geometry
3D geometry
sphere
rotation
LaTeX
calculus
integration

My Physics text derived the formulas for determining the moments of inertia for a couple of simple geometries and then it introduced a table of common geometries and their moments of inertia. Out of boredom, i tried to verify that the moment of intertia for a sphere rotated about it's central axis is I = (2/5)MR^2, as according to the text. However, i worked it out a few times and i kept getting (3/5)MR^2. Where did i go wrong? This is what i did:

First, since i don't know how to use latex, i will type S(a,b)[f(x)]dx to represent the integral from a to b of f(x) with respect to x. I apologize for the inconvenience.

Anyways, for a sphere of radius R i came up with

S(0, R)[r^2]dm = S(0, R)[Pr^2]dV where P is the density, assuming the sphere is homogeneous.

from geometry, V = (4/3)(pi)r^3 so the differential element dV = 4(pi)r^2dr. Now the integral becomes

S(0, R)[4(pi)Pr^4]dr = 4(pi)(M/V)S(0, R)[r^4]dr = 4(pi)(M/V)([r^5]/5) = [4(pi)Mr^5]/[5V]=I

substituting (4/3)(pi)r^3 in for V yields:

[4(pi)Mr^5]/[5((4/3)(pi)r^3)] which simplifies to

[12(pi)Mr^5]/[20(pi)r^3] =(3/5)Mr^2 = I

So, can someone please show me my error? I don't see it. I know this is extremely burdensome to read but i appreciate your efforts to read it.

[quote="ROLEX4life"] The moment of intertia for a sphere rotated about it's central axis is I = (2/5)MR^2, as according to the text. However, i worked it out a few times and i kept getting (3/5)MR^2.

Anyways, for a sphere of radius R i came up with

from geometry, V = (4/3)(pi)r^3 so the differential element dV = 4(pi)r^2dr. Now the integral becomes

So, can someone please show me my error? [/quote]

  Mathematically, I think your argument was corrected. I check your argument and derivate eactly the same answer as you.
  I believed you did wrong in finding dV. What you did is that you find the rate of change of volume relative to the rate of change of its radius. Nothing is relative to the moment of inertia.

   Here is what I did.
 We divide the sphere into thin disks of thickness dx (we know the moment of inertia of thin disk already). The radius of the disk is:
$r=\sqrt{R^{2}-x^{2}}$
Its volume
  $dV=\pi r^{2}dx= \pi(R^{2}-x^{2})dx$
Its mass is
$dm=\rho dV= \pi \rho (R^{2}-x^{2}) dx$
So, the moment of inertia of a disk of radius r and mass dm is:
$dI= \frac12 r^{2}dm= \frac12 (R^{2}-x^{2}) (\pi \rho (R^{2}-x^{2}) dx)$
Take intergral from $x=-R ... R$ get $I=\frac{8 \pi \rho R^{5}}{15}$
And use $V= \frac{4 \pi \rho R^{3}}{3}$ and we get $I=\frac{2}{5}MR^{2}$ :)


It's trivial, of course, since $x_1^2=x_1+1$ and $x_2^2=x_2+1$.


Prove that every even number greater than 2 can be expressed as a sum of two primes.
Thank you.

Outline:

OK so we use induction. Clearly 4 can (2+2) so that is our base case.

OK so now let n be an even number that can be expressed as a sum of primes. Note that merely doing n+2 works so our induction is complete.

oh alright
analytic bash
so an ellipse is ax^2+by^2=c
let x and y be primes

divide by ab

x^2+y^2/stuffs=other stuff

prime^2+prime^2=stuff^2

prt2=stuff

Let stuff be something even

p=stuff/2

2p=stuff

Thus all stuff can be expressed as p+p qed

yeah, thats close, so i'll give it to you.

well mewto and i just solved it so not anymore.

yeah if a mod could move to the solved problems forum it would be mucho-appreciato


modular arithmetic
MATHCOUNTS
number theory
divisibility tests

A store sold 72 decks of cards for $\$$a67.9b. Find a+b.

$\boxed{5}$!!!

EDIT:Hint to answer: Remember the requirements for divisibility by $9$.

[quote="bigboy82892"][quote="b-flat"][hide="solution"]If $a679b$ is divisible by $72$, then $a+22+b\equiv0\pmod9$ and $b=2$.

Thus, $a=3$ and $a+b=5$.[/hide]

This is probably too easy for HSB, but I won't move it...[/quote]
You don't need $b=2$ because whatever a and b are is irrelavent to the solution. You just need to know that the requirement for divisibility by $9$ is that the sum of all the digits is divisible by $9$.[/quote]
No, a+b could be 14 and the number would still work. This problem is from mathcounts target...


calculus
derivative
analytic geometry
graphing lines
slope
calculus computations

Let $ f$ be differentiable on $ [a,b]$ and $ f'(a)\equal{}f'(b)$. Prove that there exists $ c\in (a,b)$ such that $ f'(c)\equal{}\frac{f(c)\minus{}f(a)}{c\minus{}a}$.

No it is not.
The mean value theorem states that there exist $ c\in (a,b)$ such that $ f'(c)\equal{}\frac{f(b)\minus{}f(a)}{b\minus{}a}$ which is different from what I am asking.  :wink:


Let $ a,b,c\geq0$.Prove that:
$ \sqrt{\dfrac{a^3}{a^3\plus{}(b\plus{}c)^3}}\plus{}\sqrt{\dfrac{b^3}{b^3\plus{}(c\plus{}a)^3}}\plus{}\sqrt{\dfrac{c^3}{c^3\plus{}(a\plus{}b)^3}}\geq1$


function
number theory unsolved
number theory

hello everyone. I'm working in Tom Apostol's book, Introduction to Analytic Number Theory. There's this problem, prove:

$ \sum_{d^{2}| n}\mu(d) \equal{}\mu^{2}(n)$

can anyone please show me how its done? much appreciated.

It is obviosly, if exist prime $ p^{2}|n$, then $ \mu(n)=0$ and function $ f(n)=%Error. "summ" is a bad command.
_{d^{2}|n}\mu(d)$ is multiplicftive, therefore $ f(n)=\prod_{p|n}f(p^{k})=0$.
If n is squarefree, then $ \mu^{2}(n)=1$ and $ \sum_{d^{2}|n}\mu(d)=\mu(1)=1$.

[quote="Rust"]It is obviosly, if exist prime $ p^{2}|n$, then $ \mu(n) = 0$ and function $ f(n) =%Error. "summ" is a bad command.
_{d^{2}|n}\mu(d)$ is multiplicftive, therefore $ f(n) =\prod_{p|n}f(p^{k}) = 0$.
If n is squarefree, then $ \mu^{2}(n) = 1$ and $ \sum_{d^{2}|n}\mu(d) =\mu(1) = 1$.[/quote]

ah, i did think of that, but could you explain why the sum is a multiplicative function? thank you :)


which is ur fav subject guys?

this is just to analyse the trend in india (more specific, indian aopsers)

Please include another option maths n chem for me  :D  Maybe chemrock too will choose it  :D
but for now, ill vote only math  :)

hell_ever, [url=http://en.wikipedia.org/wiki/Informatics]informatics[/url] and [url=http://en.wikipedia.org/wiki/Computer_science]computer science[/url] aren't exactly the same, so i'll vote when u create an extra option "computer science" :P

I can't edit the options now...you will have to wait for the mod...

I voted in the others category madness :D

me 2 :D
and 2 % is missing again  :rotfl:

sorry that's the max options possible so others is gone :rotfl:
well i voted for Physics,chemistry and maths :)

chance illa  :P
bet pardesi voted for all 3?  :maybe:

EDIT : JUST A SEC TOO LATE IN POSTING N IT WAS REALLY A GUESS  :D

now how do i change my vote from informatics to computer science?  :lol:

Math and Computer Science (Algorithmic Computing to be precise) :)

algorithms cud get boring if u've cs as option

Anybody can learn programming languages. But as Albert Einstein said, specialized training is useless and makes you unadaptable. One can master any prog lang he/she wants when one has good logic and knows how to apply different algorithms.

On a personal note, I find algorithms more interesting than syntax.

Vendam da srini already romba peethikaran inga adhunalai avan than sonnan nu solladha da :D :rotfl:

[quote="madness"]dei pardesi, you changed others? rock's vote must have gone to computers then n he must have atleast wanted to vote for chem  :huh:   :rotfl:   :D[/quote]
I would have wanted a combination of Maths, chemistry and computers da :P  :P  :P

i wud have gone 4 but i ve to choose only 1

Thats coooooooool!!!!!!!!! :rotfl:  :rotfl:  :rotfl:

[quote="rituraj007"]that wud b awkward--there shud have been phys. in place of chem[/quote]
But I am a total [b]DUD[/b] in physics and I am much more comfortable with chem.

[quote="Cyber-Shreyas"][quote="valeriummaximum"]what the hell?? man no one has chosen physics except me...

pardesi... arent u interested in physics???[/quote]

No da. Didn't he tell you?? Pardesi wants to study Gynecology.[/quote]
Following GRS pathway eh??? :rotfl:  :rotfl:

i voted fr PCM  :D ....and gynecology is cool

But computers, math and chem is much better for me :P

[quote="rituraj007"]I have a companion- :D -who else voted 4 medical sciences?????????? :rotfl:  :blush:[/quote]

Did I???? not very sure...

Pardon me...If I made a mistake...

this place seems to have more votes than the voting for the subforum names itself :D :D :rotfl:


trigonometry
function
logarithms
algebra unsolved
algebra

If $ \frac{x}{2}$ is an odd multiple of $ \frac{\pi}{2}$ then $ f(x) \equal{} 0$, which occurs whenever $ x$ is an odd integer multiple of $ \pi$.  This immediately gives $ f(x) \equal{} 0$ whenever $ x$ is an integer multiple of $ \pi$.

Define $ \text{sinc}(x) \equal{} \frac{\sin x}{x}$ when $ x \neq 0$ and $ 1$ when $ x \equal{} 0$ and let $ f(x) \equal{} g(x) \text{sinc}(x)$.  Verify that this implies $ g(x) \equal{} g \left( \frac{x}{2} \right)$.  The general solution is as follows:  let $ \phi(x), \varphi(x)$ be any two periodic functions with period $ \ln 2$ and let

$ g(x) \equal{} \phi(\ln x) \text{ if } x > 0$
$ g(0) \equal{} 1$
$ g(x) \equal{} \varphi(\ln (\minus{}x)) \text{ if } x < 0$.


real analysis
real analysis solved
Sequence

Find  (if exists )  a strictly  increasing sequence $(x_n)_{n\ge
0} $ of real numbers such that
 $\displaystyle x_0=2, x_1 = \displaystyle r , \; r $ being rational $ > 2 $ ,and
 \[ (k+1)x_{k+2}=(k+2)x_{k}  \hspace*{0.5cm} ,\hspace*{0.5cm} \forall k=0,1,...\; .\]

If I am not mistaken, Arne posted something similar, and one had to find what is now called $r$. I think it was something related to $\pi$ ($\pi$ or $\frac {\pi}2$, I can't remember exactly right now; it can be derived with the Stirling approximation formula). Besides, you might find something useful [url=http://mathforum.org/epigone/sci.math.num-analysis/kryrbaxcen/117ac0e5.0306160456.5d9fd4b2@posting.google.com]here[/url] :).

And here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=6521&highlight=stirlings

Pierre.

[quote="harazi"]ALEX comes from Alexandru Lupas, right?[/quote]

Indeed, ALEX=A.L.


Does there exist a prime p such that,

(1) p=1 (mod 4)
(2) for all natural numbers x,
     x^2010=1 or 0 (mod p)

 :roll:

If some p exists, it has primitive roots, and by Fermat's Little Theorem, $ p \minus{} 1|2010$, and since $ p\equiv 1(mod4)$, $ 4$ divides $ \phi (n)$, but four don't divide 2010, and this prime don't exist.

I don't think this is totally correct. You just need $ p\minus{}1|2010$ if there is a primitive root, and not $ p\minus{}1\equal{}2010$.

[quote="raf92"]If some p exists, it has primitive roots, and by Fermat's Little Theorem, $ p \minus{} 1 \equal{} 2010$, so $ p$ is $ 2011$. Obviously $ 2011\equiv 3(mod4)$, and this prime don't exist.[/quote]

I don't understand why you can conclude that p-1=2010. You can have x^y=1 (mod p) even when y is not equal to p-1, although if y=p-1 it is guaranteed. For example, 2^3=1 (mod 7) but 3 not equal to 7-1=6 right?  :maybe:

[quote="Ahwingsecretagent"]I don't think this is totally correct. You just need $ p \minus{} 1|2010$ if there is a primitive root, and not $ p \minus{} 1 \equal{} 2010$.[/quote]

Couldn't it be possible that p-1 is a multiple of 2010 instead of a factor of 2010? Sure if p-1 is a factor of 2010 the stated property in question is guaranteed. But p-1=2010*k is not ruled out.

edit: Actually, I don't think you can say anything definite about the relationship between p-1 and 2010.

[quote="raf92"]Forgive my mistake. Now I think the solution is correct.[/quote]

Which solution do you mean?

[quote="raf92"]If some p exists, it has primitive roots, and by Fermat's Little Theorem, $ p \minus{} 1|2010$[/quote]

This is not right. p-1 doesn't have to divide 2010. Fermat's little theorem gives a sufficient condition for the exponent, but not a necessary one. How can you rule out the case that for some number y other than a multiple of p-1, x^y=1 mod p (as in the example I gave that 2^3=1 mod 7 even though 3 is not a multiple of 6, although FLT ensures that 2^6=1 mod 7)?

[quote="gwy"]
How can you rule out the case that for some number y other than a multiple of p-1, x^y=1 mod p (as in the example I gave that 2^3=1 mod 7 even though 3 is not a multiple of 6, although FLT ensures that 2^6=1 mod 7)?[/quote]

You misunderstood me, gwy. If $ g$ is a primitive root mod p, $ g^k\equiv 1(modp)$ if and only if $ p\minus{}1|k$. If we choose $ x\equal{}g$ in your problem, we'll get that $ p\minus{}1|k$.

[quote="raf92"]
You misunderstood me, gwy. If $ g$ is a primitive root mod p, $ g^k\equiv 1(modp)$ if and only if $ p \minus{} 1|k$. If we choose $ x \equal{} g$ in your problem, we'll get that $ p \minus{} 1|k$.[/quote]

I see your point now  :blush: . I didn't know about the theorem that for each prime there exists at least one primitive roots (I think its taught in some abstract algebra course?). Sigh... the problem becomes a pushover once you know that  :(


limit
integration
real analysis
real analysis unsolved

$ \delta >0$, prove:

$ \lim_{t\minus{}>0^\plus{}} \frac{1}{t^{n/2}} \int_{\delta}^{\infty} e^{\minus{}\frac{r^2}{16t}}r^{n\minus{}1}dr\equal{}0$

With the change of variable $ u \equal{} r/\sqrt {t}$.


Does anyone have the book Plane Trigonometry by S.L. Loney Part 1?
I can't read the second problem of the 4th chapter on mine :(

hello, i hope this will help
http://iit-dreams.blogspot.com/2008/07/download-plane-trignometery-by-sl-loney.html
Sonnhard.


MATHCOUNTS
trigonometry
geometry
parallelogram
trig identities
Law of Sines
Law of Cosines

[quote="Go Around the Tree"]I wonder if I can bring graph paper to use as scratch paper next week for a math tournament..

Anyway how about this one.

Find the median to the longest side in a triangle with sides 10,20, and 15.[/quote]
Woah, woah, woah. That problem is beyond mathcounts, (a specific case of stewart's theorem).

[quote="bpms"][quote="13375P34K43V312"]or you can apply the laws of sines and cosines a bunch of times but that's essentially what stewart's theorem is.

and these should be split into intermediate.[/quote]
No, stewart's theorem isn't law of sines, just law of cosines and $\cosh=-\cos(180-x)$[/quote]

I believe it is possible to apply the law of sines...maybe not.

Just apply the formula for a median:

$m_{a}^{2}={b^{2}+c^{2}\over 2}-{a^{2}\over 4}$ etc. in cycle.

I'm assuming $a$ is the side that it is dropped to? This is just a special case of Stewart's, right?

If you don't remember any formula, one (out of many) way to do:

If the median is $x$ Area of a triangle with sides $5,15,x$ is = equal to triangle of area with sides  $5,20,x$
($s$  (in  the formula $\ Area \ = \sqrt{s(s-a)(s-b)(s-c)}$ ) in the first case  is ((20+x)/2 etc..))  You get pretty easy form   something like

$(20^{2}-x^{2})(x^{2}-10^{2}) = (25^{2}-x^{2})(x^{2}-15^{2})$  Rest is easy.  (actually very easy as $x^{4}$ term cancels out. )
(This is same as formula given above..)

[quote="Go Around the Tree"]I wonder if I can bring graph paper to use as scratch paper next week for a math tournament..

Anyway how about this one.

Find the median to the longest side in a triangle with sides 10,20, and 15.[/quote]

Hint, try to use the fact that the sum of the squares of the diagonals of the paralleogram are equal to the sum of the squares of the sides.


This forum is DEAD.

Where are you from/living?  Though I'm living in Ontario, I was from Alberta.

I'm guessing majority is from Ontario, especially if the proportion of Ontario Math students is any indication...

Silly AoPS, it won't let me list all of the provinces and territories, so I had to condense them.

[quote="Sunny"]*cough cough*

*sigh*  Sooooo dead.

*hello?*  *echo*[/quote]

I concur... I voted though... Ontario FTW.  :D  :D

Whoa, I had people vote?  Ok, so we have lurkers, and no posters.

There goes my attempt to revive the forum a little =)

Well I live in ontario... which you already know. I would bet a million dollars for you if you were asked to correctly state what city I go to school in :)

Oh thats cool jack. May I ask whereabouts ? Are you in uni ?

:P Ontario, for sure....

an Asian in Ontario... ;)

Brit Columbia!

BC FTW! :D

Vancouver area...

Im from Nova Scotia and Nova Scotia isn't a territories!!!!!!!!!!

Im mad. :@

Hey! its the other nova scotia voter! Going to mathletes comp on Saturday? and I'm sad that all the east coast is crammed together... but I guess its ok.


superior algebra
superior algebra unsolved

First, write down all of the roots.  Then adjoin them. (actually, you only need to adjoint two elements)

Have you heard of $ \sqrt[p]{2}$?  To find the other roots, think of the complex plane.

[quote="peteryellow"]how can I find the splitting field of x^p -2.[/quote]
What do you mean? Over $ \mathbb{C}$ or $ \mathbb{Q}$,...?

Default is $ \mathbb Q$ in my oppinion... (and $ \mathbb C$ would be rather trivial  :wink: )
Nonetheless, lets wait for the original poster.


geometry
circumcircle
trigonometry
geometry unsolved

The triangle ABC has AB = 24. The median CE is extended to meet the circumcircle at F. CE = 27, and the median AD = 18. Find area ABF. :lol:

Using power, we get $CE\cdot EF=AE\cdot EB\Rightarrow EF=\frac{16}{3}$. Let $G$ be the intersection of $AD$ and $CE$. So $G$ is the centroid of $\triangle ABC$. Therefore, $AG=\frac{2}{3}AD=12$ and $GE=\frac{1}{3}CE=9$. We know that $AE=\frac{1}{2}AB=12=AG$ i.e. $\triangle AEG$ is isosceles. So $\sin\angle AEG=\frac{\sqrt{12^{2}-(\frac{9}{2})^{2}}}{12}=\frac{\sqrt{55}}{8}$. Now the height of $\triangle ABF$ from $F$ is $EF\sin\angle FEB=EF\sin\angle AEG=\frac{2\sqrt{55}}{3}$. Therefore, its area is $\frac{1}{2}AB\cdot h=\frac{1}{2}\cdot 24\cdot\frac{2\sqrt{55}}{3}=8\sqrt{55}$.


Ο Ευκλείδης πέρασε !

[color=red][b]   Ελκυστικός και ανθρώπινος !  [/b][/color]

[i](Δεν πρόλαβα να δω την Γ΄Λυκείου , αλλά εκεί τα πράγματα είναι πιο ξεκάθαρα, ό,τι θέματα και να βάλουν.Σημασία έχει ότι οι μικρές τάξεις ευχαριστήθηκαν το διαγωνισμό , από όσα άκουσα τουλάχιστον στο δικό μας εξεταστικό κέντρο).[/i]

  Καλά αποτελέσματα σε όλα τα παιδιά που συμμετείχαν !!!

 Μπάμπης

Ευχαριστούμε πολύ κ. Στεργίου

Επειδή το site της ΕΜΕ σέρνεται, ανέβασα τα θέματα και τις λύσεις εδώ για όποιον θέλει:
[code]http://rapidshare.com/files/184859711/themataeukleidi_2008-09final.pdf[/code]

[quote="Protonios"]Eγώ γενικά άκουσα ότι όλες οι τάξεις είχαν δύσκολο Ευκλείδη φέτος.

Εγώ έδωσα στην 3η Λυκείου και έλυσα σχετικά γρήγορα 1το, 3το και 4το.

Παιδεύτηκα με τη Γεωμετρία αλλά έναν-δυο το πολύ πόντους βλέπω να παίρνω γιατί είπα για το Η ριζικό κέντρο και για την καθετότητα και ότι το κέντρο του κύκλου που ψάχνουμε είναι το Ο.

Συνολικά 16-17 πόντους βλέπω να μου δίνουν 2ρο βραβείο. Θα δείξει.

Ερώτηση:

Στην Γ Γυμνασίου που έδινε ο αδερφός μου αν χρησιμοποιούσε κανείς έτοιμο το θεώρημα ότι τα συμμετρικά του ορθοκέντρου ως προς τις πλευρές βρίσκονται στον περιγεγραμμένο έβγαινε σε μία γραμμή.

Το δέχονται έτοιμο ή ήθελαν την απόδειξη???

ΚΑΛΗ ΕΠΙΤΥΧΙΑ ΣΕ ΟΛΟΥΣ!!! :)[/quote]

ennoeitai oti to dexontai etoimo. eidika an grafeis stin athina den exeis provlhma...tora an to grapsei kapoios stin eparxeia kai to diorthosei kanenas kafros,.....asta..

kosta bravo!

Thx re Nick. Polla xairetismata stoys Aggloys!!! :)

Φάση είχε το 4ο στη Γ Λυκείου. Για να βγει ήθελε να δει κανείς τις περιπτώσεις των x,y για 0, f(0), -f(0) κτλ.

την ώρα του διαγωνισμού σκέφτηκα το εξής: για x=y, αρκεί να δείξουμε ότι
η h(x)=f(f(x))-f(x) έχει πεδίο τιμών το R

Φυσικά σκέφτηκα ότι δεν ξέρω καμία θεωρία που να το λέει αυτό και έκανα την προαναφερθείσα λύση για να είμαι σίγουρος,
αλλά μήπως μπορεί να μου βρει κάποιος αντιπαράδειγμα;

Γειά σας, εγώ είμαι νέο μέλος κι έδωσα κι εγώ Ευκλείδη φέτος για πρώτη φορά (Β' Γυμνασίου)! Νομίζω ότι έλυσα και τα 4 θέματα!  :o  Δεν είμαι απόλυτα σίγουρος, αλλά τα αποτελέσματά μου βγήκαν ίδια με τις επίσημες λύσεις. Αλλά εγώ ακολούθησα άλλους δρόμους (πχ. στο 3ο αν θυμάμαι το πήγα κάπως έτσι 1,2α επί 1,3β = Ε + Ε2 και στο τέλος βρήκα ότι Ε2 = 0,56αβ, δηλαδή 56%, και στο 4ο αντί να έχω 90-80 όσον αφορά ολόκληρη τη γωνία Α, βρήκα 50-40 που αφορούσε τμήμα της γωνίας Α, αλλά χωρίς να είμαι εκτός θέματος). Λέτε να περάσω? Και αν περάσω, αξίζει να πάρω μέρος στον Αρχιμήδη ή είμαι μικρός ακόμη για Ολυμπιάδες? Από το λίγο που έχω δει στον Αρχιμήδη έχουν ασπρίσει οι τρίχες μου  :| ...

[quote="Protonios"]Μη λες ασυναρτησίες. :)  Πρώτον αν έχεις λύσει όντως 4 έχεις περάσει με Α βραβείο.

Φυσικά τώρα αξίζει να λάβεις μέρος στον Αρχιμήδη απλά και μόνο για την εμπειρία και γιατί όχι για το μετάλλιο αν κάτσει.

Μικρός δεν είσαι γιατί υπάρχει η Βαλκανιάδα junior που είναι για μικρούς οπότε cu τέλη Φεβρουαρίου χωρίς συζήτηση. :wink:[/quote]

Δεν ξέρω αν το πρόσεξες, αλλά Ασυναρτησίες είναι και το username μου.  :lol:

Επίσης δεν ξέρω αν θα μου κόψουν και κανά βαθμό επειδή δεν έχω επεξηγήσει πλήρως κλπ. Θα δούμε, τα αποτελέσματα πότε περίπου βγαίνουν?

Α, για τον Αρχιμήδη έχω αμφιβολίες επειδή έχει κάτι πράματα με συναρτήσεις, πολυώνυμα και διάφορες θεωρίες που πρέπει να ξέρω, ενώ εγώ μόλις τώρα αρχίζω συναρτήσεις. Δεν νομίζω να είμαι τόσο καλός ώστε να περάσω και στις Βαλκανιάδες, μπορεί να ήταν η τύχη μου σήμερα που γιορτάζω, μπορεί χίλια-δυό πράγματα. Δεν είμαι σίγουρος ότι νιώθω έτοιμος να βουτήξω στα 'βαθιά νερά' από τώρα, μου φτάνει να περάσω Ευκλείδη (με βραβείο, αν γίνεται!  :) )

Εγω β' λυκειου εγραψα το 1 και το 2 και λιγο απο το 4....
Ενω γεωμετρια την εφτασα μεχρι ενα σημειο αλλα δεν την καταφερα στο τελος..
Υπολογιζω γυρω στα 12-13......Ελπιζω να περασω...Τωρα για γ'βραβειο δεν το κοβω.... :(

Παιδια, βρηκα λιγο χρονο και θα ηθελα να σας κανω μερικες ερωτησεις.

Καταρχην, να πω οτι ελυσα το 1ο, το 2ο της β' λυκειου ολοκληρα. Βεβαια μου φανηκαν σχετικα ευκολα οποτε νομιζω οτι και οι ο περισσοτεροι απο τους αλλους θα τα εχουν σωστα. Παρα 'αυτα ηταν πονηρα και ισως καποιοι να εχουν χασει καποιες απο τις περιπτωσεις.

Τωρα για το 4ο εγραψα τη γενικη μορφη του πολυωνυμου και ελυσα την δοσμενη εξισωση ως προς κ εχοντας απαλλαξει το ισο του κ απο τον ορο P(x) αφου ειχα αντικαταστησει το Ρ(x) με την γενικη μορφη πολυωνυμου. Μετα ειπα θα πρεπει το ισο του κ να ειναι μεγαλυτερο ή ισο του 2 και μικροτερο ή ισο του 29. Μεχρι εκει. Και ρωτω "Θα μου δωσουν κανα μοριο? Και αν ναι ποσα?".

Με 10+μορια απο 4ο προβλημα, περναω?

Θα εκτιμουσα τις απαντησεις σας... :)

Εγω ασχοληθηκα λιγακι με αυτα της Γ Λυκειου μολις τελειωσε ο διαγωνισμος και ελυσα το 3ο και το 4ο μεσα σε ενα 20λεπτο περιπου, το 1ο τωρα ηταν υπερβολικα ευκολο για Ευκλειδη οποτε δεν ασχοληθηκα καν με αυτο... αντιθετα το 2ο ηταν αρκετα δυσκολο (Γενικευση στο 1ο προβλημα της περισυνης IMO) και θα το κοιταξω αυριο επηδη σημερα μετα το διαγωνισμο ειμουν ολη τη μερα εξω μαζι με τα υπολοιπα παιδια και δεν προλαβα να το κοιταξω καλα...

[quote="Asynartisies"]
Δεν ξέρω αν το πρόσεξες, αλλά Ασυναρτησίες είναι και το username μου.  :lol:

Επίσης δεν ξέρω αν θα μου κόψουν και κανά βαθμό επειδή δεν έχω επεξηγήσει πλήρως κλπ. Θα δούμε, τα αποτελέσματα πότε περίπου βγαίνουν?

Α, για τον Αρχιμήδη έχω αμφιβολίες επειδή έχει κάτι πράματα με συναρτήσεις, πολυώνυμα και διάφορες θεωρίες που πρέπει να ξέρω, ενώ εγώ μόλις τώρα αρχίζω συναρτήσεις. Δεν νομίζω να είμαι τόσο καλός ώστε να περάσω και στις Βαλκανιάδες, μπορεί να ήταν η τύχη μου σήμερα που γιορτάζω, μπορεί χίλια-δυό πράγματα. Δεν είμαι σίγουρος ότι νιώθω έτοιμος να βουτήξω στα 'βαθιά νερά' από τώρα, μου φτάνει να περάσω Ευκλείδη (με βραβείο, αν γίνεται!  :) )[/quote]

μα δεν είπε κανείς πως αν δε λύσεις τα θέματα του αρχιμήδη θα χάσεις τα προηγούμενα βραβεία κτλ...
δεν είναι τηλεπαιχνίδι! αντιθέτως θα κερδίσεις πολλά..
οπότε το θέμα είναι αν, σε περίπτωση που δε μένεις αθήνα, έχεις το χρόνο κτλ να έρθεις...

[quote="basted"]Εγω Β Λυκειου ελυσα 3/4. Τα δυο παντως πρωτα εβγαιναν γρηγορα και θα μπορουσαν να ειναι και θεματα σχολικου διαγωνισματος.[/quote]
Κι εγω Β. Ελυσα τα 2 πρωτα οπως οι περισσοτεροι και απο τα αλλα υλολογιζω γυρω στις 4 μοναδες.
Ευκολα αλλα οχι και για διαγωνισμα σχολειου!!!
Θα τα δειξω σε εναν φιλο μου θεωρητικη και θα φριξει :o

Τελικα η Γεωμετρια Γ Λυκειου μπορουσε να λυθει πολυ ευκολα απο καποιον ο οποιος εχει λυσει το 1ο θεμα της περσυνης ΔΜΟ με δυναμεις σημειου (Ειχα βαλει μια τετοια λυση στο συγκεκριμενο προβλημα πριν απο μερικες εβδομαδες εδω στο φορουμ)... Ενα θεωρημα Θαλη θελει για να δειχθει οτι οι διακεντροι ειναι παραληλες προς τις ευθειες που ενωνουν τα μεσα των πλευρων του ΑΒΓ (αρα και προς τις πλευρες του τριγωνου) και μετα το προβλημα αναγεται στο προβλημα τις περασμενης IMO, με την προυποθεση οτι ο λυτης σκοπευει να τη λυσει με δυναμεις σημειου...

[quote="maniopas"]
η h(x)=f(f(x))-f(x) έχει πεδίο τιμών το R

Φυσικά σκέφτηκα ότι δεν ξέρω καμία θεωρία που να το λέει αυτό και έκανα την προαναφερθείσα λύση για να είμαι σίγουρος,
αλλά μήπως μπορεί να μου βρει κάποιος αντιπαράδειγμα;[/quote]

Για αντιπαράδειγμα πάρε $ f(x) \equal{} x$. Οταν λες "προαναφερθείσα λύση" πού ακριβώς αναφέρεσαι? :!:

Btw ωραία η λύση της ΕΜΕ για το $ f(0)\equal{}0$  :wink:

sxetika me ena post mou pio pano pou grafo oti boreis na xrisimopoihseis to thoerima gia to orthokentro....

kosta tora eida to thema gia to opoio milame...th geom ths g gumnasiou.vlakeia mou pou den to eixa dei otan apantisa.tora loipon vlepo oti einai duskoli periptosi kai den ksero an tha to dextoun giati stin ousia zitane na apodeixthei to antistrofo, opote ki ego den ksero an perimeno na to dextoun.isos den perimenan oti tha kseroun to theorima ta paidia tis g.pisteuo oti tha paiksei rolo kai i genikoteri eikona tou graptou...dioti an deis ena grapto teleio to opoio lunei ti geom me to theorima auto, apla tha pei o eksetastis oti o mathitis itan polu kalos kai den ekrine skopimo na grapsei tin apodeiksi tou theorimatos diaforetika pisteuo tha parei 2-3 monades stis 5.

Για τη συναρτησιακη εμενα η πρωτη λυση που μου ηρθε οταν την πρωτοειδα ειναι:

Εστω $ u \equal{} f(x)$ και τοτε επηδη το συνολο τιμων ειναι το $ R$ ειναι: $ f(f(u) \minus{} f(y)) \equal{} u \minus{} f(f(y)) \forall u,y \in R$
Για $ u\equal{}y\equal{}0$ ειναι $ f(f(0)) \equal{} \minus{}f(0)$, για $ u\equal{}y\equal{}f(0)$ ειναι $ f(0) \equal{} f(0) \minus{} f(f(f(0))) \Rightarrow f(f(f(0))) \equal{} 0$.
Απο τις 2 τελευταιες ειναι: $ f(\minus{}f(0)) \equal{} 0$.
Για $ u\equal{}y\equal{}\minus{}f(0)$ λαμβανουμε: $ f(0) \equal{} \minus{}f(0) \minus{} f(f(\minus{}f(0)) \equal{} \minus{}f(0) \minus{} f(0) \equal{} \minus{}2f(0) \Rightarrow 3f(0) \equal{} 0 \Rightarrow f(0)\equal{}0$
Για $ u\equal{}0$ τωρα και για $ f(y)\equal{}t$ εχουμε λογο $ f(R)\equal{}R$: $ f(\minus{}t) \equal{} \minus{}f(t)$ για καθε $ t \in R$, οποτε η $ f$ ειναι περιττη.

Νικ ρώτησα στην ΕΜΕ και μου είπαν ότι αν το διατυπώνει σωστά θα θεωρηθεί σωστό.

Θα δείξει βέβαια αν το έχει διατυπώσει καλά. :maybe:  Έτσι όπως μου το είπε σχετικά καλά το έχει περιγράψει πάντως (αν και είναι λίγο τυχερός...) :)

BTW Sil θεώρημα Carnot... :rotfl:

[quote="Protonios"]
BTW Sil θεώρημα Carnot... :rotfl:[/quote]
lol :P

Να ρωτήσω εγώ κάτι άλλο... για τη συναρτησιακή!
Πρώτον, θα μπορούσαμε να αποδείξουμε ότι είναι περιττή αν δεν μας δινόταν ότι το σύνολο τιμών είναι $ R$? Εγώ το μόνο σχετικό που κατάφερα να δείξω είναι ότι $ f(\minus{}f(x))\equal{}\minus{}f(f(x))$.

Δεύτερον τί άλλο θα μπορούσαμε να βρούμε για την $ f$? Μπορούμε με κανένα τρόπο να καταλάβουμε αν απο μια τέτοια συναρτησιακή σχέση είμαστε σε θέση να βρούμε την $ f$? Π.χ. στη συγκεκριμένη απέδειξα ότι $ f(2f(x)\equal{}2f(f(x))$ ( ή $ f(2x)\equal{}2f(x)$ αν θεωρήσουμε σύνολο τιμών το $ R$). Ακουσα επίσης ότι μπορεί να αποδειχθεί ότι είναι $ 1\minus{}1$! Εχει βρει κανείς άλλος άλλες τέτοιες σχέσεις για την $ f$?

Περιμένω να ακούσω τις απόψεις σας...

Βγαινει αμεσος οτι ειναι 1-1.
Οπως γραφω και στη λυση μου στην προηγουμενη σελιδα, ειναι $ f(f(x) \minus{} f(y)) \equal{} x \minus{} f(f(y))$ για ολα τα $ x,y$, για $ x\equal{}y$ εχεις $ f(f(x)) \plus{} f(0) \equal{} x$ για ολα τα $ x$, απο εδω παιρνεις δυο αριθμους με ιδιες τιμες μεσο της $ f$ και βρησκεις οτι αυτοι ταυτιζονται...

Λογικα μπορεις να βγαλεις και αλλα πραγματα, αλλα δεν εχω χρονο αυτη τη στιγμη και ειμαι και πτωμα...

και εγω εβγαλα στον διαγωνισμο το 1-1.
επισης κατι που σκεφτομαι ειναι οτι  στην $ f( f(x) \minus{}y)\equal{} x\minus{}f(y)$
θεσουμε οπου y to 0,
$ f(f(x))\equal{}x$
kai meta opou $ x$ to $ f(x)$
$ f(x\minus{}y)\equal{}f(x)\minus{}f(y)$
kai opou $ x$ to $ \minus{}y$(einai k peritti)
$ f(x)\plus{}f(y)\equal{}f(x\plus{}y)$..na ti k i cauchy! :wink:

Παιδια εχω τρομερο αγχος ! Ειμαι Γ Γυμνασιου εδωσα αλλα μαλλον δεν θα περασω...
Ακουστε τι ελυσα :

Ελυσα το 1ο Σωστα
Απο το 2ο το ελυσα με εναν αλλο τροπο αλλα εκανα ενα λαθος στις πραξεις
Απο το 3ο Εφτασα στο τελευταιο σημειο χωρις να γραψω ακριβως τις λυσεις
Και απο την γεωμετρια εκανα το σχημα σωστο και κατι απο το πρακτικο μερος...

Δεν περναω ετσι??  :roll:  :roll:  :roll:

3erei kaneis pote anakoinwnontai oi epituxonres gia ton eukleidh?

Ο Αρχιμηδης ειναι στις 21 Φεβρουαριου οποτε τα αποτελεσματα ειναι πιθανο να βγουν μια εβδομαδα πριν οπως εγινε τα προηγουμενα χρονια.

Pws prokuptei to plhthos twn paidiwn pou pernan sthn epomenh fash? Ennow pernaei kapoio pososto p.x 1 stous 5 h kathe xrono einai enas sugkekrimenos arithmos?

kai gw auto nomizw kapou 14 Feurouariou....mporei kaneis na mou pei an pernaw estw kai sthn kaluterh periptwsh ?? plz plz plz :oops:

Φυσικα και μπορεις να περασεις αναλογα με τη βαση...Εαν η βαση ειναι γυρω στο 10 τοτε εχεις πολλες πιθανοτητες....
Αλλα μην καθεσαι και ασχολεισαι με το αν θα περασεις η οχι....Δες το απο την αλλη πλευρα......Στο κατω κατω μαθηματικα διαβαζεις για τον ευαυτο και την ευχαριστηση σου και οχι για να περασεις στον Αρχιμηδη :wink:
Μικρος εισαι ακομα και θα εχεις πολλες ευκαιριες μπροστα σου...

Δημητρης

Είναι involution και υπακούει στον Cauchy... Έχουμε τίποτα γι' αυτές??? :|

Αν όχι τότε μένει το γνησίως αύξουσα... Είμαι σχεδόν σίγουρος ότι μπορούμε να βρούμε τον τύπο... :?:

thanx dimitri X ! mallon exeis dikio  :D  :D

Βρε παιδια εμενα θα μου πειτε αν βλεπετε να παιρνω τιποτα απο τη λυση μου στο 4ο της β' :?:  :(

Εχω πιθανοτητες να "σπασω το ροδι?"

[quote="stergiu"][quote="Athinaios"]και εγω εβγαλα στον διαγωνισμο το 1-1.
επισης κατι που σκεφτομαι ειναι οτι  στην $ f( f(x) \minus{} y) \equal{} x \minus{} f(y)$
θεσουμε οπου y to 0,
$ f(f(x)) \equal{} x$
kai meta opou $ x$ to $ f(x)$
$ f(x \minus{} y) \equal{} f(x) \minus{} f(y)$
kai opou $ x$ to $ \minus{} y$(einai k peritti)
$ f(x) \plus{} f(y) \equal{} f(x \plus{} y)$..na ti k i cauchy! :wink:[/quote]

Ηλία, για δες το εξής :

 Αν δοθεί το 1-1 αλλά όχι το ''επί του IR'',δηλαδή ότι έχει σύνολο τιμών το IR , μπορεί να αποδειχθεί ότι f(0) = 0 και τελικά ότι η f είναι περιττή ; Δεν το έχω προπαθήσει και για αυτό το δίνω ως άσκηση σε σας που είστε Γ΄Λυκείου. Τώρα, το ότι είναι Cauchy δεν λέει κάτι για τον τύπο. Η άσκηση είναι μάλλον μέρος κάποιος συνθετότερης άσκησης που έχει μπει σε κάποιο διαγωνισμό και πιθανόν να μπορεί να προκύψει και ο γενικός τύπος της f. Ίδωμεν !

 Μπάμπης[/quote]

Kυριε Μπαμπη καλησπερα σας!!
το οτι ειναι cauchy το γνωριζω οτι δεν μας λεει κατι συγκεκριμενο. Το εβαλα κυριως γιατι ο Δημητρης ρωτησε αν βρηκαμε τετοιες σχεσεις(και την θεωρησα χρησιμη σχεση) καθως και διοτι αν αποδειξουμε κατι σε μονοτονια ή συνεχεια(οπως γνωριζετε σιγουρα) θα βρουμε τον τυπο!
Για την ασκηση που μας λετε παντως, νομιζω και στις λυσεις τις η ΕΜΕ δεν χρησιμοποιει οτι ειναι επι στο R για να δειξει το f(0)=0.
βρισκοντας λοιπον οτι $ f(0) \equal{} 0$
exoyme
$ f( f(f(x)) \minus{} f(y)) \equal{} f(x) \minus{} f(f(y))$ gia kathe x,y sto R
gia $ y$ to $ f(x)$
$ f(f(f(x)) \equal{} f(x) \leftrightarrow f(f(x)) \equal{} x$ και απο εκει βρισκουμε το επι.

Κωστα παντως επειδη την κοβω για την f(x)=-x (μεγαλες πιθανοτητες να κανω λαθος-διαισθητικα μιλαω)προτεινω στην συνεχεια της f να δουμε μηπως βγαζουμε τιποτα..!Παντως και εγω πιστευω οτι βρισκεται!

ΕDIT: basika kai monotonia na tis vgalume(se olo to R) logw tou epi i sinexeia einai dedomeni!  :P

Έχετε ήδη δείξει ότι η $ f$ ικανοποιεί την Cauchy και ότι είναι involution ($ f(f(x)) \equal{} x$). Ίσως να το έχετε ήδη προσέξει (δεν διάβασα προσεκτικά όλα όσα γράψατε) αλλά εύκολα μπορείτε να δείτε ότι ισχύει και το αντίστροφο. (Αν η $ f$ ικανοποιεί την Cauchy και είναι involution τότε ικανοποιεί και την συνθήκη του προβλήματος.)

Παραμένει η ερώτηση: Ποίες συναρτήσεις $ f$ ικανοποιούν την Cauchy και είναι involution;

1) Από τις "κλασσικές" λύσεις της Cauchy (της μορφής $ f(x) \equal{} ax$) εύκολα βλέπουμε ότι οι μόνες που δουλεύουν είναι η $ f(x) \equal{} x$ και η $ f(x) \equal{} \minus{}x$.

2) Αν μπορούσαμε να δείξουμε μονοτονία ή συνέχεια τότε η $ f$  θα έπρεπε να ήταν μια από τις "κλασσικές" λύσεις της Cauchy και άρα μια από τις δυο πάραπάνω λύσεις.

Αλλά...

3) Υπάρχουν μή κλασσικές λύσεις της Cauchy που ικανοποιούν αυτές τις συνθήκες. (Δεν τις έχω βρει "όλες" αλλά σίγουρα υπάρχουν και μη κλασσικές λύσεις που είναι involutions.)

Άρα σίγουρα δεν μπορούμε να δείξουμε ότι η f είναι μονότονη/συνεχής.

[quote="Athinaios"]
............................................................
ΕDIT: basika kai monotonia na tis vgalume(se olo to R) logw tou epi i sinexeia einai dedomeni!  :P[/quote]

Ηλία, αυτό δεν το ...καταλαβαίνω.Θεωρείς δηλαδή ότι η f είναι συνεχής λόγω του ''επί'' ; Αυτό δεν είναι απαραίτητα.''Αν είναι συνεχής και 1-1, τότε είναι γν. μονότονη ''. Αυτό είναι το σωστό . Αλλά δεν ξέρω αν εννοείς αυτό.

 Για το άλλο ερώτημα που είχα βάλει , η απάντηση που έδωσες είναι ικανοποιητική μεν, αλλά ... . Όντως f(0) = 0 , αλλά από έχεις ελέγξει αν βγαίνει ότι  f(f(x)) = x , αν δεν έχουμε το ''επί'' ; Ρίξτο μια ματιά γιατί ...  πνίγομαι αυτές τις μέρες στα γραπτά.

 Μπάμπης

Κυριε Μπαμπη,
προ 2 μηνων ειχα ασχοληθει, με αφορμη μια σχολικη ασκηση, με την εξης προταση: Αν μια συναρτηση ορισμενη στο R  ειναι γνησιως μονοτονη στο R και επι σε αυτο τοτε ειναι και συνεχης! (πειτε μου αν θελετε να σας δωσω την αποδειξη που βρηκα)
αυτο εννοουσα στο ποστ. Οτι αν βγαζαμε οτι ειναι γνησιως μονοτονη λογω της παραπανω προτασης η συνεχεια ειναι δεδομενη!
για το αλλο,
θεωρω γνωστο το f(0)=0,
και θετω στην αρχικη χωρις βλαβη οπου y το f(x)
και μου προκυπτει οτι ισχυει
$ f(f(f(x)) \equal{} f(x) \minus{} f(0) \equal{} f(x)$, για καθε x στο R
η ισοδυναμα αφου ειναι 1-1 $ f(f(x)) \equal{} x$..
ισως να μην βλεπω κατι αλλα νομιζω πως στην παραπανω διαδικασια δν χρησιμοποιω πουθενα το επι..

Demetres εχεις δικιο παντως σε αυτο που λες.  Η αντιστροφη διαδικασια μας δινει την απαντηση οτι δεν μπορουμε να βρουμε συνεχεια/μονοτονια!  Παντως μου φαινονται ιδιατερα ενδιαφερον οι wild functions, και επειδη ξερω ελαχιστα πραγματα για αυτες, αναρωτιομουν μηπως καποιος μπορει να μου στειλει κανενα pdf που να αναφερεται εκει..;(θα ψαξω και στο ιντερνετ παντως)

[quote="Athinaios"]
Demetres εχεις δικιο παντως σε αυτο που λες.  Η αντιστροφη διαδικασια μας δινει την απαντηση οτι δεν μπορουμε να βρουμε συνεχεια/μονοτονια!  Παντως μου φαινονται ιδιατερα ενδιαφερον οι wild functions, και επειδη ξερω ελαχιστα πραγματα για αυτες, αναρωτιομουν μηπως καποιος μπορει να μου στειλει κανενα pdf που να αναφερεται εκει..;(θα ψαξω και στο ιντερνετ παντως)[/quote]

Ηλία, έχω γράψει λίγα πράγματα για αυτές τις συναρτήσεις στο mathematica.gr. Για να μην κάνω link σε άλλο forum όμως, θα μεταφέρω αυτά που έγραψα και εδώ και αν θέλεις να εμβαθύνουμε περισσότερο με ρωτάς από εδώ.

Θα ψάξω και στο internet να δω αν υπάρχουν καλές σημειώσεις.

Νομιζω οτι χρησημοποιεις το επι Ηλια. Δεν το εχω κοιταξει, αλλα διαισθητικα νομιζω οτι χωρις το επι δεν μπορεις να δειξεις το 1-1 (πως θα εμφανισεις καπου μονο του το x η το y?).

Νικο, στην ασκηση του Κυριου Στεργιου μας δινεται το 1-1 και μας ζητειται το οτι η f  ειναι περιττη.
εγω δειχνω το οτι με το 1-1 μπορουμε να παρουμε το επι και απο εκει το οτι ειναι περιττη...

[quote="Athinaios"]Κυριε Μπαμπη,
προ 2 μηνων ειχα ασχοληθει, με αφορμη μια σχολικη ασκηση, με την εξης προταση: Αν μια συναρτηση ορισμενη στο R  ειναι γνησιως μονοτονη στο R και επι σε αυτο τοτε ειναι και συνεχης! (πειτε μου αν θελετε να σας δωσω την αποδειξη που βρηκα)
αυτο εννοουσα στο ποστ. Οτι αν βγαζαμε οτι ειναι γνησιως μονοτονη λογω της παραπανω προτασης η συνεχεια ειναι δεδομενη!
[/quote]

Ηλία εντάξει ! Είχα στο μυαλό μου ''το 1-1 και  επί του IR''  και όχι το ''γν.μονότονη και επί'' που έγραψες!

 Σε ευχαριστώ΄!
 Μπάμπης

Βγήκαν τα αποτελέσματα Ευκλείδη...πέρασα!
Ξέρει κανείς για το πότε ανακοινώνονται οι βραβευθέντες Ευκλείδη? Μπορεί να έχω πάρει αλλά δεν είμαι καθόλου σίγουρος, μιας και μου έκοψαν από 3ο θέμα 1-2 πόντους.

Καλή επιτυχία σε όλους όσους πέρασαν στον Αρχιμήδη! Θα σας δω εκεί.  :) (εξάλλου, εγώ θα πάω περισσότερο για την εμπειρία, προσωπικά δεν βλέπω να λύνω πάνω από 1 Θέμα)

[quote="Asynartisies"]προσωπικά δεν βλέπω να λύνω πάνω από 1 Θέμα)[/quote]

dn ksereis posa tha luseis. pote mi les pote :wink:


calculus
integration
real analysis
real analysis unsolved

In the book "Introduction to Calculus and Analysis. Volume2" Author: Richard Courant and Fritz John

In Chapter 4 Exercise 4.6-1(c):
calculate:
$ \int_{0}^{e}\int_{0}^{2}\frac{1}{xy}dydx$

But I think this integral is divergent. so I think it equal $ \plus{}\infty$  Right or Wrong ?


Given two circles, and their four common tangent segments (two external, two internal).

Prove that the midpoints of the four tangent segments all lie on one line.

the all have equal power with respect to the circle, so they lie on the circles' radical axis...which is a line...this is not high school basics...


$ a,b,c \geq 0$ . prove that :
$ \frac{ab}{c(c\plus{}a)} \plus{} \frac{bc}{a(a\plus{}b)}\plus{}\frac{ca}{b(b\plus{}c)}\geq \frac{a}{c\plus{}a} \plus{}\frac{b}{a\plus{}b} \plus{}\frac{c}{b\plus{}c}$

let be $ x \equal{} \frac {b}{c}$ , $ y \equal{} \frac {a}{b}$et $ z \equal{} \frac {c}{a}$

so your ineqalitie is eqivalent to

$ \frac {x}{1 \plus{} z} \plus{} \frac {z}{1 \plus{} y} \plus{} \frac {y}{1 \plus{} x } \geq \frac {1}{1 \plus{} x} \plus{} \frac {1}{1 \plus{} y} \plus{} \frac {1}{1 \plus{} z}$ and $ xyz \equal{} 1$

$ \iff$ $ (x^2 \plus{} y^2 \plus{} z^2 \plus{} xy(x \plus{} y) \plus{} yz(y \plus{} z) \plus{} xz(x \plus{} z) \minus{} 3 \minus{} (xy \plus{} yz \plus{} xz) \minus{} (x \plus{} y \plus{} z) \geq 0$

and you have by réordonement
$ xy(x \plus{} y) \plus{} yz(y \plus{} z) \plus{} xz(x \plus{} z) \geq 3 \plus{} \frac {xy(x \plus{} y) \plus{} yz(y \plus{} z) \plus{} xz(x \plus{} z)}{2} \geq 3 \plus{} \frac {(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx}{3}$

and $ xy \plus{} yz \plus{} zx \geq 3$   ....

It is equivalent to
$ \sum \frac {1}{c^2(c \plus{} a)} \ge \sum \frac {1}{bc(c \plus{} a)}$
$ \sum \frac {(a \plus{} b)(b \plus{} c)}{c^2} \ge \sum \frac {(a \plus{} b)(b \plus{} c)}{bc}$
$ \sum \frac {ab}{c^2} \plus{} \frac {a}{c} \plus{} \frac {b^2}{c^2} \plus{} \frac {b}{c} \ge \sum \frac {a}{c} \plus{} \frac {a}{b} \plus{} \frac {b}{c} \plus{} 1$
$ \sum \frac {ab}{c^2} \plus{} \frac {b^2}{c^2} \ge \sum \frac {a}{b} \plus{} 1$
But this is obvious as $ \sum \frac {ab}{c^2} \ge 3$ and $ \sum \frac{a^2}{b^2} \ge \sum \frac{a}{b}$ by Muirhead.

[quote="iqtimo"]$ a,b,c \geq 0$ . prove that :
$ \frac {ab}{c(c \plus{} a)} \plus{} \frac {bc}{a(a \plus{} b)} \plus{} \frac {ca}{b(b \plus{} c)}\geq \frac {a}{c \plus{} a} \plus{} \frac {b}{a \plus{} b} \plus{} \frac {c}{b \plus{} c}$[/quote]
my solution :
we need prove
$ \frac {ab}{c(c \plus{} a)} \plus{} \frac {bc}{a(a \plus{} b)} \plus{} \frac {ca}{b(b \plus{} c)} \plus{} \frac {c}{c \plus{} a} \plus{} \frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \geq 3$
we have :
$ \frac{ab}{c(a\plus{}c)} \plus{}\frac{a}{a\plus{}b} \equal{}\frac{a(b^{2}\plus{}c^{2}\plus{}ab\plus{}ac)}{c(a\plus{}c)(a\plus{}b)}\geq \frac{a(b\plus{}c)(\frac{b\plus{}a}{2}\plus{}\frac{c\plus{}a}{2})}{c(a\plus{}b)(a\plus{}c)}$
using AM-GM ( three number ) and
$ (\frac{a\plus{}b}{2}\plus{}\frac{b\plus{}c}{2})(\frac{b\plus{}c}{2}\plus{}\frac{c\plus{}a}{2})(\frac{c\plus{}a}{2}\plus{}\frac{a\plus{}b}{2})\geq (a\plus{}b)(b\plus{}c)(c\plus{}a)$
Done !


geometry
circumcircle
perpendicular bisector
geometry proposed

Ok, let's turn to 1996.
I hope this one has not been proposed at least 10 times here.

In the plane, we consider the triangle $ABC$ and the points $D,E,F,G,H,I$ such that $ABED,BCGF$ and $ACHI$ are squares exterior to $ABC$.
Prove that the points $D,E,F,G,H,I$ are cocyclic if and only if we are in one of the two following cases :
- $ABC$ is equilateral.
- $ABC$ is right-angled and isocele.

Pierre.

It’s clear that the center of the potential circumcircle is $O$, the same as $ABC$’s circumcenter. Draw the segments $EH,DG,FI$. The triangles $OEH,ODG,OFI$ must all be isosceles ($OE=OH$, etc.). A well-known property of this configuration is that the midpoint of $EH$ is on the perpendicular bisector of $BC$ (the same for the others), so the fact that $D,E,F,G,H,I$ are concyclic is equivalent to 1) $EH\|BC,DG\|CA,FI\|AB$ or 2) $EH\|BC$ and $DG,FI$ intersect at $O$ which must also be their common midpoint (or any of the other two analogous situations) or 3) $EH,DG,FI$ intersect at $O$ which must also be their common midpoint.

3) is impossible, and 1),2) give the two cases mentioned in the text of the problem.


Find all $ n\minus{}tuples$  $ (a_1,a_2,..,a_n)$ of positive integers such that:
                  $ (a_1!\minus{}1)(a_2!\minus{}1)...(a_n!\minus{}1)\minus{}16$ is a perfect square.

If $ a_i>3$, $ (a_i)!\minus{}1\equiv 3 \bmod{4}$
Then the number $ (a_i)!\minus{}1$ has a prime factor of the form $ p\equal{}4k\plus{}3$
then we have that $ p$ divides $ k^2\plus{}4^2$, and since $ p\equal{}4k\plus{}3$ it has to divide both $ k$ and $ 2$, so this is impossible.
If $ a_i\equal{}1$ then the expression becomes $ \minus{}16\equal{}k^2$ contradiction
If some $ a_i\equal{}2$ they don't make any difference in the product, so we can add as many $ a_j\equal{}2$ as we wish
Finally let's say that $ l$ numbers are equal to $ 3$ then:

$ 5^l\minus{}16\equal{}k^2$ clearly $ k$ is odd, then $ k^2 \equiv 1\bmod{8}$ so $ 5^l \equiv 1\bmod{8}$ implying $ l$ is even, let's say $ l\equal{}2r$
Then:

$ (5^r\minus{}k)(5^r\plus{}k)\equal{}16$
We now have to see this cases:
$ (5^r\minus{}k)\equal{}1  (5^r\plus{}k)\equal{}16$
$ (5^r\minus{}k)\equal{}2   (5^r\plus{}k)\equal{}8$
(the other cases are similar)
adding in both cases the equations we find a solution only in the second case, when $ r\equal{}1$, so $ l\equal{}2$
Then the solution of the problems are:
$ a_1\equal{}3$ $ a_2\equal{}3$ and as many as we wish of $ a_j\equal{}2$, and cyclic permutations


trigonometry inequalities geometry inradius circumcircle search

For reference, we are given a triangle $ ABC$. We also have the semiperimeter $ s$, the inradius $ r$, and the circumradius $ R$.

In a triangle $ ABC$, show that:

1. $ \sin A\plus{}\sin B\plus{}\sin C\equal{}4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$

2. $ \cos^2\frac{A}{2}\plus{}\cos^2\frac{B}{2}\plus{}\cos^2\frac{C}{2}\equal{}2\plus{}\frac{r}{2R}$

3. $ \frac{\cos B\minus{}\cos C}{s\minus{}a}\plus{}\frac{\cos C\minus{}\cos A}{s\minus{}b}\plus{}\frac{\cos A\equal{}\cos B}{s\minus{}c}\equal{}0$

4. $ (b\plus{}c)\tan \frac{B\minus{}C}{2}\plus{}(c\plus{}a)\tan \frac{C\minus{}A}{2}\plus{}(a\plus{}b)\tan\frac{A\minus{}B}{2}\equal{}0$

5. $ \frac{r}{4R}\le \frac{abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\le \frac{1}{8}$

6. $ \cos A \cos B \cos C \le \frac{1}{8}$

6. $ \cos^2 \frac{A}{2}\plus{}\cos^2 \frac{B}{2}\plus{}\cos^2 \frac{C}{2} \le \frac{9}{4}$

This definitely does not belong in basics.

1.
Since $ ABC$ is a triangle, $ A+B+C=180^{\circ}$.

$ \begin{align*}
\sin A + \sin B +\sin C &= \sin A + \sin B + \sin (180^{\circ}-(A+B))\\
&= \sin A + \sin B + \sin (A+B)\\
&= \sin A + \sin B + \sin A\cos B + \cos A\sin B\\
&=\sin A(1 +\cos B) + \sin B(1+ \cos A)\\
&=2\sin \frac{A}{2}\cos\frac{A}{2}2\cos^2\frac{B}{2} + 2\sin \frac{B}{2}\cos\frac{B}{2}2\cos^2\frac{A}{2}\\
&=4\cos\frac{A}{2}\cos\frac{B}{2}(\sin\frac{A}{2}\cos\frac{B}{2} + \sin\frac{B}{2}\cos\frac{A}{2})\\
&=4\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{A+B}{2}\\
&=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$

hello, are you sure that your second equation ist true?
By calculating the left hand site i have got:
$ \cos^2\left(\frac{\alpha}{2}\right)\plus{}\cos^2\left(\frac{\beta}{2}\right)\plus{}\cos^2\left(\frac{\gamma}{2}\right)\equal{}
\frac{s(s\minus{}a)}{bc}\plus{}\frac{s(s\minus{}b)}{ac}\plus{}\frac{s(s\minus{}c)}{ab}\equal{}
\minus{}\frac{1}{4}\frac{(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}2ab\minus{}2ac\minus{}2bc)}{abc}$.
Substituting $ r\equal{}\frac{A}{s}$ and $ R\equal{}\frac{abc}{4A}$ with
$ A\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$ and simplifying this we get for the right hand site:
$ \minus{}\frac{1}{2}\frac{a^3\plus{}b^3\plus{}c^3\minus{}2abc\minus{}a^2b\minus{}a^2c\minus{}ab^2\minus{}b^2c\minus{}ac^2\minus{}bc^2}{abc}$
This is not true.

6. $ cosAcosBcosC\leq8$

$ A\geq B\geq C$ -->

$ cosAcosBcosC\equal{}\frac{1}{2}(cos(A\minus{}B)\plus{}cos(A\plus{}B))cosC\equal{}\frac{1}{2}(cos(A\minus{}B)cosC\plus{}cos(\pi\minus{}C)cosC)\equal{}\frac{1}{2}(cos(A\minus{}B)cosC\minus{}cosC*cosC)\leq\frac{1}{2}(cosC\minus{}cos^{2}C)\equal{}\frac{1}{2}(\frac{1}{4}\minus{}(\frac{1}{2}\minus{}cosC)^{2})\leq\frac{1}{8}$

$ A\equal{}B\equal{}C\equal{}60^{o}$

hello, by substituting
$ \cos(\beta)\minus{}\cos(\gamma)\equal{}\frac{a^2\plus{}c^2\minus{}b^2}{2ac}\minus{}\frac{a^2\plus{}b^2\minus{}c^2}{2ab}$
$ \cos(\gamma)\minus{}\cos(\alpha)\equal{}\frac{a^2\plus{}b^2\minus{}c^2}{2ab}\minus{}\frac{b^2\plus{}c^2\minus{}a^2}{2bc}$
$ \cos(\alpha)\minus{}\cos(\beta)\equal{}\frac{b^2\plus{}c^2\minus{}a^2}{2bc}\minus{}\frac{a^2\plus{}c^2\minus{}b^2}{2ac}$
in the left hand side of your equation you will get your desired result.
Sonnhard.

[quote="kl2836"]For reference, we are given a triangle $ ABC$. We also have the semiperimeter $ s$, the inradius $ r$, and the circumradius $ R$.

In a triangle $ ABC$, show that:

1. $ \sin A \plus{} \sin B \plus{} \sin C \equal{} 4\cos\frac {A}{2}\cos\frac {B}{2}\cos\frac {C}{2}$

2. $ \cos^2\frac {A}{2} \plus{} \cos^2\frac {B}{2} \plus{} \cos^2\frac {C}{2} \equal{} 2 \plus{} \frac {r}{2R}$

3. $ \frac {\cos B \minus{} \cos C}{s \minus{} a} \plus{} \frac {\cos C \minus{} \cos A}{s \minus{} b} \plus{} \frac {\cos A \equal{} \cos B}{s \minus{} c} \equal{} 0$

4. $ (b \plus{} c)\tan \frac {B \minus{} C}{2} \plus{} (c \plus{} a)\tan \frac {C \minus{} A}{2} \plus{} (a \plus{} b)\tan\frac {A \minus{} B}{2} \equal{} 0$

5. $ \frac {r}{4R}\le \frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\le \frac {1}{8}$

6. $ \cos A \cos B \cos C \le \frac {1}{8}$

7. $ \cos^2 \frac {A}{2} \plus{} \cos^2 \frac {B}{2} \plus{} \cos^2 \frac {C}{2} \le \frac {9}{4}$[/quote]

[hide="1."]
we have:

$ \sin A \plus{} \sin B \plus{} \sin C \equal{} \frac {a}{2R} \plus{} \frac {b}{2R} \plus{} \frac {c}{2R} \equal{} \frac sR$

and also

$ \cos\frac {A}2\cos\frac {B}2\cos\frac {C}2 \equal{} \frac s{4R}$

$ \Rightarrow\sin A \plus{} \sin B \plus{} \sin C \equal{} 4\cos\frac A2\cos\frac B2\cos\frac C2$[/hide]
[hide="2."]
we have:

$ \cos^2\frac A2 \plus{} \cos^2\frac B2 \plus{} \cos^2\frac C2 \equal{} \frac {4R \plus{} r}{2R}$[/hide]
[hide="5."]
the right inequality is true because of AM-GM,the left inequality is posted [url=http://www.mathlinks.ro/viewtopic.php?search_id=1756613842&t=214455]here[/url][/hide]
[hide="6."]
we have:

$ \cos A \plus{} \cos B \plus{} \cos C \equal{} \frac {R \plus{} r}R$

but we know that $ 2r\leq R$ hence:

$ \cos A \plus{} \cos B \plus{} \cos C\leq\frac 32$

now by AM-GM we have:

$ \frac 32\geq\cos A \plus{} \cos B \plus{} \cos C\geq 3\sqrt [3]{\cos A\cos B\cos C}$

$ \Rightarrow \cos A\cos B\cos C\leq\frac 18$[/hide]
[hide="7."]
we have:

$ \cos^2\frac A2 \plus{} \cos^2\frac B2 \plus{} \cos^2\frac C2 \equal{} \frac {4R \plus{} r}{2R}$

hence we have to show that:

$ \frac {4R \plus{} r}{2R}\leq\frac 94$

$ \iff 2r\leq R$[/hide]

3:

$ \cos B \minus{} \cos C \equal{} 2 \sin^2 \frac{C}{2} \minus{} 2 \sin^2 \frac{B}{2}$ $ \equal{} \frac{2(p\minus{}a)(p\minus{}b)c \minus{}2(p\minus{}a)(p\minus{}c)b}{abc}\equal{}\frac{2p(p\minus{}a)(c\minus{}b)}{abc}$

Then $ \frac{\cos B\minus{}\cos C}{p\minus{}a}\plus{}\frac{\cos C\minus{}\cos A}{p\minus{}b}\plus{}\frac{\cos A\minus{}\cos B}{p\minus{}c}$ $ \equal{}\frac{2p \left [(c\minus{}b)\plus{}(a\minus{}c)\plus{}(b\minus{}a) \right ]}{abc} \equal{}0$

[color=darkblue]$ \boxed {0.}\blacktriangleright\ \|\begin{array}{c} a^2\ge 4(p - b)(p - c) \\
 \\
b^2\ge 4(p - c)(p - a) \\
 \\
c^2\ge 4(p - a)(p - b)\end{array}\|\ \bigodot$ $ \implies$ $ abc\ge 8\prod (p - a)$ $ \implies$ $ 4Rpr\ge 8pr^2$ $ \implies$ $ \boxed {\ R\ge 2r\ }$ .

$ \sum\cos A = 1 - 2\sin^2\frac A2 + 2\cos\frac {B + C}{2}\cos\frac {B - C}{2} =$ $ 1 + 2\sin\frac A2(\cos\frac {B - C}{2} - \cos\frac {B + C}{2}) =$

$ 1 + 4\sin\frac A2\sin\frac B2\sin\frac C2 =$ $ 1 + \frac {4(p - a)(p - b)(p - c)}{abc} =$ $ 1 + \frac {4pr^2}{4Rpr}\implies$  $ \boxed {\sum\cos A = 1 + \frac rR\le\frac 32}$ .

$ \boxed {1.}\blacktriangleright\ \|\begin{array}{c} \sin A + \sin B + \sin C = \sum \frac {a}{2R} = \frac pR \\
 \\
4\cos\frac {A}{2}\cos\frac {B}{2}\cos\frac {C}{2} = 4\sqrt {\frac {p(p - a)}{bc}\cdot \frac {p(p - b)}{ca}\cdot\frac {p(p - c)}{ab}} = \frac {4pS}{abc} = \frac {4pS}{4RS} = \frac pR\end{array}$ .

$ \boxed {2.}\blacktriangleright\ \sum\cos^2\frac {A}{2} =$ $ \frac 12\cdot\sum (1 + \cos A) = \frac 12\cdot (3 + \sum \cos A) =$ $ \frac 12\cdot(4 + \frac rR) = 2 + \frac {r}{2R}$

$ \boxed {3.}\blacktriangleright\ \sum \frac {\cos B - \cos C}{p - a} = \frac 1r\cdot \sum [(\cos B - \cos C)\cdot \tan\frac A2] =$ $ \frac 2r\cdot\sum[\sin \frac {B + C}{2}\sin \frac {C - B}{2}\cdot\frac {\sin\frac A2}{\cos\frac A2}] =$

$ \frac 2r\cdot\sum[\sin\frac {C - B}{2}\cos \frac {B + C}{2}] =$ $ \frac 1r\cdot\sum(\sin C - \sin B) = 0$ .

$ \boxed {4.}\blacktriangleright\ \sum (b + c)\tan \frac {B - C}{2} =$ $ 2R\cdot \sum(\sin B + \sin C)\cdot\tan \frac {B - C}{2} =$

$ 2R\cdot\sum(2\sin\frac {B + C}{2}\cos\frac {B - C}{2}\cdot\tan\frac {B - C}{2}) =$

$ 2R\cdot\sum(2\sin\frac {B + C}{2}\sin\frac {B - C}{2}) =$ $ 2R\cdot\sum(\cos C - \cos B) = 0$ .

$ \boxed {5^*.}\blacktriangleright\ \frac {r}{4R}\le \frac {abc}{(a + b)(b + c)(c + a)}\le \frac {1}{8}$ . See[/color] [url=http://www.mathlinks.ro/viewtopic.php?search_id=1756613842&t=214455][color=red][b]here[/b][/color][/url] [color=darkblue]for LHS. The RHS is evidently.

[b][u]The BaBaK Ghalebi's proof.[/u][/b] $ \frac {abc}{\prod (b + c)} =$ $ \frac {abc}{\prod (2p - a)} =$ $ \frac {4Rpr}{(2p)^3 - 2p\cdot (2p)^2 + (p^2 + r^2 + 4Rr)\cdot (2p) - 4Rpr}\implies$

$ \boxed {\ \frac {abc}{\prod (b + c)} = \frac {2Rr}{p^2 + 2Rr + r^2}\ }\ (1)$ . Since $ 4R^2 + 4Rr + 3r^2\le 8R^2 - 2Rr - r^2\Longleftrightarrow$ $ (R - 2r)(2R + r)\ge 0$

what is truly and from the [b]Gerretsen's inequality[/b] obtain $ \boxed {p^2\stackrel{(G)}{\ \le\ } 4R^2 + 4Rr + 3r^2}\le R^2 - 2Rr - r^2$ ,

i.e. $ p^2 + 2Rr + r^2\le 8R^2\ (2)$ . Thus, from the relations $ (1)$ , $ (2)$ obtain $ \frac {abc}{\prod (b + c)} =$ $ \frac {2Rr}{p^2 + 2Rr + r^2}\ge\frac {2Rr}{8R^2} = \frac {r}{4R}$ .

[b][u]My proof.[/u][/b] Denote the diameter $ [NS]$ in the circumcircle of $ \triangle ABC$ so that $ S\in AI$ , where $ I$ is the its incenter.

Since $ \max \{b,c\}\le NB=NC=2R\cos\frac A2$ obtain $ b+c\le 4R\cos\frac A2$ . Therefore,

$ \prod (b+c)\le 64R^3\prod\cos\frac A2=$ $ \frac {64R^3\cdot p^2r}{4Rpr}=$ $ 16R^2p=\frac {4R}{r}\cdot abc\ \implies\ \boxed {\ \frac {abc}{\prod (b+c)}\ge\frac {r}{4R}\ }$ .

$ \boxed {6.}\blacktriangleright\ 1 - 8\cdot\prod\cos A = 1 - 4\cos A[\cos (B + C) + \cos (B - C)] =$ $ 1 - 4\cos A[ - \cos A + \cos (B - C)] =$

$ 4\cos^2A - 4\cos A\cos (B - C) + 1 =$ $ [2\cos A - \cos (B - C)]^2 + \sin^2(B - C)\ge 0$ $ \implies$

$ \prod\cos A\le\frac 18$ with equality iff $ B = C$ and $ \cos A = \frac 12$ , i.e. $ A = B = C$ .

$ \boxed {7.}\blacktriangleright\ \sum \cos^2 \frac {A}{2} = \frac 12\cdot\sum (1 + \cos A) = \frac 12\cdot(3 + \sum\cos A) =$ $ \frac 12\cdot(4 + \frac rR) = 2 + \frac {r}{2R}\le 2 + \frac 14$ $ = \frac {9}{4}$[/color]


Let x,y,u,v are postive real numbers
please tell me the solution of the  follwoing inequality
$ \frac {u}{x} + \frac {v}{y}\geq4{\frac {uy + vx}{(x + y)^2}}$
thank you very much.

[quote="FelixD"]Please check your inequality (there are some typing mistakes)[/quote]
I got it,thanks ..

[quote="Frank-Jun"]Let $ x,y,u,v$ are postive real numbers
please tell me the solution of the  follwoing inequality
$ \frac {u}{x} \plus{} \frac {v}{y}\geq4\frac {uy \plus{} vx}{(x \plus{} y)^2}$
thank you very much.[/quote]
your inequality is equivalent as, $ (x\plus{}y)^2 \geq 4xy$ wich is true.


How many whole numbers are between $ \sqrt{8}$ and $ \sqrt{80}$?

\[ \textbf{(A)}\ 5 \qquad
\textbf{(B)}\ 6 \qquad
\textbf{(C)}\ 7 \qquad
\textbf{(D)}\ 8 \qquad
\textbf{(E)}\ 9
\]

[hide="Click for solution"]
Since $ \sqrt{8}<\sqrt{9}\equal{}3$, and $ \sqrt{80}<\sqrt{81}\equal{}9$, the desired whole numbers are $ 3,4,5,6,7,8$; there are six of them, or answer $ \boxed{\textbf{(B)}}$.
[/hide]


if we have $ x,y,z \in [\minus{}2,1]$ and $ x\plus{}y\plus{}z \equal{}0$

then $ x^2 \plus{}y^2\plus{}z^2 \leq 6$

 :D

Let $ z\equal{}\minus{}x\minus{}y$,  then inequality reduces to \[ x^2\plus{}y^2\plus{}xy\le 3\]or\[ (x\minus{}1)(x\plus{}2)\plus{}(y\minus{}1)(y\plus{}2)\plus{}(x\minus{}1)(y\minus{}1)\le 0\] or\[ (x\minus{}1)(x\plus{}y\plus{}1)\plus{}(y\minus{}1)(y\plus{}2)\le 0\] or \[ (x\minus{}1)(1\minus{}z)\plus{}(y\minus{}1)(y\plus{}2)\le 0\]  wich is true.

[quote="Hidden Scofield"]Prove it .... I know it's true  :D[/quote]
It's proved, because we can assume that $ z\leq0.$ :wink:

Nice Marius Mainea, but since Hidden Scofield appears to have trouble understanding your very elegant arrangement, and arqady appears to have the need to assume $ z\leq0$ for some strange reason, here's another proof to $ x^2\plus{}y^2\plus{}xy \leq 3$ using less elegant casework.

Note if $ x,y >0$, then each term is less than 1 and the inequality holds; the same happens when $ \minus{}1<x,y<0$. $ x\equal{}y\equal{}0$ is trivial. If $ x >0$ and $ y<0$ , then we can set $ x\equal{}1$ to maximize the left side, and we see that the remaining $ y^2\plus{}y\minus{}2 \equal{} (y\plus{}2)(y\minus{}1)\leq 0$ holds for all $ y$ in $ [\minus{}2,1]$. The same logic applies to $ y>0$ and $ x<0$. If $ x$ and $ y$ are negative, then we note that $ x,y \geq \minus{}1$, for if $ x <\minus{}1$ (or similarly for $ y$), then $ x\plus{}y <\minus{}1$, which implies $ \minus{}z < \minus{}1$, or $ z>1$, which is impossible.

With this problem,i think we only need $ (x\plus{}2)(x\minus{}1)\le 0$ hence $ x^2\le 2\minus{}x$
Similarly with$ y,z$ we have the proof
Is it similar to Marius Mainea's

I appreciate Marius Mainea solution. But , when I said "prove it" I was kidding . did u see my  :D,  .I haven't any trouble to understand Duelist  :P

My solution is like your solution Duelist.

Thank u guys .

[quote="Hidden Scofield"]if we have $ x,y,z \in [ \minus{} 2,1]$ and $ x \plus{} y \plus{} z \equal{} 0$

then $ x^2 \plus{} y^2 \plus{} z^2 \leq 6$

 :D[/quote]

 \[ \sum {(x\plus{}2)(x\minus{}1)}\le 0\]or\[ \sum {x^2}\minus{}\sum {x}\minus{}6\le 0\]or\[ \sum {x^2}\le 6\]


try to find best way to prove this very nice inequality where x,y>0 =>  (x^x)+(y^y)>=(x^y)+(y^x)

[quote="Extremal"]yes, but there is not sulution for a,b>1 because ln(a)/1-a>ln(b)/1-b is true only when 0<a,b<1  :lol:[/quote]
when a,b>1 i think it's not interesting because it's siml. Nice suluion for a,b<1


inequalities
complex numbers
absolute value
triangle inequality

Complex numbers( Z_1) ,(Z_2),.....................(Z_n) satisfy the equation

 I(Z_1 )I+ I(Z_2)I + I(Z_3)I+................ +I(Z_n)I    =  I (Z_1) +(Z_2) +(Z_3)+..................+(Z_n) I ,   whereIZI denotes absolute value or modulus  of the complex number Z

Prove the existence of a unique complex number W  such that for k= 1,2,3.......n

    (Z_k)/W is real and non-negative

Specially for Michael Niland ;) :)

Complex numbers $z_1$, $z_2$, ..., $z_n$ satisfy the equation
\[|z_1|+|z_2|+\dots+|z_n| = |z_1+z_2+\dots+z_n|,\]
where $|z|$ denotes absolute value or modulus of the complex number $z$.

Prove the existence of a complex number $w$ such that for $k= 1,2,\dots,n$ number $\frac{z_k}{w}$ is real and non-negative.

This problem was actually posted here before and solved before the temporary site shutdown.

The triangle inequality has equality iff $\text{Arg }{ z_1}=\text{Arg }{ z_2}=\ldots=\text{Arg }{ z_n}$ and $r_1,r_2,\ldots r_n\geq 0$.

Then it's enough to choose our $w$ as any of the $z_i$'s.


function
abstract algebra
algebra solved
algebra

Solve the equation

3^x + 6^x = 4^x + 5^x

<i>Folklore</i>

x=0  and  x=1  are  solutions. We  will  show that  they are  the  only solutions  of the  equation.
Let  p another  solution  with  p  not  0  and  p  not 1 .
Then  we have  6^p - 5^p = 4^p - 3^p .    (*)
Concider  the  function  f(x)=x^p . Then  we  have  f(6)-f(5) = f(4)-f(3)  so  the  mean  value theorem  gives f'(r1)=f'(r2)  where  5<r1<6  and  3<r2<4 . So  we  have  f'(r1)-f'(r2)=0  so  again  the  mean  value theorem  gives  f''(d)=0  with  r1<d<r2 , which  is  not  true  because
f''(d)=0<==>d=0 (if  p>2)  and  f''(d)=0  is wrong in  case  p<=2 , p not  0  and  p  not  1.
So  the  only solutions  are  x=0 and  x=1 .

3^x + 6^x = 4^x + 5^x

Rewrite the ecuation like so:
6^x-5^x=4^x-3^x and we consider the function f(t)=t^x to which we apply Lagrange's theorem on the intervals [5,6], respectively [3,4]. SO there exists c1 in (5,6) and c2 in (3,4) so that 6^x-5^x=x*c1^(x-1), respectively 4^x-3^x=x*c2^(x-1) and the ecuation becomes x(c1^(x-1)-c2^(x-1))=0 so
1)x1=0
2)c1^(x-1)=c2^(x-1) ==> x2=1


Let f: N -> N with f(n) = n + 10 for n < 100 and f(n) = f(f(n-11)) for n \geq 100. Determine f(2004).

Dear guest,

there are two things I do not appreciate too much:

1.) If you have something to say, please log in and post under your user-name. If you are not registered yet, you can register now. It is for free.

2.) This is the beginner's corner. If you think the problems are too easy do not care about them and leave them for the beginners. Go to the advanced section and look for some more challenging problems. I bet you will find some you cannot solve. If you can solve them all I am sure I could post problems you are not likely to solve at all. We are waiting for your contributions. Happy-Problem-Solving !!!

Here I have a slightly more general problem:

Let k \in N be a constant and f: N -> N with f(n) = n + k for n < k 2  and f(n) = f 2 (n-k-1) for n \geq k 2 . Determine an explicit formula for this recursion.

f(x)=f^2(n-k-1)
Maybe  it  is  f(x)=f(f(n-k-1))

That is no valid solution. Where is the explicit formula and its proof ?  :)


function
real analysis
real analysis solved

let f,g:R->R be to continuous functions with the properties:
i)f is periodic and nonconstant
ii)lim{x->oo}g(x)/x=oo
Prove that f o g is not periodic.

I really hope that my memory doesn't work very well (and if it is so, then I apologise). But from what I remember, this problem appeared in a contest of Gazeta Matematica in the mod of 1985 or something like this. And it was proposed by Pal Dalyai, a great problemist. It was an extraorninary problem, the most difficult in the contest. After that, a whole article was written. So, I ask friendly: how can Dan Seclaman take this problem and put his name on it?

the book's name is "1000 de probleme de matematica", editura Cardinal si Oltenia,autori:Dan Seclaman, Danut Dracea..
here doesn't write that he invented the problems, or that these are original... maybe they just pick the problems and made a set of them :)
should i write then in the source :"book"

Please write reference as precise as you can, not just book.
Well, it seems then that you have 1000 exercises to propose... :D

Pierre.

I think I can show this (asuming that f(g) is periodic of least period T, because it's onviously not constant):

For every value y of f(g) there exist a<b with a in [0,T] and b arbitrarily close to a s.t. f(g(a))=f(g(b))=y. This shows that for all values y of f(g) the set of solutions of the equation f(g(x))=y is non-countable. Can we get a contradiction only from here or do we need more?

Sorry: that stuff about a and b is true, but the set I was talking about isn't necessarily uncountable, and I think we needed more for a contradiction anyway :(.

I do not know the proof of this " theorem" (it's so nice that I think it can be called a theorem), but I'm really interested.

Here's a problem proposed by M.Krych for IMC 2002 (I think it was also chosen):

Let f:[0,1]->[0,1]
(a) Can f be monotonic s.t. for all y\in [0,1] the set of solutions of f(x)=y is uncountable?
(b) Can f be of class C1 s.t. for all y\in [0,1] the set of solutions of f(x)=y is uncountable?

The answer to both questions is no, but, as you can see, the hypotheses are, in both cases, stronger than just "f continuous". This doen't mean it's not true for f continuous, but it might be way beyond our reach.

Anyway, I'll be back soon because I might have a solution of the problem posted by Kuba.

This might work:

Let's assume that g takes all values from 0 to 00 on the interval [0,00), just for the simplicity of expression (it's clearly not crucial, since there exist c and t s.t. g takes all values from c to 00 on the interval [t,00)).

Let a be the least period of f. Then f([0,a])=f([na, (n+1)a]) for all n\in Z. Let kn be sup{x>=0|g(x)=na} for all n>=0. Then f(g([kn, kn+1]))=f(g([k0, k1]))=f([0,a]) for all n>=0. It's easy to show that for any t>0 there is an n s.t. kn+1-kn<t (try it; this is where the fact that g(x)/x->00 comes in). This means that f([0,a])=f(g([k0, k1]) can be written as f(g(In)) with In being a sequence of intervals of length decreasing to 0 s.t. the left extremity of In is in [0,T], where T is the least period of f(g).

Then there exists a subsequence formed from these left extremities which is convergent (because the sequence is in [0,T], so it's bounded). Since the lengths of the intervals decrease to 0, it means that the subsequence formed by the right extremities of the intervals whose left extremities we have considered is convergent to the same limit inside [0,T]. But we must remember that f(g(In))=f([0,a]), and from the continuity of f and g it's easy to derive that f([0,a]) is a real number, so f is constant on [0,a], so f is constant (remember that a was the period of f).

Please tell me if you think it works or not (whoever has the patience to read it :D).


calculus
geometry
conics
parabola
integration
calculus computations

Approximate the total area enclosed by the y-axis, and the curves y=tan(x) and y=(x^2)+1?

Compute the exact area between the parabola y=x(1-x) and the x-axis?

1) Use a Riemann sum with an arbitrary n.

2) Solve the x intercept and then it's an easy integration problem


If $a>1$, $b>1$, and $p=\frac{log_{b}(log_{b}(a))}{log_{b}(a)}$ then find $a^{p}$ in simplest form.

[hide]

$p=\frac{log_{b}(log_{b}(a))}{log_{b}(a)}$
$p=\log_{a}(log_{b}a)$

By the definition of a logarithm:

$a^{p}=\log_{b}a$

[/hide]


algebra
difference of squares
special factorizations

[quote="GameBot"]Find the value of $ 51^2 \minus{} 49^2 \plus{} 101^2 \minus{} 99^2$.[/quote]

$ 2 \cdot 100 \plus{} 2 \cdot 200\equal{}600$.


Eh, gotova je i IOI...

Cestitke svim tamicarima, a posebno osvajacima medalja:

 :winner_first: Goran Zuzic (Hrvatska)
 :winner_second: Ivo Sluganovic (Hrvatska)
  :winner_third: Maja Kabiljo (Srbija), Jan Bercic (Slovenija), Adrijan Satja Kurdija* (Hrvatska), Bruno Rahle (Hrvatska), Ugljesa Stojanovic (Srbija).

(E moji informaticari, da slucajno ne pogledah teletekst na HRTu, ne bih ni vidio da je IOI gotov  :lol: )

*bravo, [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=30851]djuro[/url]!!!

Hvala na cestitkama i pozdrav iz Egipta! Jako zanimljiva olimpijada, iako mi je matematicka draza :)

(Jos uvijek smo tamo, krecemo za sat vremena, a kolege iz Bosne odlaze tek sutra!)


Let $ ABC$ be an acute-angled triangle inscribed in a circle $ \omega$. The tangent to $ \omega$ at $ A$ meets the line $ BC$ at a point $ P$. Let $ M$ denote the midpoint of $ AP$, let $ R$ denote the second intersection of the line $ BM$ with $ \omega$, and let $ S$ denote the second intersection of the line $ PR$ with $ \omega$. Prove that the lines $ AP$ and $ CS$ are parallel.

Let $ X$ be the intersection of $ BR$ and $ CS$. Since $ B$, $ R$, $ S$, $ C$ all lie on a same circle, the point $ X$ lies on the polar of $ P$ with respect to $ \omega$; more precisely, the polar of $ P$ becomes $ AX$. This being said, let $ P'$ be the intersection of $ AX$ with $ BC$, thus $ \minus{}1 \equal{} (XC, XP', XB, XP) \equal{} (XC, XA, XM, XP)$. Since $ M$ is the midpoint of $ AP$, this means that $ AP \| CS$.

Not that difficult!

$ MP^2 \equal{} MA^2 \equal{} MB \times MS$

Hence, $ \angle MPS \equal{} \angle MBP \equal{} \angle SBC \equal{} 2\pi \minus{} \angle SRC$

From this, we deduce $ MP \parallel{} CS$ (QED)

Dear Mathlinkers,
1. After a well knowed theorem, the circle passing through P, B and R is tangent to AP at P
2. After the Reim's theorem applied to this circle and to the circumcircle of ABC, AMP is parallel to CS
Sincerely
Jean-Louis


function
calculus
integration
calculus computations

let caculate
                    $\sum_{i=1}^{n}i^{i}$

maybe changing $i^{i}$ to $e^{ilni}$ can help?

I managed to compute the sums for $n=1,\dots,4$ :cool: they are $1,5,32,288$. Put into Sloane's Encyclopedia and got [url=http://www.research.att.com/~njas/sequences/A001923]this[/url]. Apparently, there is no closed form known for this sequence.

[quote]There is no formula. This sum is the simplest known expression for the sum.[/quote]
why?????????????????

[quote="mathboy_cntHDVN"]why?????????????????[/quote]
You're putting the burden of proof in the wrong place. There are many functions and few closed form formulas. Start with some computation - a sum or an integral, say. The default assumption has to be that it's not likely to have a closed form formula. The burden of proof is then on anyone who wants to claim that it does have such a formula. Failure to find such a formula may merely mean insufficient cleverness, but ask if there's any good reason why there would be such a formula. The verdict most of the time? Probably not.

[quote="Kent Merryfield"]We've already answered you, although you refuse to accept that answer. Please do not post in this topic again.[/quote]

I agree with you, Kent Merryfield. This problem can't be making progresses any more.

[quote]You're putting the burden of proof in the wrong place. There are many functions and few closed form formulas. Start with some computation - a sum or an integral, say. The default assumption has to be that it's not likely to have a closed form formula. The burden of proof is then on anyone who wants to claim that it does have such a formula. Failure to find such a formula may merely mean insufficient cleverness, but ask if there's any good reason why there would be such a formula. The verdict most of the time? Probably not.[/quote]
i donnot understand


I wanna change my avatar but it seems I can't.
it writes something like this:

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and also that my image was updated, but it doesn't change

:)
 :D

I send u the avatar a while ago.
Is there a problem?

no it's just the fact that I have started moving the board to the new place and I'm not upgrading anything now

your avatar will be there :)


[quote="delegat"]If $ a$ and $ b$ are positive integers prove that: $ (\frac {a \plus{} 1}{b \plus{} 1})^{b \plus{} 1} \geq (\frac {a}{b})^{b}$[/quote]
The inequality is equivalent to
\[ \left( \frac {b(a \plus{} 1)}{a(b \plus{} 1)}\right)^{b} \ge \frac {b \plus{} 1}{a \plus{} 1}
\]
By Bernoulli's Inequality, we have
\[ \left( \frac {b(a \plus{} 1)}{a(b \plus{} 1)}\right)^{b} \ge 1 \plus{} b\left( \frac {b(a \plus{} 1)}{a(b \plus{} 1)} \minus{} 1\right) \equal{} 1 \plus{} \frac {b(b \minus{} a)}{a(b \plus{} 1)}
\]
It suffices to prove that
\[ \frac {b(b \minus{} a)}{a(b \plus{} 1)} \ge \frac {b \minus{} a}{a \plus{} 1}
\]

\[ \Leftrightarrow \frac {(b \minus{} a)^2}{a(a \plus{} 1)(b \plus{} 1)} \ge 0
\]
which is true. :)

[b]can_hang2007[/b] thanks for this nice and simple solution!


function
ratio
calculus
derivative
geometric sequence
absolute value
calculus computations

have this equation

$ x^3\plus{}7*x\minus{}11\equal{}0$

by the intermediate value theorem, the root of this equation lies between $ [1,2]$

the equation can also be expressed as a difference equation

$ x_{n\plus{}1}\equal{}\sqrt[3]{11 \minus{} 7*x_n}$

$ x_{n\plus{}1}^3\equal{}11 \minus{} 7*x_n$

I need to show that if $ x_0 \in (1,2)$ then $ x_{n\plus{}1}\equal{}\sqrt[3]{11 \minus{} 7*x_n}$ converges to the root of the equation. The root is not the whole number ($ x \equal{} 1.275196009$).

not sure how to factorise $ x^3\plus{}7*x\minus{}11\equal{}0$ to solve this problem.

would appreciate some help. many thanx.

thank you!

But how would I factorise this?

I need to prove that the difference equation converges to the limit between 1 and 2, but I don't know any other method apart from factorisation of the auxiliary equation and I am not sure how to factorise the auxiliary equation of this difference equation.

many thanx!

We're iterating a function: $ x_{n \plus{} 1} \equal{} f(x_n)$ where $ f(x) \equal{} (11 \minus{} 7x)^{1/3}.$ To assure that a function iteration converges, we need to do one of two things: either prove that the function is a contraction in the neighborhood that matters, or show that the sequence $ x_n$ is monotone and bounded.

Let's try the former. We have $ f'(x) \equal{} \minus{} 7(11 \minus{} 7x)^{ \minus{} 2/3}.$ But this is bad news. This is always negative, which pretty much rules out any argument based on monotonicity. And we really want to have $ |f'(x)| < 1$ in a neighborhood of the root - but it's not. $ f'(1)\approxeq \minus{} 2.8,f'(2)\approxeq \minus{} 3.4,$ and in between it's even worse.

The premise of this problem is [b]wrong[/b]. While it should be possible to construct some function iteration that converges to the root in question, this particular sequence does not converge.

I invite you to try that out numerically. It's easiest with a spreadsheet. Enter a value for $ x_1$ is cell A1. Enter the formula "=(11-7*A1)^(1/3)" in cell A2. Select 30 or so cells from A2 down and do a fill down. It doesn't converge, does it? Change the value in A1. It still doesn't converge, right? Not even with 1.275196 in that cell - it eventually gets further away.

Let me add that the iteration $ x_{n\plus{}1}\equal{}\frac{11\minus{}x_n^3}7$ does have a good chance.

Note that the function here, $ g(x)\equal{}\frac{11\minus{}x^3}7,$ is just the inverse of the previous function $ f$. The idea is that if $ f$ is too steep, then $ f^{\minus{}1}$ ought to be shallow, which is what we want.

In this case $ g'(x)\equal{}\minus{}\frac{3x^2}7.$ This is fine at $ x\equal{}1,$ but it's not fine at $ x\equal{}2,$ since $ \left|\minus{}\frac{12}7\right|>1.$ But let's say we start the iteration with $ x_1\equal{}1.$ Then $ x_2\equal{}\frac{10}7\approxeq 1.428.$ Say we overestimate that by $ 1.5$ We note that $ g'(1.5)\equal{}\minus{}\frac{27}{56},$ and that's comfortably smaller than $ 1$ in absolute value.

So $ g$ maps the interval $ [1,1.5]$ into itself, and $ |g'(x)|\le k$ on that interval for some $ k<1.$ This makes $ g$ a contraction, and assures that the iteration $ x_{n\plus{}1}\equal{}g(x_n)$ will converge.

Do the same thing I suggested in the post above with the spreadsheet. You'll have to run more than 30 terms to be happy about it being a good approximation, since this sequence converges a bit slowly - but if you fill down far enough, you'll get the spreadsheet approximations to stop changing. (In the language of numerical analysis, this sequence of approximations displays "first order" convergence: the error at each step is proportional to the previous error, which gives convergence at the rate of a geometric sequence or series. To a numerical analyst, that's slow. And it also depends on the constant of proportionality - the ratio for the geometric progression - which in this case isn't all that favorable.)

If you want an efficient iteration, try Newton's method: $ x_{n\plus{}1}\equal{}x_n\minus{}\frac{x_n^3\plus{}7x\minus{}11}{3x_n^2\plus{}7}.$

The whole point in using a contraction-mapping argument (or something similar) in order to argue for the convergence of the sequence is so we do [b]not[/b] depend on what aRb was hung up on: knowing the answer before we start.

[quote="Kent Merryfield"]where $ f(x) \equal{} (11 \minus{} 7x)^{1/3}.$ ... We have $ f'(x) \equal{} \minus{} 7(11 \minus{} 7x)^{ \minus{} 2/3}.$ [/quote]
Erm... Didn't you forget $ 3$ in the denominator? (Not that it changes anything :)).

Oops! Yes, divide those approximate values of the derivative by $ 3,$ so we actually have $ |f'(1)|$ just slightly less than $ 1.$ But, as fedja indicated, that still doesn't save anything, as the absolute value of the derivative gets bigger for $ x>1$ and is distinctly greater than $ 1$ near the desired root of the equation.


number theory
modular arithmetic
number theory proposed

prove that

$ \forall a,b,c \in \mathbb{N}$ we have

 $ a^2 \plus{} b^2 \plus{} c^2 not \equiv 7 [8]$

Can somebody rewrite the problem ? It`s not clear .
Thanks in advance.

There's a $ 7$ missing. And the title is just stupid (not only a writting error, but just plain useless, as it says nothing about this specific problem [believe it or not, most people here are intelligent enough to know that something posted in a forum called "Number Theory" is actually number theory]).

I'm going to assume the problem asked to prove the following situation cannot occur:
a^2+b^2+c^2=7 mod 8

Any positive perfect square must be equivalent to 0,1,4 mod 8. No sum of three of these numbers can be 7. There's the proof.

P.S. Does anyone know how to make the 3-lined = sign for modular arithmetic?


If $ABCDEF$ is a convex cyclic hexagon, then its diagonals $AD$, $BE$, $CF$ are concurrent if and only if $\frac{AB}{BC}\cdot \frac{CD}{DE}\cdot \frac{EF}{FA}=1$.

[i]Alternative version.[/i] Let $ABCDEF$ be a hexagon inscribed in a circle. Then, the lines $AD$, $BE$, $CF$ are concurrent if and only if $AB\cdot CD\cdot EF=BC\cdot DE\cdot FA$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=43719 as well as a number of different topics which I do not remember (the theorem is rather known). Note that the hexagon ABCDEF really needs to be convex - else the result is wrong.

  darij


AMC
AIME
AMC 12
USA(J)MO
USAMO
geometry
AIME I

um...your sentence is worded really wierd, especially the last sentence....

anyways, if you want to compare the 2002 AIME with the recent ones, i would say that qualifying through the floor is tougher in the years that the AIME is easier, because if the AIME is easy (like the 2005) the floor will rise very high, and there are many freshmen/sophomores whom get one or two points below the floor and don't qualify.
generally, in my opinion qualifying through floor is easiest when the AIME is niether easy (2005) nor hard (94).

2002 wasn't that hard, although the floor value was pretty low.

The floor that year may well not have dropped even if the expansion had been instituted then. Adding ~125 students doesn't guarantee a drop in the floor, although it's reasonably likely that it'll drop by one point. When it's already that low, though, I wouldn't expect it.

2002 one person who got a 150 on the AMC12 got a 6 on the AIME and qualified with the index, screwing up the floor value.

It's an outlier, don't look at it.

(if you got a 150 on the AMC12, then I guess [i]you[/i] could set the floor too!)

[quote="Iversonfan2005"][quote="Calculus"]How difficult were the 2002 aimes? The floor that year was a 6 so if the expansion was there then it would probably be a 5. That floor seems kind of low. Is qualifying through that floor in that year as difficult are qualifying like in 2005 or 2004?[/quote]

The 2002 AIME I really wasn't difficult at all. the first two problems should take you like 5 minutes to do. numbers 3,6,7, and 10 are also very doable. The rest though.... :ninja: , oh well, getting a 6 on the AIME is good...especially for me :blush:[/quote]Yeah, for me 2002 wasn't that hard...I actually thought it was similar in difficulties to the ones in 2003, which many people consider to be one of the easiest.  I actually got a higher score on the 2002 ones than 2003, because of stupid mistakes.

um..yeah..I did 2003 aime 1 today and got 5 out of the first 6 in about 1 hour...the geometry ones on the test don't seem that hard either...so yeah, i would think they are pretty easy too.

[quote="amirhtlusa"][quote="randomdragoon"]

(if you got a 150 on the AMC12, then I guess [i]you[/i] could set the floor too!)[/quote]

actually i have been thinking about this...
i don't think they just look at the absolute lowest AIME score...
so for example lets say in the 2005 AIME the index was 210, and one person gets 150/6, and qualifies for usamo, but i think the floor would still be as high as 9.

also, i know couple people who qualified for usamo last year with AIME score of 8, and relative high AMC, but the floor was still 9[/quote]

They must've taken the alternate, since the index there was in the low 220s. I'm guessing they used the higher index for both tests in that case. I would be very pissed if I got an 8 on the 2005 alternate and didn't go to USAMO because the floor was 9.

Hmm I thought the two 3D problems were pretty difficult on 2002 AIME I, I didn't solve either of them so I  solved 13 problems total(can't remember how many stupid mistakes I made), but #12-14 were really easy comparing to the normal level of #12-14's.

I was working on 2002 AIME II yesterday I solved first 11 in 1 hour 30 minutes, I'm going to finish it today.


abstract algebra
analytic geometry
superior algebra
superior algebra unsolved

Let $ \phi$ be the homomorphism from the finite generated $ R$-module to itself.
Prove that $ \phi$ is injective if $ \phi$ is surjective.

Where $ R$ is a commutatitve ring with unit.

I don't think that the case $ M\equal{}R^n$ helps in the general case.

let $ \phi : M \to M$ be a surjective homomorphism, where $ M$ is a finitely-generated $ R$-module. then we may regard $ M$ as a finitely-generated $ R[\phi]$-module, and the assumption reads as $ M \equal{} \phi M$. nakayama's lemma gives us $ a \in R[\phi]$ such that $ (1 \minus{} a \phi)M\equal{}0$. now if $ \phi(x)\equal{}0$, then $ 0 \equal{} x \minus{} a(\phi(x))\equal{}x$. thus $ \phi$ is injective.

[quote="-oo-"]I don't think that the case $ M \equal{} R^n$ helps in the general case.[/quote]Reading my own hint after some time, it is quite probable that it was not too clear. :( To clarify: If you bump into a Cayley-Hamilton like proof in coordinates of a free module, you will get the idea to use the Nakayama like proof as you gave in the general case. This is kind of a general recipe in commutative algebra.


During the softball season, Judy had 40 hits. Among her hits were 2 home runs, 1 triple and 5 doubles. The rest of her hits were singles. What percent of her hits were singles? Express your answer to the nearest whole percent.

Judy hit a total of $ 40\minus{}2\minus{}1\minus{}5\equal{}32$ singles.  Thus, $ \frac{32}{40}\equal{}\frac{8}{10}\equal{}\frac{80}{100}\equal{}\boxed{80\%}$ of Judy's hits were singles.

Judy hit 40-2-1-5= 32 singles.  Since she hit the ball 40 times, 32/40 were singles, or 80%

I wonder why the problem says "Express your answer to the nearest whole percent." ...


[quote="arqady"]Let $a,$ $b$ and $c$ are real numbers such that $a^{2}+b^{2}+c^{2}=3.$ Prove that
$\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2-c}\ge 1.$[/quote]
Solution.
By Cauchy Schwarz, it suffices to prove that
\[3 \ge-a-b-c \]
Also by Cauchy Schwarz,
\[-a-b-c \le |a|+|b|+|c| \le \sqrt{3(a^{2}+b^{2}+c^{2})}=3 \]
we are done.
Equality holds if and only if $a=b=c=-1.$ :)

if $a+b+c >3$, $ab+bc+ca > 3= a^{2}+b^{2}+c^{2}$, impossible.
Hence, $a+b+c \geq-3$ because it is now equivalent to $a^{2}+b^{2}+c^{2}\geq ab+bc+ac$.
Now it is easy to prove \[\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2-c}\geq 1+\frac{a+b+c+3}{9}.\]

Another way -
$\sum_{cyc}( \frac{1}{2-a}) \geq 1 \Leftrightarrow \sum_{cyc}( \frac{1}{2-a}-\frac{a^{2}}{3}+\frac{7(a^{2}-1)}{18}) \geq 0$
$\Leftrightarrow \sum_{cyc}\frac{(a+1)^{2}(4-a)}{18(2-a)}\geq 0$

And the last inequality is true, because a, b and c are <2. Equality holds iff a=b=c=-1.


I found this topology book on the internet and started reading it, but there's a exercize near the beginning that's bothering me.
Let R be the set of all real numbers. Which [b]three[/b] of the following collections of subsets of R are topologies?
(each of these also contains R and the empty set)

1. (a,b) a,b reals with a<b
2. (-r,r) r positive real
3. (-r,r) r positive rational
4. [-r,r] r positive rational
5. (-r,r) r positive irrational
6. [-r,r] r positive irrational
7. [-r,r) r positive real
8. (-r,r] r positive real
9. (-r,r) and [-r,r] r positive real
10. [-n,n] [-r,r] n positive integer, r positive real

I think I can show that all but example 9 aren't topologies. We can get a closed interval from an open interval by taking the intersection of all intervals > r, and we can get an open interval by taking the union of all intervals < r, and in the examples involving rationals we can take the union of all numbers less than an irrational, and vice versa for examples involving irrationals. Am I doing something wrong?

You are only allowed finite intersections for topologies. Any union of open sets is open, but only a finite intersection of open sets is open.

Ah I see. I wrongly thought unions and intersections were treated the same.


topology
inequalities
triangle inequality
superior algebra
superior algebra theorems

Let $K$ be any completion of $\mathbb{Q}$, so $K=\mathbb{R}$ or $K=\mathbb{Q}_{p}$ for some $p$ (meaning the $p$-adic numbers).

Let $\varphi$ be any automorphism of $K$ (so $\varphi(1)=1,$ $\varphi(x+y)=\varphi(x)+\varphi(y)$, $\varphi(xy)=\varphi(x)\varphi(y)$).

Show that $\varphi(x)=x$ for all $x$.

It's (very¿) well known for the real case, but the other cases are nice, too.

Let $a$ be a unit of the ring $\mathbb Z_{p}$ of $p$-adic integers. By applying Hensel's Lemma, for arbitrarily large $n\in\mathbb N$ we can find $x=x_{n}$ such that $x^{n}=a$. Then $\varphi(a)=\varphi(x^{n})=[\varphi(x)]^{n}$, i.e. for arbitrarily large $n$, we can find $\widetilde x$ such that $\widetilde x^{n}=\varphi(a)$. This implies that $\varphi(a)$ is a unit of $\mathbb Z_{p}$, so, since every element of $\mathbb Z_{p}$ is of the form $p^{k}u$ for a unit $u$ and $\varphi(p)=p$, we've shown that $\varphi(\mathbb Z_{p})\subseteq \mathbb Z_{p}$.

Now take an arbitrary $x\in\mathbb Q_{p}$. For each $n$, there is a rational $q_{n}$ such that $x=q_{n}+a_{n}$, where $a_{n}\in p^{n}\mathbb Z_{p}$. Since $\varphi|_{\mathbb Q}=\mbox{id}_{\mathbb Q}$, we have $\varphi(x)=q_{n}+\varphi(a_{n})$, and by the paragraph above, $\varphi(a_{n})\in p^{n}\mathbb Z_{p}$. This means that $\varphi(x)-x\in p^{n}\mathbb Z_{p}$ for all $n$, i.e. $\varphi(x)=x$.

Just to complete things, here's the argument for $\mathbb{R}$:
A real number is nonnegative if and only if it is a square. Therefore, every automorphism sends positive numbers to positive numbers, and must preserve order. The topology is determined by the order, so every automorphism is continuous. Every automorphism fixes the base field $\mathbb{Q}$, so every continuous automorphism fixes the completion.

A slight tweak on grobber's argument:
A $p$-adic number $a$ is an $n$-th power for arbitrarily large $n$ if and only if $|a|_{p}=1$. We have $|a|\le1$ if and only if at least one of $|a|_{p}$ and $|a+1|_{p}$ is equal to 1 by the strong triangle inequality, so $|a|_{p}\le 1$ if and only if at least one of $a$ and $a+1$ is an $n$-th power for arbitrarily large $n$. This is an algebraic condition, so the unit ball is mapped into the unit ball by any automorphism. Multiplying by $p^{k}$ (which is fixed), the balls of radius $p^{k}$ are sent to themselves, and every automorphism is continuous at zero. We can translate to any point, so every automorphism is continuous. Since the rationals are fixed, every automorphism is the identity.

This restatement emphasizes that we can generate the topology in either case with only algebraic knowledge, forcing any automorphism to be continuous. We don't even need to require the automorphism to be surjective; these arguments show that every endomorphism is the identity.


function
inequalities
trigonometry
absolute value
real analysis
real analysis unsolved

$ f$ is a differentiable function that satisfies: $ f^2 \equal{} x(f' \minus{} 1), f(0) \equal{} 0, f'(0) \equal{} 1$. Can you explicitly find, [b]without power series[/b], the maximal neighborhood of $ 0$ where $ f$ satisfies the equation? (I think it is equivalent to finding the smallest (in absolute value) infinite discontinuities)

Based (and equivalent) on the following problem I've seen:
[hide]
Let $ a_1 \equal{} 1, a_n \equal{} \frac {1}{n} \sum_{k \equal{} 1}^{n \minus{} 1} a_ka_{n \minus{} k}$.
Show that $ \frac {2}{3} \le \limsup_{n \to \infty} |a_n|^{\frac {1}{n}} < 2^{ \minus{} \frac {1}{2}}$

($ f$ is the generating function of $ a_n$, and the limit is the reciprocal of the radius of convergence of $ f$)
[/hide]
Edit (some development): [hide]After some substitutions, I found that $ f \equal{} \frac { \minus{} u'x}{u}$, where $ u \equal{} \sum_{i \equal{} 0}^{\infty} \frac {( \minus{} x)^i}{i!^2}$. So the problem is equivalent to finding the smallest root, in absolute value, of $ \sum_{i \equal{} 0}^{\infty} \frac {x^i}{i!^2}$. How does one approximate such root? [/hide]

What's wrong with the bisection or Newton methods? (I mean, I do not believe that the root can be expressed in terms of some classical constants but, due to very quick convergence of the series, you can still easily find it with any given precision using a computer). If all that you need is to show that it is between $ \sqrt 2$ and $ 3/2$, just plug the values in and compute the first 4 or 5 terms respectively. They'll have the correct sign and, moreover, the rest of the series will also have the correct sign for the trivial reason.

I am not really familiar with those methods, but I read about Newton method and it seems to be that you need to be able to calculate $ f$ (or $ u$) at certain values to approximate the root. I don't really want to use a computer for that (and I don't have the right softwares anyway), because the original problem was a competition problem (and it has a short inductive straightforward solution).
I may have misinterpreted the approximation methods - do I really need to be able to calculate the value of $ \sum \frac {x^i}{i!^2}$ at some points for that? A demonstration\link would be nice  :oops:
It would be nice to see whether this sequence has some explicit formula (it is very similar to the Catalan sequence, this is why I got the idea of differential equation), whether $ f$ or $ u$ are well-known functions, and whether $ \frac {1}{\limsup |a_n|^\frac {1}{n}}$ has also some explicit formula.

Let's say we want to show that the root is less than $ 3/2$. It is enough to show that $ u( \minus{} 3/2) < 0$. Now just plug $ x \equal{} \minus{} 3/2$ in. You'll get
\[ 1 \plus{} x \plus{} \frac {x^2}4 \plus{} \frac {x^3}{36} \plus{} \frac {x^4}{24^2} \equal{} \minus{} \frac {1}{32} \plus{} \frac {81}{16\cdot 24^2} < 0
\]
(just because $ 24^2 > 2\cdot 81$). Now the rest of the series is also negative because each term is less than the previous one in absolute value and the first term in the tail is negative.

To show that the root is greater than $ \sqrt 2$ requires just to show that $ 1 \plus{} x \plus{} \frac {x^2}4 \plus{} \frac {x^3}{36} \equal{} \left(1 \plus{} \frac x2\right)^2 \plus{} \frac {x^3}{36} > 0$ for $ \minus{} \sqrt 2\le x\le 0$. But the left hand side is increasing in $ x$ on this interval, so all you need is to plug $ x \equal{} \minus{} \sqrt 2$ and to check the inequality $ 27 > 19\sqrt 3\Longleftrightarrow 729 > 2\cdot 361$.

As to the nice expression for the root, I just don't know. I mean, if there exists one, someone will eventually find it but nobody knows how to prove that such roots are not "elementary" when they aren't and I do not think this particular equation is any better than $ x \equal{} \cos x$.

That's what I was looking for actually, thanks. I didn't realize it could be done in such simple explicit manner.
I think we also need to role out the possibility of a root smaller than $ \sqrt {2}$ in absolute value - a friend suggested Roche's Theorem but this gets kind of involved...
I am beginning to think that this is how the problem was invented - some one took a simple ODE, approximated the radius of convergence, found the linear recurrence of the coefficients and that's it.


calculus
integration
LaTeX
trigonometry
calculus computations

Hi everyone,

can someone help me for that question, where do I start, I'm doing integration at the moment.

Hence find, an implicit form, the general solution of the differential equation

     dy/dx  = [u]2y^3/2 cos x[/u]
                        (2 + sin x)^2

      The differential equation dy/dx = 2y^3/2 cos x/(2+sin x)^2,

         Has a right-hand side of the form f(x) g(y), where g(y) = 2y^3/2 cos x and
      f(x) = (2+sin x)^2

     Dividing both sides of the differential equation by g(y) means multiplying both
     sides by y. This give

∫ 2y^3/2 cos x dy = ∫(2+sin x)2 dx

        y^5/2-sin x + a = cos x^3 + b

Thus the general solution, in implicit form, is

y^5/2 = cos x^3 + c

Thanks again  :lol:

Please - either use $ \text{\LaTeX}$ or brackets because your equation is ambiguous. If you wanted to say:

$ \frac {dy}{dx} = \frac {2 y^{\frac {3}{2}} \cos x}{(2 + \sin x)^2}$

then you need to get an equation of the form

$ f(y) dy = g(x) dx$

what you got is something of the form

$ f(x,y)dy = g(x)dx$

So integrating this cannot give you a correct answer. But, multiplying the equation with $ \frac {dx}{y^{\frac {3}{2}}}$ you get

$ y^{-\frac {3}{2}} dy = \frac {2\cos x}{(2 + \sin x)^2} dx$

Integrating now, you get

$ -\frac{2}{\sqrt{y}}= - \frac {2}{2 + \sin x} + C$

$ \frac{1}{\sqrt{y}} = \frac {1}{2 + \sin x} + C$

$ \frac{1}{y}=\left(\frac{1}{2+\sin x}+C\right)^2$

$ y=\left(\frac{1}{2+\sin x}+C\right)^{-2}$

This comes from the fact that

$ \int x^n \,dx = \frac {x^{n + 1}}{n + 1} + C$

and

$ \int g'(x) f'(g(x)) \, dx = f(g(x)) + C$

where in out case

$ f(x) = - \frac {1}{x}$

$ g(x) = \sin x + 2$

Hi milin,

Thanks again for your help:lol:


algebra
polynomial
advanced fields
advanced fields solved

Merry Chr... !!!!  :D

I have some problem. Can someone help me.?

Find all polynomial P(z) so that.?

1) | P(z) | = | z | (where = complex number)
2) P(i) = i

I only find : P(z) = z  :P

Because of the rate of growing, you must have $P(z)=az+b$ with complex $a,b$ and $P(0)=0$ implies $b=0$. The rest is trivial.

what is the rate of growing. Can you write with logic of math. ? So you mean that only P(z) = z is one solution ?

If the degree of $P(z)$ is higher than 1, then $|P(z)/z|$ goes to infinity when $z$ grows, so this cannot happen because the quotient is constant. With the previous said, it follows $P(z)=az$ and because of that $ai=i$, so $a=1$.

Alternatively:

$\left|\frac{P(z)}{z}\right|=1$ on all of $\mathbb{C}$ (since the singularity at 0 is removable), so by Liouville's Theorem, $\frac{P(z)}{z}$ is a constant.

Or you can use the Schwarz Lemma generalized to the disc of radius 2, which also considers |P(z)/z| (but uses the maximum modulus principle).


function
trigonometry
limit
calculus
calculus computations

Can anyone help me with the following:

Find the limit of  sinx(1-cosx)/(x-sinx) as x approaches infinity.

I can do the problem if x were approaching 0, is there any link between the two?

Thanks!

There's no link between the behavior at zero and the behavior at infinity. At infinity the argument is simply one about size.

The numerator, $\sin x(1-\cos x),$ is bounded. (Every continuous periodic function is bounded.)

As for the denominator: with $\sin x$ being bounded and $x$ tending to $\infty,$ the whole expression, $x-\sin x,$ tends to infinity.

So we have a bounded function divided by a function that tends to infinity. That makes the whole expression tend to zero.

How fast does it go to zero? We would say it is $O\left(\frac1x\right).$

Near zero, $\frac{\sin x(1-\cos x)}{x-\sin x}=\frac{\frac{x^3}2+O(x^5)} {\frac{x^3}6+O(x^5)}=3+O(x^2).$

You did get that the limit is 3, right?

Yep, I did get 3. Thanks a lot for the help! :)

$\lim_{x\to 0} \frac{(1-\cos x)\sin x}{x-\sin x}=\lim_{x\to 0} \frac{\sin x}{x}\cdot \frac{1-\cos x}{x^2}\cdot \frac{x^3}{x-\sin x}=1\cdot \frac 12\cdot 6=3.$

[u]Remark.[/u] You use the [u]l'Hospital's rules[/u] only for $\lim_{x\to 0}\frac{x^3}{x-\sin x}=\lim_{x\to 0} \frac{3x^2}{1-\cos x}=6.$


Show that for any natural $ m$ there is natural $ n$ s.t:  $ n$ is divisble by $ m$ and all digits of $ n$ are $ 0$ and $ 1$.

Let $ m\equal{}2^{s}5^{k}t$ with $ (t,10)\equal{}1$

Let $ h$ such that $ h\phi(t)>max\{s,k\}$

Then

$ n\equal{}10^{h\phi(t)}\plus{}10^{(h\plus{}1)\phi(t)}\plus{}\ldots\plus{}10^{(h\plus{}t\minus{}1)\phi(t)}$

verifies your condition.


If $ \log 2\equal{}.3010$ and $ \log 3\equal{}.4771$, the value of $ x$ when $ 3^{x\plus{}3}\equal{}135$ is approximately:

$ \textbf{(A)}\ 5 \qquad
\textbf{(B)}\ 1.47 \qquad
\textbf{(C)}\ 1.67 \qquad
\textbf{(D)}\ 1.78 \qquad
\textbf{(E)}\ 1.63$

[hide="Click for solution"]
$ 3^x \cdot 3^3\equal{}135 \implies 3^x\equal{}5 \implies x \log 3 \equal{} \log 5 \equal{} \log \frac{10}{2} \equal{} \log 10 \minus{} \log 2$ so $ x\equal{}\frac{1\minus{}0.3010}{0.4771} \approx 1.47$ or $ \boxed{\textbf{(B)}}$.
[/hide]


As shown, a square is constructed on the hypotenuse of a right triangle whose legs have lengths 3 units and 4 units. What is the area of the pentagon $ ABCDE$, in square units?

[asy]pair a,b,c,d,e;
a=(3,7);
b=(7,4);
c=(4,0);
d=(0,0);
e=(0,3);
draw((0,0.5)--(0.5,0.5)--(0.5,0));
label("4",(2,0),S);
draw(a--b--c--e--d--c--e--a);
label("3",(0,1.5),W);
label("$A$",a,N);
label("$B$",b,E);
label("$C$",c,S);
label("$D$",d,S);
label("$E$",e,W);[/asy]

Side of the $ ABCD$ is $ 5$, so perimeter is $ 3\plus{}4\plus{}5\plus{}5\plus{}5\equal{}\boxed{22}$

The area of the pentagon is the area of the triangle: $ \frac{3\cdot 4}{2}$ or $ 6$ plus the area of the square: $ 5^2$ or $ 25$. The answer is 31

The area of the square is 5*5=25 and the area of the triangle is 3*4/2=6 so the total is 31.


[quote="archimedes1"]If $ ABCD$ is a square with area $ 4$, $ E$ is the midpoint of $ AC$, and $ F$ is on $ BE$ such that $ CF\perp BE$, then find the area of $ CDEF$.[/quote]
Aren't $ E$ and $ F$ the same point? $ E$ is the center of the square as it is the midpoint of $ AC$, a diagonal, and if $ BE\perp CF$, then $ F$ is in the same place as $ E$.

[hide="hint"]$ \triangle{ABE}$ ~ $ \triangle{BCF}$ [/hide]

[quote="archimedes1"]If $ ABCD$ is a square with area $ 4$, $ E$ is the midpoint of $ AD$, and $ F$ is on $ BE$ such that $ CF\perp BE$, then find the area of $ CDEF$.[/quote]
[hide="Solution"]$ \triangle ABE\sim\triangle FCB$ because of congruent angles.

$ BE \equal{}\sqrt{5}$ and $ BC \equal{} 2$, so $ \frac{[ABE]}{[FCB]}\equal{}\left(\frac{\sqrt{5}}{2}\right)^{2}\equal{}\frac{5}{4}$.

Since $ [ABE] \equal{} 1$, $ [FCB] \equal{}\frac{4}{5}$.

$ [CDEF] \equal{} [ABCD]\minus{}[ABE]\minus{}[FCB] \equal{} 4\minus{}1\minus{}\frac{4}{5}\equal{}\boxed{\frac{11}{5}}$[/hide]

[quote="i_like_pie"]$ \left(\frac{\sqrt{5}}{2}\right) \equal{}\frac{5}{4}$.[/quote]
I think you meant to square the LHS.
You answer is correct I think.

Wait a minute I think you mean similar not congruent.....

[hide="Alternate solution"]We have $ [EBC]\equal{}2$ and $ BE\equal{}\sqrt{5}$, so $ CF\equal{}\frac{4}{\sqrt{5}}$. Then, $ EC\equal{}\sqrt{5}$, so $ EF\equal{}\frac{3}{\sqrt{5}}$. NOw, $ [EDCF]\equal{}[EDC]\plus{}[EFC]\equal{}1\plus{}\frac{6}{5}\equal{}\frac{11}{5}$.[/hide]

@ archimedes & sonny - You're both right, edited.

@ Leet - Nice solution.


geometry
trigonometry
rotation
LaTeX
ARML
Pythagorean Theorem

[quote="Nawahd Werdna"]P is a point inside square ABCD such that PA = 5, PB = 3, and PC = 7. What is the length of the side of the square?

I just can't seem to find a solution for this problem... cosine law doesn't seem to work!

- NAWAHD WERDNA[/quote]

Try to draw a picture, and then split the square into two triangles.

Then whichever triangle has three segments within it, take that triangle separately.

Then rotate that triangle, and try to notice anything weird.

Refer to the picture. I drew perpendiculars to each of the sides, and I have the information in the square. Now, I obtain these two equations by the Pythagorean theorem:

$s^{2}-(s-x)^{2}= 16$
$(s-x)^{2}+(s-\sqrt{9-(s-x)^{2}})^{2}= 49$

Substituting gets us:
$s^{2}+(s-\sqrt{25-s^{2}})^{2}= 65$

Anyone who wants to solve this equation for $s$, the side length of the square, is free to try!

I'm not sure I did it right but using vishalarul's equation i get $18+2\sqrt{79}$ and $18-2\sqrt{79}$.

P.S. Can someone tell me how to do plus or minus in latex? Thx.

[quote="ckck"]I'm not sure I did it right but using vishalarul's equation i get 18+2\sqrt{79} and 18-2\sqrt{79}.

P.S. Can someone tell me how to do plus or minus in latex? Thx[/quote]
Here is the thing with the $\pm$:
$18 \pm 2\sqrt{79}$

Here is the source code:
[code]18 \pm 2\sqrt{79}[/code]

Could someone check my equation as well?

[quote="ckck"]Where did you get the first equation? I can't get that one.[/quote]

That is where I messed up! The equation should be:
$x^{2}+9-(s-x)^{2}= 25$

(I accidentally put an $s$ for the $x$)

So our two equations are now:
$x^{2}+9-(s-x)^{2}= 25$
$(s-x)^{2}+(s-\sqrt{9-(s-x)^{2}})^{2}= 49$
Who wants to solve that?  :stretcher:

this is not really a HSB problem...the solution involves a transformation...

[hide]rotate 90 degrees about point B...[/hide]

for some reason i think i saw this problem at arml practice last year...and i was like woa i have a cool solution...but then the grader was like omg what? thats not how we did it...so i got 4/5...then i was like wtf...and they were like don't bother us...and i was like ok