number theory
greatest common divisor
Putnam
function

A basketball player has an free throw average of less than 80% at the beginning of the season.  At the end of the season, he has an above 80% average.  Prove that at one point he must have had exactly 80% as an average.

[quote="t0rajir0u"]Generalization:  For what rational free throw averages $\frac{p}{q}$ must this hold?[/quote]

[hide="possible"] It seems like the condition is $\text{gcd}(p,q) \ge q-p$.

For this problem, suppose he never makes exactly 80%. Then there exists some point that he has tried $b$ throws and made $a$ of them such that

$\frac{a}{b}< \frac{4}{5}< \frac{a+1}{b+1}$.

But this is equivalent to $5a < 4b$ and $4b+4 < 5a+5 \Rightarrow 4b < 5a+1$. Combine those to get

$5a < 4b < 5a+1$,

an obvious contradiction.[/hide]

[quote="paladin8"][quote="t0rajir0u"]Generalization:  For what rational free throw averages $\frac{p}{q}$ must this hold?[/quote]

[hide="possible"] It seems like the condition is $\text{gcd}(p,q) \ge q-p$. [/hide][/quote]

That's right, but it's the least transparent expression for that set I've ever seen.  You might consider what exactly that means for fractions in reduced form.

The double-follow-up, of course, is what values $p$ there are such that if our basketball player starts out having made more than $p\%$ of her baskets and ends up having made less than $p\%$ of her baskets, for what values of $p$ must there have been a point when she made exactly $p\%$ of her baskets?

Also, this question is a Putnam question from (I think) 2002.

Haha, good call. I really mean any fraction in the form $\frac{n-1}{n}$.

[quote="kstan013"][hide="a start..."]Look at the number of free throws he has missed and call it x. We know that at the beginning of the season, he has made less than 4x free throws. For every one he misses, he must make four more, thus raising his total requirements to 4x+4. Notice that x is an integer, so all of the possible made/missed combinations are integers too. The only way he can "bypass" such a combination such that the number of misses is m and the number of makes is 4m, is to have one go up and one go down at the same time, and we know he can't have less misses, or to have the number of makes go up by 2 on 1 shot, which is impossible[/hide]not sure if this was too clear[/quote]
Same as above but reworded , so it  may be a little clearer ..

[hide]Let at any time:
Let s = number of free throws he was successful
Let f = number of times he missed.
Let y = s - 4f  ((y is a function of time)
Now at any throw:  y can either increase by +1 or  decrease by 4
In the beginning, y is negative, and it became positive, so it must reach zero at some point (because it can not increase  more than +1 at any point) [/hide]


real analysis
calculus
integration
function
Support
real analysis theorems

(a) Let (X,M,m) be a measure space and f belongs L1(m).Then, given e>0, there
exists g:X->S+ (simple positive measurable functions) such that
int on X |f-g|dm <e

(b)If X=R ,M=B_R(Borel),m Lebesgue measure, prove that there is a continuous function g with
compact support such that int on X |f-g|dm <e
Hint:approach using stair functions

Please show me all the technical details because this is where I have problem.
Also if this is not the right place to post this message suggest me a different one.

Thnx

Watch out, (a) is not true if $f$ is not supposed nonnegative a.e.

I'm back :D Just wanted to add: as for the second point, you can use the fact that if $f$ is integrable, then $\mu (E)=\int_{E} f$ will be a finite signed measure (where $E$ mesurable set of $R$). Also, since $H=\{x \in R, f(x) \neq 0\}$ is sigma-finite, we can write it as $H=\cup (E_{n} \cap [-n,n])$ (countable union of an increasing sequence of mesurable sets $E_{n}$, each of finite measure). But then $\mu (E_{n}) \rightarrow \mu (E)$ and there exists $N$ such that $|\mu (E-\cup_{k=1, \dots ,N} (E_{k}\cap [-k,k]))| < \epsilon$. But now on $\cup_{k=1, \dots ,n} (E_{k} \cap [-k,k])$ you can apply the first point to approximate your integral; your approximating function will have its support included in $[-N,N]$, whence it will be compact.

Sure, you can take that for granted..

Actually I want to approximate  a complex valued function f so it suffices to approximate Ref+ Ref- Imf+ Imf-

Please tell me how to approximate given f>=0 everywhere.

Classically use $f_{n}(x) = \frac{k}{2^{n}}$ whenever $\frac{k}{2^{n}} \leq f(x) < \frac{k+1}{2^{n}}$ for each $k \in [0,n-1]$, and $f_{n}(x)=n$ if $f(x) \geq n$).


Say you had u, and a function u such that $ f(u) \equal{} 2u^2 \plus{} 2u$ you want $ f(f(.....(u)$ nested k times. Can you find a formula for $ f(f(....(u))$ in terms of u and k?

Yes; it's given by $ f^k(u)$.  :P

In general, [url=http://mathworld.wolfram.com/QuadraticMap.html]quadratic maps[/url] do not have a closed form, but this one does.  Try writing the recurrence $ a_{k\plus{}1} \equal{} 2a_k^2 \plus{} 2a_k$ in a more suggestive form.

[hide="That form is..."] $ 2a_{k\plus{}1} \equal{} (2a_k \plus{} 1)^2 \minus{} 1$ [/hide]

Suppose $ f(u) \equal{} 4n^2 \plus{} 4n$, for simplicity sake.Then $ a_k \plus{} 1 \equal{} (2a_{k\minus{}1} \plus{} 1)^2$
But what would the closed form be??

"For simplicity's sake"?  That one might actually be harder.  The one you gave is very doable.

[b]mathmonster:[/b] Re-read the very end of [i]t0rajir0u[/i]'s last post. See if you can figure out what he was trying to get you to do.

[hide="Elaboration"]He noted that if you multiply by $ 2$, you can rewrite the condition as:
$ 2a_{k \plus{} 1} \equal{} (2a_k \plus{} 1)^2 \minus{} 1$

Let's add $ 1$ to both sides:
$ 2a_{k \plus{} 1} \plus{} 1 \equal{} (2a_k \plus{} 1)^2$

Define $ b_k \equal{} 2a_k \plus{} 1$
$ b_{k \plus{} 1} \equal{} b_k ^2$

Does this now look easily solvable?[/hide]
For the second recursive function you proposed, there is no linear substitution that will bring it into that form, or any useful variants of that form.

letting $ a_2 \equal{} b_k$ ( $ a_2 \equal{} 2a \plus{} 1$ not $ a_1$) then $ a_n \equal{} b_k ^{2^{n\minus{}2}}$ right??

By luck, mathemonster's recurrence also has a closed form.  Can you guys figure it out?

[b]Regarding the original problem:[/b] [i]mathmonster:[/i], I'm not exactly sure what you're talking about there in your last post, so I'm just going to take it to completion anyway.

[hide="Solution"]We want to solve the recurrence $ (2a_{k \plus{} 1}) \equal{} (2a_k \plus{} 1)^2$

To do this, we first create an auxiliary sequence $ b_k$, where each term of $ b_k$ is defined with the relationship $ b_k \equal{} 2a_k \plus{} 1$

If we write the original recurrence in terms of $ b_k$, we get $ b_{k \plus{} 1} \equal{} b_k ^2$

Solving this for $ b_k$, we get $ b_k \equal{} (b_0)^{2^k}$

Now, we simply substitute back using our definition:

$ 2 a_k \plus{} 1 \equal{} (2a_0 \plus{} 1)^{2^k}$

$ a_k \equal{} \minus{} \frac {1}{2} \plus{} \frac {1}{2} \cdot (2a_0 \plus{} 1)^{2^k}$[/hide]

[b]Regarding mathmonster's second recursion:[/b] When you figure this one out, you'll really appreciate how rare it is for tjere to be a nice solution to one of these kinds of problems.

[hide="Hint"]Double and add $ 1$ to get $ 2a_{k \plus{} 1} \plus{} 1 \equal{} 8a_k ^2 \plus{} 8a_k \plus{} 1 \equal{} 2(2a_k \plus{} 1)^2 \minus{} 1$

Substitute $ \text{c} \equal{} 2a_k \plus{} 1$ to get $ \text{c}_{k \plus{} 1} \equal{} 2\text{c}_k ^2  \minus{} 1$

Does this remind you of a well-known formula?[/hide]


trigonometry
function
calculus
derivative
inequalities
real analysis
real analysis unsolved

Prove that $\sin\sqrt{x}<\sqrt{\sin x}$ for every real $x$ such that $0<x<\frac{\pi}{2}$.

Let $f(x) = \sqrt{x}$ and $g(x) = f(\sin x) - \sin f(x)$, then on $(0,\frac{\pi}{2})$, f is increasing and f' is decreasing.
$g'(x) = \cos x f'(\sin x) - \cos f(x) f'(x)$
Since $x \le f(x)$  and $\sin x < x$ on $(0,1]$, $g'(x) > 0$ on $(0,1]$, so it's true on (0,1].

Otherwise, we have $\sin x \ge \sin \sqrt{x} \ge (\sin \sqrt{x})^2$

QED

For $1 \leq x < \frac{\pi}2$ it's ok.
For $0 < x < 1$, study the function $f(x) = \sin(x^{2})-\sin^{2}(x)$.

[quote="t_c_lhp"][quote="Soarer"] then on $(0,\frac{\pi}{2})$, f is increasing and f' is decreasing.
[/quote]

  This is not true!!![/quote]
why?

fire123, $\sqrt x$ is not less than $x$.

I've been fiddling around with the function $f(x)=\sin^2\sqrt{x},$ which is a fairly entertaining function. It's smooth. We can write its derivative as $f'(x)=\frac{\sin(2\sqrt{x})}{2\sqrt{x}}.$ We can expand out its power series:
\[f(x)=x-\frac{x^2}3+\cdots=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}2^{2n-1}x^n}{(2n)!}\]
What I haven't done yet is to turn that into a short snappy proof that $f(x)<\sin x$ on $(0,\pi/2],$ which is what we want. Although examining the first two terms of the power series does make it clear that that inequality holds on some interval $(0,\delta).$

  From $0<x<\frac {\pi}{2}$,We know that $\sin x<\sqrt {\sin x}$,and for $\sqrt {x}\le x$,which equals $1\le x\le \frac {\pi}{2}$.The inequality is obviously true.
  If $0<x<1$,we need to prove $(\sin x)^2<\sin {x^2},x\in (0,1)$.
  Which equals to $2\sin x\cos x<2x\cos {x^2}$,this is easy

I'm posting this just because I can't believe it is still unsolved here.
Using half-angle identity, we can see that inequality is equivalent to:
\[2\sin(x)+\cos(2\sqrt{x}) \ge 1 \]
Now, it is easy to see that for $t\ge 0$ following two inequalities hold:
\[ \sin(t) \ge t-\frac{t^3}{6} \quad \text{and} \quad \cos(t)\ge 1-\frac{t^2}{2}+\frac{t^4}{4!}-\frac{t^6}{6!} \]
Therefore, for $x\ge 0$ we have
\[ 2\sin(x)+\cos(2\sqrt{x}) \ge 1+\left(\frac{2}{3}-\frac{19}{45}x \right)x^2 \]
Since $\frac{\pi}{2}<\frac{30}{19}$, the problem is solved. $\blacksquare$

Uhhh

$sin^2 x<sin x$
$sinx(sin x–1)<0$
$0<sinx<1$ true for $0<x<\frac{\pi}{2}$


Πάλι στο περιοδικό GM και για τις τάξεις 7,8 ( Α΄, Β΄Γυμνασίου) προτείνεται η παρακάτω άσκηση.Βέβαια , πρόκειται για επιλεγμένες ασκήσεις που αφορούν στην προετοιμασία για το διαγωνισμό της GM, που γίνεται σε ξεχωριστή μέρα(διαφορετική από αυτή των φάσεων της ολυμπιάδας).

[color=red][b]ΑΣΚΗΣΗ[/b][/color]

Εκατέρωθεν της πλευράς [b]ΒΓ [/b]τριγώνου [b]ΑΒΓ[/b] παίρνουμε τα ίσα τμήματα [b]ΒΔ[/b] και [b]ΓΕ[/b]  (χωρίς κοινά σημεία).Πάνω στις πλευρές [b]ΑΒ, ΑΓ [/b]παίρνουμε τα σημεία [b]Ζ,Η[/b] αντίστοιχα , ώστε [b]ΖΗ // ΒΓ[/b]. Αν οι ευθείες [b]ΔΖ , ΕΗ [/b]τέμνουν  τις [b]ΑΓ,ΑΒ [/b]στα σημεία [b]Κ,Λ[/b] να αποδείξετε ότι [b]ΚΛ// ΖΗ[/b].

[i] Ελπίζω να βρεθεί και κάποια άλλη λύση , εκτός από αυτή που έχω(του περιοδικού)! [/i]

Να μια σχετικά σύντομη λύση . Από το θεώρημα Μενελάου έχουμε

$ \frac{KA}{KC}\cdot\frac{DC}{DB}\cdot\frac{ZB}{ZA}\equal{}1$

kai $ \frac{LA}{LB}\cdot\frac{EB}{EC}\cdot\frac{HC}{HA}\equal{}1$

Όμως $ \frac{ZB}{ZA}\equal{}\frac{HC}{HA}$ και $ \frac{DC}{DB}\equal{}\frac{EB}{EC}$ άρα

$ \frac{LA}{LB}\equal{}\frac{KA}{KC}$   qed  :D

[quote="silouan"]Να μια σχετικά σύντομη λύση . Από το θεώρημα Μενελάου έχουμε

$ \frac {KA}{KC}\cdot\frac {DC}{DB}\cdot\frac {ZB}{ZA} \equal{} 1$

kai $ \frac {LA}{LB}\cdot\frac {EB}{EC}\cdot\frac {HC}{HA} \equal{} 1$

Όμως $ \frac {ZB}{ZA} \equal{} \frac {HC}{HA}$ και $ \frac {DC}{DB} \equal{} \frac {EB}{EC}$ άρα

$ \frac {LA}{LB} \equal{} \frac {KA}{KC}$   qed  :D[/quote]

Ακριβώς !!!

 Με αυτό το σκεπτικό την πρότεινα :) . Περίμενα αυτή τη λύση από σας τους senior. Ας περιμένουμε μήπως κάποιος άλλος πέσει σε κάτι άλλο , επίσης απλό!

[color=red][b]    Καλά αποτελέσματα και στον Ευκλείδη ! [/b][/color]
 Μπάμπης

[quote="stergiu"][color=red][b]ΑΣΚΗΣΗ[/b][/color]
Εκατέρωθεν της πλευράς [b]ΒΓ [/b]τριγώνου [b]ΑΒΓ[/b] παίρνουμε τα ίσα τμήματα [b]ΒΔ[/b] και [b]ΓΕ[/b]  (χωρίς κοινά σημεία).Πάνω στις πλευρές [b]ΑΒ, ΑΓ [/b]παίρνουμε τα σημεία [b]Ζ,Η[/b] αντίστοιχα , ώστε [b]ΖΗ // ΒΓ[/b]. Αν οι ευθείες [b]ΔΖ , ΕΗ [/b]τέμνουν  τις [b]ΑΓ,ΑΒ [/b]στα σημεία [b]Κ,Λ[/b] να αποδείξετε ότι [b]ΚΛ// ΖΗ[/b].[/quote]
Μία απλή λύση μόνο με το [b][size=100]θεώρημα του Θαλή[/size][/b].

Έστω $ K\equiv A\Gamma\cap \Delta Z$ και $ \Lambda \in AB,$ έτσι ώστε $ K\Lambda\parallel ZH\parallel B\Gamma$ και αρκεί να αποδείξουμε ότι τα σημεία $ \Lambda,$ $ H,$ $ E,$ είναι συνευθειακά.

Αλλά, $ \frac {K\Lambda}{\Gamma E} \equal{} \frac {K\Lambda}{B\Delta} \equal{} \frac {KZ}{Z\Delta} \equal{} \frac {KH}{H\Gamma}$ και η πρόταση έχει αποδειχτεί.

Κώστας Βήττας.


function
induction
number theory proposed
number theory

Let $ \mathbb{Z}$ be the set of all integers. Define the set $ \mathbb{H}$ as follows:
(1). $ \dfrac{1}{2} \in \mathbb{H}$,
(2). if $ x \in \mathbb{H}$, then $ \dfrac{1}{1\plus{}x} \in \mathbb{H}$ and also $ \dfrac{x}{1\plus{}x} \in \mathbb{H}$.
Prove that there exists a bijective function $ f: \mathbb{Z} \rightarrow \mathbb{H}$.

[quote="pco"][quote="Raja Oktovin"]Let $ \mathbb{Z}$ be the set of all integers. Define the set $ \mathbb{H}$ as follows:
(1). $ \dfrac{1}{2} \in \mathbb{H}$,
(2). if $ x \in \mathbb{H}$, then $ \dfrac{1}{1 \plus{} x} \in \mathbb{H}$ and also $ \dfrac{x}{1 \plus{} x} \in \mathbb{H}$.
Prove that there exists a bijective function $ f: \mathbb{Z} \rightarrow \mathbb{H}$.[/quote]

There likely miss some elements in your problem since $ \mathbb H \equal{} \mathbb R^ \plus{}$ clearly respects the two conditions $ (1)$ and $ (2)$ although there is no bijection from $ \mathbb Z\to\mathbb R^ \plus{}$

If you mean that $ \mathbb H$ is the minimum set generated by successive applications of rule $ (2)$ on $ \{\frac 12\}$, then the result is obvious since all elements of $ \mathbb H$ are then positive rational numbers, and so $ \mathbb H\subseteq\mathbb Q$ is countable, hence the result.[/quote]
$ \mathbb{H}\in(0,1)$ and $ \mathbb{H}\in\mathbb{Q}$ is it?

thus by setting $ x\equal{}\frac{m}{n}$ with $ m<n$ and $ m,n\in\mathbb{N}$ we can change the form in (2) become

if  $ \frac{m}{n} \in \mathbb{H}$, then $ \dfrac{m}{m\plus{}n} \in \mathbb{H}$ and also $ \dfrac{n}{m\plus{}n} \in \mathbb{H}$.

If $ m\equal{}1$ then since $ \frac12\in\mathbb{H}$  then all numbers $ \frac{1}{x}$ are in $ \mathbb{H}$

If $ m\equal{}2$ then since we can get $ \frac23\in\mathbb{H}$ from $ \frac{n}{m\plus{}n}\equal{}\frac{2}{1\plus{}2}\equal{}\frac23$ ($ (m,n)\equal{}(1,2)$) then all numbers $ \frac{2}{x}$ that $ x\equiv_1 2$ are in $ \mathbb{H}$

If $ m\equal{}3$ we divide into 2 cases
-we can get $ \frac34\in\mathbb{H}$ from $ \frac{n}{m\plus{}n}\equal{}\frac{3}{1\plus{}3}\equal{}\frac34$ ($ (m,n)\equal{}(1,3)$) then all numbers $ \frac{3}{x}$ that $ x\equiv_1 3$ are in $ \mathbb{H}$
-we can get $ \frac35\in\mathbb{H}$ from $ \frac{n}{m\plus{}n}\equal{}\frac{3}{2\plus{}3}\equal{}\frac35$ ($ (m,n)\equal{}(2,3)$) then all numbers $ \frac{3}{x}$ that $ x\equiv_2 3$ are in $ \mathbb{H}$

And so on

so we conclude that $ \mathbb{H}\in\mathbb{Q}$ and $ \mathbb{H}\in(0,1)$

[quote="pco"]rational numbers, and so $ \mathbb H\subseteq\mathbb Q$ is countable, hence the result.[/quote]

I believe that this statement should be well-known, but can you write the details here? Thank you.

[quote="pco"][quote="Raja Oktovin"]Let $ \mathbb{Z}$ be the set of all integers. Define the set $ \mathbb{H}$ as follows:
(1). $ \dfrac{1}{2} \in \mathbb{H}$,
(2). if $ x \in \mathbb{H}$, then $ \dfrac{1}{1 \plus{} x} \in \mathbb{H}$ and also $ \dfrac{x}{1 \plus{} x} \in \mathbb{H}$.
Prove that there exists a bijective function $ f: \mathbb{Z} \rightarrow \mathbb{H}$.[/quote]

There likely miss some elements in your problem since $ \mathbb H \equal{} \mathbb R^ \plus{}$ clearly respects the two conditions $ (1)$ and $ (2)$ although there is no bijection from $ \mathbb Z\to\mathbb R^ \plus{}$

If you mean that $ \mathbb H$ is the minimum set generated by successive applications of rule $ (2)$ on $ \{\frac 12\}$, then the result is obvious since all elements of $ \mathbb H$ are then positive rational numbers, and so $ \mathbb H\subseteq\mathbb Q$ is countable, hence the result.[/quote]
I can't understand your idea  :blush:
I think that two condition of problem only to prove $ H \equal{} (0,1)\cap Q$ and we need prove that:there exist  a bijective function $ f: \mathbb{Z}\rightarrow\mathbb{H}$

@math10 :
My idea is just to show that $ \mathbb H$ is infinite (obviously) and countable (since $ \subseteq Q$). Hence the result.

@Raja Oktovin :
I think this is a basic definition of countable : an infinite set A is countable $ \iff$ $ \exists$ a bijection between A and $ \mathbb N$
And so $ \exists$ bijections between any two infinite countable sets, so between $ \mathbb Z$ and $ \mathbb H$.

[quote="pco"][quote="math10"][quote="pco"]@math10 :
My idea is just to show that $ \mathbb H$ is infinite (obviously) and countable (since $ \subseteq Q$). Hence the result.
[/quote]
We only have $ H$ is infinite and $ H\in Q$. How do you  to get the results? (sorry for my stupid  :blush: )
please write clear[/quote]
I'm sorry : I dont understand what you dont understand : isn't clear that "every infinite subset of a countable subset is countable" ?

Is this what you want I explain more ?[/quote]
I can't understand step:
*)If $ H$ is infinite and $ H\in Q$ then there exist a bijection from $ H$ to $ Z$

[quote="math10"][/quote][quote="pco"][quote="math10"][quote="pco"]@math10 :
My idea is just to show that $ \mathbb H$ is infinite (obviously) and countable (since $ \subseteq Q$). Hence the result.
[/quote]
We only have $ H$ is infinite and $ H\in Q$. How do you  to get the results? (sorry for my stupid  :blush: )
please write clear[/quote]
I'm sorry : I dont understand what you dont understand : isn't clear that "every infinite subset of a countable subset is countable" ?

Is this what you want I explain more ?[/quote][quote="math10"]
I can't understand step:
*)If $ H$ is infinite and $ H\in Q$ then there exist a bijection from $ H$ to $ Z$[/quote]

Step 1 : every infinite subset of a countable set is infinite countable.
Is step 1 OK ?

Step 2 : $ \mathbb H$ is an infinite subset of the countable set $ \mathbb Q$, and so  $ \mathbb H$ is infinite countable.
Is step 2 OK ?

Step 3 : If set A is infinite countable, there exists a bijection from A$ \to\mathbb N$
Is step 3 OK ?

Step 4 : if there exists a bijection from $ A\to C$ and a bijection from $ B\to C$, there exists a bijection from $ A\to B$
Is step 4 OK ?

Step 5 :
$ \mathbb Z$ is infinite countable, so $ \exists$ a bijection from $ \mathbb Z\to\mathbb N$ (step 3)
$ \mathbb H$ is infinite countable, so $ \exists$ a bijection from $ \mathbb H\to\mathbb N$ (step 3)
So $ \exists$ a bijection from $ \mathbb H\to\mathbb Z$
OK ?

[quote="mavropnevma"]That was a perfect piece of spam. matrix41 ! based on reading too quickly the previous post(s), then adding some completely inconsequential remarks.

pco pointed out some set(s) that fulfill both requirements, then guessed that the problem refers to the [color=red]minimal[/color] such set. What you did was some wishy-washy attempt by incomplete induction to prove $ \mathbb{H} \subseteq \mathbb{Q} \cap (0,1)$.

But this is obvious, by the fact the starting seed is $ 1/2$, while functions $ f(x) \equal{} 1/(1 \plus{} x)$ and $ g(x) \equal{} x/(1 \plus{} x)$ have both the property that they take rational values at rational points, while they move the $ (0,1)$ interval to $ (1/2, 1)$, respectively $ (0, 1/2)$.

Also, don't use $ \in$ for inclusion of sets, and fight the urge to quote entirely the post you refer to, especially if it is the one just above  :)[/quote]

Hahaha... I'm sorry my friend mavropnevma....  :)

I just want to show that $ \mathbb{H}$ is infinite countable set.

But maybe I did it in wrong way , can you please tell me what makes it wrong please, I'm just a newbie in mathematics Olympiad .

Thank you

@pco

Yeal , all OK  :P

[quote="pco"][/quote][quote="math10"][/quote][quote="pco"][quote="math10"][quote="pco"]@math10 :
My idea is just to show that $ \mathbb H$ is infinite (obviously) and countable (since $ \subseteq Q$). Hence the result.
[/quote]
We only have $ H$ is infinite and $ H\in Q$. How do you  to get the results? (sorry for my stupid  :blush: )
please write clear[/quote]
I'm sorry : I dont understand what you dont understand : isn't clear that "every infinite subset of a countable subset is countable" ?

Is this what you want I explain more ?[/quote][quote="math10"]
I can't understand step:
*)If $ H$ is infinite and $ H\in Q$ then there exist a bijection from $ H$ to $ Z$[/quote][quote="pco"]

Step 1 : every infinite subset of a countable set is infinite countable.
Is step 1 OK ?

Step 2 : $ \mathbb H$ is an infinite subset of the countable set $ \mathbb Q$, and so  $ \mathbb H$ is infinite countable.
Is step 2 OK ?

Step 3 : If set A is infinite countable, there exists a bijection from A$ \to\mathbb N$
Is step 3 OK ?

Step 4 : if there exists a bijection from $ A\to C$ and a bijection from $ B\to C$, there exists a bijection from $ A\to B$
Is step 4 OK ?

Step 5 :
$ \mathbb Z$ is infinite countable, so $ \exists$ a bijection from $ \mathbb Z\to\mathbb N$ (step 3)
$ \mathbb H$ is infinite countable, so $ \exists$ a bijection from $ \mathbb H\to\mathbb N$ (step 3)
So $ \exists$ a bijection from $ \mathbb H\to\mathbb Z$
OK ?[/quote]
For me,step 3 not clear  :blush: .

[quote="math10"][quote="pco"] Step 3 : If set A is infinite countable, there exists a bijection from A$ \to\mathbb N$
Is step 3 OK ?
[/quote]
For me,step 3 not clear  :blush: .[/quote]

I think it's the definition of infinite countable. See http://en.wikipedia.org/wiki/Countable_set for example

[quote="littletush"]since $H$ is a subset of $Q$,it's countable.
the solution is trivial......[/quote]
Ohhhh Yessssssss!
Quite new very interesting idea which solve this problem in a very nice way !
How was it possible that we miss this quite nice proof !

Thanks a lot, littletush, for your so interesting posts always adding brand new proofs :lol:


Let $\ ABC$ is a triangle. The excircle corresponding to $\ C$ touches the sides $\ BC, CA, AB$ at $\ M, N, P$ respectively. The excircle corresponding to $\ B$ touches the sides $\ BC, CA, AB$ at $\ Q, R, S$ respectively. Let $\ X=MN\cap QS, Y=MP\cap QR$. Prove that $\ A, X, Y$ are collinear.

[color=darkblue][b]Answer :[/b] the points $X$ and $Y$ belong to the $A$- altitude in the triangle $ABC\ .$

Sorry, I"ll present the my proof [u]to-morrow[/u] because here is 12:50 AM.[/color]

Let D be foot of the A-altitude of $\triangle ABC.$ Since $I_{b}Q \parallel AD \parallel I_{c}M$ (all perpendicular to BC), $\frac{DQ}{DM}= \frac{AI_{b}}{AI_{c}}.$ MP is parallel and QS perpendicular to the bisector of $\angle B,$ hence $MP \perp QS.$ Similarly, QR is parallel and MN perpendicular to the bisector of $\angle C,$ hence $QR \perp MN.$ Thus MP, QR are the M-, Q-altitudes of $\triangle XQM$ intersecting at its orthocenter Y. XY is the remaining X-altitude, $XY \perp MQ \equiv BC.$ Let D' be the foot of this altitude. Triangles $\triangle XQM \sim \triangle I_{a}I_{b}I_{c}$ are similar, having equal angles, and $I_{a}A$ is its altitude from $I_{a}.$ For example,

$\angle I_{a}I_{b}I_{c}\equiv \angle CI_{b}A = \frac{\angle C+\angle A}{2}= 90^\circ-\frac{\angle B}{2}= \angle SQB \equiv \angle XQM$

$\angle I_{a}I_{c}I_{b}\equiv \angle BI_{c}A = \frac{\angle B+\angle A}{2}= 90^\circ-\frac{\angle C}{2}= \angle NMC \equiv \angle XMQ$

Therefore, $\frac{D'Q}{D'M}= \frac{AI_{b}}{AI_{c}}= \frac{DQ}{DM},$ which means that the points $D' \equiv D$ are identical and the normals $AD,\ XD \equiv XD' \equiv XY$ to BC are also identical.

[color=darkblue][b]A metrical proof.[/b] Observe that $BM=BP=CQ=CR=p-a$ , $AN=AP=p-b$ , $AR=AS=p-c$ , $CM=CN=BQ=BS=p\ .$

$1.\blacktriangleright$ Denote the projection $D$ of the point $A$ to the line $BC\ .$ Thus, $AD\parallel MI_{c}\parallel QI_{b}$ $\Longrightarrow$ $\frac{DM}{DQ}=\frac{AI_{c}}{AI_{b}}=\frac{r_{c}}{r_{b}}=\frac{p-b}{p-c}$ $\Longrightarrow$ $\boxed{\ \frac{DM}{DQ}=\frac{p-b}{p-c}\ }\ \ (*)\ .$

$2.\blacktriangleright$ Denote $Y_{1}\in MP\cap AD$ , $Y_{2}\in QR\cap AD$ and apply the Menelaus' theorem to the transversals $\overline{MPY_{1}}$ , $\overline{QRY_{2}}$ for the triangles $ABD$ , $ACD$ respectively :

$\{\begin{array}{c}\frac{MB}{MD}\cdot\frac{Y_{1}D}{Y_{1}A}\cdot\frac{PA}{PB}=1\Longrightarrow \frac{Y_{1}D}{Y_{1}A}=\frac{MD}{p-b}\\\\ \frac{QC}{QD}\cdot\frac{Y_{2}D}{Y_{2}A}\cdot\frac{RA}{RC}=1\Longrightarrow \frac{Y_{2}D}{Y_{2}A}=\frac{QD}{p-c}\end{array}$ and $(*)$ $\Longrightarrow$ $\frac{Y_{1}D}{Y_{1}A}=\frac{Y_{2}D}{Y_{2}A}$ $\Longrightarrow$ $Y_{1}\equiv Y_{2}\equiv \boxed{\ Y\in MP\cap QR\cap AD\ }\ .$

$3.\blacktriangleright$ Denote $X_{1}\in MN\cap AD$ , $X_{2}\in QS\cap AD$ and apply the Menelaus" theorem to the transversals $\overline{MNX_{1}}$ , $\overline{QSX_{2}}$ for the triangles $ACD$ , $ABD$ respectively :

$\{\begin{array}{c}\frac{MD}{MC}\cdot\frac{NC}{NA}\cdot\frac{X_{1}A}{X_{1}D}=1\Longrightarrow \frac{X_{1}A}{X_{1}D}=\frac{p-b}{MD}\\\\ \frac{QD}{QB}\cdot\frac{SB}{SA}\cdot\frac{X_{2}A}{X_{2}D}=1\Longrightarrow \frac{X_{2}A}{X_{2}D}=\frac{p-c}{QD}\end{array}$ and $(*)$ $\Longrightarrow \frac{X_{1}A}{X_{1}D}=\frac{X_{2}A}{X_{2}D}$ $\Longrightarrow$ $X_{1}\equiv X_{2}\equiv \boxed{\ X\in MN\cap QS\cap AD\ }\ .$[/color]


AMC
AIME
USA(J)MO
USAMO
AMC 10
probability
number theory

[quote="1337h4x"]Am I allowed to write a calculator program during the test and use it to solve a problem?[/quote]
Well, they can't stop you from clicking and clacking with your calc.

still, it might be better to write some programs forehand, befor the test, such a program that finds sum of arithmatic or geometric series. those kind of problems are 99.9% guaranteed to show up in some form or another.

[quote="t0rajir0u"][quote="sapphyre571"]well, for those "find all values" or "find the sum of the values" of whatever, someone very good with a calculator could whip out a program to solve that problem I guess[/quote]

Still don't see how this would be faster than, say, making a table (with the calculator).[/quote]

or just solving the problem with pen and paper...  :D

From the AMC 12 cover page, item 5:
"No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor, erasers and calculators that are accepted for use on the SAT.  No problems on the test will [i]require[/i] the use of a calculator."

All problems are specifically constructed so that a student with a good understanding of the high school curriculum in mathematics can do the problems without lengthy or tedious calculation.  See the problem soltutions on past contests for examples.

Steve Dunbar
AMC Director

I realize that students with a good understanding of the material can solve it without a calculator. My problem is that I don't always have a very good understanding :D, so it makes sense to just write a proram.

my understanding is, you can do it, but it is possible to do without

-jorian

Go ahead and do it, but it won't help you on the AIME or USAMO, which are more important competitions.

[quote="math92"]Yesterday on the AMC 10 I made a calculator program to solve number 23.    :D[/quote]

lol I got that right by evaluating a few values and some messy algebra

[quote="t0rajir0u"]If you really think you can do it quickly enough... honestly, I can't think of a problem where writing a calculator program [b]during[/b] the test would give you a speed advantage over solving any particular problem, especially since the test is designed to be doable without a calculator.[/quote]

#25 on the AMC10 took me approximately 45 seconds to write a program to solve it.

[quote="tjhance"][quote="1337h4x"]That's what I was thinking. A loop with an if-then statement wouldn't take but a minute or two to write, so it's definitely practicle (especially if you're as bad at NT as me  :wink: ).[/quote]

What do you mean by "NT"?  :blush:[/quote]

NT=Number Theory, I think...

Are you allowed to write a program?

Well honestly, even if you're not allowed to, how would they find out if you wrote one in the middle of the test?

NT=Number Theory.

I ended up not using programs because the problems I had trouble with weren't probability or enumeration.


Determine the sum of all the divisors $ d$ of $ 19^{88} \minus{} 1$ which are of the form $ d \equal{} 2^a3^b$, $ a$, $ b$ >0

The power of 2 and 3 in the standard factorization of $ 19^88 \minus{} 1$ is 5 and 2, respectively. Hence, the sum is:
$ (2^1 \plus{} 2^2 \plus{} ... \plus{} 2^5) \times (3 \plus{} 3^2) \equal{} 744$


Duke college MIT Putnam geometry Harvard Princeton

How strong is Duke Math Department compared to Harvard's, Princeton's, MIT's, Caltech's and Chicago's? And, what is your top five best math departments in the US?

I don't believe Duke has a decent math department (as compared to MIT, CalTech, or Princeton). They have good sports teams yes, but I don't believe math is their forte.

(Could be wrong though).

Duke grabs up some IMO team members sometimes as math majors and talks up its Putnam team quite a bit in regional information sessions. I don't have a strong sense of how its first-year courses for the best prepared math majors compare to the courses at the colleges that are most often mentioned as favorites for strongly prepared math students.

These kind of questions are not only very subjective, but also misleading in suggesting that there even exists an ordering which is relevant across the wide spectrum of different types of students.

Different Universities and different math departments provide different "quality of education" to different students. One school might be a great school for person X, but a terrible school for person Y. (Not to mention that a large part of what you learn at a University is more dependent on you than on the school).

If you are going to evaluate the quality of a math department (or University), I think the best way to judge is to ask questions which are more personal to you. For example: Find out if there have been people similar to yourself - similar abilities, similar interests, similar personality, etc - who have been happy at whatever school you're looking at, and through their education at the school been able to go on to do things that you consider admirable.

There have been several students that have gone through the Duke Math department who have gone on to do outstanding things. Whether or not you could do the same is not something you can determine from other people's ad hoc rankings of schools.

@ gauss202, I understand your concern but my question is not misleading since it asked [b]YOUR[/b] top five Math Department and not [b]THE[/b] top five Math Department. Besides, I am not that naive about school selection as you might think I am. By saying strong, I mean how good are the professors there in terms of research, making themselves accessible to students, having expertise in their area and other important criteria.

That IMO/Putnam bit is based on individual recruiting. Back in 2000, they sent me a recruiting letter based on Melanie Wood's recommendation. I ultimately chose UC Berkeley, but my impression is that Duke is a fine place for top math students.

It's probably not very controversial to say that as far as faculty research reputation goes, Harvard, Princeton, MIT, and Chicago are higher up than Caltech and Duke.  However, for an undergrad education, it's not clear how important that is.

Harvard, Princeton, and MIT have the most high-powered math students, and they all offer very demanding courses (with Harvard offering the famed Math 55 course for the top first-years).  I know less about Chicago, but it's known for having a very large and inclusive math major (so that math isn't just for geniuses) and also a rigorous honors sequence.  I know next to nothing about Caltech, except that it's tiny and the math skills of the students are probably more uniformly good than any other school.

But your question is mainly about Duke, which I know quite a bit about.  Duke does very well on the Putnam partly because Duke recruits math competition stars.  They attract the best ones (e.g. Melanie Wood) by offering large academic scholarships.  There is prof there named David Kraines who does a lot of the recruiting and encourages math competition at Duke, including running a problem solving seminar for credit.  He also encourages participation in the Math Contest in Modeling, which Duke does very well in.  Because of this, if you really care about continuing competitions, then Duke is a very nurturing environment.  Also, if your college application is weak on non-math stuff, then I think Duke is more willing to give you a chance than, say, Harvard.

The top math students at Duke form a smaller, tighter community than at those other big schools, but in return they get more attention from profs (compared to, say, Harvard).  In particular, they have a PRUV fellows program which provides funds for the top students to do research with faculty over the summer and continue it during the school year.  Duke is lacking in a hardcore first-year honors sequence, but the analysis and algebra courses are tracked, so that top students take the harder courses.  The grad program at Duke is much weaker than at those other schools, so the grad classes, while still challenging, are not at the same level as at those other schools.

Teaching quality is always a tough thing to gauge.  I will say this, though:  Duke at least *considers* teaching ability (way way behind research ability) in hiring tenured or tenure-track faculty.  I'm not sure the same could be said for Harvard, MIT, and Princeton, where every faculty member is a research superstar.  At Harvard, a lot of classes for math majors are taught by postdocs, some of which are great teachers, but again, places like Harvard do not care about teaching when hiring postdocs.

Overall, where you want to go all depends on what's important to you.  You asked, "How good are the professors there in terms of research, making themselves accessible to students, having expertise in their area and other important criteria?"  It sounds as if you emphasize research quality more than most prospective college students, so the US News Graduate School Ranking should be a good stand-in for what you are looking for.  Faculty accessibility does vary some among top departments, but it's really hard to get anything more than anecdotal information about it, so you should probably mostly ignore it.  I can vouch that Duke faculty are pretty accessible.


logarithms
calculus
calculus computations

can you sum these 4 series ? :D

$ \textbf{(i)}\;\;1 \plus{} \frac {1}{1\cdot 3} \plus{} \frac {1}{1\cdot 3\cdot 5} \plus{} \frac {1}{1\cdot 3\cdot 5\cdot 7} \plus{} \frac {1}{1\cdot 3\cdot 5\cdot 7\cdot 9} \plus{} \frac {1}{1\cdot 3\cdot 5\cdot 7\cdot 9\cdot 11} \plus{} \;\ldots\;\;\;\ldots$

$ \textbf{(ii)}\;\;1 \minus{} \frac {1}{2} \minus{} \frac {1}{2^2} \minus{} \frac {1}{2^3} \minus{} \frac {1}{2^4} \minus{} \frac {1}{2^5} \plus{} \frac {1}{2^6} \plus{} \frac {1}{2^7} \plus{} \frac {1}{2^8} \plus{} \frac {1}{2^9} \plus{} \frac {1}{2^{10}} \minus{} \, \minus{} \, \minus{} \, \minus{} \, \minus{} \;\ldots\; \plus{} \, \plus{} \, \plus{} \, \plus{} \, \plus{} \;\ldots$

$ \textbf{(iii)}\;\;1 \minus{} \frac {1}{4} \plus{} \frac {1}{5} \minus{} \frac {1}{6} \plus{} \frac {1}{9} \minus{} \frac {1}{10} \plus{} \frac {1}{11} \minus{} \frac {1}{14} \plus{} \frac {1}{15} \minus{} \frac {1}{16} \plus{} \frac {1}{19} \minus{} \frac {1}{20} \plus{} \frac {1}{21} \minus{} \frac {1}{24} \plus{} \;\ldots\; \minus{} \; \plus{} \;\ldots$

$ \textbf{(iv)}\;\;1 \minus{} \frac {1}{2^2} \minus{} \frac {1}{4^2} \minus{} \frac {1}{5^2} \plus{} \frac {1}{7^2} \plus{} \frac {1}{8^2} \plus{} \frac {1}{10^2} \minus{} \frac {1}{11^2} \minus{} \frac {1}{13^2} \minus{} \frac {1}{14^2} \plus{} \frac {1}{16^2} \plus{} \frac {1}{17^2} \plus{} \frac {1}{19^2} \minus{} \, \minus{} \, \minus{} \;\ldots\; \plus{} \, \plus{} \, \plus{} \;\ldots$

[ $ \textbf{FYI}$ --- just in case it is not obvious, in $ \textbf{(iii)}$, the reciprocals of $ 2,\,3,\,7,\,8,\,12,\,13,\,17,\,18,\,22,\,23,\,27,\,28,\;\ldots\;\;\;\text{e.t.c.}$ have all been excluded, and in $ \textbf{(iv)}$, the reciprocals of the squares of multiples of $ 3$, i.e. $ 3,\,6,\,9,\,12,\,15,\,18,\,21,\,24,\;\ldots\;\;\;\text{e.t.c.}$ have been excluded. also, in $ \textbf{(iii)}$, as is probably clear already, every other term is alternating in sign, whereas  in $ \textbf{(iv)}$, every other group (of 3 [i]consecutive [/i]terms) is alternating in sign. ]

for the second

$ \sum _{i \equal{} 0}^{\infty } \frac {( \minus{} 1)^i}{\alpha ^i} \equal{} \frac {\alpha }{\alpha \plus{} 1}$

$ P \equal{} \sum _{i \equal{} 0}^{\infty } \sum _{a \equal{} 1}^5 \frac {( \minus{} 1)^{i \plus{} 1}}{2^{a \plus{} 5 i}} \plus{} 1 \equal{} 1 \minus{} \frac {31}{32} \sum _{i \equal{} 0}^{\infty } \frac {( \minus{} 1)^i}{2^{5 i}} \equal{} \frac {2}{33}$

for the third

$ \sum _{i\equal{}1}^{\infty } \sum _{a\equal{}0}^2 \frac{(\minus{}1)^{a\plus{}i}}{a\plus{}5 i\minus{}1}\plus{}1$

and after some troublesome calculations
$ \frac{1}{10} \left(\left(6\minus{}\sqrt{5}\right) \log (2)\plus{}\sqrt{5} \log \left(7\plus{}3
   \sqrt{5}\right)\right)$
but i think this result is not beutiful enougth to have a proof


geometry
geometric transformation
homothety
ratio
geometry proposed

Let O be the center of circle lying on a plane.
AB is a fix chord. If point C varies on a circle find the geometric
locus of gravity centers of the triangles ABC.
What about locus of the nine-point circles?

Considering a homothety center at $M$ (the midpoint of $AB$) and ratio $\frac 13$, we can see that the locus of the centroid $G$ is the image of the given circle under the homothety, i.e. another circle with $\frac 13$ the radius of the given circle.

Since the nine-point center $N$ is positioned s.t. $\overrightarrow{ON}=\frac 32\overrightarrow{OG}$, the locus of $N$ must also be a circle, the image of the locus of $G$ onder the homothety centered at $O$ and having ratio $\frac 32$.


Let $x$ and $y$ be $\in \mathbb{R}^+$ with harmonic mean $h = \frac{2}{\frac{1}{x} + \frac{1}{y}} = 4$ and the quadratic mean $q = \sqrt{\frac{x^2 + y^2}{2}} = 24.$ Determine the arithmetic mean $a = \frac{x +y}{2}$.

[hide="Solution"]

$h=\frac{2}{\frac 1x+\frac 1y}=\frac{2xy}{x+y}=4$

So then $2xy=4(x+y)$

$q=\sqrt{\frac{x^2+y^2}{2}}=24\Rightarrow x^2+y^2=1152$

So then we can get: $x^2+2xy+y^2=1152+4(x+y)$ or $(x+y)^2=1152+4(x+y)$  Then letting $S=x+y$ we get a quadratic:
\[S^2-4S-1152=(S-36)(S+32)=0\]

But the sum of two positive numbers is positive.  Thus $x+y=36$ and the answer is $18$.[/hide]


function
IMO
functional equation
algebra
IMO 1982

The function $f(n)$ is defined on the positive integers and takes non-negative integer values. $f(2)=0,f(3)>0,f(9999)=3333$ and for all $m,n:$ \[ f(m+n)-f(m)-f(n)=0 \text{ or } 1. \] Determine $f(1982)$.

Actually I think that we can say specifically that f(1982)=660.

By letting $m=n$ we have that $f(2m) \geq 2f(m)$.

Suppose that $n<800$ (where 800 is rather arbitary, with $4 \times 800 <3333$). Also suppose for contradiction that $f(3n+2)=n+1$.
Then $f(6n+4) \geq 2n+2$ and $f(12n+8) \geq 4n+4$.
However $f(12n+8) \leq f(12n+9) = f(3(4n+3)) = 4n+3$ (as $4n+3 < 3333$).
Contradiction, as $4n+4 > 4n+3$.

Hence for all $n<800$, $f(3n+2) \neq n+1$ and hence it equals $n$. Specifically, this is true for $n=660$ and $f(1982)=660$.

From Kalimdor's post-

Letting $m=3k$ and $n=3$ we have $f(3(k+1)) \geq f(3k)+1$. As $f(3)=1$ and $f(9999)=3333$, we must have $f(3k)=k$ for all $1 \leq k \leq 3333$.

I still don't understand... just because it works for two values, how does it work for the rest?

Suppose (for contradiction) that for some $m$, $1<m<3333$, $f(3m)>m$. Then $f(3(m+1))>m+1$ (by my previous post) and by induction $f(3(m+r))>m+r$. But for some $r$ this gives $f(3 \times 3333)>3333$, which is false.

Now suppose that for some such $m$, $f(3m)<m$. By my previous post, $f(3(m-1))<m-1$, and by induction $f(3(m-r))<m-r$. But for $r=m-1$ this gives $f(3)<1$, which is also false.

Hence $f(3m)=m$.

Clearly $f(1) \ge 1 \Rightarrow f(m+1) \ge f(m)+f(1) \ge f(m)+1$ so $f(9999) \ge 9999$.Contradiction!So $f(1)=0$.This forces $f(3)=1$.Hence $f(3k+3) \ge f(3k)+f(3)>f(3k)$ so the inequality $f(3)<f(6)<\cdots<f(9999)=3333$ forces $f(3k)=k \forall k \le 3333$.Now $f(3k+2) \ge k+1 \Rightarrow f(6k+4) \ge 2k+2 \Rightarrow f(12k+8) \ge 4k+4 \le f(12k+9)=4k+3$(Note:This is valid for $12k+9 \le 9999$ or $3k+2 \le 2499$).Contradiction!Hence the non-decreasing nature of $f$ gives $f(3k+1)=k$.Hence $f(n)=\lfloor\frac{n}{3}\rfloor \forall 1\le n \le 2499$.

So $f(1982)=\lfloor\frac{1982}{3}\rfloor=660$.

First observe that
$$3333=f(9999)\geq f(9996)+f(3)\geq  f(9993)+2f(3)\geq \cdots\geq 3333f(3)$$
Since $f(3)$ is a positive integer, we need $f(3)=1$. Next, observe that
\begin{align*}
3333=f(9999)\geq 5f(1980)+33f(3)=5f(1980)+33\quad\Longrightarrow\quad f(1980)\leq 660
\end{align*}
On the other hand, $f(1980)\geq 660f(3)=660$, so combine the two inequalities we obtain $f(1980)=660$. Finally, write
$$f(1982)=f(1980)+f(2)+0\vee 1=660\vee 661$$
Suppose that $f(1982)=661$, then
\begin{align*}
3333=f(9999)\geq 5f(1982)+29f(3)=3305+29=3334
\end{align*}
a contradiction. Hence we conclude that $f(1982)=660$.

We show that $f(n) = [n/3]$ for $n <= 9999$, where [ ] denotes the integral part.  We show first that $f(3) = 1$. $f(1)$ must be $0$, otherwise $f(2) - f(1) - f(1)$ would be negative. Hence $f(3) = f(2) + f(1) + 0$ or $1$ = $0$ or $1$. But we are told $f(3) > 0$, so $f(3) = 1$. It follows by induction that $f(3n) >= n$. For $f(3n+3) = f(3) + f(3n)$ + $0$ or $1 = f(3n) + 1$ or $2$. Moreover if we ever get $f(3n) > n$, then the same argument shows that $f(3m) > m$ for all $m > n$. But $f(3.3333) = 3333$, so $f(3n) = n$ for all $n <= 3333$.  Now $f(3n+1) = f(3n) + f(1) + 0$ or $1$ = $n$ or $n + 1$. But $3n+1 = f(9n+3) >= f(6n+2) + f(3n+1) >= 3f(3n+1)$, so $f(3n+1) < n+1$. Hence $f(3n+1) = n.$ Similarly, $4f(3n+2) = n$. In particular $f(1982) = 660$.

[b]Proof:[/b]
[b]Lemma 1: $f(mn)\geq nf(m)$[/b]
Let, $P(m,m)$ be assertion.
$$f(2m) \geq 2f(m)$$
Similarly,we can induct to get $f(nm)\geq nf(m)$.
$$f(3m) \geq f(2m)+f(m) \geq 3f(m)\implies f(nm)\geq \underbrace{f(m(n-1))+f(m)\geq .....}_{\text{n times}}\geq nf(m)$$
[i][b]Lemma proved.[b][/i]

Then we see that,
$$f(9999)\geq 3333f(3) \implies f(3)=1:f(3)>0$$
Then,
 $$f(9999) \geq 5f(1980)+33f(3) \implies 3300 \geq 5f(1980) \implies f(1980) \leq 660$$
$$f(1980) \geq 660f(3) \implies f(1980)=660$$
Then we can easily get,by assertion $P(1980,2)$
$$f(1982)=660 \space or \space f(1980)=661$$
Hence, $\boxed{f(1980)=660,661}$.And, we are done. $\blacksquare$
[b][color=red](Also, this post looks alike the previous one, but I found it genuinely myself.)

First note that $f(4) = 0 or 1$ but also $f(4) = f(3) + f(1) or f(3) + f(1) + 1 $, since $f(3) > 0$, it follows that $f(4) = 1$ and $f(1) = 0$\par
Now we will prove that $f(3k) = k$ for all integers $k$,ergo $f(m+n)-f(m)-f(n)=0$ if $3\mid m+n$,notice that this holds with $f(9999) =f(3333\cdot3) = 3333$\par
Suppose this statement holds until some $g < 9999$, then we would have that $f(3g)=g+1$, thus summing one to all multiples to 3 greater than $g$ and $f(9999) = 3334$, a contradiction\par
\begin{align*}
f(1982) &= f(1982) \\
f(1984) &= f(1982) or f(1982) + 1 \\
f(1986) &= f(1984) or f(1984) + 1 = 662 \\
f(1983) &= 661 = f(1982) + 1 or f(1982).
\end{align*}
Thus $\boxed{f(1982) = 661 or 660}$

[quote=Math.Is.Beautiful]Hey everyone, I am facing some problems while solving this problem. Kindly check my approach :
We are given that $f:\mathbb{N}\to\mathbb{N}_0$
$P(1,1)\rightarrow -2f(1)=0 or 1$
Due to the codomain of $f$, $f(1)=0$
$P(m,1)\rightarrow f(m+1)-f(m)=0 or 1$
If $f(m+1)=f(m)$ then, for $m=2$, $f(3)=0$ but this contradicts $f(3)>0$. Thus, $f(m+1)=f(m)+1$
Then, by induction, $f(n)=n-2\forall n>1$. But this contradicts $f(9999)=3333$.
So how do we resolve this problem, as I exhausted all possibilities?[/quote]

“If $f(m+1)=f(m)$ then, for $m=2$, $f(3)=0$ but this contradicts $f(3)>0$. Thus, $f(m+1)=f(m)+1$.”  That makes no sense. Since when $m=2$, we have $f(m+1)=f(m)+1$, while $m=3$, we have $f(m+1)=f(m)$.

The answer is $660$ only (which means some of the above solutions are wrong), which can be achieved by $f(x) \equiv \lfloor \frac{x}{3} \rfloor$.

Let $P(m,n)$ denote the assertion that $$f(m+n)-f(m)-f(n) \in \{0,1\}.$$ By $P(1,1)$, we have $f(1)=0$. Assume FTSOC that $f(3) \geq 2$. By $P(2,3)$, we have $P(5) \geq 2$, by $P(5,3)$, we have $P(8) \geq 4$, and so on. We eventually get that $P(9998) \geq 6664$, and by $P(9998,1)$, we get $P(9999) \geq 6664$, a contradiction. Thus, $f(3)=1$. By repeating $P(m,3)$, we get $f(x) \geq \lfloor \frac{x}{3} \rfloor$ for all positive integers $x$. Thus, $f(1982) \geq 660$. Assume FTSOC that $f(1982) \geq 661$. Then, we use $P(1982,1982)$ to get $f(3964) \geq 1322$, $P(3964,1982)$ to get $f(5946) \geq 1983$, and $P(4053,5946)$ to get $f(9999) \geq 3334$, a contradiction. Therefore, $f(1982)=660$.

[hide=Solution] It is trivial that $f(1)=0$ and $f(3)=1.$
If $f(3x)\geq x,$ then $$f(3x+3)\geq f(3x)+f(3)\geq x+1.$$
It can be trivially shown (by induction) that $f(3x)\geq x ~~\forall x\in \mathbb{N}.$
We can note that the inequality is strict for all integers greater than $x$ if it is strict for $x.$ It follows that $$f(9999)=3333\implies f(3x)=x ~~\forall x\leq 3333.$$
\begin{align*} \text{Now }3f(x)\leq f(3x) \leq 3f(x)+2 \\ \implies f(x)=\lfloor f(3x)/3 \rfloor =\lfloor x/3\rfloor ~~\forall x\leq 3333 \\ \implies f(1982)=660.\end{align*}[/hide]

There are 3 solutions that have got 2 values, but it seems their reasoning is faulty.

The answer is $f(1982)=660$. We let $P(m,n)$ denote the assertion of our FE.
-----
[b]Claim:[/b] $f(3)=1$ and $f(3n)=n$ for all integers $n\leq 3333$.
[b]Proof:[/b] Using induction one can show that for any $n$, $f(3n)=nf(3)+k$ for some $k\in \{0,1,2,...,n-1\}$. Plugging $n=3333$ into this gives $3333=f(9999)=3333f(3)+k$. Since $f(3)>0$, $f(3)=1$ must hold, but then $k=0$ and $f(3n)=nf(3)=n$ is guaranteed for all $n\leq 3333$.
-----
[b]Claim:[/b] $f(1982)=660$.
[b]Proof:[/b] Note that $f(1980)=f(3\times 660)=660$, and so $f(1980)+f(2)=660$. From $P(1980, 2): f(1982)=660$ or $661$. Assume FTSOC $f(1982)=661$, then $1982=f(1982\times 3)\geq 3f(1982)=3\times 661=1983$. Which is the contradiction. Thus $f(1982)=660$.

Note that
\[f(9999)\ge f(3)\cdot 3333\ge 3333\]
with equality case $f(3k)=k$ for all positive integers $k\le 3333$.
We know that
\[1982-f(1982)\cdot 3=f(1982\cdot 3)-f(1982)\cdot 3\in\{0,1,2\}.\]
Taking mod $3$, it must be $2$, so $f(1982)=\boxed{660}$.

Can anyone please help me to identify my mistake in the following idea .

Since f(2) = 0 , it's easy to show that f(1)=0 and f(3)=1 .

P(m,n)           : f(m+n) - f(m) - f(n) = 0,1
P(m,1)     : f(m+1) - f(m) = (0,1)+ f(1) = 0,1 ....(i)
P(m,3)     : f(m+3) - f(m) = (0,1)+ f(3) = 1,2 ....(ii)
P(m+1,2) : f(m+3) - f(m+1) = (0,1)+ f(2) = 0,1 ..(iii)

By adding (i) and (iii) and comparing that with (ii) we conclude that
Either,  f(m+3) - f(m+1) = 1
or f(m+1)- f(m) = 1 or both .
Which gives us ,
f(x+2)=f(x)+1 or f(x+1)=f(x)+1 but in this case the answers don't match.


geometry
trigonometry
trig identities
Law of Cosines

Find the area of a circle circumscribing a regular octagon whose side is 8 units in length.

[hide]The side of the octagon is 8.  If we draw a line segment from the center of the octagon to each vertex, we get 8 iscosceles triangles with legs [i]r[/i] and a base of 8.  The angle opposite the side of length 8 is 45 degrees.  By the Law of Cosines,

$r^{2}+r^{2}-2r^{2}cos45 = 8^{2}$
$2r^{2}-2r^{2}(\frac{\sqrt{2}}{2}) = 64$
$r^{2}(2-\sqrt{2}) = 64$
$r^{2}= \frac{64}{2-\sqrt{2}}= 32(\sqrt{2}+2)$

But [i]r[/i] is just the radius of the circle.  Thus,

$A = \pi(r^{2}) = 32\pi(\sqrt{2}+2)$[/hide]


geometry
3D geometry
calculus
integration
trigonometry
analytic geometry
sphere

a question i posed in class yesterday, but sadly everyone stumbled ... :(

compute the volume of a unit cube using spherical co-ordinates using the
[b]following[/b] [u]order[/u] of integration:

$ \int_{a}^{b}\;\;r^{2}\;dr\;\;\int_{f(r)}^{g(r)}\;\;d\theta\;\;\int_{k(r,\theta)}^{l(r,\theta)}\;\;\sin\phi\;\;d\phi$

where $ \boxed{x\;\equal{}\;r\cdot\cos\theta\cdot\sin\phi,\;\;\;\quad\;\;\;y\;\equal{}\;r\cdot\sin\theta\cdot\sin\phi,\;\;\;\quad\;\;\;z\;\equal{}\;r\cdot\cos\phi}$

$ V\equal{}\int_{0}^{1}r^{2}dr\int_{0}^{2\pi}d\theta\int_{0}^{\pi}\sin\phi d\phi\\ \equal{}\frac{r^{3}}{3}\Big|_{0}^{1}\theta\Big|_{0}^{2\pi}(\minus{}\cos\phi)\Big|_{0}^{\pi}\\ \equal{}\frac{1}{3}(\cdot 2\pi)\cdot (2)\equal{}\frac{4\pi}{3}$

What do you mean they stumbled? Did you set it up like that or did you just say ordered $ r$, $ \theta$, $ \phi$ and they forgot the change of coordinates needs the extra $ r^{2}\sin\phi$?

[quote="HilbertThm90"]$ V \equal{}\int_{0}^{1}r^{2}dr\int_{0}^{2\pi}d\theta\int_{0}^{\pi}\sin\phi d\phi\\ \equal{}\frac{r^{3}}{3}\Big|_{0}^{1}\theta\Big|_{0}^{2\pi}(\minus{}\cos\phi)\Big|_{0}^{\pi}\\ \equal{}\frac{1}{3}(\cdot 2\pi)\cdot (2) \equal{}\frac{4\pi}{3}$

What do you mean they stumbled? Did you set it up like that or did you just say ordered $ r$, $ \theta$, $ \phi$ and they forgot the change of coordinates needs the extra $ r^{2}\sin\phi$?[/quote]

i said volume of a unit [b]cube[/b]... not [i]sphere [/i]!!  :|

(FYI, it is a fairly tough problem i think...) the objective was to see if
one truly understood how to set up multiple integrals in polar co-ordinates...
i.e. essentially means determine $ a,\;\;b,\;\;f(r),\;\;g(r),\;\;k(r,\theta),\;\;\text{and}\;\;l(r,\theta)$

Place the cube in the first octant, so it is bounded by the coordinate planes and by $ x\equal{}1,y\equal{}1,z\equal{}1.$ Dissect it along the planes $ x\equal{}y,x\equal{}z,$ and $ y\equal{}z$ to obtain it as the union of three congruent pyramids. We will calculate the volume of the pyramid bounded by $ x\equal{}0,y\equal{}0,z\equal{}x,z\equal{}y,$ and $ z\equal{}1,$ and then triple it.

$ z\equal{}1$ becomes $ \rho\cos\phi\equal{}1$ or $ \phi\equal{}\cos^{\minus{}1}\left(\frac{1}\rho\right).$

$ x\equal{}0$ becomes $ \theta\equal{}\frac{\pi}{2}.$

$ y\equal{}0$ becomes $ \theta\equal{}0.$

$ z\equal{}x$ becomes $ \cos\phi\equal{}\sin\phi\cos\theta$ or $ \theta\equal{}\cos^{\minus{}1}(\cot\phi).$

Similarly, $ z\equal{}y$ becomes $ \theta\equal{}\sin^{\minus{}1}(\cot\phi).$

The near corners at the top have $ \rho\equal{}\sqrt{2},\phi\equal{}\frac{\pi}{4}.$ The far corner has $ \rho\equal{}\sqrt{3},\phi\equal{}\tan^{\minus{}1}\sqrt{2},\theta\equal{}\frac{\pi}{4}.$

This is really going to be an awful mess. We'll have to break the integral into several pieces, and I'm losing patience with this.

[quote="Kent Merryfield"]This is really going to be an awful mess. We'll have to break the integral into several pieces, and I'm losing patience with this.[/quote]

you are right, it is going to be a big mess, but i believe it will be less messier if you place the origin of the spherical co-ordinate system at the center of the cube and then proceed to compute the volume within the first octant, rather than place the cube entirely in the first octant and then compute the volume. :|

like you said, it will involve several pieces ($ 7$ of them as far as i can see). all i wanted to see in class were the steps leading to the setting up of $ 2$ of those $ 7$ integrals: [b](i)[/b] the octant's own "sphere-segment" integral $ \left(\; \equal{}\;\frac{1}{8}\cdot\frac{4}{3}\,\pi\,\left(\frac{1}{2}\right)^{3}\right)$, and [b](ii)[/b] $ 2$ of the rest of the $ 6$ "sphere-segment-truncated-pyramidal" integrals --- $ 1$ of them [i]near-to-the-axes[/i], and $ 1$ of them [i]far-from-the-axes[/i] ! :)

If you place the origin of the spherical co-ordinate system at the center of the cube and then proceed to compute the volume within the first octant, rather than place the cube entirely in the first octant and then compute the volume, you get exactly the same integrals but with an extra scale factor of $ \frac{1}{2}$.


What is a good score at the mental ability test at NTSE?(I will be sitting for it next year i.e. in class X)How much do people like AoPSians usually get?

Thanks a lot.I know people here are busy with the RMO discussion.

Hasnt any1 told u NTSE is now NOT held for Xth standard  :rotfl: :lol:

Ya now NTSE is not held for 10th std and it is held only for the 8th std. And I had attended that exam.

Ours was the last batch to get NTS scholarships from 10th. It was shifted to 8th to help poorer childen before they leave school. Anyways, there is always KVPY in 11th :)

who knows about KVPY beter than nuke deal!!! :D  :lol:

1. Loads of people know more. I came to know what KVPY was after I gave the exam :P
2. MY ID has no reference whatsoever to the stupid nuke deal.
3. You don't need to write 'response' in the reply topic every time.

oh!!it is out of habbit...that i write response..

Don't worry. Just live on aops from now on and you'll be thrown into the science age  :D

@ nobelium , u can see the marks of students appearing in 2nd phase on the website of NCERT. all  the marks are published there. have a look!

They do that now!  :| Nothing of that sort in our time. And I'm glad there wasn't, or my parents would have found out how screwed my SAT was?


The radius of a circular pond is 30 feet, the radius of a circular lake is 700 feet, and the center of the lake is 600 feet east and 800 feet north of the center of the pond. If a baby duck walks from the edge of the pond to the edge of the lake, then what is the shortest distance that he must walk?

I assume this is around Target Round problem... Not sure what level.

Sounds like a Chapter or a really easy State.
[hide]The distance between the centers is obviously $1000$ feet. However, since they want the distance between the edges, we subtract $730$ feet to make $270$ feet.[/hide]

I would agree on the answer and the level.

Id more say sprint though...
[hide="what I did"]Pythag triple: 3-4-5 --> 600-800-1000
So distance is 1000
minus radii: 1000-730
>>270<<[/hide]

I dunno, seems like a chapter target level to me...
[hide]Like it has been said, use the Pythagorean theorem to find that the distance between the centers is 1000 feet, and subtract the radii to get 270 ft as the answer.[/hide]

This problem is easy to do in your head, so I'd say that it should be a sprint round. (Maybe one of the last ones on school, or one of the first ones on chapter).


Let $f(x)$ are defined and continous in $[a,b](a<b)$ satisfying the condition:
\[\int^b_a f(x)dx=0\]
Prove that there exists $c \in (a,b)$ such that:
\[f(c)=2005 \int^c_a f(x)dx \]

can I prove by this method.
note $g(x)=f(x)-\int^x_af(x)dx$
assume $d$ is the smallest number $d>a$ such that $\int^d_af(x)dx=0$
we can find $g(a)=f(a),g(d)=f(d)$
but we can easily know they have different sign.
so we get there exist $c$

[quote="Moubinool"]$g(x)=e^{-2005x}\int_{a}^{x}f(t)dt$ derivable

$g(a)=g(b)=0$ then Rolle theorem gives there exist $c \in ]a,b[$
such that $g'(c)=0$

and

$g'(x)=-2005e^{-2005x}\int_{a}^{x}f(t)dt + e^{-2005x}f(x)$[/quote]
nicer :)


Here's an attachment of most of the questions and solutions proposed and solved by the carolinas community.

sorry, i would post the solutions, but i guess i reached my upload quota  :(

It's 5 MB so i don't know what to do with it.

I can try and put them on my website later and see how that works out.


Ring Theory
superior algebra
superior algebra unsolved

Am proving Z satisfies maximal condition directly.

source: unknown.

sorry.

that every nonempty set of ideals in Z has a maximal element. this is what i try show.

Take the union of all those ideals. Then take the smallest positive element of that union. Show that it generates a maximal ideal of that set.
(An exception occurs if there is no positive element. Then the only ideal in that set is $ (0)$.)


Let $ 0<a\le b \le c\le d$. Prove that:
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{d}\plus{}\frac{b}{d} \le 3\plus{}\frac{ab}{d^2}$

It's equal to prove the inequality:
\[ ac(b \minus{} a)(d^2 \minus{} b^2) \plus{} ad(c \minus{} b)(d^2 \minus{} bc) \plus{} abc(d \minus{} c)(d \minus{} b) \ge 0
\]
which is true for $ d \ge c \ge b \ge a > 0$

Nice, dear guys. I am thinking about this:

Let $ a,b,c,d$ be real numbers in $ [1,2]$, find the maximum of
\[ f\equal{}\frac{a^4\plus{}b^4\plus{}c^4\plus{}d^4}{abcd}.\]


An ant moves on the following lattice, beginning at the dot labeled $ A$. Each minute he moves to one of the dots neighboring the dot he was at, choosing from among its neighbors at random. What is the probability that after 5 minutes he is at the dot labeled $ B$?
[asy]draw((-2,0)--(2,0));
draw((0,-2)--(0,2));
draw((1,1)--(1,-1)--(-1,-1)--(-1,1)--cycle);
dot((0,0)); dot((1,0)); dot((2,0)); dot((-1,0)); dot((-2,0)); dot((0,1)); dot((0,2)); dot((0,-1)); dot((0,-2)); dot((1,1)); dot((1,-1)); dot((-1,-1)); dot((-1,1));
label("$A$",(0,0),SW);
label("$B$",(0,1),NE);[/asy]

[hide=solution]We can label each point either "even" or "odd".  The even points will be $ A$, the four corners of the square, and the four outermost points.  Then the midpoints of the sidees of the square are the odd points.  Note that when we move from an odd point we go to an even point, and vice versa.  Then there are only 4 places where the ant can be after 5 moves.  Since they all have an equal probability, we have $ \boxed{\frac{1}{4}}$.[/hide]

Lol I guessed it correctly  :rotfl:  :coolspeak:  :laugh:  :play_ball:  :trampoline:

yup got this right for like 22nd time and was at 99 now at 98

At first I didn't know what to do, and then I just put in 1/4, and it worked! :-D

$\text{This problem was really hard. I used the most tedious way and got a wrong answer . Then I looked at the answer }$ $\text{they gave me and it was so simple. It was just}$ $\boxed{\frac{1}{4}}$ :wallbash_red:  :censored:

[quote=I-_-I]I FELL FOR THE TROLL AND ACTUALLY BASHED IT >:([/quote]

This is not an appropriate post/bump. Please post only if you have an answer.

[quote=arnolde1234]When u keep getting the same problem over and over again:  :play_ball:  :wallbash_red:  :help:[/quote]

no cause you remember the answer

What a delightful problem!  [i]Probability - Think About It![/i] is probably my favorite focus on C&P Alcumus.

[hide=my full solution] In 5 minutes, you cannot be back on A (you need an even number of moves for that), on one of the corners of the square (likewise), or on one of the 4 points outside the square(likewise). You can only be on one of the 4 midpoints of the sides of the square. You are on either one of them with equal probability, so the answer is $\boxed{\frac{1}{4}}$[/hide]

haha nice problem

[hide=my sol]Note that the only possible points you can land on are the midpoints of the sides of the big square. By symmetry, the answer is 1/4.[/hide]

Richard's video should help solve this tricky problem!!  :play_ball:


[quote="silouan"]Let a function f from R to R such that
$f(x+a)=f(x)+b$  for each x in R and with $a\neq 0$ .
Prove that f has the form
$f(x)=g(x)+mx+n$  where g(x) is a periodic function[/quote]
$n=0, m=\frac{b}{a}$ :D

Prove that it has that form  :wink:
You find only such a solution .

It is obviosly. Let $g(x)=f(x)-mx, m=\frac{b}{a}$, then $g(x+a)=f(x+a)-m(x+a)=f(x)-mx=g(x)$. :D


limit
logarithms
integration
calculus
Putnam
real analysis
real analysis unsolved

Let $ a_n$ be the sequence defined by $ a_0\equal{}1$ and
\[ a_{n\plus{}1}\equal{}\frac{1}{n\plus{}1}\sum_{k\equal{}0}^{n}\frac{a_k}{n\minus{}k\plus{}2}\]

Find the limit
\[ \lim_{n\rightarrow \infty} \sum_{k\equal{}0}^n\frac{a_k}{2^k}\]
if it exists.

Someone has already asked me about this  problem (here in mathlinks by pm message  ), I just will  copy my answer  :)

if  $ f(x) \equal{} \sum_{k \equal{} 0}^{\infty}a_{k}x^{k}$ and  $ g(x) \equal{} \sum_{j \equal{} 0}^{\infty}b_{j}x^{j}$
then :
\[ f(x)g(x) \equal{} \sum_{m \equal{} 0}^{\infty}\sum_{p \equal{} 0}^{m}a_{p}b_{m \minus{} p}x^{m}\]
your condition can be rewrited as  $ (n \plus{} 1)a_{n \plus{} 1} \equal{} \sum_{k \equal{} 0}^{n}a_{k}*\frac {1}{n \minus{} k \plus{} 2}$
thus if we choose : $ g(x) \equal{} \sum_{j \equal{} 0}^{\infty}\frac {1}{j \plus{} 2}x^{j}$

and $ f(x) \equal{} \sum_{k \equal{} 0}^{\infty}a_{k}x^{k}$
 then we have
\[ f(x)g(x) \equal{} \sum_{m \equal{} 0}^{\infty}\sum_{p \equal{} 0}^{m}a_{p}*\frac {1}{m \minus{} p \plus{} 2}x^{m}\]
and since $ \sum_{p \equal{} 0}^{m}a_{p}*\frac {1}{m \minus{} p \plus{} 2} \equal{} (m \plus{} 1)a_{m \plus{} 1}$  for all ($ n\geq 0$) we have :
\[ f(x)g(x) \equal{} \sum_{m \equal{} 0}^{\infty}\sum_{p \equal{} 0}^{m}a_{p}*\frac {1}{m \minus{} p \plus{} 2}x^{m} \equal{} \sum_{m \equal{} 0}^{\infty}(m \plus{} 1)a_{m \plus{} 1}x^{m}\]
and note that
\[ f'(x) \equal{} \left(\sum_{m \equal{} 0}^{\infty}a_{m}x^{m}\right)' \equal{} \sum_{m \equal{} 0}^{\infty}(m \plus{} 1)a_{m \plus{} 1}x^{m}\]
and finaly I have :
\[ f(x)g(x) \equal{} f'(x)\]
Know we need to find $ g(x)$
we know that
\[ g(x) \equal{} \sum_{j \equal{} 0}^{\infty}\frac {1}{j \plus{} 2}x^{j}\]
but since
\[ \minus{} \ln (1 \minus{} x) \equal{} x \plus{} \frac {x^{2}}{2} \plus{} \frac {x^{3}}{3} \plus{} ...\]
then
\[ \minus{} \frac {\ln (1 \minus{} x)}{x^{2}} \minus{} \frac {1}{x} \equal{} \frac {1}{2} \plus{} \frac {x}{3} \plus{} ... \equal{} \sum_{j \equal{} 0}^{\infty}\frac {1}{j \plus{} 2}x^{j} \equal{} g(x)\]
thus we have
\[ f(x) ( \minus{} \frac {\ln (1 \minus{} x)}{x^{2}} \minus{} \frac {1}{x}) \equal{} f'(x)\]
hence (since $ f(0) \equal{} a_{0} \equal{} 1$) we have
\[ \frac {f'(x)}{f(x)} \equal{} ( \minus{} \frac {\ln (1 \minus{} x)}{x^{2}} \minus{} \frac {1}{x})\]
and
\[ \int_{0}^{t}\frac {f'(x)}{f(x)} \equal{} \int_{0}^{t}( \minus{} \frac {\ln (1 \minus{} x)}{x^{2}} \minus{} \frac {1}{x})\]
and
\[ \ln f(t) \minus{} \ln f(0) \equal{} \int_{0}^{t}( \minus{} \frac {\ln (1 \minus{} x)}{x^{2}} \minus{} \frac {1}{x})\]
and
\[ f(t) \equal{} \exp\left (\int_{0}^{t}( \minus{} \frac {\ln (1 \minus{} x)}{x^{2}} \minus{} \frac {1}{x})\right)\]
note that
\[ f(\frac {1}{2}) \equal{} \sum_{k \equal{} 0}^{\infty}a_{k}\frac {1}{2^{k}}\]
on the other hend we have
\[ f(1/2) \equal{} \exp\left(\int_{0}^{1/2}( \minus{} \frac {\ln (1 \minus{} x)}{x^{2}} \minus{} \frac {1}{x})\right)\]
try to finish yourself  :)  (I mean calulate exactly answer )

I do not know if it is too easy from here, but I got stuck precisely on this point.

There is many way how to find such integrals,
and this is what comes in my head at first time  :lol:  :
just integrate it by parts ..
\[ f(1/2) \equal{} \exp\left(\int_{0}^{1/2}( \minus{} \frac {\ln (1 \minus{} x)}{x^{2}} \minus{} \frac {1}{x})\right)\]
so :
\[ \int_{0}^{1/2}(\frac {\ln (1 \minus{} x)}{x^{2}} \plus{} \frac {1}{x}) \equal{} \lim_{\delta \to 0}\left( \minus{} \frac {\ln(1 \minus{} x)}{x}|_{\delta}^{1/2} \minus{} \int_{\delta}^{1/2}\frac {1}{x(1 \minus{} x)}dx \plus{} \int_{\delta}^{1/2}\frac {dx}{x} \right) \equal{} \lim_{\delta \to 0}\left( \minus{} \frac {\ln(1 \minus{} x)}{x}|_{\delta}^{1/2} \minus{} \int_{\delta}^{1/2}\frac {dx}{1 \minus{} x} \right) \equal{} 2\ln 2 \minus{} 1 \minus{} \ln 2 \equal{} \ln 2 \minus{} 1\]
thus answer on your question is $ e^{1 \minus{} \ln 2} \equal{} \frac {e}{2}$

I do not know. If was given in a training problem set in Colombia.... perhaps from Putnam?

I'm sure  much of mathlinker's knows where is this problem from, but no one tell us   information   :maybe: .
it was given on IMC 2003  second day, problem number 6.
for example it's possible that [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=118]Moubinool[/url]  knows more information about that, even it's possible that he know who is author of this problem  :lol:  , since he  is author of problems from [url=http://www.mathlinks.ro/viewtopic.php?mode=attach&id=9892]IMC  2003 longlist[/url] (Moubinool Omarjee, Paris) . I think he has also problems from IMC 2008 longlist, I'll be thankful, if he show these problems.


Find the integrating factors for the differential equations and then solve it.

(x-3(x^2)y)dy/dx=xy^2 + y

$ x\frac{dy}{dx} \minus{}y\equal{}3x^{2}y \plus{} xy^{2}$   u get this  after rearrangement . now the job is  done  i suppose :)

$ (x\minus{}3x^2y)\frac{dy}{dx}\equal{}xy^2\plus{}y \\ \\ \\
\frac{dy}{dx}\equal{}\frac{y}{x}\left(\frac{1\plus{}xy}{1\minus{}3xy}\right) \\ \\ \\
\text{Set: }xy\equal{}v \\ \\ \\
\frac{dv}{dx}\equal{}\left(\frac{2\minus{}2v}{1\minus{}3v}\right)\frac{v}{x}$

Now it is easy, I suppose!

Not so easy...

But the problem asks for
[hide="an integrating factor"]
\[ {1\over{\sqrt {x^3y}}
}\]
[/hide]that take us to

[hide="an solution like"]
\[ \sqrt {{y\over x}}(xy - 1)=c
\]
[/hide]


geometry
trapezoid
ratio
similar triangles

The parallel sides of a trapezoid are $ 3$ and $ 9$. The non-parallel sides are $ 4$ and $ 6$. A line parallel to the bases divides the trapezoid into two trapezoids of equal perimeters. The ratio in which each of the non-parallel sides is divided is:
[asy]defaultpen(linewidth(.8pt));
unitsize(2cm);

pair A = origin;
pair B = (2.25,0);
pair C = (2,1);
pair D = (1,1);
pair E = waypoint(A--D,0.25);
pair F = waypoint(B--C,0.25);

draw(A--B--C--D--cycle);
draw(E--F);

label("6",midpoint(A--D),NW);
label("3",midpoint(C--D),N);
label("4",midpoint(C--B),NE);
label("9",midpoint(A--B),S);[/asy]$ \textbf{(A)}\ 4: 3\qquad \textbf{(B)}\ 3: 2\qquad \textbf{(C)}\ 4: 1\qquad \textbf{(D)}\ 3: 1\qquad \textbf{(E)}\ 6: 1$

[hide="Solution?"]
Label the small non-parallel sides of the tiny long trapezoid $ a$ and $ b$, and the dividing parallel line $ x$.

$ 6\minus{}a\plus{}4\minus{}b\plus{}3\plus{}x\equal{}x\plus{}a\plus{}b\plus{}9\implies a\plus{}b\equal{}2$.

I might be misreading the question, but according to the problem, $ \frac{6\minus{}a}{a}\equal{}\frac{4\minus{}b}{b}$?

If so, we can manipulate that into $ \frac{a}{b}\equal{}\frac{3}{2}$. Putting $ a\equal{}2\minus{}b$ gives $ b\equal{}\frac{4}{5}$, and plugging this in to $ \frac{4\minus{}b}{b}$ we get the ratio as $ \boxed{\textbf{(C)}\ 4: 1}$.
[/hide]

I don't think that can be correct. It seems to me there is some extraneous information. One could just carry line segment with length 6 and line segment with length 4 to meet. Then we can see similar triangle relationships. From there using ratios and similar triangles we see that the answer has be 6/4 or 3/2.

I also get 4:1.

[hide]Extend the sides of the trapezoid to get a triangle calling where they meet point A. Starting from the top going counter clock wise with the vertices, call them A,B,C,D,E,F, and G. From similar triangles we get that $ AB\equal{}3$ and $ AG\equal{}2$. Using the same method as grn_trtle did, using $ a$ and $ b$ we get that $ a\plus{}b\equal{}2$. Also, using similar triangles with ABG and ACF, we get that $ 2a\minus{}3b\equal{}0$. Solving the systems for $ a$ gives $ a\equal{}\frac{6}{5}$. Then plugging this back into the ratio $ \frac{6\minus{}a}{a}$ gives $ 4: 1$[/hide]


modular arithmetic
number theory unsolved
number theory

Solve in positive integers:
$n^3-5n+10=2^k$

well, easy problem
if $n>2$ :

if $3 | k$
$n^3>n^3-5n+10>(n-1)^3$
contradiction

if $k \neq 0 \pmod 3$
we have contardiction by looking $\pmod 7$
because $2^k=2or 4 \pmod 7$
and $n^3-5n+10=0or1or3or5or6 \pmod 7$

if $n=2$
we have solution $k=3$
if $n=1$ we haven't solution


geometry
rectangle
combinatorics unsolved
combinatorics

for which n's it is impossible to cover a nxn board with L type elements (that include 3 squares) and at most 1 only square.
Particularly show that it is impossible to cover 99x99 board using L type elements(including 3 squares).

are you sure?because here is a tiling for a $9\times 9$ square by $L$-triominoes(i think you are talking about triominoes which are $L$-shaped):

    $\displaystyle\begin{array}{ccccccccc} 1&1&2&2&3&3&4&4&4\\ 1&5&2&6&6&3&4&4&4\\
 2&5&5&6&1&1&1&3&3\\ 2&2&4&4&1&1&1&3&3\\ 5&5&5&4&2&5&5&3&3\\ 5&5&5&2&2&5&4&4&4\\ 2&2&1&1&3&3&4&4&4\\ 2&2&1&1&3&3&6&6&6\\ 2&2&1&1&3&3&6&6&6\\
\end{array}$
 i have 'colored' the triominoes using 6 colors in this picture(it looks ok to me).
then a $99\times 99$  grid comes by placing a $9\times 9$ grid in each cell of an $11\times 11$ grid and place the triominoes this way.

i think something like this came before as well...though i can't remember exactly;probably Pierre can help me out on that...

Yes, it's been discussed before. As I recall, there's a tricky (sort of hard to find) construction for a $5\times 9$ rectangle, and, after that, we can determine the rectangles which can or cannot be covered with $L$-triominoes pretty easily.

Well, we would better find all the preceeding discussions about that stuff  ;)
Btw, we can prove the following result :

For $n \geq 2$, an $n \times n$ square can be tiled with $L$-trominoes and and most one monomino unless $n=3.$
Moreover, if $n \neq 0 \mod [3]$ then the position of the required monomino is arbitrary unless $n=5.$

Pierre.

The $m \times n$ problem (with no monomino) is here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=tiled&t=19967

Pierre.


floor function
ceiling function
function
algorithm
limit
calculus

How would you find a closed-form definition for this? :

$ f(n) \equal{} f(n \minus{} 1) \plus{} 2f(\lfloor \frac {n \minus{} 1}{2} \rfloor) \plus{} 4(3^{\lceil \frac {n \minus{} 1}{2} \rceil)} \minus{} 1) \plus{} 6$,

where f(0) = 0.

Might have a generating function that can be expressed as an infinite product (see for example [url=http://www.research.att.com/~njas/sequences/A033485]A033485[/url] at the [url=http://www.research.att.com/~njas/sequences]OEIS[/url] for a similar but simpler sequence with this property), but extremely unlikely to have a nice closed form.  What's the problem that it came from?

If you wanted to try to simplify, you might begin like this:  let $ g(n) \equal{} (f(n) \plus{} 1)/4$ and we can rewrite the recursion as $ g(n) \equal{} g(n \minus{} 1) \plus{} 2g(\lfloor(n \minus{} 1)/2 \rfloor) \plus{} 3^{\lceil (n \minus{} 1)/2 \rceil}$ with $ g(0) \equal{} 1/4$.  But it's not at all clear how to advance from here.

Thanks for the advice! I came across this while doing something for a variant of Towers of Hanoi for my math-related science fair, and the function in question was the number of moves one of my algorithms was taking, and I was stumped...  :(

I'm not very familiar with generating functions, so maybe I'll read more into it, and give it a try. Otherwise maybe I can try to find some bounds to it? What would be a good method for that?

Thanks again!

[quote="erdogankerem123"]Otherwise maybe I can try to find some bounds to it? What would be a good method for that?[/quote]  Oof ... I know good methods for this exist, but I've never learned any of them :(  I think that one could show without a huge amount of work that $ \lim_{n \to \infty} \frac {f(n)}{3^{n/2}} \equal{} 8$, which gives you a rough rate of growth (though not a strict upper bound).  More precise measurements would certainly be desirable, but you should start with that.  In any event, it's reasonably clear that the "adding powers of 3" part dominates.

Oh really? How do you get that limit? How did you get the 8, especially?

Numerically, but actually I made a silly error: the limit if $ n$ goes through even numbers is 8, but the limit if $ n$ goes through odd numbers is $ 4\sqrt {3}$.  Here's a way to determine this, assuming the limits exist.  Then $ \frac {f(n)}{3^{n/2}} \equal{} \frac {1}{\sqrt {3}} \cdot \frac {f(n \minus{} 1)}{3^{(n \minus{} 1)/2}} \plus{} 4 \cdot 3^{\lceil(n \minus{} 1)/2\rceil \minus{} n/2} \plus{} \textrm{other stuff}$, where $ \textrm{other stuff} \to 0$ as $ n \to \infty$.  Thus, if $ l_0 \equal{} \lim_{n \to \infty} \frac {f(2n)}{3^n}$ and $ l_1 \equal{} \lim_{n \to \infty} \frac {f(2n \plus{} 1)}{3^{n \plus{} 1/2}}$ exist then we have $ l_0 \equal{} \frac {l_1}{\sqrt {3}} \plus{} 4$ and $ l_1 \equal{} \frac {l_0}{\sqrt {3}} \plus{} \frac {4}{\sqrt {3}}$.  This system has the unique solutions $ (l_0, l_1) \equal{} (8, 4\sqrt{3})$.  (One still should prove those limits exist, of course.)

Thanks for the replies!

However, what are the conditions that the function must satisfy in order for the limits to exist? (We've only done continuous functions in Calculus class)


inequalities
logarithms
inequalities unsolved

Hi!everyone,I have a strange but beautiful ineq which is unsolved.
Suppose a,b,c are positive reals such that abc=1,prove that
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/b52ca6fb789a269de9b631c9dcd71697.gif[/img]

Use the inequality
$a^{a-1}-\frac{1}{a}-\ln{a} \ge 0$.


Find all positive integers x, y such that [tex] A= x^2y^4 - y^3 +1 [/tex] is a perfect square

I found two roots : [tex] ( x,y ) = ( x, 1) or ( x, 2x )[/tex] . May be there is exist other roots

Is this a contest problem? Are you sure it is solvable? (since many equations of this type can't be solved...)

Yes , I'm sure that this problem has a solution because one my friend solved it .

(0,0), (0,1),( :sqrt: A,1)

Only works when A is a perfect square.

Ok then, if you're sure it has a solution I'll start analyzing. However I still wonder if your friend found all solutions and if he can prove there are no others...

Anyway: x^2 . y^4  -  y^3  +  1 is square.

Mod 4:
Suppose y is even (2 or 0): x must be even
Suppose y is 1 mod 4: x can be both
Suppose y is 3 mod 4: no solutions

Suppose x is even (x=0 mod 4): y=1 or 0 mod 4. So the case 2 mod 4 fades away. This means [b]y is 0 or 1 mod 4[/b].

Hmm I see no next step that leads to a solution..

Can you post the solution of your friend, and if not, pm it to me? I'll verify to see if it's correct, so people know if it's worth to continue looking for the solution...


please recommend me agood reference for singular value decomposition(svd).thanks.

See this link:[url=http://www.ddj.com/204204032]Professor SVD[/url]. At the end of the page it is written "With Charles Van Loan, Golub wrote one of the best-known textbooks in the field." ( For more information google Golub-Van Loan.)

thanks for your guidance mr.ehsan khanmohammadi.


If real numbers $x$ and $y$ satisfy the condition $x^2+xy+y^2=1$, find the minimun and maximum value of:

$K=x^3y+xy^3$

So easy is it? Finish your proof then. Remember minimum-maximum problems have two-steps. You have to find values that give the minimum or maximum and then prove there exists no lesser or greater output. What are $(x,y)$ that give[list]1) $x^3y+xy^3=\frac{1}{4}$
2) $x^3y+xy^3<-2$[/list]
;)

Use polar transform, let
$x=r \cos \theta$ and $y= r \sin \theta$.
$x^2+xy+y^2=1 \Leftrightarrow r^2+r^2 \cos \theta \sin \theta = 1$
Then $\sin \theta \cos \theta = \frac 1 {r^2} - 1$
$sin {2 \theta} = 2( \frac 1 {r^2}-1) \implies -1\leq 2( \frac 1 {r^2}-1)\leq 1 \implies \frac 2 3 \leq r^2\leq 2$.---(*)

$K=(x^2+y^2)(xy)=r^2 (r^2 \cos \theta \sin \theta) = r^4 (\frac 1 {r^2} -1) = r^2(1-r^2)$.

Making use of (*) and consider the graph of [b]a part of parabola [/b]open downwards(or the function $f(t)=t(1-t)$ is strictly decreasing for $t\geq \frac 1 2$):

Maximum value of $K = (\frac 2 3)(1-\frac 2 3) = \frac 2 9$, and
mimimum value of $K= 2 ( 1-2) = -2$.


geometry
3D geometry
tetrahedron
rhombus
IMO Shortlist

$ ABCD$ is a terahedron: $ AD\plus{}BD\equal{}AC\plus{}BC,$ $ BD\plus{}CD\equal{}BA\plus{}CA,$ $ CD\plus{}AD\equal{}CB\plus{}AB,$   $ M,N,P$ are the mid points of $ BC,CA,AB.$ $ OA\equal{}OB\equal{}OC\equal{}OD.$ Prove that  $ \angle MOP \equal{} \angle NOP \equal{}\angle NOM.$

The first three conditions are much better expressed like this: two opposite sides of the tetrahedron are equal.

From here it follows quickly that $O$ lies on $MM',NN',PP'$, where $M',N',P'$ are the midpoints of $DA,DB,DC$ respectively. Moreover, $MNM'N',NPN'P',PMP'M'$ are rhombi (is this the correct plural of rhombus? :)), so the angles we have to prove equal are the angles between the diagonals of three rhombi, which are $\frac \pi 2$.

Call D,E,F,M,N,P the mid point of each side of tetrahedron.
So DE, FM, NP have the common midpoint, which is exactly O.
Besides, we have DFEN, DNEP, FNMP are lozenges.
   Thus, <MOP=<MON=<NOP=90


function
logarithms
algebra
domain
algebra proposed

Find all functions $ f: \mathbb R\rightarrow \mathbb R$ satisfying $ f (x \plus{} y \minus{}xy) \plus{} xf(y) \plus{} yf(x) \equal{} f (x) \plus{} f (y)$ for every $ x,y\in \mathbb R$

Right now I got $ f(0) \equal{} f(1) \equal{} f(2) \equal{} 0$,
\[ f(x) \plus{} f(2 \minus{} x) \equal{} 0\,,
\]
and
\[ f\left(x(2\minus{}x)\right) \equal{} 2(1\minus{}x)f(x)\,.\]
Hope it helps.

[hide]A succession of three transformations

$ f(x) \equal{} f_1(x\plus{}1)$
$ f_1(x) \equal{} xf_2(x)$
$ f_2(x) \equal{} \ln f_3(x)$

(with some random domain things like $ f_1$ odd and $ f_1(0) \equal{} 0$ thrown in) will show that this is equivalent to the Cauchy equation. Not the nicest problem.[/hide]

[hide]Considering x = -y we get

$ f(x^2) \plus{} xf(\minus{}x) \minus{}xf(x) \equal{} f(x) \plus{} f(\minus{}x)$

$ \equal{}> f(x^2) \equal{} (1\minus{}x)f(\minus{}x) \plus{} (1\plus{}x)f(x)$

If we consider, in this last expression, x = 1 then we have

$ f(1) \equal{} 2f(1) \equal{}> f(1) \equal{} 0$.

Considering now y = 1 on the given equation yields

$ f(y) \plus{} xf(1) \plus{} f(x) \equal{} f(x) \plus{} f(1)$

$ \equal{}> f(y) \equal{} (1\minus{}x)f(1) \equal{} 0$

and so f(x) = 0 is the only solution.[/hide]

Check your last step; the first term is $ f(x\plus{}1\minus{}x(1)) \equal{} f(1)$, not $ f(x)$.


A competition involving $ n \ge 2$ players was held over $ k$ days. On each day, the players received scores of $ 1,2,...,n$ points with no two players receiving the same score. At the end of the $ k$ days, it was found that each player had exactly $ 26$ points in total. Find all pairs $ (n,k)$ for which this is possible.

[hide]The total number of points awarded after $ k$ days is clearly $ \frac {kn(n \plus{} 1)}{2}$, in our case equal to $ 26n$, or $ k(n \plus{} 1) \equal{} 52 \equal{} 2^2\cdot13$, and since $ n\geq2$, then $ k\leq\frac {52}{3}$, or $ k\leq17$ must divide $ 52$.  The case $ k \equal{} 1$ is clearly impossible since all competitors would have different scores.  For $ k \equal{} 2$ and $ n \equal{} 25$, just have the $ i$-th competitor scoring $ i$ on the first day and $ 26 \minus{} i$ on the second day for any ordering of the competitors.  For $ k \equal{} 4$ and $ n \equal{} 12$, just have the $ i$-th competitor scoring $ i$, $ i$, $ 13 \minus{} i$, $ 13 \minus{} i$ on the first, second, third and fourth day, respectively, for any ordering of the $ 12$ competitors.  Remains only $ k \equal{} 13$ and $ n \equal{} 3$.  So have the following scores for each one of the competitors on each day:

$ 3,3,3,3,1,1,1,1,1,2,2,2,3$
$ 1,1,1,1,3,2,2,2,2,3,3,3,2$
$ 2,2,2,2,2,3,3,3,3,1,1,1,1$

No other values of $ k\leq17$ divide $ 52$, so the possible values of $ (n,k)$ are $ (25,2)$, $ (12,4)$ and $ (3,13)$.[/hide]


geometry
analytic geometry
graphing lines
slope

Consider $ B(5,1)$ and $ C(1,5)$. The equation of $ BC$ is:
$ (A)$ $ x\minus{}y\equal{}\minus{}4$
$ (B)$ $ x\minus{}y\equal{}4$
$ (C)$ $ x\plus{}y\equal{}5$
$ (D)$ $ x\plus{}y\equal{}6$
$ (E)$ $ 2x\plus{}y\equal{}11$

The line intercepts the y axis at $ y\equal{}6$ and has gradient -1.

The equation is therefore $ y\equal{}\minus{}x\plus{}6$ which is rearranged to give $ x\plus{}y\equal{}6$

Alternatively, plugging the values of $ (x, y)$ into each choice would give the same, and possibly more efficiently achieved, answer.

:)

the slope between (5,1) and (1,5) is -1.
this leaves only C and D because to find the slope in general form you do -A/B.
then you just plug in one of the points and you get b=6 so D is the answer.

[quote="almonkez"][quote="moldovan"]Consider $ B(5,1)$ and $ C(1,5)$. The equation of $ BC$ is:
$ (A)$ $ x \minus{} y \equal{} \minus{} 4$
$ (B)$ $ x \minus{} y \equal{} 4$
$ (C)$ $ x \plus{} y \equal{} 5$
$ (D)$ $ x \plus{} y \equal{} 6$
$ (E)$ $ 2x \plus{} y \equal{} 11$[/quote
suppose that thw equation is   y = ax + b
a is the slope of  b c
b = (5-1)/(1-5) = - 1
b the part intersected from y  axis
(5 ,1) on the line bc  then satisfi its equation
y = -x + b
1 = -5 + b     b = 6
the equation is  y = -x + 6
y + x = 6[/quote][/quote]

ok  my answer is d too
in general form
if a (x1 ,y1)   & b(x2 ,y2 ) the slope of straitline ab =(y2 -y1)/(x2-x1)
for any value of x & y
 :play_ball:


Putnam
search
number theory unsolved
number theory

Prove that if m>n are natural numbers therefore:

m divides (((GCD(m,n))*(m!)/(((m-n)!)*(n!)))

 :ninja:

Putnam 2000-A 28 :[url]http://www.mathlinks.ro/viewtopic.php?search_id=1378992668&t=150396[/url]


The webmaster of the IMO 2004 site and the webspace provider, should be shot.

The results arent online yet, the site has a random downtime...magnificent... :?

Well, I think it's the common problem :(  Remember, when were the results presented last year in Japan, and what was that site's webmaster like. Most organizers make the official webpage only because it is required by the Regulations. The best site without any doubts was that of IMO 2001 in Washington (although there was a little bug: red-eye effect on most photos).  :)
I think that the Regulations should be changed. All the organizers should at least provide the results online as they appear.

[quote="PiDeltaPhi"][quote="Valentin Vornicu"]As usual :) you can see some results on the results forum. :)[/quote]

Yes AOPS - Mathlinks is as usually faster than any news service  :D

Still, proper hosting and administration of the IMO site isnt that hard... :maybe:[/quote]Well if they would officially assign this job to us, I think we would be very happy to do it :) ;) ... maybe it will happen next year in Mexico ..  :maybe:  ;)

[quote="nonIMO"]The best site without any doubts was that of IMO 2001 in Washington.[/quote]

Well Wolfram did that one, and the stats available even rivalled some big sporting events. Hopefully next year's will be better.

Not surprisingly the IOI 2004 site is much better (eg at least it works).

If the IMO Greece committee are lazy, there are many people who can help them type the results for free, in fact I can help them type the reuslts. Why don't the IMO Greece committee just give the results to someone else to type it for them?

[quote="liyi"]and the problems and solutions are not published on the website...[/quote] they are published here :P :) ... I have just made a pdf file with the questions.

:)  :)
Mathlinks is really a fast source...

IPhO 2004 page has daily reports, video clips and results. :D
http://www.ipho2004.or.kr/home/eng/presskit/NewsLetter.asp?globalmenu=5&localmenu=1

ICho 2004 page has daily reports. :D
http://www.icho.de/english/06_catalyzer.html

IBO 2004 page has the complete results. :D
http://www.ibo2004.org.au/

But the IMO 2004 page is under construction. :? http://www.imo2004.gr/index.php?action=under_construction

The results and some graphs and some photos are online at last:

http://www.imo2004.gr/fimo/

As Russians say, "Лучше поздно, чем никогда" ("Better late than never"). The problems with solutions (official, it seems) are also available there, and I like [url=http://www.mathlinks.ro/Forum/orl/imo_2004_non_IMO_shortlist_materials.zip]that[/url] nice zipped .pdf file (although it should be "medal(s)", not "metal(s)" throughout the document).

Due to the fact that u mention  about the official IMO2004 website, the guides builded another website providing useful information... U can find it at http://www.imo2004.tk Any comments and/or photos plus additional matterial are welcome...

Thank you Nikolas for this site, I'm so glad with this site.


geometry
perimeter
combinatorics proposed
combinatorics

Prove that every triangle with area S and perimeter P could be cut into n triangles such that every of them has perimeter bigger than $ \frac {12S}{P}$ where n is arbitrarily taken natural number.

It's been long time since the problem was posted. Will somebody try to solve it?


I just found Challenging Problems in Geometry by Posamentier, Salkind.

Is this good book? Like I need some geometry problem practices so I might go get one nice geometry book. What do you think? Is it worth to buy that book?

depends on your level... i used Geo revisited, which was a good book for me. Challenges in Geometry might suit u better.


:oops:  :blush:
Maybe this is funy query but my braing has just stopped.
I need to find out: expl. [b]How many sick days goes to 1 worker [/b]in month?; if, there are 3143 workers, works 8hours/day, 22 day/month. 548 of them were on sick days; sick day is count also as 8 hours.


[quote="Interval"]A class contains 18 boys and 12 girls.  The teacher calls on students at random to recite.  What is the probability that the first student called upon to recite is:

a.  a boy?

b.  a girl?[/quote]

[hide="A"]There are 18 boys and 30 students in all. So its $\frac{18}{30}=\frac{3}{5}$.[/hide]

[hide="b"]Same thing but since its girls its just $1-\frac{3}{5}=\frac{2}{5}$.[/hide]

Or for number 2:
[hide]
You can go $12/(18+12)$[/hide]

It's good to know many ways to find the answer.

Thanks!


function
logarithms
topology
real analysis
real analysis unsolved

I am having a hard time coming up with this examples:

1. Find a function f R ---> R such that f is in LP for all finite p (>=1 or >0) but f is not in L infinity: f is actually not essentially bounded on any open set or R. (

2. Give an example of a compactly supported fucntion f R -->R such that f is holder continuous with exponent 1/2 but f is NOT in W(1,2) (The sobolev space of fucntions which are L2 and whose weak derivaties are L2).

Much thanks!

For (1), how about this:

Let $ g(x)\equal{}\minus{}\ln x$ on $ (0,1)$ and $ 0$ otherwise. Let $ r_n$ be an enumeration of all rational numbers.

Let $ f(x)\equal{}\sum_{n\equal{}1}^{\infty}2^{\minus{}n}g(x\minus{}r_n).$

For (2), how about the simplest example? Let $ f(x)\equal{}\sqrt{x}$ on $ (0,1)$ (and do something away from the origin to ramp it down to being compactly supported.) Then near the origin, $ |f'(x)|^2\equal{}\frac{c}{x},$ and that's not locally integrable.


Everyone has got a language up in games forum(okay not everyone but they will.) could we have a language forum? i know we have the different country forums but that is for math and not practical when studying dead ones.

Just wondering-
Funcia

The regional forums are not just for math, I see all kinds of stuff being posted there. And if the country forums are dead, what makes you think they are enough people to have a conversation in that specific language in a language forum?


quadratics
SFFT
algebra
polynomial
articles
number theory
special factorizations

Find all integers m so that there is an integer n such that $ 6m^2 \plus{} 2n^2 \plus{} 7mn \equal{} 5$

n = 1/4 ( -7m + (m^2 + 40)^.5)
or
   = 1/4 ( -7m - (m^2 + 40)^.5)
, so all that is needed to do is to determine when

m^2 + 40 = a^2 , for some integer a ...

But i don't know how to that, Im just beginning to learn concepts of number theory :(

[hide="Hint"]What are the factors of the right side? Of the left side?[/hide]

[hide="My Solution"]
We can use Simon's Favorite Factoring Trick to factor the left side out. (You can accomplish this by trial and error and using the coefficients of $ mn$, $ m^2$, and $ n^2$ as a hint.)

By SFFT we have $ 6m^2 \plus{} 2n^2 \plus{} 7mn \equal{} (3m \plus{} 2n)(2m \plus{} n)$.

The only factors of $ 5$ are $ 1$ and $ 5$. And because $ m$ and $ n$ are integers, we can also have $ \minus{}1$ and $ \minus{}5$. Setting up cases we have the following:

$ 3m \plus{} 2n \equal{} 5 \\
2m \plus{} n \equal{} 1 \implies 4m \plus{} 2n \equal{} 2 \\
\minus{} m \equal{} 3 \\
m \equal{} \minus{} 3 \\
n \equal{} 1 \minus{} 2m \equal{} 1 \minus{} ( \minus{} 6) \equal{} 7 \\
 \\
3m \plus{} 2n \equal{} 1 \\
2m \plus{} n \equal{} 5 \implies 4m \plus{} 2n \equal{} 10 \\
\minus{} m \equal{} \minus{} 9 \\
m \equal{} 9 \\
n \equal{} \minus{} 2m \plus{} 5 \equal{} \minus{} 18 \plus{} 5 \equal{} \minus{} 13 \\
 \\
3m \plus{} 2n \equal{} \minus{} 5 \\
2m \plus{} n \equal{} \minus{} 1 \implies 4m \plus{} 2n \equal{} \minus{} 2 \\
\minus{} m \equal{} \minus{} 3 \\
m \equal{} 3 \\
n \equal{} \minus{} 2m \minus{} 1 \equal{} \minus{} 6 \minus{} 1 \equal{} \minus{} 7 \\
 \\
3m \plus{} 2n \equal{} \minus{} 1 \\
2m \plus{} n \equal{} \minus{} 5 \implies 4m \plus{} 2n \equal{} \minus{} 10 \\
\minus{} m \equal{} 9 \\
m \equal{} \minus{} 9 \\
n \equal{} \minus{} 2m \minus{} 5 \equal{} 18 \minus{} 5 \equal{} 13$

Thus, the ordered pair solutions $ (m,n)$ are $ ( \minus{}3,7), (3, \minus{}7), (9, \minus{}13), (\minus{}9, 13)$.
[/hide]

[quote="sunehra"][hide="My Solution"]
We can use Simon's Favorite Factoring Trick to factor the left side out. (You can accomplish this by trial and error and using the coefficients of $ mn$, $ m^2$, and $ n^2$ as a hint.)

[/hide][/quote]

Isn't this just regular old factoring of polynomials?

SFFT can be used to factor the the side with the variables then you can factor 5 as $ 1*5$ and there you have your solutions :)

[quote="Smartguy"][quote="sunehra"]Yea it is. It's just a special name for it that's popular in the AoPS Community. Refer to the [url=http://www.artofproblemsolving.com/Wiki/index.php/Simon's_Favorite_Factoring_Trick]article[/url] in the  AoPS Wiki.[/quote]

ummm... i thought what was done here was considered regular factoring

i thought SFFT was something like m+n+mn=1, factor that--->

m+n+mn+1=2--> (m+1)(n+1)=2...[/quote]

Oh I think you're right. I just think of it is as SFFT for some reason.


As there is no reward system for the games played here, I am now establishing the first, and hopefully last, AoPS Bank.  Status can be viewed under the username, by adding on to the Location field in the user Profile.

To use the AoPS Bank, [b]the game you are playing on G&FF must be an AoPS Bank game[/b].

If you wish to make changes to the amount of credit(money) you have on your bank, you [b]must[/b] do the following:

[list=1][*]Make a post here.
[list][*]The post must contain a [b]reference to the game/post[/b] where the credit was earned.
[*]It must include credit earned/lost.
[*]It must include current credit total.[*]It must contain the previous total.[*]It is helpful if it is in $ \text{\LaTeX}$[/list][*]On the Bank: profile field, do not immediately change your total.  Instead, make it obvious that there needs to be a change by putting your credit earned in parenthesis.  Ex. $ \mbox{Bank: } \$0.00( + \$0.50)$[*]Wait for your change to be approved.[/list]

As for approving changes, that will be a VERY quick process (hopefully) but will be done by specific people.  Anyone who wants the job of approving changes can post here.
[b]If you are caught changing your Bank credit, there is a very high chance that you will be unable to participate in future forum games. You do NOT want that to happen.[/b]

All Bank related material belongs here.  Reference whatever you can.  You are free to sell "items" and also to give away credit to another person.

If anything is unclear, PLEASE post and save all of our lives.

[quote="MustItAlwaysBeMe"]If anything is unclear, PLEASE post and save all of our lives.[/quote]

You Got It!

What do you mean Games? Like in Mafia games, or "Comment on the person above/below you" or what?

To make it so there are more games, maybe we should either make a new topic or revive things like Stretch, Word Alchemy, etc. Doesn't seem like there are too many right now.

This is a tall request.

How do you earn money? By simply posting or creating whole new posts?

What will this money be used for?

I don't really understand. The AoPS Bank Game could be fun. Are you planning it like this?:

Every post gets $ \$.50$ . The players can't spam or something...and then they have to build up a certain amount of money to qualify for something. And then whoever has the most money at the end wins?

No, I don't think it's like this. I'm pretty sure its just a reward system for certain games, so people actually try to win. You would also be able to see who is best at GFF stuff. But I doubt it would work like by how much you post. I'm pretty sure it'll be by games with points, like possibly the rule game and stretch and that really old word alchemy thread. Stuff like that.

That would be [b][i]quite[/i][/b] difficult to do, keeping track of good posts of [b]EVERYONE[/b] who posts in Games and Fun Factory.

And for [i]every[/i] post you'll have to post in this thread too.

Talk about G&FF rising in posts quickly. I think it would double the number of posts in one time, and I'm sure 7h3.D3m0n.117 doesn't want to have extra posts to clean up!

[quote="7h3.D3m0n.117"]Similar to "Pay-Day", we'll schedule a day every week or so (most likely a Friday) where you tally up all your posts, with links, and you'll be awarded. This topic will be constantly pruned, so that this doesn't turn out to be a 50 page metathread.[/quote]

 :clap2:  :thumbup:  :coolspeak:

I think this is a great idea. But what would be main bank be for, then?

[quote="7h3.D3m0n.117"]Here's an idea:

Similar to "Pay-Day", we'll schedule a day every week or so (most likely a Friday) where you tally up all your posts, with links, and you'll be awarded. This topic will be constantly pruned, so that this doesn't turn out to be a 50 page metathread.[/quote]
Wow, that's a great idea.  That way, we won't be awarding 20 people an hour :) .  So would you just award the person who contributed most to G&FF/won most games in the week by money?

And if anyone is still wondering...it was intended to be a reward system for existing games.

What exactly is the point of this? Why would anyone want a "reward" for posting here? Why would someone want to waste 10 minutes of their life compiling links of every post they've ever made here?

[quote="7h3.D3m0n.117"]Similar to "Pay-Day", we'll schedule a day every week or so (most likely a Friday) where you tally up all your posts, with links, and you'll be awarded. This topic will be constantly pruned, so that this doesn't turn out to be a 50 page metathread.[/quote]

*is nervous* What if there is an error, which there is a high chance of having an error? Also, people*cough 1=2 cough* post a lot in G&FF.


floor function
combinatorics unsolved
combinatorics

A $ 2007\times 2007$ board is filled with pluses. In the first move, we change all signs of the [b]rows[/b] whose number is divisible by $ 1$. In the second move, we change all signs of the [b]columns[/b] whose number is divisible by $ 2$. In the third move, we change all signs of the [b]rows[/b] whose number is divisible by $ 3$, and so on. After $ 2007$ moves, is the number of pluses or minuses bigger?

The $ n$th row $ (1\leq n\leq 2007)$ will be subjected to 'sign changing operation' for $ d(n)$ times, where $ d(n)$ is the number of positive divisors of $ n$.If $ d(n)$ is even, the row will remain plus. Else if $ d(n)$ is odd, the row will become minus. We know that $ d(n)$ is odd iff n is a perfect square. Since $ \lfloor \sqrt{2007} \rfloor\equal{}44$, we have 44 perfect squares less than 2007. Hence, there are more pluses.


Asymptote
calculus
calculus computations

What does the graph of $ y^{3} \equal{} x^{3} \plus{}9$ look like? Where are its stationary points?

For an informal answer, try the site [url]http://www.wolframalpha.com/[/url]
It plots such curves and even gives some info on extrema, critical points, etc.

$ y^3\equal{}x^3\plus{}9\Rightarrow y^{\prime}\equal{}\frac{x^2}{y^2}\equal{}0\Leftrightarrow x\equal{}0, y\neq 0$.
Hence we have $ p\equal{}(0,\sqrt[3]{9})$ as a stationary point.

Much more interesting is the obvious asymptote @ y=x. We transform to polar coords then sub in $ \theta\to\theta \plus{}\pi /4$. This sets the asymptote @ y=0. This new curve has a rectangular equation given by $ x^2y\minus{}y^3\equal{}9\sqrt{2}$. It seems pointless to find that curve, but I couldn't resist--the image practically begged me to do it. :stink:


[b]The rank of an m*n matrix A is the dimension of C(A). Show that an n*n matrix A is invertible exactly when its rank is n[/b]

$A$ is inverible if $\det(A)\neq 0$ which implies that the columns are linearly independent which implies that $c_1,c_2,..,c_n$ form a basis for $\Re ^n$ so the rank is $n$.

[quote="v0id"]if the rank is r for example it means that there exist an rXr non-zero minor and all higher order minors are null. that means $det A$ is non-zero if and only if $rank A$ is non zero[/quote]

I think $rank(A)$ should be $n$ not "non zero".


show that if $ (x,y) \in [0, \frac {\pi}{2}]$ then:

 $ \displaystyle ((\cos(x))^4 (\sin(y))^6)^{\frac {1}{5}} \plus{} ((\sin(x))^4 (\cos(y))^6)^{\frac {1}{5}} \le 1$  :lol:

[edit:i added (...), infact the cos(x)^4 means (cos(x))^4..]

[quote="implex"]$ \displaystyle (\cos(x)^4 \sin(y)^6)^{\frac {1}{5}} \plus{} (\sin(x)^4 \cos(y)^6)^{\frac {1}{5}} \le (\cos(x)^4 \sin(y)^6 \plus{} \sin(x)^4 \cos(y)^6)^{\frac {1}{5}} \\
\equal{} ( \cos(A) \sin(B) \plus{} \sin(A) \cos(B))^{\frac {1}{5}}$[/quote]

scuse me, why the first and second passage?? (however the exponent is out, and the first passage if you tought cos(x^4)sin(y^6)+...need to be reversed)

hmmm, i thought power was int he angles and not in cos or sin
anyways
$ (\cos^4x \sin^6y)^{\frac {1}{5}} \plus{} (\cos^6y \sin^4x)^{\frac {1}{5}} \le 1$

Woot! I did an inequality! :D
[hide]$ \sqrt [5]{\cos^4x \sin^6y} \plus{} \sqrt [5]{\sin^4x \cos^6y}\leq \frac {2\cos^2x \plus{} 3\sin^2y}{5} \plus{} \frac {2\sin^2x \plus{} 3\cos^2y}{5} \equal{} 1$ by AM-GM.
Okay, so it wasn't that hard...[/hide]


[quote="anorithdialga"]
\[ in\_\Delta ABC
\]
prove
\[ {\rm{a}}bc < {\rm{a}}^2 (\Delta \minus{} {\rm{a}}) \plus{} b^2 (\Delta \minus{} b) \plus{} c^2 (\Delta \minus{} c) \le \frac {3}{2}{\rm{a}}bc
\]
with usual notations and
\[ \Delta
\]
=area[/quote]
is the problem correct?

okay i got it . its simple. use
\[ \Delta  \equal{} \frac{{{\bf{a}}bc}}{{4R}}
\]

can u post a detailed solution? plz?

i just can't put it together. :blush:  :D

the inequality is absolutely correct , i have solved it! :D

Would you post your solution here,please?

[quote="anorithdialga"]In $ \Delta ABC$ prove $ {\rm{a}}bc < {\rm{a}}^2 (\Delta \minus{} {\rm{a}}) \plus{} b^2 (\Delta \minus{} b) \plus{} c^2 (\Delta \minus{} c) \le \frac {3}{2}{\rm{a}}bc$ where $ \Delta$ is area of $ ABC$ .[/quote]
Maybe $ ... <\ \sum a^2\left(\sqrt {\Delta}\minus{}a\right) \le ...$ (see to dimensions !).

$ ... <\ \sum a^2\left(\sqrt {\Delta}\minus{}a\right) \le ...$

when considering the triangle with lengths 3,4,5, the supposed idea is invalid

[quote="apratimdefermat"]okay i got it . its simple. use
\[ \Delta \equal{} \frac {{{\bf{a}}bc}}{{4R}}
\]
[/quote]

Would you post your idea/solution to the inequality?

[quote="Potla"][quote="apratimdefermat"][quote="anorithdialga"]

prove                                      $ in \_ \Delta ABC$

$ {\rm{a}}bc < {\rm{a}}^2 (\Delta \minus{} {\rm{a}}) \plus{} b^2 (\Delta \minus{} b) \plus{} c^2 (\Delta \minus{} c) \le \frac {3}{2}{\rm{a}}bc$

with usual notations and

                                                      $ \Delta$

=area [/quote]
is the problem correct?[/quote]
It is from your blog itself! :P
However, I too got the solution/proof.
Thanks for your hint of using $ \Delta \equal{} \frac {abc}{4R}$
But I do not know if it is correct........... :maybe:[/quote]

would you show me some source/links to read the proof?

can anybody post the detailed solutions here?