AMC USA(J)MO USAMO email LaTeX inequalities function

I'm still looking for problems for day 2. If you have any (that are not like the day 1 problems), pm me!!! You can still participate in day 1 if you contribute to day 2.

The test is below. Please take 4.5 hours in one sitting to do these problems, and email your solutions to me at thyroldf@gmail.com (because apparently big pdfs don't like to send through pm) by midnight Monday. Title your email "solutions username"; for example "solutions joeschmoe1". I will try my best to grade them over the week and show the results hopefully by the Sunday after. I will make sure to post results before the USAMO.

If you have a question about the content of the problems, please pm me rather than posting here. Title the pm "question".

[i]Edited by mod as requested.[/i]

I'd be okay with doing it regardless. The factor of problem difficulty is important in picking between the two middle choices. The 5-problem USAMOs have much easier problems than ones in the current format.

I edited my top post with the problems. I hope you like them. :lol:

Do you think you could extend the deadline until Monday?  I'm not going to be able to work on them this weekend :(

[quote="jb05"]Umm I guess. Midnight Monday.[/quote]

Sweet thanks! :)

How should I send my solutions? Is it okay if I just send it written in like a pdf file or do I have to type to it up?

Any sort of pdf is best. All I will really ask is not to send anything massive, like (most) pictures.

Yeah yeah whatever you want. I'm not picky.

Also everyone please email me at thyroldf@gmail.com because apparently big pdfs don't like to send through pm. If a moderator is keeping up with this thread, please edit my first post.

[quote="georgiamathdude"]Ummm, I have no idea how to work the pdf files, so can I just try to do it on Microsoft word?[/quote]

I recommend downloading CutePDF and converting your Word file to PDF. This is what I do with my contest ($ \text{\LaTeX}$ takes too much of speed for my computer) and I tend to really advocate using PDF because Word can be dangerous to download.

So... are you guys procrastinating on sending in answers? :wink:

ARG

I've whittled #2 down to an inequality I'm pretty sure I've seen before.

BUT I CAN'T REMEMBER HOW TO PROVE IT.

it's driving me insane!

I know! It's insanely annoying. I spent like 1/2 and hour getting to that inequality, and spent the rest of the 4 hours trying to prove it...(I couldn't do any of the other problems...) :(

sorry jb, i wanted to give these a shot but i didn't end up having time.

The problems look cool though, I'll probably never solve them, but if I do I'll probably send you a casual sketch-up of a solution and see if it makes any sense

Hm...I don't know what you guys are trying, but there's a really short solution to #2...I'll put it up after the deadline has passed I guess if it doesn't go up.  Also #1.

EDIT: I looked at the problems, btw, but didn't take the test.  Have still to work on #3 (but yeah it definitely looks tough).

Hmm... I did not get many submissions at all. Sorry if it was hard - actually, most people are telling me it was really hard, but a few are telling me it was really easy, and nobody in between :huh: . Also, #1/2 probably should have been switched, but a certain individual convinced me otherwise. I thought I had some easier/more varied problems for the second day, but unless there is overwhelming interest, I guess I won't do that.

Altheman won, and that's about all the statistics I need to give out for this one :stink:

Before I post solutions, I'm really interested to see whether anyone can solve #3. So post away with answers.

So I take it nobody solved #3? Does your solution give a construction?

I figured out the hard part of #3, but I can't get the easy part. Well, at least I think so. I'll pm my partial to jb and see how it works.

EDIT: Whoops, I switched an inequality sign. back to the drawing board...

I am rather surprised that nobody has turned in a solution to #3, but I won't say more because I don't know if jb05 is going to extend the problem.  I'm sad because the function involved is one I came up with while thinking of problems for jb05.  Problem 2 is also not that bad, but hey.

Day 2 will be interesting...

Could we have a few days grace to work on this? (For those who were irresponsible and procrastinated on their other work  :wink: )

Sure. I guess you can send me solutions if you want, too.

I only got one more submission, and it was not from Phelpedo. Does anyone object to posting solutions?

Whooops, did the problems but forgot to tex them. I'll have that done in hopefully about an hour.

Ok I guess phelpedo has the highest score... here are solutions. My original solution to #1 is unnecessarily difficult, but I included it anyway as the second solution.

:bomb:  :censored:  :censored: Bah! I got #2 all the way down to the last step and screwed up on proving it. I forgot the x+1/x increasing part. Meh... -_-

problems have been posted in separate threads inside aops and are collected [url=https://artofproblemsolving.com/community/c3625276_2008_mock_usamo_1]here[/url]
(among others [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url], among USA Mocks/ USAMO Mocks)

Related post [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url]

I have been posting all USAJMO and USAMO Mocks, mentioned in aops in separate threads and informing by commenting here, anyone interested in this mock  where to find the related post collection, that (s)he may continue discussing if interested in the new seaprate problem threads and where to find other similar mocks inside aops


algebra
polynomial
Euler
modular arithmetic
calculus
integration
quadratics

Let $m$ be any integer. Show that there are $\infty$ many primes $p \equiv-1 \mod m$ in a way as elementary as possible (especially don't use L-series or similar things).

Some easier and already posted thing is that there are $\infty$ many primes $p \equiv 1 \mod m$.
It's also known that there is a Euklid-like proof for the existence of $\infty$ many primes $\equiv a \mod m$ iff $a^{2}\equiv 1 \mod m$, which is true for both cases given here.

Let $p_{1},p_{2},...,p_{l}$ all primes $p_{i}=-1(mod \ m)$.
Consider prime divisors q, s.t. $q|AT^{k}+B,B=\pm 1$,were $T=p_{1}p_{2}...p_{l}$. It give -AB is k-th power mod q. This method works for many m (exist $A,B=\pm 1,k$).

1)Prime divisor $q\not p_{i}$.
2) If $B=-1,m|A$, then if all divisors $q|AT^{k}-1$ are $q=\pm 1(mod \ m)$, give prime divisor $q=-1(mod \ m)$.
For example $m=4,6$ works $k=1,A=m,B=-1$.
$m=2^{a}3^{b},a>1$ works $k=\phi (m)/2,A=2^{c}3^{d},c\ge a,d\ge b,B=-1$(exist c and d ).
m=10 works $k=2,A=20,B=-1$ etc.

Why does the case $m=2^{a}3^{b}$ work¿ E.g. $a=0, b=2$ your values give the polynomial $9x^{3}-1$, but that one doesn't give the required prime factors.

I suggest using the polynomial $x^{3}+3x^{2}-1$ for primes $\equiv 1 \mod 9$ instead.

@ZetaX: you wrote that there is a Euler polynomial proof iff $a^{2}\equiv 1 \pmod{n}$. Where can I find a proof for this?

One direction is given in http://www.math.uconn.edu/~kconrad/blurbs/dirichleteuclid.pdf .
Also the converse is stated there.

My approach would be (prepare for a lot of nasty spam ;) ):

I can tell you one type of polynomial that works:
Let $a^{2}\equiv 1 \mod n$; then build pairs in the prime residue classes $\mod n$ (lol, when speaking about Dirichlet's theorem this get's another meaning :D ):
$x,y$ belong together iff $x \equiv ay \mod n$ (this is by definition of $a$ a good and working pairing).
Choose from all those pairs one of the elements and build a set S (of $\frac{\varphi(n)}{2}$ elements) of the chosen ones.
Or in short: let $S$ be a representation system for $(\mathbb{Z}/n\mathbb{Z})^{*}\mod \{1,a\}$ (seen multiplicativly).
Then an euclidean polynomial (one that up to finitely many exceptions only has prime divisores $\equiv 1, a \mod n$) is given by \[f(x)=\prod_{k \in S}\left( x-(u-\zeta_{n}^{k})(u-\zeta_{n}^{ak})\right),\] where $\zeta_{n}$ is a primitive $n$-th root of unity and $u$ some integer $\neq 0$ we can choose as we want.

This polynomial has rational coefficients because it remains the same under conjugation (via $\zeta_{n}\mapsto \zeta_{n}^{m}$, $\gcd(m,n)=1$).
It is integral since all it's roots are algebraic integers.
And it has our desired property:
Lets work in the field $\mathbb{F}_{p}$. Take $K$ to be a splitting field of $\Phi_{n}(x)$ over this prime field and let $\zeta$ be a root of $\Phi_{n}(x)$ in $K$. Then $K=\mathbb{F}_{p}[\zeta]$.
We have that the splitting field of $f(x)$ is build by that subfield that is fixed by $\{1,a\}$, and that this splitting field of $f(x)$ is already $\mathbb{F}_{p}$ itself.
Thus $K$ is equal to $\mathbb{F}_{p}$ or a quadratic extension of $\mathbb{F}_{p}$. The Frobenius isomorphism $x \mapsto x^{p}$ maps $\zeta$ either to $\zeta$ or to $\zeta^{a}$ (these are all conjugates possible). But if $\zeta^{p}= \zeta^{m}$ for some $m$, then $\zeta^{p-m}$. If we assume $p \nmid n$, then the main theorem of cyclotomic polynomials (see below) gives that $\mathrm{ord}(\zeta)=n$, so that together with the ordering lemma we get that $n|p-m$, in other words: $p \equiv m \mod n$. In our case above $m=1$ or $m=a$.
Conversly, if $p \equiv 1$ or $a \mod p$, then $f(x)$ has a root $\mod p$ by the same reasoning.

At first the lemmas etc. before we continue ($K$ is always any field, $n_{K}$ is the embedding of the integer $n$ into $K$):

Let $a \in K$, we call the smallest $k \in \mathbb{N}$ with $a^{k}=1$ the order of $a$ in $K$ and write $\mathrm{ord}_{K}(a)$ for it. For those $a$ for that there is no such $k$, we just write $\mathrm{ord}_{K}(a)=\infty$.

[b]Ordering Lemma:[/b]
If $a \in K$ and $n \in \mathbb{N}$ such that $a^{n}= 1_{K}$, then $\mathrm{ord}_{K}(a) | n$.
Proof:
Let $o=\mathrm{ord}_{k}(a)$ and take division with remainder to get $n = o \cdot q+r$ with $0 \leq r < o$. We get $a^{r}= a^{n-o \cdot q}= a^{n}\cdot \left( a^{o}\right)^{-q}= 1_{K}\cdot 1_{K}^{-q}= 1_{K}$, contradicting the minimality of $o$ again if $r>0$. Thus $r=0$ and $o | n= q o$.

Some prerequisite:
Let $n \in \mathbb{N}$, $a \in K$, and $g(x)$ a polynomial with coefficients in $K$. If $x^{n}-1 = (x-a)^{2}g(x)$ as polynomials (so if $x^{n}-1$ has a double root), then $n_{K}=0_{K}$.
Proof:
Set $y=x-a$, then $(y+a)^{n}-1=y^{2}\cdot g(y+a)$. By expanding the LHS we get $y^{n}+...+n_{K}\cdot a^{n-1}y+(a^{n}-1)$, and the RHS gives $...+0_{K}\cdot y+0_{K}$, thus $n_{K}\cdot a^{n-1}= 0_{K}$ and $a^{n}=1$, giving $n_{K}= n_{K}\cdot 1_{K}= n_{K}\cdot a^{n}= (n_{K}\cdot a^{n-1}) \cdot a = 0_{K}\cdot a = 0_{K}$.

[b]Main theorem on cyclotomic polynomials:[/b]
Let $n \in \mathbb{N}$. If there is an $a \in K$ with $\Phi_{n}(a)=0$, then $n_{K}=0_{K}$ or $\mathrm{ord}_{K}(a)=n$.
Proof:
Assume that $o : = \mathrm{ord}_{K}(a) \neq n$ (the order exists since $a^{n}=1_{K}$). Then $o|n$ by the ordering lemma, so we get that \[x^{n}-1 = \Phi_{n}(x) \cdot (x^{o}-1) \cdot \prod_{\substack{d|n , \ d \nmid o\\ d \neq n }}\Phi_{d}(x).\] Now by definition $\Phi_{n}(a)=0_{K}$ and $a^{o}-1=0_{K}$, thus $\Phi_{n}(x)=(x-a) \cdot g(x)$ and $x^{o}-1=(x-a) \cdot h(x)$ with $g(x), h(x)$ polynomials over $K$. But by this $x^{n}-1 = (x-a)^{2}\cdot j(x)$ with $j(x) = g(x) \cdot h(x) \cdot \prod_{\substack{d|n , \ d \nmid o\\ d \neq n}}\Phi_{d}(x)$.
By the prequisite above we are done.

Now if we choose for example $u=n$, then $f(0) \equiv \Phi_{n}(0)=1 \mod n$ for $n>2$. Thus if there is one prime $p \equiv a \mod n$, we take an integer $b$ such that $p|f(b)$ (exists by the above) and such that $p^{2}\nmid f(b)$ (if this doesn't hold at first, we change $b$ by adding $p$).
If there would be only finetly many primes $p_{1}, ... , p_{k}\equiv a \mod n$ (all $\neq p$), then we could take an integer $c \equiv b \mod p^{2}, \equiv , \equiv 0 \mod n$ and $\equiv a_{i}\mod p_{i}$ (here $a_{i}$ is chosen such that $p \nmid f(a_{i})$; lets skip the proof of the existence) and consider $g(x) : = f(p^{2}\cdot n p_{1}\cdot p_{2}\cdot ... \cdot p_{k}\cdot x+c)$:
Then $g(d)$ is coprime to $n$ and all $p_{i}$ for any integer $d$. But is divisible by $p$ exactly once and has only prime divisors $\equiv1 , a \mod n$, thus there must be one $q \equiv a \mod n$ with $q\neq p$ , contradiction.

Fact is: such a proof shows only: if there is one such prime, then there are $\infty$ many. Thus we can proof the existence of primes by giving one.
In total, we can (since such a prime to give alwys exists) proof it that way or any $n$ and $a$ as above.

PS: I will probably write this down in a nicer way in the future.


algebra polynomial trigonometry logarithms function

how can we use taylors polynomial to approximate $\pi$

Use the taylor polynomial for $\tan^{-1}$, plug in 1 $x$, and multiply the whole thing by four. This value is clearly $\pi$.
Now for approximations, apply lagrange error bounds to your terms :)

That's not the error bound you want- use the alternating series error bound instead, which is much easier to compute. It's also very slow convergence- there are many other much better ways of getting at $\pi$, including using arctangents of smaller numbers.

like taylor polynomial for $e^{x}$, can we make one for $\pi^{x}$?i dont liike using trignmtry stuff.

You're not trying very hard.  $\pi^{x}= e^{x \ln \pi}$, so computing a Taylor series for $\pi^{x}$ is [i]exactly the same[/i] as computing a Taylor series for $e^{x}$, with the caveat that you will need to know the value of $\ln \pi$ to actually find anything out.

The general process you want to use is to find a function $f$ and a value $x$ such that:
1) $f$ has a "nice" Taylor series which converges on some interval including $x$
2) $x$ and its powers are "easy to compute"
3) $f(x)$ involves $\pi$ in some "easy to reverse" way

I leave it to you to think about why these conditions are what they are.  Inverse trigonometric functions are, I think, the most promising.  I would have suggested $\arcsin$ rather than $\arctan$, but jmerry probably knows what he's talking about.


An example of propaganda using the logical fallacies technique:

Hypothesis 1: Bill Clinton likes guerrilla technique gun-fare (true)
Hypothesis 2:  Communist Viet Cong used guerrilla gun-fare. (true)

Our conclusion...

Bill Clinton is a Communist. :D

Yeah, whatever seeing as these posts are dumb, eh...i just randomly put this!

Not true. It actually happens to be Bill clinton like Viet Cong.


[quote="BanishedTraitor"]The units-digit of the square of an integer is $ 9$ and the tens-digit of this square is $ 0$. Prove that the hundreds-digit is even.[/quote]
[hide="Solution"]
Say that $ (10k\pm 3)^2\equiv 9\bmod 100\implies 100k^2\pm 60k \plus{} 9\equiv 9\bmod 100$.  Thus, we find that $ 60k\equiv 0\bmod 100\implies 3k\equiv 0\bmod 5\implies 5|k$.  Thus, we call $ k \equal{} 5m$.  Thus, we wish to find $ (50k\pm 3)^2\bmod 1000$.  We can do this as follows:
\[ (50k\pm 3)^2 \equal{} 2500k^2\pm 300k \plus{} 9 \equal{} 9 \plus{} 100(25k^2\pm 3k)
\]
Now, if $ k$ is odd, then $ 25k^2\pm 3k$ is even, so the $ 100$ digit is even.  If $ k$ is even, then obviously, $ 25k^2\pm 3k$ is even, so the $ 100$ digit is even.  Either way, the $ 100$ digit is even, and we are done. [/hide]

[quote="t0rajir0u"]You haven't justified that the square root ends in $ 3$ or $ 7$.  You can do so either by casework or by using Lagrange's Theorem, but it's not a trivial point (in general).[/quote]
Can't we prove it by looking at 0-9 squared?

That's what I meant by "casework."  The point is it's not obvious that the only square roots of $ 9 \bmod 10$ are $ \pm 3 \bmod 10$.  This does not hold generally.

Alternatively, use mod 8. If $ x^2\equiv 9\pmod{100}$, then $ x^2\equiv 9\pmod{200}$ or $ x^2\equiv 109\pmod{200}$. The second case fails modulo 8, so we get the first.


Triangle $ ABC$ is an isosceles triangle with $ AB\equal{}BC$. Point $ D$ is
the midpoint of both $ \overline{BC}$ and $ \overline{AE}$, and
$ \overline{CE}$ is 11 units long. What is the length of $ \overline{BD}$?  Express your answer as a decimal to the nearest tenth.

[asy]draw((0,0)--(3,112^.5)--(6,0)--cycle);
draw((6,0)--(9,112^.5)--(0,0));
label("$A$", (0,0), SW);
label("$B$", (3,112^.5), N);
label("$C$", (6,0), SE);
label("$D$", (4.75,6), N);
label("$E$", (9,112^.5), N);[/asy]

$ \Delta ADB \cong \Delta EDC$ (by SAS) $ \Rightarrow \overline{CE}\equal{}\overline{BA}\equal{}11 \Rightarrow \overline{BD}\equal{}\frac{1}{2} \overline{BC}\equal{} \frac{1}{2} \overline{BA}\equal{}\boxed{5.5}$


Five friends went to a restaurant and agreed to share the bill equally. However, one person forgot his wallet so the other four friends’ portions of the bill went up by $4 each. How many dollars was the total bill?

[hide="solution"]Since there are four other friends, each friend had to pay 16 dollars originally. The total bill was 80 dollars.[/hide]

[quote="anirudh"][quote="bpms"]Isn't this a countdown mathcounts[/quote]

I got it from MOEMS Level E.  :wink: The same problems may repeat though...[/quote]
What are the MOEMS levels, and what are the hardest problems like? Just wondering.

MOEMS Level E is grades 4-6, and it is approximately Math Counts Countdown/Sprint Questions. MOEMS Level M is grade 7-8, and is harder sprints/easy targets. Some MOEMS Level M questions, however, actually are hard targets. They are easy State Targets/hard Chapter Targets, while E is like National Countdown/State Sprints. They give you 30 minutes for 5 questions in both.

[hide][quote="anirudh"]Five friends went to a restaurant and agreed to share the bill equally. However, one person forgot his wallet so the other four friends’ portions of the bill went up by 4 each. How many dollars was the total bill?[/quote]

x/5=(x/4)-4
4x=5x-80
x=80[/hide]

[quote="kstan013"][hide]The bill is 5x, or 4(x+4)=4x+16=5x, so x=16, so 5x, the total bill, equals 5*16=[b]80[/b][/hide][/quote]
You've been here long enough to know not to revive old topics :mad: :mad: :mad:

[quote="myspacesucks"]easy, just use the formula: 5x-80=4x[/quote]

That is not really a formula...
Besides, you didn't explain it. Where did the "80" appear? Please explain.

Sorry, was being vague on the formula part.  It's actually [hide]1/4x=1/5x-4
just add the LCD and you get what I showed above
[/hide]

[quote="@nn@"]:rotfl: Hahahaha easy
$5x = 4x+(4 \cdot 4)$
$5x = 4x+16$
$x = 16$
$5 \cdot 16 = 80$

The bill is $80\$.$[/quote]
I know youre a new member, but some people feel discouraged when you say this is easy and you start laughing about it.
Also dont revive old topics.

I am sorry :( . I didn't want to hurt anyones feelings :noo: . And I apologize for saying that it was easy.


quadratics
modular arithmetic
complex analysis
number theory
arithmetic sequence
number theory proposed

Let $p \equiv 3 \pmod{4}$ be a prime. Show that the set $\{1,2,\ldots,\frac{p-1}{2}\}$ contains more quadratic residues than quadratic nonresidues.

It had not elementary solution. See Borevish, Shafarevish. "Number theory.

I wasn't sure whether the problem of finding an "elementary" solution still open (if we define "elementary" to be "avoiding complex analysis" in our situation). In this case, this thread is not appropriate for this section; can some moderator please move it to the "complex analysis" section? In any case, a classical proof using complex analysis is not hard and is rather beautiful, which is why I posted it :).

No this thread perfectly fits here....
It's number theory and not complex analysis.

Olympiad section?   :roll:  It should be "analytic number theory," but unfortunately, there is no such section.

The word olympiad is just formal.
There are also olympiad problems in Pre-Olympiad etc.  :wink:
See this as the section for serious Number Theory, please.

The class number formula normally is proven using complex analysis (it turns out to be evaluating some residues etc.).
I've never heard of a more elementary solution, but as for the prime number theorem or Dirichlet's problem on primes in arithmetic progression, there could be.

I have never seen such a proof, but the one I read in Davenport's magnifient book Multiplicative number theory is truly stunning by its simplicity. It is true it uses some analysis, but really nothing more sophisticated than the sum of $z^{n}/n$ for complex $z$. Of course, this is not elementary, but it remais however very simple.

[quote="harazi"]  the one I read in Davenport's magnifient book Multiplicative number theory is truly stunning by its simplicity. [/quote]

It is exactly the proof I had in mind (I believe, the simplest known proof so far).

[quote="Megus"]Rust, could you tell me where does it exactly appear in Shafarevich and Borevich's book? I tried to find this problem in this book today but haven't found anything  :(

I'd like to see this complex analysis proof, too :) R.K. Guy in "Unsolved Problems in Number Theory" (F4 if I'm not mistaken) says that the problem has a solution with a use of Dirichlet's class number formula, though nothing about complex analysis mentioned.[/quote]
Read shapter V. Analitic methods.

It seems to be, from a common sense point of view, due to the fact that there are more perfect squares in the range you suggest than there are perfect squares in the upper range. Obviously because those in the lower range are smaller.

Fact is that it is wrong for the primes $\equiv 1 \mod 4$, where you could use the same "reasoning".


If you drive 300 miles at 30 miles per hour and 300 miles at 20 miles per hour, what is your average rate, in miles per hour, for the 600 miles driven?


I'm interested in joining the Upstate NY ARML team this year, but I hear that there is an application that I would need to fill out. Where can I find that?

Also, does anyone have a schedule of that weekend? How do you pick everyone up?

Here is a link to a copy of the application:

http://docs.google.com/Doc?id=dcg9r97w_109fmhcf4cs&hl=en

Although the application states that the deadline was April 15, Mr. Reuter announced last Saturday (April 25) at NYSML that he was continuing to accept applications.  However, time is short, so if you want to apply, you should do so ASAP.


Jack wrote 100 positive numbers between 0 and 3, but he did not write the number 1. Jill wrote the reciprocal of every one of Jack's numbers. How many of these 200 numbers were between 0 and 1?

[hide="hint"] If Jack's numbers were $x_{1}, x_{2}, x_{3}...x_{100}$ and Jills corresponding numbers were $r_{1}, r_{2}, r_{3}...r_{1}00$, then find how many numbers are between 0 and 1 in each "corresponding pair." [/hide]
If you still can't solve it, here is my solution:
[hide="Solution"] Again, say that Jack's numbers are $x_{1}, x_{2}, x_{3}...x_{100}$ and the recipocal of  $x_{i}$ is $r_{i}$.

Now, take any random pair of $(x_{i},r_{i})$.  Either $x_{i}>1$ or $x_{i}<1$.

Now, if $x_{i}<1$, we know that $r_{i}>1$, so there is only one number between 1 and 0 in our pair (which is $x_{i}$).

Now, consider if $x_{1}>1$.  This means that $r_{i}<1$, so there is still only one number between 1 and 0 (in this case, it is $r_{i}$).

Thus, for every "pair", there is only one number that is between 1 and 0.  Since there are 200 numbers, we know that there are 100 pairs.  Thus, there are $100$ numbers between 1 and 0.  [/hide]

[quote="Quevvy"]Is the reciprocal of 1 1? and is the reciprocal of 0 undefined?[/quote]

A reciprocal of $x$ is defined as $\frac{1}{x}$ so the reciprocal of $1$ is $\frac{1}{1}=1$ and the reciprocal of $0$ is undefined because you can't divide by 0.

[quote="Quevvy"]Is the reciprocal of 1 1? and is the reciprocal of 0 undefined?[/quote]
tjhance explained this quite well, but it doesn't really matter for the problem anyways (even though it is very important).  The question said that 1 was not picked, and it also said that positive numbers were picked.  Thus, 0 could not be picked either.


geometry
3D geometry
real analysis
real analysis unsolved

A is a index set, for each a in A,  0<m(Ea)<+inf, Ea in R^n. A maximum family means there is no disjoint Ea with positive finite measure can be added into this family. Why this family has countable members? I know it can't be finite but why can't it be uncountable?

Assign to each $E_{a}$ some cube $Q_{a}$ with integer vertices such that $m(E_{a}\cap Q_{a})>0$. There is a cube $Q$ to which uncountably many sets $E_{a}$ are assigned. For sufficiently large $n$ the index set $\{a\colon m(E_{a})\cap Q>1/n\}$ is uncountable. A contradiction.


linear algebra
matrix
algorithm
combinatorics proposed
combinatorics

Given a n*n(n>=2) matrix such that aii=1,but aij=0(here i is not equal to j).
Each time you can do the following operation:
For any two rows you can add the number in each column and replace them.
It is known that one can make each row the same after a few operations.
Find all n.
(I guess n=2^M)
(For example if n=2 you have (1,0) in the first row and (0.1) in the second row,you will get (1,1) and replace both of them.Then they are the same.)

[quote="zhoraster"]I don't understand the operation.[/quote]

The algorithme is to take two rows $ X\equal{}\{x_1,x_2,...,x_n\},Y\equal{}\{y_1,y_2,...,y_n\}$ and to replace each with $ X\plus{}Y\equal{}\{x_1\plus{}y_1,x_2\plus{}y_2,...,x_n\plus{}y_n\}$


trigonometry
calculus
integration
real analysis
real analysis unsolved

(1) Show that if $0 < L < R <+\infty$, then
$\left| \int_{L}^{R}\frac{\sin x}{x}\, dx \right| < \frac{2}{L}$.

(2) Show that
$\int_{0}^{\infty}\frac{\sin t}{t^{p}}\, dt$
converges iff $0 < p < 2$.  Show that for $0 < p <2$ the integrals
$I_{n}= \int_{(n-1)\pi}^{n\pi}\frac{\sin t}{t^{p}}\, dt$
have the properties $I_{n}= (-1)^{n+1}|I_{n}|$ and $|I_{n+1}| \leq |I_{n}|$ for all $n \in \mathbb N$.  In particular, what is the significance of these properties ?

Hint: integration by parts. Within the integration by parts, take advantage of the cancellation by integrating the thing that wiggles.

If you ignored the cancellation and only looked at size, then $\sin x$ might as well be $1,$ and the antiderivative of $1$ is unbounded. But bring the cancellation back into it: $\sin x$ has a bounded antiderivative.

Special tip for integrals that start at $0:$ when taking antiderivatives, one may add any constant that is convenient or useful. If $dv=\sin x\,dx,$ there's a lot to be said for setting $v=1-\cos x.$

I'm not sure if the way I did these uses your tips, but this is what I had:

1.  $\int_{L}^{R}\frac{\sin x}{x}\, dx = \frac{\cos L}{L}-\frac{\cos R}{R}-\int_{L}^{R}\frac{\cos x}{x^{2}}\, dx$
So
$\int_{L}^{R}\frac{\sin x}{x}\, dx > \frac{\cos L}{L}-\frac{\cos R}{R}-\int_{L}^{R}\frac{dx}{x^{2}}= \frac{\cos L}{L}-\frac{\cos R}{R}+\frac{1}{R}-\frac{1}{L}\geq \frac{-2}{L}$
and
$\int_{L}^{R}\frac{\sin x}{x}\, dx < \frac{\cos L}{L}-\frac{\cos R}{R}+\int_{L}^{R}\frac{dx}{x^{2}}= \frac{\cos L}{L}-\frac{\cos R}{R}-\frac{1}{R}+\frac{1}{L}\leq \frac{2}{L}$

2.  I split the integral into $\int_{0}^{1}\frac{\sin t}{t^{p}}\, dt+\int_{1}^{\infty}\frac{\sin t}{t^{p}}\, dt$.  For the latter, integrate by parts twice to get
$\int_{1}^{R}\frac{\sin t}{t^{p}}\, dt = \left[-\frac{\cos t}{t^{p}}\right]_{1}^{R}-p \left[ \frac{\sin t}{t^{p+1}}\right]_{1}^{R}-p(p+1) \int_{1}^{R}\frac{\sin t}{t^{p+2}}\, dt$,
and this converges as $R \rightarrow \infty$ only if $p > 0$.
For the other part of the integral, since $\sin x \leq x$ on the interval
$\int_{0}^{1}\frac{\sin t}{t^{p}}\, dt \leq \int_{0}^{1}t^{-p+1}\, dt$
so by comparison this converges for at least $p < 2$.  Now if we can show the integral diverges for $p =2$, it will for all $p > 2$ by comparison.  This is the case, because
$\int_{\delta}^{1}\frac{\sin t}{t^{2}}\, dt = \left[-\frac{\sin t}{t}\right]_{\delta}^{1}+\int_{\delta}^{1}\frac{\cos t}{t}\, dt$
and as $\delta \rightarrow 0+$ the limit of $\sin \delta / \delta$ converges, but the integral on the right diverges because $\cos t > 1/2$ for $t$ close enough to $0$.  Combining the range of values for $p$ so that both pieces converge, we have $0 < p < 2$.

KM, please let me know if you see some mistake, and could you explain your hints a little more?  Also, I still don't see the significane of the part of the problem with $I_{n}$.

I think it did not follow before from what i wrote that the integral over $[1,\infty]$ diverges.  I think this works though:
Let $p =-m < 0$. Suppose, for sake of contradiction, that
$\int_{1}^{\infty}t^{m}\sin t \, dt = L \in \mathbb m R ~.$
In order to obtain the contradiction, we first estimate
$\int_{2J\pi+\pi/4}^{2J\pi+\pi/2}t^{m}\sin t \, dt$
for $J \in \mathbb N$.  Making the change of variable $t = u+2J\pi$, we have
\begin{eqnarray*}\int_{2J\pi+\pi/4}^{2J\pi+\pi/2}t^{m}\sin t \, dt & = & \int_{\pi/4}^{\pi/2}(u+2J\pi)^{m}\sin (2J\pi+u) \, du \\ & = & \int_{\pi/4}^{\pi/2}(u+2J\pi)^{m}\sin u \, du \\ & \geq & \frac{1}{\sqrt 2}\int_{\pi/4}^{\pi/2}(u+2J\pi)^{m}\, du \\ & \geq & \frac{1}{\sqrt 2}\int_{\pi/4}^{\pi/2}(2J\pi)^{m}\, du \\ & = & \frac{\pi}{4\sqrt 2}(2J\pi)^{m}> 2 \end{eqnarray*}
for $J$ sufficiently large, say $J > J_{0}$, since for $m > 0$ the limit of the final expression is infinite as $J \rightarrow \infty$.  Now by our assumption, there must exist $M \in \mathbb m R$ such that for $K > M$,
$L-1 < \int_{1}^{K}t^{m}\sin t \, dt < L+1~.$
Choose $J$ such that $2J\pi+\pi/4 > M$ and $J > J_{0}$.  Then we have
$\int_{1}^{2J\pi+\pi/4}t^{m}\sin t \, dt > L-1$
and also by our previous calculation
$\int_{2J\pi+\pi/4}^{2J\pi+\pi/2}t^{m}\sin t \, dt > 2~,$
so
$\int_{1}^{2J\pi+\pi/2}t^{m}\sin t \, dt = \int_{1}^{2J\pi+\pi/4}t^{m}\sin t \, dt+\int_{2J\pi+\pi/4}^{2J\pi+\pi/2}t^{m}\sin t \, dt > L+1~.$
But since $2J\pi+\pi/2 > M$, this contradicts our assumption; it must be that
$\int_{1}^{\infty}\frac{\sin t}{t^{p}}\, dt$
diverges for $p < 0$.

However, this seems excessive  :wink:


inequalities
function
inequalities unsolved

If $ x, y, z$ are nonnegative real numbers such that \[ xy\plus{}yz\plus{}zx\plus{}xyz\equal{}4,\]
prove that \[ x\plus{}y\plus{}z\geq xy\plus{}yz\plus{}zx.\]
(The problem is from Indian TST 1998)

let $ x \equal{}\frac{2a}{b\plus{}c},y \equal{}\frac{2b}{a\plus{}c},z \equal{}\frac{2c}{b\plus{}a}$so $ x\plus{}y\plus{}z\ge xy\plus{}yz\plus{}zx$equiv to $ (a^{2}\plus{}b^{2}\plus{}c^{2})(a\plus{}b\plus{}c)\plus{}3abc\ge 2(bc(b\plus{}c)\plus{}ca(c\plus{}a)\plus{}ab(a\plus{}b))$wich folows by shur    .

[quote="chess"]Why always are there such positive numbers $ a, b, c$ ?
Why do their existence follow from a condition of this problem? :huh:[/quote]

Plug in x=2a/(b+c) and y=2b/(a+c), then solve for z.

Here http://www.mathlinks.ro/viewtopic.php?t=163335.
Hm....actually i didn't know that this is Indian Tst when i posted this ineq.I just found this when i was doing a problem. :wink:

[quote="Cezar Lupu"]hmm, are you sure that $ x\plus{}y\plus{}z\geq xy\plus{}yz\plus{}zx$? I think that the reverse is correct. :roll:[/quote]
I was working on the first post (the problem) without reading these comments until I came across: $ (x,y,z)\equal{}(1,2,\frac{2}5)\implies (LHS,RHS)\equal{}(\frac{17}{5},\frac{18}{5})$

Let $ u\equal{}x\plus{}y,v\equal{}x\minus{}y$. Then we want to minimize \[ A\equal{}\sum(xy)\minus{}\sum(x)\equal{}u^{2}\minus{}v^{2}\plus{}2uz\minus{}z\minus{}2u\]
If we hold $ u,z$ constant then we can minimize this expression by making $ v$ as large as possible.
However, $ u,z,v$ are subject to the constraint $ (z\plus{}1)(u^{2}\minus{}v^{2})\plus{}2uz\equal{}4$
Note that by decreasing $ v$, the constraint forces us to increase $ u$. (This is because the constraint is a decreasing monovariate function in $ v$, and an increasing monovariate function in $ u$)

$ A$ is minimized when $ A\minus{}4\equal{}\minus{}z(u^{2}\minus{}v^{2}\plus{}1)\minus{}2u$ is minimized, which is precisely when $ z(u^{2}\minus{}v^{2}\plus{}1)\plus{}2u$ is maximized.
But clearly to maximize this we want to increase $ u$, decrease $ v$ which is allowed according to the constraint. Thus the minimum of $ A$ occurs when $ v\equal{}0$. Now we want to maximize $ 4\minus{}A\equal{}z(u^{2}\plus{}1)\plus{}2u$ when $ (z\plus{}1)u^{2}\plus{}2uz\equal{}4\iff z\equal{}\frac{2\minus{}u}{u}$
Notice that \[ z(u^{2}\plus{}1)\plus{}2u\equal{}(z\plus{}1)(u^{2}\plus{}1)\minus{}(u\minus{}1)^{2}\equal{}\]\[ \equal{}4\minus{}2uz\plus{}z\plus{}1\minus{}(u\minus{}2)^{2}\equal{}5\minus{}(2u\minus{}1)z\minus{}(u\minus{}1)^{2}\]
but \[ 1\leq (2u\minus{}1)z\plus{}(u\minus{}1)^{2}\iff u\leq (2u\minus{}1)(2\minus{}u)\plus{}u(u\minus{}1)^{2}\]\[ \iff 0\leq u(u\minus{}1)^{2}\minus{}2(u\minus{}1)^{2}\equal{}(u\minus{}2)(u\minus{}1)^{2}\]
Now assume $ z>1$ leads to contradiction as $ 4\equal{}\sum xy\plus{}xyz>\sum 1\plus{}1\equal{}4$
Thus $ z\leq 1$.
Assume $ x\plus{}y<2$; then $ 4\equal{}xy\plus{}z(x\plus{}y)\plus{}xyz<2xy\plus{}2\iff 4<4xy<(x\plus{}y)^{2}\implies x\plus{}y>2$ contradiction, so $ x\plus{}y\geq 2$
As $ x\plus{}y\equal{}u$, we are done
 :D

Because $ xy\plus{}yz\plus{}zx\plus{}xyz\equal{}4$,with out loss of generality we have $ (x\minus{}1)(y\minus{}1)\geq0$
$ x\plus{}y\plus{}z\minus{}xy\minus{}yz\minus{}zx\equal{}x\plus{}y\minus{}xy\plus{}(1\minus{}x\minus{}y)\frac{4\minus{}xy}{x\plus{}y\plus{}xy}$
=$ \frac{(x\minus{}y)^{2}\plus{}(4\minus{}xy)(x\minus{}1)(y\minus{}1)}{x\plus{}y\plus{}xy}\geq0$
Hence, $ x\plus{}y\plus{}z\geq xy\plus{}yz\plus{}zx$

Hmm... we've already determined that result incorrect. I do not follow the "without loss of generality" statement. Why is this?

Oh yes, I speak English not well. :(
If $ x,y,z >1$ , $ xy\plus{}yz\plus{}zx\plus{}xyz>4$
If $ x,y,z<1$ ,$ xy\plus{}yz\plus{}zx\plus{}xyz<4$
So In three numbers have two number less or great than equal to 1 together.Assum two numbers is $ x,y\Rightarrow (x\minus{}1)(y\minus{}1)\geq0$


Suppose that $ a,b,c$ are side lengths of a triangle and $ 0\leq t\leq 1$. Prove that $ \sqrt{\frac{a}{b\plus{}c\minus{}ta}}\plus{}\sqrt{\frac{b}{c\plus{}a\minus{}bt}}\plus{}\sqrt{\frac{c}{a\plus{}b\minus{}tc}}\geq 2\sqrt{1\plus{}t}$.

[quote="Γιώργος"]Suppose that $ a,b,c$ are side lengths of a triangle and $ 0\leq t\leq 1$. Prove that $ \sqrt {\frac {a}{b \plus{} c \minus{} ta}} \plus{} \sqrt {\frac {b}{c \plus{} a \minus{} bt}} \plus{} \sqrt {\frac {c}{a \plus{} b \minus{} tc}}\geq 2\sqrt {1 \plus{} t}$.[/quote]
Because $ \sum_{cyc}\sqrt {\frac {a}{b \plus{} c \minus{} ta}}\geq\sum_{cyc}\frac{2a\sqrt{t\plus{}1}}{a\plus{}b\plus{}c}\equal{}2\sqrt {1 \plus{} t}.$

[quote="Potla"][quote="arqady"][quote="Γιώργος"]Suppose that $ a,b,c$ are side lengths of a triangle and $ 0\leq t\leq 1$. Prove that $ \sqrt {\frac {a}{b \plus{} c \minus{} ta}} \plus{} \sqrt {\frac {b}{c \plus{} a \minus{} bt}} \plus{} \sqrt {\frac {c}{a \plus{} b \minus{} tc}}\geq 2\sqrt {1 \plus{} t}$.[/quote]
Because $ \sum_{cyc}\sqrt {\frac {a}{b \plus{} c \minus{} ta}}\geq\sum_{cyc}\frac {2a\sqrt {t \plus{} 1}}{a \plus{} b \plus{} c} \equal{} 2\sqrt {1 \plus{} t}.$[/quote]
Sorry but I did not understand why $ \sum_{cyc}\sqrt {\frac {a}{b \plus{} c \minus{} ta}}\geq\sum_{cyc}\frac {2a\sqrt {t \plus{} 1}}{a \plus{} b \plus{} c}$...... :huh:

Can you please elaborate a little, arquady? :oops:[/quote]
Because $ \sqrt{\frac{a}{b\plus{}c\minus{}ta}}\equal{}\frac{2a\sqrt{t\plus{}1}}{2\sqrt{b\plus{}c\minus{}ta}\sqrt{(t\plus{}1)a}} \ge \frac{2a\sqrt{t\plus{}1}}{b\plus{}c\minus{}ta\plus{}(t\plus{}1)a}\equal{}\frac{2a\sqrt{t\plus{}1}}{a\plus{}b\plus{}c}$ :)

[quote="quykhtn-qa1"][/quote][quote="Potla"][quote="arqady"][quote="Γιώργος"]Suppose that $ a,b,c$ are side lengths of a triangle and $ 0\leq t\leq 1$. Prove that $ \sqrt {\frac {a}{b + c - ta}} + \sqrt {\frac {b}{c + a - bt}} + \sqrt {\frac {c}{a + b - tc}}\geq 2\sqrt {1 + t}$.[/quote]
Because $ \sum_{cyc}\sqrt {\frac {a}{b + c - ta}}\geq\sum_{cyc}\frac {2a\sqrt {t + 1}}{a + b + c} = 2\sqrt {1 + t}.$[/quote]
Sorry but I did not understand why $ \sum_{cyc}\sqrt {\frac {a}{b + c - ta}}\geq\sum_{cyc}\frac {2a\sqrt {t + 1}}{a + b + c}$...... :huh:

Can you please elaborate a little, arquady? :oops:[/quote][quote="quykhtn-qa1"]
Because $ \sqrt {\frac {a}{b + c - ta}} = \frac {2a\sqrt {t + 1}}{2\sqrt {b + c - ta}\sqrt {(t + 1)a}} \ge \frac {2a\sqrt {t + 1}}{b + c - ta + (t + 1)a} = \frac {2a\sqrt {t + 1}}{a + b + c}$ :)[/quote]
Thanks alot, I understand now! :)


I watched Brother Bear on October 1. For the past four days, I have been in a trance! I can't remember when last a film touched me sooo much!

[img]http://www.superkids.com/aweb/pages/reviews/problem/04/brobear/bear.jpg[/img]

Oh and here is the official website
[url]http://disney.go.com/disneyvideos/animatedfilms/brotherbear/main.html[/url]

What do you think about it people? Please lemme know!

Hello Bubka,

there is already a Movies you just saw topic.
Unfortunately, even though nobody does it, movies are considered and should be "Music &Art", the confusion is selfsustaining however, everyone keeps talking about movies in Games and Fun and nobody bothers to do it in Music &Art.

Putting this in Round Table just makes it worse.

well, I can see that you...uh...well...[i][b]enjoyed[/b][/i] the movie. Maybe you can watch the $2nd$ one and [i][b]double[/b][/i] your collection of photos happiness and...uh...yeah

[quote="kheg_k"]wait... fredbel6 said that it was bubka. :huh:[/quote]

Yes, that's right.  Not only did Bubka explicitly tell me he was gonna do this under a new name "Koda", he even asked me to resize that avatar.

[quote="thatin"]so... was that the end of bubka? but like a phoenix he reborned...![/quote]
Hmm, I doubt it.  Maybe he wants to experiment with his Brother Bear love "under cover"?

Anyway, it's time we face the facts : Brother Bear isn't that good.  Okay, the images (the mountains, the rivers, the waterfalls,..) are gorgeous.  But it's all so out of place?  Mammots carrying bears on their teeth?  Bears living among humans?
The twist was VERY PREDICTABLE.
As usual, this Disney movie must have a few cheesy characters like Timon and Pumba who annoy the more mature audience.  Here, this part is done by those dear.  Their "tree" joke didn't make me laugh at all.  But unlike Timon and Pumba, they serve no purpose AT ALL.  They could have easily been left out off the entire script!

Average at best.

Tuke: I spy something green.
Rutt: Umm ... tree?
Tuke: Yeah. your turn.
Rutt: I spy something tall.
Tuke: Tree?
Rutt: Right.
Tuke: I spy something kwaark.
Rutt: Tree?
Tuke: Right.
Rutt: I spy something ... a vertical log?
Tuke: Tree?
Rutt: Yeah.
Tuke: I spy something ...
Rutt: Tree?
Tuke: Yeah.
Rutt: My turn.
Tuke: Tree.
Rutt: But I didn't spy anything.
Tuke: It counts.
Rutt: This is unfair eh?
Tuke: No it counts.
Rutt: Ok ... tree.
Tuke: Sigh ... let's play something else.

 :rotfl:

 :rotfl:

[img]http://queenbeanie.tripod.com/brotherbear4.jpg[/img]

Sorry that was a bit incorrect. I just watched the film for the fifth time amd finally made it out.

The seventh line reads:
Tuke: I spy something ... with bark?

[img]http://www.filmbuffonline.com/images/BrotherBear1.jpg[/img]

ok, that was a funny part, [i]the first time[/i]

[quote="Koda"]Tuke: I spy something green.
Rutt: Umm ... tree?
Tuke: Yeah. your turn.
Rutt: I spy something tall.
Tuke: Tree?
Rutt: Right.
Tuke: I spy something kwaark.
Rutt: Tree?
Tuke: Right.
Rutt: I spy something ... a vertical log?
Tuke: Tree?
Rutt: Yeah.
Tuke: I spy something ...
Rutt: Tree?
Tuke: Yeah.
Rutt: My turn.
Tuke: Tree.
Rutt: But I didn't spy anything.
Tuke: It counts.
Rutt: This is unfair eh?
Tuke: No it counts.
Rutt: Ok ... tree.
Tuke: Sigh ... let's play something else.
[/img][/quote]

I think that there is a similar dialogue at NEMO!

I have downloaded and watched brother bear 2. Not as good, of course, but still quite okay.

[img]http://www.geocities.jp/pochi650127_1/bb2/bb2logo.jpg[/img]

Downloading stuff is illegal. This is not the place to spread your propaganda about how they make millions of dollars anyway and how a few downloads won't affect them.

Bubka, come on, don't tell us you will make an account for each movie you see. :spam:

Here is a thing that is bothering me:

[hide="WARNING! Spoiler!"]There is some confusion about the time when Kenai decides to remain a bear. When Kenai says "He needs me", Denahi smiles. Most sites report him as saying next "You know, you did look better as a bear!" to Kenai, while what he really says next is "You know, HE did look better as a bear" to Sitka. Sitka is shown next, looking from Denahi to Kenai with a thoughtful smile. Even sites like wikiquote and imdb make this mistake.  :mad:
[img]http://asg.homelinux.org/albums/kenai/Kenai_Denahi.jpg[/img][/hide]
If you are wondering why I didn't put a picture, don't worry - it is in the hidden part!

[quote="deimos"]bubbka, or better to say koda, i remember that i've watched a part of a cartoon where an eskimo was turned into a bear by some spirit, and in the end he plays with his family bear and then spirit came with his dad but that eskimo-bear doesn't want to become human again. is that the brother bear movie?[/quote]

Not his dad - his big brother Sitka. But yes that is it.

I have reason to be elated. Yesterday evening I showed the movie to a friend. Then I gave him the CDs. Today evening I met him after my first guitar lesson. He has already watched it twice more!  :D

[img]http://images.easyart.com/i/prints/rw/lg/1/3/Disney-Koda-s-Antics-134289.jpg[/img]

I have a confusion over this point I read about in a review. It says that Kenai appears underwater twice: first when the shark upsets his boat, next in the Salmon Run. It says that there is a contrast between the frightened Kenai under hostile circumstances in the first scene, and the comfortable Kenai under friendly circumstances in the second, because in the intermediate time Kenai has learnt to love Nature. It is all very well to say that, but don't you think that there is also a sharp contrast between sharks and the salmons?  :wink: Please reply sincerely (though not necessarily seriously) everyone.

Sharks are evil  :!:  Every five year old knows that :D  They kill people... and stuff. Either that, or people watched [url=http://www.imdb.com/title/tt0073195/]Jaws[/url] too many times :o

[quote]I'd say yes, but you obviously missed the bears-hitchhiking-on-mammoth's-teeth thing, because you would have included it otherwise.  :wink:[/quote]

[img]http://www.cassioposa.net/photos/inline/o-rly.jpg[/img]

[quote="Ignite168"][quote="bubka"][quote="neelnal"]bubka, how old are you? honestly...

If you are 13+, you should be watching LOTR, or something 7 times, not BB[/quote]

I turned 8 this July.[/quote]

[img]http://files.myopera.com/drlaunch/albums/37656/no-wai001.jpg[/img][/quote]
 :rotfl:  :rotfl:  :rotfl:
8year old college student huh :wink:

[quote="junggi"][/quote][quote="Ignite168"][quote="bubka"][quote="neelnal"]bubka, how old are you? honestly...

If you are 13+, you should be watching LOTR, or something 7 times, not BB[/quote]

I turned 8 this July.[/quote]

[img]http://files.myopera.com/drlaunch/albums/37656/no-wai001.jpg[/img][/quote][quote="junggi"]
 :rotfl:  :rotfl:  :rotfl:
8year old college student huh :wink:[/quote]

Did I say 8? I meant 18 of course! Somehow missed the 1 - silly typo ... :rotfl:

[quote="neelnal"]ok so ure 18 and u go to the movies to watch brother bear.

you are scaring me right now. :rotfl:[/quote]
Not necessarily, he probably watched it on his computer... and the copy could very well be illegal.

[quote="neelnal"]ok fine then. he is risking a .0001% chance of getting caught just to watch brother bear!!!!

lol sorry bubka or koda, whomever, but it is too funny imagining an 18 year old sitting at the computer downloading brother bear  :rotfl:

note: i am just having fun, not trying to hurt anyone[/quote]
I actually watched Brother Bear on my computer as well.
I'm 21.  A 20 year old friend told me it was decent.
But it was average at best.

[quote="kstan013"]my 8 year old [i][b]sister[/b][/i] watched that movie :rotfl:[/quote]

Only proving it is a movie for all ages :rotfl:

*cough*

Where did you download it from? If you downloaded it from itunes, it is perfectly legal.

*coughs more* i thought that brother bear was a kid movie not something bubka should be making a whole new account because it wowed him so much


Ten days from Thursday, it will be Sunday. What day of the week will it be in 1,000,000 days from Thursday?

Weeks are $ 7$ days long, so we want to find the closest multiple of $ 7$ below $ 1,000,000$.

So we divide: $ \frac{1000000}{7} \equal{} \text {about} 142857$

and then multiply that by $ 7$: $ 142857 \times 7 \equal{} 999999$

Now we subtract: $ 1000000 \minus{} 999999 \equal{} 1$

So $ 1000000$ days after that Thursday is one day after Thursday, or Friday.

Our answer is $ \boxed{\text{Friday}}$

[b]or you can just just divide 1000000 by 7 and see what the remainder will be. The remainder is 1 so you add one day to Thursday: Friday[/b]


Suppose that $ a,b,c$ are pairwise coprime, and let $ \phi \equal{} ab\plus{}bc\plus{}ca$. Show that $ (a,\phi)\equal{}1$, and hence that $ (abc,\phi)\equal{}1$.

Suppose there is a prime $ p$ which divides both $ a$ and $ \phi$. Then, since $ \phi \equal{} ab \plus{} bc \plus{} ca$, we have $ p \mid bc$ (because $ p \mid a \rightarrow p \mid ab \plus{} ca$).
So $ p$ divides $ b$ or $ c$. But $ p$ also divides $ a$, a contradiction since $ a,b,c$ were pairwise coprime. Therefore there is no such prime $ p$, and $ (a,\phi)\equal{}1$.

Analogously, we see that $ (b,\phi)\equal{}1$ and $ (c,\phi)\equal{}1$. Again, if there is a prime $ q$ which divides both $ abc$ and $ \phi$, then $ q$ divides $ a$, $ b$, or $ c$, which is a contradiction. Hence $ (abc,\phi)\equal{}1$.

easy.

let $ (a,\phi)\equal{}d$. since $ d|a$, we get $ d|ab\plus{}ca$, so $ d|\phi\minus{}(ab\plus{}ca)\equal{}bc$. so $ d|(a,bc)\equal{}1$, implying that $ d\equal{}1$. similarly, $ (b,\phi)\equal{}1, (c,\phi)\equal{}1$, so $ (abc,\phi)\equal{}1$.


[hide]To write the fraction in mixed form ($a\frac{b}{c}$), divide the denominator into the numerator. The answer you get is the integeral part, $a$, and the remainder is the numerator of the fractional part, $b$. $c$ is just the original denominator.
[hide="answer"]$4149\frac{1}{3}$[/hide][/hide]

[quote="Ignite168"]Tell your teacher that is proper form.[/quote]
 Indeed it is.

[quote="Anonymous52111"]How do you solve this problem into a proper writing?

1. 12448/3[/quote]

You want common fraction or mixed?
That is already common fraction
Mixed is $4149 and 1/3$


modular arithmetic
algebra unsolved
algebra

Find the coefficient of $ x^{n}$ in the series expansion of $ \frac{1\plus{}x}{(1\minus{}x)^{2}(1\plus{}x^{2})}$, where $ \minus{}1<x<1$.

hello, about which point should the series expansion be derived?
Sonnhard.

[quote="AndrewTom"]Find the coefficient of $ x^{n}$ in the series expansion of $ \frac {1 \plus{} x}{(1 \minus{} x)^{2}(1 \plus{} x^{2})}$, where $ \minus{} 1 < x < 1$.[/quote]

$ \frac{1\plus{}x}{(1\minus{}x)^2(1\plus{}x^2)}\equal{}\frac{(1\plus{}x)(1\minus{}x^2)}{(1\minus{}x)^2(1\minus{}x^4)}$ $ \equal{}\frac{(1\plus{}x)^2}{(1\minus{}x)(1\minus{}x^4)}$

$ \frac 1{1\minus{}x}\equal{}\sum_{k\equal{}0}^{\plus{}\infty}x^k$

$ \frac 1{1\minus{}x^4}\equal{}\sum_{k\equal{}0}^{\plus{}\infty}x^{4k}$

$ \implies$ $ \frac 1{(1\minus{}x)(1\minus{}x^4)}\equal{}\sum_{k\equal{}0}^{\plus{}\infty}(1\plus{}\left[\frac k4\right])x^k$

$ \implies$ $ (1\plus{}x)\frac 1{(1\minus{}x)(1\minus{}x^4)}\equal{}$ $ 1\plus{}\sum_{k\equal{}1}^{\plus{}\infty}(2\plus{}\left[\frac k4\right]\plus{}\left[\frac {k\minus{}1}4\right])x^k$ $ \equal{}\sum_{k\equal{}0}^{\plus{}\infty}(2\plus{}\left[\frac k4\right]\plus{}\left[\frac {k\minus{}1}4\right])x^k$

$ \implies$ $ (1\plus{}x)^2\frac 1{(1\minus{}x)(1\minus{}x^4)}\equal{}$ $ 1\plus{}\sum_{k\equal{}1}^{\plus{}\infty}(4\plus{}\left[\frac k4\right]\plus{}2\left[\frac {k\minus{}1}4\right]\plus{}\left[\frac {k\minus{}2}4\right])x^k$ $ \equal{}\sum_{k\equal{}0}^{\plus{}\infty}(4\plus{}\left[\frac k4\right]\plus{}2\left[\frac {k\minus{}1}4\right]\plus{}\left[\frac {k\minus{}2}4\right])x^k$

Hence the answer : $ \boxed{\frac{1\plus{}x}{(1\minus{}x)^2(1\plus{}x^2)}\equal{}\sum_{k\equal{}0}^{\plus{}\infty}(4\plus{}\left[\frac k4\right]\plus{}2\left[\frac {k\minus{}1}4\right]\plus{}\left[\frac {k\minus{}2}4\right])x^k}$ $ \equal{}1\plus{}3x\plus{}4x^2\plus{}4x^3\plus{}5x^4$ $ \plus{}7x^5\plus{}8x^6\plus{}8x^7\plus{}9x^8\plus{}$ $ 11x^9\plus{}12x^{10}\plus{}...$

(with maybe some more details on convergences and product of series)

Looking at my previous post, I see that coefficient of $ x^n$ is $ n\plus{}1$ for $ n\equal{}0\pmod 4$ and $ n\equal{}3\pmod 4$ and is $ n\plus{}2$ for $ n\equal{}1\pmod 4$ and $ n\equal{}2\pmod 4$

So my result is $ \sum(n\plus{}1)x^n\plus{}x\plus{}x^2\plus{}x^5\plus{}x^6\plus{}x^9\plus{}x^{10}\plus{}...$ $ \equal{}\left(\sum x^n\right)'\plus{}(x\plus{}x^2)\sum x^{4n}$ $ \equal{}\left(\frac 1{1\minus{}x}\right)'\plus{}x(x\plus{}1)\frac 1{1\minus{}x^4}$ $ \equal{}\frac 1{(1\minus{}x)^2}\plus{}x(x\plus{}1)\frac 1{1\minus{}x^4}$

And so a much more simple ( :blush: ) way to get the same result : $ \frac{1\plus{}x}{(1\minus{}x)^2(1\plus{}x^2)}\equal{}\frac 1{(1\minus{}x)^2}\plus{}x(x\plus{}1)\frac 1{1\minus{}x^4}$ $ \equal{}\left(\sum x^n\right)'\plus{}(x\plus{}x^2)\sum x^{4n}$


geometry
circumcircle
search
cyclic quadrilateral
geometry unsolved

Let $ ABC$ cyclic quadrilateral and tangent line of circumcircle $ ABCD$ on $ B$ and $ D$ meet at $ P$
Prove that $ A,C,P$ collinear if and only if $ AB.CD\equal{}AD.BC$


there was a pretty easy problem on 39th IMO:
find all pairs of $(a,b)$ from positive integers, where $a^2b+a+b$ would be divisible by $ab^2+b+7$...
(please post original solutions, don't just copy it off somewhere else... :lol: )

come, on guys, this is easy,
need a hint??? :D

whats the hint ,something to do with the HCF

Let $ab^2+b+7=x, a^2b+a+b=y$. Suppose $x \mid y$. Then $x \mid by-ax=b^2-7a$, so $ab^2+b+7 \mid b^2-7a$.

Now if $b^2-7a$ is positive, $ab^2+b+7>b^2+b>b^2>b^2-7a$, contradiction. If $b^2-7a$ is zero, then $b^2=7a$, so $a$ is of the form $7k^2$, and $b^2=49k^2, b=7k$, so we have the solution $(7k^2, 7k)$ which we can see works as $7^3k^4+7k+7 \mid 7^3k^5+7k^2+7k=k(7^3k^4+7k+7)$.

Finally $b^2-7a$ may be negative. Then since $ab^2+b+7 \mid 7a-b^2$, we need $ab^2<ab^2+b+7 \leq 7a-b^2<7a$, so $ab^2 \leq 7a, b^2 \leq 7$. So $b=1$ or $b=2$. If $b=1$, then we need $a+8 \mid a^2+a+1$; since $a+8 \mid (a+8)(a-7)=a^2+a-56$, then $a+8 \mid 57$. Also $a+8$ is at least $9$ so it is $19$ or $57$, giving $a=11, a=49$, both of which work. If $b=2$, then $4a+9 \mid 2a^2+a+2, 4a+9 \mid 4a^2+2a+4, 4a+9 \mid 4a^2+9a, 4a+9 \mid 7a-4, 4a+9 \mid 3a-13$. So $3a-13$ can't be positive, but it clearly can't be zero either because it's not divisible by 3. Then it must be negative; but then $13-3a$ is at most 10, while $4a+9$ is at least 13. So there are no solutions in this case.


Can anyone post problem  Romani TST and MO 2009?


i had this problem on elias saab's website:

the obtuse angle of an isosceles triangle is bisected and each resulting angle is $ 50$ degrees larger than a base angle. how many degrees are in the obtuse angle?
i know the answer is $ 140$

if we say each resulting angle is $ x$ degrees larger than a base angle, would the obtuse angle be $ x+90$ degrees?

this is how i would do it:
let each base angle be $ x$ and since they're congruent, the base angles are both $ x$. now let each resulting, bisected angle be $ x+50$. therefore the obtuse angle is $ 2x+100$. now set up the equation $ x+x+2x+100=180$. you find that $ x=20$. substituted that back into the $ 2x+100$ and get that the obtuse angle is $ 140$ degrees.
:)

yes i know that but this was a countdown round problem(so it says) and i want to know  what the obtuse angle would be if the part about each resulting angle were 'x' degrees instead of 50.

im not really sure i understand your question..if each base angle was $ x$, then the obtuse angle would be $ 2x+100$, if each resulting, bisected angle was $ x$, then the obtuse angle would be$ 2x$. that's what i think.

ok then i will state the question the way it was given to me and i will restate it my way

website: the obtuse angle of an iscoscelles triangle is bisected and each resulting angle is 50 degrees larger than a base angle. how many degrees are in the obtuse angle?

my way: the obtuse angle of an iscoscelles triangle is bisected and each resulting angle is $ x$ degrees larger than a base angle. how many degrees are in the obtuse angle? express your answer in terms of $ x$.

im pretty sure my way's answer is $ x+90$

[quote="mountain.dew"][hide]If we have an isosceles triangle ABC with AB equal to AC, the measure of angle B equals the measure of angle C, which we'll call $ a$. Let the angle bisector of angle BAC meet BC at M. So, the measure of angle CAM = $ x+a$ = the measure of angle BAM.
$ 2(x+a)+a+a=180$
$ a = \frac{90-x}{2}$
the measure of angle BAC =$ 2a+2x=x+90$
So, you were right.  :) [/hide][/quote]

ok thanks

wait was that even the problem? :maybe:
i thought we were supposed to actually find angle measures, in degrees..

no i like to think of things that way. it helps me understand the problem bettter.


We have a circle and a line which does not intersect the circle. Using only compass and straightedge, construct a square whose two adjacent vertices are on the circle, and two other vertices are on the given line (it is known that such a square exists).

If $T$ is the projection of the center of the circle on the line, then construct a square $ABCD$ with two vertices $A,B$ on the line s.t. $T$ is the midpoint of $AB$ and s.t. $C,D$ are on the same side of the line as the circle.

Now draw lines $TA,TB$ . They intersect the circle in some points. Take $U\in TA,V\in TB$ to be two such points s.t. $UV$ is parallel to the line. Now let $P,Q$ be the projections of $U,V$ on the line. $PQVU$ is our square.


geometry
circumcircle
incenter
analytic geometry
geometric transformation
homothety
trigonometry

Let $ A^{\prime}$ be an arbitrary point on the side $ BC$ of a triangle $ ABC$. Denote by $ \mathcal{T}_{A}^{b}$, $ \mathcal{T}_{A}^{c}$ the circles simultanously tangent to $ AA^{\prime}$, $ A^{\prime}B$, $ \Gamma$ and $ AA^{\prime}$, $ A^{\prime}C$, $ \Gamma$, respectively, where $ \Gamma$  is the circumcircle of $ ABC$. Prove that $ \mathcal{T}_{A}^{b}$, $ \mathcal{T}_{A}^{c}$ are congruent if and only if $ AA^{\prime}$ passes through the Nagel point of triangle $ ABC$.

([i]If $ M,N,P$ are the points of tangency of the excircles of the triangle $ ABC$ with the sides of the triangle $ BC$, $ CA$ and $ AB$ respectively, then the Nagel point of the triangle is the intersection point of the lines $ AM$, $ BN$ and $ CP$[/i].)

[b]Lemma.- $ \alpha$ is the circumcircle of $ ABC$,Let $ D$ be an arbitrary point on the side $ BC$ of a triangle $ ABC$ . $ \beta$ is a circle tangent to $ \alpha$,$ AD$ and $ BC$, in $ D_0$,$ X$ and $ A_1$ respectively. Prove that $ I \in XA_1$where $ I$: Incenter of $ ABC$.[/b]
Dem:
(WLG-$ A_1$ between $ D$ and $ C$.)
$ IA \cap \alpha = M$, know that by the theorem Catalan $ IM=MB=MC=R$.
Let $ \phi$ cirumcircle of $ BIC$ with center $ M$,Know that $ \alpha$ and $ \beta$ They are homothetic with Homothetic transformation centre in D.
Then, $ M,A_1,D_0$ are collinear , $ AD \cap \beta =T$, $ D_0,T,A$ are collinear, moreover $ TA_1$//$ AM$.
If $ XA_1 \cap MA=L$, $ <MAD_0=<A_1TD_0=<A_1XD_0$, Then the circle $ \theta$ passes through points $ X,L,D_0,A$.
Investing on the circumference $ \phi$ .
[b]-[/b]$ \alpha$ Invested in the line $ BC$,then $ \beta$ is invested in one circle tangent to the line $ BC$ and $ \alpha$,But $ M,A_1,D_0$, are collinear, then $ \beta$ invested in $ \beta$.
if $ XM \cap \beta=T_1$, $ A_0=AM \cap BC$,
$ A,X$ is invested in $ A_0,T_1$,respectively.
Then, $ R^2=MT_1(MX)=ML(MA)$, then $ <T_1A_0A=<T_1XD=<XA_1T_1$, then the circle $ \mu$ passes through points $ T_1,A_0,A_1,L$.
Then $ \theta$ (circumcircle of $ XAD_0$), is invested in $ \mu$ (circumcircle of $ T_1A_0A_1$), Then $ L$ is invested in $ L$,  hence $ L \in \phi$, but $ \phi \cap AM=I$, then $ I=L$.
 Then $ I \in XA_1$.
                           $ Q.E.D.$
------------
In the problem.
Let $ A'$ be an arbitrary point on the side $ BC$ of a triangle $ ABC$. Denote by $ T^b_A$,$ T^c_A$ the circles simultanously tangent to $ AA'$,$ A'B$,$ \gamma$ and $ AA'$,$ A'C$ ,$ \gamma$ , respectively, where $ \gamma$ is the circumcircle of $ ABC$.
*) [b]If $ T^b_A$,$ T^c_A$ are congruent. [/b]
$ BC \cap T^b_A=B_1$,$ T^b_A \cap AA'=X$
$ BC \cap T^c_A=C_1$,$ T^c_A \cap AA'=Y$
Know by Lemma.($ I$:incenter of $ ABC$)
$ I \in B_1X$, $ I \in C_1Y$, Then , $ B_1X \cap C_1Y =I$
moreover, $ <YC_1A'=<C_1YA'$ => $ <XA'B_1=2<C_1YA'$, then $ <XB_1A'=90º-<C_1YA'$, then $ <B_1IC_1=90º$
Now, $ M$ is mind point of $ XY$, then $ <MIY=<MYI=<A'YC_1=<YC_1A'$, then $ IM$//$ BC$, If $ O_b$ and $ O_c$ are center of $ T^b_A$, $ T^c_A$ respectively.Know that $ O_a$,$ O_b$ and $ M$ are collinear, moreover, $ O_aO_b$//$ BC$(because $ O_bB_1=O_cC_1$ and $ O_bB_1$//$ O_cC_1$)
then $ I \ O_bO_c$,
$ K$ is the point of tangency of the circle inscribed to $ ABC$ with $ CB$.
then, $ IK=O_cC_1=O_bB_1$, moreover, $ <B_1IK=<O_cIC_1=<A'O_cC_1=\alpha$ (because $ <OB_1C_1=<B_1IC_1=90$)Then $ B_1K=(IK)Tang( \alpha)$, $ A'C_1=(O_cC_1)Tang( \alpha)$, (0<  $ \alpha$<90)
Then $ B_1K=A'C_1$, moreover $ BB_1=C_1C$( by simetry),
Then $ A'$ is the points of tangency of the excircles of the triangle relative to $ BC$, hence $ AA'$, passes through the Nagel point of triangle $ ABC$.
                                                              $ Q.E.D.$
[b]**)If $ AA'$, passes through the Nagel point of triangle $ ABC$[/b]
Suppose the opposite,$ T^b_A$,$ T^c_A$ aren't congruent.
Then Exist  $ X \in BC$,$ X$ different $ A'$,such that the circles tangents to $ XC,AX, \gamma$, $ BX,AX, \gamma$, are congruent,It is true, By the continuity of $ A'$ in $ BC$, (WGL-$ AB \leq AC$), then if $ A'$ is approaching $ B$,radius of $ T^b_A$ is $ r_b$ and radius of $ T^c_A$ is $ r_c$, then $ r_b \leq r_c$, but if $ A'$ is approaching $ C$, $ r_c \leq r_b$.
 Then by (*), $ AX$ passes through the Nagel point of triangle $ ABC$.(-><-)
next,  If $ AA'$, passes through the Nagel point of triangle $ ABC$, then $ T^b_A$,$ T^c_A$ are congruent.
                                       $ Q.E.D.$

1.this come easy by trilinear or baricentric coordinates
follows further easily by invertion

Let $ T_A^b$ touches $ \Gamma,\ BC,\ AA'$ at the points $ K,\ X,\ P$ respectively
and $ T_A^c$ touches $ \Gamma,\ BC,\ AA'$ at the points $ L,\ Y,\ S$ respectively
Let $ M$ the midpoint of the small arc $ BC$ . Then the homothety with center $ K$ which sends $ T_A^b$ to $ \Gamma$ sends $ X$ to $ M$ so $ K,X,M$ are collinear and in a similar way we can see that $ L,Y,M$ are also collinear .
From Casey's theorem to the 4-tuple of circles $ (A,T_A^b,B,C)$ we have
$ AP\cdot BC + BX\cdot AC = AB\cdot CX \\
\Leftrightarrow (AA' - A'X)BC + BX\cdot AC = AB(BC - BX) \\
\Leftrightarrow AA'\cdot BC - (BA' - BX)BC + BX\cdot AC = AB\cdot BC - BX\cdot AB \\
\Leftrightarrow BX(AB + BC + CA) = AB\cdot BC + BA'\cdot BC - AA'\cdot BC$ (1)

In a similar way we find that
$ CY(AB + BC + CA) = AC\cdot BC + CA'\cdot BC - AA'\cdot BC$ (2)

We will prove first that if $ AA'$ passes through Nagel point then the circles have equal radius .Then since $ BA' = p - c$ and $ CA' = p - b$ (where $ p$ is the semiperimeter) the relation (1) and (2) become $ BX(AB + BC + CA) = pBC - AA'\cdot BC$ and $ CY(AB + BC + CA) = pBC - AA'\cdot BC$
so $ BX = CY$ . From this we obtain that $ XY$ and $ BC$ share the same midpoint and $ M$ is the midpoint of arc $ BC$ so if $ O$ is the circumcenter of $ \Gamma$ then $ OM$ is the midperpedicular of $ XY$ so $ MXY$ is an isosceles triangle and so $ \angle{KXB} = \angle{LYC}$
But also $ \angle{YLC} = \angle{XKB}$ so the triangles $ KBX,\ LYC$ are equal so $ KX = LY$ (3) .
Also $ \angle{LYC} = \frac {\angle{YO_2L}}{2}$ and $ \angle{KXB} = \frac {\angle{KO_1X}}{2}$
so ${ \angle{YO_2L}} = \angle{KO_1X}$ (4) . From (3),(4) and the law of sines we get $ R_1 = R_2$

Now we will prove that if the circles have equal radius then $ AA'$ passes through Nagel point. Since the radius are equal we have that $ KX = LY$ and also $ \angle{KXB} = \angle{LYC}$
and also $ \angle{BKM} = \angle{MLC}$ so the triangles $ LYC,\ KXB$ are equal so $ BX = YC$
and from (1) and (2) we find that $ AC\cdot BC + CA'\cdot BC = AB\cdot BC + BA'\cdot BC$
and $ BA' = BC - CA'$ and so $ CA' = p - b$ so $ A'$ is the $ A$ excircle touching point , the desired.

***Note
From the Thebault's theorem we obtain that the points $ O_1,\ I,\ O_2$ are collinear so
$ R_1=R_2=r$ where $ r$ is the radius of the incircle !


[quote="Crux"]I proved following inequality but I wonder if there are some non trivial equality cases.
$ a,b,c \in \mathbb{R}$, \[ a^4\plus{}b^4\plus{}c^4\plus{}2(a^3b\plus{}b^3c\plus{}c^3a)\plus{}11(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)\geq 6(ab^3\plus{}bc^3\plus{}ca^3)\plus{}8abc(a\plus{}b\plus{}c)\]
Can you prove it ?[/quote]

$LHS-RHS=\sum [\frac{1}{2}\left ( a^2+\frac{8}{3}ab+\frac{2}{3}bc-b^2-\frac{10}{3}ca \right )^2+\frac{4}{3}a^2(b-c)^2] \ge 0 $
Equality occurs  iff $a=b=c$.

See more here
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2492966#p2492966[/url]

[quote="Crux"]I proved following inequality but I wonder if there are some non trivial equality cases.
$ a,b,c \in \mathbb{R}$, \[ a^4\plus{}b^4\plus{}c^4\plus{}2(a^3b\plus{}b^3c\plus{}c^3a)\plus{}11(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)\geq 6(ab^3\plus{}bc^3\plus{}ca^3)\plus{}8abc(a\plus{}b\plus{}c)\]
Can you prove it ?[/quote]
I can prove it, because I found a nice equation.
\[ LHS-RHS = \sum(b^2-3ab+bc+ca)^2 \ge 0.\]
So, the inequality was proved.
P/s: Sorry for my bad English :D.


geometry
MATHCOUNTS
inequalities
triangle inequality

Okay, so States aren't very far away, and I feel pretty good about Counting Problems and that type of problem. Currently, I am taking Algebra, so I have little exposure to anything that has to do with Geometry. So I was wondering if you could all help me condense a lot of the necessary things I need to know into this thread. (Posting links to websites and other threads would be great too!)

Now I know things like: Pythag. Triples, Area/Volume/Surface Area of Shapes/45-45-90 and 30-60-90 Triangles.

I see a lot of problems that involve Circles, so are there any properties of a circle/tips/tricks/formulas that could help out? Feel free to post examples too, I would GREATLY appreciate it.

Also, I know that AOPS does have a Geometry book, which I have heard is fantastic, but I think that at the place I am at right now, buying the book (although I definately will for the future) would be of no use to me with so little time left.

Anyway, I greatly appreciate any help anybody is willing to offer!

Okay. I take Geometry, but it is very hard to sum up because it has a LOT of theorems, postulates, and corollaries... :( sorry

I suggest finding a textbook about geometry and if you don't have enough time, just skim through the theorems.

Here are a few...

Polygon Inequality Postulates
    Triangle Inequality Postulate: The sum of the lengths of two sides of any triangle is greater than the length of the third side.
    Quadrilateral Inequality Postulate: The sum of the lengths of 3 sides of any quadrilateral is greater than the length of the fourth side.

I guess you could look at this site, the circle & angles one is very important.
[url]http://www.sparknotes.com/math/geometry2/theorems[/url]


Let $ m$ be a positive integer and $ p \equal{} 6m \minus{} 1$ be a prime. Prove that the numerator of $ 1 \minus{} \frac {1}{2} \plus{} \frac {1}{3} \minus{} \frac {1}{4} \plus{} \ldots \minus{} \frac {1}{4m \minus{} 2} \plus{} \frac {1}{4m \minus{} 1}$ is divisible by $ p$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=151379 or Theorem 1 in http://www.mathlinks.ro/Forum/viewtopic.php?t=150400 - it's even a bit more general.

  darij


function
functional equation
algebra
IMO
IMO 1981

The function $f(x,y)$ satisfies: $f(0,y)=y+1, f(x+1,0) = f(x,1), f(x+1,y+1)=f(x,f(x+1,y))$ for all non-negative integers $x,y$. Find $f(4,1981)$.

See
- http://www.mathlinks.ro/Forum/viewtopic.php?t=49767
- http://www.mathlinks.ro/Forum/viewtopic.php?highlight=f%280%2Cy%29&t=16384
- http://www.mathlinks.ro/Forum/viewtopic.php?highlight=f%280%2Cy%29&t=49912
etc.

We know that $f(0,y)=y+1$. We can then find formulae for $f(1,y)$, $f(2,y)$ and $f(3,y)$ in terms of $y$:
\begin{align*}
f(1,y)&=f(0,f(1,y-1))\\&=f(1,y-1)+1\\&=\cdots\\&=f(1,0)+y\\&=y+2\\
\end{align*}
and similarly:
$$f(2,y)=2y+3$$
$$f(3,y)=2^{y+3}-3$$
We know that $f(4,0)=13=2^4-3=2^{2^2}-3$. Assume, for induction, that $f(4,y)=\,^{y+3}2-3$ where $^ba$ represents $a^{a^{\iddots^a}}$ with a power tower of length $b$. We know that this holds true for $f(4,0)$ and that:
$$f(4,y)=2^{f(4,y-1)-3}-3=2^{(\,^{y+2}2)-3+3}-3=\,^{y+3}2-3$$
completing the induction. So, our answer is:
$$\boxed{^{1984}2-3}$$

Edit: I was surprised this problem was not given a solution on this site despite being an IMO problem and several years old. So, I gave it one...


calculus
integration
trigonometry
logarithms
real analysis
real analysis unsolved

These are easy integrals from my old book. But I think they may be unknown for you.
1. $ \int \frac {e^{arctgx}}{(1 \plus{} x^{2})^{\frac {3}{2}}}\,dx$.
2. $ \int \frac {xe^{arctgx}}{(1 \plus{} x^{2})^{\frac {3}{2}}}\,dx$.
3. $ \int \frac {x^{2}arcsinx}{\sqrt {1 \minus{} x^{2}}} \,dx$.
4. $ \int \frac {x}{\sqrt {x^{2} \plus{} 1}}ln(x^{2} \minus{} 1)\,dx$.
---------------------------------------------------------------
5. $ \int ln(x \plus{} \sqrt {1 \plus{} x^{2}}) \,dx$.
6. $ \int ln(\sqrt {1 \minus{} x} \plus{} \sqrt {1 \plus{} x})\,dx$.
7. $ \int xe^{x}cosx \,dx$.
If you want you may solve them and to say what is most difficult and most beautiful. Do you have any interesting integral by parts? What is the integral of any kind do you like best?

[hide="Opinion"]For 1. make $ x \equal{} \tan t$, and so

$ P \equal{} \int e^t \cos t\,dt \equal{} \frac {e^t}{2}(\sin t \plus{} \cos t) \plus{} C \equal{} \frac {e^{\arctan x}}{2}\left(\frac {1 \plus{} x}{\sqrt {1 \plus{} x^2}}\right) \plus{} C$.

For 2., let's integrate 1. by parts:

$ \int \frac {e^{\arctan x}}{(1 \plus{} x^2)\sqrt {1 \plus{} x^2}}\,dx \equal{} \frac {e^{\arctan x}}{\sqrt {1 \plus{} x^2}} \plus{} \int \frac {xe^{\arctan x}}{(1 \plus{} x^2)^{3/2}}\,dx$

and so

$ \int \frac {xe^{\arctan x}}{(1 \plus{} x^2)^{3/2}}\,dx \equal{} \frac {e^{\arctan x}}{2}\left(\frac {1 \plus{} x}{\sqrt {1 \plus{} x^2}}\right) \minus{} \frac {e^{\arctan x}}{\sqrt {1 \plus{} x^2}} \plus{} C$.

For 3., make x = sin t. For 4., make x = tan t.

For number 7. we have:

$ (xe^x \sin x)' \equal{} e^x \sin x \plus{} xe^x \sin x \plus{} xe^x \cos x$
$ (xe^x \cos x)' \equal{} e^x \cos x \plus{} xe^x \cos x \minus{} xe^x \sin x$

Now add these two relations:

$ (xe^x \sin x)' \plus{} (xe^x \cos x)' \equal{} e^x(\sin x \plus{} \cos x) \plus{} 2xe^x \cos x$

$ (xe^x \sin x)' \plus{} (xe^x \cos x)' \equal{} (e^x \sin x)' \plus{} 2xe^x \cos x$

and so

$ \int xe^x \cos x\,dx \equal{} \frac{1}{2}(xe^x \sin x \plus{} xe^x \cos x \minus{} e^x \sin x) \plus{} C$

$ \int xe^x \cos x\,dx \equal{} \frac{e^x}{2}[x(\sin x \plus{} \cos x) \minus{} \sin x] \plus{} C$.[/hide]

[hide="Number 5"]We have that

$ [\ln(x \plus{} \sqrt{1\plus{}x^2})]' \equal{} \frac{(x \plus{} \sqrt{1\plus{}x^2})'}{x \plus{} \sqrt{1\plus{}x^2}} \equal{} \frac{\frac{x \plus{} \sqrt{1\plus{}x^2}}{\sqrt{1\plus{}x^2}}}{x \plus{} \sqrt{1\plus{}x^2}} \equal{} \frac{1}{\sqrt{1\plus{}x^2}}$.

So:

$ P \equal{} \int \ln(x \plus{} \sqrt{1\plus{}x^2})\,dx \equal{} x \ln(x \plus{} \sqrt{1\plus{}x^2}) \minus{} \int \frac{x}{\sqrt{1\plus{}x^2}}\,dx$

$ P \equal{} x \ln(x \plus{} \sqrt{1\plus{}x^2}) \minus{} \sqrt{1\plus{}x^2} \plus{} C$.[/hide]

[hide="Number 6"]Start by noting that

$ (\sqrt {1 \minus{} x} \plus{} \sqrt {1 \plus{} x})' \equal{} \frac {1}{2}\left(\frac {1}{\sqrt {1 \plus{} x}} \minus{} \frac {1}{\sqrt {1 \minus{} x}}\right) \equal{} \frac {\sqrt {1 \minus{} x} \minus{} \sqrt {1 \plus{} x}}{2\sqrt {1 \plus{} x} \sqrt {1 \minus{} x}}$.

So:

$ P \equal{} \int \ln(\sqrt {1 \minus{} x} \plus{} \sqrt {1 \plus{} x})\,dx \equal{} \underbrace{x \ln(\sqrt {1 \minus{} x} \plus{} \sqrt {1 \plus{} x})}_{ \equal{} A} \minus{} \int \frac {x(\sqrt {1 \minus{} x} \minus{} \sqrt {1 \plus{} x})}{2(\sqrt {1 \minus{} x} \plus{} \sqrt {1 \plus{} x})\sqrt {1 \plus{} x} \sqrt {1 \minus{} x}}\,dx$

$ \equal{} A \minus{} \int \frac {x}{2}\left(\frac {1}{(\sqrt {1 \minus{} x} \plus{} \sqrt {1 \plus{} x}) \sqrt {1 \plus{} x}} \minus{} \frac {1}{(\sqrt {1 \minus{} x} \plus{} \sqrt {1 \plus{} x}) \sqrt {1 \minus{} x}}\right)\,dx$

$ \equal{} A \minus{} \int \frac {x}{2}\left(\frac {1}{\sqrt {1 \minus{} x^2} \plus{} 1 \plus{} x} \minus{} \frac {1}{\sqrt {1 \minus{} x^2} \plus{} 1 \minus{} x}\right)\,dx$

$ \equal{} A \minus{} \int \frac {x}{2} \cdot \frac { \minus{} 2x}{(\sqrt {1 \minus{} x^2} \plus{} 1)^2 \minus{} x^2}\,dx$

$ \equal{} A \plus{} \frac {1}{2} \int \frac {x^2}{1 \minus{} x^2 \plus{} \sqrt {1 \minus{} x^2}} \equal{} A \plus{} \frac {1}{2} \int \frac {x^2}{\sqrt {1 \minus{} x^2}(1 \plus{} \sqrt {1 \minus{} x^2})}\,dx$

$ \equal{} A \plus{} \frac {1}{2} \int \frac {x^2(1 \minus{} \sqrt {1 \minus{} x^2})}{x^2 \sqrt {1 \minus{} x^2}}\,dx \equal{} A \plus{} \frac {1}{2} \int \left(\frac {1}{\sqrt {1 \minus{} x^2}} \minus{} 1\right)\,dx$

$ \equal{} x\left[\ln(\sqrt {1 \minus{} x} \plus{} \sqrt {1 \plus{} x}) \minus{} \frac {1}{2}\right] \plus{} \frac {1}{2} \arcsin x \plus{} C$.[/hide]

Great solutions! I'm interested how you choose the order of integrals to solve? And what is your favorite integral?

Lucky 7.

$ (xe^x\cos x)' = e^x\cos x + xe^x\cos x - xe^x\sin x$
$ (xe^x\sin x)' = e^x\sin x + xe^x\sin x + xe^x\cos x$

Adding these equations gives $ 2xe^x\cos x + e^x(\sin x + \cos x) = (xe^x\cos x)' + (xe^x\sin x)'$.

$ \Longleftrightarrow 2xe^x\cos x\ dx = \{xe^x(\cos x + \sin x)\}' - (e^x\sin x)'$

$ \therefore \int xe^x\cos x\ dx = \frac {1}{2}e^x\{x(\sin x + \cos x) - \sin x\} + Const.$

This solution already has been posted by Carcul.

[url=http://www.mathlinks.ro/viewtopic.php?t=23087&search_id=1124107133]Please take a Look![/url] :)

You can also use $ e^{ix} = \cos x + i\sin x$.

Alternative Solution:

Use $ \{f(x)g(x)\}' = f'(x)g(x) + f(x)g'(x)$, we have $ \int e^x\cos x\ dx = \frac {1}{2}e^x(\cos x + \sin x) + Const.$

$ \therefore \int xe^x\cos x\ dx = \int x\cdot e^x\cos x\ dx$

$ = \frac {1}{2}xe^x(\cos x + \sin x) - \frac {1}{2}\int e^x(\cos x + \sin x)\ dx$

$ = \frac {1}{2}e^x(\cos x + \sin x) - \frac {1}{2}\int (e^x\sin x)'\ dx$

$ = \frac {1}{2}xe^x(\cos x + \sin x) - \frac {1}{2}e^x\sin x + Const.$

How about this problem?

8. $ \int (x+1)e^x\ln x\ dx$

kunny

[quote="Carcul"]Integral 6 was the most interesting for me. Now try this one:

$ P \equal{} \int \ln^2(x \plus{} \sqrt {1 \plus{} x^2})\,dx$.[/quote]

using repeated integration by parts,

$ \int\,\ln^2\,(x\plus{}\sqrt {1 \plus{} x^2\;})\;dx\equal{}\int\,\left(\sinh^{\minus{}1}\,x\right)^2\cdot 1\;dx$
$ \equal{}\;x\,\left(\sinh^{\minus{}1}\,x\right)^2\plus{}2\left(x\minus{}\left(\sinh^{\minus{}1}\,x\cdot\sqrt{1\plus{}x^2\;}\right)\right)\plus{}C$

For 6, let $ x\equal{}\minus{}t$, $ \int \ln (\sqrt{1\minus{}x}\plus{}\sqrt{1\plus{}x})\ dx\equal{}\int \ln (\sqrt{1\plus{}t}\plus{}\sqrt{1\minus{}t})(\minus{}dt)$, yielding $ \int \ln (\sqrt{1\minus{}x}\plus{}\sqrt{1\plus{}x})\ dx\equal{}0$.

It is an interesting integral. Do you composed it? I like integrals that looks easy but they are not and when you see the solution you will say..."it was so easy - how it is possible for me to be so stupid and to not solve it". Look at the file attached - what do you think for my solution?

I saw the solutions - all of them are very good.

I also have an integral for you. Calculate
\[ \int {\frac {{\sin 7x}}{{\sin x}}dx} .
\]
My solution is very short.

Is it possible to apply your method to this below
\[ \int {\frac{{\sin 2007x}}{{\sin x}}dx} .
\]

[hide="Problem number 8"]Let $ \ln x \equal{} t$, $ x \equal{} e^t$, $ x' \equal{} e^t$:

$ P \equal{} \int(1\plus{}x)e^x \ln x\,dx \equal{} \int(1\plus{}e^t)te^te^{e^t}\,dt$

$ \equal{} e^{e^t}t(1\plus{}e^t) \minus{} \int e^{e^t} (1\plus{}e^t \plus{} te^t)\,dt$

$ \equal{} e^{e^t}t(1\plus{}e^t) \minus{} \int e^{e^t}\,dt \minus{} \int e^te^{e^t}\,dt \minus{} \int e^te^{e^t}t\,dt$

$ \equal{} e^{e^t}t(1\plus{}e^t) \minus{} \int e^{e^t}\,dt \minus{} e^{e^t} \minus{} (e^{e^t}t \minus{} \int e^{e^t}\,dt)$

$ \equal{} e^{e^t}t(1\plus{}e^t) \minus{} \int e^{e^t}\,dt \minus{} e^{e^t} \minus{} e^{e^t}t \plus{} \int e^{e^t}\,dt$

$ \equal{} e^{e^t} (te^t \minus{} 1) \equal{} e^x(x \ln x \minus{}1) \plus{} C$
[/hide]

That's right!

Here is my solution:

$ (xe^x\ln x)'\equal{}e^x\ln x\plus{}xe^x\ln x\plus{}xe^x\cdot \frac{1}{x}\equal{}(x\plus{}1)e^x\ln x\plus{}e^x$.

$ \therefore \int (x\plus{}1)e^x\ln x\ dx\equal{}xe^x\ln x\minus{}e^x\equal{}(x\ln x\minus{}1)e^x\plus{}Const.$

I'm happy because of my topic is a good place for finding interesting integrals. For bookworm_vn - you may use recursion I suppose... http://www.mathlinks.ro/Forum/viewtopic.php?p=904452#904452 ... may be useful...

Problem 9. Find  $ P \equal{} \int \frac{\cos(2x)}{\sin^4x \plus{} \cos^4x}\,dx$.

[quote="borislav_mirchev"]I'm happy because of my topic is a good place for finding interesting integrals. For bookworm_vn - you may use recursion I suppose...[/quote]

I think so.  :lol:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=108624&search_id=210573473]Take a look at Fiery Hydra's solution.[/url]

[quote="borislav_mirchev"]I'm happy because of my topic is a good place for finding interesting integrals. For bookworm_vn - you may use recursion I suppose...[/quote]

I think that it is impossible!

[quote="Carcul"]Problem 9. Find  $ P \equal{} \int \frac {\cos(2x)}{\sin^4x \plus{} \cos^4x}\,dx$.[/quote]
\[ \frac {{\sqrt 2 }} {4}\ln \left( {\frac {{\tan ^2 x \plus{} \sqrt 2 \tan x \plus{} 1}} {{\tan ^2 x \minus{} \sqrt 2 \tan x \plus{} 1}}} \right)
\]
maybe?

[quote="borislav_mirchev"]What is your solution?[/quote]

I will post my solution after a while. I want to see other solutions (if possible).

Bookworm, your correct solution can be further simplyfied.

If you want you may try integrals 3 and 4. They are not difficult and are good also.
I'm wondering if to make a topic where people may put they favorite integrals. What do you think will it be a good idea?

I like your integrals borislav_mirchev and have had a lot fun doing them lately. What is the name of this old book? Post more integrals, please  :)

The name of my book is: "Sbornik ot zadachi po diferencialno i integralno smqtane" - by Georgi Bradistilov - Sofia 1947 /Collection of problems by calculus - differential and integrational/. This man was leading bulgarian mathematician - I think there is a room to Sofia's University or Technical University - Sofia by it's name. I think the book is hardly available. /There is an interesting story about this book is  - my grandmother considered it as a garbage and there were a possibilty for her to use this book for heating.../ And the bulgarian magazine where interesting integrals may be found is: "Matematicheski Forum".
There are lots of integrals - what subject exactly you want?
You may try these:
$ \int \frac {1}{x\sqrt {ax^{2n} - b}} \,dx$
$ \int \frac {1}{x\sqrt {a + bx^{n}}} \,dx$
$ \int \frac {1}{(1 + x^{2n})\sqrt {(1 + x^{2n})^{\frac {1}{n}}}}\,dx$
$ \int \frac {1}{(1 - x^{n})(2x^{n} - 1)^{\frac {1}{2n}}} \,dx$
${ \int \frac {x^{ - \frac {1}{2}}}{(1 - x^{n})(2x^{n} - 1)^{\frac {1}{2n}}}} \,dx$
They are irrational integrals ... not by parts
;-) enjoy
You my try also these two easy integrals from the magazine:
$ \int \frac {(1 + xcosx)}{x(ln^{2}x + 2lnxsinx - cos^{2}x)} \,dx$ - Alfred Ekstain, Viorel Tudoran - Arad - Romania
$ \int \frac {sin^{n}x(sin^{2n}x + 1)}{tgx(sin^{4n} + 1)} \,dx$ - Alfred Esktain, Mihaela Vizental - Romania
The last one is easiest - even I found a solution :-). First is also not difficult - but I first took a look at the solution...


Find sum:
$ P\equal{}cos{x}\plus{}2cos{2x}\plus{}...\plus{}ncos{nx}$

Well-known to check that:
$ S\equal{}sinx\plus{}sin2x\plus{}...\plus{}sinnx\equal{}\frac{cos\frac{x}{2}\minus{}cos\frac{(2n\plus{}1)x}{2}}{2sin\frac{x}{2}}$
Now $ P\equal{}S'$


Let S=1!+2!+3!+...+2007! What is the last digit of S.


symmetry
geometry
geometric transformation
reflection

[quote="13375P34K43V312"]A regular polygon has 8 lines of symmetry. What is the measure of one of its interior angles?[/quote]

[hide="Octagon"]$135^\circ$[/hide]

[quote="frost13"]And when you are done with that one :

A regular polygon has n lines of symmetry. What is the measure of one of its interior angles?  Express your answer in terms of n.[/quote]

[hide="n-gon"]$\frac{180(n-2)}{n}$[/hide]

Wait, why do the # of lines of symmetry have to equal the # of sides?

In a regular polygon, if the number of sides is even(n), then there will be n/2 lines of symmetry connecting opposite vertices, and n/2 lines of symmetry connecting the midpoints of opposite/parallel sides for a total of n lines of symmetry.

If there are an odd number of sides(n again), then there will be a line of symmetry passing through each vertex and bisecting the opposite side.  One line for each vertex = n lines of symmetry.

Try drawing the first 6-8 regular polygons and check it out for yourself, probably be more convincing that way.

[quote="13375P34K43V312"]How can we be exactly sure that there are no more?[/quote]
Well, you can think of lines of symmetry as reflections. Consider one of the vertices (and call it vertex $A$). In any of these reflections, vertex $A$ must map to another of the vertices of the regular n-gon. Thus, because it can map to any of the $n$ vertices (including itself), and all of these lines are different (a reflection that maps one point to another is uniquely defined), there are exactly $n$ lines of symmetry in a regular n-gon.


calculus
derivative
function
integration
linear algebra

i am currently in multivariable calc.. and i wanna learn some differential equations so i could learn some quantum.. am i at the level to learn differential equations? and what exactly is differential equations? and please suggest some good easy books for differentail...

Well, this is probably in the wrong forum -- I'm going to send it voer to somewhere more appropriate as soon as I'm done with this.  "Differential equations" is the study of solving equations involving the derivatives of functions.  Its interest stems from a wide range of applications and its difficulty -- given any function, it is generally quite simple to find its derivative (when one exists).  But, while we know that most functions have an integral, it tends to be very difficult to figure out what it is.  Differential equations basically amounts to trying to find ways to integrate things or solve equations where integrals will be required.

To do differential equations you really only need calculus, but to do it well you need some grasp of linear algebra and an understanding of matrices.

I unfortunately have no good book suggestions.


topology
superior algebra
superior algebra unsolved

In the definition of topological group, we set the multiplication to be continuous map together with  the inverse operator. But I want to find an example that a topological space has the structure of group together with the multiplication continuous, but inverse operator isn't continous. !
    I want to know about it very much ! Thanks !

Additive group $\mathbb R$ with the topology generated by intervals $(-\infty,a)$. This topology was [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=137293]discussed here[/url].

Ok!
Thank you very much !
That is the topological space I want to find. But I find that it isn't compact or Hausdorf . Could you find for me a space that it satisfies above properties but
be compact or Hausdorf ?.
I hope that I don't require so much from you ! Thanks !

$\mathbb{R}$ with the topology generated by $(a,b]$?

A Hausdorff example (non-elementary, but coming from RL): Take the ring of adeles over a number field. It is a topological ring, but the subspace topology on the group of units (i.e., the ideles) does not make them into a topological group because the inverse is not continuous. owk


function
real analysis
real analysis unsolved

Let for every $\alpha \in \Re$ a collection $E_\alpha$ is given.
Such that $E_\alpha \subset E_\beta \forall \alpha < \beta$, $\bigcup_{\alpha \in \Re} E_\alpha = X$ and $\bigcap_{\alpha \in \Re} E_\alpha = \emptyset$

Show that there is a measurable function $f:X \dashrightarrow\Re$
such that $f(x)\leq\alpha$ at $E_\alpha$ and $f(x)\ge\alpha$ at
$E_\alpha^c = X\setminus E_\alpha$ for all $\alpha \in \Re$

You really should say that the $E_\alpha$  are measurable sets on $X$; otherwise, this is nonsense.

Define $Y_\alpha=\bigcap_{n=1}^\infty E_{\alpha+\frac1n}\setminus\bigcup_{n=1}^\infty E_{\alpha-\frac1n}$ and let $f(x)=\alpha$ on $Y_\alpha$.
We have $x\in Y_\alpha$ when $\alpha=\sup \{t: x\in E_t\}=\inf \{t: x\not\in E_t\}$. The $\sup$ and $\inf$ exist for all $t$ and must always agree, so every $x$ is in some $Y_\alpha$.
We have $\{x: f(x)>\alpha\}=\bigcap_{n=1}^\infty E_{\alpha+\frac1n}$ and $\{x: f(x)<\alpha\}=\bigcup_{n=1}^\infty E_{\alpha-\frac1n}$, so both of these sets are measurable and $f$ is measurable. Since $E_\alpha$ lies between the two sets, $f(x)\le\alpha$ on $E_\alpha$ and $f(x)\ge\alpha$ on the complement.

This construction is unique; this is the only such function.


[quote="1/(ln x)"]My school and the local university are offering to register for the AMC A test and the B test respectively. Is it against AMC rules to take both exams?

Also just out of curiosity, is it legal for the AIME if one offers the AIME1 and the other offers the AIME2?[/quote]
You are allowed to take both the AMC A and B; however, you are not allowed to take both the AIME I and the AIME II.

For more information, see these two links:
[url]http://www.unl.edu/amc/b-registration/b1-archive/2003-2004/faq2003-04.html[/url]
[url]http://www.unl.edu/amc/e-exams/e7-aime/adminaime.html[/url]


Πώς σας φάνηκε η Μεσογειάδα και πώς τα πήγατε?
Σ' αυτό το topic μπορούμε να γράψουμε όλα τα σχόλια και τις παρατηρήσεις μας για τη φετινή Μεσογειάδα :wink:

Παραθέτω τώρα το πρόβλημα 1:

Εστω $x,y,z$ πραγματικοί αριθμοί με $x\leq y\leq z$ και $xy+yz+zx=1$
Να αποδείξετε ότι $xz<\frac{1}{2}$.
Είναι δυνατόν να βελτιώσουμε την τιμή της σταθεράς $\frac{1}{2}$ :?:

Αναμένω τις λύσεις σας :P

καλη επιτυχια σε ολους  οσους εδωσαν.οποιος μπορει ας βαλλει τα θεματα .

Το πρώτο ερώτημα προκύπτει πολύ εύκολα αν διακρίνουμε τις περιπτώσεις προσήμου των μεταβλητών.
Το δεύτερο ερώτημα ήταν ομολογουμένως αρκετά δύσκολο. Εγώ προσωπικά υπέθεσα ότι εντοπίζουμε κάποιο άλφα μηδέν όσο το δυνατόν μεγαλύτερο (για να είναι το 1/(2+α0) όσο το δυνατόν μικρότερο) και σταθεροποιούμε το χζ στην τιμή αυτή και κατέληξα σε άτοπο. Εδώ είναι και το μελανό σημείο της λύσης μου: είπα ότι μπορούμε να εκφράσουμε τα χψ, ψζ συναρτήσει της βελτιωμένης αυτής σταθεράς χωρίς να είμαι σίγουρος γι' αυτό.

Πάντως αργά ή γρήγορα θα βγουν οι λύσεις στο διαδίκτυο. Καλή επιτυχία σε όλους.

Πρόβλημα 2:

Έστω ΑBCD εγγράψιμο κυρτό τετράπλευρο του οποίου οι διαγώνιοι $AC$ και $BD$ τέμνονται στο E. Δίνεται ότι $AB=39, AE=45, AD=60, BC=56$. Να προσδιορίσετε το μήκος της πλευράς CD.

Εγώ το έλυσα ως εξής:
Βρήκα απ΄την ομοιότητα των τριγώνων $AED$ και $BED$ ότι $BE=42$
Έπειτα απ' την ομοιότητα των $ABE, DEC$ εξέφρασα τα DE, EC συναρτήσει του $DC$,
Τέλος εφαρμόζοντας το θεώρημα του Πτολεμαίου ο μόνος άγνωστος είναι το $CD$...
Το κακό ήταν ότι οι πράξεις ήταν τόσο δύσκολες που τελικά δεν έφτασα σε αποτέλεσμα :(

Το 3ο και το 4ο;Θυμάται κανείς την εκφώνηση;

3ο: Δίνεται τρίγωνο ΑΒΓ με <Α=ω και ΒΓ=sqrt(ρR), όπου R η ακτίνα του περιγεγραμμένου και ρ η ακτίνα του εγγεγραμμένου. Να προσδιορίσετε τις υπόλοιπες πλευρές (προφανώς συναρτήσει των δεδομένων).

Ήταν ιδιαίτερα δύσκολη άσκηση για την οποία έπρεπε να επιστρατεύσεις τους νόμους ημιτόνων και συνημιτόνων και τους τύπους των εμβαδών και πολύ υπομονή. Δεν κατάφερα να την ολοκληρώσω αν και έφτασα σε ικανοποιητικό επίπεδο. Έβγαλα ένα περίπλοκο σύστημα δύο αγνώστων και δεν είχα το χρόνο και το κουράγιο να προσπαθήσω να το επιλύσω. Έχει κανείς πιο εύκολη λύση?

Το 4ο είναι δύσκολο στην καταγραφή (σόρρυ) αλλά ήταν ανισοτική με ακέραια και κλασματικά μέρη.

[quote="Protonios"]Αν όμως έθετες ED=x θα γλίτωνες κάποιες πράξεις και το τριώνυμο στο τέλος θα ήταν πιο ανθρώπινο, έτσι ώστε ενδεχομένως να μη χρειαζόταν να βρεις τη ρίζα της τεράστιας διακρίνουσας, αλλά να δουλέψεις με γινόμενο άθροισμα :wink: . Το ίδιο πρόβλημα είχα κι εγώ στην αρχή. Πάντως το θέμα είχε μόνο πρακτική και όχι άλλη αξία.[/quote]

Πράγματι Kώστα έχεις δίκιο...
Το προσπάθησα λίγο σπίτι μου με αυτόν τον τρόπο και έβγαινε αρκετά πιο εύκολα  :oops: η αλήθεια είναι πως ούτε που το σκέφτηκα να θέσω το ED ως άγνωστο...

Δεν ήρθε ουρανοκατέβατο. Αφού τράβηξα όσα εσύ με τις πράξεις σκέφτηκα ότι στο θεώρημα του Πτολεμαίου εκεί που πολλαπλασιάζονται οι διαγώνιες είναι καλύτερα να έχεις 45+χ παρά 45+15/14*χ ή κάτι παραπλήσιο. Κι έτσι βγήκε... :wink:

Δεν πιστεύω ότι τα προφορικά δυσκόλεψαν κάποιον. :wink: Τα αποτελέσματα τη Δευτέρα. Καλή επιτυχία να ευχηθώ σε όλους και ας... 'κερδίσει' ο καλύτερος γιατί όλοι κερδισμένοι είναι όσοι έφτασαν έως εδώ. :wink:
Όσοι επιστρέφετε θα έχετε πολλά να πείτε είμαι σίγουρος.
Θα έχετε νεότερα.

Φιλικά,
Κώστας Παππέλης.

Γεια σας . Ρε παιδιά bilot4 , protonie ήθελα να  σας γνωρίσω από κοντά αλλά δεν μπορούσα να σας βρω ....

Silouan αν θυμάσαι στα γραπτά καθόμουν ακριβώς μπροστά σου. :)  Σόρρυ αλλά δεν είμαι τόσο εξοικειωμένος με τους υπολογιστές για να βάλω φωτογραφία αλλά φοράω γυαλιά και είμαι αρκετά ψηλός αν βοηθάει καθόλου. :wink: Δεν μπορεί, θα τα πούμε κάποτε από κοντά. Αν έχουμε περάσει δωδεκάδα σίγουρα, αν πάλι όχι στη χειρότερη περίπτωση του χρόνου, υγεία να έχουμε. :D

Ναι κατάλαβα ... Να κάθεσαι μπροστά μου και να μην πεις τίποτα ρε... :|  . Τεσπα θα τα πούμε στο μέλλον...

Σόρρυ αλλά είχα πάρα πολύ άγχος και δεν μπορούσα να αρθρώσω λέξη. :blush: Πού θα πάει όμως θα τα πούμε δεν μπορεί... :wink:

Η αλήθεια είναι ότι κι εγώ καπου εκέι καθόμουν...(δύο θέσεις μπροστά σου και διαγώνια...)αλλά δεν βρήκα την ευκαιρία να σου μιλήσω... :blush: Βέβαια εγώ σε αναγνώρισα απ' την φωτογραφία σου...  Είμαι σίγουρος ότι θα βρούμε την ευκαιρία να τα πούμε από κοντά :wink:


Let a;b;c;x;y;z be positive such that :a+b+c=x+y+z. Prove that:
                               $ a^2x\plus{}b^2y\plus{}c^2z\plus{}4(ax^2\plus{}by^2\plus{}cz^2)\ge15abc$

[quote="king.of.dragon"]Let a;b;c;x;y;z be positive such that :a+b+c=x+y+z. Prove that:
                               $ a^2x \plus{} b^2y \plus{} c^2z \plus{} 4(ax^2 \plus{} by^2 \plus{} cz^2)\ge15abc$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31344

I thank arqady, but can you solve more detail for me?I haven't understood absolutely. :(

$ 4(ax^2\plus{}by^2\plus{}cz^2)\plus{}4xyz\geq 16abc$ and $ a^2x\plus{}b^2y\plus{}c^2z\plus{}abc\geq4xyz.$
After adding we obtain your inequality.


How much does the undergraduate college you go to affect your applying to graduate and profession specific schools? [hide="3 Short Questions"]Where would be a good place to read about the order of schooling one goes through after high school?

Supposing you continue schooling after undergraduate, how much do your undergraduate years matter when you apply for?

How could I improve the format of my thread to make it easier and more attractive for people to respond? [/hide]

Offhand, I think you're thinking too hard about your future.  The plans you make now are not likely to be what you actually end up doing in a few years.  The events that shape your life and make you grow as a person are not predictable.  Just sit back and enjoy the ride :)


[quote="vishalarul"]Find all positive integer solutions to the equation:
x^2+3 = y(x+2)

[hide]y = (x^2 + 3)/x+2 = x-2 - (7/x+2)
x-y-2 = 7/(x+2)
I think I should stop at that... :P [/hide][/quote]

[hide]you basically already solved it, $\frac{7}{x+2}$ has to be an integer, so $x=5$ and it follows that $y=2$[/hide]

I didn't work it out myself, I just looked at what he did.

I know.  I was referring to his attached work. :)

from wat u gave, right side has to be 1, and's only possible wen x=5 & y must be 2.

[hide="different soln"]
$x^{2}+3=y(x+2)$

$y = \frac{x^{2}+3}{x+2}$

$y = \frac{x^{2}-4+x}{x+2}$

$y = (x-2)+\frac{7}{x+2}$

Therefore, x must be 5 and it follows that y is 4.  If you plug this into the original equation, you see that this is correct (y isn't 2)

[/hide]

[quote="NUMB3RS"]sorry,  but I don't understand how you get from one step to the next.

How do you go from
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/bdfb3fac071e08abf09537d699fe74dd.gif[/img]
to
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/288bef75f06503bdd20cf1df536be9fe.gif[/img]
to
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/c13aac75b86bf3aba800d1b4f2b4dacd.gif[/img]

I'd appreciate it if someone could explain it to me. Thanks.[/quote]

That's a typo: it should be $y = \frac{x^{2}+3}{x+2}= \frac{x^{2}-4+7}{x+2}= \frac{(x-2)(x+2)+7}{x+2}= x-2+\frac{7}{x+2}$.  :wink:

Thanks Kurt. I can follow it now.  :lol:


In a class of 20 ppl, each student is assigned an ordered pair$(m,n)$ where $m$ and $n$ are respectively his scores in written and oral exams.Each score is an integer from 1 to 10. It is known that no 2 ordered pairs are same. We say a student $A$ is better than $B$ if the 2 scores of $A$ are greater than or equal to the corresponding scores of $B$. Show that there exist 3 students $A,B,C$ such that $A$ is better than $B$ and $B$ is better than $C$. Is the same true for 19 students?

There are two posibilities. The first is when there are three students that have the same score at the same exam . Then we just need to watch the score in the other exam, they are ordered and those are the three. If the last situation doesn't occur is because every score was atained by two students in each exam. Now see both of the students which have score 10. Call them A and B. The second grades of them are ordered, so there is one which is better than the other (supose A is better than B). Now call C the student that has the same score that has B as the second score. The first score of C is less than 10 (because 10 is the maximum score) so B is better than C and we are done.

For 19 students $s_{1}, s_{2}, ... , s_{1}9$ consider the case when the students have the following scores:
$\\ s_{1}\rightarrow (1,10) \\ s_{2}\rightarrow (1,9) \\ s_{3}\rightarrow (2,9) \\ s_{4}\rightarrow (2,8) \\ s_{5}\rightarrow (3,8) \\ .\\ .\\ .\\ s_{17}\rightarrow (9,2) \\ s_{18}\rightarrow (9,1) \\ s_{19}\rightarrow (10,1)$

And for these case there are no A, B, C such that A is better than B and B is better than C.

QED


vector
linear algebra
matrix
combinatorics proposed
combinatorics

Cute.  :)

To each of the $ A_l$ associate a column vector $ a_l = \begin{pmatrix} a_{l,1} \\
\vdots \\
a_{l,n} \end{pmatrix} \in \mathcal{M}_{n,1}(\mathbb{R}),$ where $ a_{l,i}$ is one if $ i \in A_l$ and zero otherwise.

This way, we have $ n + 1$ vectors living in $ \mathbb{R}^n,$ hence they must be linearly dependent, meaning that we may find
$ (\lambda_1, \ldots, \lambda_{n + 1}) \in \mathbb{R}^{n + 1} \setminus \{(0, \ldots, 0) \}$ such that $ \sum_{l = 1}^{n + 1} \lambda_l a_l = 0.$

Now, thanks to the nonnegativity of the $ a_{l,i}$ we may find two set of indices $ P = \{i_1, \ldots, i_k \}$ and $ N = \{j_1, \ldots, j_m \}$ corresponding to the positive and negative $ \lambda_l$ so that the above equality can be rewritten as : $ \sum_{l \in P} \lambda_l a_l = \sum_{l \in N} ( - \lambda_l) a_l.$

Those sets of indices are disjoint by construction and we have that $ \bigcup_{l \in P} A_l = \bigcup_{l \in N} A_l,$ qed.

To have an idea of why the use of linear algebra is somewhat natural here, and more, I suggest reading [url=http://gowers.wordpress.com/2008/07/31/dimension-arguments-in-combinatorics/]Dimension arguments in combinatorics.[/url]

Here's an elementary solution which just uses induction. The idea is pretty simple, to act greedily, but explaining it took a lot of space.

[b]Standard stuff:[/b] We proceed by induction. $\text{Base case} =\text{Trivial case}$
Assume for $n-1$. We show for $n$.

[b]Main part:[/b] For a set $U$ of indices. define $$S_U:=\bigcup_{x \in U} A_x, V_U=\{A_x | x \in U\}$$
Note that $S_U$ is the union of the members of $V_U$.

Call a set [i]bad[/i] if it contains $n$ and [i]good[/i] otherwise.
If there exists at most one [i]bad[/i] set, then we are done by the induction hypothesis. So assume at least $2$ [i]bad[/i] sets.

Assume that there do not exist disjoint subsets $X, Y$ of indices such that $A_x, A_y$ are [i]good[/i] for all $x \in X, y \in Y$ and $S_X=S_Y$, otherwise, we are trivially done.

Choose two families of sets $A, B$ with $n$ elements each such that each of them contains at least one bad set.
[hide=Delete]
Even though we delete $n$, we still call these sets bad to distinguish them from the rest.
[/hide] the elements $n$ from all the [i]bad[/i] sets. Then by the induction hypothesis, we find two [b]disjoint[/b] subsets $X_1, Y_1$ of indices in $A$ and $X_2, Y_2$ in $B$ satisfying:
\begin{align}
S_{X_{1}}=S_{Y_{1}} \text{,} V_{X_{1}} \cap V_{Y_{1}}=\phi \\
S_{X_{2}}=S_{Y_{2}}, V_{X_{2}} \cap V_{Y_{2}}=\phi
\end{align}
By our assumption, at least one of $S_{X_{1}}, S_{Y_{1}}$ must contain a bad set, say $S_{X_{1}}$. Similarly assume that $S_{Y_{2}}$ contains a bad set.

Hence, we find $M, N$ such that
$$S_{M}=S_{X_{1}} \cup S_{X_{2}} \cup \{n\}=S_{Y_{1}} \cup S_{Y_{2}} \cup \{n\}=S_{N}$$
We can add back the $\{n\}$ to all the bad sets which exist both in the LHS and RHS, and so we are done.

Actually... it seems like we are done, but not! What if there's a set $A_k$ on both the sides? Then $M, N$ are not disjoint. Fortunately, we can fix it in the most natural way!

[b]Claim:[/b] We can remove $A_k$ from both the sides and we would still have $S_{M} \texttt{\textbackslash} A_{k}=S_{N} \texttt{\textbackslash} A_{k}.$ (here again, we do not consider the element $n$ in $A_k$)
[b]Proof of Claim:[/b] If $A_k \in V_{X_{1}}$ (note that we are talking about the family of sets $V_{X_{1}}$ and not their union $S_{X_{1}})$ then we can't have $A_k \in V_{Y_{1}}$ by (1). Hence we must have $A_k \in V_{Y_{2}} \implies A_k$ is not in $V_{X_{2}}$, by (2)

Now we know that all the [b]elements[/b] of $A_k$ are in $S_{Y_{1}}$ and $S_{X_{2}}$. Hence, even if we delete the set $A_k$ from both the sides, the [b]elements[/b] will still remain on both the sides and hence we will still have $S_{M} \texttt{\textbackslash} A_{k}=S_{N} \texttt{\textbackslash} A_{k},$ as claimed. $\square$

Hence we are done! $\blacksquare$

[b]P.S.[/b] This problem was communicated to me by [b]biomathematics[/b] on a different thread, and I recently got to know the source so posted my solution here.

[quote=Fang-jh]Let $ n$ be a positive integer, let the sets $ A_{1},A_{2},\cdots,A_{n \plus{} 1}$ be non-empty subsets of the set $ \{1,2,\cdots,n\}.$ prove that there exist two disjoint non-empty subsets of the set $ \{1,2,\cdots,n \plus{} 1\}$: $ \{i_{1},i_{2},\cdots,i_{k}\}$ and $ \{j_{1},j_{2},\cdots,j_{m}\}$ such that $ A_{i_{1}}\cup A_{i_{2}}\cup\cdots\cup A_{i_{k}} \equal{} A_{j_{1}}\cup A_{j_{2}}\cup\cdots\cup A_{j_{m}}$.[/quote]

Define the vectors $\vec{v(x)}$ for all $x \in \{1,2, \dots , n+1\}$ such that $\vec{v_i(x)} = 1 \iff i \in A_{x}$ and $0$ otherwise. Notice that we have $n+1$ vectors on a $n-$ vectorspace and hence they ought to be linearly dependent (the proof this is simple indeed). Thus there exist scalars $\epsilon_i \in \mathbb{R} \ \forall i \in \{1,2,\dots , n+1\}$ such that $$\sum_{j=0}^{n+1} \epsilon_j \cdot \vec{v(j)} = 0$$. As the components of each of these vectors are in $\{0,1\}$  and since the total is $0$, some of these must be negative. Pull them on the other side of equality and we have two disjoint non-empty subsets of $\{1,2, \dots , n+1\}$, $I$ and $J$ , where $I = \{i_1,i_2 \dots, i_m\}$ and $J=\{j_1,j_2, \dots , j_l\}$ such that $m+l=n+1$ and \\
$$ \sum_{i \in I} \epsilon_i \cdot \vec{v(i)} = \sum_{j \in J} -\epsilon_j \cdot \vec{v(j)}$$

And hence the desired result.

[quote=Wizard_32]
[b]Claim:[/b] We can remove $A_k$ from both the sides and we would still have $S_{M} \texttt{\textbackslash} A_{k}=S_{N} \texttt{\textbackslash} A_{k}.$ (here again, we do not consider the element $n$ in $A_k$)
[b]Proof of Claim:[/b] If $A_k \in V_{X_{1}}$ (note that we are talking about the family of sets $V_{X_{1}}$ and not their union $S_{X_{1}})$ then we can't have $A_k \in V_{Y_{1}}$ by (1). Hence we must have $A_k \in V_{Y_{2}} \implies A_k$ is not in $V_{X_{2}}$, by (2)

Now we know that all the [b]elements[/b] of $A_k$ are in $S_{Y_{1}}$ and $S_{X_{2}}$. Hence, even if we delete the set $A_k$ from both the sides, the [b]elements[/b] will still remain on both the sides and hence we will still have $S_{M} \texttt{\textbackslash} A_{k}=S_{N} \texttt{\textbackslash} A_{k},$ as claimed. $\square$

[/quote]

If I'm not mistaken, this is only valid for the first removal. Consider $n=7$, \[(X_1)\quad\{1,2,3,4,5\}\cup\{6,7\} \leftrightarrow (Y_1)\quad\{2,3,4,5,6\} \cup \{1,3\}\]\[(X_2)\quad\{1\}\cup \{2,3,4,5,6\} \leftrightarrow (Y_2)\quad\{1,2,3,4,5\}\cup \{4,6,7\}\]
If both pairs of repeated sets $\{1,2,3,4,5\}$ and $\{2,3,4,5,6\}$ were removed from both sides, the remaining sets \[\{1\}\cup \{6,7\} \leftrightarrow \{1,3\}\cup\{4,6,7\}\]would be irreconcilable.

One (counterintuitive) fix is to [u]only remove one copy of each repeated set[/u]. If only the repeated sets in $X_1$ and $Y_1$ are removed (while preserving their counterparts in $X_2$ and $Y_2$), we end up with disjoint set indices on both sides, but with the same union: any element in a removed set (removed from $X_1$ or $Y_1$) would still exist in both $X_2$ and $Y_2$. In the above example, this corresponds to the final configuration
\[\{1\}\cup \{2,3,4,5,6\}\cup \{6,7\} \leftrightarrow \{1,3\}\cup \{1,2,3,4,5\}\cup \{4,6,7\}.\]

The only caveat would be the removal of all sets in $X_1$ originally containing the element $n$; but this can be avoided by selecting $X_2,Y_2$ among all sets, but particularly excluding some set in $X_1$ containing element $n$.