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Diophantine equation

First of all,I am new to the forum.And I am not exactly sure whether this is right place to ask for help.Pardon me for my ignorance ,if any.
I am preparing for an entrance exam.In this entrance exam we will be given one mathematics section (with two other sections) where I have to solve as many questions as possible in 45 minutes.So my aim is to practice as much as possible and learn to approach a question in a way that I could reach a solution in the fastest way possible.

So I will be posting questions in this particular thread and would be expecting solutions from you genius guys.Please donot use any high funda concepts and solve the question in as simple way as possible.Surely I think you can spare a minute or two of your time and help me.It would be fun!

[b]Question 1:[/b]
How many ordered triples of integers (a, b, c) satisfy
|a + b| + c = 19 and ab + |c| = 97?

well i'm not exactly sure how to do this
but i think these types of problems are in AOPS intermediate algebra

It looks like a difficult Diophantine Equation to me...........

Unfortunately, I can only solve extremely basic Diophantine Equations.........................

[quote="jonathanchou711"]I'm copying this from wiki but here are a few tips:

The questions asked in Diophantine analysis include:
Are there any solutions?
Are there any solutions beyond some that are easily found by inspection?
Are there finitely or infinitely many solutions?
Can all solutions be found, in theory?
Can one in practice compute a full list of solutions?[/quote]
yes but how to solve,just form cases and find solutions?

off topic:
I want to solve it by geometry.Is it possible?Can someone solve it by geometry?
Eg
|a + b| + c = 19 and ab + |c| = 97?
CASE1
C is positive==>|a + b| + c = 19  and ab + c = 97
Case2
cIS negative==>|a + b| + c = 19  and ab -c = 97

You know, I guess you have to start to guess. That's my way. Hm... I'll try to solve t his problem. It might take me some time.

And your way of doing cases is actually pretty genius. I didn't think of it that way. I might try it that way.

Ok. uh... nvm. I give up. Sorry. But you might want to post this in a higher level forum. This is a 3 variable diophantine equation. I've only worked with 2 and did guess and c heck. Believe me. That's really hard so let alone 3 variables...

To answer your question if you can do it with geometry, I don't really know if you could. I've never heard of solving them that way. But math continues to surprise me so expect the unexpected. Good luck with your problem.

[quote="jonathanchou711"]Good luck with your problem.[/quote]
Thanks.I am also finding it quite difficult to solve  :(

ASE1
C is positive==>|a + b| + c = 19 and ab + c = 97
==>|a+b|-ab=-78
==>|a+b|=ab-78

Comparing graphs of |a+b|=0  (a pair of straight lines)and ab=78 (hyperbola)
I think we can guess the number of solutions. (or not?Do tell)
But we do have to think of c (The solution must give this value positive)
Case2
c is negative==>|a + b| + c = 19 and ab -c = 97
==>|a+b|+ab=116
==>compare |a+b|=0 (a pair of straight lines) and ab=-116(hyperbola)
Again the number of solutions can be guessed (or not?)

Is my approach right?Or am I doing something wrong here?

No way.  This should be in HSB(High school basics) at least and probably should be in High School Intermediate.

The middle school forums are ok for BASIC Diophantine equations, but this one's too hard for that.

Thanks guys,I solved it by taking help from few people in one of the other forums.Thank you all for your contributions.It helped  :)


Prove that \[\displaystyle \sum_{\{i,j,k\}=\{1,2,3\}} \csc ^{13} \frac{2^i \pi}{7}\csc ^{14} \frac{2^j \pi}{7}\csc ^{2005} \frac{2^k\pi}{7}\] is rational. Here, $(i,j,k)$ is summed over all possible permutations of $(1,2,3)$.


Suppose A, B, and C are the angles of a triangle. Show that

$cos^2$A + $cos^2$B + $cos^2$C + 2 cos A cos B cos C = 1

[hide="hint"]use the fact that C=180-A-B, and expand out with the addition identities, and use the pythagorean identity (sin^2+cos^2=1, and everything cancels but 1[/hide]

is it 2* cosA * cosB * cosC at the end?
or is it addition?

[b]Lemma[/b] $\forall \alpha \textrm{ and } \beta$:
\[ \sin^2 \beta - \cos^2\alpha = -\cos(\alpha+\beta)\cos(\alpha-\beta). \]

Proof: \[ \begin{aligned}-\cos(\alpha+\beta)\cos(\alpha-\beta) & = -(\cos\alpha\cos\beta - \sin\alpha\sin\beta)(\cos\alpha\cos\beta + \sin\alpha\sin\beta) \\ & = - (\cos^2\alpha\cos^2\beta - \sin^2\alpha\sin^2\beta) \\ & = \sin^2\alpha\sin^2\beta - \cos^2\alpha\cos^2\beta \\ & = \sin^2\alpha\sin^2\beta - \cos^2\alpha(1-\sin^2\beta) \\ & = \sin^2\alpha\sin^2\beta - \cos^2\alpha + \cos^2\alpha\sin^2\beta \\ & = \sin^2\beta(\sin^2\alpha+\cos^2\alpha) - \cos^2\alpha \\ & = \sin^2\beta - \cos^2\alpha \ \square\end{aligned} \]

[b]Problem[/b]: $\sum_{cyclic} \cos^2 \alpha + 2\prod_{cyclic} \cos \alpha = 1$

Proof: \[ \begin{aligned} \cos^2 \alpha + 2\prod_{cyclic} \cos \alpha & = 1 - \cos^2\gamma - \cos^2 \beta = \sin^2 \gamma - \cos^2 \beta \\ & \Updownarrow \textrm{Lemma} \\ \cos^2 \alpha + 2\prod_{cyclic} \cos \alpha & = - \cos(\beta+\gamma)\cos(\beta-\gamma) = - \cos(\pi-\alpha)\cos(\beta-\gamma) = \cos(\alpha)\cos(\beta-\gamma) \\ \cos \alpha \cdot \left(\cos\alpha + 2\cos\beta\cos\gamma - \cos(\beta-\gamma)\right) & = 0 \\ \cos\alpha \cdot \left(-\cos(\beta+\gamma) + 2\cos\beta\cos\gamma - \cos(\beta-\gamma)\right) & = 0 \\ \cos\alpha \cdot \left(-(\cos(\beta+\gamma)+\cos(\beta-\gamma)) + 2\cos\beta\cos\gamma \right) & = 0 \\ & \Updownarrow \textrm{Simpson} \\ \cos\alpha \cdot \left(-2\cos\beta\cos\gamma + 2\cos\beta\cos\gamma \right) & = 0,\end{aligned} \]which clearly is true. $\square$

$\cos ^2 A + \cos ^2 B + \cos ^2 C =$

$= \frac{1}{2}\left[ {3 + \cos 2A + \cos 2B + \cos 2C} \right] =$

$= \frac{3}{2} + \cos \left( {A + B} \right) \cdot \cos \left( {A - B} \right) + \frac{1}{2} \cdot \cos 2\left( {A + B} \right) =$

$= \frac{3}{2} + \cos \left( {A + B} \right) \cdot \cos \left( {A - B} \right) + \cos ^2 \left( {A + B} \right) - \frac{1}{2} =$

$= 1 + \cos \left( {A + B} \right) \cdot \left[ {\cos \left( {A - B} \right) + \cos \left( {A + B} \right)} \right] =$

$= 1 + \cos \left( {A + B} \right) \cdot 2 \cdot \cos A \cdot \cos B = 1 + 2 \cdot \cos A \cdot \cos B \cdot \cos \left( {\pi - C} \right) =$

$= 1 - 2 \cdot \cos A \cdot \cos B \cdot \cos C$

[quote="t3hggnore"]but then you would be assuming there is equality to begin with.[/quote]

If you want to, you can begin at the bottom and work yourself up until you got the equality, so you would start with $\cos\alpha \cdot 0 = 0$ and then go up. That's why there are up-down-arrows between it  ;) (despite I didn't write them all)


In quadrilateral ABCD, $\angle{ABD}=\angle{ACD}=60^{o}$, $\angle{ADB}=90^{o}-\frac12 \angle{BDC}$.
Prove: $\triangle{ABC}$ is an isoscele triangle.

[quote="M@re"]Are you sure shobber that you write this correct because, well, it is little silly that $\angle ADC=90^o$ its said in this way $\angle ADB=90^o-\angle BDC$.[/quote]
Read it more carefully, it is $\frac{1}{2} \angle{BDC}$, not $\angle{BDC}$.

After working out the angle, we can find out $\angle BAC=\angle BDC$ . Inscribe this quadilateral inside a circle and it can be seen that $\angle ABC=\angle ACB$ which implies $\triangle ABC$ is isosceles . :P

Nice one. From here you should notice that $60^{o}$ is useless. Also, the official proof did not use cyclic quadrilaterals. So I shall wait to see if any of you work it out.

[quote="probability1.01"]You can also use law of sines... $\frac{AB}{\sin 90-x/2} = \frac{AD}{\sin 60} = \frac{AC}{\sin 90+x/2}$

And yeah... 60 doesn't matter except that $\angle ABD = \angle ACD$...[/quote]

Wow , thats really a easy application of sines law ... :rotfl:  :rotfl:

Good job guys!
I shall post a more elementary way to solve it:

Extend CD to E such that DE=BD.
Since $\angle{ADB}=90^{o}-\frac12 \angle{BDC}=\frac12 (180^o- \angle{BDC})=\frac12 \angle{BDE}$.
So $EA$ bisects $\angle{BDE}$. Therefore: $\triangle{ADB} \cong \triangle{ADE}$, which gives $AE=AB$.
The congruence also tells us that $\triangle{ACE}$ is equilateral, hence $AC=AE$, so finally, $AB=AC$.

The quardilateral is inscribed in circle. We use Ptolemeus theorem. we find similar triangles and with the informations we have from the hipothesis the rest is easy


[quote="Zamfirmihai"]Aratati ca pt orice $a,b,c \geq0, a+b+c=4$ atunci:
$\sum_{ciclica}a^{2}(a-b)(b+1)\geq0$[/quote]
$\Leftrightarrow$
$\sum a^{3}b-\sum a^{2}b^{2}+\sum a^{3}-\sum a^{2}b\geq 0$
Dar $\sum a^{3}b \geq \sum a^{2}b^{2}$ si
$\sum a^{3}\geq \sum a^{2}b$
si totul este terminat.
Conditia $a+b+c=4$ inutila.  :lol:

Sunt de parere ca trebuie rectificata conditia la $a+b+c=3$.

Problema: Sa se arate ca pentru $a,b,c \geq  0$, cu $a+b+c=3$, are loc urmatoarea inegalitate:
$\sum{a(a-b)(b+1)}\geq  0$.

Astfel, o rescriem ca:

$\sum{a^{2}b}+\sum{a^{2}}\geq \sum{ab^{2}}+\sum{ab}$

Acum omogenizand, obtinem ca e echivalenta cu:

$\iff 3\sum{a^{2}b}+\sum{a^{3}}\geq 3\sum{ab^{2}}+3abc$.

Invocam urmatoarea lema: $4(a+b+c)^{3}\geq 27 \sum{ab^{2}}+27abc$.
Demonstratia lemei:
Se apeleaza la brute-force, o metoda urata, dar OK, este sa consideram $a=\min\left\{a,b,c\right\}$, si consideram $b=a+x$, si $c=a+y$, astfel inegalitatea se reduce la $9(x^{2}-xy+y^{2})a+(2x-y)^{2}(x+4y)\geq 0$, care este evident adevarata.

Astfel din lema avem ca $4\sum{a^{3}}+12\sum{a^{2}b}\geq 15\sum{ab^{2}}+3abc$.

Iar revenind la inegalitatea propusa, $15\sum{ab^{2}}+3abc \geq 12 \sum{ab^{2}}+9abc \iff \sum{ab^{2}}\geq 3abc$, care e AM-GM.

[quote="maky"]de ce $\sum a^{3}b \geq \sum a^{2}b^{2}$ ?[/quote]
Ai dreptate e gresit. :blush: .Relatia aceea e valabila doar pt suma simetrica.

scuzati-mi eroarea de redactare :oops: problema corecta este:
Aratati ca pt orice $a,b,c \geq0, a+b+c=4$ atunci:
$\sum_{ciclica}a(a-b)(b+1)\geq0$
Pusesem $a^{2}$ in loc de $a$ :oops: inegalitatea este destul de greuta dupa parerea mea. are la baza o inegalitate luata de pe forumul in engleza, restul fiind doar o mica prelucrare facuta de mine. ma scuzati si va rog sa-mi dati ratingul 1, vreau sa fiu cel mai slab cotat utlizator de pe ML :P bafta

Inegalitatea ta este echivalenta cu a arata ca daca $a,b,c\geq 0$, atunci
\[a^{3}+b^{3}+c^{3}-3abc\geq 4(a-b)(b-c)(c-a), \]
iar fiindca inegalitatea este ciclica, putem presupune ca $b$ se afla intre $a$ si $c$, si de aici vedem ca ar fi de ajuns sa demonstram inegalitatea in cazul $a\leq b\leq c$. Mai departe, ne "murdarim" un pic pe maini si notam $b=a+x$, $c=a+x+y$, si inegalitatea iese usor.

[quote]$a^{3}+b^{3}+c^{3}-3abc\geq 4(a-b)(b-c)(c-a)$
[/quote]
dap, despre aceasta iengalitatea vorbeam. de aceea am pus $a+b+c=4$


Let $ h$ be real positive real.Prove that :$ \sum_{k \equal{} 1}^{n}\frac {\sin (kh)}{k^2} < 3h$ for all positive integer $ n$

My question: Find $ c$ min such that:$ \sum_{k \equal{} 1}^{n}\frac {\sin (kh)}{k^2} < c h$  for all positive integer $ n$
(here is difficult question  :D )
I am waiting for the great solutions  :) .

[quote="math10"]Let $ h$ be real positive real.Prove that :$ \sum_{k \equal{} 1}^{n}\frac {\sin (kh)}{k^2} < 3h$ for all positive integer $ n$[/quote]
Hello math10!

I'm sorry, but this assertion is wrong (For example, with $ h\equal{}0.01$ and $ n\equal{}11$, we get $ \sum_{k \equal{} 1}^{11}\frac {\sin (\frac k{100})}{k^2}\approx 0.030187777>3h\equal{}0.03$ ). More generally :

We know that, for $ x > 0$, $ \sin(x) > x \minus{} \frac {x^3}6$

So $ \sum_{k \equal{} 1}^{n}\frac {\sin (kh)}{k^2} >$ $ \sum_{k \equal{} 1}^{n}\frac hk \minus{} \sum_{k \equal{} 1}^{n}\frac {kh^3}6$ $ \equal{} h(\sum_{k \equal{} 1}^{n}\frac 1k \minus{} h^2\sum_{k \equal{} 1}^{n}\frac {k}{6})$

So $ \sum_{k \equal{} 1}^{n}\frac {\sin (kh)}{k^2} >$ $ h(\sum_{k \equal{} 1}^{n}\frac 1k \minus{} h^2\frac {n(n \plus{} 1)}{12})$

So, choose any positive integer $ A$. Choose then $ n$ such $ \sum_{k \equal{} 1}^{n}\frac 1k > A \plus{} 1$ and $ h$ such that $ h^2\frac {n(n \plus{} 1)}{12} < 1$ and you get $ \sum_{k \equal{} 1}^{n}\frac {\sin (kh)}{k^2} > Ah$ ....

[quote="math10"]Let $ h$ be real positive real. Find $ c$ min such that:$ \sum_{k \equal{} 1}^{n}\frac {\sin (kh)}{k^2} < c h$  for all positive integer $ n$[/quote]

So no solution

[quote="pco"]
Hello math10!

I'm sorry, but this assertion is wrong (For example, with $ h \equal{} 0.01$ and $ n \equal{} 11$, we get $ \sum_{k \equal{} 1}^{11}\frac {\sin (\frac k{100})}{k^2}\approx 0.030187777 > 3h \equal{} 0.03$  [/quote]

Very very sorry for I had the wrong format  :blush: .
Problem:Let $ h$ be positive real. Prove that: $ \sum_{k \equal{} 1}^{n}\frac {\sin^2 (kh)}{k^{2}} < 3h$


Let $ x$, $ y$ and $ z$ be positive real numbers.
If for all quadrilateral with sides $ a$, $ b$, $ c$ and $ d$:
$ a^2x\plus{}b^2y\plus{}c^2z>d^2$
Prove that:
$ 1 \geq \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}$

if $ x \le 0$, take $ a \equal{} n ,b \equal{} 1 ,c \equal{} 1 ,d \equal{} n$ and get $ y \plus{} z > n^2(1 \minus{} x)$ which is contradiction. Thus $ x > 0$ and similaly $ y > 0,z > 0$.Now use a quadrilateral of side $ \frac {1}{x},\frac {1}{y},\frac {1}{z}$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \minus{} \frac {1}{n}$,where $ n$ is large.so $ \frac {x}{x^2} \plus{} \frac {y}{y^2} \plus{} \frac {z}{z^2} > (\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \minus{} \frac {1}{n})^2$
Since this hold for all $ n$,take limit as $ n \rightarrow \inf$ and get
$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \geq (\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})^2$
then
\[ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \le 1
\]


Determine limit:
\[ \lim_{n\rightarrow\infty}\sum_{i \equal{} 1}^{n}\frac {1}{i^2}.\]

$ \sum_{i \equal{} 1}^{n}\frac{1}{n^2}\equal{}n\times\frac{1}{n^2}\equal{}\frac{1}{n}$.
So, $ \lim_{n\to\infty}\sum_{i \equal{} 1}^{n}\frac{1}{n^2} \equal{} \lim_{n\to\infty}\frac{1}{n}\equal{}0$.

[quote="luckyluke8x"]Determine limit:
\[ \lim_{n\rightarrow\infty}\sum_{i \equal{} 1}^{n}\frac {1}{n^2}.\][/quote]
$ \lim_{n\rightarrow\infty}\sum_{i \equal{} 1}^{n}\frac {1}{n^2}\equal{}\frac{\pi^2}{6}$

[quote="ye109"][quote="luckyluke8x"]Determine limit:
\[ \lim_{n\rightarrow\infty}\sum_{i \equal{} 1}^{n}\frac {1}{n^2}.\]
[/quote]
$ \lim_{n\rightarrow\infty}\sum_{i \equal{} 1}^{n}\frac {1}{n^2} \equal{} \frac {\pi^2}{6}$[/quote]

I think $ \lim_{n\rightarrow\infty}\sum_{i \equal{} 1}^{n}\frac {1}{\color{red}i\color{black}^2} \equal{} \frac {\pi^2}{6}$


Hey guys,

I was just wondering how you guys study math.  For the past year I tried doing a lot of past contests for whatever contest was coming up, and although I learned some things, it never seemed like I made too much progress, and sometimes, I never even knew what was going on, by reading the Kalva solution.

Well part of the problem is that I haven't ever covered some important topics (like for example, I've never officially done any Inequalities seminar/problem set).  So i think before I do all the past contests I need to learn these basic things first.  So I was wondering what's the best way to get confortable with certain new topics (inequalities, transformational geometry, modular tables)...I don't even know what some of these topics that I need to learn may be.

I think it's a lot easier to learn them by doing a whole bunch of questions on that same topic at once instead of coming across one once over a lot of past contests.  So are there any good books/websites/services(ie.tutor) that have like lectures/whatever on these topics?  Like it teaches basic concepts, gives examples, gives problems, and gives solutions.  That's what I think is the best way to learn math.  Wondering how you guys all got so good :)

Thanks for your help.

EDIT: I also put this post in the pre-olympiad section.  No one will mind, right?

The resources page on this site has a good list - it includes problem sources, books, contests, etc.


[quote="JBL"][quote="ch1n353ch3s54a1l"]
$a^{9}-12a^{6}+48a^{3}-81a+44=0$

$f(1)=0$

$(2+\sqrt{5})^\frac{1}{3}+(2-\sqrt{5})^\frac{1}{3}=\boxed{1}$[/quote]

And what about the other 8 roots of that polynomial?![/quote]
Yes, there are 2 other real roots and 1 other positive root.

[quote="lotrgreengrapes7926"]If we are looking only at the real cube roots, then $1$ is the answer.[/quote]

Of course, but that's not the point -- there was no way to know that from the given solution!  That polynomial had a triple root at 1 and of the other six roots, two were real.  How to choose?

There is a solution in which you get a third degree with one real (1) and two imaginary.  Then you know its 1.

[quote="jhcreinhardt"]Simplify:

$(2+\sqrt{5})^\frac{1}{3}+(2-\sqrt{5})^\frac{1}{3}$[/quote]

[hide]Let $(2+\sqrt{5})^{\frac{1}{3}}+(2-\sqrt{5})^{\frac{1}{3}}=x$

Then, we cube the equation, to get

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

But we substitute the value of x to get

$x^{3}=4-3x$, so $x=1$, since we are assuming real roots.

$1$[/hide]


Let $x_1,\ldots,x_n,\delta$ and $a$ be positive reals. Then the following inequality is true: \[ (\delta-x_1-\cdots-x_n)^2+a(x_1^2+\cdots+x_n^2)\geq\frac{a\delta^2}{n+a}.  \] Equality holds iff $x_1=\cdots=x_n=\frac{\delta}{n+a}.$

it is just an application of cauchy twice  :)


There are positive integers with 0 that:
$k_1<k_2<...<k_m$
Then; $n=2^k_1+2^k_2+...+2^k_m$
How many odd coefficient have a polynomial  $P(x)=(x+1)^n$

how should I do it??

$2^k_1$ should be2^$k_1$ ( something wrong with my tex)

It's a problem from current Polish MO. Lock the topic and please do not post the solution :!:


function
integration
inequalities
calculus
probability
derivative
real analysis

[quote="perfect_radio"]Let $f : [0,1] \to \left[ 0, \infty \right)$ be a convex function such that $f(0) = 0, \, f(1) = 1$.
Furthermore, let $g : \left[ 0, \infty \right) \to \left[ 0, \infty \right), \, g(t) = (t+1) \int_{0}^{1}f^{t}(x) \, dx$.

Prove that $g(p) \geq g(q)$ for all integers $q \geq p \geq 0$.
(I suspect we can use reals here, since I don't have a solution, I'll let somebody else decide that.)[/quote]
Can we take : $f(t)=x(2-x)$? let $q=1\geq p=0$ and therefore
after substitution we have:
$1\geq \frac{4}{3}$ it's means that this inequality is wrong.

Maybe I don't know how translated from English "Convex"? when $f''(x)<0$?

[quote="didilica"]Your function is not convex.[/quote]
Thanks.  I have  not translated correct

if we substitute instead $f(x)=x^{a}$ where $a>0$ then inequality is wrong/

[quote="Extremal"]if we substitute instead $f(x) = x^{a}$ where $a>0$ then inequality is wrong/[/quote]
The inequality is correct. I guess you'll just have to live with it.

Yes you are right. I am simple was retrained also to me a little is necessary to have a rest.

$f(x)$ must be non-negative if we want to define $f^{t}(x)$, and it implies $f'(x)>0, \ x\in(0,1)$. Taking $g'(t)$ leads to a very unpleasant integral, may be there is some tricky way to show it is non-negative.

In the original post $f$ is indeed assumed to be nonnegative. However, this only implies $f'(x)\ge 0$ (where defined), not $f'(x)>0$.

Note that $f$ is a convex combination of functions $h_{r}(x)=\frac{(x-r)^{+}}{1-r}$, $0\le r\le 1$, where $h_{1}(1)=1$ and $h_{1}(x)=0$ for $x<1$. Formally, $f=\int h_{r}\,d\mu(r)$, where $\mu$ is a probability measure on $[0,1]$. The function $t\mapsto (t+1) \int_{0}^{1}h_{r}^{t}(x) \, dx$ is identically equal to $1-r$, which lends additional credibility to the statement. My guess is that the inequality is actually true for reals $s\ge t\ge 1$, and it can be proved by representing $f^{t}$ (also convex) in the above form. But I don't have a proof.  :(

OK, here is a better idea: assume that $f$ is strictly increasing, with the inverse $h=f^{-1}$. (The general case is easily handled by approximation). Substitute $x=h(y)$ and integrate by parts: $(t+1)\int_{0}^{1}f(x)^{t}\,dx=(t+1)\int_{0}^{1}y^{t}h'(y)\,dy=h'(1)-\int_{0}^{1}y^{t+1}h''(y)\,dy$. Since $h$ is concave, its follows that $t\mapsto \int_{0}^{1}y^{t+1}h''(y)\,dy$ is a nondecreasing function for $t>-1$.  :)

I like mlok's idea. Here's my somewhat tedious proof. My idea is to generalize the inequality $g(p) \ge g(p+1)$ to replace $\int_{0}^{1}$ with $\int_{0}^{X}$ and then to prove the generalized inequality by using differentiation w.r.t the variable $X$.

Suppose $f$ is a nonnegative convex function defined on the interval $[0,b]$ with $f(0)=0$. For $p \ge 0$, define $F_{p}(X) = (p+1) \int_{0}^{X}f^{p}(x) \,dx$. Let $L_{p}(X) = F_{p}(X) f(X)-F_{p+1}(X)$. We want to prove that $L_{p}(X) \ge 0$ for $X \ge 0$. (Note that if $f(1)=1$ the special case $L_{p}(1) \ge 0$ is the inequality $g(p) \ge g(p+1)$.)

Since $L_{p}(0) = 0,\quad L_{p}' = F_{p}\cdot f'-f^{p+1},\quad L_{p}'(0) = 0$ and $L_{p}'' = f'' \cdot F_{p}\ge 0$ on $[0,b]$, we have $L_{p}(X) \ge 0$ for $X \ge 0$.


I'm not a regular[b] internet[/b] user[b].This day i found out in '[/b] Young[b] Mathematician[/b]'[b] website about the recently discovered E[/b]8[b].[/b] Just what exactly do most mathematicians([u]I'm only a '[/u]Mathematics[u] Enthusiast[/u].[u]'[/u])mean[b] by[/b] E[b]8[/b]?[b].[/b] Please,help.

                                                        Respectfully,
                                                        A.Eddington


[quote="Yosh..."]If $a,b,c \in R^{+}$ and $a^{2}+b^{2}+c^{2}=9$

prove $2(a+b+c)-abc\leq 10$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122166


calculus
integration
trigonometry
calculus computations

Find the length of the curve expressed by the polar equation: $ r\equal{}1\plus{}\cos \theta \ (0\leq \theta \leq \pi)$.

the length is $ s\equal{}\int^{\pi}_0\sqrt{r^2\plus{}\left(\frac{dr}{d\theta}\right)^2}d\theta$

so substitute all those term and simplify to get

$ \int^{\pi}_02\cos\frac{\theta}{2}d\theta\equal{}4$

[quote="CyanSnow"]I got the required length of the curve answer as [b]8a[/b] ,,,[/quote]

[quote="CyanSnow"]

3. Find the perimeter of the cord $ r \equal{} a(1 \plus{} cos\theta)$[/quote]

Your answer is correct, because you will find the answer for $ 0\leq \theta \leq 2\pi$, or $ 4a\cdot 2\equal{}8a$.


[quote="#H34N1"]The sum of two numbers is 20. The first is 4 times the sum of the other two. The second is seven times the third. What is the product of all 3?

A) 28
B) 40
C) 100
D) 400
E) 800[/quote]
[hide]Make the third number x. So the second number would be 7x, and the first one would be 32x. Do you mean the sum of all three numbers is 20? Assuming you mean that, this means that 40x=20. So x=1/2. So the numbers are 1/2,7/2,and 16. So multiplying them all is going to be 28 or A.[/hide]


Show that in group $ m$ men and $ n$ women exists $ k$ women, that we can find husbands for them [b]if and only if[/b] for all $ r$ (subset of "women" set), $ r$ women like $ k \plus{} r \minus{} m$ men.

This is [url=http://en.wikipedia.org/wiki/Marriage_theorem]Hall's marriage theorem[/url].  Note that the only if direction is trivial.


[quote="mgao"]where's the class forum?
Yes, I'm logged on and I am enrolled, and can see it in My Classes.[/quote]
[url]http://www.mathlinks.ro/Forum/index.php?f=185[/url]

[quote="#H34N1"]If I click on that link, I cannot see it :maybe:[/quote]
You have to take the class

I was taknig the class.  :roll: Its back now but it disappeared for about 15 minutes or so just before the first class...


abstract algebra
Ring Theory
advanced fields
advanced fields unsolved

I'm trying to find a ring $ B$ such that $ \mbox{height } \mathfrak{p}< \infty$ for all prime ideals $ \mathfrak{p}$, but the (Krull) dimension of $ B$ is infinite.

I had a hunch that $ B\equal{}k[x_1, x_2, \ldots]$ maybe would work?  Since $ (x_1) \subset (x_1, x_2) \subset \ldots$ is a chain of prime ideals, $ \dim B\equal{} \infty$.  The height of any of these prime ideals is finite though, but I need to show it for [i] any [/i]  prime ideal.  Does anyone know how to proceed?  Or is there another example?

You might have difficulties in showing that every prime ideal in $ B$ has finite height. But take the following amendment: Define $ m_k \equal{} \sum_{n \equal{} 0}^kn$. Then $ m_k$ is increasing and $ m_{k \plus{} 1} \minus{} m_k > m_k \minus{} m_{k \minus{} 1}$. The ideals $ \mathfrak{p}_k \equal{} (X_{m_k},\ldots,X_{m_{k \plus{} 1} \minus{} 1})$, $ k\ge 0$, are prime ideals in $ B$. Then the localization $ B' \equal{} S^{ \minus{} 1}B$ with $ S \equal{} B\setminus\cup_k\mathfrak{p}_k$ will serve your needs, since you can control all the prime ideals in $ B'$.

Hint: What are the maximal ideals in $ B'$?

Additional note: $ B'$ is Noetherian, since every prime ideal of $ B'$ is a finitely generated ideal - which is a necessary and sufficient criterion for a ring being Noetherian.


Neither hard, nor easy, just trivial computations. In fact, it was already posted, as far as I remember. And "my notes" should become "Gazeta Matematica, Mihaly Bencze".


a) Find two disjoint sets $X, Y\subset \mathbb{Q^{+}}\colon X\cup Y=\mathbb{Q^{+}}$ such that $Y=\{a\cdot b|a,b \in X\}$.
b) Find two disjoint sets $X, Y\subset \mathbb{R}\colon X\cup Y=\mathbb{R}$ such that $Y=\{a+b|a,b \in X\}$.

Are you sure about the problems statement¿
For a), $0$ can't be in the set $X$ (it would then also be in $Y$), but can't be in $Y$ either (then $0$ can't be a product $ab$ with $a,b \in X$).

I don't think part a is true.

If $0 \in X$, then $0*x \in Y$, therefore $0 \in Y$, and therefore $X$ and $Y$ are not disjoint.

Now, if $0 \in Y$, then $\exists a, b \in X$ such that $a * b = 0$.
Therefore either $a = 0$ or $b = 0$.  Thus, $0 \in X$, and now $X$ and $Y$ are not disjoint.

However, if you just wanted $X \cup Y = \mathbb{Q}\backslash \{ 0 \}$, then let $X = \mathbb{Q^{-}}$, and let $Y = \mathbb{Q^{+}}$.

Now, for part b, it is easy to see that if $0 \in X$, then $\forall a \in X$, $a+0 = a \in Y$.  Therefore, $0 \in Y$.

Then, there is some $a \in X$ such that $-a \in X$ as well, since $0 \in Y$.  We also have $a+a = 2a \in Y$, and $-a+(-a) =-2a \in Y$.

In fact, $\forall x \in X$, $x+x = 2x \in Y$.

I'm not sure where to go from there...

I beg your pardon, I have forgotten that we are to consider positive rational numbers only. Then part a) becomes quite an easy one, but I don't have a solution for b).

(Written here, was one wrong proof)  :lol:

[quote="Levi"]Consider $X$ the set of real numbers such that each element has it's first digit, on the decimal representation, in the set $\{5,6,7,8,9\}$, and $Y$ is the rest. Note that if $a,b \in \{5,6,7,8,9\}$ then $a-b$ and $a+b$ have the first digit in $\{0,1,2,3,4\}$. This is the first argument to this set work out. I'll let you think about the rest of the proof.

Exemple: How to reach $e$ by the sum of two elements of $X$?
Answer: $6+e-6 = e$, where $6+e$ and $-6$ are in $X$.[/quote]

What about $912+82=994$. According to you, all three numbers belong to $X$, but this would contradict problem requirements, since then we would infer that $994$ belongs to $Y$ as well.

What a big mistake!! :)

I'm sorry, I'm still trying to solve it... I should have written this "proof" before post it, I was in a hurry for lunch...

Thank you!

There's a very simple answer to the second problem: $X$ is the set of all numbers with fractional parts in the interval $\left[\frac13,\frac23\right)$.

a. Every positive rational number can be written as $\prod p_{i}^{e_{i}}$ for some finite number of $p_{i}$, $e_{i}\ne 0$. For any such representation, consider $\sum e_{i}$. Put the given rational in set $X$ if this sum is odd; otherwise put it in set $Y$. We note that $X$ and $Y$ partition the rationals, that the product of any two elements in $X$ yields and element in $Y$, and that any element $q$ in $Y$ can be written as the product of two elements in $X$ (for example, $\frac{1}{p}$ and $qp$ for any prime $p$).

Another solution to part a):

Fix a prime number $p$ and express every rational $\frac{m}{n}$ in the form $p^{k}\cdot \frac{a}{b}$ where $a$ and $b$ are coprime with $p$. Put in $X$ all positive rationals for which $k$ is odd and all the rest in $Y$ and note that we have achieved a desired partition.


function
induction
algebra proposed
algebra

Let $ f$ be a function defined by $ f: \mathbb{N} \minus{}\minus{}> \mathbb{N}$ such that
$ f(n)$ is a square for each $ n \in \mathbb{N}$ and also
for natural numbers $ m$ and $ n$ natural numbers, $ f(m\plus{}n)\equal{} f(m)\plus{} f(n) \plus{} 2mn$.

Prove that $ f(n) \equal{} n^2$.

That's easy. We have $ f(2) \equal{} 2f(1) \plus{} 2$ and in general (can be proved by induction on $ n$): $ f(n) \equal{} nf(1) \plus{} n^2 \minus{} n \equal{} n(n \plus{} f(1) \minus{} 1) \equal{} g^2(n)$ for every $ n$ and a function $ g(n): \equal{} \sqrt { f(n)} \in \mathbb{N}$, as $ f(n)$ is a square. Hence for every prime $ p$ we have $ p \mid f(1) \minus{} 1$, which is impossible iff $ f(1)\ne 1$. Hence $ f(n) \equal{} n^2$.


First of all, I wanted to comment that I am very happy with the recent renovations within the arts and sciences forum.  I asked the moderator of the music and arts forum if topics concerning literary form or subjects would be best placed there.  He suggested that I post here to ask where the best place for such subjects would be.

I'm not sure what kind of a subject are you referring to, but try to post it in some of the Arts and Sciences forums or in the Round Table (more recommended).

I think he's talking about literature. You know...literature. Shakespeare, Hawthorne, Milton, Keats, Fitzgerald, Byron, Pepys, Austen, Shelley, Orwell...

(Yuck on the AP Lit exam.)

Yes, as that's got [b]absolutely[/b] no connection with mathematics whatsoever, he should stick the thread in the Round Table.

I don't know, Valentin. Counting how many feet Shakespeare uses could definitely fit right into the "Intermediate" forum. (It doesn't help that languages change every couple decades either. It's kind of like changing the value of 1 whenever a recognized mathematican wants to.) ;)


limit
logarithms
calculus
calculus computations

Calculate $ \lim_{x\to 0^ \plus{} }({x}^{x^x} \minus{} x^x)$

Actually $ y^x \equal{} x^{x^2}$ It is well known that $ \lim_{x \to 0^\plus{}} x\ln{x} \equal{} 0$ so the limit is 1-1 = 0

So is it ${ \lim_{x\to0^{ + }}(x^{x^x}} - x^x )$ ? If it is then just remove the y because $ y^x = x^{x^2}$ if $ y = x^x$

If this is the case then the limit is -1


Suppose $ a,b\in \mathbb{Z}$ and $ ax^{17}\plus{}bx^{16}\plus{}1$ is divisible by $ x^2\minus{}x\minus{}1$,

find the possible value of $ a .$

[hide]
Let $ f(x) \equal{} ax^{17} \plus{} bx^{16} \plus{} 1$ and the two roots of $ x^2 \minus{} x \minus{} 1 \equal{} 0$ be $ p,q$,which are distinct irrationals.

We should find $ a,b$ such that $ f(p) \equal{}f(q)\equal{} 0$.

now,$ p^2 \equal{} p \plus{} 1$,so we see that $ p^{n} \equal{} F_{n}p \plus{} F_{n \minus{} 1}$ by induction,where $ F_n$ is the Fibonacci number.

Thus,$ f(p) \equal{} ap^{17} \plus{} bp^{16} \plus{} 1 \equal{} (F_{17}a \plus{} F_{16}b)p \plus{} F_{16}a \plus{} F_{15}b \plus{} 1 \equal{} 0$ and we have

$ F_{17}a \plus{} F_{16}b \equal{} F_{16}a \plus{} F_{15}b \plus{} 1 \equal{} 0$ because $ p$ is irrational,then $ a \equal{} F_{16},b \equal{} \minus{} F_{17}$ follow,and $ f(q)\equal{}0$ is obvious.

the possible $ a \equal{} F_{16} \equal{} \boxed{987}$.[/hide]

[quote="vinskman"][quote="Bovy11"][hide]
Let $ f(x) \equal{} ax^{17} \plus{} bx^{16} \plus{} 1$ and the two roots of $ x^2 \minus{} x \minus{} 1 \equal{} 0$ be $ p,q$,which are distinct irrationals.

We should find $ a,b$ such that $ f(p) \equal{} f(q) \equal{} 0$.

now,$ p^2 \equal{} p \plus{} 1$,so we see that $ p^{n} \equal{} F_{n}p \plus{} F_{n \minus{} 1}$ by induction,where $ F_n$ is the Fibonacci number.

Thus,$ f(p) \equal{} ap^{17} \plus{} bp^{16} \plus{} 1 \equal{} (F_{17}a \plus{} F_{16}b)p \plus{} F_{16}a \plus{} F_{15}b \plus{} 1 \equal{} 0$ and we have

$ F_{17}a \plus{} F_{16}b \equal{} F_{16}a \plus{} F_{15}b \plus{} 1 \equal{} 0$ because $ p$ is irrational,then $ a \equal{} F_{16},b \equal{} \minus{} F_{17}$ follow,and $ f(q) \equal{} 0$ is obvious.

the possible $ a \equal{} F_{16} \equal{} \boxed{987}$.[/hide][/quote]

would you please explain:
In what way to solve the system of eqts:

$ F_{17}a \plus{} F_{16}b \equal{} F_{16}a \plus{} F_{15}b \plus{} 1 \equal{} 0$[/quote]

Ok,use the famous fact about Fibonacci numbers that

$ F_{n\minus{}1}F_{n\plus{}1}\minus{}F_{n}^2\equal{}(\minus{}1)^n$ for all $ n\in \mathbb{N}$.

Now, $ F_{15}F_{17}\minus{}F_{16}^2\equal{}1\ne 0$,

so we see that the system has [b]only one[/b] solution,and easily check $ a\equal{}F_{16},b\equal{}\minus{}F_{17}$ surely satisfies it.

Of course,if we solve the system about $ a,b$ directly,it would not so complicated.

[quote="Bovy11"][quote="vinskman"][quote="Bovy11"][hide]
Let $ f(x) \equal{} ax^{17} \plus{} bx^{16} \plus{} 1$ and the two roots of $ x^2 \minus{} x \minus{} 1 \equal{} 0$ be $ p,q$,which are distinct irrationals.

We should find $ a,b$ such that $ f(p) \equal{} f(q) \equal{} 0$.

now,$ p^2 \equal{} p \plus{} 1$,so we see that $ p^{n} \equal{} F_{n}p \plus{} F_{n \minus{} 1}$ by induction,where $ F_n$ is the Fibonacci number.

Thus,$ f(p) \equal{} ap^{17} \plus{} bp^{16} \plus{} 1 \equal{} (F_{17}a \plus{} F_{16}b)p \plus{} F_{16}a \plus{} F_{15}b \plus{} 1 \equal{} 0$ and we have

$ F_{17}a \plus{} F_{16}b \equal{} F_{16}a \plus{} F_{15}b \plus{} 1 \equal{} 0$ because $ p$ is irrational,then $ a \equal{} F_{16},b \equal{} \minus{} F_{17}$ follow,and $ f(q) \equal{} 0$ is obvious.

the possible $ a \equal{} F_{16} \equal{} \boxed{987}$.[/hide][/quote]

would you please explain:
In what way to solve the system of eqts:

$ F_{17}a \plus{} F_{16}b \equal{} F_{16}a \plus{} F_{15}b \plus{} 1 \equal{} 0$[/quote]

Ok,use the famous fact about Fibonacci numbers that

$ F_{n \minus{} 1}F_{n \plus{} 1} \minus{} F_{n}^2 \equal{} ( \minus{} 1)^n$ for all $ n\in \mathbb{N}$.

Now, $ F_{15}F_{17} \minus{} F_{16}^2 \equal{} 1\ne 0$,

so we see that the system has [b]only one[/b] solution,and easily check $ a \equal{} F_{16},b \equal{} \minus{} F_{17}$ surely satisfies it.

Of course,if we solve the system about $ a,b$ directly,it would not so complicated.[/quote]

Thanks,Bovy11

You are so nice and great :roll:


algebra
difference of squares
special factorizations

Hello can someone please give me one example of sum of two squares and differnce of two squares that are examples and that you can solve. I want this so I can practice. The examples on my book are to easy....thanks

I don't completely understand... do you want:

$45^2+55^2$ and $55^2-45^2$?

You mean factoring problems...?

$x^4 + 64$.  :)

Yes those the ones I want. So how would I solve for one of them?


topology
real analysis
real analysis unsolved

Suppose the real line is decomposed into two uncountable Borel sets. Prove that a suitable translated copy of the first set intersects the second in an uncountable set.

I was looking through some older problems and I found this.

Let $ A$ and $ B$ be the two disjoint Borel sets which cover $ \mathbb R$. There are two cases (we don't really need this distinction, but prefer to arrange the proof this way):

(1) Both $ A$ and $ B$ are of second category.

Then we can find non-degenerate intervals $ I$ and $ J$ such that $ A$ and $ B$ contain dense $ G_\delta$ subsets of $ I$ and respectively $ J$. Now simply translate one of the sets such that the translated $ I$ cuts $ J$.

(2) One of the sets (say $ B$) is meager.

$ A$ then contains a dense $ G_\delta$ set $ G$, and $ B$, being an uncountable Borel set, contains a perfect compact set $ K$. Now it suffices to prove that given a perfect compact subset $ K$ of the real line and a dense $ G_\delta$ real set $ G$, some translate of $ K$ cuts $ G$ along an uncountable set. Write $ G \equal{} \bigcap_{n\ge 1}G_n$, where $ G_n$ is a descending sequence of dense open subsets of $ \mathbb R$. If we find a translate $ K' \equal{} x \plus{} K$ such that all $ K'\cap G_n$ are dense in $ K'$ we're done: it then follows that $ K'\cap G$ is dense $ G_\delta$ in $ K'$, and every dense $ G_\delta$ subset of a perfect [url=http://planetmath.org/encyclopedia/PolishSpace.html]Polish space[/url] is uncountable (any such set is itself Polish and perfect, so the Baire Category Theorem implies that it can't be countable, since otherwise it could be written as the countable union of its singletons). Finding $ K'$ is easy: let $ (D_n)_{n\ge 1}$ be a basis for the topology of $ K$; the set of points $ x$ for which $ x \plus{} D_n\subseteq\mathbb R\setminus G_m$ is contained in a translate of $ \mathbb R\setminus G_m$, and so is meager. It follows that the union of all these sets (for all positive integers $ m,n$) is meager. Taking any $ x$ outside this meager set, put $ K' \equal{} x \plus{} K$.


$ a^2\plus{}b^2\plus{}c^2\equal{}1,a,b,c>0$  prove that
$ \frac{1}{1\minus{}ab}\plus{}\frac{1}{1\minus{}bc}\plus{}\frac{1}{1\minus{}ca}\le\frac{9}{2}$

It's not so easy , being a problem of Vasc , see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36873&start=40&search_id=295288265]here[/url]


Duke college Stanford AMC USA(J)MO USAMO

I made a Bracket group on Yahoo. Link at bottom. Join once you register.. The name is 'Art of Problem Solving', and the password is 'mathlinks'

[url]http://tournament.fantasysports.yahoo.com/t1/group/2702[/url]

what's the id because you need the id, not the name

It's 2702 :roll:

You can tell by way the link says group/2702

sounds awesome!!!!!!!!!although tere are only 3 brackets so far come on people join!!!

dude its not gambaling unless you sign up to for that $ \$$5,000,000 thing, its just for bragging rights, and its lots of fun! ive done it before and is tons of fun(although i lost miserably cause i dont follow college basketabll)!

im to lazy do it...so ill just make my post here...I DO acctually follow college basketball....
[u]Number 1 seeds[/u](in order)
Tenessee
Memphis
Kansas
Duke

[u]Last 4 teams in the tournament[/u](in order)
Missouri
Ohio State
S. Illinois
S. Kentucky

[quote="ASPEED"]WHAT?!- you left out UNC!!! they're gonna win it all[/quote]

Not to mention Texas, who beat both Tennessee and Kansas.

i hope its useful to you..
we provide all the stuff related to march madness like [i]2008 March madness schedule[/i], march madness picks, march madness predictions, free 2008 march madness picks , march madness betting odds , march madness handicapping , march madness bracket picks , and more ncaa tournament information. Bookmark us and visit everyday during the 2008 march madness tournament , as we are the best source for NCAA Basketball March madness schedule information, march madness picks , and [b]2008 March madness[/b] predictions .
Hey just Go through this below link
[url]http://www.thebestjerseys.com[/url]
 :lol:

Do you even know how the bracketts work people? You sound like noobs...Texas has 0 chance of getting a 1 seed, maybe a 3 seed at best, Tennessee just owned Memphis so yea. UNC well..there good but they wont get a number 1 seed.

[quote="sailingdog"]By who??? [/quote]

LOL that was a joke,  i guess you could say me, were friends and im really into college basketball so i "made" him make a bracket too.  But it was just a joke...no violence i assure you!

[quote="crazychineesmaniac"]Number 1 seeds(in order)
Tenessee
Memphis
Kansas
Duke [/quote]

Duke!!! I totally agree with that!!!!

Duke + Zolojetto = 1337 -- thats my bracket name!!!

Reporting after the ed of Day 3:

Day 1 - 16/16
Day 2 - 10/16
Day 3 - 3/8

How much worse can it get?

Note: I made the 2nd Chance Bracket Group: same password (mathlinks), ID 3902. Basically, you pick from the Sweet 16 on, knowing who is in the sweet 16.

join soon!

[quote="crazychinesemaniac"][quote="the future"]If you look at my previous post, I just mixed up Tennesse and Kansas, but thats only becuase Tennessee lost to Arkansas...(and that was a week ago)

Anyways, UCLA, Louisville, Stanford, USC (Maybe KU) final four...go Pac-10

Tennessee is overrrated, Kansas always underachieves, Xavier is underrated

And how did Arizona St. get left out over Arizona (Arizona St. beat Arizona twice, had a better record-conference and regular record-, and Arizona St. beat Stanford (3 seed), Xavier (3 seed), and USC, while Arizona barely beat Wash. St..., who cares about Strength of Schedule, its useless, I could have a really high strength of schendule even if I lose every game by 30 points to the best teams, its who you beat thats good, not who you lose to)
Arizona is possibly the worst at large to go to the tournament ever.

And how about them Bulldogs, and this could of been a great script (with the fact that they could use the two games and one day as a good thing, going from worst to first-too bad theyll lose to Xavier)[/quote]

Lol Fifty bucks say none of them make it except UCLA[/quote]

uhh 3/4 hopefully...i sitll have 7/8 elite eight teams alive...

oh well, i think i manage to qualify for USAMO so that should make up for my bracket being screwed up (its like 49th percentile but it sitll has a good chance to gain alot of points)
-atleast its an improvement from last year where i got 0/4 final four teams correct

I got 12 of my sweet 16 picks right.  Sadly, I made a lot of random upset picks in the first round, so I think I only managed 20/32 there.  I did pick Davidson over Georgetown and USD over UConn, though.

All of my elite 8 teams are still alive except Duke.  I had them going to the last game, though (losing to UNC).

On the second chance bracket i think I made Nova and WKU go to the Championship game, which means 99.9% im going to finish last or .1% ill win and everyone else loses since no one else would pick Nova, WKU

Wow. All four #1 seeds made it to the Final Four for the first time EVER.  Didn't see that coming...

And how could Texas lose to Memphis?  :mad:

haha crazychinesemaniac was right, I only got 1 right...

People that dont know basketball probably had better brackets than people that do know basketball as they will just choose the highest seeds...

err...3/4 so im happy tho only chance for getting 1/2 in championship game. Stupid Texas...nuf said

Haha, I'm near last in the regular bracket, but first in the second chance bracket. I guess I called too much random stuff happening in my regular bracket (vanderbilt in the elite eight, uconn in the final four, georgetown in the finals, etc etc)

[quote="andersonw"]I was the only one in the second chance bracket to predict davidson would beat wisconsin. :lol:[/quote]

actually i had that, but that was because i was predicting massive upsets (i had villinova in the championship game), so i really didnt "predict" it...and thats also why im last in the second chance brackets...

my first bracket might finish last also if UCLA loses, but with the way the scoring works, I can still get 3rd if UCLA wins the championship (since only 2 people other than me have that), and thats with only 1/4 final four teams

[quote="cf249"]Well, Kansas won the championship with an incredible comeback against Memphis.  Thoughts?[/quote]

Well first off, I think my main bracket finished 2nd to last or something like that...

Not really an incredible comeback, just a total meltdown in poise from Memphis.

Just made 2/5 free throws and you win the championship.  Just foul Collins or Chambers and you win the championship.

ehhhh had Memphis won I would have made a huge comeback from being in last a few days ago to being in 2nd, one point behind mustafa. But that didn't happen. On the bright side, I still won second chance  :P


a website has 20 forums. It decides to increase its forums by 10 percent a day off the previous day. After five days, how many forums does it have? For example it goes, 20, 22, 24.2 which becomes 24

EDIT: Round down every time

yes. round down every day. (A site can't have 24.8 forums  :rotfl: )

[hide]In that case, 30 forums after 5 days.[/hide]


geometry
geometric transformation
dilation

Outline a synthesis of o-allylstyrene from benzene.

[hide="answer"]
1.Gattermann hydroformylation --> Benzaldehyde
2.Claisen smidt with HCHO
3.Then sulphonate it to separate the para product
4.Now do Fridel crafts reaction with allyl chloride
5.Remove the -SO3H group
[/hide]

Sorry sorry....
Was in a bit of rush that day :oops:

You can do wittig's reaction:
$ \ce{Ph \minus{} CHO}$ + $ \ce{PPh_3 \equal{} CH_2}$  - - > $ \ce{Ph \minus{} CH \equal{} CH_2}$

Yes: that's only styrene, but the target is o-allylstyrene.

:huh: No carcul my second post where the rest of the synthesis is there :huh:

Your second post is this:

[quote="Chemrock"]Sorry sorry....
Was in a bit of rush that day[/quote]

And your first post is wrong. So...

how abt this then?
benzene + alcl3+ Ch2Cl-Ch2-CH2Cl to get indane( using high dilution technique if necessary).
indane + 1 eq. nbs  followed by treatment wid alc.KOH to get alkene foll by treatment with acid.
follow it up wid reductive ozonolysis and and then wittigs reaction.


I am doing some project, and I am in a very hury. Can someone give me code in tabular form for following pics?

$\Tex$

Assuming you are using Windows, [url=http://www.g32.org/latable/]LaTable[/url] will produce the code for the tables you want quickly and esily


function
real analysis
integration
calculus
derivative
probability and stats

Say $ X$ and$ Y$ are independent random variables with densities f and g, respectively:

$ f(x) \equal{} e^{\minus{}x}$ if $ x > 0$ and 0 if $ x < 0$

$ g(y) \equal{} 2e^{\minus{}2y}$ if $ y > 0$ and 0 if $ y < 0$

Calculate a density function for $ Y/(X\plus{}1)$

Ideas I thought of were change of variable, or convolution. I know convolution lets you find the addition of densities, but dividing them is something altogether new for me, and I am not sure how to even approach it.

We begin by considering the cumulative distribution function of Z = Y/(X+1):

$ F_Z(z) \equal{} \Pr\left[\frac {Y}{X \plus{} 1} \le z\right]$

$ \equal{} \Pr[Y \le z(X \plus{} 1)]$

$ \equal{} \int_{x \equal{} 0}^\infty \Pr[Y \le z(X \plus{} 1) | X \equal{} x] f_X(x) \, dx$

$ \equal{} \int_{x \equal{} 0}^\infty F_Y(z(x \plus{} 1)) f_X(x) \, dx$.

Since $ Y$ is exponentially distributed with mean 1/2, it follows that the CDF of Y is simply $ F_Y(y) \equal{} 1 \minus{} e^{ \minus{} 2y}$.  Hence

$ F_Z(z) \equal{} \int_{x \equal{} 0}^\infty (1 \minus{} e^{ \minus{} 2z(x \plus{} 1)}) e^{ \minus{} x} \, dx \equal{} 1 \minus{} \frac {e^{ \minus{} 2z}}{1 \plus{} 2z}$.

Therefore the density of Z is the derivative of the CDF, or

$ f_Z(z) \equal{} \frac {4e^{ \minus{} 2z}(1 \plus{} z)}{(1 \plus{} 2z)^2}$.

one classical way to do this kind of exercice is to consider a nice function $ f$ (e.g: positive, bounded) and to compute $ E[f(\frac{Y}{X\plus{}1})]$:
$ E[f(\frac{Y}{X\plus{}1})]\equal{}\int_{y \geq 0} \int_{x \geq 0} f(\frac{y}{x\plus{}1}) e^{\minus{}x} 2 e^{\minus{}2y} dx dy$
and to to some computations to write it as
$ E[f(\frac{Y}{X\plus{}1})] \equal{} \int_{u \in \mathbb R} f(u) \psi(u) du$
then one can conclude that the density of $ \frac{Y}{X\plus{}1}$ is $ \psi$. If you have difficulties, do not hesitate to ask questions ..


Find the maximum and minimum values of $\sqrt{3}\sin \theta \cos \theta+\cos^{2}\theta \ (0^\circ \leq \theta \leq 90^\circ).$


A farmer want to fence in a pigpen along a stone wall which will serve as part of a boundary. The rest of the pen is circular with a radius of 2 yards and with the center of the circle 1 yard away from the wall and within the pen. Find the amount of fence required for the circular part of the boundary.

Wait I don't understand...is it a semi circle? then the solution is

[hide]

$4 \pi$

[/hide]

but since the center is 1 yard away from the wall then this should definately be in HSB...

[quote="Art of Owna"]I get 8pi/3..but I need someone to confirm this, considering, I thought for about 3 mins before drawing the picture, which I might have drawn wrong..[/quote]
You are right. So now you can show your solution.
[quote="worthawholebean"] No need to triple post, and this should probably go in Classroom Math. [/quote]
This was in a hoover geo test. I really don't think this should belong in classroom math.

Big hint
[hide]you can find that its a right triangle, with hypotenuse of length 2, a side of length 1 (what kind of special triangle has this), then you can find the central angles and the arc length is pi (diameter)* (central angle)/360[/hide]

[quote="max_tm"]Yes, that's what I get too.

hint:
[hide]
Draw a circle of 2 units then draw a chord that is 1 unit away from the center. How would you find the length of this newly formed arc?

[/hide]

EDIT: on a totally unrelated note, how do you make the hide tag something other than hide?[/quote]
In order to do that:
[code][hide=blah blah]something[/hide][/code] and that gets you
[hide="blah blah"]something[/hide]


function
real analysis
real analysis theorems

Let X be an uncountable set. Describe the smallest sigma-algebra that
contains all the one point subsets of X. What is the corresponding
sigma-algebra of real functions?

So if we define P = { {x} | x \in X } (P is the collection of all one
point subsets of X) then we need to describe S(P) (the sigma algebra
generated by P)

This is by definition the intersection of every sigma-algebra containing
P.

If X were countable then it seems plainly clear the S(P) = powerset(X)
since given any H \subset X \UNION{{h} | h \in H} is a countable union of
elements in the generating family P, and so H is in S(P) so powerset(X) \subset S(P)
but clearly S(P) \subset powerset(X) so they are equal. But what about
in the case of an uncountable set X?

I know (from the above reasoning) any countable subset of X is in the sigma
algebra generated by P. Also X\{x} \in S(P) for each x \in X. Using this reasoning
we can construct any uncountable subset missing only a countable number
of elements from X. By intersecting X\{x_1} \intersect X\{x_2} \intersect ... X\{x_n} ...
But what about uncountable sets Y, with uncountably many elements of X not in Y??

As for the corresponding sigma-algebra of real functions I am clueless.

Thanks for your help.

Define F={C \subset X | C is countable} \union {U \subset X | X\U is countable}. Then I am almost positive this is a sigma algebra. Closed under complements is trivial. The empty set is countable so its in there, and the union of any countable collection is either countable or has a countable complement so F is closed under countable unions. Therefore its a sigma algebra. It countains P since {x} is countable for each x in X. And any sigma algebra which contains P can be shown to contain F.

I am still a little confused about the corresponding real function sigma algebra what do you think?


What ways are there of finding the equations of the tangents at the pole for a curve given in polar coordinates?

In particular, I am interested in the curve $ r\equal{}\sin^{3}\theta$ for $ 0\le\theta\le\pi$.

Well you could convert it to rectangular coordinates by multiplying both sides by $ r^3$. Then you just have to use implicit differentiation.

Thanks sunehra.

That is what I did in another post called reduction formula as part of another question: College Playground, Calculus Computations and Tutorials, reduction formula.

I want(ed) to know if my answer there is correct and also wanted to find out about other ways of doing this which is why I've sort of double-posted.

Can anyone explain any (other) ways of approaching this problem?


Hay les va este problema, k a mi gusto, esta divertido:

Hallar todos los primos distintos P,Q,R,S tales que:
1) P+Q+R+S sea primo y
2) P^2+QR y P^2+QS sean ambos cuadrados perfectos.

Espero sus soluciones :P

Las unicas soluciones son $(p, q, r, s): (2, 7, 3, 11)$ y $(2, 7, 11, 3)$

Dado que $p+q+r+s$ es primo, entonces forsosamente uno de los cuatro es $2$.

Si $q=2$ o $r=2$ o $s=2$, dado que  $p^{2}+qs$ y $p^{2}+qr$ son cuadrados perfectos, tendriamos la siguiente ecuación:
$p^{2}+2k=n^{2}$ con $k$ uno de los primos.
$2k=(n-p)(n+p)$ pero $(n-p,n+p)=1$ ó $2$, entonces $2$ divide a $(n-p)$ y a $(n+p)$ o a ninguno de los dos, ambos casos imposibles.
Entonces $p=2$.

Entonces obtenemos:
$qr=(n-2)(n+2)$  y  $qs=(m-2)(m+2)$
Si $r=(n-2)$ entonces $q=(n+2)=(m-2)$ y $s=(m+2)$
$m=n+4$ entonces $s=n+6$

De donde $q$, $r$, $s$ dejan residuos $0$, $1$, $2$ modulo $3$, en algun orden, por lo tanto uno de los primos es $3$. Como el menor es $r=n-2$ entonces $r=3$ asi $q=7$ y $s=11$.

Analogamente si $r=(n+2)$ obtenemos $s=3$, $q=7$ y $r=11$.

En ambos casos se cumplen las condiciones del problema.

 :ninja:  :huh:


hi

im faced with a math problem in my program

here is the deal :

there is a point on the circle , the circle is moving to right by X amount only in horizontal direction
at the same time the point is rotating counterclock wise by "alpha" radians.

now when the circle has moved by X amount , what is the new location of this point on the circle?

obviously both its position in x,y has changed .

how do you find the location of the new point

please , any help is appreciated and its very urgent ;-)

ps. radius is known

the new position of the point is given by this transformation

$\left( \begin{array}{cc} x' \\ y' \\ \end{array} \right) =r\cdot \left( \begin{array}{cc} x+X \\ y \\ \end{array} \right) \cdot \left( \begin{array}{cc} cos(\alpha) & -sin(\alpha) \\ sin(\alpha) & cos(\alpha) \\ \end{array} \right)$


Hi all,

May I know how to insert background pictures to a page only?  I do not want it to appear on subsequent pages.

Thanks!

Cheers,
Ruifang

The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=watermark]watermark[/url] or the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=wallpaper]wallpaper[/url] packages will do this.


search Harvard college MIT HMMT ARML Stanford

[quote="redragon104"]...I was wondering if anyone here knows any math competition we can do, and/or how we should study for them...[/quote]

Hi...I would like to point you toward a new mathematics contest, the American High School Internet Mathematics Competition.

Check us out at http://www.ahsimc.com or http://www.americanmathcompetition.com and sign up!  It will make a great competitive mathematics learning experience, and will help sharpen you for future competitions.

Try a search of this website -- people have brought related questions up before.  To see what kind of material to study, you can just look through these forums (particularly Intermediate and AMC, I would say).  I know that there's a list of competitions around somewhere as well, if you're willing to go find it.

I'll have more substantive advice later, particularly if you have more specific questions.

[quote="MithsApprentice"][quote="redragon104"]I want to start a math team at my school, mostly because i want to be in a math team and my school doesnt have one.  My school is in new jersey, I was wondering if anyone here knows any math competition we can do, and/or how we should study for them.  We already participate in the AMC/AIME, but we dont really prep at all.  Any help would be greatly appreciated.[/quote]

lun posted a lot on this a few weeks back...you could PM her (lunmu) or Caroline (Caroline) or any other AASTer...[/quote]

I'll take that as an invitation for me to post ;)

Yep, I go to AAST in New Jersey.  We participate in a lot of competitions.. The only one I can think of that's distinctive to NJ is the NJML (New Jersey Math League - http://www.mathleague.com/ ), though I'm pretty sure most states have one of these.  There's also the Atlantic-Pacific something or other (we all refer to it as AtPac), and of course the Mandelbrot (http://www.mandelbrot.org).  Each of the above consist of a few rounds of individual, 6 or 7 question exams.  The Mandelbrot also encompasses a team round.  I'm pretty sure there's at least one other in-house competition I'm not thinking of.

As for the out-of-school annual competitions, there's the Harvard-MIT Math Tournament (HMMT - http://web.mit.edu/hmmt/ ), which usually takes place late February/early March, as well as American Regions Math League (ARML.. most New Jersey people go to the Penn State competition.. http://www.arml.com/ ).  HMMT switches between Harvard and MIT, and takes teams of 8 members, while ARML takes teams of 15.

This is a really really brief description of the competitions.. if you have specific questions about any of them, you can PM me, or IM me (loro46mpnorp).

Oh yes.. as far as preparation goes.. most of the competitions offer practice exams and stuff through their website or by mail.  It's good to get those and practice with them, and help each other by explaining different topics.

My school has math team every weekend.. complete with soda, donuts, candy, cheese balls, etc.  Helps to get people motivated ;)

[quote="cats..."]how many hrs/weekend?  Do you meet on weekdays too?[/quote]

Well, AAST is unique in that Mr. Holbrook is our math team coach, and not an official teacher of anything.  We have unique scheduling, as well.. our school days go from 8 til 4:30, and they're split into 24 20-minute "mods".. So during the day, when we have free mods, sometimes we hang out in Mr. Holbrook's room and do math and stuff.  He's usually there from 10 AM - 3:30 PM on Saturdays and Sundays, but most people arrive between 12 and 1:00.

When we have "math team" .. since there are so many of us.. it's not really a meeting so much as a time for us to go and do math.  We're allowed to do pretty much whatever we want, though Mr. Holbrook provides us with some direction based on our grade and level.

Again, AAST is very unique - it'll be pretty difficult to replicate our math team practice patterns at a standard high school, especially if there aren't thaaat many interested students.

Good luck =)

[quote="Caroline"]Well, AAST is unique in that Mr. Holbrook is our math team coach, and not an official teacher of anything.  We have unique scheduling, as well.. our school days go from 8 til 4:30, and they're split into 24 20-minute "mods".. So during the day, when we have free mods, sometimes we hang out in Mr. Holbrook's room and do math and stuff.  He's usually there from 10 AM - 3:30 PM on Saturdays and Sundays, but most people arrive between 12 and 1:00.

When we have "math team" .. since there are so many of us.. it's not really a meeting so much as a time for us to go and do math.  We're allowed to do pretty much whatever we want, though Mr. Holbrook provides us with some direction based on our grade and level.

Again, AAST is very unique - it'll be pretty difficult to replicate our math team practice patterns at a standard high school, especially if there aren't thaaat many interested students.

Good luck =)[/quote]

oh nice...it was kinda like dat back in middle school minus the weekends...

we'd just hang out at math club for bout an hour after school everyday and being mr. president and all, i got to boss ppl around to get snacks and stuff (yes, i know...abusing executive rights) and we'd order pizza sometimes...didn't get much math done though...

[quote="MithsApprentice"]oh nice...it was kinda like dat back in middle school minus the weekends...

we'd just hang out at math club for bout an hour after school everyday and being mr. president and all, i got to boss ppl around to get snacks and stuff (yes, i know...abusing executive rights) and we'd order pizza sometimes...didn't get much math done though...[/quote]

The snacks can be somewhat distracting.. but it's nice if you just sort of need to get up and stop looking at a problem for a second to walk over, get a cup of soda, take a sip, sit down, and try again with a possibly fresh perception ;)

[quote="ComplexZeta"]That reminds me of Stanford Putnam practices. An hour after the practices began, pizza would arrive, and then everyone would stop doing math and get some food and then return to the math with some new inspiration (hopefully).[/quote]

I think that's the general hope.. but it doesn't always happen.  If it's a weekend and we take a Mandelbrot or some other competition, no one wants to do more math after the competition, so we all just get donuts and soda and sit around and talk ;)

[quote="ComplexZeta"]That reminds me of Stanford Putnam practices. An hour after the practices began, pizza would arrive, and then everyone would stop doing math and get some food and then return to the math with some new inspiration (hopefully).[/quote]

the food part sounds about right...new inspiration?...nah...

[quote="redragon104"]My school is in new jersey, I was wondering if anyone here knows any math competition we can do, and/or how we should study for them.[/quote]

And I would like to point you to my [url=http://goodlymaths.0catch.com/contests/directory.htm]growing list of math contests[/url].  There are a couple of math contests in New Jersey on my list, plus a bunch of national ones (denoted by [O] or [S]).

Wow. I didn't know all these existed. Are they all open to high schoolers?

On the note of starting a math team-- I started a math team this year at my school. We have about 6 to 8 members. What I found most difficult is getting new people to join because my math department is not exactly the most popular. Actually, I am pretty sure that it is the least popular of all the departments. However, most of the people on the team are people I know. I found it really difficult to provide incentives for people to join.

We did the AMC12, AIME (two of us), and HMMT this year. However, we are currently not having any meetings because there are no math competitions left this year that we could do. Most of them occur in the beginning of the school year or start then. You should definitely try to catch some of these.

I was wondering, what do people generally do at math team meetings? I taught what I knew and gave out problems but honestly even I would be bored if I had to listen to myself speak. How do keep people engaged?

[quote="mathnovice"]Wow. I didn't know all these existed. Are they all open to high schoolers?[/quote]

Yes, I'm pretty sure.

[quote]On the note of starting a math team-- I started a math team this year at my school. We have about 6 to 8 members. What I found most difficult is getting new people to join because my math department is not exactly the most popular. Actually, I am pretty sure that it is the least popular of all the departments. However, most of the people on the team are people I know. I found it really difficult to provide incentives for people to join.[/quote]

I've had the same experience at my school.  The only people who joined were people in my grade who I know pretty well.  Everyone else said "math...aaah!" or something to that effect when I invited them to join.

But we don't even have food at math club meetings, so it's really hard to get people to come.

[quote]I was wondering, what do people generally do at math team meetings? I taught what I knew and gave out problems but honestly even I would be bored if I had to listen to myself speak. How do keep people engaged?[/quote]

I think I gave one lecture on the Pigeonhole Principle, then gave up and just handed out photocopied AMC10s.  It's a considerable bit of work to prepare a presentation, and everyone gets really bored.   Most of the time, we do math competitions during the time we meet (mandelbrot individual and team take up two meetings, etc.).  A lot of times we don't even meet.

If I were returning to the same school, I'd probably plan something more formal - more discussions of topics, maybe write a contest and give out prizes from the prize money I won, etc.  Something like that.  But I'm not, and this year is almost over.

[quote="mathnovice"]I was wondering, what do people generally do at math team meetings? [/quote]

We have around 10 people at most meetings (out of 13 on the roster). We meet for 2 hours once a week. Some days we have contests, like Math League, or Mandelbrot. Otherwise, I have books of problems from various competitions and will bring something to work on each time. Everyone sits around a big table and works on them together, asking each other for help or hints. Then students volunteer to present solutions on the board. We do have snacks, but something simple like a couple of packages of cookies. Everyone chips in $5 each at the beginning of the semester, and one family volunteers to bring the snacks. Previously we tried taking turns, but too many people would forget. Oh, and sometimes we do practice problems from various science competitions.

Mmm... the snacks and stuff idea sounds good.

One problem I have is that the math team doesn't have an advisor. There are several reasons for this. I had to contact the head of the math department to get our school involved in the first place. However, unfortunately, she is one of the most despised teachers of all. Most of the other math teachers are not liked very much either. My AP Calculus teacher seems to be the most amicable, but, of course, I can't ask him to be the advisor after all the other teacher has done for me. I don't know why, but most of the math teachers at my school are honestly some of the strictist and meanest people I know. All the other departments have teachers who actually are someone nice and talk to their students.

I have another year at this high school. : (

your sponsor doesn't have to be a math teacher. The students can explain the math to each other, referring to the solutions when needed. A non-math teacher, or even a parent, can be a math team sponsor. The job is mostly organizational/administrative.

I attend Skyline High School which is in Utah. The math team is sort of combined with the chess team. Its more of a self motivation and self study concept if you want to do good in math. The chess team does very well, probably because there is chess practices once every week. There is a competition called the Utah State Math Contest. Skyline took 2nd in 10th grade and 11th grade. Plus, the junior high that feeds into Skyline took 1st in 7th, 8th, 9th.

[quote="churchilljrhigh"]Its more of a self motivation and self study concept if you want to do good in math. [/quote]

Seems like that's the only way to get real results.. if someone wants to learn something, they can study it - the math team's job is to provide them with the proper resources/atmosphere.

[quote="redragon104"]I want to start a math team at my school, mostly because i want to be in a math team and my school doesnt have one.  My school is in new jersey, I was wondering if anyone here knows any math competition we can do, and/or how we should study for them.  We already participate in the AMC/AIME, but we dont really prep at all.  Any help would be greatly appreciated.[/quote]

So, people have already given you a good list of competitions.  However, you should be warned: most of the good contests (AMC an exception) require you to register in the Fall.  That means that if you want to do this, you should probably work on getting it organized now and over the summer, so that you can start as soon as possible next year.

As for teaching, if you own problem books, look at the kinds of things that are necessary to know for the problems.  [u]The Art of Problem Solving[/u] book is a truly excellent resource in this regard, in that it's essentially a textbook for competitions.  It is quite possible to have students teach math team, but it's hard.  One way to do it is to set up a lesson plan (this is what we'll do this week, etc.), and if you have 5 or more people, everyone can prepare to teach a lesson a week in advance.  Obviously, this requires a fair amount of dedication, but it's certainly doable.  Alternatively, one person could teach all the time, which is a heck of a lot of work.  And then you can go the "lets just do and present a lot of problems" approach.

I'd think geometry should definitely be taught if there's a math competition class.  And I mean advanced geometry not the normal geometry taught in class of course.

[quote="JBL"][quote="redragon104"]It is quite possible to have students teach math team, but it's hard.  One way to do it is to set up a lesson plan (this is what we'll do this week, etc.), and if you have 5 or more people, everyone can prepare to teach a lesson a week in advance.  Obviously, this requires a fair amount of dedication, but it's certainly doable.  Alternatively, one person could teach all the time, which is a heck of a lot of work.  And then you can go the "lets just do and present a lot of problems" approach.[/quote][/quote]

Not necessairily.  I know enough stuff that the kids on my math team don't that I could teach them off the top of my head.

[quote="jehr314pi"]Our school has a large math team but we don't practice, we've been lucky for the past few years because we have naturally talented people.  Unfortunately our strong seniors are leaving so we want to run actual practices now.  Any advice on how to run practices and prepare people?[/quote]

1. Keep doing problems:
Do a lot of problem solving... say if it is AMC coming up, get a bunch of those tests and start practicing. Then grade it. It is not important that you get all of them right, but it is more important that you understand what you missed and finish off it. You can do that by discussion in groups, looking at solutions and trying to carve your own method, ask advisor, etc. There are a lot of methods for that.

2. Study
You dont improve on something you do not know. Even if you keep doing 1., if you don't know what you are doing, then you wont improve. Study on your own, study in math meet. Select a day where everyone can learn one thing. Good, useful books are AOPS book or Art and Craft. Some of the easier books can also be used.

3. Do not memorize, derive!
You will come across a lot of formulas and such, and when you learn them, you want to know HOW things are the way they are. Memorizing might be good for many multiple-choice type or easy fill-in competitions, such as AMC, Mu Alpha Theta, Math League, etc., but it barely does any good for prestigious competitions, such as ARML, USAMO, AIME... and such.

hope these things helped.

The main problem in my school is the motivation to participate/study in math contests.  Some are discouraged, while others just plain don't care.  However, most of them are reasonably talented.  Anyone have any ideas for helping this situation?

yeah, we had the same situation in our school this year, there are always some people that really care, and some who don't. well i guess, you just have to deal with it.
oh, by the way, one thing that i did for my team was that i made a website, for the team, that i could post files and other stuff on there, so if people don't wanna go the meetings to learn stuff, they could get files from the web...


geometry
circumcircle
geometric transformation
homothety
ratio
trigonometry
cyclic quadrilateral

Let $ H$ be the orthocenter of triangle $ \bigtriangleup ABC$, $ O$ be the midpoint of segment $ BC$.
$ AD$ is a line penpendicular to $ OH$ and $ D$ be the intersection of line $ OH$ and $ AD$.
Prove that $ A,B,C,D$ are in a circle.

$ P,Q,R$ are feet of the altitudes onto $ BC,CA,AB.$ Negative inversion with center $ H$ an power $ HA \cdot HP$ maps the circumcircle $ (O)$ of $ \triangle ABC$ into its nine-point circle $ (N).$ From the cyclic quadrilateral $ ADPM$ on account of the right angles $\angle ADP$ and $\angle AMP,$ we have $ HA \cdot HP \equal{} HM \cdot HD$ $\Longrightarrow$ $M$ is the inverse image of $ D$ through $(H, \minus{} k^2),$ since $ M \in (N).$ Then $D$ lies on $(O),$ as desired.

Let $ O$ be the circumcenter of triangle $ ABC$. It is well known that $ AH \equal{} 2.OM$ (The homothety with center $ G$(the centroid) and ratio $ \minus{} 1/2$ carries $ OM$ into $ HA$ )
Let $ AO$ and $ HM$ meet at point $ P$ . Since $ AH$ and $ OM$ are parallel and $ AH \equal{} 2.OM$, $ AO \equal{} OP$ whiCh implies that $ P$ is on the circumcenter and $ AP$ is a diameter of the circumcircle. Since $ \angle ADP$ is right, we're done

Let $ P$,$ Q$,$ R$ be the feet of the perpendiculars to $ BC$,$ CA$, $ AB$. Let $ S$ be the foot of the perpendicular from $ H$ to $ AO$. Then since $ \angle HSO \equal{} \angle HPO \equal{} 90$, HPOS is cyclic. Also, since $ \angle ADH \equal{} \angle ARH \equal{} \angle ASH \equal{} \angle AQH \equal{} 90$, $ ADRHSQ$ is cyclic. Next, $ P$,$ O$,$ Q$, and $ R$ are on the nine-point circle, so $ POQR$ is cyclic. So by the radical axis theorem, $ PO$, $ HS$ and $ QR$ concur at a point $ X$. Also, since $ \angle ADO \equal{} \angle APO \equal{} 90$, $ ADPO$ is cyclic. Thus, the radical axes of $ ADPO$, $ HSPO$, and $ ADRHSQ$ concur, so $ AD$ passes through $ X$. But since $ AD$, $ HS$, and $ PO$ concur, by the converse of the radical axis theorem, $ A$,$ B$,$ C$, and $ D$ are concyclic (end of problem).

Note that since $ \angle BQC \equal{} \angle BRC \equal{} 90$, $ BQRC$ is cyclic and since $ QR$,$ HS$, and $ BC$ concur, by the converse of the radical axis theorem, $ B$,$ H$,$ S$ and $ C$ are concyclic. So we have 7 cyclic quadrilaterals!!!!! (see picture) :)  :D  :lol:  :P
[geogebra]8b3d249524d402aa7591fd1eda1b5d9849bc1b55[/geogebra]

But we also have $ ARPC$,$ CPRA$, $ BRHP$, $ DHRX$ and $ PHQC$ are cyclic, so I guess you could make that 12.

[hide]Let $ E$ and $ F$ be points on sides $ AC$ and $ AB$ respectively such that $ BE\perp AC$ and $ CF\perp AB$. Let $ EF$ and $ BC$ intersect at $ G$. From [url=http://www.mathlinks.ro/viewtopic.php?t=296444]here[/url], we know that $ GH\perp AO$. Consider triangle $ AOG$. Line $ GH$ and $ AH$ are two altitudes of this triangle, hence $ OH$ is also the altitude of triangle $ AOG$. Thus $ OD$ is perpendicular to $ AG$, which implies that $ A,D,G$ are collinear.

Let the circumcircle of $ ABC$ and the circumcircle of $ ADFHE$ intersect at $ A$ and $ D'$ ($ D'\neq A$). Consider the circumcircles of $ ADFHE$, $ BFEC$, $ AD'BC$. The three radical axes are $ AD',EF,BC$. By the radical axis theorem, these lines are concurrent. Since $ EF$ and $ BC$ meet at $ G$, then $ AD'$ also passes through $ G$, which implies that $ D'\equal{}D$. Thus the cirumcircle of $ ABC$ passes through $ D$, and hence $ A,B,C,D$ lie on a circle.[/hide]

[quote="tang zy"]Nice solution of all of you! :)
Let $ P,Q,R$ be the feet of the perpendiculars to $ BC,CA,AB$ .Let $ G$ be the intersection of line $ BC$ and $ QR$,We can also prove that $ A,D,G$ are in a line.[/quote]I used this fact to prove the original problem. :D

A very simple solution.

Let $ AK$ be the diameter of the circumcircle of $ \triangle ABC$ then $ KB \perp AB$ and $ CH \perp AB$ so $ KB \parallel CH$. Analogously, $ KC \parallel BH$ so $ KBHC$ is a parallelogram. This means that $ KH$ passes through $ O$. Let $ D'\equiv KH \cap (ABC)$ then $ AD' \perp OH$ and so $ D' \equiv D$. Therefore $ A,B,C,D$ are concylic, QED.

Why is the map $ (O)$ under $ I(H,k)$ is $ (N)$?

[hide="Another solution inspired by my friend:"]It is well-known that $ BH\equal{}2R\cos B$. Note that $ \angle BCK\equal{}90^{\circ}\minus{}B$ and $ \angle CKB\equal{}\angle CAB\equal{}A$, so $ BK\equal{}\frac{BC\sin(90^{\circ}\minus{}B)}{\sin A}\equal{}2R\cos B\equal{}BH$. Since $ BM\perp HK$, then $ BHMK$ is a kite, which gives $ HM\equal{}MK$. Let $ CL$ be a diameter of the circumcircle of $ ABC$. Note that $ \angle CKL\equal{}90^{\circ}$. So $ HKL$ is a right triangle. Since $ HM\equal{}MK$, then $ HM\equal{}MK\equal{}ML$. Extend line $ LMH$ until it cut the circle at $ D'$. Since $ \angle MD'C\equal{}90^{\circ}$, then $ D'\equal{}D$, which implies that $ A,B,C,D$ lie on a circle.[/hide]


How many different numbers we can make from digits 1,2,3,4,4,5,5,5?
Edit: We [b]also[/b] can make 1-digit, 2-digit, 3-digit, ... , 7-digit number not just 8-digit number  :)

[quote="juki"][hide] 8!/2!3! all permutations if they were distinct are 8! and all permutations between the same gitis 3! for 5 and 2! for 4[/hide][/quote]

I think this is correct! :lol:
PS: $ \frac {8!}{2!3!} \equal{} 3360$   :razz

[quote="Bachukas"]Has somebody any ideas?  :maybe:[/quote]
My idea, and probably the obvious one, is brute-force it by casework. We just have to do 8 cases though :P

[quote="ghjk"]
My idea, and probably the obvious one, is brute-force it by casework. We just have to do 8 cases though :P[/quote]

The problem is that we actually have a lot more than 8 cases. For example, when we have seven digits the number could contain three 5's and two 4's, with $ \frac{7!}{3!2!}$ permutations, or it could have two 5's and two 4's giving $ \frac{7!}{2!2!}$ permutations...and so on.

So the case work could take a while...

[quote="Attila the Hun"]you can solve such problems only by casework.  :D

PS :[hide] I'm having login problems. whenever I log in to post something, it automatically logs off again. I have to log in some 3 or 4 times to post something. why is this happening.[/hide][/quote]

I think you should change your password......

[quote="Bachukas"]You didn't understood my problem. We also can make 1-digit,2-digit,3-digit,...,7-digit numbers.[/quote]
Yeah, I forgot about it.....
Sorry. :oops:

I'm thinking about 12 cases:

1) There are three 5's and two 4's
2) There are three 5's and one 4
3) There are three 5's and zero 4's
4) There are two 5's and two 4's
5) There are two 5's and one 4
6) There are two 5's and zero 4's
7) There is one 5 and two 4's
8) There is one 5 and one 4
9) There is one 5 and zero 4's
10) There are zero 5's and two 4's
11) There are zero 5's and one 4
12) There are zero 5's and zero 4's

What do you think?  :whistling:

It's good! I think we have another nicer way to do this problem.Let me see if I can find it!

[hide="Solution"] Pretend that the $ 4$s and $ 5$s are distinct and let $ S$ be the set of all such numbers obtained in this way (that is, out of $ 8$ distinct digits), including the empty string.  The group $ G \equal{} S_2 \times S_3$ acts by permutation on the two $ 4$s and the three $ 5$s and has order $ 12$.  We now wish to count the number of orbits $ X/G$.  By [url=http://en.wikipedia.org/wiki/Burnside%27s_lemma]Burnside's lemma[/url], we just need to count the fixed points of every element of $ G$.  (These are not the same $ 12$ cases as Bachukas:  here the number of cases is $ 2! 3!$, whereas in his division the number of cases is $ (2 \plus{} 1)(3 \plus{} 1)$.)  $ G$ has six conjugacy classes, which we will write in cycle notation using the embedding $ S_2 \times S_3 \to S_5$; we will also multiply each answer by the number of elements in the conjugacy class.

[b]The identity[/b] (1 element):  fixes everything.  $ \sum_{i \equal{} 0}^{8} {8 \choose i} i!$.
[b](12)[/b] (1 element):  fixes every number that uses both $ 4$s.  $ \sum_{i \equal{} 0}^{6} {6 \choose i} (i \plus{} 2)!$.
[b](34)[/b] (3 elements):  fixes every number that uses two of the three $ 5$s.  $ 3 \sum_{i \equal{} 0}^{5} {5 \choose i} (i \plus{} 2)!$.
[b](345)[/b] (2 elements):  fixes every number that uses every $ 5$.  $ 2 \sum_{i \equal{} 0}^{5} {5 \choose i} (i \plus{} 3)!$.
[b](12)(34)[/b] (3 elements):  fixes every number that uses both $ 4$s and two of the three $ 5$s.  $ 3 \sum_{i \equal{} 0}^{3} {3 \choose i} (i \plus{} 4)!$.
[b](12)(345)[/b] (2 elements):  fixes every number that uses every $ 4, 5$.  $ 2 \sum_{i \equal{} 0}^{3} {3 \choose i} (i \plus{} 5)!$.

Summing and dividing by $ 12$ gives the answer.  I don't know if this sum simplifies.  If it does, you might want to evaluate it using [url=http://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem]Polya's enumeration theorem[/url]. [/hide]

Your solution is much harder than my  :lol:

I'm only in 10th grade now.  :)


There are several ways to simplify this. For example, you don't have to look at the whole board; only one fourth of it is relevant, i.e. you only have to look at a 4x4 board.

i think you first find the total number of rectangles, then subtract those with only 3,2, or 1 (complementary counting)

[hide="hint"]
Total number of rect's in a m x n grid:
$\frac{m \cdot n \cdot (m+1) \cdot (n+1)}{4}$
[/hide]

[quote="CircleSquared"][hide="hint"]
Total number of rect's in a m x n grid:
$\frac{m\cdot n\cdot(m+1)\cdot(n+1)}{4}$
[/hide][/quote]

Theres no way that formula's correct..(try for 2x2 case)
I think you mean mn(m+1)(n+1)/4

OMG what happened?

....hmm, must have messed up with the $\LaTeX$.

You're completely right, madskillz, I've gone back and fixed it.

So isn't it just (9C2)*(9C2) - w/e... i really don't feel like getting up from my sofa and working it out

[hide]
I'm not sure, but would it have any thing to to do with the fact that the area is at least 8? This is still semi- brute force, but it's quicker than drawing.
1*8
2*4 2*5 2*6 2*7 2*8
3*3 3*4 3*5 2*6 3*7 3*8
4*4 4*5 4*6 4*7 4*8
5*5 5*6 5*7 5*8
6*6 6*7 6*8
7*7 7*8
8*8
So, 27?
[/hide]

[hide]
I think the best way is to count the ones with width 1, width 2, width 3, and so on.  Therefore it would be:
8(1(with height 7)+1(with height 8))+         :Width 1
7(5(with height 4)+4+3+2+1)+                 :Width 2
6(6+5+4+3+2+1)+                                 :Width 3
5(7+6+5+4+3+2+1)+                             : ...
4(7+6+5+4+3+2+1)+
3(7+6+5+4+3+2+1)+
2(4(discounting the weird ones which don't actually have four black squares because of their placement)+7+6+5+4+3+2+1)+
1(8+7+6+5+4+3+2+1)=
8(2)+7(15)+6(21)+5(28)+4(28)+3(28)+2(32)+1(36)=
16+105+126+140+112+84+64+36=
676
Wow, now that I write it out it seems kind of like brute force.
It still seems easier than drawing 676 rectangles though.
[/hide]

Is it just coincidence that $676=26^{2}$?

The solution manual says that the answer is 683. I read the explanation, and what they did was brute force it with really complex casework depending on width.

Platypi_Rule! did everything right, except he added wrong. 16+105+126+140+112+84+64+36=683, not 676.

:blush: Wow, that was very stupid.  I dropped seven on the third line for some reason.  So yes it is just a coincidence that 676=26^2

Thanks, but after reading these solutions, I've decided that I will skip this type of problem until the end of the contest. Thanks anyways.

[quote="alkjash"]There is an 8 by 8 checkerboard with alternating black and white squares. How many distinct rectangles, with sides on the grid lines of the checkerboard and containing at least 4 black squares, can be drawn on the checkerboard?

Is there an easy way to do this, or must you brute force it?[/quote]

[hide="@mathcrazed"]You might want to then subtract the smaller ones, since the problem asks for 4 or over ;)[/hide]


function
calculus
derivative
integration
quadratics
complex analysis

Find all analytic functions  for every $ z \in \mathbb{C}$ s. t.
$ z \equal{} 0$ is a zero of order 2, $ |f'(z)| \le 6|z|$ and $ f(i) \equal{} \minus{} 2$.

$ f(z)\equal{}2z^2$ is the only solution in this case.

...because we may use the cauchy formulae to express third-or-higher derivatives of $ f$ at any point in terms of an integral of $ f'(z)$ over an appropriate large power of $ z \minus{} a$ that dies out as we take the contour to infinity, so $ f$ is quadratic and is a multiple of $ z^2$


We put a small charge of $Q$ and another charge of $-4Q$ in distance $L$ from each other.
Where will the electrostatic potential be $0$ (zero)?
(The potential is zero in infinity)

And $\frac{4L}{3}$ from the charge -4Q. :)
So?

There should be an entire equipotential plane. I'll try to solve it, hang on :)

let the charge $Q$ be at origin.and choose  line joining $Q$ and $-4Q$ as $+X-axis$
for $0$ potential $Y_{-4Q}=4X_{Q}$
let $(x,y,z)$ be coordinates of points satisfying the condition this gives

$16(x^{2}+y^{2}+z^{2})=[(x-l)^{2}+y^{2}+z^{2}]$

Hey guys,I think you mustn't think so deeply on this problem.Didn't you see,Thaakisfox posted it as an [b]'Easy Electrostatic Problem[/b] :lol:  :D

Well namiqphysic I meant easy, compared to the other problems posted here on the ADVANCED physics forum. :wink:
Btw. I was thinking you should do what Pardesi did. :D


for b=(a)^(1/n),a,n are natural numbers.prove that there is not exist a polynomial whose degree is  <= n-1,sitasfy b is the root of this polynomial.
generally,for b=(a_1)^(1/k1)+(a_2)^(1/k2)+??+(a_n)^(1/k_n),there is not exist a polynomial whose degree is  <= k_1*k_2??*k_n-1,sitasfy b is the root of this polynomial.
BTW,I'm not sure it is ture.

To the second one: you must add more conditions. What if $a_1=a_2=\ldots=a_n$ and $k1=k2=\ldots=k_n$?
Also in some cases whan not all of them are equal you can find a polynomial with smaller degree.

As for the first one, you also need some extra conditions. What if $a=u^n$ for some natural $u$? Let's assume $a$ has no prime factor whose exponent is at least $n$ (i.e. $a$ is sort of "$n$'th power free", a generalization of the concept of square-free).

$\theta=a^{\frac 1n}$ is a root of $P(x)=x^n-a$. We have to show that $P$ is, in fact, the minimal polynomial (with rational coefficients, of course) of $\theta$. Let $\psi$ be a primitive root of $x^n-1$. The roots of $P$ in $\mathbb C$ are precisely $\theta,\psi\cdot \theta,\psi^2\cdot\theta,\ldots,\psi^{n-1}\cdot\theta$. It's easy now to obtain a contradiction from the assumption that $P$ is reducible, because the free term of any proper divisor of $P$ would have $a^{\frac mn}$ as its modulus, for some $0<m<n$, which is clearly impossible, since $a^{\frac mn}$ is not rational.

sorry for my unstrict problem.and thank you for you improvement.
but dear grobber,I'm afraid I didn't understand your solution compeletly
how do you know "a contradiction from the assumption that  is reducible"

Assume $P=Q\cdot R$, where $Q,R$ have degrees strictly smaller than that of $P$. I wrote there that in this case, the free terms of $Q$ and $R$ would not be rational. I think it's pretty clear. :?


Positive integers $a,b,c,d$ satisfy $a+c=10$ and \[\displaystyle S=\frac{a}{b} + \frac{c}{d} <1.\] Find the maximum value of $S$.

I think the answer is $S=\frac{6}{7}+\frac{4}{29}$.

I'm sure the maximum value of $S$ for any given $a$ is $\frac{a}{a+1}+\frac{10-a}{(10-a)(a+1)+1}=1-\frac{1}{(a+1)((a+1)(10-a)+1)}$, but I can't prove this.


Let $ a,b,c$ be positive real numbers which satisfy $ 5(a^2\plus{}b^2\plus{}c^2)<6(ab\plus{}bc\plus{}ca)$. Prove that these three inequalities hold: $ a\plus{}b>c$, $ b\plus{}c>a$, $ c\plus{}a>b$.

If we suppose the claim is not true and take $ a\equal{}b\plus{}c\plus{}x ; 0 \leq x$ after computations the inequality at beginning is equvivalent to: $ 4(b\minus{}c)(b\minus{}c)\plus{}x(5x\plus{}4b\plus{}4c)<0$ which is obviously false...

Suppose $ c>\equal{}b>\equal{}a$. Then we can choose a=a, b=a+x and c=a+x+y where $ x, y>\equal{}0$.
We can see that we just need to prove that $ a\plus{}b>c$, it means, $ a>y$. Then, we suppose that $ a<\equal{}y$.
Now after substituying the expresion given we have:
$ 0<3a^2\plus{}4ax\plus{}2ay\minus{}4xy\minus{}4x^2\minus{}5y^2<\equal{}3y^2\plus{}4yx\plus{}2y^2\minus{}4xy\minus{}4x^2\minus{}5y^2\equal{}\minus{}4x^2<\equal{}0$
which is an absurd.
Then we can conclude $ a>y$ and the problem is over.

[quote="Johan Gunardi"]Let $ a,b,c$ be positive real numbers which satisfy $ 5(a^2 \plus{} b^2 \plus{} c^2) < 6(ab \plus{} bc \plus{} ca)$. Prove that these three inequalities hold: $ a \plus{} b > c$, $ b \plus{} c > a$, $ c \plus{} a > b$.[/quote]

For any $ a,b,c$ the following identity is true:

$ (a \plus{} b \minus{} c)\,(a \minus{} b \plus{} c) \equal{} \frac {1}{4}\left(\,6\,(ab \plus{} ac \plus{} bc) \minus{} 5\,(a^2 \plus{} b^2 \plus{} c^2) \plus{} (b \plus{} c \minus{} 3\,a)^2\right)$

So, if it is given that  $ 6\,(ab \plus{} ac \plus{} bc) \minus{} 5\,(a^2 \plus{} b^2 \plus{} c^2) > 0$, then $ (a \plus{} b \minus{} c)\,(a \minus{} b \plus{} c) > 0$. If the product $ (a \plus{} b \minus{} c)\,(a \minus{} b \plus{} c)$ is positive, then either both multipliers are positive, or both are negative. But if both $ a\plus{}b\minus{}c$ and $ a\minus{}b\plus{}c$ are negative, then their sum will also be negative. But their sum is $ 2\,a \equal{} (a\plus{}b\minus{}c) \plus{} (a\minus{}b\plus{}c)$. Now it's time to remember that it is given that $ a > 0, b > 0$ and $ c > 0$, so $ 2\,a$ cannot be negative, and the problem is solved.

Thank you, Johan Gunardi, for the nice inequality! Here is my reply to your inequality:

Let $ a,b,c$ be positive real numbers which satisfy $ 5\,(a^2 \plus{} b^2 \plus{} c^2) < 6\,(ab \plus{} bc \plus{} ca)$. Prove, that $ 11\,a \plus{} b > 5\,c$.

Sergey Markelov
markelov@mccme.ru


AMC
USA(J)MO
USAMO
trigonometry
inequalities
geometry
power of a point

This is probably already on the forums somewhere, but it's hard to search.

Say you've made some good progress on a USAMO problem, or perhaps have a near-complete solution with a gap, but have given up the endeavor or do not have enough time to continue working on the problem. Should you
1) Write down your progress, and leave it at that.
2) Write down what you have, and acknowledge that the solution is incomplete
3) Write down your work, and put "QED" or "as desired" or "and we're done" as if you actually solved the problem
4) Write down what you've done, and continue the solution with nonsense as if it solves the problem
5) or another alternative...

Personally, I'd write down whatever I had, then make a note of exactly what I would need to do to finish the proof.  It may turn out that the gap is a lot smaller than you think.

[quote="jb05"]If you pretend to have solved the problem, you run the risk of annoying the grader. However, if you turn out to be missing just a little, you could get a 5 or 6. If you pretend to have solved the problem and write nonsense, you WILL annoy the grader and get a low score. You don't have to acknowledge that the solution is incomplete, but you shouldn't try to be tricky about it. All of the USAMO graders are VERY smart. I would say that (1), (2), (3) are all fine, depending on the situation.[/quote]

What if the situation a student thinks he solved the problem but really forget about some cases? (as I forget about the possibility of equal adjacent term in the sequence this year) Will grader regard him as trying to do trick and get annoyed? How to differentiate these situations?

Well then there's nothing you can do about it. I would imagine it is easy to tell the difference between a genuine attempt at a solution and a deceitful solution.

[quote="JRav"]Definitely don't don't do 4, that's for sure.[/quote]
I agree. My friend got #1, except he didn't know the radical axis theorem; instead of just stopping there, he said that "this result [radical axis theorem] can easily be shown by trig Ceva." He got a 0 overall.

[quote="JRav"]Definitely don't don't do 4, that's for sure.[/quote]

Oops.  :oops:  So much for trying to get partial credit.

Zuming once told a class personally that if you reached something you knew was obviously true but you couldn't prove, to put QED and bluff it or just write Muirhead's Inequality, as in (3).

Personally, I do (1) more, but (3) can actually get you a major point boost in case of luck.

1.) looks to be the right choice.
2.) might hurt you if you [i]were[/i] actually just one trivial step away from solving it but didn't see the step. so don't write that you don't know how to solve it
3.) is a bit icky, and I agree the graders wouldn't be too happy. I'd rather leave it at one with an [i]air[/i] of completeness, not a declaration.
4.) is icky, you might lose all your points

When I was taking the USAMO, if I could reduce a problem to a single claim I would usually cite that claim as known and proceed to prove the problem from there (or just state the claim without any proof and hope the graders thought it was obvious). On the problem from 2007 with the circles, I stated that some bound followed by Descarte's circle theorem and think I got at least a 5 on that problem (it did in fact follow from it, fairly straightforwardly, but I didn't know Descarte's circle theorem at the time so I couldn't actually prove it).

I did something similar to (4) on #1 (pointed out two random triangles from the diagram that were similar, from which the conclusion follows). I think that cost me any partial credit I might have gotten.

Yeah, the thing to keep in mind is that any incorrect math on your paper anywhere can cost you points.  I remember one of the speakers at the 2002 USAMO in Boston told us that if we wrote up a completely correct solution to a problem, and then put "By the way, 0=1" at the bottom, they could still dock points for that.

"Yeah, the thing to keep in mind is that any incorrect math on your paper anywhere can cost you points."

^Haha, I always resort to using method number 4 on USAMOs if I haven't completely solved a problem, no wonder I don't pick up much partial credit whatsoever.

Gah, I'm still annoyed at my number 6.

I did get most of the second official solution (i showed that for any prime power p^n that was in the denominator of a t_i - t_j, p^n was in the numerator of every s_i - s_j)
Sadly, i missed the fact that there can be only a finite number of primes in r, making my proof incomplete.

From my scores, I got about a 1 on this problem, when a 5 would get me into blue mop. Is this really how harshly the USAMO is graded (where missing a simple observation that can be quickly proven after my work can cost you 6 points?)


function
limit
real analysis
real analysis unsolved

Let $ f: (0,\infty) \rightarrow (0,\infty)$ be a bounded function and $ p>0$ a fixed real number. If $ \displaystyle\lim_{x\to 0,x>0}(f(x)\minus{}\alpha f^p (\alpha x))\equal{}0$, for any $ \alpha >0$, prove that $ \displaystyle\lim_{x\to 0, x>0}f(x)\equal{}0$.

We use the same procedure as in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=276623.

We produce a sequence $ x_1 > x_2 > \ldots$ converging to $ 0$ with $ f(x_i) > \epsilon_0$

We again conclude from $ \lim_{x\to 0,x > 0}(f(x) \minus{} \alpha f^{p}(\alpha x)) \equal{} 0$, that
\[ \frac {f(x_i) \minus{} \epsilon}{\alpha} < f^{p}(\alpha x_i) < \frac {f(x_i) \plus{} \epsilon}{\alpha}.
\]

Now we send $ \alpha$ to $ 0$, such that:
$ \infty \equal{} \frac {f(x_i) \minus{} \epsilon}{\alpha} < f^{p}(\alpha x_i)$

Since $ {\alpha x_i}$ converges to $ 0$ and $ f(\alpha x_i) > 2\epsilon_0$, $ \epsilon_0$ is not $ \sup{\Phi}$. Therefore $ \Phi$ is empty and conclusion follows.

Comment: All the notations are the same as in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=276623.

We do not need to send $ \alpha$ to 0. We choose $ \alpha$ small enough, such that $ f(\alpha x_i) > 2\epsilon_0.$


a) Which number(s) less than a 1000 have the most factors?

b) What is the smallest natural number n having exactly 100 factors?