trigonometry
inequalities
geometry
circumcircle
area of a triangle
Heron\u0027s formula
inequalities proposed

$ A\plus{}B\plus{}C\equal{}180$,Prove that;

$ 4(sin^{2}Asin^{2}B\plus{}sin^{2}Bsin^{2}C\plus{}sin^{2}Csin^{2}A)\leq(sinA\plus{}sinB\plus{}sinC)^{2}$

[quote="cnyd"]$ A \plus{} B \plus{} C \equal{} 180$,Prove that;

$ 4(sin^{2}Asin^{2}B \plus{} sin^{2}Bsin^{2}C \plus{} sin^{2}Csin^{2}A)\leq(sinA \plus{} sinB \plus{} sinC)^{2}$[/quote]

substitution   $ a \equal{} x \plus{} y,b \equal{} y \plus{} z,c \equal{} z \plus{} x$

$ \sin{A} \equal{} \frac {4(x \plus{} y \plus{} z)xyz}{(y \plus{} z)(z \plus{} x)}$

$ \sin{B} \equal{} \frac {4(x \plus{} y \plus{} z)xyz}{(z \plus{} x)(x \plus{} y)}$

$ \sin{C} \equal{} \frac {4(x \plus{} y \plus{} z)xyz}{(x \plus{} y)(y \plus{} z)}$

The following inequality is equivalent to

$ 64\frac {(x \plus{} y \plus{} z)^2x^2y^2z^2}{(x \plus{} y)^2(y \plus{} z)^2(z \plus{} x)^2}\sum{\frac {1}{(x \plus{} y)^2}}\le 4(x \plus{} y \plus{} z)xyz(\sum{\frac {1}{(x \plus{} y)(y \plus{} z)}})^2$

$ \Longleftrightarrow 4xyz\sum{\frac {1}{(x \plus{} y)^2}}\le x \plus{} y \plus{} z$

true due to AM-GM

If $ a,b,c$ are the sides of a triangle with angles $ A,B,C$, $ R$ is the circumradius and $ K$ is the area, then we know from geometry that $ a \equal{} 2R \sin A$, $ b \equal{} 2R \sin B$, and $ c \equal{} 2R \sin C$, and that $ 2R \equal{} abc / 2K$.  Therefore, we need to show that
$ 4(a^2 b^2 \plus{} b^2 c^2 \plus{} c^2 a^2) \leq (a \plus{} b \plus{} c)^2 a^2 b^2 c^2 / 4K^2$.
Now, we can use the substitution $ a \equal{} y\plus{}z$, $ b \equal{} z \plus{} x$, $ c \equal{} x \plus{} y$, and by Heron's Formula $ K^2 \equal{} xyz(x\plus{}y\plus{}z)$, the inequality then is equivalent to proving that for non-negative $ x,y,z$,
$ 4 \sum_{cyc} (x\plus{}y)^2 (x\plus{}z)^2 \leq (x\plus{}y\plus{}z)(x\plus{}y)^2 (y\plus{}z)^2 (z\plus{}x)^2 / xyz$.
However, this can be simplified to
$ \sum_{cyc} \frac{1}{(x\plus{}y)^2} \leq \frac{1}{4} \sum_{cyc} \frac{1}{xy}$
which finally becomes
$ \minus{}\sum_{cyc} \frac{(x\minus{}y)^2}{(x\plus{}y)^2} \leq 0$,
clearly true.


[quote="2424"]with $ a,b,c > 0$ Prove that
   $ \frac {b(a \plus{} b \minus{} 2c)}{b^2 \plus{} c^2 \plus{} ab \plus{} ac} \plus{} \frac {c(b \plus{} c \minus{} 2a)}{a^2 \plus{} c^2 \plus{} ab \plus{} bc} \plus{} \frac {a(c \plus{} a \minus{} 2b)}{a^2 \plus{} b^2 \plus{} bc \plus{} ca} \geq0$[/quote]
no one interested in this ineq?


function
real analysis
trigonometry
real analysis unsolved

I am working on the following problem  :

let $ f : \mathbb{R}\minus{}\minus{}\minus{}\minus{}>\mathbb{R}$ be a bounded (above and below) Lebesgue measurable function. Prove that $ S: \equal{}\{ a : m(f^{\minus{}1}(a;\infty)\}\equal{}0$ has a smallest element . Where m is Lebesgue measure

Here is my "proof" :

Let $ E_{a} : \equal{}\{x : f(x)>a\}\equal{}f^{\minus{}1}(a;\infty)$ then $ E_{a}$ is measurable sine $ f$ is measurable.  Since $ m\leq f\leq M$ then I have $ a\geq M$ (?) ,   so $ S$ is bounded from below so there exists $ \alpha\equal{}inf(S)$

I have to prove that $ \alpha\equal{}inf(S)\in S$, I choose $ (a_{n})$ st : $ a_{n}\leq a_{n\plus{}1}$    (?)
since $ \alpha\equal{}inf(S)$ then there exist $ a_{n}\in S$ st : $ a_{n}\rightarrow \alpha$

consider $ E_{a_{n}}\equal{}\{x : f(x)>a_{n}\}$ then since $ a_{n}\leq a_{n\plus{}1}$, I have $ E_{a_{n\plus{}1}}\subset E_{a_{n}}$, since $ a_{n}\rightarrow \alpha$ I have $ \bigcap_{n\geq 1} E_{a_{n}}\equal{}E_{\alpha}$ so  $ m(E_{a_{n}})\rightarrow m(E_{\alpha})$ so choose $ n$ big enough I have $ m(E_{a_{n}})>0$ , but $ a_{n}\in S$ therefore $ m(E_{a_{n}})\equal{}0$ so I have a contradiction
Thus $ \alpha \in S$

Could you please give me some comments?
$ f$ is given bounded from above and below but I did use the fact that $ f$ is bounded from below. Is there some thing missing?

many thanks

Please give me some comment

is my "proof" ok?

many thanks

[quote="1234567a"] I choose $ (a_{n})$ st : $ a_{n}\leq a_{n \plus{} 1}$    (?)[/quote]
I have problem only here. Ok it's true that since $ \alpha \equal{} inf(S)$  there exist $ a_{n}\in S$ st : $ a_{n}\rightarrow \alpha$. How would you justify that there exists an increasing sequence?

I can choose such $ (a_{n})_{n\geq 1}$ since $ \alpha \equal{} inf(S)$ and $ \alpha\geq s\,\,\forall s\in S$

Am I right? thanks

What if your $ S$ happens to be $ \{1, \frac{1}{2}, ...\}$?


Find all set of non-negative numbers $ (a,b,c)$ such that
$ 2^{a\plus{}1}\equal{}3^{2b\plus{}1}\plus{}5^{2c\plus{}1}$


geometry
circumcircle
perpendicular bisector
geometry proposed

$\triangle ABC\ : \ \boxed {\ 4Rm_a=b^2+c^2\Longleftrightarrow b=c\ \vee\ A=90^{\circ}\ }\ .$

$b^2+c^2 = 2m^2_a + \frac{a^2}{2}$

$4Rm_a = 2m^2_a + \frac{a^2}{2} \Leftrightarrow$

$2Rm_a = m^2_a + \frac{a^2}{4} \Leftrightarrow$

$\frac{a^2}{4} = 2Rm_a - m^2_a \Leftrightarrow$

$\frac{a^2}{4} = m_a (2R - m_a) \ \ (1)$

Let $M$ be the midpoint of $BC$. The ray $AM$ intersects the circumcircle $(O,R)$ at the point $N$.
The $(-)$ power of point $M$ with respect to the circumcircle is

$-pow(M) = R^2 - OM^2 = AM \cdot MN = BM \cdot MC$

But $BM \cdot MC = \frac{a^2}{4} = m_a (2R - m_a)$ from $(1)$

So $AM \cdot MN = m_a (2R - m_a) \Rightarrow$

$MN = 2R - AM \Rightarrow$
$AM+AN =2R \Rightarrow$
$AN =2R$

So, the segment $AN$ is the diameter of the circle.

from $R^2 - OM^2 = MC^2$ we have two cases:

1) if $OM=0$ then $AM=R=MC = \frac{a}{2}$, so $A=90^o$ ( a triangle with $m_a = \frac{a}{2}$)

2) if $OM>0$ then the triangle $OMC$ is right $(OM \perp BC) \Rightarrow  AM \perp BC$

So the point $A$ is on the perpendicular bisector of $BC \Rightarrow AB=AC$

[color=blue]For the inverse:[/color]
- If $A=90^o$ then $m_a=R\Rightarrow 4Rm_a=4R^2$
and $b^2+c^2 = a^2 = (2R)^2$

- if $b=c$ then $AM \perp BC\Rightarrow AN$ is diameter $\Rightarrow  BM \cdot MC = AM(2R-AM)\Rightarrow$

$\frac{a^2}{4} = m_a(2R-m_a)$, which is the equation $(1)$ and we can go inversely to prove that $b^2+c^2=4Rm_a$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=73562
Generally, $\boxed {\ b^2+c^2\le 4Rm_a\ }\ .$


email
MATHCOUNTS
AwesomeMath
summer program
geometry

Wasn't such an ugly way to spend 3 weeks of summer vacation after all. Hope yall had fun as well. Maybe I'l see some of you later elsewhere.

peace
 :bye:

IVAN!!! *claps for 5 minutes*

anyone from track 2 here?  thanks for the great times guys.  ill post some pictures maybe

[quote="Fanatic"]Uh, there was also a russian from New York at the camp named Ivan... I think this is him, not Ivan from track 2, judging by his other post.[/quote]
That Ivan is prowler.

[quote="Altheman"]woot track 2

on another note, my rating went down like 1.5 points when i was away[/quote]

gosh alex you should stop spamming by posting math on a math forum...

track 2 is beast bc we had chuck norris as a teacher

Yes I believe that i_v_a_n is the New York Russian from track 4

k hoo!

wtf...somebody really hates u alex.

No this is not the Chuck norris ivan.

this is the most hardcore ganstah!!!...

of the math camp =(

It should have been me!!!!!!!!!!!!!!!

 :mad:

 :lol:

Fun fun..I just came home like 5 hours ago..
Johnny I can`t understand crap from your yearbook message :P  Same for ivan.

PS. I`m Andrei :)

Andrei!!!

most of the stuff is about my roomates.  just making sure everone knows im not samsung.

dude i was totally going to like pull one of their tails and/or kick them in the areas but those bodyguards are pretty sketchy

[quote="FightTheTide"]hmmm. My signature hasn't been showing up lately. Maybe this will help.

Apparently sarcasm doesn't pass well over the internet.[/quote]
huh what
huh
what huh huh what
huh

[quote="FightTheTide"]
Apparently sarcasm doesn't pass well over the internet.[/quote]

Gosh, really?

[quote="mustafa"][quote="pianoforte"]wait, is that sarcasm?[/quote]

 :D  :D  :D[/quote]

 :D
:D [hide=":D"]no[/hide]

AMP is not over.
AMP is not over.
AMP is not over.
AMP is...
sigh

apparently repeated denial doesnt help.

God exists
God exists
God exists
God e...
sigh

apparently repeated denial doesn't help.

[quote="DavidL"]Wear you're such a spammernoob  :rotfl:[/quote]

ur m0m 1s such 1 5pamm3r n00b

lolz!!!!1!one!!!!

[quote="bowei"]Hi Alex! HI! HIIII!
HEllo!
I just discovered this forum!

Hi Andrei!

Does the Ivan from track two even post here?

Where's ccy?
Hi mustafa!

go amp!

Hi alex! plz dont mark this as spam[/quote]
OMG ITS BOWEI
OMG


Which is the smallest natural number such that all of the conditions are met:
-Its digits consist only of 3's, 5's, and 7's
-The sum of its digits is divisible by 3, 5, and 7
-The number itself is divisible by 3, 5, and 7

Side note: This was a question set during the scavenger hunt of University of Waterloo's math orientation. None of the people in our team (consisting of all the advanced math people) could do it, so someone wrote something in Java to bruteforce it, and it took quite a while since the answer was so large.

[hide="solution"]
1. 3 (trivial)
3. lcm(3,5,7) = 105
2. Since the sum of the digits must be a multiple of 105, the number is the smallest number whose digits sum to 105.  And since we want to maximize the number of 9s to get the fewest number of digits possible, and 105 = 9*11 + 6, the smallest number is 699999999999 (a six followed by 11 nines)

[/hide]

[quote="fshiu"]maybe i wasnt clear enough, see edit
(otherwise this would be too easy)[/quote]

Either way this will indeed be a fairly large number, since the sum of the digits is at least 105. This would probably take quite some time....

[hide] OK, so let's assume there is such a number with the sum of its digits equal to $ 105$. To make it small, we should maximize the number of 7's. If there are 15 7's, the number will not be divisible by 5. If there are 14 or 13 7's, the sum cannot be 105. If there are 12 7's, there must be 3 5's and 2 3's. Trying out a few numbers, I believe $ 33577577777777775$ is the smallest possible. It is easy to show that no numbers with digital sum 105, 210, etc. are smaller. [/hide]

[quote="lotrgreengrapes7926"]If there are 14 or 13 7's, the sum cannot be 105.[/quote]

Hmm, actually 13 7s, 1 5, and 3 3s works, but that's still 17 digits and the minimum number possible (I think at least) from that is 37377737777777775.


I don't quite understand the chinese remainder theorem. Can someone explain it to me?

The Chinese Remainder Theorem states that all simultaneous congruences can be solved, and gives a way to find the solutions.

However, further discussion about this would probably go beyond the Basics forum.

[quote="ch1n353ch3s54a1l"][quote="t0rajir0u"]The statement needs to be modified.

Basically, the Chinese Remainder Theorem states that, if you have a system like

$x \equiv a_1 \bmod m_1$
$x \equiv a_2 \bmod m_2$
...
$x \equiv a_k \bmod m_k$

Where (and this is very important) [i]all of $m_1, m_2, ... m_k$ share no common factors with one another[/i], then a unique solution

$x \equiv x_0 \bmod (m_1 m_2 m_3 ... m_k)$

Is guaranteed.  For example, the Chinese Remainder theorem will tell you that

$x \equiv 1 \bmod 2$
$x \equiv 2 \bmod 3$
$x \equiv 4 \bmod 5$

Uniquely exists $\bmod 30$.[/quote]

Does the Chinese Remainder Theorem say anything about $x_0$ in $x \equiv x_0 \bmod (m_1 m_2 m_3 ... m_k)$...?[/quote]

Well, yes, it does, but the expression is really annoying.

Define $M = m_1 m_2 m_3 ... m_k$.

Define $b_i$ as integers such that $b_i \frac{M}{m_i} \equiv 1 \bmod m_i$.  Then

$x_0 \equiv a_1 b_1 \frac{M}{m_1} + ... + a_k b_k \frac{M}{m_k} \bmod M$.


geometry
circumcircle
parallelogram
number theory
least common multiple
perpendicular bisector
geometry proposed

Points $ K,L,M$ and $ N$ lie on sides $ AB,$ $ BC,$ $ CD$ and $ AD$ of the convex quadrangle $ ABCD.$ Let $ ABLMD$ and $ KBCDN$ be inscribed pentagons and $ KN\equal{}LM.$ Prove that $ \angle BAD\equal{}\angle BCD.$

Let $ (O_a, R_a), (O_c, R_c)$ be circumcircles of the cyclic pentagons $ ABLMD, CDNKB.$ The $ \triangle CML \sim \triangle CDB$ are oppositely similar, having equal angles. Let $ P_c$ be circumcenter of the $ \triangle CML.$ Because of the opposite similarity, $ CP_c$ is isogonal of $ CO_c$ with respect to the angle $ \angle BCD,$ hence $ CP_c \perp BD$ and similarly, $ CO_c \perp LM.$ Thus $ P_c$ is the intersection of the C-altitude of the $ \triangle CDB$ with the perpendicular bisector of $ LM$ passing through $ O_a$ on the perpendicular bisector of $ BD.$ $ CP_c \parallel O_cO_a$ (both $ \perp BD$), $ P_cO_a \parallel CO_c$ (both $ \perp LM$), $ CP_cO_aO_c$ is a parallelogram, so that the circumradius of the $ \triangle CLM$ is equal to $ P_cC \equal{} O_aO_c.$ In exactly the same way, we can find that the circumradius of the $ \triangle AKN$ with circumcenter $ P_a$ is equal to $ P_aA \equal{} O_cO_a.$ If the bases $ LM \equal{} NK$ and the circumradii $ P_cC \equal{} P_aA \equal{} O_aO_c$ of the $ \triangle CLM, \triangle AKN$ are equal, the angles $ \angle DAB \equiv \angle NAK \equal{} \angle LCM \equiv \angle BCD$ are also equal.

[quote="yetti"]$ CP_c \parallel O_cO_a$ (both $ \perp BD$), $ P_cO_a \parallel CO_c$ (both $ \perp LM$), $ CP_cO_aO_c$ is a parallelogram, so that the circumradius of the $ \triangle CLM$ is equal to $ P_cC \equal{} O_aO_c.$ [/quote]

This argument works only when $ LM\nparallel BD.$
If $ LM\parallel BD$ then points $ P_c,C,O_a,O_c$ are collinear and it is not difficult to verify that $ P_cC \equal{} O_aO_c$ also holds true. :)

If $ LM \parallel BD,$ then $ P_cC \equal{} O_aO_c$ by the continuity principle.


search function email algorithm algebra polynomial \/closed

So for everysingle post, under my username there is some kind of stars

then it says P verus NP and I don't think I made this option...
what is it?

Your post count deterines your rank of stars.  The ranks are:

0 - 19 New Member (Zero Stars)
20 - 49 P versus NP (Half Star)
50 - 99 Hodge Conjecture (One Star)
100 - 249 Poincare Conjecture (Two Stars)
250 - 499 Riemann's Hypothesis (Two and Half stars)
500 - 999 Yang Mills Theory (Three Stars)
1000 - 2499 Navier-Stroke's Equation (Four Stars)
2500 - $\infty$ Birch & Swinnerton Dyer. (Five Stars)

These are the [url=http://www.claymath.org/millennium/]Clay Mathematics Institute's[/url] unsolved "Millenium Problems."

This question has been posted several times.  See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24794]this thread[/url] for more.  Also, remember that the search function of AoPS can be very useful!  Look for the 'Search" button on the toolbar above.  Cheers!

Why there is no "Goldbach's Conjecture"?  :(

Ask the Clay Mathematics Institute people :)

unfortunatelly I have not their emails and I am not going to search for them :) Do you know why I cannot see my message at the forum but only at the topic review?

The P versus NP problem is to determine whether every language accepted
by some nondeterministic algorithm in polynomial time is also accepted by
some (deterministic) algorithm in polynomial time.
See also http://www.claymath.org/millennium/ P_vs_NP/Official_Problem_Description.pdf


looks kinna ugly. lemme' translate it :)
 . .   _________
3 \sqrt  -1 + 3 \sqrt 2   versus 3 \sqrt 1/9 - 3 \sqrt 2/9 + 3 \sqrt 4/9

This exercise is based on a famous identity of Ramanujan.

Letting x=(1/3)^(1/3) and y=(2/3)^(1/3) we have only to prove

(x+y)(2^(1/3)-1)^(1/3)=x^3+y^3=1  or equivalently

((2)^(1/3)+1)^3(2^(1/3)-1)=3

so the two numbers are equal.


Suppose a stadium with $11$ rows for spectators, each row consisting of $199$ seats. We want to invite some students there, but at most $39$ students from each school are allowed. We must place students of each school in one row.
Find the maximum number of students, so that we can be sure that they are placed properly.

Can anyone prove or disprove this conjecture? Since $199$ is a prime number, we consider $200$.
$25$ is the greatest divisor of $200$ which is less than $39$. Seven $25$-student groups can be placed in one row. Hence, the answer must be strictly less than $a= 10\times (7\times 25) + 8 \times 25$. In my opinion the answer is $a-1$. What do you think?


Find all positive $x$ such that

$\sqrt[5]{x}=[\sqrt[5]{3x}]$, where the bracket is the floor function


Λοιπόν αγαπητοί μου φίλο θέλω να δώ πόσο πιστεύουμε στην εθνική μας
Με το χέρι πρώτα στο μυαλό κι μετά στην καρδιά ψηφίστε μπορείτε να ψηφίσετε κι κατά την διάρκεια του παιχνιδιού

ετσι ειναι δεν πειραζει.συγχαρητηρια .αργυρο μεταλιο :)

[quote="ambros"]ΟΣΗ ΠΙΚΡΑ ΚΙ ΑΝ ΕΧΟΥΜΕ ΠΡΕΠΕΙ ΝΑ ΔΩΣΟΥΜΕ ΤΟ ΧΕΡΙ ΣΤΟΥΣ ΙΣΠΑΝΟΥΣ ΚΙ ΝΑ ΤΟΥΣ ΔΩΣΟΥΜΕ ΣΥΓΧΑΡΗΤΗΡΙΑ ΟΠΩΣ ΜΠΡΑΒΟ ΚΙ ΣΤΗΝ ΕΛΛΑΔΑ ΓΙΑ ΤΗΝ ΚΑΛΗ ΠΟΡΕΙΑ ΣΤΟ ΤΟΥΡΝΟΥΑ ΤΗΣ ΕΛΕΙΠΕ Η ΨΥΧΡΑΙΜΙΑ
ΔΕΝ ΠΕΙΡΑΖΕΙ ΕΙΜΑΣΤΕ ΥΠΕΡΗΦΑΝΟΙ ΓΙΑ ΤΗΝ ΕΘΝΙΚΗ ΜΑΣ Η ΟΠΟΙΑ ΕΧΕΙ ΜΕΛΛΟΝ ΚΙ ΣΤΟ ΜΕΛΛΟΝ ΘΑ ΞΑΝΑΝΕΒΕΙ ΤΟΣΟ ΨΗΛΑ[/quote]

Ακριβώς   :)   :winner_second:

... Η Ισπανία! :)

Σοβαρά πάντως, καλό και το αργυρό.

προτίνω να κλείσουμε τη δημοσκόπηση...


limit
calculus
derivative
algebra
binomial theorem
calculus computations

:huh:  :huh:  :huh:  :huh:  what is a LIMIT?

have you taken (pre)calculus? He wants you to find $\lim_{h\rightarrow 0} \frac{1}{h} \left( \frac{1}{\sqrt{1+h}}-1 \right)$. If you don't know what that is, it's just the value that $f(h)$ gets closer and closer to...

[quote="Derek"]Evaluate the limit as h approaches 0:

$ \frac {1}{h} (\frac {1}{\sqrt {1+h}}-1) $[/quote]

If u ask L'Hospital he would tell u that the limit is $L= \lim_{h \to 0} \left ( -\frac{1}{2}(1+h)^{-\frac{3}{2}} \right ) = -\frac{1}{2}$
 ;)

Let's treat this as an exercise somewhere near the boundary between precalculus and first-semester calculus. In other words, we never heard of L'Hôpital, and maybe don't yet know what a derivative is.

$\lim_{h\to0}\frac1h\left(\frac1{\sqrt{1+h}}-1\right).$

Let's see - if $h$ is very small, then $\sqrt{1+h},$ and its reciprocal, must be very near 1. Subtract 1 and you get something very small. Now divide by $h$ - uh-oh, what's a very small number divided by a very small number? The answer to that is "It depends - how small and how small?" We can't settle the issue without doing some more work. (By the way, that's what " $\frac00$ " means: not literally zero over literally zero, but very small over very small.)

Since we don't know any calculus yet, we must fall back on what we do know: algebra.

$\frac1h\left(\frac1{\sqrt{1+h}}-1\right)= \frac1h\left(\frac{1-\sqrt{1+h}}{\sqrt{1+h}}\right)$

At this point, note that the $\sqrt{1+h}$ in the denominator is harmless - it might as well be 1. We don't need to do anything about it. We [i]do[/i] need to do something about the $1-\sqrt{1+h}$ in the numerator, which is far from harmless, being a non-obvious version of "very small." So we attack that - we choose to rationalize the numerator.

$\frac1h\left(\frac{1-\sqrt{1+h}}{\sqrt{1+h}}\right)= \frac1h\left(\frac{1-\sqrt{1+h}}{\sqrt{1+h}}\cdot\frac{1+\sqrt{1+h}} {1+\sqrt{1+h}}\right) = \frac{1-(1+h)}{h\sqrt{1+h}\,(1+\sqrt{1+h})}$

$=\frac{-h}{h\sqrt{1+h}\,(1+\sqrt{1+h})}= \frac{-1}{\sqrt{1+h}\,(1+\sqrt{1+h})}$

Now let $h$ tend to zero. All the parts of the denominator are now harmless and predictable, and we can see that the limit must be $-\frac12.$

What in that solution was the least obvious and required the most insight? The choice to rationalize the numerator rather than the denominator. Can you see why I did that?

Of course, if you free me from the "boundary of precalculus and Calculus I" constraint and let me use all of my favorite tools, I'll use Newton's Binomial Theorem for this one:

$\lim_{h\to0}\frac1h\left((1+h)^{-\frac12}-1\right)= \lim_{h\to0}\frac1h\left((1-\frac h2+O(h^2))-1\right)$

$=\lim_{h\to0}\left(-\frac12+O(h)\right)=-\frac12.$

Besides, this is just the derivative of $(1+x)^{-\frac 12}$ at $x=0$; using L'Hopital would be circular.

Yes we assume that the person has never heard of L'Hopital rule before, although it would be convenient to use it.

Thanks guys.


Here are two problems -

Here's what the paper gave as given information:
Let X, Y, Z be three points on a circle. Let Z' be a point outside the circle, on the same side of XY as Z.
Then angle XZ'Y < angle XZY. Now, if Z' is inside the circle, then angle XZ'Y > XZY.

Problems: (ABCD is a quadrilateral)
Prove that if $ \angle ABD$ = $ \angle ACD$, ABCD is cyclic.
Prove that if $ \angle ABC$ = $ \angle ADC$, ABCD is cyclic.

I kind of need help. Help?  :(

[hide="hints"]
The Theorems on [url=http://regentsprep.org/regents/mathb/5A1/CircleAngles.htm]This[/url] website will be helpful

For the second two, I would assume that in the first case, B and C are on the same side of AD
and in the second one, B and D are on the same side of AC.
[/hide]

a)consider the circle (ABD).
proof:direct consequence of the given information.
b)the second affirmation is wrong.


I'd just like to thank the coordinators of this site for taking the time to set it up and invite me to it. It seems very promising, especially the "Math Jams." I was reading the about section, I think, and I was very pleased to see that this site aims to answer that question that I have had so many times over: "How would I have come up with that answer?" Anyway, I think this is a great site, and I bet just about anyone else here would say the same thing that I am saying here.


Prove or disprove that if x,y,z>0 then:

y^4+z^4+x^2 yz+xy^2 z+ xyz^2+x^2 y^2 >= 2y^2 z^2 + 2 z^2 x^2 + 2x y^3.

Put $f(x, y, z) = LHS - RHS$ then

$ f(x, y, x) = y^4 - x^4 + 2x^3y - 2xy^3$

So, it is easy to see that initial ineq does not hold. For example
$f(2, 1, 2) = -3 < 0.$

There is general rule to check if such kind of ineq holds.

Namdung

[quote="Namdung"]
There is general rule to check if such kind of ineq holds.
[/quote]

Would you please describe the "general rule" in more detail?

It is long history. But in short, the rule is "It must be true for all hyperspaces". For example, in our example, this is x = z.

Namdung


modular arithmetic
IMO Shortlist
number theory
Divisibility

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Assume $n$ is not a power of $2$. In this case there is an odd number $k\ge 3$ s.t. $2^k-1|2^n-1|m^2+9$. $2^k-1$ is of the form $4t+3$, so it has a prime factor of the form $4t+3$ and $\ne 3$, because $3\not |\ 2^k-1$, so it can't divide $m^2+9$ because if a prime of the form $4t+3$ divides $a^2+b^2$, then it divides $a,b$. This means that $n$ must be a power of $2$.

On the other hand, assume $n=2^k$. In this case, the only prime of the form $4t+3$ which divides $2^n-1$ is $3$. Indeed, let $p$ be such a prime. We then have $p|(2^{2^k}-1,2^{p-1}-1)=2^{(2^k,p-1)}-1=2^2-1=3$ (because $p-1=4t+2$). At the same time, $9\not |2^n-1$ (because $9|2^n-1\iff 6|n$), so $2^n-1$ is of the form $3\alpha$, where $\alpha$ has only divisors of the form $4t+1$, so there is some $m'$ s.t. $\alpha|m'^2+1$. All we need to do now is to take $m=3m'$, and we have found $m$ s.t. $2^n-1|m^2+9$.

And what do you want to tell us by this¿

i think he wants the proofs for this facts

davron

Hm...my solution is much much longer and uglier than grobbers but since ML is a kind of archieve of solutions I've found (and they are precious few), I think I'll post it:

[hide]

[b]Case 1: [/b] $n$ is odd and $n>1$.
Clearly $3\nmid2^n-1$
Then there is a prime $p\neq 3$ dividing $2^n-1$ with $p\equiv 3 \mod{4}$ .
Let $p=4l+3$.
We have
\[ \left(\frac{-9}p\right) = \left(\frac{-3}p\right)\left(\frac{3}p\right)  \]
If $-9$ is a quadratic residue $\mod {2^n-1}$, it must be a quadratic residue $\mod{p}$, implying that $\left(\frac{-3}p\right) = \left(\frac{3}p\right)$.
But we have
\begin{eqnarray*} \left(\frac{3}p\right) &\equiv& \left(\frac{-3}p\right) \mod p \\ \Leftrightarrow 3^{\frac{p-1}2} &\equiv& (-3)^\frac{p-1}{2} \mod p \\ \Leftrightarrow 3^{2l+1} &\equiv& (-3)^{2l+1} = -3^{2l+1} \mod p\\ \Leftrightarrow 2\cdot 3^{2l+1} &\equiv& 0 \mod p \end{eqnarray*}
Which is impossible for $p>3$.

[b]Case 2:[/b] $n$ is even.
Let $n = 2^{k+1}u$ with $k\ge 0$ and $u$ odd.
Hence $2^n-1$ is divisible by $3$, so $m$ must be divisible by $3$. Let $m=3r$.
Also, notice that $2^{2^{k+1}u} - 1 = (2^u-1)(2^{1u}+1)\ldots (2^{2^{k}u}+1)$
So we have $2^u - 1 \mid 9(r^2+1)$ which is impossible for $u \ge 3$ because then $2^u - 1$ would have at least one prime divisor $p>3$ and with $p\equiv 3 \mod 4$ and it is well known that $-1$ is not a quadratic residue mod such a $p$.
Hence $u=1$ and $n$ is a power of $2$.
Now we can rewrite the divisibility to the form
\[ (2^2+1)(2^4+1)\ldots (2^{2^k}-1)\mid 3(r^2+1)  \]
but since all brackets on the left side are coprime to $3$, it's equivalent to
\[ (2^2+1)(2^4+1)\ldots (2^{2^k}-1)\mid (r^2+1)  \]
Notice that the numbers in the brackets on the left side are the Fermat-numbers. It is well known that they are pairwise coprime and (except for 3) not divisible by any prime $\equiv 3 \mod 4$.
Hence the left side only consists of prime divisors $\equiv 1\mod 4$ and it is well known that $-1$ is quadratic residue mod $p\in P$ iff $p=2$ or $p\equiv 1 \mod 4$.
Hence, by chinese remainder theorem and by the well known fact that $a$ is q.r. $\mod {p\ge 3}$ iff it is a q.r. $\mod p^\alpha$ for some $\alpha>0$, we are done...

Conclusion: Such an $m$ exists iff $n$ is a power of $2$.

[/hide]

[quote="abdurashidjon"][quote="grobber"]Assume $n$ is not a power of $2$. In this case there is an odd number $k\ge 3$ s.t. $2^k-1|2^n-1|m^2+9$. $2^k-1$ is of the form $4k+3$, so it has a prime factor of the form $4t+3$ and $\ne 3$, because $3\not |\ 2^k-1$, so it can't divide $m^2+9$ because if a prime of the form $4t+3$ divides $a^2+b^2$, then it divides $a,b$.
[/quote]
 Hi here grobber has used that if an number is of the form $4k+3$ then it will a prime divisor of the same form, and also, if a number  of the form $4k+3$ divides $a^2+b^2$ then it have to divide $a, b$
 Abdurashid[/quote]

Seven years ago, but for future reference, here is the proof. Consider a number of the form $4k+3.$ Clearly, all its prime factors must be of the form $4m+1$ or $4m+3.$ Since $4k+3 \equiv 3\pmod 4,$ the factors cannot all be $1\pmod 4,$ so there must be at least one prime factor of the form $4m+3.$

Now suppose that a prime $4m+3$ divides $a^2+b^2.$ Let $a^2 \equiv j\pmod{4m+3}.$ Assume for sake of contradiction that $j \neq 0.$ Then $b^2 \equiv -j\pmod{4m+3}.$ So we obtain $a^{4m+2} \equiv j^{2m+1}\pmod{4m+3}$ and $b^{4m+2} \equiv -j^{2m+1}\pmod{4m+3}.$ However, by FLT, we must have $j^{2m+1}=-j^{2m+1}=1,$ a contradiction. Therefore, $a \equiv b \equiv 0\pmod{4m+3}.$

I shall be using the fact that if any prime $p$ of the form $4k-1$ divides $m^2+9$ then $p=3$.
Firstly as $2^n-1\equiv -1\pmod{4}$ for $n>1$ we have $n=2s$ for some integer $s$, otherwise $3\nmid 2^n-1$.

If $n$ has any prime factor $q\neq 2$, then $2^q-1\nmid 2^s-1$. But $2^q-1$ has a prime factor of the form $4k-1$ which is greater than $3$. Hence $n=2^t$ for some integer $t$.

The rest is easy, let $2^{2^t}-1=3p_1^{\alpha_1}\cdots p_n^{\alpha_n}$.
By Chinese Remainder Theorem there exists a solution to
\[\begin{cases}X\equiv x_1\pmod{p_1^{\alpha_1}}\\ \cdots\\X\equiv x_n\pmod{p_n^{\alpha_n}}\end{cases}\] Where $p_i^{\alpha_i}\mid x_i^2+1$. Easy to see such $x_i$ exists.
Now $3X$ suffices.

SO the answer is $\boxed{n=2^t}$ for some non-negative integer $t$.

[hide="Alternative Solution with Legendre's Symbol"]I claim that the desired values for $n$ are the power of $2.$ We must find values of $n$ such that $-9$ is a quadratic residue of $2^n-1.$ Now, suppose for sake of contradiction that $n$ has an odd factor $k>1.$ Then $-9$ is a quadratic residue of $2^k-1.$ So consider a prime factor of $2^k-1$ that is greater than $3,$ say $p.$
\[\left( \frac{-9}{p} \right) = \left( \frac{-1}{p} \right) \left( \frac{9}{p} \right) = (-1)^{\frac{p-1}{2}}\left( \frac{3}{p}\right) ^2 = (-1)^{\frac{p-1}{2}}.
\] Hence, $(p-1)/2$ must be even, so $p \equiv 1\mod 4.$ However, since $k$ is odd, $3 \not{|} 2^k-1,$ so all prime factors of $2^k-1$ must be $1\mod 4,$ which is impossible since $2^k \equiv 3\mod 4.$

Thus $n$ must be a power of $2,$ and the only prime factor of $2^n-1$ that is $3\mod 4$ is $3.$ From above, we see that $-9$ is a quadratic residue for all other prime factors $q.$ From the well-known fact* that all quadratic residues of $q$ are quadratic residues of $q^{\alpha},$ $-9$ is a quadratic residue of $2^n-1$ by the Chinese Remainder Theorem. We are done.

*[u]Proof[/u]: We proceed by induction. Suppose $r$ is a q.r. of $p^n$ for some positive integer $n.$ There there exists an integer $a<p$ such that $a^2 = kp^n+r.$ Then consider $(a+bp^n)^2=kp^n+r+2abp^n+b^2p^{2n}.$ It suffices to show that there exists a value of $b$ such that this value is $r\mod p^{n+1}.$ We have
\[kp^n+r+2abp^n+b^2p^{2n} = p^n(2ab+k)+b^2p^{2n}.
\] Clearly, there exists $b$ such that $p | 2ab+k,$ so we are done. (This results also follows directly from Hensel's Lemma.)[/hide]

$2^n - 1 \mid m^2 + 9$

Let's assume $n$ has a prime divisor $p$ different than $2$

Therefore $2^p-1\mid2^n-1$, and $2^p-1$ has a prime divisor which is different than $3$, and it's equivalent $3$ in $\mod4$. Assume it's $q$

Since $q\mid 2^p-1\mid m^2+9$, $-9$ is a quadratic residue in $\mod q$

By Legendre's Symbol, we have

$1=\left(\dfrac{-9}{q}\right)=\left(\dfrac{-1}{q}\right) \cdot \left(\dfrac{9}{q}\right)$

Since $\left(\dfrac{9}{q}\right)$ is always equal to $1$, $\left(\dfrac{-1}{q}\right)$ must also be equal to $1$

Hence, $q \not\equiv 3 \mod4$. Contradiction.

Therefore, $n$ can't have a prime divisor different than $2$. So $n=2^k$

Now, we will use induction to prove $n=2^k$ always satisfies.

We will be done if $2^n-1$ doesn't have any prime divisor different than $3$, and equivalent $3$ in $\mod4$. Because if it really doesn't, $-1$ and so $-9$ will be a quadratic residue in all of it's prime divisors modulo, so it also will be in modulo $2^n-1$. And that's what we want.

For $k=1$, $2^n-1=3$ satisfies the condition. Assume also $2^{2^k}-1$ satisfies.

$2^{2^{k+1}}-1=\left(2^{2^k}+1\right)\left(2^{2^k}-1\right)$. We know the second factor satisfies. So we have to show $2^{2^k}+1$ also satisfies.

$r\mid 2^{2^k}+1 \Longrightarrow 2^{2^k} \equiv -1 (\mod r)$. So $-1$ is a quadratic residue in $\mod r$. Hence $r$ can't be equivalent $3$ in $\mod 4$. Therefore, we're done.

I have corrcted my solution. Thank you @TheOverlord

First of all note that if $2^n-1$ has a prime divisor $p \equiv 3\pmod{4}$ other than $3$ then $-9$ is not a quadratic residue modulo $2^n-1$,because $(-9)^{\frac{p-1}{2}}=-3^{p-1} \equiv -1\pmod{p}$ and this follows at once by Euler's criterion and composite quadratic residue law.

Now if $n$ has an odd prime divisor $j$ then $2^j-1$ is a factor of $2^n-1$ so the latter has a prime divisor $p \equiv 3\pmod{4}$(other than $3$ since $j$ is odd) so our condition is not satisfied.

So it remains to check for $n=2^k$ ($n=1$ case is obvious).

The crucial fact is that if an odd prime $p \equiv 3\pmod{p}$ other than $3$ divides $2^{2^k}-1$ then since it also divides $2^{p-1}-1$ it must also divide $2^{\text{gcd}(2^k,p-1)}-1=2^2-1=3$.This is a contradiction.

Also see that $2^{2^k}-1=(2^{2^{k-1}}-1)(2^{2^{k-1}}+1)$ inducting backwards we see that the highest power of $3$ dividing it cannot exceed the highest power of $3$ dividing $2^8-1=255$,i.e, $2$.(note that the latter factor is not divisible by $3$ for $k \ge 2$)

Finally we see that all the primes of $2^{2^k}-1$ other than $3$ are $\equiv 1\pmod{4}$ hence $-9$ is a quadratic residue of each of them(one may see this by applying Euler's criterion: $(-9)^{\frac{p-1}{2}}=3^{p-1} \equiv 1\pmod{p}$).Also we may choose $l$ so that the highest power of $3$ in $2^{2^k}-1$ divides $l^2+9$.Now applying CRT we may find $m$.

Thus the answer to our problem is $n=1,2^k \forall k \in \mathbb{Z^+}$

An easy proof that $n$ is a power of $2$.

Let $n=2^a b$ where $a\ge 0$ and $b$ is odd and let $p$ be a prime factor of $2^b-1$, hence $p\mid 2^b-1\mid 2^n-1\mid m^2+9$ and so $m^2\equiv -3^2\pmod p$. Since $b$ is odd we have $3\nmid 2^b-1$ giving $p\ne 3$, therefore the last congruence implies that $-1$ is a quadratic residue of $p$, implying that $p\equiv 1\pmod 4$. It follows that $2^b-1\equiv 1\pmod 4$, thus $b=1$. So $n=2^a$.

Used the 20% hint on ARCH.

The answer is when $n$ is a power of $2$(including $1$). Obviously $n=1,2$ works.

Otherwise, $2^n-1$ is $3$ mod $4$, and by Fermat's Christmas Theorem on $m^2+9$, any prime $p$ that is $3$ mod $4$ and divides $2^n-1$ has to be $3$.

Now if $\nu_3(2^n-1)\ge 3$, we get a contradiction by Fermat's Christmas Theorem. If it is $2$, that means that $2^n-1$ has some other $3$ mod $4$ divisor, which doesn't work. So $\nu_3(2^n-1)=1$.

[b]Claim.[/b] Assuming $n>2$, a value $n$ works if and only if $\dfrac{2^n-1}{3}$ is an integer and only has $1$ mod $4$ factors.
[b]Proof.[/b] We have proven the necessity already. For the sufficiency, simply apply CRT to find a working $m$.

Now assume FTSOC that $n$ is not a power of $2$. Then let $x=\nu_2(n)$ and $k=\dfrac{n}{2^x}$. We have that $2^k-1$ divides $2^n-1$. But $2^k-1$ is $3$ mod $4$ and not a multiple of $3$, contradiction.

If $n$ was a power of $2$, then
\[2^{2^a}-1=\left(2^{2^{a-1}}+1\right)\left(2^{2^{a-2}}+1\right)\cdots\left(2^{2^0}+1\right).\]
So if a $3$ mod $4$ prime that isn't $3$ were to divide into $2^{2^b}+1$ for some positive integer $b$, then we get a contradiction by Fermat's Christmas Theorem. QED.

Notice that all divisors of $m^2+9$ are of the form $k$ or $3k$ where $k \equiv 1 \pmod 4$. As $2^n -1 \equiv 3 \pmod 4$ unless $n=1$, we only have to look at the latter case.

The condition implies that no $3 \bmod 4$ primes divide $2^n - 1$ except for 3. Obviously $n$ must be even. But now if $p \mid n$ for some $p \geq 3$, then $2^p - 1 \mid 2^n-1$, and $3 \nmid 2^p - 1$. But then as $2^p - 1 \equiv 3 \pmod 4$, some $3 \bmod 4$ prime not $3$ must divide $2^p-1$, contradiction.

Thus $n$ must be a power of two. It is easy to check that such $n$ indeed work by Fermat theorem, say inductively as $$2^{2^n} - 1 = (2^{2^{n-1}} - 1)(2^{2^{n-1}} + 1)$$ where the latter factor is $1 \bmod 4$.

Only powers of 2 work. If $p\mid n,$ then $2^p-1 \mid 2^n-1 \mid m^2+9$. By Fermat's Christmas theorem, all the prime factors of $m^2+9$ are $1$ mod $4$, other than possibly $3$. This means $2^p-1$ is a power of $3$ and this implies $p=2.$ This proves that the power of $2$ condition is necessary. To prove it's sufficient, first we see that $3\mid 2^{2^k}-1, 9\nmid 2^{2^k}-1$. Assume $p\equiv 3 \pmod 4, p\neq 3$ divided $2^{2^k}-1$. Then $\text{ord}_p(2)\mid \gcd(2^{2^k}, p-1) = 2,$ but this means $p=3,$ contradiction. We are done.

Solved with [b]RaceCar21[/b]

For any prime $p \mid 2^n-1$, we have that $p \mid m^2+3^2$. If $p \neq 3$, then by Fermat's Christmas Theorem we know that $p \equiv 1 \pmod 4$. So, we know that $2^n-1$ must be expressed as the product of $3$'s and $1 \pmod 4$ primes. Let $n = 2^k \cdot m$ where $m > 1$ is odd. Note that $2^m-1 \mid 2^{2^k \cdot m}-1$ and $3 \nmid 2^m-1$ because $m$ is odd. Since $2^m-1 \equiv 3 \pmod 4$, we know that there must exist a prime $q \equiv 3 \pmod 4$ that is not $3$. This means that a prime $3\pmod 4$ divides $2^{2^k \cdot m}-1$, contradiction. Therefore, $m = 1$ and we will prove that all $n = 2^k$ work.

Note that $$2^{2^k}-1 = (2^{2^{k-1}}+1)(2^{2^{k-2}}+1) \dots (2^2+1)(2+1)$$ Denote $a_i = 2^{2^i}+1$ for simplicity. Note that all $a_i$ are coprime. Assume otherwise. WLOG $i>j$. If a prime $p \mid a_i$ and $p \mid a_j$ then $2^{2^i} \equiv -1 \pmod p$ and. $2^{2^j} \equiv -1 \pmod p$ but $$2^{2^i} \equiv  (2^{2^j})^{2^i-2^j} \equiv (-1)^{2^i-2^j} \equiv 1 \pmod p$$ contradiction.

Also, for each $a_i$ we can find an $m_i = 3 \cdot 2^{2^{i-1}}$ such that $a_i \mid m_i^2+9 = 9 \cdot a_i$. To construct $m$, take $m \equiv m_i \pmod {a_i}$ for all $1 \leq i \leq k-1$. By the Chinese Remainder Theorem, there exists such an $m$. To prove this construction suffices, let $m = m_i+ca_i$ for some constant $c$. Note that since $a_i \mid m_i^2+9$ we also have $a_i \mid (m_i+ca_i)^2+9 = m^2+9$ for all $i$. Again using the fact that all $a_i, a_j$ are coprime,  we have $2^{2^k}-1 = a_{k-1}a_{k-2} \dots a_2a_1 \mid m^2+9$ done!

Notice that we must have $m^2 = -(3^2) \pmod{2^n-1}$. Let $p$ be a prime dividing $2^n-1$. Notice that if $p \neq 3$ and $p \equiv 3 \pmod 4$ then the statement is impossible. Therefore, the prime factors of $2^n-1$ must be either $1 \pmod 4$ or $3$. Let $n = 2^a \cdot b$, where $b$ is the odd part of $n$. Notice that $b$ must be $1$, which implies that $n$ must be a power of $2$. It suffices to prove all powers of $2$ work. Notice it suffices to prove that $2^{2^b}+1$ has only prime factors that are $1 \pmod 4$, because induction will clearly finish after this. This is true because $2^{2^{b+1}-1}-1 = (2^{2^{b}}-1)(2^{2^b}+1)$. Now notice that -1 is a quadratic residue $\pmod{2^{2b}+1}$, so all prime divisors must be $1 \pmod{4}$. Now it is easy to finish with induction.

[quote=orl]Determine all positive integers $n$ for which there exists an integer $m$ such that ${2^{n}-1}$ is a divisor of ${m^{2}+9}$.[/quote]

Good. The answer is powers of 2 only.

If $n$ has an odd factor $f > 1$, then $d = 2^f-1\mid2^n-1\implies\exists d_0\mid d:d_0\equiv 3 \pmod 4,d_0\not\equiv 0 \pmod 3$, which means there's a $3 \pmod 4$ prime other than $3$ dividing $m^2+9$, contradiction by Fermat's Christmas Theorem (FCT).

This works because for $i=p^e$ be any prime power dividing $2^n-1$, either $i= 3$, or again by FCT $p \equiv 1 \pmod 4$; irregardless, there exists $m$ such that $i\mid m^2 + 9$, since in second case i=4j+1 with 4j-8 dividing m squared, and there's such an m since any multiple of 4 divides a perfect square. $\blacksquare$
\end{soln}
\begin{remark}
    In my opinion very easy to fall for the $2^n-1=4i-1,m^2+9=4j-c\implies m^2=4k$ trick. (joke; it comes from the fact that 1. i misread as $m^2+9=4i-1\implies m^2=4i-8$ instead of minus 10, which i thought trivialized it. 2. it's a dumb fallacy so i mimicked when ihatemath123 said in that JMO diophantine about "very easy to fall for 2+11+2022=2024" to sarcastically give mathmax 1 pt. for claiming the trivial answer n=1 and mathnerd101 fell for it
\end{remark}

Note that all priem factors are either $1\pmod 4$ or $3$ by fermats christmas theorem. Then if $o | n$ for odd $o$, then $3\nmid 2^o - 1$ and if $o\geq 1$ then $2^o - 1 \equiv 3 \pmod 4$, so it has a $3\pmod 4$ prime factor that isnt 3, a contradiction. Thus $2^{2^i}-1$ is forced. Note that this factors into $$(2^{2^0} + 1)(2^{2^1} + 1)(2^{2^2} + 1)\dots.$$
Note that each $3\pmod 4$ prime must have an even power in this expression, but $p^2 \nmid 2^{2^i} + 1$ because otherwise $$ord_{p^2}(2) = 2^{i+1} | p(p-1) \implies p \equiv 1 \pmod 2^{i+1}.$$ Since $2^{2^i} + 1 \equiv 1 \pmod 4$ so all of its factors must be $1 \pmod 4$ primes, with the exception of simply $2 + 1 = 3$. Thus by FCT it is possible for all $n = 2^i$.

The answer is when $n$ is a power of 2.\\

The problem is just saying "find all $n$ such that $-9$ is a quadratic residue mod $2^n-1$".\\

Claim 1: For an odd prime $p\neq 3$, $-9$ is a quadratic residue mod $p$ if and only if $p\equiv 1\pmod{4}$. If $p\equiv 1\pmod{4}$, then $\frac{p-1}{2}$ is even, so $$(-9)^{\frac{p-1}{2}}\equiv 3^{p-1}\equiv 1\pmod{p},$$ so by Euler's criterion we have that $-9$ is a quadratic residue. On the other hand, if $\frac{p-1}{2}$ is odd, then it will evaluate to $-1$, so it is not a quadratic residue, hence proven.\\

Claim 2: If $r$ is a quadratic residue mod $p$ for an odd prime $p$, then it is also a quadratic residue mod $p^m$ for all $m$. We will induct. Suppose that $$r\equiv k^2\pmod {p^m}.$$ Then, note that $$(k+sp^m)^2\equiv k^2+2ksp^m+p^{2m}\equiv r+cp^m+p^m(2ks)=r+p^m(2ks+c)\pmod{p^{m+1}}$$ for some unknown constant $c$. However, since $p$ is relatively prime to $2k$, we can choose $s$ so that $2ks+c\equiv 0\pmod{p}$, hence $r$ is a quadratic residue since some number of the form $(k+sp^m)^2$ works, shown by induction.\\

Combining Claims 1 and 2 as well as the Chinese Remainder Theorem, $n$ works if and only if all prime divisors of $2^n-1$ are either 1 mod 4 or equal to 3.\\

Now, if an odd prime $q$ divides $n$, then $2^q-1$, which is 3 mod 4, divides $2^n-1$, and it's also not divisible by 3, contradiction, so $n$ must be a power of 2.\\

Now, we will just show that all powers of 2 work. Suppose FTSOC that $3\neq v\equiv3\pmod{4}$ is a prime that divides $2^{2^k}-1$. Then, by repeated difference of squares factorization, we have that $v\mid 2^{2^s}+1$ for some positive integer $s<k,$ since $v\neq 3.$ However, for any prime $v$ that divides $2^{2^s}+1$, the order of 2 mod $v$ has to be exactly $2^{s+1}$ (it has to be a power of 2, but if its a smaller power of 2 then $2^{2^s}$ is already 1 mod $v$). Hence, $v\equiv 1\pmod{2^{s+1}}$. However, since $s\geq 1$, this implies $v\equiv 1\pmod{4}$, contradiction, hence done.

Average FCT problem.

The answer is all positive integers $n$ that are a power of $2$.

[b][color=#00f]Proof that $n$ must be a power of $2$:[/color][/b] Let $k$ be the largest odd divisor of $n$. Assume for contradiction that $k>1$. By FCT, all prime divisors of $m^2+9$ are either $3$ or $1 \bmod{4}$, and since $2^k - 1$ is a divisor of $2^n - 1$, all prime divisors of $2^k-1$ are either $3$ or $1 \bmod{4}$. Since $3 \nmid 2^k-1$ and $2^k-1 \equiv 3 \pmod{4}$, it must have a prime divisor that is $3 \pmod{4}$, a contradiction. Thus $k=1$, and $n$ is a power of $2$.

[b][color=#00f]Powers of $2$ work:[/color][/b] Let $n=2^j$. We prove the desired via induction on $j$, with the base cases $j=1, 2$ trivially true. For the inductive step, suppose $j$ works. Then factor \[ 2^{2^{j+1}}-1 = (2^{2^j}-1)(2^{2^j}+1). \] By inductive hypothesis the former factor works, and the latter works as it is $1 \bmod{4}$, so by the multiplicative nature of FCT we conclude.

We will show that the answer is all $n$ that are powers of $2.$ First suppose $p \mid n$ and $p \ne 2.$ Then there must be some prime $q \mid 2^p-1$ and $q \ne 3$ and $q \equiv 3 \pmod 4$ since $2^p-1 \equiv 3 \pmod 4.$ By Fermat christmas we have $q \mid m^2+9 \implies q \mid 9,$ contradiction. Now if $n=2^k$ we may write $2^{2^k}-1=(2+1)(2^2+1)(2^4+1)\dots(2^{2^{k-1}}+1).$ We notice that besides the first term, each term is a sum of two squares, so by Fermat christmas we have that all factors of this except for $3$ are $1 \pmod 4.$ Let one of these primes be $p.$ If we let $g$ be a primitive root $\pmod p,$ we may see that when $m\equiv 3g^{\frac{p-1}{4}} \pmod p$ we have $p \mid m^2+9,$ and when $m\equiv 0 \pmod 3$ we have $3 \mid m^2+9.$ Now taking all possible $p$ we get a system of congruences that must have a solution by CRT. Thus there exists such an $m,$ and we are done.

The answer is when $n$ is a power of $2$, including $n = 1$.

[b][color=#00f]Proof n must be a power of 2:[/color][/b]
Assume $n \geq 3$ is not a power of $2$. Then there exists an odd integer $k$ such that $$2^k-1 \mid 2^n - 1 \mid m^2 + 3^2.$$ Since $2^k -1 \equiv 3 \pmod 4$, $2^k -1$ must have a $3 \pmod 4$ prime factor. From Fermat's Christmas Theorem, $2^k -1$ cannot have any $3 \bmod 4$ prime factors other than $3$. However, $$ 2^k-1 \equiv (-1)^k -1 \equiv 1 \pmod 3,$$ so this is impossible. Therefore, $n$ must be a power of $2$.

[b][color=#00f]Proof n is a power of 2 works:[/color][/b]
Let $n = 2^k$, and $p$ be a prime such that $p \mid 2^{2^k} - 1$. Let the $a$ be the order of $2 \pmod p$. Then $a \mid \gcd(2^k, p-1).$ If $p \equiv 3 \pmod 4$, we must have $a = 2$, so $p = 3$. Therefore, $3$ is the only $3 \pmod 4$ prime factor of $2^{2^k} -1$, and all other prime factors are $1\pmod 4$. If $3^2 \mid 2^{2^k} - 1$, then we get $6 \mid 2^k$, since the order of $2 \pmod 9$ is $6$. This is clearly impossible, therefore $9 \nmid 2^{2^k}$. Thus, from Fermat's Christmas Theorem, there exists $m$ such that $2^{2^k} - 1 \mid m^2 + 9$.

  nice :blush:   $$\boxed{\text{The answer are all powers of 2.} }$$[color=blue][b] Claim. [/color] $n$ must be a power of $2$
 take a odd divisor $k>1$ of $n$, then $2^k-1\mid 2^n-1\mid m^2+9$ by Fermat Christmas Theorem all prime divisors of $m^2+9$ are $1\pmod 4$ or $3$, but $2^k-1\not\equiv 3\pmod 3$ as $k$ odd, nor $2^k-1\equiv 1 \pmod 4$, a contradiction.

[color=blue][b] Claim. [/color] all powers of $2$ work.
Let $n=2^k$, write $$2^{2^k}-1=(2+1)(2^2+1)(2^{2^2}+1)\cdots (2^{2^{k-1}}+1)$$
Note that each term except for the first one, by Fermat Christmas Theorem have only $1\pmod 4$ prime divisors, so choose $g$ a primitive root $\pmod {p^\theta}$  then by Chinese Remainder Theorem exists $m$ such that $$m\equiv 3 g^{\frac{\phi({p^\theta})}{4}}\pmod {p^\theta} \hspace{0.4cm}\forall\hspace{0.2cm} p\mid 2^{2^j}+1 \hspace{0.5cm}\text{and} \hspace{0.5cm} m\equiv 0\pmod 3$$ then $p^\theta\mid m^2+9$  , and $3\mid m^2+9$, as $\text{Fermat numbers}$ are relative primes we would have that $2^{2^n}-1\mid m^2+9$, as desired.

The answer is $\boxed{\text{powers of 2}}$.

Let $d$ be the largest odd prime divisor of $n$. For the sake of contradiction, assume $d>1$. Fermat's Christmas Theorem states that the factors of $m^2+9$ are either $1 \pmod{4}$ or $3$. Since $2^d-1$ divides $2^n-1$, all the factors of $2^d-1$ are either $1 \pmod{4}$ or $3$. Then, as

\[2^d-1 \equiv 3 \pmod{4},\]

it must contain a factor that is $3 \pmod{4}$, contradicting our assumption that $d>1$.

To prove the powers of $2$ work, induction with difference of squares easily works.

Our answer is $\boxed{n=2^b, b \ge 0}$. The most convenient method to show each works is by setting $m=3a$. Our condition is reduced to proving there exists such an $a$ with
\[2^{2^b}-1 = 3\left(2^2+1\right)\left(2^{2^2}+1\right) \ldots \left(2^{2^{b-1}}-1\right) \mid 3(a^2+1) \mid (3a)^2+9.\]

Since the Fermat primes are pairwise relatively prime, and we have the solution $a \equiv 2^{2^{t-1}} \pmod{2^{2^t}+1}$ for each $t$, there must exist such an $a$.

Otherwise, suppose $p>1$ is an odd prime divisor of $n$. Then neither 2 or 3 are factors of $2^p-1 \equiv 3 \pmod 4$, so Fermat's Christmas Theorem tells us we must have $p \equiv 1 \pmod 4$ for all prime divisors of $n$, contradiction. $\blacksquare$

Since it's not specified that $m$ has to be an integer, the answer is obviously all positive integers $n$! [color=#9a00ff](Just kidding! Real solution below)[/color] \\\\
I claim that the answer is all $n$ that can be expressed in the form of $2^k$ for some nonnegative integer $k$. \\\\
First, note that by Fermat's Christmas Theorem, if a prime $p$ divides $m^2+3^2$, then it is either $1$ mod $4$ or it divides $\gcd(m,3)$. Using this, I now claim that $n$ cannot have any prime factor $p$ that is larger than $2$. FTSOC, assume that $p\mid n$, where $p>2$. This then implies that
\[2^p-1\mid 2^n-1,\]
and since $p>2$, we have that $2^p-1$ must be $3$ mod $4$. Additionally, since $p$ is a prime $>2$, this means that $p$ is odd, implying that
\[2^p-1 \equiv 2-1\equiv 1\mod 3,\]
so if $2^p-1$ is $3$ mod $4$, some other prime $q\neq 3$ divides $2^p-1$, a contradiction to the Christmas Theorem statement. Therefore $n$ cannot have any prime factor $p>2$, meaning that $n$ must be in the form of $2^k$ in order for $2^n-1$ to have a multiple in the form of $m^2+9$. \\\\
I now claim that for all $n=2^k$, $2^n-1$ has a multiple of the form $m^2+9$. I will prove this using induction. Suppose that $2^{2^k}-1$ for $k\geq 1$ has a multiple of the form $m^2+9$. This implies that $-9$ is a quadratic residue mod $2^{2^k}-1$. Now, note that
\[2^{2^{k+1}}-1=(2^{2^k}-1)(2^{2^k}+1),\]
and that $\gcd(2^{2^k}-1,2^{2^k}+1)=1$. Therefore, since we already know that $-9$ is a quadratic residue mod $2^{2^k}-1$, we just need to prove that $-9$ is also a quadratic residue mod $2^{2^k}-1$. Since the two modulos are relatively prime, by CRT, this will also prove that $-9$ is a quadratic residue mod $2^{2^{k+1}}-1$, meaning that a multiple of $2^{2^{k+1}}-1$ in the form of $m^2+9$. \\\\
Using this, notice that,
\[(2^{2^{k-1}})^2 \equiv -1 \mod (2^{2^{k}}+1) \iff (3*2^{2^{k-1}})^2 \equiv -9 \mod (2^{2^{k}}+1),\]
which proves that $-9$ is indeed a quadratic residue mod $2^{2^{k}}+1$. Therefore, if there exists a multiple of $2^{2^{k}}+1$ in the form of $m_1^2+9$, then there exists a multiple of $2^{2^{k+1}}+1$ in the form of $m_2^2+9$!\\\\
Finally, to complete our induction, we need to cover the base case of $n=2$ and the external case $n=1$. The latter is covered by $m=0$ and the former also by $m=0$. Therefore, there exists a multiple of $2^n-1$ in the form of $m^2+9$ if and only if $n$ is a power of $2$, finishing the problem.


geometry
3D geometry
number theory proposed
number theory

$ a\in\mathbb{N}$ and for all $ n\in\mathbb{N}$ , $ 4(a^n\plus{}1)$ is a perfect cube,prove that $ a\equal{}1$

Check Out Here: http://www.mathlinks.ro/viewtopic.php?t=201851  :)


abstract algebra
group theory
superior algebra
superior algebra theorems

Let $ <G, . >$ be an abelian group. Let $ S \equal{} \{x \in G|x^3 \equal{} e\}$. Prove that $ S$ is a subgroup of $ G.$

[quote="t0rajir0u"]Which of the group axioms are you having trouble verifying?[/quote]

Im having trouble with the inverse part.

Since $ G$ is a group you have inverses in $ S$: Clearly $ (x^{\minus{}1})^3\equal{}(x^{\minus{}1})^3x^3\equal{}e$ since $ G$ is abelian..

existence of inverses does not have anything to do with $ G$ abelian.

Indeed.  Ottem, you don't need $ G$ abelian to show that $ x^{\minus{}1}$ commutes with $ x$.

Eh, of course not , and I never really  used commutativeness in the argument... thank you both for pointing that out though..  :roll:

But note that you do need the assumption that $ G$ is abelian. If you drop that hypothesis, the result would be false. Oh, you'd still have inverses in $ S;$ that's not the problem. The problem is that the set $ S$ wouldn't be closed under multiplication.

Just as an example, consider $ G\equal{}S_n,$ the permutations on $ n$ elements, with $ n\ge 4.$ The set of elements such that $ x^3\equal{}e$ consists of the identity and all the 3-cycles. The product of two different but overlapping 3-cycles will not be a 3-cycle and will not have order 3.

I said "different but overlapping." The word "overlapping" was meant to exclude that particular example.

allright but your description for the $ x$ such that $ x^3\equal{}1$ is wrong.


limit
function
number theory
number theory theorems

Does anyone know an elementary proof of the prime number theorem?

Prime number theorem:

 \lim  \pi (x)lnx/x = 1 when x goes to infinity., where  \pi (x) is the number of primes not exceeding x and lnx is the logarithmus with base e.

I have looked at proof of this theorem in an AMM article, but it uses some facts in complex analsys. It is said that Erdoes and Selberg has found an elemtary proof of this theorem.

Yes, Erdos and Selberg found an elementary proof of this theorem in 1948. A very readable reconstruction of it can be found in the book [u]A Course in Number Theory[/u] by H.E. Rose. The discovery of an elementary proof surprised many who did not think that an elementary proof could be found.

An interesting bit of history behind it can be found in the book [u]The Music of the Primes[/u] by Marcus du Sautoy. There was quite a bit of controversy surrounding who should be given credit for it.

While this proof is famous, I think the proof using the zeta function is much more interesting.

I have a pretty good reference for this theorem!
The book
 "Introduction to Number Theory", Hua Loo Keng, Ed. Springer Verlag
has a complete proof of this theorem. And it is really strange!

Bye-bye, guys!
Ass.:Johann


induction
inequalities solved
inequalities

Yes, it works to prove by induction that a_1>=...>=a_n then (a_1)^2+...+(2n-1)(a_n)^2 \leq (a_1+...+a_n)^2. The inductive step comes down to 2a_1+...+2a_n>=2na_(n+1), true,.


geometry
geometric transformation
rotation
real analysis
real analysis unsolved

Show that the torus is a 2-fold covering of the klein bottle. Compute pi1(Klein bottle)

I assume you know the model of the Klein bottle as a square with one pair of opposite sides identified in the same orientation, and the other pair identified with opposite orientation; tile the plane with copies of this model, reflecting or translating as necessary.
Note that adjacent pairs of squares combine into the standard model of a torus.

This picture is a universal covering of both the torus and the Klein bottle, so the fundamental groups are isomorphic to the groups of rotations and translations of the plane which are the identity after taking quotients. These groups of rotations and translations are easy to understand; we can write their elements as $3\times 3$ matrices with integer entries.


Back in the summer of 2003, I gathered up some Web links about how to study math. Any reader of this who has suggestions for additional links would surely be doing several readers of this thread a favor by sharing them, or by commenting on the links I post here.

[url=http://www.cde.ca.gov/ta/tg/hs/mathguide.asp]Preparing for the California High School Exit Examination (CAHSEE): A Mathematics Study Guide[/url]

[url=http://www.sunytccc.edu/instruct/sbrown/math/succeed.htm]How to Succeed in Math[/url]

[url=http://euler.slu.edu/Dept/SuccessinMath.html]Success in Mathematics[/url]

[url=http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/]College Algebra[/url] (includes Tutorial 1 "How to Succeed in a Math Class")

[url=http://www.math.clemson.edu/~mjs/courses/misc/study.pdf]How to Survive Your College Math Class (and Take Home Something of Value)[/url] (.PDF file)

[url=http://www.iwu.edu/~lstout/HowToStudy.html]How to Study Mathematics[/url] (Illinois Wesleyan)

[url=http://www.math.utah.edu/~alfeld/math.html]Understanding Mathematics[/url]

[url=http://www.math.ohio-state.edu/students/how_to_study.html]How to Study Mathematics[/url] (The Ohio State University)

[url=http://www.math.fau.edu/Niederhausen/HTML/SurvivalGuide.htm]Survival Guide for Students Taking University Courses in Mathematics[/url]

[url=http://math.ucsd.edu/~ebender/proofs.html]Some Help on Reading Mathematics and Creating Proofs[/url]

[url=http://www.math.jhu.edu/~sz/Education/]Education[/url] (articles about undergraduate math study by Steven Zucker of Johns Hopkins U.)

[url=http://www.math.ohiou.edu/~mjm/goodproblems/goodstudent.html]Student's Guide to Good Problems[/url] (a great guide to mathematical writing)

[url=http://www.maths.cam.ac.uk/undergrad/inductionday/studyskills/]Study Skills in Mathematics[/url] (Cambridge University)

[url=http://www.maths.ox.ac.uk/current-students/undergraduates/study-guide/]Study Guide[/url] (Oxford University)

[url=http://www.math.purdue.edu/academic/undergrad/credit/undergrad]Credit Exams for Undergraduates[/url] lots of study guides and practice exams for undergraduate math courses

[url=http://www.math.cornell.edu/~hubbard/readingmath.pdf]Reading Mathematics[/url] (Professor Hubbard at Cornell)

[url=http://www.cargalmathbooks.com/]Books in the Mathematical Sciences[/url] (James M. Cargal) more than just a book list, a guide full of opinions on how to learn different levels of mathematics

[url=http://www.ocf.berkeley.edu/~abhishek/chicmath.htm]Chicago undergraduate mathematics bibliography[/url] a book list and an introduction to studying math at the U of Chicago

[url=http://server1.fandm.edu/departments/Mathematics/writing_in_math/guide.html]Guide to Writing in Mathematics Classes[/url] (Franklin and Marshall College)

[url=http://www.esotericka.org/math/guide.html]Guide to being a successful math student[/url]

[url=http://math.berkeley.edu/~vojta/survival.html]Survival Guide to Calculus[/url] (UC Berkeley)

[url=http://web.mit.edu/jrickert/www/mathadvice.html]Writing a Research Paper in Mathematics[/url] (Ashley Reiter)

[url=http://tex.loria.fr/typographie/mathwriting.pdf]Mathematical Writing[/url] (Donald E. Knuth, Tracy Larrabee, and Paul M. Roberts)

[url=http://www.dartmouth.edu/~acskills/success/index.html]Learning Strategies: Maximizing Your Academic Experience[/url] (Dartmouth College Academic Skills Center)

[url=http://www.wadsworth.com/colsuccess_d/special_features/weblinks.html]Student Web Links[/url] (Wadsworth Publishing, links to many sites about study tips for all subjects)

[url=http://math.arizona.edu/ugprogram/mcenter/gradprep.html]How to prepare for a graduate program in the mathematical sciences[/url] (U. of Arizona)

Here are some famous pages on how to avoid errors in mathematics study:

[url=http://www.math.vanderbilt.edu/~schectex/commerrs/]THE  MOST  COMMON  ERRORS  IN UNDERGRADUATE MATHEMATICS[/url]

[url=http://www.swarthmore.edu/NatSci/smaurer1/WriteGuide/write_errors.pdf]Common Errors in Writing Mathematics[/url] (.PDF file)

Hope this helps!

[quote="JS1527"]Wow thanks tokenadult.[/quote]

You're welcome. Your recent thread reminded me that I had already gathered those links last summer, before I joined this site.

very informative.  I don't suppose you have any good links on effectively studying for other subjects as well, because they would be very useful.

[quote="zscool"]I don't suppose you have any good links on effectively studying for other subjects as well, because they would be very useful.[/quote]

I know of one [url=http://learninfreedom.org/languagebooks.html]bibliography on language learning[/url] (my field) that is not too bad. Generalizing the subjects covered by this FAQ is an interesting suggestion. After all, learning all those other subjects more efficiently leaves more time for math.  :D

Many people know this already, but I found it very useful.
[url=http://docs.google.com/View?docid=df5w5j9q_5gj6wmt]So You Want To Be A Physicist[/url]

Also, I have a general question. How do we make the most efficient use of modern learning tools like [url=http://en.wikibooks.org/wiki/Main_Page]Wikibooks[/url] and [url=http://en.wikiversity.org/wiki/Wikiversity:Main_Page]Wikiversity[/url]? How far do you think such tools can influence/improve education?

It is great to have free sources that are just a mouse click away but you should understand that they are just sources of information. The most important thing is some "education program" that guides you step by step from simple things to more and more sophisticated ones and is adjusted to your personal abilities, learning speed, preferences, etc. The difficulty with education is exactly in creating such a program (indeed, how can you do it without a priori expert knowledge of the subject?) and forcing yourself to follow it. Once this is done, I do not see much difference between reading online books and ordinary ones. So, the only influence the mentioned sources can have is making education slightly more accessible to those who know how and are willing to educate themselves.

How to use these tools effectively? The same way as you use the library, this forum, etc. effectively. Essentially everything boils down to finding the right books to read, finding the right people to ask questions and learn from, and having enough willpower to continue the studies when it is not easy. This works today the same way it did in China and Egypt thousands of years ago and will work the same way in the foreseeable future with new tools or without them. Everything else (like most modern education theories or fancy online programs) is secondary. All that is just my humble opinion, of course :)


algebra polynomial calculus derivative MIT college

If $x$  is a real number such that:

$1 + x +\frac {x^2}{2! } + \frac {x^3}{3!}+\frac {x^4}{4!} ......+\frac {x^{2004}}{2004!} = 0  $

Prove that   $[x^{2004}] \equiv 1 \mod 2005$

Where  $[x^{2004}] $ , of course means an integer  less than or equal to  $x^{2004} $

Edited later: Thanks JBL  for editing
P.S. . Please put answers in spoilers ...

That should be, $[x^{2004}] \equiv 1 \mod 2005$.

Hmmmm, not my kinda question, but yet intriguing... any hints that dont give away too much?

[b]radagast[/b] is right.Look if $n$ is even $e^{x}>1+x+\frac{x^{2}}{2!}+\ldots+\frac{x^{n}}{n!}$ for $x>0$ and $e^{x}<1+x+\frac{x^{2}}{2!}+\ldots+\frac{x^{n}}{n!}$ for $x<0$.So $1+x+\frac{x^{2}}{2!}+\ldots+\frac{x^{n}}{n!}>0$ for all $x$.That means that the equation
$1+x+\frac{x^{2}}{2!}+\ldots+\frac{x^{n}}{n!}=0$ does not have real solutions for even $n$-s.For odd $n$-s the equation above has one real root because it is a polynomial of odd degree and its derivative is the polynomial $1+x+\frac{x^{2}}{2!}+\ldots+\frac{x^{n-1}}{(n-1)!}>0$.

[quote="radagast"]Gyan, maybe i misunderstood something but why would that polynomial have any real roots at all?

[hide]
First its clear $x<0$.  Also By taylor's theorem, there exists $c\in\ (x,0)$ such that $e^x=\frac{x^{2005}}{2005!}e^c+\sum_{i=0}^{2004}\frac{x^i}{i!}$  Now if the summation is 0 then
$e^x=\frac{x^{2005}}{2005!}e^c$ .... but the right side is negative and the left is positive...so whats wrong here??
[/hide][/quote]

[hide]No reason at all! But don't you see the beauty of that?    :D   Since there are no real roots :D  the "theorem" is proved ~ :lol:  (  for all x the second part is true since there are/is  no such x ..)   (Yes I could have  put anything there, but   was  just  looking for some  funny looking stuff)

:lol: Okay,  ducking for cover now! :lol:
BTW there was ":)" in the subject line.
PS  Random House, if you don't know is a MIT dorm  where all the nerdy math students live ..
[/hide]

Heh yes i was kinda suspecting something was afoot when seeing the source... :)


The COPSD (COP for middle schoolers ONLY) will start in August.

Anyone interested?

Sign-up here!

Can I sign up? Do you have to qualify? I would like to join.

??? Everyone in middle school can sign up. I'll sign up as well. I forgot to the the COP 2nd round last time.  :blush:

[quote="236factorial"]lol, I want to start this earlier. I've never done this and I can't wait!

There's still 30+ days till August!!!!![/quote]

I heard this from someone.. :D

[quote="orl"]Patience is a virtue as Valentin prefers to say[/quote]

Longer you wait, the better it will be.  :)

Now it's August  :D

What date will the problems be posted?

I'll sign up, because it sounds like fun.

Umm... what does COP stand for?

[quote="Drunken_Math"]Umm... what does COP stand for?[/quote]

I think it's Contest of Opportunity.

For the answers, do we have to show work or do you only want the answers?

The questions arent' out yet...

I think you have to show work... or not.

I would want to join, except for I'm not in middle school anymore.   ;)

[quote="math92"]wait, if you are going into 9th can you still join?[/quote]

I don't think so.

oh, i was just wondering becuase i thought 4everwise was going into 9th.

probably since nobody will post here again.


induction
inequalities proposed
inequalities

Let $x_1,x_2,\cdots x_n$ be real numbers such that $x_1^2+x_2^2+\cdots+x_n^2=1$.

Find the maximum and minimum values of $x_1^2+2x_2^2+\cdots+nx_n^2$.

We make the substitution $a_i=x_i^2$. Then $a_i$ are non-negative and add up to $1$. Now I prove by induction that the minimum is attained for $a_1=1$ and $a_2=...=a_n=0$ and the maximum for $a_1=...=a_{n-1}=0$, $a_n=1$.
For $n=1$ it is trivial. Now we consider the case $n+1$. For every fixed $a_{n+1}$ the minimum is attained for $a_1=1-a_{n+1}$, $a_2=...=a_n=0$ and similiar for the maximum (because of induction hypothesis). Thne the term gets $(n-1)a_{n+1}+1$ in the minimum case and $n-1+a_{n+1}$ for the maximum case. So to achieve the minimum we have to choose $a_{n+1}=0$ and for the maximum $a_{n+1}=1$. The values of the extrem points are $1$ and $n$.

Misha

Thank you for your reply, Misha.
You are right.

kunny


ratio
logarithms
geometric series
geometric sequence

Hi everyone,

   I am a new member just recently joined.  Didn't know exactly where to post this question.  This is not for school so I hope no one thinks I am cheating by asking outside sources.  But I am a paintballer and came across this problem.  Wasn't sure how to solve it.  I hope someone here can lend me a hand.

Suppose I have an 80cf (cf = cubic foot) tank at 3000psi. It is filled with air. I also have an empty 68ci (ci = cubic inches) tank. The procedure is the following: I will fill the small tank by attaching a hose to the larger tank. The air will go into the smaller tank till both pressures are equalized (thus the pressure in the larger tank will drop since air is being shared in a larger volume of space now). Then I will empty the smaller tank and repeat this step. Assume room temperature (but a small factor is that when pressure starts to build, temperature rises. Thus, when the small tank is being filled, the temperature will rise since it goes from 14.4psi to whatever psi it equalizes at. Therefore, if you assume constant room temperature, it will throw off the calculations a little, but I don’t think it will be THAT off.)

How many times can I repeat the above step before the pressure in the larger tank gets below 800psi?

I tried solving, then realized air is not an ideal gas and couldn't use the ideal gas law. Anyone know of what formula or method to solve this?

Thanks guys for any help!!

This is a physics question, so I'm sending it over to the physics forum.  (A link to your post from the Round Table will continue to exist for a while.)

Welcome to the forums.

If you allow the reaction to occur slowly enough, it is sufficient to assume that the temperature will stay fairly constant.

spoon: your equation assumes that he empties the small tank to 0 pressure, (a physical impossibility). i would work with moles of gas instead. with the ideal gas law: $PV=nRT$

i dont think the temperature changes.. (ideally) i think that the decrease in pressure is reciprocated by a loss of moles of gas.

here is an example. i am going to assume that you empty the small tank to 1 atm, and not a vacumm. if you really do empty the small tank to close to a vaccum, you should use spoons equation.

[hide="a really long attempt at an answer"]
first, you have the big tank of volume $V_1$. It is filled with a pressure of $P_1$.

from the ideal gas law, you have $n_1=\frac{P_1 V_1}{RT}$ moles in the large tank

now, in the small tank, of $V_2$ you have $n_2=\frac{V_2}{RT}$

therefore, your total moles: $n_1+n_2=\frac{P_1 V_1+V_2}{RT}$

now, back to the ideal gas law $P_{new} (V_1+V_2)=P_1 V_1+V_2$

therefore, $P_{new}=\frac{P_1 V_1+V_2}{V_1+V_2}$

from this, we can see a recursive sequence defined as:

$a_o=P_1$

$a_n=\frac{a_{n-1} V_1+V_2}{V_1+V_2}$

let $\alpha=\frac{V_1}{V_1+V_2}$, and $\beta=\frac{V_2}{V_1+V_2}$ (these are easily calculated)

therefore, $a_n=\alpha \,a_{n-1} + \beta$

the general formula i found for this is:

$a_n=\alpha^n P_1 + \beta(\alpha^{n}+\alpha^{n-1}+\alpha^{n-2}...+1)$

that last part is the sum of the first n terms of a geometric series of ratio $\alpha$, and initial value $1$.

that is equal to $\frac{1-\alpha^n}{1-\alpha}$

therefore, $a_n=\alpha^n P_1 + \beta\frac{1-\alpha^n}{1-\alpha}$

solving this, i get $n=3$ which doesnt make sense to me...

i have a feeling i made a stupid mistake

i think thats the general idea though.
[/hide]

[hide="Read"]You're right, PV=K isn't going to hack it. I was considering 0 volume, oops. Not trusting your answer of 3 (for obvious reasons) I did the complete problem on my own, my approach being to calculate $n_n$ rather than $P_n$. After about as much work as you, I came to the answer of 2713 refills. Also, I changed all units to Pascals and Cubic Meters so that PV=nRT was satisfied.

Now that is a bit of a descrepency, so I went through your work carefully. It took me a few seconds to realize that you left out $P_2$ because you made it in atm. That is perfectly fine except I don't think you converted your volumes to liters!!! Hehe.

I converted them to liters, and then used your recurrsion relation, except rather than using the closed form, I used an excel spread sheet to run down the list. This way, any error in that closed form wouldn't matter. Guess what iteration the pressure drops to below 800 psi (54.43 atm). 2713. That is the final answer.

Your work is right, good job. Now I need sleep.[/hide]

[quote="jsjaedon"]
I tried solving, then realized air is not an ideal gas and couldn't use the ideal gas law. Anyone know of what formula or method to solve this?
[/quote]

Most basic mixing problems can be made under the assumption that the gas is ideal. This one included.

[quote="jsjaedon"]... Assume room temperature (but a small factor is that when pressure starts to build, temperature rises. Thus, when the small tank is being filled, the temperature will rise since it goes from 14.4psi to whatever psi it equalizes at. Therefore, if you assume constant room temperature, it will throw off the calculations a little, but I don’t think it will be THAT off.)...[/quote]

This is not correct, the temperature of the small tank will not rise, because the gas does not perform any work. For an ideal gas, it would remain exactly the same. Most real gases cool down when released under pressure through a small openning. For example, we blow air on hot food to cool it down. Or recall how refrigeration or air conditioning work.

Let $p_A \doteq 14.4\ \text{psi}$ be the atmospheric pressure, $p_0 = 3000\ \text{psi}$ the initial pressure in the pressurized tank and $p_n = 800\ \text{psi}$ the pressure after equalizing the tank pressures n-times. Let $V_1 = 80\ \text{cf},\ V_2 = 68\ \text{cf}$ be the volumes of the 2 tanks. Imagine that air from the pressurized tank 1 flows into the evacuated tank 2 and simultaneously, air from tank 2 at the atmospheric pressure flows into the evacuated tank 1. The pressure $p_1$ after the 1st equalization is given by

$p_0V_1 + p_AV_2 = p_1(V_1 + V_2)$

$p_1 = \frac{p_0V_1 + p_AV_2}{V_1 + V_2}$

Similarly, the pressure $p_2, ..., p_n$ after the 2nd, ..., n-th equalization will be

$p_2 = \frac{p_1V_1 + p_AV_2}{V_1 + V_2} =  \frac{ \frac{p_0V_1 + p_AV_2}{V_1 + V_2}\ V_1 + p_AV_2}{V_1 + V_2} =$

$= \frac{p_0V_1^2}{(V_1 + V_2)^2} + \frac{p_AV_1V_2}{(V_1 + V_2)^2} + \frac{p_AV_2}{V_1 + V_2}$

$p_n = \frac{p_0V_1^n}{(V_1 + V_2)^n} + \frac{p_AV_1^{n-1}V_2}{(V_1 + V_2)^n} + \frac{p_AV_1^{n-2}V_2}{(V_1 + V_2)^{n-1}} + ... + \frac{p_AV_2}{V_1 + V_2}$

The terms with the atmospheric pressure $p_A$ form a finite geometric sequence with n terms, the 1st term $\frac{p_AV_2}{V_1 + V_2}$ (the last term in the equation) and with the quotient $q = \frac{V_1}{V_1 + V_2}$ Adding this geometric sequence,

$p_n = \frac{p_0V_1^n}{(V_1 + V_2)^n} + \frac{p_AV_2}{V_1 + V_2} \cdot \frac{1 - q^n}{1 - q}$

$\frac{1 - q^n}{1 - q} = \frac{(V_1 + V_2)^n - V_1^n}{(V_1 + V_2)^n} \cdot \frac{V_1 + V_2}{V_1 + V_2 - V_1} = \frac{(V_1 + V_2)^n - V_1^n}{(V_1 + V_2)^{n-1}V_2}$

$p_n = \frac{p_0V_1^n}{(V_1 + V_2)^n} + \frac{p_AV_2}{V_1 + V_2} \cdot \frac{(V_1 + V_2)^n - V_1^n}{(V_1 + V_2)^{n-1}V_2} =$

$= \frac{p_0V_1^n}{(V_1 + V_2)^n} + \frac{p_A[(V_1 + V_2)^n - V_1^n]}{(V_1 + V_2)^n} = p_A + (p_0 - p_A) \left(\frac{V_1}{V_1 + V_2}\right)^n$

Calculating n from the last equation (1st, invert the fractions on both sides in order to get positive logarithms) and substituting the values:

$\frac{p_n - p_A}{p_0 - p_A} = \left(\frac{V_1}{V_1 + V_2}\right)^n$

$\frac{p_0 - p_A}{p_n - p_A} = \left(\frac{V_1 + V_2}{V_1}\right)^n$

$\log{\frac{p_0 - p_A}{p_n - p_A}} = n \log{\frac{V_1 + V_2}{V_1}}$

$n = \log{\frac{p_0 - p_A}{p_n - p_A}} : \log{\frac{V_1 + V_2}{V_1}} =$

$= \log{\frac{3000 - 14.4}{800 - 14.4}} : \log{\frac{80 + 68}{80}} = \log{\frac{2985.6}{785.6}} : \log{\frac{148}{80}} =$

$= \frac{\log{3.80041}}{\log{1.85}} = \frac{0.579830}{0.267172} \doteq 2.17$

It is kind of obvious that n must be approximately 2 and since it came out slightly larger than 2, equalizing the pressure 2-times is not sufficient, it must be done at least 3-times to get below $800\ \text{psi}$.

[quote="yetti"]
Let $p_A \doteq 14.4\ \text{psi}$ be the atmospheric pressure, $p_0 = 3000\ \text{psi}$ the initial pressure in the pressurized tank and $p_n = 800\ \text{psi}$ the pressure after equalizing the tank pressures n-times. Let $V_1 = 80\ \text{cf},\ V_2 = 68\ \text{cf}$ be the volumes of the 2 tanks. Imagine that air from the pressurized tank 1 flows into the evacuated tank 2 and simultaneously, air from tank 2 at the atmospheric pressure flows into the evacuated tank 1. The pressure $p_1$ after the 1st equalization is given by
[/quote]

$V_2$ is cubic Inches... 68 cf is slightly larger than 68 cubic inches.

Being a long time SCUBA diver, I should have realized this earlier!

By 80cf tank, do you mean that the tank is actually 80cf, aka the volume enclosed by a twin bed three feet off the ground, or is it a typical 80cf scuba tank? The "80cf" size rating on a tank like that isn't the actual size of the tank, rather it's the amount of actual air at 1 atm that will fit into the tank at a given pressure. Meaning, I have several 80 cf tanks, that we fill to around 3000-3200 psi, yet they are only about 2 cf in size. It means that they have 80 cf of air at 1 atm once they are filled. If this IS the case, then we are all wrong, and some extra conversions need to be done before the problem can be attempted.

spoon: what you said makes sense

those numbers are rediculous otherwise. in his post he says that he is a paintballer. i would assume the little container is for a gun?? this would make 68 cf (at 1 atm).. WAY too big.

i noticed i do have an error .. i think:

$\sum_{i=0}^n x^i = \frac{1-x^{n+1}}{1-x}$

i used:

$\sum_{i=0}^n x^i = \frac{1-x^{n}}{1-x}$

from that i get:

$a_n=\alpha^n P_1 + \beta\frac{1-\alpha^{n+1}}{1-\alpha}$

yetti if i understand your work, i think you did that part correctley.

spoon.. your seggustion of spreadsheets is a good one.. (i dont really trust that equation... )

[quote="Spoon"]$V_2$ is cubic Inches... 68 cf is slightly larger than 68 cubic inches.[/quote]
Sorry, I did not notice the different units.

$V_1 = 80\ \text{cf} = 80 \cdot 12^3\ \text{ci} = 138240\ \text{ci}$

$n = \log{\frac{p_0 - p_A}{p_n - p_A}} : \log{\frac{V_1 + V_2}{V_1}} =  \log{\frac{3000 - 14.4}{800 - 14.4}} : \log{\frac{138240 + 68}{138240}} =$

$= \log{\frac{2985.6}{785.6}} : \log{\frac{138308}{138240}} \doteq \frac{\log{3.80041}}{\log{1.000491898}} \doteq \frac{0.579830}{0.000213576} \doteq 2715$

This number checks with a spreadsheet calculation.

we still have the problem of misleading information. I do not believe the large tank is actually 80cf at 3000 psi. We have to wait to hear back from the original poster to clarify this issue.

i tried to look up paintball tanks to get some sort of idea, but the stores i saw didnt say..


When do you expect last year's scores to be released? It has been so long and I am getting impatient...

According to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=178011]this topic[/url], I received my results mailer on:

March 18, 2005

March 30, 2006

March 22, 2007

March 13, 2008

The mailing I'm talking about goes to the contest supervisor at each college (which I am). It's mailed from the West Coast, so I probably get it in Long Beach a day or two earlier than it arrives on the East Coast. The process of disseminating the information from there depends entirely on how the local contest supervisor chooses to handle it.


Find min value of $ k>0$,$ x,y,z>0$
$ \sum\frac{x}{\sqrt{x\plus{}y}}\le k\sqrt{x\plus{}y\plus{}z}$

[quote="wxydx00"]Find min value of $ k > 0$,$ x,y,z > 0$
$ \sum\frac {x}{\sqrt {x \plus{} y}}\le k\sqrt {x \plus{} y \plus{} z}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1459
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88454
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90923
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=152848


function
trigonometry
calculus
calculus computations

Prove these aren't periodic functions:
[b]1[/b] $f\left( x \right) = \sin x^2$
[b]2[/b] $f\left( x \right) = \sin \sqrt x$
[b]3[/b] $f\left( x \right) = \sin \left| x \right|$

$g(x)=\sin f(x).$
Supose  $T>0$ -is a period. Then $g(x+T) - g(x) = 0 \forall x$
Then $\sin f(x) - \sin f(x+t) = 2\sin\left(\frac{f(x)-f(x+T)}{2}\right)\cos\left(\frac{f(x)+f(x+T)}{2}\right)$.
So $\forall\; x$ must be $f(x)-f(x+T) = 2k(x)\pi$ or $f(x)+f(x+T) = 2l(x)\pi+\pi$ where $k,l: R \to Z$.
But for a)b)c) $f(x)-f(x+T)$  and $f(x)+f(x+T)$ is non-constant functions, but continuous.
So the RHS functio must be not-constant and continuous. Contradiction, because obvious that $k,l$is not continuous if they are non-constant.  :)


calculus
integration
parameterization
geometry
linear algebra
matrix
calculus computations

If $ S$ is the helicoid: $ r(u,v)\equal{}ucos(v){i}\plus{}usin(v){j}\plus{}v{k}$ with $ 0\le u\le 1,$ and $ 0\le v\le 2\pi$
how do you evaluate $ \displaystyle\int_S \sqrt{1\plus{}x^2\plus{}y^2}dS$

[b]Edit:[/b] Oops! I thought $ u$ was a given parameter somewhere in $ [0,1]$.

Two things are unclear about that answer.

1. Why are you factoring $ u,$ which is a variable, out of the integral sign?

2. Since $ S$ is a two-dimensional surface, talking about its arclength doesn't make a lot of sense.

Been a little while since I did this, but if I recall correctly: Plug in x and y in terms of u and v, and transform the area element dS via the determinant of the Jacobian:

http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant

dS = |J|dudv

After that, it should be a straightforward double integral with the given u,v limits.

I get:

$ \frac{2\pi}{3}(2\sqrt{2}\minus{}1)$


Sorry for being paranoid, but it is 11:20 eastern time...

My question is, if you live in the east coast, is the deadline 3 AM? I have all my solutions typed except half of one, and I would rather turn in a partial answer in one instead of 0/25, but I don't want to lose the chance to get them all because I worried to much. If possible, someone answer in the next 40 minutes.

I was just trying to send my solutions and the page kept giving me "Sorry, there was an unknown error while processing your upload. Please try again."
And the alternate page gave me "FPDF error: Unable to find xref table - Maybe a Problem with 'auto_detect_line_endings'".

Is there any way to fix this? Thanks! (I just e-mailed usamts@usamts.org)

EDIT: just successfully submitted. It seemed the error occurred when the pdf is generated from the PDF Merge. I later re-scanned every page directly into one single pdf and then it worked.

I'm done now (I really felt like proofreading; presubmission jitters I guess). Thanks again worthawholebean; if I get a 5 on a certain problem it is thanks to you.


A piece of yarn is 60 cm long. The yarn is cut so that one piece is five times the length of the other piece. How many centimeters long is the shorter piece?

We have that $ y\equal{}5x$ and $ x\plus{}y\equal{}60$, so that $ x\plus{}5x\equal{}6x\equal{}60$, so $ x\equal{}10$.  Hence this is the shorter length.


inequalities
real analysis
integration
calculus
calculus computations

For each $ x,x'\in\mathbb R$ is $ \Big|\big|f(x)\big|\minus{}\big|f(x')\big|\Big|\le\big|f(x)\minus{}f(x')\big|.$

This inequality is used to prove that if $ f$ is integrable on $ [a,b]$ then so does $ |f|,$ but no idea how that inequality helps to prove the integrability.

are you talking about Lebesgue integrable ? if so , come back to the definition of integrable of $ f$ : $ f$ is integrable iff $ \int_{E}f^{\plus{}}$ and $ \int_{E}f^{\minus{}}$  are finite and define $ \int_{E}f\equal{}\int_{E}f^{\plus{}} \minus{}\int_{E}f^{\minus{}}$

and $ \int_{E}|f|\equal{}\int_{E}f^{\plus{}} \plus{}\int_{E}f^{\minus{}} <\infty$  so $ |f|$ is integrable

I'm pretty sure the OP meant Riemann integration. The key point is that $ f$ is Riemann integrable if and only if for every $ \varepsilon>0$, there exists a partition $ P\equal{}\{I_1,\dots I_n\}$ of $ [a,b]$ into subintervals $ I_j$ such that $ \sum_j |I_j|\sup_{x,x'\in I_j}|f(x)\minus{}f(x')|<\varepsilon$. The rest should be clear.

Interestingly enough, the same approach can be used for the Lebesgue integration. The only difference is that we'll have to consider not finite partitions into subintervals, but countable partitions into measurable subsets.

An other point of vue  :D  ( for the fun  :lol: ).  Let $ \mu$ be the associated measure .  Let $ D_{(f)}$ be the set of discontinuity points of $ f$ .
$ f$ is Riemann_integrable iff $ f$ is bounded and $ \mu(D_{(f)})\equal{}0$ . Manifestly $ |f|$ is bounded ,  $ D_{(|f|)}\subset D_{(f)}\Longrightarrow \mu(D_{(|f|)})\equal{}0$, and so on ... .  :cool:


[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107262]IMO 2007 HKTST 1[/url]
Problem 5

The sequence $\{a_{n}\}$ is defined by $a_{1}=0$ and $(n+1)^{3}a_{n+1}=2n^{2}(2n+1)a_{n}+2(3n+1)$ for all integers $\geq 1$. Show that infintely many members of the sequence are positive integers.


What's the Squirty Chicken formula?
ahaha.

hello,this is what i found
http://www.playworkz.com/swingset/maoforums/index.php?act=Print&client=printer&f=3&t=1143507547
Sonnhard.


I have two terms that I can't think in my class. It's from crossword :)

A. An independent candidate.

It's five-lettered word with like this:

_ _ r _ t

B. Progressive candidate for president

It's nine-lettered word like this:

_ _  _ _ e _ _ _ _

Anyone knows this? :lol:

The answers are both names of particular people, not terms.  I think the first was from 1992, and the second roughly 1912?

Hehe, I know the first one from watching SNL's Presidential Debates last Monday. Good show.


topology
linear algebra
linear algebra unsolved

[b]Source: Homework[/b]

Let $M_{k}(n,n)$ be the set of matrices of $rank=k$ in $M(n,n)$, and $M^{*}_{k}(n,n)$ be the set of matrices with $rank\leq k$.
Show that $M_{k}(n,n)$ is open and dense in $M^{*}_{k}(n,n)$.

Use, respectively, the continuity of the determinant and the density of $GL_{n}$ in $M_{n}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=115505]This[/url] should also solve the problem.

The assumption that $k$ is algebraically closed is only used for proving that the $\leq i$ rank matrices are a non-empty set which is otherwise obvious wihtout this hypothesis.

Of course there is no problem in passing from the Zariski topology to the euclidean one.


function
calculus
integration
ARML
algebra
functional equation

1) Consider a family of functions $f_{b}(x)$ such that $f_{b}(0)=b$ and $f_{b}(x)=2^{a}f_{b}(x-a)$. Find an expression for $f_{c}(2x)$ in terms of $f_{b}(x)$. (MA$\theta$ 1992)

2) If $f(n+1)=(-1)^{n+1}n-2f(n)$ for integral $n\ge 1$, and $f(1)=f(1986)$, compute $f(1)+f(2)+f(3)+\dots+f(1985)$. (ARML 1985)

3) Find all solutions to the functional equation $f(x+y)-f(y)=x/y(x+y)$. (M&IQ 1991)

[quote="ch1n353ch3s54a1l"]1) Consider a family of functions $f_{b}(x)$ such that $f_{b}(0)=b$ and $f_{b}(x)=2^{a}f_{b}(x-a)$. Find an expression for $f_{c}(2x)$ in terms of $f_{b}(x)$. (MA$\theta$ 1992)

um for number one (I hope):
[hide]$f_{b}(x)=2^{a}f_{b}(x-a)$

$f_{b}(x)=2^{x}f_{b}(x-x)$

$f_{b}(x)=2^{x}b$[/hide]

3) Find all solutions to the functional equation $f(x+y)-f(y)=x/y(x+y)$. (M&IQ 1991)

I hope
[hide]if y=-x, then

$f(x+y)-f(y)=x/y(x+y)$

$f(0)-f(-x)=-(0)$

$f(0)-f(y)=0$

so f(x)=f(y) for any y or x
let f(x)=a
then a-a=x/y(x+y)

0=x^2/y+x

which isn't true for all x and y, say x=1 and y=2

[/hide]


Prove that if $a,b,c$ are positive distinct reals:

$a^2+b^2+c^2+2abc+1 \geq 2ab+2ac+2bc$.

$ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}2abc\plus{}1\geq 2ab\plus{}2ac\plus{}2bc$ , $ a, b, c$ positive reals.
Solution: From the following numbers: $ 1\minus{}a, 1\minus{}b, 1\minus{}c$ suppose WLOG that $ (1\minus{}a)(1\minus{}b)\geq 0$. As $ c\geq 0$, we have :
$ 2c(1\minus{}a)(1\minus{}b)\geq 0$ $ \leftrightarrow 2c\minus{}2ac\minus{}2bc\plus{}2abc\geq 0$
Actually the former inequailty is equivalent to: $ (c\minus{}1)^{2}\plus{}(a\minus{}b)^{2}\plus{}2c(1\minus{}a)(1\minus{}b)\geq 0$
I think this ineq is well known, I think I have seen it somewhere before... maybe in Vasc's book, but I'm not sure...


Let $S$ be a set containing $n^{2}+n-1$ elements, for some positive integer $n$. Suppose that the $n$-element subsets of $S$ are partitioned into two classes. Prove that there are at least $n$ pairwise disjoint sets in the same class.

Isn't a complete solution, only parcial, because i'm not certain if is completely right!

Let G a graph with $C(n^{2}+n-1,n)$ vertices (each vertice assigned to one of the sets with n elements).
Two vertices are linked by one edge if the intersection of your sets are empty.

We easily see that each vertice have degree $C(n^{2}-1,n)$.

If we cut G in two graphs X and Y (erasing the edges from X to Y, only considering the edges from one vertice of X to X, and the same for Y), i found that, using turan, if the two sets don't have a $K_{n}$, we will get a contradction!

My doubt is about my computations, but i guess that my idea isn't wrong!

[quote="N.T.TUAN"]Let $S$ be a set containing $n^{2}+n-1$ elements, for some positive integer $n$. Suppose that the $n-$ element subsets of $S$ are partitioned into two classes. Prove that there are at least $n$ pairwise disjoint sets in the same class.[/quote]
I think maybe it should be $|S|=n^{2}+n-1$ in the title of the topic.
Let us prove the stronger problem by induction:
Proposition:Let $S$ be a set containing $nk+k-1$ elements, for some positive integer $n,k$. Suppose that the $n-$ element subsets of $S$ are partitioned into two classes. Prove that there are at least $k$ pairwise disjoint sets in the same class.
[b]Proof:[/b] Induction base: For $k=1$, the statement is obvious.
Induction step: assume  that for all integer not greater than $k$, we have proved the proposition, Let us prove the $k+1$ case:
Let us say $A$ and $B$ (with $|A|=|B|=n$ and $A,B\subset S$) are adjacent if only if $\left|A\cap B\right|=n-1$.
Separate into two cases:
Case 1: If there exist $X,Y\subset S$ such that $X,Y$ are adjacent, and $X,Y$ are not in the same class, then Let us consider the set $S'=S-X\cup Y$ then $|S'|=n(k+1)+k+1-1-(n+1)=nk+k-1$,
By the assumption of the induction, there exist $k$ pairwise disjoint sets in the same class,then these $k$ sets and one of $X,Y$ are pairwise disjoint and in the same class. (This is because $X,Y$ are not in the same class, and $(X\cup Y)\cap S'= \emptyset$.)
Case 2: If for all $X,Y\subset S$ with the property that $X,Y$ are adjacent, $X,Y$ are in the same class, then we can prove easily that all the sets are in the same class. (Because we find some subset $A_{1}$ of $S$ with $|A_{1}|=n+1$; then all the subsets of $A_{1}$ must be in the same class, and for any element $u$ of $S-A_1$ and any $n-1$-elements set $P,P\subset A_{1}$, the subset $P\cup \{u\}$ is in the same class with all the subsets of $A_{1}$, then let $A_{2}=A_{1}\cup \{u\}$, then $A_{2}$ has the property that all the subsets of $A_{2}$ are in the same class; in the same way, we can find $A_{3}$, $A_{4}$,........., $A_{t}=S$.)
I hope my solution is correct. I always commit some stupid and small mistakes, I am not so confident now....... (Er, Maybe  I am so confident that always ignore the details :blush: )

I didn't check all the details, but you've got the right idea (I tried to solve it for quite a few hours, but then I finally... snapped... and I looked over the official solution).

[quote="Hawk Tiger"]Proposition:Let $S$ be a set containing $nk+k-1$ elements, for some positive integer $n,k$. Suppose that the $n-$ element subsets of $S$ are partitioned into two classes. Prove that there are at least $k$ pairwise disjoint sets in the same class.[/quote] What is the motivation for this? Is there some heuristic that might lead you to believe the number of sets ($k$) and their sizes ($n$) are actually unrelated?

(Sorry for the massive revive.)

[quote="v_Enhance"][quote="Hawk Tiger"]Proposition:Let $S$ be a set containing $nk+k-1$ elements, for some positive integer $n,k$. Suppose that the $n-$ element subsets of $S$ are partitioned into two classes. Prove that there are at least $k$ pairwise disjoint sets in the same class.[/quote] What is the motivation for this? Is there some heuristic that might lead you to believe the number of sets ($k$) and their sizes ($n$) are actually unrelated?[/quote]
I don't really like this problem, but I think the main motivation for generalizing the problem is that the original problem doesn't allow you to look at small cases. (Also, it's not initially clear where the $n^2+n-1$ comes from.)

And pretty much the simplest way to get lots of similarly-flavored small cases is to start with $k=2,3$ in "Find the smallest $N(n,k)$ such that when we partition the $n$-subsets of a $\ge N(n,k)$-set into $2$ classes, we can find some $k$ pairwise disjoint sets in the same class." One way to see that it might be helpful: If you first look at the obvious bound $N(n,n)\le (2n-1)n$ (by considering the $2n-1$ pairwise disjoint sets $[1,n],[n+1,2n],\ldots,[(2n-2)n+1,(2n-1)n]$), it's fairly natural to think about the analogous bound $N(n,k)\le (2k-1)n$. (On the other hand, changing the number of classes just complicates the situation.)

Of course, it's not immediately obvious that the $k$ and $k+1$ cases are related---at best you might be able to build off from $k$ to $k+1$, but a priori you might only expect it to give a better understanding of how $N(n,k)$ grows.

[quote]Proposition:Let $S$ be a set containing $nk+k-1$ elements, for some positive integer $n,k$. Suppose that the $n-$ element subsets of $S$ are partitioned into two classes. Prove that there are at least $k$ pairwise disjoint sets in the same class.[/quote]

Proof of easier proposition:

Induct on $k$. If $k=1$, we are considering $n$-element subsets of an $n$-element set, which makes the statement trivial.

Now suppose the statement holds for $k-1$, for some $k>1$, and AFSOC that we lose for $k$. Suppose that $A$ and $B$ are two $n$ element subsets of $S$, with $n-1$ elements in common. We claim that they are in the same class. Indeed, $S-A\cup B$ has size $(n+1)(k-1)-1$, so by the inductive hypothesis, one of the two classes has a collection $C$ of $k-1$ size $n$ subsets of $S-A\cup B$. So if $A$ and $B$ are in different classes, we can choose one of them appropriately to add to $C$, yielding a new collection of size $k$, as desired.

So we've shown that replacing any element in an $n$ element subset of $S$ keeps its class constant, which easily implies that all $n$ element subsets of $S$ must be in the same class, which implies what we want.

[hide=Solution]
We prove the following statement below, by induction on $m$:
[quote]
Let $S$ be a set containing $mn + m - 1$ elements, for some positive integers $m$ and $n$. Suppose that the $n$-element subsets of $S$ are partitioned into two classes. Prove that there are at least $m$ pairwise disjoint sets in the same class.
[/quote]
When $m = 1$, the result is obvious. Now, suppose the result is true for $m$, and consider $m \to m + 1$. For a contradiction, suppose neither class contains $m + 1$ disjoint subsets. We have $S = (mn + m - 1) + (m + 1)$. Take an $mn + m - 1$-element $T$ of $S$, by the inductive hypothesis, there exist $m$ pairwise disjoint subsets in the same class (call it "red") with union a subset of $T$. Now, consider the remaining $m + 1$ elements. None of the $m$-element subsets of this set can be in the red class, so all of these subsets are in the other class. Doing this for all $mn + m - 1$ element subsets of $S$, we see all subsets are in the same class, which is a contradiction, and the inductive step is complete.

Setting $m = n$ finishes the problem.
[/hide]

[hide="A Question"]
Shouldn't it be $S = (mn+m-1) + (n+1)$?
[/hide]

EDIT: Oh wait I see now.

if m=2 S has 2n - 1 elements and please show me there are at least n-elements $2$ pairwise disjoint sets in the same class.

Induction argument supplied by [b]awang11[/b].

We show a generalization of this result: let $S$ be a set of $k(n+1)-1$ sets. Let the subsets of $S$ with cardinality $n$ be partitioned into two classes. Show that one of the classes contains $k$ pairwise disjoint sets.

The proof is by induction on $k$. If $k=1$, the result is obvious. For $k\ge 2$, we use the following argument to achieve the result. Let $A$ and $B$ be sets with cardinality $n$ and $|A\cap B|=n-1$. We claim that $A$ and $B$ lie in the same class or we are done, which clearly suffices. Suppose that $A$ and $B$ lie in separate classes. By the inductive hypothesis on $T=S\setminus (A\cup B)$, one of the classes restricted to $T$ contains $k-1$ pairwise disjoint sets, so the result holds for $k$ by considering those $k-1$ pairwise disjoint sets by using the appropriate choice of $A$ and $B$ as the $k$-th set.

We prove the more general statement: in a set $S$ of $m(n+1)-1$ elements, if we partition the $n$-element subsets of $S$ into two classes, then there are at least $m$ pairwise disjoint sets in the same class. This is by induction on $m$, with the base case of $m=1$ being trivial. For the inductive step, consider any $n+1$ elements in $S$ and temporarily hide them; among the remaining $(m-1)(n+1)-1$ elements, we can find $m-1$ $n$-element subsets in the same family which are disjoint. Then, if we cannot find $m$ pairwise disjoint sets in the same class, no $n$-element subset formed from the $n+1$ hidden elements can be in the same family, so they must all be in the same family—in other words, any two $n$-element subsets which overlap in $n-1$ positions are in the same family. But this clearly implies that every $n$-element subset is in the same family, in which we can obviously find $m$ pairwise disjoint sets, as desired. $\blacksquare$

[b]Remark:[/b] Perhaps a possible way to motivate the idea that the number of sets and their size is unrelated is to discover a class of "equality cases" where the size of $S$ is "tight": for some fixed $1 \leq k \leq n$, take $kn-1$ elements and color them red. Put every set with at least $k$ red elements in one family and the rest in another. We can find $n$ disjoint subsets in the second family by forming each subset using $k-1$ red elements and $n-k+1$ non-red elements, and we will have enough non-red elements to do this, but for $|S|=n^2+n-2$ this is no longer the case.

However, the fact that the number of sets and their size is the same is not relevant to this "equality case". For the more general case we can take $km-1$ elements instead and essentially employ the same construction. Furthermore, it seems that the generalized statement is true in general by checking small cases, so it is motivated to pursue it instead, with induction now seeming viable.

    Solved with hint 40\% to consider general. when generalizing makes it easier skull

    We induct on k with \# of elements k(n+1)-1, with base case k=1 evident; ignore an arbitrary n+1 elements. AWLOG the first family has by inductive hypothesis n-1 pairwise disjoint sets, FTSOC exactly that amount, then the other n+1 sets formed by the n+1 elements choosing n elements are all in the second family; in particular, any n sets that overlap in n-1 elements are in the same family, which by varying the such n+1 elements that are ignored, implies all sets are in the same family, so we can find n disjoint sets in the same family. We conclude. :surf:

We generalize to showing $k$ pairwise disjoint sets for $k(n+1)-1$ elements.
The base case of $k=1$ holds.
Now suppose it holds for $k$ and consider a set $S$ of $(k+1)(n+1) - 1$ elements and a partition.
FTSOC there are not $k+1$ pairwise disjoint sets.
By the inductive hypothesis, considering the $n$ element subsets of a $k(n+1) - 1$ subset $S' \subset S$ which can vary, one of the partitions of $S$ contains $k$ pairwise disjoint subsets with elements in $S'$. This implies that all subsets only containing elements in $S \setminus S'$ must be in the other partition.
Since $S \setminus S'$ can be any subset of $n+1$ elements, repeating this as $S'$ varies implies that all elements are in the same partition, contradiction.

[hide="Key word: Generalize!:"]

We will prove that there are at least $i \leq n$ pairwise disjoint sets in the same class for $S$ containing $ni+i-1$ elements.\\
Induction on $i$:\\
For $i =1$ : trivial\\
Now suppose the statement holds for $i=k-1 \leq n-1$.Assume FSOC that the statement doesn't hold for $i=k$\\
$Claim:$ All the $n$-elements subsets of $S$ are in the same class.\\
$Proof:$ Assume the contrary. Let $A$ and $B$ be two $n$ elements subsets of $S$ such that $|A\cap B|=n-1$. Notice that $|S$ $\symbol{92}$ $A\cap B|= n(k-1)-(k-1)-1$, thus by the induction hypothesis there is $k-1$ $n$-elements subsets in the same class. \\
If $A$ and $B$ are in different classes the result follows immediately.
If $A$ and $B$ are in the same class, this means that all the $n$-elements subsets of $S$ with $n-1$ elements in common are in the same class, hence all of the $n$-elements subsets of $S$
are in the same class and this leads to a contradiction .

$$\mathbb{Q.E.D.}$$

      [/hide]


trigonometry
geometry
number theory
greatest common divisor
ratio
trig identities
Law of Sines

In $\triangle ABC$, $E$ is the midpoint of side $BC$ and $D$ is on side $AC$. If the length of $AC$ is $1$ and $\angle BAC=60^{\circ}$, $\angle ABC = 100^{\circ}$, $\angle ACB=20^{\circ}$ and $\angle DEC=80^{\circ}$, then the area of $\triangle ABC$ plus twice the area of $\triangle CDE$ equals

(A) $\frac{1}{4}\cos 10^{\circ}$
(B) $\frac{\sqrt{3}}{8}$
(C) $\frac{1}{4}\cos 40^{\circ}$
(D) $\frac{1}{4}\cos 50^{\circ}$
(E) $\frac{1}{8}$

Tough problem.
[hide="Solution"]Using the Law of Sines, we get $AB=2\cos80^{\circ}$ and $BC=\frac{\sin60^{\circ}}{\sin100^{\circ}}$. Then we use the formula for a triangle to get

$[ABC]=\sin60\cos80$
$[CDE]=\frac{3\cos80}{16\sin80}$.

$S=[ABC]+2[CDE]=\sin60\cos80+\frac{3\cos80}{8\sin80}$
in which we get
$S=\frac{\sqrt{3}}{8}\frac{2\sin60\cos80+4\sin80\cos80}{\sin80}$
after combining the above expressions into one single expression and then simplifying terms.

Changing the sine into cosines and using $2\cos\alpha\cos\beta=\cos(\alpha+\beta)+\cos(\alpha-\beta)$, we can get

$S=\frac{\sqrt{3}}{8}\frac{\cos110+\cos50+2\cos70}{\cos10}$
$=\frac{\sqrt{3}}{8}\frac{\cos50+\cos70}{\cos10}$
$=\frac{\sqrt{3}}{8}\frac{2\cos60\cos10}{\cos10}$
$=\frac{\sqrt{3}}{8}$,

so our answer is $\boxed{B}$[/hide]

project $C$ onto $AB$ at $F$ and project $C$ onto $DE$ at $G$

$\angle EDC=80=\angle DEC$ so $DC=EC$, then by HL, $\triangle CGE\cong \triangle CGD$, also $\angle CBF=180-100=80$,

note $\angle CGE=90=\angle BFC$, so $\triangle GCD\sim \triangle FCB$ (by AA) with a side ratio of $1: 2$ (due to the midpoint), hence an area ratio of $1: 4$

hence $[ABF]=[ABC]+[CFB]=[ABC]+4[GCE]=[ABC]+2[CDE]$

now $\triangle ABF$ is a 30-60-90 triangle so $[ABF]=\frac{\sqrt{3}}{8}\implies \boxed{B}$


calculus
integration
real analysis
real analysis unsolved