calculus
integration
ratio
induction
algorithm
algebra unsolved
algebra

Prove that every positive integer can be written as a finite sum of distinct integral powers of the golden ratio.

Can you add and subtract distinct powers of $\phi$? Because otherwise I do not believe this is true

Can someone please tell me which one of the two values is considered golden ratio: $\frac{\sqrt{5}+1}{2}>1$ or $\frac{\sqrt{5}-1}{2}<1$?

Besides, regarding orl's question - it's possible that negative powers do the work.

It's the positive, I believe.

It doesn't really matter, because the other one is just the negative power of the first.

I really don't see how one can do it by simply adding...

Assuming we can subtract:

[hide]

Lemma 1:  is an integer for all integers .

Proof: Let . We know that  and . Moreover, it also satisfies the characteristic equation , so then we get a reccurrence relationship for  which is , , and . We're only doing linear operations on the values, and since our starting values are positive integers, all values are positive integers. Moreover, define .

Next, each term gives us a pair of distinct powers of , so if we can prove that all positive integers can be formed through sums or differences of distinct , then we can show that each positive integer can be formed by distinct (and even!) powers of .

Lemma 2: We will show  or  by induction on .

Proof: For , we have  or , which is true. For our inductive step, we can assume that

Adding  to both sides, we get that

 and our inductive step is complete.

Finally, we will show that every integer less than or equal to  can be written as a sum or difference of distinct terms from the set . We will do this by induction. For , we must express all numbers less than  as sums of  or .

Next, assume we can express all integers less than or equal to  using sums or differences of distinct terms from . Clearly, we can express all integers between  and  by simply adding  to every integer that we've already assume we've formed elss than or equal to . However,  by our second lemma, and thus we can subtract every integer less than or equal to  from  and be able to obtain every one of these. But the smallest number in this interval of consecutive integers, , is less than the the largest number in the interval , and thus we can express every integer in the interval  as a finite sum or difference of distinct terms in the sequence , and thus are inductive step is complete.

As a result of our second induction, we know that there is no integer left out, and therefore every integer can be expressed as a finite sum of distinct integral powers of the golden ratio.

[/hide]

You can (must) use negative powers.

[hide]Let t be the golden ratio. We know that t^n = t^(n-1) + t^(n-2) so we can deduce that t^n + t^n = t^(n+1) + t^(n-2). We want to writte an integer k as a sum of powers of t so we write k = k * t^0.
Then every step we erase t^n + t^n and write t^(n+1) + t^(n-2) for an integer n that is repeated in the sum. Every step we reduce the average of the exponents by 1/k. Supose that this algorythm never ends, after 3k^2 steps the average of the exponents will be -3k so the lower exponent may be lower than -3k. Now we divide the integers from -3k to 0 in k groups of three ((-1, -2, -3), (-4, -5, -6)...(-3k+2, -3k+1, -3k)). Before we reach a group we might have been in all the upper groups before, and once we reach a group we can't leave it because t^n leads to t^n-2, t^n-2 leads to t^n-1 and t^n-1 leads to t^n.
Now we have at least one power of t in each group but we have k powers and k groups so there are not two equal powers of t [/hide]

Elemennop, i think "sum" means addition only....otherwise note that the sequence {a_n}n, if n is even, a_n = phi^n + phi^(-n); if n is odd, a_n = phi^n - phi^(-n); then {a_n} gives the Lucas sequence (2, 1, 3, 4, 7, 11, 18...) , then a simple induction solves the problem (this is easy to see because if you can get all numbers from 1-7 as sums of distinct powers, add 11 to each give you 12-18 while still using distinct powers; similarly, adding 7 to 1-4 gives 8-11, so no overlap and every integer is covered.)

Can we just use the fibonacci series general formula, which include $1$ and $2$ thus we are done? :O

[quote=tenniskidperson3]Sir, as an example to prove your ideas, can you show me how to write $10$ as the sum of distinct powers of $\frac{1+\sqrt{5}}{2}$ using the Fibonacci sequence?[/quote]

if golden ratio = t , then 10 = t^4 + t^2 + t^(-2) + t(^-4)

Use fibonacci and Lucas sequences, note that $\phi^n =\frac{f_n+L_n \sqrt{5}}{2}$ (where $f_n$ is the sequency of fibonacci and $L_n$ is Lucas' sequency),  only see a recurrency with the coefficents to prove this

The key is to use the property $\tau^2 = \tau + 1$.

[b][color=#f00]Claim.[/b][/color] Suppose that a positive integer $n$ can be written as the sum of powers of $\tau$. Then it can be written as a sum of powers of $\tau$ with no two exponents consecutive.

[i]Proof.[/i] Consider the largest $\tau^k$ in the expansion of $n$ such that $\tau^k$ and $\tau^{k+1}$ both appear in the expansion. Then, replacing $\tau^k$ and $\tau^{k+1}$ with $\tau^{k+2}$ yields another valid expansion, thus eliminating exactly one consecutive pair. We may repeat this process until no two powers of $\tau$ are consecutive. $\blacksquare$

Now we can induct upwards. Notice that $1 = \tau^0$, so the base case is obvious. For the inductive step, we have three scenarios.

If $\tau^0$ is not in the expansion of $n$, then simply adding $\tau^0$ yields a valid expansion for $n+1$. If $\tau^0$ is in the expansion of $n$ and nor is $\tau^{-2}$, then removing $\tau^0$ and replacing it with $\tau + \tau^{-2}$ yields a valid expansion for $n+1$. (This is because an expansion of $n$ exists without consecutive powers by our first claim.)

Finally, if $\tau^0$ and $\tau^2$ are both present, add $1 = \tau + \tau^{-2} - \tau^0$ to the expansion. Then, replace one of the $\tau^{-2}$ terms with $\tau^{-3}$ and $\tau^{-4}$.

In general, if we have a $2\tau^{-2n}$ term, then we may replace one of the $\tau^{-2n}$ terms with $\tau^{-2n-1} + \tau^{-2n-2}$. We may repeat this process because the presence of a $\tau^{-2n}$ term means that a $\tau^{-2n-1}$ term cannot exist. Let $-2N$ be the least even power in the expansion of $n$. Then, the process must terminate and we have a valid expansion when we reach a $\tau^{-2N}$ term, as needed. Thus, our inductive step is complete, and every positive integer $n$ is representable.


function
trigonometry
real analysis
real analysis solved

For a continuous function $f:\mathbb R\to\mathbb R$, we say that it crosses the axis $Ox$ at $a\in\mathbb R$ if $f(a)=0$, and, in every neighbourhood of $a$, we can find $b,c$ s.t. $f(b)<0,f(c)>0$. Characterize those subsets $M$ of the real line for which there is a continuous function which crosses $Ox$ in every point of $M$.

The problem in the contest asked first to exhibit a function which crosses the axis infinitely many times, and then asked whether it was possible for a function to cross the axis uncountably many times (and they worked with $f:[0,1]\to\mathbb R$, which we can do too, if we want).

I guess $M$ is closed and $\overline{\mathbb R\setminus M}=\mathbb R$.
For each $a\in \mathbb R\setminus M$ define the largest interval $[a_1,a_2]\in \mathbb R\setminus M$, $a\in (a_1,a_2)$, and construct  something like "hat" over $[a_1,a_2]$.
That's all.

Something like that (and that is indeed the answer: closed sets with void interior), but I think that construction requires more work. For instance, you need to choose carefully which hats you take to be negative and which to be positive.

Oh, I disabled images on pages long time ago and sometimes I incorrectly recognize conditions. :(

For reason you pointed we just need to construct something like $x\sin 1/x$ on the endpoints $a_1$ and $a_2$. Then everything is ok.


If $ f: M\to N$ is a smooth map and at $ p\in M$ rank$ (f) \equal{}$dim$ N$,show that
there is a cordinate chart at $ p$  such that rank of $ f$ is constant whole of this cordinate chart.

Do you have some sort of "Rank Theorem" to work with, or must this be done by scratch. You have a constant rank smooth map whose differential is surjective, so this is a submersion. Do you have any theorems about submersions to work with?

When I use this result, I usually just call these "submersion charts" and consider it a consequence of the Rank Theorem.


This is a question on one of my textbooks:

Why can't Aluminium be produced through the electrolysis of an acid & an aqueous solution of Al 3+ ions?

I think the answer they are looking for is that the cathodic deposition of Aluminium would be limited by the electrolytic decomposition of water.

Is my answer correct?

Also, is the question correct?

Because I think that Aluminium CAN be produced by this method, just not very much of it.

Thanks everyone.

Dave.  :)

yes, it is correct.
it is all because of electro potential differences.$E^{\circ}$

Thanks amirhtlusa,

I obviously need to study a bit harder!

Thanks again mate.

Dave.  :D

Overvoltage also comes into play; I'm not exactly sure how, for Aluminum.


geometry
inequalities
quadratics
number theory
IMO Shortlist
algebra
quadratic formula

[url=http://www.artofproblemsolving.com/Forum/download/file.php?id=26628]IMO 2008 Shortlist Download (.pdf file)[/url]

[size=150][b]Algebra[/b][/size]

[url=http://www.artofproblemsolving.com/viewtopic.php?p=1191683#1191683]Problem A1[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1190551#1190551]Problem A2[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287852]Problem A3[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287853]Problem A4[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287855]Problem A5[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287857]Problem A6[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287856]Problem A7[/url]

[size=150][b]Combinatorics[/b][/size]

[url=http://www.artofproblemsolving.com/viewtopic.php?t=287859]Problem C1[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1472110#1472110]Problem C2[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287860]Problem C3[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1191679#p1191679]Problem C4[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287861]Problem C5[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287863]Problem C6[/url]

[size=150][b]Geometry[/b][/size]

[url=http://www.artofproblemsolving.com/viewtopic.php?p=1190553#p1190553]Problem G1[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287864]Problem G2[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287865]Problem G3[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287866]Problem G4[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287867]Problem G5[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287868]Problem G6[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1191671#p1191671]Problem G7[/url]

[size=150][b]Number Theory[/b][/size]

[url=http://www.artofproblemsolving.com/viewtopic.php?t=287869]Problem N1[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287870]Problem N2[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287871]Problem N3[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287872]Problem N4[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?t=287873]Problem N5[/url]
[url=http://www.artofproblemsolving.com/viewtopic.php?p=1190546#p1190546]Problem N6[/url]

Title says it all. Post comments on problems that have appeared on IMO Shortlist 2008. If [url=http://www.mathlinks.ro/viewtopic.php?t=156803]the discussion of 2006 ISL[/url] could be done, why not also 2008?

http://www.mathlinks.ro/viewtopic.php?t=287876
This is the link where you can find the problems.

I'll post some more comments later, for now just:
I like N3, G1, G4, C2, A1, A5 and A7.

I haven't tried to solve all of them, but I like the number theory problems. The algebra problems are quite easy though. :D

I'll post my comments on the most of:

A1: IMO4, nice and (almost) easy problem, good choice
A2: IMO2, I don't like it at all, the inequality is solved in almost every way you could think of, and b) is straightfoward after inequality
A4: Ugly
A5: Very nice inequality!
A7: Another nice inequality, but too standard

C1: Good, standard problem
C2: Nice one, includes divisibility also
C4: IMO5, this is a standard, and good problem (although not the best combinatorics)

G1: IMO1, nice problem, and very easy - perfect no. 1
G2,3: Also computational problems, don't like'em so
G4: Very nice!
G7: IMO6, very nice, but very hard

N1: Easy, straightfoward, and it would also be a good 1/4
N2, N3: both nice problems
N6: Ugly, very ugly, and familiar

If I could choose the IMO problems, I would take out 2. and 3.

Algebra part was nice, Combinatorics could be better(I love the Combinatorics part in ISL'07), Geometry was OK, and NT was the best probably.

[b]A[/b]
1 nice choice for problem 4
2 I have yet to see a solution of this that does not involve expanding or the quadratic formula, easier (in my opinion) than A1
5 splendid, although AM-GM kills it easily  :D
6 nice
7 nice... but not for problem 3/6

[b]C[/b]
1 nice
4 an okay problem, though the use of "turning on and off" is a little turn-off  :D
6 a better choice for problem 3

[b]G[/b]
1 too easy if you know radical axis...
7 good choice for problem 6

[b]N[/b]
1-4 nice, if only at least one of them were chosen...
7 messy

I'm basing this mostly on beauty... :lol:

Personally, I think

A1. simply nice but not so challenging.
A2. people say that this is easier than A1 but in fact i solved A1 more easily   :)  in fact, part b is so stupid but if they only pick part a, it'll be more stupid.
A3. nice but doesn't give any surprise.
A4. i think there must be nicer problem than this! i found nothing interesting of problem. even if someone solve this problem he doesn't need MUCH inspiration. just sit down and break this problem down.
A5. wew, very nice and challenging! need no special technique but algebra creativity!
A6. aw, very good-looking but i can't solve to the end.
A7. no idea.

C1. standard but it is fun.
C2. very very easy, no need much inspiration.
C3. no idea
C4. another easy but nice problem. we need an upgrade!
C5. no idea
C6. wew

G1. nice and beautiful. but i think this is not a good problem for imo because when you find "something" once, you tackle the problem! so i think this is too simple.
G2. just like G1, once you find "something" you tackle the problem. i get bored!
G3. bery nice!
G4. this is what i call a beautiful liar. very beautiful.
G5. gee! i hate this combinatorial-look... considering all points on plane? blah.
G6. there are two things to be proved here, one of them is useless and the other one is very crazy. too far...
G7. no idea

N1. bad, cases, need to be careful.
N2. bery nice. the observation is standard and some ideas really work!
N3. the face is innocent, i love it, but unfortunately i failed to solve this one and read the solution.
N4. seems familiar so i just open the solution. crazy.i don't like this one.
N5. very nice property, but i haven't try to tackle this one.
N6. nice, fun, familiar.


geometry
trapezoid
analytic geometry
graphing lines
slope

Prove that if the sides of trapezoid ABCD (with AB parallel to CD) satisfy the condition AC squared + BD squared = (AB+CD)^2, then ABCD is orthodiagonal.

[quote="mr. math"]Prove that if the sides of trapezoid ABCD (with AB parallel to CD) satisfy the condition AC squared + BD squared = (AB+CD)^2, then ABCD is orthodiagonal (ABCD has diagonals that are perpendicular to each other).[/quote]

My solution:
[hide]
Let the following points be placed on the xy-plane.

Let A = (a,e)
Let B = (b,e)
Let C = (0,0)
Let D = (d,0)

The condition AC squared + BD squared = (AB+CD)^2 translates to:

$ [a^2 \plus{} e^2] \plus{} [(b \minus{} d)^2 \plus{} e^2] \equal{} [(a \minus{} b) \plus{} d]^2$

$ a^2 \plus{} e^2 \plus{} b^2 \minus{} 2bd \plus{} d^2 \plus{} e^2 \equal{} a^2 \minus{} 2ab \plus{} b^2 \plus{}2(a \minus{} b)d \plus{} d^2$

$ 2e^2 \minus{} 2bd \equal{} \minus{}2ab \plus{} 2ad \minus{} 2bd$

$ 2e^2 \equal{} \minus{}2ab \plus{} 2ad$

$ e^2 \equal{} \minus{}ab \plus{} ad$

$ (e)(e) \equal{} \minus{}1(a)(b \minus{} d)$.

$ (\frac{e}{a})(\frac{e}{b \minus{} d}) \equal{} \minus{}1$.

The slopes of AC and BD are worked out below**

Because the product of the two slopes is negative one,
AC and BD are perpendicular to each other.

So ABCD is orthodiagonal.

 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

**The slope of $ AC \equal{} \frac {e \minus{} 0}{a \minus{} 0} \equal{} \frac{e}{a}$.

The slope of $ BD \equal{} \frac{e \minus{} 0}{b \minus{} d} \equal{} \frac {e}{b \minus{} d}$.

[/hide]

Draw line $ CK \parallel{} BD$. (K is on AB).


geometry
trapezoid
angle bisector
geometry unsolved

Two circles are tangent externaly at a point $B$. A line tangent to one of the circles at a point $A$ intersects the other circle at points $C$ and $D$. Show that $A$ is equidistant to the lines $BC$ and $BD$.

Let $E,F$ be points on $BD,BC$ respectively such that $AE \perp BD$ and $AF \perp BC$. As AD is a tangent and B is an external tangential point, so \[ \angle ABF = \angle ABG + \angle CBG = \angle BAD + \angle BDA = 180^{\circ} - \angle ABD = \angle ABE.  \] Therefore $\triangle ABE \equiv \triangle ABF$ and hence $AE=AF$.

I will solve this nice problem by inversion.

Consider an inversion with center $A$ and an arbitrary radius. and suppose points $C,B,D$ changes to

$C',B',D'$ respectively. It is clear that the problem changes to this one:

there are two parallel lines $L_1, L_2$, a circle is tangent to $L_1$ at $B'$, and intersects$L_2$ at $C',D'$

prove that: circum circles of $AB'D',AB'C'$ have equal diameters.

To prove this easy fact, u should only know that a trapezoid can inscribed in a circle if and only if it is isosceles.

Ok………………………………………………………………………………………………………………….

Draw the circum circle of $AC'B'$, and let it intersect $L_1$ at $E'$.

Then$AC'B'E'$ should be an isosceles trapezoid.and because of $C'B'D'$ is an isosceles triangle,

$AE$ is parallel to $BD$.

And it shows us that $AEBD$ is parallelogram. Then triangles$AEB,ABD$ are equal, and we can conclude

That their circum circles have equal diameters too.

This problem is easy and nice :∠ABC=∠CMB+∠BCM=∠NBM+∠BDC=∠ABP+∠BDA;Because l and AC are tangent lines to Γ1;so PA=PB;PAB=<PBA;SO :∠ABC=∠PAB+∠BDA=∠HBA:So, BA is the angle bisector of <HBG; SO:AH=GA;---Q.E.D :lol:

Let $ BD$ intersect the first circle again at $ E$.

$ \angle ABC \equal{} \angle AED \plus{} \angle ADE \equal{} 180^o \minus{} \angle EAD \equal{} \angle EBA$. Thus $ AB$ is the bisector of $ EBC$ and the result follows.

Define $E = BB \cap AC$ and let $BD$ meet the circle containing $A$ again at $F$. Since $EA = EB$, we have $$\angle ABC = \angle ABE + \angle EBC = \angle BAE + \angle BDC = \angle BAD + \angle BDA = 180^{\circ} - \angle ABD = \angle ABF$$ so $A$ lies on the external bisector of $\angle CBD$, which finishes. $\blacksquare$


I have found difficulties in solving the following problem. :)
$ \{X_{n}\}$ is a Markov chain which is irreducible, with invariant distribution $ \pi$. Show that,\[ P(\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k\equal{}0}^{n\minus{}1}\textbf{1}_{\{X_k\equal{}i,X_{k\plus{}1}\equal{}j\}}\equal{}\pi _{i}p_{ij})\equal{}1\]

what if you consider the Markov chain $ Y_n\equal{}(X_{n\minus{}1}, X_n)$ ?


I'm trying to upload my computer documents (wps., rtf., .pdf) into my Gmail because my computer has been recently attacked by Trojan viruses, Spyware, and whatever else that people like to make. But I uploaded many files and came across this error for last two:

Ooops. The system was unable to perform your operation (error code 008). Please try again in few seconds.

Okay well I've tried in few seconds, minutes, hours, and even days and it still doesn't work. Anyone has idea what's wrong?

[url]https://mail.google.com/support/bin/static.py?page=troubleshooter.cs&problem=bugflow&selected=send6[/url]

But I had this error message when I tried to save my files that were 10+ MB, which is over the limit.


Where can I download a free version of Gabri geometry? thanks for help

I'm not sure what Cabri geometry is.  I have it on my TI-89 though. :?


geometry
trigonometry
trig identities
Law of Cosines

In triangle ABC, side a = 2 radical 5, angle A = 45 degrees, and b + c = 6 + 2 radical 2. Find the exact value of the area of triangle ABC without finding the exact values of b and c.

[hide="hint"]
Law of Cosines, then square the value of b+c
Manipulate the equation to find the value of bc
[/hide]

EDIT: more hints

[hide="bigger hint"]
Use heron's, but keep semiperimeter as $\frac{b+c+a}{2}$
For the $\frac{b-c+a}{2}$ and $\frac{-b+c+a}{2}$ terms,
[/hide]
[hide="EVEN bigger hint"]
change them to $\frac{a}{2}-\frac{b-c}{2}$ and $\frac{a}{2}+\frac{b-c}{2}$
[/hide]
[hide="This is turning into a solution"]
When you multiply the 2 terms from the previous hint it becomes
$\frac{a^{2}}{4}-\frac{(b-c)^{2}}{4}$
and $(b-c)^{2}=(b+c)^{2}-4bc$
[/hide]
I'm just too lazy to do everything.

Oh wow, I forgot that you could use sines to find the area of a triangle.


Let $0<u_1<u_2<....<u_n<.....$such that:$ \displaystyle\sum_{i=1}^{\infty}\frac {1}{u_i}$ is bounded.Let f(x) denote the number of pairs of integers (i,j) for which : $\sum_{i=1}^j {u_k}< x$
Prove that: $\displaystyle\lim_{x\to\infty}\frac{f(x)}{x}=0 $

...beautiful! if i see it correctly this is actually the best possible bound for $f(x)$!

Peter

Do you mean

"the pairs $(i,j)$ for which $\sum_{k=i}^j u_k<x$" ?

Yes,I do. "The pairs $ (i,j)$ for which : $\sum_{k=i}^j{u_k}< x$" is right.I'm sorry! ;)


if $ 39p\plus{}1$ is a perfect square then find the p (where p is prime number).

Here is my solution:
$ 39p \plus{} 1 \equal{} k^2$
$ 39p \equal{} (k \minus{} 1)(k \plus{} 1)$
so with the last condition p is 41 or 37.
$ 39p \equal{} 3*13p$
and with the last equation p equals to 5 (since 5*3=15 and since 11/3 is not prime)


One of the factors of $ x^4\plus{}2x^2\plus{}9$ is:

$ \textbf{(A)}\ x^2\plus{}3 \qquad
\textbf{(B)}\ x\plus{}1 \qquad
\textbf{(C)}\ x^2\minus{}3 \qquad
\textbf{(D)}\ x^2\minus{}2x\minus{}3 \qquad
\textbf{(E)}\ \text{none of these}$


There are 2009 boxes numbered from 1 to 2009, some of which contain stones. Two players, $ A$ and $ B$, play alternately, starting with $ A$. A move consists in selecting a non-empty box $ i$, taking one or more stones from that box and putting them in box $ i \plus{} 1$. If $ i \equal{} 2009$, the selected stones are eliminated. The player who removes the last stone wins
a) If there are 2009 stones in the box 2 and the others are empty, find a winning strategy for either player.
b) If there is exactly one stone in each box, find a winning strategy for either player.

An odd box is, for example, the box 17, the box 35,...

For the $ a$ part, the player $ B$ win because he allways can  play such that all the stones, after he play,  will be in even boxes, and then the player $ A$ cannot win, and so like the game is finite, the player $ B$ must win.

For the $ b$ part, I use the same strategy, it is that the player who can play such that, after the other player plays, the number of boxes odd and nonempty will be odd, and then is at least one, so he cannot lost. I will prove that the player $ A$ win.
Let $ k$ be the number of odd boxes wich ar nonempty in any moment.
In the begining of the game there are 1005 boxes that are nonempty and odd. The first play of the player $ A$ is select one stone of the box 2009. Now $ k$ is even. Then the player B can play of two ways:
1- If he select $ x$ stones from the box $ 2y$, then in the next play the player $ A$ select $ x$ stones from the box $ 2y \plus{} 1$ (it is easy to see that it allways is possible).
2- If he select $ 1$ stone from the box $ 2y \plus{} 1$, then in the next play the player $ A$ select $ 1$ stone from the box $ 2z \plus{} 1$.
We see that like in the begining there is one stone in each box, after $ B$ plays, the number $ k$ is allways odd if $ B$ played 2 (so $ z$ allways exist).
After $ A$ plays the number $ k$ is even and in each odd box the number of stones is 0 or 1. So the player $ A$ allways can play, and then he must win.

[quote="gilcu3"]
Let $ k$ be the number of odd boxes wich ar nonempty in any moment.
In the begining of the game there are 1005 boxes that are nonempty and odd. The first play of the player $ A$ is select one stone of the box 2009. Now $ k$ is even. Then the player B can play of two ways:
1- If he select $ x$ stones from the box $ 2y$, then in the next play the player $ A$ select $ x$ stones from the box $ 2y \plus{} 1$ (it is easy to see that it allways is possible).[/quote]if the first play of B is 1., then k is even after the play of A   :|

Yes, I proved that after each play of $ A$, the number $ k$ is even.

Let a stone in box $i$ have value $i$ and let the sum of all the values of the stones be denoted as $E$ (assume that there is some box of value 2010 for which no stone can be removed). Notice that after each move, $E$ increases by $1$ and thus changes parity.

a) Notice that $E$ is initially even. Thus when $A$ operates, $E$ goes from even to odd and $B$ goes from odd to even. Clearly, $A$ cannot perform the operation of putting the last stone in box 2010 since such a move would be odd to even.

b) Notice that the initial $E$ is now odd, thus when $A$ operates, $E$ goes from odd to even and $E$ goes from even to odd. Making $A$ win this time.


I don't ask you, if you found a satellite photo of yourself, this is nearly impossible so far :D But did someone find such pic of her/his home at Google Maps/MSN Virtual Earth?

Above my home is a large cloud so far  :(

The addresses for Google Earth are a bit off near where I live. And they're at least two years old. Good resolution, at least.
Some areas like the US Capitol are purposely blotted out, for security I think.

[quote="mathclass"]I can't use Google Earth on my creaky Windows 98 machine :([/quote]
It doesn't work on my slow proc too  :( So I'm forced to use [url=http://maps.google.com/]Google Maps[/url]

yep, i can see my house on google maps. i guess 25mi out of DC is far enough for it to not be blotted out

On the free version of Google Earth, the non densely populated areas. (such as maybe places in Nevada) don't have as good as a resolution as those places that are densely populated. I live in Metro-Detroit, so I could see my home.  :)

That thing is so cool!
i saw my home and a bunch of other places.

there are too many programs on my computer for math stuff my parents use so i cant download like anything so i use maps

google earth is pretty cool. i live just a little too far in the country to get a close picture of my house

[quote="zeb"][quote="noneoftheabove"]google maps is messed up when i searched my house they gave me like a middle school[/quote]
Does this mean that you live in a middle school?[/quote]
no, it means the the google systme isn't perfect. it also messes up my house by putting it down the road half a mile. i guess they have some fine tuning to due
and for some reason i want to use this emoticon:  :arrow:

I have been using google earth since it was in beta. The addresses are a bit off for individual homes. Large buildings, such as schools and stadiums, are mostly accurate. Google earth now has new imaging options for areas devistated by hurricane Katrina. Go to the google homepage and click on the katrina link. then go to the google earth link on that page. It has all the instructions. Washington D.C. is not blocked. It is in fact included in the google earth 3D buildings option. I flew around mount everest and went to moscow yesterday. If anyone knows the lattitude and longitude of area 51/ groom lake air force base please post it. It doesn't register on google earth.


Princeton college Harvard Yale MIT inequalities geometry

Well, you got into Princeton. Congratulations! 9.25%, eh? Not as low as Harvard and Yale, but that's pretty intense.

So, you've probably gotten tons of info from Princeton telling you about Princeton Preview. I'm not going to tell you anything new about that. I just want to say that [b]if you are admitted to Princeton and to another school, going to the prospective student event (for each school) is a good idea![/b]

You can read what you want online. You can look at the US News rankings. You can see which one gives you more hits on Google and you can look at the school websites and see what there is to see. But you're not going to get an idea of what the school is actually like from these things. If you go to the hosting event, you can meet lots of different students, all doing different things, and see what they have to say about the school. You can see the sort of activities that are generally available during the year. You can see how students eat and how good the food is, you can see what to do off campus, you can see how the place looks and how easy it is to get around and how nice the dorms are.

Do keep in mind, though, that these programs are meant to impress you! Princeton Preview is not "a day in the life of a Princeton student." It is, however, something that I recommend for you to experience if you are still not sure that you actually want to go to Princeton.

Princeton Preview is April 10-12 and 17-19. It's possible to request a particular host, so if you want to be hosted by a Princeton AoPSer, post here! I will be hosting for both weekends, and I also know a number of other AoPSers at Princeton (maybe even including a girl!) who will be hosting as well.

I'm also happy to answer any questions about Princeton. Post here or PM me.

Although I'm already a student at Hopkins and I never applied to Princeton, I really envy those who got in and it is my goal to go there for grad school.  Congrats to those who got in.

Aah, I hate how it conflicts with MIT CPW.

Here's what I'd most like to know:  [i]what kind of people go to Princeton?[/i]  (Especially as opposed to, say, Harvard.)  Describe some of the most interesting people you know for me.  I realize this is highly subjective, but it's the only kind of information I'm interested in at this point.

Yeah, alot of the preview conflict with each other.  :|  There are other opportunities for similiar programs, right?

I agree with t0rajir0u , describe some interesting people.  :D  Though I do have to say you are very interesting yourself, Xevarion.

[quote="t0rajir0u"]Aah, I hate how it conflicts with MIT CPW.[/quote]
That's part of the reason that our admissions office decided to offer two different times for Princeton Preview this year. If you want to go to MIT's, you can now go to Princeton's too! (Unless you have to do something else on the other Princeton date, and in that case,  :( )

[quote="t0rajir0u"]Here's what I'd most like to know:  [i]what kind of people go to Princeton?[/i]  (Especially as opposed to, say, Harvard.)  Describe some of the most interesting people you know for me.  I realize this is highly subjective, but it's the only kind of information I'm interested in at this point.[/quote]
This is a great question. Unfortunately, as I am only a student at Princeton, I can't say anything at all about what it's like [i]relative to somewhere else[/i]. (Be wary of anyone who says something like "A is better than B, and here's why:" -- most likely they are actually only familiar with A!)

I'll tell you about some people I know. I'm going to keep it (to some extent) anonymous; some of these people are close friends of mine, some of them are friends but not very close, and some are people I hardly know. Note that these people are NOT going to be representative of the "typical" Princeton student.

[hide="One"] This guy went to high school in New Jersey. (NJ is home for more Princeton students than any other state.) He is majoring in the Woodrow Wilson school -- it's like a souped-up politics degree. He's interesting because he likes building and maintaining organizations. It's kind of a strange thing to enjoy, wouldn't you think? He likes to think of more efficient ways to run things. He's always coming up with ideas for how the university or a residential college or some student group could accomplish one of its goals more effectively. He is involved in lots of student groups, but unlike most people, it's not just because he likes the thing that the student group is about. It's because he likes to help run the organization! He'll probably end up working for an NGO eventually, but he wants to go to graduate school first.

This guy is also interesting because he is very efficiency-minded. He is an RCA (residential advisors) and he teaches his advisees never to leave a free food event unpillaged. (Unpillaged should totally be a word. "The Vikings left no town unpillaged!") He goes on "scavenging trips" after conferences to see if anything good has been left. He's also kind of OCD -- he gets really mad if anyone messes with his stuff, or even goes into his room without permission when he's not around, and he is paranoid about cleanliness. [/hide]

[hide="Two"] This guy is Chinese, but he spent the last few years of high school in London living with a host family. So he has an English accent, even though he grew up in China! He is also very interesting because he plans to major in math, but his alternate interest is theater. You don't see math majors acting much, and you don't see many Chinese acting either (at least in the USA), so he's quite unique! In addition, he also likes ballroom dance! Last I heard, he was planning on starting a Chinese theater group on campus -- they'll actually perform traditional Chinese works! It sounds pretty cool, although I'll have no idea what's going on unless they provide pretty detailed programs.

This guy also interesting because he's such a great entertainer. He's not at all a typical math major. He's always full of energy. He is great at impersonating people, he's very passionate and melodramatic about everything, and he comes up with crazy and hilarious ideas (e.g. rewriting The Phantom of the Opera as The Phantom of Fine Hall (the math dept building at princeton), a story about a crazy old mathematician who visits some undergrad, and having the Math Club perform it). He's also very into sharing Chinese culture with other people, which I think is admirable -- there are plenty of Chinese people at Princeton, but very few of them are actively trying to show everyone else the greatness of the art, ideas, etc. from China. [/hide]

[hide="Three"] Okay, so I talked about some people who break stereotypes, how about someone who fits them? This guy is really socially awkward, but he's very nice (isn't this the stereotypical math nerd? poor guy who's totally awkward and doesn't interact well with people, but he's actually a great guy when you get to know him). Anyway, he's brilliant at math and physics, but he has really idiosyncratic ways of thinking about things that are hard to understand. I used to give him inequality problems (back when I still did olympiad inequalities) and he would always solve them using some weird variational methods that I couldn't understand.

He loves playing games. I mean [i]loves[/i]. He is extremely good at DDR and Super Smash Brothers and some other games of that type. But he is also pretty good at bridge and chess, frighteningly good at SET, and plays a lot of other games of skill as well. In fact, he has done some research about the game SET and I think he even published a paper about it! Hardcore!

He is also interesting because he makes a really valiant effort to get to know people. I often feel ashamed when I'm around him, because I am often too lazy to try to learn people's names, and he's really good about it. He also tries to break out of his stereotypical role as the loner math geek by meeting new people and trying new stuff all the time. [/hide]

[hide="Four"] Okay, this guy is way more interesting than any of the previous people. He's also a math major. (Hmm, I guess even when I talk about people I know that I am not close friends with, most of them do math or physics.) He likes to explore. He likes to get into places. If he were at MIT, he probably would be the culprit behind all of the craziest and most daring hacks. Unfortunately, there's no hacking culture at Princeton, so he has to do everything himself. I've seen him climbing random buildings. He showed me how to get on top of the roof of the math building one time (it requires going through a trap door in the ceiling!), and we've talked about getting onto the roof of other buildings too.

He's also responsible for the most hilarious prank I've ever heard of at Princeton. (Well, the [url=http://en.wikipedia.org/wiki/Princeton_Branch#The_Great_Dinky_Robbery]Great Dinky Robbery[/url] might be more hilarious, but at least in recent memory his is the best.) So, there's one street where almost all of the eating clubs are located. This is where most upperclassmen eat, and it's the social center of Princeton. Most of the dancing, partying, drinking, and other social activity goes on here. The street is called Prospect Avenue, but everyone refers to it as "The Street." So, one time this guy got some green and white construction paper, and he changed the street sign from "Prospect Ave" to "The Street". Classic. It was more than a week before Public Safety noticed and fixed the sign. [/hide]

[hide="Five"] Well, I have talked about only guys, so how about a girl? I don't know her that well, so I don't have too much to say about her. She is from Singapore. She is an evangelical Christian. She studies psychology. The thing that's interesting about her is that she listens to industrial music and likes to cosplay! (I hope cosplay doesn't carry a negative connotation for you; she isn't one of those weird types who lives their lives wishing they were characters in their favorite anime.) I have gone to a couple of shows with her. It's really quite unexpected -- you meet her and she is a quiet little Chinese girl. Then you go to her facebook page and the picture is of her in a very professional-looking cosplay type costume. Then you go to an Assemblage 23 show with her and she is singing along and headbanging. She also introduced me to Eisbrecher, who are now one of my favorite bands. [/hide]

Well, I'm getting tired and don't want to think of more people to write about right now. Also, I've written a lot and I have no idea if this was the kind of thing you were looking for. So I'm just going to finish off with...

[hide="Some anecdotes"] One time, I was in the Math Dept common room playing bridge with some people, and a couple of the math grad students had a Rubik's cube and they had written down the group of motions of the Rubik's cube and the group elements that were side rotations for each side, and they were trying to use group theory to solve it! I also heard that they wrote a program to play [url=http://en.wikipedia.org/wiki/Phutball]Conway Football[/url] and that they've published at least one paper about it. The math grad students are awesome.

A lot of people play pick-up basketball in the gym. I used to play with some chem grad students, and occasionally with math grad students. My roommate used to play with some professors.

One time I saw a guy biking through campus with a cat sitting on his shoulder. (This is possibly the most bizarre and amazing thing I've ever seen.)

There's a student who goes around on a unicycle from time to time. I don't know him. Princeton's on a hill -- how do you control a unicycle on a hill?  :o

One time a guy I know kind of pressured me and a couple of my friends to go to The Street with him. We played a few games of beer pong, and ended the guy's 6 month long winning streak (none of us had ever played before). The guy was really mad. (I've never done it again, so my perfect 3-0 record still stands.)

Somebody has a tight rope. I often see it strung up between two trees. The people who own it are pretty good by now, and sometimes you can see them teaching people how to do it. It looks really hard to me, but I bet it's great way to improve your balance. I don't know if they teach random people; I've never talked to them.[/hide]

I hope some of this was, uh, interesting. :)

Indeed interesting...thanks for the info and help!

I will be there on the 17th-19th weekend, if people want to meet up...

Congratulations to you all who got into Princeton  :)

I am not sure if this is relevant as you make your choice, but I would just like to say that I am really amazed by the quality of the math department. The profs are really eager to help you and work with you. Courses are really great, though quite difficult. In every course I have taken so far (which is only 3...) the prof made an effort to go beyond the textbook into interesting topics and cool stuff. People who take the courses have varied levels of experience, but everyone is intelligent and the learning environment features multiple levels of support (p-set groups, problem sessions, prof office hours, grad students, etc.) so everyone catches up quickly.

Non-math courses are also pretty amazing and quite rewarding, especially if you choose to put lots of time into them. There are lots of distribution requirements, but no worries, they are quite interesting (and if you really want to do lots of math, many of the math courses can be difficult enough that they take up >50% of your workload even if you're doing only one math course).

If you have any general or specific questions, post them here or pm me and I'll be happy to answer them.

[quote="Xevarion"]One time a guy I know kind of pressured me and a couple of my friends to go to The Street with him. We played a few games of beer pong, and ended the guy's 6 month long winning streak (none of us had ever played before). The guy was really mad. (I've never done it again, so my perfect 3-0 record still stands.) [/quote]

lol, there are many things that could be said about a 6 month long beer pong winning streak :D

[quote="Xevarion"]If he were at MIT, he probably would be the culprit behind all of the craziest and most daring hacks.[/quote]

heh, perhaps.   I personally value daring, exciting adventures pretty highly.  Being at MIT has made getting onto a roof and changing a sign almost commonplace.  It's interesting to consider the vast cultural difference in the hacking aspect.

[quote="Xevarion"]We played a few games of beer pong, and ended the guy's 6 month long winning streak (none of us had ever played before). The guy was really mad. (I've never done it again, so my perfect 3-0 record still stands.) [/quote]The first time I ever played beer-pong I went undefeated 6-0 and the keg ran out :)

[quote="stupidkid"]Meh, I'm kind of disappointed by my Princeton fin aid. It's about $ \$$6k/yr less than Harvard's.[/quote]
Really? Do you mean you'd be paying 6K more per year to go to Princeton, or do you mean Harvard is giving you 6K more? Remember that these universities don't give awards based on tuition, they give awards based on "estimated financial need", which includes tuition, room & board, allowances for travel and books, fees, etc. So if Harvard gave you 6K more, it may be because Harvard estimates that your total financial need is more, or they give allowance for more stuff. (If you're flat out paying 6K more for Princeton, then you're probably in a high income bracket, and thus benefiting from the new Harvard "give lots of money even to pretty high income upper-middle-class families" policy.)

I've never understood complaints about financial aid. These universities are incredibly generous. They are pretty good at ensuring that no one will have to take out loans in order to go to their school. So what's the big deal? If you can afford to go to one of the top universities in the world, and you won't even have to be in debt as a result, you should choose what you would enjoy, learn the most at, whatever! Your education is worth FAR more than 24,000. Do what you want. (I also find it quite depressing that some of my friends are doing advanced standing and spending only 3 years at Princeton for financial reasons.)

Seriously, if money is a problem, talk to the financial aid office! They're open to negotiation.

If only 9.25 % get in, how big are their math programs? How come so many students apply for math? Is mathematics popular in America, or are there simply not a lot of universities? In Sweden, there are almost no students in mathematics, so search pressure is non-existant. At my university, a majority of the students at higher level classes in mathematics are exchange students

[quote="Kalle"]If only 9.25 % get in, how big are their math programs?[/quote]
I don't understand your question. In the USA, people don't apply to university for a specific subject. At Princeton, you don't specify your subject for at least 1 year, 2 if you don't do engineering. The 9.25% is overall, not for math.

The Math Dept here is actually very small. A lot of people come thinking they'll do math and then switch because it's too hard, too abstract (the applied math dept is small and still pretty theoretical), or because they think they can get a better salary after graduation if they take a different major. Typically, there have been 15-20 math majors each year; my year (I'm 3rd year) there are just over 30.

I mean that I have to pay $ \$$6k more per year for Princeton. Is it possible that Princeton's financial aid office will match Harvard's fin aid?

I'm kind of curious, Pton accepted 1976 applicants this year but will only have room for about 1250. So they're expecting a matriculation rate of 63%?? That seems kinda low.

Yeah, matriculation has been getting lower at ivies because they are accepting a lot of the same people. That's why the overadmit. (Also, they probably don't know how many people have Princeton as first choice, due to no more ED.) But generally they actually try to make low estimates for that -- it's worse to overfill than to underfill, because if they underfill they can just let in waitlisted students.

I don't know if you can get the FinAid office to match Harvard's aid, but you can try. I have definitely heard of them upping people's aid either due to competing offers or extenuating circumstances, but not by that much. Good luck with that!

Xevarion, you should introduce me to student #4 sometime.

[quote]The Math Dept here is actually very small. A lot of people come thinking they'll do math and then switch because it's too hard, too abstract (the applied math dept is small and still pretty theoretical)...Typically, there have been 15-20 math majors each year; my year (I'm 3rd year) there are just over 30.[/quote]

Do people come wanting to do math and get "forced out" because it's too hard? I'm pretty sure about wanting to do math and I think I've had enough experience with "real math" (research, summer stuff, self study, etc.) to know that I actually do want to be a math major, so I'm sort of worried that I won't be able to hack it. I'm not one of the crazy USAMO/IMO contest people that frequent AoPS.

[quote="piper"]Do people come wanting to do math and get "forced out" because it's too hard? I'm pretty sure about wanting to do math and I think I've had enough experience with "real math" (research, summer stuff, self study, etc.) to know that I actually do want to be a math major, so I'm sort of worried that I won't be able to hack it. I'm not one of the crazy USAMO/IMO contest people that frequent AoPS.[/quote]
Nobody gets "forced" out. If you really want to do math, you can do it. It just might take a lot of work to stay afloat.

Also, it sounds like you'll be okay. The sort of experience you have with math is exactly the kind of thing you need to be able to survive a rigorous math program. I never even qualified for USAMO when I was in high school, and I'm doing fine in the math dept here.

The people who leave the dept because it's too hard are either not willing to spend enough time doing math, or they discover that what they thought was "math" in high school is quite different from what math actually is.

This did happen to a few students when I was there (mid 90s), but I'd say they were students who, for the most part, weren't [i]completely[/i] committed to being math majors at the outset. They were students who may have been wavering between math and some other subject (usually science, engineering, or comp sci) and chose the other subject when they either didn't like or had difficulty with the math department's intro courses.

When I was there, pretty much the only way to become a math major was to do MAT 217/218 that first year, which was difficult for students who hadn't had multivariable calculus as well as an introduction to proof-based mathematics already. Those classes started with about 40 students and maybe 10 became math majors (plus 2 more who came from MAT 203/204 and 3 who started in 300 or 400 level courses).

Now they have more points of entry into the math department. In particular they have courses that introduce you to proof-based mathematics (MAT 214-215). They also have more options for what you can take once you're in the department, so you have more opportunity to choose classes that fit your interests and abilities.

Most of the math majors at Princeton are not former Olympiad contestants. Some are, but certainly not the majority. And at least while I was there, the best students weren't necessarily the former Olympians. They were generally students who had taken some college level courses or had done some research before they got there. They were also the ones who were extremely tenacious workers!

The most important thing (and this is true no matter where you go to school) is that you don't try to compare yourself to other students. Just stay focused on your own learning and growth, and the success will follow. I'll add that I also never made it to USAMO in high school and did fine in the math department.

The idea of being "forced out" simply arises because people come into the "intro" math courses with significantly different levels of experience. You're bound to be much less experienced than some people as well as much more experienced than others. The most important thing to keep in mind is that the less experienced people really do catch up if they just stay in the math class. The truth is, an IMO gold medal alone will not give you an A, while hard work will pay off. And I am not talking of ridiculous amounts of hard work: maybe 5-10 hrs a week on average (this might be more than for some of the other classes you would take here). An important observation is that you'll not catch up if you alienate yourself from others. The support system is quite extensive here, but you have to make an effort to use it.

Judging from what you describe as your experience, I doubt you'll need to do much catching up.

I was one of the people in the middle of the pack when I came, but I went to a summer program which taught me that it's possible to learn things when there are people smarter than myself (i.e. taught me to compete against myself and not others).

Another thing to keep in mind is that for USAMO problems, you usually have to figure out the thing that "makes it tick," so you sometimes need to have incredible intuition so that you are able to start them in the right way and solve them in a couple of hours. In the math you do here, you have to build up concepts, generalize, solve "longer" problems which have the solution outlined over multiple exercises because you'd never be able to solve them otherwise, etc. all done in basically unlimited time. In other words, being an IMOer is not required; hard work is required.


geometry
analytic geometry
number theory unsolved
number theory

Given a triangle with sides $ a,b,c$ and area $ K$, where $ a^2,b^2,c^2,2K$ are integers and $ \gcd (a^2,b^2,c^2)\equal{}1$. Prove that the triangle is congruent to a triangle in the plane whose vertices are lattice points.

It is not difficult to prove that a triangle is $ \mathbb{Z}^2$-inscriptible if and only if the $ tan(\theta)$ is rational for each angle $ \theta \neq \frac {\pi}2$ of the triangle.

Pierre.

Thanks for your help, but what is $ \mathbb{Z}^2$-inscriptible ?

It means that you may find three points of the plane, with integer coordinates, which form a triangle isometric to the given one.

Pierre.


Given integer $ n$ with $ (n,3)\equal{}1$, prove that there exists an integer $ m$ with the same sum-of-digit as $ n$ and $ (m,n)\equal{}1$

Let $ n\equal{}\overline{a_na_{n\minus{}1}\ldots a_1000\ldots0}$, take $ m\equal{}\overline{a_na_{n\minus{}1}\ldots (a_1\minus{}1)1}$. So $ (m,n)\equal{}(\overline{a_na_{n\minus{}1}\ldots a_10},\overline{a_na_{n\minus{}1}\ldots (a_1\minus{}1)1})\equal{}\overline(9,\overline{a_na_{n\minus{}1}\ldots a_10})\equal{}1$. Even if $ n$ does not end with 0, there's no problem, for example for $ n\equal{}1244923$, take $ m\equal{}12449221$, then $ (m,n)\equal{}(m,10n)\equal{}(m,10n\minus{}m)\equal{}(m,9)\equal{}1$.


If I have a prepaid cell phone, and wish to call someone in California (that's long distance since I'm in Maryland), will it just take off minutes off will there actually be a bill? Also does this apply to text messages? I'm not too good with cell phones  :blush: .

In general, you will only have a bill when you decide you need to add to your account. Other than that- they don't even have the information needed to bill you for anything. All calls/messages will simply deduct from your account; they may be charged at different rates for long distance, but that's all.


What is the sum of the digits of 99999^2 (without a calculator)

99999=10^5-1, which means that 99999^2=10^10-2.10^5+1, easy to find that the solution is 9*4+8+1=45.
In general, the sum of the digits of 999..99^2(n digits) is 9n

That's right, especially the last line is important.

How about this: compute the sum of the digits of x99..99^2 where x is a digit ?


Kalispera , simera psaxnontas mia askisi mou genithikan 2 apories kai tha ithela tin voithia sas...Epsaksa ligo alla den mporesa na vrw kati na mporesei na me voithisei akomi....

1. Mporei kapoios na mou vrei mia sinartisi f me sinexi prwti paragogo tetoia wste otan to x--->+oo   limf =0 kai limf'<>0    ???  (diesthitika mou fenete oti den mporei..ipoxreotika otan Limf=0 tote kai Limf'=0 , alla diesthitika panta..skeftika dld kai mia xazomara ws tekmiriosi alla den me kaliptei!)

2. Mporoume na xrisimopoiisoume ton orismo tou oriou tis paragogou(kai pws??) gia na vroume  tin timi tis f' sto apiro??( to mono pou skeftika einai na grapsw LimXo--->+00(LimX--Xo F-Fo/X-Xo) , (epeidi den kserw latex elpizw na katalavenete ti ennow))

[quote="Durandal"][quote="SorrowMan"]1. Mporei kapoios na mou vrei mia sinartisi f me sinexi prwti paragogo tetoia wste otan to x--->+oo   limf =0 kai limf'<>0    ???  (diesthitika mou fenete oti den mporei..ipoxreotika otan Limf=0 tote kai Limf'=0 , alla diesthitika panta..skeftika dld kai mia xazomara ws tekmiriosi alla den me kaliptei!)[/quote]

$ f(x) \equal{} e^{ \minus{} x}\cos(e^x)$.

Αν όμως το όριο της $ f'$ υπάρχει... άσκηση :)

[quote]2. Mporoume na xrisimopoiisoume ton orismo tou oriou tis paragogou(kai pws??) gia na vroume  tin timi tis f' sto apiro??( to mono pou skeftika einai na grapsw LimXo--->+00(LimX--Xo F-Fo/X-Xo) , (epeidi den kserw latex elpizw na katalavenete ti ennow))[/quote]

Δεν καταλαβαίνω τί εννοείς εδώ πέρα...

Cheerio,

Durandal 1707

PS: Μήπως να το μεταφέρουμε στις σχολικές ασκήσεις? Είναι ενδιαφέρουσα ερώτηση που δε χρειάζεται ύλη πιο προχωρημένη από Γ' Λυκείου...[/quote]

signwmi gia to prwto paralipsis mou , ennow an to orio tis f' otan x--->+00 iparxei tote ti ginete otan kserw pws limf=0... (gia na pw tin alithia kolisa pros stigmin oti me to na pw oti i prwti paragogos einai sinexis kaliftika , pou einai lathos fusika)

Gia to deftero ti lew...ama x anikei se ena diastima tote i F'(xo) = Lim F(x)-F(xo)/x-xo , otan to x-->xo , swsta? Ama egw omws thelw na vrw tin timi tis F'(x) otan x--->+00 ti kanw?pws mporw na xrisimopoiisw ton orimo tou oriou tis paragogou?Elpizw na to eipa ligo pio kala twra....To mono pou skeftika einai na xrisimopoiisw diadoxika oria alla den mou poli aresei i alithia einai...dld na pw oti gia xo-->+oo kai , x--->xo LimLim F(x) - F(xo)/x-x0   (to prwto Xo-->+oo antistixi sto prwto lim kai to X->x0 antistixi sto deftero Lim) ...

P.S. A den kserw ama thelete metafereteto den exw prob...Emena pantws se kati panepistimiakes askiseis mou dimiourgithikan aftes oi apories opote protimisa edw ...Alloste de nomizw kapoion mathiti likiou na ton endiaferoun tetoiou eidous "leptes" enies...

[quote="Durandal"][quote="SorrowMan"]signwmi gia to prwto paralipsis mou , ennow an to orio tis f' otan x--->+00 iparxei tote ti ginete otan kserw pws limf=0... (gia na pw tin alithia kolisa pros stigmin oti me to na pw oti i prwti paragogos einai sinexis kaliftika , pou einai lathos fusika)[/quote]

Ένα hint: αν $ \lim_{x\to\infty} f'(x) \equal{} L > 0$ τότε $ f'(x) > \frac {L}{2}$ για μεγάλα $ x$. Τι σου λέει αυτό?

[quote]Gia to deftero ti lew...ama x anikei se ena diastima tote i F'(xo) = Lim F(x)-F(xo)/x-xo , otan to x-->xo , swsta? Ama egw omws thelw na vrw tin timi tis F'(x) otan x--->+00 ti kanw?pws mporw na xrisimopoiisw ton orimo tou oriou tis paragogou?Elpizw na to eipa ligo pio kala twra....To mono pou skeftika einai na xrisimopoiisw diadoxika oria alla den mou poli aresei i alithia einai...dld na pw oti gia xo-->+oo kai , x--->xo LimLim F(x) - F(xo)/x-x0 (to prwto Xo-->+oo antistixi sto prwto lim kai to X->x0 antistixi sto deftero Lim) ... [/quote]

Το μόνο πράγμα για το οποίο μπορείς να μιλήσεις (εκτός και αν δουλέψεις στο συμπαγοποιημένο $ \overline{\mathbb{R}}$) ειναι το όριο της $ f'$ στο $ \infty$. Ακόμα δεν καταλαβαίνω όμως τί ακριβώς ψάχνεις... :(

Cheerio,

Durandal 1707

[quote]P.S. A den kserw ama thelete metafereteto den exw prob...Emena pantws se kati panepistimiakes askiseis mou dimiourgithikan aftes oi apories opote protimisa edw ...Alloste de nomizw kapoion mathiti likiou na ton endiaferoun tetoiou eidous "leptes" enies...[/quote]

Μπα, νομίζω ότι μια χαρά θα ενδιαφέρονται για κάτι τέτοιο και είναι κρίμα να χαθεί στο φοιτητικό forum σε περίπτωση που δεν το διαβάζουν οι πιο μικροί, γι'αυτό το πρότεινα, that's all :)[/quote]

Oti i f einai gnisios afksousa se ena diastima (x,+00) me x poli megalo...??

Tha prospathisw na to ksana eksigisw ti pasxnw...Loipon an erthw kai sou pw estw i sinartisi F apo ton R ston R , kai einai C^00 (gia efkolia afto) kai zitisw na mou vreis me ton orismo tin paragogo sto simio x=3 ti tha kaneis?tha grapseis limF(x)-F(3)/X-3 kathos to x---->3 sosta??? E twra ama sou pei kapoios vres mou tin paragogo tis F kathos to x--->00 , tote pws to doulevw afto me ton orismo????Twra elpizw na itan pio sosta diatipomeni i erwtisi!!!!

[quote="SorrowMan"][quote="Durandal"]Ένα hint: αν $ \lim_{x\to\infty} f'(x) \equal{} L > 0$ τότε $ f'(x)>\frac{L}{2}$ για μεγάλα $ x$. Τι σου λέει αυτό? [/quote]
Oti i f einai gnisios afksousa se ena diastima (x,+00) me x poli megalo...??[/quote]

Much better: η $ f(x) \minus{} \frac {L}{2}x$ είναι αύξουσα ;)

[quote]Tha prospathisw na to ksana eksigisw ti pasxnw...Loipon an erthw kai sou pw estw i sinartisi F apo ton R ston R , kai einai C^00 (gia efkolia afto) kai zitisw na mou vreis me ton orismo tin paragogo sto simio x=3 ti tha kaneis?tha grapseis limF(x)-F(3)/X-3 kathos to x---->3 sosta??? E twra ama sou pei kapoios vres mou tin paragogo tis F kathos to x--->00 , tote pws to doulevw afto me ton orismo????Twra elpizw na itan pio sosta diatipomeni i erwtisi!!!![/quote]

Ok. Όπως έγραψες κι εσύ, με βάση τον ορισμό, το όριο της παραγώγου καθώς $ x\to\infty$ είναι: $ f'(\infty) \equiv \lim_{x\to\infty} f'(x) \equal{} \lim_{x\to\infty} \lim_{\xi\to x} \frac {f(x) \minus{} f(\xi)}{x \minus{} \xi}$.

Υπό προϋποθέσεις (ποιές? :P) μπορείς να εναλλάξεις τα όρια και να γράψεις ότι $ f'(\infty) \equal{} \lim_{x\to\infty} \frac {f(x)}{x}$. Αλλιώς... βλ. παραπάνω ;)

Cheerio,

Durandal 1707


Given an integer $ m\geq 2$. Find the smallest integer $ n\geq m$ such that for every partition of the set $ \{m,m\plus{}1,...,n\}$ into two subsets, one of the subsets contains three numbers $ a,b,c$ (not necessarily distinct) such that $ ab\equal{}c$.


Determine of there exist four polynomials such that the sum of any three of them has a real root while the sum of any two of them does not.


a) Simplify: $ (1 \plus{} tg\alpha)(cos\alpha)$
b) If $ cos\alpha \plus{} sin\alpha \equal{} m$  find $ cos\alpha *sin\alpha$

What's $ tg \alpha$?  :huh:

LaTeX commands for trig functions:

[code]\sin \alpha[/code]

produces $ \sin \alpha$. I'm sure you can guess the other codes.

$ tg\alpha$ I mean tangent or $ tan\alpha$

yeah actually I think tg is also a correct way of expressing tangent or tan.

Anyways.....

[hide="1"]$ \sin \alpha \plus{} \cos \alpha$[/hide]

[hide="2"]$ (\sin \alpha \plus{} \cos \alpha)^{2} \equal{} \sin^{2} \alpha \plus{} 2\sin \alpha \cos \alpha \plus{} \cos^{2} \alpha \equal{} 1 \plus{} 2\sin \alpha \cos \alpha \equal{} m^{2}$
So $ \sin \alpha \cos \alpha \equal{} \frac{m^{2} \minus{} 1}{2}$
[/hide]


In a math contest, there are $ 20$ students, and each of them solved at least one problem. Prove that there is a nonempty subset $ S$ of problems such that the number of students who solved [i]every[/i] problem of $ S$ is even.

Count the number of pairs $ (s, S)$ such that $ s$ is a student who solved all of the problems in (non-empty) set $ S$ (and possibly others).  A student who solves $ n > 0$ problems is in $ 2^{n}-1$ such pairs, an odd number; as there are 20 students, this means we have in total an even number of pairs.  If there are $ p > 0$ problems in total, there are $ 2^{p}-1$ total subsets, an odd number.  Thus, some subset must appear in an even number of pairs, as desired.

I don't understand the question.
If  we have to prove that there exists a subset S st the number of the students who solved at least problem of S is  even,we can choose S is the set of all the problems
And if we want S to satisfy: For each problem of S, the number of the students who solved that problem  is even ,then we can easily see that the case  each student solved exactly a problem (distinct) isnt satisfying the question .
Do we need S satisfy : the number of the students who solved all of the problems of S is even ? :)

No: we want $S$ to be nonempty and satisfy, "The number of students who solved [i]every[/i] problem in $S$ is even."


geometry
rectangle
area of a triangle
Heron\u0027s formula

I don't know if this is right, but you can use Heron's formula 3 times...

If the sides of the triangle are a, b, c, and the semiperimeter is s (s = (a+b+c)/2), then the area is sqrt[s*(s-a)*(s-b)*(s-c)].

Plug in the values for a, b, and c, and solve from there...is this right?

If someone could make a picture, that would be great.  In the meantime, try and follow my thinking.  Let's call the bottom left-hand corner $ A$, and, going clockwise, it is Pentagon $ ABCDE$.  From point $ B$, draw a segment parallel with segment $ AE$ and perpendicular to segment $ DE$.  This will make a rectangle ($ 20 \text{by} 24$), and another quadrilateral.  That quadrilateral can be divided into two right triangles using line $ BD$.  This would give one triangle $ 15 \text{and} 20$ unit legs, and another one $ 7 \text{and} 24$ unit legs.  Plugging into area formulas, we get $ 20\cdot24\plus{}\frac{1}{2}\cdot15\cdot20\plus{}\frac{1}{2}\cdot7\cdot24\equal{}480\plus{}150\plus{}84\equal{}\boxed{714}$.


geometry
geometric transformation
homothety
ratio
symmetry
absolute value

Is there a general method for finding the largest square that can be inscribed in any triangle? If there isn't, what about an isosceles triangle?

[hide]
In order to inscribe a square in a triangle $ ABC$ such that the square shares more than a point with, say, side $ BC$, construct a square $ BCDE$ to $ ABC$'s exterior Then, perform a homothety that takes $ BCDE$ to a square with $ D'E'$ on $ BC$ with center $ A$. This is a square inscribed in the triangle.
Now suppose that the triangle $ ABC$ has area $ K$
Then, the distance from $ DC$ to A is $ h_a \plus{} BC$ where $ h_a$ is the length of the altitude from a. This is equal to $ \frac {2K}{BC} \plus{} BC$
Then, the homothety that takes $ BCDE$ to the inscribed square has ratio $ \frac {\frac {2K}{BC}}{\frac {2K}{BC} \plus{} BC}$
$ BCDE$ has side length $ BC$
Therefore, the inscribed square's side length is $ BC \cdot \frac {\frac {2K}{BC}}{\frac {2K}{BC} \plus{} BC} \equal{} \frac {2K}{\frac {2K}{BC} \plus{} BC}$
By symmetry, we know the lengths of the sides of the other 2 possible inscribed squares.
We want to maximize the expression $ \frac {2K}{\frac {2K}{x} \plus{} x}$
Whis is equivalent to minimizing $ \frac {2K}{x} \plus{} x$
My AM-GM, we want these to be as close as possible (absolute value is minimized) since the 2 terms have a constant product for a given triangle.

So, the length of the largest inscribed square is $ \frac {x^3}{2K \plus{} x^2}$ where x is the side such that $ \left|\frac {2K}{x} \minus{} x\right|$ is minimized where $ K$ is the area.
The first term is just the height to the side of length $ x$
This is constructed by the homothety given
[/hide]


inequalities
LaTeX
inequalities unsolved

c>a>=b>0,c>a+b

a*sqrt[b/(a+c)]+b*sqrt[c/(a+b)]+c*sqrt[a/(b+c)]<5/4*sqrt(ab+bc+ca)

Sorry.I don't know how to type it in the formula format. :oops:

$c\geqq\geq b\geq 0,c\geq a+b$

$a\sqrt{\frac{b}{a+c}}+b\sqrt{\frac{c}{a+b}}+c\sqrt{\frac{a}{b+c}}<\frac{5}{4}\sqrt(ab+bc+ca)$

[quote="andy_tok"]:) THANKS.
BTW, is there any link in the forum that shows how to type those formulas?[/quote]
See [url]http://www.mathlinks.ro/LaTeX/AoPS_L_About.php[/url]. You can ask questions about the typesetting program $\LaTeX$ in [url=http://www.mathlinks.ro/Forum/index.php?f=123]this forum[/url].

[quote="vardanch"]$Problem:$
$c>a\geq b>0$,$c>a+b$
$a\sqrt{[b/(a+c)]}+b\sqrt{[c/(a+b)]}+c\sqrt{[a/(b+c)]}<5/4\sqrt{(ab+bc+ca)}$

$Solution:$
<=> $\sum{\sqrt{ab}\sqrt{\frac{a}{a+c}}}<\frac{5}{4}\sqrt(ab+bc+ca)$
$\sqrt{\frac{a}{a+c}}<1/2$; $\sqrt{\frac{b}{b+a}}<1/2$; $\sqrt{\frac{c}{c+b}}<1$;
<=> $2\sqrt{ab}+2\sqrt{bc}+4\sqrt{ac}<5\sqrt{(ab+bc+ca)}$
By $A(M)\geq G(M)$
$(2\sqrt{ab}+2\sqrt{bc}+4\sqrt{ac})^{2}\leq 12ab+12bc+16ac+16a\sqrt{bc}+16c\sqrt{ab}<25ab+25bc+25ac=(5\sqrt{(ab+bc+ca)})^{2}$
$Q-E-D$[/quote]
I'm not sure about this part :P !
[quote]$\sqrt{\frac{a}{a+c}}<1/2$[/quote]
Can you prove it  :P  :P ?

I think there's something wrong.
c is not necessary bigger than 3a


Solve the equation : $ x^4\minus{}4\sqrt{3}x\equal{}5$


limit
ratio
integration
logarithms
calculus
calculus computations

hi!!

Let $ (a_{n})_{n>0}$ an arithemtic sequence such that: $ a_{1},a_{2}....,a_{n}>0$ and:

$ lim_{n\to\infty}\frac{a_{n}}{n}\equal{}1$

we consider $ (x_{n})_{n>0}$ such that:

$ x_{n}\equal{}\sum_{i\equal{}1}^{n}\frac{1}{a_{n\plus{}i}}$

Prove that $ (x_{n})$ is convergent.

thanx.

It may help you to notice that $ \lim_{n \to \infty} \frac {a_n} {n} \equal{} \lim_{n \to \infty} \frac {a_1 \plus{} (n\minus{}1)r} {n} \equal{} r$, so the ratio is $ 1$.

yes,this is the way to find the solution!

so $ x_n\equal{} \sum_{k\equal{}n}^{2n\minus{}1} \frac{1}{a_1\plus{}k}$

$ \frac{1}{ \max(0,a_1\minus{}1) \plus{}2}\le \frac{n}{(a_1\minus{}1)\plus{}2n}\le x_n \le \frac{n}{a_1\plus{}n}\le 1$

so its bounded..

finding a monotone subsequence seems a little harder

[quote="kenn4000"]so $ x_n \equal{} \sum_{k \equal{} n}^{2n \minus{} 1} \frac {1}{a_1 \plus{} k}$

[/quote]

Right. And this sum is bounded above by $ \int_{n}^{2n} \frac{dx}{a_1 \plus{} x}$ with a similar estimate from below. Looks to me like $ x_n\rightarrow \ln 2$.


induction
MIT
college
pigeonhole principle

this is sort of a conjecture that i have made, and i am pretty sure it is true, but could anybody help rigorously prove it?

Prove that: $\prod^{n}_{k=1}(x-k) \mod{n!} \equiv 0$ for any integer $n>0$ and any integer $x$

Consider the numbers x-k for k from 1 to n.  They are consecutive integers.  How do these numbers look mod 2, mod 3, mod 4, etc?

i dont really understand what you would do, i only know basic mods, i can see proving for specific cases, it is almost like double induction, where 1 part for n then one part for x (also with the x you would need to show that $S_{x+1}$ implies $S_{x}$ (i.e. the opposite for all negative x)

You can do it this way . Since the product has $n$ consecutive number , we know from pigeon hole that , it can be divide by 2,3,4...n .

Example :

How we know the product can be divided by 2 ? Well since the factor of $2$ repeat in every two cycle , and there are $[\frac{n}{2}]$ cycle in the product so clearly it can be divided by 2.

Lets see a bigger number say $19$ . Since we know that the factor containing $19$ repeats every 19th cycle , so there are $[\frac{n}{19}]$ cycle in the product which shows that it can be divided by 19 .

Similarly work out for other case varying from $2$ to $n$ , which you can see they can be divided by $2\times 3\times 4\times 5...\times n=n!$  :D

these numbers are residue system modulo m(it is true for all $x$).
another solution is combinotric if $x>n$ so it is $\binom{x}{n}$ so $n!|(x)(x-1)(x-n+1)$.

It's like if you had 2 consecutive integers.  One of them must be divisible by 2.
If you have 3 consecutive integers, one must be divisible by 2, another by 3.
You just extend that to as many as needed.

Seems like you could just pigeonhole the mods since they are consecutive.

thank you for the help...i understand what to do now :)


AMC
AIME
USA(J)MO
USAMO
AMC 10
AMC 10 B
AMC 12

Please poll your AIME score * 10 + AMC score.
This is for both AIME I and AIME II takers.
If you are not sure of your score, please poll what you think you got, based on the answers people have been posting.

i got 4 on the AIME II and 135 on the AMC 10 (if that counts).

oh well , it's only my first year...

I'm pretty sure it does ...
I got 270, I think (150 AMC 10, 12 (? i'm pretty sure) AIME II)
If it doesn't count, then 252 (132 AMC 12B)

Well, it technically does, but it'll only matter if there are a whole lot of people in 10th or under who have the floor score. The floor is determined by the indices with the AMC 12.

Not that it'll matter for the floor, but $144+10\cdot13=274$ is my index. I was a binomial expansion and two addition/subtraction errors away from 300. Oh well, there's always next year.

while a 300 is a good goal, if you're doing that well on the AMC/AIME i think you would benefit from USAMO practice  :lol:

well... counting 12B it would be 209.5 but counting 10A it would be 238 so I say 238

136.5 AMC10B
8 AIME II
Not bad for me... considering I'm in seventh...
I still haven't gotten over missing #1 in AIME, by means of stupid undercounting.

114 AMC 12A
117 AMC 12B
11 AIME II

So 227, which I'm relatively happy with.

136.5 (12b) + 10*10 (AIME II) = 236.5. Much better than my 189  (129+60) in 7th grade last year.

Question: for 10th grade and below, are USAMO indices used only to break ties between people who scored exactly the AIME floor?  This is important to me because, since I took the AMC12 as a 10th grader, scored a 123, and got an 11 on the AIME II, I am almost certainly above the floor but not necessarily above the 160 highest indices.

oh well why the hell am i worrying about this, i'll probably get in anyway.

[quote="Teki-Teki"]Question: for 10th grade and below, are USAMO indices used only to break ties between people who scored exactly the AIME floor?  This is important to me because, since I took the AMC12 as a 10th grader, scored a 123, and got an 11 on the AIME II, I am almost certainly above the floor but not necessarily above the 160 highest indices.

oh well why the - am i worrying about this, i'll probably get in anyway.[/quote]

It's impossible for you to not get in, stop worrying.

[quote="Teki-Teki"]Question: for 10th grade and below, are USAMO indices used only to break ties between people who scored exactly the AIME floor?  This is important to me because, since I took the AMC12 as a 10th grader, scored a 123, and got an 11 on the AIME II, I am almost certainly above the floor but not necessarily above the 160 highest indices.

oh well why the - am i worrying about this, i'll probably get in anyway.[/quote]
233 is way more index points than you will need by any standards

206.5 in 7th.
my earlier post was wrong, i got a 7 not 8

Erg... lame...

150+ 9*10 = 240.

You AIME II kids... ugh!

187 =(...7 on AIME(bash head) and 117 on amc 12b but  144 on amc 10...so technically uhh 214

221.  141 (12A) +8

I am amazed at this score, and I didn't believe I would get so high a score on the AMC12!

Wow.

126 AMC10A + 4 AIME 1 = 166 index

 :(

Shouldn't this thread read "USAMO index" instead of AIME index?

145.5 (AMC12) + 10*14 (AIME2)=285.5

103.5 AMC 12A + 6 AIME II = 163.5

Ouch.

Wish I had taken the 10.

OMG my index is lower than most of your amc 10 and 12 scores i feel really retarted right now.

Its okay, you're only in 7th grade. While some of these other kids are too, you just have to work harder to do better, pretty much. Practice is the best way to improve. Do lots and lots of problems, and then ask questions about the ones you don't really know how to do.

[quote="Zenith Xan"][quote="vishalarul"][quote="davidyko"]There was already a thread like this.
112.5 (12A)+50 (AIME I)=162.5.
And I don't really think the AMC 10 counts.[/quote]

Hehehe:

102.5 (12B) + 60 (AIME I) = 162.5

Even though you pwned me on the 12, I caught up on the AIME.[/quote]

A 102.5 on the AMC10/12 isn't possible...[/quote]

Oops, I got 102... Missed by .5 of a point...
Oh well, there's always next year...

I got a 185... Had I not made several stupid mistakes, I would have gotten in. Perhaps next year I'll make it to USAMO.


Prove for every natural n (n > 2) :

\[\sum_{i = 1}^{n-1}{i\left( \begin{array}{l}n \\ i+1 \\ \end{array}\right)(n-1)^{n-i-1}}= (n-1)^{n}\\ \]


function
algorithm
number theory open
number theory

Is it possible to construct a sequence $a_{n+1} = f(a_n, a_{n-1}, ..., n)$ that only takes on prime values $\forall n \in \mathbb{N}$?  (Not necessarily every prime.)

If so, how strong can the restrictions on $f$ be?  Only algebraic?  Algebraic allowing floors?  Allowing functions such as $\phi(n)$?

If not, can a definite upper bound be found on the maximum number of possible consecutive prime values for $f$ an algebraic function?  $f$ allowing floors?  Etc.

http://mathworld.wolfram.com/PrimeFormulas.html is a nice startup.

what are algorithms? curiousity haunts me

[quote="now a ranger"]what are algorithms? [/quote]

An algorithm, in the most basic sense, is just a procedure for doing something.

[quote="now a ranger"]what are algorithms?[/quote]
I'm not an expert on computer science, but I believe that an algorithm is a finite set of statements $\{w_{1},w_{2},\cdots,w_{n}\}$ such that every $w_{i}$ can be simulated by the Turing's machine and the Turing machine ends processing every statement after finitely many steps, no matter about the input values. That was only a guess.

sounds interesting, I want to learn more about that  :wink:


Spain is champion  on Euro 2008!  :!:   :lol:

I also like Spain s' team  :lol:  Because I like Cesc Fabregas  He is a good player but........

Yes. A shame we couldn't have been there for the final. It must have been a crazy night :D (I remember as I was in Athens in 2004 when Greece won the semifinal with France watching the game with a friend of mine).

[quote="Naphthalin"][quote="lchserious"]Fernando Torres  :lol:  :first:

It makes my trip to Spain more interesting.[/quote]
mine too^^

Naphthalin[/quote]You wanted Spain to win. Otherwise people in Madrid wouldn't have been very happy with you being German and all that :D

Spain  - this's my dream, but i can't go  in Spain.
 And  congratulation all nembers Vietnamese team!!!
 i wish all success

I was a fan of Germany and Russia. Spain kicked them both out...

damn you nick, you had to share what we discussed!!! spain are brilliant though, and if anyone has something against torres, they shud strictly reconsider, or i'll prove them wrong.... with mathematical means

[quote="bnick17"]Greece won Euro 2004 when IMO was in Athens

Italy won World Cup in 2006 when IMO was in Slovenia (IMO excursion was in Italy!)

Spain win Euro 2008 when IMO is in Madrid

And in 2010 who will win the World Cup? Could it be The Netherlands? We will have to see  :lol:[/quote]The Netherlands IMO is in 2011, so sorry, no it won't happen :D

[quote="Inequalities Master"][b]Fire_dragon wrote[/b]:
[quote]I also like Spain s' team  Because I like Cesc Fabregas He is a good player but........[/quote]

I think Netherlands was a much better team than Spain,
but they got out of competition because of low morale after what has happened to Boulahruz.
Actually, I'm a fan of Italy. :)[/quote]
i'm a fan of  Netherland and  Russia.
 But what's  pity, they  killed at  1:4 Euro  champion.

Isn't IMO 2010 in Kazahstan?
So, Kazahstan has to be world champion 2010... I respect every team, but NO WAY THAT KAZAHSTAN WILL BE THE CHAMPIONS!!

:D  i m fan of spain cos this year imo in spain. next year i ll turn to germany

[quote="silouan"]Maybe Kazahstan will won the cup . For those who remember at the BMO this year the Kazahstan team with also Adrei Frimu
and me won all the football matches  :wink:  :lol:[/quote]Hey if I find enough coordinators that play soccer, we can make a match at the end "students vs coordinators" in soccer ;)

[quote="Valentin Vornicu"]Hey if I find enough coordinators that play soccer, we can make a match at the end "students vs coordinators" in soccer ;)[/quote]

That would be great :P

the coordiantors might have to play a couple of matches, because of the large number of teams that "may" be willing to play

[quote="silouan"]Maybe Kazahstan will won the cup . For those who remember at the BMO this year the Kazahstan team with also Adrei Frimu
and me won all the football matches  :wink:  :lol:[/quote]
I remember that too,as I played in those matches with Silouan and Andrei :wink:In Barcelona uniform,if you remember...
We played football at IMO 2008 too,Kazakhstan team with two guys from Germany,and we beat Bulgarians and Andrei...


In $\triangle{ABC}$, $2B=A+C$, Find the minimum of $\cos^2{A}+\cos^2{B}+\cos^2{C}$.

hello, from $ A+B+C=\pi$ and $ 3B=A+C$ follows $ B=\frac{\pi}{3}$ and $ C=\frac{2\pi}{3}-A$, setting this in $ \cos^2(A)+\cos^2(B)+\cos^2(C)$ we get $ f(\alpha)=\cos^2(\alpha)+\cos^2(\frac{\pi}{3})+\cos^2(\frac{2\pi}{3}-\alpha)$, differentiating $ f(\alpha)$ with respect to $ \alpha$ we get
$ f'(\alpha)=-\cos(\frac{\pi}{6}+2\alpha)-\sin(2\alpha)$
$ f^{''}(\alpha)=-2\cos(2\alpha)+2\sin(\frac{\pi}{6}+2\alpha)$
we get $ f^'(\alpha)=0 \Leftrightarrow \alpha_1=\frac{\pi}{6}$ or $ \alpha_2=\frac{5\pi}{6}$, further
we have $ f^{''}(\alpha_1)=2$ and $ f^{''}(\alpha_2)=-2$, hence our minimum is $ f(\alpha_1)=\frac{3}{4}$.
Sonnhard.

After various elementary trigonometric transformations, we get that the desired expression is equal to $ {5\over 4}\plus{}{1\over 2}\cos\left(2A\plus{}{\pi\over 3}\right)$, which is obviously minimal for $ 2A\plus{}{\pi\over 3}\equal{}\pi\iff A\equal{}{\pi\over 3}$ (equilateral triangle), and its value is $ {3\over 4}$.

$ 2B \equal{} A \plus{} C$
$ 180 \minus{} B \equal{} A \plus{} C$
$ B \equal{} 60$
$ A \plus{} C \equal{} 120$.

minimize $ \cos^2A \plus{} \cos^2(120 \minus{} A)\ge 2|\cos A \cdot\cos(120 \minus{} A)|$
with equality when $ A \equal{} 120 \minus{} A$ or $ A \equal{} 60$

@Farenhajt: What trig transformations did you use?

[quote="mathwizarddude"]@Farenhajt: What trig transformations did you use?[/quote]

Standard addition formulas, product-to-sum and sum-to-product formulas.

[quote="mathwizarddude"]$ \cos^2A \plus{} \cos^2(120 \minus{} A)\ge 2|\cos A \cdot\cos(120 \minus{} A)|$
with equality when $ A \equal{} 120 \minus{} A$ or $ A \equal{} 60$[/quote]

True, but this inequality doesn't say anything other than that the LHS is non-negative, so unfortunately its meaningless.  :wink:

Yeah I know but I just couldn't think up of a solution using ineq bashing...  :(
(the other solutions are kinda messy)

$ \begin{array}{ccc} & A \equal{} \phi\ <\ 90^{\circ}\mathrm{\ (constant)} & \\
 \\
\mathrm{min.\ } & \left(\cos^2A \plus{} \cos^2B \plus{} \cos^2C\right) & \\
 \\
\mathrm{min.\ } & \left(\cos^2B \plus{} \cos^2C\right) & \\
 \\
\mathrm{max.\ } & \left(\sin^2B \plus{} \sin^2C\right) & \frac 12\left(1 \minus{} \cos 2B \plus{} 1 \minus{} \cos 2C\right) \\
 \\
\mathrm{min.\ } & \left(\cos 2B \plus{} \cos 2C\right) & \minus{} 2\cos A\cos (B \minus{} C) \\
 \\
\mathrm{max.\ } & \cos (B \minus{} C) \ \le\ 1 & \\
 \\
& B \equal{} C \equal{} 90^{\circ} \minus{} \frac {\phi}{2} & \end{array}$


Put the angle a on line r and through point P a line paralell to the angle side

last year i  proved that if we have a half of circle with center we can do any thing with ruler because we can draw parallel and draw perpendicular from apoint to a line.


[quote="ComplexZeta"]It's too long.[/quote]

What's the limit on the length?

I think it's 500 or 750 characters (can't remember which), but that one was probably tripped up on image size, which is 500 pixels wide as the limit (wide images make the board a mess).


calculus
mathleague.org
Princeton
college
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inequalities
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\/closed

I tried to get to the forum and this happened:
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DEBUG MODE

SQL Error : 1062 Duplicate entry '14436-fjbox' for key 1

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After I closed the window and opened the link again, this stopped happenning.