AMC
AIME
AMC 12
logarithms
algebra
polynomial
Vieta

Well I barely made it to the AIME this year. :D I'm a HS junior and I took the AMC12A.

I haven't actually done any preparation for math contests and such ever, but I am quite good in all the math classes I've taken.  Since I have the chance to do the AIME this year, I want to do the best I can given my limited experience and prep.

So...how should I best prepare?  Is there any optimal starting point for me now?  The first half of the AMC12 seemed quite routine to me, but after that I hit a roadblock and only got one or two.  Should I try my best to tackle AMC12 problems now, or should I practice on some AIME problems?

Thanks very much ahead of time!

Well, to tell you the truth, I don't think there's terribly much you can do with 9 days before the contest. But with the limited time you have, if you don't have any math books like AoPS II on you, I'd advise looking at some old AIMEs at the "Contests" link you see above the forum. At the very least, you'll get acquainted with the sort of problems that show up.

What if you are taking the AIME II on April 2, and qualified just barely for the AIME through AMC 10?

is that you???i didnt make AIME (*sobbing*) even though i prepared alot for the AMC...but im still in Junior high!!!(darn 109.5 incase anyone was wondering)  er...i would suggest trying a few of the AIME tests...either from the contest box above or go to AOPS Wiki and type verbatum "AIME problems and solutions".

Hope this helps!

Well, thanks very much everyone!

I figured I was pretty much at the end of the line anyway, I'll check out the AIME problem sets and see how it goes.

Well this helped me a lot, since I didn't know logs much at all.

Prove that $ \log_{a}b \plus{} \log_{a}c \equal{} \log_{a}bc$
Prove that $ \log_{a}b^c \equal{} c\log_{a}b$
Prove that $ \log_{a}b \minus{} \log_{a}{c} \equal{} \log_{a}\frac {b}{c}$
Prove that $ (\log_{a}b)(\log_{c}d) \equal{} (\log_{a}d)(\log_{c}b)$
Prove that $ \frac {\log_{a}b}{\log_{a}c} \equal{} \log_{c}b$
Prove that $ \log_{a^n}b^n \equal{} \log_{a}b$

Past problems can be found here: http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45

These are probably your best way to practice. A couple years ago I was pretty noob at AIME (although I did decently on the AMC 10/12 with scores of 150/124). I did a bunch of old tests over two weeks and managed to pull the floor.

As for the standards to look for: if you can get more than two or three right, you're doing very well.
You have a lot more time for AIME problems; there are 12 minutes per question, and effectively more if you give up on some of them quickly. It's probably a good idea to go through at least one old test real time, so you can get used to spending three hours at once on the contest.

2 problems or so are always bruteforce able I believe. My plan is to do around 6-7 and brute force the 1-3 problems that are able. Then guess for the rest

I spned like 5 minutes checking bubbles incase I dont have a bubbling error or I dont skip a question (also so you wont be stressed out later worrying about it)

I also do that for the AMC and I havent messed up a bubble or skipped a question yet

Plus AIME is 3 hours, do it when you get bored

I have noticed that olympiad practice doesn't help my AIMEs very much... due to my lack of speed :wink:

So... know:

algebra - basic log/exponent rules; AM-GM; factoring; polynomials (roots, Vieta, blah); very basic functional equation stuff, sometimes

geometry - Pythagorean, areas, Power of a point, SIMILAR TRIANGLES, circle stuff, trig, cyclic quads; perhaps coordinates

combo - hmm... add/product rule, bijections, geometric interpretation :P I am pretty bad at combo. But maybe should also know Fibonacci/Catalan? (more former than latter)

NT - modular arithmetic, Euler/Fermat's little theorem, divisibility stuff, factoring stuff (# of factors, etc.), maybe Euclidean algorithm, -maybe- Wilson's theorem

yeah and probably some other stuff that I can't think of

also, think! bad things happen when you don't

[quote="ZhangPeijin"]My prediction:
I do the first 6 problems, have no idea how to do 7-15, finish the test in 1hr, spend 2 hrs going bored out of my mind and doodling on the desk or something.[/quote]
You can generally get at least 2 problems by coordinate bashing or similar :)

or check your 6 problems, unlike my 8th grade year :wink:

hmm ive really havent seen that many NT theory problems (they are more modified to become more algebric) on the AIME, i really havent seen fermats or euclidean used on it...

well something like "find the remainder when you divide $ 2^1000$ by 11" :D

Of course, pattern finding would work too, but why do that when you have Fermat?

[quote="moogra"][quote="aspiringchesslete"]well something like "find the remainder when you divide $ 2^{1000}$ by 11" :D

Of course, pattern finding would work too, but why do that when you have Fermat?[/quote]

that would be extremely easy for aime (not to mention for guessing)[/quote]
It'd probably be more like

Blah blah blah blah blah blah blah equals $ x$.
Find the remainder when $ x$ is divided by $ 1000$.

Yeah, they usually do that, especially with combinatorial questions.

[quote="distracted523"][quote="moogra"][quote="aspiringchesslete"]well something like "find the remainder when you divide $ 2^{1000}$ by 11" :D

Of course, pattern finding would work too, but why do that when you have Fermat?[/quote]

that would be extremely easy for aime (not to mention for guessing)[/quote]

Yep, especially if you know mods. This is #1-5 AMC 10/12 material.

I think old tests help too, but it's frustrating when you don't know how to do something (and even more if you screw up in computation). I'm hoping to get 10 (I got two 9's when I did the 2005 I and 2005 II).

EDIT: Like I said, randomdragoon, if you know mods then that sort of problem isn't hard (unless maybe something like a brute-force kind such as #24 AMC 10A this year).

EDIT2: Unless you're talking about a general problem crunched into the 000-999 grid?[/quote]

Sorry, slightly off topic, but how would you solve that problem with mods? ><
Thanks guys =]

And on topic: AOPS vol. 1 / 2 and past tests are good.  :lol:

no... that method doesn't generalize, although it happens to work.

instead, use Fermat's little theorem

11 is prime, so $ 2^{10}\equiv 1 \pmod{11}$

so $ 2^{1000}\equal{}(2^{10})^{100}\equiv 1^{100} \equal{} 1 \pmod{11}$

I dont think Ive ever seen an AIME number theory problem that used fermat's little theorm (I think its pretty useless anyways since most of the important pairings of n^power divided by p most people memorize)

I dont think NT is that important on the AIME when compared to the other 3 subjects...

Most problems that you think NT would be important end up being counting problems (like find the number of numbers less than 2^32*3^45 but divide 2^64*3^90-something like that or find the number of divisors of 2004^2004 that have exactly 2004 factors) or algebric problems (like last year's number 7 on AIME II with those cube roots) or just by logic/algebra (like the one where its like find the sum of all number a/b where a and b are relatively prime and a and b are factors of 1000)

The only number theory technique you need to know is how to calculate the number of factors a number has, how to write even numbers, how to write odd numbers, how to change bases, 2 is the only even prime, what primes are, and that 2*5=10

[quote="ZhangPeijin"]My prediction:
I do the first 6 problems, have no idea how to do 7-15, finish the test in 1hr, spend 2 hrs going bored out of my mind and doodling on the desk or something.[/quote]
In 8th grade (2006 II) I just spent an hour and a half brute forcing #15. I got it right.

[quote="mcsquared"]Yeah so theres 8 days till the AIME.

I'm really bad at geometry as a whole because I don't know when/where to use geometric theorems. Most of my points come from algebra and combinatorics because they don't have as many theorems. I've done several old AIMEs now and I usually get about 6-7, but that's still kind of dangerous since I'm planning on making USAMO.

Any tips on how to improve?[/quote]

in eight days you are not going to improve drastically

just chill with friends and stuff cuz it is what i did last year

and it worked (qualified for USAMO), lol

Don't do any math 1 or 2 days before the contest. For AIME Geometry, I recommend Aops Intro to Geom. or Geometry Revisited.


Err...on the Course Introduction PDF it says there are 12 classes, but [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=11]here[/url] it says there are 18.  Which one is right?

[quote](Note: This course used to be called "Intermediate Trigonometry/Complex Numbers." This course contains seven weeks of material covering two- and three-dimensional vector spaces that the old course did not include.) [/quote]

I think they forgot to update. But I am not sure...

Will the Precalc book be available by then?

Yes, it will be 18 weeks.  I don't know if we'll have the book ready by the time the course starts, but if it isn't, it will be out shortly thereafter.


I sent in my round 2 solutions before the deadline, but all of the files were compressed in a .zip file. I didnt think it would be a problem before, but now looking at the usamts.org website it says "We will only accept the following file types: DVI, PDF, RTF, TEX, TXT, TIF, GIF, BMP, JPG, EPS".

So... am I in a predicament..?

I've sent my files in a zip in the past, and it's been fine.

.zip files are NOT acceptable.  We may decide to accept them anyway, but don't plan on it.  If your file type is not on the list cited above, then we cannot guarantee that we will accept them.

Ideally, all we want to receive is 1 PDF file, plus a .tex source file (if you used LaTeX) and any image files you used.


function
limit
logarithms
real analysis
real analysis unsolved

$F(n)=1$ if $n$ is prime, $0$ if not.
Find an equivalent of $f(x)=\sum_{\mathbb{N}} p(k)x^k$ when $x$ tends to $1^-$

Let me understand: you are looking for a function $g(x)$ such that
\[\lim_{x\to 1^-}g(x)\cdot(x^2+x^3+x^5+x^7+\ldots)=1.\]

i'm looking for a function $g$ such that:
$lim_{x\to 1^-} \frac{x^2+x^3+x^5+x^7+..}{g(x)} = 1$

I think that it is equivalent to find an equivalent of $\sum \frac{x^k}{ln(k)}$ .

What is your conjecture? Maybe
\[g(x)=-\frac{1}{(1-x)\log(1-x)}.\]

[quote="alekk"]how can one find an equivalent of $\sum \frac{x^n}{ln(n)}$ ?[/quote]
There are several ways based on different ideas. The simplest approach is to notice that $\ln n$ is a very slowly changing function that stays almost constant (about $\ln\frac 1{1-x}$) when $\frac 1{(1-x)^{a}}=A<n<B=\frac 1{(1-x)^{b}}$ where $a<1<b$ are two real numbers very close to $1$. Now split the sum into 3 parts: $\sum_{n\leq A}$, $\sum_{A<n<B}$ and $\sum_{n\geq B}$. The first sum does not exceed $A=O((1-x)^{-a})=o\left(\frac 1{1-x}\frac 1{|\ln(1-x)|}\right)$ (because $a<1$), the last sum does not exceed $\frac 1{1-x}\frac 1{|\ln(1-x)|}x^{(1-x)^{-b}}=o\left(\frac 1{1-x}\frac 1{|\ln(1-x)|}\right)$ because the last factor tends to $0$ as $x\to 1-$. It remains to note that the middle sum is almost the same as $\frac 1{|\ln(1-x)|}$ times $\sum_{A<n<B}x^n\approx \frac {1}{1-x}$ (because $x^A\to 1$ and $x^B\to 0$ as $x\to 1-$. So the answer is, indeed, $\frac 1{(1-x)|\ln(1-x)|}$. This is also the answer to the original question. BTW, I had to use the Prime Number Theorem to switch from the original sum to that with logatrithms. I still wonder whether it can be avoided.

Thanx a lot. Of course I also used the Prime number Theorem to switch to the other expression. It would be amazing if we could solve that problem without that theorem .


Given triangle $ ABC$.Bisectors of $ <A$ and $ <B$ intersect $ BC$ and $ CA$ in  in $ D$ and $ F$ respectively and  line parallel to $ AB$ passing thrugh $ C$ in $ E$ and $ G$ . Prove that if $ FG\equal{}DE$ then $ CA\equal{}CB$.

I can't solve it without trigonometry -_-..

$ CF\frac{sinA}{sin\frac{B}{2}}\equal{}GF\equal{}DE\equal{}CD\frac{sinB}{sin\frac{A}{2}}$
$ \rightarrow sinAsinB(sin\frac{A}{2}\minus{}sin\frac{B}{2})\plus{}sinC(sinAsin\frac{A}{2}\minus{}sinBsin\frac{B}{2})\equal{}0$
Suppose $ \angle A \prec \angle B$ or $ \angle A \succ \angle B$
Then, Left side of equality is bigger than 0 or smaller than 0. so contradiction.
Therefore $ \angle A\equal{}\angle B$ and it means $ CA\equal{}CB$


I don't understand what you asked. Do you mean: What's the purpose of this problem? If [i]yes[/i], you should ask those who created, else i'm sorry ;)

[quote="Slizzel"]Let $x,y\in\mathbb{R}$ such that $x^2-x+1=\frac{3}{y^2-2y+5}$. Find $\lambda=x+y$.[/quote]

Considering the given equation as a quadratic in x,
we need to have its discriminant non negative as x is given real.
we should have,

$1-4(1-\frac{3}{y^2-2y+5})>=0$
simplifying we get
$-(y-1)^2>=0$
which can happen only when y=1.

substituting $y=1$ in the given equation, we get $x=1/2$
hence $\lambda=3/2$

[quote="socrates"]We have $x^2-x+1\geq\frac{3}{4}$ for all $x\in R$ and the equality exists only for $x=\frac{1}{2}$

Similarly we get $y^2+2y+5\geq4$ for all $x\in R$ and the equality exists only for
$y=-1$

So we obtain $(x^2-x+1)(y^2+2y+5)\geq\frac{3}{4}\cdot 4=3$ and the equality exists only for $y=-1$ and $x=\frac{1}{2}$.

However $\displaystyle x^2-x+1=\frac{3}{y^2-2y+5}$ so it is obvious that $y=-1$ and $x=\frac{1}{2}$ and $x+y=-\frac{1}{2}$[/quote]

Theres a typo in "y^2 + 2y + 5"   ;)  ;)

vidyamanohar's solution looks ok and it's a simple ideea (anyway I didn't found it :blush:). As for the other solution, is same as in my book but is a little complicated ... Thanks a lot :)


Anyone watch it or listen to it? Very surprising race. A 50-1 odds horse finishing first! And a 70-1 horse finishing second! Huge surprises. I was rooting for Bandini, but he totally sucked out there. Good thing I'm too young to have a bookie.  :D

For those who didn't watch, here's the top 3:

Win- Giacomo (50-1)
Place- Closing Argument (70-1)
Show- Afleet Alex (4-1, I think)

i totally would have bet on that horse. or tried, anyways. i have a history of being talked out of winning gambles.

ha, steinbrenner loses AGAIN!!! his horse was the favorite to win :D

What a week for George. His $200 million team loses 4 straight games, three games to Tampa Bay, falls to last place....And now his favored horse loses the Kentucky Derby. He's going to be furious. :P


According to [url=http://www.research.att.com/~njas/sequences/A002467]this page[/url], the exact value is $ \left\lfloor\frac{2009!(e\minus{}1)}e\plus{}\frac12\right\rfloor$.

[quote="Johan Gunardi"]According to [url=http://www.research.att.com/~njas/sequences/A002467]this page[/url], the exact value is $ \left\lfloor\frac {2009!(e \minus{} 1)}e \plus{} \frac12\right\rfloor$.[/quote]

Very nice, indeed !  :)

Thanks for this remark.


$n$ is positive integer
prove that if $3n+1$ is a perfect square then $n+1$ is the sum of three perfect square


function
trigonometry
vector
limit
calculus
calculus computations

For $ x > 0$, let the function $ f$ be defined by:

$ f(x, y) \equal{} (x^{y}, 3x^{2}y)$.

What value could be assigned to $ f(0, 0)$ such that $ f$ is continuous for all $ y$ and $ x\geq 0$?

I was thinking of this example when I remembered that $ g(x) \equal{}\frac{\sin{x}}{x}$ is continuous if $ g(0) \equal{} 1$.

What if we tried something similar with vector functions?

[quote="jmerry"]You need the existence of a limit at the origin, and $ x^{y}$ doesn't have a limit.[/quote]

Does $ 0^{0}\equal{} 1$ hold?  :oops:

Formally, a function of two variables $ x, y$ only has a limit defined at a point if that limit is the same regardless of the path taken to get to that point.

So along the path $ x \equal{} 0, y\to 0$ we have $ \lim_{(x, y)\to (0, 0)}x^{y}\equal{} 0$, yet along the path $ x\to 0, y \equal{} 0$ we have $ \lim_{(x, y)\to (0, 0)}x^{y}\equal{} 1$; hence the limit does not exist.


algebra
polynomial
inequalities unsolved
inequalities

[quote="caubetoanhoc94"][quote="hedeng123"][quote="caubetoanhoc94"]Let:$ x;y;z > 0$ such that:$ x \plus{} y \plus{} z \le 18;y \plus{} z \le 9;z \le 1$.Find maximum:
$ A \equal{} \sqrt {x} \plus{} \sqrt [3]{y^2} \plus{} \sqrt [4]{z}$[/quote]

maximum  is a root of the following polynomial

$ \minus{} 21414132145800 \plus{} 15989048930124k \minus{} 9229391305254k^2 \plus{} 3541364067564k^3$

$ \minus{} 1061105161833k^4 \plus{} 244414734336k^5 \minus{} 42127589376k^6 \plus{} 5435817984k^7$

 $ \minus{} 452984832k^8 \plus{} 16777216k^9$

$ 8.0140856123620763141.....$[/quote]

Can you post full solution ?. I like a solution simple  :)[/quote]

sorry  I got the answer with the computer.


geometry
geometric transformation
reflection
circumcircle
Euler
homothety
geometry proposed

Let an acute triangle $ ABC$ with curcumcircle $ (O)$. Call $ A_1,B_1,C_1$ are foots of perpendicular line from $ A,B,C$ to opposite side. $ A_2,B_2,C_2$ are reflect points of $ A_1,B_1,C_1$ over midpoints of $ BC,CA,AB$ respectively. Circle $ (AB_2C_2),(BC_2A_2),(CA_2B_2)$ cut $ (O)$ at $ A_3,B_3,C_3$ respectively.
Prove that: $ A_1A_3,B_1B_3,C_1C_3$ are concurent.

In fact, the concurent point is centroid $ G$ of triangle $ ABC$.

Yes, of course it is :lol:. By the way, are you from VN??

[quote="mathVNpro"]This problem is really nice. Here is my solution to this problem:
First I want to introduce you 2 lemmas that I use:

[b]Lemma 1[/b]: Denote $ O_1$, $ O_2$, $ O_3$ the center of circumcircle of triangle $ AB_2C_2$, $ BC_2A _2$, $ CA_2B_2$. Then $ O_1$, $ O_2$, $ O_3$, respectively, are the reflections of the midpoints of $ BC$, $ CA$, $ AB$ through the center $ O$ of circumcircle of triangle $ ABC$.

[i]Proof[/i]: It is very easy, I will leave it for the readers :lol: .

[b]Lemma 2[/b]: Let $ ABC$ be triangle inscribed in circumcircle $ (O)$. The line through $ A$, $ B$, $ C$ intersect each other by the points $ A_4$, $ B_4$, $ C_4$ ($ A_4$ is opposite to $ A$,...). Then A, B, C will be midpoints of $ B_4C_4$, $ C_4A_4$, $ A_4B_4$, respectively.Further, (O) will be the Euler circle of triangle $ A_4B_4C_4$.

[i]Proof[/i]: This is also very easy to prove, a junior student in my country can "kill" it easily. :D.

Back to our problem.
According to Lemma 1, we can get this result: $ AA_3$, $ BB_3$, $ CC_3$, respectively, parallel to BC, CA, AB. Let $ A_4$, $ B_4$, $ C_4$ be the intersections of $ AA_3$, $ BB_3$, $ CC_3$, pairwise ($ A_4$ is opposite to $ A$,...). Then according to the lemma 2, we have triangle $ A_3B_3C_3$ is the orthic triangle wrt triangle $ A_4B_4C_4$. Hence there exist a homothety that maps triangle $ A_3B_3C_3$ into triangle $ A_1B_1C_1$. Therefore $ A_1A_3$, $ B_1B_3$, $ C_1C_3$ are concurent.
Our proof is completed :blush:[/quote]
I don't need so many lemmas :D

[hide="Solution"]Let I be the circumcenter of triangle $ AB_2C_2$
We have $ AC_1.AB\equal{}AB_1.AC$ or $ BC_2.BA\equal{}CB_2.CA$
Then $ BI\equal{}CI$
So $ IO\perp BC$. Let M be the midpoint of $ BC, AM\cap A_1A_3\equal{}\{G\}$
Since $ IO\perp AA_3$, we get $ AA_3\equal{}2A_1M$
Using Thales's theorem, we have $ \frac{AG}{GM}\equal{}\frac{AA_3}{A_1M}\equal{}2$
Thus G is the centroid of triangle ABC
Similarly, $ B_1B_3, C_1C_3$ pass through G -> q.e.d[/hide]

Oh, very nice one, thank you so much, I always think complicatedly, bad me, bad me :D  :blush:
And by the way, there is no need to write out the lemmas like me in my solution.
Actually, these lemmas are the steps that I have followed while I was saving this problem,
Write out the lemma so that the solution will be clear and understandable, that's all. :blush:


Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:

$ \frac{a^{2}\plus{}b^{2}\minus{}c^{2}}{a\plus{}b}\plus{}\frac{a^{2}\plus{}c^{2}\minus{}b^{2}}{a\plus{}c}\plus{}\frac{b^{2}\plus{}c^{2}\minus{}a^{2}}{b\plus{}c}\leq\frac{a\plus{}b\plus{}c}{2}.$

I have spent a while on this problem...I have been trying various new techniques on it...I have come up with a pretty messy proof [I basically left all the computations off...].

Wlog $ x\ge y\ge z$. Let $ f(a,b,c)$ be the RHS minus the LHS. Note that:
\[ f(x,y,z)\ge f\left(\frac{x\plus{}y}{2},\frac{x\plus{}y}{2},z\right)\iff\]

\[ (x\minus{}y)^{2}(2(x^{3}\plus{}y^{3}\minus{}z^{3})\plus{}5(xy\minus{}z^{2})(x\plus{}y)\plus{}3(x^{2}\plus{}y^{2})(z)\plus{}4xyz)\ge 0\]
which is obviously true.

Applying that, $ f(a,b,c)\ge f\left(\frac{a\plus{}b}{2},\frac{a\plus{}b}{2},c\right) \equal{}\frac{\left(\frac{a\plus{}b}{2}\minus{}c\right)^{2}*c}{(a\plus{}b)\left(\frac{a\plus{}b}{2}\plus{}c\right)}\ge 0$, as desired.

Equality occurs when $ a \equal{} b$ and $ \frac{a\plus{}b}{2}\equal{} c\implies a \equal{} b \equal{} c$.

--

Sorry that the solution is relatively ugly. Perhaps there is a nicer solution? I always am reluctant to use weaker methods because I end up with many false inequalities. Perhaps there is a much easier way to attack this problem? Do you have a proof arqady?

I didn't  search  the nice proof. Maybe it exists. My proof it's expanding:
$ \sum_{cyc}\frac{a^{2}\plus{}b^{2}\minus{}c^{2}}{a\plus{}b}\leq\frac{a\plus{}b\plus{}c}{2}\Leftrightarrow\sum_{cyc}(2a^{4}\minus{}a^{3}b\minus{}a^{3}c\minus{}2a^{2}b^{2}\plus{}2a^{2}bc)\geq0\Leftrightarrow$
$ \Leftrightarrow2\sum_{cyc}(a^{4}\minus{}a^{3}b\minus{}a^{3}c\plus{}a^{2}bc)\plus{}\sum_{cyc}ab(a\minus{}b)^{2}\geq0,$ which true by Schur.

[quote="arqady"][color=darkred]Let $ a$, $ b$, $ c$ are positive numbers. Prove that $ \frac{a^{2}\plus{}b^{2}\minus{}c^{2}}{a\plus{}b}\plus{}\frac{a^{2}\plus{}c^{2}\minus{}b^{2}}{a\plus{}c}\plus{}\frac{b^{2}\plus{}c^{2}\minus{}a^{2}}{b\plus{}c}\leq\frac{a\plus{}b\plus{}c}{2}.$[/color][/quote]
[color=darkblue][b]Proof.[/b]I"ll use only the identity $ \sum a(b^{3}\plus{}c^{3}) \equal{}\sum a^{3}(b\plus{}c)$ and the inequality $ \frac{1}{b\plus{}c}\le\frac{b\plus{}c}{4bc}$. Therefore,

$ \sum\frac{b^{2}\plus{}c^{2}\minus{}a^{2}}{b\plus{}c}\le$ $ \sum\frac{(b\plus{}c)(b^{2}\plus{}c^{2}\minus{}a^{2})}{4bc}\equal{}$ $ \sum\frac{bc(b\plus{}c)\plus{}(b^{3}\plus{}c^{3})\minus{}a^{2}(b\plus{}c)}{4bc}\equal{}$ $ \frac{1}{4}\sum (b\plus{}c)\plus{}\sum\frac{a(b^{3}\plus{}c^{3})\minus{}a^{3}(b\plus{}c)}{4abc}\equal{}$

$ \frac{a\plus{}b\plus{}c}{2}\plus{}\frac{1}{4abc}\cdot\left[\sum a(b^{3}\plus{}c^{3})\minus{}\sum a^{3}(b\plus{}c)\right] \equal{}\frac{a\plus{}b\plus{}c}{2}$ $ \implies$ $ \frac{a^{2}\plus{}b^{2}\minus{}c^{2}}{a\plus{}b}\plus{}\frac{a^{2}\plus{}c^{2}\minus{}b^{2}}{a\plus{}c}\plus{}\frac{b^{2}\plus{}c^{2}\minus{}a^{2}}{b\plus{}c}\leq\frac{a\plus{}b\plus{}c}{2}$.

[b]Remark.[/b] You can use directly the well-known identity $ \boxed{\ \sum a(b\plus{}c)(b^{2}\plus{}c^{2}\minus{}a^{2}) \equal{} 2abc(a\plus{}b\plus{}c)\ }$. Thus,

$ \sum\frac{b^{2}\plus{}c^{2}\minus{}a^{2}}{b\plus{}c}\le$ $ \sum\frac{(b\plus{}c)(b^{2}\plus{}c^{2}\minus{}a^{2})}{4bc}\equal{}$ $ \frac{1}{4abc}\cdot\sum a(b\plus{}c)(b^{2}\plus{}c^{2}\minus{}a^{2}) \equal{}\frac{2abc(a\plus{}b\plus{}c)}{4abc}\equal{}\frac{a\plus{}b\plus{}c}{2}$.[/color]

[quote="Virgil Nicula"]Therefore,

$ \sum\frac{b^{2}\plus{}c^{2}\minus{}a^{2}}{b\plus{}c}\le$ $ \sum\frac{(b\plus{}c)(b^{2}\plus{}c^{2}\minus{}a^{2})}{4bc}$[/quote]
You assumed that $ b^{2}\plus{}c^{2}\geq a^{2}$ etc. That can fail and one must bring another argument to handle that case.
[I don't have the patience (or am I willing) to check the other case right now, so I don't know if it's trivial.]

Yes, you are right, Perfect_radio ! I didn't carefully. Thank you. I corrected above. Must think more over it !

[hide="Hope I didn't make any mistakes"]
$ \sum_{cyc}\frac{a^{2}\plus{}b^{2}\minus{}c^{2}}{a\plus{}b}\leq\frac{a\plus{}b\plus{}c}{2}$
$ \Longleftrightarrow\sum_{cyc}\frac{a^{2}\plus{}b^{2}}{a\plus{}b}\minus{}(a\plus{}b\plus{}c)\leq\sum_{cyc}\frac{c^{2}}{a\plus{}b}\minus{}\frac{a\plus{}b\plus{}c}{2}$
$ \Longleftrightarrow\sum_{cyc}\frac{(a\minus{}b)^{2}}{2(a\plus{}b)}\leq\sum_{cyc}(a\minus{}b)(\frac{2a\plus{}b\plus{}c}{4(b\plus{}c)}\minus{}\frac{2b\plus{}c\plus{}a}{4(a\plus{}c)})$
$ \Longleftrightarrow\sum_{cyc}\frac{(a\minus{}b)^{2}}{2(a\plus{}b)}\leq\sum_{cyc}(a\minus{}b)^{2}\frac{a\plus{}b\plus{}c}{2(b\plus{}c)(c\plus{}a)}$
$ \Longleftrightarrow\sum_{cyc}\frac{a^{2}\plus{}b^{2}\plus{}ab\minus{}c^{2}}{2(a\plus{}b)(b\plus{}c)(c\plus{}a)}(a\minus{}b)^{2}\geq 0$
Now let $ S_{a}\equal{}\frac{b^{2}\plus{}c^{2}\plus{}bc\minus{}a^{2}}{2(a\plus{}b)(b\plus{}c)(c\plus{}a)}$
and $ S_{b}\equal{}\frac{a^{2}\plus{}c^{2}\plus{}ac\minus{}b^{2}}{2(a\plus{}b)(b\plus{}c)(c\plus{}a)}$
and $ S_{c}\equal{}\frac{a^{2}\plus{}b^{2}\plus{}ab\minus{}c^{2}}{2(a\plus{}b)(b\plus{}c)(c\plus{}a)}$
WLOG we assume that $ a\geq b\geq c>0$
That gives $ S_{b}\geq 0$
Moreover, we can easily have $ S_{b}\plus{}S_{a}\geq 0$ and $ S_{b}\plus{}S_{c}\geq 0$
So, the inequality is proved. [/hide]


given AB = 4, CD = 10   and AB//CD

find the radius of the circle

 :(  :(
I cannot solve this problem.........
anybody who knows how to solve this problem please help me.....
or give me any advice
thankyou very much

chaktp@yahoo.com

[quote="jackoff"]given AB = 4, CD = 10   and AB//CD

find the radius of the circle[/quote]

The lengths of AB and CD are not enough information to find the radius of the circle. You need another data, for instance the distance between the parallel lines AB and CD.

  Darij

thankyou for your help....

I try to solve this problem for sometime.......

after that i realise that the information is missing .....
thankyou very much again for your help.....


for the first drawing, 2/3 chance, and for the second, a 1/2 chance, so a total of a (2/3)(1/2)=1/3 chance.

[quote="Iron Fist"]Albatross has three socks in his drawer - two red, one blue. What is the probability of him picking two blue socks, assuming he picks two socks at random?[/quote]

[hide] Umm...$0$? [/hide]

[quote="Arvind_sn"]Wait...since he only has one blue sock, wouldn't it be zero??? :huh:  :what?:[/quote]

I think thats a typo and its 2 blue 1 red  ;)
[hide="answer"]1/3[/hide]
if it wasn't a typo ur right

I think that daermon actually is right...thanks. But then there would still be two cases! You see, the problem does not specify how many socks Albatross picks at one time. You see, the problem [i]said[/i] that he picks two socks at random. This might be wrong, but in that case, the answer is [hide]$\frac{2}{3}$[/hide].

In the case where Albatross only picks one at a time, the answer is, of course, [hide]$\frac{1}{3}$[/hide].

[quote="Karth"]I think that daermon actually is right...thanks. But then there would still be two cases! You see, the problem does not specify how many socks Albatross picks at one time. You see, the problem [i]said[/i] that he picks two socks at random. This might be wrong, but in that case, the answer is [hide]$\frac{2}{3}$[/hide].[/quote]

[hide]are you sure?
there are 3 combinationsof choosing two socks out of three. and one of these is the two blue socks, so $\frac{1}{3}$[/hide]

[hide="if there is no typo"]0[/hide]

[hide="if there is a typo"]
1/3[/hide]

[quote="Ignite168"][hide="If there is a typo"]
1, he's guranteed to pick 2 blue socks out if Treething meant 6 blue socks.[/hide][/quote]

How did you just jump to that reasoning, there are still some red socks.

[hide]Um... if there isn't any typo, it's just 0 (because you can't pick 2 blue socks if there's only 1).[/hide]


[quote="MCrawford"]Of the 2400 jellybeans in a jar, $\frac{1}{6}$ are purple.  Lucia then removes 150 purple jellybeans from the jar.  What fraction of the jellybeans left are purple?  Express your answer as a common fraction.[/quote]

[hide="Solution"]
400/2400
250/2250
[b]1/9[/b]
[/hide]

[quote="MCrawford"]Of the 2400 jellybeans in a jar, $\frac{1}{6}$ are purple.  Lucia then removes 150 purple jellybeans from the jar.  What fraction of the jellybeans left are purple?  Express your answer as a common fraction.[/quote]

[hide]there are 2400*1\6=400 purple jellybeans.

so $\frac{400-150}{2400-150}=\frac 19$[/hide]

[quote="math92"][hide]
1/6 of 2400 are purple so 400 are purple
she took out 150 purples so there are 250 left
the answer is 250/2400.
Simplify that and you get $5/48$[/hide][/quote]

Remember, if 150 purples are taken out, then 150 beans are taken out of the whole thing too. So it wouldnt be 2400 left but....

oh i get it now... thanks GoBraves :lol:


floor function
real analysis
real analysis unsolved

let

$ f_1(x)\equal{}x$

$ f_{n\plus{}1}(x)\equal{}f_n(x)\plus{}n\lfloor x\rfloor \plus{}\frac{\pi^2}{6}\cdot\frac{x}{2^n}$ for $ x\in\mathbb{Q}$

$ f_{n\plus{}1}(x)\equal{}f_n(x)\plus{}\frac{x}{n^2}$ for $ x\notin\mathbb{Q}$

find the set on which $ (f_n)$ converges pointwise and the set on which $ (f_n)$ converges  uniformly.

Find also the limit function.

The second question is ill-posed. It's not like the sets on which it converges uniformly is totally ordered. You can always add finite sets of pointwise convergence, so I honestly don't tknow what it means.


Does anyone here know about nomography?

There isn't too much in the internet about it.

The only interesting links I found over internet were:

http://www.archive.org/details/firstcourseinnom00broduoft
http://www.projectrho.com/nomogram/intro.html
http://myreckonings.com/wordpress/2008/01/09/the-art-of-nomography-i-geometric-design/
http://mathforum.org/library/drmath/view/63338.html

Etc...

You have 3 collinear points, make 3 scales, join 2 points with a line, the intersection of the other scale with the line gives the third point.

All links I put the Math is very poor (the books are intended to engineers).

If someone have some lost papers about monography, one could share this will us.

I were more interested in the transformations of nomograms to make them more efficient and accurate.

Thanks.

From my limited understanding, the operation of a nomograph consists of connecting two points with a line, extending the connector, and finding the intersection of this line with the line containing neither original point. I'm not sure how it could be made more efficient to use, and the cost to create isn't very high either. Accuracy seems to me that it's going to be more accurate than Cartesian coordinates, what say you?

There is infinitely many ways to make a nomogram of a equation.
Some ways are better than others ways.

For example: There is a way to solve $ x^2 \plus{} a x \plus{} b \equal{} 0$ with 2 perpendicular scales and a circle, there is other way using a parabola, you can even make a elliptic curve to solve this if you want.

So there is a set of "solutions", differents monograms for the same thing. Don't you agree that there need to be a least one optimal solution, where you could make more efficient calculations with this? (efficient here means how many significant figures you have)

A precision of 30 digits is impossible, I know that alread. But I read in some site (don't remember where) that in some cases we can have a precision of 2 to 3 digits (in a A4 paper more digits would mean that the scale is shorter and a monogram to solve only for 1.00 to 1.01 is not good at all)

I could've do this by myself, but to re-invent something that Alread exist is just a loss of time.

Do you got the point? In math there is always some easy way to do things, but sometimes is more difficult to find this way.

And since this is just a hobbie, don't have any pratical purpose, is more loss of time, that's why I asked some material or others stuff about nomography.


[quote="Alisher"]If $x$, $y$, $z$ are positive reals, prove the inequality

$\frac{x}{\sqrt{yz+4x\left(y+z\right)}}+\frac{y}{\sqrt{zx+4y\left(z+x\right)}}+\frac{z}{\sqrt{xy+4z\left(x+y\right)}}\geq 1$.[/quote]

Let $x+y+z=1$.

Then using Generalised Jensen inequality for convex function $f(x)=\frac{1}{\sqrt{x}}$ we have:

$x\frac{1}{\sqrt{yz+4x(y+z)}}+y\frac{1}{\sqrt{zx+4y(z+x)}}+z\frac{1}{\sqrt{xy+4z(x+y)}}\geq \frac{1}{\sqrt{3xyz+4x^{2}(y+z)+4y^{2}(z+x)+4z^{2}(x+y)}}$

Now we have to prove:

$1\geq3xyz+4x^{2}(y+z)+4y^{2}(z+x)+4z^{2}(x+y)=\sum xyz+4\sum_{sym}x^{2}y$

$1=(x+y+z)^{3}=\sum x^{3}+2\sum xyz+3\sum_{sym}x^{2}y$

So our inequality becomes:

$\sum x^{3}+2\sum xyz+3\sum_{sym}x^{2}y\geq\sum xyz+4\sum_{sym}x^{2}y$

$\sum x^{3}-\sum_{sym}x^{2}y+\sum xyz\geq0$

$\sum x^{3}-x^{2}y-x^{2}z+xyz=\sum x(x-y)(x-z)\geq0$ which is true by Schur.


Find all functions $ f : \mathbb R^* \to \mathbb R^*$ such that \[ f(x^2 \plus{} y) \equal{} f^2(x) \plus{} \frac {f(xy)}{f(x)}\] holds for all $x, y \in R^*$ with $y \neq -x^2$.

[quote="hophinhan"]Let $ \mathbb{R^{*}}$ be the set of non-zero real numbers. Find all functions $ f : R* \minus{} > R*$ such that : $ f(x^2 \plus{} y) \equal{} f^2(x) \plus{} \frac {f(xy)}{f(x)} (1)$
for all $ x, y \in \mathbb{R^{*}}, y \neq \minus{} x^{2}$[/quote]

 [i]  I think i have a solution : :lol:

Change $ y \equal{} 1$ into $ (1)$ , we have : $ f(x^2 \plus{} 1 ) \equal{} (f(x))^2 \plus{} 1 (2)$

 Change $ x \equal{} 1$ into $ (1)$ : $ f(y \plus{} 1) \equal{} \frac {f(y)}{f(1)} \plus{} (f(1))^2 \equal{} \frac {f(y)}{a} \plus{} a^2 \ \forall \ y > 0 \ \ ( f(1) \equal{} a)$

$ \longrightarrow f(2) \equal{} a^2 \plus{} 1 ; f(3) \equal{} \frac {f(2)}{a} \plus{} a^2 \equal{} \frac {a^3 \plus{} a^2 \plus{} 1}{a}$

  Continue that work , we can calculate $ f(5)$ as a function of $ a$

   And , we can calculate $ f(5)$ in the other way by using  $ (2)$ ( $ f(5) \equal{} (f(2))^2 \plus{} 1$ )

  Thus , we will have a equation of $ a$ , this equation seems to be hard but actually very easy

 $ \iff (a \minus{} 1) \left( a^4 (a^2 \plus{} a \plus{} 1) \plus{} (a \plus{} 1)^2 (a^2 \minus{} a \plus{} 1) \plus{} 2a^2 \right) \equal{} 0$

The unique root of it is $ a \equal{} 1$

   So $ f(y \plus{} 1) \equal{} f(y) \plus{} 1 \ \forall \ y > 0$

  $ \longrightarrow f(y \plus{} n) \equal{} f(y) \plus{} n \forall \ n \in \ \mathbb{N} , \forall \ y > 0 (3)$

  Put $ y \equal{} \frac {1}{n} ; x \equal{} n \in \ \mathbb{N}$ into $ (1)$ , we have : $ f(1/n) \equal{} 1/n$

Put $ y \equal{} \frac {1}{n} ; x \equal{} m \in \ \mathbb{N}$ into $ (1) \longrightarrow f( m/n) \equal{} m/n$

    So $ f(x) \equal{} x \forall x \ \in \ \mathbb{Q^{ \plus{} }}$

   By $ (2) ; (3) : f(x^2) \equal{} (f(x))^2 \forall \ x > 0 \longrightarrow f(x) > 0 \forall \ x > 0$

   $ f(x^2 \plus{} y) > (f(x))^2 \forall \ x;y > 0$ , change $ x$ to $ \sqrt {x}$  we can prove $ f(x) < f(x \plus{} y) \forall \ x;y > 0$

$ \longrightarrow f$ is stricly increasing in $ (0; \plus{} \infty)$ ; $ f(x) \equal{} x \forall x \ \in \ \mathbb{Q^{ \plus{} }}$

For each  fixed positive real number $ x$ .Choose $ 2$ sequences of rational numbers $ (u_n) ; (v_n)$ such that :
               $ \lim u_n \equal{} \lim v_n \ \equal{} \ x ; u_n \ \leq x \ \leq \ v_n \forall \ n \in \ \mathbb{N}$

$ \longrightarrow f(u_n) \ \leq \ f(x) \ \leq f(v_n) \forall \ n \in \ \mathbb{N} \longrightarrow u_n \ \leq \ f(x) \ \leq v_n \forall \ n \in \ \mathbb{N}$

Setting $ n \to \infty$ , we can find : $ f(x) \equal{} x \ \forall \ x > 0$
         With $ y < 0$ , choose a positive real number $ x$ such that : $ x^2 \plus{} y > 0$
From $ (1)$ , we have $ f(xy) \equal{} xy$ . Note that , $ xy$ can take all value in $ ( \minus{} \infty ; 0)$
so   $ f(x) \equal{} x \ \forall \ \ x < 0$   And $ f \ \equiv \ x \ \forall \ x \ \in \ \mathbb{R^{*}}$ is the unique solution :lol:

     [/i]

[quote="hophinhan"]Let $ R*$ be the set of non-zero real numbers. Find all functions $ f : R* \minus{} > R*$ such that : $ f(x^2 \plus{} y) \equal{} f^2(x) \plus{} \frac {f(xy)}{f(x)}$
for all x, y ∈ R*, y ≠ −x^2[/quote]
Same demo as nguyenvuthanhha's one (congrats!  :) ), a litlle bit quicker in the beginning :

Let $ P(x,y)$ the assertion $ f(x^2\plus{}y)\equal{}f^2(x)\plus{}\frac{f(xy)}{f(x)}$
Let $ a\equal{}f(1)$

$ P(x,y)$ $ \implies$ $ f(x^2\plus{}1)\equal{}f^2(x)\plus{}1$ and so $ f(x)\equal{}f(\minus{}x)$. As a consequence, $ f(\minus{}1)\equal{}\epsilon a$ where $ \epsilon\in\{\minus{}1,\plus{}1\}$

Comparing then $ P(1,x)$ and $ P(\minus{}1,x)$, we get $ f(\minus{}x)\equal{}\epsilon f(x)$ $ \forall x\neq 0$
$ P(1,1)$ $ \implies$ $ f(2)\equal{}a^2\plus{}1$
$ P(1,\minus{}2)$ $ \implies$ $ f(\minus{}2)\equal{}\epsilon a^2\minus{}a^3$

And since $ f(\minus{}2)\equal{}\epsilon f(2)$, we get $ a\equal{}\minus{}\epsilon$ and so $ f(\minus{}1)\equal{}\minus{}1$

Then $ P(1,x^2)$ $ \implies$ $ f(x^2\plus{}1)\equal{}1\minus{}\epsilon f(x^2)$ and $ P(x,1)$ $ \implies$ $ f(x^2\plus{}1)\equal{}f^2(x)\plus{}1$ and so $ f(x^2)\equal{}\minus{}\epsilon f^2(x)$

$ P(1,1)$ $ \implies$ $ f(2)\equal{}2$
$ P(1,2)$ $ \implies$ $ f(3)\equal{}1\minus{}2\epsilon$
$ P(1,3)$ $ \implies$ $ f(4)\equal{}3\minus{}\epsilon$ But $ f(4)\equal{}f(2^2)\equal{}\minus{}\epsilon f^2(2)\equal{}\minus{}4\epsilon$ and so $ \epsilon\equal{}\minus{}1$

So $ f(1)\equal{}1$ and $ f(\minus{}x)\equal{}\minus{}f(x)$ and $ f(x\plus{}1)\equal{}f(x)\plus{}1$ and $ f(x^2)\equal{}f^2(x)$ and so $ f(x)\equal{}x$ $ \forall x\in\mathbb Z^*$
$ P(n,x)$ $ \implies$ $ f(nx)\equal{}nf(x)$ $ \forall n\in\mathbb Z^*,x\in\mathbb R^*$ and so $ f(x)\equal{}x$ $ \forall x\in\mathbb Q$

From $ f(x^2)\equal{}f^2(x)$ and $ f(\minus{}x)\equal{}\minus{}f(x)$, we get that $ f(x)>0$ $ \forall x>0$ and $ f(x)<0$ $ \forall x<0$
Then $ P(x,y)$ with $ y>0$ shows that $ f(x^2\plus{}y)>f(x^2)$ and so $ f(x)$ is increasing for $ x>0$, so $ f(x)\equal{}x$ $ \forall x>0$
And, since $ f(\minus{}x)\equal{}\minus{}f(x)$ : $ \boxed{f(x)\equal{}x\text{   }\forall x\in\mathbb R^*}$

hi!
see here: [url]http://www.mathlinks.ro/viewtopic.php?p=1599126#1599126[/url]

[quote="pco"][quote="hophinhan"]Let $ R*$ be the set of non-zero real numbers. Find all functions $ f : R* \minus{} > R*$ such that : $ f(x^2 \plus{} y) \equal{} f^2(x) \plus{} \frac {f(xy)}{f(x)}$
for all x, y ∈ R*, y ≠ −x^2[/quote]
Same demo as nguyenvuthanhha's one (congrats!  :) ), a litlle bit quicker in the beginning :

Let $ P(x,y)$ the assertion $ f(x^2 \plus{} y) \equal{} f^2(x) \plus{} \frac {f(xy)}{f(x)}$
Let $ a \equal{} f(1)$

$ P(x,y)$ $ \implies$ $ f(x^2 \plus{} 1) \equal{} f^2(x) \plus{} 1$ and so $ f(x) \equal{} f( \minus{} x)$. As a consequence, $ f( \minus{} 1) \equal{} \epsilon a$ where $ \epsilon\in\{ \minus{} 1, \plus{} 1\}$

Comparing then $ P(1,x)$ and $ P( \minus{} 1,x)$, we get $ f( \minus{} x) \equal{} \epsilon f(x)$ $ \forall x\neq 0$
$ P(1,1)$ $ \implies$ $ f(2) \equal{} a^2 \plus{} 1$
$ P(1, \minus{} 2)$ $ \implies$ $ f( \minus{} 2) \equal{} \epsilon a^2 \minus{} a^3$

And since $ f( \minus{} 2) \equal{} \epsilon f(2)$, we get $ a \equal{} \minus{} \epsilon$ and so $ f( \minus{} 1) \equal{} \minus{} 1$

Then $ P(1,x^2)$ $ \implies$ $ f(x^2 \plus{} 1) \equal{} 1 \minus{} \epsilon f(x^2)$ and $ P(x,1)$ $ \implies$ $ f(x^2 \plus{} 1) \equal{} f^2(x) \plus{} 1$ and so $ f(x^2) \equal{} \minus{} \epsilon f^2(x)$

$ P(1,1)$ $ \implies$ $ f(2) \equal{} 2$
$ P(1,2)$ $ \implies$ $ f(3) \equal{} 1 \minus{} 2\epsilon$
$ P(1,3)$ $ \implies$ $ f(4) \equal{} 3 \minus{} \epsilon$ But $ f(4) \equal{} f(2^2) \equal{} \minus{} \epsilon f^2(2) \equal{} \minus{} 4\epsilon$ and so $ \epsilon \equal{} \minus{} 1$

So $ f(1) \equal{} 1$ and $ f( \minus{} x) \equal{} \minus{} f(x)$ and $ f(x \plus{} 1) \equal{} f(x) \plus{} 1$ and $ f(x^2) \equal{} f^2(x)$ and so $ f(x) \equal{} x$ $ \forall x\in\mathbb Z^*$
$ P(n,x)$ $ \implies$ $ f(nx) \equal{} nf(x)$ $ \forall n\in\mathbb Z^*,x\in\mathbb R^*$ and so $ f(x) \equal{} x$ $ \forall x\in\mathbb Q$

From $ f(x^2) \equal{} f^2(x)$ and $ f( \minus{} x) \equal{} \minus{} f(x)$, we get that $ f(x) > 0$ $ \forall x > 0$ and $ f(x) < 0$ $ \forall x < 0$
Then $ P(x,y)$ with $ y > 0$ shows that $ f(x^2 \plus{} y) > f(x^2)$ and so $ f(x)$ is increasing for $ x > 0$, so $ f(x) \equal{} x$ $ \forall x > 0$
And, since $ f( \minus{} x) \equal{} \minus{} f(x)$ : $ \boxed{f(x) \equal{} x\text{ }\forall x\in\mathbb R^*}$[/quote]

I think you forgot the condition $ f : R\plus{} \rightarrow R\plus{}$  :)

[quote="ll931110"] I think you forgot the condition $ f : R \plus{} \rightarrow R \plus{}$  :)[/quote]

I think you misread the problem : $ f : \mathbb R^*\to\mathbb R^*$ and not $ \mathbb R \plus{} \to\mathbb R \plus{}$ :)

[quote="pco"]Same demo as nguyenvuthanhha's one (congrats!  :) ), a litlle bit quicker in the beginning :

Let $ P(x,y)$ the assertion $ f(x^2 \plus{} y) \equal{} f^2(x) \plus{} \frac {f(xy)}{f(x)}$
Let $ a \equal{} f(1)$

$ P(x,y)$ $ \implies$ $ f(x^2 \plus{} 1) \equal{} f^2(x) \plus{} 1$ and so $ f(x) \equal{} f( \minus{} x)$. As a consequence, $ f( \minus{} 1) \equal{} \epsilon a$ where $ \epsilon\in\{ \minus{} 1, \plus{} 1\}$

Comparing then $ P(1,x)$ and $ P( \minus{} 1,x)$, we get $ f( \minus{} x) \equal{} \epsilon f(x)$ $ \forall x\neq 0$
[/quote]

pco, I don't understand why you can put $ P(\minus{}1,x)$ while $ f: R* \rightarrow R*$, and from the problem description, $ R*$ is the set of non-negative real numbers  :?:

[quote="ll931110"] pco, I don't understand why you can put $ P( \minus{} 1,x)$ while $ f: R* \rightarrow R*$, and from the problem description, $ R*$ is the set of non-negative real numbers  :?:[/quote]

I'm sorry, but from the problem description, $ R*$ is the set of non-[u][b]zero[/b][/u], (and not non-negative) real numbers  (and it is a rather classical notation)

This example team selection test for 46.IMO(2005) has been given :lol:


calculus
derivative
geometry
trigonometry
graph theory
calculus computations

First problem is:

Use implicit differentiation to find the following:

y+sin y=x

I think the answer is  1/(cos(y)+1

is this correct?

Second problem is a related rate problem:

Assume that oil spilled from a ruptured tanker spreads in a circular pattern whose radius increases at a constant rate of 2ft/s. How fast is the area increasing when the radius of the spill is 60ft?

I dont really know where to begin with this one.

Third problem I have is:

Sketch a complete graph of y=(x-4)^(2/3)

That 2/3's is an exponent.

find the x and y intercepts:
find the critical points for f and give the intervals where f is increasing or decreasing:
find the inflection points for f, and give the intervals where f is concave up or down:
Make an accurate graph using the above:

I can't seem to get by the 2/3's exponent. I know i have to use the exponent rule.

Any hints are welcome

Thanks for your help,
Sailguy

[quote="sailguy"]First problem is:

Use implicit differentiation to find the following:

y+sin y=x

I think the answer is  1/(cos(y)+1

is this correct?[/quote]
If you meant $\frac{1}{\cos{y} +1}$, that's correct.
[hide]
$y+\sin y=x$

Using implicit differentiation,

$\frac{dy}{dx}+(\cos{y}) \frac{dy}{dx}=1$
$\frac{dy}{dx} (1+\cos{y})=1$
$\therefore \frac{dy}{dx}=\frac{1}{1+\cos y}$[/hide]

[quote="sailguy"]
Second problem is a related rate problem:

Assume that oil spilled from a ruptured tanker spreads in a circular pattern whose radius increases at a constant rate of 2ft/s. How fast is the area increasing when the radius of the spill is 60ft?

I dont really know where to begin with this one.

[/quote]

First you know that $\textsl{dA}=2$
then $\textsl{dA}=2{\pi}rdr$
then solve for $\textsl{dr}$
$\textsl{dr}=\frac{2}{2{\pi}r}$
insert  $\textsl{r}=60$
and you get $\textsl{.00530516}sqft/sec$


conics
parabola
hyperbola
vector
calculus
calculus computations

1. find equation to plane which is parallel to plane x+7y-5z+3=0 and distance between them is 10
2. find distance between lines (x-1)/1=(x+2)/-2=(z-3)/4 and (x+2)/3=(y+1)/-2=(z+2)/4
3. find equation to parabola if directrix is x+y-4=0 and focus (5, 7)
4. find equation to tangent of hyperbola (x^2)/5-(y^2)/4=1 which is parallel to line 7x+y-6=0

Please give me some clue how can I calculate these!!!

Thank You!

1.[1,7,-5] is the vector perpendicular to the plane.The line [1,7,-5]*k+[0,0,3/5] is perpendicular to the plane.Absolute value of[k,7k,-5k] must be =10.So k=2*3^.5/3 and -2*3^.5/3.The plane will be x+7y-5z+50*3^.5+3=0


G finite sub-group of $GL_n(R)$
cardinal of $G$ is denoted by g

Let $S=\sum_{M\in G}M$

1)prove g divides $tr(S)$

2) prove for any $r$ positive integer the sum $\sum_{M \in G}tr(M)^{r}$
is divisible by g

for the first we can show that S²-gS=0 , the coclusion follows immediatly ( Tr(S)=rank(S)g)
for the second we my use the operator * given in the exam and reapply the 1) question and using that Tr(M*r)=Tr(M)^r


trigonometry
inequalities
geometry
geometry proposed

Let's prove the stronger inequality : $ \boxed{\ \sum\frac{1}{\cos B\cos C}\ge \frac{36R^2}{s^2\plus{}r^2\minus{}4R^2}\ge 12\ }$ .

[b][u]Proof.[/b][/u]

First let's prove the identity $ \boxed{\ \sum\cos B\cos C\equal{}\frac{s^2\plus{}r^2\minus{}4R^2}{4R^2}\ }\ (*)$ .

Since $ \left(\sum\cos A\right)^2\equal{}\sum\cos^2 A\plus{}2\sum\cos B\cos C$ $ \implies\ 2\sum\cos B\cos C\equal{}\left(\frac{R\plus{}r}{R}\right)^2\minus{}\sum\left(1\minus{}\sin^2 A\right)$

$ \Longleftrightarrow\ 2\sum\cos B\cos C\equal{}\left(\frac{R\plus{}r}{R}\right)^2\minus{}3\plus{}\frac{a^2\plus{}b^2\plus{}c^2}{4R^2}\equal{}$ $ \left(\frac{R\plus{}r}{R}\right)^2\minus{}3\plus{}\frac{2s^2\minus{}8Rr\minus{}2r^2}{4R^2}$

$ \Longleftrightarrow\ 2\sum\cos B\cos C\equal{}\frac{4R^2\plus{}8Rr\plus{}4r^2\minus{}12R^2\plus{}2s^2\minus{}8Rr\minus{}2r^2}{4R^2}\equal{}$ $ \frac{2s^2\plus{}2r^2\minus{}8R^2}{4R^2}$

$ \Longleftrightarrow\ \boxed{\ \sum\cos B\cos C\equal{}\frac{s^2\plus{}r^2\minus{}4R^2}{4R^2}\ }$ .

Coming back to the inequality we have : $ \sum\frac{1}{\cos B\cos C}\ \stackrel{\mbox{C.B.S}}{\ge}\ \frac{9}{\sum\cos B\cos C}\stackrel{(*)}{\equal{}}\frac{36R^2}{s^2\plus{}r^2\minus{}4R^2}$

Finally let's prove that $ \frac{36R^2}{s^2\plus{}r^2\minus{}4R^2}\ge 12$ . This inequality is equivalent to : $ 3R^2\ge s^2\plus{}r^2\minus{}4R^2\ \Longleftrightarrow\ s^2\plus{}r^2\le 7R^2$

From [b]Gerretsen[/b]'s inequality $ s^2\le 4R^2\plus{}4Rr\plus{}3r^2$ we only have to show that $ 4R^2\plus{}4Rr\plus{}4r^2\le 7R^2\ \Longleftrightarrow\ 4Rr\plus{}4r^2\le 3R^2$

$ \Longleftrightarrow\ (R\minus{}2r)(3R\plus{}2r)\ge 0$, which is true .

[quote="Mateescu Constantin"]Let's prove the stronger inequality : $ \boxed{\ \sum\frac {1}{\cos B\cos C}\ge \frac {36R^2}{s^2 \plus{} r^2 \minus{} 4R^2}\ge 12\ }$ .

[b][u]Proof.[/b][/u]

First let's prove the identity $ \boxed{\ \sum\cos B\cos C \equal{} \frac {s^2 \plus{} r^2 \minus{} 4R^2}{4R^2}\ }\ (*)$ .

Since $ \left(\sum\cos A\right)^2 \equal{} \sum\cos^2 A \plus{} 2\sum\cos B\cos C$ $ \implies\ 2\sum\cos B\cos C \equal{} \left(\frac {R \plus{} r}{R}\right)^2 \minus{} \sum\left(1 \minus{} \sin^2 A\right)$

$ \Longleftrightarrow\ 2\sum\cos B\cos C \equal{} \left(\frac {R \plus{} r}{R}\right)^2 \minus{} 3 \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{4R^2} \equal{}$ $ \left(\frac {R \plus{} r}{R}\right)^2 \minus{} 3 \plus{} \frac {2s^2 \minus{} 8Rr \minus{} 2r^2}{4R^2}$

$ \Longleftrightarrow\ 2\sum\cos B\cos C \equal{} \frac {4R^2 \plus{} 8Rr \plus{} 4r^2 \minus{} 12R^2 \plus{} 2s^2 \minus{} 8Rr \minus{} 2r^2}{4R^2} \equal{}$ $ \frac {2s^2 \plus{} 2r^2 \minus{} 8R^2}{4R^2}$

$ \Longleftrightarrow\ \boxed{\ \sum\cos B\cos C \equal{} \frac {s^2 \plus{} ^2 \minus{} 4R^2}{4R^2}\ }$ .

Coming back to the inequality we have : $ \sum\frac {1}{\cos B\cos C}\ \stackrel{\mbox{C.B.S}}{\ge}\ \frac {9}{\sum\cos B\cos C}\stackrel{(*)}{ \equal{} }\frac {36R^2}{s^2 \plus{} r^2 \minus{} 4R^2}$

Finally let's prove that $ \frac {36R^2}{s^2 \plus{} r^2 \minus{} 4R^2}\ge 12$ . This inequality is equivalent to : $ 3R^2\ge s^2 \plus{} r^2 \minus{} 4R^2\ \Longleftrightarrow\ s^2 \plus{} r^2\le 7R^2$

From [b]Gerretsen[/b]'s inequality $ s^2\le 4R^2 \plus{} 4Rr \plus{} 3r^2$ we only have to show that $ 4R^2 \plus{} 4Rr \plus{} 4r^2\le 7R^2\ \Longleftrightarrow\ 4Rr \plus{} 4r^2\le 3R^2$

$ \Longleftrightarrow\ (R \minus{} 2r)(3R \plus{} 2r)\ge 0$, which is true .[/quote]

Very nice my friend....  :)

Dear friend [color=red][b]Constantin[/b][/color] who is the author of the inequality? where is published? thanks

Dear Lingouras,
your problem I can solve by Mixing variable:
WLOG, A>=B>=C=> A>=60;
your equality equivaleteral with: cosA+cosB+cosc>=12cosAcosBcosC
we have: f(a,b,c)-f(a,(b+c)/2,(b+c)/2)=(1-cos(b-c)/2)(12cosA+12cosAcos(b-c)/2-2cos(b+c)/2)>=0
so we only to prove: cosA+2cos(b+c)/2>=cosA(cos(b+c)/2)^2 (this so easy by analysis)
 (I think you should make all your geometry inequalities to one box for everyone easy to review and make our forum more clever :D)

\[ \sum\frac {1}{\cos B\cos C}\ \stackrel{\mbox{C.S}}{\ge}\ \frac {9}{\sum\cos B\cos C} \ge \frac {9}{\frac {1}{3} (\sum \cos A)^2} \ge \frac {9}{\frac {1}{3}.\frac {9}{4}} \equal{} 12\]
@Ligouras : could you plz post your ineqs in a topic

[quote="Ligouras"]Very nice my friend....  :)

Dear friend [color=red][b]Constantin[/b][/color] who is the author of the inequality? where is published? thanks[/quote]

Thank-you, [b]Ligouras[/b] . This stronger ineq. is mine . :)

[quote="Ligouras"][color=darkred]Prove that if $ \triangle ABC$ is acute, then $ \frac {1}{\cos A\cos B} \plus{} \frac {1}{\cos B\cos C} \plus{} \frac {1}{\cos C\cos A}\geq12$ .[/color] [/quote]

[color=darkblue][b][u]Proof I[/u].[/b] $ \boxed {a \equal{} b\cdot\cos C \plus{} c\cdot\cos B}$ $ \implies$ $ a^2\ge 4bc\cdot \cos B\cos C$ $ \implies$ $ \sum\frac {1}{\cos B\cos C}\ge 4\cdot \sum \frac {bc}{a^2}\ge 12\cdot \sqrt [3]{\frac {bc}{a^2}\cdot \frac {ca}{b^2}\cdot\frac {ab}{c^2}} \equal{} 12$ .

[b][u]Proof II[/u].[/b] $ \sum \frac {1}{\cos B\cos C} \equal{}$ $ \sum \frac {2}{\cos (B \minus{} C) \plus{} \cos (B \plus{} C)}\ge$ $ \sum \frac {2}{1 \minus{} \cos A} \equal{}$ $ \sum\frac {1}{\sin^2\frac A2} \equal{}$

$ \sum\frac {bc}{(s \minus{} b)(s \minus{} c)} \equal{}$ $ \frac {abc}{pr^2}\cdot \sum \frac {s \minus{} a}{a} \equal{}$ $ \frac {4R}{r}\cdot\left(s\cdot\sum \frac 1a \minus{} 3\right)\ge$ $ \frac {4R}{r}\cdot\left(\frac 92 \minus{} 3\right) \equal{}$ $ \frac {6R}{r}\ge 12$ .

Observe that $ \boxed {\ \sum \frac {1}{\cos B\cos C}\ \ge\ \frac {36R^2}{s^2 \plus{} r^2 \minus{} 4R^2}\ \ge\ \frac {6R}{r}\ \ge\ 12\ }$ . The fiest inequality belongs to [u]my dear[/u] [b]C. Mateescu[/b].[/color]

[quote="Ligouras"]Prove that
$ \frac{1}{\cos A\cos B}\plus{}\frac{1}{\cos B\cos C}\plus{}\frac{1}{\cos C\cos A}\geq12$[/quote]
$cosAcosBcosC\le \frac{1}{8}\implies$
$ \frac{1}{\cos A\cos B}\plus{}\frac{1}{\cos B\cos C}\plus{}\frac{1}{\cos C\cos A}\ge\frac{3}{\sqrt[3]{(cosAcosBcosC)^2}}\geq12.$


Let $ a$, $ b$, and $ c$ be real numbers, all different from $ \minus{} 1$ such that $ abc \equal{} 1$. Prove that
\[ \frac {a^2}{(a \plus{} 1)^2} \plus{} \frac {b^2}{(b \plus{} 1)^2} \plus{} \frac {c^2}{(c \plus{} 1)^2} \plus{} \frac {4}{(a\plus{} 1)(b\plus{} 1)(c \plus{} 1)} \ge 1.
\]
btw, this ineq is true too

Let $ a$, $ b$, and $ c$ be real numbers, all different from $ \minus{} 1$ and $ 1$, such that $ abc \equal{} 1$. Prove that
\[ \frac {a^4 \minus{} a^2}{(a^2 \minus{} 1)^2} \plus{} \frac {b^4 \minus{} b^2}{(b^2 \minus{} 1)^2} \plus{} \frac {c^4 \minus{} c^2}{(c^2 \minus{} 1)^2} \plus{} \frac {2}{(a \plus{} 1)(b \plus{} 1)(c\plus{} 1)} \ge 1.
\]

To the 1st ineq:

As $ (\frac{a}{a\plus{}1} \plus{}\frac{b}{b\plus{}1} \plus{} \frac{c}{c\plus{}1} \minus{} 1)^2 \geq 0$
$ \sum \frac{a^2}{(a\plus{}1)^2} \plus{} \sum \frac{2ab}{(a\plus{}1)(b\plus{}1)} \minus{} 2 (\sum \frac{a}{a\plus{}1}) \plus{}1 \geq 0$
$ \sum \frac{a^2}{(a\plus{}1)^2} \plus{} \sum \frac{2ab}{(a\plus{}1)(b\plus{}1)} \minus{} 2 (\sum \frac{2ab\plus{}a\plus{}b}{(a\plus{}1)(b\plus{}1)}) \plus{}1 \geq 0$
$ \sum \frac{a^2}{(a\plus{}1)^2} \minus{} \sum \frac{a\plus{}b}{(a\plus{}1)(b\plus{}1)} \plus{}1 \geq 0$
$ \sum \frac{a^2}{(a\plus{}1)^2} \minus{} \frac{2(ab\plus{}bc\plus{}ca\plus{}a\plus{}b\plus{}c)}{(a\plus{}1)(b\plus{}1)(c\plus{}1)} \plus{}1 \geq 0$
$ \sum \frac{a^2}{(a\plus{}1)^2} \plus{} \frac {4}{(a\plus{} 1)(b\plus{} 1)(c \plus{} 1)} \plus{}1 \geq \frac{2(ab\plus{}bc\plus{}ca\plus{}a\plus{}b\plus{}c\plus{}2)}{(a\plus{}1)(b\plus{}1)(c\plus{}1)}$
$ \sum \frac{a^2}{(a\plus{}1)^2} \plus{} \frac {4}{(a\plus{} 1)(b\plus{} 1)(c \plus{} 1)} \plus{}1 \geq \frac{2(abc\plus{}ab\plus{}bc\plus{}ca\plus{}a\plus{}b\plus{}c\plus{}1)}{(a\plus{}1)(b\plus{}1)(c\plus{}1)}$
$ \sum \frac{a^2}{(a\plus{}1)^2} \plus{} \frac {4}{(a\plus{} 1)(b\plus{} 1)(c \plus{} 1)} \plus{}1 \geq 2$
So. $ \sum \frac{a^2}{(a\plus{}1)^2} \plus{} \frac {4}{(a\plus{} 1)(b\plus{} 1)(c \plus{} 1)} \geq 1$


algebra
polynomial
trigonometry
absolute value
real analysis
real analysis unsolved

Let $P(x)=\sum_{i=o}^n a_ix^i $such that:$|P(x)|\leq 1\forall x \in [-1;1];G(x)= \sum_{i=o}^n a_{n-i}x^i$.Prove that:$|G(x)|\leq 2^{n-1}\forall x\in [-1;1]$

As far as I remember, this was an open question of Walther Janous, published in Crux. I managed to prove it, but the proof is quite long and difficult. In fact, for x=0 it reduces to a well-known deviation theorem. For $ x$ non-zero we can prove even more, that is $ K( G(x))\leq K( x^n\cdot T_n(\frac{1}{x})) $ where K is the absolute value function.

The competition deadline has passed so I'll post my solution. (which is probably the one Harazi intended, seeing his comment)

First, check the internet to see that $T_n(x)=\cos(n\arccos(x))=\frac12\left((x+\sqrt{x^2-1})^n +(x-\sqrt{x^2-1})^n\right)$ is a real polynomial, its set of extremal values on $[-1,1]$ obviously being $S=\left\{c_k|k=0..n\right\}$, where $c_k=\cos\left(\frac{k\pi}{n}\right)$.

Now taking lagrange interpolants of this $T_n(x)$ and any $n$-th degree polynomial with $\max_{\xi\in[-1,1]} |f(\xi)|\le1$ on $S$, we obtain: \[ f(x)=\sum_{k=0}^n f(c_k)\prod_{i\not=k}\frac{x-c_k}{c_i-c_k}\le \sum_{k=0}^n \left|f(c_k)\prod_{i\not=k}\frac{x-c_k}{c_i-c_k}\right|=\sum_{k=0}^n |f(c_k)|\prod_{i\not=k}\frac{|x-c_k|}{|c_i-c_k|},  \] \[ T_n(x)=\sum_{k=0}^n T_n(c_k) \prod_{i\not=k}\frac{x-c_k}{c_i-c_k}=\sum_{k=0}^n (-1)^k \prod_{i\not=k}\frac{x-c_k}{c_i-c_k}=\sum_{k=0}^n \prod_{i\not=k}\frac{x-c_k}{|c_i-c_k|},  \] where the latter follows from the sign of $\prod_{i\not=k}(c_i-c_k)$ being $(-1)^k$.

Since for all $|x|\ge1$ all $(x-c_k)$ have the same sign, and since $c_k\in[-1,1]$ thus $|f(c_k)|\le\max_{\xi\in[-1,1]}|f(\xi)|=1$, we obtain Harazi's statement.

From there on it's easy to tackle this problem: Since $g(x)=x^n f\left(\frac{1}{x}\right)$ we have $|g(x)|\le|x^n| \left|T_n\left(\frac1x\right)\right|$, and since $T_n(x)=\frac12\left((x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n\right)$ we have $x^n T_n\left(\frac{1}{x}\right) = \frac12\left((1+\sqrt{1-x^2})^n+(1-\sqrt{1-x^2})^n\right)$, which is convex on $[0,1]$ in $\sqrt{1-x^2}$, obtaining its extremal values at the endpoints, which yields $x^n T_n\left(\frac{1}{x}\right)\le 2^{n-1}$, QED.


calculus
integration
real analysis
real analysis solved

Study the convergence of the following improper integral:

\int _0^{+\infty} (e^(sin x) sin 2x / (x^p)) dx

Well, here is the easy bit ...

For positive x, then
0<=e^(sin x)<=e
|sin 2x|<=1

so

|Integral| <= e  \int  (0 to \infty) x^(-p) dx = e/(p-1) if p>1

so integral certainly converges for all p>1

gandalf61, don't you think that your solution is a little indiscreet?

have a look at the interval again...it's 0 to \infty, not 1 to \infty

well...No one answers this post?  :(

My answer is
The improper integral diverges when p\geq 2 or p\leq 0,
and it is absolutely convergent when 1<p<2;
and it is conditional convergent when 0<p\leq 1.


Princeton
college
real analysis
complex analysis
number theory

[quote="moby"]I'm a pure math major (finishing up my sophomore year; the only math courses I've taken so far are "pure" ones), and I do want to go to grad school (in the pure math track) and academia after that.

However:
I'm also kind of nervous and doubtful about the prospect.  I'm leaning heavily towards it, but I feel like I should also explore other options before committing myself to the ideal of academia.

I'm just wondering what kind of courses/activities I should do in the event that I should eventually go into industry (if I do, it will probably be after grad school, although who knows, I might even give grad school a pass, or go for the applied track in grad school?)

There is also something else on my mind - the possibility of taking fall semester off senior year to expand my math horizons (and for a change of environment) although I'm not sure what form this will/should take.  Does anyone have suggestions? Internships in industry; or just math research opportunities at other universities (how would I secure them?!); etc.

Thanks a lot.

P.S.  I don't know if this is the best section to get lots of (good) advice, but I certainly hope for it.[/quote]

What courses did you take? I think 'pure' math is a quite general expression. What kind of pure math do you prefer?

I've taken Real Analysis (2 sems), Abstract Algebra (2 sems), Complex Analysis (1 sem), Diff. Geom (1 sem), Logic (1 sem)... My grades are OK but not great.  Not counting the A's for my year long honors multivariable class in freshman year, I have 3 A's and 2 B+'s so far (and 2 more grades to come).

I guess I'm leaning towards Algebra and Number Theory academically.
From an applied math perspective, although I've not taken any classes in these subjects, but I feel that combinatorics and discrete math can really get me excited.  (I guess this straddles applied and pure though?)
And yeah, I'm hoping to go to one of the top grad schools (princeton, etc.) although it's a crapshoot.
I'd be elated with a place that is just off the very top tier, like UCLA.

[b] Digression [/b]

I personally feel that almost nobody in this forum wants to post or participate in the discourse (with regards to a thread created by someone) here :( . I really can't understand the reason behind this behavior (which appears weird). Almost no one is interested in creating threads or replying to others' queries or whatever.

I feel this section of AoPS deserves more attention than it is being given now.

Well I'm not surprised. The major focus on AoPS is high school (and even middle school) math competitions. If you want to know more about career paths, you can always talk to your college teachers.

:lol:  You must be kidding! I'm not from the USA, and I bet I'm the only person in my school who is an 'AoPSer' ; I really can't consult my teachers on a career in Math because they don't have the requisite knowledge regarding such questions, so, I have to decided to ask people in this section of AoPS. I hope this section of AoPS becomes active again.


inequalities
trigonometry
inequalities unsolved

$a,b,c$ are the length of a triangle
prove or disprove:
$2\sum_{cyc}a^2 \sin^2 \frac{B-C}2 \ge \sum_{cyc} (b-c)^2$

and by calculation:
I find this inequality is equal to:
$\sum_{cyc}(ab^2+ac^2-a^3)(b-c)^2 \ge 0$
or
$\sum_{cyc}\cos A (b-c)^2 \ge 0$
surprised me...
and can anybody find a solution,then?

No one answer me,I just answer myself :) .
and finally I find a solution:
use the fact:
$\cos \frac{B-C}2=\frac{b+c}{2a} \sqrt{\frac{(a-b+c)(a+b-c)}{bc}}$
we can get:
$2\sum_{cyc}a^2 \sin^2 \frac{B-C}2 \ge \sum_{cyc} (b-c)^2 \iff \sum_{cyc} \cos A(b-c)^2 \ge 0$
wlog $A \ge B \ge C$
case (i) $A \le \frac{\pi}2$,obvious
case (ii) $A > \frac{\pi}2$,we have $\cos B,\cos C>0,and \cos A+\cos B >0$
then by the method SOS,we can get:
$\sum_{cyc} \cos A(b-c)^2 \ge 0$,QED.
hope it is right.and enjoy it. :)

Cool inequality, Zhaobin!

[quote="zhaobin"]$\sum_{cyc}(ab^2+ac^2-a^3)(b-c)^2 \ge 0$[/quote]

In fact, in this form, your inequality is true for any three nonnegative reals a, b, c (not necessarily sides of a triangle).

In my proof of this inequality, I will denote cyclic sums simply by the $\sum$ sign. Then, we have to show that the inequality

$\sum\left(ab^2+ac^2-a^3\right)\left(b-c\right)^2\geq 0$

holds for any three nonnegative reals a, b, c. In fact, since $\left(b-c\right)^2=\left(b-c\right)\left(b-a\right)+\left(c-a\right)\left(c-b\right)$, we have

$\sum\left(ab^2+ac^2-a^3\right)\left(b-c\right)^2=\sum\left(ab^2+ac^2-a^3\right)\left(\left(b-c\right)\left(b-a\right)+\left(c-a\right)\left(c-b\right)\right)$
$=\sum\left(ab^2+ac^2-a^3\right)\left(b-c\right)\left(b-a\right)+\sum\left(ab^2+ac^2-a^3\right)\left(c-a\right)\left(c-b\right)$
$=\sum\left(ca^2+cb^2-c^3\right)\left(a-b\right)\left(a-c\right)+\sum\left(bc^2+ba^2-b^3\right)\left(a-b\right)\left(a-c\right)$
$=\sum\left(\left(ca^2+cb^2-c^3\right)+\left(bc^2+ba^2-b^3\right)\right)\left(a-b\right)\left(a-c\right)$
$=\sum\left(a^2-\left(b-c\right)^2\right)\left(b+c\right)\left(a-b\right)\left(a-c\right)$
$=\sum a^2\left(b+c\right)\left(a-b\right)\left(a-c\right)-\sum\left(b-c\right)^2\left(b+c\right)\left(a-b\right)\left(a-c\right)$
$=\sum a^2\left(b+c\right)\left(a-b\right)\left(a-c\right)+\sum\left(b-c\right)^2\left(b+c\right)\left(a-b\right)\left(c-a\right)$
$=\sum a^2\left(b+c\right)\left(a-b\right)\left(a-c\right)+\left(b-c\right)\left(c-a\right)\left(a-b\right)\underbrace{\sum\left(b-c\right)\left(b+c\right)}_{=\sum\left(b^2-c^2\right)=0}$
$=\sum a^2\left(b+c\right)\left(a-b\right)\left(a-c\right)$,

so that it remains to prove that $\sum a^2\left(b+c\right)\left(a-b\right)\left(a-c\right)\geq 0$. In fact, WLOG assume that $a\geq b\geq c$. Then, $a\geq b$ yields $ac\geq bc$, so that $ab+ac\geq ab+bc$, what becomes $a\left(b+c\right)\geq b\left(c+a\right)$, and thus, upon multiplication with $a\geq b$, we obtain $a^2\left(b+c\right)\geq b^2\left(c+a\right)$. Hence, if we denote $x=a^2\left(b+c\right)$, $y=b^2\left(c+a\right)$, $z=c^2\left(a+b\right)$, then $x\geq y$. Thus, according to the Vornicu-Schur inequality (Theorem 1 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=387639#p387639]http://www.mathlinks.ro/Forum/viewtopic.php?t=65385 post #2[/url]), we get $\sum x\left(a-b\right)\left(a-c\right)\geq 0$, what rewrites as $\sum a^2\left(b+c\right)\left(a-b\right)\left(a-c\right)\geq 0$. The proof is complete.

  Darij

Cool solution.
I can't believe a inequality seems use SOS
then  vornicu-schur.
Thanks very much


inequalities
algebra
polynomial
absolute value
inequalities proposed

Let  $P\in {\mathbb R}[x]\; ,\; P(x)=x^n+a_1x^{n-1}+\ldots+a_n\; ,$ and $-\infty<a<b<\infty .$  If $|P(x)|\le 1\; ,\; \forall x\in [a,b] ,$ prove that
  \[ b-a\le \frac{4}{\sqrt[n]{2}} .  \]
This result is optimal  : when  $b-a= \frac{4}{\sqrt[n]{2}} \;, \;$ there is a monic polynomial $P^*$ of degree $n$   such that  $\left|P^*(x)\right|\le 1$  for all $x\in [a,b] .$

Until now ALEX collection= a long list with problems
[of course,not all original].
Regards,Alex

[quote="flip2004"]...
 $  -\infty<a<b<\infty . $  If $ P(x)\le 1\; ,\; \forall x\in [a,b] ,$ prove that
  \[ b-a\le     \frac{4}{\sqrt[n]{2}} . \]
[/quote]

Is there any absolute value in $P(x) \leq 1$ ?

Pretty nice, but not that difficult. Just consider the polynomial $ Q(X)=P(\frac{x(b-a)+a+b}{2})$ and apply Cebasev minimal deviation theorem. This will also give you the optimal polynomial.

What do call Cebasev minimal deviation theorem ?

I think he means following:
if $|P(x)|\leq 1$ for all $x\in[-1,1]$ and $\deg P=n$ then leading coefficient of $P$ is at most $2^{n-1}$.

Yes, that is the statement. Sorry for using this name, but I just saw in some papers this name.

Hi Moubinool,
Thanks for your remark. Indeed, in hypothesis the right condition is $|P(x)|\le 1 $ .


If there exists a regular n-gon with its vertices at lattice points in a cartesian plane, prove that n=4

[hide="solution"]
Suppose we had regular polygon with latice point vertices.

The area of a regular n-gon with side length $ s$ is given by $ A \equal{} \frac {n}{4}\cdot s^2\cdot \cot \left(\frac {180}{n}\right)$

Note that the side length squared, $ s^2$, is an integer because $ s \equal{} \sqrt {(x_1 \minus{} x_2)^2 \plus{} (y_1 \minus{} y_2)^2}$ where $ (x_1,y_1)$, $ (x_2,y_2)$ are two adjacent vertices

Also $ \cot \left(\frac {180}{n}\right)$ is irrational when $ n\not \equal{} 4$

Thus we conclude that our polygon has an irrational area when $ n\not \equal{} 4$

However, by the [url=http://www.artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem]Shoelace Theorem[/url], a polygon with lattice point vertices has a rational area. Contradiction $ \blacksquare$

[/hide]


quadratics
function
algebra unsolved
algebra

Let $ x,y,z$ be real numbers such that

$ |x \plus{} y \plus{} z|\leq 1,|x \minus{} y \plus{} z|\leq 1,|4x \minus{} 2y \plus{} z|\leq 8,|4x \plus{} 2y \plus{} z|\leq 8$.

Prove that $ |x| \plus{} 3|y| \plus{} |z|\leq 7$

I think something is wrong with the problem, try $ x\equal{}y\equal{}z\equal{}0$ or $ y\equal{}0, x\equal{}z\equal{}\frac{1}{2}$ or $ x\equal{}y\equal{}z\equal{}\frac{1}{3}$ for example.

...

...

??

 :)

Yeah, whatever you try it will be wrong.

I've reversed the sign  :)

Just use Lagrange's Interpolation formula for the quadratic functions with x,y,z coeficients.!


The product of a 3-digit base 7 number and the base 5 number, 43 equals
50,435 in base 6. What is the base 7 number?

Solution
[hide]562[/hide]

[quote="shentang"]Good Job! When converting 289 back to base 10, did you use remainders?[/quote]
I think you meant to say converting 289 back into base 7?

When you are converting base 10 into something else, you use the greatest power. I don't know how to explain this.. If you have a AoPS book 1, take a look at page 40.

[quote="ZhangPeijin"]Wow everyone's using AoPS books these days...
But to me, they get boring after awhile....
I get bored and stop reading.
And then I forget what I read,
So I never really got past chapter 6...[/quote]

HOW COULD YOU???

The remainder method is something I don't use as much as I don't see how it relates to the idea of bases.

[quote="ZhangPeijin"]Wow everyone's using AoPS books these days...
But to me, they get boring after awhile....
I get bored and stop reading.
And then I forget what I read,
So I never really got past chapter 6...[/quote]
Keep on trying. I was like that too, but I changed and decided to finish it. And I am almost there.


In a triangle ABC ,BD is internal bisector of angle CBA. m(CBA)= 80 . And AD=BD+BC. Find the angle CAB.

fro $ AB \le BC$ we have $ BC > \frac{AC}{2} > AD$ is it'a ipossible. For $ AB>BC$ let 2 triangle with angle C respectively X and Y with X>Y, then BC on the first is bigger than BC on the second and the seme for BD. Then there is only a solution and it's simple to prove that $ \angle CAB \equal{} 20$ it's a solution because
$ \frac{BD \sin60}{\sin80} \plus{} BD\equal{} \frac{BD \sin40}{\sin 20} \Longleftrightarrow \sin 60 \plus{} \cos 10 \equal{} 2 \cos20\cos 10 \Longleftrightarrow \sin 60 \plus{} \cos 10 \equal{} \cos 30 \plus{} \cos 10$


Let $a,b,c$ be positive reals. Prove that $\frac{ab+bc+ca}{a+b+c}\geqq\frac{3abc}{ab+bc+ca}$. When does equality hold?

This is a direct result of Newton's inequality.

What's Newton's inequality? I can't find in Mathworld.

Is there a shorter way to do this?

The inequality is homogeneous, that is, we can replace a, b, c with ka, kb, kc (k > 0) and we get an equivalent inequality.  Thus, assume without loss of generality that a + b + c = 1.  Then cross multiplication yields that we must prove

(ab + bc + ca) 2  >= 3abc.

We multiply out to get [LHS means "left-hand side"]

LHS = a 2 b 2  + b 2 c 2  + c 2 a 2  + 2a 2 bc + 2ab 2 c + 2abc 2 .

The right three terms of this expression factor to 2abc (a + b + c) = 2abc because a + b + c = 1.  We use AM-GM to get an inequality out of the three left terms:

Three left terms = (1/2)(a 2 b 2  + b 2 c 2) + (1/2)(b 2 c 2  + c 2 a 2) + (1/2)(c 2 a 2 + a 2 b 2)  >=  ab 2 c + abc 2  + a 2 bc = abc (a + b + c) = abc.

Thus the whole LHS >= abc + 2abc = 3abc, which is what we wanted.  We have equality when the AM-GM we used has equality.  This is when a = b = c, which is a = b = c = 1/3 for the case we are considering.  But by our first argument about homogeneity we get equality whenever a = b = c (and a, b, c > 0).

Multiplying out, we get:

$\displaystyle \sum_{sym} a^2b^2+2\sum_{sym}a^2bc\geq 3\sum_{sym} a^2bc$, which is an immediate consequence of Muirhead.

For cn, if you don't know Muirhead, we can use

$\displaystyle \frac {1}{2}\left(\sum_{sym} a^2b^2+a^2c^2\right)\geq \sum_{sym} a^2bc$ by AM-GM, and the rest proceeds as above.

For Newton's inequality, see page 10 of Kiran Kedlaya's [url=http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/ineqs-080299.ps]A < B[/url].

Thank you very much,[b] billzhao[/b].

It's great!

kunny

See my second solution, it's just multiplication and AM-GM for two numbers.

The proof of inequality is basically [b]taking the difference [/b]and examining it's sign.


To $ x\ge 0$,$ 0\ge (x \minus{} 2)(x \plus{} 1)^2 \equal{} x^3 \minus{} 3x \minus{} 2$
To $ 0\ge x$,$ x^3 \minus{} 3x \plus{} 2\ge0$
Then……easy


integration
calculus
derivative
real analysis
real analysis unsolved

At some point in a discussion of Laplace's equation, Evans writes 'we cannot just compute
\[\Delta u(x)=\int_{\mathbb{R}^{n}}\Delta_{x}\Phi(x-y)f(y)dy=0 \]
Anyways, $\Delta$ is just the Laplacian, but what is $\Delta_{x}$?  Does it just mean that it is independent of $y$ since $y$ is a 'dummy' variable?

$\Delta_{x}$ is the sum of second partial derivatives in $x_{1},\dots,x_{n}$. The clarification is necessary here because $\Phi(x-y)$ depends on $x$ and $y$. Also, this notation is used when considering the heat equation $U_{t}(x,t)=\Delta_{x}U(x,t)$.

[quote]Notation is a nightmare[/quote]
(L.C. Evans)


If you take both A and B tests, but 1 score is higher than the other, do you get to choose which score to use with your AIME score?

I don't think you get to choose. I think they just choose your higher score.

http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2006-aa/AIME%20I.pdf

Note the section to the right of your name.

You fill in your choice of AMC scores, but if, for instance, you took both AMC 10 and AMC 12, they'll consider both.

We use the higher numerical score, wherever it came from,
whatever test it was acheived on.

Steve Dunbar
AMC Director.

[quote="life=tennis+math"]What if you are in 10th grade. Say you took the AMC 10 and barely qualified for the AIME, as it was a bad day.
Then let's say, on the B date, you took the AMC 12 and got a higher score than on your AMC 10.
Since your AMC 12 score is higher, would you have to qualify by the index or, since you are in 10th grade, could you qualify by the AIME floor score?[/quote]
floor

Hmm, I didn't realize you could take both the A and B tests.  If you had a really off day on the A test, how would you go about taking the B one?

Just like you would normally take the B one.  The only issue you might have is where the B test is offered.  For example, my school doesn't offer the B test, so I'm probably not going to take it...


Let $ P$ be a point outside a circle $ \omega$,  draw a line through $ P$ intersect $ \omega$ at $ Q$ and $ E$ ($ E$ is between $ P$ and $ Q$ ).
$ A$ and $ C$ are points of tangent from $ P$ such that $ A$ is on the minor arc of $ QE$.
$ B$ is a point on $ AC$ such that $ QB$ bisects $ \angle AQC$ and intersects $ EC$ at point $ G$.
$ F$ is a point on $ AQ$ such that $ BF\perp BE$.
$ FG$ intersects $ AC$ at point $ H$.
Prove that the quadrilateral $ BHFQ$ is cyclic.

No one? This problem was built by my friend and I dont know whether it is true. But by my rough computer drawing it's true.


Οι δύο παράλληλες πλευρές ενός τραπεζίου έχουν μήκη $a,b$ .Ένα ευθύγραμμο τμήμα μήκους $m$
παράλληλο προς τις πλευρές το χωρίζει σε $2$ τραπέζια με ίσα εμβαδά .Αν οι $a,b,m$ είναι σχετικά
πρώτοι θετικοί ακέραιοι δ.ο. ούτε το $2$ ούτε το $3$ είναι παράγοντες των $a,b,m$ .

 :)

Γειά σας και από μένα. Έχω φτάσει στο εξής αποτέλεσμα:
a^2+b^2=2m^2. Συγνώμη, για τον τρόπο γραφής αλλά προσπάθησα να κάνω copy από τον Equation Editor της Microsoft και δεν ήταν εφικτό. Είμαι σίγουρος ότι από αυτό το σημείο μπορεί να φτάσει κανείς στην λύση, αλλά είναι ώρα για formula και θα επανέλθουμε σε 2 ωρίτσες.
Κώστας

Καλως ηρθες Κωστα στην παρεα. :) η εξισωση που βρισκεις δινει το αποτελεσμα.ενδιαφερουσα  εξισωση. :!:


abstract algebra
superior algebra
superior algebra solved

Find three finite groups G, H, and K, along with surjective homomorphisms g: G --> K, h: H --> K such that:
1) ker(g) is isomorphic to ker(h)
2) G is not isomorphic to H

$G=\mathbb{Z}/6\mathbb{Z},H=S_{3},K=\mathbb{Z}/2\mathbb{Z}$ with $g$ given by further reduction $\mod 2$ and $h$ being the signum.
It's clear that the kernels have three elements each, so they are isomorphic, whereas $G,H$ are not.

But is h surjective?  Can you please explain that to me, because I don't understand how you can do signum(x), where x is an element of S_3.

"signum" here is the sign of a permutation. For $\mathbb{S}^{3}$ mapping to $\mathbb{Z}/2\mathbb{Z}$, $\text{sgn}(p)=\begin{cases}1&p\text{ is a transposition}\\0&\text{otherwise}\end{cases}$.


What are you gonna do about it????
Does it matter???
Why do you get so angry when I congratulate someone????

Now that I am noticing...Math154 also did a pretty good job in replying to my posts.

If you want to congratulate someone, the best way to do it is to PM them, unless they did a great feat. Then I guess you could say it in public. Answering a question isn't very great, therefore you should have just PMed him, or done nothing.

[quote="USAMO13"]Out of all the people who have gave solutions to my problems, I think XPmath gave the best solutions.
Thanks xpmath! You are the best. Also, to others who reply to my posts please dont spam.[/quote]

first of all, if you're gonna congratulate someone (which is complete spam), at least do it in the same thread.

and while you're telling others to not spam by replying to your posts, why can't others find alternate solutions and post them, inform you that your question was far above national mc level, or discuss the true difficulty in finding the answer? i don't believe that any of those are spam.

why did you post the same question in three different threads? if you're trying to challenge someone, do you realize that they could go back and find the answer? if you're not, do you realize that posting one question in three threads will return just as many, if not less responses than posting it in one thread.

although i see that you are new on AoPS, your multiple posting, spamming, (and stealing of avatars... :mad: ) is absolutely unacceptable. please stop.

BTW, are all of the following your multis?

 mathgenius123

 megamath

 AIME

[quote="Dojo"]You guys don't need to so severely criticize him. USAM013: just don't do it again.

Also, can you change the avatar? I made that one my self specially for nikeballa.[/quote]

he claims to have gotten it from another "friend" who supposedly told USAMO13 that i stole it from him. and that's why there are nike signs.

Could you just change the avatar and location?
Cause that seems to be the main problem that nike doesn't like about this topic.


By the way, did anyone ever find two digital numbers whose sum was also a digital number?  (A digital number is a nine-digit number which includes each of the digits from 1 to 9 exactly once.)  Just wondering... I would love to see how it could be done!

I have a list of a whole lot of them, but let me just give you the first one:

123456789+123456789=246913578

Awesome.  Hard to improve on that answer, I must admit.  Thanks!

Want all the ones I have so far?

(It would take my computer days to find the rest)


inequalities
LaTeX
number theory unsolved
number theory

$\text{\LaTeX} \text{'d}:$

If $p_n$ denote the $n\text{th}$ prime, so $p_1 = 2, p_2 =3 \ldots$. Show that:

a) $2 \times 3 \times \ldots \times p_n > p_{n+1}$
b) $2 \times 3 \times \ldots \times p_n > p^2_{n+1}$ for all $n > 4$.

Part a) is easy, $p_1p_2 \ldots p_n - 1$ is coprime to all $p_i < p_{n+1}$ and so must be prime or a product of two primes $> p_n$. $\therefore 2 \times 3 \times \ldots \times p_n > p_{n+1}$.


function
number theory unsolved
number theory

Prove that for all $ k \in N$, $ k>2$ , always exist $ n \in N$ such that : $ \phi(n\plus{}k)\equal{}\phi(n)$.

[quote="conan_naruto236"]Prove that for all $ k \in N$, $ k > 2$ , always exist $ n \in N$ such that : $ \phi(n \plus{} k) \equal{} \phi(n)$.[/quote]
$ k\equal{}1$ work $ n\equal{}3$.
Let $ (k,6)\equal{}1$, then work $ n\equal{}3k$ $ \phi (3k)\equal{}2\phi(k)\equal{}\phi(4k)\equal{}\phi(3k\plus{}k)$.
If $ k\equal{}2^lm, (m,6)\equal{}1$ work $ n\equal{}4k$.
If $ k\equal{}3^lm, (m,6)\equal{}1$ work $ n\equal{}k$,
If $ k\equal{}2^b3^am, (m,6)\equal{}1, (m,91)\equal{}1$ work $ n\equal{}13k$ $ \phi(13k)\equal{}12\phi(k)\equal{}\phi(14 k)$,...

Rust wrote:
[quote]If $ k\equal{}2^l m$,$ (m,6)\equal{}1$ work $ n\equal{}4k$[/quote]
This is wrong when 5|m , check k=10 .

Let $ p$ be the smallest prime that doesn't divide $ k$. Then:
$ \phi((p\minus{}1)k)\equal{}(p\minus{}1)\phi(k)\equal{}\phi(pk)$

The first equality is true because all the prime factors of $ p\minus{}1$ already appear on $ k$, and the last is $ p\perp k$ and $ \phi(p)\equal{}p\minus{}1$.


function
trigonometry
geometry
3D geometry
sphere
calculus
integration

Suppose the gradient of $ f(x,y,z)$ is parallel to $ <x,y,z>$ everywhere.  Then show that $ f(0,0,a) = f(0,0,-a)$.

By the multivariable chain rule, the function $ g(t)=f(0,a\sin t,a\cos t)$ has zero derivative.

Another take on the same idea, which generalizes to $ n$ dimensions with $ n\ge 2$:
Choose two points $ A,B$ on a sphere centered at the origin, and let $ \gamma(t)$ be a path contained in the sphere, with $ \gamma(0)=A$ and $ \gamma(1)=B$. By the FTC for line integrals, $ f(B)-f(A)=\int_{0}^{1}\nabla f(\gamma(t))\cdot \gamma'(t)\,dt=\int_{0}^{1}0\,dt=0$, and $ f$ is a radial function.


LaTeX
inequalities
search
inequalities unsolved

Because I have not LAtex and this is better. You need 1 sec to download in word

I will download it when I get to Bucharest,because here in Constantza I don't have Microsoft Word or Acrobat Reader. ;)

Please post in $\LaTeX$ here, it's not hard, and it's the custom.

The problem: Find the rational number $a/b$ ($a$, $b$ positive integers) with the minimum $b$.

We know that
\[\frac{52}{303} < \frac{a}{b} < \frac{16}{91}.\]

No answers for second day. Try again it is very difficult

What is strange is not the inequality, but the way you talk, silouan. If [b]I[/b] have been able to find a solution, then it is certainly not that difficult  :mad:

The answer to your problem (which should have been posted in the Arithmetic section) is : $\frac{4}{23}$. Here are some details : let $r_1=\frac{52}{303}$ and $r_2=\frac{16}{91}$ ; with the help of a calculator, search for the first $b\geq 1$ such that there exists an integer $a$ between $br_1$ and $br_2$. In other words, stop the search loop as soon as you have

$[br_1] + 1 = [br_2]$

where [.] denotes the integer part.


In the unbalanced chemical equation $ \ce{BrO3 \minus{} } \plus{} \ce{Br \minus{} } \plus{} \ce{H \plus{} } \rightarrow \ce{H2O}\plus{}\ce{Br2}$, find the ratio of bromate to bromide.

Try and balance the equation.

[hide=" Chemical Reaction"]

$ 2 \ce{BrO3\minus{}} \plus{} 10\ce{Br\minus{}} \plus{} 12\ce{H\plus{}} \rightarrow 6 \ce{Br2} \plus{} 6 \ce{H2O}$

[/hide]

One of the common technique to doing this is the ion-electron method. You can also work with oxidation numbers etc.

[quote="BanishedTraitor"]One of the common technique to doing this is the ion-electron method. You can also work with oxidation numbers etc.[/quote]

Could you explain these various methods other than trial-and-error?

use the half reaction method:

Basically that states that :

$ BrO_3^\minus{} \plus{} Br^\minus{} \leftarrow Br_2$

The rules are as follows:

1. Add water to balance oxygens
2. Add Hydrogen on the other side to balance hydrogens
3. Add Electrons to balance charge
4. If in basic medium, add equal amt of OH's to each side as number of H+'s and convert H+ + OH --> H2O, then cancel H2O's

I might just elaborate on that further so it becomes clearer how it works. Okay let's start with balancing:

$ \ce{BrO3\minus{}} \rightarrow \ce{Br2}$

First because there are two atoms of $ \ce{Br}$ on the right so there should be two on the left as well:

$ 2 \ce{BrO3\minus{}} \rightarrow \ce{Br2}$

Now we have 3 atoms of oxygen on the left and to balance that out we add molecules of water to the right:

$ 2 \ce{BrO3\minus{}} \rightarrow \ce{Br2} \plus{}3\ce{H2O}$

However adding water has created 6 atoms of H on the right so to balance this out we add hydrogen ions to the left:

$ 2 \ce{BrO3\minus{}} \plus{}6 \ce{H\plus{}} \rightarrow \ce{Br2} \plus{}3\ce{H2O}$

Finally we must conserve the charge and we have $ \plus{}4$ chrage on the left and none on the right so we add electrons:

$ 2 \ce{BrO3\minus{}} \plus{}6\ce{H\plus{}} \plus{} 4\ce{e\minus{}} \rightarrow \ce{Br2} \plus{}3\ce{H2O}$

u can do without balancing :D  :P
just use equivalent concept....... :roll:


function
combinatorics unsolved
combinatorics

Let $M_{1}, M_{2}, M_{3}, ..., M_{n}$ be sets and let the union of $a$ such sets have at least $a$ elements. Set $M=\{M_{1}, M_{2}, M_{3}, ...,M_{n}\}$. Proof that there is an injective transformation $f: M\to \bigcup_{i=1}^{n}{M_{i}}$, so that $f(M_{k})\in M_{k}$ for every natural $k\leq n$.

I think it's itself Marriage Principle or Hall's theorem.
   Solution:Induct on $n$


logarithms calculus integration inequalities

Find the smallest x such that $9^x<10^{x-1}$.

 :P

[hide]Dividing by $10^{x-1}$ on both sides we have:
$\frac{9^x}{10^{x-1}}<1$.  We can rewrite this as $9\left(\frac{9}{10}\right)^{x-1}<1$, so $\left(\frac{9}{10}\right)^{x-1}<\frac{1}{9}$.  If set this as an equality we will find the smallest value.  So we have $\log_{\frac{9}{10}}\frac{1}{9}=x-1$.  Using a calculator we find $x-1\approx 20.85$ so $x\approx 21.85$.[/hide]

Rearrange the equation as follows:

$9^x < 10^{x - 1}$
$9^x < \frac {10^x}{10}$
$10 < \frac {10^x}{9^x} = (\frac {10}{9})^x$

I shall assume you want the smallest integral $x$. We can find the solution by solving the equation $10 = (\frac {10}{9})^x$. Taking logs (to base ten) on both sides, $1 = \log (\frac {10}{9})^x$, $x \log \frac {10}{9} = 1$, $x = \frac {1}{\log \frac {10}{9}} \approx 21.85$

So the answer the the inequality is $x = 22$

[quote="shadysaysurspammed"]Hmmmmmm...cant this question be done without logs??Just azking,i dont know the method without logs..if somebody does....then plz tell so.[/quote]
you know logs though???? :huh:

Yup i do and i do know its easy to solve it with logs but i was just askin for another way :lol:


Constructive Counting Part 2

We first explore into how many regions we can divide a rectangle with 6 fences, and then we determine the number of additional regions we can form if we also have a circular fence.

What happens if the circle we draw is tangent to two of the outer fences? What if it is tangent to three of them?

i sort of agree with your thinking of the tangent circles, but i drew some diagrams, and it looks like there are still the same number of regions in the sqaure.  Also, for the formula of n lines, it is 1+ (n(n+1))/2
i have a theory for m circles, but none for n lines and m circles

I have found THE FORMULA!  :rotfl:  :rotfl:  :rotfl:  The formula is for the number of regions created by n fences. It is ((n(n+1))/2)+1.

The formula for regions with m circles is m(m-1) +2 (assuming the circles don't intersect the original fences)
The formula for regions with n lines and m circles is a combination of the two formulae:
m(m-1) + 2mn + ((n(n+1)/2)+1

I think that the number of regions into which you can divide the rectangle into with $n$ line is $1+1+2+3... + n = 1 + \dfrac{n(n+1)}{2} = \dfrac{n^2 + n + 2}{2}$


quadratics
algebra
polynomial
invariant
function
combinatorics proposed
combinatorics

Is it possible to obtain the array (5,13,42) starting from (1,21,42) by repeating the following steps: It is allowed to swap any two numbers in an array of three numbers (a,b,c), or replace the array by (a,b,2a+2b-c)

There is a quadratic polynomial invariant under those operations:
[hide]
$a^2 + b^2 + c^2 - 2ab - 2bc - 2ca$ [/hide]

We want a quadratic function symmetric in $a,b,c$ whose roots, for fixed $a$ and $b$, are $c$ and $2a+2b - c$.  That is, it looks like $c^2 - (2a+2b)c$ plus terms without $c$, and this fixes the function by symmetry.


Provide a laboratory synthesis of catechol from phenol.

Method 1. a. Use Riemer Tiemann reaction to get CHO in ortho position .
               b. Use H2O2 and NaOH to get Catechol .

Method 2. a. Use fuming H2SO4 to get SO3H in ortho position .
               b. Fuse with NaOH and add H+ to get Catechol .

Method 3. Perform Elbs Persulphate reaction to get catechol as a by product .

P.S. Is any of these right .

[quote="Madness"]Method 3. Perform Elbs Persulphate reaction to get catechol as a by product.[/quote]

From the Elbs persulphate oxidation we only get the para-dihydroxy product.

[quote="Madness"]Method 2. a. Use fuming H2SO4 to get SO3H in ortho position.[/quote]

A specific temperature range is needed if we are to obtain the ortho product: otherwise, we can obtain the para product or a mixture of the two.

I think 2 name reactions are enough.

1.Duff reaction- Treating phenol with urotropine((CH2)6N4), glycerol and boric acid to give salicylaldehyde.

2.Dakin Reaction- salicylaldehyde on oxidation with alkaline hydrogen peroxide is converted into catechol.

C6H5-OH + (CH2)6N4 / H+ --> HO-C6H4-CHO (ORTHO) + H2O2 /OH- -->HO-C6H4-OH (CATECHOL)


geometry
geometric transformation
dilation

Ok I don't know about everyone else, but I find the USNCO Free Response q. 5 (the reactions problem) to be a lot of fun but rather difficult.  I also have a ton of prior problems for which I have no answer keys (http://dbhs.wvusd.k12.ca.us/webdocs/NChO/NChO-Menu.html).  So I figured if there is interest we could start a Question 5 of the Day thread (until next weekend's national exam), and anyone who wants to can post in their ideas or corrections, with the intention of getting practice for this question.  I'll give it a start with the 1997 reactions:

a) One mole of ammonia is bubbled into a liter of one molar sulfuric acid.
b) Excess sodium hydroxide solution is added to a solution of nickel(II) chloride.
c) Solutions of potassium permanganate and hydrogen peroxide are mixed in acid solution.
d) Chlorine water is added to a solution of sodium iodide.
e) Acetic acid is added to solid sodium hydrogen carbonate.
f) Excess concentrated ammonia is added to a solution of zinc nitrate.
g) Ethanol is heated with concentrated sulfuric acid.

[hide]a) $NH_{3 (g)} + H_3O_{(aq)}^+ \rightarrow NH_4^+ + H_2 0 _{(l)}$

b) $OH^-_{(aq)} + Ni^{2+}_{(aq)} \rightarrow Ni(OH)_{4 (aq)}^{2-}$

c) Umm... aren't both of these strong oxidizing agents??  I can't think of anything here, except maybe decomposition of the hydrogen peroxide.  And of course, I suppose it's possible there's a typo there.  Anyone who can get this one is officially amazing.

d) $Cl_{2 (aq)} + I^-_{(aq)} \rightarrow Cl^-_{(aq)}+ I_2_{(g)}$

e)  $CH_3 COOH _{(aq)}+ NaHCO_3_{(s)} \rightarrow CH_3 COO^-_{(aq)}+ Na^+ + H_2 0 _{(l)} + CO_2_{(g)}$

f) $NH_3_{(conc)} + Zn^{2+}_{(aq)} \rightarrow Zn(NH{}_3){}_4^{2+}_{(aq)}$ ... am not sure about that number of ligands, but I think it's right.  I know some things like to have just 2.

g) $CH_3 CH_2OH \rightarrow CH_3 CH{}_2OCH{}_2CH{}_3$
[/hide]

I am sooooo bad at reactions!
c) Solutions of potassium permanganate and hydrogen peroxide are mixed in acid solution.
Complete guess:
[hide]MnO4(-) + H2O2 --> Mn(2+) + H2O + O2[/hide]

I like that... it should probably use $H^+$ as a reactant though, to use up the $O^{2-}$ generated from the permaganate.

for part A, you might want to just write H+, since you can cancell the H2O from the RHS,

balanced "c"
[hide]
$5O_2^{2-}+16H^++2MnO_4^-\rightarrow 2Mn^{2+}+8H_2O+5O_2$[/hide]

amirhtlusa -- Thanks for the idea about H+.  That's probably a better idea; it just gets cumbersome.  I think your reaction is right, except in acid solution shouldn't $O_2^{2-}$ be protonated?

Ok here's day 2:  (1996)

a) A piece of calcium is added to water.
b) Concentrated hydrochloric acid is added to manganese(IV) oxide.
c) Solutions of lead acetate and dilute sulfuric acid are mixed.
d) Sodium cyanide is added to water.
e) Silver is added to dilute nitric acid.
f) Excess aqueous sodium hydroxide is added to aqueous aluminum nitrate.
g) A mixture of acetic acid and excess methanol is refluxed in the presence of a catalyst.

(I'll let someone else post their answers this time.)

b) (guess mainly)  $H^+$ + $MnO_2 \rightarrow$ $Mn^{2+}$ + $O_2$ + $H_2 O$ (since something has to be oxidized, and oxygen is more easily oxidized than chlorine I believe).

c)  Make sure to write Pb(C2H3O2)2 as Pb and C2H3O2-, and H2SO4 as H and HSO4- (or even SO4-, since pKa of hydrogen sulfate is very low).

e)  I think it might be redox... $Ag$ + $H^+$ + $NO_3^- \rightarrow$ $Ag^+$ + $H_2 O$ + $NO$  (and of course, don't forget that nitric acid is a strong acid and will be completely deprotonated!)... makes me wonder why it is "dilute" though

f) Make sure to write Al(NO3)3 as Al3+ (and then NO3- is just a spectator ion)

g)  Fischer esterification (whenever you see something with alcohols or acids, just get rid of an OH group from one, and the H off the other hydroxyl, and stick them together, and 9 times out of 10 you'll get it)  $CH_3COOH + HOCH_3 \rightarrow CH_3COOCH_3 + H_2O$

[quote="saturnlife"]Alright, here is my attempt...

a) Ca + H2O --> Ca(2+) + OH- + H2
b) H+ + MnO2 --> Mn (2+) + something
c) Pb(C2H3O2)2 + H2SO4 --> HC2H3O2 + PbSO4
d) NaCN + H2O --> HCN + Na(+) + OH-
e) Ag + HNO3 --> Ag(NO3)2 (-) + H+
f) OH- + Al(NO3)3 --> Al(OH)4 - + NO3 -
g) ugh, i can't remember..[/quote]

a) should only produce ${Ca(OH)_{2}}$, a metal + water= base, similarily a nonmetal + water=acid

also, (f) contains (in it's net ionic) only $OH^{-}+Al^{+2} \Longrightarrow Al(OH)_{2}$ I believe.

I assume we're only doing net ionic for practice of the AP Chem Exam

BTW, I NEVER KNEW AOPS HAD A CHEM SECTION :blush:

You're definitely right it should be $Ca(OH)_2$ (a solid) (nice catch)... however, since it's redox it has to have something reduced, and I think it will be $H^+ + e^-\rightarrow H_2$

[quote="gregx007"]Oh, and for the Al problem, be careful -- Al forms $Al^{3+}$ (not $Al^{2+}$), which I believe does indeed complex with $OH^-$ to form $Al(OH)_4^-$[/quote]

thanx for the catch, i didn't have a periodic table w/ me :)

Thanks for the suggestions! I always forget to write the salts in solution as ions  :blush:

And here (ok a couple hours in advance, but I won't be online until late tomorrow) is the third round:  (1995)

a) Concentrated phosphoric acid is added to solid sodium bromide.
b) Solutions of barium hydroxide and copper(II) sulfate are mixed.
c) Boron trichloride is bubbled through excess water.
d) Lead sulfide is burned in air.
e) A solution of potassium dichromate is added to an acidic solution of tin(II) chloride.
f) Excess aqueous ammonia is added to a solution of cobalt(II) chloride.
g) Solid aluminum nitrate is added to water.[/hide]

err...at first glance:

b. $Ba^{+2}+OH^{-}+Cu^{+2}+SO_{4}^{-2} \longrightarrow BaSO_{4}+Cu(OH)_{2}$

e. $Sb^{+2}+CrO_{7}^{-2} \longrightarrow SbCrO_{7}$

a. $H_{3}PO_{4}+NaBr \longrightarrow H_{2}O+Br_{2}+Na^{+}+PO_{4}^{3-}$

geez...these reactions are tough :blush: , well, atleast I can get the reactants to all of em :P

a) Concentrated phosphoric acid is added to solid sodium bromide.
b) Solutions of barium hydroxide and copper(II) sulfate are mixed.
c) Boron trichloride is bubbled through excess water.
d) Lead sulfide is burned in air.
e) A solution of potassium dichromate is added to an acidic solution of tin(II) chloride.
f) Excess aqueous ammonia is added to a solution of cobalt(II) chloride.
g) Solid aluminum nitrate is added to water.

a) NaBr --> Na+ + Br-
b) Ba (2+) + OH- + Cu (2+) + SO4 (2-) --> BaSO4 + Cu(OH)2
c. BCl3 + H2O --> I think it's lewis acid/base reaction, but I'm not too sure of reactants
d. PbS + O2 --> Pb + SO2
e. Cr2O7 (2-) + Sn (2+) + H+ --> Sn (4+) + Cr (3+) + H2O
f. NH3 + Co (2+) --> Co(NH3)4 (2+)
g. Al(NO3)3 + H2O --> Al(H2O)4 (3+) + NO3 (-)

Here's what I got:

a)  seems like there should be more... NaBr is soluble, so why go through the trouble of adding concentrated H3PO4 if its just dissolving?  (I think).  Here's what I put (not sure if it's right though)
H3PO4 + NaBr --> Na+ + H2PO4- + HBr(g)
Could it be redox?

b)  same

c)  BCl3 + H2O --> B(OH)3 + H+ + Cl-      (fairly certain, note they say excess)

d)  PbS + O2 --> PbO + SO2 (don't think lead is reduced here; O2 is more easily reduced I believe)

e)  same

f)  same (according to your ligand rule, it has four right?)

g)  Al(NO3)3 + H2O --> Al(H2O)6 (3+) + NO3 (-)  (99% certain that Al3+ has a coordination number of 6.  Not sure why.  Ligand field theory :) ?)

And the next day:

a) Hydrochloric acid is added to solid zinc sulfide.
b) Solutions of barium chloride and phosphoric acid are mixed.
c) Phosphorus is burned in air.
d) An acidic solution of potassium permanganate is added to one of the iron(II) sulfate.
e) Solid sodium acetate is added to water.
f) A solution of hydrogen peroxide is warmed gently.
g) Excess concentrated ammonia is added to a copper(II) chloride solution.

You're right, I was thinking of Al(OH)4 - when doing the equation. Although Al(OH)6 3- is also acceptable as a product in a reaction, I think.

Good to know.  Ok here's what I got:

[hide]a)  $H^+ + ZnS \rightarrow H_2 S_{(g)} + Zn^{2+}$  (would there be complexing here?  Probably not, that'd be too complicated, but I'm not sure)

b)  $Ba^{2+} + H_3 PO_4 \rightarrow Ba_3 (PO{}_4){}_2 + H^+$ (not sure about that... it'd be nice to have some kind of redox... does phosphoric acid participate in redox reactions?)

c) $P_4 + O_2 \rightarrow P_4 O_{10}$

d)  $MnO_4^- + H^+ + Fe^{2+} \rightarrow Mn^{2+} + H_2 O + Fe^{3+}$

e) $NaOOCCH_3 \rightarrow Na^+ + CH_3COO^-$

f)  $H_2 O_2 \rightarrow H_2 O + O_2_{(g)}$

g) $NH_3 + Cu^{2+} \rightarrow Cu(NH{}_3){}_4^{2+}$[/hide]

Well I had all my answers typed up, and then pushed refresh:(...     Here are some more:

(A) Sulfuric acid is added to a sodium nitrite solution.
(B) A solution of potassium dichromate is added to a solution of sulfurous acid.
(C) Solid sodium phosphate is added to water.
(D) A solution of concentrated aqueous ammonia is added to solid silver chloride.
(E) Copper metal is added to concentrated nitric acid.
(F) Solutions of barium chloride and potassium carbonate are mixed.
(G) Solutions of iron(III) sulfate and potassium iodide are mixed.
(H) Strontium-90 decays by the loss of a beta particle.

Enjoy.


Let p be a prime, k \geq 2 and n \in N. Determine all solutions of the equation 2p + 1 = n^k.

created by Ralf Kleinschmidt

I think this problem is not yet solved.
The case that needs considering is 2p = 3^k - 1, where k is odd prime, which divides p-1.
for example one solution is p = 13; k = 3.

I think it's not a problem for beginners, as the equation, which the last guest pointed out, seems to be very hard. maybe someone can come up with a solution ?

OK, as far as I know the problem is not solved yet or does anybody have different knowledge ?

 1.) n = 1 => p = 0: contradiction; for even n => nk  is even and 2p+1 is odd; hence n \geq 3 must be odd; for p = 2 => 2p + 1 = 5 = nk  for k \geq 2 is not valid => therefore p is an odd prime; from the decomposition 2p = nk -1 = (n-1)* \sumk-1 i = 0  ni  follows with n \geq 3:

a.) n - 1 = 2 => n = 3 or
b.) n - 1 = p => due to n odd is p even: contradiction or
c.) n - 1 = 2p => sum = 1; hence k = 1: contradiction => hence n = 3.

2.) suppose k is not prime, e.g. k = rs with r,s \in N, r,s > 1; then
2p = 3k  - 1 = 3rs  - 1 = (3r ) s  - 1 = (3r -1)* \sum s-1 i = 0     3ri  with 3r  - 1 \geq 8 even, which means p has non-trivial divisors: contradiction

Therefore the original equation can have solutions at most in the case if k is prime. (e.g. for 2,5,11,17,19 1/2*(3^k-1) is no prime but for k = 3,7,13).

So the partial result: 2p + 1 = nk  with p prime; k \geq 2 is solvable at most if n = 3 and p,k odd primes; but it is not solvable for odd primes k. Does anybody know whether there are finitely or infinitely many solutions ?

PS: Why don't the guest do not register or if registered yet then log in please.  :)


Does anyone play connect 5 (gomoku)?

I mean ya the rules are pretty simple so you guys may not think it takes much strategy, but if you're really good at it, it is a pretty fun strategy game, and it doesn't take hours like chess, or a specialized board and game pieces.

Yeah, I did use to play connect 5.
Unfortunately, it seemed to me that the first player has a substantial advantage.

So now I mostly play Connect6, sometimes connect 5 agains classmates.


Let $f: E\to E$ be an isometry with $E$ complete and $f(E)$ totally bounded. Prove that $f$ is a bijection and that $E$ is compact.
[hide]Injection is trivial. I can think of two approaches:1.(the way I tried!) First we show that $f$ is surjective. Then according to a theorem $E$ being complete and totally bounded, will be compact.
2.We prove $E$ is compact and according to a problem on the forum $f$ will be surjective.[/hide]

(2) makes more sense. Since $f(E)$ is totally bounded, so is $E$, because $f$ is an isometry. Hence $E$ is compact.


Is it possible to list the non-negative integers as $n_1$, $n_2$, $n_3$, ..., with each non-negative integer appearing precisely once in the list, in such a way that for each $i = 1$, 2, 3, ..., we have $n_{i + 1} = n_i + 2$, $n_{i + 1} = n_i - 2$, $n_{i + 1} = 2n_i$, or $n_{i + 1} = n_i/2$?

to start with note that from $0$ one has only 1 option; $0\rightarrow 2$;likewise from 2 there are only 2 options; $2\rightarrow 1,2\rightarrow 0$,so the following is forced on us: $a_0=0,a_1=2,a_2=1,a_3=3$.
now we make the following observation: from $2n+1$ the following sequence of 'moves' is possible:
 \[2n+1\rightarrow 4n+2\rightarrow 4n\rightarrow 4n-2\rightarrow\cdots\rightarrow 2n+2\rightarrow 4n+4\rightarrow 4n+6\rightarrow 2n+3\rightarrow 2n+5\rightarrow\cdots\rightarrow 4n+5\rightarrow 4n+7\] which implies that if we have a sequence containing all the integers $\leq 2n+1$,with $2n+1$ as the lastin sequence, we can extend it to contain all the integers till $4n+7$, which is again odd and so this can be extended further again.
since we start out having $0,1,2,3$ as the first 4 terms with $a_3=3$, it follows that we can get such a sequence.