geometry topology function algebra polynomial ratio calculus

What do you guys think would be an interesting (competition or non-competition) math topic to self-study during the summer? Something fun and not so dense and difficult that I'd get very easily frustrated (although it's good if it's challenging)? I've taken math up to multivariable calculus, plus linear algebra and number theory.

Diff Eq is fun.  I personally have been meaning to getting around to learning basic abstract algebra, you could try that.

And, if all else fails, solve the Riemann Hypothesis.

[quote="Magnara"]I've found the perfect compromise.  Differential Geometry![/quote]

Differential geometry is fun, but it is heavy in computation, at least at first.

Why do people care about proofs so much? You can't really "prove" anything.

We can prove things given basic axioms. Sure, we can't prove the axioms. But they're interesting nonetheless, and the properties we derive from them are interesting and useful.

I'm not sure what you mean -- within the structure of mathematics, you can prove quite a lot.

[quote="nr1337"][quote="Magnara"]And, if all else fails, solve the Riemann Hypothesis.[/quote]

I agree with Magnara.  Sadly, it might be a bit difficult to solve one of the Millenium Problems.[/quote]

Meh, they can't be [i]that[/i] hard.  I'm sure if I tried hard enough I could do it.

:-P

What does it mean to try hard enough? If you mean that you will first learn for 25 years and then work for 10 on nothing else, then you would have a small chance of solving one of the problems. If you mean that after a month of getting nowhere you'll give up, you won't solve one.

I definitely believe that all proofs can be written down in terms of the axioms. If it can't be done, it isn't a proof. If you have a proof, then what you have proven is clearly true. If you have a wrong proof, then, well that isn't a proof. An argument that convinces yourself is nothing like a proof.

[quote="mathnovice"]Assuming that the Riemann Hypothesis and all the other unsolved math problems were proved, what difference would that make in the world? They don't generalize anything for us which might allow us to solve numerical problems. I can see the value of proving, say, the Pythagorean Theorem, but what good will deciding P v. NP do?[/quote]

Proving the Riemann Hypothesis would help in finding properties of the distribution of primes on the number line.

[quote="mathnovice"][quote="ComplexZeta"]If you can't prove something, how do you know it is true?[/quote]

Even if you can prove something, how do you know it is true? You may have convinced yourself that it is true, but this doesn't necessarily make it true. Also, even if the axioms were true, most proofs cannot be written in terms of the axioms. Can all proofs be written without words, i.e. only with symbols and mathematical notation? I don't think so.[/quote]

To prove is too mathematically prove it, something proven is true within the mathematical world. And by the way most proofs have plenty of words in them.

[quote="mathnovice"]Can all proofs be written without words, i.e. only with symbols and mathematical notation? I don't think so.[/quote]

Well, you're wrong.  Every single mathematical proof tool can be written in purely symbolic terms.  However, it is an enourmously long and difficult procedure to do.  People don't write math like that for the same reasons that people don't write computer programs in 1's and 0's -- they know it is possible, and they know everything they do is based on the principles of that realm, but you get just as valid results if you do it the other way, and it's a lot easier and more fun.  Mathematics, at its core, is entirely axiomatized -- everything we can prove can be expressed purely in terms of symbols and the axioms of the system we are working in.  (Go read [u]Godel's Proof[/u] if you'd like to know more about this.  It's quite interesting, although somewhat rough going in places.  It talks all about the limitations of the axiomatic system.)

As for whether or not the axioms are true, that is not a mathematicians problem.  Every theorem of math has the implicit sentence, "Assuming the following axioms, [u]insert axioms here[/u], we have the following result, [u]insert theorem here[/u]."  The proof of the theorem then demonstrates the validity of that statement.  The truth of the axioms is really a problem for physicists -- they take the empirical data that they have and try to use it to figure out what the "fundamental nature of the universe" is.  What that means, give or take, is to figure out what axioms are true.  Then, they use mathematics to figure out what the results would have to be.  Experimental data contradicting some of these theoretical results indicates that there is something wrong with what they thought were the axioms.  And so on.  None of this really concerns mathematicians, however -- their work is all done assuming the validity of axioms.  Whether these axioms are true or not doesn't matter to them, at least not in a professional sense.

Finally, you've picked a bad bunch of problems to attack as unimportant in the Millenium Problems.  The Navier-Stokes Equations, for example, arise out of trying to solve the physical problem of fluid flow.  Solving those equations would translate almost immediately into improvements in the aerodynamics of every single product we use.  The most obvious example of this being useful is in the airplane industry.  Airplanes would almost instantaneously become faster, more fuel efficient, and require lower takeoff speeds.  It would pretty much do away with the need for air-tunnel experimentation.  All in all, it would save billions of dollars of unnecessary costs.  P vs. NP tells you something significant about the limitations of computers.  If P = NP, you immediately know that the potential exists for efficient solutions to a wide variety of problems.  If P [tex]\neq[/tex] NP, we know that computers are inherently limited.  Yang Mills theory will help in figuring out, as I discussed above, exactly which axioms really apply in our world by giving a solid mathematical basis to something that works for reasons that people don't understand.

[quote="mathnovice"]Also, even if the axioms were true, most proofs cannot be written in terms of the axioms. Can all proofs be written without words, i.e. only with symbols and mathematical notation? I don't think so.[/quote]

Most proofs are written with "according to so and so's theorem concerning blah-dee-blah," and surely the proof of that theorem had an equivelent statement, and so on and so forth, but, if you look far enough, they all made reference to basic theorems that were proven axiomatically.  During the...late 19th century?...a bunch of logicians went crazy and decided that everything really did have to be proven axiomatically (not with all of the axioms in the publishing, but with reference to theorems that were proven axiomatically).  And ever since, everything in math is proven with the utmost rigor.
1+1 is 2, because it has been proven.  And, with our axioms, it will always be two.  There will never, ever be a situation where 1+1=3.  Or even 2.1.  Or even 2.01.  :rt2: and :pi: have been proven to be irrational, so no one will ever find two whole numbers whose ratio is equal to either of those.

We should bring inconsistency into this discussion.

Well, as for math self-study, this is what I will be doing this summer:

1. Learning to Tex
2. Reviewing multivariable calc and linear algebra
3. Frontloading for my analysis course next year
4. Working through Larson's Problem Solving through Problems book

[quote="mathnovice"]Does anyone have tips on HOW to self-study? I tend to be extremely inefficient at self-study, especially math and science. This is especially true when it comes to proofs, which I usually have to read about three to five times before I understand it. However, this process takes about 10 to 15 minutes for me because I have to take out a sheet of paper and try to write things down and then try to visualize the problem and sometimes look things up in other books. So, I usually end up, after a couple hours of studying, realizing that I have gotten practically nowhere. This also true when I am working on a problem books; 3 hours = 3 pages. This is so annoying...[/quote]

This is how I self-study new topics.

I read and memorize the theorems and some introductory material on the topic.
Then I try to prove the theorems myself.  If this fails, I look up the proof of the theorem and do not proceed until I understand it completely.  I find that this gives me an insight into the actual subject material.
Next, I read some solutions to problems involving the topic at hand.  This gives me more insight as to how the topic is actually applied to problems.
Next, I do all the problems I can.

I have two methods I use for studying mathematics. They work for me, but they might not work for other people.

Method 1: Read the book aloud and don't write anything down. If you have trouble understanding something, try to get a geometric picture of what is going on. Don't go on until you can visualize everything and you can see [i]why[/i] the results are true. (Understanding a proof is not the same as understanding why a result is true.) This method works well for geometry-like topics.

Method 2: Go through all the results very meticulously, writing down every step in the book. If something is non-obvious, work out all the intemediate steps until everything becomes obvious to you. This works better when there isn't very much to visualize (problems dealing with polynomials, for example).

I agree with both of the two above suggestions. To those I would add, if possible try to read the same thing in two (or more) different books. Seeing the same thing explained in two different ways can often be very helpful.

Also, don't be in a rush. Reading mathematics takes time. So don't expect to master a whole subject in a week, or even a month or two. Take your time to get to know the subject and to absorb it.

[quote="gauss202"]Also, don't be in a rush. Reading mathematics takes time. So don't expect to master a whole subject in a week, or even a month or two. Take your time to get to know the subject and to absorb it.[/quote]

Over Spring Break, I read all of AoPS Volume One and did over half of the problems.  :lol:

I forgot to mention my method for learning problem-solving. I usually went through a book once very quickly, only solving a few problems on each topic, and then after I had done that, I went through the book more slowly and carefully, solving many problems. I think it is nice to get a feel for many different ideas before focusing on them in-depth.

I'm puzzled as to how everyone says "take your time" or something of the sort, but some people here still manage to study for an entire AP exam in one month or less! How do you guys do it?!

Studying for an exam is different than studying to learn something for fun.  If you're doing the second, you (a) get to go about it in a much more relaxed way and (b) should spend more time to appreciate fully the beauty of what you're learning.

I don't think anyone has ever applied the word "beauty" to AP economics ;)

[quote="ComplexZeta"]I plan to study topology, elliptic curves, and modular functions over the summer.[/quote]

How much longer until you read the proof of Taniyama-Shimura? ;-)

I probably won't get to that until I'm in graduate school at the earliest.


limit
quadratics
algebra
polynomial
calculus
calculus computations

this is something that i am not sure if it is true, but it seems logical,

consider a sequence $a_n$ that converges as $n \rightarrow \infty$, and is defined by the recursion $f(a_n)=a_{n+1}$
just a few restirictions, because this was inspired by a real world problem

$f: \mathbb{R} \rightarrow \mathbb{R}$ and $a_0 \in \mathbb{R}$

let $x$ be the only (real) solution to $f(x)=x$

prove (or disprove, i am not sure) that $\lim_{n \rightarrow \infty} a_n = x$

i don't know if there are any other restrictions, but this relates to modeling frequencies in biology, so there are all real numbers,

if the sequence converges, then $a_{n+1}\rightarrow a_n$ as $n\rightarrow \infty$

then $f(a_n)\rightarrow a_n\Rightarrow a_n\rightarrow x$.

The only problem is, need to show that $a_n$ converges first.

This doesn't always happen (try to construct counterexamples; think of quadratic polynomials $f(x)$ such that $f(x)-x$ is a perfect square...)

As any numerical analyst will tell you, (assuming $f$ is differentiable), what you should be particularly concerned about is whether $|f'(x)|$ is greater or less than 1 at that fixed point.

The convergence need some conditions as Kent quoted. The result known as fixed point theorem.


quadratics
complex numbers
algebra
system of equations
polynomial
Rational Root Theorem

Solve the following system for real $x,y,z$
\[ \{ \begin{array}{ccc} x+ y -z & =& 4 \\ x^2 - y^2 + z^2 & = & -4 \\ xyz & =& 6. \end{array} \]

y = 4+z-x from 1

x^2 + z^2 +4 =  (16 + z^2 + x^2 + 8z - 8x -2zx) from 2

implying $0 = (6 + 4z - 4x -zx)$ or $6 = 4x+zx-4z$ (*)

Also, $xz(4+z-x) = 6$

So $(6-4x+4z)(4+z-x) = 6$

Let k = z-x.  Then $(3+2k)(4+k) = 3$ is a quadratic in k, with 2 solutions.  From (*) we determine zx, and together with k we can determine z and x.  From this we determine y, then we have to check that the (x,y,z) things work.  [I am too lazy to do it.]

we have:
x+y-z=4 so(x-y)^2=(4-y)^2 so (x^2+y^2)-2xy=16+y^2-8y (1)
and we have also x^2+z^2=-4+y^2 and xz=6/y.
so (1) become : y^2-4-12/y=16+y^2-8y
equivalent to:y=3 or y=-1/2
1/ y=3:
so z=1 and x=2
2/ y=-1/2:
it impliques that x^2+y^2=-17/4 (it's impossible)
the only one couple to this system is (x,y,z)=(2,3,1)

Re-arrange equation 2 to get

$(x - z)^2 + 2xz = y^2 - 4$

Substituting from equations 1 and 3, we get

$(4 - y)^2 + 12/y = y^2 - 4$

which simplifies to

$16 - 8y + 12/y = -4$.

Multiplying by $y$ and solving the resulting quadratic yields $y = 3, - 1/2$.

Now we get from the third equation $z = 2/x, -12/x$ for the first and second values of $y$, respectively.  Plugging these two solutions into the first equation, multiplying by $x$, and solving the resulting quadratics leads to $x = -1, 2$ for the first solution, and complex numbers for the second solution, which must be discarded.  Hence, we have the two solutions $(x,y,z) = (-1, 3, -2)$ and $(x,y,z) = (2,3,1)$.

Note that it is valid to multiply any equation by one of the variables, because equation 3 implies that all variables are nonzero.

Also, I think that the solution on the official problems page should be a complete solution - it looks sloppy to have a solution that's not finished (esp. since it doesn't give the answer!).

Please correct me if I'm wrong(or tell me if i'm right),can we put x,y,z as roots of a cubic equation with coefficients as the original system of equations and then use the rational root theorem?

Here is my solution
It's obvious that ; $x,y,z,$ are not $0$
From (2) $x^2 + z^2 = y^2 -4$
From (1) $x^2 + z^2 = 2xz +16 -8y+y^2$
So $-4 = 2xz + 16 -8y$
   And we get that ;
 $y = 3 or -\frac{1}{2} (by use 2xz = \frac{12}{y})$

[hide=my proof]we have $x-z=4-y$\\
$x^2+z^2=y^2-4\\
xz=\frac{6}{y}$\\
notice $(x-z)^2=(4-y)^2\implies y^2-4-\frac{12}{y}=16+y^2-8y$\\
which gives $2y-\frac{3}{y}=5\implies y=3$ or $\frac{-1}{2}$\\
we get $x=1+z$ or $\frac{9}{2}+z$ and $xz=2,-12$\\
notice case of $x=\frac{9}{2}+z$ don't works as upon solving it don't gives a real solution for $z$\\
so solving for case of $x=1+z$ we get possible $x,y,z$ as $(2,3,-1)$ or $(1,3,-2)$ so\\
$\boxed {\ (x,y,z)\in \left\{ \ (\ 2,\ 3,\ 1\ ),\ (\ -1,\ 3,\ -2\ )\ \right \}\ }.$ $\blacksquare$[/hide]


[hide=" What have I missed?"]

$ 4p\equal{}5q$

$ 2p\equal{} \frac{5q}{2}$

$ 2p\plus{}q\equal{} \frac{7q}{2}$

[/hide]

hello, from $ \frac{p}{q}\equal{}\frac{5}{4}$ we get
$ p\equal{}\frac{5}{4}q$
plugging this in the term above we get
$ 2\frac{5}{4}q\plus{}q\equal{}\frac{7}{2}q$.
Sonnhard.


[quote="jeanvaljean"]\documentclass{article}
\usepackage{latexsym}

\begin{document}

Va rog sa ma ajutati cu solutia la problema 3/pag 283, culegere Ganga, clasa
a xi-a:

Fie $f: [0,1]\to{\rm R},\;f(x)=\left\{{\begin{array}{l}0,\;x\in{\rm R}-{\rm Q}\\ \frac{1}{q},\;\mbox{daca}\;\mbox{x=}\frac{\mbox{p}}{\mbox{q}},\;(p,q)=1,\;p,q\in{\rm N}^\ast \\ \end{array}}\right.$. Sa se arate ca $f(x)$ are limita zero \^{\i}n orice punct
$a\in [0,1]$.

Va multumesc foarte mult.

\end{document}[/quote]

Nu cred ca este adevarat. $x_{n}=\frac{n}{2n+1}, lim_{x\to \frac{1}{2}}f(x)=lim_{n\to \infty}f(x_{n})= \frac{1}{2}$, daca $x_{n}=\frac{n\sqrt{2}}{2n\sqrt{2}+1},lim_{x\to \frac{1}{2}}f(x)=lim_{n\to \infty}f(x_{n})=0$

Enunţul este corect. Fie $x_{n}=\frac{p_{n}}{q_{n}}$ cu $(p_{n},q_{n})=1$ un şir cu limita $a\in [0,1]$ astfel ca $x_{n}\ne a, \forall n\in \mathbb{N}.$ Atunci rezultă că $p_{n}\rightarrow+\infty,q_{n}\rightarrow+\infty,$ şi deci $f(x_{n})=\frac{1}{q_{n}}\rightarrow 0.$
Dacă $x_{n}\rightarrow a,$ cu $x_{n}\in \mathbb{R}-\mathbb{Q},$, evident $f(x_{n})\rightarrow 0.$
P.S. Faptul că $p_{n}\rightarrow+\infty,q_{n}\rightarrow+\infty,$ rezultă din următoarea lemă: un şir de numere naturale care nu are limita $+\infty$ conţine un subşir constant.

O problemă legată de asta, propusă de Radu Gologan la finala din 1978:
Să se arate că funcţia $f: \mathbb{R}\rightarrow\mathbb{R},$
\[f(x)=\left\{\begin{array}[c]{l}0,x\in\mathbb{R}-\mathbb{Q}\\ \frac{1}{q^{3}},x=\frac{p}{q},p\in\mathbb{Z},q\in\mathbb{N},\gcd(p,q)=1 \end{array}\right. \]
este derivabilă în punctele iraţionale de forma $\sqrt{k}$, cu $k\in\mathbb{N}.$

si cum se face?  :wink: eh, de fapt... cu ideile de dinainte nu mai e chiar asa de greu

Deci, daca $p_{n}$ nu ar avea limita infinit, ar contine un subsir constant $p{}_{n}{}_{k}$. Se ia subsirul $\frac{p{}_{n}{}_{k}}{q{}_{n}{}_{k}}$ care trebuie sa tinda la $a$.

Mai departe nu stiu....

Dacă $(q_{n})$ nu are limita infinită, atunci conţine un subşir constant $(q_{k_{n}}),$ şi fie $q_{k_{n}}=q.$ Atunci şirul $\frac{p_{k_{n}}}{q_{k_{n}}}=\frac{p_{k_{n}}}{q}$ are limita $a$ doar dacă $p_{k_{n}}$ e constant de la un anumit rang, dar atunci $\frac{p_{k_{n}}}{q_{k_{n}}}=a$ pentru orice n de la un anumit rang, ceea ce contrazice ipoteza ($x_{n}\ne a$).

Bine ca exista acest forum si acesti oameni minunati!


A square piece of paper is folded in half vertically. The resulting figure has a perimeter of 21 inches. What is the number of square inches in the area of the original square?

The resulting paper's width is half of the original width, but the length stays the same. Call the length $ x$.

$ 0.5x \plus{} 0.5x \plus{} x \plus{} x \equal{} 21$

$ 3x \equal{} 21$

$ x \equal{} 7$

Since it's a square, our answer is $ \boxed{49}$


geometry
geometric transformation
reflection
trigonometry
\/closed

Mathematics and problem solving has a wide range of applications.  For discussion of these, we have created a set of forums for discussion of [url=http://www.artofproblemsolving.com/Forum/index.php?f=328]various sciences and arts[/url].  Specifically, we now have a forum for each of the following topics:

[url=http://www.artofproblemsolving.com/Forum/index.php?f=329]Biology[/url]
[url=http://www.artofproblemsolving.com/Forum/index.php?f=330]Chemistry[/url]
[url=http://www.artofproblemsolving.com/Forum/index.php?f=331]Computer Science and Informatics[/url]
[url=http://www.artofproblemsolving.com/Forum/index.php?f=332]Physics[/url]
[url=http://www.artofproblemsolving.com/Forum/index.php?f=333]Economics[/url]
[url=http://www.artofproblemsolving.com/Forum/index.php?f=334]Graphical Design[/url]
[url=http://www.artofproblemsolving.com/Forum/index.php?f=335]Music & Art[/url]

Please note that the Music & Art forum is intended for the musicians and artists among us to discuss techniques and their work.  If you want to talk about your favorite band, please do it in [url=http://www.artofproblemsolving.com/Forum/index.php?f=139]Games & Fun Factory[/url].

Thank you!  This really helps a lot since now I don't have to sift through "Other US Contests & Programs" or "International Contests & Programs" just to find the threads about computer science and informatics.  I hope that these new forums will get a lot of use.  Also, the range of topics in the forums outside of the art & sciences forums should become more focused.  Great job!  :)

Yes that is exactly what we intended. Have fun :D

w00t w00t

( now where does graph theory fit? Combinatorics or informatics? ) :D

Great idea.

I assume each level will be open to questions of all levels?

I'm going to be starting university this september, so will college level chem/bio/physics questions be ok?

Or is it geared more towards the middle school/high school area?

Either way, good stuff.

For now all levels. It would be nice if the author of the thread would specify which level is the topic addressing in the subject.

[quote="manuel"]Valentin do you think it should be a good idea to add a Philosophy forum?? or no people would be intersted in that..??[/quote]I think the Round Table is good enough for Philosophy forum ;)

While we are talking about new forums, I think an area for formulas would be good.  It could be moderated by someone who would make stickys with all the formulas for each subject (Algebra, Geometry, Trig, etc.) that had been posted.  Or do we already have an area like that?  I couldn't find one.

There is no need for such an area for "pre-olympiad" since all the forumulaes can be easily googled. In plus there are already such stickies in the high-school forums. In the Olympiad and College forums there is, under each subject, a "Theorems and Formulaes" subforum.

can we post chem and physics and biology problems there??

[quote="rrusczyk"]Mathematics and problem solving has a wide range of applications.  For discussion of these, we have created a set of forums for discussion of [url=http://www.artofproblemsolving.com/Forum/index.php?f=328]various sciences and arts[/url].  Specifically, we now have a forum for each of the following topics:

[url=http://www.artofproblemsolving.com/Forum/index.php?f=329]Biology[/url]
[url=http://www.artofproblemsolving.com/Forum/index.php?f=330]Chemistry[/url]
[url=http://www.artofproblemsolving.com/Forum/index.php?f=331]Computer Science and Informatics[/url]
[url=http://www.artofproblemsolving.com/Forum/index.php?f=332]Physics[/url]
[url=http://www.artofproblemsolving.com/Forum/index.php?f=333]Economics[/url]
[url=http://www.artofproblemsolving.com/Forum/index.php?f=334]Graphical Design[/url]
[url=http://www.artofproblemsolving.com/Forum/index.php?f=335]Music & Art[/url]
[/quote]

This is excellent news, rrusczyk! AoPs is definitely growing. :) Who knows, maybe someday AoPS will be the largest (and best) online intellectual community!

[quote="bubala"]Who is going to moderate the Arts and Sciences Forums?[/quote]The moderators of those forums of course :D
If you are asking who will they be, probably we'll gather some of the current mods, and maybe appoint one or two new ones from the old members of the site. Although this is not of big importance at this very moment.

[quote="shobber"][quote="bubala"]Who is going to moderate the Arts and Sciences Forums?[/quote]
i think Yetti can be the moderator of Physics Forum. :D[/quote]

We welcome highly experienced and interested potential moderators to PM us.  We will certainly need a few.

Does discussion in the Chemistry forum have to be math-related, or can we go into more theoretical, non-mathematical discussion? Basically, is it meant to be about Chemistry in its own right, or Chemistry insofar as it relates to Mathematics?

Discussion does not have to be math-related in any of these forums (though we would like the discussion in Art & Music to be about creation and performance - talk about your favorite bands in Fun & Games).

[quote="nr1337"]Now is Fractal Art to be put in a math forum or an arts+sciences forum?[/quote]Yes, I would go for the Graphical Design forum to discuss about fractals.

[quote="mysmartmouth"][quote="bubala"]Isn't it already?

I wouldn't know since I haven't heard of any others...[/quote]

AoPS is the best math forum by far, but there are better, larger forums for other subjects.[/quote]

Would it be considered impolite to ask for some specific forums that are in the same league with AoPS in subjects like Computer Science and Chemistry?

If it is impolite, pretend I never asked  :blush:

Not at all impolite - part of the purpose of thtis forum is to share information,

However do not discuss it in [b]this[/b] thread. Do it in the specific forum.

Hey Valentin, while your creating new forums... perhaps you may check this thread of yours again :)

http://www.mathlinks.ro/Forum/viewtopic.php?t=5771

[quote="Valentin Vornicu"]Ok I will reflect on that.[/quote]

I.Q. i can be moder of it
u have all without it i think in this forum
exuse me if i am not write

[quote="DON FAVOR"]exuse me if i am not write[/quote]I think one of the main needs of a moderator is to have good communication skills ;)

Ok enough off topic. Everyone knows the news, it's time to end this :)


floor function
geometry
rectangle
induction
number theory proposed
number theory

prove that $\gcd(x,y) = 2 \sum_{k=1}^{x-1} \left [ \frac{ky}{x} \right ] + x + y - xy.$ for all integers $x,y$

Consider all lattice points $(i,j)$ with $0 \leq i \leq x$ and $0\leq j \leq y$.
Then $\gcd(x,y)+1$ is the number of such points on the line through $(0,0)$ and $(x,y)$.
Now $x+y+1+\sum_{k=1}^{x-1} (\left\lfloor \frac{ky}{x} \right\rfloor ) = \sum_{k=0}^{x} (\left\lfloor \frac{ky}{x} \right\rfloor +1)$ is the number of such points on or strictly under that line, so (since it's symmetric) $2x+2y+2+2\sum_{k=1}^{x-1} \left\lfloor \frac{ky}{x} \right\rfloor$ is the number of such points on that line added to the total number of points.
Since there is a total number of $(x+1)(y+1)$ points in that rectangle, we get that $\gcd(x,y)=2x+2y+2+2\sum_{k=1}^{x-1} \left\lfloor \frac{ky}{x} \right\rfloor - (x+1)(y+1)-1=2\sum_{k=1}^{x-1} \left\lfloor \frac{ky}{x} \right\rfloor +x+y-xy$ is already the result.

Yep! This is my solution. Induction on $m+n$.

If $m+n=2$ then $m=n=1$ , both equal $1$.

Suppose it true for all $m+n\in\{1,2,...,k\}(k\geq 1)$.

Now, if $m+n=k+1$, write $n=mq+r$ with $0\leq r<m$. Cases follow occur:

i)If $r=0$ then $(n,m)=m$ and $2\sum_{k=0}^{m-1}[\frac{kn}{m}]+m+n-mn=2q(1+2+...+m-1)+m+n-mn=qm(m-1)+m+n-mn=n(m-1)+m+n-mn=m$.

ii)If $r>0$ then $2\sum_{k=0}^{m-1}[\frac{kn}{m}]+m+n-mn=2\sum_{k=0}^{m-1}[\frac{kmq+kr}{m}]+m+n-mn=2q(1+2+...+m-1)+2\sum_{k=0}^{m-1}[\frac{kr}{m}]+m+n-mn=qm(m-1)+m+n-mn+(r,m)-m-r+mr=(n-r)(m-1)+m+n-mn+(r,m)-m-r+mr=(r,m)=(m,n)$. (Because $r+m<m+n$ and $r,m>0$).

QED.

[Comment by $ \zeta X$: I moved the solution in the thread mentioned below into this thread (see the post before)]

Nice solution! There are two other solutions, one of them I posted here
http://www.mathlinks.ro/Forum/viewtopic.php?t=129155

Now, I will post 2nd solution:

[b]Problem[/b]. Show that for positive integers $ m,n$ we have
$ \gcd{(m,n)} \equal{} 2\sum_{k \equal{} 0}^{m \minus{} 1}[\frac {kn}{m}] \plus{} m \plus{} n \minus{} mn$.

[b]Solution.[/b] Let $ g \equal{} \gcd{(m,n)}$ and $ m \equal{} m'g,n \equal{} n'g,\gcd{(m',n')} \equal{} 1$. For any $ k \equal{} 1,2,...,m \minus{} 1$ we have $ m|kn$ iff $ m'|k$, which happents $ [\frac {m \minus{} 1}{m}] \equal{} g \minus{} 1$ times. Note that $ [\frac {kn}{m}] \plus{} [n \minus{} \frac {kn}{m}] \equal{} n \minus{} 1$ if $ m\not | kn$ and equal $ n$ if $ m|kn$.

So $ \sum_{k \equal{} 1}^{m \minus{} 1}([\frac {kn}{m}] \plus{} [n \minus{} \frac {kn}{m}]) \equal{} mn \minus{} m \minus{} n \plus{} g$.

Hence $ 2\sum_{k \equal{} 0}^{m \minus{} 1}[\frac {kn}{m}] \equal{} 2\sum_{k \equal{} 1}^{m \minus{} 1}[\frac {kn}{m}] \equal{} \sum_{k \equal{} 1}^{m \minus{} 1}[\frac {kn}{m}] \plus{} \sum_{k \equal{} 1}^{m \minus{} 1}[\frac {(m \minus{} k)n}{m}] \equal{} \sum_{k \equal{} 1}^{m \minus{} 1}([\frac {kn}{m}] \plus{} [n \minus{} \frac {kn}{m}]) \\
\equal{} mn \minus{} m \minus{} n \plus{} g$. QED

[b]References[/b]
Titu Andreescu and Zuming Feng,Mathematical Olympiads 1998-1999 Problems and solutions from around the world.


The sum of two positive integers is $2310$. Show that their product is not divisible $2310$

[hide]
Suppose that there are two positive integers $x_{1},x_{2}$ which are defined as follows:
$x_{1}+x_{2}=2310$
$x_{1} \cdot x_{2}=2310k$ while k is also a positive integer.

$x_{1}$ and $x_{2}$  must be solutions of  the square equation $x^2-2310x+2310k=0$.
So we obtain $x_{1/2}=1155 \pm \sqrt{1155^2-1155 \cdot 2k}$
As $x_{1}$ and $x_{2}$ are positive integers $1155^2-1155 \cdot 2k=1155(1155-2k)$ has to be a square.
So $1155-2k$ has to be of the form $3 \cdot 5 \cdot 7 \cdot 11 \cdot i^2$, where $i$ is a positive integer.
So we obtain $1155-2k \geq 1155$ CONTRADICTON!!!!
[/hide]

This problem has been discussed in
http://www.mathlinks.ro/Forum/viewtopic.php?t=26298


I can't log into the classroom and I'm confused why.  Please help.

Yes. I did some logging in and out-and it ended up working, yay!


linear algebra
matrix
invariant
linear algebra unsolved

How do I show that for any matrix $A \in \mathbb{C}^{n \times n}$, there exists a matrix $P \in \textrm{GL}_n{\mathbb{C}}$ such that $A^t = P^{-1}AP$? That is, how do I show that a complex matrix and its transpose are similar?

you just have to prove it for Jordan blocks. And it is not that hard : just change a little bit the basis ...

It is true for arbitrary fields.(proof with invariant subspace for example)


Prove that no matter how we choose 5 points in the plane (no three collinear), we can pick 4 of them to form a convex quadrilateral.

[hide]ANy 3 noncollinear points makes a triangle.  Extend the three lines of the triangle.  Four points define a convex polygon when the points are in regions B,D,E.  Therefore, when we add the fifth point, we don't need to consider that region.  In regions A,C,F, when we add point 4, we create another triangle with one of the original points stuck inside that triangle.  (The case for region G is the same, though we can't use the word original and another.)  Since all these cases are equivalent in terms of point placement, proving that this works for five points in any one sector proves it for all the others.  Let's prove it for G.  When we have 4 points in G, we create 3 triangles.  We apply the line extensions for all three triangles.  Notice that if the fifth point is in the triangle, the only extension area that's in the triangle is the extension of the lines that connect to the point in the interior of the original triangle's area.  Such extensions' areas never intersect meaning that any point drawn in the triangle will always make a convex polygon with 4 points.  Now, if the fifth point is outside the triangle in regions A,C,F, we have the same case before with 5 points in a triangle except this time, the triangle is bigger but there are still 2 points in the interior of the triangle.  Such a triangle encompasses the original triangle, so there is no way a point in the original 4 points can be outside the new, big triangle.[/hide]


calculus
linear algebra
calculus computations

Can someone post a proof of the product rule?

By product rule, I guess you mean product rule for differentiation. [url=http://tutorial.math.lamar.edu/AllBrowsers/2413/ProductRuleProof.asp]Here[/url] is one.

There are also quite good notes on Calculus I, II and III, Linear Algebra, Algebra and Differential Equations on the same site.


inequalities
function
vector
trigonometry
geometry
geometric transformation
reflection

Let$ a,b,c>0, a\plus{}b\plus{}c\equal{}3$. Prove that:
$ \sum {\frac{1}{{\sqrt {a^2  \minus{} 3a \plus{} 3} }}}  \le 3$ :help:

[quote="TRAN THAI HUNG"]Let$ a,b,c > 0, a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ \sum {\frac {1}{{\sqrt {a^2 \minus{} 3a \plus{} 3} }}} \le 3$ :help:[/quote]

You must have a mistake in your problem.

You don't have any variables b and c in the expression.

[hide="solution"]

Observe that by Am-Gm $ a^2 -3a +3 \geq 2-a$

This way we just need to prove that :
\[ \sum {\frac{1}{\sqrt{2-a}} \leq 3}\]

Function $ f(x)=\frac{1}{\sqrt{x}}$ is concave in positive real numbers , so using Jensen's inequality we obtain :

\[ \frac{1}{3}\sum{\frac{1}{\sqrt{2-a}} \leq f\left(\frac{2-a +2-b +2-c}{3}\right) =f(1)=1}\]
[/hide]

You reversed Jensen's inequality. Notice that as $ a$ approaches 2, the fraction blows up.

[quote="enndb0x"][hide="solution"]

Observe that by Am-Gm $ a^2 - 3a + 3 \geq 2 - a$

This way we just need to prove that :
\[ \sum {\frac {1}{\sqrt {2 - a}} \leq 3}\]
Function $ f(x) = \frac {1}{\sqrt {x}}$ is concave in positive real numbers , so using Jensen's inequality we obtain :
\[ \frac {1}{3}\sum{\frac {1}{\sqrt {2 - a}} \leq f\left(\frac {2 - a + 2 - b + 2 - c}{3}\right) = f(1) = 1}\]
[/hide][/quote]
Take a=1, b=1.5 and c=0.5
$ \frac {1}{3}\sum{\frac {1}{\sqrt {2 - a}}=3.23}$ :D
So, you have make mistake.

[quote="ocha"]Let $ x \equal{} 3 \minus{} 2a, y \equal{} 3 \minus{} 2b, z \equal{} 3 \minus{} 2c$ such that $ x \plus{} y \plus{} z \equal{} 3$ and $ x,y,z \in \mathbb{R}$, we have

$ \sum \sqrt {\frac {1}{x^2 \plus{} 3}} \le \frac {3}{2}$

Now it suffices to prove the case where $ x,y,z\ge 0$ because if say $ z < 0$, and $ x,y > 0$ then we can replace $ z$ with $ 0$ and decrease $ x$ or $ y$ by $ z$ so $ x \plus{} y \plus{} z \equal{} 3$ and we increase each term of the sum.

Let $ f(x) \equal{} \sqrt {\frac {1}{x^2 \plus{} 3}}$

$ f'(x) \equal{} \frac { \minus{} x}{(x^2 \plus{} 3)^{\frac {3}{2}}} < 0$ so $ f$ is concave

$ f(x) \plus{} f(y) \plus{} f(z) \le 3f\left(\frac {x \plus{} y \plus{} z}{3}\right) \equal{} 3f(1) \equal{} \frac {3}{2}$ as desired[/quote]
you should prove $ f''(x)<0$ to prove it is concave. :)

I have checked f(x).It is not always concave :( .Is there any idea? :)

ocha : $ f'(x) \leq 0$ doesn't mean that $ f$ is a concave fonction, it means only that $ f$ is an decreasing fonction  :)

you should have prouved that $ f''(x) \leq 0$ to conclude that $ f$ is a concave fonction !!

in this cas $ f''(x) \ge 0$ $ \implies$ $ f$convex fonction

so Jensen won't work in this inequalitie  :)

The same problem is also posted here:
http://www.mathlinks.ro/viewtopic.php?t=312336

Just an idea:(I don't know if this can actually work or not :S)

[hide]   Consider the vectors: $ u \equal{} (a \minus{} \frac {3}{2},\frac {\sqrt {3}}{2}),v \equal{} (b \minus{} \frac {3}{2},\frac {\sqrt {3}}{2}),w \equal{} (c \minus{} \frac {3}{2},\frac {\sqrt {3}}{2}),s \equal{} (1,1)$.

We have: $ |u| \equal{} \sqrt {a^2 \minus{} 3a \plus{} 3}$,$ |v| \equal{} \sqrt {b^2 \minus{} 3b \plus{} 3}$,$ |w| \equal{} \sqrt {c^2 \minus{} 3c \plus{} 3}$.
We can therefore write the desired inequality in terms of the magnitudes of those vectors on the left-hand side
and on the-right hand side we could express 3 as the sum of the dot products of the vectors u,v,w and s.
Sorry if this doesn't help at all.I will try to figure out more.[/hide]

yes i see i made a silly mistake, this inequality seems sharper than jensen or cauchy

I did notice that $ \sqrt{a^2\minus{}3a\plus{}3} \equal{} \sqrt{a^2 \plus{} (\sqrt{3})^2 \minus{} 2\sqrt{3}\cos\frac{\pi}{6}}$
which is the cosine law, so perhaps there is some geometric interpretation

Put $ x \equal{} a \minus{} 1, y \equal{} b \minus{} 1, z \equal{} c \minus{} 1$ so that $ x \plus{} y \plus{} z \equal{} 0$ and we need
\[ \sum{\frac {1}{\sqrt {x^2 \minus{} x \plus{} 1}}} \le3\]
Not all $ x,y,z$ are positive.
Case 1:  Suppose $ x,y \ge0$. Since $ c\ge0, z\ge \minus{} 1, x \plus{} y \le1$, so that at most one of $ x,y \ge1/2$
If $ x\ge1/2$, since $ f(x) \equal{} \displaystyle\frac {1}{\sqrt {x^2 \minus{} x \plus{} 1}}$ is symmetrical about $ x \equal{} 1/2$, we can replace $ x$ with its reflection in $ x \equal{} 1/2$ and use the surplus to increase $ z$ (still negative), so increasing the overall sum since $ f(z)$ is strictly increasing for negative $ z$. Hence we can assume that $ x,y \le1/2$

$ f(x) \equal{} \displaystyle\frac {1}{\sqrt {x^2 \minus{} x \plus{} 1}} \Longrightarrow f'(x) \equal{} \minus{} \frac {2x \minus{} 1}{2(x^2 \minus{} x \plus{} 1)^{3/2}}$ and $ f''(x) \equal{} \displaystyle\frac {8x^2 \minus{} 8x \minus{} 1}{4(x^2 \minus{} x \plus{} 1)^{5/2}}$

$ f''(x) \equal{} 0$ when $ x \equal{} 1/4(2\pm\sqrt {6})$ and $ f''(0) \equal{} \minus{} \frac {1}{4}$, so $ f(x)$ is concave throughout $ 0\le x\le1/2$.

By Jensen's inequality, $ f(x) \plus{} f(y)\le2f(\displaystyle\frac {x \plus{} y}{2})$. Put $ m \equal{} \frac {x \plus{} y}{2}$ so that $ z \equal{} \minus{} 2m$, since $ x \plus{} y \plus{} z \equal{} 0$.

We now have $ f(x) \plus{} f(y) \plus{} f(z)\le 2f(m) \plus{} f( \minus{} 2m) \equal{} P(m)$, say, and $ 0\le m \le1/2$

For turning points, $ P'(m) \equal{} 0$.

$ P'(m) \equal{} 2f'(m) \minus{} 2f'( \minus{} 2m) \equal{} 0 \Longrightarrow f'(m) \equal{} f'( \minus{} 2m)$

But for $ 0\le m \le1/2, f'(m) < 1/2$ and $ f'( \minus{} 2m)\ge1/2$ so this is only possible when $ m \equal{} 0$. $ P''(m) \equal{} 2f''(m) \plus{} 4f''( \minus{} 2m)$ and so $ P''(0) \equal{} 6f''(0) < 0$, giving a maximum value for $ P(m)$ in this range of $ 3f(0) \equal{} 3$

Case 2: Suppose $ x\ge0$ and $ y, z\le0$. As before, we may suppose $ x\ge1/2$
Let $ g(x) \equal{} 1/2x \plus{} 1 \minus{} \displaystyle\frac {1}{\sqrt {x^2 \minus{} x \plus{} 1}}$.
Solving $ g(x) \equal{} 0$,
\[ (1/2x \plus{} 1 )^2(x^2 \minus{} x \plus{} 1) \equal{} 1 \Longrightarrow \displaystyle\frac {1}{4}x^2(x^2 \plus{} 3x \plus{} 1) \equal{} 0\]
This gives solutions $ x \equal{} 0$, (repeated, tangent), $ x \equal{} 1/2( \minus{} 3 \plus{} \sqrt {5}) \equal{} \minus{} \theta$, say, and $ x \equal{} 1/2( \minus{} 3 \minus{} \sqrt {5})$, (spurious, from squaring).

We have $ g(x) \ge0$ for $ x \ge \minus{} \theta$ and strictly increasing for $ x\ge0$. $ g(x)$ is also strictly increasing for $ x \le \minus{} \theta$ and so for $ \minus{} 1/2 \le x \le \minus{} \theta, g(x)\ge g( \minus{} 1/2) \equal{} \frac {3}{4} \minus{} \frac {2}{\sqrt {7}}$

Suppose $ z \equal{}$min$ \{y,z\}$

If $ z\ge \minus{} \theta$,both $ z$ and $ y$ are $ \ge \minus{} \theta$ and $ g(x) \plus{} g(y) \plus{} g(z)\ge0$

If $ z\le \minus{} \theta$, then $ \minus{} 1/2 \plus{} \theta\le y \le 0$, $ x \ge \theta$ and $ g(x) \plus{} g(y) \plus{} g(z)\ge g( \minus{} 1/2) \plus{} g(\theta)\ge0$, since $ g(\theta) > 0.04 > g( \minus{} 1/2)$

In either case, $ g(x) \plus{} g(y) \plus{} g(z)\ge0$, i.e.
\[ \displaystyle\frac {1}{2}x \plus{} 1 \minus{} \displaystyle\frac {1}{\sqrt {x^2 \minus{} x \plus{} 1}} \plus{} \displaystyle\frac {1}{2}y \plus{} 1 \minus{} \displaystyle\frac {1}{\sqrt {y^2 \minus{} y \plus{} 1}} \plus{} \displaystyle\frac {1}{2}z \plus{} 1 \minus{} \displaystyle\frac {1}{\sqrt {z^2 \minus{} z \plus{} 1}}\ge0\]
and summing with $ x \plus{} y \plus{} z \equal{} 0$ gives us the result we needed.

Alternatively, to prove the second case without using numbers, we can use the fact that $ f'(x)$ is concave to show that $ h(x) \equal{} x \minus{} f(x) \plus{} f( \minus{} x) \equal{} g(x) \minus{} g( \minus{} x)$ is non-decreasing which is equivalent to the result above.


geometry
rectangle
search
calculus
integration
function
combinatorics open

For each $k=1,2,\ldots,$ consider the rectangle $I_k := \frac{1}{k} \times \frac{1}{k+1}$. The sum of the areas of these rectangles is $1$; can they tile the unit square?

I think that you could search the arXiv site, I once read about tilings in infinity!
There was also something about tiling a rectangle of $\displaystyle \frac{\pi}{\sqrt{6} }$ with rectangles of sides $\frac{1}{n}$. It is a well known limit. I think the idea was with double integrals, that is the sum of integrals on each rectangle should be equal to the entire integral for any function. Take some particular functions and see, maybe you find a solution.....

Oh, I just learned that this is an open question  :). Perhaps a moderator should move it to the 'open problems' section of the forum.

DusT, the approach with the double integrals is quite standard; however, I don't think that the answer to this question should be negative. For example, Paulhus has come close by showing that the squares can be packed into a square of side $1.0000000001$.

--Vesselin


geometry
circumcircle
incenter
geometry proposed

I would like to tell me if the following fact about the mixtilinear circle is known . Here is the problem : Known or not I hope you will enjoy it  :)

Let a triangle $ ABC$ and its circumcircle $ C_1(O,R)$ . Let the circle that is tangent at the same time to $ AB$ at $ P$ to $ AC$ at $ Q$ and to $ C_1$ at $ T$
and let also the circle which is passing through the incenter $ I$ of the triangle $ ABC$ and it is tangent (internally) to $ C_1$ at $ T$ .
This circle meets $ PQ$ at the point $ S$ . Prove that the points $ S,T,A$ are collinear .

Denote K is intersection of TP and ⊙O   h is the tangent line of ⊙O and ⊙O1
IK^2=BK^2=KP*KT  => △IPK∽△TIK  => ∠KIP=∠ITP=∠CIQ=∠AQP-∠ICA=
90-1/2∠A-1/2∠C=1/2∠B    => ∠KIP=∠IBP  => I,P,B,T are concyclic
∠SIT=∠PBT=∠ABT=∠ITH
So h is tangent to the circumcircle of △IST

Dear Mathlinkers,
a proof of this result can be seen on
http://perso.orange.fr/jl.ayme    vol. 4  A new mixtilinear incircle adventure I p. 59
Sincerely
Jean-Louis

$ AT$ cut $ PQ$ at $ S'$
We prove that $ (TIS')$ tangent $ (O)$$ (1)$
$ TI$ cut $ (O)$ at $ J$ we have $ PQ//AJ$
then $ (1)$ true.
done.


algorithm
combinatorics unsolved
combinatorics

Question:
http://www.cartalk.com/content/puzzler/transcripts/200520/index.html

Answer:
http://www.cartalk.com/content/puzzler/transcripts/200520/answer.html

I don't particularly like their answer... or at least their method for finding an answer.  It seems that they just kept testing new combinations until they decided there probably wasn't a higher unreachable number.

How would you go about solving a problem like this?

Thanks...
Gideon

You're right, that's no solution. However, obviously if you can get $x$ you can also get $x+6$... so if 43 is indeed the answer, it will suffice to check 44, 45, 46, 47, 48, 49 by hand. Which shouldn't take too long.

I've already done the work to verify that 43 is indeed not reachable... but that doesn't prove whether or not a greater number exists that is also not reachable.  That's my issue with it.  There's got to be some definitive way to know when to stop looking higher.... they just seemed to give up after awhile and decided it must be 43.

It makes sense to me that the higher you go, the fewer gaps there should be until eventually there aren't any gaps, but I can't figure out how to find that point.

As you said, given any reachable number $x$, we know that $x+6$, $x+9$, and $x+20$ are all reachable.  Furthermore, we know that any multiple of $x$ is reachable ($2x$, $3x$, etc).  I can't really come up with any other generalizations that might help, though.

Thanks for looking at it...
Gideon

Yes, you're right. But checking 44-49 does in fact prove it. Because then by adding 6 you get 50-55, 56-61, 62-67 and so on. So if 44-49 are reachable, every number greater than 43 is reachable.

Ahhh... I see what you mean, now!  That makes complete sense to me.

Thanks for explaining it to me.
Gideon

Alright, well, I was satisfied for a minute with the knowledge of how to prove the answer... however, I still wonder now if there is a process by which to find the answer in the first place (something other than brute force would be nice).

When I first looked at the problem, I had in mind to write a program that would determine reachable numbers up to a given limit to look for a greater unreachable number, but I got stuck.  I can figure out in my head the set of reachable numbers, but I couldn't come up with a general algorithm for doing it...

$6=6$
$9=9$
$12=6+6$
$15=6+9$
$18=6+6+6$ or $9+9$
$20=20$
etc...

It's like trying to count in a numbering system with only 6, 9, and 20... but since the numbers don't evenly divide into each other, it's not very straightforward.

The other option would be to work backwards... for each number $x$ up to whatever limit, try and factor $x$ using only 6, 9, and 20.  That might be easier, but probably not as effecient, since you'd have to try every number.

Any idea what the best generic approach would be for any given increments?  Thanks!
Gideon

There is a more systamaitac way of doing it.

First a theorem called Sylvesters result: The largest non-repersentable number in the form $ax+by$ with $\gcd(a,b)=1$ and $x,y \geq 0$ is $ab-a-b$

Now, $6x+9y+20z=3(2x+3y)+20z$

Here, $2x+3y$ takes all values from $(3)(2)-3-2+1=2$  upwards So let it equal $2+t$

$3(2+t)+20z=3t+20z+6$ This equation assumes all values from $(3)(20)-3-20+6=43
$ upwards. So 42 is the greatest non-repersentable number of chicken nuggets.


percent
analytic geometry
geometry
perimeter
algebra
polynomial
ratio

i have several problems that i need clarification on from a chapter and state competition...please provide full explanations not just answers....

1. the salary of a salesperson was a 4% commission on all sale. her commission rate is then raised to 4.5%. what will the percent of increase in her salary be, assuming she sells the same amount of merchandise?

2. how many different, positive three-digit numbers can be made using any of the three following five digits: 1,2,2,3,3?

3. in order to direct observers to a school exhibit, the students are building a road of paper sheets that measure 18" by 24" each. the path is 18" wide. they need to tape around the outside of each sheet of paper completely, either to join the sheet to another or to stick the paper to the floor. (when two sheets meet at an edge, only one piece of tape is used to hold the sheets together). what is the least number of feet of tape they need for a 100-foot long path? express your answer as a decimal to the nearest tenth.

4. if b and 7 are solutions to the equation: x^2 -4x=a, find the value of a+b.

5. a triangle has vertices with coordinates A(0,15), B(0,0), and C(10,0). find the coordinates of point D on AC so that the area of triangle ABD is equal to the area of triangle DBC.

6. if an iron ball 8 in. in diameter weighs 80 pounds, how many pounds does an iron ball 12 inches in diameter weigh?

7. find the length of the line segment with endpoints (a,a) and (9a, 7a) and a>0.

8. a manufacturing company identifies its products by codes consisting of a letter (A-Z) followed by 3 digits (0-9) followed by another letter. how many different products acn the company manufacture before it uses up all possible codes?

9. the eight grade basketball team averaged P points per game in the X games it won this season. the team scored a total of T points during the season. express the total number of points scored in the games it lost in terms of P, X, and T.

10. the area of a semicircular region is 49/8*Pi square inches. express the number of inches in the perimeter of the semicircular region to the nearest whole number.

Thanks.

[hide="5"]

The area of triangle ABC = 15*10/2 = 75
So the area of each of the other two triangles is 75/2
So, AB * height of ABD = 75 so hieght = 5 = abscissa of D
Also, CB * height of CBD = 75 so hieght = 7.5 = ordinate of D
D = (5,7.5)

[/hide]

circlesquared...i don't know if what you did is right....the way i thought about it is as follows:
since the area of the semicircular region is 49/8 pi...the area of the entire region is 98/8*pi...dividing pi out, you get that the radius squared is 98/8...find the square root of that and you get that the radius is (7*root 2)/8...multiply that by 2 to get the radius which is (7*root2)/4....multiply this by pi and divided by 2 to get the perimeter as being (7*root2)/2 pi...you can estimate the rest...

[quote="CircleSquared"][hide="1"]
The salary starts off as a 4% commission.
It increases by .5%.
.5% is $\frac{1}{8}=12.5\%$ of 4%.
$12.5\%$ is the answer.
[/hide]
[hide="2"]
Semi brute force:
122 x 3
133 x 3
123 x 6
223 x 3
332 x 3
Which gives a total of $18$.
[/hide]
[hide="3"]
Put together 5 18x24 sheets.
The total amount of tape needed is $18\cdot6+10\cdot24=348$.
[/hide]
[hide="4"]
Given the information, $x^{2}-4x-a=x^{2}-(b+7)x+7b$
Two polynomials are equal if their corresponding terms are congruent soooo...
$4=b+7$, or $b=-3$
$a=-7b$, or $a=21$
$a+b=18$
[/hide]
[hide="5"]
Draw a line parallel to AC through B, and you can see that no matter where D is, the height of DBA and DBC will always be equal.  (Perpendiculars between 2 parallel lines, at any points, are always congruent.)  So all we have to do is find the midpoint of AC, and AD and DC will be equal as well.  Using midpoint formula on A and C, D is $(5,7.5)$.[/hide]
[hide="6"]
These balls have a ratio of diameters (or scale factor) of 2:3.  The ratio of volumes is the scale factor cubed, or 8:27.  The diameter 8 ball weighs 80 lbs, so the diameter 12 ball weighs $270$ lbs.
[/hide]
[hide="7"]
Distance formula.
$\sqrt{64a^{2}+36a^{2}}$
$=\sqrt{100a^{2}}$
$=10a$
[/hide]
[hide="8"]
_ _ _ _ _
There are 26 letters for the 1st blank
26 _ _ _ _
There are $10\cdot10\cdot10$ digits for the 2nd, 3rd, and 4th blanks.
26 10 10 10 _
There are 26 letters for the last blank.
26 10 10 10 26
So the answer is $26^{2}\cdot1000$
$=676000$
[/hide]
[hide="9"]
The total number of points the team earned in the games it won is PX.  The total number of points the team earned in total is T.  So the number of points the team earned in games it lost is $T-PX$.
[/hide]
[hide="10"]
The area of the total circle is $\frac{49\pi}{16}$.
Reverse the area formula for the circle, and the radius is $r=\frac{7}{4}$.
The total perimeter is $2r+r\pi$
To the nearest whole number, that is $9$.[/hide]

Owwww that hurt
I'm not surprised if I got any wrong.

EDIT: fixed #7.  I'm tired.[/quote]

For number 10, shouldn't the radius be $\frac{7}{2}$, not $\frac{7}{4}$?

[quote="magixter"][quote="CircleSquared"][hide="1"]
The salary starts off as a 4% commission.
It increases by .5%.
.5% is $\frac{1}{8}=12.5\%$ of 4%.
$12.5\%$ is the answer.
[/hide]
[hide="etc"][hide="2"]
Semi brute force:
122 x 3
133 x 3
123 x 6
223 x 3
332 x 3
Which gives a total of $18$.
[/hide]
[hide="3"]
Put together 5 18x24 sheets.
The total amount of tape needed is $18\cdot6+10\cdot24=348$.
[/hide]
[hide="4"]
Given the information, $x^{2}-4x-a=x^{2}-(b+7)x+7b$
Two polynomials are equal if their corresponding terms are congruent soooo...
$4=b+7$, or $b=-3$
$a=-7b$, or $a=21$
$a+b=18$
[/hide]
[hide="5"]
Draw a line parallel to AC through B, and you can see that no matter where D is, the height of DBA and DBC will always be equal.  (Perpendiculars between 2 parallel lines, at any points, are always congruent.)  So all we have to do is find the midpoint of AC, and AD and DC will be equal as well.  Using midpoint formula on A and C, D is $(5,7.5)$.[/hide]
[hide="6"]
These balls have a ratio of diameters (or scale factor) of 2:3.  The ratio of volumes is the scale factor cubed, or 8:27.  The diameter 8 ball weighs 80 lbs, so the diameter 12 ball weighs $270$ lbs.
[/hide]
[hide="7"]
Distance formula.
$\sqrt{64a^{2}+36a^{2}}$
$=\sqrt{100a^{2}}$
$=10a$
[/hide]
[hide="8"]
_ _ _ _ _
There are 26 letters for the 1st blank
26 _ _ _ _
There are $10\cdot10\cdot10$ digits for the 2nd, 3rd, and 4th blanks.
26 10 10 10 _
There are 26 letters for the last blank.
26 10 10 10 26
So the answer is $26^{2}\cdot1000$
$=676000$
[/hide]
[hide="9"]
The total number of points the team earned in the games it won is PX.  The total number of points the team earned in total is T.  So the number of points the team earned in games it lost is $T-PX$.
[/hide]
[hide="10"]
The area of the total circle is $\frac{49\pi}{16}$.
Reverse the area formula for the circle, and the radius is $r=\frac{7}{4}$.
The total perimeter is $2r+r\pi$
To the nearest whole number, that is $9$.[/hide]

Owwww that hurt
I'm not surprised if I got any wrong.[/hide]

EDIT: fixed #7.  I'm tired.[/quote]

For number 10, shouldn't the radius be $\frac{7}{2}$, not $\frac{7}{4}$?[/quote]

Yes; the area of the total circle is $2 \cdot \frac{49}{8}\pi = \frac{49}{4}\pi$. Then the answer would be exactly two times what CircleSquared got.


calculus
integration
trigonometry
calculus computations

Hi can someone please help me with this.  Thank you.

Question: Evaluate $ \displaystyle\int_{t\equal{}0}^{\infty}\int_{u\equal{}0}^t\frac{e^{\minus{}t}\sin u}{u}dudt$

Hi :) !!

I think that

$ \int_{t \equal{} 0}^{ \plus{} \infty}\int_{u \equal{} 0}^t \frac {e^{ \minus{} t}sin(u)}{u}dudt \equal{} \sum_{n \geq 0}\frac {( \minus{} 1)^n}{(2n \plus{} 1)(2n \plus{} 1)!}\Gamma(2n \plus{} 2)$

than $ \int_{t \equal{} 0}^{ \plus{} \infty}\int_{u \equal{} 0}^t \frac {e^{ \minus{} t}sin(u)}{u}dudt \equal{} \sum_{n \geq 0}\frac {( \minus{} 1)^n}{2n \plus{} 1}\equal{}\frac{\pi}{4}$

without error :)

thanks

[quote="gauss202"]Where is your justification for these results?[/quote]

Hi :)

did you know that $ sinc_1(x)\equal{} \sum_{n \geq 0}\frac{(\minus{}1)^n x^{2n\plus{}1}}{(2n\plus{}1)!}$

than we remplace and we use the theorem of convegence .....

PS: if you don't understand we complete this after

Hi I think I figured out another method.

I did not realize that I could have done the problem like this:

First, consider $ \displaystyle I \equal{} \int_0^{\infty}e^{ \minus{} st}\left(\int_0^t\frac {\sin u}{u}du\right)dt \equal{} \mathcal{L}\left\{\int_0^t\frac {\sin u}{u} du\right\}$
Since $ \displaystyle\mathcal{L}\left\{\frac {\sin u}{u} \right\} \equal{} \tan^{ \minus{} 1}\frac {1}{s}$
Then $ \displaystyle I \equal{} \mathcal{L}\left\{\int_0^t\frac {\sin u}{u} du\right\} \equal{} \frac {1}{s}\tan^{ \minus{} 1}\frac {1}{s}$.  When $ s \equal{} 1$, we have the desired integral.  Hence the value is $ \displaystyle\frac {1}{1}\tan^{ \minus{} 1}(1) \equal{} \frac {\pi}{4}$.


Given$ a,b,c$ be real numbers .Prove that:
$ |ab(a^2\minus{}b^2)\plus{}bc(b^2\minus{}c^2)\plus{}ca(c^2\minus{}a^2)| \le \frac{9\sqrt{2}(a^2\plus{}b^2\plus{}c^2)^2}{32}$
Sorry, if anybody posted it.

[quote="darij grinberg"]Shortlist is a bit of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=101294]understatement[/url] ?

  darij[/quote]
Given$ a,b,c$ be real numbers,and$ ab\plus{}bc\plus{}ca \ge 0$.Prove that:
$ |a^2b^2(a\minus{}b)\plus{}b^2c^2(b\minus{}c)\plus{}c^2a^2(c\minus{}a)| \le \frac{4\sqrt{(a^2\plus{}b^2\plus{}c^2)^5}}{25\sqrt{5}}$

[quote="quykhtn-qa1"][quote="vo thanh van"]You can see in this file[/quote]
Thank you :)
@ anh biet ket qua thi quoc gia chua?anh dc giai gi?[/quote]
Chưa em.Chắc là tuần sau có kq rồi

[quote="Mathias_DK"][quote="quykhtn-qa1"][quote="darij grinberg"]Shortlist is a bit of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=101294]understatement[/url] ?

  darij[/quote]
Given$ a,b,c$ be real numbers,and$ ab \plus{} bc \plus{} ca \ge 0$.Prove that:
$ |a^2b^2(a \minus{} b) \plus{} b^2c^2(b \minus{} c) \plus{} c^2a^2(c \minus{} a)| \le \frac {4\sqrt {(a^2 \plus{} b^2 \plus{} c^2)^5}}{25\sqrt {5}}$[/quote]
$ a^2b^2(a \minus{} b) \plus{} b^2c^2(b \minus{} c) \plus{} c^2a^2(c \minus{} a) \equal{} (a \minus{} b)(b \minus{} c)(c \minus{} a)(ab \plus{} bc \plus{} ca)$.
Let $ 3u \equal{} a \plus{} b \plus{} c, 3v^2 \equal{} ab \plus{} bc \plus{} ca, w^3 \equal{} abc$. Assume $ v \neq 0$, since then the ineq is trival. Set $ u^2 \equal{} tv^2$, so $ t \ge 1$.
Then notice $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 9u^2 \minus{} 6v^2$. So we are to prove:
$ (a \minus{} b)(b \minus{} c)(c \minus{} a)3v^2 \le \frac {36\sqrt {3}}{25\sqrt {5}} \sqrt {(3u^2 \minus{} 2v^2)^5}$.
Notice that $ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \equal{} 27( \minus{} (w^3 \minus{} (3uv^2 \minus{} 2u^3))^2 \plus{} 4(u^2 \minus{} v^2)^3) \le 108(u^2 \minus{} v^2)^3$, so it suffices to prove:
$ 108(u^2 \minus{} v^2)^3 \cdot 9v^4 \le \frac {36^2 \cdot 3}{5^5}(3u^2 \minus{} 2v^2)^5 \iff$
$ 5^5(u^2 \minus{} v^2)^3v^4 \le 8(3u^2 \minus{} 2v^2)^5 \iff$
$ 5^5(t \minus{} 1)^3 \le 8(3t \minus{} 2)^5$
$ 1944t^5 \minus{} 6480t^4 \plus{} 5515t^3 \plus{} 3615t^2 \minus{} 7455t \plus{} 2869 \ge 0 \iff$
$ (t \plus{} 1)(1944t^4 \minus{} 8424t^3 \plus{} 13939t^2 \minus{} 10324t \plus{} 2869) \ge 0$
$ 1944t^4 \minus{} 8424t^3 \plus{} 13939t^2 \minus{} 10324t \plus{} 2869 \ge 0$ is true, but it is very hard to show in my opinion :oops:[/quote]
You can prove the last inequality.Your solutions very ugly with me :wink:
My solution very nice :)

[quote="Mathias_DK"][quote="quykhtn-qa1"]You can prove the last inequality.Your solutions very ugly with me :wink:
My solution very nice :)[/quote]
That wasn't very polite ;) ! You should post your own solution before you say that mine is ugly.

Btw. (A bit off-topic) Why is a solution which involves a bit of calculations, considered "ugly" by so many people ? :roll: I hate when people act like there solution is special, because it only uses AM-GM, cauchy-schwartz, or something like that.. I would appreciate it if everyone would look at the ideas in a solution, instead of the techniques which are used. :!:[/quote]
I speak it ugly because my solution only use AM-GM :oops:

@Mathias:
[quote]
$ 5^5(u^2 \minus{} v^2)^3v^4 \le 8(3u^2 \minus{} 2v^2)^5 \iff$
[/quote]
It should be
$ 5^5(u^2 \minus{} v^2)^3v^4 \le 4(3u^2 \minus{} 2v^2)^5 \iff$  :wink:
Then we get the equality case at $ t\equal{}\frac32$.  :P

[quote="Mathias_DK"][quote="quykhtn-qa1"][quote="darij grinberg"]Shortlist is a bit of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=101294]understatement[/url] ?

  darij[/quote]
Given$ a,b,c$ be real numbers,and$ ab \plus{} bc \plus{} ca \ge 0$.Prove that:
$ |a^2b^2(a \minus{} b) \plus{} b^2c^2(b \minus{} c) \plus{} c^2a^2(c \minus{} a)| \le \frac {4\sqrt {(a^2 \plus{} b^2 \plus{} c^2)^5}}{25\sqrt {5}}$[/quote]
$ a^2b^2(a \minus{} b) \plus{} b^2c^2(b \minus{} c) \plus{} c^2a^2(c \minus{} a) \equal{} (a \minus{} b)(b \minus{} c)(c \minus{} a)(ab \plus{} bc \plus{} ca)$.
Let $ 3u \equal{} a \plus{} b \plus{} c, 3v^2 \equal{} ab \plus{} bc \plus{} ca, w^3 \equal{} abc$. Assume $ v \neq 0$, since then the ineq is trival. Set $ u^2 \equal{} tv^2$, so $ t \ge 1$.
Then notice $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 9u^2 \minus{} 6v^2$. So we are to prove:
$ (a \minus{} b)(b \minus{} c)(c \minus{} a)3v^2 \le \frac {36\sqrt {3}}{25\sqrt {5}} \sqrt {(3u^2 \minus{} 2v^2)^5}$.
Notice that $ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \equal{} 27( \minus{} (w^3 \minus{} (3uv^2 \minus{} 2u^3))^2 \plus{} 4(u^2 \minus{} v^2)^3) \le 108(u^2 \minus{} v^2)^3$, so it suffices to prove:
$ 108(u^2 \minus{} v^2)^3 \cdot 9v^4 \le \frac {36^2 \cdot 3}{5^5}(3u^2 \minus{} 2v^2)^5 \iff$
$ 5^5(u^2 \minus{} v^2)^3v^4 \le 8(3u^2 \minus{} 2v^2)^5 \iff$
$ 5^5(t \minus{} 1)^3 \le 8(3t \minus{} 2)^5$
$ 1944t^5 \minus{} 6480t^4 \plus{} 5515t^3 \plus{} 3615t^2 \minus{} 7455t \plus{} 2869 \ge 0 \iff$
$ (t \plus{} 1)(1944t^4 \minus{} 8424t^3 \plus{} 13939t^2 \minus{} 10324t \plus{} 2869) \ge 0$
$ 1944t^4 \minus{} 8424t^3 \plus{} 13939t^2 \minus{} 10324t \plus{} 2869 \ge 0$ is true, but it is very hard to show in my opinion :oops:[/quote]
your proof can solve the inequality ???
Determine the smallest number M such that the inequality:
$ |a^2b^2(a\minus{}b)\plus{}b^2c^2(b\minus{}c)\plus{}c^2a^2(c\minus{}a)| \le  M\sqrt{(a^2\plus{}b^2\plus{}c^2)^5}$
with the same condition :oops:

[quote="FelixD"]@Mathias:
[quote]
$ 5^5(u^2 \minus{} v^2)^3v^4 \le 8(3u^2 \minus{} 2v^2)^5 \iff$
[/quote]
It should be
$ 5^5(u^2 \minus{} v^2)^3v^4 \le 4(3u^2 \minus{} 2v^2)^5 \iff$  :wink:
Then we get the equality case at $ t \equal{} \frac32$.  :P[/quote]
Of course! Thank you! :)
Then we have:
$ 5^5(t \minus{} 1)^3 \le 4(3t \minus{} 2)^5 \iff$
$ ( ((243t \plus{} 405)(t \minus{} 1) \plus{} 76)(t \minus{} 1) \plus{} 4 )(2t \minus{} 3)^2 \ge 0$
Which is obviously true :) So equality when $ t \equal{} \frac {3}{2}$, $ u^2 \equal{} tv^2$ and $ w^3 \equal{} 3uv^2 \minus{} 2u^3$.
Setting $ u \equal{} m$ we get $ 3u \equal{} 3m$, $ 3v^2 \equal{} 2m^2$ and $ w^3 \equal{} 0$. So roots in $ x^3 \minus{} 3mx \plus{} 2m^2x \equal{} (x \minus{} 0)(x^2 \minus{} 3mx \plus{} 2m^2) \equal{} (x \minus{} 0)(x \minus{} m)(x \minus{} 2m)$. So equality if and only if $ (a,b,c)$ is a permutation of $ (0,m,2m)$ for some $ m \in \mathbb{R}$.

[quote="quykhtn-qa1"]
your proof can solve the inequality ???
Determine the smallest number M such that the inequality:
$ |a^2b^2(a \minus{} b) \plus{} b^2c^2(b \minus{} c) \plus{} c^2a^2(c \minus{} a)| \le M\sqrt {(a^2 \plus{} b^2 \plus{} c^2)^5}$
with the same condition :oops:[/quote]
But we have both proven that $ M \equal{} \frac{4}{25\sqrt{5}}$ :lol:

[quote="quykhtn-qa1"][/quote][quote="Mathias_DK"][quote="quykhtn-qa1"][quote="darij grinberg"]Shortlist is a bit of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=101294]understatement[/url] ?

  darij[/quote]
Given$ a,b,c$ be real numbers,and$ ab + bc + ca \ge 0$.Prove that:
$ |a^2b^2(a - b) + b^2c^2(b - c) + c^2a^2(c - a)| \le \frac {4\sqrt {(a^2 + b^2 + c^2)^5}}{25\sqrt {5}}$[/quote]
$ a^2b^2(a - b) + b^2c^2(b - c) + c^2a^2(c - a) = (a - b)(b - c)(c - a)(ab + bc + ca)$.
Let $ 3u = a + b + c, 3v^2 = ab + bc + ca, w^3 = abc$. Assume $ v \neq 0$, since then the ineq is trival. Set $ u^2 = tv^2$, so $ t \ge 1$.
Then notice $ a^2 + b^2 + c^2 = 9u^2 - 6v^2$. So we are to prove:
$ (a - b)(b - c)(c - a)3v^2 \le \frac {36\sqrt {3}}{25\sqrt {5}} \sqrt {(3u^2 - 2v^2)^5}$.
Notice that $ (a - b)^2(b - c)^2(c - a)^2 = 27( - (w^3 - (3uv^2 - 2u^3))^2 + 4(u^2 - v^2)^3) \le 108(u^2 - v^2)^3$, so it suffices to prove:
$ 108(u^2 - v^2)^3 \cdot 9v^4 \le \frac {36^2 \cdot 3}{5^5}(3u^2 - 2v^2)^5 \iff$
$ 5^5(u^2 - v^2)^3v^4 \le 8(3u^2 - 2v^2)^5 \iff$
$ 5^5(t - 1)^3 \le 8(3t - 2)^5$
$ 1944t^5 - 6480t^4 + 5515t^3 + 3615t^2 - 7455t + 2869 \ge 0 \iff$
$ (t + 1)(1944t^4 - 8424t^3 + 13939t^2 - 10324t + 2869) \ge 0$
$ 1944t^4 - 8424t^3 + 13939t^2 - 10324t + 2869 \ge 0$ is true, but it is very hard to show in my opinion :oops:[/quote][quote="quykhtn-qa1"]
your proof can solve the inequality ???
Determine the smallest number M such that the inequality:
$ |a^2b^2(a - b) + b^2c^2(b - c) + c^2a^2(c - a)| \le M\sqrt {(a^2 + b^2 + c^2)^5}$
with the same condition :oops:[/quote]
mathias_DK,you can prove this inequality,by your proof?? :oops:

[quote="Mathias_DK"][quote="quykhtn-qa1"][quote="quykhtn-qa1"]
your proof can solve the inequality ???
Determine the smallest number M such that the inequality:
$ |a^2b^2(a \minus{} b) \plus{} b^2c^2(b \minus{} c) \plus{} c^2a^2(c \minus{} a)| \le M\sqrt {(a^2 \plus{} b^2 \plus{} c^2)^5}$
with the same condition :oops:[/quote]
mathias_DK,you can prove this inequality,by your proof?? :oops:[/quote]
As I said before: We have both proved $ M \equal{} \frac {4}{25\sqrt {5}}$.[/quote]
Yes, it is true :wink: .But I don't know this inequality before :wink:

[quote="quykhtn-qa1"][/quote][quote="Mathias_DK"][quote="quykhtn-qa1"][quote="quykhtn-qa1"]
your proof can solve the inequality ???
Determine the smallest number M such that the inequality:
$ |a^2b^2(a - b) + b^2c^2(b - c) + c^2a^2(c - a)| \le M\sqrt {(a^2 + b^2 + c^2)^5}$
with the same condition :oops:[/quote]
mathias_DK,you can prove this inequality,by your proof?? :oops:[/quote]
As I said before: We have both proved $ M = \frac {4}{25\sqrt {5}}$.[/quote][quote="quykhtn-qa1"]
Yes, it is true :wink: .But I don't know this inequality before :wink:[/quote]
What do you mean? Do you ask if I could find $ M$, if you hadn't posted it? If that's the question, then the answer is yes..

[quote="Mathias_DK"][/quote][quote="quykhtn-qa1"][/quote][quote="Mathias_DK"][quote="quykhtn-qa1"][quote="quykhtn-qa1"]
your proof can solve the inequality ???
Determine the smallest number M such that the inequality:
$ |a^2b^2(a - b) + b^2c^2(b - c) + c^2a^2(c - a)| \le M\sqrt {(a^2 + b^2 + c^2)^5}$
with the same condition :oops:[/quote]
mathias_DK,you can prove this inequality,by your proof?? :oops:[/quote]
As I said before: We have both proved $ M = \frac {4}{25\sqrt {5}}$.[/quote][quote="quykhtn-qa1"]
Yes, it is true :wink: .But I don't know this inequality before :wink:[/quote][quote="Mathias_DK"]
What do you mean? Do you ask if I could find $ M$, if you hadn't posted it? If that's the question, then the answer is yes..[/quote]
you can solve by your proof?Please post solution

[quote="quykhtn-qa1"][quote="Mathias_DK"]
What do you mean? Do you ask if I could find $ M$, if you hadn't posted it? If that's the question, then the answer is yes..[/quote]
you can solve by your proof?Please post solution[/quote]
Let $ a \equal{} m,$ $ b \equal{} 2m,$ where $ m\neq0,$ $ c \equal{} 0$  and see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1435325#1435325
 :wink:

[quote="quykhtn-qa1"][/quote][quote="Mathias_DK"][/quote][quote="quykhtn-qa1"][/quote][quote="Mathias_DK"][quote="quykhtn-qa1"][quote="quykhtn-qa1"]
your proof can solve the inequality ???
Determine the smallest number M such that the inequality:
$ |a^2b^2(a - b) + b^2c^2(b - c) + c^2a^2(c - a)| \le M\sqrt {(a^2 + b^2 + c^2)^5}$
with the same condition :oops:[/quote]
mathias_DK,you can prove this inequality,by your proof?? :oops:[/quote]
As I said before: We have both proved $ M = \frac {4}{25\sqrt {5}}$.[/quote][quote="quykhtn-qa1"]
Yes, it is true :wink: .But I don't know this inequality before :wink:[/quote][quote="Mathias_DK"]
What do you mean? Do you ask if I could find $ M$, if you hadn't posted it? If that's the question, then the answer is yes..[/quote][quote="quykhtn-qa1"]
you can solve by your proof?Please post solution[/quote]
Using my method it comes down to finding the max value of:
$ \frac {(t - 1)^3}{(3t - 2)^5}$. From here we can easily find $ M$ ;)
But this is easy. Set $ f(t) = \frac {(t - 1)^3}{(3t - 2)^5}$. Then:
$ f'(t) = \frac {3(t - 1)^2(3t - 2) - 15(t - 1)^3t}{(3t - 2)^5} = \frac {3(5 - 1)^2(3 - 2t)}{(3t - 2)^5}$. So the max is obtain when $ 3 - 2t = 0 \iff t = \frac {3}{2}$. Then we can easily find the max value and thus easily find $ M$. So it actually seems a lot easier to solve the problem by finding the least constant, than using that you already know it ! :)

[quote="Mathias_DK"][/quote][quote="quykhtn-qa1"][/quote][quote="Mathias_DK"][/quote][quote="quykhtn-qa1"][/quote][quote="Mathias_DK"][quote="quykhtn-qa1"][quote="quykhtn-qa1"]
your proof can solve the inequality ???
Determine the smallest number M such that the inequality:
$ |a^2b^2(a - b) + b^2c^2(b - c) + c^2a^2(c - a)| \le M\sqrt {(a^2 + b^2 + c^2)^5}$
with the same condition :oops:[/quote]
mathias_DK,you can prove this inequality,by your proof?? :oops:[/quote]
As I said before: We have both proved $ M = \frac {4}{25\sqrt {5}}$.[/quote][quote="quykhtn-qa1"]
Yes, it is true :wink: .But I don't know this inequality before :wink:[/quote][quote="Mathias_DK"]
What do you mean? Do you ask if I could find $ M$, if you hadn't posted it? If that's the question, then the answer is yes..[/quote][quote="quykhtn-qa1"]
you can solve by your proof?Please post solution[/quote][quote="Mathias_DK"]
Using my method it comes down to finding the max value of:
$ \frac {(t - 1)^3}{(3t - 2)^5}$. From here we can easily find $ M$ ;)
But this is easy. Set $ f(t) = \frac {(t - 1)^3}{(3t - 2)^5}$. Then:
$ f'(t) = \frac {3(t - 1)^2(3t - 2) - 15(t - 1)^3t}{(3t - 2)^5} = \frac {3(5 - 1)^2(3 - 2t)}{(3t - 2)^5}$. So the max is obtain when $ 3 - 2t = 0 \iff t = \frac {3}{2}$. Then we can easily find the max value and thus easily find $ M$. So it actually seems a lot easier to solve the problem by finding the least constant, than using that you already know it ! :)[/quote]
Thanks :lol: .But i think my solution than nice

[quote="quykhtn-qa1"]Thanks :lol: .But i think my solution than nice[/quote]
Of course, it's nicer. But you aren't the first people which find this form neqs and this form solutions.  :P
[quote="nguoivn"][quote="nguoivn"][quote="quykhtn-qa1"]Determine the smallest number M such that the inequality:
$ |a^2b^2(a \minus{} b) \plus{} b^2c^2(b \minus{} c) \plus{} c^2a^2(c \minus{} a)| \le M\sqrt {(a^2 \plus{} b^2 \plus{} c^2)^5}$
with the same condition :oops:[/quote]
It isn't a new form ineq. I have found many ineq like that for 2 years ago. See it here:
http://www.maths.vn/forums/showthread.php?t=9244
( problem 1,15,31,32,33...) and in our another book of MIC, this form ineq is in the topic "The strange ineqs"  :oops:[/quote]

Indeed, I also don't like the proofs above. I had created and solved my similar problems which I mention above by Am-Gm  :)[/quote]

[quote="nguoivn"][/quote][quote="quykhtn-qa1"]Thanks :lol: .But i think my solution than nice[/quote]
Of course, it's nicer. But you aren't the first people which find this form neqs and this form solutions.  :P
[quote="nguoivn"][quote="nguoivn"][quote="quykhtn-qa1"]Determine the smallest number M such that the inequality:
$ |a^2b^2(a - b) + b^2c^2(b - c) + c^2a^2(c - a)| \le M\sqrt {(a^2 + b^2 + c^2)^5}$
with the same condition :oops:[/quote]
It isn't a new form ineq. I have found many ineq like that for 2 years ago. See it here:
http://www.maths.vn/forums/showthread.php?t=9244
( problem 1,15,31,32,33...) and in our another book of MIC, this form ineq is in the topic "The strange ineqs"  :oops:[/quote]

Indeed, I also don't like the proofs above. I had created and solved my similar problems which I mention above by Am-Gm  :)[/quote][quote="nguoivn"][/quote]
Sorry,I don't know this inequality ,before :wink:


logarithms
limit
calculus
calculus computations

Analyze the convergence of:

1.  $ \sum_{n=1}^\infty (-1)^{n}\left( \frac{\pi}{2}-\arctan ( \log n) \right)$

2.  $ \sum_{n=2}^{\infty}\frac{(-1)^{n}}{(n+(-1)^{n})^{s}}$

3.  $ \sum_{n=1}^\infty a_{n}$, $ a_{n}= 1/n$ if $ n$ is a square, $ 1/n^{2}$ otherwise.

4.  $ \sum_{n=1}^\infty a_{n}$, $ a_{n}= 1/n^{2}$ if $ n$ is odd, $ -1/n$ if $ n$ is even.

Determine all $ z \in \mathbb C$ such that the series converges:

1. $ \sum_{n=1}^{\infty}\frac{(-1)^{n}}{z+n}$

2.  $ \sum_{n=1}^\infty \left( 1+\frac{1}{1+5n}\right)^{n^{2}}|z|^{17n}$

3.  $ \sum_{n=1}^\infty \frac{n}{n+1}\left( \frac{z}{2z+1}\right)^{n}$

1. $ \sum_{n=1}^{+\infty}(-1)^{n}\left[\frac{\pi}{2}-\arctan(\ln n)\right]$.

Let's study the convergence of the series of the modulus by comparing with the divergent series having $ a_{n}= \frac{1}{\ln n}$:

$ \lim_{n\to\infty}\frac{\frac{\pi}{2}-\arctan(\ln n)}{\frac{1}{\ln n}}= \lim_{n\to\infty}\frac{-\frac{1}{n(1+\ln^{2}n)}}{-\frac{1}{n \ln^{2}n}}= \lim_{n\to\infty}\frac{\ln^{2}n}{1+\ln^{2}n}= 1$

and so the series of the modulus also diverges. Let's now see if we are in the conditions of the Leibniz Test. If $ a_{n}= \frac{\pi}{2}-\arctan(\ln n)$ then:

- $ a_{n}$ is always positive;
- $ \lim_{n\to\infty}a_{n}= 0$;
- $ a_{n}$ is monotone decreasing, because $ (a_{n})' =-\frac{1}{n(1+\ln^{2}n)}$ is strictly negative for $ n \ge 1$.

So, our initial series is simply convergent.


linear algebra
matrix
calculus
calculus computations

Consider the equation $ x'' \plus{} tx' \minus{} x \equal{} 0$.

Find the Wronskian $ W(t)$ for the pair of solutions $ (\phi,\psi)$ where $ \phi(t) \equal{} t$ and $ \psi$ is a solution satisfying $ \psi(0) \equal{} 1, \psi'(0) \equal{} 1$.
Using this Wronskian, find $ \psi(t)$.

[hide="Hint"]$ W(t) \equal{} \begin{vmatrix}\phi(t)&\psi(t)\\\phi'(t)&\psi'(t)\end{vmatrix} \equal{} \begin{vmatrix}t&\psi(t)\\1&\psi'(t)\end{vmatrix} \equal{} t \psi'(t) \minus{} \psi(t)$. So, $ W(0) \equal{} 0 \cdot 1 \minus{} 1 \equal{} \minus{}1 \neq 0$, and this means those two solutions are linearly independent, and so $ t \psi'(t) \minus{} \psi(t)$ is always different from zero.[/hide]

If $ \phi$ and $ \psi$ are solutions of $ y''\plus{}p(t)y'\plus{}q(t)y\equal{}0,$ then the Wronskian $ W(\phi,\psi)(t)$ satisfies $ W'(t)\plus{}p(t)W\equal{}0.$

Hence in this case, $ W'(t)\plus{}tW(t)\equal{}0,$ from which we deduce $ W(t)\equal{}Ce^{\minus{}t^2/2}.$

The other conditions given are sufficient to determine that

$ W(0)\equal{}\begin{vmatrix}0&1\\1&1\end{vmatrix}\equal{}\minus{}1.$

Hence, $ W(t)\equal{}\minus{}e^{\minus{}t^2/2}.$

Now, that gives us that

$ \begin{vmatrix}t&\psi\\ 1&\psi'\end{vmatrix}\equal{}t\psi'\minus{}\psi\equal{}\minus{}e^{\minus{}t^2/2}.$

This last equation - a first-order linear differential equation for $ \psi$ - does not have a closed-form elementary solution.


calculus
derivative
calculus computations

Find the derivation of:

$ f(x) \equal{} \sqrt {x \plus{} \sqrt {x \plus{} \sqrt {x \plus{} ...........\sqrt {x \plus{} \sqrt x } } } }$ (N radical)

I'm think that's very difficult!  :wink:  :wink:

This should be in the Calculus forum.
[hide="HINT"]
It is just the chain rule over and over.  Find the "pattern" and simplify.
[/hide]

We have $ f(x)\equal{}\sqrt{x\plus{}f(x)}$ and solving for $ f(x)$ gives $ \frac{1\plus{}\sqrt{1\plus{}4x}}{2}$.  Then taking the derivative should be pretty easy.

jli, the number of radicals is fixed, not infinite.  Personally, I would try a recursion in terms of $ N$.


floor function
number theory unsolved
number theory

Let $x_1$ be a given positive integer. A sequence $\{x_n\}_ {n\geq 1}$ of positive integers is such that $x_n$, for $n \geq 2$, is obtained from $x_ {n-1}$ by adding some nonzero digit of $x_ {n-1}$. Prove that

a) the sequence contains an even term;

b) the sequence contains infinitely many even terms.

If we assume the contrary, we get that each of the $x_i$ must have at least one even digit.  But since the $x_i$ are strictly increasing, and can increase by at most $8$ at each increment, $\left\lfloor\frac{x_i}{10}\right\rfloor$ increases in steps of one and zero without bound, and hence must pass through every positive integer greater than $\left\lfloor\frac{x_1}{10}\right\rfloor$.  But at least one of these will have all odd digits, and since the last digit of the corresponding $x_i$ is also odd, the next number in the sequence must be even.

For part (b), assume the contrary and let $x_k$ be the largest even value.  Then apply part (a) to $y_i=x_{i+k}$ to get a contradiction.

Bassically the same.

Suppose for the sake of condraction that there exists $ m\in\mathbb{Z}^ \plus{}$ such that $ \forall\; n\in\mathbb{Z}\cup[m,\infty]$, $ x_n$ is odd.
Then $ x_m$ must be odd and must have at least one even digit.
Suppose that $ x_m \equal{} \overline{a_k...E..a_0}$ (where $ \overline{ab} \equal{} 10a \plus{} b$) where $ E$ is first even digit counting from left to right in $ x_m$.
As $ \{x_n\}$ is increasing and it increases in no more than $ 8$ in each step there exists a positive integer $ l$ such that $ \overline{a_k...(E \plus{} 1)1...1}\le x_l\le\overline{a_k...(E \plus{} 1)1...19}$ by assumption $ x_l$ is odd, but in that case $ x_{l \plus{} 1}$ can't be odd, contradiction.
Then such an $ m$ does not exist, proving that there are infintely many even terms in the sequence.

:)


after i deleted my blog, i can't make a new one

Lesson? Don't delete your first blog!  :wink: .

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190223[/url]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=173209[/url]


Two persons are playing the following game. They have a pile of stones and take turns removing stones from it, with the first player taking the first turn. At each turn, the first player removes either 1 or 10 stones from the pile, and the second player removes either m or n stones. The player who can not make his move loses. It is known that for any number of stones in the pile, the first player can always win (regardless of the second player's moves). What are the possible values of m and n?


Solve for $ a$:

$ |a \minus{} 3|^2 \equal{}  8 \plus{}9|a \minus{} 3|$

I know its easy but I think its a nice problem.

[hide]Notice that the minimum value of the LHS is 0, and the maximum value of the RHS is -8.  Therefore, there are no solutions.[/hide]

I'm really sorry but there is an error in my original question. So I've edited it.[/hide]

[quote="JZambrano201"]Solve for $ a$:

$ |a \minus{} 3|^2 \equal{} 8 \plus{} 9|a \minus{} 3|$

I know its easy but I think its a nice problem.[/quote]

let $ x\equal{}|a\minus{}3|$

$ x^2\minus{}9x\minus{}8\equal{}0$

Then use quad equation to solve for x

are you sure its not $ \minus{}8$ instead of $ 8$?

do casework, first case let a >= 3
second let a < 3
i would do it however i do not understand the coding of this forum

[quote="jli"][quote="JZambrano201"]Solve for $ a$:

$ |a \minus{} 3|^2 \equal{} 8 \plus{} 9|a \minus{} 3|$

I know its easy but I think its a nice problem.[/quote]

let $ x \equal{} |a \minus{} 3|$

$ x^2 \minus{} 9x \minus{} 8 \equal{} 0$

Then use quad equation to solve for x

are you sure its not $ \minus{} 8$ instead of $ 8$?[/quote]

remember, x cannot return a negative value after you quadradic it.

[quote="JZambrano201"]Solve for $ a$:

$ |a \minus{} 3|^2 \equal{} 8 \plus{} 9|a \minus{} 3|$[/quote]
[hide="Solution"]Substituting $ u \equal{} |a \minus{} 3|$ and bringing everything over to the LHS, we get\[ u^2 \minus{} 9u \minus{} 8 \equal{} 0
\]Using the quadratic formula, we solve for $ u$:\[ u \equal{} \frac {9\pm\sqrt {81 \plus{} 32}}{2}
\]Since $ u$ cannot be negative,\[ u \equal{} \frac {9 \plus{} \sqrt {113}}{2}
\]Now, we solve for $ a$:\[ a \equal{} 3\pm\frac {9 \plus{} \sqrt {113}}{2}\quad\Longrightarrow\quad a \equal{} \left\lbrace\frac { \minus{} 3 \minus{} \sqrt {113}}{2},\ \frac {15 \plus{} \sqrt {113}}{2}\right\rbrace
\][/hide]


i)Let be $ A\equal{}\{y\in \mathbb{Z}\ |\ y\equal{}x\plus{}x^2\ ,\ x\in \mathbb{Z}\}$ and $ B\equal{}\{y\in \mathbb{Z}\ |\ y\equal{}x\plus{}x^4\ ,\ x\in \mathbb{Z}\}$ . Find $ A\cap B$ .

ii)Let be $ M\equal{}\{x\in \mathbb{Z}\ |\ (\exists)n\in \mathbb{Z}\ ,\ x^2\plus{}5x\plus{}4\equal{}n^2\}$ . Prove that $ M \subset [\minus{}5,0]$ .

ii) $ x^2 \plus{} 5x \plus{} 4 \equal{} (x\plus{}1)(x\plus{}4)$ so if $ n\equal{}x\plus{}1$ we need to find all $ n$ such that $ n(n\plus{}3)$ is a perfect square.
This is clearly true if $ n\equal{}0$ of $ n\equal{}\minus{}3$. If this is not the case, then $ |n|$ and $ |n\plus{}3|$ are both perfect squares, or are both a perfect square multiplied by $ 3$.

In the first case, we know that $ |n|$ and $ |n\plus{}3|$ are two perfect squares that differ by 3. Hence they are $ 1$ and $ 4 \rightarrow n\equal{}\minus{}4$ or $ n\equal{}1$.

In the second case, we know that $ |n|$ and $ |n\plus{}3|$ are both a perfect square multiplied by $ 3$, since they differ by 3, those perfect squares must differ by 1. Hence they are $ 0$ and $ 1$. But we've already checked that case.

So $ n$ can only take the values $ \minus{}4, \minus{}3, 0, 1$, then $ x$ can only take the values $ \minus{}5, \minus{}4, \minus{}1, 0$, and $ M \subset [\minus{}5,0]$, as wanted.

hello

alex2008 , you can give me the proof of  i)  ? thanks

[b]i)[/b] WLOG, $ x$ in $ A$ is non-negative (because $ x^2\minus{}x\equal{}(x\minus{}1)[(x\minus{}1)\plus{}1]$)
Then we write $ B$ like these: $ y\equal{}k^2\plus{}\sqrt{k}$ or $ k^2\minus{}\sqrt{k}$.

So Case 1:$ x^2\plus{}x\equal{}k^2\plus{}\sqrt{k}$, then we have if $ k<x$, $ R.H.S.<L.H.S.$.
If $ k \ge x\plus{}1$, then $ R.H.S. \ge (x\plus{}1)^2 >L.H.S.$.
So $ k\equal{}x$ gives solutions $ y\equal{}2$, which is a solution to the original question after checking.

Case 2:$ x^2\plus{}x\equal{}k^2\minus{}\sqrt{k}$, then we have if $ k<x$, $ R.H.S.<L.H.S.$.
If $ k \ge x\plus{}1$, we have $ R.H.S. \ge k^2\minus{}k \ge x^2\plus{}x$,
equality holds only when $ k\equal{}0$ or $ 1$, which gives solutions $ y\equal{}0$
If $ x\equal{}k$,then $ k\equal{}0$ gives the same solution again.

So $ A \cap B\equal{}\{0,2\}$.

P.S. Sorry for my wrong solution before.

[quote="stephencheng"][b]i)[/b]
or $ \begin{cases} |x \minus{} z^2| \le |z \minus{} x|.......(1) \\
|x \plus{} z^2| \le |z \minus{} x|......(2) \end{cases}$

$ (1)^2 \plus{} (2)^2$, we get $ x^2 \plus{} z^4 \le x^2 \plus{} z^2$ [/quote]

I think this should be $ x^2 \plus{} z^4\le x^2 \minus{} 2zx \plus{} z^2$

Hello,

thanks stephencheng , but tell me please why $ z^4 \le z^2\minus{}2xz \Rightarrow z \equal{} \minus{} 1,0,1$