Assume that $f: [a,b]\to{\mathbb R}$ is derivable on its domain and moreover $f^{\prime}\ne 0$ on $[a,b].$ Suppose that equation $f(x)=0\; ,\; x\in [a,b]$ has a solution, say ${\mathbf x}$. To solve this equation, some approximations $x_{1},x_{2},...,x_{k},...$ of ${\mathbf x}$ are furnished by [b]Newton method[/b], i.e.
\[(1)\; \; \; \; \; x_{n+1}: =x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}\; ,\; \; n\in \{0,1,...,\}, \]
the "start-point"  $x_{0}$ being selected in $[a,b].$

[b]
I propose following question: let $A>1$  and $f_{1}(x): =x^{2}-A\; ,\; x\in \left[1,\frac{A+1}{2}\right]\; ,\; x_{0}=x_{0}^{[1]}: =\frac{A+1}{2}$.
Using (1) you rediscover a well-known method   (Heron-formula,"Babyloonian method").
 Let us denote by $(x_{n}^{[1]})$ the general term of the sequence given by (1).

Then, choose an integer $p\; ,\; p\in \{3,4,...\}$ and consider an equivalent equation, namely
\[f_{2}(x): = \frac{x^{2}-A}{\left((x+\sqrt{A})^{p-1}-(x-\sqrt{A})^{p-1}\right)^{\frac{1}{p-1}}}=0\; ,\; x\in \left[1,\frac{A+1}{2}\right]\; ,\]
with $\; x_{0}=x_{0}^{[2]}: =\frac{A+1}{2}\; .$
 By means of (1) one finds a sequence, say $\left(x_{n}^{[2]}\right)$.

[color=blue]
Prove/verify/that $\left(x_{n}^{[2]}\right)$ converges  to $\sqrt{A}$ much faster than $\left(x_{n}^{[1]}\right)$ . [/color]

For instance, consider $A\in (1,4)$ and take $p=5\; .$

[/b]

[b]Hint:[/b] Let $\varepsilon: =\left(\frac{\sqrt{A}-1}{\sqrt{A}+1}\right)^{2}$ and observe that $\varepsilon \in [0,1) .$ Try to prove that there exists positive constants $m_{1},m_{2},M_{1},M_{2}$ such that for all $n$
\[m_{1}\cdot\varepsilon^{2^{n}}\le \left|x_{n}^{[1]}-\sqrt{A}\right|\le M_{1}\cdot\varepsilon^{2^{n}}\; . \]
At the same time
\[m_{2}\cdot\varepsilon^{p^{n}}\le \left|x_{n}^{[2]}-\sqrt{A}\right|\le M_{2}\cdot\varepsilon^{p^{n}}\; . \]
For instance, when $A\in (1,4),p=5,n=5 ,$ we give $\left|\sqrt{A}-x_{5}^{[2]}\right|\le \frac{1}{{10}^{2944}}\; .$


limit
function
algebra
polynomial
search
calculus
calculus computations

Prove that $ \lim_{x \to 4}{(x^2\plus{} x \minus{} 11)} \equal{} 9$.

Let $ \delta =\min \left\{ 1,\frac{\epsilon }{10} \right\},$ then

\begin{eqnarray*}
   \left| x^{2}+x-11-9 \right|&=&\left| x^{2}+x-20 \right| \\
 & =&\left| x-4 \right|\left| x+5 \right| \\
 & <&10\left| x-4 \right| \\
 & <&10\cdot \frac{\epsilon }{10} \\
 & =&\epsilon,
\end{eqnarray*}

and we're done.

You would be more well-served by showing that the sum and product of continuous functions is continuous.  After showing that $ x$ and constant functions are continuous, you've then proven that polynomials are continuous.

[url=http://mathrefresher.blogspot.com/2009/02/polynomials-are-continuous.html]Here[/url] is a proof that polynomials are continuous.  It's not particularly difficult but it's quicker to search for a proof on google than actually type it out myself. :wink:


Prove that
\[ (\sin\gamma \plus{} a\cdot \cos\gamma)(\sin\gamma \plus{} b\cdot \cos\gamma)\leq 1 \plus{} (\frac {a \plus{} b}{2})^2\]

We are to prove that  $ \sin^2 \gamma \plus{} ab \cos^2 \gamma\plus{} \sin \gamma \cos \gamma (a\plus{}b) \le 1 \plus{} \left(\frac{a\plus{}b}{2} \right)^2$

or $ (ab\minus{}1) \cos^2 \gamma \plus{} \sin \gamma \cos \gamma \ (a\plus{}b) \le \left(\frac{a\plus{}b}{2} \right)^2$

Since $ ab \le \left(\frac{a\plus{}b}{2} \right)^2$ we attempt to prove that

$ \left[ \left(\frac{a\plus{}b}{2} \right)^2\minus{}1 \right] \cos^2 \gamma \plus{} \sin \gamma \cos \gamma \ (a\plus{}b) \le \left(\frac{a\plus{}b}{2} \right)^2$ which is equivalent to proving

$ t^2\minus{}t\plus{}1 \ge 0$ where $ t \equal{} \left(\frac{a\plus{}b}{2} \right) \tan \gamma$ which is obviously true


geometry
perimeter
inequalities unsolved
inequalities

If $ A$, $ B$, $ C$ are values of angles in triangle $ \triangle ABC$, $ a$,$ b$,$ c$ its sidelengths, $ s$ perimeter and $ x$,$ y$,$ z$ arbitrary real numbers prove that:

$ \frac{y\plus{}z}{x}\frac {A}{s(s\minus{}a)}\plus{} \frac{z\plus{}x}{y}\frac{B}{s(s\minus{}b)}\plus{}\frac{x\plus{}y}{z}\frac{C}{s(s\minus{}c)} \geq \frac{9}{s^{2}}$

[quote]I think the right hand side may be missed $ \pi$[/quote]

This is most probably true - can anyone give a proof__


[quote="STARS"]Find the greatest possible value of $ sin(cos x) \plus{} cos(sin x)$ and determine all real
numbers x, for which this value is achieved.[/quote]
$ Max(Sin(Cos(x))) \equal{}  Sin(1)$ -> is for $ x \equal{} 2{\pi}N$ ( N - integer )
$ Max(Cos(Sin(x))) \equal{}  1$   ->    is for $ x \equal{} {\pi}K$ ( K - integer )
So : $ Max(Cos(Sin(x))\plus{} Sin(Cos(x)))\equal{} 1 \plus{} Sin(1)$ -> is for $ x \equal{} 2{\pi}N$ ( N - integer ) :)

let $ f : x\equal{}\equal{}>sin(cos(x))\plus{}cos(sin(x))$. $ f$ is periodic with period $ 2\pi$ .and $ f(\minus{}x)\equal{}f(x)$.
in $ [0,\pi]$ we have $ f$ is non increasing (if $ x<y$ then $ f(x)>f(y)$) in $ [0,\pi/2]$ so $ f(x) \le f(0)\equal{} sin(1)\plus{}1$ and for $ [\pi/2,\pi] f(x) <1$.so the maximal value is $ 1\plus{}sin(1)$ which is achieved when $ x\equal{}0$.

see here

http://www.mathlinks.ro/viewtopic.php?t=215123


We consider that we have a $ 2$x$ 2$ matrix. Could we say that $ a^2\minus{}b^2\equal{}(a\minus{}b)(a\plus{}b)$ and why?

You mean $ a, b$ matrices?  No.  $ (a \minus{} b)(a \plus{} b) \equal{} a^2 \minus{} b^2 \plus{} (ab \minus{} ba)$, so the two are equal if and only if $ a, b$ commute.


ARML
geometry
calculus
integration
trigonometry
3D geometry
circumcircle

1.
In right triangle $ABC$, $\angle C = 90^\circ$, $c=3$, and $a+b=\sqrt{17}$.  Compute the area of the triangle.

2.
If $(a+b):(b+c):(c+a) = 6:7:8$, and $a+b+c=14$, compute the value of $c$.

3.
Let $z$ be a root of $z^5-1=0$, with $z \not= 1$.  Compute the value of

\[z^{15} + z^{16} + z^{17} + \cdots + z^{50}. \]

4.
A regular polygon of $2n$ sides is inscribed in a circle of radius 5.  The sum of the squares of the distances from any point on the circle to the vertices of the polygon is 600.  Compute the value of the integer $n$.

5.
The circles, each of radius 5, lie in perpendicular planes.  They intersect in two points, $X$ and $Y$, thus determining a common chord $\overline{XY}$.  If $XY = 8$, compute the distance between the centers of the circles.

6.
How many positive integers less than 54 are equal to the product of their proper divisors?

7.
Find all positive integers $n$, less than 17, for which $n! + (n+1)! + (n+2)!$ is an integral multiple of 49.

8.
In triangle $ABC$, $a \ge b \ge c$.  If $\displaystyle \frac{a^3+b^3+c^3}{\sin^3{A} + \sin^3{B} + \sin^3{C}} = 7$, compute the maximum possible value of $a$.

Are these supposed to be answered?

[hide="Problem 1"]We are given that $a+b=\sqrt{17}$ and that (from pythagorean theorem) that $a^2+b^2=9$.  We seek the area of the triangle, or $\frac12ab$.  With some manipulation:

$a^2+b^2=9$
$a^2+b^2+2ab=9+2ab$
$(a+b)^2=9+2ab$
$\displaystyle \left(\sqrt{17}\right)^2=9+2ab$
$17=9+2ab$
$ab=4$
$\frac12ab=\boxed{2}$
[/hide]

[hide="Problem 2"]We have a proportion, so we let the proportional constant be k.  Therefore, we have:
$a+b=6k$
$b+c=7k$
$c+a=8k$

We are also given that $a+b+c=14$.  Adding all three equations, we get:

$2(a+b+c)=21k$
$k=\frac{4}{3}$

Substituting $a+b=6k$ into our given: $a+b+c=14$, we get:
$6k+c=14$
$6\displaystyle \left(\frac{4}{3}\right)+c=14$
$c=\boxed{6}$[/hide]

For problem #6:

[hide="Problem #6 Solution"]The only way a number can be equal to the product of their proper divisors is when they only have $1$ or $3$ proper divisors (excluding $1$).  These are just all the prime, all the perfect squares of primes, and numbers who are products of two primes...which are:

$2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53$

$1^2=1$

$3^2=9$

$5^2=25$

$7^2=49$

$6,10,14,22,26,34,38,46,15,21,33,39,52,35$

This is a total of $\boxed{24}$ integers less than 54 are equal to the product of their proper divisors.  I think I may be missing a few.  But I believe this approach is correct.
[/hide]

[quote="MCrawford"]
3.
Let $z$ be a root of $z^5-1=0$, with $z \not= 1$.  Compute the value of

\[z^{15} + z^{16} + z^{17} + \cdots + z^{50}. \]
[/quote]
$\#3$ is challenging for me.  How can $\sqrt[5]{1}\not=1$?

[quote="Jason27603"]Complex roots. For example, $i$ is a $4th$ root of $1$.[/quote]
Yeah, I was thinking about that.  But how might you solve for which complex root it is?

[hide="Problem 8, not exactly sure about"]

By the Law of Sines, $\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}}$, so their cubes must also be the same.  Hence, $\frac{a^3}{\sin^3{A}} = \frac{7}{3}$.

From here, we use the extended Law of Sines, that $\frac{a}{\sin{A}} = 2R$, $R$ being the circumradius.  So $8R^3 = \frac{7}{3}$.  Thus, $R^3 = \frac{7}{24}$.  If we inscribe the triangle in a circle, then the circle will have circumradius $\sqrt[3]{\frac{7}{24}}$.  So $\boxed{2\sqrt[3]{\frac{7}{24}}}$.

[/hide]

[hide="Actually..."] By the Extended Law of Sines, note that $a=2R\sin A$, $b=2R\sin B$ and $c=2R\sin C$.

Hence,

$\displaystyle\frac{a^3+b^3+c^3}{\sin^3A+\sin^3B+\sin^3C}$

$=\displaystyle\frac{8R^3\sin^3A+8R^3\sin^3B+8R^3\sin^3C}{\sin^3A+\sin^3B+\sin^3C}$.

$\implies 8R^3=7 \implies R=\frac{\sqrt[3]{7}}{2}$.

Consequently, the maximal value of $a$ is $a=2(\frac{\sqrt[3]{7}}{2})=\sqrt[3]{7}$, achieved when $A=\frac{\pi}{2}$.[/hide]

[hide="#5 Solution"]Let the center of one circle be $C$.  Extend a line segment from $C$ to $X$ and another line segment from $C$ to a point on the circle so that the line segment is a perpendicular bisector to $\overline{XY}$.  Call the point that intersects $\overline{XY}$ point $P$.

Now we have a right triangle with leg one $4$ and hypotenuse $5$.  That means the other leg, or $\overline{CP}$ is $\sqrt{5^2-4^2}=\sqrt{25-16}=\sqrt{9}=3$ units long.

The line connecting the two circles is a hypotenuse of a right triangle with right angle at point $P$ and other two vertices at the centers of the circles.  Then the legs of the right triangle is $4$ units.

Therefore, the distance between the centers of the circles is the hypotenuse of the right triangle.  Since the two legs are of the same length, this is a $45-45-90$ right triangle.  Then the hypotenuse, or the distance between the two centers is $\boxed{4\sqrt{2}}$ units.[/hide]


Thought some of you might find this interesting.  I went to MOP with the author in 1987,88,89.

[url]http://slate.msn.com/id/2103903/[/url]

[quote="Kyptonite"]I found this article to be quite interesting.  Good problem-solvers seem to be good writers too.[/quote]

The correlation seems self-evident if we are considering specifically those who make it to the IMO, where ostensibly the selection process is at least partially designed to assess one's skill in writing..

I don't know about the correlation, but I do know Jordan has a published novel and has a masters in creative writing, which I think he wedged between undergrad and his math PhD.

And might I add that he is one of the heads of the Princeton Department of Mathematics (or is that another Jordan Ellenberg?) which I noticed while snooping around today.

[quote="darij grinberg"]Actually, in order to achieve full marks on a problem, you should write the solution in a long and boring way, what is just opposite to the skill of a good writer.
[/quote]

As a member of my school's Academic Decathalon team, the AD idea (at least at the RI competition) of a good writer isn't someone who answers the essay in a boring way, but definitely in a long way.  Same with most standardized tests that require essays.  In comparison, winning math solutions would probably be very short and elegant.

[quote="darij grinberg"][quote="farmertim"]The correlation seems self-evident if we are considering specifically those who make it to the IMO, where ostensibly the selection process is at least partially designed to assess one's skill in writing..[/quote]

I am not really convinced of this correlation. Actually, in order to achieve full marks on a problem, you should write the solution in a long and boring way, what is just opposite to the skill of a good writer.

  Darij[/quote]

I disagree.  I think a good proof has three qualities: (1) it convinces those who already understand, (2) provides an understandable explanation for interested readers with the appropriate background by making clear step-by-step leaps of logic, and (3) is concise.

I think great essays are highly similar in the qualitative sense.

[quote="MCrawford"]I disagree.  I think a good proof has three qualities: (1) it convinces those who already understand, (2) provides an understandable explanation for interested readers with the appropriate background by making clear step-by-step leaps of logic, and (3) is concise.[/quote]

I was speaking about olympiads, not about good proofs in general. And I do have the experience that the more detailed the proof is (actually, even if the details are completely trivial), the better are the chances to achieve a good score. My solution of problem 1 in the second round of the BWM (Bundeswettbewerb Mathematik) was not incomplete, but it was written too sketchily (at least, in the eyes of the correctors). And it's for that reason I got a third prize, not a first one. This is not just limited to "homework" olympiads; actually, I had the same difficulties on an exam, where, however, I suceeded to get the full score back (actually, the problem was that I squared the inequality $k - \frac12 \leq \sqrt{i} < k + \frac12$, where k and i are positive integers, as an equivalence transformation, without saying that everything was $\geq 0$).

  Darij


geometry
3D geometry
number theory proposed
number theory

Given two positive integers $ n,k$ such, that
                                                 $ (n\minus{}1)^{3}\plus{}n^{3}\plus{}(n\plus{}1)^{3}\equal{}k^{3}$.
Prove, that $ n$ is divisible by $ 4$.

It was on Croatia TST 2007, also.

$ k^{3}\equal{} 3n(n^{2}\plus{}2)$

If $ n$ is even, it must be divisible by 4 because $ 3n(n^{2}\plus{}2)$ is divisible by $ 8$ (cube!).

If $ n$ is odd then $ n$ and $ n^{2}\plus{}2$ are relatively prime. Two cases:
1) $ 3n \equal{} a^{3}$, $ n^{2}\plus{}2 \equal{} b^{3}$. Now $ n$ must is divisible by $ 9$ but then $ b^{3}\equiv 2$ (mod $ 9$) - contradiction.
2) $ 3(n^{2}\plus{}2) \equal{} a^{3}$, $ n \equal{} b^{3}$. Now $ n^{2}\plus{}2$ is divisible by $ 9$ so $ n\equiv 4, 5$ (mod $ 9$). But then $ n$ cannot be a cube.
In both cases we used the fact: $ x^{3}\equiv 0, 1,\minus{}1$ (mod $ 9$).
So $ n$ cannot be odd.

$ LHS \equal{} (\frac{(n\plus{}1)(n\plus{}2)}{2})^{2}\minus{}(\frac{(n\minus{}1)(n\minus{}2)}{2})^{2}\equal{} 3n(n^{2}\plus{}2)$

So the equation is $ k^{3}\equal{}3n(n^{2}\plus{}2)$.
If n is even, the result is trivial because 4 will not divide $ n^{2}\plus{}2$. It remains to show that n cannot be odd.

Let n=2m+1. It is also clear we can let k=3l
$ (2m\plus{}1)(4m^{2}\plus{}4m\plus{}3)\equal{}9l^{3}$.
The two brackets are both coprime (neither being divisible by 2), so we have 2 cases.

Case 1:
$ 2m\plus{}1\equal{}a^{3}, 4m^{2}\plus{}4m\plus{}3\equal{}9b^{3}$
So $ 9b^{3}\equal{}a^{6}\plus{}2$, but this is absurd, because $ a^{6}\equiv 0,1\mod 9$.

Case 2:
$ 2m\plus{}1\equal{}9a^{3}, (2m\plus{}1)^{2}\plus{}2\equal{}b^{3}$, but again, we get a contradiction because $ b^{3}\not\equiv 2\mod 9$.

So n must be even, and we are done.


Here are a few fun math problems that I found:
Note... You can look at the answer, but it will spoil the fun.

1. This equation is true with Roman Numerals:

       X L I V * X =C D X L

Make it true using numbers...keep it in base 10, folks...
THERE ARE 2 SOLUTIONS... FIND THEM BOTH
[hide]the two are: 2864*2=5728 and 2814*2=5628[/hide]

 :play_ball:

2. Sue was born on Friday the 13th.  She recently celebrated a birthday on Friday the 13th and found that when she added her birthday of that year (ex. 26th birthday), it was five more than when she multiplied them together.  Which birthday did she celebrate? (you have to show both parts)
HINT:
[hide]leap years count[/hide]

ANSWER:
[hide]50th birthday[/hide]

THESE ARE SOME FUN ONES...
I LIKE #2 BETTER...

 :starwars:

This is me:

 :ddr:

I WILL DANCE AWAY WHILE YOU ALL ARE PUZZLED!
(except the guys here that are good at math...which is pretty much all of them)
(and girls...)

Your wording is a bit weird...
[quote="thacheezinator"]
1. This equation is true with Roman Numerals:

X L I V * X =C D X L

Make it true using [b]numbers[/b]...keep it in base 10, folks...
[/quote]

What sort of numbers? You should specify that they represent digits.

[quote="thacheezinator"]
2. Sue was born on Friday the 13th. She recently celebrated a birthday on Friday the 13th and found that when she added her [b]birthday of that year[/b]...
[/quote]
You could just say age.....

sorry, DIGITS...
I was trying to keep it as specific as possible, and it backfired...
people like to use other bases when figuring out these things...

[quote="istos"]Could you please post a solution
to question 1? I got the first
solution... by trial and error.  :blush:[/quote]

I think he was asking a question, not just testing us  :maybe:

maybe..

[quote="ajain"]I think there are multiple answers for #2, true 50 works, but 34 and 43 works also!!! 12-7=5!!

27 and 72 work also!!14-9=5!!!![/quote]

Ummm....
READ IT...
ADDITION is five more than MULTIPLICATION
not MULTIPLICATION if five more than ADDITION

[quote="istos"]Could you please post a solution
to question 1? I got the first
solution... by trial and error.  :blush:[/quote]

Sure...
I have a book that says an answer:  Here it is, in direct quotation:
(By the way, I really wasn't testing you, i just wanted to see an alternate version to what the book was saying, because i wanted some OUT OF THE BOX answers to help me think...) :blush:

HERE IS THE SOLUTION:
[hide]Because X*X<10, X must be 1, 2, or 3.  However, X=1 is obviously impossible, because then X L I V would have to equal C D X L.  It can also be shown that X=3 doesn't work: L would then have to equal 1 or 2, and either choice produces a contradiction.  If L=1, then V=7, but then I would also have to equal 7 to produce the X in CDXL.  Similarly, if L=2 then V=4, but I would also have to be 4.

It follows that X=2, and it turns out that there are 2 solutions: (the ones i said...)[/hide]

I just wanted to see if there was an easy way to get the other numbers... :yoda:

YAY!
:ddr:


Hello Everyone!

I am searching for a DVI viewer for Windows, but in all searches I have made, this is already included in an TeX installation.
There exists an alternateive to this (an DVI viewer only)?

Thanks!
Johann

First of all, it shouldn't really matter whether you download the whole tex. You ccould just go to the miktex installation and download miktex and install it. Leave all of the settings the same. Then alfter you finisthed installing, just go to start>programs or all programs>miktex>dvi viewer. :)

Yes, but he need only dvi viewer. To see documents on another computer (exp. friend's).

Click on this link:

[url=http://prdownloads.sourceforge.net/miktex/small-miktex-2.4.1705.exe?use_mirror=ufpr]DVI Viewer[/url]


geometry
probability
probability and stats

Let $ X$ be the number of car accidents in a certain area every day. We know that the average is $ 0.5$.

What would be a reasonable pdf? Because I need to compute the probability that no accidents happen today.

We have three pieces of information here:

- The random variable is a count. That is, it takes values in the nonnegative integers.

- There is no immediately obvious upper bound. Either that, or if there is an upper bound, it is very large compared to any reasonably likely values of the random variable.

- We are given exactly one piece of quantitative information, and we are given that in the form of a [i]rate[/i]: the given occurance happens, on average, a certain number of times per time period or per spatial sample of a given size.

There is a probability distribution which has these features, and it is always the most natural distribution to think of when we do see precisely these features. That is the [b]Poisson distribution.[/b]

Model it by a Poisson distribution with mean $ 0.5$ and we get that the probability of no accidents today is $ e^{ \minus{} 0.5},$ which is about $ 61\%.$

Note that we [i]must[/i] have $ P(Y \equal{} 0)\ge\frac12$ for any $ Y$ taking values in the nonnegative integers such that $ E(Y) \equal{} \frac12,$ no matter what the specifics of the distribution. Do you see why?


Prove that $\frac{1}{h_{1}}+\frac{1}{h_{2}}+\frac{1}{h_{3}}=\frac{1}{r}$, where a,b,c are the altitudes of a triangle and r is the inradius.

Today has been a bad day. 3 problems posted incorrectly in a row! The right problem is up now.

As I said, a very bad day. I meant altitudes, not sides.

[hide]multiply each side of the equation by [ABC] and simplify[/hide]

I think this is too easy for Pre-Olympiad.

[hide]Put $h_{a}=\frac{2P}{a},h_{b}=\frac{2P}{b},h_{c}=\frac{2P}{c},r={P\over s}$ where $P$ is the area and $s$ the semiperimeter and you get it immediately.[/hide]

You're right. I originally thought the problem was for side lengths, so I said wow! this would be really useful if I could prove it. THen I reread it after splashD said it didn't work and...


function
limit
number theory proposed
number theory

Find all $ a\in R$ to
$ 4[an]\equal{}1\plus{}[a[an]],\forall n\in N$

[quote="TTsphn"]Find all $ a\in R$ to
$ 4[an] \equal{} 1 \plus{} [a[an]],\forall n\in N$[/quote]
Obviosly, from your cjnditioon followed $ a^2\equal{}4a$, but $ a\equal{}0,4$ are not solutions.

[quote="Rust"][quote="TTsphn"]Find all $ a\in R$ to
$ 4[an] \equal{} 1 \plus{} [a[an]],\forall n\in N$[/quote]
Obviosly, from your cjnditioon followed $ a^2 \equal{} 4a$, but $ a \equal{} 0,4$ are not solutions.[/quote]
I think you have a mistake when you write solution.

$ an\minus{}1<[an]\le an\to 4an\minus{}4<4[an]\le 4an, a^2n\minus{}a\minus{}1<[a[an]\le a^2n$, therefore
$ a(4\minus{}a)n\minus{}5<4[an]\minus{}[a[an]]\minus{}1\equal{}0<a(4\minus{}a)n\plus{}a$.
If $ a(4\minus{}a)\not \equal{}0$ it is false for big n.  :D

A good solution.
This is my solution:
Let $ n\equal{}1$ then
$ [a]\equal{}1\plus{}[a[a]]>a[a]$
Imply that $ a<4$
Easy to check $ a>0$
$ [an]\equal{}k$ so
$ 4k\equal{}1\plus{}[ak]$
$ 4k\minus{}1\equal{}[ak]$
$ 4k\minus{}1\leq ak\leq 4k$
$ 4\minus{}\frac{1}{k}<a<4$
But $ \lim_{n\to\infty} 4\minus{}\frac{1}{k}\equal{}4$ so $ a\equal{}4$ (tradition)


Take a checker-board with $2007 \times 2007$ squares.

On each square, write a real number whose absolute value is less than $1$ such that the sum of numbers in ANY $2 \times 2$ matrix is zero.

Now add up all the $4,028,049$ numbers.

What is the maximum value of the total that you can get?

[quote="t0rajir0u"][hide="Solution"] In a $2006 \times 2006$ subsquare the sum of the values is clearly $0$; hence we only need to consider the border left over, of length $4013$.  It's not hard to realize that the values in the border can be set arbitrarily, and that they uniquely determine the values in the rest of the board, so our maximum total is $\boxed{4013-\epsilon}$ (since you added the condition absolute value strictly less than $1$).  [/hide][/quote]

How do you prove the statements in your solution?

(I'm going to assume the absolute value is less than or equal to 1, because else the answer must contain

I'm not getting how you can get 4013. What about the corner of the subspace you were talking about? If that space and the two next to it are 1s, then the 2x2 matrix there can't add up to 0. :maybe:

However, I can't even find a way to make 4011. Making all the 4012 non-corner spaces 1s and the corner -1 forces the following:

[code].?.-1..1
-1.-1..1
.1..1.-1[/code]
I have to go now, but I think you know what I mean.

MellowMelon: They can't all be 1 and -1 because the problem states their absolute value is less than one. Even with that, however, I don't think t0rajir0u's way works.

Right.   :oops:  The problem with my method is that I force an awful lot of squares to have an absolute value greater than $1$...

[hide="But my lemma is correct"] [b]Lemma:[/b]  The values in a strip of length $4013$ uniquely determine the values in the rest of the checkerboard.

Proof:  Denote the checkerboard by the lattice $(x, y), 1 \le x, y \le 2007$, where we will be considering lattice points for simplicity.  The points $(1, y)$ and $(x, 1)$ uniquely define the rest of the lattice as follows:

$(2, 2) =-\left( (1, 1)+(1, 2)+(2, 1) \right)$
$(3, 2) =-\left( (2, 1)+(2, 2)+(3, 1) \right)$
...
$(k+1, 2) =-\left( (k, 1)+(k, 2)+(k+1, 1) \right)$
...
$(2, k+1) =-\left( (1, k)+(2, k)+(1, k+1) \right)$

We have now defined the points $(2, y)$ and $(x, 2)$.  We can continue in this way until the rest of the lattice is filled.[/hide]

Of course, this might not be the best way to arrive at a solution...


[quote="Chinaboy"]1. Convert .23232323... into a fraction in simplest terms. What about .234234234... ?

2. In my Mathcounts class, 4 of the 6 girls are 8th graders while 4 of the 5 boys are 8th graders. What is the probability that my 4-person team consists of...
3 8th grade girls and 1 8th grade boy?
2 8th grade girls and 2 8th grade boys?
2 7th grade girls and 2 8th grade boys?
1 7th grade girl, 2 8th grade boys, and 1 7th grade boy?

3. What is the value of ...
1/2 - 1/3
1/3 - 1/4
1/4 - 1/5
1/(n) - 1/(n+1)

So what is the value of 1 + 1/2 + 1/6 + 1/12 + 1/20 + ... + 1/90?
What is the value of 1 + 1/2 + 1/6 + 1/12 + ... + 1/(n*(n+1)) ???[/quote]
Tell me if I did something wrong...

[hide="#1"]Might not be all the way simplified..
$ \ 0.\overline{23}=\frac{23}{99}$
$ \ 0.\overline{234}=\frac{234}{999}$[/hide]

[hide="#2"]I was too lazy to do this...but here are the equations.

Total: $ \binom{11}{4}$

1. $ \frac{\binom{4}{3}\cdot \binom{4}{1}}{\binom{11}{4}}$

2. $ \frac{\binom{4}{2}\cdot \binom{4}{2}}{\binom{11}{4}}$

3. $ \frac{\binom{2}{2}\cdot \binom{4}{2}}{\binom{11}{4}}$

4. $ \frac{\binom{2}{1}\cdot \binom{4}{2}}{\binom{11}{4}}$[/hide]

[hide="#3"]1.$ \frac{1}{6}$

2.$ \frac{1}{12}$

3.$ \frac{1}{20}$

4.$ \frac{1}{n^2+n}$
-----------------------------------------------------------------------------
$ 1 + \frac12 + \frac16 + \frac1{12} + \frac1{20} + \cdots + \frac1{90}=$

$ 1 + \frac12 + (\frac12-\frac13) + (\frac13-\frac14)+\cdots +(\frac19-\frac1{10})=$

$ 2-\frac1{10}=\boxed{1\frac9{10}}$

$ 1 + \frac12 + \frac16 + \frac1{12} + \cdots + 1/(n*(n+1))=\boxed{2-{\frac{1}{n+1}}}$[/hide]

Good job DK!

How about this?

Express .121212...(base 6) in base 10 as a fraction.

And on 3, can't 1/2 also be broken down into 1/1 - 1/2?

So what if I said I wanted 1/2 + 1/6 + 1/12... + 1/110 (just more practice)?

Good job!

I know very little about bases... but I'll give it a try..

Let $ x$ be what we want.(Since it's base 10.. it should work out.)

$ \frac{x}{6}\equal{}\frac{4}{33}$

$ \ x\equal{}\frac{4}{33}\cdot6\equal{}\frac{8}{11}$

$ \boxed{x\equal{}\frac{8}{11}}$

[quote]And on 3, can't 1/2 also be broken down into 1/1 - 1/2? [/quote]

Yes, that also works, too.

[quote]So what if I said I wanted 1/2 + 1/6 + 1/12... + 1/110 (just more practice)? [/quote]
$ 1\minus{}\frac{1}{11}$

$ \boxed{\frac{10}{11}}$

[quote="Chinaboy"] It's the same principle for any base. How can you move the decimal point in base 6?[/quote]

I don't quite get what you are talking about...

Isn't base determined by the reminders from a division??

[quote="Chinaboy"]Do you remember how we dealt with repeating fractions on Thursday? It's the same principle for any base. How can you move the decimal point in base 6?[/quote]

[quote]Express .121212...(base 6) in base 10 as a fraction. [/quote]

Hmm.. I still have this in mind.

To remove decimal point, wouldn't you multiply it by 99?

$ \frac{12_6}{99}$ and $ 12_6\equal{}8_{10}$

So $ \frac{8_{10}}{99}\equal{}.\overline{08}$???

[quote="DonkeyKong"][quote="Chinaboy"]Do you remember how we dealt with repeating fractions on Thursday? It's the same principle for any base. How can you move the decimal point in base 6?[/quote]

[quote]Express .121212...(base 6) in base 10 as a fraction. [/quote]

Hmm.. I still have this in mind.

To remove decimal point, wouldn't you multiply it by 99?

$ \frac {12_6}{99}$ and $ 12_6 \equal{} 8_{10}$

So $ \frac {8_{10}}{99} \equal{} .\overline{08}$???[/quote]

Again, why am I here?

From here on, all numbers are in base six except for when it's a subscript.

let $ x \equal{} .\overline{12}$

then $ 100x\equal{}12.\overline{12}$.

Subtracting, we get $ 55x\equal{}12$, so $ x\equal{}\frac{12}{55}$.  Now we convert to base ten and we get $ \frac{9_{10}}{35_{10}}\approx .257_{10}$.

[quote="alanchou"][quote="DonkeyKong"][quote="Chinaboy"]Do you remember how we dealt with repeating fractions on Thursday? It's the same principle for any base. How can you move the decimal point in base 6?[/quote]

[quote]Express .121212...(base 6) in base 10 as a fraction. [/quote]

Hmm.. I still have this in mind.

To remove decimal point, wouldn't you multiply it by 99?

$ \frac {12_6}{99}$ and $ 12_6 \equal{} 8_{10}$

So $ \frac {8_{10}}{99} \equal{} .\overline{08}$???[/quote]

Again, why am I here?

From here on, all numbers are in base six except for when it's a subscript.

let $ x \equal{} .\overline{12}$

then $ 100x \equal{} 12.\overline{12}$.

Subtracting, we get $ 55x \equal{} 12$, so $ x \equal{} \frac {12}{55}$.  Now we convert to base ten and we get $ \frac {9_{10}}{35_{10}}\approx .257_{10}$.[/quote]
IT says Jenks :wink:

Yeah, that's what I meant, so the decimal isn't correct either

[quote="Chinaboy"]You meant 8/35 right?

DK, what he did is move the decimal to the right 2 spots by multiplying by 6^2 (or as he wrote it, 100_6). Then, the stuff to the right of the decimal is the same as the original x. Go from there.[/quote]
Wait.. is 8/35 right??

But by what you said, it gives you different answer..

You multiply $ x$ by 100, which is $ 36_6$, and subtract to get $ 99x\equal{}12$ and new $ x\equal{}\frac{12}{99}$.

Then you convert that into base 10, which would be $ \boxed{\frac{8}{63}}$

Wouldn't it be $ \frac{8}{63}$, not $ \frac{8}{35}$

@xpmath: It's fine with me that alanchou posted the answer.. I was really anxious to know how to do this.. and of course, you guys may think that this is piece of cake....

If we assume we're always working in base six, then we have $ 100\minus{}1\equal{}55$, since this is base six, when borrowing, you don't borrow a ten or hundred or whatever, you're borrowing powers of sixs.

This is the point where I go back to base 10

Just remember that 9 isn't a valid base 6 digit

[quote="alanchou"]If we assume we're always working in base six, then we have $ 100 \minus{} 1 \equal{} 55$, since this is base six, when borrowing, you don't borrow a ten or hundred or whatever, you're borrowing powers of sixs.

This is the point where I go back to base 10

Just remember that 9 isn't a valid base 6 digit[/quote]

Okay, I get why 100-1=55.

Irrelevant to the problem:
But on $ \frac {12}{99}$, can't I reduce it to $ \frac {4}{33}$ and go from there?

I don't think it will work out because it's in base 6, not base 10...


I remember the following problem of VasC
\[ \sum \frac {a^2 \minus{} bc}{2pa^2 \plus{} p^2b^2 \plus{} c^2} \ge 0
\]
for all $ a,b,c \in mathbb R, p > 0$.
It received a very nice proof of VasC for this by Cauchy Schwarz.
In here, I will present my proof for it, it is totally long but it is the way for us to go more with six degree inequality for real numbers :)
Here is it
After expanding, we can rewrite the inequality as
\[ \sum (pb \minus{} c)^2 [ (pab \plus{}ca\minus{} b^2 \minus{} pc^2)^2 \plus{} 3p (a^2 \minus{} bc)^2] \ge 0
\]
which is obviously true.
And we have done.
:)

[quote="can_hang2007"]I remember the following problem of VasC
\[ \sum \frac {a^2 \minus{} bc}{2pa^2 \plus{} p^2b^2 \plus{} c^2} \ge 0
\]
for all $ a,b,c \in mathbb R, p > 0$.
It received a very nice proof of VasC for this by Cauchy Schwarz.
In here, I will present my proof for it, it is totally long but it is the way for us to go more with six degree inequality for real numbers :)
Here is it
After expanding, we can rewrite the inequality as
\[ \sum (pb \minus{} c)^2 [ (pab \plus{} ca \minus{} b^2 \minus{} pc^2)^2 \plus{} 3p (a^2 \minus{} bc)^2] \ge 0
\]
which is obviously true.
And we have done.
:)[/quote]
Nice Can, Congratulation!  :lol:

[quote="Allnames"]Very nice ,can_hang2007,in my memory,we can prove it by Chebysev.can't we?[/quote]
Hm, I dont think so, Dang. Just notice at the case of equality $ (a,b,c)\equal{}(1,p,p^2),(1,1,1)$.

[quote="can_hang2007"]I remember the following problem of VasC
\[ \sum \frac {a^2 \minus{} bc}{2pa^2 \plus{} p^2b^2 \plus{} c^2} \ge 0
\]
for all $ a,b,c \in mathbb R, p > 0$.
It received a very nice proof of VasC for this by Cauchy Schwarz.
In here, I will present my proof for it, it is totally long but it is the way for us to go more with six degree inequality for real numbers :)
Here is it
After expanding, we can rewrite the inequality as
\[ \sum (pb \minus{} c)^2 [ (pab \plus{} ca \minus{} b^2 \minus{} pc^2)^2 \plus{} 3p (a^2 \minus{} bc)^2] \ge 0
\]
which is obviously true.
And we have done.
:)[/quote]

Very nice can_hang2007 :lol:


A coin is flipped 10 times. What is the expected number of pairs of consecutive tosses that come up heads? (For example, $ THHTTTHHTT$ has 2 pairs of consecutive heads)

Expected no. of consecutive tosses can be given by sum of all (P(x)*X) where P(x) represents the probability of getting x consecutive tosses and x represents the no. of consecutive tosses .
In the given problem x can vary from 0 to 9 (case of HHHHHHHHHH) .
Hence E(x) also called as expectations of X is given by (P(0)*0)+(P(1)*1)+..........+(P(9)*9) which can be found out :wink:

A much nicer idea is to use
[hide]linearity of expectation: you can consider each pair of consecutive flips as if they were independent.  Thus the answer is just $ \frac {9}{4}$.[/hide]


algorithm
inequalities
function
number theory proposed
number theory

Hi,

I don't know if it's the right place to post.
However, I'm trying to find some documentation about a largest prime divisor (LPD) of any integer.
I googled it but without useful results.

Example :

n=39 LPD(39)=13
n=17 LPD(17)=17
n=125 LPD(125)=5
and so on

Thank you for any help

As far as I know, finding the largest prime divisor of an integer is not any easier than [url=http://en.wikipedia.org/wiki/Integer_factorization]completely factoring it[/url].

Right, because if you had a fast algorithm to find the largest prime divisor of a number, you could iterate it to quickly find the complete factorization of a number.

Thank you for your answers.
What I meean by documentation is studies about LPD. Some rules or theorems.
Example

a,b integers

LPD(ab)=max(LPD(a),LPD(b))

or LPD(a^k)=LPD(a)

LPD(a+b)<=LPD(a)+LPD(b)

and so on

[quote="Akashnil"][quote="Matharith"]LPD(a+b)<=LPD(a)+LPD(b)[/quote]
a=12,b=14 :huh:
 :?:[/quote]

You are right. I was wrong.

I did it quickly but my aim was :
Are there any rules that we can build which could be useful.

But if a and b are both prime
the inequality LPD(a+b)<=LPD(a)+LPD(b) hold.

Hi,

I tried to see how the function LPD (Largest prime divisor) behave.
I solved quickly some values.
Here some examples

LPD(x)+LPD(y)=LPD(a)
such as x+y=a

LPD(20)=LPD(2)+LPD(18)=5
LPD(20)=LPD(8)+LPD(12)=5
LPD(21)=LPD(5)+LPD(16)=7
LPD(22) has no solution
LPD(23) has no solution
LPD(24) has no solution
LPD(25)=LPD(9)+LPD(16)=5

All the LPD(p) have no solution when p is prime. It can be proved.

My first conclusion is that if we can not split LPD(n) as sum of two primes it has no solution.
How many composite numbers have no solutions?
Is there a way to use the LPD as sieve for primes?

Thank you for any help.


geometry
3D geometry
tetrahedron
analytic geometry
linear algebra

what is the formula for a tetrahedron in the x,y,z plane???

In my opinion, the best way to figure this out is to construct it in $ 4$-space on $ (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)$, and then do a coordinate substitution into the subspace formed by $ x \plus{} y \plus{} z \plus{} w \equal{} 1$. This takes linear algebra, but it works for a [url=http://en.wikipedia.org/wiki/Simplex]simplex[/url] of any dimension.


wait until friday to post scores/mop qualifiers?

or is there a specific reason why the notification process is done by phone and not email or both?

because i'm literally not getting sleep at night wondering/worrying over this >_<

it really would be nice if i could just take the shock now and spend the rest of the week in jubilation or lamentation.

thanks

I think they say Friday so that in case they are running behind the schedule they would like to have things done they will still be on time for what they said.  They most likely want to let the MOPpers know that they qualified for MOP before they release all the results.  After this happens they will release the results as soon as possible.  (Of course this is just my understanding of things...it might be totally wrong though...some of this was based on kedlaya's post [url=http://artofproblemsolving.com/Forum/viewtopic.php?p=215624#p215624]here[/url])


Adam bought a number of ping-pong balls in a store in Tennessee, where a 5% sales tax (we wish) is added to every purchase. If he did not have to pay tax, he could have bought 3 more balls for the same amount of money. How many balls did he buy?

a) 30 b) 45 c) 57 d) 60 e) none of these

[hide="Solution"]If removing the sales tax meant that he could buy $ 3$ more balls, then if we scale the amount saved by $ 20$, we could buy $ 60$ more balls. However, $ 0.05 \times 20 \equal{} 1$, so he bought $ 60$ balls. Answer: $ \boxed{D}$.

[/hide]


function
superior algebra
superior algebra unsolved

let $ C$ be a linear spaces of all continous function $ C\equal{}\lbrace f| f: R \to R \rbrace$ Let $ M\equal{} \lbrace f \in C | f(1)\equal{}0 \rbrace$ Find $ B$ -subspace in $ C$ which realize $ C\equal{} B \oplus M$ Show that B is truly subspace in $ C$

$ B \equal{} \{\lambda \textrm{id}_{\mathbb{R}} \ \mid \ \lambda \in \mathbb{R} \}$, of $ \dim B \equal{} 1$.

Or $ B \equal{} \{\lambda \ \mid \ \lambda \in \mathbb{R} \}$.

... or the constant multiples of any one function which is nonzero at the origin.


linear algebra
matrix
search
linear algebra unsolved

At the beginning of the semester, my professor assigned prob 12 at the end of 1.6 in Hoffman and Kunze.  Nobody in the class solved it, and my professor came to class on the day that it was due and confessed he couldn't solve it either.  He said that he assigned the problems before doing them, and later realized how hard it was.  Now I'm curious if anyone can solve it.  It's at the end of chapter 1, so theoretically it should be solvable with just the information in chapter 1.

Prove the following matrix

$ A \equal{} \begin{bmatrix} 1 & 1/2 & \cdots & 1/n \\  & & &  \\ 1/2 & 1/3 & \cdots & 1/(n\plus{}1) \\ \vdots & \vdots & & \vdots \\ 1/n & 1/(n\plus{}1) & \cdots & 1/(2n\minus{}1) \end{bmatrix}$

is invertible and $ A^{\minus{}1}$ has integer entries.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1357476973&t=163506]This topic[/url] has a post in it with links to several other useful topics.

I think it's safe to say that the link KM posted and the links and solution therein all point to analysis type proofs with Hoffman and Kunze do not assume the reader does not have knowledge of (especially for a first chapter problem).

I am also of the private suspicion that the problem was thrown in there for good measure; there are no 'elementary' proofs based on row-reduction etc (ie. what is covered in the first chapter of that book). I showed it to my TA and he knew no determinant free analysis free elementary proofs either.

It is the only unsolvable problem I encountered in that book so don't fret :)


real analysis
topology
advanced fields
advanced fields unsolved

Let a neighborhood basis of a point $ x$ of the real line consist of all Lebesgue-measurable sets containing $ x$ whose density at $ x$ equals $ 1$. Show that this requirement defines a topology that is regular but not normal.

[i]A. Csaszar[/i]

In other words, the open sets are those measurable sets which have density $ 1$ at all their points.

Regular:

Let $ F$ be a set closed in the given topology, and which does not contain the origin $ 0$ of the real line. Let $ (x_n)_n$ be a sequence of positive reals converging to $ 0$ such that the measure of $ F\cap( \minus{} t,t)$ is at most $ \frac 1n$ times the measure of $ ( \minus{} t,t)$ for all $ 0 < t\le x_n$; it's obviously possible to find such a sequence because the complement $ \mathbb R\setminus F$ has density $ 1$ around $ 0$. Also assume that $ 2x_{n \plus{} 1} < x_n,\ \forall n$ (the factor $ 2$ is not important here). It follows from these conditions that it's possible to cover $ F\cap(( \minus{} x_n,x_n)\setminus( \minus{} x_{n \plus{} 1},x_{n \plus{} 1}))$ with an open set $ D_n$ of measure $ \le\frac 2n$ times the measure of $ ( \minus{} x_n,x_n)\setminus( \minus{} x_{n \plus{} 1},x_{n \plus{} 1})$. The union $ D$ of all $ D_n$ is an open set containing $ F$ but not $ 0$, and its complement $ U$ has density $ 1$ at $ 0$. By the Lebesgue Density Theorem we can eliminate a set of measure zero from $ U$ so that the remainder has density $ 1$ at all its points, and we have the disjoint open sets containing $ x$ and $ F$ that we were looking for.

Not normal:

Let $ A$ and $ B$ be disjoint dense $ G_\delta$ sets of measure zero (constructing such sets should be easy enough). $ A$ and $ B$ are disjoint closed sets in our topology; I claim that there they cannot have disjoint neighborhoods. Assume otherwise, and let $ U\supset A$ and $ V\supset B$ be disjoint open sets. For each positive integer $ n$, let $ A_n$ ($ B_n$) be the subset of $ A$ ($ B$) consisting of those points $ x$ with the property that the measure of $ (x \minus{} t,x \plus{} t)\cap U$ (resp. $ (x \minus{} t,x \plus{} t)\cap V$) is at least $ \frac 23$ times the measure of $ ( \minus{} t,t)$ for all $ 0 < t\le\frac 1n$. $ A_n$ are closed subsets of $ A$, and $ A \equal{} \bigcup A_n$ (similarly for $ B$). Because $ G_\delta$'s in Polish spaces are Polish, we can apply Baire's Theorem to conclude (using also the density of $ A$ and $ B$) that we can find an $ n$ and an interval of the real line in which both $ A_n, B_n$ are dense. But this is absurd: we would be able to find two intervals of length $ \frac 2n$, arbitrarily close (i.e. their centers are arbitrarily close), such that $ U$ has density $ \ge\frac 23$ in one of them, while $ V$, which is disjoint from $ U$, has density $ \ge\frac 23$ in the other one.


Now I'm learning Russian, and I'm having trouble with grammar. I know that "HET" and "HE" are basically different forms of "no" or "not", but when should I use which?

I. "Net" either means "no" (when you answer a question)  or "don't have" (present tense):
1)
- Ты был в Бостоне?
- Нет, не был.

2) У меня нет дома. (c.f., future: У меня не будет дома. past: У меня не было дома.)

3) However, in Russian, you can't say "I got no response from him" (but rather you say "I do not get any response from him."/"Я не получил от него ответа.").

II. "Не" means "not".


limit
function
real analysis
real analysis unsolved

There is a value of $ c$, where the limit $ \lim_{x\to\infty}{\lfloor(x^5\plus{}7x^4\plus{}2)^c \minus{} x\rfloor}$ is finite and nonzero.
Determine $ c$ and find the limit.

[hide]$ \left(7 x^5 \plus{} 2 x^5 \plus{} x^5\right)^c \minus{} x\geq \left(x^5 \plus{} 7 x^4 \plus{} 2\right)^c \minus{} x\geq \left(x^5 \plus{} 0 \plus{} 0\right)^c \minus{} x$

$ 10^c x^{5 c} \minus{} x\geq \left(x^5 \plus{} 7 x^4 \plus{} 2\right)^c \minus{} x\geq x^{5 c} \minus{} x$

The function is only bounded if $ c \equal{} \frac {1}{5}$

And this limit is not a difficult one. Try it.

[hide]
Or proove that.
$ \lim_{x\to \infty } \, \left(\left(x^n\plus{}\sum _{i\equal{}0}^{n\minus{}1} \alpha _i
   x^i\right)^{1/n}\minus{}x\right)\equal{}\frac{\alpha _{n\minus{}1}}{n}$
[/hide]

[/hide][/hide]


$ n \in N$,
Let $ p_{1},p_{2},...$ be all primes numbers smaller than $ n$. Proof, that
$ p_{1} \cdot p_{2} \cdot ... \le 4^{n}$


algebra
polynomial
symmetry
geometry
geometric transformation
inequalities
algebra proposed

$f(x), g(x), h(x)$ are polynpmials all with degree $2$. Is it possible that the numbers $1, 2, 3, \cdots ,8$ be all the roots of equation $f(g(h(x))) = 0$ ?

I don't think it is possible.
Suppose $\{ 1,2,3,4,5,6,7,8 \}$ are all roots of the equation $f(g(h(x)))=0$; consider that for any polynomial $P$ of degree 2, given a set $\{ a_{i}\}$ of numbers, $|\{ a_{i}\}|\leq 2|\{ f(a_{i}) \}|$, with equality only if for each $a_{i}$ in the set there is another $a_{j}\neq a_{i}$ in the set such that $x=\frac{a_{i}+a_{j}}{2}$ is the axis of symmetry of $P$. In our case we must have equality with each one of the three polynomials (for $f$ and $g$ the set of arguments are not $i\in \mathbb N$  $1\leq i\leq 8$, but their images), because otherwise  $|\{f(g(h(i)))|i\in \mathbb{N}, 1\leq i\leq 8 \}|>1$, and so the starting numbers would not be all roots.
This means the axis of $h$ must be $x=\frac{9}{2}$, and so $h(1)>h(2)>h(3)>h(4)$ or $h(1)<h(2)<h(3)<h(4)$  is the new set of numbers which will enter as arguments in $g$. However this set doesn't admitt an axis of symmetry, since the concavity of $h$ implies $|h(1)-h(2)|>|h(3)-h(4)|$.

[Editing: uups; I forgot the case when $h$ has negative first coefficient. ]

You may find a solution in french here :
http://www.animath.fr/tutorat/dossier_02036sol.html
Exercice 4

Pierre.

Dude, do you have an English translator for http://www.animath.fr/tutorat/dossier_02036sol.html ? :|

[hide="Translation"]We'll prove that such polynomials don't exist. The whole reason is based on the following statement: the graph of a second-degree polynomial has a symmetrial axis.

Reduction to absurdity. Numbers $h(1),\dots,h(8)$ are the roots of fourth-degree polynomial. On the other hand, $h$ is of the second degree, hence we must have $h(1)=h(8), h(2)=h(7), h(3)=h(6), h(4)=h(5)$. Besides that, we must have $h(1)<h(2)<h(3)<h(4)$ or $h(1)>h(2)>h(3)>h(4)$

Now we regard the numbers $g(h(1)), g(h(2)), g(h(3)), g(h(4))$. They are the roots of a second-degree polynomial, which, according to the previous inequality, implies $g(h(1))=g(h(4))$ and $g(h(2))=g(h(3))$. From there we deduce $h(1)+h(4)=h(2)+h(3)$, which means that the leading coefficient in $h$ is zero. That's a contradiction.[/hide]


find all positive integers x such that [x/5]-[x/7]=1 , where , for any real number t , [t] is the greatest integer less than or equal to t.


The pattern in the diagram can be folded along the dashed lines to form a cube. The shaded triangle has an area of 12 square units. How many square units are in the surface area of the cube?
[asy]fill((1,2)--(1.5,2.5)--(1,3)--cycle,gray(.7));
draw((1,0)--(2,0)--(2,2)--(3,2)--(3,3)--(2,3)--(2,4)--(1,4)--(1,3)--(0,3)--(0,2)--(1,2)--cycle);
draw((1,1)--(2,1),dashed);
draw((1,2)--(2,2)--(2,3)--(1,3)--cycle,dashed);
draw((1,2)--(2,3));
draw((1,3)--(2,2));[/asy]

There are four of those little triangles in each side of the cube, so the area of one cube is $ 12 \cdot 4 \equal{} 48$. When we fold it up to make a cube, there are six sides, each with area 48, so the answer is $ 6 \times 48 \equal{} \boxed{288}$.


Let $ a,b,c\geq0$,no two of which are zero.Prove that:
$ \frac{a^2\minus{}bc}{\sqrt{a^2\plus{}bc}}\plus{}\frac{b^2\minus{}ca}{\sqrt{b^2\plus{}ca}}\plus{}\frac{c^2\minus{}ab}{\sqrt{c^2\plus{}ab}}\geq0$

by  ineq CHebyshev's
we purpose that : $ a \geq b$ then : $ (a^2 - bc)(b + c) \geq (b^2 - ca)(c + a) \leftrightarrow (a - b)(ab + c^2) + c(a^2 - b^2) \geq 0$  and
$ \sqrt {a^2 + bc}(b + c) \leq \sqrt {b^2 + ca}(c + a) \\ \leftrightarrow (a - b)\left(c^3 + c^2 (a + b) + c(a^2 + b^2) - abc \right) \geq 0$
 Hence  $ LHS \geq 0$
( note that:$ \sum \limits_{cyc} (a^2 - bc)(b + c) = 0$)


A point P belongs to the interior of a square with 10 cm side. At most, how many points the edge of the square may be at a distance of 6 cm from the point P?

A) 1
B) 2
C) 4
D) 6
E) 8

Portuguese question:

Um ponto P pertence ao interior de um quadrado com 10 cm de lado. No máximo, quantos pontos da borda do quadrado podem estar a uma distância de 6 cm do ponto P?

Here's my solution - I think it's correct.
[hide]
Let P be the center of the square, and construct a circle with radius 6 cm centered at P. Since $ 5\sqrt{2}>6$, the circle will intersect the square twice for every side of the square, or E) 8 times.

[/hide]


inequalities
geometry
inequalities unsolved

[quote="gemath"]We can prove that
$\sum\frac{1}{(b+c)^{4}}\ge\frac{9}{16abc(a+b+c)}\forall a,b,c>0$
But maybe we need a stronger following:
$\forall x,y,z>0,\forall a,b,c>0$ prove that
$\frac{x^{4}}{(b+c)^{4}}+\frac{y^{4}}{(c+a)^{4}}+\frac{z^{4}}{(a+b)^{4}}\ge \frac{3xyz(x+y+z)}{16abc(a+b+c)}$
Please help me to prove it or to disprove it! Thanks!
 :)[/quote]
I think it is not true.try $,b=c=1,a\rightarrow{0}$

sorry two inequalitys need a condition:
$(a+b+c)abc\ge max\{b^{2}c^{2},c^{2}a^{2},a^{2}b^{2}\}$

I see mathlinkers don't like discussing about geometric inequalities in inequalities box therefore I write it in algebraic form, but really it has a nice geometric form (I think you known about them,too)
Given acute triangle $ABC$ area $S$ prove that
$\frac{1}{a^{4}}+\frac{1}{b^{4}}+\frac{1}{c^{4}}\ge \frac{9}{16S^{2}}(*)$ and that problem has source form Iran 96 inequality.
But My problem is find a form of $(*)$ with weight that is (it is only a conjecture)
$\frac{x^{4}}{a^{4}}+\frac{y^{4}}{b^{4}}+\frac{z^{4}}{c^{4}}\ge \frac{3xyx(x+y+z)}{16S^{2}}$ for all acute triangle
maybe we can get stronger
$\frac{x}{a^{4}}+\frac{y}{b^{4}}+\frac{z}{c^{4}}\ge \frac{3\sqrt{3(xy+yz+zx)}}{16S^{2}}$ for all acute triangle
but all are only conjectures, it really hard with me, hope you are interested in!  :)  :)


ARML email AMC AIME USA(J)MO USAMO geometry

Hi all!

Unfortunately, I can't make ARML this year. So I won't be seeing any of you there ;(

However, since ARML practice are (hopefully) coming up, I have thoughtfully created a topic to discuss possibly ARML-related issues.

Since registration entails signing up like 7 times in different places (including 2 times from your school), I cannot participate.  Plus I didn't meet the deadline.

:(  I swear, they need to make it so only the individuals have to sign up, and only once at that.

I'm almost certain I won't be going to ARML this year either :/

Agreed, the registration process is a pain.

I'm fairly sure I won't be going to the NorCal Championship thingy either. Is it too late to register now?

SFBA isn't dead though, just the ARML team

I forgot to do the application too, but they sent out an email to all the A people from last year, that didn't yet apply (presumably).

RIP?
Nonetheless we shall still have a spammy thread.

lol i think i got the email 1 day before the deadline.. there goes arml for me

WOW. Totally missed the application due date.
Why don't they email us a couple of days before the deadline?

If you haven't filled out the application yet, you can still do it.  I don't know what Tim will do with late applications, but it couldn't hurt.

Also, first practice will be in two weeks.  Stay tuned for the full schedule once we finalize it.

I heard that Tim is kinda loose on the deadline.

I applied on time!  Woot!

Yeah, I filled it in yesterday...
I couldn't remember my AMC 12 score though, so I put down a figure that seemed close to it...  :oops:

Anyone know of a way to get your scores, short of asking the teacher who has them?

And holy crap, AP exams and practices are clashing...

For what it's worth, I have your scores, since the AMC office made them available to ARML coaches. They were probably also available to Tim, although I don't know that for sure. The reason I'd have your scores is that the AMC office doesn't split California into regions, but simply gives us scores for the whole state. The only names on this list are those of AIME qualifiers. It came as two separate lists, and everyone who took the AMC twice and got qualifying scores twice was listed twice. I've taken the trouble to try to match up the names and get down to unique individuals.

I would agree with bookaholic - many people commit to ARML and come to practices in hopes of making it onto A team or B team. Now that the A and B teams were announced ahead of time, there wasn't the same drive. The people who felt strongly about making A team but were placed on B team didn't commit or come. The people who felt strongly about making B team but did not, did not commit or participate.

To explain some of this: We didn't intend to have an increased emphasis on competition.  We wanted a way to select teams objectively before the start of practices so that practice time could be spent working with your teams, which was something that you, the students, had been clamoring for.  And if you ask me, it shouldn't matter what team you're on.  ARML should be a fun experience whether you're on the A team or the Z team.  When I was a freshman in high school, I got picked as an alternate for the Minnesota team.  I worked my butt off in practices, and was absolutely thrilled the day I earned a spot on the B team.  I went to ARML two weeks later, and it was one of the highlights of my year.  Coming from that perspective, in a region where just making the team is a huge honor (and you don't turn down your spot without a very good reason), this business of "I won't go unless I'm on the A team" is completely mystifying to me.

So how do we fix this problem?  It seems that to get people to participate with enthusiasm, they have to have a chance at making A or B team.  Our goals this year were to select teams early and objectively, which has not worked.  So it seems one or both of those might have to go.  Selecting teams early was a good goal, but given how much it killed enthusiasm among non-A-team students, and given that any benefit it might have had was lost to an extent due to poor attendance at practices, maybe we won't do that.  Lots of other ARML teams (including my old team) announce team placement the night before the competition, and they do just fine.

Then there's also the matter of objectively picking teams.  We've always used an objective formula before, to deflect the possibility of students or parents hassling us about team placement.  But it's also possible that next year's team placement could take into account things like attendance at practices, and how much you contribute to the team during team and power rounds in practice.

[quote="junggi"][quote="Sly Si"]And if you ask me, it shouldn't matter what team you're on.  ARML should be a fun experience whether you're on the A team or the Z team.  When I was a freshman in high school, I got picked as an alternate for the Minnesota team.  I worked my butt off in practices, and was absolutely thrilled the day I earned a spot on the B team.  I went to ARML two weeks later, and it was one of the highlights of my year.  Coming from that perspective, in a region where just making the team is a huge honor (and you don't turn down your spot without a very good reason), this business of "I won't go unless I'm on the A team" is completely mystifying to me.
[/quote]
agree with this
at first i thought blarghiwanttobeontheateam
then i was like hmm wait it doesnt really matter if im on the c team because arml is just awesome

which is why i havent showed up to many practices/tryouts <.<[/quote]

but this also might be why SFBA doesn't win ARML

[quote="serialk11r"]the way things are going now, i doubt sfba will ever win arml in the near future[/quote]
A slow year in terms of participation is never any reason to doubt! We haven't even competed yet! D:

I think the early try-outs are fine.  It gives the students more chances of practices. Also it gives the coaches more objective measure of selecting the teams.

But we should not announce the teams too early. The early team announcement really hurts the students' enthusiasm of participation in the later practices.

I remember that when I participated ARML practices two years ago (as a 5th grader), I had two goals in mind:
1. get some practices in high school math so that it may help me do better in math in general.
2. hope to get onto A or B team as pride.

Now I would add another reason:
3. it's fun to meet other mathletes in the practices and during the trip to UNLV.

Well to be perfectly frank, I would think the early tryouts are a decent idea, but Mathleague contests tend to take up a lot of time out of the day, and since its more contest oriented, it might be less conducive to learning than actual practices.
A complaint I know some of the otherwise A-team people have is that the earlier mathleagues were on the easy side, and some (especially busy sophomores and juniors) viewed them as being extraneous to their mathematical development. Coupled onto the long time commitment, this discouraged some of the upper-level participation. Naturally of course, contests should be reasonably accessible, but to a point.

If i may dare to suggest, perhaps make the target round problems at mathleagues similar to those at the NorCal championships or an actual ARML round?

On a more jovial note, Fantasy ProLeague signups are from now running for four days. BackHo is only 3 points!

[quote="bookaholic"]On the subject of attitudes toward team placement, there is also the parental factor. If my parents are going to approve of me spending so much time in preparation for an event and during the event weekend itself, it had darn well better sound at least a little like something that would look good on a college application. (But I might not be the best example since I couldn't even convince them that going on A team would be brag-worthy enough...)

Hmm, and now that I've written that, I'm thinking about how silly I'd feel if it turned out to be inapplicable to absolutely everyone else, which is entirely possible.

For me personally, it also seems like ARML would be an...inefficient...use of time and effort if ultimately my participation didn't contribute to the strongest Bay Area team. This is somewhat irrational, and I probably wouldn't actually be bothered by it if given the chance, but it's there.

Also, for people who have never previously been to ARML, "you should go even if you're on a low-ranking team because ARML is AWESOME" is a little unpersuasive.[/quote]
Well, this won't be solved by earlier or later tryouts, although it probably wasn't intended to comment on that.

Yeah, in retrospect announcing the teams before getting commitments was a bad idea - i hadn't counted on A-team-or-bust being such a big factor in so many people's decisions, so clearly my greater emphasis on transparency backfired this year.  i guess i've learned that SFBA can't rely on students going to ARML for the love of the game, so we'll have to figure something else out for next year.  Meanwhile, i don't anticipate that this will affect our A team's performance this year (they weren't the ones who bailed), but there's a lot to be said for bringing a full complement of developmental teams in terms of preparing us for future years - something i will try to find a solution to for next year.  Meanwhile, bookaholic is right that we can't expect insanely strong students to exhibit the same enthusiasm on a secondary team, but unfortunately that's the curse of having the strongest region in the country, as i clearly can't put all 60 or so people who deserve it on the A team.  As always, constructive suggestions sent privately are very welcome, and feel free to continue using the forum to vent complaints if you just need to rant.  :)  Thanks to all of you who agree with me and Silas that ARML is not just about being on the A team, and i look forward to an awesome time in Vegas with you this week..

Yeah, I'll see you all in... 15 hours!  Get psyched!

woohoo Vegas!
for those of you on the trip, let me know if you get the QR code on the back of your nametags to work.  apparently the camera on my phone doesn't focus well enough to read it..  :(

Darn, I suspect you guys are not going to see this until ARML is over, but I might as well say it anyway--good luck, people!


I just learned about Menelaus' theorem, and I have some questions about it. First of all, I'm not exactly understanding what ratio of segments does the theorem specify? For example, in the standard form where its the product of 3 ratios equalling 1, I look at a diagram and I don't see any particular pattern in the ratios. It's hard to explain, but in some of the ratios the numerator is the vertex to the point on the opposite side (extended if necessary) and in some the numerator is a smaller segment.
I guess it would really help if someone could explain the statement of the theorem in words. "Menelaus' theorem says that the ratio of the extend segment..." (particularly the ratios)
Thanks!

btw, i already tried aops wiki.

[url]http://en.wikipedia.org/wiki/Menelaus_theorem[/url]

Have you used Ceva's Theorem? The order in which you use the segments is just like in Ceva's Theorem, except now the three points don't have to lie on the triangle now.

[quote="kunny"]I hope this help you.[/quote]
lol. cute. thanks!


The mean of five consecutive odd integers is 17. What is the greatest of the five integers?

Let $ 2k\plus{}1$ be the first integer in the sequence, so that $ 2k\plus{}1\plus{}2k\plus{}3\plus{}2k\plus{}5\plus{}2k\plus{}7\plus{}2k\plus{}9\equal{}10k\plus{}25\equal{}5(17)\Rightarrow 2k\plus{}5\equal{}17\Rightarrow k\equal{}6$, so the greatest integer in the sequence is $ 2(6)\plus{}9\equal{}21$.


Dear everyone.

find the value of $ 1\plus{}\frac{1\cdot 2}{2\cdot 3}\plus{}\frac{1\cdot 2\cdot 3}{3\cdot 4\cdot 5}\plus{}\frac{1\cdot 2\cdot 3\cdot 4}{4\cdot 5\cdot 6\cdot 7}\plus{}\cdots$.
please help me with your interests and answers~ :(  thanks.

your sum is:

$ \displaystyle\sum_{n\equal{}1}^{\infty}\frac{1}{C^{n}_{2n\minus{}1}}\equal{}2\displaystyle\sum_{n\equal{}1}^{\infty}\frac{1}{C^{n}_{2n}}\equal{}\frac{2}{27}(2\pi\sqrt{3}\plus{}9).$  :)

$ C^{n}_{m}\equal{}\frac{m!}{n!(m\minus{}n)!}.$

I gave a hand waving approach to proving what F-o-u-r-i-e-r has given [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=249135]here[/url]


x,y,z>0
α).  If  x+y+z \geq 3 , is  it  necessarily  true  that  1/x + 1/y + 1/z \leq 3 ?
β). If  x+y+z \leq 3 , is  it  necessarily  true  that  1/x + 1/y + 1/z \geq 3 ?

(a) No: take x = 3, y = 1/2, z = 1/2, then x + y + z > 3 and 1/x + 1/y + 1/z > 3.
(b) Yes: (x + y + z)(1/x + 1/y + 1/z) \geq 9 for all x, y, z > 0 (Cauchy).
Since x + y + z \leq 3, we have 1/x + 1/y + 1/z \geq 9/(x + y + z) \geq 3.


Vjerojatno tema ne zanima većinu na forumu, ali očekivao sam od Melkiora neku najavu. Vidim da su pingvini mrtvi hladni što se MEMO-a tiče.

Molim, ne vrijeđati pingvine :lol: samo zato što su neki od njih malo usporeni i toliko izgubljeni u upornim pokušajima da dobiju svoju broncu s IMC-a i rješavanju (manje ili više trivijalnih, da budem istovremeno i iskren i skroman) zadataka na podforumu Real Analysis da više ne znaju gdje im je glava, a gdje rep. :D

Još jednom, sretno svima!

Evo, da se iskupim za neotvaranje topica, barem da budem prvi na Mathlinksu koji će staviti neke rezultate:

[b]Ekipno natjecanje[/b] [i](maksimum 8x8=64 boda)[/i]
Poljska 60 (već tri godine zaredom, dakle, od početka natjecanja  :lol: , Poljska je prva)
Mađarska 55
Njemačka 52
Hrvatska 52 (Njemačka je imala više potpuno riješenih zadataka, što je kriterij za rješavanje izjednačenog rezultata, tako da je treća)
Češka 42
Slovenija 36 (Slovenija je (jedina od svih ekipa?) imala tri ovogodišnja IMO olimpijca u svojoj ekipi)
Austrija 34
Slovačka 34 (ne znam koji je i postoji li uopće poredak između Austrije i Slovačke)
Švicarska 30
Litva 24

[b]Pojedinačno natjecanje[/b] [i](maksimum: 4x8=32 boda)[/i]
Hrvatska je osvojila dva srebra (Grgur Valentić, Matija Bucić) i tri bronce (Kristijan Kilassa Kvaternik, Ivana Antoliš, Ognjen Stipetić), a Anti Malenici su nedostajala dva boda do bronce. Po ukupnom broju bodova u pojedinačnom natjecanju, Hrvatska je četvrta.

Koliko vidim na austrijskom forumu (makar moj njemački i nije najjači, pa ćete me ispričati ako griješim :blush: ), pragovi su 8 za broncu, 12 za srebro i 16 za zlato... dakle, dosta niski s obzirom na to da je maksimalni broj bodova 32. Na istoj stranici (oemo.at) na forumu se mogu naći i zadaci (zaprepašćujuće, ali opet na njemačkom  :D ).

Čestitam još jednom, rezultat je zbilja izvrstan!

Upravo sam se vratio iz Poljske, pa ću kratko komentirati. Rezultat je, za hrvatsku ekipu izvrstan, osobito u ekipnom natjecanju.

Što se Slovenije tiče, oni su imali prijavljena tri sudionika IMO-a (uz suglasnost ostalih), ali, kako se jedan od njih prije puta razbolio, doputovali su ukupno s petoricom, od čega dva IMO-vca. No, u pojedinačnom su dijelu ta petorica osvojili (također kao i hrvatska ekipa) dvije srebrne i tri brončane medalje.


Hi, im having some trouble understanding what they are asking me here.....

Demonstrate that for natural numbers m and n that the modulo 3^m residue of n can be determined from its last m digits alno when expressed in base 12

I cant understand what their asking about the residue also what does it mean to
"express a number in base 12.

For example does it mean that if n=100 then n expressed in base 12 :arrow: is 144 or is n expressed in base 12 =84 base 12 or anything else??

i spent a whole day trying to figure this out  :rotfl: , any help would be nice.

[quote="mackdonnalds"][quote="matt276eagles"]The "modulo $ 3^{m}$ residue of $ n$" means $ n\pmod{3^{m}}$.  If you don't know what mods are, $ a\pmod{b}$ is basically the remainder when $ a$ is divided by $ b$.  Mods are very useful in many math competition problems.  Take a look at this link from the AoPS wiki:

http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic/Introduction

If $ n$ is expressed in base 12, that means convert $ n$ to base 12 and then look at its digits.  So 100 expressed in base 12 is 84.  Hope this helps![/quote]

hmmm.so are they saying that since for example 55=55(mod 3^m) then 55=47(mod 3^m)??[/quote] There was also a problem in the Intro to NT book that said"

Let the product of (12)(15)(16) each factor written in base b equals 3146 in base b. Let S=12+15+16 each term expressed in base b. Find the value of S in base b. Well once they find that b=9 they proceed to say that the answer is 12 base(9) +15(base 9)+16(base9)=44(base 9)..... shouldnt S=13(base9)+16(base9) + 17(base9) since it says that "each term is expressed in base b"?? :?:  :maybe:

Oh maybe my last post wasnt very clear..i just need an example of a 3^m modulo which leaves the same remainder by looking at its last m digits...so maybe the meaning of that can clear up for me :(

 PS i noticed i misspelled the title  :blush:   :oops:


does anyone ahve previous years INChO question papers .My friend needs them  :)

[quote="Cyber-Shreyas"][quote="amaron"]Does anybody have  papers of Indian National Astronomy Olympiad?
I have been selected to write it this year<senior category>,But I do not satisfy the eligibility criterion.
That is, I was born in 1990.What options do I have now?<I am in XII grade>[/quote]

I think you have Astrophysics olympiad na?? Is India's participation in it confirmed?[/quote]

I have no idea about all that.This is the first time I've written this exam.
I read in their website(HBCSE) about this Astrophysics olympiad ,but no clarifications are given.
It is just written that I am eligible for further camps......It does not state clearly if I can write the INAO.

[quote="amaron"]
I have no idea about all that.This is the first time I've written this exam.
I read in their website(HBCSE) about this Astrophysics olympiad ,but no clarifications are given.
It is just written that I am eligible for further camps......It does not state clearly if I can write the INAO.[/quote]
u will be writing the INAO but For IOAA .that's astrophysics and Astronomy.u can probably go through the questions of 2006 INAO Senior [url=http://www.mathlinks.ro/viewtopic.php?t=78566]here[/url]
by [b]udi_s37[/b] + there will be some objective questions
But alst years quetsions were much much much simpler than these :)

[quote="Cyber-Shreyas"]Luckily I was the one who set these papers, so I happened to have a spare copy in my drive.

INCHO 2005 - http://www.hbcse.tifr.res.in/olympiads/olychem/INChO05pdf
INCHO 2007 - http://www.hbcse.tifr.res.in/olympiads/olychem/incho07pdf[/quote]

do u happen to have INPhO papers by any how , or can give me any reference whatsoever where i can get hands on it
plss its imp

that should be easily available ask ur regional coordinator or look for it in the magazine published by HBCSE.

PS:why didn't u post ur marks in NSEP in the post results discussion thread  :|

Are class 12 students allowed to try for astronomy olympiad???? I thought no...

Ofcourse when class 9 students (like pardesi's girlfriend sowmya) get medals, job of class 12 students become useless.

no acually we will be appearing for Internation Olympiad on Astronomy and Astrophysics.
and shreyas that looked nice  :blush:

on a note these INChO questions are really good  :omighty:
some seem better than a INPho paper.

[quote="amaron"]Does anybody have  papers of Indian National Astronomy Olympiad?
I have been selected to write it this year<senior category>,But I do not satisfy the eligibility criterion.
That is, I was born in 1990.What options do I have now?<I am in XII grade>[/quote]

Being an IOAA (International Olympiad on Astronomy and Astrophysics) participant in 2007 at Chiang Mai, Thailand I can assure you that participation of India in Bandung , Indonesia in IOAA - 2008 (august 29-??) is confirmed...

And the age limit is 20 years for IOAA and you shouldn't have entered into university by the time of the olympiad...

only for astronomy olympiad (IAO- International Astronomy Olympiad) age limit is 18 years...(and you are not eligible for this nor am I)

And this year's pattern for INAO will be different for XII class people..
you may have to sit for an extra hour for an extra test...
(not very sure ...our team leader said this)

[quote="Cyber-Shreyas"]Are class 12 students allowed to try for astronomy olympiad???? I thought no...

Of course when class 9 students (like pardesi's girlfriend sowmya) get medals, job of class 12 students become useless.[/quote]

I am his boyfriend  :P  :oops:

In IAO being a 9ther and a 10ther , you can get medals...
(as only ingenuity and visualisation is required...
no need of higher physics or mathematics (except spherical trigo))

But in IOAA, you can't expect a 9ther or 10ther to compete and win a medal..
as knowledge of astrophysics is required for which lot of physics is required for understanding the concepts...
(I won a gold, and I was a 11ther)

so try for IOAA this year..
Hopefully India will send a full team of 5 people
unlike in 2007 when we had a 3-member team...

seems no indian book will be sufficient to answer the questions from organic here.
Morrison and Boyd will be the resort.Our books rae really informative but they hardly tach u what organic is .What the teach u is to synthesize benzene from substances like Calcuim Carbide which is hardly of any use :rotfl:
u have questions from NMR spectro and isotope exchange and racemization studies.

[quote="pardesi"][quote="madness"]How inconvenient
Can anybody tell me where to find solns to the incho papers[/quote]
[url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=25503]here[/url] are the solutions. :D[/quote]

Dei i know that the solns are there but do ya think i am soooooooooo jobless to type each and every question.
Its highly irritatin you know  :D

Anyway much talk is here in chennai abt ur girlfriend ( Soumya  :wink: )
People were shocked initially when i suddenly in a spurt of thought told them that you are a girl  :rotfl:

They attained adhirchi and said nooooooooooooooooooo  :o    it cant beeeeeeeeee
I liked him soooooooooooooo much  :(
 :rotfl:

Oh my, topic going out of hand... Pardesi you are blushing :D

None of us here have ever written any astro related olympiad except this year for the first time some 11th'ers wrote it. I thought 12th weren't allowed. In 9th & 10th, all we heard of was math olympiad.

@pardesi: Why the sudden talk about M n B (Sidenote: Madness has read it 4 times).... M n B is a brillo book to LEARN organic. Most of our other books focus on only the JEE aspect and are narrow minded.

because i love it and ad been reading from it then suddenly i remembered that indians understood a completely diff organic chemistry :rotfl:


The ratio of cats to dogs to squirrels in Dogpatch is 4:5:12, while the ratio of squirrels to racoons to opossums is 10:3:6. What is the ratio of dogs to opossums?

If the ratio of dogs to squirrels is 5:12, and the ratio of squirrels to racoons is 10:3, then the ratio of dogs to racoons is:
\[ \frac {5}{12} \cdot \frac {10}{3} \equal{} \frac {25}{18}
\]
This works because "squirrels" cancels out. If the ratio of dogs to racoons is 25:18, and the ratio of racoons to opossums is 3:6, then the ratio of dogs to opossums is
\[ \frac {25}{18} \cdot \frac {3}{6} \equal{} \boxed{\frac{25}{36}}
\]
as desired.


function
integration
real analysis
real analysis theorems

Prove that $ f: [a,b]\rightarrow\mathbb{R}$ is Lipschitz if and only if $ f(x)\equal{}C\plus{}\int_{a}^{x}g$ for some constant $ C$ and some $ g\in L^{\infty}[a,b]$.

I know Lipschitz implies absolutely continuous, and I know that $ f\in AC[a,b]\Longrightarrow f(x)\equal{}C\plus{}\int_{a}^{x}g$ for some constant $ C$ and some $ g\in L^{1}[a,b]$.

$ g \equal{} f'$ (almost everywhere) which is of course (essentially) bounded.

What about the other direction ($ f(x)\equal{}\cdots\Rightarrow f$ is Lipschitz)?

$ \left | \int_x^y g \, dt\right | \leq \lVert g \rVert_\infty |y\minus{}x|$


[quote="And 1"]lol change the year  [code]©Copyright 2003-4, AoPS Incorporated. All rights reserved.[/code][/quote]

Jajajajaj!!!!

[quote="robinhe"]Copyright doesn't mean what year it is. :roll:[/quote]

That's why I laughed

Jajajajaja is supposed to be a laugh?

Sorry if I offended you :(

I didn't know it was a laugh either. jajajajajaj

Actually, here in Argentina "jejejejej" or "jjajjajaajaja" is like laughing

The way "j" is pronounced in Spanish is like "h" in English, I believe.

So "jajaja"(Spanish) = "hahaha" (English)


calculus
integration
Euler
trigonometry
real analysis
real analysis unsolved

Solve these without using Euler Substitution

1. $ \int \frac {xdx}{({\sqrt {7x \minus{} 10 \minus{} x^2})}^3} dx$

2. $ \int \frac {dx}{x \plus{} \sqrt {x^2 \minus{} x \plus{} 1}} dx$

Second I think is slightly easier.

1. $ \lambda$   $ \mu$ theorem da shre :D
2. Rationalise Then solve.

1. Oops, its power 3/2. My bad
2. Solve it when you have time, it involves nice manip.

1. still $ \lambda$   $ \mu$ theorem only. :)

[hide="Number 1"]$ P \equal{} \int \frac{x}{(7x \minus{} 10 \minus{} x^2)^{3/2}}\,dx \equal{} \int \frac{x}{\left[\frac{9}{4} \minus{} \left(x \minus{} \frac{7}{2}\right)^2\right]^{3/2}}\,dx$,

and now we are to make the substitution $ x \minus{} \frac{7}{2} \equal{} \frac{3}{2} \sin t$: the result is

$ P \equal{} \int \frac{\frac{3}{2} \sin t \plus{} \frac{7}{2}}{\left(\frac{9}{4} \minus{} \frac{9}{4} \sin^2t\right)} \cdot \frac{3}{2} \cos t\,dt \equal{} \frac{2}{9}\int (3 \cos^{\minus{}2}t \sin t \plus{} 7 \sec^2t)\,dt$

and you can finish from here.[/hide]

Carcul:  :thumbup:  :10:

Euc: Lambda Mu theorem is basically splitting up the numerator as Lambda(Differential of Den) + Mu (a constant).

However rakesh It doesn't work here.

Oh god! Please call it a METHOD, not a theorem. it just happens to be a procedure isnt it  :?:

Number 2 can also be solved by substitution $ t \equal{} x \plus{} \sqrt{x^2 \minus{} x \plus{} 1}$ which yields
\[ I \equal{} 2 \int \frac{t^2 \minus{} t \plus{} 1}{t(2t \minus{} 1)^2} \, \mbox{d}t \equal{} \ldots\]
which is fairly easy.


Show that \[ \frac{1}{5}x^5+\frac{1}{3}x^3+\frac{7}{15}x  \]
is entire number for every entire number $x$ .(AuMO 1994) Is this to hard for this place ?

[color=darkred][size=75][Yes it is too hard. Split from the basics forum][/size][/color]

[quote="t0rajir0u"][hide="Solution"]
The expression can be rewritten $\frac{3x^5 + 5x^3 + 7x}{15}$, so we wish to show $3x^5 + 5x^3 + 7x \equiv 0 \bmod 15$.

We do this by evaluating $\bmod 3, \bmod 5$.  By Fermat's Little Theorem, $x^3 \equiv x \bmod 3$ and $x^5 \equiv x^3 \equiv x \bmod 3$, so $3x^5 + 5x^3 + 7x \equiv 3x + 5x + 7x \equiv 15x \equiv 0 \bmod 3$.  Next, $x^5 \equiv x \bmod 5$ so $3x^5 + 5x^3 + 7x \equiv 3x + 7x \equiv 10x \equiv 0 \bmod 5$.  Hence the expression is always divisible by $15$.  Q.E.D.
[/hide][/quote]

how do you conclude $x^5 \equiv x^3 \equiv x \bmod 3$???

[quote="pkerichang"]

how do you conclude $x^5 \equiv x^3 \equiv x \bmod 3$???[/quote]

Sorry, the logic I used was slightly incorrect there :blush:

[hide="It should go something like this..."]
For $x \not \equiv 0 \bmod 3$ we can conclude that $x^2 \equiv 1 \bmod 3$.  Squaring, $x^4 \equiv 1 \bmod 3$ and hence $x^5 \equiv x \bmod 3$.  For $x \equiv 0 \bmod 3$ obviously $x^5 \equiv x \equiv 0 \bmod 3$.
[/hide]

ok, that's better.  Before I thought there's a second "Femat's Little Theorem"...


Does anyone know where I can find past WPI Invitational exams?  And/Or can someone tell me what I need to know to do well on it?

Thank you very much! :lol:

Can someone please respond?  The meet is coming up soon and I want to be ready!  Thanks :-)


For each parallelogram, one side of length $ 20'$ is part of the perimeter. For $ 4$ of the side parallelograms, one side of $ 62'$ is counted. Thus, the total perimeter is $ 48\cdot20'\plus{}4\cdot62'\equal{}960'\plus{}248'\equal{}\boxed{1208'}$.

No, because it is made up of 42 total pieces, meaning 24 on each side. Thus, it would actually be $24\times20\times2$, which is equal to $48\times2$.


How many integers are in the range of the function y (x) = $ (4x^2\plus{}75) / (2x^2\plus{}3)$

[hide]Let $ k$ be in the range of $ y(x)$. Then there exists $ x$ for which $ y(x) \equal{} k$ or $ \frac {4x^2 \plus{} 75}{2x^2 \plus{} 3} \equal{} k$. This can be written as $ x^2 \equal{} \frac {75 \minus{} 3k}{2k \minus{} 4}$. Thus $ \frac {75 \minus{} 3k}{2k \minus{} 4}\ge 0$. Hence we need to consider two cases:

Case 1: $ 2k \minus{} 4 > 0,75 \minus{} 3k\ge 0$. These are equivalent to $ 2 < k\le 25$. Thus $ k$ can be any of the $ 23$ integers $ 3,4,5,\dots,25$.

Case 2: $ 2k \minus{} 4 < 0,75 \minus{} 3k < 0$. These imply $ k < 2\wedge k > 25$, which is absurd.

Thus the answer is $ \boxed{23}$.[/hide]


calculus
derivative
analytic geometry
graphing lines
slope
calculus computations

The tangent line to y=f(x) at (6,7) passes through the point (0,6).  find f'(6).

Can anyone show me how to do this problem step by step?

Thanks in advance.

Draw a picture.

Remind yourself of this: "the derivative is the slope of the tangent line."

What do you know about tangent lines and their slopes in this problem?

I still have no idea where to start.  Do I use the two point to find a equuation.  Then find it's derivative.  Is taht  what you mean?

is the answer 1/6? can anyone verify it.


geometry
circumcircle
geometric transformation
homothety
ratio
geometry proposed

Well I don't know if it's interesting or not but it was ehat I have found with some calculations (it's a easy result)

Given $\triangle ABC$, let $P_{1}$, $P_{2}$, $P_{3}$ on its circumcircle and let $H$ be its ortocenter. Let $\mathfrak{F}$ its Feuerbach circle and let $X_{i}=HP_{i}\cap \mathfrak{F}$ the, the triangles formed by the tangents to $\mathfrak{F}$ at $X_{1}$, $X_{2}$, $X_{3}$ and the tangents to its circumcircle at $P_{1}$, $P_{2}$, $P_{3}$ are homotetic at $H$.

that's simply because H is the outer homotety center of the circumcenter and the feuerbach circle :)
the ratio is 1/2 !


Really, an inequality in classroom math?  :|

proof + inequality $\neq$ middle school classroom math.

Uhh...A little help? I'll move it to Intermediate I guess...

[hide]let a = 1+u and b = 1+v, and a, b real and greater than 1.

a^6 + b^6 + 3a^2b^2 > 2a^3b^3 + 2a^3 + 2b^3

a^6 - 2a^3b^3 + b^6 > 2a^3 + 2b^3 - 3a^2b^2

(a^3 - b^3)^2 > 2(a^3 + b^3) - 3a^2b^2

did i do something wrong with my math? the last line doesn't necessarily seem to hold for all a, b > 1....

for example, let a = b = some number slightly bigger than 1. then lhs is approximately 0 while rhs is approximately 1.

what did i do wrong?  :? [/hide]

just plug in 1 to $u$ and $v$ to get $176>160$.  :|

No. I mean it. You should move on to the pre-olympiad, stancioiu.

[quote="mbabbitt2003"]just plug in 1 to $u$ and $v$ to get $176>160$.  :| [/quote]

This doesn't constitute a proof. In order for it to be a proof, it has to hold for all positive reals u and v.

Lasagnaman is correct, the inequality is wrong.  If $u = v = \frac13$ we have equality, but if $u = v < \frac13$ the inequality is the wrong direction.

Sorry, I corrected it. :oops: I don't think it's necessary to move it to Pre-Olympiad. ;)


number theory
least common multiple
inequalities
number theory unsolved

Here is an Erdös' problem which he proposed to Louis Posa.
Posa succeded to solve it when aged of $12$. :wink:

Is there someone with a proof?

Show that the sum of the serie of the $\frac{1}{a_{n}}$ is irrational, where $a_{n}= lcm (1,2,...,n).$

Pierre.

I don't understand the problem..
Clearly, $a_{n}$ is an integer.. So, the sum of the inverses must be rational..  :maybe:
How can it be irrational ?

So what? You have to be more precise...
This kind of argument works for $a_{n}= 2^{n}$ for which the sum is rational. The fact that $a_{n}$ divides $a_{n+1}$ is not sufficient to conclude.
There are many deep works about irrationality of this kind of series (try google for Ahmes series), but I'm hoping this one could avoid heavy machinery :wink:

Pierre.

If $\sum_{k}\frac{1}{a_{k}}=\frac{p}{q}$ we get $q|a_{q}$. Let m is prime, suth that $a_{m}=ma_{m-1},m>q$.
Then $\frac{1}{a_{m}}+\frac{1}{a_{m+1}}+...<\frac{1}{a_{m-1}}$, therefore from
$\frac{p}{q}-\sum_{k=1}^{m-1}\frac{1}{a_{k}}=\frac{b_{m-1}}{a_{m-1}}>\frac{1}{a_{m-1}}$ we get contradition.
I think it is obviosly. :lol:

[quote="Rust"]
because $m_{1}<\frac{3}{2}m$. [/quote]

Ok, I get it.
Do you have easy proof for this inequality, or did you use the prime number theorem?

Pierre.

[quote="Rust"]It proof Chebechev by elementary method.[/quote]
Would you mind making it more clear?


probability
function
analytic geometry
graphing lines
slope

----1
On the disk shown above, a player spins the arrow twice. The fraction [i]a/b[/i] is formed, where [i]a[/i] is the number of the sector where the arrow stops after the first spin and [i]b[/i] is the number of the sector where the arrow stops after the second spin. On every spin, each of the numbered sectors has an equal probability of being the sector on which the arrow stops. What is the probability that the fraction a/b is greater than 1?

(The disk is a circle with 3 lines making 6 equal pieces. Each sector is numbered 1-6.)

----2
The figure above shows the graph of y = g(x). If the function [i]h[/i] is defined by h(x) = g(2x) + 2, what is the value of h(1)?

The picture looks like: y= -|x| + 3 but the right line of the equation is stopped at x-axis and starts a new line from that point and is parallel to the left line of  y= -|x| + 3 and going into a positive direction. (What I mean by left and right is by the y-axis separating left and right.) Basically there are 2 slopes with 1 and one slope with -1

----3
Let the operations "triangle" and "square" be defined for all real numbers [i]a[/i] and [i]b[/i] as follows.

a"triangle"b = a+3b
a"square"b = a+4b

If 4"triangle"(5y) = (5y)"square"4, what is the value of y?

("triangle" and "square" are symbolic shapes)

----4
In the xy-coordinate plane, the graph of x = y^2 -4 intersects line [i]l[/i] at (0,p) and (5,t). What is the greatest possible value of the slope of [i]l[/i]?

----5
Esther drove to work in the morning at an average speed of 45 miles per hour. She returned home in the evening along the same route and averaged 30miles per hour. If Esther spent a total of one hour commuting to and from work, how many miles did Esther drive to work in the morning?

#1

so the problem is asking for 'what is the probability that a>b'

if a=6, b=5, 4, 3, 2, 1
if a=5, b=4, 3, 2,1
if a=4, b=3, 2, 1
if a=3, b=2, 1
if a=2, b=1
so total 15 outcomes satisfy the condition.
and there are total 6*6=36 outcomes.

answer : $ \frac {5}{12}$

#2

...i have no idea what ur saying...

#3

4"triangle"(5y)=5+15y
(5y)"square"4=5y+16

so 10y=11

answer : y=1.1

#4

$ x \equal{} y^2 \minus{} 4 \implies y^2 \equal{} x \plus{} 4 \implies y \equal{} \sqrt {x \plus{} 4}$

$ p \equal{} \sqrt {0 \plus{} 4} \implies p \equal{} 2$
$ t \equal{} \sqrt {5 \plus{} 4} \implies t \equal{} 3$

slope : $ \frac {3 \minus{} 2}{5 \minus{} 0} \equal{} \frac {1}{5}$

answer : $ \frac {1}{5}$

#5

$ \frac {x}{45} \plus{} \frac {x}{30} \equal{} 1 \implies 5x \equal{} 90 \implies x \equal{} 18$

answer : 18 miles

If I understood well...
#2: $ h(1)\equal{}g(2)\plus{}2\equal{}\minus{}|2|\plus{}3\plus{}2\equal{}3$

I didn't bother finding the graph for #2, but here's how to solve it.

x=1, so plug in 2x = 2 into g. Then add 2 to get h(1). Assuming Bugi got the graph correct, his solution is right.

Isn't there a problem with #4? The graph of x = y^2 - 4 is not linear. Therefore, p could be 2 or -2 and t could be 3 or -3. Maximizing the slope would yield 1.

However, we have a contradiction. Line / cannot pass through those 4 points becuse it is linear. There is a problem with the question.


Let $ x,y,z \in [0,1]$ satisfy the condition $ (x\plus{}y\plus{}z\minus{}2)^2\equal{}2xyz$.
Prove that:
$ \frac{x^2}{(x\minus{}1)^2}\plus{}\frac{y^2}{(y\minus{}1)^2}\plus{}\frac{z^2}{(z\minus{}1)^2} \geq 3$

Let $ a\equal{}x\plus{}y\plus{}z\in[0,3]$. Then the given condition says that $ (a\minus{}2)^2\equal{}2xyz\leq2(\frac{a}{3})^3$
$ \iff2a^3\geq27a^2\minus{}108a\plus{}108\iff(2a\minus{}3)(a\minus{}6)^2\geq0 \iff a\geq\frac{3}{2}$. We have $ a\equal{}\frac{3}{2}$ iff $ x\equal{}y\equal{}z$.

Now since $ x,y,z\in[0,1]$, $ x\geq y \iff(x\minus{}1)^2\leq(y\minus{}1)^2. \therefore(x,y,z)$ and $ (\frac{1}{(x\minus{}1)^2},\frac{1}{(y\minus{}1)^2},\frac{1}{(z\minus{}1)^2})$ are oppositely ordered, so that from Chebyshev's Inequality,

$ \frac{x^{2}}{(x\minus{}1)^{2}}\plus{}\frac{y^{2}}{(y\minus{}1)^{2}}\plus{}\frac{z^{2}}{(z\minus{}1)^{2}}\equal{}3\plus{}\frac{2x\minus{}1}{(x\minus{}1)^2}\plus{}\frac{2y\minus{}1}{(y\minus{}1)^2}\plus{}\frac{2z\minus{}1}{(z\minus{}1)^2}$
$ \geq(2a\minus{}3)\left(\frac{1}{(x\minus{}1)^2}\plus{}\frac{1}{(y\minus{}1)^2}\plus{}\frac{1}{(z\minus{}1)^2}\right)\geq0$ with equality iff $ x\equal{}y\equal{}z$.

[quote="bakerbakura"]Let $ a \equal{} x \plus{} y \plus{} z\in[0,3]$. Then the given condition says that $ (a \minus{} 2)^2 \equal{} 2xyz\leq2(\frac {a}{3})^3$
$ \iff2a^3\geq27a^2 \minus{} 108a \plus{} 108\iff(2a \minus{} 3)(a \minus{} 6)^2\geq0 \iff a\geq\frac {3}{2}$. We have $ a \equal{} \frac {3}{2}$ iff $ x \equal{} y \equal{} z$.

Now since $ x,y,z\in[0,1]$, $ x\geq y \iff(x \minus{} 1)^2\leq(y \minus{} 1)^2. \therefore(x,y,z)$ and $ (\frac {1}{(x \minus{} 1)^2},\frac {1}{(y \minus{} 1)^2},\frac {1}{(z \minus{} 1)^2})$ are oppositely ordered, so that from Chebyshev's Inequality,

$ \frac {x^{2}}{(x \minus{} 1)^{2}} \plus{} \frac {y^{2}}{(y \minus{} 1)^{2}} \plus{} \frac {z^{2}}{(z \minus{} 1)^{2}} \equal{} 3 \plus{} \frac {2x \minus{} 1}{(x \minus{} 1)^2} \plus{} \frac {2y \minus{} 1}{(y \minus{} 1)^2} \plus{} \frac {2z \minus{} 1}{(z \minus{} 1)^2}$
$ \geq(2a \minus{} 3)\left(\frac {1}{(x \minus{} 1)^2} \plus{} \frac {1}{(y \minus{} 1)^2} \plus{} \frac {1}{(z \minus{} 1)^2}\right)\geq0$ with equality iff $ x \equal{} y \equal{} z$.[/quote]
Very nice!
In fact, we can even prove that $ xyz \geq (1\minus{}x)(1\minus{}y)(1\minus{}z)$


[quote="hainam-sp"]a,b,c>0 such that abc<1.Prove that:
$\frac{a}{c}+\frac{b}{a}+\frac{c}{a}\ge a+b+c$[/quote]

Let a, b, c = a, b, c/abc^1/3

then abc=1

when LHS reminds still, the value of RHS increases. so we can only prove that when abc=1

then,

2a/c + c/b >=3a,

2b/a+ a/c>=3b,

2c/b +b/a>= 3c,

so,   a/c+b/a+c/b>=a+b+c

[quote="hainam-sp"]a,b,c>0 such that abc<1.Prove that:
$\frac{a}{c}+\frac{b}{a}+\frac{c}{a}\ge a+b+c$[/quote]
   I think you mean b/c+c/a. ;)

[quote="pvthuan"][quote="hainam-sp"]a,b,c>0 such that $abc\leq1$.Prove that
$\frac{a}{c}+\frac{b}{a}+\frac{c}{a}\ge a+b+c$[/quote]

These are known inequalities

$\frac ab+\frac bc+\frac ca\geq\frac{a+b+c}{\sqrt[3]{abc}},$ and

$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{d^2}+\frac{d^2}{a^2}\geq\frac{a+b+c+d}{\sqrt[4]{abcd}}.$[/quote]
I have found the solution of the inequalitiy:
$\frac ab+\frac bc+\frac ca\geq\frac{a+b+c}{\sqrt[3]{abc}},$
we have
$(\sum\frac ac)(\sum \frac{1}{ac})\geq (\sum\frac{1}{a})^2$
$(\sum\frac ac)(\sum{ac})\geq (\sum a)^2$
thus:$(\sum\frac ac)^2*(\sum\frac ac)\geq \frac{(\sum\frac{1}{a})^2}{\sum \frac{1}{ac}}* \frac{(\sum a)^4}{(\sum{ac})^2}=\frac{(\sum a)^3}{abc}$

Is $\frac{a}{c}+\frac{b}{a}+\frac{c}{b} \ge a+b+c$ equivalent to $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \ge a+b+c$?

[quote="Lovasz"]Is $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\ge a+b+c$ equivalent to $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge a+b+c$?[/quote]

If we put $\;a'=c,\;b'=b,\;c'=a$ we have \[\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\ge a+b+c \;\;\;\Leftrightarrow\;\;\;\frac{a'}{b'}+\frac{c'}{a'}+\frac{b'}{c'}\ge a'+b'+c'.\] and every symmetric relation is satisfied by both or none of $(a,b,c)$ and $(a',b',c').$

Yes, I thought it was true, too. From this point, we'll find some more solutions for many problem.
Wait.

Let $abc=k^3\leq 1$ so that $0<k\leq 1$. Then letting $a=\frac{kx}y,b=\frac{ky}z,c=\frac{kz}x$ then the inequality become

$\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}\geq k\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)$

$\Longleftrightarrow x^3+y^3+z^3\geq k(xy^2+yz^2+zx^2)$

But $x^3+y^3+y^3\geq 3xy^2$ , so we can prove that $x^3+y^3+z^3\geq xy^2+yz^2+zx^2\geq k(xy^2+yz^2+zx^2)$


geometry
probability
combinatorics proposed
combinatorics

Every phone number in an area consists of eight digits and starts with digit $ 8$. Mr Edy, who has just moved to the area, apply for a new phone number. What is the chance that Mr Edy gets a phone number which consists of at most five different digits?

We can count the number of 8-digit phone numbers starting with 8 and with exactly $ k$ other digits using standard inclusion-exclusion techniques. We get
\[ {9\choose k}\sum_{i\equal{}0}^k(\minus{}1)^{k\minus{}i}{k\choose i}(i\plus{}1)^7.\]
So the number of such phone numbers with at most 5 distinct digits is
\[ \sum_{k\equal{}0}^4\left[{9\choose k}\sum_{i\equal{}0}^k(\minus{}1)^{k\minus{}i}{k\choose i}(i\plus{}1)^7\right]\equal{}4103200.\]
The total number of possible phone numbers is clearly $ 10^7$, so the desired probability is $ 0.41032$.


I'm not in charge of this but Nukular is doing this.  He's going to do the discussion through AIM which is going to be preparing for MOP or any USAMO leveled discussion.  In the discussion, people will talk about the variety way of problem solving and trust me, it is something you do and you'll like it.  As I said it, I'm not in charge of it but the reason I'm posting is to get people something preparing for MOP so they can do well and in future, they're the people who's going to teach "us."
So, if you're interested, please post here.

I'd be interested; when will these be?  Also, is it OK if I'm not really MOP level?  I can usually make significant progress on at least one out of four USAMO problems though.

Still deciding on a time.

But It's not going to be all MOP level. That would be pretty silly. If you can do 1 USAMO question that should be more than enough.

I think that if you can do USAMO(which I can't btw)  you should be fine because it would be ridiculous for only moppers(or those at that that level) to participate.  The number of them on this site is too small to have a discussion like that.

I will go but first I have to create a screen name

You might want to PM me your screen names, so I know who to invite to the chat.

Just to make sure, it's free, right?
[edit]- never mind, it's free.

Of course. It's just some discussion for fun and learning.

Sounds like fun, count me in AIM: zscool

p.s. I suppose college students can participate, no?

sure... why not?

Does anyone have any topics they might want to discuss?

(That way, I can find some problems relating to those topics...)

[quote="Fierytycoon"]How are the chats going to be run? Will someone just post an olympiad problem and the rest of us type in ideas that we have, etc...?[/quote]

That would be the general format.

Also, I think it would be important to analyze how solutions are arrived at, and identify patterns among different problems.

For example, the Well Ordering Principle is often used in problems involving integers, and the Smoothing principle can be used in inequalities.

I think this would be good to come together as a group and share develop ideas and techniques for olympiad problem solving.

Yeah, I think that it would be best if we all voiced our ideas and lthought process along the way to the solution instead of having one person just abruptly type in the complete solution.

If everyone else doesn't mind, could people explain known theorems used in proofs?  I personally know few, so I might get totally baffled when names of theorems are flying around.  For example, "By Schur" means absolutely nothing to me.  Just a brief explanation for each would be fine; I'll probably look into them more later.

"Schur," we'll go over commonly used theorems when they come up.

sounds like fun; aim = mj2forever

weekend is best for me, if possible.

[quote="Silverfalcon"][quote="Syntax Error"]Just keep in mind that math isn't necessarily all competition oriented.  Have fun with it; moreover, if one does not, most likely succeeding will be a bit tougher.[/quote]

Nice saying, but not necessarily related to topic. So, keep in mind that when you post something, it has to relate to what someone originally posted.[/quote]

I will worry about keeping threads on topic.  Let me do my job.

I'll try and be around, but it will depend on what day the session takes place.  Any ideas on what time and place?

Oh yeah, my screenname is soccerdude9869.

It seems that a weekend night or something like that would be best.

We'll decide on the details soon.

tetrahedr0n, i was there for the last one.

I would be here for this weekend(in fact, i have to leave now)

Sounds like fun, although I'll likely mainly be a spectator - not too familiar with most of the theorems and such that are required in many of these level problems.

AIM - Murray65536 (original, ain't I?)

Hi, I would like to come even though I am not really at this level.  I am always up for a challenge.  SN=js1527k. I'll PM you too.

Let's decide on a time now, since we have enough people now.

I want to pick a time where everyone is going to be available.

I propose either friday night or saturday night.

I would be interested. I'm still working mostly on AIME level things, but I would be interested in giving a go at, or at least seeing the thought processes for USAMO problems. My screenname is scm200x

Friday or Saturday sounds fine.

Fri and Sat should be fine for me. We could even do a little bit tonight if enough people are around.

I'll be around. What is the planned chatroom for these meetings? "USAMO mathchat"?

AIM = Haitao55555
Sat. night sounds good.
How long will this be?

So... any finalized times for this event?

[quote="JS1527"]I would suggest this saturday evening.[/quote]

I agree... probably about 9PM EST (so that ppl further west wont have problems)

If by Saturday you mean tomorrow, 5/22, then Saturday evening sounds good to me.
So how is this going to work? Are you going to contact each one of us at that time?

I can't make this Saturday.  (Grr, a Relay for Life fundraiser.)  See y'all next time, though.

Yeah I'd appreciate it if sessions were posted here (or if you'd rather communicate privately, PM me).


$ a,b,c \in R$
$ \left( {a \plus{} b \minus{} c} \right)^2 \left( {a \plus{} c \minus{} b} \right)^2 \left( {c \plus{} b \minus{} a} \right)^2  \ge (a^2  \plus{} b^2  \minus{} c^2 )(a^2  \plus{} c^2  \minus{} b^2 )(c^2  \plus{} b^2  \minus{} a^2 )$ :lol:
I have solved this stuff but with a huge solution. :wink:

[hide="solution"]
[b]Case 1[/b]: let $ a,b,c$ be sides of a triangle

make the substitution $ a \equal{} x \plus{} y, b \equal{} y \plus{} z, c \equal{} z \plus{} x$ and we get

$ (2y)^2(2x)^2(2z)^2 \ge ( 2y^2 \plus{} 2yx \plus{} 2yz \minus{} 2xz)(2x^2 \plus{} 2xy \plus{} 2xz \minus{} 2yz)(2z^2 \plus{} 2zx \plus{} 2zy \minus{} 2xy)$

By AM-GM notice that $ \dfrac{x^2 \plus{} y^2}{2} \ge xy$ so sub this into the RHS to find

RHS is a maximum at $ (4y^2)(4x^2)(4z^2) \rightarrow$ LHS=RHS when $ x \equal{} y \equal{} z$ and the inequality follows

[b]Case 2[/b]: $ a,b,c$ are not sides of a triangles $ \longrightarrow |a|\plus{}|b|<|c| \Rightarrow a^2\plus{}b^2<c^2$

Now clearly the left hand side of the inequality is positive and the right and side is negative because $ (a^2\plus{}b^2\minus{}c^2) < 0$ so the inequality holds

[/hide]

EDIT:  Thanks Peine, i think I've fixed it

[quote="ocha"][hide="solution"]
make the substitution $ a \equal{} x \plus{} y, b \equal{} y \plus{} z, c \equal{} z \plus{} x$
[/hide][/quote]
why?they are not sides of a triangle.

You can't use AM GM with -2yz;-2xz;-2xy.By the way you don't need to divide in case because we don't need x,y,z>0 . :lol:

Does anybody have any other ideas? :lol: