inequalities
calculus
derivative
inequalities unsolved

Let a,b,c be positive real numbers such that
a^2+b^2+c^2=3
Prove that
2(a+b+c)+ab+bc+ac≥3(a*(b^2)+b*(c^2)+c*(a^2))
It seems to be true but how to prove or disprove it?
Thanks a lot  :lol:

Anyone can give me the answer or some hint? =)

[quote="enndb0x"]$ 2(a \plus{} b \plus{} c) \plus{} ab \plus{} bc \plus{} ca \equal{} (a \plus{} b \plus{} c) \plus{} (a \plus{} b \plus{} c) \plus{} ab \plus{} bc \plus{} ca \geq 3 \sqrt [3]{(a \plus{} b \plus{} c)(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}$

Since $ (a \plus{} b \plus{} c)^2 \geq 3(ab \plus{} bc \plus{} ca)$

(since $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$)

We have that $ 3 \sqrt [3]{(a \plus{} b \plus{} c)(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)} \geq 3\sqrt [3]{(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)(ab \plus{} bc \plus{} ca)}$

Now using Holder's inequality we have

$ 3\sqrt [3]{(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)(ab \plus{} bc \plus{} ca)} \geq 3(b^{4/3}a^{2/3} \plus{} c^{4/3} b^{2/3} \plus{} a^{4/3}c^{2/3})$

we have to prove that $ 3(b^{4/3}a^{2/3} \plus{} c^{4/3} b^{2/3} \plus{} a^{4/3}c^{2/3})\geq 3 (ab^2 \plus{} bc^2 \plus{} ca^2)$

I don't think the last inequality is true  !![/quote]

This is a conclusion I got when I try to prove another much stranger ineq  :lol: I wrote a program to grnerate 10^4 sets of random testdata and 10^10 equal-gap data(0<a<sqrt(3),0<b<sqrt(3),a += GAP,b += GAP,GAP = 1E-5) ,no counterexample is found,so the ineq is likely to be true but is really hard to me :blush: anyway thanks to enndb0x and I'll try it later  :D

[quote="chengdi"]Let a,b,c be positive real numbers such that
$ \sum_{cyc}a^2 \equal{} 3$prove $ 2\sum{a} \plus{} \sum{ab}\geq 3\sum_{cyc}ab^2$[/quote]
$ \iff\frac {2}{3}(\sum{a^3} \plus{} \sum{a^2(b \plus{} c)}) \plus{} \sum{ab}\geq 3\sum_{cyc}ab^2$
$ 6\sum{a^3} \plus{} 6\sum{a^2(b \plus{} c)} \plus{} 3\sqrt {3}\sum{ab\sqrt {(a^2 \plus{} b^2 \plus{} c^2)}}\geq 27\sum{ab^2}$
use global derivative
$ f'\equal{}(6\sum{a^3} \plus{} 6\sum{a^2(b \plus{} c)} \plus{} 3\sum{ab(a \plus{} b \plus{} c)} \minus{} 27\sum{ab^2})'\plus{}(3\sum{ab}(\sqrt{3a^2\plus{}3b^2\plus{}3c^2}\minus{}a\minus{}b\minus{}c))'\equal{} 18\sum{a^2} \plus{} 12\sum{ab} \plus{} 12\sum{ac} \plus{} 6\sum{a^2} \plus{} 6\sum{a^2} \plus{} 6\sum{ab} \plus{} 3\sum{b^2} \plus{} 3\sum{bc} \plus{} 3\sum{a^2} \plus{} 6\sum{ab} \plus{} 3\sum{ac} \plus{} 3\sum{ab} \minus{} 27\sum{b^2} \minus{} 54\sum{ab}\plus{}(2\sum{a}\minus{}\frac{3\sum{ab}}{\sqrt{3a^2\plus{}3b^2\plus{}3c^2}})(\sqrt{3a^2\plus{}3b^2\plus{}3c^2}\minus{}a\minus{}b\minus{}c)$
$ \geq  9(\sum{a^2} \minus{} \sum{ab})\geq 0$
when a=0,it becomes
$ \frac {2}{3}(b \plus{} c)(b^2 \plus{} c^2) \plus{} bc\sqrt {\frac {b^2 \plus{} c^2}{3}}\geq 3bc^2$
$ \iff\frac {2}{3}(t \plus{} 1)(t^2 \plus{} 1) \plus{} t\sqrt {\frac {t^2 \plus{} 1}{3}}\geq 3t$
which is true
hold when a=b=c


function
integration
calculus
probability and stats

I have a piece of question, that completely stumped me. Preemptive thanks for any help, suggestions, hints, solutions.

For an univariate distribution function $ F$, define a transformation
$ \beta_F(t) \equal{} \int\limits_{ \minus{} \infty}^\infty e^{ \minus{} |x \minus{} t|}dF(x)$, $ t\in\mathbb{R}$.

Show that,
1. $ \beta(t)$ is continuous and $ \int_{ \minus{} \infty}^\infty \beta(t)dt \equal{} 2$.
2. Does $ \beta_{F_1} \equal{} \beta_{F_2}$ imply $ F_1 \equal{} F_2$?
3. Show that if $ F_n\to F$, then $ \beta_{F_n}\to \beta_F$ pointwise.
4. Conversely, if  $ \beta_{F_n}\to \beta_F$ pointwise, then what conditions should we impose on $ \beta$ to conclude $ F_n\to F$.

literally nothing :blush:

We can exchange the order of integration (by Tonelli's thm)
\[ \int_{\minus{}\infty}^{\infty} \beta(t)dt \equal{} \int_{\minus{}\infty}^{\infty} \int_{\minus{}\infty}^{\infty} e^{\minus{}|x\minus{}t|}dF(x)dt \equal{} \int_{\minus{}\infty}^{\infty}\int_{\minus{}\infty}^{\infty} e^{\minus{}|x\minus{}t|}dt dF(x)\]
On the other hand
\[ \int_{\minus{}\infty}^{\infty}e^{\minus{}|x\minus{}t|}dt \equal{} \int_{\minus{}\infty}^{\infty} e^{\minus{}|t\minus{}x|}d(t\minus{}x) \equal{} \int_{\minus{}\infty}^{\infty} e^{\minus{}|t|}dt \equal{} 2 \int_{0}^{\infty}e^{\minus{}t}dt \equal{} 2\]
Since $ F$ is distribution function,
\[ \int_{\minus{}\infty}^{\infty} \beta(t)dt \equal{} \int_{\minus{}\infty}^{\infty} 2 dF(x) \equal{} 2.\]
This is what I have done so far.


[quote="mdk"]Find all natural numbers n that divide 1^n +2^n +···+(n−1)^n[/quote]

If $ n$ is odd then we can write,
$ [1^{n}+(n-1)^{n}]+[2^{n}+(n-2)^{n}]+...+[t^{n}+s^{n}]$

The reason why I wrote $ s,t$ is because Im to lazy to determine in term of $ n$ what $ s$ and $ t$ are, but you get the idea.

Now you can use the identity for $ x^{n}+y^{n}$ when $ n$ is odd, you know the factoring identity.

Then we see that, each is divisible by $ n$. Now, I guess we should try to show it fails when $ n$ is even.


inequalities
induction
function
integration
calculus
geometry
geometric transformation

let $ x > 0$ then prove that for every positive integer $ n$ that:

$ e^x > \sum_{i \equal{} 0}^{n} \frac {x^i}{i!}$

[hide="N.B :"]There are two nice solutions for this problem  :P[/hide]

The solution I know is induction on n.
If $ n\equal{}1$ then it is true.
Suppose it is true for $ k$.
We will prove it it true for $ k\plus{}1$
$ F_{k\plus{}1}(x)\equal{}e^x\minus{}\sum_{i\equal{}1}^{k\plus{}1}\frac{x^i}{i!}$
$ F'_{k\plus{}1}(x)\equal{}F_{k}(x)>0$ so the function increasing on $ [0,\plus{}\infty)$
But $ F(0)\equal{}0$ so $ F(x)>0,\forall x>0$

[quote="TTsphn"]
But $ F(0) \equal{} 0$ so $ F(x) > 0,\forall x > 0$[/quote]

Ok this is one, but $ F(0)\equal{}1$

There's another one very nice.

Other is consider the number :
$ (1\plus{}\frac{1}{n})^{nx}$

[quote="TTsphn"]Other is consider the number :
$ (1 \plus{} \frac {1}{n})^{nx}$[/quote]

TTsphn, could you please give a hint?(detail)

[quote="0714446459923"]wouldn't it be a lot easier just to use the taylor series expansion of e^x?[/quote]

of course

Here is the solution for Japanese high school students, because they don't study Taylor expansion.

$ e^x \minus{} e^0 \equal{} \left[e^x\right]_0^x \equal{} \int_0^x e^xdt$

$ \equal{} \left[(t \minus{} x)e^x\right]_0^x \minus{} \int_0^x (t \minus{} x)e^xdt$

$ \equal{} x \minus{} \left(\left[\frac {(t \minus{} x)^2}{2}e^x\right]_0^x \minus{} \int_0^x \frac {(t \minus{} x)^2}{2!}e^xdt\right)$

$ \equal{} x \plus{} \frac {x^2}{2!} \minus{} \frac {1}{2}\left(\left[\frac {(t \minus{} x)^3}{3}e^x\right]_0^x \minus{} \int_0^x \frac {(t \minus{} x)^3}{3}e^xdt\right)$

$ \equal{} x \plus{} \frac {x^2}{2!} \plus{} \frac {x^3}{3!} \plus{} \int_0^x \frac {(t \minus{} x)^3}{3!}e^xdt$

Repeating this procedure gives $ e^x \minus{} e^0 \equal{} x \plus{} \frac {x^2}{2!} \plus{} \frac {x^3}{3!} \plus{} \cdots \plus{} \frac {x^n}{n!} \plus{} \int_0^x \frac {(t \minus{} x)^n}{n!}e^x dt > 0\ (n\in\mathbb{N})$ for $ x > 0.$

Maybe it depends on the definition of e. It is possible to go from $ \left( {1 \plus{} \frac {1}{n}} \right)^n \to e$ to what you are looking for.

The key is to use Newton's binomial theorem on that and simplify it cleverly. I would try to help you but it's quite a lot to put it in Latex.

Very shortly, $ \left( {1 \plus{} \frac {1}{n}} \right)^n \equal{} \sum\limits_{k \equal{} 1}^n {\frac {{n!}}{{k!(n \minus{} k)!n^k }}} \equal{} \sum\limits_{k \equal{} 1}^n {\frac {1}{{k!}}} \prod\limits_{i \equal{} 0}^{k \minus{} 1} {\left( {1 \minus{} \frac {{i \minus{} 1}}{n}} \right)}$

If you take the first k terms of this sum, you'll see that its less than the entire sum. Write down this inequality and take the limit as n tends to infinity and you'll get your inequality, basically. It can be generalized by replacing 1/n with x/n in the original sequence, which would then tend to e^x, but the idea is the same.

You'll have to write this out painstakingly to see what I'm talking about. It's much easier to show on paper than on a forum.

There are lots of mistakes in your solution, please check it again.
[quote="kunny"]
$ e^x \minus{} e^0 \equal{} \left[e^x\right]_0^x \equal{} \int_0^x e^xdt$
[/quote]
This equality doesn't occur:  $ \int_0^x e^xdt\equal{}xe^{x}$

[quote="kunny"]
$ \equal{} \left[(t \minus{} x)e^x\right]_0^x \minus{} \int_0^x (t \minus{} x)e^xdt$
[/quote]
This one also: $ \left[(t \minus{} x)e^x\right]_0^x \minus{} \int_0^x (t \minus{} x)e^xdt\equal{}xe^{x}\plus{}\frac{1}{2}x^{2}e^{x}$

Sorry for that I made some typos, I can't edit it now.

Here is the corrected version.

$ e^x \minus{} e^0 \equal{} \left[e^x\right]_0^x \equal{} \int_0^x e^tdt$

$ \equal{} \left[(t \minus{} x)e^t\right]_0^x \minus{} \int_0^x (t \minus{} x)e^tdt$

$ \equal{} x \minus{} \left(\left[\frac {(t \minus{} x)^2}{2}e^t\right]_0^x \minus{} \int_0^x \frac {(t \minus{} x)^2}{2!}e^tdt\right)$

$ \equal{} x \plus{} \frac {x^2}{2!} \minus{} \frac {1}{2}\left(\left[\frac {(t \minus{} x)^3}{3}e^t\right]_0^x \minus{} \int_0^x \frac {(t \minus{} x)^3}{3}e^tdt\right)$

$ \equal{} x \plus{} \frac {x^2}{2!} \plus{} \frac {x^3}{3!} \plus{} \int_0^x \frac {(t \minus{} x)^3}{3!}e^tdt$

Repeating this procedure gives $ e^x \minus{} e^0 \equal{} x \plus{} \frac {x^2}{2!} \plus{} \frac {x^3}{3!} \plus{} \cdots \plus{} \frac {x^n}{n!} \plus{} \int_0^x \frac {(t \minus{} x)^n}{n!}e^t dt > 0\ (n\in\mathbb{N})$ for $ x > 0.$

[quote="kunny"]Here is the solution for Japanese high school students, because they don't study Taylor expansion.[/quote]
If that's not the Taylor formula with the remainder in the integral form (together with its standard derivation), I don't know what it is :roll:.

Fedja wasn't commenting on your use of language.  [hide="Although"] as long as we're discussing it, the proper phrasing for the first part of your last post would have been, "I don't know, either."  The construction you were aiming at was "Neither do I," but I don't believe it's used unless the verb (in this case, "know") is implicit (so one wouldn't write "Neither do I know.")  I hope this is all accurate :).[/hide]  He was commenting that the integral formulas you used (and the final result you arrived at) are a derivation of the Taylor expansion (in integral form), so if Japanese high school students know them then they essentially know Taylor series (at least for the exponential function).

I meant that the expression "Neither I know" may be wrong, that's why I said, Is my expression correct?

Anyway, few students know or have heared the name Taylor, although the problem related to this has been posed in university entrnace exam so far. If Japanese high school students use Taylor expansion in test, the will gain zero marks, let alone l'Hopital's theorem. Thank you for your appointment about my mistakes, JBL. :)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19548&search_id=1853829453]Please take a look[/url]

Alternative Solution:

$ e^t>t\plus{}1$ for $ t>0$, $ \int_0^x e^t\ dt>\int_0^x (t\plus{}1)\ dt\Longleftrightarrow e^x>1\plus{}x\plus{}\frac{x^2}{2!}$,

similarly repeating gives $ e^x>1\plus{}x\plus{}\frac{x^2}{2!}\plus{}\cdots \plus{}\frac{x^n}{n!}\plus{}\cdots.$

[quote="kunny"] If Japanese high school students use Taylor expansion in test, the will gain zero marks, let alone l'Hopital's theorem. [/quote]
Hmmm... This doesn't sound fair. I mean, if somebody knows the theorem, can state it without a mistake, use it correctly, and, in addition, can do just as much as to name exactly the textbook in which it can be found, I think he must get full credit for his solution. Of course, I don't have any power to change Japanese grading policies :( but, generally speaking, I have a very negative attitude towards the exam problems with restrictions on what you are allowed to use. In my humble opinion, the only thing they reflect is the lack of imagination of those who put them on the exams. One can always find a good problem with "no hands barred" that tests any particular skill :)

Of course, if a sudent solve in using Taylor and l'Hopital with proof, he would get full marks, generally speaking, few people could do that even if Tokyo and Kyoto university's applicants. In my opinion, it is impossible to teach high school students Calculus without teaching the principle, say, epsilon-delta, particularly the volume rotated by the figure about $ y$ axis, you would use the formula $ V \equal{} \int_a^ b 2\pi f(x)\ dx$ and Pappus-Guildin's theorem, which are not allowed to use in test.

[quote="kunny"] In my opinion, it is impossible to teach high school students Calculus without teaching the principle, say, epsilon-delta, [/quote]
With that I agree 100% (I had some doubts about it before I came to the USA, but after 4 years of teaching business calculus, they all evaporated :P). Of course, full proofs deserve full credit but what I meant was that they should not be required. I would settle exactly for what I said: the ability to correctly state, use, and reference the theorem (after all, that is the standard in the professional mathematical writing).

As far as I remember, in Russia we had something like "integration and differentiation techniques" during the last two grades in school and we learned how to prove inequalities using derivatives (the main theorem being, of course, that if $ f(a)\ge g(a)$ and $ f'(x)\ge g'(x)$ for all $ x\ge a$, then $ f(x)\ge g(x)$ for $ x\ge a$) but it is true that I learned the $ \varepsilon\minus{}\delta$ definition of continuity only in the university (in defence of the old Russian educational system, I should say that it happened on the second day of lectures). I cannot tell what exactly was the policy on using the theorems outside the school curriculum on the entrance exams, but I remember that none of the problems I saw would become any easier (not to say "trivial") if I knew the entire university program. Later, when I talked to people who made exams, they told me that they always tried to find problems that wouldn't give any advantage to anybody who "knows too much" or a serious disadvantage to anybody who "knows too little". The reason  was that, despite we had a standard school curriculum that all schools in the country had to comply with, the level of school education you could get in Moscow and Leningrad was far above the one you could get in rural areas (though I should mention here that my uncle was a physics teacher in a small village school and, looking back, I should admit that he knew more physics than I ever learned in my life :)). I suspect that today's Japan is not much different in this respect, so it would be natural to expect the same attitude.

Anyway, those who know the Taylor formula in the high school usually are clever enough to use it without making an impression that they do, so I shouldn't, probably, worry about them too much :lol:.


For a,b,c positive reals, we have
\[ \frac{(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\left[\frac{ab}{c(c\plus{}a)}\plus{}\frac{bc}{a(a\plus{}b)}\plus{}\frac{ca}{b(b\plus{}c)}\right]\leq\frac{27}{16}.\]

[quote="hitek"]For a,b,c positive reals, we have
\[ \frac {(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\left[\frac {ab}{c(c \plus{} a)} \plus{} \frac {bc}{a(a \plus{} b)} \plus{} \frac {ca}{b(b \plus{} c)}\right]\leq\frac {27}{16}.
\]
[/quote]
Try $ a\rightarrow0^\plus{}.$ :wink:


email
trigonometry
geometry
circumcircle
3D geometry
Stanford
college

Did anyone get an e-mail saying they got an award? I probably got an honorable mention again...

Maybe we should reserve a corner of the lecture room for AoPS members :D then everyone can really see how fat, ugly, stupid, and mean I am!

What did people think about the test, anyway? I thought that 1 was too easy, 3 was way too easy for a 3, 2 was really confusing (well it was really easy but I just failed :( :( ), 4a was really easy, 4b was really hard, and 5 was beastly. I remember most of the solutions to the problems in case anyone is interested (although I'm pretty sure solutions are already up on the website).

[quote="Ubemaya"]Did anyone get an e-mail saying they got an award? I probably got an honorable mention again...
[/quote]
I got one...w00t
[quote]
Maybe we should reserve a corner of the lecture room for AoPS members :D then everyone can really see how fat, ugly, stupid, and mean I am!
[/quote]
you need to be less emo online.
[quote]
What did people think about the test, anyway? I thought that 1 was too easy, 3 was way too easy for a 3, 2 was really confusing (well it was really easy but I just failed :( :( ), 4a was really easy, 4b was really hard, and 5 was beastly.
[/quote]
I thought 1 was trivial, 2 was interesting, yet easy, 3 was dumb(as in I was being a retard and couldn't do so wrote up a bogus solution), 4a was trivial, and 4b,5 were beastly. Wrote up another bogus to 4b(should've written 'lol' at the end of that proof...).
[quote]
(although I'm pretty sure solutions are already up on the website).[/quote]
From Paul Zeitz's email:
[quote](Sorry, we will not divulge any scores or tell you anything about
prizes, so don't ask.  That's what the awards ceremony is for.  After
March 11, we will put the solutions on the web and I will respond to
score requests.)[/quote]
which means solutions aren't up yet

[quote="Ubemaya"]Did anyone get an e-mail saying they got an award? I probably got an honorable mention again...[/quote]
im willing to bet 5 dollars you got 1st place in 8th grade.

I'm willing to bet my chances of ever getting a girlfriend that I won't place.

(I have a 99% chance of winning this bet, but just in case I somehow got lucky by copying off Quatto during the test, I have a negative chance of getting a girlfriend so I'm not losing anything).

bookaholic, I seriously doubt you did worse than everyone here (namely, me). I made stuff up for every problem but 2, and my proofs SUCK. I think I got a 11-12 total.

[quote="i_like_pie"]I'm going!

I agree with you two on the difficulty of the problems.[/quote]

And junggi, there's proof that someone owned me. If he agrees with us two, then he solved 1-3 and 4a, for a score of about 22.

Oh yeah, does anyone want to see the super-elegant, no trig way for #3?

[quote="bookaholic"]I'm going. I didn't do very well, probably worse than everyone else in this thread, but whatever.[/quote]
oh no
that probably means you pwned us all

@ubemaya: uh.....alright then. wait isn't pie in 9th grade?

@ Ubemaya - I didn't do #2. :ninja:

@ junggi - I'm in 8th grade. :)

[quote="i_like_pie"]
@ junggi - I'm in 8th grade. :)[/quote]
oh nevermind...
wow what is up with all these 8th graders doing as well as me, if not much better
you people make me feel bad

Dude wtf I'm stupid.

I_like_pie - Congrats on winning 2nd! (I'm pretty sure Moly Mol won first, as usual).

ooh i got an email
im gonna try and wear my AoPS shirt so that you guys all know who I am
i solved 1-3 and 4a and didn't use trig on 3, just circumcircles and stuff
i thought my diagram for 3 was pretty good
i actually think i did pretty well on 2, though i made a (probably) unjustified assumption
im in 7th grade

O_o okay, you are now the new 8th grade and below winner.

How exactly did you do circumscribed stuff for #3?

[quote="Ubemaya"]
Maybe we should reserve a corner of the lecture room for AoPS members :D then everyone can really see how fat, ugly, stupid, and mean I am![/quote]

wow...you sure are proud... :lol:

Hey wait a minute... I just realized that my email had NO hint that I had won something, it just invited me to the Awards Ceremony... :( I think that everyone who put down a email address got that email which means that I didn't even get an honorable mention :( :( Oh well I didn't expect anything better anyway.

Wow, me too.  Rather large club we have here.

Today on AIM talking to Andy Tsao and random other people about my AIME score, I said the f-word about twice as much as I've ever swore in my life. (I consider swearing to be d***, h***, etc.)

Oh yeah, and... WOAH! SAME HERE!!

hmm i got a 3 even though i solved 9 problems
i guess i just made a bunch of mistakes
I COUNTED POWERS INSTEAD OF MULTIPLES FOR #1 AAUUGGHH
but im happy :D
wasn't really expecting to do extremely well anyways and i still have 5 more years :)

i get from the discussion here that you guys all did pretty badly on the AIME? :P

[quote="Ubemaya"]DUDE I FAILED SO BADLY DON'T EVEN REMIND ME!!!!![/quote]
shut up you !@^%&@W$E!(\#$%!%D

Ubemaya, you did not fail by any definition of "fail", reasonable or otherwise. Don't complain.

Daermon--that's great for a 7th grader. I wish I could solve 9 of the problems on an AIME when I was in 7th grade. (I could solve maybe around 4-6 back then, and never got them all right either.)

Yeah I kinda failed the AIME. What is the best way to prepare? Oh yeah and don't listen to Ubemaya, even a 15 is a failing score for him.

In my experience, the only really useful way to prepare is to do lots and lots of old tests and carefully go over the solutions to any problems you didn't solve. (This is assuming you already know all the background math--if you don't, learn it first.)

Hehe I sure wish that was the case.  The aime this year was so un-aimelike.  Seriously, what is up with the score distribution.  USAMO honorable mentions getting 5s and random mathcounts people getting 8s???  There is seriously something wrong with that.

[quote="Ubemaya"]HA! SEE, I'M A RANDOM PERSON!!!!

I think it's people with prep in stuff like MC got good scores, while people who actually studied for the AIME were unprepared and failed.

My definition is "fail" is doing worse than I could have done. If I had gotten a 10 (no stupid mistakes), I wouldn't have failed.[/quote]
you got an eight?  well, if you're so sure that you failed, I'd be happy to switch with you :P
dude don't be so hard on yourself you're in eighth grade and you're already really good

Not good enough.
Just watch, I won't get my goal of 1 for the USAMO.

[quote="Ubemaya"]HA! SEE, I'M A RANDOM PERSON!!!!

I think it's people with prep in stuff like MC got good scores, while people who actually studied for the AIME were unprepared and failed.

My definition is "fail" is doing worse than I could have done. If I had gotten a 10 (no stupid mistakes), I wouldn't have failed.[/quote]
The number of stupid mistakes i made is the same as the number of problems you got right, so I really suggest you shut up or you might get a punch from someone(me possibly)  :mad: :P

and really, there is something very very wrong with this AIME. adrian, steven, altheman, tony liu person, etc didn't make usamo. that is just wrong.

stephen Ge(krustyteklown).

there happy stephen? i spelled your name right =P


Take the number 1100 . Increase it by 70%, then decrease the result by 70%. The number you obtain is

you should hide the answer
[hide]1100*1.7=1870
1870*.3=561[/hide]

There is a second possibility. If we consider $70\%$ as a normal number $0.7$, not as a $70\%\cdot1100$, then the answer will be $1100$.
We add $70\%$ to $1100$.
$1100+70\%=1100.7$
Next we decrease this number by $70\%$.
$1100.7-0.7=1100$

[color=darkred][size=75][Edited for Latex][/size][/color]

[quote="dondigo"]There is a second possibility. If we consider $70\%$ as a normal number $0.7$, not as a $70\%\cdot1100$, then the answer will be $1100$.
We add $70\%$ to $1100$.
$1100+70\%=1100.7$
Next we decrease this number by $70\%$.
$1100.7-0.7=1100$
[/quote]

Good thinking, but that's not the problem  :P


quadratics
ratio
algebra
quadratic formula

Two different positive numbers $ a$ and $ b$ each differ from their reciprocals by $ 1$. What is $ a\plus{}b$?

$ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ \sqrt{5} \qquad
\textbf{(D)}\ \sqrt{6} \qquad
\textbf{(E)}\ 6$

[hide="Solution"]One of the numbers is the positive root of $ x \minus{} \frac {1}{x} \equal{} 1$; the other is the positive root of $ \frac {1}{x} \minus{} x \equal{} 1$.

Solving the two equations, we get $ a \equal{} \frac {1 \plus{} \sqrt {5}}{2}$ and $ b \equal{} \frac { \minus{} 1 \plus{} \sqrt {5}}{2}$.

$ a \plus{} b \equal{} \sqrt {5}\quad\Longrightarrow\quad\text{C}$[/hide]
Thanks for catching my error, 123456789.

@i_like_pie, that's wrong...one of the roots is negative...

[hide]We have two possibilities:
1. $ x \minus{} \frac {1}{x} \equal{} 1$
2. $ \frac {1}{x} \minus{} x \equal{} 1$
Solving each equation for its positive roots, we find that the two solutions are $ \displaystyle \frac {1 \plus{} \sqrt {5}}{2}$ and $ \displaystyle \frac { \minus{} 1 \plus{} \sqrt {5}}{2}$. The sum of these two numbers is $ \sqrt {5} \implies \boxed{C}$.[/hide]

[hide="Solution"]
If a number satisfies the condition, its reciprocal satisfies the condition as well. Let the two numbers be $ (b\plus{}1)/2$ and $ (b\minus{}1)/2$, which sum to $ b$ and differ by $ 1$. Then, $ (b\plus{}1)(b\minus{}1)/4 \equal{} 1 \implies b^2 \minus{} 1 \equal{} 4 \implies b \equal{} \sqrt{5}$[/hide]

[hide="asdfasdfasdf"]
We can form the equation

$ x\equal{}\frac{1}{x}\pm1$

This is because it be either greater or less than its reciprocal by 1.
Now we have 2 cases:

$ x\equal{}\frac{1}{x}\plus{}1$

Rewrite the 1 as $ \frac{x}{x}$

$ x\equal{}\frac{x\plus{}1}{x}$

Multiply both sides of the equation by x then subtract x+1. We get the equation

$ x^{2}\minus{}x\minus{}1\equal{}0$

Using the quadratic formula and the fact that we are looking for the positive numbers, we have

$ x\equal{}\frac{1\plus{}\sqrt{5}}{2}$.

Similary, for the -1 case, we end with the equation

$ x^{2}\plus{}x\minus{}1\equal{}0$

Again using the quadratic formula, we have

$ x\equal{}\frac{\minus{}1\plus{}\sqrt{5}}{2}$

Now, we just find

$ \frac{\minus{}1\plus{}\sqrt{5}}{2}\plus{}\frac{1\plus{}\sqrt{5}}{2}$

The -1 and 1 cancel out, leaving us with

$ \frac{2\sqrt{5}}{2}$

The 2s cancel out, finally leaving us with our final answer, $ \sqrt{5}$. Hence, the answer is

$ \boxed{C}$[/hide]

The numbers that satisfy this requirement are the [url=http://en.wikipedia.org/wiki/Golden_ratio]Golden Ratio[/url] and its reciprocal.

Why is this wrong.
Case 1.
a+1=1/a. $a^2+a=1$. b+1=1/b. $b^2+b=1$. $b^2+b=a^2+a$. $a^2-b^2=b-a$. $(a+b)(a-b)=b-a$. $a+b=-1$. But there is no -1 answer choice.
Case 2.
a+1=1/a. $a^2+a=1$. b=1+1/b. $b^2-b=1$. $b^2-b=a^2+a$. $a^2-b^2=a-b$. $(a+b)(a-b)=a-b$. $a+b=1$. So answer can be A.
Case 3
a=1+1/a. $a^2-a=1$. b+1=1/b. $b^2+b=1$. $b^2+b=a^2-a$. $a^2-b^2=a+b$. $(a+b)(a-b)=a+b$. $a-b=1$. NO solution?.
Case 4
a=1+1/a. $a^2-a=1$. b=1+1/b. $b^2-b=1$. $b^2-b=a^2-a$. $a^2-b^2=a-b$. $(a+b)(a-b)=a-b$. $a+b=1$. So answer can be A.
Therefore I got A, can someone explain what I did wrong.
I understand the right answer (i.e. NerdOfTheAges) and everything its just confusing as to why this solution is wrong.
Thanks

@ above, in cases 2 and 4 you finished by dividing by $a - b$. However, in those cases the equations for $a$ and $b$ were symmetric, thus making $a - b = 0$. Also in case 1 you realized that $-1$ is not an answer choice; however, this is because the question asks for distinct [b] positive [/b] numbers. Lastly, Case 3 yields: \begin{align*} a^2 - a - 1 &= 0 \\ b^2 + b - 1 &= 0 \\ a &= \frac{ 1 \pm \sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2} \\ b &= \frac{ -1 \pm \sqrt{5}}{2} = \frac{-1 + \sqrt{5}}{2} \\ a + b &= \frac{1 + \sqrt{5} - 1 + \sqrt{5}}{2} = \frac{2\sqrt{5}}{2} = \boxed{\sqrt{5}} \end{align*}


If $x \neq 0$, $\frac x{2} = y^2$ and $\frac{x}{4} = 4y$, then $x$ equals

$ \textbf{(A)}\ 8\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 32\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 128 $

[hide="Answer"]$x$ must be of the form $2^{2n+1}$ for $\frac{x}{2}$ to be a perfect square. For $y^2>y$, we must have a large value of $x$, so our answer is $128\Rightarrow \boxed{E}$.[/hide]

[hide]Well if you simply solve it and put $x=2y^2$ in the second equation,then you get y=8 hence x=$2*64$=128[/hide]

[hide="long way"]$x-2y^2=0$.
$x-16y=0$.

Subtracting we have the quadratic $2y^2-16y=0$, $y=8$. y can't be 0 because then x will also be 0.

Thus x=128. [/hide]

[hide]
$\\\frac{x}{2}=y^2\\ x=2y^2$
y goes into equation number 2:
$\\\frac{y}{2}=4\\ y=8$
As result, we have
$\\\frac{x}{2}=64\\ x=128$

The answer is $\boxed{E}$.
[/hide]

Masoud Zargar


Let n be a natural number and $ n\equal{}2^{2007}k\plus{}1$,such that k is an odd number.Prove that $ n\not|2^{n}\minus{}1\plus{}1$


limit
calculus
derivative
trigonometry
LaTeX
calculus computations

hello
find
 $ \lim_{x\rightarrow \frac {\pi}{4}}\frac {tanx \minus{} 1}{x \minus{} \frac {\pi}{4}}$

$ lim_{x \rightarrow 1} (x\minus{}1)tan(\frac{\pi}{2}x)$
 $ \lim_{x\rightarrow \frac {\pi}{4}}\frac {tan2x}{cosx \minus{} sinx}$

The first is a derivative. Do you know the derivative of $ \tan$?

The second is a $ \frac {\pm\infty}{0}$ form, which is very far from indeterminate. After checking that it's positive on both sides, we get that the limit is $ + \infty$.

$ \text{\LaTeX}$ note: use \sin, \cos, \tan and so on for "standard" functions. It looks better.

hello
thx but I must not use derivative ^^

It is precisely the limit that appears in the definition of the derivative. You're not going to avoid that completely.
Look at how we found the derivatives of $ \sin$ and $ \cos$ in the first place, and mimic that for $ \tan$. You need an angle addition formula, along with some basic limits such as $ \lim_{x\to 0}\frac{\sin x}{x}\equal{}\lim_{x\to 0}\frac{\tan x}{x}\equal{}1$.

Here's the solution for the first, without derivatives:

$ L \equal{} frac{\frac{\sin x}{\cos x}\minus{}\frac{\sin (\pi/4)}{\cos (\pi/4)}}{x\minus{}\pi/4} \equal{} \frac{\frac{\sin x \cos(\pi/4) \minus{} \cos x \sin(\pi/4)}{\cos x \cos(\pi/4))}}{x \minus{} \pi/4} \equal{} \frac{\sin(x \minus{} \pi/4)}{(x \minus{} \pi/4)(\cos x \cos(\pi/4)} \equal{} \frac{1}{\cos x \cos(\pi/4)} \equal{} 2$

(Used sint/t=1, t->0).

EDITED, Sorry...

[quote="hsiljak"][Edited for readability and including the limits]

$ L \equal{} \lim_{x\to\frac {\pi}{4}}\frac {\frac {\sin x}{\cos x} \minus{} \frac {\sin (\pi/4)}{\cos (\pi/4)}}{x \minus{} \frac {\pi}{4}} \equal{} \lim_{x\to\frac {\pi}{4}}\frac {\frac {\sin x \cos(\pi/4) \minus{} \cos x \sin(\pi/4)}{\cos x \cos(\pi/4)}}{x \minus{} \frac {\pi}{4}}$
$ \equal{} \lim_{x\to\frac {\pi}{4}}\frac {\sin(x \minus{} \frac {\pi}{4})}{(x \minus{} \frac {\pi}{4})(\cos x \cos\frac {\pi}{4}) } \equal{} \lim_{x\to\frac {\pi}{4}}\frac {1}{\cos x \cos\frac {\pi}{4}} \equal{} 2$

(Used $ \lim_{t\to 0}\frac {sint}{t} \equal{} 1$).[/quote]

You really should pay more attention to how these things look; fractions are much better than / in most cases. To see the details of what I did, quote my post.

This calculation also works to find the derivative of $ \tan$ in general.

the given limit's value is -2..........

multiply and divide by 1+$ \ce{tanx}$ and the numerator becomes....$ \ce{tan(pi/4 \minus{} x)}$.........
and we know the limit as limtanx/x=1..........

Ps..learning to use latex..........

hello, the second, it is
${ \lim_{x\rightarrow 1}(x-1)\tan\left(\frac{\pi}{2}x\right)}=-\frac{2}{\pi}$
Sonnhard.

hello,
$ \lim_{x\rightarrow \frac {\pi}{4}}\frac {tan2x}{cosx \minus{} sinx}\equal{}\infty$.
Sonnhard.

[quote="chemrock"]For the third write $ \cos{x} \minus{} \sin{x} \equal{} \sqrt {2}\times\cos{(\pi/4 \plus{} x)}$
then put $ t \equal{} \pi/4 \minus{} x$
But I am getting Infinity as the answer :o[/quote]

Ofcourse you do: it is because the form of the limit isn't indeterminate. Actually, just by putting in the value for $ x$, you get $ \infty/0\equal{}\infty$ (as jmerry and dr Sonnhard already pointed out).

You can use the [u]method of substitution[/u] : $ \boxed {\ \lim_{x\to x_0}f(x)\stackrel {\left(t\equal{}x\minus{}x_0\right)}{\equal{}\equal{}\equal{}\equal{}}\lim_{t\to 0}f(t\plus{}x_0)\ }$ , where $ x_0\in \mathbb R$ .


inequalities
complex numbers
inequalities unsolved

Let $a,b,c$ be three positive reals such that $a+b+c=1$. prove that
$(ab)^\frac{5}{4}+(bc)^\frac{5}{4}+(ac)^\frac{5}{4}<\frac{1}{4}$.

just wandering. is "inequalities" the appropriate section for posting inequalities with absolute values of complex numbers?

[quote="hungkhtn"]It was post in Mathlinks before.
The General inequality:

$\sum(ab)^{k} \le max(\frac{1}{3^{2k-1}},\frac{1}{4^k})$[/quote]

where can i find it?


Does anybody know what tests the ti-89 titanium is allowed on? (ie. Sats, things like that)

It's allowed on the AP exams too, but not on the ACT exams.

I'm moving this to the Round Table; it'll probably get more replies there.

It is only allowed on certain parts of the Calculus, Statistics, and Physics AP exams i think.

I'm pretty sure though for the AP Calculus exam, there's a calculator part and a non-calculator part, so it's not just the TI-89 that's banned.

[quote="mysmartmouth"]TI-89 (vs. TI-84):

[u]Allowed[/u]:
MathCounts
SAT
AMC 8/10/12

[u]Forbidden[/u]:
IB Exams[/quote]
Which calculator are you talking about the TI-84 or 89? And what are IB exams?

I was talking about the TI-89.  And IB exams are International Baccalaureate exams (it's a popular international diploma program).

Threads like this don't exactly belong here because they tend to clog up the top of the board. But before I lock it, math92, just note that the TI-89 Titanium is accepted on just about every test and exam that allows graphing calculators. The SAT seems to set the standard in regards to those regulations and it is allowed on the SAT.


Prove that if for every point P in the interior of a triangle ABC one can construct a triangle having sides equal to PA, PB, and PC, then ABC is equilateral.

Assume $BC<AB$. Then make $P\to B$. In this case $PB+PC\to BC,\ PA\to AB$, so when $P$ is really close to $B$ (in the disc centered at $B$ and having radius $\varepsilon>0$ as small as we need it to be) we will have $PB+PC<PA$, so $PA,PB,PC$ won't be the sides of a triangle. This means that $AB=BC$, so each two sides are equal, and we're done.


calculus
integration
calculus computations

Maybe the following five formulas might help:

$ \int(f(x) \plus{} g(x))dx \equal{} \int{f(x)}\,dx \plus{} \int{g(x)}\,dx$

$ \int C f(x)\,dx \equal{} C\int f(x)\,dx$

$ \int x^n\,dx \equal{} \frac {x^{n \plus{} 1}}{n \plus{} 1} \plus{} K$

$ \int \,dx \equal{} x \plus{} K$

If $ F^{'}(x) \equal{} f(x)\,$ then $ \int^{b}_{a}f(x)\,dx \equal{} \int{f(x)}\,dx\Big|^{b}_{a} \equal{} F(b) \minus{} F(a)$


If $q$ is an integer that can be expressed as the sum of two integer squares, then prove that $3q$ cannot be so expressed.

[hide="Partial answer"]$q=m^2+n^2$
If $q\equiv 1\bmod{4}$, then $3q\equiv 3\bmod{4}$. Since all perfect squares are $\equiv 0,1\bmod{4}$, then $3q$ cannot be expressed as the sum of two integer squares.

I'm not immediately sure how to do the second part, and don't have time to figure that out at the moment. =\[/hide]

[hide]Well, it works for $q=0$. I'm guessing it means $q\in \mathbb{N}$.

Suppose there exists $q$ such that both $q$ and $3q$ are sums of two squares. Consider the least $q$ with this property and write $q=a^2+b^2,\ 3q=c^2+d^2\Rightarrow 3(a^2+b^2)=c^2+d^2$.

Reduce mod 3 to get rid of the left side. $c^2+d^2\equiv 0\pmod{3}$. Since squares are 0 or 1 mod 3, the only way this can happen is if $c$ and $d$ are both divisible by 3. Let $c=3j$, $d=3k$.

Then $3(a^2+b^2)=(3j)^2+(3k)^2\Rightarrow 3(j^2+k^2)=a^2+b^2$. But $j^2+k^2=\frac{q}{3}<q$ so $q$ is not the least number such that $3q$ is a sum of two squares. Contradiction.[/hide]

[hide]We know that any square must be $1\mod{3}$ or $0\mod{3}$.  Either way, $3q$ must be $3\mod{3}$.  Hence, if it is the sum of two squares, then each square is $0\mod{3}$, so each square is divisible by 3, and hence also by 9.  Then, if $3q=9m^2+9n^2$, then $q=3m^2+3n^2$.  Thus, $q$ is divisible by 3, and can only be the sum of two squares, each of which is divisible by 3.  So by similar reasoning, $\frac{q}{3}$ is the sum of two squares, each of which is divisible by 3, as is $\frac{q}{9}$, and so on.  At some point, we'll run out of 3's, and $\frac{q}{3^x}$ will not be an integer for some x.  Thus, our original hypothesis was false.[/hide]


calculus
integration
function
algebra
polynomial
floor function
real analysis

${\sum_{i=a}^{b}f(i)=\int_{a}^{b}G(f(t)) \, dt}$

There exist a function $G$ such as this is true for all nice function $f$ ?

I was thinking, $G$ is a some kind of combinations of floor function and other function.

No way. On the left we have a linear functional of $f$. This should imply that $G$ is a linear function, i.e. $G(x)=cx$ for some constant $c$. And this $G$ doesn't do what you want.

$G$ is not linear!! no way!

${f(x)=x^{n}}$

${G(x)=-n \zeta \left(1-n,x^{1/n}+1\right)-\frac{a^{n}}{a-b}}$

${b>a>0}$

It could be showed that:

${\sum_{i=a}^{b}f(i)=\int_{a}^{b}G(f(t)) \, dt}$

an example that $G$ can be non-linear.

you didn't proof that $G$ must be linear, we can have a simple counter example

No, it's hopeless. Look at the cases $f(x)=c$. That gives you $G(c)=\frac{b-a+1}{b-a}c$. This doesn't work in general.

for what $n$ you transform magically ${f(x)=x^{n}}$  into $f(x)=c$?

i really don't get this

I made another for everybody become happy.

${f(x)=\alpha x^{n}}$

${G(x)=-n \alpha \zeta \left(1-n,\left(\frac{x}{\alpha }\right)^{1/n}+1\right)-\frac{\alpha a^{n}}{a-b}}$

${a>b>0\land \alpha >0}$
${(n\in \mathbb{R})\land (n\neq 0)}$
${\sum_{i=a}^{b}f(i)=\int_{a}^{b}G(f(t)) \, dt}$

Note "see that the function is undefinied on $n=0$ :)"

I don't. I just think that constant functions are "nice".

This is a very silly way of approaching things. Asking for exactness is unreasonable, and your $G$ is in a nonsensical place.

silly? you post 2 non-sense posts and i'm the murder?

oh man.. c'mon...

go to orkut to say those things for someone. here is just math.

I just want a methode to find $G$ if you don't know this methode, just dont post here. because you don't gain point every post you made

c'mon.... don't insult persons.

It's not personal. It's just math- this has no chance of working. The class of "nice" functions is too large, no matter what you do. I can always add something that's zero at the points in the sum- that's almost certain to change the right hand side.
Also, sums and integrals are linear objects- introducing nonlinearity in just one place is doomed. If we choose a class of "nice" functions that includes the constants (such as every class I've ever seen on a closed interval), all choice is eliminated just looking at them, by the structure you've chosen.

i just asked for a Indirect Way to find $G$

for some kind of $f$ , $G$ can be easy to calculate.

think this as a Functional Diferential equation.

I want to take G out the integral.

and let's talk about unreasonable things.

a lot things in math dont make sense.

Why should I solve a ODE since it dont have any physical aplication? just for fun?!

So, if you ask, why should I find G?

i will ask for you. just for fun. a challenge. the result can be beautiful and useful, maybe yes maybe not. how knows? but give up unless you try

There is no $G$.

At least, that's the answer if you want a $G$ that works for any nice function $f$ on a fixed interval. If you want something else, you haven't done a very good job of asking the question clearly.

i will stop

there is no logical reason to ask good problem to people that cant understant it

just because you dont know how to solve, you bully me

silly

I'm afraid that yagaron means a different order of quantifiers: for every $f$ there exists $G$ (depending on $f$) such that the corresponding equality holds (at least, all his examples are like that). This is, of course, true (just take any $G$ you want and add an appropriate constant) but completely non-interesting. As to "universal" function "G", there is none, of course (not in the sense that I or jmerry do not know how to find one but in the sense that for any given $G$, there is a polynomial $f$ such that the identity fails (I hope that all polynomials are "nice")).


geometry
3D geometry
algebra unsolved
algebra

Positive Integers are written on all faces of a cube, one on each. At each vertex, the product of the numbers on the faces that meet at the vertex are written. The sum of the numbers in the corners is 2004. If T denotes the sum of numbers on all faces, find the maximum value of T.

[hide]Labeling each of the faces a,b,c,d,e,f, we obtain the factoring of T, (a+e)(b+f)(c+d), where a,e b,f and c,d are on opposite faces to each other.

The factoring of 2004 is 2,2,3,167.

Since the product of those 3 sums=2004, one of the three factors must contain 167 as a factor. As well, since these are positive integers, each of the valuse must contain a factor. So, since the 2's and 3's would attribute to the sum very little in the non 167 factor, we should maximize that factor, by adding the 3, and, since we need every factor to contain a number, put the 2's in the other 2 factors. So we have:

WLOG
a+e=167*3
b+f=2
c+d=2

So the maximum value of T=167*3+2+2=505[/hide]


MATHCOUNTS
LaTeX
geometry
geometric transformation
ratio
geometric series

[quote="vamathletes"]This is from a high school competition, but I thought it was Mathcounts level.

What are the roots of the equation x^11+x^10...x^1+x^0?

If anyone knows how to put this in LaTeX, please do.[/quote]
Translation:$x^{11}+x^{10}+x^{9}....+x^{1}+x^{0}$.
I have no idea if this is mathcounts level or not.

technically since the greatest exponent is 11 there should be at max 11 different roots..i think...or something like that.

use the $\frac{p}{q}$ method where $p=\pm 1$ and $q=\pm 1$.

this is as far is I bothered to pay attention in math class  :blush:

but I believe the possible rational roots are $\pm 1$ but I don't know how to find the imaginary ones

[quote="Glen Mann"][hide="possible solution"]
to find the roots we must make the equation equal to zero
there are 12 powers of x, and in order for them to equal zero they must cancel each other out, the only way for this to happen is for x to be a negative number and in order for them to cancel out x must be -1

odd exponents keep the negative, even ones lose it.

so essentially we have -1 + 1 occur 6 times leaving zero which makes our root of the equation -1
[/hide][/quote]

[hide="answer"]
-1 is the answer.
[/hide]

[quote="vamathletes"]This is from a high school competition, but I thought it was Mathcounts level.

What are the roots of the equation x^11+x^10...x^1+x^0?

If anyone knows how to put this in LaTeX, please do.[/quote]
[hide="intro"]
Multiply the equation by (x-1).
$x^{12}=1$
[/hide]
Now I am going to show two different methods, both using the intro:
[hide="beyond HSB solution"]
This is the 12th roots of unity.
$\frac{\pm \sqrt{3}}{2}+\frac{\pm i}{2}$
$\frac{\pm 1}{2}+\frac{i\sqrt{\pm 3}}{2}$
$\pm i$
-1
[/hide]
[hide="solution that takes less knowledge"]$x^{12}-1=0$
$(x^{6}+1)(x^{6}-1)$
$(x^{2}+1)(x^{4}-x^{2}+1)(x^{3}+1)(x^{3}-1)$
For the first term, the roots are i and -i.
For the third, we have $-1$, $\frac{1 \pm i\sqrt{3}}{2}$
For the 4th it's $\frac{-1 \pm i \sqrt{3}}{2}$
For the second, let $x^{2}=n$
$n^{2}-n+1$
$n=\frac{1 \pm i\sqrt{3}}{2}$
$x=\frac{\sqrt{3}\pm i}{2}$
[/hide]

[quote="ckck"][quote="max_tm"]Um.... am I missing something?:

$x^{0}=1$

so we check $\pm1$

$f(-1)=0$

so we conclude $-1$ is a root.

Simple as that?[/quote]
Yea but its not as simple as that. Since it has a degree of 11, theres going ot be at least 11 roots, rational or irrational.[/quote]
Not only at least, but exactly 11 complex roots. Like bpms showed above, only -1 is the other real root.

[quote="mysmartmouth"]How do we see so easily that this equation can be rewritten?

[hide="Answer"]
We can use the formula for the sum of a finite geometric series:

$S = \frac{a(1-r^{n})}{1-r}$

where a is the first term, r is the common ratio, and n is the number of terms.

Now consider the given equation:

$x^{11}+x^{10}+x^{9}....+x^{1}+x^{0}$

$a=x^{0}=1$
$r=x$
$n=12$

So
$S = \frac{1(1-x^{12})}{1-x}$
$S = \frac{1-x^{12}}{1-x}$

So that means $x^{11}+x^{10}+x^{9}....+x^{1}+x^{0}= \frac{1-x^{12}}{1-x}=0$

$\frac{1-x^{12}}{1-x}=0$

Obviously the only roots are 1 and -1, except 1 gives us a 0 in the denominator, so -1 is the only (real) solution.

[/hide][/quote]
Or just multiply by (x-1).

[quote="1=2"]uuhh.. this is way above mathcounts level. Just put it into the intermediate forum.[/quote]
It can be solved without methods above intermediate.

Assuming only real roots to be checked for this is how I did it:

[hide]
Just 'see' the root at -1, then:

$(x+1)(x^{10}+x^{8}+x^{6}+x^{4}+x^{2}+1)=0$

Notice how the right hand side is always greater than zero so it can't possibly contain another (real) root.
[/hide]

Another problem from a competition was: Find the number of real roots of $x^{8}-x^{7}+2x^{6}-2x^{5}+3x^{4}-3x^{3}+4x^{2}-4x+\frac{5}{2}= 0$

That one can be solved using similar logic as in my solution to this problem.

[quote="perfect628"]Real roots are pretty trivial with RRT.[/quote]Wrong. RRT only guarantees [i]rational[/i] roots, not real roots...  :ninja:

oops-wasn't paying attention to what I was saying... :blush:


Billy Bob, back from the dead, decides to build a bridge across the river next to his house.  Being dead didn't do him any good, so he asks his evil twin, Billy Bobo, to help.  He tells Billy Bobo to meet him at the river at 8:00 a.m.  When the day comes, however, Billy Bobo, being evil, decides to eat a meatspin steak sandwich instead of helping Billy Bob.  Billy Bob decides to build the bridge anyway, and finishes at 1:00 p.m.  Soon after he finishes, Billy Bobo finishes his meatspin steak sandwich and absentmindedly wanders over to the river.  Billy Bob yells at him, "Where where you?  If you had come when you were supposed to, we could have finished by 10:55 a.m.!"  How long would it have taken Billy Bobo to make the bridge by himself?

[hide]It takes Billy Bob 5 hours to finish the bridge, so his rate is $\frac15$.  If they worked together, it would have taken $\frac{35}{12}$ hours.  Call Billy Bobo's rate $x$.  Now, we have an equation $\frac15+x=\frac{12}{35}$.  Solve for $x$:

$\frac15+x=\frac{12}{35}$
$x=\frac{5}{35}$
It would have taken Billy Bobo $7$ hours to build the bridge by himself.
[/hide]

[quote="b-flat"][hide]It takes Billy Bob 4 hours to finish the bridge, so his rate is $\frac14$.  If they worked together, it would have taken $\frac{23}{12}$ hours.  Call Billy Bobo's rate $x$.  Now, we have an equation $\frac14+x=\frac{12}{23}$.  Solve for $x$:

$\frac14+x=\frac{12}{23}$
$x=\frac{25}{92}$[/hide][/quote]
[hide]So it would've taken him $\frac{92}{25}$ hours to build the bridge [/hide]:wink:

Thanks, Klebian.

[quote="superwizard188"]Wait, isn't it 5 hours from 8 - 1? o-0

[hide]Billy Bob's rate is 1/5. Together, their rate is 12/35. 12/35-7/35 = 5/35. So... It would take 35/5 or 7 hours?[/hide][/quote]

You are also right about the amounts of time.  Thanks again  :blush:
 :wallbash:

[hide]It takes Billy Bob 5 hours to build the bridge, and it takes $\frac{35}{12}$ hours for the two to build it together.

We can set up the equation $\frac{35}{12}\left(\frac{1}{5}+\frac{1}{x}\right)=1$. Solving we find that Billy Bobo's time alone would be [b]7 hours[/b].[/hide]

[hide]7[/hide]hours is correct :w00t:  :thumbup:  :w00tb:


Determine all even numbers $n$, $n \in \mathbb N$ such that \[{ \frac{1}{d_{1}}+\frac{1}{d_{2}}+ \cdots +\frac{1}{d_{k}}=\frac{1620}{1003}}, \]
where $d_1, d_2, \ldots, d_k$ are all different divisors of $n$.

Really, is it possible to have a more pointless question than this?

Since $1003 = 17 \times 59$, we observe that $17 | n$ and $59 | n$, otherwise the denominator of the sum will not be divisible by $17$ and/or $59$.  Since $n$ is even, $2006 | n$.  It remains to observe that $\sum_{d|2006} \frac{1}{d} = \frac{1620}{1003}$, so all the solutions for $n$ are simply the multiples of $2006$.

The hardest part of the question seems to be factorising $1003$. :P

Valiowk is wrong. Because he wrote all solution of n are simply multiples of 2006.
but if n is a certain solution of the equation then any multiplication of n(assume p*n and  p>1) will not hold the equality. Because then the L.H.S. of the given equation will become larger as there will be additional divisiors which will work as denominator. That means if solution exists, it will be unique

But as Valiowk said if n is a solution then n will be divisible by 2006.
but for n=2006,
L.H.S.= 1/1 + 1/2+ 1/17+ 1/59+ 1/2.17+ 1/2.59+1/17.59+ 1/2.17.59=1620/1003

So n=2006 is the only solution

(Sorry for not being able to write in latex. i've forgot how to use it.. i'll learn it again as soon as possible) :)


inequalities
function
inequalities unsolved

can anyone help me with this one?
Given a+b+c=1, a,b, c > 0,

Prove that

(a+1)(b+1)(c+1)>=8(1-a)(1-b)(1-c)

After using homogeneous method, it turns out to prove that

$ \prod_{cyc}(2a+b+c) \ge 8\prod_{cyc}(b+c)$

Let

$ \{\begin{array}{ccc}
x=a+b\\ \\
y=b=c\\ \\
z=c+a \end{array}$

The inequality becomes

$ \sum_{cyc}z(x-y)^2 \ge 0$ which is obviously true

It can be easily proved if you use Am-Gm and $ a\plus{}1\equal{}(1\minus{}b)\plus{}(1\minus{}c)$  :wink:

[quote="gai"]can anyone help me with this one?
Given $ a\plus{}b\plus{}c\equal{}1$, $ a,b, c > 0$,

Prove that

$ (a\plus{}1)(b\plus{}1)(c\plus{}1)\ge 8(1\minus{}a)(1\minus{}b)(1\minus{}c)$[/quote]

$ \Leftrightarrow (a\plus{}1)(b\plus{}1)(c\plus{}1)\ge (2\minus{}2a)(2\minus{}2b)(2\minus{}2c)\Leftrightarrow$ $ xyz\ge (x\plus{}y\minus{}z)(x\minus{}y\plus{}z)(\minus{}x\plus{}y\plus{}z)$ , where $ x\equal{}a\plus{}1,...$

[quote="Stephen"]It can be easily proved if you use Am-Gm and $ a \plus{} 1 \equal{} (1 \minus{} b) \plus{} (1 \minus{} c)$  :wink:[/quote]
$ a\plus{}1\equal{}(1\minus{}b)\plus{}(1\minus{}c) \ge 2\sqrt{(1\minus{}b)(1\minus{}c)}$
$ b\plus{}1\equal{}(1\minus{}a)\plus{}(1\minus{}c) \ge 2\sqrt{(1\minus{}a)(1\minus{}c)}$
$ c\plus{}1\equal{}(1\minus{}b)\plus{}(1\minus{}a) \ge 2\sqrt{(1\minus{}b)(1\minus{}a)}$
Multiplying and we're donne.
$ (a\plus{}1)(b\plus{}1)(c\plus{}1)\ge 8(1\minus{}a)(1\minus{}b)(1\minus{}c)$

[hide="Solution"]Expanding, we are to prove that $ abc \plus{} ab \plus{} bc \plus{} ca \plus{} a \plus{} b \plus{} c \plus{} 1\ge 8 \minus{} 8a \minus{} 8b \minus{} 8c \plus{} 8ab \plus{} 8bc \plus{} 8ca \minus{} 8abc$  $ \implies 2\ge 7ab \plus{} 7bc \plus{} 7ca \minus{} 9abc$.

Note that since $ (7 \minus{} 9a)(7 \minus{} 9b)(7 \minus{} 9c) \equal{} 81(7ab \plus{} 7bc \plus{} 7ca \minus{} 9abc) \minus{} 98$, we have $ 7ab \plus{} 7bc \plus{} 7ca \minus{} 9abc \equal{} \frac {(7 \minus{} 9a)(7 \minus{} 9b)(7 \minus{} 9c) \plus{} 98}{81}$.

Hence, it suffices to prove that $ 2\ge \frac {(7 \minus{} 9a)(7 \minus{} 9b)(7 \minus{} 9c) \plus{} 98}{81}$.

Applying AM-GM on $ (7 \minus{} 9a)(7 \minus{} 9b)(7 \minus{} 9c)$, we obtain $ \left(\frac {21 \minus{} 9(a \plus{} b \plus{} c)}{3}\right)^3\ge (7 \minus{} 9a)(7 \minus{} 9b)(7 \minus{} 9c)\implies 64\ge (7 \minus{} 9a)(7 \minus{} 9b)(7 \minus{} 9c)$.

Hence, we have $ \frac {(7 \minus{} 9a)(7 \minus{} 9b)(7 \minus{} 9c) \plus{} 98}{81}\le \frac {64 \plus{} 98}{81} \equal{} 2$, which is what we wanted to prove. $ \blacksquare$[/hide]

[quote="FantasyLover"]Applying AM-GM on $ (7 \minus{} 9a)(7 \minus{} 9b)(7 \minus{} 9c)$,
[/quote]
You can't do this as $ 7\minus{}9a$ might be negative.

[hide="@ gai"]It is Indian RMO 1994 Prob 8.. Are you from India? :) [/hide]

Thanks a lot guys.. I stumbled upon the paper

17-th All-Russian Mathematical Olympiad 1991

http://imomath.com/othercomp/Rus/RusMO491.pdf

you can go check out the problems

jensen inequality for the function $ ln(\frac{1\plus{}a}{1\minus{}a})$! :wink:


Suppose $ I$ is an interval in $ \mathbb{R}$ and $ f: I\rightarrow \mathbb{R}$ is continuous and one-to-one (i.e. $ f(x_1)\not \equal{} f(x_2)$).  Show that $ f$ must be strictly increasing or strictly decreasing on $ I$.

If $ f$ is not monotone, then you can find three points $ a<b<c$ such that $ f(b)$ is either smaller or greater than both $ f(a)$ and $ f(c)$. Use the Intermediate Value theorem on $ [a,b]$ and $ [b,c]$ to prove that $ f$ is not one-to-one.


Consider k and n positive integers and A1, A2, . . . Ak real n x n square matrices. Prove that: det (A1At1 + A2At2 + . . . + AkAtk ) >= 0.
Ati is the transpose of Ai.

A sum $U=\sum U_k$ of symmetric positive semidefinite matrices $U_k$ is symmetric and positive semidefinite, since $x^tUx=\sum x^tU_kx\ge 0,\ \forall x\in\mathbb R^n$, so its determinant is $\ge 0$.


i know it should be somewhere in the latex tutorial but i never found it. :oops:

how do you start a new line in latex  :lol: i no you need to use two line if you want two separate lines but i want a space between two separate lines

You can use \vskip with a distance measurement, such as \vskip 3mm.

Or you could use \newline if you just want a new line.


Let $a_n>0$ such that $\sum_{n\geq 1}a_n$ is convergent
$u_n=(\prod_{k=1}^{n}a_k)^\frac{1}{n}$
Prove that for any $n\in N*$
$\sum_{k=1}^{n}u_k\leq e.\sum_{k=1}^{n}a_k$

AKA Carleman's Inequality:
[url]http://mathworld.wolfram.com/CarlemansInequality.html[/url]

If only they gave me this problem at an oral!

i guess there is some solution in the "RMS" magazine of "grandes ecoles" 98!!
i m not sure i'll go to check it out!
bye


calculus
integration
SFFT
special factorizations
algebra
system of equations

Here are some "challenge" problems from my pre-cal textbook...they don't require much beyond basic algebra, though.  :D

1. Determine the value $x+y$ if $2^{x}=8^{y-1}$ and $9^{y}=3^{x-8}$.

2. Determine the pairs of integers $(a,b)$ that satisfy the equation $a+b=ab$ and prove that these are the only ones.  (The prove part isn't in the original problem, I added it in.)

3. What is the length of each side of a regular hexagon if opposite sides are $18$ inches apart?

[quote="eryaman"]2. Determine the pairs of integers $(a,b)$ that satisfy the equation $a+b=ab$ and prove that these are the only ones.
[/quote]

$0=ab-a-b$, so $1=(a-1)(b-1)$ from Simon's favorite factoring trick  :lol: $1=1 \cdot 1$ and $-1 \cdot -1$.

\[a-1=1\]
\[a=2 \Rightarrow b=2\]

\[a-1=-1\]
\[a=0 \Rightarrow b=0\]

They are the only ones because there are only 2 distinct integral factors of 1. (+ and -)

3.
[hide] This means that the apothem is $9$, and use your knowledge of 30-60-90 triangles to see that each side is

$2 \cdot 3\sqrt3$
=
$6\sqrt3$

Note: If you don't know how I came up with using the 30-60-90 triangles to solve this problem, use your knowledge of interior angles of a polygon and note that the radius of a regular polygon bisects the interior angle.[/hide]

it'd be easier to note that the apothem bisects the central angle.

3.
[hide="Did you mean this Hamster1800?"]
the apothem bisects the central angle, forming a a 30-60-90 triangle where the apothem meets half the side at right angles

So, the length of the side is
$%Error. "display" is a bad command.
2(9  tan  30)= 18(\frac tion{1}/{\sqrt3})$
                  $=  6\sqrt3$
[/hide]

Ooh, never looked at it like that before, Hamster. Cool stuff!

Um...I did problem number 1)

2^x=8^y-1 and 9^y=3^x-8

Then you know that 2^x=2^3y-3, so x=3y-3

Then for the other you know that 3^2y=3^x-8, so 2y=x-8

Now you gotta system of equations:

So then you can solve that and get y=11 and x=30, so x+y=[b]41[/b]

For number 2, when you have a+b=ab, you can find out that a=ab-b, which is a=b(a-1) so b=a/(a-1), and so....whoah....am I doing this right?


geometry
analytic geometry
angle bisector

If we apply an inversion with arbitrary radius center $ M$, then $ \angle MA'P'\equal{}\angle APM\equal{}\angle BPM\equal{}\angle MB'P$ so $ A'P'\equal{}B'P'$ so $ P'$ lies on a line. If we do an inversion back, $ P$ lies on a circle so we're done.

This locus is common; it is the appolonius' circle of the triangle $ APB$. I think that it is amusing that I came up with this problem before I went to sleep last night.

[quote="BanishedTraitor"]Let $ A, M$ and $ B$ be three collinear points on a line in that order such that $ M$ is not the midpoint of $ AB$. Let $ P_{i}$ where $ i \in \{ 1,2,3,4 \}$ be co-planar points such that $ P_{i}M$ is the angle bisector of $ \angle{AP_iB}$. Prove that $ P_1P_2P_3P_4$ is a cyclic quadrilateral.[/quote]

But the locus of all such points $ P$ such that $ PM$ is the bisector of $ \angle APB$ is the Apollonius circle of $ AMB$. Are you looking for a proof of the original theorem? An easy proof exists [url=http://www.mathlinks.ro/weblog_entry.php?t=166655]using coordinates[/url] and it's not hard to derive.

This is what i tried to write yesterday...but my net line was down  :mad:.

[quote="BanishedTraitor"]Let $ A, M$ and $ B$ be three collinear points on a line in that order such that $ M$ is not the midpoint of $ AB$. Let $ P_{i}$ where $ i \in \{ 1,2,3,4 \}$ be co-planar points such that $ P_{i}M$ is the angle bisector of $ \angle{AP_iB}$. Prove that $ P_1P_2P_3P_4$ is a cyclic quadrilateral.[/quote]
I think that it is simple. As we know that, the locus of the points $ P_i$ s.t. $ \frac{AP_i}{P_iB}\equal{}k$ is a circle, the problem is solved.
Here, $ \frac{AP_i}{P_iB}\equal{}\frac{AM}{MB}$

Precisely, this circle is the Apollonian circle as [b]Altheman[/b] says.


induction
inequalities
number theory unsolved
number theory

Show that for all positive integer n$ \ge$2, prove that the "multiplication" of all primes that are less or equal than n is smaller than $ 4^n$.

for example, if we take n=4, then $ 2*3\le 4^4 \equal{} 256$

Thanks.

The standard trick, if I recall correctly, is to note that $ 4^n \equal{} 2^{2n} > {2n \choose n}$ and induct, first showing that $ {2n \choose n}$ is divisible by the primes in $ (n, 2n]$.  It's a little tedious.

Some specific solution of the induction?

Thanks

We use induction on $ n$.

If $ n\equal{}2$, the inequality is trivial. Let's assume that the inequality holds for all $ k\equal{}1,2,..,n\minus{}1$

Denote by $ P(a,b)$ the product of all the primes in the closed interval $ [a,b]$.
The induction hypothesis is that $ P(1,k) < 4^k$ for all $ k\equal{}1,2,..,n\minus{}1$.

If $ n$ is even, then $ P(1,n) \equal{} P(1,n\minus{}1) \leq 4^{n\minus{}1} < 4^n$.

If $ n$ is odd, let $ n\equal{}2t\plus{}1$.

Observe that all primes in the interval $ [t\plus{}2, 2t\plus{}1]$ divide the value $ {2t\plus{}1 \choose t}$.
This implies $ P(t\plus{}2, 2t\plus{}1) \leq {2t\plus{}1 \choose t}$
Since $ 2^{2t\plus{}1} \equal{} {2t\plus{}1 \choose 0} \plus{} {2t\plus{}1 \choose 1} \plus{} .. \plus{} {2t\plus{}1 \choose t} \plus{} {2t\plus{}1 \choose t\plus{}1} \plus{} .. \plus{} {2t\plus{}1 \choose 2t\plus{}1} \geq 2{2t\plus{}1 choose t}$
Therefore, $ 4^t \equal{} 2^{2t} \geq {2t\plus{}1 \choose t} \geq P(t\plus{}2,2t\plus{}1)$.

By the induction hypothesis, $ 4^{t\plus{}1} > P(1,t\plus{}1)$.
Multiplying the two above inequalities, one arrives at the conclusion $ 4^{2t\plus{}1} \geq P(1,2t\plus{}1)$, which completes the induction hypothesis.


rotation
combinatorics unsolved
combinatorics

Assume that $ A_1, A_2, \ldots, A_8$ are eight points taken arbitrarily on a plane. For a directed line $ l$ taken arbitrarily on the plane, assume that projections of $ A_1, A_2, \ldots, A_8$ on the line are $ P_1, P_2, \ldots, P_8$ respectively. If the eight projections are pairwise disjoint, they can be arranged as $ P_{i_1}, P_{i_2}, \ldots, P_{i_8}$ according to the direction of line $ l.$ Thus we get one permutation for $ 1, 2, \ldots, 8,$ namely, $ i_1, i_2, \ldots, i_8.$ In the figure, this permutation is $ 2, 1, 8, 3, 7, 4, 6, 5.$ Assume that after these eight points are projected to every directed line on the plane, we get the number of different permutations as $ N_8 \equal{} N(A_1, A_2, \ldots, A_8).$ Find the maximal value of $ N_8.$

I'm not sure, but the answer seems to be $2 {{n} \choose 2}$

Just rotate the $l$, then whenever it becomes perpendicular with some line $P_i P_j$, i and j switches their position on the permutation. When $l$ is rotating, it becomes perpendicular with $P_i P_j$ 2 times for specific $i, j$, so new permutation occurs $2 {{n} \choose 2}$ times in total.

For two parallel and directed lines with the same direction, the order of projections of $A_1 , A_2 \dots A_8$ must be the same. So, we only need to look at all the directed lines passing through a fixed point, say $O$

If a directed line is taken perpendicular to a line joining two given points, then the projections of these two points must coincide,
and a corresponding permutation will not be produced. When the directed lines taken are not perpendicular to any line joining two given points, any two projections of $A_1 , A_2 , \dots, A_8$ must not coincide. There is therefore a corresponding permutation.

(1) Suppose that the number of lines through point $O$ that are also perpendicular to a line joining two given points is $k$. Then $k\leq \dbinom{8}{2}= 28$. Then there arise $2k$ directed lines placed anticlockwise. Assume that they are in order of $l_1 , l_2 , \dots, l_{2k}$. For an arbitrary directed line $l$ (different from $l_1 , \dots, l_{2k}$ ) , there must be two consecutive directed lines $l_j$ and $l_{j+ 1}$ such that $l_j , l , l_{j+ 1}$ are placed anticlockwise. We can see that for a given $j$, the corresponding permutations obtained from such $l$ must be the same.

(2)For any two directed lines $l$ and $l'$ different from $l_1, \dots l_{2k}$, if we cannot find $j$ such that both $l_j, l, l_{j+1}$ and $l_j, l', l_j+1$ satisfy (1) then there must be $l$ so that $l’, l_j , l$ are placed anticlockwise. Assume that $l_j$ is perpendicular to the line joining $A_{j_1}$, and $A_{j_2}$;  the orders of the projections of points $A_{j_1}$, and $A_{j_2}$ on the directed lines $l$ and $l’$ must be different, so the corresponding permutations must also be different.

It follows from (1) and (2) that the number of different permutations is $2k$. Note that $k=\dbinom{8}{2}$ is attainable. so the maximal value of $N_8 = 56$.


Positive integers $ a,b,c$ satisfy $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c} \le 1.$ Prove that: $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c} \le \frac{41}{42}$.

It seems that the problem is wrong, but this is the form of the exercise in the pdf source. Maybe something must be changed. I don't know, try for example $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}<1$ instead of $ \le$.

If $ a,b,c\ge 3$, then $ \frac1a\plus{}\frac1b\plus{}\frac1c\le\frac13\plus{}\frac13\plus{}\frac14<\frac{41}{42}$. Assume $ a\equal{}2$. If $ b>3$, then $ \frac1a\plus{}\frac1b\plus{}\frac1c\le\frac12\plus{}\frac14\plus{}\frac15<\frac{41}{42}$. Assume $ a\equal{}2,b\equal{}3$. We have $ \frac1c<1\minus{}\frac12\minus{}\frac13\equal{}\frac16$ so $ c>6$. Therefore $ \frac1a\plus{}\frac1b\plus{}\frac1c\le\frac12\plus{}\frac13\plus{}\frac17\equal{}\frac{41}{42}$, as desired.

I found a Finnish source where the condition is $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} < 1$.


Let $ a$, $ b$, $ c$ be positive real numbers such that $ a\plus{}b\plus{}c\equal{}1$. Prove that \[ \frac{\sqrt{a^2\plus{}abc}}{c\plus{}ab}\plus{}\frac{\sqrt{b^2\plus{}abc}}{a\plus{}bc}\plus{}\frac{\sqrt{c^2\plus{}abc}}{b\plus{}ca}\le\frac{1}{2\sqrt{abc}}\]

This might help: $ \frac{\sqrt{a^2\plus{}abc}}{c\plus{}ab}\equal{}\frac{\sqrt{(a)(a\plus{}b)(a\plus{}c)}}{(c\plus{}a)(c\plus{}b)}$

Can you give me a full solution? Thank you.

That wasn't supposed to be a full solution, just a hint.

Make sure you can figure out why that's true before you use it though.  :D

Use a permutation $ (b,c)\rightarrow(c,b)$ and see the can_hang2007's post here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204969


Okay, the following equation's solution is obvious $(2, -1)$ intuitively, but can someone post an algebraic solution?

$\frac{(x-2)^2}{2} + \frac{(y+1)^2}{5} = 0$

I'd argue that IS the algebraic solution; if you have a big polynomial that you want to find the root of, you'd factor it, it's just yours is already factored. So it's done. A factored polynomial is far more useful than an expanded one.

Note that $(x-2)^2\ge 0$ and $(y+1)^2 \ge 0$ for all $x$ and $y$.
Therefore, RHS=0 only if $(x-2)^2=0$ and $(y+1)^2=0$.
Then obviously the only solution is $(x,y)=(2,-1)$.

Trivial inequality shows that both LHS terms must be 0.


algorithm
number theory
greatest common divisor
Euclidean algorithm
Diophantine equation

Was just thinking about this the other day; using bezout's lemma + the extended euclidean algorithm it's fairly easy to generate an initial solution to a linear diophantine equation, but I always seem to forget the general form of the solution set since it has always been presented to me out of the blue and then proven that it satisfies all solutions. Is there some motivated way to help "remember" the general solution set?

See here: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202353&start=3]click here[/url]

Basically, say you have $ Ax\plus{}By\equal{}C$. Divide through by $ \text{GCD}[A,B]$ to get another equation $ A'x\plus{}B'y \equal{} C'$. Clearly, $ C$ must be an integer or there is no solution.

Now, if we have an initial solution $ (x_0,y_0)$, then the general solution is $ (x_0 \plus{} B' n, \; y_0 \minus{} A' n)$


Ok, so basically I would like to start a discussion about the chapter that just happened in February in the AoPS classroom. I think this would be a great oppurtunity for people to get a better understanding of every problem they either missed or had a bad way of solving.

This would be at 8:00 p.m. EST today (7 for central, 6 for mountain, and 5 for pacific)

Thanks for helping others like me,
Jorian Hoover

No, we do not open the classroom for things such as this.  (Basically, it is too time-consuming for us to monitor.)

Please use this message board for your discussion -- that's what it's here for!


hello, your inequality is equivalent with
$ 2\plus{}2x\plus{}2y\plus{}x^2\plus{}y^2\minus{}2\sqrt{x}y\minus{}2x\sqrt{y}\plus{}2x^{\frac{3}{2}}\plus{}2y^{\frac{3}{2}}\minus{}8\sqrt{x}\sqrt{y} \geq0$.
We observe that the following inequalities hold
(1)$ x\plus{}y \geq2 \sqrt{xy}$
(2)$ x\plus{}y^2\geq 2y\sqrt{x}$
(3)$ y\plus{}x^2\geq 2x\sqrt{y}$
(4)$ 2\plus{}2y^{\frac{3}{2}}\plus{}2x^{\frac{3}{2}}\geq6\sqrt{xy}$.
Adding (1),(2),(3),(4) we get the desired rsult.
Sonnhard.

[b]limes123[/b] can you explain me first part of your proof?

$ (1\plus{}xy)(1\plus{}\frac{x}{y})\geq (1\plus{}x)^2\iff \frac{1}{(1\plus{}x)^2}\geq \frac{1}{1\plus{}xy}\cdot \frac{y}{x\plus{}y}$ and analogously $ \frac{1}{(1\plus{}y)^2}\geq \frac{1}{1\plus{}xy}\cdot \frac{x}{x\plus{}y}$ hence result ;)

[quote=vo thanh van]Problem 2. Prove that for $ x$ and $ y$ positive reals,
$ \frac {1}{(1 \plus{} \sqrt {x})^2} \plus{} \frac {1}{(1 \plus{} \sqrt {y})^2} \ge \frac {2}{x \plus{} y \plus{} 2}$[/quote]
[quote=limes123]Or
$ \frac {1}{(1 \plus{} \sqrt {x})^2} \plus{} \frac {1}{(1 \plus{} \sqrt{y})^2}\geq \frac {1}{1 \plus{} \sqrt {xy}}\geq \frac {2}{x \plus{} y \plus{} 2}$ ;)[/quote]

 Prove that for $ x$ and $ y$ positive reals,
$$ \frac {1}{(1 \plus{} \sqrt {x})^2} \plus{} \frac {1}{(1 \plus{} \sqrt {y})^2} \ge \frac {1}{\sqrt{xy}\plus{} 1}\ge \frac {2}{x \plus{} y \plus{} 2}$$
[url=https://artofproblemsolving.com/community/c6h1830243p12253317]here[/url]


linear algebra
matrix
algebra
polynomial
absolute value
linear algebra unsolved

Let $A \in M_n(Z) , 25A^{25} = 9A^9 + 16I_n$. Show that :
1. $\exists M > 0: |(A^m)_{i,j}| \leq M, \forall m \in N, \forall i, j = 1,...,n$
2. $|det(A)| = 1$

[My problem is that i proved $A = I_n$, but i can be wrong, can you check this beautiful problem ?]

You have great Linear Algebra problems, I'd defiantely want to get my hands on that source of yours  :lol:

All the eigenvalues of $A$ are $1$ because i think that the only real solution to $25\lambda_i^{25}=9\lambda_i^9+16$is $1$ ... so $A\in\mathcal l M(\mathcal l Z)$ is a matrix with the eigenvalues $\lambda_i=1$ shouldn`t $A=I_n$?   :?

First, not any matrix with eigenvalues 1 is the identity. Secondly, A may have very well non real eigenvalues.

Anyway, the problem is very simple, though nice. The first point is clear: if one puts $N(A)$ the square root of the sum of the squares of the absolute values of all elements of $A$, this will be a norm on $M_n(C)$ and since $25N(A^{k+25})$ is  smaller than $9N(A^{k+9})+16N(A^k)$, we conclude immediately that $N(A^n)$ is bounded and since all norms are equivalent  in finite dimension, we are done. The second question is now immediate, since $A$ is clearly invertible and the powers of $A$ have bounded values, thus two powers of $A$ will be identical and conclusion. What is nice is that therefore $A$ is of finite order. Note that the second question is also immediate without a), since it is clear that $A$ is invertible with all eigenvalues in the unit disk, thus its determinant is a non zero integer with absolute value at most 1. Now, nice question in order to elucidate ther structure of these matrices: all eigenvalues of $A$ are roots of unity (note that this could also be obtained without a), by using Kronecker"s theorem) and roots of $25X^{25}-9X^9-16$. What values can they take?  ;)

i agree the problem is not very difficult, but nice ;), but ... i wonder if there exists one A different from $I_n$ that satisfies the initial conditions, because all eigenvalues are 1, $(A-I_n)^n = O_n$ and from here by several computations, $A=I_n$  :oops: [or i have done some stupid things...]

Yes, of course, each eigenvalue has modulus 1 and the only root of modulus 1 of this polynomial is 1, this $A$ is unipotent and diagonalizable (since it has finite order) thus it must be the identity.


how do I write (a sub 0) i can't figure out how to do it

Did you mean $ a_{0}$ ?
[code]$a_{0}$[/code]

Please check the [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:About]LaTeX[/url] wiki, next time.


geometry AMC USA(J)MO USAMO AIME inequalities

I can only do about half the problems in this book with the half I can't do being the ones requiring extra segments to be drawn.  What's the best way to learn to do these problems?  Should I spend a few hours on them or just look at the solution when I can't solve it?

I think you should spend hours on it. If/when you find the solution, you will learn much more. That said, there are some instances where this may be a waste of time as well. Eventually look at the solution after about an hour [i]at least[/i] (meaning it may be better to work on it more).

I have this book also, and can someone tell me if it is harder than [i]Geometry Revisited[/i]? The first chapter seemed really simple, and I don't know if I should spend more time on it.

There are a few problems that are USAMO level. Most of the problems are Medium AIME level. Geometry Revisited covers USAMO material (but doesn't have a lot of hard geometry problems for practice.)

I agree. What is the best geometry book for practice?

[quote="archimedes1"][quote="Anaxerzia"]
I have this book also, and can someone tell me if it is harder than [i]Geometry Revisited[/i]? The first chapter seemed really simple, and I don't know if I should spend more time on it.[/quote]
Geometry Revisited wins, no contest.  CPIG problems seem like mostly one-step problems, whereas Geometry Revisited takes you throguh a lot of material.  That said, there are only about 10-15 olympiadish problems there.

For olympiad geometry I would recommend Kiran Kedlaya's free [i]Geometry Unbound[/i].  It has helped a lot.[/quote]

Same from me too! Also Bottema and co can also help if you are interested in inequalities related to geometry!

I just have to add two of my favourites: Prasolov and Yaglom. Prasolov's Problems in plane geometry is probably the best collection of geometry problems I've seen in my life-and it's so rich when it comes to ideas and methods. Same holds for Yaglom's Geometric transformations (it's not so general as Prasolov's plane geometry, but still interesting and educative, even for those who aren't much of a geometry fans-I include myself in that class  :) )

My suggestion:
1) First take theoretical books with some problems that might help - Geometry Revisited
2) Go to books with lots of problems of different level - Prasolov
3) Pass to olympiad level - decide on your own or take all


Let ABCD be a rectangle such that AB= 46 cm and BC= 22 cm . Let M be a point on AB such that AM=18 cm , N be a point on BC such that BN=8 cm , P be a point on CD such that CP=18 cm , and Q be a point on DA such that DQ=8 cm . Let H be the intersction of MP and NQ . What is the number of the square centimeters in the area of the quadrilateral MHNB ?

[hide]246[/hide]

hard to explain without a diagram but....

[hide]start by drawing every single thing that is given to you....
Connect point M with point N, point N with point P, and etc....
It will create a rhombus.....
Now if you already drew everything, you'll see that MHNB is actually
made up of 1/4 of the rhombus + triangle MNB!!!!
The rhombus has an area of 46*22-2((14*18)/2)-2((8*28/2) which is equal to
46*22-14*18-8*28=536... 1/4 of that is 134.
Triangle MNB has an area of 28*8/2=112.
112+134=246![/hide]


calculus integration trigonometry algebra polynomial function limit

Help required

Prove that
$\int_0^{\infty} {\frac{dx}{1 + x^n}}=\int_0^1{\frac{dx}{{(1 - x^n)}^{\frac 1{n}}} (n > 1)}$

Make a silly substitution $\frac{1}{1 + x^n} = 1 - y^n$. Then

$1 + x^n = \frac{1}{1 - y^n},\ \ x^n = \frac{1}{1 - y^n} - 1 = \frac{y^n}{1 - y^n},\ \ x = \frac{y}{\left(1 - y^n\right)^{\frac 1 n}}$

$y = 0$ when $x = 0$ and $y \longrightarrow 1$ when $x \longrightarrow \infty$. This takes care of the integration limits. Furthermore,

$\frac{\text dx}{\text dy} = \frac{\left(1 - y^n\right)^{\frac 1 n} - y \cdot \frac 1 n \left(1 - y^n\right)^{\frac 1 n - 1} \cdot \left(-ny^{n - 1}\right)}{\left(1 - y^n\right)^{\frac 2 n}} =$

$= \frac{1 + \frac{y^n}{1 - y^n}}{\left(1 - y^n\right)^{\frac 1 n}} = \frac{1}{(1 - y^n) \left(1 - y^n\right)^{\frac 1 n}}$

Consequently,

$\int_0^\infty \frac{\text dx}{1 + x^n} = \int_0^1 (1 - y^n) \cdot \frac{\text dy}{(1 - y^n)(1 - y^n)^{\frac 1 n}} = \int_0^1 \frac{\text dy}{(1 - y^n)^{\frac 1 n}}$

A more interesting problem is how to actually calculate this definite integral. The answer is $\int_0^\infty \frac{\text dx}{1 + x^n} = \frac{\frac \pi n}{\sin{\frac \pi n}}$.

It belong to the complex calculus. First, assume that n = 2m is even. The polynomial $1 + z^n = 1 + z^{2m}$ has n complex roots $z_k = \exp \left(\frac{(2k - 1)\pi i}{n}\right),\ k = 1, 2, ..., n$. Half of these roots are in the upper half of the complex plane. Using the residue theorem for the function $f(z) = \frac{1}{1 + z^{2m}} = \frac{1}{g(z)}$, we choose the closed integration path $C_R$ to be a segment of the real axis from -R to +R and a semicircle |z| = R in the upper half of the complex plane, R sufficiently large (R > 1), so that all m roots $z_k = \exp \left(\frac{(2k - 1) \pi i}{2m}\right),\ k = 1, 2, ..., m$ are inside of this closed curve:

$\oint_{C_R} \frac{\text dz}{1 + z^{2m}} = \int_{-R}^{+R} \frac{\text dx}{1 + x^{2m}} + \int_{|z| = R} \frac{\text dz}{1 + z^{2m}} =$

$= 2 \pi i\ \sum_{k=1}^m \text{Res} f(z_k) = 2 \pi i\ \sum_{k=1}^m \frac{1}{g'(z_k)} = 2 \pi i\ \sum_{k=1}^m \frac{1}{2m z_k^{2m - 1}} =$

$= \frac{2\pi i}{2m}\ \sum_{k=1}^m \frac{1}{\exp \left(\frac{(2k - 1)(2m - 1) \pi i}{2m}\right)} = \frac{2\pi i}{2m}\ \sum_{k=1}^m \exp \left(-\frac{(2k - 1)(2m - 1) \pi i}{2m}\right) =$

(using $\exp(2\pi k i) = 1$ and substituting a new index $l = m - k + 1$)

$= \frac{2\pi i}{2m}\ \sum_{k=1}^m \exp \left(-\frac{[2(m - k) + 1] \pi i}{2m}\right) = \frac{2\pi i}{2m}\ \sum_{l=1}^m \exp \left(-\frac{(2l - 1) \pi i}{2m}\right) =$

(because of $\cos(\pi - z) = -\cos z$, the cosine terms cancel out)

$= \frac{2\pi}{2m}\ \sum_{l=1}^m \sin \left(\frac{(2l - 1) \pi}{2m}\right)$

Now we use the following trigonometric formula:

$2 \sin\left(\frac{\pi}{2m}\right) \sum_{l=1}^m \sin \left(\frac{(2l - 1) \pi}{2m}\right) =$

$= \sum_{l=1}^m \left[\cos\left(\frac{2(l - 1) \pi}{2m}\right) - \cos\left(\frac{2l \pi}{2m}\right)\right] = \cos(0) - \cos(\pi) = 2$

Substituting this back into the sum of residues,

$\int_{-R}^{+R} \frac{\text dx}{1 + x^{2m}} + \int_{|z| = R} \frac{\text dz}{1 + z^{2m}} = 2 \cdot \frac{\frac{\pi}{2m}}{\sin \left(\frac{\pi}{2m}\right)}$

The right side does not depend on R, hence, the last equation holds in the limit transition $R \longrightarrow \infty$. But in this transition, the integral over the semicircle vanishes and we get

$\int_0^{\infty} \frac{\text dx}{1 + x^{2m}} = \frac 1 2 \left(\int_{-\infty}^{+\infty} \frac{\text dx}{1 + x^{2m}} + 0\right) =$

$\frac 1 2 \left(\lim_{R \rightarrow \infty} \int_{-R}^{+R} \frac{\text dx}{1 + x^{2m}} + \lim_{R \rightarrow \infty} \int_{|z| = R} \frac{\text dz}{1 + z^{2m}}\right) = \frac{\frac{\pi}{2m}}{\sin \left(\frac{\pi}{2m}\right)}$

This completes the calculation of the integral for n = 2m even.
____________________________________________________

Unfortunately, without a modification, this approach does not work for n = 2m + 1 odd. The modified approach not only works for both even and odd n, but it is also simpler. Choose the closed integration path to be the perimeter of a circular sector $C_R$ centered at the origin z = 0, i.e., the segment of the real axis from 0 to R, then the circular arc from z = R to $z = R \exp \left(\frac{2 \pi i}{n}\right)$ and then the remaining segment to the origin z = 0. This closed path contains only the root $z_1 = \exp\left(\frac{\pi i}{n}\right)$, hence, it is necessary to calculate only one residue.

$\oint_{C_R} \frac{\text dz}{1 + z^n} = \int_0^R \frac{\text dx}{1 + x^n} + \int_{|z| = R} \frac{\text dz}{1 + z^n} + \int_{R \exp \left(\frac{2 \pi i}{n}\right)}^0 \frac{\text dz}{1 + z^n} =$

(substitute $z = x \exp \left(\frac{2 \pi i}{n}\right)$ into the last integral and use $\exp(2 \pi i) = 1$)

$= \int_0^R \frac{\text dx}{1 + x^n} + \int_{|z| = R} \frac{\text dz}{1 + z^n} - \exp \left(\frac{2 \pi i}{n}\right) \int_0^R \frac{\text dx}{1 + x^n} =$

$= 2 \pi i\ \text{Res} f(z_1) = \frac{2 \pi i}{g'(z_1)} = \frac{2 \pi i}{n \exp\left(\frac{(n - 1)\pi i}{n}\right)} = -\frac{2 \pi i}{n} \exp\left(\frac{\pi i}{n}\right)$

In the limit transition $R \longrightarrow \infty$, the integral over the circular arc vanishes and we obtain an equation for the given integral:

$\left[1 - \exp \left(\frac{2 \pi i}{n}\right) \right] \int_0^{\infty} \frac{\text dx}{1 + x^n} = -\frac{2 \pi i}{n} \exp\left(\frac{\pi i}{n}\right)$

$\int_0^{\infty} \frac{\text dx}{1 + x^n} = -\frac{2 \pi i}{n} \cdot \frac{\exp\left(\frac{\pi i}{n}\right)}{1 - \exp \left(\frac{2 \pi i}{n}\right) \right} = -\frac{\pi}{n} \cdot \frac{2i}{\exp \left(-\frac{\pi i}{n}\right) - \exp \left(+\frac{\pi i}{n}\right)} = \frac{\frac{\pi}{n}}{\sin\left(\frac{\pi}{n}\right)}$

[quote="yetti"]$\ldots$
But in this transition, the integral over the semicircle vanishes and we get
$\ldots$
In the limit transition $R \longrightarrow \infty$, the integral over the circular arc vanishes and we obtain an equation for the given integral:
$\ldots$[/quote]

Why? Sorry if it's a dumb question.

The arc length is of order $R$. The function on the arc is of order $\frac{1}{R^n}$. Obviously the integral is divergent when $n=1$, and the integral over the arc goes to 0 when $n > 1$.

[quote="perfect_radio"]how do you solve the last problem?[/quote]
see http://www.mathlinks.ro/Forum/viewtopic.php?p=465691#p465691


function
parameterization
integration
algebra
partial fractions
probability and stats

first of all my professor uses $ \xi(t)\equal{}E[e^{tX}]$ seemingly interchangably with $ \phi(t)\equal{}E[e^{itX}]$ without the i, we get distinctly different functions..

but im wondering if (like $ \phi$) there is a correspondence between $ \xi(t)$ and $ X$ density function,  how i would go about getting the inverse..

specifically.. i am working on a problem where the simplified model has n exponential distributed variables of equal parameter s

$ E[e^{tX_1}]\equal{}\int_0^\infty e^{tr\minus{}sr}dr\equal{}\frac{s}{s\minus{}t}$ when $ t<s$

so $ E[e^{t(X_1\plus{}... \plus{}X_n)}]\equal{}\prod_{k\equal{}1}^n E[e^{tX_k}]\equal{}\left(\frac{s}{s\minus{}t}\right)^n$ means $ Y\equal{}\sum_{k\equal{}1}^n X_n$ has density $ \frac{s^n}{\Gamma(n)}x^{n\minus{}1}e^{\minus{}sx}$

but as soon as i release the restriction that they are all the same parameter, i get completely stuck

$ E[e^{t(X_1\plus{} ... X_n)}]\equal{}\prod_{k\equal{}1}^n E[e^{tX_k}]\equal{}\prod_{k\equal{}1}^n \frac{s_k}{s_k\minus{}t}$

so when i look for a function $ g(x)$ such that $ \int_0^\infty e^{tx}g(x)dx\equal{}\prod_{k\equal{}1}^n \frac{s_k}{s_k\minus{}t}$ i feel like its impossible...? but there SHOULD exist such a function.. shouldnt there? any help would be appreciated.

The texts I know use the name "moment generating function" for $ E[e^{tX}].$

The moment generating function is a Laplace transform; the characteristic function is a Fourier transform.

Here's an idea for your specific problem, and it's very much a Laplace transform idea: can't you use partial fractions to expand your product into a sum?


ABCD is convex quadrilateral. AC is bisector of an angle BAD. <BAD=64, <BCA=18 ,<ACD=40 then find <ADB=?

Take $ E$ on $ (AC)$ so that $ \angle ADE \equal{} \angle ABC \equal{} 18^\circ$; obviously, $ DE \perp CD$ and $ \triangle ABE \sim \triangle ACD$, hence $ \angle ABE \equal{} 40^\circ$ and $ BE\perp BC$, consequently $ BCDE$ is cyclic, $ \angle EBD \equal{} \angle ECD \equal{} 40^\circ$, so $ \angle ABD \equal{} 40^\circ$.

Best regards,
sunken rock

[geogebra]66975d881033c89ed8ee93a510eeeb8550cfbe3c[/geogebra]
I hope this may help.
MOVE THE TEXT IN CASE <ADB =36 degree IS NOT VISIBLE


function
limit
real analysis
real analysis unsolved

Let $ f_j: [a,b]\to \mathbb{R}$, $ j\equal{}1,2,3,\ldots$ be a sequence of functions defined on the interval
$ [a,b]$. Assume that for every $ \epsilon>0$ there exists $ N>0$ s.t. for all $ m>N,n>N$ we have
\[ |f_m(x)\minus{}f_n(x)|<\epsilon
\]
Show that the sequence $ f_j$ converges uniformly to a function $ f: [a,b]\to\mathbb{R}$

For each $ x$ fixed, $ \{ f_n(x)\}_n$ is a Cauchy sequence. This allows us to construct a function $ f$ such that $ f_n(x) \to f(x)$ for each $ x$ as $ n \to \infty$.

Now, uniformly convergence means that
\[ \mathop {\lim }\limits_{n \to  \plus{} \infty } \mathop {\sup }\limits_{a \leqslant x \leqslant b} \left| {f_n \left( x \right) \minus{} f\left( x \right)} \right| \equal{} 0.
\]
This can be done very easy.


superior algebra
superior algebra unsolved

If we have a finite group with an even number of elements, prove that there must exist some element in the group $ a$ not equal to the identity $ e$ such that $ a^2\equal{}e$.

Look at inverses. Every element has one. Note that $ a^2\equal{}e$ just means that $ a$ is its own inverse. Suppose that every element has an inverse different from itself and count whether there are an even or odd number of elements.


What is the largest even integer that cannot be written as the sum of two odd composite numbers?

[hide="Solution"] $6$
Because any even number greater than $6$ is either a multiple of $4$, in which case it can be represented as $\frac{m}{2}+\frac{m}{2}$ where, because $m=4k$ then each $\frac{m}{2}$ is even and thus composite. For those that are not a multiple of $4$, they can be represented as $\frac{m}{2}-1+\frac{m}{2}+1$ and becase $\frac{m}{2}$ is odd (because $m$ is not a multiple of $4$) $\frac{m}{2}+1$ and $\frac{m}{2}-1$ must be even, and thus composite. $6$ works because in the case that it is represented as $\frac{m}{2}-1+\frac{m}{2}+1 \Rightarrow 2+4$ the first even number is $2$, the only even prime.[/hide]

The original problem was actually [b]odd[/b] composite integers.

Well that adds a new level of difficulty now doesn't it?
I'll try to come up with a non-even approach.

[hide]If $2n$ is a positive even integer that can be expressed as the sum of two composite odd integers, then there exist positive integers $a,\ b,\ c,\ d$ such that $(2a+1)(2b+1)+(2c+1)(2d+1)=2n$.

Expanding and simplifying, $n=2(ab+cd)+(a+b+c+d)+1$. Then $2n$ can be the sum of two composite odd integers if:

i. $n=3a+6$ ($b=c=d=1$)
ii. $n=3a+14$ ($b=1,\ c=d=2$)
iii. $n=3a+19$ ($b=1,\ c=2,\ d=3$)

6, 14, and 19 are all different mod 3, so any integer $n>19$ can be expressed in one of the forms above. Then the largest possible $n$ such that $2n$ might not be the sum of two composite odd integers is 19, which gives $2n=38$.

After some trial and error, it turns out that indeed 38 is [b]not[/b] a sum of two composite odd integers $\Rightarrow \boxed{38}$.[/hide]

[hide="Another proof"]
For any sufficiently large even integer $2k$, one of the three numbers $2k - 15, 2k - 25, 2k - 35$ is divisible by three and not equal to three, so $2k = 3a + 5(3, 5, 7)$ for some odd integer $a$.  We find that $a = 1$ for $2k = 38$, so for any $2k > 38$, it is clear that $a > 1$ and $2k$ can be written as the sum of two composite odd integers.  $\boxed{38}$ itself cannot.  Q.E.D.
[/hide]

Can someone please explain this with a little more detail? Forgive me for my stupidity, but I am having some difficulty understanding the solutions. Thank you

Assuming that this number would be less then 100
(I would like to see proof for this if its possible)
By Goldbach's conjecture we can get odd number by sum of 2 primes (not actually proven, but been run through to really big numbers, so for numbers this small we can use that)
So this numbers would have to be sum of 2 primes $ p_{1}$ and $ p_{2}$ (not necessarily distinct).
Therefore $ p\pm 2k$ has to be prime (at least one). Otherwise it could be written as 2 composite numbers.

By process of elimination (for primes to 100) (Notice this is kinda big hole in my solution. Hopefully someone can fill it.) we get that only 3,5,7,11,13, and 19 satisfy this.
So biggest number we could get is $ 19\plus{}19\equal{}38$

[quote="scorpius119"][hide]If $ 2n$ is a positive even integer that can be expressed as the sum of two composite odd integers, then there exist positive integers $ a,\ b,\ c,\ d$ such that $ (2a \plus{} 1)(2b \plus{} 1) \plus{} (2c \plus{} 1)(2d \plus{} 1) \equal{} 2n$.

Expanding and simplifying, $ n \equal{} 2(ab \plus{} cd) \plus{} (a \plus{} b \plus{} c \plus{} d) \plus{} 1$. Then $ 2n$ can be the sum of two composite odd integers if:

i. $ n \equal{} 3a \plus{} 6$ ($ b \equal{} c \equal{} d \equal{} 1$)
ii. $ n \equal{} 3a \plus{} 14$ ($ b \equal{} 1,\ c \equal{} d \equal{} 2$)
iii. $ n \equal{} 3a \plus{} 19$ ($ b \equal{} 1,\ c \equal{} 2,\ d \equal{} 3$)

6, 14, and 19 are all different mod 3, so any integer $ n > 19$ can be expressed in one of the forms above. Then the largest possible $ n$ such that $ 2n$ might not be the sum of two composite odd integers is 19, which gives $ 2n \equal{} 38$.

After some trial and error, it turns out that indeed 38 is [b]not[/b] a sum of two composite odd integers $ \Rightarrow \boxed{38}$.[/hide][/quote]

Based on what did you list all those possibilities?

[quote="t0rajir0u"]Those are the three smallest ones.[/quote]

But how do you know only to try these three smallest ones, not 4 or 5?

[quote="matt276eagles"]Not as elegant as the other solutions, but hopefully more understandable.  I tried to explain this how I would solve it on the test.  :)

[hide="Solution"]Start by listing some odd composite integers.  First we have all numbers of the form $ 3a$ for some odd $ a\ge 3$.  This gives us $ 9, 15, 21, \ldots$, all integers that are $ 3\pmod{6}$ and greater than or equal to $ 9$.  Clearly this doesn't help us much on its own since adding two of these numbers always results in an integer that is $ 0\pmod{6}$, so we can't get anything that is $ 2\pmod{6}$ or $ 4\pmod{6}$.

We need some more odd composite integers, so consider the numbers of the form $ 5a$ for some odd $ a\ge 5$.  This gives us $ 25, 35, 45, \ldots$.  Since $ 25\equiv 1\pmod{6}$, we can add one of our numbers of the form $ 3a$ to get any integer that is $ 4\pmod{6}$ and greater than or equal to $ 34$.  Similarly, since $ 35\equiv 5\pmod{6}$, we can also get any integer that is $ 2\pmod{6}$ and greater than or equal to $ 44$.  We can already get all integers that are $ 0\pmod{6}$ and greater than or equal to $ 18$ by using only numbers of the form $ 3a$.

The greatest even number remaining is $ 38$, and we can verify that $ 38$ can't be written as the sum of two odd composite numbers.[/hide][/quote]

how did you prove the numbers that are inclusive if they are 1 or 5(mod6)?  Since 25 and 35 are only two starting case, we can't prove later 31 or 41 are also 1,5(mod6)

I'm not sure I understand your objection.  What part of the solution do you think is incomplete?

The main tactic being used here is not a complete consideration of all cases:  one only needs to eliminate enough cases to quickly arrive at the number $ 38$ and then show that it works.  Because this is a weaker condition than actually eliminating every case, there are a variety of solutions.

[quote="t0rajir0u"]Three of them gets you an answer.  Any larger possibilities get you a larger answer, which can't be the minimum.[/quote]

But why don't you try b=1,c=1,d=2 or b=1,c=1,d=3 or b=3,c=2,d=1?

You could, but that's not the point. To prove the answer, it suffices to show that we have a bound, and then show that our best case works; that was already done successfully. We could try to 'tighten' our result using the cases you mentioned, but they're superfluous.

Note that for many of the solutions posted, they might not be the way that one would try during an actual competition. However, the solutions do contain some of the necessary intuition to build a path to a solution.

[hide="low tech solution"]
$ 2n$ is the number we seek:
\[ 2n\equal{}(2a\plus{}1)(2b\plus{}1) \plus{} (2c\plus{}1)(2d\plus{}1) \\ 2n \equal{} 4ab \plus{} 2a \plus{} 2b \plus{} 1 \plus{} 4cd \plus{} 2c \plus{} 2d \plus{}  1 \\ n \equal{} 2ab \plus{} a \plus{} b \plus{} c \plus{} d \plus{} 2cd \plus{} 1\] Attempting to get the terms to be linear: we plug in $ a \equal{} c \equal{} 1:$
\[ n \equal{} 2b \plus{} 1 \plus{} b \plus{} 1 \plus{} d \plus{} 2d \plus{} 1 \equal{} 3b \plus{} 3d \plus{} 3\]
This is not very helpful, since it gives only $ n$ that are multiples of 3, so now we try $ a \equal{} 1, c \equal{} 2$:
\[ n \equal{} 2b \plus{} 1 \plus{} b \plus{} 2 \plus{} d \plus{} 4d \plus{} 1 \equal{} 3b \plus{} 5d \plus{} 4.\] However, we want the terms $ 2a\plus{}1, 2b\plus{}1$ etc to be composite, and so we want $ a,b,c,d$ to be positive, so we substitute $ f \equal{} b \plus{} 1, g \equal{} d \plus{} 1$ to get
\[ n \equal{} 3f \plus{} 5g \plus{} 12 \Longleftrightarrow n \minus{} 12 \equal{} 3f \plus{} 5g\]
Now the right side is linear, and chicken mcnugget tells us that $ n \minus{} 12$ is minimized at $ n \minus{} 12 \equal{} 3 \times 5 \minus{} 3 \minus{} 5 \equal{} 7$. so $ n \equal{} 19$ and the answer is $ 2n \equal{} 38$[/hide]

[hide=@#19][url=aops.com/community/user/69906][b]Eternal[/b][/url] · Mar 15, 2017, 8:57 PM [url=aops.com/community/p7865402](view)[/url][color=transparent]helo[/color]
Here's arguably the easiest solution.

[hide=Solution]
Note $24 = 3\cdot 3 + 3\cdot 5$, and so if $2k$ has a representation as the sum of even multiples of 3 and 5, say $2k = e_3\cdot 3 + e_5\cdot 5$, we get a representation of $2k+24$ as a sum of odd composites via $2k+24 = (3+e_3)\cdot 3 + (5+e_5)\cdot 5$. But by the Frobenius coin problem every number $k > 3\cdot 5 -3-5 = 7$ has such a representation, hence every number $2k > 14$ has a representation as the sum of even multiples of 3 and 5. Thus every number $n > 24+14=38$ has a representation as the sum of odd composites. Checking, we see that $\boxed{38}$ has no representation as a sum of odd composites.
[/hide]

-----------
[color=#5b7083][aops]x[/aops] 19[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]
It comes from a wanted "consistency" in the expression. Honestly, when I saw the "largest" and "cannot be expressed" in one problem, my brain flicks to Chicken McNugget. After that, if you manipulate through the consistency, you're inclined to think that the two smallest distinct non-1 numbers (3, 5) would be sufficient to express the numbers.

Indeed, the bound is optimal so our motivated assumption works out.

My solution for this wasn't that nice. It is likely that this number is less than $50$, testing we get that the answer is $\boxed{38}.$ Don't do this solution, it took me 20 minutes to do this.


integration
real analysis
real analysis theorems

.... one recognizes that

$ \int^1_0\int^1_0 \frac {(x^r)(y^s)}{1 \minus{} xy} dxdy \equal{} \frac1{r \minus{} s}(\frac1{s \plus{} 1} \plus{} ... \plus{} \frac1r), r > s,$ represents a rational number, the denominator of which is certainly contained in the square V(r)^2 of the lowest common multiple V(r) of all numbers 1,..,r. Note: r and s are non-negative integers

What is V(r)?  And isn't it obvious by RHS of the equation $ \frac1{r \minus{} s}(\frac1{s \plus{} 1} \plus{} ... \plus{} \frac1r)$ must be rational.  [/code]

In the above text,
[quote]the lowest common multiple V(r) of all numbers 1,..,r[/quote]

but then why is that statement related to the denominator and V(r) true.  I don't see what lcm(1,..., r) has to do with the denominator of the rational number.  I don't see der anschluß!

oh ok, never mind, i think i figured it out


Euler
limit
logarithms
integration
calculus
calculus computations

i searched google for eulers constant and got approximately
0.577215665
But i always thought that euler's constant was
[i]e[/i]? (lim x->infinity (1+1/x)^x)

can someone explain what euler's constant is?

There's two of them:

[url=http://mathworld.wolfram.com/Euler-MascheroniConstant.html]The Euler-Mascheroni Constant :approx: 0.5772...[/url]
e :approx: 2.718281828459045

Euler has the distinction of being (maybe?) the only person with two constants named after him.  The first, $e$ , is
\[e=\lim_{x\to\infty}\left(1+1/x\right)^x\approx 2.718281828\]

The second, also known as the 'Euler-Mascheroni Constant' and commonly symbolized by $\gamma$ is
\[\gamma=\lim_{x\to\infty}\left(\sum_{k=1}^x {\frac{1}k} - \ln x\right)\approx0.577215665 \]

In my experience, $e$ is generally referred to as the "euler number" rather than as his constant.

[quote="Magnara"]Mascheroni sounds more like a noodle than a constant.

And I've usually seen $e$ referred to as the exponential constant.  That's what the $e$ stands for, not "Euler."[/quote]

That's probably not true either.  I read that Euler used "e" because it was the next unused letter of the alphabet (I'm not sure what the previous letters were taken up for)  , and it stuck.

Actually, $e$ is more often known as the Napier's constant.

I hadn't heard $e$ referred to as Napier's constant before, but if Napier deserves his name on any constant, it would be $1/e$ rather than $e$ itself. (His logarithms were essentially in base $1/e$, except he used 10 million for $\infty$.)

[quote="Danbert"]Euler has the distinction of being (maybe?) the only person with two constants named after him.  The first, $e$ , is
\[e=\lim_{x\to\infty}\left(1+1/x\right)^{x}\approx 2.718281828 \]
The second, also known as the 'Euler-Mascheroni Constant' and commonly symbolized by $\gamma$ is
\[\gamma=\lim_{x\to\infty}\left(\sum_{k=1}^{x}{\frac{1}k}-\ln x\right)\approx0.577215665 \]
[/quote]

How would you prove that this sequence converges:

\[\gamma=\lim_{x\to\infty}\left(\sum_{k=1}^{x}{\frac{1}k}-\ln x\right)\approx0.577215665 \]

$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}-log(n)$ ?

To start, I know that

$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}$ diverges as n gets very large, and so does log(n)

But together, how how you prove that the difference of these two converges?

well...

$\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\log(n)$

is really the same as

$(\frac12-\log\frac21)+(\frac13-\log\frac32)+...+(\frac1n-\log\frac{n}{n-1})$

just speculating, the mathworld link probably has more in-depth and more correct information but i'm too lazy to decipher math language into english

$\sum_{k=1}^{x}\frac{1}{k}< 1+\int_{1}^{x}\frac{dt}{t}$ so we know the limit of the difference is less than $1$ (and also that it is positive by a similar argument).