[quote="amir2"]Prove that if $a, b, c, d$are positive integers such that $ad = bc$, then $a+b+ c + d$cannot be a prime number.[/quote]
it is equal to prove that it is a composite number.(AKHE VASE CHI SOALO MIPICHOONI?) :D

Suppose $a+b+c+d=p$ is prime. Then $a(p-a-b-c)=bc$, so $ap=bc+ab+ac+a^2=(b+a)(c+a)$. So $p \mid b+a$ or $p \mid c+a$. Since $p>0$ this implies $p \leq b+a$ or $p \leq c+a$. But $p=a+b+c+d>a+b, a+c$, contradiction.

$ad=bc \Rightarrow \frac{a}{b}=\frac{c}{d}$ So we can imagine $a>c,b>d \Rightarrow a=ck,b=dk$ and $k \in N$.Now we have $a+b+c+d=ck+dk+c+d=(c+d)(k+1)$ (We know that $k+1>1$).By this way of proving we can say  $ad=bc \Rightarrow$ there exists $p$ that divides $a^n+b^n+c^n+d^n$.


inequalities
real analysis
probability and stats

$ \{X_n\}$ is a sequence of random variables, $ \{\mathcal{F}_n\}$ is a sequence of $ \sigma$-algebra. If $ \mathbb{E}(|X_n|\Big|\mathcal{F}_n)$ converges to 0 in $ \mathbb{P}$, then prove $ \{X_n\}$ converges to 0 in $ \mathbb{P}$.

Thanks

Hi,
as I didn't found the answer I asked a friend of mine on another forum and he gave me the tricks that solve the issue.

Here we are
First suppose $ X_n$ are positive and let's take $ g(x)\equal{}min(\epsilon, x)$ wich is concave
So applying Jensen we get
$ E[g(X_n)|F_n]\leq g(E[X_n|F_n])\equal{}g(Y_n)$ (posing $ Y_n\equal{}E[X_n|F_n]$)
so we get by taking expectation on both sides:
$ E[g(X_n)]\leq E[g(Y_n)]$
Now we have
$ P(X_n>\epsilon)<1/\epsilon E[g(X_n)]\leq 1/\epsilon E[g(Y_n)]$
Let's decompose the $ E[g(Y_n)]$
we have for any integer $ p>0$

$ E[g(Y_n)]\leq E[g(Y_n)1_{Y_n>\epsilon/p}]\plus{}  E[g(Y_n)1_{Y_n\leq\epsilon/p}]$
then it comes :
$ E[g(Y_n)1_{Y_n>\epsilon/p}]\leq  \epsilon P(Y_n>\epsilon/p)$ and the second term

$ E[g(Y_n)1_{Y_n\leq\epsilon/p}]\leq \epsilon /p$

reintagrating both terms we have then :
$ P(X_n>\epsilon)\leq E[g(Y_n)]/\epsilon \leq P(Y_n>\epsilon/p)\plus{}1/p$

To finish letting $ n\rightarrow \plus{}\infty$, we get  $ limsup P(X_n>\epsilon) \leq 1/p$
letting now go to infinity we get
$ limsup P(X_n>\epsilon) \equal{}0$
so $ Lim P(X_n\geq \epsilon)\equal{}0$

For the general case you just need to decompose $ X_n$ into positive and negative part and treat them as done earlier.

see ya


Inside a square of sidelength $ 1$ there are finitely many disks that are allowed to overlap. The sum of all circumferences is $ 10$. Show that there is a line intersecting or touching at least $ 4$ disks.

Did I miss something or is it quite easy?

Each circle, with diameter $ d_i$, is projected onto a given side of the square as a segment of length $ d_i$. Thus, the sum of these projections is $ \sum d_i \equal{} \frac {10}{\pi} > 3$.
It follows that some point on the side of the square, say $ M$, is covered at least $ 4$ times by these projections. Therefore, the line through $ M$ and perpendicular to the side intersects at least $ 4$ of the circles.

Pierre.

You missed nothing. But the time it remained unsolved shows that a lot of people have problems with this one (in the contest itself, it had a surprisingly low rate of solutions). Maybe nobody thinks of the trivial things...

The funny thing is that this same problem and solution is in Engel's book [i]Problem-Solving Strategies[/i] (Chapter 4, Problem 62).  I figured that students in Germany would have seen this book, since Engel is from Germany.


integration
function
inequalities
geometry
rectangle
calculus
real analysis

For $n\geq 1$ we define the following sequences:

$u_{n}=\sum_{k=0}^{n-1}\sqrt{1+\frac{k}{n}}$  and   $v_{n}=\sum_{k=1}^{n}\sqrt{1+\frac{k}{n}}$
Demonstrate that for all $n\geq 1$ and $k\epsilon\{0,...,n-1\}$
$\frac{1}{n}\sqrt{1+\frac{k}{n}}\leq\int_{\frac{k}{n}}^{\frac{k+1}{n}}\sqrt{1+x}dx\leq\frac{1}{n}\sqrt{1+\frac{k+1}{n}}$ is true,
next demonstrate that
$u_{n}\leq\int_{0}^{1}\sqrt{1+x}dx\leq v_{n}$
Give me at least an idea, please.

I believe you've left a $\frac1n$ out of the definition of $u_{n}$ and of $v_{n}.$

This is a Riemann sum argument, and Riemann sums are inherently visual. Note that the function $f(x)=\sqrt{1+x}$ is monotone increasing; you need that for the inequality arguments.

Draw pictures. Draw pictures that compare the area under a curve to the area of a rectangular box (or of a sum of such boxes.)

[quote="Kent Merryfield"]I believe you've left a $\frac{1}{n}$ out of the definition of $u_{n}$ and of $v_{n}.$

This is a Riemann sum argument, and Riemann sums are inherently visual. Note that the function $f(x)=\sqrt{1+x}$ is monotone increasing; you need that for the inequality arguments.

Draw pictures. Draw pictures that compare the area under a curve to the area of a rectangular box (or of a sum of such boxes.)[/quote]
Thanks for the hints!
No, I didn't leave
 $\frac{1}{n}$
out of the definition of $v_{n}$ and $u_{n}$. I've double checked.

There's an $n$ somewhere. Since $\sqrt{1+t}>1$ for $t>0,$ we have that $u_{n}>n$ and $v_{n}>n,$ while $\int_{0}^{1}\sqrt{1+x}\,dx<2.$

Is perhaps then the inequality

$u_{n}\le n\int_{0}^{1}\sqrt{1+x}\,dx\le v_{n}?$

I'm trying to do the index shift: would this be correct
$v_{n}=\sum_{k=1}^{n}\sqrt{1+\frac{k}{n}}=\sum_{k=0}^{n-1}\sqrt{1+\frac{k+1}{n+1}}$  OR more likely equal to this
...$=\sum_{k=0}^{n-1}\sqrt{1+\frac{k+1}{n}}$  ?

[quote="Kent Merryfield"]If an index shift is confusing you, don't try to replace $k$ with $k.$ Use a different letter - say, $j.$ $k=j+1$ or $k=j-1$ or whatever you need. Follow that, and it should make sense.

But, overall, don't let yourself be a prisoner of notation, especially sigma notation. Write out the sums: first few terms, last term or last couple of terms, ... in the middle. Don't see things as symbols, see them as numbers; better yet, see them as rectangles of a certain area.[/quote]
Yes, at the beggining it is confusing. Fortunatelly, as soon as I get the answer to a particular question( illustrating, let's say, the method of index shifting at this point or any actual problem of the moment on my mind), the same question(or type of problems) won't ever make me confused again in the future.
So,please :)
if I want to have k=0 instead of k=1 how this series would look like
$v_{n}=\sum_{k=1}^{n}\sqrt{1+\frac{k}{n}}=???$

All this time you're spending fussing around with the formalities of index shifting is time spent not solving the problem. Back away, take a breath, and look at what this is about.

A basic property of integrals: if $m\le f(x)\le M$ for all $x\in [a,b]$, then $m(b-a)\le \int_{a}^{b}f(x)dx\le M(b-a)$. Use this property with $f(x)=\sqrt{1+x}$, $a=k/n$, and $b=(k+1)/n$. You will find $m$ and $M$ by observing that $f$ is increasing.

Thank you all for the hints :)
P.S.: Dear M.Merryfield, yes, I'll be having "that shifting" in the background of my mind :) up to the point where somebody actually shows me how my series with the shifted index would look like :lol: (I need it in order to solve the second half-which I haven't posted here- of the problem). On the other hand, at the same time I'm familiarising myself with Herr B.Riemann's ideas on the integrals .


geometry
circumcircle
inequalities
inequalities proposed

2)

Let M the midpoint of BC and t the line tangent to the circuncircle of ABC trought A.
Let X the foot of the perpendicular, trought M, to t. Let B' and C' the foots of perpendiculars, trought B and C, to t.
We have $m_{a}\geq MX = \frac{(BB'+CC')}{2}= \frac{b.sinB+c.sinC}{2}$.

Suming the analogous inequalities, we find $m_{a}+m_{b}+n_{c}\geq a.sinA+b.sinB+c.sinC = \frac{a^{2}+b^{2}+c^{2}}{2R}$

Nice geometric solution there are two other ways to prove $b^{2}+c^{2}\le 4Rm_{a}$ is $GA\cdot OA\ge\vec{OA}\cdot\vec{GA}$ or $AA'\le 2R$ with $A'$ is intersection of median and circumcircle. :)


number theory
greatest common divisor
relatively prime

#1 If k=[2004^4+4(2004^3)x2+6(2004^2)(2^2)+4(2004)(2^3)+2^4]/[2008^3-3(2008^2)x2+3x2008(2^2)-2^3]
Compute k.
#2 In Pasculand, there is only 5 dollars and 11dollars bills. Compute the largest sum of money that cannot be made using these bills. (Assume that there exists an endless supply of each ty of bill)

[quote="lotrgreengrapes7926"][quote="superwilson"]#1 If $k=\frac{2004^{4}+4(2004^{3})\cdot2+6(2004^{2})(2^{2})+4(2004)(2^{3})+2^{4}}{2008^{3}-3(2008^{2})\cdot2+3\cdot2008(2^{2})-2^{3}}$
Compute k.
#2 In Pasculand, there is only 5 dollars and 11 dollars bills. Compute the largest sum of money that cannot be made using these bills. (Assume that there exists an endless supply of each type of bill)[/quote][hide="Hint"] Factor it. $(x+y)^{4}=x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^{4}$ and $(x-y)^{3}=x^{3}-3x^{2}y+3xy^{2}-y^{3}$[/hide]
[hide="#2"] Using Chicken McNugget, since $(5,11)=1$ the answer is $5\cdot11-5-11=39$ [/hide][/quote]

Please explain the logic behind this operation
Please explain wat wd happen in a case when the GCD is not 1...like if we have bills of 4 and 8 dollars...then th largest sum that cant be created is???????????

[quote="Varun Nakra"]Please explain the logic behind this operation
Please explain wat wd happen in a case when the GCD is not 1...like if we have bills of 4 and 8 dollars...then th largest sum that cant be created is???????????[/quote]
[hide] The first gives $\frac{(2004+2)^{4}}{(2008-2)^{3}}$ :wink:
In your example, 4 and 8, any number that is not divisible by 4 cannot be made. For example, 40000000000001 or 40000000000002 or 40000000000003. :D [/hide]

[quote="mysmartmouth"]Conjecture:
If the GCD of $x$ and $y$ is not $1$, you can only form amounts that are multiples of the GCF of $x$ and $y$.[/quote]
It's quite obvious :D . Just put $x= x'\cdot \gcd(x,y)$, $y=y'\cdot \gcd(x,y)$. Then you can for only numbers like $ax+by= \gcd(x,y)\cdot\left(ax'+by'\right)$.
It can be proved that if $\gcd(x,y)=1$, then$\left\{ax+by: a,b\in\mathbb{Z}\right\}= \mathbb{Z}$.

Note:
McDonalds sold chicken McNuggets in a variety of different (relatively prime) sizes back in the day.  People were trying to find out the largest number of Chicken McNuggets one couldn't buy, and thus this name was created.

[quote="Rzeszut"][quote="Tim_Lou"][quote="lotrgreengrapes7926"]Using Chicken McNugget, since $(5,11)=1$ the answer is $5\cdot11-5-11=39$[/quote]

what is this "Chicken McNugget"? is that the name of a theorem or is it just Chicken McNugget?[/quote]
I don't know what's Chcken McNugget, but it's a well - known fact that if $\gcd(x,y)=1$, then any number $n> xy-x-y$ can be written as $ax+by$, where $a,b\geqslant 0$.[/quote]

Is there any place where I can get a formal proof for this result?

[quote="smohanram"][/quote][quote="Rzeszut"][quote="Tim_Lou"][quote="lotrgreengrapes7926"]Using Chicken McNugget, since $(5,11)=1$ the answer is $5\cdot11-5-11=39$[/quote]

what is this "Chicken McNugget"? is that the name of a theorem or is it just Chicken McNugget?[/quote]
I don't know what's Chcken McNugget, but it's a well - known fact that if $\gcd(x,y)=1$, then any number $n> xy-x-y$ can be written as $ax+by$, where $a,b\geqslant 0$.[/quote][quote="smohanram"]

Is there any place where I can get a formal proof for this result?[/quote]
See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=105458[/url].


Let $ a,b,c>0$ with $ abc\equal{}1$. Prove that

$ \sum\frac{a^7\plus{}c}{a^7c(b\plus{}c)}\geq\frac{3}{2}$

i think there is something wrong with your inequality because only this term
$ \sum_{cycl}\frac{1}{a^{7}(b\plus{}c)}\geq3/2$
 :blush:

[quote="saif"]i think there is something wrong with your inequality because only this term
$ \sum_{cycl}\frac {1}{a^{7}(b \plus{} c)}\geq3/2$
 :blush:[/quote]

SAIF you are sure?

yes sure because if we put $ x\equal{}\frac{1}{a}$,$ y\equal{}\frac{1}{b}$,$ z\equal{}\frac{1}{c}$
$ \sum_{cycl}\frac{1}{a^{7}(b\plus{}c)}\equal{}\sum_{cycl}\frac{x^{6}}{y\plus{}z}$
using chebychev inequality we can easily get the result.

[quote="Ligouras"]Let $ a,b,c > 0$ with $ abc \equal{} 1$. Prove that

$ \sum\frac {a^7 \plus{} c}{a^7c(b \plus{} c)}\geq\frac {3}{2}$[/quote]

Dear Friend ARQADY that you say? is it wrong?


if this equation has natural solutions , determine all natural n?
$ x^3 \plus{} y^3 \plus{} z^3 \equal{} nx^2y^2z^2$

Maybe $ \mathbb N$ is the set of positive integers.  :huh:
[quote]Find all positive integer $ n$ such that there exits positive integers x,y,z, satisfying $ x^3 \plus{} y^3 \plus{} z^3 \equal{} nx^2y^2z^2$  (*)[/quote]
[hide="Answer"]1 or 3[/hide]
[hide="Solution."]WLOG, assume that $ x\ge y\ge z$.
$ 3x^3\ge x^3 \plus{} y^3 \plus{} z^3 \equal{} nx^2y^2z^2\Rightarrow 3x\ge ny^2z^2$.
In (*), see that $ x^2|(y^3 \plus{} z^3)$, so $ 2y^3\ge y^3 \plus{} z^3\ge x^2$
$ \Rightarrow 18y^3\ge 9x^2\ge n^2y^4z^4\Rightarrow 18\ge n^2yz^4$.
Then, $ n\le 4$.
For $ n \equal{} 4$, $ y \equal{} z \equal{} 1$, but the equation $ x^3 \plus{} 2 \equal{} 4x^2$ don't have solution in $ \mathbb N$.
For $ n \equal{} 3$, $ x \equal{} y \equal{} z \equal{} 1$ satisfy (*).
For $ n \equal{} 2$, $ z \equal{} 1$, $ y\in \{1,2,3,4\}$, but not found $ x\in \mathbb N$.
For $ n \equal{} 1$, $ (x,y,z) \equal{} (1,2,3)$ satisfy (*).
Hence all possible n are 1;3.[/hide]

Hello

Do u have Mongolian Matematical olimpiad's problems of previous years?

If u have it please post it here

Thanking u


For WOOT classes, are they the same every year?  Because I wouldn't want to pay for them this year and then take the same classes again if I make it to Red MOP next year (in which case the class will be free).

Hm...I'm not sure if the lectures or the handouts are the same, but the practice AIMEs and Olympiads are different each year, I think...

The classes and handouts and tests from year n will not be used in year n+1.  A very small portion of them may be used in year n+2, and definitely some of the class material will be used in year n+3, but not in year n+1.


Online Math Open
algebra
polynomial
Vieta

1)Ν.δ.ο. δεν υπάρχουν θετικοί ακέραιοι $ x$, $ y$, $ z$, ώστε ο αριθμός $ 4xyz \minus{} x \minus{} y$ να είναι τέλειο τετράγωνο.

2)Αν $ a,b,c$ είναι θετικοί ακέραιοι με $ 0 < a^2 \plus{} b^2 \minus{} abc < \equal{} c$, ν.δ.ο. ο αριθμός $ a^2 \plus{} b^2 \minus{} abc$ είναι τέλειο τετράγωνο.

    [hide]Το 1 είναι γενίκευση της γνωστής άσκησης: Αν $ x,y$ θετικοί ακέραιοι ν.δ.ο. ο αριθμός $ 4xy \minus{} x \minus{} y$ δεν είναι τέλειο τετράγωνο.
                Το 2 είναι γενίκευση του προβλήματος 6 της ΙΜΟ του 1988. Let α and b be two positive integers such that $ ab \plus{} 1$  divides $ a^2 \plus{} b^2$ . Show that ($ a^2 \plus{} b^2$)/$ (ab \plus{} 1)$ is a perfect square. [/hide]

GEIA S FILE M PANO!!!VAZO PROS T PARON TN LYSI M G T DEYTERO...

$ a^2 \plus{}b^2\minus{}abc\equal{}d$  TOTE $ a^2 \minus{}abc\plus{}b^2 \minus{}d\equal{}0$(1)     me c$ \geq$d>0 ..profanos an $ d\minus{}b^2\equal{}0$ teleiosame..logo summetrias os pros a,b upotheto w.l.o.g. pos $ a\geq$$ b$ kai pairnoume tora 2 periptoseis:

1) $ b^2 \minus{}d>0$:esto (A,B) to zeugari pou epalitheuei tin(1) kai elaxistopoiei to $ a\plus{}b$.i (1) omos exei 2 lyseis os pros $ a$ afou einai deuterou vathmou.(tin A kai tn K)tote apo vieta:$ K\equal{}Bc\minus{}A\equal{}(B^2\minus{}d)/A <B^2/A<A$ afou ypothesame pos A$ \geq$B.
Ara apo tis sxeseis g t K vriskoume oti o K einai fysikos k ikanopoiei tn (1),k to zeugari (K,B) einai lysi me K<A diladi K+B<A+B atopo apo tin upothesi..

2) $ b^2 \minus{}d<0$:prepei s auti tn periptosi n einai $ a^2 \minus{}abc>0$ dld $ a>bc$ ara $ a(a\minus{}bc) \plus{}b^2\minus{}d\geq a(bc\plus{}1\minus{}bc)\plus{}b^2\minus{}d>bc\plus{}b^2\minus{}d \geq b^2\plus{}bc\minus{}c>0$ atopo...

ara prepei upoxreotika $ b^2\minus{}d\equal{}0$....Ara o d einai teleio tetragono!!! :wink:  :o

elpizo n min m diefyge kt.....

[quote="lefmast"]ena mikro hint.....pollaplasiazoume me 4z kai prosthetoume monada!!!!!! :wink:[/quote]

Afti einai i prwti skepsi, nomizw omos oti moni tis den boithaei kai poli... meta apo afti ti skepsi i askisi teleionei an gia kapoio prwto diaireti tis morfis p = 4k+3 tou 4zk^2 + 1 (iparxei toulaxiston enas tetoios prwtos diairetis logo tis morfis twn paragontwn tou allou melous) o z einai tetragoniko kataloipo (modp), se alli periptwsi omws den mporo na skeftw kati allo. Opoios mporei as voithisei dioti einai poli kalo kai diskolo thema kai exei minei poli kairo alyto.


Euler
modular arithmetic
geometry
geometric transformation
rotation
More Sequences

An integer sequence $\{a_{n}\}_{n \ge 1}$ is given such that \[2^{n}=\sum^{}_{d \vert n}a_{d}\] for all $n \in \mathbb{N}$. Show that $a_{n}$ is divisible by $n$ for all $n \in \mathbb{N}$.

Use the moebius inversion formula to obtain $a_n = \sum_{d|n} \mu (d) 2^{\frac{n}{d}}$.
For an arbitrary prime $p$, let $p^k$ be the highest power of the prime $p$ dividing $n=p^k m$.
Then $\sum_{d|n} \mu (d) 2^{\frac{n}{d}} = \sum_{p\nmid d|n} \mu (d) 2^{\frac{n}{d}} + \sum_{p|d|n} \mu (d) 2^{\frac{n}{d}} = \sum_{p\nmid d|n} \mu (d)( 2^{\frac{mp^k}{d}} - 2^{\frac{mp^{k-1}}{d}})$.
But $2^{\frac{mp^k}{d}} - 2^{\frac{mp^{k-1}}{d}}= 2^{\frac{mp^{k-1}}{d}} \cdot \left( (2^m)^{\frac{p^{k-1}(p-1)}{d}} -1\right)$ is divisible by $p^k$ by the theorem of Euler-Fermat (in case of $p=2$, we still have $p^k=2^k|2^{2^{k-1}} |2^{\frac{mp^{k-1}}{d}}$), so the whole sum is divisible by $p^k$.
But since $p$ was arbtrary, the sum is divisible by $n$.

Alternatively, let $b_n$ be the number of aperiodic sequences $\left(x_1,x_2,\ldots,x_n\right) \in \{1,2\}^n$ of length $n$ (i.e., there exists no $d \mid n$ such that $x_k = x_l$ whenever $k \equiv l \pmod{d}$).  It is obvious that $n \mid b_n$ for every $n$ since the rotation \[\left(x_1,x_2,\ldots,x_{n-1},x_n\right) \mapsto \left(x_n,x_1,x_2,\ldots,x_{n-1}\right)\] preserves aperiodicity.  The rest is showing that $a_n=b_n$ by observing that $\left\{b_n\right\}$ obeys the same identity that defines $\left\{a_n\right\}$.  Indeed, the integer $2$ in the definition may be replaced by any other integer and the result still holds.


function
real analysis
real analysis unsolved

Let $ M$ be the set of the points at which a bounded variation function is [url=http://mathworld.wolfram.com/RegularFunction.html]regular[/url].
Show that the set $ [0,1] \minus{} M$ is at most enumerable set.

Are you sure that's the kind of "regularity" that you want? A function of bounded variation may be discontinuous on a dense set.


calculus
integration
calculus computations

Evaluate $ \int_0^1\int_0^1  \cdots \int_0^1 \left(\sum_{k\equal{}1}^n x_k\right)^2\ dx_1dx_2\cdots dx_n$.

$ \frac {n}{3} \plus{} \frac {{n(n \minus{} 1)}}{4}$

I hope  :maybe:

I found the same using $ I_{n} \equal{} I_{n\minus{}1} \plus{} \frac{n\minus{}1}{2} \plus{} \frac{1}{3}$ where $ I_{n}$ is the given integral.

[quote="Ramanujan"]I found the same using $ I_{n} \equal{} I_{n \minus{} 1} \plus{} \frac {n \minus{} 1}{2} \plus{} \frac {1}{3}$ where $ I_{n}$ is the given integral.[/quote]

A recursion is hardly necessary, as
\[ \left(\sum_{1\leq j\leq n} x_n\right)^2 \equal{} \sum_{1\leq j\leq n} x^2_j \plus{} \sum_{1\leq j\not \equal{} k\leq n} x_jx_k
\]
The squared terms integrate to $ 1/3$ and there are $ n$ of them.  The mixed terms integrate to $ 1/4$ and there are $ n^2 \minus{} n \equal{} n(n \minus{} 1)$ of them, so...


Can you suggest me some good book for practising topology ?
 With nice and hard problems  :) ?
( level of 3rd and 4th year at university )

Dugundji ["Topology"], Willard ["General Topology"]?


A standard six-sided die is rolled, and P is the product of the five numbers that are visible. What is the largest number that is certain to divide P?

[hide="answer"]There, clearly, is more than 1 # divisible by. 1. the same follows for 2 and 3. It will always be divisible by 4, as if 4 isn't one of them, 2 and 6 will be. 5 isn't guaranteed, or 9, so the # is 12.[/hide]


What do I need to bring? I'm a little clueless. I have a binder, pens, pencils, paper, etc. Anything else I need for the classes at AMSP?

Also, what about six flags? Do I need to bring money for it? I keep hearing about it.

bring calculator, ruler, compass, graph paper, whatever you use for math.
i didnt bring anything except a notebook last year and i had to buy everything at the bookstore which ...is ridiculously overpriced [yeah stupid me.]
anyways, bring supplies that you would use in school math classes or whatever.

and as for six flags, bring money if you want to get drinks there [one of those big cups that you can get unlimited refills?]
and if you want any stuff from gift shops and stuff, bring moneyy. [AND SUNSCREEN]

Don't go to six flags. If you lose your plane ticket (I almost did), it costs a fortune to get another one.

Off topic, but just telling you, it's spelled stationery, not stationary. That's another word. :)

yeah you should bring a backpack/knapsack or whatever else you can use to carry binders and other material in, unless you want to carry each item by hand everyday.


Find all prime pairs $ (p|n)$ such that $ 3p+4=n^{2}$.

[hide="Is not it (p,n)?"]
$ 3p+4 = n^{2}$
$ 3p = (n+2)(n-2)$
$ n+2=3 \Rightarrow n=1 \ , \ p=-1$
$ n-2=3 \Rightarrow n=5 \ , \ p=7$
$ (-1,1)$ Is the negative prime still considered a prime?
$ (7,5)$.
[/hide]

why do we have to solve the equations $ n+2=3$ and $ n-2=3$? i mean why $ =3$?

Here it is with a few more comments.

[hide]$ n \in P \implies n > 0 \implies n+2 > 0$

$ 3p = 3 \times p = (-3) \times (-p) = 1 \times 3p = (-1) \times (-p)$

Those are the only possible factorizations of $ 3p$ into integers, since both $ 3$ and $ p$ are prime.

However, since $ 3p=(n-2)(n+2)$ and $ n+2>0$, $ n-2 >0$ as well.

So, $ 3p = 3 \times p = 1 \times 3p$

Since $ n+2$ and $ n-2$ are integers, one of the two above factorizations must hold.

So, either; $ n-2=3$ and $ n+2 = p$ or $ n-2 = 1$ and $ n+2 = 3p$
In the first case, we get $ n=5$ and $ p=7$

In the second case, we get $ n=3$ and $ 3p=5$, which doesn't work.

So the only solution is $ (5,7)$[/hide]

thx. just one question:

[quote="The Zuton Force"]So, either; $ n-2=3$ and $ n+2 = p$ or $ n-2 = 1$ and $ n+2 = 3p$[/quote]

but aren't there also the possibilities $ n+2=3$ and $ n-2=p$ or $ n+2=1$ and $ n-2=3p$? i know that they doesn't work, but do we have to check them?

They trivially don't work because $ (n+2)-(n-2)>3>1$, so it would force a number to be negative. But yes, they are cases.


Let $w=C(I,r)$, $w_a=C(I_a,r_a)$, $w_b=C(I_b,r_b)$, $w_c=C(I_c,r_c)$ be the incircle and the exincircles of the triangle $ABC$ (standard notations). Prove that the middlepoints of the sides of the triangle $I_aI_bI_c$ belong to the circumcircle $e=C(O,R)$ of the triangle $ABC$.
[hide="Indication"]
[u]Indication.[/u] I note the second intersections $A'$, $B'$, $C'$ of the lines $AI$, $BI$, $CI$ respectively with the circumcircle $e$ and the second intersections $A''$, $B''$, $C''$ of the lines $OA'$, $OB'$, $OC'$ respectively with the same circumcircle $e$. Then $A''\in I_bI_c,\ A''I_b=A''I_c$ a.s.o.
[/hide]

Virgil this problem is easy get to the well know problem about [b]Nine point circle(Euler's circle)[/b]

Yes. This problem has a indication. I received it from a young in a private message and I think is interestingly for more. Using the above indication, the solution of this problem is very simply.

Tiks, you report on the Feuerbach's theorem (the Euler's circle is tangent to the incircle and to the exincircles) ? I don't see how you use the Euler's circle in prove of this problem. Give me some details.


Finally just about a week left
And i m feeling really vetti :rotfl:

I'll start with a simple integral (not simple for me :( )

$ \int {\sqrt{\frac{sin(x \minus{} a)}{sin(x \minus{} b)}}}$

Good but not very. 3 mechanisms in one 3 marker is too much to ask.

and add to it some tough calculations..

i messed one calculation (Os Pres one)

What is the product when benzene diazonium chloride reacts with ethanoyl chloride???????????????
I wrote acetophenone ( guess work). I did not find this reaction in ncert,abc,arihant and even in i.l.finar :maybe: !!!!!!!!!!!!!

how was physics??

and wat did u write in the capacitor  ques
(pot diff b/w X and Y) :maybe:

Hey it went well.
p.s.: it was all india paper

Could you ask deepak vasishth how much he is getting in INChO?

srry...i dont hav his no.

And i think the result will be out  by the time i get to kno
RITE  NOW..

anyways dont worry...he is not such a beast in Chem....(lik other vmc'ians)

congrats u got rank 12 in NSTSE  :D . sadly i didn't give (would hav failed anyway) and what what do u mean by other vmc ians  :?: (try and find out about their marks plzzzzzz.)

thanks...

and suppose 2 capacitors(20 and 5 F) r in series with a 12V battery

wat will be pot diff accross them???

[hide]i've written 12V[/hide]

how were the papers :P

now pls solve the integral in first post :)

Is anyone aware when the CBSE 12 Board results will be out! :roll:
 I couldn't find any info at their site....
Just wondering cuz I,ve nothing else to do

I hope after the 25th cuz my boards were pathetic , all because of one subject: comp sc . And guess what I want to do in college. :D

Let me guess ... Comp sci.

So is no one aware of when the CBSE Board ( Both 10 and 12 ) results will be coming  :?:

Seeing how many people have posted in this thread would give someone the idea that very few ppl. on AoPS have written the CBSE board examinations this year  :D . (or maybe no one really cares about their board marks; Even the few ppl. who have posted here don't seem to be interested in their marks)


In set $S$, there is an operation $'' \circ ''$ such that $\forall a,b \in S$, a unique $a \circ b \in S$ exists. And

(i) $\forall a,b,c \in S$, $(a \circ b) \circ c = a \circ (b \circ c)$.

(ii) $a \circ b \neq b \circ a$ when $a \neq b$.

Prove that:

a.) $\forall a,b,c \in S$, $(a \circ b) \circ c = a \circ c$.

b.) If $S = \{1,2, \ldots, 1990\}$, try to define an operation $'' \circ ''$ in $S$ with the above properties.

Observe first that $(a\circ a)\circ a=a\circ (a\circ a)$ implies $a\circ a=a$, for all $a$. Now, we have
\[
a\circ\left(  a\circ\left(  b\circ a\right)  \right)  =\left(  a\circ
a\right)  \circ\left(  b\circ a\right)  =a\circ\left(  b\circ a\right)
\]
and%
\[
\left(  a\circ\left(  b\circ a\right)  \right)  \circ a=\left(  a\circ
b\right)  \circ\left(  a\circ a\right)  =\left(  a\circ b\right)  \circ
a=a\circ\left(  b\circ a\right)
\]

Again, we deduce that $a\circ\left(  b\circ a\right)  =a,$ for all $a,b.$

Finally,%
\[
\left(  a\circ c\right)  \circ\left(  a\circ\left(  b\circ c\right)  \right)
=\left(  a\circ\left(  c\circ a\right)  \right)  \circ\left(  b\circ c\right)
=a\circ\left(  b\circ c\right)
\]
and%
\[
\left(  a\circ\left(  b\circ c\right)  \right)  \circ\left(  a\circ c\right)
=\left(  a\circ b\right)  \circ\left(  c\circ\left(  a\circ c\right)  \right)
=\left(  a\circ b\right)  \circ c,
\]
therefore%
\[
a\circ\left(  b\circ c\right)  =a\circ c
\]
For the second part, a trivial example is $a\circ b=a,$ for all $a,b$ in $S.$
A less trivial example can be obtained by considering a set $S^{\prime
}=\left\{  \left(  x,y\right)  |1\leq x\leq n,1\leq y\leq m\right\}  ,$ with
$nm=1990,$ defining a one-to-one function $f: S\rightarrow S^{\prime}$ and
transporting through $f$ the structure defined on $S^{\prime}$ as follows:
$\left(  x,y\right)  \circ\left(  x^{\prime},y^{\prime}\right)  =\left(
x,y^{\prime}\right)  ,$ which clearly satisfy the requested conditions. The
previous trivial example is obtained for $n=1990$ and $m=1$


vector
analytic geometry
induction
linear algebra
matrix
combinatorics proposed
combinatorics

Let $ A_1,A_2,...,A_n$ be $ n$ distinct subsets of $ \{1,2,...,n\}$. Prove that there exists a number $ x$ such that
\[ A_1/\{x\},A_2/\{x\},...,A_n/\{x\},\]
are also distinct.

I have known one solution only.

This is a theorem from A. Bondy.
Induced subsets, J. Combin. Theory Ser. B 12 (1972), 201- 202.

There are also three proofs in the book "combinatorics" about set systems by B. Bollobas.
D.

I knew that this problem belongs to A. Bondy, (and the solution I got based on one of this idea). I remembered I read this problem in "Problems & Solution of National Mathematical Competition Around the World" (series of Titu. Adresscu) and it seems that there is a nice solution in this book, but I can't remember it now.

Could you post three solutions, Sclormu? Thank you very much,

[quote="hungkhtn"]

Could you post three solutions, Sclormu? Thank you very much,[/quote]

Please, i have other things to do, and i don't have the book with me. But I'll post the one I remember... perhaps it is the one you know.

The fact that an element $ x$ of the ground set cannot be suppressed (so that the traces of the $ n$ sets would be distinct) means that there exists two sets A and B in the family such that $ A\equal{}B \cup \{x\}$. Say that the pair $ (A,B)$ witness the fact that $ x$ cannot be suppressed. Consider the graph on $ n$ vertices, which are the sets of the family, and for each $ x$ that cannot be suppressed add one edge labelled $ x$ between one pair $ A,B$ that witness for $ x$ (only one). Then this graph cannot have a cycle, because on a cycle the number of edges labelled $ x$ must be even. So this graph is a forest, and its number of edges is at most $ n\minus{}1$.

The main idea is the same as the solution I know, but the fact that the graph does not have a cycle is not true. Consider
\[ \{1\},\{1,2\},\{1,2,3\},\{1,2\}.\]
However, we can remove some edges so that the graph does not have a cycle (remove so that the number of $ x$ with $ A/\{x\}\equal{}B/\{x\}$ does not change).

Anyway, nice for your idea, though I wait for another solution (if exists).

A Corollary of next Problem:
In an n*n array of numbers,all rows are different(two rows are different if they differ in at least one entry).Prove there is a column which can be deleted in such a way that the remaining rows are still different.
aij=1 for j∈Ai, aij=0 if not.

[quote="sclormu"][for each $ x$ that cannot be suppressed add [b]one[/b] edge labelled $ x$ between [b]one[/b] pair $ A,B$ that witness for $ x$ (only one). [/quote]

In my version there is only one edge labelled x so there cannot be any cycle. Someone knows another proof ?

Consider the $ A_i$ as vectors in $ V \equal{} \mathbb{R}^n$ with basis $ x_i$, where the $ x_i$ coordinate of a set is $ 1$ if the set contains the number $ i$ and $ 0$ otherwise. The span $ W$ of the $ n\minus{}1$ vectors of the form $ A_{i\plus{}1} \minus{} A_i$ contains all vector differences $ A_i \minus{} A_j$. $ W$ is $ n\minus{}1$ dimensional, so it can not contain all the $ x_i$; therefore, we can find a projection $ \pi_k$ along at least one of the coordinate axes $ x_k$ so that $ W$ gets mapped injectively, so that $ A_i \minus{} A_j \ne 0 \implies \pi_k(A_i) \minus{} \pi_k(A_j) \ne 0$. This means that the $ \pi_k(A_i)$ are distinct, so that the corresponding $ A_i \minus{} \{k\}$ are also distinct.

My solution based on the ide of induction:(Probably the easiest one I think)
Consider a matrix nXn filled with entries 0 and 1 corresponding to the elements not contained and contained in the reflecting set!
First,let's delete one arbitrary column.In the case the remaining matrix contains distinct rows ,we're done! Otherwise,we partition the present rows into cliques of identical rows and relace each clique with one representing row.Now,induction!!! :lol:

$ \mathbb{R}^n$?  Offhand, it seems more natural to work in $ \mathbb{F}_2^n$, although I don't think it changes the argument either way.


Let $ a,b,c>0$ and $ a^2\plus{}b^2\plus{}c^2\equal{}3$ and $ ab\plus{}bc\plus{}ca\equal{}3\minus{}t^2$ and $ abc\equal{}1\minus{}r^2$ find the maximum value of  $ \frac{r}{t}$


function
modular arithmetic
number theory
divisibility tests
graph theory
combinatorics unsolved
combinatorics

In  a school class with $ 3n$ children, any two children make a common present to exactly one other child. Prove that for all odd $ n$ it is possible that the following holds: For any three children $ A$, $ B$ and $ C$ in the class, if $ A$ and $ B$ make a present to $ C$ then $ A$ and $ C$ make a present to $ B$.

the total number of members is odd. take a member $ x$ notice that there are an even amount of members left. pair them. for any pair $ y,z$ have $ x,y$ give a present to $ z$ and $ x,z$ give a present to $ y$. all presents from $ x$ satisfy the condition now. next we take another member $ q$ and we do the same (*except we have given a present from $ x,q$ already), but now we are left with some $ w$ which is unpaired. put $ q,w$'s present aside and repeat for a different member (not $ w$), again we can pair everyone. now take a different member $ r$, make $ w$ the reminder again. and now pair remainder presents $ qw\minus{}>r$ and $ rw\minus{}>q$. so far, every single present from the first 4 members has been given and satisfies the condition. now we see we can repeat this to the end beacuse only the even numbered members leave a remainder present. we also see that this is independent of divisibility by 3, therefore it is all wrong.  :oops: ? can some1 tell me where i went astray..

Let the children be $ X_i, Y_i, Z_i$ for $ 0 \leq i \leq n\minus{}1$; we will be considering the subscripts mod $ n$. The key is that $ 2$ is invertible when $ n$ is odd.

Also note that $ (A,B) \to C$ and $ (A,C) \to B$ implies $ (B,C) \to A$, so we basically just need to get each pair of children into a unique triple.

First, make the triples $ (X_i, Y_i, Z_i)$ for each $ i$.
Then for $ i \neq j$, if $ k$ is equal to $ \frac{i\plus{}j}{2}$ in mod $ n$, make the following triples: $ (X_i, X_j, Y_k), (Y_i, Y_j, Z_k), (Z_i, Z_j, X_k)$.

This pairs off any children labelled with the same $ X,Y,Z$ letter. Given say $ X_a, Y_b$ (with $ a \neq b$ or the triple is obvious), we know that the third child is $ X_{2b\minus{}a}$, and the pair $ Y_b, X_{2b\minus{}a}$ also means $ X_a$ was third by the same rule, so the grouping is well defined.

wow somebody actually replied... thank you MellowMelon.
im new to olympiad type questions so i am wondering how did you know to look for an answer in specifying the triples instead of say graph theory?

I don't see a good way to get graph theory to work, because what you're dealing with is a function from pairs of kids to kids. Graph theory works better when you have a symmetric relation; I don't see an obvious one here.

The triple idea came straight from the observation that $ (A,B) \to C$ implies $ (A,C) \to B$ and $ (B,C) \to A$, so that takes care of the gift given by every pair in the triple $ (A,B,C)$, and that's what suggests grouping them off.

The hardest part of the solution was probably just thinking of the $ \frac{i\plus{}j}{2} \pmod{n}$ construction. Everything else, like the triple thing, is just toying around with the problem and getting intuition about it. Really, this is how solving most olympiad problems goes. You spend some time developing a basic idea for how things work and find out what makes the problem difficult, and then you try to overcome those barriers.

Asking for the existence of a Steiner triple system of order $ 6n \plus{} 3$ (see also http://www.mathlinks.ro/Forum/viewtopic.php?t=240152 ) is soooo original...

  darij


http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168612&start=120

There is a weird blue stripe cutting across the page, moving all the messages to the side. What happened?

I do! It's cutting right down the middle of my page.

Please always specify the browser and OS when submitting bugs.

It happens in IE browser and ML style, regardless of OS, as far as I can tell.

I didn't see if it happens for my other computer which uses mozilla, but on the computer it happened on:

Internet Explorer
Windows XP

Edit: I also just noticed that it also happnes on the link that jmerry gived for his currently stickied topic in the support forum about the tildes. Maybe we're actually reporting the same error and someone messed up their LaTeX with a tilde? I don know...

Update: I'm trying the pages on mozilla (still windows xp), and the "blue stripe" doesnt appear in either one of them. So I'm pretty sure it's something to do with internet explorer

this only happened to me in one topic that was:
MATHCOUNTS Problems of the Day

[quote="archimedes1"][url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168612&start=120[/url]shows up with the blue stripe for me as well.  I am using IE 7.0.5730.11on XP with AoPS skin[/quote]

Uh, why does archimedes' post not show up? ML, FF

See http://www.mathlinks.ro/Forum/viewtopic.php?t=179090 for the bug.

I've checked the Mac side- ZetaX's link shows up as a blank post in both Firefox and Safari. That's still a bug, but the subject of this thread is definitely IE only.

Zeta's bug will show up in any browser, regardless of skin or OS. I thought the content might be somehow contained into an HTML tag by accident, but viewing the source proves that it's not sent to the client at all.

[quote="archimedes1"][url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168612&start=120[/url]shows up with the blue stripe for me as well.  I am using IE 7.0.5730.11on XP with AoPS skin,[/quote]
That's what I meant.

Incidentally, the trigger there is a simple missing space error- \pir is not a valid command. For me, this means it just shows up blank (probably the intended behavior, although standard $ \text{\LaTeX}$ generates an error). For the IE users, it seems to do weird things to the whole page.

Am I talking about the same bug?

[url=http://img405.imageshack.us/my.php?image=bugmz0.jpg][img]http://img405.imageshack.us/img405/5622/bugmz0.th.jpg[/img][/url]

I have appended the source code in the RAR file, but I think it should not be of much help, since the bug  is not reproducible in a logical way (when I press F5, it disappears, but can reappear in a rather arbitrary way when viewing any topic).

Firefox 2.0.0.11 under Windows.

  Darij


linear algebra
matrix
Gauss
algorithm
linear algebra theorems

Hi there, first of all, I would like to tell you guys that I love these forums, and thank you in advance for any help you can give me.

My question is simple, I know how to, given a matrix, calculate the adjugate. The thing is, I know there's another method, using Gauss-Jordan to get it, and I do not know it. I cannot find it using internet, so if anyone could explain it to me OR provide me with a link it will be very appreciated.

Oh, and by the way, I just want to get the adjugate so I can get the inverted matrix.

Sorry for my maths vocabulary, English is not my primary language and, even though I can speak it fluently, I'm missing some vocabulary.

Thanks again.

You've got the vocabulary right. The adjugate is sometimes called the (classical) adjoint, but that's a terrible name since it conflicts with another operation, and adjugate is better.

I don't know a natural way to get the adjugate that way, but I do know an efficient algorithm for the inverse using Gaussian elimination. Run the full elimination process on the augmented matrix $ \begin{bmatrix}A&I\end{bmatrix}$, including normalizing the pivot entries and clearing the columns above the pivots. The endpoint of this process is $ \begin{bmatrix}I&A^{\minus{}1}\end{bmatrix}$ if $ A$ is invertible.
You can extract the adjugate out of that; if you kept track of all operations that multiplied a row by a constant, that gets you the determinant, which you can multiply the inverse by. This is ridiculous for your purposes, of course.

The adjugate, or classical adjoint, provides a formula for the inverse of a matrix.

Gaussian elimination can also be used to compute the inverse of a matrix. If you do this, the adjoint never appears in the computation - it's a process for getting the inverse. This differs fundamentally from the whole notion of using the adjoint. The adjoint-based formula for the inverse is a [b]formula.[/b] Using Gaussian elimination to find the inverse is an [b]algorithm.[/b]

Suppose $ A$ is an $ n\times n$ matrix. Form the $ n\times 2n$ partitioned matrix $ [\,A\,|\,I\,].$ Perform row elimination on this $ n\times 2n$ matrix, aiming towards bringing the left half equal to the identity. There are two possibilities:

(1) The process fails, because we cannot get $ n$ pivot elements into the left half; the left half arrives at a state in which there is a row of zeros across the bottom. Once we see that, we declare that $ A$ is not invertible.

(2) The process succeeds, and we can reduce the left half to the identity. The whole $ n\times 2n$ matrix is now $ [\,I\,|\,A^{\minus{}1}\,]$ and we can read off the right hand half as $ A^{\minus{}1}.$

Why does this work? Each row step is the same as left-multiplying by an elementary matrix. If we can row-reduce $ A$ to $ I,$ we have $ E_kE_{k\minus{}1}\cdots E_2E_1A\equal{}I.$ In that case, we have that $ E_kE_{k\minus{}1}\cdots E_2E_1\equal{}A^{\minus{}1}.$ But what we've done to the right hand half gives us $ E_kE_{k\minus{}1}\cdots E_2E_1I\equal{}E_kE_{k\minus{}1}\cdots E_2E_1,$ which we've just declared to be $ A^{\minus{}1}.$

At worst, this process takes something like $ \frac43n^3$ multiplication/division steps and a similar number of addition steps. I'm not completely sure of the coefficient, but I am sure that it's $ O(n^3).$ For large $ n,$ the classical adjoint is less efficient.

The vocabulary reaaaally gets in my way.

But ooh, thanks very much, cause I just searched for "Gaussian elimination", and I found that in order to use Gauss-Jordan to find the adjugate, you have to first augment (I dunno again if that's the correct vocabulary) the matrix with the Matrix Identity.

After that, It says that you have to perform "elementary row operations" on the new matrix ( [AI] ) until A reaches reduced row echelon form.

... and while I'm posting this, I'm re-reading what you said, and it's the exact same thing you said, but explained "noob-friendly".

And now I understand yes, it's ridiculous to extract the adjugate out of that, when my purpose is just to get the inverse matrix. It's like going to steps forward, and then one step back.

I will now do further research to know what's a row echelon form, and what are those "elementary row operations"

Thank you!

****EDIT: Kent you posted while I was posting. I understood the most of it, and I loved the way in which you differed between a formula and an algorithm.

The only part I do not get is "Row elimination"

I just use regular Gauss-Jordan right? for example, with a 3x3 matrix, I will get a 3x6 matrix, and the 3 pivots would be a11, a22 and a33?

As a vocabulary note, "Gauss-Jordan", "Gaussian elimination", and "row elimination" are all names for the same thing. It's the fundamental algorithm of linear algebra, so everybody learns it- but not necessarily by the same name.

Oooooooooh I see now.

Your post saved my day. Seriously.

Thanks!


vector
linear algebra
matrix
linear algebra unsolved

$ W$ is a finite dim vector space, an inner product $ <,>$ is defined on $ W$. $ v \in Endomorphism(W)$.
If $ <v \bullet w_1, w_2>\equal{}\minus{}< w_1,v \bullet w_2>$ for all $ w_1,w_2 \in W$, is it equiverlent to say that
the matrix of $ v$ is skew symetric?

Try $ w_1\equal{}e_i$ and $ w_2\equal{}e_j$, where $ e_i,e_j$ are elements of an orthonormal basis.


I was going through some old posts and the word " centrally symmetric" appeared quite a no of times. could someone tell me what that means? they use it in the context of " A is centrally symmetric with respect to O " and stuff like that. any help appreciated

Two points $P$ and $P'$ are symmetric about $O$ if $O$ is the midpoint of line segment $PP'$.

A set $A$ is (centrally) symmetric about $O$ if for any $P\in A$ the point $P'$ (defined as above) is also contained in $A$.


inequalities
induction
function
calculus
derivative
inequalities unsolved

[quote="mffmartinelli"]If we make $x=\frac{a^p}{a^p+b^p+c^p}$, $y=\frac{b^p}{a^p+b^p+c^p}$ and $z=\frac{c^p}{a^p+b^p+c^p}$, we´ll have that $x+y+z=1$ and our inequalitie is equivalent to $1\geq x^{u}+y^{u}+z^{u}+(3^{u}-3)\sqrt[3]{x^{u}y^{u}z^{u}}$, where $u=\frac{3}{p}$ and we can notice that $u\geq 1$. In fact we have to prove that $(x+y+z)^{u}\geq x^{u}+y^{u}+z^{u}+(3^{u}-3)\sqrt[3]{x^{u}y^{u}z^{u}}$, and if we make $m=\frac{y}{x}$ and $n=\frac{z}{x}$, we have to prove that ${(1+m+n)^{u}\geq 1+m^{u}+n^{u}+(3^{u}-3)\sqrt[3]{m^{u}n^{u}}}$, for all reals $m,n$ and $u$, such that $m,n>0$ and $u\geq 1$. And it´s more easily prove this last ineq.[/quote]
can you tell me how to prove the last easy ineuqlity as you said:
${(1+m+n)^{u}\geq 1+m^{u}+n^{u}+(3^{u}-3)\sqrt[3]{m^{u}n^{u}}}$
thanks :)

I said that this ineq is more easily. Just it! Let´s prove it!

First I´ll prove by induction that ${P(u)= (1+m+n)^{u}\geq G(u) = 1+m^{u}+n^{u}+(3^{u}-3)\sqrt[3]{m^{u}n^{u}}}$ for all natural $u$ such that $u\geq 1$.

For $u=1$, the ineq is obvious true and now we can assume that it´s true for $u=k$. So we have

${P(k) = (1+m+n)^{k}\geq G(k) = 1+m^{k}+n^{k}+(3^{k}-3)\sqrt[3]{m^{k}n^{k}}}$.

So we can aplly our ineq to obtain the following:

${P(k+1) \geq (1+m+n)(1+m^{k}+n^{k}+(3^{k}-3)\sqrt[3]{m^{k}n^{k}}})$. And so we have:

$P(k+1) \geq 1+m^{k+1}+n^{k+1}+f(m,n)+(1+m+n)(3^k-3) \sqrt[3]{m^kn^k}$, where $f(m,n)=m^k+n^k+mn^k+nm^k+m+n$

And now, we can notice that $(1+m+n) \geq 3\sqrt[3]{mn}$. So we have

$P(k+1) \geq G(k+1)+g(m,n)$, where $g(m,n)=f(m,n)-6\sqrt[3]{m^{k+1}n^{k+1}}$

But now we can prove that:

$g(m,n)\geq 0$ and this just because $m^{k}+n^{k}+mn^{k}+nm^{k}+m+n \geq 6 \sqrt[6]{m^{2k+2}n^{2k+2}} = 6\sqrt[3]{m^{k+1}n^{k+1}}$.

To extend our result for all real numbers let $u$ a real number such that $k<u<k+1$ ($k$ is a natural number) and so considere $a$ such that $u=k+a$, and now use the previous result to extend this ineq.

can you just explain your last step, how you deal with the $0<a<1$ you introduce? I can't see how you prove e.g. for $u=1.5$.

Anyway, note that the ineq can be written in a nice way as an ineq invoving just AM&GM means:

$(x+y+z)^{u}-(3\sqrt[3]{xyz})^{u}\geq x^{u}+y^{u}+z^{u}-3\sqrt[3]{x^{u}y^{u}z^{u}}$

What about generalizations?
I think the obvious generalization to more than three variables can be handled the same way you do (for natural $u$). But more interesting:
[b]If the AM and GM of $x,y,z$ and $x^u,y^u,z^u$, taken as power means $S_1$ and $S_0$, are replaced by other power means $S_n$ and $S_m$, $n>m$, can the corresponding ineq be proved as well?[/b] Maybe with Jensen??

[quote="mffmartinelli"]

To extend our result for all real numbers let $u$ a real number such that $k<u<k+1$ .[/quote]
Can you explain how can we extend???(sorry for my my stupidness :( )

to spanferkel:
I think this problem can be general for n viarables :)
do you have another solution,if yes,please post for details thanks

So, here we are again!!! In fact $0<a<1$ is much valious. Let´s see.

(I´ll try short my notation)

First we have $(1+m+n)^{u} = (1+m+n)^{k}(1+m+n)^{a}$. And we want to show that

$(1+m+n)^{u} \geq p(m,n)+q(m,n)$, where we have:

$p(m,n) = 1+m^km^{a}+n^kn^{a}$ and $q(m,n) = (3^{a}3^{k}-3)\sqrt[3]{m^{a}m^{k}n^{a}n^{k}}$.

And thus we have to show that $(1+m+n)^k \geq \frac{p(m,n)}{(1+m+n)^{a}}+\frac{q(m,n)}{(1+m+n)^{a}}$.

And now we can note that if we consider a function $f$ such that $f(x) = x^{a}$, it´s second derivative is negative since $0<a<1$ (!!!).

So we can write by Jensen´s inequalitie (with pounds):

$\frac{p(m,n)}{(1+m+n)^{a}} \leq (1+m^k+n^k){h(m,n)}^{a}$, where we have:

$h(m,n) = \frac{1+m^{k+1}+n^{k+1}}{(1+m+n)(1+m^k+n^k)}$.

And we have to notice that we must be able to prove that

$(1+m+n)^k \geq (1+m^k+n^k){h(m,n)}^{a}+\frac{q(m,n)}{(1+m+n)^{a}}$.

At this point we have to notice that if we make $a\to 0+$, our ineq is true just because the previous induction. And if we make $a\to 1-$, our ineq it´s true so, by the same reason.

Now to extend our result it´s just we prove that $z(a) = (1+m^k+n^k){h(m,n)}^{a}+\frac{q(m,n)}{(1+m+n)^{a}}$, where $z: (0,1) \to R$ is a continuous one and it´s first derivative is never $0$. So it´s finally extend!

Nice problem, no? Zhaobin, how did you do this question?

[quote="mffmartinelli"]Prove that the first derivative of $z$ is never $0$ isn´t hard becuase we can write $z$ as exponential functions.[/quote]
sorry for being late,I don't know this why $z$ is exponential functions than  that the first derivative of $z$ is never $0$???

But very sorry,and unforturenaly that:I find the nice inequality:
$(x+y+z)^{u}-(3\sqrt[3]{xyz})^{u}\geq x^{u}+y^{u}+z^{u}-3\sqrt[3]{x^{u}y^{u}z^{u}}$
is not ture....just try $u=1.00002,x=1,y=z=1.1$(if my caculation is right)...that my inequality is aslo not ture.
and I think we there exist a minimum of $u$,But I think it must be a very hard problem.

[quote="spanferkel"]I think fortunately your calculation is wrong :). After plotting some graphs, it looks like the minimum of LHS-RHS is always below u=1, for your choice of x,y,z for example it is near u= 0.9116365. Whew.

For my proposed generalisation, see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=401083]here[/url][/quote]
No,I am sure it is not right for all $u>1$
if my calculation is wrong,I think there will other ones.

and if you said it is ture ,do you have a solution.
anyway,I think mffmartinelli is incorrect.

I'm sorry that I try to find a counterexample.but failed.
but I still trust this problem is not true for all $u>1$.
and when $n=2$ in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=401083]spanferkel's topic[/url],it is easy to find a counterexample.


There seems to be an evil mod somewhere. See
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1010355#1010355

That's not really a problem... just a moderator having some fun. It is the Test Forum, after all.

Mods are allowed to practice "modding" in the Test Forum.

^ Now that that has been brought up, I shall go try it.  :lol:

BTW, I thought I've heard of that elsewhere, as in other forums?

Oh. Ok.

 The message is too small. Please make the message longer before submitting.


Not that Im gonna be in it, but how do they choose the people to be in it?
My best guess is 1st countdown, then 1st written, 2nd written, 3rd written, 4th written (if necessary)
Not sure though.

I read somewhere, it's top 4 individuals.

It's the top two written round scorers, and the top two after countdown. If some people are the same, then they'd take 3rd and (if needed) 4th in the written round.

I thought it would be top 4 after countdown, oh well.

[quote="XxHarryxX"]^ Well, at least that's what it was last year. I'd know, I was there.[/quote]
Well it was both what you said and what I said.
There was Neal Wu (1st countdown), Mark Zhang (2nd countdown 3rd written), Sergei Bernstean (1st written), Nathan Benjamin (2nd written)
So either works, it depends whether Mark Zhang made it because of his written or countdown performance.

I'm pretty sure it was his countdown performance, but not 100% sure.  :P

definitely, becuase nathan and sergei each did not do well in countdown.

I heard somewhere that one year only one person out of all the 228 students accepted the masters round, so he immediately won. Of course, that does sound pretty unlikely though.

I'm sure there would be more people willing to make fools of themselves or try and succede.

Masters round is #1 and #2 in Countdown, and #1 and #2 in written. If there are overlaps or people decline, they use the next highest in written. All Countdown people are told not to leave the stage untill they have 4 participants.

How would a masters round run and be graded?  (I know what it is)

There are judges who listen to what each has to say about the problem and then ask challenging questions. Then they deliberate after all are done to decide the winner which is announced at the awards banquet.

[quote="Xantos C. Guin"]Each participant is alloted 30 mins to work on the problems.
Each participant starts at a different time so they can present immediatly when their time is up.
A participant is allowed 11 mins to show solutions to a problem and 4 mins for the judges to ask questions.[/quote]
Also, I think each participant has 15 minutes to prepare. Is that right?

You can't really be stuck and then suddenly make it up on the spot:

Well, when I first looked at this problem...obviously, of course, then you must consider

[quote="Xantos C. Guin"]I believe that the 30 mins is for preparation also, but I could be wrong.

Then again, a participant could be working the problems he didin't solve for "preparation" and "Ad-lib" the presentation.[/quote]

It is 30 minutes.


Find all $k$ such that :
 $2(x^3+y^3+z^3) +3(1+3k)xyz \geq (1+k)(x+y+z)(xy+yz+xz)$ , $\forall x,y,z >0$

$k \leq 1$. For this we can use schur to show. It remains to see that k =1 is the best constant.
Rewrite the inequality as $\sum (x+y - kz)(x-y)^2 \ge 0$. If $k > 1$, we try to take $x = y$ with z not fixed yet. Then the ineq becomes $(y+z - kx)^2 (y-z)^2 + (x+z-ky)(x-z)^2 \ge 0$. Now it's clear that this is false, for if $z < (k-1)x = (k-1)y$, the left hand side is smaller than 0


1200$ \$$ dar sandooghe pasandaz darim. bad az yek sal 240$ \$$ an ra migirim va bad az yek sale digar 1701$ \$$ dar sandoogh darim. nerkhe bahre ra hesab konid   :)

:wink: 1 sal ra barabar ba 365 rooz begirid

salam
nerkhe bahre ra tarif konid!ya begid che no nerkhe bahreie made nazaretune!tarifaye mokhtalefi baraye nerkhe bahre vojud dare,be ebarate dige nerkhe bahraro be raveshaye mokhtalefi hesab mikonan,ke javabaye mokhtalefi bedast miad!

fek konam manzooresh in bashe ke masalan age nerkhe bahraro $ r$ begirim oonvakh age $ n$ TOMAN pool berizim too hesab,bade yesal mishe $ n \plus{} nr$.

sohbate aghaye [b]BaBaK Ghalebi[/b] kamelan doroste  :)

2 ta moadele ehtiaj dare
ma nerkhe bahre va poole baghimandeye sal av'al ro nadarim  :idea:

vaght tamam!!!
shayad baratoon in soal gong bood vali hich masalei nadere alan man javabesho midam!!  :)

ma $ 1200\$$ dar sandoghe pasandaz gozashtim va bad az yek sal 1yr=365days, $ 240\$$ pas gereftim vali ma pooli ke baghi mimanad ra nadarim. in meghdar ra barabare $ y$ gharar midahim pas darim:
$ y\equal{}1200\plus{}365x\minus{}240$
haman tor ke dar masale molaheze kardeid nerkh bahre niz majhool mibashad pas in meghdar ra ham barabar ba $ x$ gharar midahim. haman tor ke edameye masale gofteshode, yek sale digar $ 1071\$$ darim yani:
$ y\plus{}365x\equal{}1071$
moadeleye aval ra dar moadeleye dovom gharar midahim:
$ 1200\plus{}365x\minus{}240\plus{}365x\equal{}1071$
va in yek moadeleye yek majhooli ke nerkhe bahre ra be ma midahad.  movafagh bashid :)


Find the probability of getting the following in a 5-card hand:
1.  4 of a kind
2.  Full house (3 of one kind, 2 of another)
3.  Flush (5 of one suit)
4.  Straight (5 numbers in a row)
5.  3 of a kind
6.  2-Pair (2 pairs)
7.  Pair (2 of a kind)
8.  None of the above
Then find the probability with one wild card.
Then two.
Hide all solutions.
[hide="If you are trying straight..."]Remember that Aces are high or low, but you can't go K, A, 2.[/hide]

[quote="breez"][hide="Straight"]Since A, 2, 3, 4, ..., J, K, Q, A is 14, and we want 5 card hand straight, there are $14-5+1 = 10$ consecutive 5 card sets. Also, for each consecutive card, there are 4 choices, as there are 4 suits. Hence:

$P = \frac{(10)4^{5}}{\binom{52}{5}}= \frac{128}{32487}\approx .00394$
[/hide][/quote]
There's one problem:
You're counting those cases in which all the cards are of the same suit, such as 3h-4h-5h-6h-7h.  That is technically NOT a straight, but a straight flush.

Well, in that case, just subtract 40 from the numerator.

I'm sorry to revive an old thread, but does anybody want to do the remaining hands?

[hide=" One-Pair"]
\[\frac{ \dbinom{13}{1}\dbinom{4}{2}\cdot \dbinom{12}{3}\dbinom{4}{1}^{3}}{ \dbinom{52}{5}}= \frac{352}{833}\approx 0.423 \] [/hide]
[hide=" Two-Pair"]
\[\frac{ \dbinom{13}{2}\dbinom{4}{2}^{2}\cdot \dbinom{11}{1}\dbinom{4}{1}}{ \dbinom{52}{5}}= \frac{198}{4165}\approx 0.0475 \] [/hide]
[hide="Three-of-a-Kind"] \[\frac{ \dbinom{13}{1}\dbinom{4}{3}\cdot \dbinom{12}{2}\dbinom{4}{1}^{2}}{ \dbinom{52}{5}}= \frac{88}{4165}\approx 0.0211 \] [/hide]
Calculating the probability of getting absolutely nothing is just a lot of casework, I think. And I don't feel like doing casework right now. :P

Good, but don't you have to make sure that the pair is not a 2-pair, 3-of-a-kind, 4-of-a-kind full house, or flush and that the 2-pair is not a full house or a flush and that the 3-of-a-kind is not a 4-of-a-kind or a full house?

Look again. I've already done that.  :)

(A note: why are the "hide" and "quote" features on AoPS not working?)

[quote="cincodemayo5590"](A note: why are the "hide" and "quote" features on AoPS not working?)[/quote]
[hide="I dont know why"]That happened to me too.[/hide]
 :rotfl:

Forgive me for reviving an old topic, but I don't get cincodemayo's answer for a two pair.

[hide="I got"]

$ {4\choose 2}^{2}\cdot{52\minus{}8\choose 1}$

Since there are 4 cards with a certain value (ex/ 4 aces, 4 kings, 4 10's, etc...), we must pick 2 out of 4 of a kind to get a pair. Do this again with 4 different cards, then just pick your remaining card as whatever you want.

And of course the probability is just what I mentioned divided by $ {52\choose 5}$, so

$ \frac{{4\choose 2}^{2}\cdot{52\minus{}8\choose 1}}{{52\choose 5 }}\approx 0.06\%$

[/hide]

Does anyone agree or am I overlooking something?

[quote="Prime factorization"]Does anyone agree or am I overlooking something?[/quote]

[hide="Yes, you forgot to include the fact that"]
there are 13 such groups of 4 to choose your pair from.[/hide]

[quote="Prime factorization"]So should it be [hide="this?"]

$ {13\choose 1}\cdot{12\choose 1}\cdot{4\choose 2}^{2}\cdot{52\minus{}8\choose 1}$

Since you are choosing 1 set of 4 of a kind, then another.

[/hide][/quote]

<<quoted post was deleted, I believe>>

[hide="Close, but"]
you shouldn't be picking the two groups of four in order. E.g. it doesn't matter if you pick Jacks then Queens, or Queens then Jacks, you'll get the same thing. So, it's $ \dbinom{13}{2}$ and not $ \dbinom{13}{1}\dbinom{12}{1}\equal{} 2\dbinom{13}{2}$.[/hide]


analytic geometry
graphing lines
slope
algebra
linear equation

The points $ (x, y)$ represented in this table lie on a straight line.
The point $ (28, t)$ lies on the same line. What is the value of $ t$?

\begin{tabular}{c|c}
$ x$ & $ y$ \\ \hline
1 & 7 \\
3 & 13 \\
5 & 19 \\
\end{tabular}

The pattern is that as $ x$ increases by 1, $ y$ increases by 3.  So as $ x$ increases by 23, $ y$ increases by 69 so $ (28,88)$ is the point.

straight line$ \implies$linear equation
we see that as $ x$ increases $ 2$ $ y$ increases 3 so the slope is $ 3$
now substituting $ x\equal{}1$ and $ y\equal{}7$ into point-slope form of line $ y\equal{}mx\plus{}b$ $ 7\equal{}3\cdot1\plus{}b$ $ b\equal{}4$
so now substituting $ 28$ for $ x$ we find $ y$.
$ y\equal{}3\cdot28\plus{}4\equal{}84\plus{}4\equal{}88$
the answer is 88.

I think you mean "as $ x$ increases $ 2$, $ y$ increases $ \it{6}$ so the slope is $ 3$."


linear algebra
matrix
linear algebra unsolved

find all the matrixes

A=\a  b\
           \c   d\
wich apply that
A^k=\a^k    b^k\
                  \c^k    d^k\

regards

squaring the matirx A gives the system:

a^2+bc=a^2
b(a+d)=b^2
c(a+d)=c^2
d^2+bc=d^2

Now, either b,c=0 which will give a solution for every a,d, or one of b,c equals 0, both cannot be non-zero because from equation 1, bc=0.

WLOG, b=0. Then from equation 3, a+d=c. Note that this matrices squared will also have to satisfy the intial problem conditions. So for this matrix,

/a^2          0  /
/(a+d)^2  d^2/ we have either a+d=0, which is covered by the case b=c=0, or a^2+d^2=(a+d)^2, 2ad=0, a=0, or d=0. This gives the remaining solutions, making the solutions of the form,

A row, or column of zeroes, or b=c=0.


inequalities
geometry
inequalities unsolved

Prove that

$ abc < a^2(A \minus{} a) \plus{} b^2(A \minus{} b) \plus{} c^2(A \minus{} c)\le \frac {3}{2}abc$

where $ A$ is the area of a trangle which has sides $ a,b,c$

 :o

Are you sure? Consider equilateral triangle with $ a\equal{}b\equal{}c$, then $ A\equal{}\frac{\sqrt{3}}{4}a^2$ and must be $ a^3<3a^2(\frac{\sqrt{3}}{4}a^2 \minus{}a)\le \frac{3}{2}a^3\Rightarrow a< 3(\frac{\sqrt{3}}{4}a^2 \minus{}a)\le \frac{3}{2}a$ which is obviously wrong, take for example $ a\equal{}1$ or $ a\equal{}100$.

[quote="nilenbert"]Are you sure? Consider equilateral triangle with $ a \equal{} b \equal{} c$, then $ A \equal{} \frac {\sqrt {3}}{4}a^2$ and must be $ a^3 < 3a^2(\frac {\sqrt {3}}{4}a^2 \minus{} a)\le \frac {3}{2}a^3\Rightarrow a < 3(\frac {\sqrt {3}}{4}a^2 \minus{} a)\le \frac {3}{2}a$ which is obviously wrong, take for example $ a \equal{} 1$ or $ a \equal{} 100$.[/quote]

 Could it hold when the triangle is not equilateral ?

Yes, it could, but I don't see clear conditions for this. The problem is that this inequality is not homogeneous.


trigonometry
inequalities unsolved
inequalities

Given reals $ x$ and $ y$ such that $ xy(x^{2} \minus{} y^{2}) \equal{} x^{2} \plus{} y^{2}$. Find minimal possible value of $ x^{2} \plus{} y^{2}$.

you need also some condition
because I can say $ x^{2}\plus{}y^{2}\equal{}0$
or let $ x\equal{}y\equal{}0$

Sorry I forgot the condition that $ x$ isn't equal to $ 0$.

Put  $ x\equal{}r\cos t ,y\equal{}r\sin t, r\neq 0$
$ xy(x^{2} \minus{} y^{2}) \equal{} x^{2} \plus{} y^2 \ \Longrightarrow \ r^4 \cos t \sin t (\cos^2 t \minus{} \sin^2 t) \equal{} r^2$$ \ \Longrightarrow\  r^2 \sin (4t) \equal{}4 \ \Longrightarrow \ r^2_{min} \equal{}4$
 :cool:

I don't think that this problem is so difficult. It is first question of British M.O. held on 31. January 2006.

Please see also link:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=420940&search_id=764050530#420940[/url]


In a metropolis, there are n underground lines (n>4). At most three lines meet at any station, and for any two lines there is a third line such that one can make a transfer to it from each of the two lines. Prove that there are at least 5/6*(n-5) underground stations.

[i]Kmal, A. 292, March, 2002[/i]

The statement is incorrect : Let L be a metro line with metro stations S_1,...,S_p.
For each i, let L_i and L_i' be two metro line which only have S_i as metro station.
We then have n = 2p+1 metro lines ans p metro stations. But, for n large enough, 5(n-5)/6 > (n-1)/2, thus a problem....

Pierre.

I don't understand your example, Pierre!


conics
ellipse
analytic geometry
graphing lines
slope
calculus
calculus computations

Let $a,b$ be postive numbers. In $xy$ plane, let $C$ be the part of the ellipse $E: \ ax^2+by^2=1\ (x\geq 0,\ y\leq 0),$
and $l_t$ be the half-line $y=tx\ (x\geq 0)$ with slope $t\geq 0.$ When the segment $PP'$ joining the point on $l_t$ and the point $P'$ on $C$ moves parallel to $y$-axis, denote $P_t$ be $P$ which maximize the length of $PP'$ and denote $C'$ be the curve ploted by $P_t.$ Let $Q(\alpha,\ \beta)$ be the intersection point of $E$ and $C'.$

(1) Find the equation $y=f(x)$ of  the curve $C'.$
(2) Find $\alpha$ and $\beta.$
(3) Let $D$ be the region enclosed by the line $y=\beta$ and the curve $C'.$ Find the volume $V$ formed by the revolution of $D$ about $y$-axis.

Frankly speaking, I could not understand this problem. There must be something wrong with the translation. Please check.


trigonometry
geometry
3D geometry
summer program
MathPath
similar triangles

Prove that for all integer n, $\sin n$ has a calculable exact value.

Please clarify.

1) Is n in radians or degrees?
2) What do you mean by "a calculable exact value"? Do you mean that the number is rational? (This is false.) That the number can be expressed in terms of radicals, i.e. algebraic? (This is also false.) That it's transcendental, but has a closed form expressed in terms of something else? (In terms of what?)

After all, you can't even calculate the value of $\frac{1}{3}$ if you define "calculate" as "express using a finite number of digits".

[quote="lingomaniac88"][hide]ALL VALUES ARE IN DEGREES.

Since $\sin 15$ and $\sin 18$ are both calculable, then so are $\cos 15$ and $\cos 18$.  So $\sin 3$ is also calculable (simply rearrange to get $\sin (18-15)$).
Since $\sin 3x=3\sin x-4\sin^{3}x$, $\sin 3=3\sin 1-4\sin^{3}1$.  Rearranging gets $-4\sin^{3}1+3\sin 1-\sin 3=0$.  Since this is a cubic, we can compute the exact values of $\sin 1$ by the cubic formula.  We can then compute $\cos 1$.  Repeated usage of $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$ reveals a way to compute $\sin x$ for all $x\in\mathbb{N}$.  Finally, using the fact that $\sin-\theta=-\sin\theta$, we can then calculate $\sin x$ for all $x\in\mathbb{Z}$.  But some of these values will be really ugly.

$\mathbb{QED}$[/hide][/quote]
Correct.

"o i c" :blush: Sorry...

Here's a related problem: Which values are constructible (or equivalently, which values are surds?)

[quote="lingomaniac88"]Since $\sin 15$ and $\sin 18$ are both calculable, [/quote]

How would you find $\sin 18$?

It was discussed at MathPath 2005...

...basically, you find $\sin 72$ using some really amazing similar triangles in a regular pentagon.

I think.

$x=18^\circ$

$\cos{2x}=\sin{3x}$ (better because you don't need to derive sin5x in terms of sinx, also because cos2x can be put in terms of sinx)

or $\sin{5x}=1$

[quote="t0rajir0u"][quote="lingomaniac88"]Since this is a cubic, we can compute the exact values of $\sin 1$ by the cubic formula. [/quote]

Incorrect.

Have you taken a look at the cubic formula?  One step (for this particular problem) involves finding the cube root of $e^{i \frac{\pi}{60}}$ (or something along those lines) which is [b]equivalent[/b] to calculating $\sin 1^{\circ}$, so this is in fact circular logic.[/quote]
:ewpu:
I just realized that the problem statement was horribly wrong, I made this same mistake.
If the question was true then the 20 degree angle would actually be constructable. It also explains the multiple of three for angle construction. Gah.

Anyways, to calculate sin18, my favorite method is taking an equiangular pentagon, drawing a diagonal, and using ptolomey's theorem.

No, the problem statement is correct, if you define "calculable" as "algebraic".

$\sin 20$ is algebraic; it's just a solution to a cubic. On the other hand, it is not constructible. The condition for constructability is slightly stronger: to be constructible, a number must be a surd (i.e. expressible using square roots only).


Let $ u\equal{}e^x\plus{}\frac{1}{e^x}\plus{}1$ then $ du\equal{}e^x\minus{}\frac{1}{e^x}$

$ \int {\frac {{e^{2x} \minus{} e^x }} {{(e^x \plus{} 1)\sqrt {e^{3x} \plus{} e^{2x} \plus{} e^x } }}} dx\equal{}\int {\frac {{(e^{x} \minus{}\frac{1}{e^x })dx}} {{(e^x \plus{}2\plus{} \frac{1}{e^x })\sqrt {e^{x}\plus{}\frac{1}{e^x }\plus{}1 } }}}\equal{}\int{\frac{du}{(u\plus{}1)\sqrt{u}}}$

Rest is easy...


[b]Problem.[/b] [i]In the parliament of Zargaria, there are $7$ non-empty parties.  A dictator, creating multiple parties to fool its people, is to choose $13$ people (pretty small parties) from $17$ of the most intelligent politicians of Zargaria and place them into the distinct parties.  In how many ways can this dictator create parties?  Find its last nonzero digit (if you want).[/i]

Instead of counting all those arrangements, you can just find the number of positive integer solutions to $a_{1}+\hdots+a_{m}=n$.  If $n=1+1+...+1$, we have to place $m-1$ $)+($ in the sequence $1\ \_\ 1\ \_\ 1\hdots\_\ 1$ with $n-1$ $\_$.  So there are $\binom{n-1}{m-1}$. solutions.  In this case, we have $n=13$ and $m=7$, which gives us $\binom{12}{6}$.  You miscounted the arrangements.

Nice solution, but I dont get how you do the distribution. Could you explain it a little clearer? Thank you!

@ i_like_pie: did you use a calculator?  :D

[hide]Clearly, there are $\binom{17}{13}$ ways to choose the initial 13 elites. Assuming the elites to be 'pearls,' we can place these pearls into boxes, or parties, $a_{1}, a_{2}, \ldots, a_{7}.$

We have:

$\sum_{i=1}^{7}a_{i}= 13 | a_{i}\ge 1.$

A bijection exists between the number of solutions to the above equation and the following:

$\sum_{i=1}^{7}a_{i}=6 | a_{i}\ge 0.$

The number of solutions to this is simply $\binom{7+6-1}{6-1},$ so our final answer is:

$\binom{17}{13}\binom{12}{6}$

If we take this number modulo $10$ after eradicating all factors of $10,$ we have to evaluate:

$7\cdot 8 \cdot 17 \cdot 7 \cdot 3 \cdot 11 \bmod{10}.$

This number is clearly: $7^{3}\cdot 4 \equiv-108 \equiv 2 \bmod{10},$ so our first non-zero digit is $2.$

[/hide]

[quote="Karth"]@ i_like_pie: did you use a calculator?  :D[/quote]
Only for the last step (multiplying the three numbers). :)

[quote="boxedexe"]You have to multiply your answer by $13!$ since scientists cannot possibly be clones at this period of time.[/quote]
So I [i]was[/i] right (besides miscounting the arrangements)!


Vieta
quadratics
complex numbers
algebra
quadratic formula

For real numbers $ x,y,z$:

\[ \sqrt{x} \plus{} \sqrt{y} \plus{} \sqrt{z} \equal{} 5
\\
\sqrt{xy} \plus{} \sqrt{xz} \plus{} \sqrt{yz} \equal{} 8
\\
\sqrt{xyz} \equal{} 4\]

Find all triples $ (x,y,z)$ that satisfy above systems of equations.

Just use Vieta's formula $ \{\sqrt {x},\ \sqrt {y},\ \sqrt {z}\} \equal{} \{1,\ 2,\ 2\}$.

[quote="kunny"]Just use Vieta, $ \{\sqrt {x},\ \sqrt {y},\ \sqrt {z}\} \equal{} \{1,\ 2,\ 2\}$.[/quote]

kunny, I believe your solution set's incomplete. Furthermore, I want to see at least ways you have done it (I know what you mean by Vieta's, I'm saying this in sake of others who don't).

More harder Problem created by kunny.

Find complex numbers $ a,\ b,\ c$ such that:

$ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \equal{} 3$

$ \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \equal{} 4$

$ \sqrt {abc} \equal{} 2$

[quote="kunny"]I can't understand what you say. Maybe you don't understand what $ \{\sqrt {x},\ \sqrt {y},\ \sqrt {z}\} \equal{} \{1,\ 2,\ 2\}$ does mean.

Anyway this problem is very easy.

$ \sqrt {x},\ \sqrt {y},\ \sqrt {z}\in{t^3 \minus{} 5t^2 \plus{} 8t \minus{} 4 \equal{} 0}$

$ \Longleftrightarrow (t \minus{} 1)(t \minus{} 2)^2 \equal{} 0$

$ \Longleftrightarrow t \equal{} 1,\ 2(double\ root)$.

$ \therefore (x,\ y,\ z) \equal{} (1,\ 4,\ 4),\ (4,\ 1,\ 4),(4,\ 4,\ 1)$. Done![/quote]

Yes, now you have it correctly.

By incomplete, I meant that your solution set (three of them) weren't there (you only had one), and I wasn't sure if you meant brackets to be symmetric for $ x,y,z$. And people are more than welcomed to try kunny's problem (which is definitely more fun and challenging!)  :)

I have to say, there's only one system of equations there

kunny:

$ \sqrt {a}, \sqrt {b}, \sqrt {c} \in t^{3} \minus{} 3t^{2} \plus{} 4t \minus{} 2 \equal{} 0 \Longleftrightarrow$

$ (t \minus{} 1)(t^{2} \minus{} 2t \plus{} 2) \equal{} 0 \Longleftrightarrow$

$ t \equal{} 1, 1 \pm i$  (quadratic formula) $ \Longrightarrow$

$ \{a, b, c \} \equal{} \{ 1, 2i , 2 \minus{} 2i \}$ ?

[quote="t0rajir0u"][quote="Silverfalcon"]I meant that your solution set (three of them) weren't there (you only had one)[/quote]
kunny was using set notation and sets are unordered.[/quote]

Oh okay. I wasn't sure what that meant.  :oops:

So, if the answer was like $ \{ 4, 5, 6 \} \equal{} \{ r, s, t\}$ then it means there are 6 ordered triples of $ (r,s,t)$?

Yes, as long as the equations are symmetric; your problem had only $ 3$ ordered triples, because $ 4$ appeared twice in the set of possible values.


inequalities
function
three variable inequality

This is a hard inequality. And also note that from this one we can find another tough inequality, posted by me long time ago: if $ a^2+b^2+c^2=3 $ then $ a^3(b+c)+b^3(c+a)+c^3(a+b)\leq 6 $.

We may consider that $a,b,c\leq1$. Consider the functions $g(a,b,c)=f(a)=(\sum a^2)^2-3\sum a^3b$. Then $g(a,b,c)$ has a minimum.
$f'(a)=4a^3-9a^2b+4a(b^2+c^2)-4c^3$ and $f''(a)=12(a^2-\frac32ba+\frac13(b^2+c^2))\geq0$
So $f(a)$ is minimal when $f'(a)=0$. Analogously for $b$ and $c$, $g(a,b,c)$ is minimal when
$4a^3-9a^2b+4a(b^2+c^2)-3c^3=4b^3-9b^2c+4b(c^2+a^2)-3a^3=4c^3-9c^2a+4c(a^2+b^2)-3b^3=0$
Summing we get $\sum (a^3+4ab^2-5a^2b)=0$ which holds iff $a=b=c$.

I don't think the solution is good. How do you know that the minimum when considered as a function in a is attained at the same point as the minimum when considered as a function in b or c. Am I missing something.

Let g(a,b,c) have minimum in (a0,b0,c0). Assume that f(a) doesn't have minimum in a0 then we can a1 such as f(a1)>f(a0). Then g(a1,b0,c0)>g(a0,b0,c0). Contradiction. So a0 is the minimum of f(a). Consider h(b)=s(c)=g(a,b,c). In the same way we get that h(b) has a minimum in b0 and s(c) has a minimum in c0. Then f'(a0)=h'(b0)=s'(c0)=0.
Is it good now? :)

I see, you use partial derivatives. Well, in this case it works with the condition to explain very well the existence of the minimum and so on. Anyway, do not take into account what I say now, I'm very tired and I'm surely idiot!

I don't know this method of partial derivatives, but I knew that for proving that it is a minimum it doesn't suffice just to compute the first partial derivatives and then to solve the system. I knew you have to compute the second partial derivatives and to show that some determinants are positive or negative.

Can anyone help me and tell if this method is corect? It sem\ems to easy to be true!

I think that this method isn't correct because to find the min and max of a function of more variables you must use Sylvester theorem which involves the calculus of three determinants composed with partial derivatives.

Sasha is only partially right.
If $f_1(a)=g(a,b,c)$ is minimal at $a=1$, $f_2(b)=g(a,b,c)$ is minimal at $b=1$ and $f_3(c)=g(a,b,c)$ is minimal at $c=1$, then $g(a,b,c)$ is minimal at $(1,1,1)$.
Indeed, we have $g(a,b,c\geq g(1,b,c) \geq g(1,1,c) \geq g(1,1,1)$.
But the obtained inequality $g(a,b,c)\geq g(1,1,1)$ is trivial.
In our case, $f_1(a)=(\sum a^2)^2-3\sum a^3b$ is not minimal at $a=1$ for all $b$ and $c$. Moreover, the conditions $f_1'(1,1,1)=f_2'(1,1,1)=f_3'(1,1,1)=0$ and $(f_1)''(1,1,1)>0$, $(f_2)''(1,1,1)>0$, $(f_3)''(1,1,1)>0$
dont imply $g(a,b,c)$ is (local)minimal at (1,1,1). Consequently, the Sasha solution is not correct.
I have to notice that the inequality proposed by Manlio with unknown source was published in G.M 7-8/1992 (problem 22695, pg 287, author Vasc, sorry) in the form

$a^2(a-b)(a-2b)+b^2(b-c)(b-2c)+c^2(c-a)(c-2a)\geq0$.

This inequality is not trivial, is a very strong inequality. It follows from the identity

$4(a^2+b^2+c^2-ab-bc-ca)[(a^2+b^2+c^2)^2-3(a^3b+b^3c+c^3a)]=$
$=[(a^3+b^3+c^3)-5(a^2b+b^2c+c^2a)+4(ab^2+bc^2+ca^2)]^2+$
$+3[(a^3+b^3+c^3)-(a^2b+b^2c+c^2a)-2(ab^2+bc^2+ca^2)+6abc]^2$.

I have also a very nice solution for this inequality, but I will not post it, because it is a part of a problem proposed by me to Crux.

I have just one simple question: how on earth did you deduce this identity? I mean, I just can't imagine deducing such thing. It is absolutely unbelievable!

This identity follows from the proof.
This one can be an little indication about my proof. :P

Vasc, this little clue makes the search of your VERY NICE

proof a fantastic challenge for all very clever solvers of inequalities of the forum.

(unfortunately I am not among them, but I will try it anyway)

again this problem depends on the sign of S=(a-b)(b-c)(c-a).if S is greater than 0 ,then the inequality is proved by what i've done but if S is less than zero then taking a= max{a,b,c} and c<b is th true
the substitution b=b,a=b+ :lambda,c=b- :gamma:   kills me with computations
any1 has cum up with anything different ,not vasc's identity,it's a bolt from the blue

My last dream (well, maybe not exactly the last, but close!) is to see a beautiful 3 lines solution without identities. But somthing tells me that such solution can't exist. The inequality is far too strong. Keep searching. After all, for that problem in Crux I spent more than 2 months.

[b]Leonard[/b] and [b]Kittyw[/b], you are very special mathlinkers, please don't hurt each other!

There is a related version of it posted under some "I Get it" collection here in this forum on AoPS by Grotex. But can't find it now due to poor searching.

[quote=TaHongQuang][b]THQ202[/b] (Inspired by VASC and Sladjian Stankovik)
Let $ a, b, c   $ be real numbers and $s = \sqrt {\frac{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}{7}}  $. Prove

$$
 \left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{2}
\ge 3\left( {a}^{3}b+{b}^{3}c+{c}^{3}a \right) +{\frac {7\,{s}^{2} \left( a+b+c-7\,s
 \right) ^{2}}{9}}\
 $$[/quote]
After a few moves (not easy, but not difficult either) we get:
$$\min\left[\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{2}-3\left( {a}^{3}b+{b}^{3}c+{c}^{3}a \right)\right]=\frac {7\,{s}^{2} \left( a+b+c-7\,s
 \right) ^{2}}{9}.$$
(of course, by fixing first $a+b+c\neq0$ and $ab+bc+ca$).
Note that if $a+b+c=0,$ then LHS=RHS. Interesting identity.

[quote=arqady][quote=TaHongQuang][b]THQ202[/b]
Let $ a, b, c   $ be real numbers and $s = \sqrt {\frac{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}{7}}  $. Prove

$$
 \left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{2}
\ge 3\left( {a}^{3}b+{b}^{3}c+{c}^{3}a \right) +{\frac {7\,{s}^{2} \left( a+b+c-7\,s
 \right) ^{2}}{9}}\
 $$[/quote]
The $uvw$'s technique should kill it.[/quote]

I wrote in October 2017 an article, where I show through different methods that
$$\min\left[\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{2}-3\left( {a}^{3}b+{b}^{3}c+{c}^{3}a \right)\right]=\frac {7\,{s}^{2} \left( a+b+c-7\,s
 \right) ^{2}}{9},$$
for $a+b+c\neq0$ and $ab+bc+ca,$ both fixed. It will appear this week in GMB (Gazeta Matematica).

[quote=arqady][quote=manlio]For $a,b,c$ positive real numbers prove that

 $ a^2+b^2+c^2 \geq \sqrt {3(a^3b+b^3c+c^3a)} $[/quote]
Let $a=x+2\cos\frac{\pi}{7}y$, $b=y+2\cos\frac{\pi}{7}z$ and $c=z+2\cos\frac{\pi}{7}x$.
Hence, $\sum_{cyc}a^3b=\sum_{cyc}\left(x+2\cos\frac{\pi}{7}y\right)^3\left(y+2\cos\frac{\pi}{7}z\right)=$
$=\sum_{cyc}\left(x^3+6\cos\frac{\pi}{7}x^2y+12xy^2\cos^2\frac{\pi}{7}+8\cos^3\frac{\pi}{7}y^3\right)\left(y+2\cos\frac{\pi}{7}z\right)=$
$=\sum_{cyc}\left(Ax^4+Bx^2y^2+Cx^2yz+\left(1+16\cos^4\frac{\pi}{7}\right)x^3y+\left(2\cos\frac{\pi}{7}+12\cos^2\frac{\pi}{7}\right)x^3z\right)$,
where $A$, $B$ and $C$ are constants.
But $1+16\cos^4\frac{\pi}{7}=2\cos\frac{\pi}{7}+12\cos^2\frac{\pi}{7}$.
Thus, we need to prove a fourth degree symmetric inequality.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative and $xyz=w^3$.
Id est, we need to prove that $f(w^3)\geq0$, where $f$ is a linear function.
Hence, $f$ gets a minimal value for an extremal value of $w^3$, which happens, when two variables are equal.
Since we have a fourth degree inequality, we can assume $y=z=1$.
Thus, it remains to prove that $\left(\left(x+2\cos\frac{\pi}{7}\right)^2+\left(1+2\cos\frac{\pi}{7}\right)^2+\left(1+2\cos\frac{\pi}{7}x\right)^2\right)^2\geq$
$\geq3\left(\left(x+2\cos\frac{\pi}{7}\right)^3\left(1+2\cos\frac{\pi}{7}\right)+\left(1+2\cos\frac{\pi}{7}\right)^3\left(1+2\cos\frac{\pi}{7}x\right)+\left(1+2\cos\frac{\pi}{7}x\right)^3\left(x+2\cos\frac{\pi}{7}\right)\right)$,
which is $(x-1)^2\left(x-8\cos^2\frac{\pi}{7}-2\cos\frac{\pi}{7}+1\right)^2\geq0$. Done! :)[/quote]

beautiful!

[quote=mihaig]It's wonderful. But you have also access to GMB and you'll soon read my article[/quote]

Thanks, I look forward to it!

See my article on this fascinant result of the professor in the below link. By courtesy of the Romanian Mathematical Society (the article is in the official site)
[url]https://ssmr.ro/gazeta/gmb/2021/9/articol.pdf[/url]

[quote=manlio]For $a,b,c$ positive real numbers prove that

 $ a^2+b^2+c^2 \geq \sqrt {3(a^3b+b^3c+c^3a)} $[/quote]

We gef

 $$ a^2+b^2+c^2 \geq \sqrt {3(a^3b+b^3c+c^3a)} $$
$$\implies a^2(a^2+2b^2-3ab) +b^2(b^2+2c^2-3bc) + c^2 (c^2 +2a^2 - 3ac) \geq 0 $$

WLOG we can suppose $a= min(a,b,c)$ so:

$$\implies a^2(a^2+2b^2-3ab) +b^2(b^2+2c^2-3bc) + c^2 (c^2 +2a^2 - 3ac) \geq 3a^2(a^2+b^2+c^2 -ab-ac-bc) \geq 0$$

with equality if $a=b=c$

[quote=teomihai]indeed  splendid articol.[/quote]

Thank you, dear professor. Not only solves the problem, but it refine it. I will mention again that the article was sent in 2017. I'm sure that meanwhile new results appeared

[url=https://artofproblemsolving.com/community/c6h536442_a2b2c22__3a3bb3cc3a__f_f__]Some other identities.[/url]
[url=http://tieba.baidu.com/p/7717458615][/url]Something stranger: For \(a,b,c>0\), show that \[7\sum_{\rm cyc}{a^4bc}+\sum_{\rm cyc}{a^3b^3}-29\sum_{\rm cyc}{a^3b^2c}+12\sum_{\rm cyc}{a^3bc^2}+9\sum_{\rm cyc}{a^2b^2c^2}\ge0\] and \[\sum_{\rm cyc}{a^5}-\sum_{\rm cyc}{a^4b}-2\sum_{\rm cyc}{a^4c}+3\sum_{\rm cyc}{a^3b^2}-\sum_{\rm cyc}{a^3c^2}+2\sum_{\rm cyc}{a^3bc}-2\sum_{\rm cyc}{a^2b^2c}\ge0\text.\]


Let $ABC$ be a triangle with othocentre$H$. let $M$ be the midpoint of $BC$.
Let $\ell$ be a  line passing through $A$ and prependicular to $MH$.
Let $A`=\ell \cap BC$.
Same definations for $B`,C`$.
Prove that $A`B`C`$ is a line prependicular to euler line of $ABC$.

Hint: Prove that the $A'B'C'$ is the orthic axis of triangle $ABC$.

Just a quick sketch:

Let $D,E,F$ be the feet of the altitudes from $A,B,C$. It's easy to see now that $A',B',C'$ are the harmonic conjugates of $D,E,F$ wrt $(B,C),(C,A),(A,B)$ respectively, so $A'=EF\cap BC,B'=FD\cap CA,C'=DE\cap AB$.

Now perform the inversion of pole $H$ which takes $A$ onto $D$. It turns $A',B',C'$ into the perpendiculars from $H$ to the medians through $A,B,C$ respectuvely, which means that it turns the line $A'B'C'$ into the circle having $GH$ as diameter ($G$ is the centroid). The angle between this circle and the Euler line is $\frac\pi 2$, so the same must hold for the images of these two, which are the line $A'B'C'$ and the Euler line respectively.

Edit: the post above wasn't there when I started typing this :)


Consider the expression 1 ÷ 2 ÷ 3 ÷ 5 ÷ 7 ÷ 11 ÷ 13 ÷ 17 ÷ 19. If you are allowed to put as many parentheses as you want, how many different values can be obtained for this expression?

[hide]$ \boxed{128}$[/hide]

Well... $ 128\equal{} \dbinom{7}{0}\plus{}\dbinom{7}{1}\plus{}\hdots\plus{}\dbinom{7}{7}$

Maybe err.......  xD

well you don't have 7,7 because that is 7!/0

[quote="mathcrazed"]Well... $ 128 \equal{} \dbinom{7}{0} \plus{} \dbinom{7}{1} \plus{} \hdots \plus{} \dbinom{7}{7}$

Maybe err.......  xD[/quote]

How did you get this expression?  Thanks.

I think by his explanation he just got it randomly from the Hockeystick Identity...

That's not hockey stick :wink:

Try the problem for 1 number, 2 numbers, 3 numbers, etc.  try to explain your results, and note that the numbers being used are prime.

[hide="Possible Solution?"]There are a total of 7 possible ways to put parentheses (after that, the answers just repeat), like around (1/2) or (1/2/3),...Each parentheses can be there, or not be there, so there are 128 ways.[/hide]

Instead of looking at where the parentheses can go, which makes the problem much harder, look at the final answer.  You know that 2 can't be in the numerator, so the other 7 odd primes, namely 3, 5, 7...19 can either be in the denominator, or the numerator, for a total of 2^7 possibilitites.  2 has been taken care of, since it has to be in the denominator, and 1 has also been taken care of, since when all of the numbers are in the denominator, 1 is in the numerator.  Hence, the answer is 2^7=[u]128[/u]


trigonometry
geometry
circumcircle
radical axis
geometry unsolved

Let $ AB$ be a diameter of a circle $ S$, and let $ L$ be the tangent at $ A$. Furthermore, let $ c$ be a fixed, positive real, and consider all pairs of points $ X$ and $ Y$ lying on $ L$, on opposite sides of $ A$, such that $ |AX|\cdot |AY| \equal{} c$. The lines $ BX$ and $ BY$ intersect $ S$ at points $ P$ and $ Q$, respectively. Show that all the lines $ PQ$ pass through a common point.

[asy]defaultpen(fontsize(8));
pair A=(0,0), B=(10,0), X=(0,10), Y=(0,-5), H, P, Q;
path S = Circle(midpoint(A--B),abs(A-B)/2);
P = intersectionpoints(S, X--B)[1];
Q = intersectionpoints(S, Y--B)[1];
H = extension(P,Q,A,B);
draw(S);
draw(A--Y--B--X--A--P--Q--A--B);
dot(P^^Q^^A^^B^^X^^Y^^H);
label("A",A,(-1,0));label("B",B,(1,0));label("P",P,(1,1));label("Q",Q,(1,-1));label("X",X,(-1,0));label("Y",Y,(-1,0));label("H",H,(1,-1));[/asy]Let $ H\equal{}PQ\cap AB$.
Claim: As $ X$ and $ Y$ vary on $ L$ such that $ AX\cdot AY \equal{} c$, $ H$ remains fixed on $ AB$.
It is sufficient to show that $ \frac{[PAQ]}{[PBQ]}$ remains constant.
$ \frac{AX}{AB}\equal{}\frac{AP}{BP}$, $ \frac{AY}{AB}\equal{}\frac{AQ}{BQ}$
$ \implies\frac{c}{AB^2}\equal{}\frac{AX\cdot AY}{AB^2}\equal{}\frac{AP\cdot AQ}{BP\cdot BQ}\equal{}\frac{\frac{1}{2}AP\cdot AQ\sin{PAQ}}{\frac{1}{2}BP\cdot BQ\sin{PBQ}}\equal{}\frac{[PAQ]}{[PBQ]}$

Let $ X',\ Y'$ be, the points on the tangent line $ L$ of the given circle $ (O)$ at point $ A,$ as the problem states and also such that $ AX' \equal{} AY'$ $ ,(1)$

So, we have that $ (AX)\cdot (AY) \equal{} u^{2} \equal{} (AX')^{2} \equal{} (AY')^{2}$ $ ,(2)$

We denote the points $ P'\equiv (O)\cap BX',\ Q'\equiv (O)\cap BY'$ and let be the point $ H\equiv AB\cap P'Q'.$

We will prove that the line-segment $ PQ,$ always passes through the point $ H.$

From the right angled triangle $ \bigtriangleup ABX',$ we have that $ (BP')\cdot (BX') \equal{} (AB)^{2}$ $ ,(3)$

Similarly from the right angled triangle $ \bigtriangleup ABY',$ we have that $ (BQ')\cdot (BY') \equal{} (AB)^{2}$ $ ,(4)$

From $ (3),$ $ (4),$ we conclude that the points $ X',\ P',\ Q',\ Y'$ are concyclic, with their circumcircle so be it $ (O'),$ which intersects the line-segment $ AB,$ at points $ M,\ N$ such that $ (BP')\cdot (BX') \equal{} (BM)\cdot (BN) \equal{} (AB)^{2}$ $ ,(5)$ and $ (AM)\cdot (AN) \equal{} u^{2}$ $ ,(6)$

By the same way as before, we can say that $ (BP)\cdot (BX) \equal{} (AB)^{2} \equal{} (BQ)\cdot (BY)$ $ ,(7)$ from the right angled triangles $ \bigtriangleup ABX,\ \bigtriangleup ABY.$

From $ (5),$ $ (7),$ we conclude that the points $ X,\ P,\ M,\ N$ are concyclic, with circumcircle so be it $ (O_{1}).$

Because of $ (AM)\cdot (AN) \equal{} u^{2} \equal{} (AX)\cdot (AY)$ $ ,(8)$ we conclude that the point $ Y,$ lies on $ (O_{1}).$

The point $ Q$ now, lies also on the circle $ (O_{1}),$ because of $ (7).$

That is, the circle $ (O_{1}),$ always passes through the constant points $ M,\ N$ and then, we conclude that the line segment $ PQ,$ always passes through the fixed point $ H\equiv AB\cap P'Q',$ as the radical center of the circles $ (O),\ (O'),\ (O_{1})$ and the proof is completed.

Kostas Vittas.


probability
function
integration
calculus
derivative
expected value
probability and stats

I don't really understand expected value in accordance with probability distribution function. I looked up on Google and stuff but the explanations seem to be vague to me. Like, I don't know understand how they work. Can anyone clarify this for me?

OK, let's start with probability distribution function, also known as density function. Assume that $ \xi$ is a real valued random variable, which merely means that for every "reasonable" set $ E\subset\mathbb R$ we can tell the probability $ P\{\xi\in E\}$. Reasonable here really means "Borel", but, since I do not expect you to know any measure theory yet (if you do, say so!), think just of sets you can easily integrate over like finite unions of intervals. Now, it may happen that there exists a "reasonable" function $ f$ (again, the right word is "measurable" but for the purposes of this explanation, just think of "piecewise continuous") such that
\[ P\{\xi\in E\}\equal{}\int_E f(x)\,dx \qquad(*)\] for all reasonable $ E$. In this case $ f$ is called the density function of $ \xi$. That is the bare definition.
Question 1: do you have any problem with the formal side of it?

Now let's see what the density function allows you to do and what its basic properties are.
1) If $ f$ exists, it must be non-negative. Indeed, since the integral over every interval is non-negative (it is just the probability that $ \xi$ is in that interval), we can conclude that the value at every point where $ f$ is, say, continuous, is non-negative.
2) $ \int_{\minus{}\infty}^{\infty}f(x)\,dx\equal{}1$. Indeed, this is just the probability that $ \xi\in \mathbb R$, which is $ 1$.
Actually, given any reasonable function $ f$ with properties 1 and 2, we can design a random variable with density $ f$: just use $ (*)$ to define the probabilities and check that the resulting probabilities satisfy (countable) additivity property, which is just the additivity property of integral (the integral over the union of disjoint sets equals the sum of the integrals over the sets).

By definition, the density function $ f$ of $ \xi$ allows you to compute the probabilities of the events $ \xi\in E$ as the corresponding integrals. What can the value $ f(x)$ allow you to tell? It can tell you the approximate probability of the event $ \xi\in I$ where $ I$ is a very short interval containing $ x$: the integral then is almost the same as $ f(x)|I|$ because $ f$ doesn't change too much within $ I$ and the smaller $ I$ is, the better the approximation is. So, if you split the entire real line into very short intervals $ I_j$, you can approximate your random variable $ \xi$ with continuous distribution by the random variable with discrete distribution, namely, the random variable taking the value $ c_j$ (where $ c_j$ is some point in $ I_j$, say the center) with probability about $ f(c_j)|I_j|$.
Question 2. Does everything make sense so far?

I glanced through this, and it pretty much confirmed what I knew.  The notation was bit confusing (since obviously, it wasn't as same as what it was in the book) but mostly, you explained it really well!

I finally got some chance to write down some questions that I have so here they are!

Let's say that the function I'm considering is $ f(x) \equal{} \frac{3x^2}{8}$ on interval $ 0 \leq x \leq 2$. Now, I want to review/ask for finding mean, median, mode, standard deviation, interquartile range, mean deviation, and coefficient of skewness, $ \alpha^{\text{*}}_{3}$.

Below are my ideas (I hid them so it's easier to view).

[hide="Mean"]

$ \mu \equal{} \int^2_0 x\left(\frac{3x^2}{8}\right) dx$ [/hide]

[hide="Median"]

$ F(x) \equal{} \int^x_0 \frac{3x^2}{8} dx \equal{} \frac{x^3}{8}^{x}_0 \equal{} \frac{x^3}{8}$ (This was subtracting value of $ 0$ and $ x$ ---> I don't know how to put bracket and initial and final value as in integral for brackets so I just used subscripts and superscripts). [/hide]

[hide="Mode"]

Is this taking derivative (that's what my teacher suggested)? In my example in the book, this is what they wrote (the function was different; it was $ f(x) \equal{} \frac{1}{2} e^{\minus{}\frac{x}{2}}$ for $ x > 0$ and 0 for elsewhere).
"By examination, it can be seen that the pdf approaches a maximum when you approach zero from the right.  Therefore, we would say that the mode is 0 in the limit (Taking derivatives and setting to zero will not help in this case)." ---> WHY?!
 [/hide]

[hide="Standard Deviation"]

$ \sigma \equal{} \sqrt{\sigma^2} \equal{} \sqrt{E(X^2) \minus{} E(X)^2}$ [/hide]

[hide="Interquartile Range"]
$ x_{0.75} \minus{} x_{0.25} \equal{} Q_3 \minus{} Q_1$

For ex: $ \{1,3,5,7,9,11,13,15\}$
$ Q_3 \minus{} Q_1 \equal{} 11\minus{}3 \equal{} 8$, right? [/hide]

[hide="Mean Deviation"]
$ E|X \minus{} \mu|$

Question: In this case, is $ E(X) \equal{} \mu$ [I know I can't separate $ E(X)$ in $ E|X\minus{}\mu|$ but still, it confuses whether they are always treated like same or not. [/hide]

[hide="Coefficient of Skewness"]

$ \alpha^{\text{*}}_{3} \equal{} \frac{\mu_3}{(\sigma)^3}$

I don't understand this at all.. What it means and how to find it, its importance, etc.... [/hide]

1) You are completely correct about the mean. In general, if you want to find the expectation $ E G(X)$ where $ G$ is any function, you can do it by computing the integral $ \int_{ \minus{} \infty}^\infty G(x)f(x)\,dx$ where $ f$ is the density function. So, now you should be able to set the integrals for $ EX^2$, $ E|X \minus{} \mu|$ and such yourself.

2) Be careful: the cumulative distribution function is $ \frac {x^3}8$ only on $ [0,2]$. To the left of $ 0$ it is $ 0$ and to the right of $ 2$ it is $ 1$ (the reason is that $ f$ is given by your formula only on that interval and is extended by $ 0$ outside it). Now, the only thing you need to find the median and the quartiles is to solve $ F(x) \equal{} \frac 12, \frac14,\frac 34$. Your understanding of the interquartile range is correct except the quartiles like the medians are not always well-defined (in general, the upper quartile, say, is the value $ a$ such that $ P(X>a)\equal{}\frac 14$ or, which is the same, $ F(a)\equal{}\frac 34$ but, since the cumulative distribution fnction $ F$ may have jumps and intervals on which it is constant, this definition doesn't always make sense. Jumps are OK, of course, because jumping over $ \frac 34$ at $ a$ is as good as having the value $ \frac 34$ at $ a$, but flat intervals on the graph give you a headache. Some people prefer to think that there is more than one median/quartile in this case, some others prefer to define it as the middle point of that interval:
[asy]size(200);
D((-2,0)--(4,0),black);
D((0,2)--(0,0),black);
D(MP("\frac 34",(-0.05,3/4),8,W)--(1,3/4),dashed);
MC("F(x)",D((-2,0.1)..(-0,0.3)..(1,3/4)--(2,3/4)..(3,0.85)..(4,0.92),green+linewidth(1)),1);
for(int k=0;k<5;++k)
{
draw((1,-1)--D((1+k/4,3/4)),blue,Arrow);
}
MP("\mbox{Which one is the quartile here?}",(1,-1),8);[/asy]
You should just check the convention your book follows on that issue.

3) The mode is any point where $ f(x)$ attains its maximum. For smooth $ f$, the maxima can be found using the derivative. But in your case $ f$ is not only not smooth but even discontinuous at $ 2$. That's why you should check what happens at this point separately. Strictly speaking, the mode does not exist here but, since it is clear that to get the largest possible value of $ f$, you should approach $ 2$ as close as possible from the left, you can say what your teacher said: "the mode is $ 2$ in the limit".

4) Yes, $ \mu\equal{}E(X)$ in the mean deviation formula.

5) Here is a quote from Mathworld (adjusted to your notation):
[quote]
Skewness is a measure of the degree of asymmetry of a distribution. If the left tail (tail at small end of the distribution) is more pronounced that the right tail (tail at the large end of the distribution), the function is said to have negative  skewness. If the reverse is true, it has positive  skewness. If the two are equal, it has zero skewness.

Several types of skewness are defined, the terminology and notation of which are unfortunately rather confusing. "The" skewness of a distribution is defined to be
$ \alpha_3^*\equal{}\frac{\mu_3}{\sigma^3}$
where $ \mu_3\equal{}E(X\minus{}\mu)^3$ is the third central moment and $ \sigma$ is the standard deviation.
[/quote]
This tells you how to compute it. As to how to use it, it is not terribly useful as far as I can tell, so just learn how to determine it and skip over the rest.

I just printed what you typed up there. Right now, my computer is not working (I'm at local library) so I won't be able to type up my response until it gets fixed (hopefully soon...).

But thank you! Hopefully I can progress bit further with your explanation now  :)


[quote="TheAmazingOne"]I used PNG images to put graphics into my solutions where needed, because JPGs reduced the quality and it wasn't recognizing gifs or bmps. Am I not allowed to send PNG files? Will I have to convert them to JPGs?[/quote]
I don't see why you wouldn't be allowed to send PNG files.

It wasn't listed on this page
http://www.usamts.org/Problems/U_ProblemsSoln.php
As one of the accepted filetypes. But I'm wondering if that's a mistake, because PNG is just another common image format.

Well, I sent the PNG files, but I also sent a copy of the images as GIF files... I hope that this will be okay.

It's OK to send PNG files...we'll add them to the list for the next round.


calculus
derivative
calculus computations

How can we find the general expression of the derivative

$\frac{d^{n}\sqrt{f(x)}}{dx^{n}}~~~~\forall n\in \{0,1,2,\ldots\}$

By [url=http://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula]Faà di Bruno's formula[/url].


function
symmetry
algebra unsolved
algebra

find all continuous functions $f:R->R$ satisfy:
                $f(x+y+f(xy))=f(x+y)+xy$ for all $x,y \in R$

If we put $y=0$ then $f(x+y+f(xy))=f(x+f(0))$, but for $y=0$ $f(x+y)+xy=f(x)$ so $
f(x+f(0))=f(x)$ so if $f(0) \neq 0$ then $f(x)$ is a periodic function and has a finite $min$ and a $max$ in $R$ - from the Weierstrass theorem. But putting $x=-y \neq 0$ we obtain $f(f(-x^2))=-x^2$ (1) so $f(x)$ obtains every negative number what contradicts with the fact that $f(x)$ has a $min$. Thus $f(0)=0$.
In addition $f(x)$ satisfies $f(x) \neq f(y)$ for $x,y \leq 0 : x \neq y$.
That means that in $(-\infty,0]$ $f(x)$ is strictly decreasing or strictly increasing.
Case 1: So for $t \leq 0 $ we have
$f(f(t))=t$, so if $f(t) \neq t$ and $f(t)<0$ for $t<0$ and if $f(t)<t$ then $f(f(t))=t>f(t)$ so $f(x)$ is strictly decreasing but it has also a negative value so it can't have the continuousness property in $x=0$ - contradiction.
If $f(t)>t$ and $f(t)<0$ for $t<0$ then $f(f(t))<f(t)$ so we use the same argumentation as above - also contradiction. Thus for all $t \leq 0$ we have $f(t)=t$ if $f(t)<0$ for some $t<0$. It's trivial to show that $f(x)=x$ holds then for all real $x$.
Case 2:If $f(t)>0$ for some $t<0$ then $f(t)>0$ for all $t>0$ so $f(x)$ is strictly decreasing and $f(x)$ has a symmetry axis $y=x$ but putting $y=-\frac{1}{x}$ we get $f(x-\frac{1}{x}+f(-1))=f(x-\frac{1}{x})-1$ but
$x-\frac{1}{x}$ can obtain every real value so for all $t \in R$  $f(t+f(-1))=f(t)-1$ thus
points $(-f(-1),1),(0,0),(f(-1),-1)$ lie on a line so if $f(x)$ has a symmetry axis $y=x$ then the line on which the points lie
is $y=-x$ so obviously $f(-1)=1$ and $f(t+1)=f(t)-1$ for all $t \in R$. Now let's note that if we would replace
$y=x-\frac{1}{x}$ with $x-\frac{1}{kx}$ for $k$ positive then we would receive $f(-\frac{1}{k})=\frac{1}{k}$ solving the same way as before so $f(x)=-x$ for all $x \leq 0$. That implies-from the symmetry axis -that
$f(x)=-x$ for all real $x$.

So the answer is : $f(x)=x$ for all $x \in R$ or $f(x)=-x$ for all $x \in R$.