AMC USA(J)MO USAMO AIME trigonometry geometry rectangle

Yes, Lassiter signed up.  I think it's an excellent thing, because I think Round 2 will provide an opportunity for more people to do something closer to the USAMO than the AIME than just USAMO qualifiers.

When are we allowed to talk about results and answers?

My school entered... I did... eh, we'll see.

Hmm, maybe we should wait 'til Saturday or something, just to be sure.  I'm looking forward more to Round 2 than I was to Round 1, though.

I found a very nice solution to the $\sin x= \sin x^{2}$ question.

I didn't, I had to leave it blank.

I messed up #7 since It was a [i]squared rectangle[/i] not a [i]squared square[/i].

M, I got the sin x = sin x^2 one also; just a bit of mods.

And I got the square one too.

But I guessed on the birthday one.

And missed the 5-digit-number-div-by-11-one- I solved it with 5 minutes left, and forgot to count the permutations of 79999. :(

What was the square one again?

I think I got them all except 21 and 23.  I liked 25.

For number 25, I took my fingers and compared the line segment to  the one that was 12 long. I thought that it was about 1 and 2/3 as long, so I guessed 12.

I'm guessing that I got 23 right, 1 blank, and 1 wrong for a total of 140.

For 25, you were supposed to draw the equilateral triangle that was begging to be drawn.  Then you notice that the perpendicular from the center of the circle to the midpoint of that one chord is parallel to the a median of the equilateral triangle that you just drew, and you use ratios of segments...
[hide="Problem Statement (for those who do not remember)"](I might be changing some notation) Let $P$ be the center of a circle, and let $A,B,C$ be on the circle.  $PA=8$, $AB=12$, and $PAB = ABC = \frac{\pi}{3}$.  Find the length of $BC$.

(Okay, so the equilateral triangle that's begging to be drawn is made by extending $AP$ to intersect $BC$, and you get an answer of something like 20.)[/hide]

And what was the square one?

What's the point of that median stuff? Add the equilateral triangle, drop the perpendicular from P, and angles show that the length from the third vertex of the triangle to the foot of the perpendicular from P is 2, so the other part of the triangle is 10, and since it's a perpendicular intersection through center it's bisected, so the full length is 20.

[quote="Elemennop"]What's the point of that median stuff?[/quote]
It doesn't take significantly more effort, it's more elegant and general than trig, some of us don't trust our own knowledge of trig tables, and it's easier to communicate without putting in the effort of drawing a diagram. :P

westside school signed up and we took the first one 2 weeks ago i believe or 3

How does everything think they did on round 2? I think I got three of them, but made a small mistake on number four (that didn't affect my answer, but affected the validity of my proof). I know other super ARML geniuses probably got all five of them in ten minutes, but I've never really done proofs before, so I think I did pretty well for myself. :maybe:

Our school scheduled it from 3:30 to 5:30.
I rode the bus home by mistake and didn't get back to the school until 4:30.

With less than one hour left, I wrote proofs to 1, 4, and 5. I solved #3 20 mins after time was called while riding back home :mad: . Atleast, I did better than the others at my school, atleast I think so.

I recall #1 from part of a Power Round at an ARML practice.

I'm not sure we're supposed to comment on it yet? Thus I refuse to comment on how I did.

There are no "official" restrictions, but I recommend that we withold our comments until Saturday.

Hope everybody did well!


superior algebra
superior algebra unsolved

How can I show that there are infinitely many primes congruent to 5,7, 11 modulo 12?

Look t prime divisors of numbersof type $ x^2\plus{}1$, $ x^2\plus{}3$, $ x^2\minus{}3$ repectively.

The problems are a special case of Dirichlet's theorem.


1.  The USAMTS "My Scores" page contains a link to the "Grading Subforum" for USAMTS.  If such a subforum exists, it isn't visible to my registered students.  In any case, the link on that page is misspelled (http://www.artofproblemsolving.com/Fourm/viewforum.php?f=164).

2.  Erin Schram posted some insightful comments on the grading of one of the Round 1 Problems.  I read the post in the USAMTS Forum the day he made the posting.  Now, the post shows as "Moved".  Depending on how one tries to link to the post, I get an error saying the post does not exist or an error saying you need to be granted special access to view the post.  Should this be visible to USAMTS participants?

Thanks in advance!

We put all those grading posts in a forum that is only visible to participants.  I'll take a look at the link and fix it.  If you have a registered participant who can't see that forum when she/he is logged in, let me know.

The link is fixed, and my student was able to view the subforum after we closed the browser and partially cleared the browser's cache.  Thanks for the quick resolution!


inequalities
search
inequalities unsolved

please post if anyone has a nice solution. i solved this with muirhead and its not nice at all.
$ a,b,c>0$

$ (ab\plus{}bc\plus{}ca)(\frac{1}{(a\plus{}b)^2}\plus{}\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}) \geq \frac{9}{4}$

this is from iran 96,posted many many many times before,for example:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1973718867&t=3547[/url]


Do there exist two sequences of real numbers $ \{a_i\}, \{b_i\},$ $ i \in \mathbb{N},$ satisfying the following conditions:

\[ \frac{3 \cdot \pi}{2} \leq a_i \leq b_i\]

and

\[ \cos(a_i x) \minus{} \cos(b_i x) \geq \minus{} \frac{1}{i}\]

$ \forall i \in \mathbb{N}$ and all $ x,$ with $ 0 < x < 1?$

The problem had in book ( The IMO Compendium ) is
Do there exist two sequences of real numbers $ \{a_i\}, \{b_i\},$ $ i \in \mathbb{N},$ satisfying the following conditions:
\[ \frac {3 \cdot \pi}{2} \leq a_i \leq b_i\]
and
\[ \cos(a_i x) \plus{} \cos(b_i x) \geq \minus{} \frac {1}{i}\]
$ \forall i \in \mathbb{N}$ and all $ x,$ with $ 0 < x < 1?$

So I don't know which will true.
Who can help me to solve the problem, please?
Thank you.

Help............
I can't solve the problem.
Thank


calculus
logarithms
real analysis
real analysis unsolved

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125085[/url]

Shouldn't this be in the calculus forum?

[hide="hint"]

Let $ g(x) \equal{} \frac{f'(x)}{f(x)}$ Then $ g'(x) \equal{}0$

[/hide]

[quote="hsbhatt"][hide="hint"]

Let $ g(x) \equal{} \frac {f'(x)}{f(x)}$ Then $ g'(x) \equal{} 0$

[/hide][/quote] who do we know that f(x) isnt 0 ??

Hmm, we will need help here but it seems to me that these will be point discontinuities

plz i need to solve this exercise ! :maybe:

Don't know if this helps but about that $ \frac {f'(x)}{f(x)}$,

$ \frac {f'(x)}{f(x)} \equal{} \frac {f''(x)}{f'(x)}$

Integrate both sides.

$ \ln f(x)\equal{}ln(f'(x)$.

$ f(x)\equal{}f'(x)$

Also, $ f(x) \equal{} e^x$ satisfies all of those conditions.

[quote="Stank"]Not really a solution, but $ f(x) \equal{} e^x$ satisfies all of those conditions.[/quote] yeah i know that :/ (we dont know that f(x),f'(x),...f'''''''''..''(x) isnt 0 ...

Hello,

Since we are only interested in finding the value at zero we can just use the continuity of $ f$
to deduce that $ f$ is nonzero sufficiently close to zero. Therefore, with $ g\equal{}f'/f$ (which is
well defined close to zero), we get
\[ g'(x) \equal{}0 \qquad x\approx 0.\]
From this we have $ g(x)\equal{}c$, $ x\approx 0$ so $ f'(x)\equal{}cf(x)$. Differentiate this equality $ f''(x)\equal{}cf'(x)\equal{}c^2f(x)$ and so on.
we differentiate further and get $ f^{(2009)}(x)\equal{}c^{2009}f(x)$. For $ x\equal{}0$ we deduce that $ c\equal{}1$. The conclusion
is that $ f'(0)\equal{}1$.

I think that this proof is valid since, as I said, we have a neighborhood of 0 where $ f$ is nonzero.

Regards,
Hormander


The vertices of square $ EFGH$ lie on the edges of square $ ABCD$. $ \frac{AE}{EB} \equal{} \frac{1}{2}$. What is the ratio of the area of square $ EFGH$ to the area of square $ ABCD$?

[asy]pair A,B,C,D,F,G,H,I;
A=(0,10); B=(10,10); C=(10,0); D=(0,0);
I=(3.3,10); F=(10,6.7); G=(6.7,0); H=(0,3.3);
draw(A--B--C--D--cycle,linewidth(0.7));
draw(I--F--G--H--cycle, linewidth(0.7));
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$E$",I,N);
label("$F$",F,E);
label("$G$",G,S);
label("$H$",H,W);[/asy]

Assuming that $ AB \equal{} 3$ (any number in form of $ 3k$ works),

By Pythagorian Theorem we have that

$ EB^2 \plus{} BF^2 \equal{} EF^2$

$ 2^2 \plus{} 1^2 \equal{} EF^2$

$ EF \equal{} \sqrt {5}$

So the ratio of the areas is $ \frac {(\sqrt {5})^2}{3^2} \equal{} \boxed{\frac {5}{9}}$


What should the roght balance be between checking over problems already solved and tackling unsolved ones? Just to let you know, I'm not somebody very good. Basically, looking for a score from 105-120 AMC10, 90-102 AMC12, so I'm not someone who will, like, get everything done.

Depends on what you are taking and what your goal is. I you are taking 10 and your goal is to make the AIME I suggest that you do at least 21 problems and start checking your work because you can get 2 wrong and still pass with 120.

The sweet spot for qualifying for the AIME in the AMC 12 is to  answer 18 questions. That way you need to get 15 correct, as opposed to the minimum qualifying score which requires you to answer 14 correctly and skip the rest. Answering [15,17] questions means you still need to get 15 correct, so there is no reason to answer less. After attempting 18 then that's when I go back to check my work. This means you can still score up to 118.5 without taking too much of a risk.


first of all ...even if you're not writing in LATEX(neither am I :oops: ) you should have written it like this:
(a+1)/4 + (2a-3)/4 = a/2 - 2. if you multiply each member of the equality by 4(and sum up the left part) you'll get : 3a-2=2a-8 ...the answer is obviously a=-6.  :)

Oh, its (a+1)/4 not a+ 1/4, that really changes it, so *thinks in head for 1 second*, yep -6

[quote="Cyberman"]a+1/4 + 2a-3/4 = a/2 -2

i worked this problem out.. and all i could come up with was

1a-2/4 = -2

i looked in the back of the book. the answer is -6.. i don't know how to get the answer.. i don't know if the order of what i did was right.[/quote]

$a+\frac{1}{4}+2a-\frac{3}{4}=\frac{a}{2}-2$

$\text{Multiply both sides by 4}$

$4a+1+8a-3=2a-8$

$\text{Rearrange so a is all on one side}$

$10a=-6$

$a=-\left(\frac{3}{5}\right)$

[quote="Cyberman"]a+1/4 + 2a-3/4 = a/2 -2

i worked this problem out.. and all i could come up with was

1a-2/4 = -2

i looked in the back of the book. the answer is -6.. i don't know how to get the answer.. i don't know if the order of what i did was right.[/quote]

The first step is to remove fractions. They make your life hard. So multiplying thru by 4....
$4a+1+8a-3=2a-8$
Now make all your a's belong to R.S (right  side) and all your constants belong to L.S (left side) to get...
$10a=-6$
$a=-\frac{3}{5}$

DONE


$ Ito$  $ ang$ $ problem:$

A Pentagon is inscribed in a Hexagon. If they share a common side of length $ 2$ inches, find for the area within the Hexagon but outside of the Pentagon.

i think pag regular lang yung polygon then you can solve it

its easy to see that the area of hexagon is 6 times area of equilateral triangle with side 2
so 6(sqrt(3))

then yung area of pentagon is 5 times area of triangle with sides 72, 54,54 where the base is 2.
so the area would be 5 x 1/2 x 2(base) (tan 54)(height)..

so you just subtact the 2 from above..

if you are ask to solve what is tan54 without any calculator. you can start by sin 2a = cos 3a,  use identities to solve sina, you will have a 3 degree in sin a here.   :D

regards


What do you do if you've been studying a list of french words (it could be anything) to just memorize them, not the definitions, and you can't get all of them. Just so you know, I'm doing the "Dr. & Mrs. Vandertramp" verbs for french. My teacher will just write down:[code]
D
R
M
R
S
V
A
N
D
E
R
T
R
A
M
P[/code]
and you have to put the correct verb next to each letter.

What should I do?

-jorian

I took French for two years.  It wasn't bad considering how I'd already had one for Latin.  The mnemonic jhredsox mentioned is to remember a list of verbs that are conjugated irregularly.  I'm not taking it this year, so I forgot some of them, but a couple like "venir" (to come), and "rentrer" and some others.  redsox, they're conjugated differently in the past tense, right?  What's better is to always try to understand [i]why[/i] they're different or what they have in common that sets them apart, (a strategy that can be applied to languages, math, and sciences) so it's easier to remember them.  For sure, the meanings of the words are important; it's a foreign language.  Speaking of differences, don't all those verbs involve some kind of motion, or something dynamic going on?  I think that's what it is.

I did the Vandertrampp verbs last year. There are basically two ways to memorise them that my teacher presented - the mneumonic DR MRS VANDERTRAMPP (we had an extra P - passer, which is only a vandertrampp sometimes, if I remember correctly...) and the House of Être model, if you can do it in a house it is a Vandertrampp verb.  I found the house method vauge and it didn't work for me.  The mneumonic is how it got stuck in my head.

The best way for me to memorise it to write it out.  Looking at the original to make sure that you get it right, copy it out 4 or 5 times.  Then test yourself on it - just write out the vertical DR MRS VANDERTRAMPP and see if you can fill it in properly.  Do this until you get it, then say it out loud.  My weakest learning style is verbal, so if I can master something verbally then I can do pretty well on it.  Figure out how you learn best, learn it that way first, then master it in your weaker learning styles.

If you just have tonight, write it out a gazillion times until you can recreate it.  Good luck!


topology
function
real analysis
real analysis unsolved

define A Continuous function $ f$ as $ f: E\rightarrow\mathbb{R}$ .

Let $ (x_n)$ be a caushy sequnce of E. then :

if $ E$ is closed then $ f(x_n)$ is a caushy sequnce. .Also, give an counter example if $ E$ is not closed.

Thanx.  :)

This should be in the analysis section. Also, the name is Cauchy (always capitalized, two "c").

A set being "closed" has no independent meaning; in the metric space $ (\minus{}1,1)$, the full set $ (\minus{}1,1)$ is closed. Given this, your statement is false.
The way to fix it is to substitute "complete" for "closed". That means that Cauchy sequences converge, and the problem is easy. In general, continuous functions take convergent sequences to convergent sequences, while uniformly continuous functions take Cauchy sequences to Cauchy sequences. For the counterexample, look for a function that's continuous but not uniformly continuous.

Making it a function to $ \mathbb{R}$ is irrelevant; any metric space would work equally well there.

[b]Hello Jmerry :

you're right that must be in the Analysis section , and don't notice the C in Caushy I wrote it quickly. :wink:

Okay , you said that In full metric the $ (\minus{}1,1)$ is closed and that's give us false statement.

Did you mean by $ [ \minus{} 1,1]$ . , because I meant by closed set [definition from Topology].

did I misunderstanding? . Yes the problem is so easy if we write Complete instead of Caushy.

Clarify more please.? :)

[/b]

If $ E\equal{}(\minus{}1,1)$, then as a subset of $ \mathbb{R}$, this is not closed, but in the metric space $ E$, the set $ (\minus{}1,1)$ is closed (remember in any topological space the whole space is both open and closed along with the empty set).

[b]Thx HilbertThm90 .

I didn't study these subjects and I tried to study it alone .  good Information .

Again I'm so thankful guys  :)

[/b]


induction
algebra
polynomial
inequalities
strong induction
absolute value
triangle inequality

The sequence $ x_n$ is defined as follows:
$ x_1\equal{}x_2 \equal{}..\equal{}x_{1996}\equal{}0; x_{1997}\equal{}1;x_{n\plus{}1997}\equal{}\frac{x_n\plus{}x_{n\plus{}1}\plus{}..\plus{}x_{n\plus{}1996}}{1997}\forall n\geq1$
Prove that it has a limit and find it.

[quote="tdl"]The sequence $ x_n$ is defined as follows:
$ x_1 \equal{} x_2 \equal{} .. \equal{} x_{1996} \equal{} 0; x_{1997} \equal{} 1;x_{n \plus{} 1997} \equal{} \frac {x_n \plus{} x_{n \plus{} 1} \plus{} .. \plus{} x_{n \plus{} 1996}}{1997}\forall n\geq1$
Prove that it has a limit and find it.[/quote]

Not entirely convinced I'm right, still.

We claim the limit is 1.

Principle of Strong induction.

Base cases (it is given for the first 1997

Assume it is true $ \forall n \le k$

$ 1 \equal{} \frac {\underbrace{1 \plus{} 1 \plus{} \dots \plus{} 1\plus{}1}_{1997 \text{ 1s}}}{1997} \ge x_{k \plus{} 1}$

Thus it holds for all k by princple of strong induction

The limit is not $ 1$. It's strictly between zero and one, and it's tricky algebra to find it.

Since $ 1997$ is clearly just playing the role of a large integer here, call it $ N$. We have the recurrence $ Nx_{k \plus{} N} \minus{} x_{k \plus{} N \minus{} 1} \minus{} \cdots \minus{} x_{k \plus{} 1} \minus{} x_k \equal{} 0$, with associated polynomial $ Nt^N \minus{} t^{N \minus{} 1} \minus{} \cdots \minus{} t \minus{} 1 \equal{} (t \minus{} 1)(Nt^{N \minus{} 1} \plus{} (N \minus{} 1)t^{N \minus{} 2} \plus{} \cdots \plus{} 2t \plus{} 1)$. Since the two factors don't share roots, we can separate them; the solutions $ x_k$ to the original recurrence are of the form $ a_k \plus{} b_k$, where $ a_{k \plus{} 1} \minus{} a_k \equal{} 0$ and $ Nb_{k \plus{} N \minus{} 1} \plus{} (N \minus{} 1)b_{k \plus{} N \minus{} 2} \plus{} \cdots \plus{} 2b_{k \plus{} 1} \plus{} b_k \equal{} 0$. I claim that all such $ b_k$ tend to zero.
To prove this, we apply the general theory of constant-coefficient difference equations again: the solutions are of the form $ \sum_i p_i(k)r_i^k$ where $ r_i$ are the roots of the characteristic polynomial and $ p_i$ are polynomials with degree less than the multiplicity of $ r_i$. If all roots are less than $ 1$ in absolute value, this always goes to zero. By a simple triangle inequality estimate, $ |Nr^N|\ge|r^{N \minus{} 1} \plus{} \cdots \plus{} r \plus{} 1|$ if $ |r|\ge 1$, with equality only if $ |r| \equal{} 1$ and $ r,r^2,\dots,r^{N \minus{} 1}$ are all positive real multiples of $ 1$. The only root of the original polynomial on or outside the unit circle is $ 1$ with multiplicity 1; when we divide by $ t \minus{} 1$, the remaining polynomial has no such roots. This proves the claim.

We have proved existence of the limit; we must now calculate it by finding the constant sequence $ a_k \equal{} a$ with the initial conditions $ x_1 \equal{} x_2 \equal{} \cdots \equal{} x_{N \minus{} 1} \equal{} 0,x_N \equal{} 1$. We have $ x_k \minus{} a \equal{} b_k$, so
$ N(x_{N} \minus{} a) \plus{} (N \minus{} 1)(x_{N \minus{} 1} \minus{} a) \plus{} \cdots \plus{} 2(x_{2} \minus{} a) \plus{} (x_1 \minus{} a) \equal{} 0$
$ N \minus{} Na \minus{} (N \minus{} 1)a \minus{} \cdots \minus{} 2a \minus{} a \equal{} N \minus{} \frac {N(N \plus{} 1)}{2}a \equal{} 0$, and $ a \equal{} \frac {2}{N \plus{} 1}$. In the original problem with $ N \equal{} 1997$ (ten years old!), this is $ \frac1{999}$.

This method would work with any initial conditions; the limit is $ \frac2{N(N \plus{} 1)}\sum_{k \equal{} 1}^N kx_k$.


geometry
geometric transformation
rotation

Can somebody awesome help me find solutions to these?

1. Find the sum of the first 30 terms of the sequence 1,5,6,11,17,28,... if the 30th term is 2888956 and the 31st term is 4564429.

2. Each petal of a six-petal flower is to be painted one of two colors. In how many different ways can this be done? (Two flowers are the same only if one can be obtained from another by rotating the flower).

1. Think about how the sequence is constructed.

1+5+6=12=17-5
17-5+11=28-5.

Conjecture: Sum of first n terms is (term n+1) - 5.

Proven by induction.

4564429-5=4564424.

[hide]
Let's say the colors are red and blue.
6 reds: 6 ways
5 reds 1 blue: 6 ways
4 reds 2 blues: 6*5=30 ways
now double it to get 84.
3 reds 3 blues 6*5*4=120 ways.
120+84=204/6 (rotations)=34 ways.[/hide]

It's incorrect because of some rotation stuff that gets tricky in the "4 reds 2 blues" and "3 reds 3 blues" cases, at first glance.

I believe [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1474481#1474481]this solution[/url] is correct.


[hide] It's just traveling from the origin, to two points, then to the point $(5,6)$. We all know the shortest distance between $(5,6)$ and the origin will be a straight line, i.e. $\frac{b}{a}= \frac{c}{d}= \frac{6}{5}$. The shortest distance is then $\sqrt{ 5^{2}+6^{2}}= \boxed{ \sqrt{ 61 }}$ [/hide]

[hide="Solution"] a=b=c=d=0? so the answer would be $\sqrt{5^{2}+6^{2}}= \sqrt61$

[/hide]

I know it's wrong....

[quote=kohjhsd]Find the minimum value of
\[\sqrt{a^{2}+b^{2}}+\sqrt{(a-c)^{2}+(b-d)^{2}}+\sqrt{(5-c)^{2}+(6-d)^{2}}\]
for real numbers $a$, $b$, $c$, $d$.[/quote]
??


While I was using TeXnicCenter, I put the command \choose, but I saw a warning message, that said it was an undefined control sequence, but when I viewed it as a pdf, the thing showed up perfectly fine.

Why would it say undefined control sequence for \choose?

It also said that I actually typed \atopwithdelims, but I did no such thing.

\choose is an old plain-TeX command, and is not officially part of the current version of LaTeX.  It still works (to provide backward compatibility, so that old documents will still compile), but it will give a warning that you're using an outdated command.

You should use \binom{5}{3} rather than 5 \choose 3 to get a binomial coefficient.


Find all primes $ p$ such that both $ 4p^{2} \plus{} 1$ and $ 6p^{2}\plus{}1$ are primes.

If $ p \equal{} 1 or \minus{}1 (mod 5)$

$ 4p^{2}\plus{}1 \equal{} 0 (mod 5)$

If $ p \equal{} 2 or \minus{}2 (mod 5)$

$ 6p^{2}\plus{}1 \equal{} 0 (mod 5)$

Therefore one of the three numbers must be divisible by 5.

$ \boxed {p\equal{}5}$ is the only solution.


Mods: if you move this post, please leave the URL intact, if possible.  Otherwise, please pm me with the new URL.  I think that this is more of a college-level discussion, but it didn't seem to fit into any of the categories.

This topic is devoted to discussion on the nature of infinity, and is primarily for the users the math forum listed as my web site, which does not have $\LaTeX$, but anyone here should feel free to post as well.

Well, here's some trivial  :wink:  information that everybody already knows:

Where $n$ is a finite number, $\infty/n$, $\infty*n$, $\infty+n$, and $\infty-n$ all $=\infty$

Some less trivial information is that the properties attributed to $\infty$ vary based on the field of mathematics one is working in, that is, there is no overarching definition of $\infty$. :huh:  :o

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=aleph&t=43524

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=aleph&t=12962

Here are two pretty good topics devoted to the concept of infinity.

So, basically, we have two old topics that we can sift through :sleeping: , or we can have our own conversation.

Well what do you want to discuss. There isnt much really. most of these forums have some sort of question that gets ansered 20 times by everyone and then just left.
Any ideas?

Well, those two old topics do cover many points, at least.

Well, I'd like to say a few things.

First we have the ordinals. These are 1, 2, 3, 4, 5, 6, 7, 8, 9, ... or in words, first, second, third, forth, fifth, sixth, seventh, ... The first infinite ordinal is $\omega$. I could also go into a game theory aspect of these infinities, but I won't unless someone asks.

We also have the cardinals. These are also 1,2,3,4,5,6,7,8,9,... but in words they are one two three four five six seven etc. The first infinite cardinal is $\aleph_{0}$. We also have $\aleph_{1}, \aleph_{2}$ and so on. Now suppose we keep going up until we have the $\omega$th infinite cardinal. We'd have something like $\lim_{n\to\infty}\aleph_{n}= \aleph_{\omega}$. But then we can keep going up again and get $\aleph_{\omega+1}$ and $\aleph_{omega+2}$ and so on. Now suppose we perform another one of our limit operations. We can then get $\aleph_{2\omega}$. We can also get $\aleph_{\omega^{2}}$ and so on if we do stuff like applying our limit operations to our limit operations and doing crazy stuff like that.

But this isn't the extent of it! We can have even BIGGER cardinals!!!! These are the so called "unreachables" that can never be obtained by these addition and limit operations. Let's call the first unreachable $K$ or $K_{0}$. We can then apply our addition and limit operations AGAIN to get stuff like $K_{1}$ or $K_{\omega}$ or even $K_{3\omega^{5}+2\omega+56}$. But there's still MORE unreachables!!!!!!!!!!!!!!! We can continue this making bigger and bigger infinities!

So, basically what we have is that the infinities can get infinitely large, and infinitely larger an infinite number of times, right? :roll:  :surrender:  :wallbash:


linear algebra
matrix
Extremal combinatorics
combinatorics
IMO Shortlist

For what values of $ n$ does there exist an $ n \times n$ array of entries -1, 0 or 1 such that the $ 2 \cdot n$ sums obtained by summing the elements of the rows and the columns are all different?

Hi ! Can somebody prove that it`s impossible for odd $ n$ ?
I have a solution for even numbers (all even $ n$ have this property :wink: ) . I will post my solution.

[quote="rahulakaneo"]I have seen the solution from an external source. Shall I post it ?[/quote]
Please post it . :blush:
Thanks in advance .

Here you go. :)

Consider a matrix M[1..N][1..N] that satisfies the given property and let the arrays Row[1..N] and Col[1..N] denote the Row sums and the Col sums. Since the maximum value a row sum or a col sum can take is N and the minimum value is -N and there are only 2N+1 values that all the sums can take, exactly one value from the set {-N,-N+1,...,N-1,N} is missing. Since there are N+1 non-negative values and N+1 non-positive values in the set {-N,-N+1,...,N-1,N}, the set of 2N sums contain at least N non-negative and N non-positive values.

Now, note that exchanging any two rows or any two columns does not affect the set of distinct 2N sums. By permuting the rows and columns, we can obtain a matrix for which the row sums R[1],R[2],...,R[k] and C[1],C[2],...,C[N-k] are non-negative.

Clearly,

(Sum for i=1 to N) |R[i]| + (Sum for j=1 to N) |C[j]| >= (Sum for k=-N to N) |k| - N = N^2

But, the same summation can be rewritten as,

(Sum for i=1 to N) |R[i]| + (Sum for j=1 to N) |C[j]|

= (Sum for i=1 to k) R[i] - (Sum for i=k+1 to N) R[i] + (Sum for j=1 to N-k) C[j] - (Sum for j=N-k+1 to N) C[j]

= (Sum for all i such that i<=k) M[i][j] - (Sum for all i such that i>k) M[i][j] + (Sum for all j such that j<=N-k) M[i][j] - (Sum for all j such that j>N-k) M[i][j]

= 2.( (Sum for i=1 to k) (Sum for j=1 to N-k) ) M[i][j] - 2.( (Sum for i=k+1 to N) (Sum for j=N-k+1 to N) ) M[i][j] <=4k(N-k)

This yields N^2 <= 4k(N-k) ie (N-2k)^2 <= 0 which proves that N has to be even and k exactly half of it.

The latter part of the solution contains more mathematical signs and it isn't going to be easier for me to write it continuing in this way. Could someone please tell as to from where do I learn about Latex etc to be able to post with mathematical signs properly.

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 and [[LaTeX]].  (Note that you don't need to download anything to use LaTeX in the forums -- just enclose your code between dollar signs.)

I have an inductive proof for the existence for even $n$.I will prove a stronger result:

For any even $n$ there exist a $n*n$ matrix so that the sum of all its columns and rows produce numbers $-n,-(n-1),\dots , n-1$.The base of induction is trivial just check the below matrix:

$\begin{bmatrix}1&0\\-1&-1\\ \end{bmatrix}$

For the inductive step put the $n+2$-th row all $-1$'s.For the first column of $n+1$-th row put $0$ and for other entries of $n+1$-th row put all $1$'s.For the remaining entries of first column put $-1$'s and for the remaining entries of the $n+2$-th column put $1$'s.Now put the $n*n$ matrix obtained by induction in the remaining part.Here is an example of a constructed matrix for $n=4,6$ using induction:

$\begin{bmatrix}-1&1&0&1\\-1&-1&-1&1\\0&1&1&1\\-1&-1&-1&-1\\ \end{bmatrix}$

$\begin{bmatrix}-1&-1&1&0&1&1\\-1&-1&-1&-1&1&1\\-1&0&1&1&1&1\\-1&-1&-1&-1&-1&1\\0&1&1&1&1&1\\-1&-1&-1&-1&-1&-1\\  \end{bmatrix}$

I have no proofs for impossiblity for odd numbers I will post it as soon as I find it.

I couldn't find a solution for the second part myself so I just put @rahulakaneo solution in $LATEX$

[quote=rahulakaneo]Here you go. :)

Consider a matrix $M(n*n)$ that satisfies the given property and let the arrays $R_1,R_2 \dots R_n$ and $C_1,C_2,\dots C_n$ denote the Row sums and the Col sums. Since the maximum value a row sum or a col sum can take is $n$ and the minimum value is $-n$ and there are only $2n+1$ values that all the sums can take, exactly one value from the set $\{ -n,-n+1,\dots ,n-1,n \}$ is missing. Since there are $n+1$ non-negative values and $n+1$ non-positive values in the set $\{ -n,-n+1,\dots ,n-1,n \}$, the set of $2n$ sums contain at least $n$ non-negative and $n$ non-positive values.

Now, note that exchanging any two rows or any two columns does not affect the set of distinct $2n$ sums. By permuting the rows and columns, we can obtain a matrix for which the row sums $R_1,R_2,\dots ,R_k$ and $C_1,C_2,...,C_{n-k}$ are non-negative.

Clearly,

$\sum\limits_{i=1}^n |R_i| + \sum\limits_{j=1}^n |C_j| \ge  \sum\limits_{k=-n}^n |k| - n = n^2$

But, the same summation can be rewritten as,

$\sum\limits_{i=1}^n |R_i| + \sum\limits_{j=1}^n |C_j|= \sum\limits_{i=1}^k  R_i - \sum\limits_{i=k+1}^n R_i + \sum\limits_{j=1}^{n-k} C_j - \sum\limits_{j=n-k+1}^n C[j]= \sum\limits_{i \le k} M_{ij} - \sum\limits_{i>k} M_{ij} + \sum\limits_{j \le n-k} M_{ij} - \sum\limits_{j>n-k} M_{ij}= 2*\sum\limits_{i=1}^k(\sum\limits_{j=1}^{n-k} M_{ij}) - 2*\sum\limits_{i=k+1}^n(\sum\limits_{j=n-k+1}^n M_{ij}) \le 4k(n-k)$

This yields $n^2 \le 4k(n-k)$ ie $(n-2k)^2 \le 0$ which proves that $n$ has to be even and $k$ exactly half of it.

[/quote]

Well I have another construction for even n.
Maybe it's actually the same as Taha1381's(just swapping some rows and columns or mutiply -1 to everything)
but i think it's easier to understand.
for $n=2$ we take
$\begin{bmatrix}1&-1\\1&0\end{bmatrix}$
for $n=4$ we take
$\begin{bmatrix}1&0&-1&-1\\1&1&-1&-1\\1&1&0&-1\\1&1&0&0\end{bmatrix}$
for $n=6$ we take
$\begin{bmatrix}1&0&0&-1&-1&-1\\1&1&0&-1&-1&-1\\1&1&1&-1&-1&-1\\1&1&1&0&-1&-1\\1&1&1&0&0&-1\\1&1&1&0&0&0\end{bmatrix}$
and so on.
generally,for $n=2k$ for cell $(i,j)$ we put $-1+[j\le i]+[j\le k]$ here $[P]$ is $1$ if $P$ is true and $0$ otherwise.
then we can easily get sum of the columns are $2k,2k-1,\cdots,k+1,-k,-k-1,\cdots,1-2k$ and sum of the rows are $1-k,2-k,\cdots,0,1,2,\cdots,k$.
and they are pairwise distinct so we are done.

The answer is $n$ even only. Here is a construction for $n=6$ which readily generalizes:
$$\begin{bmatrix}\textcolor{red}{1}&-1&-1&-1&-1&-1\\1&\textcolor{red}{0}&-1&-1&-1&-1\\1&1&\textcolor{red}{1}&-1&-1&-1\\1&1&1&\textcolor{red}{0}&-1&-1\\1&1&1&1&\textcolor{red}{1}&-1\\1&1&1&1&1&\textcolor{red}{0}\end{bmatrix}$$
It's not hard to verify that this works.

Now we prove that $n$ even fail. By multiplying every element of the matrix by $-1$ if necessary we may assume that the missing row entry is $m\geq 0$; note that $n \in [-n,n]$.
We may readily swap whole rows and columns, so WLOG assume that the $k$ leftmost columns have nonnegative sum, and the $n-k$ rightmost columns have negative sum. W may further assume that the $n-k$ top columns have nonnegative sum and the $k$ bottom columns have negative sum. Truthfully, there's not an actual reason we do this, but it makes the rest of this solution easier to think about.
Split the grid into four regions:
[list]
[*] A top-left $n-k \times k$ region with $A$ with sum $a$
[*] A top-right $n-k \times n-k$ region $B$ with sum $b$
[*] A bottom-right $k \times k$ region $C$ with sum $c$
[*] A bottom-left $k \times n-k$ region $D$ with sum $d$
[/list]
Then double-counting over regions and rows,
$$2a+b+d=\frac{n(n+1)}{2}-m\text{ and } 2c+b+d=-\frac{n(n+1)}{2}.$$
Hence we require
$$2a-2c=n^2+n-m.$$
On the other hand, the LHS is bounded above by $4k(n-k)$ because the total area covered by regions $A$ and $C$ is $2k(n-k)$. Since $n$ is odd, this quantity is at most
$$4\cdot \frac{n-1}{2}\cdot \frac{n+1}{2}=n^2-1,$$
but $n^2+n-m\geq n^2$ which is a contradiction. Thus all $n$ odd fail. $\blacksquare$


Simplify: $ \frac{2^{n\plus{}4} \minus{} 2(2^n)}{2(2^{n\plus{}3})}$. Express your answer as a common fraction.

Question...
How did you get the numerator to 2^n +1(2^3 - 1)?
That part doesn't make sense...

[quote="flyingpanda321"]
How did you get the numerator to 2^n +1(2^3 - 1)?[/quote]

In the numerator, we have $2^{n+4}-2(2^n)$.

Let's rewrite this as $2^{n+4}-(2^1)(2^n)$.

We know that when we multiply exponential expressions with the same base, we add the exponents.
Therefore, the numerator becomes $2^{n+4}-2^{n+1}$

Now consider this. $2^{n+4}$ can be written as $(2^{n+1})(2^3)$. (See why?)

So now, we have $(2^{n+1})(2^3)-2^{n+1}$.

From here, we just use the reverse distributive property to take out that $2^{n+1}$ to get $2^{n+1}(2^3-1)$.

Hope that cleared things up a bit!


[b]1.[/b]Sa se demonstreze ca pentru diferite alegeri ale semnelor + si - din expresiile
\[ \pm 1 \pm 2 \pm 3 \pm ... \pm(4n \plus{} 1)
\]
se obtin numere impare mai mici sau egale cu $ (2n \plus{} 1)(4n \plus{} 1)$

si inca o problema poate chiar mai interesanta decat prima
[b]2.[/b] Fie n un numar natural arbitrar si k un numar intreg.Sa se demonstreze ca daca k admite o reprezentare de forma
\[ \pm 1^n \pm 2^n \pm ... \pm m^n
\]
, m natural, pentru o anumita alegere a semnelor + si -, atunci admite o infinitate de reprezentari de aceasta forma.

It's roumain...I hope you will understand...those problems are very nice

Sorry, I don't understand them !!!???
Please rewrite it in English !

1.Prove that for any diffrent chioces of the signs + and - from the expresions \[ \pm1 \pm3 \pm... \pm (4n\plus{}1)\], we obtain all odd numbers $ <\equal{}$ with (2n+1)(4n+1).

2.Let n be a nutural number and $ k \in Z$. Prove that if k admits o representation with the form \[ \pm 1^n \pm 2^n \pm ... \pm m^n\], with m a natural number, for a diffrent choice of the signs + and -, then admites an infinite number of solutions.

My English is not so good but I hope you will understand

[hide="1."]  If we set the signs to be all +, then we get $ 1 \plus{} 3 \plus{} \cdots \plus{} (4n \plus{} 1) \equal{} \sum_{i \equal{} 0}^{2n} 2i \plus{} 1 \equal{} (2n \plus{} 1)(2n) \plus{} (2n \plus{} 1) \equal{} \left(2n\plus{}1\right)^2$.  This is obviously odd for $ n \in \mathbb{Z}$.  But now we change the sign of some $ a_k$ from + to -.  This decreases the expression by $ 2a_k$, which is always even.  Hence the sum will always be odd.[/hide]

2. I've seen this one before, but alas the solution is lost to the abyss that is my memory.  I think induction on $ m$ is the way to go.


algebra
polynomial
linear algebra
matrix
complex analysis
complex analysis theorems

Is there a theorem that states given a Hurwitz polynomial $ p(z)$, and you truncate its low order tail and shift all the terms by the same order, you still get a Hurwitz polynomial?

Example:

$ p(z) \equal{} z^2 \plus{} 4z \plus{} 1$ All poles are on the OLHP, so it is Hurwitz.
$ q(z) : \equal{} z \plus{} 4$, all poles are still on the OLHP, so it is also Hurwitz.

It is not hard to prove, but does this property have a name?

I doubt it. The statement immediately follows from the general Routh-Hurwitz stability criterion in the matrix form (you just need to check fewer determinants), so I'll be extremely surprised if it has some other name associated with it.

What about chopping the higher order terms?

$ p(z) \equal{} z^{2} \plus{} 4z \plus{} 1$
$ q(z) : \equal{} 4z \plus{} 1$

Is there a common term that means this type of property like "telescoping", etc.

Well, if $ p(z)$ is Hurwitz, then so is $ z^{\text{deg\,}p}p(1/z)$ and vice versa, so it doesn't matter from which end to chop.

Suppose I have two polynomials $ p(z)$ and $ q(z)$, is there a standard way to tell whether the norm of $ \frac{p(z)}{q(z)}$ equals $ \frac{p(0)}{q(0)}$? (Assuming that the degree of $ q$ is greater than or equal to the degree of $ p$ and $ q$ is Hurwitz.)

i.e. Under what condition does:

$ \left\|\frac{p(z)}{q(z)}\right\| \equal{} \frac{p(0)}{q(0)}$ ?

$ H_{\infty}$ norm, as in the one on the right:
[img]http://www.mathworks.com/access/helpdesk/help/toolbox/robust/hincon72.gif[/img]


Note: Please keep this topic flame-free, and don't attack each other for their music choice.

Anyway, what are the songs you absolutely hate?

amen.

uhhh...

anything by sean kingston...i want to cringe every time i hear it

Ah so you're a classical geek. How can you tolerate metal but not rock? What I really hate is 99% of country. But I like rock. The most. Because it sounds good.

Soulja boy

That thing is played like every other second at my school.  Once or twice, I can ignore.  But it's like constant on the news each morning and stuff...  (Well, that's a hyperbole, but it still annoys me)

[quote="serialk11r"]Any rock, pop, rap, hip hop, country...metal is tolerable, so is techno. Other than that, I only like Liszt, Rachmaninoff, Chopin, etc. :)[/quote]Strange, a classical & metal fan.  Hmm...

I don't like anything hardcore, it's primarily just pointless noise.
I hate almost everything from the last 5 years, though the Black Eyed Peas weren't awful.
I like some modern-ish synthetic-type stuff, but not as much as classic 60s-70s rock.

No no no I don't like metal. I don't think its worth listening to, but it is more so than other stuff...by a little.
If I had an iPod (GASP) then I would fill it with my favorite classical stuff :D (but I HATE mozart, bach, hadyn or however its spelled, schumann).

Metal, Rock, Pop, Hip Hop, basically most of Mainstream music, but there are a few exceptions. A spoof on Soulja Boy is good, a spoof on Glamorous, White and Nerdy, and that's it for now.

I like Electric World.

[quote="ZhangPeijin"][youtube]oHg5SJYRHA0[/youtube][/quote]
The song you hate is "sorry this page isn't avaliable"...interesting music preferences.

Wow...
It works on YouTube, why not here?

Sorry, but I just hate [size=34]Fergie.[/size]

Hannah Montana.
Chris Brown.
Jonas Brothers.
And one more I forgot.

this topic is funny

[quote="miyomiyo"]
I don't like anything hardcore, it's primarily just pointless noise.
[/quote]
though hardcore sucks, i dont think thats the reason; actually go listen to some. generally, if you begin to think a certain genre is just "pointless noise," you should reconsider. people tend to not like listening to pointless noise; there definitely are reasons why some people like it
[quote]
I hate almost everything from the last 5 years, though the Black Eyed Peas weren't awful.[/quote]
i think you are assuming that the only music to be released in the last 5 years are from the US pop charts

[quote="hunter34"]Linkin fart.

And bands trying to be awesome but then failing (Example:Dragonforce).[/quote]
agreed
[youtube]jdMIQ1jfJm0[/youtube]

[quote="infinity4ever"]I like rock and metal.

I hate 110% of country.

Chris Brown, Sean Kingston, Justin Timberlake, Rihanna, etc. just plain tick me off.  I hate R&B.

umm, that's it? 1% of classical.

YEAH GO ROCK AND CLASSICAL FANS

techno:  annoying.  they all sound the same
classical:  annoying.  they all sound the same.
hiphop:  hate about 75% of it.  annoying.  they all sound the same.
country:  don't even ask.
rock:  WOOT
R&B:  eccckk.  i hate about 80% of it.
rap:  okay. gets annoying sometimes.  hate about 85% of it.

I like all of Avril Lavigne's old albums... back then, she actually knew how to SING.

basically, Linkin Park and Avril Lavigne.  I don't tolerate anything else except for maybe Three Days Grace... and some others..

Ironically most of my favorite songs are rock and rap.  like Walk it Out by DJ Unk ^^[/quote]
THIS POST IS FUNNY
especially the part about "hiphop is annoying but rap is okay"
also i get the feeling that you think avril lavigne and linkin park are good rock artist/bands

I'm sorry if this offends anyone but: BREAK AWAY FROM MAINSTREAM MUSIC!!!! THERE ARE WAY BETTER SONGS OUT THERE!!!

Dudes. I hate:
Country (But not southern rock. There is a difference)
Any band they show performances of on Disney Channel (Hannah Montana, Jonas Brothers, etc.)
Rap (Music is like candy: it's good as long as you throw out the w[rapper])
"You're Beautiful" by James Blunt
Any song that curses (besides anything by MCR, Papa Roach's "Last Resort", and Seether's "Fake It").

I like classic rock. THE BEATLES ROCK!

Candy can be good if you don't throw out the wrapper...but I do agree, rap is pretty terrible.

[quote="perfect628"]rap is pretty terrible.[/quote]

fo shizzle

Some raps are hilarious but I generally don't listen to them for pure entertainment.

i prefer hard metal or techno or classical rock. I hate rap...

i hate the song see you again by cyrus, and sailing away by styx.

[sarcasm]Turn your volume way up.[/sarcasm]

[youtube]atjByPSKTRA[/youtube]

I have a high tolerance level for different varieties/genres of music. Even...*gasp*...pop fluff! And even [i]mainstream[/i]. I do rather detest the mainstream music and concept has a whole, but some isn't half bad the first few times around.

I even have a few strands of tolerance for Han--no, I can't type the name. It would scortch my fingertips. And I rather like my fingerprints. And the Jona--. Ow. That hurt.*blows on fingertips*

I like stuff that's been listed.
'Cept a lot of country. I only listen to Taylor Swift and Rascl Flatts. Everything else is just a blur...

I haven't heard a lot of techno and opera. I hear those are quite good, though. I also haven't heard a lot of the American oldies stuff.

Musicals can be good. Some can be quite scary and traumatizing.....
Does anyone else here like the musical from certain musicals?

I like classic rock, especially Rush/Boston/Styx/etc.

I absolutely despise:
Disney Channel music (ick), country (sorry, but my sisters listen to it, so naturally I think it's evil), rap, pop, hip-hop (+ any other "modern"/"normal" kind of music)

I also like most jazz, and some classical music.  (only the more interesting pieces, though.  Some are really cool and different, and others just repetitive)

Oh, and Lemon Demon is awesome.  Same goes for any other kind of "weird" music (including Weird Al)

My philosophy is that excluding entire genres without any regard to the actual quality of specific songs is stupid.

I currently have the 1812 Overture, Rhapsody in Blue, the full Moonlight Sonata, Under the Sea, Holy Orders (Be Just or Be Dead), Bohemian Rhapsody, Do You Hear the People Sing?, Eruption, Gonna Make You Sweat, Legend of the Snake 2, Hello Muddah/Hello Faddah, Cottoneye Joe, Clocks, and My Heart Will Go On on my mp3 player. I'm looking for country, electronic, and hip-hop songs with clear lasting appeal; I have some in mind to go check out.

Expand your horizons, you might be surprised at what you like.

Oh, there are so many :roll:
[u]
Newest on hate-list:[/u] Pharoahe Monch - Simon Says

I was just talking to someone about this rather recently.
He was talking about how he liked...I think it was Smash Mouth before they signed with Disney. Proof that underneath it all, some Disney isn't entirely...

Are there other bands like this?


abstract algebra
superior algebra
superior algebra unsolved

Why $ \mathbb{Z}$ in not a semisimple ring?

We have a property that every ring $ R$ is (left) semisimple if and only if all of its left modules are projective.

But $ \mathbb{Z}\minus{}$module $ \mathbb{Q}$ is not projective (it is flat).


Let $ a,b,c$ be real numbers in the interval $ (0,1]$

Prove that
$ \frac{a}{bc\plus{}1}\plus{}\frac{b}{ca\plus{}1}\plus{}\frac{c}{ab\plus{}1}\le 2$

[quote="Rijul saini"]Let $ a,b,c$ be real numbers in the interval $ (0,1]$

Prove that
$ \frac {a}{bc \plus{} 1} \plus{} \frac {b}{ca \plus{} 1} \plus{} \frac {c}{ab \plus{} 1}\le 2$[/quote]

$ a,b,c$ symmetrical  ----> $ 0<a\leq b\leq c\leq 1$ ---> $ (1\minus{}a)(1\minus{}b)\geq 0$  <-->$ a\plus{}b\leq 1\plus{}ab$---> $ a\plus{}b\leq 1\plus{}ab<1\plus{}2ab$ --->
$ a\plus{}b\plus{}c\leq a\plus{}b\plus{}1< 1\plus{}2ab\plus{}1\equal{}2\plus{}2ab$-> and $ 1\plus{}ab\leq 1\plus{}ac\leq 1\plus{}bc$--->

$ \frac{a}{bc\plus{}1}\plus{}\frac{b}{ac\plus{}1}\plus{}\frac{c}{ab\plus{}1}\leq\frac{a}{1\plus{}ab}\plus{}\frac{b}{1\plus{}ab}\plus{}\frac{c}{1\plus{}ab}\equal{}\frac{a\plus{}b\plus{}c}{1\plus{}ab}<\frac{2\plus{}2ab}{1\plus{}ab}\equal{}2$

[quote="Rijul saini"]Let $ a,b,c$ be real numbers in the interval $ (0,1]$

Prove that
$ \frac {a}{bc \plus{} 1} \plus{} \frac {b}{ca \plus{} 1} \plus{} \frac {c}{ab \plus{} 1}\le 2$[/quote]

$ \frac {a}{bc \plus{} 1} \le\frac{2a}{a\plus{}b\plus{}c}$ so... :o


Prove that if I select any 10 positive 2-digit integers, then there will always exist two disjoint subsets having equal sums.

Here's my solution:

[hide] There are 2^10 total subsets.  The total number of possible sums is 990-100= 890, and this is less than 1024, so there are always going to be two subsets with the same sum. If these are not disjoint, then simply remove the members they have in common (so the sums will stay the same). [/hide]

Did i miss anything?


MIT
college
Harvard
Yale
videos
University of Chicago
number theory

Hi. Since yesterday was the results day, you should know by now if you were admitted.. so, which of you were accepted at Caltech?
I was not, but I'm still curious who was and who were'nt accepted, so I'm waiting for your responses.

I know two people who applied, both of them didn't get in I believe  :(

[quote="E^(pi*i)=-1"]I was accepted Early Action.

EDIT: Also, for those who were accepted, are you going to Prefrosh Weekend?  I'll be there.[/quote]

I'm going. :)

I got into Caltech EA. It was pretty exciting. However, my first choice is Harvey Mudd and I have yet to hear from them (first week of April). I see myself going there if it turns out to be cheap and I get rejected by HMC and MIT :D.
And also, I really, really want to go there for graduate school.

In what respect? How do you know there isn't some compelling reason [i]for that person[/i] to go to Harvey Mudd instead?

[quote="jvenezuela716"][quote="mathemonster"]yea i guess if HM is cheaper[/quote]

Hmm, I think you're making college seem like a black and white option. It's really not.[/quote]
yes your right, but i bet your going to CalTech

[quote="TZF"]Harvey Mudd is a liberal arts college.[/quote]
Not really. The Claremont Colleges have several colleges placed together on the same campus, with fairly free cross-enrollment opportunities. Each of the colleges has its own style and emphasis. The college that comes closes to what most people mean when they say "liberal arts college" is Pomona College. Harvey Mudd is the one college of the Claremont group which has an engineering school, and it has a heavy science/engineering emphasis.

[quote="mathemonster"]
yes your right, but i bet your going to CalTech[/quote]

That's quite a presumption from someone who I've never even met. And of course your "bet" is completely worthless since I couldn't even profit from it should you be wrong.

It's certainly reasonable for someone to prefer Harvey Mudd over Caltech for various personal reasons, but I'm actually curious about what those reasons actually are.  I only ask because as far as head-to-head comparisons go, these schools are more easily comparable than most pairs of schools, and for the most part, the comparison seems to favor Caltech.  Sort of how Harvard and Yale are comparable, and the comparison seems to favor Harvard.  http://www.instantrimshot.com/

[quote="E^(pi*i)=-1"]As Dr. Merryfield said, it is very easy to cross-enroll from Mudd at the other Claremont Colleges, meaning that Mudd would probably allow for a much stronger humanities education than Caltech would.[/quote]
but his is aops haha so......

You sound like me, which is why I currently prefer UChicago to MIT and Caltech (although I'm still very much undecided).

Don't know if anyone cares, but I feel like completing my story. The financial aid info came in, and Caltech is much cheaper (as in $ \$$10,000 cheaper, per year), then either UChicago or Harvey Mudd. So guess where I am going next year :) My parents sure are much happier, and if I made it seem like Caltech was totally not for me, that was unintentional. I should end up being very happy there (and actually, when I visited, it just seemed perfect, especially when I saw the dry erase boards with Maxwell's equations scribbled on them).

[quote="stuck"]The financial aid info came in, and Caltech is much cheaper (as in $ \$$10,000 cheaper, per year), then either UChicago or Harvey Mudd.[/quote]

Yeah, I'm not sure why people on collegeconfidential and even teachers at my school bash all the "good" schools for being so expensive. My Caltech financial aid package also has the cost of attendance far below that of UChicago, WPI (by a long shot), and UConn. And MIT is even cheaper than Caltech.

[quote="stuck"]So... why? I'll try to break it down.

I'm a very well-rounded student. My desk has books on number theory and Kierkegaard, quantum mechanics and Kafka. I like to read, and I think that having a school that encourages such a personality would be perfect for me. To some, it may seem like a burden and even be frowned upon (Erdos apparently never read a book). But I think I have become too enmeshed in philosophy and literature to satisfyingly ignore it.

There is also the undergrad vs. grad distinction. Caltech places more of an emphasis on graduate students (and understandably so). But Harvey Mudd has its professors focus solely on the undergrad students for both teaching and research. This has both benefits and drawbacks, depending on how you play it. I honestly don't know how this will turn out.

I visited both schools, and I liked the environments at both. Having stayed overnight at Harvey Mudd, I know more about it, so maybe my view is a bit biased, or at the least incomplete. The dorm structure is wonderful there, each has its own personality, and I'm a person of multiple personalities. I can play video games at the nerd dorm, party at the party dorm, play sports with with jock dorm, or shut myself away and work on a physics problem without any difficulty.

So there you are. Three solid reasons as to why Harvey Mudd works for me. I know I would still fit in well at Caltech, but I sincerely don't expect anyone to share my view. Different strokes for different folks ;)

And also, apparently the hardest humanities classes are at Mudd itself. ;)[/quote]

What athletes at HM do you consider jocks rofl


[quote="espark52"]There are posite inteers k, n, and m such that $\frac{19}{20}< \frac1k+\frac1n+\frac1m < 1$. what is the smallest possible value of $k+n+m$?[/quote]
Um... one is 1/2, 1/3, and 1/7. Aren't there a lot?

yes, there are a lot. the smallest oout of the many

hehe cognos stole the answer from mathmaster GOOD JOB!

[quote="espark52"]There are posite inteers k, n, and m such that $\frac{19}{20}< \frac1k+\frac1n+\frac1m < 1$. what is the smallest possible value of $k+n+m$?[/quote]

[hide]For convenience, let the numbers $k,n,m$ be
$k \le n \le m$. Then $\frac1k \ge \frac1n \ge \frac1m$.

1)$k=2$.

Then in order for the sum to be greater than $\frac{19}{20}$, other two needs to have sum of $\frac{9}{20}$ but less than $\frac{1}{2}$.
--If $n=3$, then from inequality, smallest possible value of $m$ is $m=7$. This yields the sum of $12$. (Other possible value of $m$ is $8$ only.)
--If $n=4$, then $m \ge 4$, but the sum of fractions equal $1$ when $m=4$ and $\le \frac{19}{20}$ when $m \ge 5$.
--If $n \ge 5$, then the greatest possible value of $\frac1k+\frac1n+\frac1m$ (when $m=5$) is less than $\frac{19}{20}$ and thus impossible.

2)$k=3$.

  --If $n=3$, then it is impossible since there are no such $m$ that satisfy the original inequality. (try $m=3,4$)
  --If $n \ge 4$, then it is also impossible since $m=4$ yields sum less than $\frac{19}{20}$.

3)$k \ge 4$. Then because $\frac{3}{4}$ is less than $\frac{19}{20}$, it is impossible.

Therefore the only possible ordering of $(k,n,m)=(2,3,7)$ and $(2,3,8)$ and the smallest sum possible is $2+3+7=\boxed{12}$.[/hide]

[quote="cognos599"]correct me if im wrong but is the answer[hide]12[/hide][/quote]

that is correct


function
limit
integration
inequalities
probability
real analysis
real analysis unsolved

Let $ \phi$ be a finite valued non-negative increasing function on $ [0,\infty)$ such that $ \lim_{t\to\infty}\frac{\phi(t)}{t}\equal{}\infty$.

Let $ M\equal{}\{f\in L^1[0,1] : \int_0^1\phi( |f|)dt\leq 1\}$.

Show that for every $ \epsilon>0$ there exists $ \delta>0$ such that $ \int_E |f|dt<\epsilon$ for all $ f\in M$ and $ m(E)<\delta$ (i.e. M is uniformly integrable).

My instinct tells me that Jenson's inequality is involved.

i think it's a result due to de la vallee poussin or something.  i don't remember the proof but i think it's in a lot of probability texts.  searching google for 'poussin uniform integrability' should yield some results.

See e.g. http://www.math.gatech.edu/~carlen/6327F06/uniformint2.pdf (theorem 5).


geometry
circumcircle
parallelogram
trigonometry
cyclic quadrilateral
geometry solved

in an arbitrary triangle ABC, denote by O the circumcenter, H the orthocenter, M,N,P the midpoints of BC, AC, and respectively AB. Let A'' be the midpoint of AH. Prove that

a) A''M congruence to AO;

b) the quadrilateral APON is cyclic

c) the circumcircle of APON has the same radius as Euler's nine-point circle in triangle ABC

[color=blue][b](a)[/b][/color] Put a line from the vertex $C$ through the circumcenter $O$ intersecting the circumcircle at another point $D$. The lines $BD \parallel AH$ are parallel (both perpendicular to $BC$) and the lines $BH \parallel AD$ are also parallel (both perpendiculr to $CA$). Hence the quadrilateral $ADBH$ is a parallelogram and $AH = BD$. The angles $\measuredangle CAB = \measuredangle CDB$ are equal (spanning the same arc $BC$ of the circumcircle $(O)$. But from the right angle triangle $\triangle BCD$, $BD = 2R \cos{\widehat{CDB}}$. As a result, $AH = 2R \cos {\widehat{CAB}}$

The lines $AH \equiv A''H \parallel OM$ are parallel, because they are both perpendicular to the side $BC$. In addition, the segments $A''H = OM$ are equal, because

$A''H = \frac{AH}{2} = R \cos {\widehat{CAB}}$

$OM = OB \cos{\widehat{OMB}} = R \cos{\frac{\widehat{BOC}}{2}} = R \cos{\widehat{CAB}}$

The quadrilateral $A''HMO$ is therefore a parallelogram. Hence, its diagonals cut each other in half at the midpoint of the segment $OH$, which is identical with the center of the 9-point circle. Since the points $A'', M$ both lie on the 9-point circle, $A''M$ is a diameter of the 9-point circle, i.e., $A''M = R$ is equal to the circumradius $R$ (see (c) for details).

[color=blue][b](b)[/b][/color] The circumcenter is the intersection of the perpendicular bisectors $OM, ON, OP$ of the triangle sides. The angles $\measuredangle APO = \measuredangle ANO = 90^o$ are both right, their sum is $180^o$, so the quadrilateral $ANPO$ is cyclic.

[color=blue][b](c)[/b][/color] The 9-point circle is the circumcircle of the medial triangle $\triangle MNP$, which has half the sides of the triangle $\triangle ABC$. So, the 9-point circle radius is half the circumradius $R$. The segment $AO = R$ is the circumradius and also the diameter of the circumcircle of the quadrilateral $ANPO$. Hence, both the quadrilateral circumcircle and the 9-point circle have radii equal to $\frac R 2$.

[quote="pestich"]It would seem that before the notion of  COS  came into the open and became
    officially accepted on  Aug. 14, 1347,  there were doubts about congruency and
    cyclicity of the  items discussed in this topic.

    Maj.  Pestich[/quote]

Hopefully, any doubts still lingering in your mind have now been cleared, major.

Before this discussion gets moved into Solved Problems, let me show a synthetic solution of the problem:

Being the circumcenter of the triangle ABC, the point O must lie on the perpendicular bisectors of its sides CA and AB. Thus, $ON\perp CA$ and $OP\perp AB$, since the points N and P are the midpoints of the segments CA and AB, respectively. Hence, < ONA = 90° and < OPA = 90°; consequently, the points N and P lie on the circle with diameter OA. Thus, the quadrilateral APON is cyclic. Problem [color=blue][b]b)[/b][/color] is solved.

Since the points P and M are the midpoints of the sides AB and BC of triangle ABC, we have PM || CA. In other words, PM || AN. Similarly, NM || AP. Thus, the quadrilateral APMN is a parallelogram. This yields that the triangles APN and MNP are congruent. Hence, the circumcircles of these triangles have equal radius. But the circumcircle of the triangle APN is the circumcircle of the cyclic quadrilateral APON, and the circumcircle of triangle MNP is the nine-point circle of triangle ABC (since the points M, N and P lie on the nine-point circle of triangle ABC). Hence, the circumcircle of the cyclic quadrilateral APON and the nine-point circle of triangle ABC have equal radius. This solves problem [color=blue][b]c)[/b][/color].

Since the points A" and P are the midpoints of the sides AH and AB of triangle AHB, we have A"P || BH. But since the point H is the orthocenter of the triangle ABC, the line BH is an altitude of triangle ABC, and thus $BH\perp CA$. Consequently, $A^{\prime\prime}P\perp CA$. Finally, since we know that PM || CA, this becomes $A^{\prime\prime}P\perp PM$. Thus, < A"PM = 90°. Similarly, < A"NM = 90°. This implies that the points P and N lie on the circle with diameter A"M. In other words, the circumcircle of triangle MNP is the circle with diameter A"M. On the other hand, we know that the points N and P lie on the circle with diameter OA, so that the circumcircle of triangle APN is the circle with diameter OA. Since, as seen above, the triangles APN and MNP are congruent, their circumcircles have equal diameters, and thus, it follows that the diameter OA of the circumcircle of triangle APN is equal to the diameter A"M of the circumcircle of triangle MNP. So we have proven that OA = A"M. And problem [color=blue][b]a)[/b][/color] is solved.

  Darij


projective geometry
Euler
geometry proposed
geometry

Let $ ABC$ be a triangle. $ M$ and $ N$ are the midpoints of the sides $ AB,AC$, respectively. $ D$ and $ E$ are the feet of the perpendiculars from $ C$ and $ B$ to $ AB, BC$, respectively. Denote by $ A'\equal{}DN \cap EM$. $ B'$ and $ C'$ are defined similary. Prove that the points $ A'$, $ B'$, $ C'$ are collinear.

apply pappus on the feuerbach circle and the thesis follow.

It's also well-know that this 3 points lies on the euler line of ABC.

Dear Mathlinkers,
a proof (like Gabriel) but with not the Feuerbach's circle, and a developpment can be seen on my website : http://perso.orange.fr/jl.ayme  vol.2 Les deux points de Schroeter.
Sincerely
Jean-Louis

[quote="moldovan"]Let $ ABC$ be a triangle. $ M$ and $ N$ are the midpoints of the sides $ AB,AC$, respectively. $ D$ and $ E$ are the feet of the perpendiculars from $ C$ and $ B$ to $ AB, BC$, respectively. Denote by $ A' \equal{} DN \cap EM$. $ B'$ and $ C'$ are defined similary. Prove that the points $ A'$, $ B'$, $ C'$ are collinear.[/quote]
[hide][img]http://i209.photobucket.com/albums/bb172/Ineq/Pascal.jpg[/img][/hide]
Fig
Apply Pascal on the Euler circle we have desired!
:D


Prove that 4 perfect squares(all distinct) cannot form an arithmetic progression!!!!

pls help me!!

Solution:

Lets say that there exist a^2,b^2,c^2 and d^2 that verify the conditions with a<b<c<d. Then we have:

a^2+c^2=2b^2 and b^2+d^2=2c^2

WLOG we can say that a,b,c,d are are odd. We have
a^2(2c^2-b^2)=d^2(2b^2-c^2) which gives:
2(a^2c^2-b^2d^2)=a^2b^2-c^2d^2.

we can now denote: ac=x, bd=y, ab+cd=2z, ab-cd=2w. with these notations we have:

x^2-y^2=2zw and xy=z^2-w^2 which gives us:

x^4-x^2y^2+y^2=(z^2+w^2)^2.

Lemma: the natural solutions of the equation x^4+x^2y^2+y^2=alpha^2 is given by (x,y,alpha)=(k,0,k^2) and (0,k,k^2) and (k,k,k^2) with k from Z+. And from it we get that a) xy=0 or b)x=y.

a) is impossible
b) x=y implies that w=0 which gives: ab=cd which again is impossible cause: a<b<c<d.

Hope it is correct!!! cheers every1  :D


The altitude of an equilateral triangle is $ \sqrt6$ units.  What is the area of the triangle, in square units? Express your answer in simplest radical form.

What's the formula when it's a 30-60-90 triangle? I forgot :D

[hide][asy]draw((0,0)--(.5,sqrt(3)/2)--(.5,0)--cycle);
label("$x$",(.25,0),S);
label("$x\sqrt{3}$",(.5,.4),E);
label("$2x$",(.25,.5),W);[/asy]

Sorry my asymptote stinks.

Anyway, the side opposite the 30 degree angle is x, the side opposite the 60 degree angle is $ x\sqrt{3}$, and the hypotenuse is 2x.[/hide]

Whenever you forget this, you can always use Pythagorean Theorem to rederive it.

[hide]Drawing an altitude of an equilateral triangle splits it into two 30-60-90 right triangles: [asy]
unitsize(0.6inch);
pair A, B, C, F;
A = (0,1);
B = rotate(120)*A;
C = rotate(120)*B;
F = foot(A,B,C);
draw(A--B--C--A,linewidth(1));
draw(A--F);
label("$A$",A,N);
label("$B$",B,S);
label("$C$",C,S);
label("$M$",F,S);
[/asy]

The altitude is the longer leg of each 30-60-90 triangle, so $AM = \sqrt{3}\cdot BM$ in the diagram above. Since $AM = \sqrt{6}$, we have \[BM = \frac{AM}{\sqrt{3}} = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{\frac63} = \sqrt{2}.\] Therefore, we have $BC = 2BM = 2\sqrt{2}$, so the area of the triangle is $(BC)(AM)/2 = (2\sqrt{2})(\sqrt{6})/2
=\sqrt{12} = \boxed{2\sqrt{3}}$ square units.[/hide]

[hide][quote=AIME15][asy]draw((0,0)--(.5,sqrt(3)/2)--(.5,0)--cycle);
label("$x$",(.25,0),S);
label("$x\sqrt{3}$",(.5,.4),E);
label("$2x$",(.25,.5),W);[/asy]

Sorry my asymptote stinks.

Anyway, the side opposite the 30 degree angle is x, the side opposite the 60 degree angle is $ x\sqrt{3}$, and the hypotenuse is 2x.[/quote][/hide]

Also, isn't the side opposite to the $30$ degree angle, the smallest.


USAMTS inequalities trigonometry algebra polynomial

For those of you who sent the proof (Jesusfreak197, dk2, and Andreas), you may talk about the problems. The result of the proof will be out around sometime this weekend.

After careful grading which took about 2 hours, I found that all solutions I received deserve 5/5 for both questions so the result is that we'll have three people qualifying to Championship Round. Now, here are ways to do the problems.

Number one:

I personally think in any Olympiad-looking problems, you should first see if expansion helps. If you know Holder's or other things but sometimes, simple AM-GM can do the job (I remember seeing on USAMTS problem inequality problem that can be solved with simple AM-GM --> a lot faster than using Cauchy if I remember correctly).

First of all, $r$ is just fancy looking thing in this problem. If you expand and cancel, the inequality is actually:

$\left(1 + \frac{a}{b} \right)\left(1 + \frac{b}{c}\right)\left(1 + \frac{c}{a}\right) \geq 2 + \frac{2(a+b+c)}{\sqrt[3] {abc}}$

Now, here is my proof but I'm not *quite* sure if this will work out.

Expand LHS (you should get 2 1's) and you should get $\frac{a}{b} + \frac{b}{c} + \frac{a}{c} + \frac{c}{a} + \frac{c}{b} + \frac{b}{a}$. From here, we can directly approach AM-GM to get the minimum is 6 when $a=b=c$. Now, to prove the given inequality, we consider RHS, which is $\frac{2(a+b+c)}{\sqrt[3] {abc}}$.

From AM-GM and writing the expression as three terms (we'll leave two out for moment), we have $\frac{a}{\sqrt[3] {abc}} + \frac{b}{\sqrt[3] {abc}} + \frac{c}{\sqrt[3] {abc}} \geq 3(1) = 3$ so the whole RHS will be at least 6.

So, if you connect those, it works out but I'm not quite confident. Maybe someone can tell if this is rigorous? Thanks.

On the other hand, here is another proof by actual contestant.

[quote="JesusFreak197"][b]1.[/b] $r+\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=r+2+\frac{a}{b}+\frac{b}{a} +\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}$
$2\left(1+\frac{a+b+c}{\sqrt[3]{abc}}+\frac{r}{2}\right)=2+r+\frac{2(a+b+c)}{\sqrt[3]{abc}}$
Replacing these into the inequality, we simplify to $\frac{a}{b}+\frac{b}{a}+\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+ \frac{c}{b}\ge \frac{2(a+b+c)}{\sqrt[3]{abc}}$. The LHS simplifies to $\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}$. Multiply both sides by $abc$ to get $a^2b+ab^2+a^2c+ac^2+b^2c+bc^2\ge 2(\sqrt[3]{(abc)^2})(a+b+c)$. AM-GM of the LHS yields $a^2b+ab^2+a^2c+ac^2+b^2c+bc^2\ge 6abc$. We must have $6abc\ge 2(\sqrt[3]{(abc)^2})(a+b+c)$, which gives $3\sqrt[3]{abc}\ge a+b+c$, which is only possible with equality. However, this does not apply to the original equation; rather, this means that $6abc=2(\sqrt[3]{(abc)^2})(a+b+c)$, so we have $a^2b+ab^2+a^2c+ac^2+b^2c+bc^2\ge 2(\sqrt[3]{(abc)^2}(a+b+c)$ as desired. [/quote]

Number two:

This problem is a trig problem that I accidentally found while punching calculator.  :D I thought it would be nice to show the proof though. The "key" idea to this problem to me was to use De Moivre's to get angle for $\cos (3 \theta)$ but again, it can be solved through trig simplification.

[quote="Andreas"]
2) a scolastic approach
Rewrite the expression as:

$\frac{\cos 160\cos 80 + \cos 160\cos 40 + \cos 80\cos 40}{\cos 160\cos 80\cos 40}$

We know that $\cos a\cos b = \frac{1}{2}(\cos (a + b) + \cos (a - b)$.

$\frac{\frac{1}{2}(\cos 240 + \cos 80) + \frac{1}{2}(\cos 200 + \cos 120) + \frac{1}{2}(\cos 120 + \cos 40)}{\cos 160\cos 80\cos 40} =$

$= \frac{-\frac{3}{4} + \frac{1}{2}(\cos 80 + \cos 40 + \cos 200)}{\cos 160\cos 80\cos 40}$

We know that $\cos a + \cos b = 2\cos \frac{a + b}{2}\cos \frac{a - b}{2}$.

$\frac{-\frac{3}{4} + \frac{1}{2}(2\cos 60\cos 20 + \cos 200)}{\cos 160\cos 80\cos 40}$

$\frac{-\frac{3}{4} + \frac{1}{2}(\cos 20 + \cos 200)}{\cos 160\cos 80\cos 40}$

$\frac{-\frac{3}{4} + \frac{1}{2}(2\cos 110\cos 90)}{\cos 160\cos 80\cos 40}$

$\frac{-\frac{3}{4}}{\cos 160\frac{1}{2}(\cos 120 + \cos 40)}$

$\frac{-\frac{3}{2}}{-\frac{1}{2}\cos 160 + \frac{1}{2}(\cos 200 + \cos 120)}$

$\frac{-\frac{3}{2}}{\frac{1}{2}(\cos 200 - \cos 160) - \frac{1}{4}}$

We know that $\cos a - \cos b = -2\sin \frac{a + b}{2}\sin \frac{a - b}{2}$.

$\frac{-\frac{3}{2}}{-(\sin 180\sin 20) - \frac{1}{4}}$

$\frac{-\frac{3}{2}}{-\frac{1}{4}} = 6$.[/quote]

Below method is same as what I did using polynomial. Go polynomial!  :lol:

[quote="JesusFreak197"]
[b]2.[/b] $\cos 120=\cos 240=\cos 480=-\frac{1}{2}$, and $\cos 120=\cos 3(40)=4\cos ^340-3\cos 40=-\frac{1}{2}\Rightarrow 8\cos ^340-6\cos 40+1=0$, so, by the same method, $\cos 80$ and $\cos 160$ are also roots of the given cubic. From there, we have $\frac{8}{\sec ^340}-\frac{6}{\sec 40}+1=0\Rightarrow \sec ^340-6\sec ^240+8=0$, and the sum of the roots is the negative of the coefficient of the $\sec ^2$ term, so our answer is $\boxed{6}$, and we are done.[/quote]

Congratulation again to dk2,Jesusfreak197, and Andreas for qualifying to Championship Round. The Championship Round problems will be posted sometime later (I'm working on this) but it will be posted in Pre-Olympiad forum due to couple problems that will be too hard for Intermediate.

Silverfalcon


I dunno.. im new to $LaTeX$.. however you write that symbol out.. but Texnic wont work for me anymore. I cant build or do much of anything except write code that i cant use in it. Any ideas on what i should do?

Judging from the word "anymore," am I correct in saying that you were working on something and all of a sudden, when you tried to compile it as usual, it had a fatal error and wouldn't work anymore? If so, close texnicenter and reopen it. It will work fine. (This has happened to me, so don't worry too much.)

BTW, it is typed as $\text{\LaTeX}$ (or $\text{\TeX}$ or $\text{\LaTeX}e$)


Let $ n\geq 2\in \mathbb{Z}$ and let $ k$ be any constant such that $ 0<k<n(n\minus{}2)/2$. If $ 0<a_{1},a_{2},\ldots ,a_{n}<1$ and $ \sum_{1\leq i\leq y\leq n}a_{i}a_{j}\equal{}k$, then prove that $ \sum_{i\equal{}1}^{n}\frac{a_{i}}{1\minus{}a_{i}^{2}}\geq \frac{n\sqrt{2kn(n\minus{}1)}}{n(n\minus{}1)\minus{}2k}$. Also, find when the equality holds.

I suppose you meant the sum $ \sum a_ia_j$ to be taken over all $ i,j$ satisfying $ 1\leqslant i<j\leqslant n$, not $ i\leqslant j$, i.e., we have to prove
\[ \sum_{i\equal{}1}^{n}\frac{a_{i}}{1\minus{}a_{i}^{2}}\geq\frac{n\sqrt{2n(n\minus{}1)\sum_{1\leqslant i<j\leqslant n}a_ia_j}}{n(n\minus{}1)\minus{}2\sum_{1\leqslant i<j\leqslant n}a_ia_j}\]
with $ a_i \in (0;1)$.

MacLaurin-Inequality states that
\[ \frac{\sum a_i}{n}\geqslant\sqrt{\frac{\sum_{1\leqslant i<j\leqslant n}a_ia_j}{\binom{n}{2}}}\mbox{.}\]
From this, we get
(1) $ (n\minus{}1)\sum a_i\geqslant\sqrt{2n(n\minus{}1)\sum_{1\leqslant i<j\leqslant n}a_ia_j}$ and
(2) $ \frac{(n\minus{}1)\left(\sum a_i\right)^2}{n}\geqslant 2\sum_{1\leqslant i<j\leqslant n}a_ia_j$.
Plugging this into the right-handed side of the inequality, we have to prove
\[ \sum \frac{a_i}{1\minus{}a_i^2}\geqslant \frac{\sum a_i}{1\minus{}\left(\frac{\sum a_i}{n}\right)^2}\mbox{,}\]
or eqivalently
\[ \sum\frac{\frac{a_i}{1\minus{}a_i^2}}{n}\geqslant \frac{\frac{\sum a_i}{n}}{1\minus{}\left(\frac{\sum a_i}{n}\right)^2}\mbox{,}\]
which is clear from Jensen's Inequality because $ f(x)\equal{}\frac{x}{1\minus{}x^2}$ is convex on $ [0;1]$.
Equality occurs if and only if all $ a_i$ are equal.
(all sums not specified are taken over all $ i$ satisfying $ 1\leq i\leq n$)
 :)

Job well done Mr. Martin N.!...I've proved this inequality for n = 3 using simple trigonometric substitutions and then made a generalization..c",)


Princeton
college
email
MIT
Harvard
IMC
college contests

Princeton is sending a team! I am going! :D:D

What other AoPSers might I meet this summer?  :lol:

Also I get 1st prize on IMC 2006 but it seems I can't be in MSU team this year.

[quote="puuhikki"]I would like to go but I don't know how can I take part in that competition. My university has not given any information about that competition.[/quote]
But you can do it yourself! Princeton's team was arranged by a student. He only went to the Math Dept when he wanted some money from them.
http://www.imc-math.org/
http://www.ucl.ac.uk/~ucahjej/

Has anybody already received an official invitation from organizers this year?

I can't participate. I have studied five years in university.[quote="IMC"]
The competition is planned for students completing their first, second, third or fourth year of university education[/quote]

I have taken part in IMC 2 times 2005 and 2006. Although I am in the 4th year of university but I am not comming this year.
I bet you'll enjoy this trip!
Have fun
Ahmad

[quote="ZetaX"]I will come together with NoName and Qwerwulff  :)[/quote]
Good luck,Zetax.
But it seems no Chinese go to IMC. :(

sharif university of technology selected a team and send it to compete:1)Erfan Salavati 2)Mohammad Gharakhani 3)Ali Akbar Daemi 4)Nima Ahmadipour 5)Naser TalebiZadeh 6)Behzad Mehrdad 7)Rouzbeh Farhoudi 8)Sam Nariman

University of Bucharest is sending the following team:

   1. Claudiu Raicu
   2. Alexandru Petre Tache
   3. Cezar Lupu
   4. Dragos Fratila
   5. Diana Stan

From Princeton:

1. Kenneth Tay
2. Eve Waddington
3. Andrei Negut
4. Andrei Ungureanu (?)
5. Aleksander Lishkov (?)
6. Mihnea Stangescu-Petre
7. Simion Filip
8. Nathan Savir  :ninja:

I believe some of you may be friends with a few of these people. ;)

---
I am actually not 100% sure whether Andrei Ungureanu or Alex Lishkov are coming; according to the last email I got from the organizer (which was like 10 weeks ago), they were both awaiting approval from their summer employers.

Our team leader is Balin Fleming, a grad student.

how come K. is not coming?

colombians, are you coming?

Hi,  I have a question.

I heard that IMC is an Individual event, is it possible to register personally? (without team leader).

thx.

[quote="Beginner"]I heard that IMC is an Individual event, is it possible to register personally? (without team leader).[/quote]
I think this is possible. But you will have to pay the registration and travel fees yourself then.

You could try contacting that guy Dr. Jayne, see the link I posted above.

We have not yet solved all questions with visas but team of Kyiv Taras Shevchenko National University seems to be:

Team leader: Dr. Andriy Bondarenko
Students:
Ivan FESHCHENKO
Iaroslav ZHURBA
Oleksandr KRAVETS
Sergii SLOBODIANIUK
Ivan IURCHENKO
Larisa HOMENKO
Taras TYMOSHKEVYCH
Ivan LIVINSKYI

Belarussian State University sent a team which contain  7 participant:
Mikhail Blotski
Alexandr Medvedev
Ivan Bliznets
Uladzimir Yahorau  other participant i don't know how to write names in English.

HI, University of Tehran is also sending a team, i know two of them my friends but none of them are AoPSers!

Naser Golestani, and Esmaeeli Zadeh :)

i hope i'll be in Tehran's universitys Team in IMC 2008 :D

[quote="dmitin"][quote="sinajackson"]Naser Golestani, and Esmaeeli Zadeh :) [/quote]
Could I have seen them at [url=http://olympiad.sanjesh.org/EnglishV/News/E8604231124.aspx]ISOM[/url] in Tehran University in July?[/quote]

i dont think so, actually both of them have a Silver Medal from the Iranian Mathematical Society(IMS2007) which was:

http://www.mathlinks.ro/Forum/resources.php?c=2&cid=109&year=2007

and know they are in Tehran University's Team to come to IMC2007 :)

i think they got medals from IMS 2006 too! :)

[quote="Peter"]There is no inherent limitation, but of course your university decides for how many they want to pay. ;)[/quote]

Well, if hosting country is not too far from yours then even if your university has not enough money to pay for you or does not like you enough to do this then nevertheless you can go for your own money. Surely if you are real olympiader and have a little money. In this sense there is no limitation. :wink:

Which is what I am doing for the third year in a row now. ;)

But from MIT (USA) to eastern Europe is not what I call 'not too far'... ;)

[quote="PTynan89"][quote="dmitin"][quote="PTynan89"]How many are allowed to go from each university?[/quote]

As far as I know there is no limitation. :wink:[/quote]

That's good to know, as I will be attending MIT next year.

=)[/quote]

Hey,

MIT hasn't participated before to my knowledge.

In terms of cost, the competition fee is 175 Euros, or 238 dollars.
Airfare from Newark to E-Europe is around 2000, but you can sometimes get it down to 1200.

So you're looking at 1400 to 2200 bucks, which is quite a sum.

If you give me an email (jytang@princeton.edu), I can get back to you on a couple other contests that may be available and for which your school may be able to send a team for.

-Jeff

[quote="Xevarion"]This past summer, we (Princeton) contacted MIT and Harvard to see if their students were interested in forming a USA team to send to the IMC. However, apparently nobody was interested. We'd probably be willing to do this again if people at Harvard/MIT wanted to in the future.[/quote]

Well, I am certainly interested in going.

I'm willing to bet a few others would be interested as well (my roommate happens to be a former IMOer, so I could ask him).


Suppose $f(x)$ is a polynomial with integer coefficients such that $f(0) = 11$ and $f(x_1) = f(x_2) = ... = f(x_n) = 2002$ for some distinct integers $x_1, x_2, . . . , x_n$. Find the largest possible value of $n$.

[quote=scorpius119][hide]
For each $x_k$, an $x_k$ can be factored out of $f(x_k)-f(0)=1991=11\cdot 181$. So $x_k$ must be a divisor of 1991, and there exist 8 such divisors: $\pm 1,\pm 11,\pm 181,\pm 1991$.

In addition, $P(x)=2002+Q(x)\prod_k (x-x_k)$ for some polynomial $Q$, and $Q$ will also have integer coefficients. Substituting 0 in for $x$ gives
\[ Q(0)=(-1)^{n+1}\frac{1991}{\prod_k x_k} \]
which must be an integer. So only one $x_k$ can be divisible by 19 and only one can be divisible by 181. After using up 1 and -1 that means at most 4 for $n$.

The polynomial $P(x)=2002+(x+1)(x-1)(x-19)(x-181)$ does satisfy the conditions for $n=4$, so the maximum is $\boxed{4}$.
[/hide][/quote]

check once i don't think this gives P(0)=11, i guess you mean 11 in place of 19

[hide]we can set $f(x)=\prod_{i=1}^{n}(x-x_{i})Q(x)+2002$ for some $Q(x)\in \mathbb Z[X]$\\

$f(0)=(-1)^{n}\cdot \left(\prod_{i}^{n} x_{i} \right)Q(0)+2002$\\

clearly this gives $(-1)^{n}\cdot \left(\prod_{i}^{n} x_{i} \right)Q(0)=-1991$\\

[color=#f00][b]Claim:-[/b][/color] $n\le4$\\

[color=#00f]pf:-[/color]since we have $x_{i}\neq x_{j}$ we observe we can't have $(-1)^{n}\cdot \left(\prod_{i}^{n} x_{i} \right)=1$ and $Q(0)=-1991$\\

so we have either $(-1)^{n}\cdot \left(\prod_{i}^{n} x_{i} \right)=-1 , Q(0)=1991$ or $(-1)^{n}\cdot \left(\prod_{i}^{n} x_{i} \right)=-1991$ and $Q(0)=1$\\
or $(-1)^{n}\cdot \left(\prod_{i}^{n} x_{i} \right)=1991$ and $Q(0)=-1$ , and some more cases\\

clearly from above cases we get only $n=4$ to be the maximal  case working  ( for $(x_{1},x_{2},x_{3},x_{4}) \in \{-1,1,11,181\})$\\

hence we have $n\le 4$ as claim follows $\blacksquare$\\

hence maximal $n$ that works is $\boxed{4}$[/hide]

Who knows merthods to prove surjectivity


A Thief steals four gallons of liquid soap kept in a bathroom. He then fills the container with water to avoid detection. Unable to resist temptation he repeats the process again. The final ratio of the liquid soap when checked is 36:13 what was the initial amount of liquid soap in the container.( the container was initially full of liquid soap and no one used it during the entire course of action)
a.	7 Litres
b.	14 Litres
c.	21 Litres
d.	28 Litres

what would be the simplest way to solve this in under a minute? what should be the thought flow. I know how to use brute force but that would consume more time. I'm looking for a solution that can be arrived at in under a minute.

The given solution is 28 litres.(i'm not sure if the gallon <=> litre conversion has been done or misinterpreted!)

Let there be $s$ liters of soap. Then after the first step we have $s-4$ liters of soap and $4$ liters of water. The proportion factor of soap to total is is $\frac{s-4}{s}$, hence after removing another $4$ liters of the mixture and pouring in another $4$ liters of water, we have $\frac{(s-4)^{2}}{s}$ liters of soap and $4\frac{s-4}{s}+4=\frac{8s-16}{s}$ liters of water. Then the given ratio is $\frac{(s-4)^{2}}{8s-16}=\frac{(s-4)^{2}}{8(s-2)}$. Now just plug the given choices in and see which one gives the denominator divisible by $13$. That's $D$ since $8(s-2)=8\cdot 26$, and indeed the result is $\frac{24^{2}}{16\cdot 13}={36\over 13}$

In my opinion, this shouldn't take much more than a minute.


calculus derivative inequalities induction

Without using any limit or derivative, prove the existance of A, satisfying the inequality such that $ n > 0$ and $ 2 \le (1 \plus{} \frac {1}{n})^n \le A$  ,  $ n \in\mathbb{N}$

I might use hints, as I can't really think of a way to approach this problem to prove the "existence" of something.

I assume you know how to prove the lower bound of $ 2$.  Now, you want to show that the sequence $ s_n \equal{} \left( 1 \plus{} \frac{1}{n} \right)^n$ is bounded above.  Are you aware of elementary results on bounds on quantities of the form $ (1 \plus{} x)^n$?  Does induction look promising?

I wasn't aware till you said that >.<

I'll try induction, thanks.


If $ \frac35 \equal{} \frac{M}{45} \equal{} \frac{60}{N}$, what is $ M \plus{} N$?

$ \textbf{(A)}\ 27\qquad
\textbf{(B)}\ 29\qquad
\textbf{(C)}\ 45\qquad
\textbf{(D)}\ 105\qquad
\textbf{(E)}\ 127$

[hide]

Let us set up two separate proportions:

$ \frac35 \equal{} \frac{M}{40} \implies 135 \equal{} 5M \implies M\equal{}27$

$ \frac35 \equal{} \frac{60}{N} \implies 300 \equal{} 3N \implies N\equal{}100$

Adding reveal the answer is $ \boxed{\textbf{(E)}\ 127}$[/hide]

[quote="cgyao15"]Are you writing questions and solving them just to get posts?[/quote]

Adding to resources.  Turns out there are only like 2 AMC 8's there.


Should it read like this? (because such as it is, it doesn't seem totally sensible):

$\frac{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\dots}}}}{2+\frac{1}{1+\frac{1}{2+\frac{1}{1+\dots}}}}$

no, it's correct the way it is
why does it not look sensible?

In that case, the dots aren't clear enough - what's the pattern to be repeated?

look at the symmetry across the largest "fraction" line. I think it would be called horizontal symmetry. I tried making the pattern go on further, but the thing got very large, and it had a weird shape. how do you made text larger?

[hide="Hint 1"] Let the numerator be $B$.  Can you write $A$ in terms of $B$?  [hide="Hint 2"] Once you have done that, can you find an equation that will let you solve for $B$? [/hide] [/hide]

Hope I am reading this problem correctly.

[hide]
Assume that the numerator equals x. Then,
x = 1+1/(2+x/2)
This can be written as,
x(2+x/2) = 2+x/2+1
2x+(x^2)/2 = 3+x/2
(1/2)x^2+(3/2)x-3 = 0
Which can also be written as,
x^2+3x-6 = 0
Solving the quadratic equation, we get
x = (-3+(33)^(1/2))/2

Now, assume that the denominator equals y. Then,
y = 2+(1+1/y)/2
This can be written as,
2y = 5+1/y
Multiplying both sides by y, we get
2y^2 = 5y+1
2y^2-5y-1 = 0
Solving the quadratic equation, we get
y = (5+(29)^(1/2))/4

Finally, dividing x by y, we get
[b](-6+2(33)^(1/2))/(5+(29)^(1/2))[/b][/hide]

[quote]
I'm fairly certain the problem was not intended this way. It should be .
[/quote]
Okay, then, here is the new solution:

[hide]
Solving the new quadratic equation we get
x = (-1+(13)^(1/2))/2
The value for the denominator is
y = (5+(29)^(1/2))/4
Dividing x by y, we get
[b]A = (-2+2(13)^(1/2))/(5+(29)^(1/2))[/b][/hide]

Your values for $x$ and $y$ should not be so different.

[hide="The problem intends..."] $y = 2+x$ (can you see why?) [/hide]

[quote="t0rajir0u"][quote="echatterjee"]
x = 1+1/(2+x/2)[/quote]

I'm fairly certain the problem was not intended this way.  It should be $x = 1+\frac{1}{2+x}$.[/quote]

Confer my statement of the problem.

the way echatterjee originally interpreted the problem is correct. i know its unusual, but i was trying to get an unusual problem

Here's how I did it (the one in AIME class was much friendlier :) )

[hide]Given $a = \frac{n}{d}$,

$n = 1+\frac{1}{d}$ , $d = 2+\frac{n}{2}$

So
$d = 2+\frac{1+\frac{1}{d}}{2}\Rightarrow 2d^{2}-5d-1=0$
In quadratic, d must be plus because the denominator is obviously positive :)

$d = \frac{5+\sqrt{33}}{4}$

So $n=1+\frac{4}{5+\sqrt{33}}$

This can be simplified to $\frac{48}{12+4\sqrt{33}}\Rightarrow \frac{12}{3+\sqrt{33}}$

So you're answer is:
$\frac{12}{3+\sqrt{33}}\ast \frac{4}{5+\sqrt{33}}$

Or better put $12-2\sqrt{33}$[/hide]

This is pretty much the same as minsoen's.... but oh well :)


function
inequalities
algebra
binomial theorem

For $ r$ element of N*, let $ a_r$ be the number of integer solutions to the inequality $ x_1 + x_2 + x_3 + x4 <=r$ where $ 3 <=x_1<=9$, $ 1 <=x_2<=20$, $ x_3>=9$ and $ x_4>=0$. Find the generating function for $ a_r$ and $ a__{20}.$

[hide="Start"]
$ x_a \equal{} x_1 \minus{} 3,\;x_a \le 6$
$ x_b \equal{} x_2 \minus{} 1,\;x_b \le 19$
$ x_c \equal{} x_3 \minus{} 9$
$ x_a \plus{} x_b \plus{} x_c \plus{} x_4 \le r \minus{} 13$

We're looking for the coefficient of the term with power $ 20 \minus{} 13 \equal{} 7$
$ (1 \plus{} x \plus{} x^2 \plus{} ... \plus{} x^6)(1 \plus{} x \plus{} ... \plus{} x^{19})(1 \plus{} x \plus{} ...)^2$
$ \equal{} (1 \minus{} x^7)(1 \minus{} x^{20})(1 \minus{} x)^{ \minus{} 4}$

Then do some case work, and use newton's binomial theorem on $ (1 \minus{} x)^{ \minus{} 4}$
[/hide]


calculus
derivative
real analysis
superior algebra
superior algebra solved

Let $A$ is associative ring in which every element is idempotent. Id est $\forall a\in A$  $a^2=a.$
Prove that $A$ is commutative ring.

$x^2=0\Rightarrow x=0$ (1)
$x,y\in A$
Then $(xy-xyx)^2=0$ and $(yx-xyx)^2=0$ (very easy, use that $a^2=a, \forall a\in A$).
From (1) we get that $xy=xyx$ and $yx=xyx$. So $xy=yx$, so $A$ is commutative.

Nice solution, Kuba!
My proof:
$(a+ab)^2=a+ab.$ Hence, $ab=-aba.$
$(a+ba)^2=a+ba.$ Hence, $ba=-aba.$

[quote="enescu"]$1+1=(1+1)^2=1+1+1+1$, hence $1+1=0$, implying $x=-x$ for all $x$.
$a+b=(a+b)^2=a^2+ab+ba+b^2=a+ab+ba+b$, hence $ab=-ba=ba$ as desired.[/quote]
Why $1\in A?$ ;)

[quote="enescu"]You may replace $1$ with some $x$. It still works.
I assumed $1\in {A}$ because in Romania high school curriculum includes only unitary rings.[/quote]

High school books include ring theory in Romania? :oops:

[quote="fredbel6"]High school books include ring theory in Romania? :oops:[/quote]

Currently, the high school curriculum is in perpetual change, but in my time (4-8 years ago), what most of the world considered to be elementary mathematics, stoped around the 10th grade and from 11th to 12th we got to know groups, rings, fields, matrices in algebra and real analysis in one variable (differentiation and integration).


function
calculus
derivative
real analysis
real analysis unsolved

Prove or disprove:
Let f be a differentiable function on R.Then,the derivative,df/dx is a measurable function.

Can anyone help me?

If $ f$ is differentiable, then it it continuous, and continuous functions are measurable.

Let $ h_n$ be some particular sequence of nonzero numbers that tends to zero. Let $ g_n(x)\equal{}\frac{f(x\plus{}h_n)\minus{}f(x)}{h_n}.$

Each $ g_n$ is a continuous function, hence measurable.

If $ f$ is differentiable, then $ g_n(x)\to f'(x)$ pointwise. And the pointwise limit of measurable functions is measurable.

As a general plan, when you're trying to show that a function is measurable, it's a good idea to write that function as a limit. The reason we used all those countable unions and countable intersections in the definition of measure is so that the resulting theory would be stable under the taking of limits.


number theory
relatively prime
number theory unsolved

Let $p$ be an odd prime, and let $S = \{n_{1},n_{2}, \cdots, n_{k}\}$ be an arbirtrary set of perfect squares relatively prime to $p$. Compute the smallest $k$ such that there necessarily exists a subset of $S$ with the product of its elements one more than a multiple of $p$.

That's nice. We will show that $k=\frac{p-1}{2}$. First, we prove that $k \geq \frac{p-1}{2}$. Consider the set $S = \{g^{2}, g^{2}, ..., g^{2}\}$ of $k$ elements with $k < \frac{p-1}{2}$ and $g$ being primitive root $\mod{p}$. Then it's obvious that product of any subset of $S$ is of a form $g^{2a}$ where $a<\frac{p-1}{2}$ so it can't be congruent to $1$ mod $p$, since $g$ is a primitive root. Now, let's prove that $k \leq \frac{p-1}{2}$. Indeed, suppose that $k \geq \frac{p-1}{2}$, $S=\{n_{1}, n_{2}, ..., n_{k}\}$ and write $n_{i}$ = $g^{2\alpha_{i}}$ for each $i$ (as before $g$ is a primitive root mod $p$). Product of any subset of $S$ has a form $g^{2A}$ where $A$ is a sum of some subset of the set $\{\alpha_{1}, \alpha_{2}, ..., \alpha_{k}\}$. So we are left with proving that for any set of at least $\frac{p-1}{2}$ elements we can find a subset of it with sum of its elements being divisible by $\frac{p-1}{2}$. But that's classic. Consider the numbers $\alpha_{1}, \alpha_{1}+\alpha_{2}, \alpha_{1}+\alpha_{2}+\alpha_{3}, ..., \alpha_{1}+\alpha_{2}+...+\alpha_{\frac{p-1}{2}}$. There are exactly $\frac{p-1}{2}$ of them, so if there is no number which is divisible by $\frac{p-1}{2}$ then there are two numbers, which are congruent $\mod \frac{p-1}{2}$. Now, we just have to take their difference and we are done.

Is every odd prime number has a primitive root?
How we prove it?

Thx..

@Yosh... : Is this YO (Oscar) ? If yes, you can read Elementary Number Theory page 154 - 157 for the details about primitive roots..

@TomciO : Nice solution, pal..  :wink:


calculus
integration
logarithms
trigonometry
calculus computations

Here's a couple of integral that I'm stuck with them now. I used substitution but it doesn't help much on these problems. Probably someone can point out other useful methods to help me solve these problems. Anyway, here they are:
1.$ \int(x/{x^4 + 1}))dx$
2.$ \int{tan^( - 1)x/{x^2 + 1})dx}$
3.$ f(x) = \int(x^2.sin(x^2))dx$ with $ f(1) = 0$

You can solve the problems in using this.

Let $ \int f(x)\ dx \equal{} F(x) \plus{} \text{Const}.$,

$ \boxed{\int f(g(x))g'(x)\ dx \equal{} \int f(g(x))\ dg(x) \equal{} F(g(x)) \plus{} \text{Const}.}$

hello
Is the first question is :$ \int\frac{x}{x^2\plus{}1}dx$
[hide="If it so then answer"]
put $ x^2\plus{}1$=y
$ \int \frac{dt}{2t}$
=$ \frac{log[x^2\plus{}1]}{2}\plus{}C$ or $ log\sqrt{x^2\plus{}1}\plus{}C$
[/hide]
thank u

$ \int \frac {x}{x^2 \plus{} 1}\ dx \equal{} \frac {1}{2}\int \frac {1}{x^2 \plus{} 1}(x^2 \plus{} 1) '  \ dx$ follows your answer.

$ \boxed{\int \frac{f'(x)}{f(x)}\ dx\equal{}\ln |f(x)|\plus{}C}$

hello
Second question::$ \int \frac{tan^{\minus{}1}x}{x^2\plus{}1}dx$
[hide="Answer"]
put $ tan^{\minus{}1}x\equal{}y$
$ \int ydy$
=$ \frac{(tan^{\minus{}1}x)^2}{2}$
[/hide]
thank u

[quote="kunny"]$ \int \frac {x}{x^2 + 1}\ dx = \frac {1}{2}\int \frac {1}{x^2 + 1}(x^2 + 1) '\ \,dx$ follows your answer.[/quote]
Oh I made a typo in this problem. It should be $ x^4+1$. Sorry kabi and Mr. Kunny!
@Mr. Kunny: Your formula restates the substitution rule, doesn't it?? But I can't deal with these problems by using substitution(like let $ u=x^4+1$ but then it doesn't work well).

[quote="ghjk"][quote="kunny"]$ \int \frac {x}{x^2 + 1}\ dx = \frac {1}{2}\int \frac {1}{x^2 + 1}(x^2 + 1) '\ \,dx$ follows your answer.[/quote]
Oh I made a typo in this problem. It should be $ x^4 + 1$. Sorry kabi and Mr. Kunny![/quote]

No problem.  :)  That integration is difficult, probably we will first factorize $ x^4+1$.

[quote]@Mr. Kunny: Your formula restates the substitution rule, doesn't it?? But I can't deal with these problems by using substitution(like let $ u = x^4 + 1$ but then it doesn't work well).[/quote]

Yes, it seems that foreigners are apt to use substitution, don't you? Japanese highschool students do integeration for level of your problem directly without using substitution, but They don't study arc tangent e.t.c.  :lol:

hello
$ \int \frac{x}{x^4\plus{}1}dx$
use the same substitution that i have given in my post.
$ \int \frac{dy}{2(y^2\plus{}1)}$
=$ \frac{1}{2}tan^{\minus{}1}x^2 \plus{}C$
thank u

[quote="kunny"]3. $ \int x^2\sin ^ 2 \ dx \equal{} \frac {1}{2}\int x^2(1 \minus{} \cos 2x)\ dx$

$ \equal{} \frac {1}{6}x^3 \minus{} \frac {1}{2}\left(\frac {1}{2}x^2\sin 2x \plus{} \frac {1}{2}x\cos 2x \minus{} \frac {1}{4}\sin 2x\right) \plus{} C$.[/quote]
Kunny
Sorry but I think  ghjk has given us $ sin(x^2)$. :?:

in the last integral there is no any closed form.
I mean you can't calculateso as to express it by elementary functions.
for example
by the substitution $ x^{2} \equal{} t$ you will get:
$ \int t^{1/2}sin(t)dt$
or it integral is of this type : $ \int t^{\alpha}e^{it}dt$ where $ \alpha < 1$
it can be calculated exactly for some intervals, for example when we integrate it on the interval $ (0,\infty)$ it can be expressed as $ \Gamma$ function.
generraly there is no some "good answer"


You and a friend are trying to perform the following "magic" trick.  Your friend leaves the room.  You hand a standard deck of 52 cards to someone in the audience.  The audience member chooses any five cards from the deck and hands those five cards to you. Now, you must select one of these cards to keep concealed from your friend, and you must display the other four cards side-by-side. Your friend now enters the room, examines the only the four displayed cards, and then tells the audience what's on the concealed card. Can you and your friend come up with a scheme so that, based solely on the four displayed cards, the fifth card can always be determined?

Hint[hide]This is an advice given to all beginers when playing agains notump at bridge: "Play the longest suit you hold! :wink: [/hide]

Solution[hide]
Obviously from any 5 cards we have at least 2 with the same color. The [b]first card to be shown [/b]is one of this two. Let the Ace count for 1, J=11,Q=12,K=13. For any card we call its group the 6 higher cards counted clockwise. Example: the group of 2 are (3,4,5,6,7,8) and the group of 10 are (J,Q,K,A,2,3).
 It is clear that for any 2 cards in a suit one is in the group of the other. The [b]first card we show determins the group [/b](if the cards were 2 and 10 we show the 10, if 2 and 7 we show 2).
We have 3 more cards = 3!=6 to identify one of the 6 cards from the group.
If we forget about cards we see that with 4 numbers we can code any number from 1 to 52.
[b]Is it possible to code more?
How many numbers one can code with 5 numbers? [/b][/hide]


[color=red][size=150]How many distinct three letter sequences w/ at least one "T" can be formed by using 3 of the 6 letters of TARGET :?:  Of one such squences "T-R-T"[/size][/color]

There is a total of $ 6\cdot5\cdot4\equal{}120$ ways for a 3 word to be made. To make this simpler, you count off the ones with no Ts. This is$ 4\cdot3\cdot2\equal{}24$. Thus there are $ 120\minus{}24\equal{}\boxed{96}$

how did u get 6*5*4???
and 4*3*2?

thanks

Doesn't hunter34's solution not account for having two T's in a sequence? I mean, doesn't it count TRT or ATT twice?

I might be wrong, but:

6*5*4-4*3*2=96

Then subtract half of the sequences with two T's... 1*1*4*3/2=6
96-6=90

?

I would say that the safest way to solve the problem here would be through casework.

[hide]CASE 1:  There is exactly 1 T in the sequence.  There are $ \binom42\equal{}6$ ways to choose the other two letters that appear in the sequence.  In addition, there are 3 distinct letters in the sequence, so there are $ 3!\equal{}6$ ways to rearrange the letters, for a total of $ 6\cdot6\equal{}36$ sequences.

CASE 2:  There are exactly 2 T's in the sequence.  There are $ 4$ choices for the remaining letter and $ 3$ ways to rearrange the letters, so there are $ 4\cdot3\equal{}12$ total sequences.

Because there can be only one or two Ts, all cases have been considered, for a total of $ 36\plus{}12\equal{}48$ sequences.[/hide]

I think that iin is right.
it would be 6*5*4/(2!) because there are 2 "T"s    which equals 60

Then you subtract the ones without any "T"s in them whichi is

4*3*2= 24 so 60-24=36

The 6,5,4 and 4,3,2 thig show how many possibilities can be put in that spot

___   ___   ___

We use Casework.

The first case is that there is only $ 1$ T. Then,

We have 4 choices for the next letter, and 3 for the next.

So, $ \frac{4 \cdot 3}{2} \cdot 6\equal{}36$. This is because 4 choices,

then 3

choices, then divide by 2 since AR is the same as RA, then

mulitply by 6 for the ordering.

The second case is that there is $ 2$ T's. Then, we get

4*3=12, 4 choices for the other letter, 3 for ordering.

So, $ 36\plus{}12\equal{}48$.


calculus
integration
function
trigonometry
calculus computations

Given a continuous function $ f(x)$ such that $ \int_0^{2\pi} f(x)\ dx \equal{} 0$.

Let $ S(x) \equal{} A_0 \plus{} A_1\cos x \plus{} B_1\sin x$, find constant numbers $ A_0,\ A_1$ and $ B_1$ for which $ \int_0^{2\pi} \{f(x) \minus{} S(x)\}^2\ dx$ is minimized.

[hide="Solution"]$ \int_0^{2\pi} f(x)\, dx \equal{} 0$

$ F(a,b,c) \equal{} \int_0^{2\pi}[f(x) \minus{} a \minus{} b\cos x \minus{} c\sin x]^2\, dx$

$ \frac {dF}{da} \equal{} \frac {dF}{db} \equal{} \frac {dF}{dc} \equal{} 0$

1. $ \frac {dF}{da} \equal{} 0$

$ \frac {d}{da} \int_0^{2\pi}[f(x) \minus{} a \minus{} b\cos x \minus{} c\sin x]^2\, dx \equal{} 0$

$ \int_0^{2\pi} \frac {\partial}{\partial a} [f(x) \minus{} a \minus{} b\cos x \minus{} c\sin x]^2\, dx \equal{} 0$

$ \int_0^{2 \pi } (2 a \plus{} 2 b \cos x \minus{} 2 f(x) \plus{} 2 c \sin x) \, dx \equal{} 0$

$ a \equal{} 0$

2. $ \frac {dF}{db} \equal{} 0$

$ \frac {d}{db} \int_0^{2\pi}[f(x) \minus{} a \minus{} b\cos x \minus{} c\sin x]^2\, dx \equal{} 0$

$ \int_0^{2\pi} \frac {\partial}{\partial b} [f(x) \minus{} a \minus{} b\cos x \minus{} c\sin x]^2\, dx \equal{} 0$

$ \int_0^{2 \pi } (2 a \cos x \plus{} 2 b \cos^2 x \minus{} 2 \cos x f(x) \plus{} 2 c \cos x \sin x ) \, dx \equal{} 0$

$ b \equal{} \frac {\int_0^{2 \pi } [\cos x f(x)] \, dx}{\pi }$

3. $ \frac {dF}{dc} \equal{} 0$

$ \frac {d}{dc} \int_0^{2\pi}[f(x) \minus{} a \minus{} b\cos x \minus{} c\sin x]^2\, dx \equal{} 0$

$ \int_0^{2\pi} \frac {\partial}{\partial c} [f(x) \minus{} a \minus{} b\cos x \minus{} c\sin x]^2\, dx \equal{} 0$

$ \int_0^{2 \pi } \left(2 a \sin x \plus{} 2 b \cos x \sin x \minus{} 2 f(x) \sin x \plus{} 2 c \sin^2 x \right) \, dx \equal{} 0$

$ c \equal{} \frac {\int_ 0^{2\pi} [\sin x f(x)]\, dx} {\pi}$[/hide]

That's right. But Japanese highschool students don't study partial differentiation.


Find the value of the constant k in $f(x) =-x^{2}-8x+k$ if the maximum value of $f(x)=17$.

[hide="ans"]
put the equation in vertex form
$f(x) =-x^{2}-8x+k$
$f(x) =-(x^{2}+8x+16-16)+k$
$f(x) =-(x+4)^{2}+16+k$

thus the maximum value of the function is 16 + k,

knowing that the max value of the function is 17,
$16+k = 17$

thus $k = 1$
[/hide]