weaker result: "show that $p\equiv1\pmod{12}$"
One can see that clearly if $r$ is even then $3|p$ and since $r>1$ we must have $r$ even.If $r$ is even then $p=(2^{2k}-1)^2+(2^k)^2$ and thus $p$ can be expressed as the sum of 2 squares $\Rightarrow p\equiv1\pmod{4}$ .Now $4\equiv-2\pmod{6}$ and since $r$ is even $4^r\equiv2^r\pmod{6}$ so $6|4^r-2^r=p-1$ and thus $12|p-1\Rightarrow p\equiv1 \pmod{12}$.
Is it possible to involve $r$ in the modular using elementary methods?
(as usual sorry for bumping)


Let $ p$ be a prime and $ g$ an integer such that the period of $ g (mod p)$ is $ p\minus{}1$. Let all $ d_k$ be divisors of $ p\minus{}1$. Prove that there is an element with period $ d_k$ for all $ d_k$

It says in the book that since g has period p-1, then $ g^{(p\minus{}1)/(d_1)}, g^{(p\minus{}1)/(d_2)}...$ all have periods $ d_1, d_2...$, respectively.

Why is this?

For $ d \mid p\minus{}1$, let $ k$ be the multiplicative order (not period) of $ g^{\frac {p\minus{}1} {d}}$ modulo $ p$. This means  $ \left (g^{\frac {p\minus{}1} {d}}\right )^k \equal{} g^{k\frac {p\minus{}1} {d}} \equiv 1 \pmod{p}$, so $ p\minus{}1 \mid k\frac {p\minus{}1} {d}$, i.e. $ d \mid k$. But $ \left (g^{\frac {p\minus{}1} {d}}\right )^d \equal{} g^{p\minus{}1} \equiv 1 \pmod{p}$, so $ d$ is the multiplicative order of $ g^{\frac {p\minus{}1} {d}}$.


this is from 1997-

Call a positive real number special if it has a decimal representation that consists entirely of digits 0 and 7. For example, 700/99=7.070707070707....and 77.007 are special numbers. What is the smallest n such that 1 can be written as the sum of n special numbers?

(A) 7  (B) 8  (C) 9  (D) 10  (E) 1 can't be represented as the sum of finitely many special numbers

how do you do this? btw the answer is [hide] B [/hide]

Oooh, this was a fun one :).  Just try and make 1 from lots of 7's!  (Although 1 will have to look slightly different.)

ah thanks, i was thinking there was a better way than just writing it out.

Here's my solution

[hide]

We want to build up (digit by digit) a set of special numbers that will sum to 1=.99999...   So:

.7 = 1x0.7

.98=.7+.07x4

.994=.98+.007x2

.9996=.994+.0007x8

.99995=.9996+.00007x5

.999999=.99995+.000007x7

Since the highest coefficient was 8, it is easy to piece together 8 special numbers that sum to .999999.  Then if we turn each of those numbers into repeating decimals with period 6, we have 8 special numbers that sum to .999999...=1.

It isn't clear to me how to prove that 8 is the minimum necessary.

One more thought.  For some reason, when I saw this problem, I almost immediately thought to make use of the fact that 1=.999999...  After that, everything is relatively straightforward.  If I'm trying to teach someone how to do this, how do I explain the jump to the idea of using the fact that 1=.999999...?

[/hide]

--Dan

Your proof includes the fact that it's a minimum:  if you were to do it in fewer than 8, you'd need to move some 7 up one place.  Basically, there isn't any room for larger numbers.  This is terribly explained, but it's true.  The fact that you need 8 7's in one place won't change no matter what and there's no way you can make it change.

You explain it based on the fact that you want to get as close to 1 as you can, and .9999... is as close to 1 as you can get (that is, it's equal).  Your algorithm works fine, even if you start out trying to get 1.000..., you just have to recognize that once you have .999999, you slap down a repeating decimal and are done with it.

[quote="wyra"]The way I learned .999... = 1 was by writing out fractions:

1/9 = .111...
2/9 = .222...
...
9/9 = .999... = 1[/quote]
There are other cool ways. Such as x=.99999...  10x=9.99999 9x=9 x=1

[quote="Ragingg"]Hmm... I don't get it... why are we multiplying .07 and .007 and such.[/quote]

Any special number between 0 and 1 is of the form .7a+.07b+.007c+.0007d+..., where each of a,b,c,d,... is either 0 or 1 depending on whether that digit in the special number is 0 or 7.

Say we have k special numbers that sum to 1:
.7a_1 +.07b_1+.007c_1+..., .7a_2+.07b_2+.007c_2 +..., up to
.7a_k+.07b_k+.007c_k+...

Then if we let a = a_1 + a_2 + ... +a_k, b = b_1 + b_2 + ... + b_k, etc,
we get 1 = .7a + .07b + .007c + .0007d + ...
where a, b, c, d, ... are all integers between 0 and k.

So what I did above was find values for a, b, c, d, ...
After that it is easy to pick values for a_i, b_i, c_i, d_i, ... and reconstruct the k special numbers (but we don't need to do that for the problem).  The answer to the problem is max(a,b,c,d,...) = 8.

Hope that is a bit clearer.
--Dan

Take a look at the decimal expansion of 1/7.  See if that will get you a solution.


$n \ge 2$ ... needn't be integer
We want: $\prod_{k=1}^{\infty}\frac{n^{k}-1}{n^{k}}$


ARML
probability
geometry
trigonometry
quadratics
algebra
quadratic formula

1.
When the sum of the first $k$ terms of the series

\[ 1^2 + 2^2 + 3^2 + \cdots + n^2 + \cdots \]
is subtracted from the sum of the first $k$ terms of the series

\[ 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1) + \cdots, \]
the result is 210.  Find the numerical value of $k$.

2.
Three coplanar circles are concentric, and the lengths of their respective radii are 10, 20, and 30.  A square is circumscribed about the smallest circle, and squares are inscribed in the other two circles.  Find the length of the shortest line segment that could contain a point on each square.

3.
Find the numerical value of $k$ for which

\[ \frac{7}{x+y} = \frac{k}{x+z} = \frac{11}{z-y}. \]

4.
Spender Sam always carried only dollars and pennies, but he never carried more than 99 pennies.  When leaving home one day, Sam noted that the number of dollars he had was 150% of the number of pennies he had.  After spending 10 dollars and 78 cents, Sam discovered that the number of dollars he had remaining was 1 less than the number of pennies he then had.  With how much money did Sam leave home that day?

5.
Two cards are randomly selected from a standard 52 card deck of playing cards, without replacement.  Find the probability that at least on king and at least one heart are selected (though not necessarily on different cards).

6.
Find the ordered pair of positive integers $(a, b)$, with $a<b$, for which
\[ \sqrt{1+\sqrt{21+12\sqrt{3}}} = \sqrt{a} + \sqrt{b}. \]

7.
The length of the base of an isosceles triangle is 20.  In the plane of the triangle, a point 4 units from this base is 10 units from each leg.  Find the area of the triangle.

8.
If Arc denotes the principle value, find all values of $x$ which satisfy

\[ \sin{(4\text{Arc}\tan{x})} = \frac{24}{25}. \]

Anyone gonna try any of these?  I'll start this off by doing one of them.  I would post a solution to #6 but then furious wouldn't try doing it and I know he will like that problem ;) .  Instead I'll do #1.

[hide="Number 1"]
Guess what?  You don't even have to know the formula for the second sum or even the formula for the sum of the first $n$ squares for this problem :!:

Using sigma notation the first sum is:

\[\sum_{j=1}^k{j^2}.\]

The second sum is:

\[\sum_{j=1}^k{j(j+1)}=\sum_{j=1}^k{j^2}+\sum_{j=1}^k{j}.\]

Thus the desired difference is:

\[\sum_{j=1}^k{j^2}+\sum_{j=1}^k{j}-\sum_{j=1}^k{j^2}=210.\]

So then:

\[\sum_{j=1}^k{j}=1+2+\cdots+k=210.\]

Which means:

\[\frac{k(k+1)}{2}=210\Rightarrow k=20,-21.\]

$k$ cannot be negative so thus: $k=\boxed{20}.$[/hide]

yay. i did one.

[hide="number 6"]
(3,1)
$\displaystyle\sqrt{3}+1$
just used the quadratic trick to solve it.[/hide]

Err... the quadratic trick? Do tell. Oh, and it says a<b. So (1,3).

Sorry; it's hard to clearly solve a geometry problem without drawing anything, but I'll try...

[hide="7"]Since the point is equidistant from the legs, it is on the bisector of the angle formed by the legs. If the point were inside the triangle, its length of 10 from the leg and the right angle and shared angle would imply that it was the same as half of the original triangle, thus implying that the point was 0 units from the base. So the point is outside of the triangle. Again the triangle that is half of the quadrilateral formed is equal to the triangle that is half of the original triangle. Denote the value we want as n. Then :sqrt: $n^2-100$+4=n (because they are congruent). Skipping a couple steps, $n^2-8n+16$=$n^2-100$, and 116=8n. Therefore, [b]n=29/2=14.5[/b].[/hide]

[quote="jb05"]Err... the quadratic trick? Do tell. Oh, and it says a<b. So (1,3).[/quote]

i'm dislexic so i saw the sign going to other way. o well.

the quadratic trick (i'm sure it has a real name):
for $\displaystyle\sqrt{m+\sqrt{n}}$,
let $a^2+b^2=m$ and $2ab=\sqrt{n}$
then create $(ab)^2$ from 2ab, which equals $(\frac{\sqrt{n}}{2})^2$ or $\frac{n}{4}.$
then you can make a quadratic formula with the value of $a^2+b^2$ as bx and the value of $(ab)^2$ as c.
then factor the quadratic and take the square root of your solutions. usually one of the solutions will be a perfect square so you end up with $k+\sqrt{j}$

Bumped for people who haven't tried these yet.

[hide="3"]$\frac{7}{x+y} = \frac{k}{x+z} = \frac{11}{z-y}$

After cross multiplying, we have:

$(1)$  $7(x + z) = k(x + y )$
$(2)$  $11(x + z) = k(z - y)$

$(1)+(2)$ gives $18(x+z)=k(x+z) \Longrightarrow k=\boxed{18}$
[/hide]


Find all numbers of the form $ 5x^2 \minus{}4$ such that are a perfect square

There are infinite numbers like that unless $ x$ is a integer

[quote]There are infinite numbers like that unless  is a integer[/quote]
that is obvious.you need not say it :o

Can someone provide the solution please? I don`t even understand the notation of Kouichi Nakagawa.

Let　$ \alpha \equal{} \frac {1 \plus{} \sqrt {5}}{2}$, $ \beta \equal{} \frac {1 \minus{} \sqrt {5}}{2}$.
So $ (\alpha^n \plus{} \beta^n)^2 \equal{} (\alpha^n \minus{} \beta^n)^2 \plus{} 4 (\alpha \beta)^n$.
Now $ L_n \equal{} \alpha^n \plus{} \beta^n$, $ F_n \equal{} \frac {\alpha^n \minus{} \beta^n}{\sqrt {5}}$, $ \alpha \beta \equal{} \minus{} 1$.
Thus $ L_n^2 \equal{} 5F_n^2 \plus{} ( \minus{} 1)^n \times 4$

For Example
$ L_{2n \minus{} 1}^2 \equal{} 5F_{2n \minus{} 1}^2 \plus{} ( \minus{} 1)^{2n \minus{} 1} \times 4 \iff 5F_{2n\minus{}1}^2 \minus{} 4 \equal{} L_{2n\minus{}1}^2$
$ L_{2n}^2 \equal{} 5F_{2n}^2 \plus{} ( \minus{} 1)^{2n} \times 4 \iff 5F_{2n}^2 \plus{} 4 \equal{} L_{2n}^2$

Excuse me Kouichi Nakagawa :oops: , i continue don't understanding where do $ \alpha \equal{}\frac{1\plus{}\sqrt{5}}{2}$ came out from. I search on wikipedia for solving it as a Pell equation, but i didn't understood the notation. I am begginer in those subjects, so the notation is hard for me.

[quote="Chalao"]Excuse me Kouichi Nakagawa :oops: , i continue don't understanding where do $ \alpha \equal{} \frac {1 \plus{} \sqrt {5}}{2}$ came out from. I search on wikipedia for solving it as a Pell equation, but i didn't understood the notation. I am begginer in those subjects, so the notation is hard for me.[/quote]

This comes from the fact that X^2=X+1 has that root, and that is the characteristic equation of the Fibonacci Sequence recursion.

And what does the fibonancci sequence has to do with the problem?

Can someone explain me how to solve pell equations?

[url=http://www.imomath.com/tekstkut/pelleqn_ddj.pdf]This[/url] is pretty good.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Is it true that given are $n$ numbers which add up to 0 (consider them reals, complex,...) we can always find a $nxn$ nilpotent matrix which has these numbers as main diagonal?

Yes, it is true.
There is more general result:For evry polynomial $f(x)=x^n+c_1x^{n-1}+...+c_n$ and every fixed numbers $a_1,...,a_n$ with addition condition $a_1+...+a_n=-c_1$ there exists matrix $n\times n $ which have $f(x)$ as it's characteristic polynomial and $a_1,...a_n$ as its main diagonal numbers

Sure.
The polynomials $u_0=1,u_1=x-a_1,...,u_n=(x-a_1)...(x-a_n)$ forms a base in the space of polynomials with degree $\leq n$, thus our polynomial $f$ we can represent in the form $f=u_n+\lambda_1u_{n-1}+...+\lambda_nu_0$. Now let's take the matrix $A$ to be $A= \left(   \begin{array}  {cccccc} a_1 & 1 & 0 & ... &0 & 0 \\ 0 & a_2 & 1 & ... &0 & 0 \\ ... & ... & ... & ... &... & ... \\ 0 & 0 & 0 & ... &a_{n-1} & 1 \\ -\lambda_n & -\lambda_{n-1} & -\lambda_{n-2} & ... &-\lambda_2 & a_n  \end{array} \right)$.
Expanding the determinant of the matrix $xI-A$ by the last row we get $|xI-A|=u_n+\lambda_1u_{n-1}+...+\lambda_nu_0=f$, thus $A$ is our requested matrix

Great, that's also what I found. Too bad I considered only nilpotent matrices. And please, a moderator erase the source "created by harazi" and replace it with "classical".


In a game, letter "B's" are placed in an arrangement shown below
BBBBB
BBBB
BBB
BB

The object of this game is to cross out a certain number of B's in one row  at one time; then the other person can do the same. A player can only cross out the B's in one row, and can't cross inbetween rows. The object of the game is to leave the other player with only one letter left.  A sample game  is shown below (the bolding of letters represents crossing the letters):

Player 1's first move:
B[b]BB[/b]BB
BBBB
BBB
BB

Player 2's first move:
[b]BBB[/b]BB
BBBB
BBB
BB

Player 1's second move:
[b]BBB[/b]BB
[b]BBBB[/b]
BBB
BB

Player 2's second move:
[b]BBB[/b][b]BB[/b]
[b]BBBB[/b]
BBB
BB

Player 1's third move:
[b]BBB[/b][b]BB[/b]
[b]BBBB[/b]
BB[b]B[/b]
BB

Player 2's third move:
[b]BBB[/b][b]BB[/b]
[b]BBBB[/b]
BB[b]B[/b]
[b]BB[/b]

Player 1's fourth move:
[b]BBB[/b][b]BB[/b]
[b]BBBB[/b]
B[b]B[/b][b]B[/b]
[b]BB[/b]

Player 1 wins because he has left player 2 with only one B.

Here is the question:
[i]If the game is replayed, what first move could Player 1 do to win the game?[/i]

Nim game, cross out 4 Bs, then 3 Bs, 2 Bs, etc.

http://www.ebaumsworld.com/pearl.shtml

this is a flash game of it
arrangement is similar, but lowest is 3 balls


factorial
Ring Theory
superior algebra
superior algebra unsolved

The ideal class group of $ \mathbb Z [i]$ is trivial because it is a factorial ring.

It's euclidean for the norm $ N(a \plus{} bi) \equal{} a^2 \plus{} b^2$ (in practical, given $ a$ and $ b \neq 0$, you find $ q,r$ such that $ a \equal{} bq \plus{} r$ and $ N(r) < N(b)$ by considering $ \frac ab$ as a complex number and taking the closest number in $ \mathbb Z[i]$).

And $ \mathbb Q(i)$ is a field (trivially a UFD, but that's not interesting). You still seem to talk about $ \mathbb Z[i]$.


[quote="warmachine10"]Given triangle ABC with a= 12, b=10, and c=8 find the angles in minutes[/quote]
[hide]
$12^2 = 10^2 + 8^2 -2\cdot8\cdot10\cdot\cos C$
$10 = 80\cos C$
$\cos C = \frac{1}{8}$
$C \approx 83^o$

$\frac{\sin 83}{12} = \frac{\sin B}{8} = \frac{\sin A}{10}$
$12\sin B = 8\sin 83$
$B \approx 41^o$
$A \approx 56^o$[/hide]

You can't without a calculator or trigonometric tables.

[quote="robinhe"]It said the angles in minutes, not degrees.

But, I have no idea how to solve it.  :help:[/quote]

How do you solve angles with $\sin$ and $\cos$ while using minutes and stuff?? I misunderstood what he meant by "in minutes"  :oops:

Minutes is in DMS which stands for degrees-minutes-seconds.  I learned it last semester but forgot it.  All I know is that it's a way of representing a degree.

it is just you have 60 min in 1 degree and 60 sec in 1 min, just take the integer part, that is the degrees, take the fractional part, and multiply by 60, the integer part of that is the minutes, fractional part of that multiply by 60, that is the seconds


Let $ a_{0},a_{1},\ldots,a_{n}$ be integers, one of which is nonzero, and all of the numbers are not less than $ \minus{} 1$. Prove that if  \[ a_{0} \plus{} 2a_{1} \plus{} 2^{2}a_{2} \plus{} \cdots \plus{} 2^{n}a_{n} \equal{} 0,\] then $ a_{0} \plus{} a_{1} \plus{} \cdots \plus{} a_{n} > 0$.

Also induction.

$2| a_0$, so $a_0\geq 0, a_0/2\geq 0$. And we have $(a_0/2 + a_1) + 2a_2 +\ldots + 2^{n-1}a_n = 0$, $a_0/2 + a_1\geq -1$. So by induction hypothesis we get that $a_0/2 + a_1+a_2+\ldots + a_n\geq 0$ and so ($a_0\geq 0$) $a_0+\ldots + a_n\geq 0$. Done

[quote=BG Yoda]Induction:

1. $ a_{0} \plus{} 2a_{1} \equal{} 0$ $ \Leftrightarrow (a_0,a_1) \equal{} (2, \minus{} 1)$. (If $ a_1\ge 0$, then $ a_0$ is at least $ \minus{} 1$ but $ \textrm{even} \minus{} 1$ can't be $ 0$.)

2. $ a_0 \plus{} 2a_1 \plus{} 2^2 a_2 \plus{} \cdots \plus{} 2^n a_n \equal{} 0$ $ \Leftrightarrow a_0 \plus{} 2(a_1 \plus{} 2a_2 \plus{} \cdots \plus{} 2^{n \minus{} 1}a_n) \equal{} 0$
Thus $a_0$ is even, so $ a_0 \equal{} 2k$ for some integer $ k\geq 0$ and therefore $ (a_1 \plus{} k) \plus{} 2a_2 \plus{} \cdots \plus{} 2^{n \minus{} 1}a_n \equal{} 0$.

Obviously, $ a_1 \plus{} k \ge \minus{} 1$ and then by the induction hypothesis we have $ k \plus{} a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n > 0$, but $ 2k \geq k$ $ \Rightarrow a_{0} \plus{} a_{1} \plus{} \cdots \plus{} a_{n} > 0$, too :flex:[/quote]
the same solution i had  :blush:


When will the Round 3 LaTeX source be out?

I find it useful when typesetting solutions.

Sorry -- this was an oversight on our part.

It's on the website now.


Hey everyone.

I know there are lots of music fans on here; but I doubt that many of you have had the pleasure of experiencing some genuine, cheese-free, non-commercial trance music.  Trance differs from techno and house music in that the music is rich in melodies and harmonies rather than sound effects, beepy noises, and annoying, repetitive percussive loops.  As a proud member of tranceaddict.com for 4 years, it is my pleasure to present you guys with some aural pleasure.

The standard method by which trance is enjoyed is when several tracks are mixed live by a DJ into one big set (livesets are also free and legal to distribute).  I made this mix a few weeks ago; I think the tunes are pretty sweet.  It's about 77 minutes long; just right for burning to a CD.

Enjoy (this should be up for about 1 week; get it before the link dies!)

[url=http://s4.yousendit.com/d.aspx?id=0DV6MVV8YANNY0E66T4CFONWH7]http://s4.yousendit.com/d.aspx?id=0DV6MVV8YANNY0E66T4CFONWH7[/url]

For those interested, I can provide a tracklisting.  A few of the tracks in the mix were produced by myself, and the rest are some of my favourite tunes from the past year or so.

~ Elyot

Glad you liked it.  Last weekend I went to a party at my friend's house where 5 DJs mixed for 15 hours.  It was amazing.  A 1 hour liveset is really nothing.  I have thousands of hours worth of this stuff on my hard drive.

Tracklisting below:

00) 00:00 Elyot Grant - Intro
01) 00:28 Powerplant - Turn It Into Gold (Probspot Remix)
02) 04:13 PQM - You Are Sleeping (PQM vs Luke Chable Vocal Pass Remix)
03) 09:23 Gabriel & Dresden - Arcadia
04) 13:08 Above and Beyond & Andy Moor - Air For Life (Original Mix)
05) 19:15 DT8 featuring Andrea Britton - Winter (Max Graham Sidechain Mix)
06) 24:10 The Scarab - Monday P.M.
07) 29:48 Purple Haze - Adrenaline (Original Mix)
08) 36:04 Bart Claessen - Playmo (2nd Play)
09) 41:23 Sunny Lax - PUMA
10) 48:18 Mike Folye vs Signalrunners - Love Theme Dusk (Signalrunners Mix)
11) 53:51 Elyot Grant - Risen
12) 60:19 Robert Nickson - Twisted By Design
13) 64:29 Young Parisians featuring Ben Lost - Jump The Next Train (Kyau vs Albert Remix)
14) 70:16 Adam White & Andy Moor present White Room - White Room

The intro and the track at 53 mins were done by me.  Sunny Lax - PUMA has got to be my favourite track of 2005.  The 3 minutes from 45:00 to 48:00 are just insane.

[quote="Elyot"]Sunny Lax - PUMA has got to be my favourite track of 2005.  The 3 minutes from 45:00 to 48:00 are just insane.[/quote]

r[b]Amen[/b]

[quote="rzsolt"]I enjoyed it enough that I tuned out most of the room I'm in. When I relized it was on its second run through, it was like 2am, on a school night.

Yeah, thats pretty nice stuff to listen too. Do you have a single torrent that could get a few gigs of the stuff in my poessesion in one go?[/quote]

I don't really do torrents.  I have some nice connections with private DC++ hubs that I really can't let you guys in on.

However, there are tons of great places online to go to get trance livesets.  Tranceaddict has a few available for download (for non-members, you might have to sign up or get bombarded by ads or something).

[url=http://www.tranceaddict.com/livesets.shtml]http://www.tranceaddict.com/livesets.shtml[/url]

I also have a few other locations that I'll divulge but I'm not gonna post them for the public's eyes to see.  Private message me if you really want to know.  I might post another of my own sometime next week.

[quote="claralex"]That stuff is pretty awesome, I  must say I didn't think I'd like it cuz I hate house music, but its good. :)[/quote]

To tell you the truth, I don't like house music much either, and I think it's too bad that electronic music gets a bad name because of all of the rotten, boring, dull, annoying, or cheesey "techno songs" that people get exposed to.  It just kinda puts a sour taste in some people's mouths and really turns them off it.  And I'll admit, about 90% of electronic music is CRAP.  But there are enough different kinds that everyone should be able to find something that they'll eb able to enjoy.

Trance and house are two different entities.  It's like comparing ice cream to lasagna.

House music is like... fine when it's in the background; when you're doing something else.  They play house music in clubs where people just go to meet up, guys go to meet women, etc.  It's just the background noise that sets the mood.

Trance music is for listening to.  You put that stuff on, and it just carries you away.  You don't go to trance clubs to pick up girls.  You go to trance events to listen to the music, jump around like an idiot, and just have a blast.  It's got the emotion and the energy that house music just lacks.

[quote="Mathgod"]I kinda like hip-pop, rap, r&b better than those classic musics and house musics.  :P[/quote]

We have almost compleatly opposite music tastes. Hip-hob, rap, r&b and popy stuff, makes up the only music I have never really enjoyed.

I'm enjoying this trance stuff so far, before this the only electric I ever listened too was the chemical brothers and the crystal method.

I like a very wide spectrum, but I lean more heavily to 'face melting' music.

[quote="rzsolt"]I like a very wide spectrum, but I lean more heavily to 'face melting' music.[/quote]

Trance melts my whole body.  Wait till you guys here the next thing I'm cooking up... I'm working on a remix of a classic Jean-Michelle Jarre piece called Souvenir De Chine.  I promise that it'll be awesome.

(yea yea... I know the link is dead by now, I'll upload again soon!)

OK.  I made a new mix this afternoon.  It's more of a progressive-style trance set, with more mellow, darker tunes than the previous one.  Hope you all enjoy it!

Click below to download it:
[url=http://s56.yousendit.com/d.aspx?id=1CVRXQNQEIMTC2PQAOVS7KI4ZC]http://s56.yousendit.com/d.aspx?id=1CVRXQNQEIMTC2PQAOVS7KI4ZC[/url]

[quote="Sunny"]Definitely liking the new set.  I dunno, I like the darker tone better than the first one.  Hehehe thanks.[/quote]

Glad you liked it.  I started out darker, but it gets a bit more uplifting towards the end.  Track 2 is by me!

Tracklisting:

1) 00:00 Atmos - Ordinary Weekend
2) 08:12 Elyot Grant - Majesty
3) 14:51 Solar Stone - Release
4) 21:69 Sasha - Xpander
5) 26:58 Probspot - Blueberry
6) 32:53 Perry O'Neil - Numb
7) 39:30 Vadim Zhukov - Exit (Robert Nickson Remix)
8) 45:49 Aly & Fila - Eye Of Horus (Ronski Speed Remix)
9) 53:35 Sonourous - Last Sunday (Positive Ways Remix)
10) 58:37 Solid Globe - Sahara
11) 65:20 George Hales & Ant Williams - Ocean Drive
12) 72:57 Modulation - Spirits (Cosmicman Mix)

I way prefure the first set over the second one. The second one is not a push over, but I really enjoyed the first one.


Prove that the only prime $p$ such that $p^2+2^p$ is prime, is 3.

[hide]Clearly, 3 works and 2 fails.  For a prime $p>3$, we have $p^2\equiv 1\pmod{3}$, and, since $p$ is odd, $2^p\equiv 2\pmod{3}$, so $3|p^2+2^p$, a contradiction, since $p^2+2^p>p>3$.[/hide]

Right. And note that $p^2 \equiv 1\pmod{3}$ since all primes $p$ greater than 3 can be written in the form $p=6k \pm1$. Squaring this gives you the result. This leads to my next question:

Prove that all primes $p>3$ can be written in the form $p=6k\pm1$


If $a,b,c,x,y,z,t>0$ and $x \geq y \geq x \geq t$ then we have:

$ \frac{x}{ay+bz+ct} + \frac{y}{az+bt+cx} + \frac{z}{at+bx+cy} + \frac{t}{ax+by+cz} \geq \frac{4}{a+b+c}$.

x/(ay+bz+ct) + y/(az+bt+cx) + z/(at+bx+cy) + t/(ax+by+cz)  \geq 4/(a+b+c)
<=> x 2 /(axy+bxz+cxt) + y 2 /(ayz+byt+cyx) + z 2 /(atz+bxz+cyz) + t 2 /(axt+byt+czt)  \geq  4/(a+b+c)
We note that :
x 2 /(axy+bxz+cxt) + y 2 /(ayz+byt+cyx) + z 2 /(atz+bxz+cyz) + t 2 /(axt+byt+czt)  \geq (x+y+z+t) 2 /((a+c)(xy+yz+zt+tx)+b(2xz+2yt))
If xy+yz+zt+tx  \geq 2xz+2yt then :
(x+y+z+t) 2 /((a+c)(xy+yz+zt+tx)+b(2xz+2yt))
 \geq (x+y+z+t) 2 /((a+b+c)(xy+yz+zt+tx)) (1)
And : (x+y+z+t) 2  \geq 4(xy+yz+zt+tx) (2)
(2) <=> (x-y+z-t) 2  \geq 0 (obvious)
(1)(2) => q.e.d
If xy+yz+zt+tx \leq 2xz+2yt then :
(x+y+z+t) 2 /((a+c)(xy+yz+zt+tx)+b(2xz+2yt))
 \geq (x+y+z+t) 2 /((a+b+c)(2xz+2yt)) (3)
And : (x+y+z+t) 2   \geq 4(2xz+2yt) (4)
(4) <=> x 2 +y 2 +z 2 +t 2 +2xy+2yz+2zt+2tx-6xz-6yt  \geq 0
We denote :
u=x/t , m=y/t , n=z/t
u \geq m \geq n \geq 1
4) <=> u 2 +m 2 +n 2 +1+2um+2mn+2n+2u-6un-6m  \geq 0
We fix u,n and let f(m) = m 2 +m(2u+2n-6)+u 2  +n 2 +1+2n+2u-6un
f'(m) = 2m+2u+2n-6 \geq 0 for u \geq m \geq n \geq 1
So : f(m)  \geq f(n) = 4n 2 +u 2 -4un-4n+2u+1 = (2n-u-1) 2  \geq 0 (obvious)
(3)(4) => q.e.d


algebra
polynomial
calculus
integration
induction
symmetry
binomial theorem

This is really tedious problem in competition:
Assume that $R$ is a ring which has no nilpotent except zero in the ring. If two elements $a$ and $b$ of $R$ and two positive integers $m$ and $n$ with $(m,n)=1$ satisfy $a^m=b^m$ and $a^n=b^n$, prove that $a=b$.

[quote="sam-n"]This is really tedious problem in competition:
Assume that $R$ is a ring which has no nilpotent except zero in the ring.if for $a,b\in R$ we have $a^m=b^m$ and $a^n=b^n$ ,$(m,n)=1$ prove that $a=b$.[/quote]

I can prove the result is true if the ring is commutative.  We first observe that since $(m,n)=1$,

$\forall i\ge (m-1)(n-1) \exists p,q\ge 0: pm+qn=i$

From this it follows easily that

$a^{i}=a^{pm+qn}=(a^{m})^{p}(a^{n})^{q}=(b^{m})^{p}(b^{n})^{q}=b^{i}$

Now we let $T=2(m-1)(n-1)$.  Using the binomial theorem and the fact that the ring is commutative, we can compute:

$\displaystyle (a-b)^{T}=\sum_{i=0}^{T} (-1)^{i} {T \choose i} a^{T-i}b^{i}$

Given that $T=2(m-1)(n-1)$, it follows that either $T-i\ge (m-1)(n-1)$ or $i\ge (m-1)(n-1)$.  Consider these two subcases.

In the $T-i\ge (m-1)(n-1)$ case we get:

$a^{T-i}b^{i}=b^{T-i}b^{i}=b^{T}$

And in the $i\ge (m-1)(n-1)$ case we get:

$a^{T-i}b^{i}=a^{T-i}a^{i}=a^{T}=b^{T}$

So in either case, we can substitute in to the binomial expansion to get:

$\displaystyle (a-b)^{T}=\sum_{i=0}^{T} (-1)^{i} {T \choose i} b^{T}$

This may be rewritten as:

$\displaystyle (a-b)^{T}= b^{T}\sum_{i=0}^{T} (-1)^{i} {T \choose i}=0$

Since the ring has no nonzero nilpotent, it follows that $a-b=0$ or $a=b$.  QED

I haven't been able to prove the noncommutative case.  In fact consider the ring consisting of the following:

$R=\{p_1 (b)+babp_2 (ab)+ap_3 (ba)+abp_4 (ab)+bap_5(ba)\}$

where $a^{2}=b^{2}$, $a^{3}=b^{3}$, $ab\ne ba$, and $p_{1}$, $p_{2}$, $p_{3}$, $p_{4}$, and $p_{5}$ are polynomials with integer coefficients.  I would conjecture that this is a counterexample in the noncommutative case.  The key, of course, would be to prove that this ring has only zero as a nilpotent--so far I haven't been able to prove it but would conjecture that this is true.

[quote="amfulger"][quote="NoYoo-hooForMe"]
I haven't been able to prove the noncommutative case.  In fact consider the ring consisting of the following:

$R=\{p_1 (b)+babp_2 (ab)+ap_3 (ba)+abp_4 (ab)+bap_5(ba)\}$

where $a^{2}=b^{2}$, $a^{3}=b^{3}$, $ab\ne ba$, and $p_{1}$, $p_{2}$, $p_{3}$, $p_{4}$, and $p_{5}$ are polynomials with integer coefficients.  I would conjecture that this is a counterexample in the noncommutative case.  The key, of course, would be to prove that this ring has only zero as a nilpotent--so far I haven't been able to prove it but would conjecture that this is true.[/quote]

Are you sure $R$ is a ring? Keep in mind non-commutativity.[/quote]

It's definitely a ring.  Basically $R$ is a set of 5-tuples of polynomials with integer coefficients:

$R=\{(p_1(x),p_2(x),p_3(x),p_4(x),p_5(x))\}$

Addition is just defined by adding corresponding elements.  Multiplication is defined by letting:

$a=(0,0,1,0,0)$

and

$b=(x,0,0,0,0)$

The above identity can be used to uniquely define multiplication.  I won't work through all the cases but, to give some examples:

$(0,0,1,0,0)(0,0,1,0,0)=a^{2}=b^{2}=(x^{2},0,0,0,0)$

$(x,0,0,0,0)(x,0,0,0,0)=b^{2}=(x^{2},0,0,0,0)$

$(0,0,0,p_{4} (x),0)(0,0,0,q_{4} (x),0)=abp_{4} (ab)abq_{4} (ab)=ababp_{4} (ab)q_{4}(ab)$ $=(0,0,0,xp_{4} (x)q_{4} (x),0)$

$(0,0,0,p_{4} (x),0)(0,0,0,0,q_{5} (x))=abp_{4} (ab)baq_{5} (ba)=$ $(x^{4} p_{4} (x^{2} )q_{5} (x^{2} ),0,0,0,0)$

I am quite confident that I can work through all the subcases and establish that $R$ is a ring.  I can also prove that in $R$, for all positive integers $T$, $(a-b)^{T} \neq 0$.  Hence $R$ is behaving substantially differently than in the commutative case.  If we consider the following variant of the original problem:

Assume that $R$ is a ring.  If for $a,b\in R$ we have $a^m=b^m$ and $a^n=b^n$ ,$(m,n)=1$ prove that $a-b$ is nilpotent in $R$.

Essentially I have proved that for $R$ commutative, the revised problem is true.  For the noncommutative case, I can work through and complete my proof that $R$ is a ring.  I can also show (this is nontrivial but when I get a little time I will post it) that $a-b$ is not nilpotent in $R$.  So the revised problem is not, in general, true for $R$ noncommutative.

What I do not claim to be able to prove so easily is that this potential counterexample to the original problem has no nilpotent elements except zero.  I conjecture that that is true but I do not think it is as easy to prove.

[quote="NoYoo-hooForMe"]Theorem:  $R$ is defined as a ring consisting of the set of all 5-tuples of polynomials with integral coefficients:

$R=\{(p_1(x),p_2(x),p_3(x),p_4(x),p_5(x))\}$

Addition is defined by adding corresponding elements.  Multiplication is defined by letting $a=(0,0,1,0,0)$, $b=(x,0,0,0,0)$ and then setting $a^{2}=b^{2}$,  $a^{3}=b^{3}$ and interpreting any given element of $R$ as follows:

$(p_1(x),p_2(x),p_3(x),p_4(x),p_5(x))=$ $p_1 (b)+babp_2 (ab)+ap_3 (ba)+abp_4 (ab)+bap_5(ba)$

Then $a-b$ is not nilpotent in $R$.

Proof:  ...[/quote]

The above theorem and proof is definitely correct, and the associated object $R$ is definitely a ring.  However, it seems I was mistaken in my earlier conjecture that this specific noncommutative ring $R$ would have no nonzero nilpotent elements.  In particular, we observe:

$(b^{3}-bab)^{2}=(b^{3}-bab)(b^{3}-bab)=b^{6}-b^{4}ab-bab^{4}+bab^{2}ab$
Hence $(b^{3}-bab)^{2}=b^{6}-b^{2}a^{3}b-ba^{3}b^{2}+ba^{4}b$
or $(b^{3}-bab)^{2}=b^{6}-b^{6}-b^{6}+b^{6}=0$

Thus $(b^{3}-bab)$, which is nonzero in the specific ring $R$ under consideration (since it corresponds to the 5-tuple $(x^{3},-1,0,0,0)$), is nilpotent in $R$.

However, this observation may be used to attack the noncommutative case of the original problem with $m=2$ and $n=3$.  In the original problem, it is a given that there are no nonzero nilpotents, so it follows that $b^{3}-bab=0$ or $b^{3}=bab$.  We next observe that:

$(b^{2}-ba)^{2}=(b^{2}-ba)(b^{2}-ba)=b^{4}-bab^{2}-b^{3}a+baba$
Hence $(b^{2}-ba)^{2}=b^{4}-b^{4}-b^{3}a+b^{3}a=0$

The requirement that there be no nonzero nilpotents then forces $b^{2}-ba=0$ or $b^{2}=ba$.  By symmetry it is easily shown that $a^{2}=b^{2}=ab=ba$.  From this it readily follows that $(b-a)^{2}=0$ and hence $b-a=0$.  This proves the noncommutative case of the original problem for $m=2$ and $n=3$.

I presume it is possible to generalize this argument for all $m$ and $n$ with $(m,n)=1$ but have not done so yet.

I believe I can generalize this argument for the general noncommutative case for all $m$, $n$ with $(m,n)=1$.  We first observe that since $(m,n)=1$,

$\forall i\ge (m-1)(n-1) \exists p,q\ge 0: pm+qn=i$

where $p$ and $q$ are nonnegative integers.  From this it follows easily that

$a^{i}=a^{pm+qn}=(a^{m})^{p}(a^{n})^{q}=(b^{m})^{p}(b^{n})^{q}=b^{i}$ (Eq 1)

for all $i\ge (m-1)(n-1)$.  Now we let $T=(m-1)(n-1)+1$.  We observe that:

$(b^{T}-bab^{T-2})^{2}=(b^{T}-bab^{T-2})(b^{T}-bab^{T-2})$
Hence $(b^{T}-bab^{T-2})^{2}=b^{2T}-bab^{2T-2}-b^{T+1}ab^{T-2}+bab^{T-1}ab^{T-2}$
or $(b^{T}-bab^{T-2})^{2}=b^{2T}-ba^{2T-1}-a^{T+2}b^{T-2}+ba^{T+1}b^{T-2}$
or $(b^{T}-bab^{T-2})^{2}=b^{2T}-b^{2T}-b^{2T}+b^{2T}=0$

by repeated application of Eq 1.  Since $R$ has no nonzero nilpotents, it follows that $b^{T}-bab^{T-2}=0$ or $b^{T}=bab^{T-2}$.  We next show that for $2\le t\le T$:

$b^{t} =bab^{t-2}$ (Eq 2)

We proceed by mathematical induction.  We have already shown the basis case $t=T$.  We assume that Eq 2 holds for $t=k$ and consider the case where $t=k-1$.  We observe that:

$(b^{k-1}-bab^{k-3})^{2}=(b^{k-1}-bab^{k-3})(b^{k-1}-bab^{k-3})$
Hence $(b^{k-1}-bab^{k-3})^{2}=b^{2k-2}-bab^{2k-4}-b^{k}ab^{k-3}+bab^{k-2}ab^{k-3}$
or $(b^{k-1}-bab^{k-3})^{2}=b^{2k-2}-(bab^{k-2})b^{k-2}-b^{k}ab^{k-3}+b^{k}ab^{k-3}$
or $(b^{k-1}-bab^{k-3})^{2}=b^{2k-2}-b^{2k-2}=0$

by repeated application of Eq 2 for $t=k$.  Since $R$ has no nonzero nilpotents, it follows that $b^{k-1}-bab^{k-3}=0$ or $b^{k-1}=bab^{k-3}$.  Hence Eq 2 is now established for $t=k-1$, and proceeding by mathematical induction, it follows that Eq 2 holds for $t=2$.  Thus $b^{2}=ba$ and by arguments of symmetry it can easily be shown that $b^{2}=a^{2}=ba=ab$.  From this it follows that $(b-a)^{2}=0$ and thus $b-a=0$ or $b=a$.  QED

I believe this now completes the proof for the original problem for all cases.


At a party, a guest asks the age of the hosts 3 children. The host tells the guest the following information:

Assume each kid is a counting number age;
The product of their ages is 72
The sum of their ages is the same as my house number.

The guest goes outside and looks at the house number. Upon returning, the guest asks for additional information. The host then says "Oops, I forgot to tell you that the oldest likes strawberry ice cream."

You now have all the information to determine the ages of the kids. How old are they, and why?

so i read a problem like this before.

first factor them up.
72-8,9

so
age age age sum
1 1 72 74
1 2 36 39
1 3 24 28
1 4 18 23
1 6 12 19
1 8 9   18
2 2 18 22
2 3 12 17
2 4 9   15
2 6 6   14
3 3 8   14

then he can't tell if they are 2,6,6 or 3,3,8 so he gets told theres an oldest.  338 has an oldest but not 266 so there  3,3,8 years old


I am a bit confused about the official version of Holder's and Minkowski's Inequalities.

Here is Mathworld's entry for Minkowski:
http://mathworld.wolfram.com/MinkowskisInequalities.html

Here is Planetmath's entry for Minkowski:
http://planetmath.org/encyclopedia/MikowskiInequality.html

Here is Planetmath's entry for Holder:
http://planetmath.org/encyclopedia/HolderInequality.html

However, Kedlaya gives a more generalized version of these two inequalities:
http://caltizzle.caltech.edu/explications/ineq/node14.html

Which version is correct? If I state the more generalized version of either on the USAMO, will the judges take points off?

Thanks.

Mathworld and planetmath have the same inequality for Minkowski.  I can't imagine ever using the generalized versions of either of those inequalities, but you could always just reference the inequalities packet and you'd be fine.  ("By the generalized Minkowski inequality, as found in Kiran Kedlaya's inequality packet, ...")

[quote]("By the generalized Minkowski inequality, as found in Kiran Kedlaya's inequality packet, ...")[/quote]

Oh, students can do that? That is, students can cite theorems from obscure books/magazines/publications?

Yeah, I know these two inequalities aren't used very much on olympiads, but I wanted to know just in case. It took me a while to understand their proofs, and I don't really think that I'd be comfortable writing them up again on an olympiad.

Well, maybe you can't quote obscure books/magazines/publications, but Kiran's Inequalities book is not exactly obscure. At the very least, AMC should recognize a publication that is available for download on their own website.

[quote="Fierytycoon"]Oh, students can do that? That is, students can cite theorems from obscure books/magazines/publications?[/quote]

Absolutely, although generally they discourage you from citing results which trivialize a problem or of which a problem is a special case.

Although I guess Simon is right and you probably should avoid quoting things that are really obscure :).  Anyhow, I doubt you'll have an opportunity to do so, so it's kind of a moot point.

[quote="tokenadult"]Not related to this inequality, isn't there a story about Reid Barton citing some obscure theorem for a one-line proof at the Hong Kong IMO?[/quote]

There have been many similar occasions at the IMO, but remember that at the IMO, the team leaders get to fight for points with the judges and argue their students' cases or provide clarification for anything written, something that does not happen on the USAMO.

I cannot believe that if you indeed cite a correct theorem which trivializes the problem you will not get points. The problem asks you for a proof and you provide it.

[quote="JBL"] ("By the generalized Minkowski inequality, as found in Kiran Kedlaya's inequality packet, ...")[/quote]

Given that Kiran Kedlaya is one of the graders, that would probably work  :D

I have a question... how often can you get away with just saying "by a well-known theorem"?  I did that for #1 this year and I think I got the credit....but that was something simple (a+c=b+d for circumscribed quadrilaterals).

Also, if I cite a theorem, do I have to state its full form and then apply it?  Or can I just say "By ....., [special case needed for the problem]"?

Thank you!

        -OR

Reid Barton was not at the Hong Kong IMO: he was a fifth grader at the time and not quite yet up for making the US IMO team.  However, Walter Mientka did say at MOP last year that something of the kind did happen at the Hong Kong IMO.

Maybe someone should make a math competition urban legends page...

[quote="tokenadult"]Not related to this inequality, isn't there a story about Reid Barton citing some obscure theorem for a one-line proof at the Hong Kong IMO?[/quote]

This particular legend is usually attached to Lenny Ng; I can't attest to its accuracy, but it wouldn't surprise me.


Seven people are sitting in a theater watching a show. The row they are in contains seven seats. After intermission, they return to the same row but choose seats randomly. What is the probability that neither of the people sitting in the two aisle seats were previously sitting in an aisle seat?

Umm...[hide]

There are 4 combinations of the aisle seats:

The 2 guys who had the aisle seats last time gets them again

(AA)

The left seat is picked up by a person who had an aisle seat before but the right isn't (or vice versa)

(AI)/(IA)

Neither aisle seat is held by someone who was in one before.

(II)

The odds of AA are 1/21: the first guy has 2/7 chance of getting an aisle seat, and the second guy has a 1/6 chance if the first guy makes it.

By similar reasoning, the odds of AI and IA are 5/21, and their sum is 11/21. The odds of II are 21/21-11/21=10/21.[/hide]

Different method ...

[hide]5 out of 7 people are not in aisle seats originally. So the probability that one of these 5 is in the left-hand aisle seat after the intermission is 5/7.

Given that one of these 5 non-aisle seaters is now in the left-hand aisle seat, then 4 of the remaining 6 people were originally not in aisle seats. So the probability that one of these 4 is now in the right-hand aisle seat is 4/6=2/3.

Joint probability is 5/7x2/3 = 10/21.[/hide]

... same answer.

Different way to solve it: [hide]there are 7 people and two aisle seats,  2  of the 7 people have to sit in the aisle seats, there are  7C2 ways. Out of these, 2 of the 5 people that were not sitting in the aisle seats before have to sit in the aisle seats, there are 5C2 ways. 5C2/7C2 =10/21[/hide]


I would have posted a link to a proof now (it is on ML) but your behaviour of bumping up topics after a few hours annoys me. Find it on your own.

  darij

[quote="mathboy_cntHDVN"]for the positive real number $a;b;c$ such that $a\geq b \geq c$
prove that : $x(a-b)(a-c)+y(b-a)(b-c)+z(c-a)(c-b)\geq 0$ if $ax\geq by$ or $cz\geq by$[/quote]
It's here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66660

Fpr n number ; this inequality is true ?


What is the positive difference between the $ 2000^{\mathrm{th}}$ term and the $ 2005^{\mathrm{th}}$ term of the arithmetic sequence -8, -2, 4, 10, $ \ldots$?

We could actually find the 2000th and 2005th term and find the difference, but there's a much faster way. Note that the common difference is 6, and the 2000th and 2005th term are five terms apart, so our answer is $ 6 \times 5 \equal{} \boxed{30}$.

I put in 30, but it said I was wrong! I don't know why because your solution seems right.


geometry
trigonometry
analytic geometry
conics
ellipse
angle bisector
geometry solved

Given angle A < 180  and segment  length d.  Find 3 points:
      K and M on sides of the angle  and  L inside the angle  so that
      KL + LM  = d   and area AKLM  is max.

      Col. Pestich.

Is that area $\frac{d^2}{8}$?

BTW: if this value is true, then don't you think, that angle A is not necessary? I just've worked with maximalization of area of KLM and it's clearly only when $|KL|=|LM|$ and KL, LM being perpendicular, or not?

Generalissimo Franco broadcasted (along with the statement that the sky is cloudless
   over Spain)   that he was sure that the
   max area could not be independent of the angle measure.
   But the question was to find locations of the 3 points, not to
   estimate the area.  You have to provide at least some
   justification of the area being the maximum.

   Col. Pestich.

Denote $\alpha = \angle A$. Assume first that the angle $\alpha \le 90^o$ is acute (or right). The locus of points $L$ with a constant sum $d  = LK + LM$ of distances from the lines $AK, AM$ is the segment $BC$ of a line perpendicular to the bisector of the angle $\alpha$ and intersecting the lines $AK, AM$ at points $B, C$, such that $AB = AC = \frac{d}{\sin\alpha}$. Let $D$ be the midpoint of the segment $BC$, i.e., $AD$ is a bisector of the angle $\alpha$. The lines $LK, BC$ are perpendicular to the lines $AB, AD$, respectively, and they form the same angles. Hence, $\angle BLK = \angle BAD = \frac \alpha 2$ and similarly, $\angle CLM = \frac \alpha 2$.

$LK = LB \cos{\frac \alpha 2}$, $LM = LC \cos{\frac \alpha 2}$

$LK + LM = (LB + LC) \cos{\frac \alpha 2} = BC \cos{\frac \alpha 2} = AB \sin\alpha = d$

regardless of the position of the point $L$ on the segment $BC$. The area $S(AKLM)$ of the quadrilateral  $AKLM$ is clearly maximum, when the point $L$ coincides with the midpoint $D$ of the segment $BC$ and its value is then

$S(AKDM) = S(\triangle ABC) - [S(\triangle BKD) + S(\triangle CMD)] = S(\triangle ABC) - 2 S(\triangle BKD)$

$S(\triangle ABC) = \frac{AB^2}{2} \sin\alpha = \frac{d^2}{2 \sin \alpha}$

$S(\triangle BKD) = \frac{DK\cdot DB}{2} \sin{\frac \alpha 2} = \frac{DK^2}{2} \frac{\sin{\frac \alpha 2}}{\cos{\frac \alpha 2}} = \frac{d^2}{8} \tan{\frac \alpha 2}$

$S(AKDM) = \frac{d^2}{2 \sin\alpha} - \frac{d^2}{4} \tan{\frac \alpha 2} = \frac{d^2}{4} \frac{1 - 2 \sin^2{\frac \alpha 2}}{2 \sin{\frac \alpha 2} \cos{\frac \alpha 2}} =$

$= \frac{d^2}{4} \frac{\cos^2{\frac \alpha 2} - \sin^2{\frac \alpha 2}}{2 \sin{\frac \alpha 2} \cos{\frac \alpha 2}} = \frac{d^2}{4} \frac{\cos\alpha}{\sin\alpha} = \frac{d^2}{4 \tan\alpha}$

Calculating the area $S(AKLM)$ for an arbitrary position of the point $L$ on the segment $BC$:

$S(\triangle BKL) + S(\triangle CML) = \frac 1 2 (LK \cdot LB + LM \cdot LC) \sin{\frac \alpha 2}  = \frac 1 2 (LB^2 + LC^2) \sin{\frac \alpha 2} \cos{\frac \alpha 2} = \frac 1 4 [LB^2 + (BC - LB)^2] \sin\alpha = \frac 1 4 [x^2 + (a - x)^2] \sin\alpha$

where $a = BC$ and $x = BL$. For a maximum area $S(AKLM)$, $\frac{dS(AKLM)}{dx} = 0$:

$\frac{dS(AKLM)}{dx} = \frac{d}{dx} \left[\frac{d^2}{2 \sin\alpha} - \frac{x^2 + (a - x)^2}{4} \sin \alpha \right] = -\frac{2x - 2(a - x)}{4} \sin \alpha = \frac{2x - a}{2} \sin\alpha = 0$

i.e., $x = \frac a 2$ or $BL = \frac{BC}{2}$, as was claimed before. The case, when the angle $\alpha > 90^o$ is obtuse is handled in the same way, the only difference being that the locus of the points $L$ is not the entire segment $BC$, but only the segment $B'C'$ cut on the segment $BC$ symmetrically with respect to the midpoint $D$ by a right angle with the same vertex $A$ and the same angle bisector as the angle $\angle A$.

Small trouble is that LK and LM are not distances to the angle sides from
   point L, all we know is len(KL) + len(LM) = d, and K and M are on
   the sides.

All right, let me think about it. BTW, say hello to your friends.

For a given value x of MK in (0,d), the maximum value of area(AMK) is x^2*cot(A/2)/4 and the maximum value of area(LMK) is x*root(d^2-x^2)/4; now f(x)=x^2*cot(A/2)/4+x*root(d^2-x^2)/4 is maximum for x0 = d cos(A/4) and f(x0) = d^2*cot(A/4)/8. Equivalently AM = AK = AL = d/4/sin(A/4).
Kind regards

JPE does not write much, but very nice and to the point.
  Thanks.

  Maj.  Pestich

  BTW, hello to my friends.

Denote $\alpha = \angle KAM, \gamma = \angle KLM, a = AK, b = AM, c = KL$. The given distance is $d = KL + LM$. The area of the quadrilateral $AKLM$ is

$S(a, b, c, \gamma) = (ab\ sin\alpha + c(c-d)\ cos\gamma)/2$

We have to find the maximum area subject to the constraint

$KM^2 = a^2 + b^2 - 2ab \cos \alpha = c^2 + (d - c)^2 - 2c(d - c) \cos\gamma$
$G(a, b, c, \gamma) = a^2 + b^2 - 2ab \cos\alpha - [c^2 + (d - c)^2 - 2c(d - c) \cos\gamma]$

Using Lagrange muliplier $\lambda$,

$F(a, b, c, \gamma) = S(a, b, c, \gamma) + \lambda G(a, b, c, \gamma)$

For the maximum area,

$\frac{\partial F}{\partial a} = \frac{\partial F}{\partial b} = \frac{\partial F}{\partial c} = \frac{\partial F}{\partial \gamma} = 0$ and $G(a, b, c, \gamma) = 0$

$\frac{\partial F}{\partial a} = \frac b 2 \sin\alpha + 2\lambda (a - b \cos\alpha) = 0$

$\frac{\partial F}{\partial b} = \frac a 2 \sin\alpha + 2\lambda (b - a \cos\alpha) = 0$

$\frac{\partial F}{\partial c} = \frac{d - 2c}2 \sin\gamma + 2\lambda  [d - 2c + (d - 2c) \cos\gamma] = 0$

$\frac{\partial F}{\partial b} = \frac{c(d - c)}2 \cos\gamma - 2\lambda \frac{c(c - d)}2 \sin\gamma = 0$

From the last equation (assuming $c > 0$ and $d - c > 0$), $\lambda = \frac{1}{4 \tan\gamma}$. Substituting this $\lambda$ to the 3rd equation,

$\frac{d - 2c}2 (\sin\gamma + \frac{1 + \cos\gamma}{\tan\gamma}) = 0$
${\frac{d - 2c}2 \frac\sin^2{\gamma} + \cos^2{\gamma} + \cos\gamma}{\sin\gamma} = \frac{d - 2c}2 \frac{1 + \cos\gamma}{\sin\gamma} = 0$

Assuming $\gamma \# \pi, \cos\gamma \# -1$ and we get $c = d/2$. Substituting $\lambda$ to the first 2 equations,

$b \sin\alpha + \frac{a - b \cos\alpha}{\tan\gamma} = 0$

$a \sin\alpha + \frac{b - a \cos\alpha}{\tan\gamma} = 0$

Adding these 2 equations, reducing by $a + b  \# 0$ and rearranging,

$tan\gamma = -\frac{1 - \cos\alpha}{\sin\alpha} = -\frac{2 \sin^2{\frac\alpha 2}}{2 \sin{\frac\alpha 2} \cos{\frac\alpha 2}} = -\tan{\frac\alpha 2}$

Hence, $\gamma = \pi - \frac\alpha 2$. Substituting this to either of the above 2 equations leads to $a = b$:

$a \sin\alpha - \frac{b - a \cos\alpha}{\tan{\frac\alpha 2}} = 0$

$2a \sin{\frac\alpha 2} \cos{\frac\alpha 2} \tan{\frac\alpha 2} - (b - a \cos\alpha) = 0$

$a (1 - \cos\alpha) - (b - a \cos\alpha) = 0$
$a = b$

Together with $a = b,\ c = d/2$, the relation  $\gamma = \pi - \frac\alpha 2$ implies that the triangles $\triangle AKL, \triangle AML$ are both isosceles and $AL = AK = AM$. Consequently,

$a = b = \frac{d}{4 \sin\frac\alpha 4} = \frac{d}{2\sqrt{2(1 - \cos{\frac\alpha 2}})}$

${S(AKLM) = a^2 sin\frac\alpha 2 = \frac{d^2}{8} \frac{\sin{\frac\alpha 2}}{1 - \cos{\frac\alpha 2}}}$

But what about using just geometrical considerations, like an average
   school student would have done then ( or how would it be done before
   the French contaminated geometry with coordinate system), starting with pretty
   obvious conclusion that point L is on the angle bisector?

   Maj. Pestich

Well, I did not use any coordinate system anywhere. I started with "pretty obvious" conclusion that the angles $\angle AKL, \angle AML$ are both right, i.e., that the angle $\angle KLM = \pi - \angle KAM$, but the Lagrange multiplier was giving me $\angle KLM = \pi - \frac{\angle KAM}{2}$. I could not believe it and kept searching for an error. After that, nothing was obvious to me.

yetti

Step 1: prove KL=ML=d/2.
If M and K are fixed, then so is [KAM], so we need to maximize [KLM]. The locus of points L so that KL+LM is fixed is an ellipse with foci at K and M, and in order to maximize [KLM] we need to maximize the distance from L to KM. This distance is easily seen to be maximized when L is at the vertex of the ellipse, i.e. KL=ML.

Step 2: prove KA=MA.
If we consider triangle KLM fixed, we need to maximize the area of KAM. The locus of points A so that <KAM is fixed is an arc of a circle. The area of KAM is maximized when the distance from A to KM is maximized, i.e. when A is at the midpoint of that arc. When this happens, MA=KA.

Step 3: Show KAL is isosceles.
Since everything is symmetric with respect to the angle bisector of <A, we need to maximize the area of triangle KAL. If we consider K and L fixed, then <KAL=<A/2 is also fixed. The locus of positions of A is thus an arc of a circle, and the area of triangle KAL is maximized when A is at the midpoint of the arc.

So the maximum area occurs when AK=AL=AM with L on the angle bisector.