find all pairs $ (x,y)$ of positive integers such that $ \frac{x^2\plus{}y^2}{x\minus{}y}$ is an integer divisible by 1995.

[quote="om256"]find all pairs $ (x,y)$ of positive integers such that $ \frac {x^2 \plus{} y^2}{x \minus{} y}$ is an integer divisible by 1995.[/quote]
It's rather easy to show that  $ x^2\plus{}y^2\equal{}k(x\minus{}y)$ has general solution $ (x,y,k)\equal{}(\frac{\lambda v(u\plus{}v)}2,\frac{\lambda v(u\minus{}v)}2,\frac{\lambda(u^2\plus{}v^2)}2)$
[hide="How ?"]=================== begin ==================
$ x^2\plus{}y^2\equal{}k(x\minus{}y)$ $ \iff$ $ x(k\minus{}x)\equal{}y(k\plus{}y)$ and so :

$ x\equal{}uv$, $ k\minus{}x\equal{}wt$, $ y\equal{}uw$ and $ k\plus{}y\equal{}vt$ with the constraint $ uv\plus{}wt\equal{}vt\minus{}uw$ $ \iff$ $ u(v\plus{}w)\equal{}t(v\minus{}w)$ and so :

$ u\equal{}ab$, $ v\plus{}w\equal{}cd$, $ t\equal{}ac$ and $ v\minus{}w\equal{}bd$ and so $ (u,v,w,t)\equal{}(ab,d\frac{b\plus{}c}2,d\frac{c\minus{}b}2,ac)$

So $ (x,y,k)\equal{}(abd\frac{b\plus{}c}2,abd\frac{c\minus{}b}2,ad\frac{b^2\plus{}c^2}2)$
=================== end ====================== [/hide]
And so we just have to choose $ u,v,\lambda$ such that $ \lambda(u^2\plus{}v^2)$ is  divisible by $ 3990$. And so $ \lambda \equal{} \alpha \frac{3990}{\gcd(u^2\plus{}v^2,3990)}$

Hence a general solution (with obviously infinitely many pairs) :

$ (x,y)\equal{}\left(k \frac{1995v(u\plus{}v)}{\gcd(u^2\plus{}v^2,3990)},k \frac{1995v(u\minus{}v)}{\gcd(u^2\plus{}v^2,3990)}\right)$ for any $ u,v,k\in\mathbb N$, $ u> v$

[hide="Some examples of such pairs"]
$ (u,v,k)\equal{}(2,1,1)$ $ \implies$ $ (x,y)\equal{}(1197,399)$
$ (u,v,k)\equal{}(3,1,1)$ $ \implies$ $ (x,y)\equal{}(798,399)$
$ (u,v,k)\equal{}(3,2,1)$ $ \implies$ $ (x,y)\equal{}(19950,3990)$
$ (u,v,k)\equal{}(4,1,1)$ $ \implies$ $ (x,y)\equal{}(9975,5985)$
$ (u,v,k)\equal{}(4,2,1)$ $ \implies$ $ (x,y)\equal{}(2394,798)$
$ (u,v,k)\equal{}(4,3,1)$ $ \implies$ $ (x,y)\equal{}(8379,1197)$
$ (u,v,k)\equal{}(5,1,1)$ $ \implies$ $ (x,y)\equal{}(5985,3990)$
$ (u,v,k)\equal{}(5,2,1)$ $ \implies$ $ (x,y)\equal{}(27930,11970)$
$ (u,v,k)\equal{}(5,3,1)$ $ \implies$ $ (x,y)\equal{}(23940,5985)$
$ (u,v,k)\equal{}(5,4,1)$ $ \implies$ $ (x,y)\equal{}(71820,7980)$
$ (u,v,k)\equal{}(6,1,1)$ $ \implies$ $ (x,y)\equal{}(13965,9975)$
$ (u,v,k)\equal{}(6,2,1)$ $ \implies$ $ (x,y)\equal{}(3192,1596)$
$ (u,v,k)\equal{}(6,3,1)$ $ \implies$ $ (x,y)\equal{}(3591,1197)$
$ (u,v,k)\equal{}(6,4,1)$ $ \implies$ $ (x,y)\equal{}(39900,7980)$
...
[/hide]

Or, simpler again, (but equivalent), we have :

$ x\equal{}\lambda u$ and $ y\equal{}\lambda v$  and $ 1995|\lambda \frac{u^2\plus{}v^2}{u\minus{}v}$ $ \iff$ $ 1995(u\minus{}v)|\lambda(u^2\plus{}v^2)$ and so $ \lambda\equal{}k\frac{1995(u\minus{}v)}{\gcd(u^2\plus{}v^2,1995(u\minus{}v))}$

Hence another trivial general solution (easier to get) :

$ (x,y)\equal{}(k\frac{1995u(u\minus{}v)}{\gcd(u^2\plus{}v^2,1995(u\minus{}v))},k\frac{1995v(u\minus{}v)}{\gcd(u^2\plus{}v^2,1995(u\minus{}v))})$ $ \forall u,v,k\in\mathbb N$, $ u>v$


Να μια ωραία έξυπνη και σύντομη ασκησούλα .
Να δείξετε ότι δεν υπάρχει μη μηδενικός ακέραιος $ n$ ώστε οι αριθμοί
$ 2n^{2}\plus{}1,\ 3n^{2}\plus{}1,\ 6n^{2}\plus{}1$ να είναι και οι τρεις τελεια  τετράγωνα .

Τι θα λέγατε αν :
Να βρείτε τους θετικούς ακεραίους $ n$ ώστε οι $ 2n\plus{}1,\ 3n\plus{}1,\ 6n\plus{}1$ να είναι και οι τρεις τέλεια τετράγωνα ...

Καλησπέρα Silouan,

Καταρχήν συγχαρητήρια για την εύφημο μνεία στην Ολυμπιάδα! Είναι ισάξια ενός μεταλλίου ειδικά αφού πλησίασες τόσο κοντά!!

Επίσης θέλω να σου θυμίσω την περίοδο πριν από τα Χριστούγεννα η οποία ήταν παρα πολύ νεκρή! Θυμίσου όμως μετά τι έγινε αφότου ξεκινήσαμε και πάλι!! Είναι λογικό με τόσα πολλά post να έχουν κουραστεί οι περισσότεροι... Είμαι όμως σίγουρος ότι την νέα χρονιά θα είμαστε και πάλι δυναμικοί και με νέο αίμα να προσπαθεί!! Ας είμαστε λίγο ανακτικοί με τους νέους οι οποίοι λόγω έλλειψης εμπειρίας δεν μπορούν να ακολουθήσουν με τον ίδιο ρυθμό κάποιους παλαιότερους. Ας μην σας κουράζω όμως! Να μπω στο θέμα που είναι οι ασκήσεις που πρότεινες και συγκεκριμένα η πρώτη. Την δεύτερη δεν μπόρεσα να την ολοκληρώσω!Κάπου κολάω και θα την ξαναδω!

Για την πρώτη:

Έχουμε $ (3n^{2}\plus{}1)(6n^{2}\plus{}1)\equal{}18n^{4}\plus{}9n^{2}\plus{}1 \equal{} 9n^{2}(2n^{2}\plus{}1)\plus{}1\equal{}(3n)^{2}(2n^{2}\plus{}1)\plus{}1$
δηλαδή
$ (3n^{2}\plus{}1)(6n^{2}\plus{}1)\minus{}(3n)^{2}(2n^{2}\plus{}1)\equal{}1$, δηλαδή έχουμε μία διαφορά τετραγώνων ίση με $ 1$. Αυτό είναι εφικτό μόνο εαν $ n\equal{}0$ το οποίο δεν είναι θετικό.

Αλέξανδρος

Γεια σε όλους. Βάζω μια λύση στη δεύτερη. Πιστέυω βέβαια ότι μάλλον υπάρχουν και άλλες πιο σύντομες και όμορφες λύσεις, αλλά... ίδωμεν  :wink: .

Εστω $ 2n\plus{}1\equal{}a^{2}$, $ 3n\plus{}1\equal{}b^{2}$ και $ 6n\plus{}1\equal{}c^{2}$. Θα είναι $ 4b^{2}\equal{}3a^{2}\plus{}c^{2}$, δηλαδή $ (2b\minus{}c)(2b\plus{}c)\equal{}3a^{2}$, οπότε $ b\equal{}\frac{y^{2}\plus{}3x^{2}}{4}$, $ c\equal{}|\frac{y^{2}\minus{}3x^{2}}{2}|$ και $ a\equal{}xy$, αφού οι $ a,b,c$ είναι περιττοί και ανα δύο πρώτοι μεταξύ τους, οπότε $ (2b\minus{}c,2b\plus{}c)\equal{}1$ και άρα
$ \{2b\minus{}c, 2b\plus{}c\}\equal{}\{3x^{2}, y^{2}\}$. Τώρα από τη σχέση π.χ. $ c^{2}\plus{}1\equal{}2b^{2}$ παίρνουμε
$ 9x^{4}\minus{}18x^{2}y^{2}\plus{}y^{4}\plus{}8\equal{}0$, που είναι τριώνυμο ως προς $ x^{2}$ με διακρίνουσα $ D\equal{}144(2y^{4}\minus{}2)$. Για να είναι η διακρίνουσα τέλειο τετράγωνο πρέπει $ y^{4}\minus{}1\equal{}2z^{2}$, οπότε αν
$ t\equal{}\frac{y\minus{}1}{2}, r\equal{}\frac{z}{2}$ θα έχουμε $ t(t\plus{}1)[2t(t\plus{}1)\plus{}1]\equal{}r^{2}$ και συνεπώς $ t\equal{}r\equal{}0$, αφού πρέπει οι $ t,t\plus{}1,2t(t\plus{}1)\plus{}1$ να είναι τέλεια τεράγωνα. Άρα $ x\equal{}y\equal{}1$, δηλαδή $ n\equal{}0$.

[quote="savvas_michael"]Γεια σε όλους. Βάζω μια λύση στη δεύτερη.[/quote]

Όχι που δεν θα έβρισκες λύση . Το πίσμα νίκησε  :wink:
Λοιπόν να η λύση μου στο 1 και αύριο θα βάλλω και τη λύση μου στο 2
Dedicated to Milano  :wink:
Λοιπόν αφού 2*3=6 πολλαπλασιάζοντας και τις τρεις θα έχουμε ότι ο συντελεστης του μεγιστοβάθμιου όρου θα είναι τελ τετράγωνο .Άρα έστω $ A\equal{}(2n^{2}\plus{}1)(3n^{2}\plus{}1)(6n^{2}\plus{}1)$ Tote
$ (6n^{3}\plus{}3n)^{2}<A<(6n^{3}\plus{}3n\plus{}1)^{2}$  QED

Η λύση του Αλέξανδρου είναι προφανώς πιο κομψή αλλά ...
Καλά ξεκινήσαμε παντως  :)

Για τη 2 π.χ. δε μπορούσαμε να πάρουμε τις περιπτώσεις: $ n\equiv0\pmod{10}, n\equiv1\pmod{10}$ κ.λπ. και λαμβάνοντας υπόψη ότι ένας αριθμός είναι τέλειο τετράγωνο αν και μόνο αν το τελευταίο ψηφίο του είναι $ 0,1,4,5,6,9$, να τις απορρίψουμε μία-μία ? Ή χάνω κάπου ?

Δε ξέρω αν βγαίνει όντως έτσι γιατί δεν κάθησα να υπολογίσω όλους τους συνδιασμούς... Μία πρόταση έκανα!

A! Αλέξανδρε η λύση σου μου άρεσε πολύ!  :lol:

Στέλιος

Στέλιο δεν είμαι σίγουρος αλλά νομίζω ότι η άσκηση δεν λύνεται όπως λες :|

Για παράδειγμα αν πάρουμε την απλή περίπτωση του $ n\equiv 0\pmod{10}$ το τελευταίο ψηφίο και των τριών αριθμών είναι 1!!!

Να κι από μένα ίνα πισκέσι :rotfl:
Λύση
Γνωρίζουμε ότι
Τελευταίο ψηφίο του αριθμού ν^2={0,1,4,5,6,9,}
Αν τελευταίο ψηφίο ν^2= {1,6}
Τελευταίο ψηφίο 2*ν^2+1={3,3}
Αν τελευταίο ψηφίο ν^2= {4,5,9}
Τελευταίο ψηφίο 3*ν^2+1={3,6,8}
Αν τελευταίο ψηφίο ν^2=0      ν^2=100μ (*)
Αν υποθέσουμε ότι υπάρχει ν που αντίκειται στο ζητούμενο τότε
2ν^2+1=χ^2
3ν^2+1=ψ^2
6ν^2+1=ω^2
Κι με βάση τις σχέσεις αυτές καταλήγουμε σε αυτή την σχέση

   (χ*ψ)^2 =ω^2+5ν^2

Κι με αντικατάσταση της  (*) σε αυτήν κι μετά τις πράξεις έχουμε
100μ^2+1-μ=0  η οποία μέσω διακρίνουσας   διαπιστώνουμε  ότι δεν έχει λύσεις άρα δεν είναι δυνατόν να υπάρχουν ν  τέτοια ώστε να ισχύει η  υπόθεση

Υ.Γ το δεύτερο σκέλος το παλεύω

Hmm... μάλλον έχεις δίκιο Δημήτρη.

Χωρίς να είμαι σίγουρος, ας δούμε λίγο την περίπτωση $ n \equal{} 0\pmod{10}$.

Έχουμε:

Έστω $ 2n\plus{}1 \equal{} a^{2}$ και $ 3n\plus{}1 \equal{} b^{2}$

Για $ n \equal{} 0\pmod{10}$, θα έχουμε ότι ο $ a^{2}$ παίρνει τη μορφή $ 2\ldots1$ και ο $ b^{2}$ τη μορφή $ 3\ldots1$ (ή $ a^{2}\plus{}10^{n}$)

Δηλαδή $ 10^{n}\equal{} b^{2}\minus{}a^{2}\equal{} (b\minus{}a)(b\plus{}a)$.

Προφανώς ($ 10 \equal{} 2\cdot5$ , $ 10^{n}\equal{} 2^{n}\cdot5^{n}$ (αδύνατο να γραφτεί έτσι η πρώτη, και οι δύο είναι άρτιοι κ.λπ.)

Σίγουρα αυτό δε σημαίνει πως έχουμε αποκλείσει εντελώς την περίπτωση αυτή. Πάντως, πάνω-κάτω αυτές είναι οι σκέψεις που έκανα για το $ n \equal{} 0\pmod{10}$, με κάθε επιφύλαξη. Θα το ψάξω λίγο ακόμα. Προφανώς μόνο η περίπτωση $ n \equal{} 0\pmod{10}$ είναι η "ελαττωματική", αλλά έχω την πεποίθηση ότι βγαίνει και έτσι  :P

Στέλιος


AMC
AIME
modular arithmetic
number theory

How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?

[hide]
2n-1 = n^2 - (n-1)^2.
So we can make any odd integer from 1...1000.  That's 500.
4n = (n+1)^2 - (n-1)^2
So we can make any multiple of 4 from 1...1000.  That's a total of 750.

All squares are 0 or 1 mod 4, so their difference must be 0,1, 3 mod 4.  So we can't make any integer of the form 4n+2.  So our final answer is 750.[/hide]

[hide]The number can be written in the form $N=x^2-y^2=(x+y)(x-y)$.

If $x$ is even and $y$ is even, then both $x+y$ and $x-y$ are even, so $N$ is the produce of two even numbers.

If one of $x$ and $y$ odd, then both $x+y$ and $x-y$ are odd, so $N$ is the product of two odd numbers.

If both $x$ and $y$ are odd then $x+y$ and $x-y$ are even, so $N$ is the product of two even numbers.

If $N$ is the product of two even numbers, say $2m$ and $2n$, then $2*2*mn\equiv 0 \pmod4$.  There are $250$ multiples of $4$ from $1$ and $1000$.

If $N$ is the product of two odd integers, say $2m+1$ and $2n+1$ then $(2m+1)(2n+1)=4mn+2m+2n+1\equiv 1 \pmod2$.  This is the same as all odd integers, of which there are $500$.

So final answer $250+500=750$.[/hide]

Haha, everyone knows Tim... what a beast...

different solution(although stupider and longer, this is god for people like me who dont know much about number theory):

[hide]
well, we know that a perfect square can be express as the sum of consecutive odd integers starting from one. so the difference of 2 perfect squares will be sum of 1 or more consecutive odd integers. Immediately we see that all odd number works. for even number with the form $2(2n)$, we can express it as $(2n - 1) + (2n + 1)$, which is the sum of two consecutive odds. but integers with the form of $2p$, where $p$ is odd, can never be express as sum of consecutive odds. so the answer is $500 + 250 = 750$
[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

If $ G$ is a group with $ |G|\equal{}2009$ find the number of solutions of the equation $ x^{49}\equal{}e$ in $ G$.

What can you say about the number of Sylow 7-subgroups of $ G$?


For $ a,b,c$ positive reals prove that

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

[quote="manlio"]For $ a,b,c$ positive reals prove that

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

I think you create it from

You are right!!

Write it as

$ \sum \frac{ca}{b}(\frac{a}{c}\plus{}\frac{b}{a})$ and apply CRUX 3076 to obtain directly the ineq.

[quote="arqady"][quote="manlio"]You are right!!

Write it as

$ \sum \frac {ca}{b}(\frac {a}{c} \plus{} \frac {b}{a})$ and apply CRUX 3076 to obtain directly the ineq.[/quote]
Are you sure?[/quote]
$ \sum \frac {ca}{b}(\frac {a}{c} \plus{} \frac {b}{a}) \geq 2 \sqrt {(\sum \frac {ac}{b}\frac {ab}{c} )(\sum \frac {a}{c}.\frac {b}{a})}$
   ineq proved completey

[quote="arqady"]Ohh, thank you, Hong Quy! My mistake.
Very nice proof, shfdfzhjj![/quote]

Thank you，arqady.I think you are a very kind man. :)
Can you help me? http://www.mathlinks.ro/Forum/viewtopic.php?t=188479
I am very eager to konw how to slove it. :wink:
I think you can! :lol:


calculus
integration
trigonometry
calculus computations

uhh

$ \tan x \in [0, 1]$ as x ranges over $ [0, \frac{\pi}{4}]$
and $ \frac{\pi}{4}<1$ too, so..

$ x \tan x \le 1$ for each of these x, and

$ (x\tan x)^{n\plus{}1} \le (x\tan x)^n$ for all x in there, so the integrals should be in descending order..?


Consider the dark square in an array of unit squares, part of which
is shown. The first ring of squares around this center square
contains 8 unit squares. The second ring contains 16 unit squares.
If we continue this process, then what is the number of unit squares in the
$ 100^{th}$ ring?

[asy]fill((2,2)--(7,2)--(7,7)--(2,7)--cycle,gray(0.8));
fill((3,3)--(6,3)--(6,6)--(3,6)--cycle,gray(0.6));
fill((4,4)--(4,5)--(5,5)--(5,4)--cycle,black);
for (int i=0; i<10; ++i) {
draw((0,i)--(9,i));
draw((i,0)--(i,9));
}[/asy]

For the first part of the solution, an example of the four rectangles that are spoke of could be given to allow students to see them,

For the second part, an example of the larger square minus the smaller square would be a nice visual to help understand the solution

A third visual way to show would be an applet showing the rings breaking up and filling in the next ring,

the top section could rise one, the bottom section fall one, and the left and right sections spread out showing the 8 vacant unit squares in the next ring, fill that ring in with 8 red squares, fade to the normal color, and repeat. Students will see that the next ring always needs 8 more units to be completed.

In this way, the answer is $ 8 \times 100$, or $ \boxed{800}$.

I have a slightly alternative proof

Imagine four equal rectangles composed of unit squares that are put in a ring around the other rings

Then the square units in the rings are always $ 2n$

And there are four rings, therefore the total square units in the ring $ 4\cdot(2n)$

$ n\equal{}100, \text{ therefore } 800.$

Is this valid? Or am I just reiterating the solution...

Instead of subtracting number of unit squares from previous rectangle we can also find total no of units in every rectangle.
The number of unit squares along any rectangle will be 2n + 1

So number of units will be :
= 2(2n + 1) - 2(2n - 1)
= 8n

That's what I thought (almost the same as the 2nd solution).

The nth square has the same a "perimeter" equal 4 times the side of the (n-1)th square plus 4, that is, 4(2n-1)+4=8n.

[hide=AMC 10 Solution ]We can partition the $n^\text{th}$ ring into $4$ rectangles: two containing $2n+1$ unit squares and two containing $2n-1$ unit squares.

There are $2(2n+1)+2(2n-1)=4n+2+4n-2=8n$ unit squares in the $n^\text{th}$ ring.

Thus, the $100^\text{th}$ ring has $8 \times 100 = \boxed{ 800}$ unit squares.[/hide] Last post was 2010 :what?:  :oops_sign:

This problem is AMC 10 2001, Problem 11.

[hide=Solution]Given that the first ring contains $8$ squares, and the second ring contains $16$ squares, we can see that there is a $8:1$ ratio from the amount of squares to rings. So, using this fact, this means there must be $800$ squares in order to satisfy the condition.[/hide]


FInd all  $ f : \mathbb{R} \rightarrow \mathbb{R}$
$ f( f(x) \minus{} f(y) ) \equal{} (x\minus{}y)^2 \cdot f(x\plus{}y)$.

change th position of x,y get that $ f(x)\equal{}f(\minus{}x)$ and put y->-y and get easily that $ f(x)\equal{}x^2$ ..if f is constant then f is equal to zero...... :o

[quote="lefmast"]change th position of x,y get that $ f(x) \equal{} f( \minus{} x)$[/quote]

How do you know every real is representable as f(x)-f(y)?

dgreenb801 is right, and for that reason, lefmast neglected the solution $ f(x)\equal{}\minus{}x^2$.

To prove $ f(x)\minus{}f(y)$ takes on all values, it is sufficient to show that $ f$ takes on either all nonnegative or all nonpositive values (it is easily found that $ f(0)\equal{}0$).  If $ f(x\plus{}y)\ne0$ for some value of $ x\plus{}y$, we use this and we may get any value for $ x\minus{}y$, thus achieving all nonnegative or nonpositive values for $ f$ from $ f( f(x)\minus{}f(y) ) \equal{} (x\minus{}y)^{2}\cdot f(x\plus{}y)$.


geometry
circumcircle
projective geometry
geometric transformation
rotation
geometry proposed

let ABC and BCD be two different equilateral triangles. If N and M are the points where a line r through D cuts AB and AC respectively, let P = BM^CN. Find the locus of possibles position of P when r varies.

It's the circumcircle of $ABC$. We look at it backwards: for any point $M$ on $AC$, we can take $P$ to be $BM\cap(ABC)$, $N=AB\cap PC$. In this case the line $MN$ is the polar of $AP\cap BC$, which lies on $BC$, so it must pass through the pole of $BC$ wrt $(ABC)$, which is $D$.

Really impressive proof, grobber. A part from be situation independent, also generalize the results which is valid even when ABC is a generic triangle and D is the common points to the tangents to (ABC) on B and C.

here another way (which is situation dependent) and makes use of all hypothesis given (for this reason, I think, is not valid for a more general case).

I describe the case where M is internal to the segment AC (the other cases are similar).

Since APM and BCD are perpective wrt N, by Desargues theorem, the intersections of the corresponding sides are collinear. This means that if A'=AP^BC and B'=BP^DC, as AM//BD, that A'B'//BD//AM. In other words A'B'C is equilateral. The two triangles ACA' and BCB' are congruent, as one can obtain one from the other by a rotation of 60. This means, in particular, that <PAC = <PBC, i.e. ABCP is ciclic.

Can some one find a completely classic euclidean proof of that results?


In professional football it is possible to score $ 6$ points(a touchdown), $ 3$ points(field goal), or $ 2$ points(a safety). If a touchdown is scored it is possible to score an additional point (the point after). What is the largest total score that cannot be achieved in football?

you can make anything except 1 with 3's and 2's
so 1

[hide]if you can have every units digit #, hten you can have every number from thereon because 5 safeties=10pts
you can have

safety=2
FG=3
2 safeties=4
FG+safety=5
TD=6
2 safeties+FG=7
TD+safety=8
TD+FG=9
TD+2 safeties=10
TD+FG+safety=11

thats every units digit

so the smallest impossible score=1[/hide]

[quote="abacadaea"]thats what everyone else just said[/quote]

yea, but they didnt explain it thoroughly

yeah. that was thurough enough....for me atleast


Maybe it is well known,but if you try to prove it by yourself it  is so interesting:

Prove ther AM_GM by the rearrangement inequailty.

Well, the solution that i know for this is beautiful but hard too reach.
I will give some guidelines for anyone who is interested.
[hide]

Let $C=\sqrt[n]{x_1...x_n}$

We should prove :

$\displaystyle \frac{\displaystyle \sum_{i=1}^n x_i}{n} \ge C$

Or:

$\sum_{i=1}^n \frac{x_i}{C} \ge (1+1+...+1)(n$ times$)$
.

We want to use rearrangement so we sould build two sequences like

$\{a_i\}_{i=1}^n,\{b_i\}_{i=1}^n$ such that :

$a_i \le a_j \leftrightarrow b_i \ge b_j$

To have :

$a_1b_{n}+a_2b_{n-1}+...+a_nb_1 \ge a_1b_1+a_2b_2+...+a_nb_n$

for changing it to :

$\sum_{i=1}^n \frac{x_i}{C} \ge (1+1+...+1)(n$ times$)$

We should have $b_i=\frac{1}{a_i}$(which means$a_ib_i=1,\sum a_ib_i=n)$

So what remains to  do is just to build the $\{a_i\}_{i=1}^n$ such that :
$a_1b_n+a_2b_{n-1}+...+a_nb_1=\frac{a_1}{a_n}+\frac{a_2}{a_{n-1}}+...+\frac{a_n}{a_1}=\displaystyle \sum_{i=1}^n \frac{x_i}{C}$[/hide]
[hide] What is your idea about the $\{a_i \} _{i=1}^n$? :D  :D [/hide]

You can compare your method of proof by the proof given here:
http://matholymp.com/TUTORIALS/Rear.pdf

Interesting article.
That method is not mine.I read it in book.
As oldman has said:
 the sequance shall be made in this way:

$a_i=\frac{x_1...x_i}{c^i}$
I think this is a interesting problem.


Find a real number $x$ such that $\sum^{1996}_{k=0}|x-\sqrt{k}|$ attains its maximum

Could you not just stick in a massive $x$ ???

No, you can't.  Now stop saying stupid things about the power of calculus, and solve it if you can (and yes, the question should be asking about minima)

This isn't too bad.

[hide]x should always be increased if there are more values of $\sqrt{k}$ above it than below to lower the sum, and vice versa.  Therefore x = $\sqrt{998}$.  I don't see calculus coming here anywhere...[/hide]

Considering $\sum_{i=0}^{998} |x - \sqrt{i}| + |x - \sqrt{1996-i}|$, the (i+1)th term is minimized obviously when x is between $\sqrt{i}$ and $\sqrt{1996-i}$.  All of these minimums can be realized when x is between $\sqrt{998}$ and $\sqrt{999}$, thus, this is where the minimum occurs.


I've just learned about rank of matrices, and here is a problem after this lesson, they asked me using properties of rank of matrices to solve it, but I don't know how  :( . Here is the problem:

Find $ \lambda$ for which vector $ X$ is a linear combination of the other vectors.
a) $ X\equal{}(7, \minus{}2, \lambda)$, $ X_1\equal{}(2, 3, 5)$, $ X_2\equal{}(3, 7, 8)$ and $ X_3\equal{}(1, \minus{}6, 1)$
b) $ X\equal{}(1, 3, 5)$, $ X_1\equal{}(3, 2, 5)$, $ X_2\equal{}(2, 4, 7)$ and $ X_3\equal{}(5, 6, \lambda)$

Thanks in advance

Since in this particular case we are working in $ \mathbb{R}^3$, vector X will be a linear combination of the other three vectores iff X belongs to the plane generated by X1, X2, X3. In other words, we just have

$ \left | \begin{array}{cccc} 7 & \minus{} 2 & \lambda & 1 \\
2 & 3 & 5 & 1 \\
3 & 7 & 8 & 1 \\
1 & \minus{} 6 & 1 & 1 \end{array} \right| \equal{} 0$ and $ \left | \begin{array}{cccc} 1 & 3 & 5 & 1 \\
3 & 2 & 5 & 1 \\
2 & 4 & 7 & 1 \\
5 & 6 & \lambda & 1 \end{array} \right| \equal{} 0$.

However, observe that the vectors X1, X2, X3 from (a) are linearly dependent, since

$ \left | \begin{array}{ccc} 2 & 3 & 5 \\
3 & 7 & 8 \\ 1 & \minus{}6 & 1 \end{array} \right| \equal{} 0$,

and so Span(X1,X2,X3) is a line. Therefore, for (a) we must assure that X belongs to the same line generated by, for instance, X1 and X2. In matrix language, we are to demand the rank of the matrix

$ \left [ \begin{array}{ccc} 7 & \minus{} 2 & \lambda \\
2 & 3 & 5 \\
3 & 7 & 8 \\
1 & \minus{} 6 & 1 \end{array} \right]$

to be 2.


Brandon has an 8-inch square pan. Barbara has a 9-inch square pan. Both pans have the same volume. What is the ratio of the height of Barbara's pan to the height of Brandon's pan? Express your answer as a common fraction.

Let's say that the volume of the pans are $ 72$.  Then, Brandon's pan would have height $ 9$ and Barbara's height $ 8$.  Thus, the answer is $ \boxed{\frac89}$.

BTW, you can just choose a volume because for any volume x the hieght are $ \frac{x}{8}$ and $ \frac{x}{9}$, so the x's cancel out.

However, the area is found by squaring the side lengths so it's the square of both.  Therefore, the answer is $\boxed{\frac{64}{81}}$.

Wait, what do you mean by area?  The question is asking for the ratios of the heights...or am I missing something here?

Well, the volume is found by multiplying the area of the bottom and the height together.  Let the height of Brandon's pan be $y$ and the height of Barbara's pan by $x$, and the volumes are the same so:
\begin{align*}8^2x&=9^2y \\ 64x&=81y \\ \frac{y}{x}&=\boxed{\frac{64}{81}}\end{align*}


algebra
polynomial
trigonometry
inequalities
calculus
integration
superior algebra

Superior algebra $ \neq$ algebra with superior computer power. ;)

  dg

Could somebody please send this "thing"  where it belongs: inequalities forum, the place where all computational horrors and nightmares become true... ? I suggest creating a whole book with such kind of mathematical creatures, just to prove how uninteresting mathematics can become when playing around with a computer in a totally useless way.  :(

2 Darij and Harazi: I am not sure that this question is so much senseless. The general problem to determine when a non-negative poynomial is a sum of squares is of much interest (it's intimately related with the Hilbert's 17th problem), and some particular machineries, which lead to such representations, may be of interest.

As for computer's help, it looks like these numbers are just roots of some explicit polynomial with integral coefficients.

My question is: are those the only polynomials such that the equality takes place? (i.e. is this representation as sum of squares of real polynomials unique?)

You should be careful when asking this: the group of orthogonal transformations in $ \mathbb R^4$ is quite large and I don't think you are interested in solutions differing from the given one by an orthogonal transformation. If you allow more than 4 polynomials, you have obvious extra options in the same venue. Also, if you have the product of two sums of 2 squares, you can represent it as a sum of 2 squares in two seemingly different ways: $ (ac\minus{}bd)^2\plus{}(ad\plus{}bc)^2\equal{}(ac\plus{}bd)^2\plus{}(ad\minus{}bc)^2$ and I do not see how these 2 representations can be obtained from each other by a simple orthogonal transformation when $ a,b,c,d$ are arbitrary polynomials. On the other hand, it is quite plausible that the set of all possible representations has some algebraic structure, at least, "generically"; I just don't know :(.

[quote=Ji Chen]Exist real polynomials $ p_1,p_2,p_3,p_4$ in $ a,b,c$ such that

$ \left(a^2 \plus{} b^2 \plus{} c^2\right)^3 \minus{} 8\left(b^3c^3 \plus{} c^3a^3 \plus{} a^3b^3\right) \equal{} p_1^2 \plus{} p_2^2 \plus{} p_3^2 \plus{} p_4^2.$

[hide="Proof."]$ \left(a^2 \plus{} b^2 \plus{} c^2\right)^3 \minus{} 8\left(b^3c^3 \plus{} c^3a^3 \plus{} a^3b^3\right)$

$ \equal{} r(b \minus{} c)^2(c \minus{} a)^2(a \minus{} b)^2$

$ \plus{} \left[sa^3 \plus{} ta^2(b \plus{} c) \plus{} ua\left(b^2 \plus{} c^2\right) \plus{} vabc \plus{} w(b \plus{} c)(b \minus{} c)^2\right]^2$

$ \plus{} \left[sb^3 \plus{} tb^2(c \plus{} a) \plus{} ub\left(c^2 \plus{} a^2\right) \plus{} vabc \plus{} w(c \plus{} a)(c \minus{} a)^2\right]^2$

$ \plus{} \left[sc^3 \plus{} tc^2(a \plus{} b) \plus{} uc\left(a^2 \plus{} b^2\right) \plus{} vabc \plus{} w(a \plus{} b)(a \minus{} b)^2\right]^2,$

where $ r \equal{} \frac {49}{3} \minus{} \frac {10\sqrt {65}}{3}\sin{\left(\frac {\pi}{6} \minus{} \frac {1}{3}\arccos{\frac {13049}{1625\sqrt {65}}}\right)} \equal{} 3.5946\cdots,$

$ s \equal{} \sqrt { \minus{} \frac {2}{9} \plus{} \frac {16\sqrt {5}}{9}\sin{\left(\frac {\pi}{6} \minus{} \frac {1}{3}\arccos{\frac {1}{\sqrt {5}}}\right)} \minus{} \frac {40}{9}\sin^2{\left(\frac {\pi}{6} \minus{} \frac {1}{3}\arccos{\frac {1}{\sqrt {5}}}\right)}}$

$ \plus{} \frac {2\sqrt {5}}{3}\sin{\left(\frac {\pi}{6} \minus{} \frac {1}{3}\arccos{\frac {1}{\sqrt {5}}}\right)} \minus{} \frac {2}{3} \equal{} 0.096086\cdots,$

$ t \equal{} \sqrt { \minus{} \frac {2}{9} \plus{} \frac {22\sqrt {5}}{9}\sin{\left(\frac {\pi}{6} \minus{} \frac {1}{3}\arccos{\frac {1}{\sqrt {5}}}\right)} \minus{} \frac {40}{9}\sin^2{\left(\frac {\pi}{6} \minus{} \frac {1}{3}\arccos{\frac {1}{\sqrt {5}}}\right)}}$

$ \minus{} \frac {2\sqrt {5}}{3}\sin{\left(\frac {\pi}{6} \minus{} \frac {1}{3}\arccos{\frac {1}{\sqrt {5}}}\right)} \plus{} \frac {2}{3} \equal{} 1.1540\cdots,$

$ u \equal{} \minus{} \sqrt { \minus{} \frac {11}{18} \plus{} \frac {10\sqrt {5}}{9}\sin{\left(\frac {\pi}{6} \minus{} \frac {1}{3}\arccos{\frac {1}{\sqrt {5}}}\right)} \minus{} \frac {10}{9}\sin^2{\left(\frac {\pi}{6} \minus{} \frac {1}{3}\arccos{\frac {1}{\sqrt {5}}}\right)}}$

$ \minus{} \frac {2\sqrt {5}}{3}\sin{\left(\frac {\pi}{6} \minus{} \frac {1}{3}\arccos{\frac {1}{\sqrt {5}}}\right)} \plus{} \frac {2}{3} \equal{} \minus{} 0.54628\cdots,$

$ w \equal{} \minus{} \sqrt { \minus{} \frac {1}{18} \plus{} \frac {4\sqrt {5}}{9}\sin{\left(\frac {\pi}{6} \minus{} \frac {1}{3}\arccos{\frac {1}{\sqrt {5}}}\right)} \minus{} \frac {10}{9}\sin^2{\left(\frac {\pi}{6} \minus{} \frac {1}{3}\arccos{\frac {1}{\sqrt {5}}}\right)}}$

$ \plus{} \frac {2\sqrt {5}}{3}\sin{\left(\frac {\pi}{6} \minus{} \frac {1}{3}\arccos{\frac {1}{\sqrt {5}}}\right)} \minus{} \frac {2}{3} \equal{} \minus{} 0.70383\cdots,$

$ v \equal{} 2\sqrt {5}\sin{\left(\frac {\pi}{6} \minus{} \frac {1}{3}\arccos{\frac {1}{\sqrt {5}}}\right)} \minus{} 3 \equal{} \minus{} 2.3115\cdots.$[/hide][/quote]

If more than four polynomials are allowed, SOStools can give exact SOS as follows (In matrix form,
but it is easy to derive the sum of squares of $7$ polynomials):
$$\left(a^2 + b^2 + c^2\right)^3 - 8\left(b^3c^3 + c^3a^3 + a^3b^3\right) = \frac{1}{4}Z^\mathsf{T}QZ$$
where
$$Z = \left[\begin{array}{cccccccccc} a^3, & a^2\, b, & a\, b^2 ,& b^3 ,& a^2\, c ,& a\, b\, c ,& b^2\, c ,& a\, c^2 ,& b\, c^2 ,& c^3 \end{array}\right]^\mathsf{T}$$
and
$$Q = \left(\begin{array}{rrrrrrrrrr} 4 & 0 & -3 & -1 & 0 & 1 & 1 & -3 & 1 & -1\\ 0 & 18 & -15 & -3 & -1 & -1 & 0 & 0 & 2 & 1\\ -3 & -15 & 18 & 0 & 0 & -1 & -1 & 2 & 0 & 1\\ -1 & -3 & 0 & 4 & 1 & 1 & 0 & 1 & -3 & -1\\ 0 & -1 & 0 & 1 & 18 & -1 & 2 & -15 & 0 & -3\\ 1 & -1 & -1 & 1 & -1 & 12 & -1 & -1 & -1 & 1\\ 1 & 0 & -1 & 0 & 2 & -1 & 18 & 0 & -15 & -3\\ -3 & 0 & 2 & 1 & -15 & -1 & 0 & 18 & -1 & 0\\ 1 & 2 & 0 & -3 & 0 & -1 & -15 & -1 & 18 & 0\\ -1 & 1 & 1 & -1 & -3 & 1 & -3 & 0 & 0 & 4 \end{array}\right).
$$

[quote=dragonheart6]
option.solver = 'sedumi';
[Q, Z] = findsos(p,'rational',option);[/quote]
I type same, but my result is different.
[img]https://i.imgur.com/gnVFr6O.png[/img]

[quote=SBM][quote=dragonheart6]
option.solver = 'sedumi';
[Q, Z] = findsos(p,'rational',option);[/quote]
I type same, but my result is different.
[hide][img]https://i.imgur.com/gnVFr6O.png[/img][/quote][/hide]

Strange. But your result is also simple (the numbers are not large).


When do we receive our AIME scores? Will we have enough time to ask for a hand grading if necessary? (I had to erase some stuff on my answer form, and I'm not sure if the machine will read my erased marks). Also, when hand grading, do the bubbles or the numbers written over the bubbles take precedence?

I heard that the bubbles are all that matter; you can even leave the boxes over the bubbles blank.

I believe you can request a hand-grading, but there is a fee involved.  Contact the AMC.

"A discrepency between the numbers written and the bubbled numbers is grounds for ruling a question incorrect" (or something to that effect on the AIME problem booklet). But I think they'd just go with the bubbles.

There is still a chance they will count answers that you have written correctly but misbubbled; however, if it makes the difference between USAMO or no USAMO I doubt they'll give it to you.

[quote="JSteinhardt"]But then why not write $004$ and bubble $003$ if you are unsure of which answer it is?[/quote]

You have a good point.  They wouldn't know if you were doing that or if it was an honest mistake.


probability
combinatorics proposed
combinatorics

The player has $ {k}(0<k<100)$ coins. For one course it or loses one coin, or wins. The probability to lose a coin at each course is equal $ {1/3}$. What probability to win 100 coins before to lose for all?

Up to errors on the order of $ 10^{\minus{}30}$, the answer is $ 1 \minus{} \frac{1}{2^k}$.  (These are the probabilities that you never visit the origin in an infinite walk of this sort with no absorption at 100.  By the way, where can I play this game?)


Last year's Canadian IMO team had no Canadian flag(not a single one). It was quite sad since many countries had so many flags or an enormous flag. :D

I have this small Canadian flag, which I'm planning to bring. Please bring a Canadian flag if you(team members) have any. :)

[quote="blackegg2"]aww too bad CMO team has to leave before canada day. or else you can get a ton of free mini canadian flags!!![/quote]
We are going to be in Canada on Canada day though. :?  We are going to leave on 11th.

[quote="billzhao"]Richard says that he'll bring his huge flag.[/quote]
Yeah!!!! :)  We have a huge flag now!!!

[quote="Sunny"]lol, I should get you this pin wheel with maple leaves all over it.  David, you'll love it.  I could so see you standing there like a 5 year old with the pin wheel making it spin, lol.[/quote]
Hmm...  :huh: I wonder...  :!: LMAO!  :rotfl:

Someone should  dye a maple leaf flag blue, for the Toronto Maple Leafs!  :D

[quote="Sunny"]lol, I should get you this pin wheel with maple leaves all over it.  David, you'll love it.  I could so see you standing there like a 5 year old with the pin wheel making it spin, lol.[/quote]
What is 'pin wheel'? :?

[quote="Sunny"][quote="lightrhee"]What is 'pin wheel'? :?[/quote]

 :o !!!!  You don't know!?  I'm so bringing you one now!!![/quote]
You still didn't tell me what it is!! What is it? :?

[quote="Sunny"][quote="lightrhee"][quote="Sunny"]Wait and see.   :D[/quote]
Thank you for the pin wheels. :D
They are fun!!!![/quote]

Hahaha awesome.  Did you fix that broken one yet?[/quote]
Not yet :( . I need to get a tape somehow.

[quote="Sunny"][quote="lightrhee"]
Not yet :( . I need to get a tape somehow.[/quote]

Considering how the nice the place you're staying at is, shouldn't be too hard, they probably have everything![/quote]
I agree. Even though I heard how good this place is from many people, this place is even better than I expected. I might get fat(for the first time in my life) by staying here for 10 days or so(the food is so good :D ). Me being fat is hard to imagine. :rotfl:

[quote="lightrhee"]
I agree. Even though I heard how good this place is from many people, this place is even better than I expected. I might get fat(for the first time in my life) by staying here for 10 days or so(the food is so good :D ). Me being fat is hard to imagine. :rotfl:[/quote]

*Sarah falls out of chair laughing when picturing a pig with David's head*

Good thing that we have many Canadian flags this year. We have at least
3 huge flags
2 small flags
2 pin wheels.

Comparing to the last year(no flag), this is very good. :)

[quote="Sunny"][quote="d.lam.86"]Don't get too wrapped up with the team "look" of Canada. You're trying to win medals and self-honour (I assume) and I tend to have these superstitions that whenever I get too wrapped up into the preparation, I fail the purpose of the whole journey/endeavour. GOOD LUCK.  :)[/quote]

You guys would look better if you all wore the flags, lol.  Only need 3 more!  I personally think Good luck, but have fun.  Hehehe...enjoy the problems...if you worry too much about getting medals....kinda defeats the spirit...well, that's only my two cents.[/quote]

Well... you know my attitude toward that... ;)

[quote="Elyot"][quote="Sunny"][quote="Elyot"]

Well... you know my attitude toward that... ;)[/quote]

Hahahaha, spray paint all the way.[/quote]

Instructions for winning IMO gold medal:

1) Win IMO bronze medal:  :winner_third:
2) Buy can of gold spray paint:  :harhar:
3) Spray medal gold:  :first:[/quote]
Are the medals even worth any sum of money?  :roll:

[quote="d.lam.86"][/quote][quote="Elyot"][quote="Sunny"][quote="Elyot"]

Well... you know my attitude toward that... ;)[/quote]

Hahahaha, spray paint all the way.[/quote]

Instructions for winning IMO gold medal:

1) Win IMO bronze medal:  :winner_third:
2) Buy can of gold spray paint:  :harhar:
3) Spray medal gold:  :first:[/quote][quote="d.lam.86"]
Are the medals even worth any sum of money?  :roll:[/quote]

Would anyone in their right mind sell one???  If you won at the olympics or something, you wouldn't wanna sell it...

*hee* just curious, that's all. I haven't really considered the IMO as part of the Olympics. It just doesn't seem to fit in my head because I tend to think the Olympics are only for sports and mainly nothing else, but I guess I'm mistaken, eh?

[quote="Elyot"]Well... guys like me would like to consider it pretty close to an equivalent... I mean, it's not as big because it doesn't get the TV coverage that the olympics get (kinda hard to make an interesting TV series out of kids writing math proofs) but it's still very prestigious and well-recognized throughout the math community.[/quote]
I'd be interested in watching the IMO but you're right—not a large enough population of the world has such interests. Who knows? They already broadcast spelling competitions and MathCounts on television. The International Mathematics Olympiad is most likely next.

[quote="d.lam.86"]
I'd be interested in watching the IMO but you're right—not a large enough population of the world has such interests. Who knows? They already broadcast spelling competitions and MathCounts on television. The International Mathematics Olympiad is most likely next.[/quote]

I doubt it...i mean...in those competitions, you can see them doing something.  Spelling...or racing against one another.  That's not the point of the IMO, all you'll see is them writing proofs.

[quote="TheDreamer"]

I'm still convinced that if they can make marathon coverage exciting, then they can make writing math proofs exciting.  Like they can have cameras hovering over you so that ppl can see how the pro's solve problems...and a skilled commentator who can build suspense.  In marathons the commentator kinda just repeats "oh, he's fallen out of the picture, he has used it all on the first uphill" which is comparable to "Elyot has discovered that AQRT is cyclic!! omg omg, he is so close to solving the problem..he has 15 minutes to realize that the areas are the same and he will win a gold medal.." etc.

I'm sure I'd watch that and cheer you guys on in front of the TV.  :D[/quote]
HAhaha they filmed us trying math problems on tuesday when the media came.  It was really funny, they got a couple of us to sit down and solve problems, so I gave them my own geometry problem.  Too bad we didn't have a commentator...

[quote="TheDreamer"][quote="Elyot"]Well... guys like me would like to consider it pretty close to an equivalent... I mean, it's not as big because it doesn't get the TV coverage that the olympics get (kinda hard to make an interesting TV series out of kids writing math proofs) but it's still very prestigious and well-recognized throughout the math community.[/quote]

I'm still convinced that if they can make marathon coverage exciting, then they can make writing math proofs exciting.  Like they can have cameras hovering over you so that ppl can see how the pro's solve problems...and a skilled commentator who can build suspense.  In marathons the commentator kinda just repeats "oh, he's fallen out of the picture, he has used it all on the first uphill" which is comparable to "Elyot has discovered that AQRT is cyclic!! omg omg, he is so close to solving the problem..he has 15 minutes to realize that the areas are the same and he will win a gold medal.." etc.

I'm sure I'd watch that and cheer you guys on in front of the TV.  :D[/quote]
That would be awfully strange and awkward.  :rotfl:

[quote="TheDreamer"][quote="Elyot"]Well... guys like me would like to consider it pretty close to an equivalent... I mean, it's not as big because it doesn't get the TV coverage that the olympics get (kinda hard to make an interesting TV series out of kids writing math proofs) but it's still very prestigious and well-recognized throughout the math community.[/quote]

I'm still convinced that if they can make marathon coverage exciting, then they can make writing math proofs exciting.  Like they can have cameras hovering over you so that ppl can see how the pro's solve problems...and a skilled commentator who can build suspense.  In marathons the commentator kinda just repeats "oh, he's fallen out of the picture, he has used it all on the first uphill" which is comparable to "Elyot has discovered that AQRT is cyclic!! omg omg, he is so close to solving the problem..he has 15 minutes to realize that the areas are the same and he will win a gold medal.." etc.

I'm sure I'd watch that and cheer you guys on in front of the TV.  :D[/quote]

Best.  Post.  Ever.
 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

[quote="TheDreamer"][quote="Elyot"]Well... guys like me would like to consider it pretty close to an equivalent... I mean, it's not as big because it doesn't get the TV coverage that the olympics get (kinda hard to make an interesting TV series out of kids writing math proofs) but it's still very prestigious and well-recognized throughout the math community.[/quote]

I'm still convinced that if they can make marathon coverage exciting, then they can make writing math proofs exciting.  Like they can have cameras hovering over you so that ppl can see how the pro's solve problems...and a skilled commentator who can build suspense.  In marathons the commentator kinda just repeats "oh, he's fallen out of the picture, he has used it all on the first uphill" which is comparable to "Elyot has discovered that AQRT is cyclic!! omg omg, he is so close to solving the problem..he has 15 minutes to realize that the areas are the same and he will win a gold medal.." etc.

I'm sure I'd watch that and cheer you guys on in front of the TV.  :D[/quote]
 :rotfl:


The qn was of physics , but here's the problem ->
A particle of mass 2kg is subjected to a 2D conservative force given by

Fx = - 2x + 2y
Fy = 2x - y^2

Now just give me the eqn of U (potential energy)

[hide="my attempt"]
$ U\equal{}\minus{}\int(F_x)dx \minus{}\int(F_y)dy \minus{}\int(F_z)dz \plus{}C$

so,$ U\equal{}\minus{}\int(\minus{}2x\plus{}2y)dx \minus{}\int(2x\minus{}y^2)dy \plus{}C$
so,$ U\equal{}\minus{}(\minus{}x^2\plus{}2xy)\minus{}\left (2xy\minus{}\frac{y^3}{3} \right) \plus{}C$
so,$ U\equal{}x^2\minus{}4xy\plus{}\frac{y^3}{3} \plus{}C$

but when I partially differentiated the expression,it yielded $ F_x\equal{}\minus{}2x\plus{}4y$ and $ F_y\equal{}4x\minus{}y^2$
so there is an error in my way, :(
the answer would be $ U\equal{}x^2\minus{}2xy\plus{}\frac{y^3}{3}$(It yields right expressions for Fx and Fy)

Can you please give the answer and please tell where i went wrong?(i am not sure about integration part)
thanks[/hide]

It helps if you have some knowledge of differential equations. Basically, the mistake skand has committed is that the constant of integration when integrating f(x,y) with respect to x is a function of y.

The actual solution would go something like this.

$ \frac {d(\Delta U)}{dx} \equal{} 2x \minus{} 2y \Rightarrow U \equal{} x^2 \minus{} 2xy \plus{} \phi (y)$
$ \Rightarrow \frac {d(\Delta U)}{dy} \equal{} \minus{} 2x \plus{} \phi ' (y) \Rightarrow \minus{} 2x \plus{} \phi ' (y) \equal{} \minus{} 2x \plus{} y^2 \Rightarrow \phi (y) \equal{} \frac {y^3}{3} \Rightarrow \Delta U \equal{} x^2 \minus{} 2xy \plus{} \frac {y^3}{3}\plus{}C$

couldn't understand your second line :(

probably I need to learn differential equations

i had exactly the same doubt as u skand  :wink:
i'll wait for differential eqns to get over then  :D
sorry hell , couldn't follow that , haven't studied that yet

[hide]Till then , i'll just follow this rule ->
IF
 [img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/6/d/c/6dcdd3cc5f891e6ff75ea1662132a3d6b98426c9.gif[/img]

That is ,the common part(2xy) wouldn't be added while adding the components of U,this worked in another qn too :)[/hide]

good method :)  :)
you have brilliant brain :)

well that was the method that our physics teacher told us last yr :P
refering that "u bio ppl can never understand maths :P  :rotfl: "

see the method is pretty simple and has already been shown by hell_ever
but i want to mention thsi imp thing about mathematical opertaors ...
it wud be really good if u guys/gals(mythili  :P  :P  :rotfl: ) readabt mathematical operators before hand.
check out this wiki site about [url=http://en.wikipedia.org/wiki/Gradient]gradient[/url]
there are otehr operators too like curl and divergence and their respective combo's
if u know these operators then ur lyf with theoritical physics becomes easier
see here by definition $ \vec{F} \equal{} \minus{} \nabla U$ and now use the definition of grad and compare teh coefficients and we are dun

I knew about curl (our teacher told us,i am his fan :) )but he told us about curl only to check whether any force is potential(comservative)or not
I was unaware of these applications
probably we'ld be told about gradient later(he mentioned it while teaching curl)

Can we use these in subjective examinations(i know,i can't use in school examination)but like boards or other subjective exams???

hi skand
why i toldthese things is because u shud know certain stuffs ....i cant tell u where this will help u or where u can apply them but knowng them surely makes u better.

hi par

seems you were active for a longer duration yesterday :)

i wanted to know,whether I can use these in subjective exams(i asked my teacher(school)and he asked me to refrain from these methods :(

you can surely use them in jee if it becomes subjective but dont use them in skool and all :P


LaTeX
superior algebra
superior algebra unsolved

Hey, the link above is broken, where we can go in order to get latex help.

thanks


It is possible to earn 0, 1, 3, 4 or 10 points with each shot in the game of Blippy. How many positive scores less than 30 cannot be achieved in three shots?

[hide] Running through general tests, with the knowledge of a few patterns, the scores that can not be obtained are $ 19,22,25,26,27,28,29$. This is a total of $ \boxed{7}$ scores.[/hide]

You just go do guess and check..
[hide] So then the scores that cannot be obtained are 19,22,25,26,27,28,and 29. So the answer to the problem is 7.[/hide]

I should've read the question more carefully. Thanks.

Ok.No problem, I see that you have edited your post!


$ a_i\ge0,\sum_{i \equal{} 1}^n a_i \equal{} 1,a_{n \plus{} 1} \equal{} a_1,a_{n \plus{} 2} \equal{} a_2$
$ f(n) \equal{} Max{\sum_{i \equal{} 1}^n a_ia_{i \plus{} 1}a_{i \plus{} 2}} \equal{} ?$
$ x1,x2,x3,x4,x5\ge0, x1 \plus{} x2 \plus{} x3 \plus{} x4 \plus{} x5 \equal{} 1$
$ 2(x1x2 \plus{} x2x3 \plus{} x3x4 \plus{} x4x5 \plus{} x5x1) \plus{} x1x3 \plus{} x2x4 \plus{} x3x5 \plus{} x4x1 \plus{} x5x2 < \equal{} 3/5$
$ x1,x2,x3,x4,x5,x6\ge0,x1 \plus{} x2 \plus{} x3 \plus{} x4 \plus{} x5 \plus{} x6 \equal{} 1$
$ 2(x1x2 \plus{} x2x3 \plus{} x3x4 \plus{} x4x5 \plus{} x5x6 \plus{} x6x1) \plus{} x1x3 \plus{} x2x4 \plus{} x3x5 \plus{} x4x6 \plus{} x5x1 \plus{} x6x2 < \equal{} 2/3$

[quote="arqady"][quote="shfdfzhjj"]
$ x1,x2,x3,x4,x5\ge0, x1 \plus{} x2 \plus{} x3 \plus{} x4 \plus{} x5 \equal{} 1$
$ 2(x1x2 \plus{} x2x3 \plus{} x3x4 \plus{} x4x5 \plus{} x5x1) \plus{} x1x3 \plus{} x2x4 \plus{} x3x5 \plus{} x4x1 \plus{} x5x2 < \equal{} 3/5$
[/quote]
The following inequality is also true.
Let $a,$ $b,$ $c,$ $d$ and $e$ are real numbers such that $a+b+c+d+e=1.$ Prove that:
\[2(ab+bc+cd+de+ea)+ac+bd+ce+da+eb\leq\frac{3}{5}\][/quote]
 Let $x:=a-b$ etc. Homogenizing yields (all sums are cyclic)
\[Ineq \Longleftrightarrow 8\sum ab\le 6\sum a^2+2\sum ac \\ \\
\Longleftrightarrow \sum (a-c)^2\le 4\sum (a-b)^2 \\ \\
\Longleftrightarrow \sum (3a-2b-c)(a-2b+c)\ge 0  \\ \\
\Longleftrightarrow \sum (3x+y)(x-y)\ge 0  \\ \\
\Longleftrightarrow \sum (2x^2-2xy)\ge 0   \\ \\
\Longleftrightarrow \sum (x-y)^2\ge 0  \]

In fact, this means simply \[Ineq \Longleftrightarrow \sum (a-2b+c)^2 \ge 0  \] :D

[quote="arqady"]The following inequality is also true.
Let $a,$ $b,$ $c,$ $d$ and $e$ are real numbers such that $a+b+c+d+e=1.$ Prove that:
\[2(ab+bc+cd+de+ea)+ac+bd+ce+da+eb\leq\frac{3}{5}\][/quote]
Using the Cauchy-Schwarz Inequality, we have
$4=\left[ \sum (b+e)\right]^2\le 5 \sum (b+e)^2=10 \left( \sum a^2+\sum ac\right).$
It follows that
$\sum a^2+\sum ac \ge \frac{2}{5}.$
But
$\sum a^2=\left( \sum a\right)^2 -2\sum ab -2\sum ac =1-2\sum ab -2\sum ac.$
Combining this with the above inequality, we get
$1-2\sum ab -\sum ac \ge \frac{2}{5},$
or
$2\sum ab +\sum ac \le \frac{3}{5}.$
[quote]
$ x_1,x_2,x_3,x_4,x_5,x_6\ge0,x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} x_5 \plus{} x_6 \equal{} 1$
$ 2(x_1x_2 \plus{} x_2x_3 \plus{} x_3x_4 \plus{} x_4x_5 \plus{} x_5x_6 \plus{} x_6x_1) \plus{} x_1x_3 \plus{} x_2x_4 \plus{} x_3x_5 \plus{} x_4x_6 \plus{} x_5x_1 \plus{} x_6x_2 < \equal{} 2/3$[/quote]
This inequality is true according to the inequality
$\sum_{i=1}^6 x_i^2 \ge \sum_{i=1}^6 x_ix_{i+1}.$

[quote="spanferkel"][quote="arqady"]
The following inequality is also true.
Let $a,$ $b,$ $c,$ $d$ and $e$ are real numbers such that $a+b+c+d+e=1.$ Prove that:
\[2(ab+bc+cd+de+ea)+ac+bd+ce+da+eb\leq\frac{3}{5}\][/quote]
In fact, this means simply \[Ineq \Longleftrightarrow \sum (a-2b+c)^2 \ge 0  \] :D[/quote]
Nice! :lol:
My poof is very ugly: $\frac{3}{5}-2(ab+bc+cd+de+ea)-ac-bd-ce-da-eb=$
$=\frac{(4a-b-c+4d-6e)^2}{60}+\frac{(a-b+2c-2d)^2}{12}+\frac{(a-b)^2+(b-c)^2}{4}.$
[quote="shfdfzhjj"][quote="can_hang2007"][quote="shfdfzhjj"]$ a_i\ge0,\sum_{i \equal{} 1}^n a_i \equal{} 1,a_{n \plus{} 1} \equal{} a_1,a_{n \plus{} 2} \equal{} a_2$
$ f(n) \equal{} Max{\sum_{i \equal{} 1}^n a_ia_{i \plus{} 1}a_{i \plus{} 2}} \equal{} ?$[/quote]
In my proof, I proved that $f(n)=\max\left\{ \frac{1}{n^2},\; \frac{1}{27}\right\}.$[/quote]

That's the answer.[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=112980

[quote="can_hang2007"]
[quote]
$ x_1,x_2,x_3,x_4,x_5,x_6\ge0,x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} x_5 \plus{} x_6 \equal{} 1$
$ 2(x_1x_2 \plus{} x_2x_3 \plus{} x_3x_4 \plus{} x_4x_5 \plus{} x_5x_6 \plus{} x_6x_1) \plus{} x_1x_3 \plus{} x_2x_4 \plus{} x_3x_5 \plus{} x_4x_6 \plus{} x_5x_1 \plus{} x_6x_2 < \equal{} 2/3$[/quote]
This inequality is true according to the inequality
$\sum_{i=1}^6 x_i^2 \ge \sum_{i=1}^6 x_ix_{i+1}.$[/quote]
Yes, it is very easy for non-negative $x_i$.

For [b]real [/b]$x_i$ and $n=6$, we only have $\sum ab+\sum ac\le \frac 13$.
And I think that for $n\ge 7$ and [b]real [/b]$x_i$, there is no maximum $M$ such that $\lambda\sum ab+\mu\sum ac\le M$ for given $\lambda,\mu\ge 0$.

[quote="can_hang2007"]The following inequality is also interesting:

Let $a_i$ $(i=1,\; 2,\; \ldots, \; 6)$ be [b]real [/b]numbers such that $\sum_{i=1}^6 a_i=0$ and $\sum_{i=1}^6 a_i^2=1.$ Denote $a_{7}=a_1,$ prove that
$\sum_{i=1}^6 a_ia_{i+1} \le \frac{1}{2}.$

This inequality can be proved by Cauchy-Schwarz. Can someone find it?[/quote]

It is the special case of Fan inequality.


logarithms
number theory unsolved
number theory

If $ 0<a<b$, there exists an $ x_0$ such that for $ x\geq x_0$, there is at least one prime between $ ax$ and $ bx$.

Suppose otherwise.  Then $ \pi(x)$ grows asymptotically at most as fast as $ \log_{b/a} x$, but this contradicts the prime number theorem.


[quote="BOGTRO"]I asked this because I put 29 and it said it was wrong, then gvae 29 as the answer.[/quote]

Can't you clikc "report error with question or solutions"?

Also, that happened to me before, but it was just a typo.

[quote="BOGTRO"]I asked this because I put 29 and it said it was wrong, then gvae 29 as the answer.[/quote]

Seems like they fixed it then. 29 worked fine for me.

I tripped myself up on the first part. XD
Well, they said, How many natural numbers greater than 6 but less than 60 are relatively prime to 15?
So I assumed:
6 < {set of numbers relatively prime to 15} < 60
It wasn't inclusive, so I subtracted 60 from 6, and I got 54.
They did it from 7 to 59 inclusive, which I understand, and it got 53 for the first step.
I think I've got my inclusive exclusive rules wrong, or something...
can someone explain?

Every time I solve a problem correctly, the answer given is always different than the one I used to solve the the problem. Is this a problem, or is this just a new, creative way to solve a problem?

Well, there isn't a solution here yet, so:
[hide=Solution]
We can use complementary counting by subtracting the numbers that share a common divisor greater than 1 from the total number of natural numbers in the set $6<n<60$, which means that the least possible value of $n$ is 7 and the greatest possible value of $n$ is 59 (Relatively prime means that the greatest common factor of two numbers is 1). We know that any numbers not relatively prime to 15 must have a factor of 3,5, or 15. The smallest number in the set divisible by 3 is 9, and the greatest number divisible by 3 in the set is 57. The number of total multiplies of 3, inclusive, will be $\frac{57-9}{3}+1=\frac{48}{3}+1=16+1=17$. We can repeat the same process with 5, where the smallest number divisible by 5 in the set is 10, and the greatest is 55. $\frac{55-10}{5}+1=\frac{45}{5}+1=9+1=10$. $10+17=27$. But wait, there are numbers that are multiplies of 15. There are 3 in the set-15, 30, and 45. We subtract this from 27 to get 24. Subtracting this from all the numbers in the set gives $(59-7+1)-24=(52+1)-24=53-24=\boxed{29}$.

I made a list but accidentally miscounted to get 30

[hide=Solution]There are $17$ multiples of $3$ between $7$ and $59$ and there are $10$ multiples of $5$ between those numbers. Subtracting both these numbers from $53$ gives us $26$, and we then add $3$ to account for the multiples of $15$ giving us $\boxed{29}$.[/hide]

[quote=YT2019]It didn't specify if you should include 6 and 60... But I got it right on the second try.[/quote]

"greater than" and "less than" as opposed to "greater than or equal to" and "less than or equal to" imply that it is exclusive.

[quote=fucongcong][quote=BOGTRO]I asked this because I put 29 and it said it was wrong, then gvae 29 as the answer.[/quote]

*gave

you should report this error.[/quote]

That post was from 2008.

We can use complementary counting;
# of integers that are relatively prime to 15= (total # of numbers greater than 6, less than 60)-(# of integers that are not relatively prime to 15).

The # of numbers greater than 6, less than 60 is 60-6=54 since we are excluding 6 itself. The # of integers that are not relatively prime to 15 are multiples of the prime factorisation of 15- namely, 3, 5, or 3 and 5. Thus, the # of integers that are not relatively prime to 15 is: $\lfloor \frac {60} {3} \rfloor + \lfloor \frac {60} {5} \rfloor - \lfloor \frac {60} {15} \rfloor=20+12-4.$ However, since the numbers must be greater than 6, we counted more numbers than we needed to! The multiples of 3 less than 6 are 3 and 6. The multiples of 5 less than 6 is just 5 and there are no multiples of 15 that are less than 6. So we've counted 3 more numbers than we needed to; thus, the answer is $28-3.$

feel free to ask me any part that you don't get and sry if I made it more confusing than it needs to be :/


"From point $A$ located on a highway, one has to get by car as soon as possible to point B located in the field at a distance $\eta$ times slower on the highway.  At what distance from point $D$ must one turn off the highway?"

So it looks kind of like this

(point D)--------------highway-------------------------(point A)

(point B)

and the distance from $D$ to $B$ is $l$

Im gonna call that strange charachter x.

in general, the time taken when travelling at a speed v on the highway length h, and the distance from D to B is l:

first part: speed is v, and distance travelled is ah, where 0<a<1 (a is the fraction of highway travelled), so the time here is t = ah/v

second part: the car leaves the road with a distance h(1-a) to go, and travels along a diagonal, of length, by pythagoras, sqrt[l^2 + (h-ha)^2], at speed v/x, takng a time of t = xsqrt[l^2 + (h-ha)^2]/v

total time is therefore:

T = {x sqrt[l^2 + (h-ha)^2] + ah }/v

you want to find a minimum value for the total time, so differentiate (w.r.t. a) and set to zero,

dT/da = how convenient, my pizza is ready

yeah, I get that solution.  I just wanted to find something more imaginative- I tried making a Snell's Law analogy, but I'm not so proficient with geometry(or maybe my drawing was just wrong).  Can anyone come up with an "intuitive" solution?

And,
What kind of pizza is it?  :P

This problem is very similar to the lifeguard problem. You can use an analogy with light - the car is a lightwave, the field is a material with refractive index $\eta$ and the highway is a boundary between the material and air. The lightwave is traveling almost parallel to the boundary and when it hits the boundary it refracts into the material at the critical angle. Therefore the optimal path is a path with angle between the field part of the path and a line perpendicular to the highway equal to the critical angle for a material with refractive index $\eta$.

if you consider, as x (that weird charachter) tends to infinity, the optimal distance from D tends to zero, and as x tends to zero, the optimal distance tends to l.

I burnt my pizza a bit. I put garlic, oregano and extra chese on it and it was very tasty. but the crust was burnt! that's what you get with calculus... burnt pizza.


quadratics
number theory proposed
number theory

Let $n$ be a positive integer. Show that if $x^2+xy+y^2=n$ has rational solutions then it also has integer solutions.

I just love this problem. I know thieir solution uses some rings...or even Minkowski theorem.  I don't know how to use those, so I gave an elementary solution. First of all, it's not very difficult to prove that if a prime p of the form $ 3k+2 $ divides a number of the form $ a^2+ab+b^2$ then it divides a and b (a,b are supposed integers) . This means that the exponent of any prime of the form 3k+2 in the number $ a^2+ab+b^2$ is even. Now, if the equation has a solution in the rationals, then the equation $ a^2+ab+b^2=c^2n$ has a solution in integers. But again the exponent of any prime of the form 4k+3 in n must be even. And now we will prove that if the exponent of any prime  of the form 3k+2 in n is even, then n can be written in the form $ x^2+xy+y^2$ with x,y integers. Let A the set of those numbers which can be written as $ x^2+xy+y^2$ with x,y integers. It is not difficult to show that for any a and b in A we have ab in A. Thus, it is suficient to prove that any prime of the fom 3k+1 is in A. Let p such a prime. Then p is 1 mod 6. Then using the quadratic reciprocity law we find that -3 is a quadratic residue mod p. This means that we can find an integer k such that $ p|k^2+3 $. But from Thue theorem there exists integers x and y such that $ 0\geqq,y<sqrt{p} $ and $ p|k^2x^2-y^2$. This means that $ p|3x^2+y^2 $.But using the bounds for x and y we immediately find that p is of the form $ u^2+3v^2$. But in this case $ p=(u-v)^2+2v(u-v)+(2v)^2 $ and the problem is solved.

Wow! that was fast! :) My idea was exactly the same, but I did use rings in order to show that if $p=3k+1$ then $p=x^2+xy+y^2$. I used the fact that the integer ring of $Q[i\sqrt3]$ is Euclidean (the integer ring is the ring made up of integer algebraic numbers and it can be shown that for $Q[i\sqrt 3]$ this ring must be $Z[\frac{1+i\sqrt 3}2]$).

Thue's lemma usually replaces the usage of rings in cases like this one.


calculus
integration
trigonometry
real analysis
real analysis unsolved

Calculate

$ \int^{\pi /6}_0 \frac{tgx}{1\plus{}x^2} \,dx$

It's an alternate abbreviation for "tangent" used in China and Russia and some other countries ;)

Writing $ \tan{x}$ would have made it a lot clearer.

We cen't calculate this integral elementary. Only by serie I think.


Let ABC be any triangle right angled at A. If D and E trisect the hypotenuse BC then prove that,
$ AD^2 \plus{} AE^2 \equal{} \frac{5}{9} \cdot BC^2$

Another approach using vectors:

Let [b]AB[/b]=[b]b[/b], [b]AC[/b]=[b]c[/b]. Then [b]b[/b] $ \cdot$ [b]c[/b]=0

$ L.H.S. \equal{} (\frac {(2b \plus{} c) \cdot (2b \plus{} c)}{3})^2 \plus{} (\frac {(2c \plus{} b) \cdot (2c \plus{} b)}{3})^2$
$ \equal{} \frac {5}{9}(b \cdot b \plus{} c \cdot c)$
$ \equal{} BC^2$


(ABCD)^2=(s-a)(s-b)(s-c)-abcd cos^2((B+D)/2)
when AB=a BC=b CD= c AD=d in a qudrilateral.

how do you prove this. look p 50 on AoPS Vol 2

It doesn't have a proof (or at least an explanation) in the book?

Try searching for "Brahmagupta" on the forum...

http://www.mathlinks.ro/Forum/viewtopic.php?t=15973

  Darij


Euler
number theory unsolved
number theory

Prove that it doesn't exist a n-odd and n>1 such as n|(3^n + 1).

By Euler Theorem, if $n$ is an odd prime then 4=0 mod $n$   (contradiction) and if $n$ is composite, if we take a prime of $n$, namely $t$, then as gcd(n,t-1)=1 and $\varphi(t)=t-1$ then 4=0 mod $t$ then t is even (contradiction).

Best regards.

RX the Crazy Math

[quote="RobertuX"]...if $n$ is composite, if we take a prime of $n$, namely $t$, then as gcd(n,t-1)=1 and $\varphi(t)=t-1$ then 4=0 mod $t$ then t is even (contradiction).

Best regards.

RX the Crazy Math[/quote]

I didn't really get this part, could you explain it again please?

Yes, if $n$ is composite then we consider a prime $t>2$ (divisor of $n$) then $gcd(n,t-1)=1$ ¿ok? then

\[3^n+1=0 \ \ mod \ n \quad \Rightarrow \quad 3^{2n}=1 \ \ mod \ t \quad {\rm and }\quad  3^{t-1}=1 \ \ mod \ t.\]

Hence $3^2=1 \ mod \ t$ then $t=2$ which is a contradiction.

Remember, if $a^b=1 \ \ mod \ t \quad {\rm and} \quad a^c=1\ \ mod\ t \quad {\rm with } \quad gcd(b,c)=1\quad
\Rightarrow \ a=1\ \ mod \ t$.

Best regards.

RX the Crazy math

[quote="kueh"]icic... just choose thesmallest prime divisor.[/quote]
why we should choose the smallest prime divisor?I think everyone is OK. :)
am I wrong?

It's needed to state that gcd(n, t-1) = 1.

(Otherwise, counterexample: n = 105. Take prime = 7. Then gcd(n, p-1) = 3.)

[quote="StRyKeR"]It's needed to state that gcd(n, t-1) = 1.
[/quote]

Sorry, I forgot that $t$ is a prime divisor of $n$, see my before topic, so if $t|n$ ($t>2$) then clearly $gcd(n,t-1)=1$.

Best regards.

RX the Crazy Math


A standard deck of 52 playing cards consists of four different suits and 13 cards of different ranks in each suit. Four cards will be drawn at random without replacement from this deck. The probability that the
four cards will be all the same suit is k times the probability that they will be all the same rank. What is the value of k?


(If we cannot discuss the test yet, please lock this thread until we can).

How did those taking the exam feel about the questions?

They were harder than the open round last year; in fact, some were probably the level of the invitational problems. The fifth one (the one with the newspaper summary) was an absolute puzzle to me, although I think most of the numbers did not change, so nearly an entirely blank sheet could earn some major points  :D
Overall, I cannot imagine myself getting over a 65 on the test.

Also, I noticed that there were not many "explain your answers" problems. Only about 25% of the test was "theoretical", while the rest was just giving an answer. Were we supposed to give elaborate explanations anyway, like in the margins, back of the sheets, or something? If so, then I am done for, especially on problem number 1.

Aw, I missed it this year again. Must remember next year. Back to math...

[quote="236factorial"]
Also, I noticed that there were not many "explain your answers" problems. Only about 25% of the test was "theoretical", while the rest was just giving an answer. Were we supposed to give elaborate explanations anyway, like in the margins, back of the sheets, or something? If so, then I am done for, especially on problem number 1.[/quote]

My suspicion is that there were fewer "explain your answer" problems on the open round this year because they have an increasing number of participants (somewhere between 1100 and 1200, compared to about 800 last year) and they have enough trouble getting the grading done in time for the invitational round as it is.  Compare with the AMC, in which they do [increasingly difficult] multiple choice for the first two rounds, and only start grading by hand once they're down to 500 or so exams. The goal for the first round is to be fun, and to winnow the participant pool down to 100 or so,  and THEN ask the the more difficult-to-grade questions.

Of course, I'm totally out of the loop on this specific contest [though I do know a bit about it second-hand] and am basing this on my own logistical experiences with other competitions.

added: 2 :30 PMish, pacific time: scores for those who qualified for the invitational round are now up:
[url]http://tangra.si.umich.edu/~radev/clo/naclo09/NACLO09R1scores.txt[/url]

I did far better than I could have ever imagined with a 78.5. I found the "Summer Eyes" problem (the newspaper one) wholly ridiculous, and, well, that definitely brought me down a bit. But in all honesty I didn't even think I would make it, so this is definitely a pleasant surprise. :D

I kind of wish there had been some more explanation as well, such as the first problem which I did quite thoroughly--but I suppose that TMA is right, that the harder stuff gradingwise comes later.


email
summer program
Mathcamp
MathPath
MATHCOUNTS
articles
calculus

I sent in my application but it hasn't come back.
People are probably saying "wtf did this kid [u]read[/u] the questions?"

*posting in hope of people coming back to this forum now that we aren't allowed to discuss the quiz*

So...is there anyone new who is going?

I might be, but I bet I won't qualify.  :(

Do try though. Supposedly if you just get 3 it's alright. :)

I tried X, X, and X.

Better than I expected.

[quote="rd5493"]did you get in or not, alanchou.  what was your score?  can we post that or talk about that? sorry, if we can't...

can we ask what score you need to get in?[/quote]
no, i posted my score a while ago and they got mad :(

[quote="daermon"][quote="funcia"]*cough cough* were we supposed to get our scores too? i think i either threw that away or missed reading it if so...[/quote]
um i dunno if we were supposed to
i just know mine cuz i called them and asked if theyd gotten my application
cuz they were a bit slow in grading them at the beginning
and they mentioned my score[/quote]
oh i see, clearly i should have called them instead of email  :P
well okay, that makes me feel better
i thought it might be the case i got the minimum and....
yeah i can just be quite now.
anyway, i hope to see you there(as i got in which i am not sure if i mentioned.)

I applyed, and I'm pretty sure I'll get in...

mustafa- did you post in the right thread & all b/c i thought you were in 9th *gets more confused*

I'm pretty sure that as long as you don't turn 15 before June this year, you can go to camp. A camper I know from last year is returning, and he's in 9th grade.

anyone going should pm someone who frequents this thread(and said they were going) with their email address if they have the time to read there inbox everyday [/warning]

 i definitely recommend pming daermon unless he disappears or dies or decides he hates math or...well we don't need to be rigourous here...

woah, val, why do you have a new account?   :maybe:
ohhhhhh i get it
your other account is leet now

and no, i dont think i'm gonna die or decide i don't like math
and you forgot 'abducted by aliens'
uh oh, i just jinxed myself :ninja:
or 'dies of overspazziness'
its probably not a good idea to get me, meena, and val in the same place
it might explode because of the extreme amounts of energy :(
or 'die because he can't read the harry potter 7 book at camp'
well actually i wont die
i'll just be kinda pissed off
but val will definitely die
i'll come to your funeral :D

oh yeah, does anybody know piggyinc's email?  its not listed, and its probably no use pming her because she only has 16 posts, so probably isnt active

[quote="pianoforte"]perhaps mustafa fulfills the age requirement?
[quote]Age Requirement: Students should be in the age range of 11-14 years. Assuming that the first birthday of a person is the day they complete the age of one year, the student should not be 15 before the end of the camp, whereas the student should be 11 by the start date of camp. Since the camp begins on either the last Sunday of June or the first Sunday of July and ends on the fourth Sunday after, we use the following formula as the age requirement: The applicant's birthdate must fall on a date from July 30, Y-15, to June 30, Y-11, where Y means the year of the camp. The birthdate can fall exactly on July 30, Y-15, or June 30, Y-11. Exceptions are considered on a case-by-case basis, but no exception will be granted for students born earlier than July 30, Y-15.[/quote]
(heh, i'm a month exactly too old (june 30th birthday))[/quote]

WOAH that means I could have gone as a fifth grader... :ninja:

[quote="daermon"]
but val will definitely die
i'll come to your funeral :D
[/quote]

YOU FORGOT ONE IMPORTANT FACT:
I will have spent all my money going to mathpath so how the hack will I afford for a
funeral? :(

and umm huh i am really bad at math(as we all know) but wouldn't most people be able to go in 5th grade(by age)?

I'm going too.

Last year, there was a 10 year old at mathpath (Ryan Yoo was his name I think), but he was in 7th grade because he skipped two grades.

[quote="jhredsox"]Can we get a list going?

Even if you're just applying

jorian[/quote]

I'm applying...

Uh, this is from last Mathpath...you shoouldn't respond to this, maybe a new thread?But I think it's too early to get a list going.

I find it very hard to believe they gave out the score to Anyone -- who did you speak to? My mom called and spoke to one of the top directors and he wouldn't even Hint as to my score - so I can't believe someone right out gave it .. not saying it didn't happen, -- just saying it's hard to believe

I use to be a director at MathPath, and as far as I know, none of us ever gave out scores. Only one of us had the tests before the camp started, the rest of us just thumb through them for a few minutes sometime during camp. We don't bother memorizing anyone's score.

I uber phailed that one. :) wait... :huh:

[quote="Fermat1601"]Wow...

On the first day of camp all of us sleepy campers had to do this test, and if we got high scores we had an advantage in getting classes.[/quote]

I think that's simply a misconception. As far as I know, the test basically tests your computational and/or MathCounts-type skills. These skills basically give no indication of your interest in math, and how well you will do in the classes (which with the exception of MathCounts and 1-2 star problems solving have no relevance to the types of problems on that test). For the most part, if you want to be in a class, you will be put in that class, unless there are too many people.

As an example, when I went to MathPath, I was basically at MathCounts robot, and had little exposure to real problem solving and real math (disregard the cliche...). I scored 31/35 on the test anyway, which, for bragging's sake, apparently was the first time someone under the age of 13 got less than 5 problems wrong.

[quote="Fermat1601"]On the first day of camp all of us sleepy campers had to do this test, and if we got high scores we had an advantage in getting classes.[/quote]
I'm going to disagree with some of the implications of this statement.  Everyone who attends MathPath gets a full set of classes.  Everyone who attends MathPath gets a set of classes customized to the student's abilities and to the student's preferences.  Sometimes a particular class is popular, and cannot accomodate all who request the class.  In this case, some students will get their second choice.  Wildly popular classes are sometimes offered again, later, during the camp.  Some classes are offered primarily to returning students, so they don't have to repeat a class they took the previous year.  Occasionally, the staff doesn't think a student has the necessary background for a class.

The camp does give an assessment to every student at the start of camp.  The assessment is primarily of your problem solving skills.  I hope you won't disagree the assessment might be useful in initial placement of students in the best problem-solving course.  Realize the staff doesn't know you on the first day of MathPath.  They need to use some criteria to distinguish the level of participants for the first round of problem-solving classes.  The assessment isn't the only criteria, but it does play a role.

However, once camp gets going, instructors and counselors quickly get to know students.  Students often move from one problem-solving group to another.  For most students, your goal shouldn't be to rush to the highest-level classes you can get.  (Read Richard Rusczyk's article on "The Calculus Trap".)  Your goal should be to learn as much as you can, while enjoying the learning process.  If you aren't having fun, you should discuss things with the staff.

[size=75]I have no official connection with MathPath.  I have a child who attended in '04 and '05.  I have another child who was a counselor in '07 and has been hired for '08.  These comments are based on my own personal observations through my visits to the camp and on conversations with numerous students, parents, counselors and MathPath staff.[/size]


[quote="Thao"]Let A,B,C be the points outside circle with centre O. AD,BE,CF are tangents of circle respectively. Prove that AD.BC = BE.CA + CF.AB (where AD=max{AD,BE,CF}.[/quote]

This is, generally, wrong. We have $AD\cdot BC=BE\cdot CA+CF\cdot AB$ if and only if the circle through the points $A$, $B$, $C$ touches the circle $O$. At least one direction of this fact is a degenerate case of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=146125]Casey's theorem[/url].

  darij


How do you make the latex displayed in the beginning of the line, not in the center
For example it always come out as  "\[/diplaystyle i got 3 dollars\]"

You typoed "displaystyle" there, but you don't need it on this forum. Everything you put in $\LaTeX$ is automatically displaystyle. As for not centering- enclose your code in dollar signs.

You made your slash go the wrong way. It should be a backslash(\) rather than a frontslash(/).

Why cant i type spaces? I typed $i have three dollars$, but no spaces came out

in math mode, latex doesn't really care of the spaces you put. You need to use specific command to spaces

[quote]Here are just a few examples of various horizontal spaces.

Code:
a \  b
a \, b
a \: b
a \; b
a \! b
a \quad b
a \qquad b
a \hspace{1cm} b
a \hspace{1in} b

\quad = 1 space according to current type size
\qquad = 2 quads
\, = small space =3/18 of a quad
\: = medium space = 4/18 of a quad
\; = large space = 5/18 of a quad
\! = negative space = -3/18 of a quad
\hspace can be any size both positive and negative and can use as units
cm, mm, in, pt, bp (=1/72 in), pc (12 pt), dd (1238/1157 pt), cc (12 dd), em (width of M), ex (height of x)[/quote]

nice list but I have no credit for it, stevem has in one of his post

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=spaces&t=13028[/url]

I think this should answer your question

ledurt

Or you could just do this:
\[\text{I have three dollars.}\]

:blush:
well, yeah....but the way, there's more commands, so I look better, like if I knew a lot of things... :P

anyway, your way is much better

ledurt

please [b]DO NOT PUT TEXT INTO LATEX[/b] formulaes, unless really really but really necessary! [img]http://www.mathlinks.ro/Forum/images/icons/icon_exclaim_blue.gif[/img] [img]http://www.mathlinks.ro/Forum/images/icons/icon_exclaim_blue.gif[/img]


There is a circuit containing 3 capacitors with capacity $C$ and two inductors with inductance $L_{1}$ and $L_{2}$.At the beginning th capacitor in the left side is charged.Find the frequence of oscilations of the system.


Prove that there are functions $ f: \mathbb{Z} \rightarrow \mathbb{Z}$ so that $ f(f(x))\plus{}f(x)\plus{}x\equal{}0$ for any $ x \in \mathbb{Z}$.

[quote="moldovan"]Prove that there are functions $ f: \mathbb{Z} \rightarrow \mathbb{Z}$ so that $ f(f(x)) \plus{} f(x) \plus{} x \equal{} 0$ for any $ x \in \mathbb{Z}$.[/quote]

Here is a simple way to create one :

Construct the following sequences of natural numbers $ a_n,b_n,c_n$ ($ n > 0$)

$ a_1 \equal{} 1$
$ b_1 \equal{} 2$
$ c_1 \equal{} 3$

For $ n > 1$ : Let $ A_n \equal{} \bigcup_{1\leq k < n}\{a_k,b_k,c_k\}$. Then :
Let $ a_n \equal{}$ fewest positive integer not in $ A_n$
Let $ b_n \equal{}$ fewest positive integer such that $ b_n\notin A_n\bigcup\{a_n\}$ and $ (a_n \plus{} b_n)\notin A_n\bigcup\{a_n\}$
Let $ c_n \equal{} a_n \plus{} b_n$

[hide="What these sequences look like"]
This will construct the following $ (a_n,b_n,c_n)$ :
$ (1,2,3),(4,5,9),(6,7,13),(8,10,18),(11,12,23),$
$ (13,14,27),(15,16,31),(17,19,36),(20,21,41),$
$ (22,24,46) ...$
[/hide]

It's immediate to see that all $ a_i,b_j,c_k$ are pairwise different and that $ \bigcup_k\{a_k,b_k,c_k\} \equal{} \mathbb N$

Then define $ f(x)$ as :
$ f(0) \equal{} 0$
For any $ n\neq 0$, $ |n|\in\mathbb N$ and $ \exists! k > 0$ such that $ |n|\in\{a_k,b_k,c_k\}$. Then :
If  $ a_k \equal{} |n|$, then $ f(n) \equal{} sign(n)b_k$
If  $ b_k \equal{} |n|$, then $ f(n) \equal{} \minus{} sign(n)c_k$
If  $ c_k \equal{} |n|$, then $ f(n) \equal{} \minus{} sign(n)a_k$

And it is easy to see that $ (x,f(x),f(f(x))$ is :
either $ (0,0,0)$
either $ (a_k,b_k, \minus{} c_k)$
either $ (b_k, \minus{} c_k,a_k)$
either $ (c_k, \minus{} a_k, \minus{} b_k)$
either $ ( \minus{} a_k, \minus{} b_k,c_k)$
either $ ( \minus{} b_k,c_k, \minus{} a_k)$
either $ ( \minus{} c_k,a_k,b_k)$
And in all cases $ x \plus{} f(x) \plus{} f(f(x)) \equal{} 0$


trigonometry
ratio
Pythagorean Theorem
geometry
similar triangles
power of a point

$ \triangle{ABC}$ is right angled at A. D is a point on AB such that CD = 1. AE is the altitude from A to BC. If BD = BE = 1, what is the length of AD?

[hide="Did I mess up"]
Umm... doesn't E=D, so D is the midpoint of AD, so AD=BD=$ \boxed{1}$?
Yeah... so I think I messed up...
[/hide]
EDIT: DARN!!

Yeah, I messed up.  I misread the problem.  I thought that D was on CB.
[hide="Revised Solution"]
Yeah, okay.  So, we have that BD=1=CD=BE.  Now, call AD=x.  Thus, we can say that, by the Pythagorean Theorem, $ AC \equal{} \sqrt {1 \minus{} x^2}$.  Again, by the Pythagorean Theorem, we have that $ BC \equal{} \sqrt {2 \plus{} 2x}$.  Thus, $ CE \equal{} \sqrt {2 \plus{} 2x} \minus{} 1$.  So, $ AE^2 \equal{} AC^2 \minus{} CE^2 \equal{} 1 \minus{} x^2 \minus{} (3 \plus{} 2x \minus{} 2\sqrt {2 \plus{} 2x}) \equal{} 2\sqrt {2 \plus{} 2x} \minus{} 2 \minus{} x^2 \minus{} 2x$.  Thus, $ AE^2 \plus{} BE^2 \equal{} AB^2$, so $ 1 \plus{} 2x \plus{} x^2 \equal{} 1 \plus{} 2\sqrt {2 \plus{} 2x} \minus{} 2 \minus{} x^2 \minus{} 2x \equal{} 2\sqrt {2 \plus{} 2x} \minus{} 1 \minus{} x^2 \minus{} 2x$.  Thus, $ 2\sqrt {2 \plus{} 2x} \equal{} 2 \plus{} 2x^2 \plus{} 4x$, giving $ \sqrt {2 \plus{} 2x} \equal{} x^2 \plus{} 2x \plus{} 1$.  So, $ 2(1 \plus{} x) \equal{} (x \plus{} 1)^4$, giving $ (x^3 \plus{} 3x^2 \plus{} 3x \minus{} 1)(x \plus{} 1) \equal{} 0$.  Yeah, so x cannot be -1, so it must one of the roots of $ x^3 \plus{} 3x^2 \plus{} 3x \minus{} 1 \equal{} 0$.  This equation gives that $ x^3 \plus{} 3x^2 \plus{} 3x \plus{} 1 \equal{} 2$, so $ (x \plus{} 1)^3 \equal{} 2$, so $ x \plus{} 1 \equal{} \sqrt [3]{2}$, and thus, $ x \equal{} \boxed{\sqrt [3]{2} \minus{} 1}$ (note that the other two roots of this equation are imaginary.)  [/hide]

[hide]
Quicker method: After you got $ AC$ and $ BC$, use similar triangles (or trig ratio) that $ \frac {BE}{AB} \equal{} \frac {AB}{BC} \equal{} cosA$, or the power-of-a-point formula $ BE*AC \equal{} AB^2$, which is just the same thing.[/hide]

Let $ a\equal{}BC, b\equal{}CA, c\equal{}AB$. Then

$ b^2\plus{}c^2\equal{}a^2$

$ (c\minus{}1)^2\plus{}b^2\equal{}1^2$

$ BE\equal{}{c^2\over a}\implies a\equal{}c^2$

From the second equation we get

$ a^2\minus{}2c\equal{}0\implies a^2\equal{}2c$

Plugging the third equation in we get

$ c^4\equal{}2c\implies c\equal{}\sqrt[3]{2}$

Hence $ AD\equal{}c\minus{}1\equal{}\sqrt[3]{2}\minus{}1$

[quote="Farenhajt"]Let $ a \equal{} BC, b \equal{} CA, c \equal{} AB$. Then

$ b^2 \plus{} c^2 \equal{} a^2$

$ (c \minus{} 1)^2 \plus{} b^2 \equal{} 1^2$

$ BE \equal{} {c^2\over a}\implies a \equal{} c^2$

From the second equation we get

$ a^2 \minus{} 2c \equal{} 0\implies a^2 \equal{} 2c$

Plugging the third equation in we get

$ c^4 \equal{} 2c\implies c \equal{} \sqrt [3]{2}$

Hence $ AD \equal{} c \minus{} 1 \equal{} \sqrt [3]{2} \minus{} 1$[/quote]

Yup all of the answers are correct.


I tossed that tag in because there weren't any links to other articles.  Generally, "wiki style" means proper grammar, formatting, and links to other articles where appropriate.


calculus
integration
calculus computations

Nea you are left to perform the substitutions:

1. $x^5 = u$
2. $x^2+1 = u$


inequalities
logarithms
calculus
calculus computations

For $x>0,$ prove the following inequality.

\[ \frac{1}{2}<\frac{1}{x}\ln \frac{e^x-1}{x}<1. \]


superior algebra
superior algebra solved

Let $\pi=\pi_{1}\cdot \pi_{2}\cdot ... \cdot \pi_{k}\in S_{n}$ such that order of $\pi_{i}$ is $n_{i}$ and all the $n_{i}$ are reletivly prime.
Prove that order of $\pi$ is $lcm(n_{1},...,n_{k})$.

I think the $\pi_{i}$ should be cycles(disjoint)?

Yes, probably disjoint permutations are meant, otherwise take $(12)(123)$ in $S_{3}$.

But use all $n_{i}$ are reletively prime.
Then $lcm(n_{1},n_{2},\cdots,n_{k})=\prod_{i=1}^{k}n_{i}$？

Look at my counterexample ;)
Probably they are meant to commute pairwise or something like that (special case: they permute desjoint sets).

sorry i was wrong. the $n_{i}$ are not necceserely  reletivly prime.
I think i have solved it.
thanx.

I used that they are disjoint.

rustam, would you mind sharing the solution?

I used that this cycles are disjoint and any product ${n_{i}}^{k_{i}}$ from i=1 to n will not be identity unless $k_{i}| \text{order}(\pi_{i})$.


geometry
advanced fields
advanced fields theorems

I want to read the work of Field medallists but their papers is for graduate students. It is far beyond my math level.

A) Is there any math book that explain the work of a Field medallist to a high school student?
(a high school student can understand the work of a Field medallist after reading this book)

B) Is there any book that explain the work of a Field medallist to undegraduate?
(an undergraduate can understand the work of a Field medallist after reading this book)

Unfortunately, the answer is no.  Fields Medals are given out for cutting edge research, and it is simply not accessible to high school students or even undergraduates, in any meaningful sense.  I'll put it this way: Most of the time there is no book that will explain the work of a recent Fields Medalist to a *mathematician with a PhD*.

At best you might read about the *results* that the Fields Medalists proved rather than the work they actually *did*.  For example, there are many descriptions of the Poincare Conjecture that are accessible to high school students, but one must take a graduate course in Riemannian geometry before one can approach Perelman's work.  Similarly, the Green-Tao Theorem can be explained to anyone who knows what a prime number is, but to understand the proof requires an education.  This year is actually atypical in this respect; usually, even the biggest *results* of Fields Medalists are essentially impossible to explain to a lay audience.  (Again, they are often difficult to explain to professional mathematicians.)

Sorry about the bad news, but math is a very mature and diverse subject.


Support
linear algebra
matrix
number theory proposed
number theory

Let $ (a, b) \equal{} \{a, b \in \mathbb{N}|7a \plus{} 8b, 8a \minus{} 7b \in \text{positive square number}\}$

(1) Show that there are pair of $ (a, b)$ innumerably.
(2) Find the minimum of $ a$.

[quote="Kouichi Nakagawa"]Let $ (a, b) \equal{} \{a, b \in \mathbb{N}|7a \plus{} 8b, 8a \minus{} 7b \in \text{positive square number}\}$

(1) Show that there are pair of $ (a, b)$ innumerably.[/quote]

$ (4u^2,u^2)$ is a solution $ \forall u$

[quote](2) Find the minimum of $ a$.[/quote]

$ a\equal{}1$ and $ 8a\minus{}7b>0$  would imply $ b\in\{1\}$ but then $ 7a\plus{}8b\equal{}15$ is not a square
$ a\equal{}2$ and $ 8a\minus{}7b>0$  would imply $ b\in\{1,2\}$, so $ b\equal{}1$ (else $ 8a\minus{}7b>0$ is not a square) but then $ 7a\plus{}8b\equal{}22$ is not a square
$ a\equal{}3$ and $ 8a\minus{}7b>0$  would imply $ b\in\{1,2,3\}$ but then $ 8a\minus{}7b\in\{3,10,17\}$ cant be a square.
$ a\equal{}4$ and $ b\equal{}1$ matches the requirement and so minimum value of $ a$ is $ 4$.

Let
$ 7a \plus{} 8b \equal{} x^2,$
$ 8a \minus{} 7b \equal{} y^2,x,y \in N_ \plus{} .$
Then
$ a \equal{} \frac {7x^2 \plus{} 8y^2}{113},$
$ b \equal{} \frac {8x^2 \minus{} 7y^2}{113}.$
So
$ 113|7x^2 \plus{} 8y^2,113|8x^2 \minus{} 7y^2.$
Thus
$ 113|(y \minus{} 18x)(y \plus{} 18x).$
However,when $ (x,y)$ is which $ 113|(y \minus{} 18x)(y \plus{} 18x)$,then it is easy to prove that $ a,b \in N_ \plus{}$.

So $ y \equal{} 113k\pm18x,k \in Z.$
$ a \equal{} 8*113k^2\pm16*18kx \plus{} 23x^2$
$ a \equal{} 23(x\pm\frac {16*9}{23}k)^2 \plus{} \frac {56}{23}k^2$
Then we prove that $ a_{min}\equal{}4$.
When$ |k|>\equal{}2,a\ge\frac{224}{23}>9>4$
When$ k\equal{}\pm1$,if$ x\equal{}\mp6,a\equal{}a_{min}\equal{}4$,else $ a\ge4$
When $ k \equal{} 0,a \equal{} 23x^2\ge23 > 4$.

Thus $ a_{min} \equal{} 4$

Thank you for your nice solution.
I think that your solution can support a generalized problem.

[color=blue][b]Memo[/b][/color]

\begin{eqnarray*}
\begin{pmatrix} 7 & 8 \\ 8 & -7 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} x^2 \\ y^2 \end{pmatrix} \iff \begin{pmatrix} a \\ b \end{pmatrix}
& = & \begin{pmatrix} 7 & 8 \\ 8 & -7 \end{pmatrix}^{-1}\begin{pmatrix} x^2 \\ y^2 \end{pmatrix} \\
& = & \frac{1}{\det\begin{pmatrix} 7 & 8 \\ 8 & -7 \end{pmatrix}}\begin{pmatrix} -(-7) & 8 \\ 8 & -7 \end{pmatrix}\begin{pmatrix} x^2 \\ y^2 \end{pmatrix} \\
& = & \frac{1}{113}\begin{pmatrix} 7 & 8 \\ 8 & -7 \end{pmatrix}\begin{pmatrix} x^2 \\ y^2 \end{pmatrix} \\
\end{eqnarray*}

Background is $ A^2=(\det A)I$?
But, I don't know a name of Matrix $ A$ :(
If $ A^2=I$, $ A$ is [b](matrix) involution[/b].


Let $p(x)= x^{10}+x^{8}+x^{6}+x^{4}+x^{2}+1=0$. Find the area of the polygon enclosed by the roots of $p(x)$.

[quote="kunny"]$(1-x^{2})p(x)=1-x^{12}=0$ gives the roots of regular dodecagon with side-length 1 on complex plane, thus the sired area is $\left(\frac{1}{2}\cdot 1\cdot 1\cdot \sin 30^\circ \right)=3.$[/quote]

The area is not that of a regular dodecagon, remember that $p(x)$ only has 10 roots.

[hide="That is to say"] $p(x)$ is a decagon but not regular. $(1,0), (-1,0)$ are missing.[/hide]

Of course, what you did is pretty much the first step I took. :D

[quote="kunny"]You are right. I made a mistake. :blush:

$p(x)=(x^{4}+x^{2}+1)(x^{6}+1)=(x^{2}+x+1)(x^{2}-x+1)(x^{6}+1)=0$ gives the roots $x=\frac{-1\pm \sqrt{3}i}{2},\ \frac{1\pm \sqrt{3}i}{2},\ \frac{\sqrt{3}\pm i}{2},\ \frac{-\sqrt{3}\pm i}{2},\ \pm i$.

Thus desired area is $\left\{\left(\frac{1}{2}\cdot 1\cdot 1 \cdot \sin 30^\circ \right) *2+\frac{1}{2}\cdot 1\cdot 1 \cdot \sin 60^\circ\right\}*4=2+\sqrt{3}$.[/quote]

Interestingly, my answers don't agree with yours. And then again, I don't have the actual answers and won't get them until next week.

My method involved $x^{12}-1=(x^{2}-1)p(x)$. First construct the dodecagon enclosed by a unit circle. Label $(1,0)$ as $A$. Then go clockwise and label the next vertex $B$, then go counterclockwise from $(1,0)$ and label the vertex $C$. Let $O(0,0)$ be the origin.

Clearly to find the area enclosed by $p(x)$ we need to subtract $2| \triangle ABC|$ from the area of the dodecagon. As you have shown, the area of the dodecagon is $3$.

$| \triangle ABC|= 2| \triangle OAB|-| \triangle OBC|$.

Eventually, I got that the area is $\frac{4+\sqrt{3}}{2}$.


[hide="Click to reveal hidden content"]Long division gives $32$ and a fraction that has a numerator smaller than its denominator. Thus, r must be greater than $1$.[/hide]

[hide]for any "large"  $n$  if $a>b$  then $a^n$ is "much larger" than $b^n$ ... so much so that you can forget/ignore $29^n$  (when compared to $32^n$) and ...would not miss much..
Answer would be pretty close (as if $29$ wasn't there at all) to $32^{65}/32^{64}$ or  about $32$ (just a little less)
(In fact, the answer  would not change much (would hover around 32) even if you change "+" to "-" (or "-" to "+") in numerator (or denominator))
(Or replace 29 with any other positive number, as long as  it is less than 32 :) - )
[/hide]

Has anyone found a clever factoring approach?

i have a pretty good start, it's just the end that i'm having trouble with
[hide]
Rewrite the $32^{65}-29^{65}$ as $(30\frac{1}{2}+1\frac{1}{2})^{65}-(30\frac{1}{2}-1\frac{1}{2})^{65}$
Then after doing the binomial therom on this you can reduce it to...
$2\binom{65}{1}30\frac{1}{2}^{64}1\frac{1}{2}+2\binom{65}{3}30\frac{1}{2}^{62}1\frac{1}{2}^3...$
if you do the same thing to the bottom, you can reduce the bottom to...
$2\binom{64}{0}30\frac{1}{2}^{64}+2\binom{64}{2}30\frac{1}{2}^{62}1\frac{1}{2}^2...$
Now you can factor out a $1\frac{1}{2}$ from the top, then cancel the terms and reduce it to
$\frac{3(\binom{65}{1}+\binom{65}{3}+\binom{65}{5}...)}{2(\binom{64}{0}+\binom{64}{2}+\binom{64}{4}...)}$
this is where i'm stuck, if anyone can help that would be great, or maybe i didn't even go in the right directions?
[/hide]
Hopefully someone can help...

[quote="eminem"]altheman ,how did you get R>3 or how did you say that long 2nd polynomial in the numerator is greater than denominator??[/quote]

Well that is pretty obvious since the numerator is totally positive and has $32^64$ and $29^64$ added together along with mannnyyyy other +ve terms,so its >denominator.

Felipesa's method seems pretty clever and right for people ho don't know the factoring of $x^n-y^n$

good explanation, but I think you mean $29^64\cdot3$ and $32^64\cdot3$

[quote="shadysaysurspammed"][quote="eminem"]so altheman , the formula is $(x-y)(x^n*y^0+x^{n-1}*y^1............x^0*y^n)$[/quote]

Nope,its...

$x^n-y^n=(x-y)(x^{n-1}*y^0+..........+x^0*y^{n-1})$[/quote]

ok  :)

is that derived using binomial theorem??  :?

As far as my vague knowledge aboput binomial theorem is concerned,No....B.theorem is usually for $(x-y)^2$,i.e whole squares,i guess.


Consider a a positive integere a>2 and p_1<..p_k be all primes less than a.
Prove that A=sum{[a/p_a1*p_a2*p_a3*..p_al]} and a have opposite parity.
the sum is taken over all non-empty subsets {a_1..a_l} of {1..k}

yes this one is really nice ... however did you think of something like this?
first we rephrase: let q1<q2<...<q(k)<=n be squarefree numbers less than n (including q1=1). we want to prove [n/q1]+[n/q2]+...+[n/q(k)] is always odd. (this is same as the problem, since if the product of primes is always squarefree, occurs once, and if it exceeds n then [n/product of primes]=0. the product does not tho include 1, and since [n/q1]=n, its equivalant).
next let f(n) = [n/q1]...+[n/q(k)], g(n) be the number of squarefree divisors of n (not including n and 1). write out f(n) like so:
5  2 1 1
6  2 1 1
7  3 2 1 1 1
8  4 2 1 1 1
9  4 3 1 1 1
105 3 2 1 1 1
etc.
now its obvious that f(n) 'increases' from f(n-1) for every squarefree divisor of f(n-1), 1 since [n/1]=[n-1/1]+1, and a further 1 if n if squarefree, so:
f(n) = f(n-1) + g(n) + 1 if n is not squarefree
      = f(n-1) + g(n) + 2 if n is squarefree
but its also obvious that g(n) is even iff n is not squarefree. so thus f(n) - f(n-1) is always even. so the parity of f(n) is always same. Checking for some n :P we find f(n) is always odd, which is what we wanted to prove. [sorry if i didn't explain well].

Your proof is true vinoth and very nice.
My proof is combinatorial however.
I will post it soon.

We prove B=sum{(-1)^l [a/p_a1*..*a_l]^2} has opposite parity with a.
Let's count The number of pairs (x,y) with 1<=x,y<=a and gcd(x,y)>1.
Let S_i be the set of pairs that satisfy p|gcd(x,y).
We must compute how many elements has the union of S1..Sk.
Applying Inclusion-Exclusion principle this number is B.
However if (x,y) is a pair then (y,x) is likewise.
So they group in pairs unless x=y.
In this case we have a-1 pairs:(2,2),(3,3)..(a,a).
So B has same parity with a-1 which implies conclusion


function
algorithm
AMC
AIME
algebra
binomial theorem
number theory

Could someone give me a basic rundown on how the combinatorics sequence progresses?

In high school:
Counting, permutations, binomial theorem, etc...

Advanced high school:
Special cases, Generating functions, restrictions, etc.

College:
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Please include the details.

[quote="1234567890"]College:
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Nothing unless you are really interested in the topic. I mean, after the level you descrive, it is very hard to do a distinction between combinatorics and the rest of discrete mathematics, like recurrences, number theory, graph theory, algorithms and complexity theory and a lot of other things.

[quote="1234567890"]How much knowledge is needed to read Stanley's books.[/quote]

I don't have any idea, I have heard that there are some Stanley-Whatever (Whatever= theorem, identity, conjecture, etc) but I have no idea about the books that guy have written.

I am far from being a Combinatorics specialist (very very very far away of it :P ), but if you give me a title, I could give it a look (assuming that there is some information in the web, or it is in the library in my university) or ask some specialist!

He's written Eunemerative Combinatorics, Volume 1 and Volume 2.

Off to "Other Problem Solving Topics" you go.

I have been reading the Stanley books lately. They're not easy! People usually use them for textbooks in a graduate class in combinatorics. An easier book that is also very good is [i]Concrete Mathematics[/i] by Graham, Knuth, and Patashnik.

Having said that, if you can get through the first chapter of Stanley's first book (and specifically section 1.4), you will never have any trouble on combinatorics problems on contests up to AIME level at least -- section 1.4 solves every problem that ever appears, more or less. (Well, you might want to read section 2.1 as well to be certain.)


$Question 1$ show that for every prime $p>3,24\mid p^{2}-1$
$Question 2$ for which $n$, $(n+1,n+11,n+111)$ are all primes?
$Question 3$ how many natural solutions have the equation $x^{2}-y^{2}=2006$?
$Question 4$ prove: for every natural numbers there is a multiple that is made of only zeroes and ones (in any base), for example: 3-111,4-100, 5-10. are there any generalization here?
$Question 5$ find all natural numbers that are squares, and all their digits are odd.
$Question 6$ you need to complete a magic square 3X3 with natural numbers such that in every row, column and diagnosal, the sum of the numbers is 100.  generalise for other sums! what about other magic squares?
$Question 7$ the letters A-I represent the natural numbers 1-9, so that ABC is a represent of 3 digit number. prove that if ABC+DEF=GHI, $9\mid GHI$.
$Question 8$ find how many solutions does the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$ have (for constant $z$,and natural numbers).
$Question 9$ find the numbers between 20 to 60 that aren't the sum of $n$ consecutive numbers

[hide="1"]Every prime $p > 3$ is expressible as $6m \pm 1$. So $p^{2}-1 = 36m^{2}+1 \pm 12m-1 = 12m(3m \pm 1)$. Clearly one btw $m$ and $3m \pm 1$ is even, and the result follows.[/hide]

[hide="2"]$n+1$, $n+11$, and $n+111$ belong to different classes $\bmod 3$, so one of them is divisible by $3$. Clearly it must be $n+1 = 3$ from which $n = 2$. $3, 13, 113$ are all primes.[/hide]

[hide="hint for 3"]$(x-y)(x+y) = 2 \cdot 17 \cdot 59$. Now, several cases.[/hide]

[hide="5"]Let the number be $(10^{n}a_{n}+10^{n-1}a_{n-1}+...+10a_{1}+a_{0})^{2}$ For $n=0$, $a_{0}=1, 3$ works. For all $n\geq1$, when we expand it, the tenth digit of the number is also the tenth digit of $20a_{1}a_{0}+a_{0}^{2}$. It's easy to see that the tenth digit will always be even. The only numbers are 1 and 9.[/hide]

[hide="For 3rd"]
$(x-y)(x+y)=2\cdot17\cdot59$
Since this is not possible for natural numbers, hence no natural solutions
[/hide]

[hide="4"]
There are two cases here:

case 1: if the number $n$ has only the factors that the base $b$ has, then a sufficiently large power of this base will be divisible by this number $n$.  So $100000\dots$ can be divisible by $n$, given enough zeros.

case 2: other numbers.  Then take $1, b, b^{2}, \dots$ mod by the number $n$, and we form a sequence.  Since there's only $n$ possible values, it must repeat somewhere.  Then let's say the repeating block has total sum of $S$.  We know that $nS$ must be divisible by $n$, then we take bunch of $1$s on those repeating blocks $n$ times and put $0$ in the rest of them.

In case you don't get what I'm doing in case 2, remember how we get the divisibility of 3 in base ten: the number $10^{n}a_{n}+10^{n-1}a_{n-1}+\dots$ mod by $3$ is $a_{n}+a_{n-1}+\dots$.  So if the latter is divisible by 3, then the original number will be divisible by 3.  I just use this method and generalize to any number $n$ and any base $b$.
[/hide]

[hide="2"]Since the only consecutive primes are 2 and 3, $\boxed{\boxed{\boxed{\boxed{\boxed{2}}}}}$ is the answer.[/hide]

[hide="8"]I think it was posted before, but I can't find where.
Anyway, the number of solutions equals the number of factors of $z^{2}$.[/hide]

[hide="Full Solution to 3"] $(x+y)(x-y)=2\cdot17\cdot59$.
The 2 numbers $x+y$ and $x-y$ must both be either even or odd. Because $x+y+x-y=2x\equiv0\pmod2$. Clearly, $x+y$ cannot be $0\pmod2$ while $x-y$ is $1\pmod2$, and vice versa. $2\cdot17\cdot59$ can be expressed as the product of 2 numbers 3 ways: $34\cdot59$, $118\cdot17$, and $2\cdot1003$. In none of these pairs do the numbers have the same parity. [/hide]

you have'nt finish everything! I think all the solutions until now are right  :)

[hide="#8"]The given equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$ is equivalent with $yz+xz=xy$, so $xy-xz-yz+z^{2}= z^{2}$, so $x(y-z)-z(y-z)=z^{2}$, i.e. $(x-z)(y-z)=z^{2}$. Clearly, $x>z$, otherwise we would have that $\frac{1}{x}+\frac{1}{y}= \frac{1}{z}\leq\frac{1}{x}$, which would mean that $\frac{1}{y}\leq 0$, which is impossible. Same for $y>z$. Therefore, $x-z$ and $y-z$ must both be non-zero positive divisors of $z^{2}$. Therefore, there are $d(z^{2})$ solutions, where $d(n)$ denotes the number of divisors of $n$. [/hide]