Lately, I've been comparing math to cello practice a lot, and I noticed that the components of cello practice are far more defined (this is going somewhere related to AMC) . It is basically comprised of three things:

1)Scale/Double Stops/Bowing/Other Excercises(basics of playing)

2)Etudes (applying and concentrating on technique)

3)Repertoire (the real deal)

As I've been planning this year's AMC circuit training I'm trying to organize a better plan. I was wondering if anyone would like to share components of their practice regiment. I'll start with what I've come up with so far.

1)Memorizing Formulas, Patterns, Computation Tricks (basics you need to use).

2)Find/do problems with the same formulas,patterns, and tricks you learned (applying and concentrating on technique).

3)Work on a longer and more challenging problem. This does not have to be finished in one sitting (much like learning a piece), and your work for the day may just be making your plan of attack. Pay special attention to current concepts you are learning and how you can apply them. (This is the real deal part).

I guess its a nice idea that you have going.

But spending time memorizing formulas is likely not the best step. I would suggest fusing your steps 1 and 2. Not only will you be doing problems then, but through repetition, you will not only be learning the formulas, but also to just what extent their usefulness is (an imperative step).

I like the challenging problem part. Even as you do simple problems, they get quite easy after a while. Challenging problems (and spending a long time on them after numerous failures) is one of the best ways, in my opinion, to learn more fully a mathematical concept/method.

I think I might borrow from this  :D

Yeah, the challenging problems thing came from me thinking about how I could spend weeks or even months on a piece, but I got impatient if I spent more than 20 min. on a problem. I was reading "How to Solve It" and there were things about doing parts of a problem and putting it together and I realized that one problem could last me for days.   Thanks for your ideas about fusing 1 and 2. That makes a lot more sense now that I think about it. I think that the AoPS books I bought will help me in all of this since they often have a concept plus challenge problems directly related to the topic, and all the problems I found on here should give me no end to challenges. One flaw I see in my practice components though, is that I have not decided how to divide my practice time up.

If you have the AOPS books, just learn concepts + problems from there, then do some other challenging problems.

You need to know the basics - the formulas - but they aren't something you can practice with anything equivalent to scales. In music, you don't learn how to apply the scales in new ways - they're the scales, pretty unchanging. Compare that with, for example, the Pigeonhole Principle, which is applied in an entirely different way in every problem. Practicing with almost trivial problems doesn't help you understand how to apply the principle to more difficult ones. In math, you just have to dive in.

[quote="worthawholebean"] In music, you don't learn how to apply the scales in new ways - they're the scales, pretty unchanging. [/quote]

While I am pretty sure that if it were not two in the morning, I could come up with a slightly more nuanced argument to refute this point for classical music as well, but for now, in jazz, at least, it is absurd to say that you don't learn how to apply the scales in new ways.

As a simple example, take all the notes of a C major scale other than the root: D,E,F,G,A, and B.

If you are playing that partial scale over a C maj 7 chord, it lends the very consonant sound of a major scale without the root.

If, however, you play the same notes in a different context, over say a C# dominant 7 chord, it provides the sometimes-desired dissonance of an altered-dominant scale without the root--What is essentially the same scale can be applied in a more creative way in a less obvious situation, just as, as you say, the pigeonhole principle can be applied more creatively to a less obvious problem.

[quote="cellolad"]Well, I think the point we can all agree on is that it is impossible to solve problems without knowing your basics no matter how much of a creative thinker you are. For example, if you did not anything about triangles, you would have a hard time naming the length of the hypotenuse in a right triangle with legs of 3 and 4. Also, many formulas/patterns can save you a lot of time, especially if it's related to computation. It is also essential to know when and how to apply these basics. I do agree that solving trivial problems does not always equal solving the big ones, but I often need to have a simple problem to learn a new concept, and I do think that a hard problem can be broken down into one or many simple ones.[/quote]

This can be put in one sentence.
"Everyone starts off as a beginner."

It is certainly a good idea to solve some simple problems to solidify your understanding of an idea before moving on to harder ones. Many AoPSers have probably gotten away without doing this so far because they have good natural intuition for mathematics, but even in this case, by the time you get to higher-level math you will not yet have the intuition to continue without some basic examples.

Terence Tao said something along the lines that you don't know how powerful a technique is until you have found cases where it doesn't work. This is at least one reason why it's important to do hard problems.

I also like the idea of coming up with a plan of attack. My favorite thing to do when solving problems is coming up with a good plan, although this has been my downfall sometimes when the real thing to do is just to dive in and try examples.

[quote="worthawholebean"]I'm not a musical expert, so don't trust what I say about music. If I'm wrong, I'm wrong. My point was that regardless of situation, the scales are still the same notes in the same order up to rotation.[/quote]
I think he was referring to the technical aspect. Anyone can pick out the right notes, but playing them clearly, fast, and with correct technique takes some degree of practice.

In classical music, a scale is technique wise pretty much the same anywhere on any instrument, most likely. Like he said in jazz, it may not be the case.

As to jazz, I'm pretty sure he was talking about jazz improvisation.

[quote="JSteinhardt"]
I also like the idea of coming up with a plan of attack. My favorite thing to do when solving problems is coming up with a good plan, although this has been my downfall sometimes when the real thing to do is just to dive in and try examples.[/quote]

I think that even diving in and trying examples can be planned. You think about what kind of examples relate to your problem, and if one example doesn't help, you think about what other examples you can use.

[quote="JSteinhardt"]
Terence Tao said something along the lines that you don't know how powerful a technique is until you have found cases where it doesn't work. This is at least one reason why it's important to do hard problems.[/quote]

This is a really state something great you shared. I also think that you discover many new things about a technique when it is put to the test in a difficult situation.


function
integration
calculus
algebra
domain
derivative
calculus computations

I am very confused. If a function is conservative its curl F=0, but stokes thereom says:
$\int\int_{S}curl\vec{F}\cdot\,dS$

Does this mean that Stoke's Theory allows us to find line intregrals for surfaces that are piecewise-smooth and bounded by a simple, closed, piecewise-smooth boundary of curve C as long as it's [b]not[/b] conservative???

As long as the regularity conditions are met (the curve and the surface are "nice enough," the closed curve bounds the surface in the appropriate way, and $\vec{F}$ is differentiable on the whole surface), then Stokes's Theorem is true, period, whether the field is conservative or not.

If $\vec{F}$ happens to be a conservative field (i.e. $\exists\,f$ such that $\nabla f=\vec{F}),$ then Stokes's theorem tells you that the integral over any closed curve of $\vec{F}\cdot d\vec{r}$ is zero.

Of course, you knew that anyway, from the fundamental theorem of calculus for line integrals: if the curve $C$ starts at $\vec{p_0}$ and ends at $\vec{p_1},$ then

$\int_C \vec{F}\cdot d\vec{r}=\int_C\nabla f\cdot d\vec{r}=f(\vec{p_1})-f(\vec{p_0}).$

For a closed curve, $\vec{p_1}=\vec{p_0}$ and this is zero.

So curl F is only zero if it is conservative and a closed? I think I missed that!

We've got to be careful here:

$\vec{F}$ is conservative implies $\text{curl }\vec{F}=\vec{0}$ everywhere.

If $\text{curl }\vec{F}=\vec{0}$ everywhere on a simply connected domain, then $\vec{F}$ is conservative.

But the condition of "simply connected" matters. Here's the classic example:

$\vec{F}=\frac{-y\vec{i}+x\vec{j}}{x^2+y^2}$

$\text{curl }\vec{F}=\vec{0}$ everywhere except the $z$ axis, where the field and its derivatives are undefined. However, the set $\mathbb{R}^3$ with the $z$ axis deleted is not simply connected. If we take a closed curve $C$ that loops around the $z$ axis - most simply the unit circle in the $xy$ plane, once counterclockwise - then

$\int_C \vec{F}\cdot d\vec{r}=2\pi.$ Since this is not zero, the field is not conservative.

(Roughly speaking, $\vec{F}=d\theta,$ where $\theta$ the the angle defined in cylindrical coordinates. The problem is that we can't define $\theta$ as a continuous function all the way around.)

And fields are not "closed." I only used the word closed to refer to a curve. A curve is closed if its endpoint is the same as its beginning point.

A field $\vec{F}$ is conservative if and only if $\int_C \vec{F}\cdot d\vec{r}$ for every closed curve $C$ in the domain of $\vec{F}.$


inequalities
real analysis
triangle inequality
calculus
calculus computations

Can someone give me a small hint to show the triangle inequality of $ \|. \|_{\infty}$

For almost all $ x,$ $ |f(x)|\le \|f\|_{\infty}$ and for almost all $ x,$ $ |g(x)|\le\|g\|_\infty.$ Stay away from the set where either one of those inequalities is false - as the union of two sets of measure zero, it's a set of measure zero. Then we can say that

$ |f(x)\plus{}g(x)|\le |f(x)|\plus{}|g(x)|$ (by the triangle inequality for $ \mathbb{C}$ or $ \mathbb{R}$)

You finish it from there.


geometry
circumcircle
number theory proposed
number theory

Let $ P$ be a regular polygon. A regular sub-polygon of $ P$ is a subset of vertices of $ P$ with at least two vertices such that divides the circumcircle to equal arcs. Prove that there is a subset of vertices of $ P$ such that its intersection with each regular sub-polygon has even number of vertices.

Just use induction for number of prime divisors of n and you can easily find a non empty subset for the question


ok so I have a function y = f(x), and I need to find the points where f(x) = f(f(x)) so I figure this implies that x = f(x) so the solutuions must be on the like y = x.  Is this right? It seems to make sense to me but I'm not really sure if my logic is right

You are correct, but there may be some places in which the inverse function is undefined.

This isn't necessarily correct.  Your statement that $x = f(x)$ relies on the assumption that $f(x)$ is one-to-one; that is, if $f(x) = f(y)$, then $x = y$.   Otherwise, there could be some $x \neq f(x)$ such that $f(x) = f(f(x))$, and your assumption would be invalid.  An example is $f(x) = x^2$.  In this case $f(x) = f(y)$ if $x = - y$.  This would then yield the result $f(x) = f(f(x)) \Rightarrow x^2 = (x^2)^2 \Rightarrow x = 0, x = 1, x = -1$, and we see that the solution $x = -1$ violates the assumption that $x = f(x)$.

Well, you can make the assumption $x=f(x)$ if you know it's a linear function...

And if not, then it should be $|x|=f(x)$, I think...


pigeonhole principle
modular arithmetic
combinatorics unsolved
combinatorics

Let A consist of 16 elements of the set {1, 2, 3 ..., 106} so that no two elements of A differ by 6, 9, 12, 15, 18 or 21. Prove that two elements of A should differ by 3.

[quote="Albanian Eagle"]Look at things $ \pmod{3}$
at least 6 of your chosen numbers are going to be congruent to each other. Their difference must be a multiple of 3. In case it can't be 3,6,9,12,15,18,21 it has to be 24 or higher for any two of them so the difference between the biggest and the largest is at least 120.
Contradiction![/quote]

Thanks so much! :)


An exciting new ride in a theme park is described as follows: "The vehicle plus riders takes only a few seconds to accelerate to a spine-tingling 53m/s. Then (with motor off) it shoots vertically upwards to the height of a forty-two storey building and pauses."

Calculate a value for the average resistive force acting on the vehicle during the climb.

Thank you!

Why don't you try showing us some work and beginning steps?

[quote="Bobtheball"]calculate using energy.
Find the difference in energy between the kinetic and potential energies and then the energy lost is due to the resistive forces on the rail (neglecting air resistance). Then divide the energy lost by the length of the rail and you have your answer.[/quote]

thanks very helpful!

Calculate the time for the vehicle plus riders to make the vertical climb to 126m. State an assumption that you make.

(Power needed here at all? The only information given is what's in the question I gave before. Assumption= air resistance negligible?)


What book should I use to prep for the Physics Olympiad? I am looking to get a headstart on it. Thanks!

Once you know the basics, the rest is just natural ability (in my opinion).  I would pick up any introductory college physics textbook.  Halliday/Resnick is good, and I've heard the book they use as MIT is really hard (I can't remember the authors).

Yes.  Volume 1 is Mechanics and Volume 2 is E&M, although both books are expensive.

does ne1 think 'university physics' is a nice one.....

There's about 100 books called "University Physics" so you'll have to be a little more specific.

I have both books in pdf, but personally I think University Physics is too easy. I can do all the problems, including the challenge ones. The text in too small, but that is all the bad about it. Fundamentals of Physics, however has pretty cool problems and better examples to help you understand. I am not sure but Fundamentals of Physics alone can get you into the semifinal, maybe?

If you are only interested in Olympiad physics, I would buy Irodov Problems in general Physics.  Those problems are VERY hard though, and won't help you too much unless you are fluent in calculus (some multivariate calc would help on some problems).  Oh, this book has only problems and solutions.   Might not be what you look for if you just want to learn.

A book to learn:

An Introduction to Mechanics, by Kleppner and Kolenkow.  It's hard, and the problems are very similar to olympiad problems, but it's helpful to know some vector calculus (there are sample problems in the advanced physics forum, posted by me)

E&M: I like to study from Tipler's Physics for scientists and engineers for E&M, and all other topics.  If it's too easy for you, then you're pretty much set to do well.

So do we have to learn vector calculus for the olympiad? If so, that is unfair!

You don't need vector calculus.

And the Kleppner and Kolenkow is the book used at MIT (I think).

Phew! Yeah, but don't you have to learn the basics first to use that book, or does that book teach stuff that is understandable to someone new in physics?

It would help to know some physics going in since that will make it easier to understand the second time around, but you don't need to know any physics.

the UNIVERSITY PHYSICS i mentioned is authored by HUGH.D.YOUNG

These books are not going to teach you how to solve Physics Olympiad problems.  They will teach you the basic physics, and then expect you to understand the concepts well enough that you can apply the methods you learned.  Make sure you understand this.


Let $c$ be a given fixed real number. Let $M$ be a infinite number of integers, including 0. And the distance among neighboring elements in $M$ is not greater than $c$ (assumed that you look at the set in as the numbers would be sorted increasingly). Consider the following step: Let $X$ be a given set of integers, then generate the sets:

\[X \pm M = \{x \pm s | x \in X, s \in M\}.\]

Starting with $S_0 = 0$ generate consecutively $S_1, S_2, S_3, \ldots$ by choosing for $X$ consecutively $S_0, S_1, \ldots$. Prove that after a finite steps, there will no newly generated set, e.g. there is no number $k_0$ such $S_k = S_{k_0}$ $\forall k \geq k_0$ and $k, k_0 \in \mathbb{N}$.

[hide="an example"] Let $M$ be composed of the number 1 and all non-negative numbers being divisible by 5. Let $\bar{i} = 0,1,2,3,4$ be the residue class $i$ modulo 5 in $\mathbb{Z}$. Then we can represent $M$ as $M = \bar{0} \cup \{1\}$.

Let $S_0 = \{0\}$. Then we get for $S_1 = S_0 \pm M = \{0\} \pm (\bar{0} \cup \{+1\}) = \bar{0} \cup \{1,-1\}$, for $S_2 = S_1 \pm M = \bar{0} \cup \bar{1} \cup \bar{4}$.  For $S_3$ we get $S_3 = S_2 \pm M = (\bar{0} \cup \bar{1} \cup \bar{4}) \pm (\bar{0} \cup \{1\}) = \bar{0} \cup \bar{1} \cup \bar{2} \cup \bar{3} \cup \bar{4} \cup \bar{5} = \mathbb{Z}$. Thus we also get $S_4 = S_3 \pm M = \mathbb{Z} \pm (\bar{0} \cup \{1\}) = \mathbb{Z}$.

Hence $\forall k \geq 3$ we have $S_k = S_3 = \mathbb{Z}$. That means for $k_0 = 3$ our existence property is satisfied.
 [/hide]

Let $M=\{a_1<a_2<a_3<...\}$, $d=\liminf_{n\geq 1} (a_1,...,a_n)$.
Then $d=(a_1,...,a_m)$ and we can express $d$ as a linear integer combination of $a_1$, ..., $a_m$. Since $a_{i+1}-a_{i}\leq c$ and $d\mid a_i$, we conclude that $S_k=d\mathbb{Z}$ for all $k\geq k_0$.


probability
MATHCOUNTS
complementary counting

I got this question in saab:

We select  3  numbers at random, with replacement, from the set of integers from  1  to  600  inclusive. What is the probability that the product of the  3  numbers is  even  ? Express your answer as a commom fracton in lowest terms.

I worked it this way:

The possibilities for it to be even is OOE, OEE, or EEE, where O stands for an odd number and even stands for an even integer. Then, it is just $1/8$ for each of those resulting in an answer for $3/8$.

However, the saab person used complementary counting, and said hte only possible way for it was $OOO$ for it to be not even. So, the answer was $1-1/8=\fbox{7/8}$

What did I do wrong :(

If you count it that way, order matters. So you're missing the other permutations.

the possibilities for it to be even are:
OOE
OEO
EOO
OEE
EOE
EEO
EEE

so it's 7/8..

EDIT: ehh yeah.. what davidli said :lol:

i think i know what u did. the probability of getting EEO or EOO is not the same as getting EEE. think about heads and tails. the probability of getting HHH is not the same as the probability of getting HHT or TTH.

But complimentary counting is the way i would do it. it is way easier, but if u like to do things differently, u can do it the other way. i just wouldn't advise doing it in a mathcounts test unless ur insanely fast and have time to spare...

edit: oops, what the other two said.... :oops:


Let $N$ be the number of positive integers that are less than or equal to 2003 and whose base-2 representation has more 1's than 0's.  Find the remainder when $N$ is divided by 1000.

[hide]1s; There is 1 number where the # of 1s exceed the # of 0s and 0 numbers where the # of 1s doesn't exceed the # of 0s.
10s: There is 1 number where the # of 1s exceed the # of 0s and 1 number where the # of 1s doesn't exceed the # of 0s.
100s:  There are 3 number where the # of 1s exceed the # of 0s and 1 number where the # of 1s doesn't exceed the # of 0s.
1000s:  There are 4 number where the # of 1s exceed the # of 0s and 4 numbers where the # of 1s doesn't exceed the # of 0s.
...  And so on.
One quickly notices that the formula for the number of numbers where the the 1s exceed the 0s between $2^n$ and $2^{n+1}-1$ is $\binom{n}{n}+\binom{n}{n-1}+...+\binom{n}{\lceil {n/2} \rceil}$.  The formula for when the opposite occurs is $\binom{n}{0}+\binom{n}{1}+...+\binom{n}{\lceil {n/2} \rceil-1}$.  These formulas are easily proven when one tries to list all the possiblities for the 1s, 10s, 100s, and so on, and then tries to group them into their appropriate catagories.  When n is odd, these two are equal.  Therefore, the # of numbers between {2,3}{8,15}{32,63}{128,255}{512,1023} that satisfy this condition are 1+4+16+64+256=341.
However, when n is even, these two aren't equal, the first equation is greater than the second because it has an additional $\binom {n}{n/2}$.  But since n is known, these values are $\frac{2^n-\binom{n}{n/2}}{2}+\binom{n}{n/2}$.  This means that the # of numbers between 1+{4,7}{16,31}{64,127}{256,511}{1024,2047} that satisfy this condition are 1+3+11+42+163+638=858.  However, one must realize that the problem only goes to 2003, so the last 45 numbers are discarded.  Those last 45 numbers all satisfy this condition because 2003 in binary is 11111010011.  Notice that there's 11 digits and that the first 5 digits are 1s.  The only way for a number with those requirements to not follow the problem's original requirements is if the rest of the digits are zeroes, which would then be less than 2003.  Therefore, the remainder when divided by 1000 is 858+341-45-1000=154.[/hide]

I used the exact same method. I believe your answer is 1 off though.

[hide]At the end of the problem, you said there are $45$ numbers from the interval $2004$ to $2048$ which need to be subtracted.

However, $2048$ was not counted in the casework process. (Note that it has 12 digits, not 11.) Hence, 44 numbers should've been subtracted, for an answer of $155.$[/hide]

I got 155 too. http://www.rolyata.com/?p=53 (I would rather not rewrite it as the writing is not exactly "cross-platform".

[quote="dengmi"][hide]1s; There is 1 number where the # of 1s exceed the # of 0s and 0 numbers where the # of 1s doesn't exceed the # of 0s.
10s: There is 1 number where the # of 1s exceed the # of 0s and 1 number where the # of 1s doesn't exceed the # of 0s.
100s:  There are 3 number where the # of 1s exceed the # of 0s and 1 number where the # of 1s doesn't exceed the # of 0s.
1000s:  There are 4 number where the # of 1s exceed the # of 0s and 4 numbers where the # of 1s doesn't exceed the # of 0s.
...  And so on.
One quickly notices that the formula for the number of numbers where the the 1s exceed the 0s between $ 2^n$ and $ 2^{n \plus{} 1} \minus{} 1$ is $ \binom{n}{n} \plus{} \binom{n}{n \minus{} 1} \plus{} ... \plus{} \binom{n}{\lceil {n/2} \rceil}$.  The formula for when the opposite occurs is $ \binom{n}{0} \plus{} \binom{n}{1} \plus{} ... \plus{} \binom{n}{\lceil {n/2} \rceil \minus{} 1}$.  These formulas are easily proven when one tries to list all the possiblities for the 1s, 10s, 100s, and so on, and then tries to group them into their appropriate catagories.  When n is odd, these two are equal.  Therefore, the # of numbers between {2,3}{8,15}{32,63}{128,255}{512,1023} that satisfy this condition are 1+4+16+64+256=341.
However, when n is even, these two aren't equal, the first equation is greater than the second because it has an additional $ \binom {n}{n/2}$.  But since n is known, these values are $ \frac {2^n \minus{} \binom{n}{n/2}}{2} \plus{} \binom{n}{n/2}$.  This means that the # of numbers between 1+{4,7}{16,31}{64,127}{256,511}{1024,2047} that satisfy this condition are 1+3+11+42+163+638=858.  However, one must realize that the problem only goes to 2003, so the last 45 numbers are discarded.  Those last 45 numbers all satisfy this condition because 2003 in binary is 11111010011.  Notice that there's 11 digits and that the first 5 digits are 1s.  The only way for a number with those requirements to not follow the problem's original requirements is if the rest of the digits are zeroes, which would then be less than 2003.  Therefore, the remainder when divided by 1000 is 858+341-45-1000=154.[/hide][/quote]

How did you get the formula $ \binom{n}{n}\plus{}\binom{n}{n\minus{}1}\plus{}...\plus{}\binom{n}{\lceil {n/2} \rceil}$.  I don't get how the ceiling functions were made.  Thanks!

When $ n$ is odd, you must choose more than half in order to have more ones, which results in the ceiling, but if $ n$ is even, half will suffice.

[quote=E^(pi*i)=-1][hide="A Different Approach"]
We seek the number of allowed numbers which have $ k$ 1's, not including the leading 1, for $ k\equal{}0, 1, 2, \ldots , 10$.

For $ k\equal{}0,\ldots , 4$, this number is

$ \binom{k}{k}\plus{}\binom{k\plus{}1}{k}\plus{}\cdots\plus{}\binom{2k}{k}$.

By the Hockey Stick Identity, this is equal to $ \binom{2k\plus{}1}{k\plus{}1}$.  So we get

$ \binom{1}{1}\plus{}\binom{3}{2}\plus{}\binom{5}{3}\plus{}\binom{7}{4}\plus{}\binom{9}{5}\equal{}175$.

For $ k\equal{}5,\ldots , 10$, we end on $ \binom{10}{k}$ - we don't want to consider numbers with more than 11 digits.  So for each $ k$ we get

$ \binom{k}{k}\plus{}\binom{k\plus{}1}{k}\plus{}\binom{10}{k}\equal{}\binom{11}{k\plus{}1}$

again by the Hockey Stick Identity.  So we get

$ \binom{11}{6}\plus{}\binom{11}{7}\plus{}\binom{11}{8}\plus{}\binom{11}{9}\plus{}\binom{11}{10}\plus{}\binom{11}{11}\equal{}\frac{2^{11}}{2}\equal{}2^{10}\equal{}1024$.

The total is $ 1024\plus{}175\equal{}1199$.  Subtracting out the $ 44$ numbers between $ 2003$ and $ 2048$ gives $ 1155$.  Thus the answer is $ 155$.
[/hide][/quote]

Could someone explain this solution a bit more? I am really confused on how this person got the first expression. Thanks (I get the rest after that).

@above I am also confused, lol. I found a neat solution which I'll put in wiki and here in detailed:

We will count the number of it $< 2^{11}=2048$ instead of $2003$ (In other words, the length of the base-2 representation is at most $11$. If there are even digits, $2n$, then the leftmost digit is $1$, the rest, $2n-1$, has odd number of digits. In order for the base-2 representation to have more $1$'s, we will need more $1$ in the remaining $2n-1$ than $0$'s. Using symmetry, this is equal to
$\frac{2^9+2^7+..+2^1}{2}$
Using similar argument where there are odd amount of digits. The remaining even amount of digit must contains the number of $1$'s at least as the number of $0$'s. So it's equal to
$\frac{\binom{10}{5}+2^{10}+\binom{8}{4}+2^8+\binom{6}{3}+2^6+...+\binom{0}{0}+2^0}{2}$
Summing both cases, we have $\frac{2^0+2^1+..+2^{10}+\binom{0}{0}+..+\binom{10}{5}}{2} = 1199$. There are $44$ numbers between $2004$ and $2047$ inclusive that satisfy it. So the answer is $1199-44=1\boxed{155}$


function
real analysis
real analysis unsolved

hi
construct a function f:R -> R that takes every real value on every interval - in other words, for every a < b and every c there is an x with a < x<b such that f(x)=c.
thanks

There are lots of ways of constructing such a function.

[b]Method 1.[/b] Assuming AC, it is not so hard.

Construct a function $ g : \mathbb{Q} \to \mathbb{Q}$ such that for any $ a < b$, the image of $ (a, b)$ under $ g$ is $ \mathbb{Q}$.

Let $ \{x_\alpha \in \mathbb{R} \}$ be a set of representatives of the quotient group $ \mathbb{R}/\mathbb{Q}$.

Then define $ f(x) \equal{} x_{\alpha} \plus{} g(r)$ if $ x \equal{} x_{\alpha} \plus{} r$ for some $ r \in \mathbb{Q}$.

Every real number assigns a unique such decomposition, so $ f$ is well-defined.

Finally, it is an easy task to check that $ f$ does satisfy the desired property.

One can ask if such $ g$ exists, but it is not hard to construct such a function. Here is an example of a construction:

[hide="Construction"]Consider a rational number of the form $ r \equal{} 2^{ \minus{} m}q \plus{} p^{ \minus{} n}$, where $ q$ is an odd integer, $ m, n \geq 0$ and $ p$ is an odd prime satisfying $ p > \max ( |q|, 2^m )$.

Clearly, any rational number can be represented as at most one such form.

Let $ \{ r_k \}_{k \equal{} 1}^{\infty}$ be an enumeration of $ \mathbb{Q}$, and for a fixed $ m, q$, list all such primes in increasing order and write them as $ p_1, p_2, p_3, \cdots$.

Then assign $ g(r) \equal{} r_k$ for each $ r \equal{} 2^{ \minus{} m}q \plus{} p_k^{ \minus{} n}$. For a rational number $ r$ which cannot be written as this form, just let $ g(r) \equal{} 0$.[/hide]

[b]Method 2.[/b] It suffices to construct a function $ f : [0,1] \to [0,1]$ such that for any $ 0 \leq a < b \leq 1$, the image of $ [a, b]$ under $ f$ is $ [0, 1]$.

For each $ x \in (0, 1]$, there is a unique non-terminating binary representation $ x \equal{} 0.a_1 a_2 a_3 \cdots$.

Then define $ f(x) \equal{} \limsup_{n\to\infty} \frac{\# \{ k \, | \, 1 \leq k \leq n \text{ and } a_k \equal{} 1 \} }{n}$ and $ f(0) \equal{} 0$. Then $ f$ satisfies the following properties:

1. $ f([0, 1]) \equal{} [0, 1]$

2. $ f(x) \equal{} f(2^{\minus{}m}x)$ if $ x \in [0, 1]$ and $ m \geq 0$

3. $ f(x) \equal{} f(x\plus{}2^{\minus{}m})$ if both $ x, x\plus{}2^{\minus{}m}$ are in $ [0, 1]$.

This ensures that $ f$ satisfies the desired property.


http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14696

http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14697

 :(

[color=darkblue]Let $ \omega$ be the circumcircle and $ O$ its center.
By Leibnitz $ \boxed{PA^2 \plus{} PB^2 \plus{} PC^2 \equal{} GA^2 \plus{} GB^2 \plus{} GC^2 \plus{} 3PG^2}$.
So $ P \equal{} P_{min}$ when $ PG \equal{} min$ and $ P \equal{} P_{max}$ when $ PG \equal{} max$.

We only draw a line through points $ O,G$.  Then $ \boxed{[GO) \cap \omega \equal{} P_{max}}, \boxed{[OG) \cap \omega \equal{} P_{min}}$.

I hope my solution is correct. Isn't it?[/color]

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14700[/img]

 :thumbup:


solve equation :
       $x^{2005} +y^{2005} =2007^{2005}$
 where$x,y$  is positive integers

Fermat's Last Theorem implies no solution.  Or are we supposed to prove no solution by more elementary methods?   :huh:

no we have solution x=0 and y=2007 or x=2007 and y=0 i think it's the only solution

Excuse me, sir, but... 0 is not supposed to be a positive integer. :?

We'll first show that $2<(\frac{2007}{2006})^{2005}$.

It is easy to show that $a_n=(\frac{n+1}{n})^{n+1}$ is a decreasing sequence with limit $e$, so any term of this sequence is $>e$. Then $a_{2006}>e$. Now, it would be enough to show that $2<e \times \frac{2006^2}{2007^2}$, which is easy (just do very rough bounds). Thus, our assertion is proven, and from it, it follows that $2007^{2005}>2 \times 2006^{2005}$. Then one of $x, y$ must be $\geq 2007$. But then the other must be $\leq 0$, absurd.


geometry
circumcircle
symmetry
angle bisector
geometry unsolved

Let $ ABC$ be an acute triangle. Consider all equilateral triangles $ XYZ$ in the plane, such that the points $ A,B,C$ lie on segments $ YZ,ZX,XY$ respectively. Find the locus of the centers of triangles $ XYZ$. (The centers of triangles are defined as the point where the medians intersect)

I am surprized it is from Polish 2003, since first time I saw this problem 15 years ago.

Construct 3 arcs on sides of $ ABC$ s.t. all of them have measure $ 4\pi/3$. Choose arbitrary point $ X$ on arc $ AB$, draw rays $ XA$ and $ XB$, let $ Y$ and $ Z$ be their intersections with another 2 arcs. It is easy to show that $ Y$, $ C$ and $ Z$ are collinear. It means that we have showed how to construct all possible equilateral triangles $ XYZ$ with desired property.

Now let use some complex numbers. Suppose $ O_1$ be circumcenter of $ ABX$, $ O_2$ be circumcenter of $ BCZ$, $ r_1$ and $ r_2$ are their radii, $ \phi$ is an argument of p.$ X$.
It is a good exercise to show that angle between $ O_1X$ and $ O_2Z$ is const, denote it as $ \alpha$. Using it we obtain that $ Z\equal{}O_2\plus{}r_2e^{i(\phi\plus{}\alpha)}$. Thus $ Z\minus{}X\equal{}O_2\minus{}O_1\plus{}(r_2e^{i\alpha}\minus{}r_1)e^{i\phi}\equal{}O_2\minus{}O_1\plus{}a\cdot e^{i(\phi\plus{}b)}$ for some fixed $ a$ and $ b$. Let $ O$ be center of $ XYZ$, then $ O\equal{}X\plus{}\frac{\sqrt3}2(Z\minus{}X)e^{i\pi/6}\equal{}c_1\plus{}c_2e^{i(\phi\plus{}c_3)}$.
Since $ \phi$ varies on some segment $ [\phi_1,\phi_2]$, we conclude that locus $ O$ is an arc.

Thanks, Myth. I appreciate your superb depth of knowledge.

Anyway, are there any nice synthetic solutions?

[quote="Altheman"]I remember that this problem had a very difficult synthetic solution.[/quote]
could you please post the link?

Let $ P$ be the fermat point of $ ABC$, let $ \omega_1$, $ \omega_2$, and $ \omega_3$ be the circumcircles of $ PYZ$, $ PZX$, and $ PXY$, respectively, let $ X_1$, $ Y_1$, and $ Z_1$ be the midpoints of $ YZ$, $ ZX$, and $ XY$, respectively, let $ XX_1$, $ YY_1$, and $ ZZ_1$ intersect $ \omega_1$, $ \omega_2$, and $ \omega_3$ again at $ X_2$, $ Y_2$, and $ Z_2$, respectively, let $ G$ be the centroid of $ XYZ$.

Clearly, $ X$, $ Y$, and $ Z$ lie on $ \omega_1$, $ \omega_2$, and $ \omega_3$, respectively. Observe that $ XX_2$ is the angle bisector of $ \angle YXZ$, so $ X_2$ is the midpoint of minor arc $ AB$ of $ \omega_3$. By symmetry, $ Y_2$, and $ Z_2$ are fixed.

By considering convenient cases, we can determine that $ X_2Y_2Z_2$ is equilateral (when $ X_2G$ is tangent to the circumcircle of $ \triangle X_2Y_2Z_2$ we easily find $ Y$, then draw lines to intersect circles, and show $ B$, $ X$, and $ Z$ are collinear, then $ \angle X_2Y_2Z_2 \equal{} \angle XGZ_1 \equal{} 60^\circ$.)

When $ G$ is on segments $ Y_2Y$ and $ Z_2Z$, $ \angle Y_2GZ_2 \equal{} 120^\circ$ when $ G$ is on the circumcircle of $ X_2Y_2Z_2$. We can repeat that for other cases to show that $ G$ is always on that circumcircle.

Now we have to show that all of those points lie on the locus but that isn't actually true because $ P$ is not on the locus...There is an arc that is not on the circle also. The restriction that $ A$ lie on segment $ YZ$ makes this problem difficult because it sorta cuts the circle in a difficult place to measure, etc.

1987 ISL #12 (POL): Given non-equilateral triangle $ ABC$, the vertices listed counterclockwise, find the locus of the centroids of the equilateral triangles $ A'B'C'$ (the verticies listed counterclockwise) for which the triples of points $ A,B,C'$; $ A',B,C'$; and $ A',B,C$ are collinear.

--

This particular problem is different because it stipulates that $ A,B,C$ must lie on the segments $ XY$, etc. I think this makes the problem harder and more ugly.

The locus is an arc, but I think that it needs to be specified better...


algebra
polynomial
inequalities
geometry
geometric transformation
dilation
algebra unsolved

I'm sorry if this problem is too easy. :(  I think that it is famous problem,but i can't find a link.
Let $ f(x)\equal{}x^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\dots\plus{}a_{1}x\plus{}a_{0}$ be a polynomial.Let $ c\equal{}max\{|f(x)|\},x\in (a,b)$.Prove that $ c\geq 2(\frac{b\minus{}a}{4})^{n}$.

It is indeed wellknown that the monic polynomial of degree n which maximises the maximal deviation in $ [\minus{}1,1]$ is the n-th Chebyshev Polynomial which has maximal deviation $ 2^{1\minus{}n}$. To get your result just use translation and dilation of your polynomial  :)


what is the first Fibonacci number that is divisible by 1000000?

can we generalize for any n?

[hide="Hint"]
You may use the fact that $a|b \implies F_{a}| F_{b}$
[/hide]

i dont think that helps much?
anyone can solve the first part?

[hide="Some concepts"] Consider pairs of consecutive Fibonacci numbers $(F_{i}, F_{i+1}) \bmod m$ for an arbitrary positive integer $m$.  Since there are only at most $m^{2}$ such distinct pairs, and since such a pair determines any successive pairs, the Fibonacci sequence must be periodic $\bmod m$ for any $m$.

In another thread a poster posted an exact expression for the period that I have forgotten. [/hide]


Let $ B$ and $ C$ be the endpoints and $ A$ the midpoint of a semicircle. Let $ M$ be a point on segment $ AC$, and $ P,Q$, the feet of the perpendiculars from $ A$ and $ C$ to line $ BM$, respectively. Prove that $ BP\equal{}PQ\plus{}QC$.

Continue $ BQ$ to $ C'$ such that $ QC'\equal{}QC$. We just need to show that $ P$ is the midpoint of $ BC'$. Let the center of the semicircle A.K.A the midpoint of $ BC$ be $ O$. Then it suffices to show that $ OP$ is parallel to $ CC'$. Let $ \angle PAM\equal{}\theta$. Then, since $ APM\sim CQM$, $ \angle MCQ\equal{}\theta$, so $ \angle BCC'\equal{}\theta \plus{}\angle OCA\plus{}\angle QCC'\equal{}\theta\plus{}90$. Now, since $ \angle BAC\equal{}90, BAM\sim APM$. So $ \angle ABP\equal{}\theta$. Now notice that $ \angle APB\equal{}\angle AOB\equal{}90$. That means $ APOB$ is a cyclic quad, so $ \angle AOP\equal{}\angle ABP\equal{}\theta$. Hence, $ \angle BOP\equal{}\angle BOA\plus{}\theta\equal{}\theta\plus{}90$. Hence $ OP$ and $ CC'$ are parallel and $ P$ is the midpoint of $ BC'$ as required.

[quote="TheHeinrich"]Let $ B$ and $ C$ be the endpoints and $ A$ the midpoint of a semicircle. Let $ M$ be a point on segment $ AC$, and $ P,Q$, the feet of the perpendiculars from $ A$ and $ C$ to line $ BM$, respectively. Prove that $ BP \equal{} PQ \plus{} QC$.[/quote]
[hide="Solution"]
Let $ S$ be the point on segment $ BP$ such that $ PS \equal{} QC$ and $ R$ be the point such that $ \triangle PSR$ is congruent to $ \triangle CQP$.  This means that $ PR \equal{} PC$ and $ \angle RPC \equal{} 180 \minus{} \angle SPR \minus{} \angle QPC \equal{} 180 \minus{} \angle SPR \minus{} \angle SRP \equal{} 180 \minus{} (180 \minus{} \angle PSR) \equal{} 90$.  Thus, $ \triangle PCR$ is an isoceles right triangle as is $ \triangle ABC$, so $ \triangle ABC\sim \triangle PRC$.  Hence, a spiral similarity maps $ \triangle CAP\sim \triangle CBR\implies \angle CBR \equal{} \angle CAP$.  Now, let $ AP$ and $ BR$ meet at $ X$, so $ \angle XBC \equal{} \angle XAC$, so $ XBAC$ is cyclic, which means that $ X$ lies on the circle.  Since $ \angle APQ \equal{} 90\implies \angle BPX \equal{} 90 \equal{} \angle BSR\implies SR\parallel PX$, so $ \angle BRS \equal{} \angle AXB \equal{} \angle ACB \equal{} 45$, which means that $ \triangle BSR$ is a $ 45$-$ 45$-$ 90$ triangle, so $ SB \equal{} SR \equal{} PQ\implies PB \equal{} PS \plus{} SB \equal{} QC \plus{} PQ$, as desired. [/hide]

As in first solution, take $C'$ on $BQ$, produced so that $QC'=QC$ we just need to show that $AC'=AB$; also note that $Q$ lies on the semicircle ($BQ\bot CQ$).
It is quite obvious that $\angle AQC'=\angle AQC=135^\circ$, hence $\Delta AQC'\equiv\Delta AQC\implies AC'=AC$, but $AC=AB$ and our proof is completed then.

Best regards,
sunken rock


i think this topic has been posted before, but here it is again..
the following are two questions from old mathcounts test:
[b]Round $ \sqrt{55}$ to the nearest tenth[/b]
[b]Round $ \sqrt{91}$ to the nearest tenth[/b]

How do u do problems like this?

use a calculator!!! lol

the sqrt of 55 is between 7 and 8 or something

1) this is a sprint/countdown round question
2) knowing what two integers a square root is between doesn't really help me round it to the tenth :wink:

[quote="mihail911"]knowing what two integers a square root is between doesn't really help me round it to the tenth[/quote]
Actually it does, but only a little.

[hide="Solution"]Pick some number $ n$, and calculate $ 55\div n$. If possible, try to determine what the square root could be close to to avoid having to waste a lot of time. Whatever you get, average it with $ n$. Now take this number, and repeat the process. When you get the same tenths digit twice, you have evaluated $ \sqrt{55}$ correctly to $ 2$ place values.[/hide]

[quote="stupidityismygam"][hide]In order to determine the $ \sqrt{55}$ to the nearest tenth is the same as determining the $ \sqrt{5500}$ to the nearest whole number. So, the rest is just calculations.[/hide][/quote]

that is really slick! :)
good solution!


algebra
polynomial
analytic geometry
calculus
integration
graphing lines
slope

We are given the graph of a polynomial with integer coefficients. We choose two points in the graph with integer coordinates and such that their distance is an integer too. Prove that the segment joining these to points is parallel to the x-axis.

I think I have it:
[hide]Let the polynomial in question be $P(x)$.  If the two vertices are $(x_1,P(x_1))$ and $(x_2,P(x_2))$ then the integral distance $d$ between these points satisfies
$(P(x_1)-P(x_2))^2+(x_1-x_2)^2=d^2$  It is fairly well known that for a polynomial $P$ that $x_1-x_2$ divides $P(x_1)-P(x_2)$ which means there exists an integer $k$ such that $P(x_1)-P(x_2)=k(x_1-x_2)$.  Substituting and factoring the LHS yields $(x_1-x_2)^2(k^2+1)=d^2$.  This also implies $d$ is divisible by $(x_1-x_2)$ so there exists another integer $m$ such that $d=m(x_1-x_2)$.  So either $x_1=x_2$ (the degenerate case) or $k^2+1=m^2$.  However it is an easy number theory practice to show that the only integral solutions to the above equation are $(k,m)=(0,-1)$ or $(0,-1)$.  In either case $k=0$.  Notice though that $k=\frac{P(x_1)-P(x_2)}{x_1-x_2}$ which is numerically equal to the slope connecting these points.  Since this slope is zero it is therefore parallel to the x-axis.[/hide]


Find the general soln. of the equation $ \cos4\theta=\frac{1}{2}$

Use DeMoivre's Theorem to show that $ cos4\theta=8cos^{4}\theta-8cos^{2}+1$

Show that the eqn. $ 16x^{4}-16x^{2}+1 = 0$ has roots $ x=\cos\frac{\pi}{12}$ $ x=-cos\frac{\pi}{12}$, $ x=cos\frac{5\pi}{12}$ and $ x=-cos\frac{5\pi}{12}$

By considering this eqn as quadratic eqn in $ x^{2}$, show that $ cos\frac{\pi}{12}$ = $ \frac{\sqrt2+sqrt3}{2}$

Well, I won't exactly follow the directions...

$ \cos 4\theta=\frac{1}{2}=\cos \frac{\pi}{3}\implies 4\theta=\pm \frac{\pi}{3}+2k\pi$.

Take $ \Re [(\text{cis}\, 4\theta)^{4}]$ to establish your identity.

$ 16x^{4}-16x^{2}+1=0\implies (8x^{2}-1)^{2}-8x^{2}=0\implies 8x^{2}\pm 2\sqrt{2}x+1=0$


ratio
geometry
trapezoid
trigonometry
similar triangles
geometry proposed

A line through the orthocenter H of an acute △ABC meets CA,BC at points D,E respectively.
A line through H,perpendicular to DE,meet AB at a point F.
Prove that DH:HE=AF:FB.

Is it possible to solve this problem with a projective geometric method?
I want to know the method particularly,feeling that it is so beautiful.
Please give your cherishful answers to me~.Frankly I don't have any solution. :)

[quote="77ant"] [color=darkred]A line through the orthocenter $ H$ of an acute $ ABC$ meets $ CA$ , $ CB$ at points $ D$ , $ E$ respectively.

A line through $ H$ perpendicular to $ DE$ meet $ AB$ at a point $ F$ . Prove that $ \boxed {\ \frac {HD}{HE}=\frac {FA}{FB}\ }$ .[/color] [/quote]

[color=darkblue][b][u]Proof I (synthetical).[/u][/b] Denote $ \{\begin{array}{c}
U\in BH\cap CA\\\\
V\in AH\cap CB\end{array}$ and $ \{\begin{array}{c}
X\in HA\ ,\ DX\perp DE\\\\
Y\in HB\ ,\ EY\perp DE\end{array}$ . Observe that

the quadrilaterals $ AUVB$ , $ XDVE$ , $ YEUD$ are cyclically. Thus, $ \{\begin{array}{c}
\widehat {VUB}\equiv\widehat {VAB}\\\\
HD\cdot HE=HX\cdot HV\\\\
HD\cdot HE=HY\cdot HU\end{array}\ \ \implies$

$ HX\cdot HV=HY\cdot HU$ , i.e. the quadrilaterals $ XUVY$ is cyclically. Therefore, $ \widehat {HXY}\equiv\widehat {VUB}\equiv\widehat {VAB}$ $ \implies$

$ \widehat {HXY}\equiv\widehat {HAB}$ $ \implies$ $ XY\parallel AB$ . Denote $ Z\in HF\cap XY$ . Then in the right trapezoid $ XDEY$

we have $ HZ\parallel DX\parallel EY$ and $ \frac {FA}{FB}=\frac {ZX}{ZY}=\frac {HD}{HE}$ , i.e. $ \boxed {\  \frac {HD}{HE}=\frac {FA}{FB}\ }$ .

[b][u]Proof II (metrical).[/u][/b] $ \{\begin{array}{c}
\widehat {FHA}\equiv\widehat {CED}\ ,\ \widehat {FHB}\equiv\widehat {CDE}\\\\
HA=2R\cos A\ ,\ HB=2R\cos B\end{array}$ $ \implies$

$ \{\begin{array}{c}
\frac {HD}{HE}=\frac {CD}{CE}\cdot\frac {\sin\widehat {HCD}}{\sin\widehat {HCE}}=\frac {\sin\widehat {CED}}{\sin\widehat {CDE}}\cdot\frac {\cos A}{\cos B}\\\\
\frac {FA}{FB}=\frac {HA}{HB}\cdot\frac {\sin \widehat{FHA}}{\sin\widehat {FHB}}=\frac {2R\cos A}{2R\cos B}\cdot\frac {\sin \widehat {CED}}{\sin\widehat {CDE}}\end{array}\|$ $ \implies$ $ \boxed {\ \frac {HD}{HE}=\frac {FA}{FB}\ }$ .[/color]

:blush: Thank you very much.I'm ashamed of myself poor at math  :ewpu:

How can I get the insight for math problems? ToT
Math doesn't admit me into its beautiful world.ToT

We denot as $ M,\ N,$ the intersection points of the sidelines $ BC,\ AC$ respectively, from the line segment $ HF.$

From the triangle $ \bigtriangleup CDE,$ with transversal $ NHM,$ by [b][size=100]Menelaus theorem[/size][/b],

we have $ \frac {HD}{HE}\cdot \frac {ME}{MC}\cdot \frac {NC}{ND} \equal{} 1$ $ \Longrightarrow$ $ \frac {HD}{HE} \equal{} \frac {MC}{ME}\cdot \frac {ND}{NC}$ $ ,(1)$

Similarly from the triangle $ \bigtriangleup ABC,$ with transversal $ NFM,$

we have $ \frac {AF}{FB}\cdot \frac {MB}{MC}\cdot \frac {NC}{NA} \equal{} 1$ $ \Longrightarrow$ $ \frac {AF}{FB} \equal{} \frac {MC}{MB}\cdot \frac {NA}{NC}$ $ ,(2)$

From $ (1),$ $ (2),$ it is enough to prove that $ \frac {MC}{ME}\cdot \frac {ND}{NC} \equal{} \frac {MC}{MB}\cdot \frac {NA}{NC}$ $ \Longrightarrow$ $ \frac {NA}{ND} \equal{} \frac {MB}{ME}$ $ \Longrightarrow$ $ \frac {NA}{AD} \equal{} \frac {MB}{BE}$ $ ,(3)$

But $ (3)$ is true, based on the below Lemma.

Hence, we conclude that $ \frac {HD}{HE} \equal{} \frac {AF}{FB}$ and the proof is completed.

[b][size=100][color=DarkBlue]LEMMA. - A triangle $ \bigtriangleup ABC$ is given and let $ H$ be, its orthocenter. We draw an arbitrary line through $ H,$ which intersects the sidelines $ AC,\ BC,$ at points $ D,\ E,$ respectively. The line through also $ H$ and perpendicular to $ DE,$ intersects the sidelines $ BC,\ AC,$ at points $ M,\ N,$ respectively. Prove that $ \frac {NA}{AD} \equal{} \frac {MB}{BE}.$[/color][/size][/b]

[b][size=100]PROOF[/size][/b]. – From the similar triangles $ \bigtriangleup ANH,$ $ \bigtriangleup BHE,$ we have $ \frac {NA}{HB} \equal{} \frac {AH}{BE}$ $ \Longrightarrow$ $ (NA)\cdot (BE) \equal{} (AH)\cdot (HB)$ $ ,(1)$

From the similar triangles $ \bigtriangleup AHD,$ $ \bigtriangleup BMH,$ we have $ \frac {AD}{HB} \equal{} \frac {AH}{MB}$ $ \Longrightarrow$ $ (AD)\cdot (MB) \equal{} (AH)\cdot (HB)$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac {NA}{AD} \equal{} \frac {MB}{ME}$ and the proof is completed.

Kostas Vittas.

Thank you so much,vittasko.  :blush:
I'm happy to get many good answers. :)


Let $ R_{\plus{}}$ be the set of all  positive real numbers and $ f\colon \mathbb{R_{\plus{}}}\to\mathbb{R_{\plus{}}}$ be a function such that    $ y f(x^{2})\plus{}x f(y^{2})\equal{}2xy$. Find all such function   :?:

_______________________________________
Azerbaijan Land of Fire
$ \Rightarrow\bigstar\Leftarrow$
 :rotfl:

[quote="Dimitris X"]Setting $ x \equal{} y \equal{} 1$ we take:

$ f(1) \equal{} 1$.

Setting $ x \minus{} \minus{} > \sqrt {x},y \minus{} \minus{} > \sqrt {y}$ we take:

$ \sqrt {y}f(x) \plus{} \sqrt {x}f(y) \equal{} 2\sqrt {xy}$.

Now for $ y \equal{} 1$ we take:

$ f(x) \plus{} \sqrt {x} \equal{} 2\sqrt {x} \Longleftrightarrow f(x) \equal{} \sqrt {x}, \forall x \in \mathbb{R_ \plus{} }$.[/quote]

In order to be quite Ok, you need to add a line saying that this necessary condition ($ f(x)\equal{}\sqrt x$) is sufficient (easy to verify just by plugging this value in the original equation).

[quote="pco"]

In order to be quite Ok, you need to add a line saying that this necessary condition ($ f(x) \equal{} \sqrt x$) is sufficient (easy to verify just by plugging this value in the original equation).[/quote]

Yes i know but it is easy to see it....


inequalities
algebra
polynomial
vector
inequalities theorems

Is there a variant\generalization of Muirhead that deals with conditions on $ \vec{a}\equal{}(a_1,\cdots,a_{2n}),\vec{b}\equal{}(b_1,\cdots,b_{2n})$ such that the following inequality holds for all $ x,y \in R^\plus{}$? :
$ \sum_{sym} x^{a_1}y^{a_2} \plus{} \cdots \plus{} \sum_{sym} x^{a_{2n\minus{}1}}y^{a_{2n}} \ge \sum_{sym} x^{b_1}y^{b_2} \plus{} \cdots \plus{} \sum_{sym} x^{b_{2n\minus{}1}}y^{b_{2n}}$

$ \sum a_i \equal{} \sum b_i$ is probably necessary...

[quote="bambaman"]Is there a variant\generalization of Muirhead that deals with conditions on $ \vec{a} \equal{} (a_1,\cdots,a_{2n}),\vec{b} \equal{} (b_1,\cdots,b_{2n})$ such that the following inequality holds for all $ x,y \in R^ \plus{}$? :
$ \sum_{sym} x^{a_1}y^{a_2} \plus{} \cdots \plus{} \sum_{sym} x^{a_{2n \minus{} 1}}y^{a_{2n}} \ge \sum_{sym} x^{b_1}y^{b_2} \plus{} \cdots \plus{} \sum_{sym} x^{b_{2n \minus{} 1}}y^{b_{2n}}$

$ \sum a_i \equal{} \sum b_i$ is probably necessary...[/quote]
Well, first of all, I think it only makes "really" sense if the polynomials are homogenous, i.e. there is $ m\in\mathbb N$ such that $ a_{2i}\plus{}a_{2i\minus{}1}\equal{}b_{2i}\plus{}b_{2i\minus{}1}\equal{}m$. So, dividing by $ x^m$, you are left with \[ \sum_{i\equal{}1}^n t^{a_{2i}}\ge \sum_{i\equal{}1}^n t^{b_{2i}}\] for all $ t>0$. Try to figure that one out. ;)

Oops, I missed half of each sum. :blush: Forget the above. Here is a better approach.

First an example for the inhomogeneous case (which of course can make sense :blush: )

Put $ a \equal{} (9,0 ;3,3)$ and $ b \equal{} (7,1 ;5,2)$, then the inequality can be easily proven by AM-GM as follows :)
 ‘Homogenizing’ the inequality
\[ \sum (x^9 \plus{} x^3y^3)\ge \sum (x^7y \plus{} x^5y^2)
\]
by putting $ w : \equal{} \sqrt [3]{xy}$, we can write it as
\[ \sum (x^9 \plus{} w^9)\ge \sum (x^6w^3 \plus{} x^3w^6)
\]
which is obviously true.

Inspired by this idea, here is the most general sufficient (and maybe necessary :) ) condition I can find:
[b]Suppose there are vectors $ \vec u, \vec v \in\mathbb R^2$ and scalars $ \lambda_1,\dots, \lambda_n\in\mathbb R$ and $ \mu_1,\dots, \mu_n\in\mathbb R$ such that for all $ i \equal{} 1,\dots,n$ we have $ (a_{2i}, a_{2i \minus{} 1}) \equal{} \vec u \plus{} \lambda_i \vec v$ and $ (b_{2i}, b_{2i \minus{} 1}) \equal{} \vec u \plus{} \mu_i \vec v$.[/b] [hide]
This means things are "affinely homogenous" in a sense, i.e. the inequality can be homogenized similarly as above.[/hide]

[b]If $ (\lambda_1,..., \lambda_{n})\succ(\mu_1,..., \mu_{n})$, we get easily by Karamata that
\[ \sum_{i \equal{} 1} ^n \sum_{sym} x^{a_{2i \minus{} 1}}y^{a_{2i}}\ge \sum_{i \equal{} 1} ^n \sum_{sym} x^{b_{2i \minus{} 1}}y^{b_{2i}}\ \ (*)
\]
as wanted.[/b]

I have the impression that this may already cover all possibilities, apart from permuting $ a_{2i \minus{} 1}$ with $ {a_{2i}}$ and/or $ b_{2i \minus{} 1}$ with $ {b_{2i}}$ for certain $ i$’s, which doesn’t change $ (*)$.

Note that $ \sum a_i \equal{} \sum b_i$ is indeed necessary:
Let’s fix $ y \equal{} 1$ and put $ f(x): \equal{} LHS \minus{} RHS \equal{} \sum_{i \equal{} 1} ^n x^{a_{i}} \minus{} \sum_{i \equal{} 1} ^n x^{b_{i}}$. So we have by assumption $ f(x)\ge0$ for all $ x > 0$ and $ f(1) \equal{} n \minus{} n \equal{} 0$, thus a local minimum with $ 0 \equal{} f'(1) \equal{} \sum a_i \minus{} \sum b_i$ qed.

Note that the inequality $ (*)$ and the sufficient condition generalize to $ k > 2$ variables in the obvious way, but it is far from being a necessary one then (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=56369]here[/url]). :roll:


calculus
integration
geometry
3D geometry
sphere
analytic geometry
real analysis

[quote="PeterPan123"]For which $ \alpha \in \mathbb{R}$ converge the Integrals

$ I_\alpha \equal{} \int_{B^n} {\Vert x \Vert}^\alpha dx$ und $ J_\alpha \equal{} \int_{\mathbb{R}^n \setminus B^n} {\Vert x \Vert}^\alpha dx$

with $ B^n$ being the $ n$-dimensional Sphere.[/quote]

Hint: The polar coordinates formula for $ R^n$ is $ \int_{R^n} f(x)dx \equal{} \int_0^{\infty} r^{n\minus{}1} \int_S f(r\zeta) d\sigma(\zeta)$, where $ \sigma$ is surface area measure on the unit sphere $ S$.


The answer is [hide]1992[/hide]

For the solution, (with skimping on the details) set $1992 = n$ and let $x_i - x_{i+1} = d_i$ and then [hide]$|y_i-y_{i+1}| = \frac 1 {i(i+1)} |\sum_{j=1}^i j\cdot d_j| \leq \sum_{j=1}^i \frac j{i(i+1)} |d_j|$ where the first equality results from just writing the thing out and appropriate adding/subtracting, while the inequality is the triangle inequality.  Summing this for $i$ from 1 to $n$ gives us $|y_1 - y_2| + |y_2 - y_3| + \ldots + |y_n - y_{n+1}| \leq \frac n{n+1} |d_1| + b_2|d_2| + \ldots + b_n|d_n|$ where $0 < b_i \leq \frac n{n+1}$ for each $i > 1$.  But this expression is at most $\frac n{n + 1}(|d_1| + \ldots |d_n|)$.  But the second part of this product is just $n + 1$, so this shows that the maximum is at most $n$.  We can achieve $n$ by setting $x_1 = n, x_2 = x_3 = \ldots = x_{n+1} = 0$.[/hide]


analytic geometry
inequalities unsolved
inequalities

Let $ n$ be a given positive integer. In the coordinate set, consider the set of points $ \{P_{1},P_{2},...,P_{4n\plus{}1}\}\equal{}\{(x,y)|x,y\in \mathbb{Z}, xy\equal{}0, |x|\le n, |y|\le n\}.$

Determine the minimum of $ (P_{1}P_{2})^{2} \plus{} (P_{2}P_{3})^{2} \plus{}...\plus{} (P_{4n}P_{4n\plus{}1})^{2} \plus{} (P_{4n\plus{}1}P_{1})^{2}.$

we can find out that zhe points are (0,i),(0,-i),(0,0),(i,0),(-i,0),i from 1 to n.

$16 n - 8 $ seems the least possible.

For the corresponding problem with the lattice points of the axis in $\mathbb R^d$, i.e. \[ \{P_{1},P_{2},...,P_{2dn+1}\}=\Bigl\{(x_1,...,x_d)\in\mathbb{Z}^d\ \biggl| \ \sum_{i<j}x_i^2x_j^2=0, |x_i|\le n\ \forall i\Bigr\}, \]I think the minimum is $4d(2n-1)$ for $d\ge2$ and $8n-2$ for $d=1$. :)

Indeed, $16n-8$ is the minimum value.

For $n = 1$ it's trivial to check, so WLOG assume that $n \ge 2.$

Let $S$ be the set of $4n+1$ points described in the problem.

[hide = Construction]

For $n = 2$, if we let $P_1, P_2, \cdots, P_9$ be $(2, 0), (1, 0), (0, -2), (0, -1), (-2, 0), (-1, 0), (0, 1), (0, 2), (0, 0)$ then we obtain the desired value of $16 \cdot 2 - 8 = 24$ for $\sum (P_iP_{i+1})^2.$

For $n = 3$, we can let $P_1, P_2, \cdots, P_{13}$ be $(3, 0), (1, 0), (0, 2), (0, 3), (0, 1), (-2, 0), (-3, 0), (-1, 0), (0, 0), (0, -2), (0, -3), (0, -1), (2, 0)$ in order to obtain the desired value of $16 \cdot 3 - 8 = 40$ for $\sum (P_iP_{i+1})^2.$

Assume now that $n > 3.$ We will describe a way to connect pairs of points in $S$ such that the resulting edges comprise a $(4n+1)-$cycle with the sum of squares of lengths of edges being $16n-8$. This would clearly upper bound our answer by $16n-8.$ Start by adding edges between the following $8$ pairs of vertices:

$$((0, n), (0, n-1)), ((0, n), (0, n-2)), ((n, 0), (n-1, 0)), ((n,0), (n-2, 0)), ((0, -n), (0, -(n-1))), ((0, -n), (0, -(n-2))), ((-n, 0), (-(n-1), 0)), ((-n, 0), (-(n-2), 0)).$$

The sums of the squares of the lengths of these edges is $20.$

Now, for each $3 \le i < n,$ add edges between the following $4$ pairs of vertices:

$$((0, i), (0, i-2)), ((i, 0), (i-2, 0)), ((0, -i), (0, -(i-2))), ((-i, 0), (-(i-2), 0)).$$

The sums of squares of the lengths of these edges is $16$, and so our total sum so far is $20 + (n-3) \cdot 16 = 16n - 28.$

Finally, add edges between the following $5$ pairs of vertices:

$$((-2, 0), (0, 1)), ((0, 2), (1, 0)), ((2, 0), (0, -1)), ((0, -2), (0, 0)), ((0, 0), (-1, 0)).$$

The sum of the squares of these edges is $5+5+5+4+1 = 20$, and so our grand total is $16n-28 + 20 = 16n-8,$ as desired. It's not hard to see that the $4n+1$ edges we've now drawn indeed comprise a $(4n+1)-$cycle on the vertices of $S$, and so the construction is complete.

[/hide]

[hide = Proof of Optimality]

We will show that $(P_1P_2)^2 + (P_2P_3)^2 + \cdots + (P_{4n}P_{4n+1})^2 + (P_{4n+1}P_1)^2 \ge 16n-8$ for an arbitrary labeling $P_1, P_2, \cdots, P_{4n+1}$ of the points of $S.$

Let $x_1, x_2, \cdots, x_{4n+1}$ be the $x-$coordinates of $P_1, P_2, \cdots, P_{4n+1}.$ We will show that:
$$\sum_{i=1}^{4n+1} (x_i - x_{i+1})^2 \ge 8n-4, \qquad (1)$$
 where $4n+2 = 1$ here. This, together with an analogous result for the $y-$coordinates, would solve the problem with the Pythagorean Theorem.

So let's just show this. Let $a_1, a_2, \cdots, a_t$ be the sequence which is just $x_1, x_2, \cdots, x_{4n+1}$ after compressing blocks of $0$'s into a single $0$. For example, in the construction for $n = 3$ above, $(x_1, x_2, \cdots, x_{13}) = (3, 1, 0, 0, 0, -2, -3, -1, 0, 0, 0, 0, 2)$ and $(a_1, a_2, \cdots, a_{13}) = (3, 1, 0, -2, -3, -1, 0, 2).$ Observe that every nonzero number appears in the sequence $a_1, a_2, \cdots, a_t$ exactly one time. Now, WLOG assume that $a_1 = -n$ and $a_b = n$, for some $2 \le b \le t.$ Notice that $\sum_{i=1}^{b-1} (a_{i+1}-a_i)^2$ is minimized when $a_1, a_2, \cdots, a_b$ is nondecreasing. Hence, for the purposes of proving $(1)$, we can assume that $a_1, a_2, \cdots, a_b$ is strictly increasing. Analogously, we may assume that $a_b, a_{b+1}, \cdots, a_t, a_1$ is strictly decreasing. In this way, we know that $0$ appears at most two times among the $a_i$'s (once among $\{a_2, a_3, \cdots, a_{b-1}\}$ and once among $\{a_{b+1}, a_{b+2}, \cdots, a_t\}$). Therefore, as all other values appear exactly once, we obtain that $t \in \{2n+1, 2n+2\}.$ From the Cauchy-Schwarz Inequality, we know that

$$\sum_{i=1}^{b-1} (a_{i+1} - a_i)^2 \ge (\sum_{i=1}^{b-1} (a_{i+1}-a_i))^2/(b-1) = \frac{4n^2}{b-1}.$$

Analogously we get that

$$\sum_{i = b}^{t} (a_{t+1}-a_t)^2 \ge \frac{4n^2}{t+1-b}.$$

Summing the above two inequalities, we obtain

$$\sum_{i = 1}^{t} (a_{i+1} - a_i)^2 \ge \frac{4n^2}{b-1} + \frac{4n^2}{t+1-b} = \frac{4n^2t}{(b-1)(t+1-b)} \ge \frac{4n^2t}{\frac{t^2}{4}} = \frac{16n^2}{t}.$$

If $t = 2n+1$, then we have $\frac{16n^2}{t} = \frac{16n^2}{2n+1} > 8n-4,$ and so we're done. Else, suppose that $t = 2n+2.$

Then, we know that there exists $1 < c < b$ with $a_c = 0$ and $b < d \le t$ with $a_d = 0.$

[b] Lemma. [/b] For any permutation $\sigma$ of $[n]$, we have $\sum_{i = 1}^{n} (\sigma(i) - \sigma(i+1))^2 \ge 4n-6.$

[b] Proof. [/b] We proceed by induction on $n$. For $n = 2, 3$, the result is easily verified. Suppose that the lemma holds for $n = k.$ When $n =k+1,$ let $a, b$ be the two numbers on either side of $k+1$ in the permutation. In other words, if $\sigma(i) = k+1,$ then $\{a, b\} = \{\sigma(i-1), \sigma(i+1)\}$ with indices modulo $k+1.$ Consider the effect of removing $k+1$ from the permutation entirely. Then, from the inductive hypothesis, it suffices only to show that $(k+1-a)^2 + (k+1-b)^2 - (a-b)^2 \ge 4.$ Notice that $c = k+1-a$ and $d = k+1-b$ are distinct positive integers, and so therefore $c^2 + d^2 - (c-d)^2 = 2cd \ge 2 \cdot 1 \cdot 2 = 4,$ as desired.
$\blacksquare$

Now, apply the lemma on the permutations $(a_c+1, a_{c+1}+1\cdots, a_{d-1}+1)$ and $(a_d+n+1, a_{d+1}+n+1, \cdots, a_t+n+1, a_1+n+1, \cdots, a_{b-1}+n+1)$ of $[n+1].$ Summing the two resulting expressions and using that $a_c = a_d = 0$ gives us that:

$$\sum_{i=1}^{t} (a_{i+1}-a_i)^2 \ge 4(n+1)-6 + 4(n+1)-6 = 8n-4,$$

as desired.

$\square$

[/hide]


Can anyone provide a full proof to this problem. We had it on a previous test and I got it wrong. I am studying for the final and I really need to know the correct method:

Suppose $f_{n}: R \to R$ is a sequence of bounded functions (ie: $|f_{n}(x)| \le M_{n}$ for all $x \in R$) and that $f_{n}$ converges to $f$ uniformly on $R$. Prove that $f$ is bounded on $R$.

Thanks.

Your problem says that the uniform limit of a sequence of bounded functions is a bounded function.
Here is an ideea:

Let $f_{n}$ be a sequence of functions bounded on a set $A$ which converges uniformly to a function $f$. Then
let $\epsilon =1$. It follows that there is $N=N(1)$ such that for all $x\in A$ we have that

$|f(x)-f_{N}(x)|\leq 1$, hence $|f(x)|\leq |f_{N}(x)|+1$.
But $f_{n}$ is bounded therefore there is $M$ such that $|f_{N}(x)|\leq M$.

It follows that $|f(x)|\leq M+1$.

Why do you use the $\le$ sign rather than the $<$ sign? I though the definition for uniform convergence was $|f_{N}(x)-f(x)|< \epsilon$.


probability
expected value
probability and stats

I'm stuck on this problem from a practice final.

Let $ S$ be the set of permutations of $ n \geq 100$ obejcts.
Consider the experiment of choosing an element of $ S$ randomly with equal probabilities. Let $ f$ be the random variable defined by $ f(\sigma)\equal{}|\{j\colon \sigma(j)\equal{}j\}|$, so $ f(\sigma)$ is the number of fixed points of $ \sigma \in S$.

Compute the variance $ V(f)$, and show that a randomly chosen permutation of $ n \geq 100$ objects is 99% certain to leave fewer than 11 objects fixed.

So far I've been able to compute that the expected value $ E(f)\equal{}1$, and I know $ V(f)\equal{}E(f^2)\minus{}{E(f)}^2$, but I don't know how to directly compute it.

Many thanks for any help.

This problem is relatively straightforward by the following reasoning:

Let $ X_i$ be equal to $ 1$ if $ \sigma(i)\equal{}i$ and zero else. Let $ X\equal{}X_1\plus{}...\plus{}X_n$

Then $ Var X\equal{}\sum Var(X_i)\plus{}2\sum_{i<j} Cov(X_i,X_j)$
$ \equal{}nVar(X_1)\plus{}(n)(n\minus{}1)Cov(X_1,X_2)$.

Now $ Cov(X_1,X_2)\equal{} E( (X_1\minus{}EX_1)(X_2\minus{}EX_2))\equal{}E(X_1X_2)\minus{}(EX_1)(EX_2)$

$ EX_1\equal{}\frac{1}{n}$.

Now $ Var(X_1)\equal{}E(X_1^2)\minus{}E(X_1)^2\equal{}\frac{1}{n}\minus{}\frac{1}{n^2}$

$ E(X_1X_2)\equal{}P(X_1\equal{}1 \cap X_2\equal{}1)\equal{}\frac{1}{n}\cdot \frac{1}{n\minus{}1}$

So the answer is:

$ n\cdot \left(\frac{1}{n}\minus{}\frac{1}{n^2}\right)\plus{}(n)(n\minus{}1)\cdot \left(\frac{1}{n}\cdot \frac{1}{n\minus{}1}\minus{}\frac{1}{n^2}\right)\equal{}1$

Also $ EX\equal{}1$.

Also see: http://en.wikipedia.org/wiki/Random_permutation_statistics#Moments_of_fixed_points

I'm not sure how to finish your problem...

Thanks for your explanation, Altheman, as well as the link. The second part of the question is now clear to me with your help.

Here goes,
To find the probability that a random permutation leaves 11 or more objects fixed, I must have $ 11 \leq f(\sigma) \rightarrow 10 \leq f(\sigma)\minus{}E(f)$. By Chebyshev's equality,

$ p(|f(\sigma)\minus{}E(f)|\geq10)\leq \frac{V(f)}{{10}^2}\equal{}.01$. Hence the probability that a random permutation leaves fewer than 11 objects fixed is about $ 1\minus{}.01\equal{}.99$.

Again, thank you.


Let $ x,y,z\geq 0$,prove that:
$ \frac{x^2}{x^2\plus{}2yz}\plus{}\frac{y^2}{y^2\plus{}2zx}\plus{}\frac{z^2}{z^2\plus{}2xy}\geq 1$
BQ

[quote="xzlbq"]Let $ x,y,z\geq 0$,prove that:
$ \frac {x^2}{x^2 \plus{} 2yz} \plus{} \frac {y^2}{y^2 \plus{} 2zx} \plus{} \frac {z^2}{z^2 \plus{} 2xy}\geq 1$
BQ[/quote]

$ \sum_{cyclic}\frac {x^2}{x^2 \plus{} 2yz}\ge \sum_{cyclic}\frac {x^2}{x^2 \plus{} y^2\plus{}z^2}\equal{}1$ :wink:

Summary　and collecting experience：

１。I just through a concrete example，have been so many，Lenovo is very important to note。
２。old and new.
old<=>new,
Let [size=200]old[/size] problems output [size=200]new [/size]shine ......

[quote="xzlbq"]Let we look it a while,the cases has changed.....

n=3;

1.$ \frac {x}{x \plus{} 2y} \plus{} \frac {y}{y \plus{} 2z} \plus{} \frac {z}{z \plus{} 2x} \geq 1;$

2.$ \frac {x^2}{x^2 \plus{} 2yz} \plus{} \frac {y^2}{y^2 \plus{} 2zx} \plus{} \frac {z^2}{z^2 \plus{} 2xy} \geq 1;$

(I guess，You have guessed my mean)
BQ
(n-1 inequality)[/quote]

1.$ \frac {x}{x \plus{} \frac {1}{2}y} \plus{} \frac {y}{y \plus{} \frac {1}{2}z} \plus{} \frac {z}{z \plus{} \frac {1}{2}x} \le 2;$

2.$ \frac {x^2}{x^2 \plus{} \frac {1}{2}yz} \plus{} \frac {y^2}{y^2 \plus{} \frac {1}{2}zx} \plus{} \frac {z^2}{z^2 \plus{} \frac {1}{2}xy} \le 2;$


Salut la toti,

Puteti va rog sa ma ajutati cu ceva informatii despre modul de desfasurare a pregatirii? Mi-ar fi utile si informatii din anii trecuti. Multumesc mult!

In principiu va duceti la bucuresti, si faceti lectii cu diversi profi (si studenti). Cel mai bine este sa lucrezi toate problemele date la baraje in anii trecuti (e o carte de D. Branzei care le contine pe toate), si mai e o carte de Dinu Serbanescu si Panaitopol cu probleme de teoria numerelor si combinatorica pentru juniori. Daca le faci pe toate alea nu ar trebui sa ai probleme.

[url=http://www.mathlinks.ro/resources.php?c=142&cid=78]JBMO TST[/url]
[url=http://www.mathlinks.ro/resources.php?c=1&cid=21]JBMO Questions[/url]

Poate cineva sa-mi spuna cand si unde va avea loc pregatirea lotului largit de juniori (daca se mai tine :D )? Thx

Vad ca nu prea exista informatii publice despre acest subiect. Miercuri am primit prin inspectoratul judetean o instiintare pentru perioada 2-13 iunie, la colegiul Nichita Stanescu in Bucuresti. Daca cineva are mai multe detalii il rog sa le posteze aici.

Aveti testele de selectie date in anii trecuti si la seniori?


inequalities
Cauchy Inequality
inequalities proposed

Let $ a_1,a_2,...,a_n$ be positive real numbers, such that $ a_1\cdot a_2\cdot...\cdot a_n\equal{}1$.
Prove that the following inequality holds
$ \left.\frac{a^2_1}{a_2(a^2_1\plus{}a_1a_2\plus{}a^2_2)}\plus{}\frac{a^2_2}{a_3(a^2_2\plus{}a_2a_3\plus{}a^2_3)}\plus{}\cdots\plus{}\frac{a^2_n}{a_1(a^2_n\plus{}a_na_1\plus{}a^2_1)}\geq \frac{n}{3}\right.$

Nice inequality my friend.

Having observed that we possess the following inequality

\[ \frac{a_{i}^2}{a_{i\plus{}1}(a_{i}^2\plus{}a_ia_{i\plus{}1}\plus{}a_{i\plus{}1}^2)} \ge \frac{2}{3a_{i\plus{}1}} \minus{} \frac{1}{3a_i} \, \, \forall i \in {1,2,3,..,n}\]

The last step is to use Cauchy inequality for n variables which completes our proof.

You are great, [b]great math[/b]! This inequality, between $ a_i$ and $ a_{i\plus{}1}$, is essential for solving the problem. I think that not many people know how to find it! It's easy to prove it, but it's not trivial to reach to it. First of all, you must "guess" that you need an inequality like that for your proof. Congrats!  :)

[quote="great math"]Nice inequality my friend.

Having observed that we possess the following inequality
\[ \frac {a_{i}^2}{a_{i \plus{} 1}(a_{i}^2 \plus{} a_ia_{i \plus{} 1} \plus{} a_{i \plus{} 1}^2)} \ge \frac {2}{3a_{i \plus{} 1}} \minus{} \frac {1}{3a_i} \, \, \forall i \in {1,2,3,..,n}\]
[/quote]
Could you prove it, please ?

For your interest, Dijkschneier. This 'small' inequality is equivalent to

\[ (a_i\minus{}a_{i\plus{}1})^2(a_i\plus{}a_{i\plus{}1}) \ge 0\]


In triangle $ ABC$, $ \overline{CA} \equal{} \overline{CB}$. On $ CB$ square $ BCDE$ is constructed away from the triangle. If $ x$ is the number of degrees in angle $ DAB$, then

$ \textbf{(A)}\ x\text{ depends upon triangle }ABC \qquad\textbf{(B)}\ x\text{ is independent of the triangle}$
$ \textbf{(C)}\ x\text{ may equal angle }CAD \qquad\textbf{(D)}\ x\text{ can never equal angle }CAB$
$ \textbf{(E)}\ x\text{ is greater than }45^{\circ}\text{ but less than }90^{\circ}$

Well this is easy.
Let $ <CBA\equal{}a$.  Therefore, $ <ACB\equal{}180\minus{}2a$, and $ <ACD\equal{}270\minus{}2a$.  From isosceles triangle $ \triangle{ACD}$, $ <CAD\equal{}a\minus{}45$.  Hence, $ <DAB\equal{}x\equal{}a\minus{}(a\minus{}45)\equal{}45^{\circ}$.  Thus, $ x$ is independent of triangle $ ABC$.  $ \boxed{(B)}$. (A quick check at the other answers confirms that this is the only possibility given our finding).


probability
inequalities
function
combinatorics unsolved
combinatorics

From the set of all permutations $f$ of $\{1, 2, ... , n\}$ that satisfy the condition:
$f(i) \geq i-1$   $i=1,...,n$
one is chosen uniformly at random. Let $p_n$ be the probability that the chosen permutation $f$ satisfies
$f(i) \leq i+1$  $i=1,...,n$
Find all natural numbers $n$ such that $p_n > \frac{1}{3}$.

They are $1,2,3,4,5,6.$

Let $A_n$ be the number of permutations satisfying the first inequality.
Let $B_n$ be the number of permutations satisfying both inequalities.

1) We have $A_{n+1}=2A_n$ so that$A_n =2^{n-1}.$
From the first inequality, we have $f(n+1)=n$ or $n+1$.
- If $f(n+1) = n+1$, the rest of the permutation is one satisfying the inequality for $\{1, \cdots , n \}.$
Conversely, any of these restriction leads to a permlutation for $n+1$. Thus there are exactly $A_n$ permutations in that case.
- $f(n+1)=n.$
There exists $i_0 \in \{1, \cdots , n\}$ such that $f(i_0) = n+1.$
Let $g$ be the permutation of $\{1, \cdots , n \}$ such that $g(k)=f(k)$ for all $k$ excpet $g(i_0)=n.$
Let $\Phi$ be the function such that $\Phi(f) = g.$
It is easy to verify that $\Phi$ is a bijection from the set of the permutations of $\{1, \cdots , n+1 \}$ satisfying the two inequalities and such that $f(n+1)=n$ to th set of the permutations of $\{1, \cdots , n \}$ satisfying both inequalities.
Thus there are exactly $A_n$ permutations in that case.
Then, we have $A_{n+1}=2A_n$, and since $A_1 =1$, we deduce $A_n =2^{n-1}.$

2) We have $B_{n+1} = B_n + B_{n-1}.$
- If $f(n+1) = n+1$ we use the same resoning as above, and there are exactly $B_n$ permutations in that case.
- If $f(n+1) = n$ then $f(k) \leq n$ for all $k \leq n-1$ so that $f(n) = n+1.$
Thus, $f$ is determined by its restriction to $\{1, \cdots ,n-1 \}$ which staisfies both inequalities. It follows that there are exactly $B_{n-1}$ permutations in that case.
Then $B_{n+1} = B_n + B_{n-1}$, as claimed.

Since $B_=1$ and $B_2 = 2$ we may solve this linear recurrence and find $B_n = \frac{3a+2} {4a+3} a^n + \frac{a+1} {4a+3} (-\frac 1 a)^n$ where $a = \frac{1 + \sqrt 5} 2$.

Then $p_n = \frac{B_n} {A_n} = 2 \frac{3a+2} {4a+3} (\frac a 2)^n + 2 \frac{a+1} {4a+3} (-\frac 1 {2a})^n$.
It is straightforward to see that $p_n \leq 2 (\frac a 2 )^n$. It follows that $pn < \frac 1 3$ for $n > 8.$
Direct checking shows that $n$ such that $p_n > \frac 1 3$ are $1,2,3,4,5,6.$

Pierre.


The paths of two particles for t >= 0 are given by

r_1(t) = (t-3)[b]i[/b] + (t-3) 2 [b]j[/b],
r_2(t) = (3t/2 - 4)[b]i[/b] + (3t/2 - 2)[b]j[/b].

a. Determine the exact time(s) at which the particles collide.

For this, do I just set the x-components of the vectors equal to each other and the y components equal to each other?

ie: t-3 = 3t/2 - 4, and (t-3) 2  = (3t/2 - 2).

b. Find the direction of motion of each particle at the time(s) of collision.

Since v/|v| is the direction of motion, do I just differentiate each r(t) and divide that by the magnitude of the velocity (plugging in the values at which they intersect)?

Helped is needed and appreciated. Thanks.

Doing exactly what you described:

$i$-component:

$t - 3 = \frac 3 2\ t - 4,\ \ t = 2$

$j$-component:

$(t - 3)^2 = \frac 3 2\ t - 2,\ \ 2t^2 - 15 + 22 = 0$

$t = \frac{15 \pm \sqrt{225 - 168}}{4} = \frac{15 \pm \sqrt{49}}{4} = \frac{15 \pm 7}{4} = 2$ [color=white].[/color] or [color=white].[/color] $\frac{11}{2}$

The particles collide at $t = 2\ \text{sec}$. If the solutions were not the same, the particles would miss each other.

$\vec v_1(t) = \frac{\text d\vec r_1(t)}{\text dt} = \vec i + 2(t - 3) \vec j$

$\vec v_2(t) = \frac{\text d\vec r_2(t)}{\text dt} = \frac 3 2 (\vec i + \vec j)$

At the time of the collision $t = 2$, the velocities are:

$\vec v_1(2) = \vec i - 2 \vec j$

$\vec v_2(2) = \frac 3 2 (\vec i + \vec j)$

$v_1(2) = |\vec v_1(2)| = \sqrt{1^2 + 2^2} = \sqrt 5$

$v_2(2) = |\vec v_2(2)| = \frac 3 2 \sqrt{1^2 + 1^2} = \frac 3 2 \sqrt 2$

and the directions of motion are:

$\vec n_1(2) = \frac{\vec v_1(2)}{v_1(2)} = \frac{1}{\sqrt 5}(\vec i - 2 \vec j) = \frac{\sqrt 5}{5} (\vec i - 2 \vec j)$

$\vec n_2(2) = \frac{\vec v_2(2)}{v_2(2)} = \frac{1}{\sqrt 2} (\vec i + \vec j) = \frac{\sqrt 2}{2} (\vec i + \vec j)$


trigonometry
inequalities
geometry
geometric inequality
inequalities proposed

$ABC$ is an triangle. Prove that
\[\cos{A}.\cos{B}.\cos{C}\le \frac{r^{2}}{3R^{2}-2Rr}\]
:)

[color=purple]Easy? Can you post ypur solution? I know it's very hard.I've solve it in the case triangle  ABC have two angles less than 60.How about you Can?[/color]

[quote="evarist"][color=purple]Easy? Can you post ypur solution? I know it's very hard.In the case triangle  ABC have two angles less than 60 ? How about you Can?[/color][/quote]
Actually, i have solved this problem. And i will post my solution next week because i am too busy now. :(
Please wait. :)

@: you don't to consider any cases, evarist! :)

[quote="gemath"]Hey it is quite simple !  :D
we well know $\prod\cos A=\frac{s^{2}-(2R+r)^{2}}{4R^{2}}$ thus we need to prove
$\frac{s^{2}-(2R+r)^{2}}{4R^{2}}\le \frac{r^{2}}{3R^{2}-2Rr}\Leftrightarrow s^{2}\le (2R+r)^{2}+\frac{4Rr^{2}}{3R-2r}$
but we know the inequality
$s^{2}\le R^{2}+10Rr-r^{2}+2(R-2r)\sqrt{R^{2}-2Rr}$ (W.J.Blundon 1965)
thus we need to prove
$R^{2}+10Rr-r^{2}+2(R-2r)\sqrt{R^{2}-2Rr}\le (2R+r)^{2}+\frac{4Rr^{2}}{3R-2r}$ we can prove this quite easy by let $t=\frac{r}{2R}\Rightarrow t\in [0,1]$ and prove $f(t)\ge 0$ by computer.
@ I see Vietnamese like inequality with $s,r,R$ very much, why?  :D[/quote]
yeah, this is also my solution. :) So i don't need to post it. :D

@gemath: sorry, but not all Vietnameses like $s,R,r$ inequality, such as me :D, this is a problem of a student in VN (not me) :D, only some Vietnameses like it :P

[quote="gemath"]
@ I see Vietnamese like inequality with $s,r,R$ very much, why?  :D[/quote]
[color=purple]oh yeah iam very like R,r,s and Bui Viet Anh is too.How about you gemath?
  Can , gemath kill the following problem!
  $\sum\frac{1}{a^{2}}\geq \frac{27}{a^{2}+b^{2}+c^{2}+8\sqrt{3}S}$
    Before i have an interesting proff for this problem applying Iran 96.What a pity it have an error.So i don't  think it true but"here"is true
   ${\sum\frac{1}{a^{2}}\geq\frac{63}{3(a^{2}+b^{2}+c^{2})+16\sqrt{3}S}}$
[/color]

I don't like methods killed the beauty of geometry,$s,R,r$ is one of them but it is a strong method. I like geometric inequality and only like geometric method for them, but I like the application of them in algebra, too. :D  :D  :D
ok I will try your problem and I will post it if I have time!  :)

[quote="evarist"]
Can , gemath kill the following problem!
  $\sum\frac{1}{a^{2}}\geq \frac{27}{a^{2}+b^{2}+c^{2}+8\sqrt{3}S}$
    Before i have an interesting proff for this problem applying Iran 96.What a pity it have an error.So i don't  think it true but"here"is true
${\sum\frac{1}{a^{2}}\geq\frac{63}{3(a^{2}+b^{2}+c^{2})+16\sqrt{3}S}}$
[/quote]

ok now I will kill it for new year 2007 :D  :D
we see $\sum\frac{1}{a^{2}}\geq \frac{27}{a^{2}+b^{2}+c^{2}+8\sqrt{3}S}(\star)$ is stronger than ${\sum\frac{1}{a^{2}}\geq\frac{63}{3(a^{2}+b^{2}+c^{2})+16\sqrt{3}S}}$
we will prove $(\star)$ but note that $(\star)$ is only true for acute triangle
We well know the Iran inequality
$\sum\frac{1}{(y+z)^{2}}\ge \frac{9}{4(xy+yz+zx)}\forall x,y,z>0$
By Ravi subtitude $x=s-a,y=s-b,z=s-c$ and $\sum (s-b)(s-c)=r(4R+r)$ we have the Iran in triangle
$\sum\frac{1}{a^{2}}\ge \frac{9}{4r(4R+r)}$
and now we will show
$\frac{9}{4r(4R+r)}\ge \frac{27}{a^{2}+b^{2}+c^{2}+8\sqrt{3}S}\Leftrightarrow a^{2}+b^{2}+c^{2}+8\sqrt{3}S\ge 12r(4R+r)$
by $a^{2}+b^{2}+c^{2}=2(s^{2}-4Rr-r^{2})$ so we need to prove
$s^{2}+4\sqrt{3}sr\ge 28Rr+7r^{2}\Leftrightarrow (s+2\sqrt{3}r)^{2}\ge 28Rr+19r^{2}\Leftrightarrow s \ge \sqrt{28Rr+19r^{2}}-2\sqrt{3}r$
and now using Blundon inequality for all triangle
$s^{2}\ge 2R^{2}+10Rr-r^{2}-2(R-2r)\sqrt{R^{2}-2Rr}$
we need to prove the following for all acute triangle
$2R^{2}+10Rr-r^{2}-2(R-2r)\sqrt{R^{2}-2Rr}\ge (\sqrt{28Rr+19r^{2}}-2\sqrt{3}r)^{2}$
by let $t=\frac{r}{2R}$ we can prove it, but really I know it true by computer not by hand  :D
Now I have a problem for you if you like that method:
Prove that
$\frac{27}{a^{2}+b^{2}+c^{2}+144\frac{r^{2}}{R}}\ge\sum\frac{1}{a^{2}}$
It is pity that the inequality
$\frac{27}{a^{2}+b^{2}+c^{2}+72r^{2}}\ge\sum\frac{1}{a^{2}}$ isn't true for all acute triangle, it need more stronger conditions about angle!
Note that we also solve some problems in acute traingle by subtitude
$a'=a^{2},b'=b^{2},c'=c^{2}$ then $a',b',c'$ is sides of a triangle too, and
$S_{ABC}$ can be calculate by $16S^{2}=16R'r'+4r'^{2}$ :D

[color=purple]Ok i'll solve it.please read my message i've send to you.Thanks[/color]


calculus
limit
algebra
function
domain
calculus computations

:(

hi,

i took calc in high school and now i am retaking it again as a prereq for college and i am finding out very quickly that i have been taught cookbook calculus :(

so, basically i've been doing the homework using what i know... but i really do want to learn the "real" way to go about solving a problem... but i dunno if its the way my professor explained it or what, but i simply don't get the relation between epsilon and delta and how they are applied to limiits and how they are applied thoughout calculus..

i've poured through the previous pages here and didn't find an answer, so now i am forced to pose this thread at the risk of looking like an idiot :(

thanks in advance
tb

There's no way I could teach you this in a few on-line posts. Talk to your instructor, or get help from someone else who knows what he's doing.

I do have one insight to offer. Let's look at that definition again:

$\lim_{x\to a}f(x)=L\iff\,\forall\,\epsilon>0,\,\exists\,\delta>0$ such that if $x$ is in the domain of $f$ and $0<|x-a|<\delta,$ then $|f(x)-L|<\epsilon.$

Note that to use this definition, you must already know $L.$ In other words, this, the official definition of limit, cannot itself be used to find an unknown limit. It's not a technique for finding limits; it is a standard of proof. You use it to verify a number of useful theorems: the limit of a sum is the sum of the limits (provided both exist); the limit of a product is the product of the limits (provided both exist); the limit of a quotient is the quotient of the limits, provied both exist and the denominator limit is not zero; the squeeze theorem; the composition of continuous functions is continuous. And those theorems are what you use in practice to find limits.

so as it pertains to limits, epsilon and delta are just variables(essentially) that are used to represent arbitrary values that exist near the point of c????

I've talked to him and he offers up the same stuff he threw out in class  :mad:

so it's not a way to solve limit equations then, just to prove the theorums that are used to solve limits....

so, if i may(i'm probably getting ahead of myself, but i'll ask anyway), how are these applied to computer science and such?

and, sorry for this last question, but im reading more as im typing... but is one supposed to be looking for a coorelation b/w |f(x) - L| and |x-c| ??? or something else....

sorry for the randomness.... i'll try talking to my instructor again, hopefully i can force him to explain this to a point where i get it...

thanks for the insight and any further insight,
tb


number theory
greatest common divisor
number theory proposed

Find all $d$ such that

$d\mid n^2+1$

and

$d\mid (n+1)^2+1$

for fix $n\in \mathbb{Z}$


Hi, My name is Catherine and I live in Scarborough area, Toronto. My school, Dr. Norman Bethune C.I. is so not a math school and I basically cannot find anyone in my school on the same level as I do. I think it is much more fun and progressive to work with partners on math problems than totally alone, especially after going to the national camp last week. My goal for next year is to go to CMO, USAMO, APMO and possibly (if I can make it) the winter camp. I'm not a pro in my age, but I think I'm ok. If I work somewhat hard this summer, I'll be able to achieve my goal, and It will be much more interesting to study with a buddy, so if anyone is interested in this idea, contact me at czcat@yahoo.ca

P.S. I live near Warden and McNicoll, so as long as you don't live really far away, like Etobicoke or north of Markham or ext., it'll be easy to work out a place to meet.

Just out of curiousity, what grade level are you at?

Grade 11 next year, not many is as young as you do

Yeah, I am young for this forum. I'm very fortunate to have discovered this site in record time or I wouldn't be as determined to succeed. Witnessing other brilliant minds helps a lot. Being younger does give me an advantage. In my high school years, I'll probably be the expert of this site since everyone else will be in university/college. LOL, besides the G-Unit guy. Have you considered the AoPS books? What type of materials are you planning to explore on?

[quote="G-UNIT"]who is this G-UNIT person you speak of?  he sounds like quite the handsome fellow from your description..........[/quote]
You're weird. I gave no description at all; I only referred to your username.  :P

[quote="p.c"]I believe that is a great idea to have partners. I would like to help u but i live past Etobicoke in Mississauga.[/quote]

I suggest you private message Catherine so she knows who you are, or you can introduce yourself in the Introduction sticky.

[quote="TheDreamer"]Hmm, I've usually had study groups instead of study buddies.  Like just join one of the math forums on this site devoted to problem solving and you can pose questions on there and answer their questions :)

But certainly, it helps to discuss problems with others.[/quote]

Sometimes, I believe it's easier to have actual guidance from someone who you can rely on for math help. Typically, at my level—definitely, problems can be answered quite easily by your fellow peer members. It's harder to actually explain solutions on MathLinks, which is what I've noticed lately. I tend to stick with the human-explaining solutions rather than the ones written on MathLinks. *ha* These are just my random thoughts that I've decided to share.  :lol: (*-__-)

[quote="d.lam.86"]Have you considered the AoPS books? What type of materials are you planning to explore on?[/quote]If she want's to reach USAMO and do well on that (and also on CMO) I would suggest she takes some of the AoPS Olympiad classes and maybe [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=40675]WOOT[/url]   in addition to spending more time in the [url=http://www.mathlinks.ro/Forum/index.php?f=217]Olympiad Forums[/url] ;)

Ok, I've split the WOOT part of the topic, and you can continue discussion about that here: http://www.mathlinks.ro/Forum/viewtopic.php?t=45258


In $\triangle ABC$, $D$ is on $\overline{AB}$ and $E$ is on $\overline{BC}$. Let $F = \overline{AE} \cap \overline{CD}$. $AD = 3, DB = 2, BE = 3$ and $EC = 4$. Find $EF:FA$ in lowest terms.

I used Mass Point but I'm not sure what the answer is.

[hide="Mass Point on my answer"]

$\frac{EF}{FA} = \frac{1}{7}$

Assign weight of 2 to A and 3 to B then because of $nP + mQ = (n+m)R$ for $R \in PQ$, we get that D has weight of 5. Similarly, assign weight of 4 to B and 3 to C and get E to have weight of 7.

Let F has weight of k. Then:

$mA + 7E = kF$ and $nC + 5D = kF$.

Substitution yields:

$mA + 7E = nC + 5D$ and for them to have same weight of $k$, $m+7 = n+5$. I'm not sure if this is allowed but since $n = 3$ and $m = 1$ is the smallest case that satisfy the equation, we will then have:

$1A + 7E = 8F = 3C + 5D$. Since $1A + 4B = 5D$ and putting $4B$ for other equation have $4B+3C = 7E$ is also true, we confirm that our values are correct. Now, since answers are now given, all we have to do is read the answer in ratio.

Since $A:E = 1:7$ for weight, the segments $AF:FE = 7:1$ or $\frac{EF}{FA} = \frac{1}{7}$[/hide]

Is my answer right? Also, what I did there on guessing $m$ and $n$, is that allowed?

You can find the answer without what I did though. I have to see if my answer is right or not first so.

[quote="tarquin"]I got something different
[hide]Apply a pressure of 12 on B and 8 on A. This lets C have a weight of 9. A total of 21 is on E so the ratio of EF to FA is $\frac 8{21}$[/hide][/quote]

same here.

I don't know what I did wrong.

Perhaps $m$ and $n$ stuff was wrong.. I'll have to get pencils to see where I screw up. :(

ive seen similiar problems in Cayley and Fermat competitions.... and they used ratios and stuff to figure out, which took  me a long time to understand. I did it that way and the mass point way.

Use Menelaos' theorem, $\frac{AD}{DB}\cdot \frac{BC}{CE}\cdot \frac{EF}{FA}=1$, yielding $\frac{EF}{FA}=\frac{21}{8}$.

Lots of questions like that can be solved if you change the ratio of segments into the ratio of area of triangles.
In this one, we can do it like:
[hide]
$\frac{FE}{FA}=\frac{\triangle{DEC}}{\triangle{ADC}}=\frac{\triangle{DEC}}{\triangle{BDC}}\cdot \frac{\triangle{BDC}}{\triangle{ADC}}=\frac{4}{7}\cdot \frac{2}{3}=\frac{8}{21}$.[/hide]


geometry
calculus
rotation
LaTeX
3D geometry
sphere
integration

Okay, so I'll just copy the stuff down and post it, and edit in my answers and explanations because they are on about 10 different sheets of paper lying around my room.

I'm self studying Calculus AB because there are not that many new topics between my regular calc class and calc ab, and this way I'll get to take BC next year in school (senior year). My math teacher at school emailed me some problem sets she had, and gave me a book and asked  me to work through them.

1. Find the volume of the solid formed when the region bounded by the curves y = x^3 + 1, x = 1 and y = 0 is rotated about the x-axis.
My answer: int(pi(x^3+1)^2,x,-1,1)
(how do i use a latex input or something like that?)

2. Find the volume of the solid of revolution obtained by revolving the region bounded by y = 1/x and the lines x = pi/8 and x = pi/2 around the x axis the x-axis.
My answer: int(pi(1/x)^2,x,pi/2,pi/8)

3. Find the area bounded by the curves f(x) = x^3 + x^2 and g(x) = 2x^2 + 2x.
int((x^3+1)-(2x^2+2x),x,-1,0) + int((2x^2+2x)-(x^3+1),x,0,2)

4. Find the area bounded by the curves y = sqrt(x), y =  (5-x)/4, and y = (3x-8)/2
int(sqrt(x)-(5x-4)/4,x,1,3) + int(sqrt(x) - (3x-8)/2,x,3,4)

5. Let R be the region bounded by: y = x^3, the tangent to y = x^3 at (1,1), and the x-axis. Find the area of R integrating: a) with respect to x, and b) with respect to y.
Solving for tangent yields tangent = 3x-2
integrating for x yields
int(x^3-(3x-2),x,0,1)
b) Help me with this, not sure how to do at all.

6. Find the volume of the solid generated by revolving the region bounded by the graphs of y = x^2 - 4x + 5 and y = 5 - x about the line y = -1.
disk: int(pi((5-x)^2-(x^2-4x+5)^2),x,0,2)
shell: int(2pix(-x^2+3x),0,3)

7. Use cylindrical shells to find the volume of the solid obtained by revolving around the y-axis the region bounded by the curves y^2 = 8x and x = 2.
int(2pix(2sqrt(2x),x,0,2)

8. A cylindrical hole is drilled through the center of a sphere of radius R. Use the method of cylindrical shells to find the volume of the remaining solid, given that the solid is 6 cm high.
I believe the answer is 36 pi in^3, but can not seem to do this problem with the shell method.

9. Find the volume of the solid generated by revolving about the line x = -1, the region bounded by the curves y = -x^2 + 4x - 3 and y = 0.
10. Consider the region in the xy-plane between x = 0 and x=pi/2 bounded by y = 0 and y = sin x. Find the volume of the solid generated by revolving this region about the x-axis.
11. Let R be the region bounded by y = 1/x, y = x^2, x = 0, and y = 2. Suppose R is revolved around the x-axis. Set up but do not evaluate the integrals for the volume of rotation using: a) the method of cylindrical shells; b) the method of circular disks.

I have just checked your solution.

Japanese Highschool students are not allowed to use the shell method as formula, I dare not to give a solution by the method.

The unexplained problems are left to the others.

1.

$ V_1 \equal{} \pi \int_{ \minus{} 1}^1 (x^3 \plus{} 1)^2\ dx$.

2.

$ V_2 \equal{} \pi \int_{\frac {\pi}{8}}^{\frac {\pi}{2}} \frac {1}{x^2}\ dx$.

3.

$ f(x) \minus{} g(x) \equal{} x^3 \plus{} x^2 \minus{} (2x^2 \plus{} 2x) \equal{} x(x \plus{} 1)(x \minus{} 2)$.

$ \therefore S_3 \equal{} \int_{ \minus{} 1}^2 |f(x) \minus{} g(x)|\ dx$

$ \equal{} \int_{ \minus{} 1}^0 \{f(x) \minus{} g(x)\}\ dx \minus{} \int_0^2 \{f(x) \minus{} g(x)\}\ dx$.

4.

The coordinate of $ x$ of  the irrational function $ y \equal{} \sqrt {x}$ and the line $ y \equal{} \frac {5 \minus{} x}{4}$ can be obtained by $ \sqrt {x} \equal{} \frac {5 \minus{} x}{4}$

$ \Longleftrightarrow (\sqrt {x} \minus{} 1)(\sqrt {x} \plus{} 5) \equal{} 0$, yielding $ x \equal{} 1$, then the coordinate of $ x$ of  the irrational function $ y \equal{} \sqrt {x}$ and the line $ y \equal{} \frac {3x \minus{} 8}{2}$ can be obtained by $ \sqrt {x} \equal{} \frac {3x \minus{} 8}{2}$

$ \Longleftrightarrow (\sqrt {x} \minus{} 2)(3\sqrt {x} \plus{} 4) \equal{} 0$, yielding $ x \equal{} 3$.

$ \therefore S_4 \equal{} \int_1^3 \left(\sqrt {x} \minus{} \frac {5 \minus{} x}{4}\right)\ dx \plus{} \int_3^4 \left(\sqrt {x} \minus{} \frac {3x \minus{} 8}{2}\right)\ dx$.

5.

$ S_5 \equal{} \int_0^1 (x^3 \minus{} 3x \plus{} 2)\ dx$.

then, $ y \equal{} 3x \minus{} 2\Longleftrightarrow x \equal{} \frac {y \plus{} 2}{3}$ and $ y \equal{} x^3\Longleftrightarrow x \equal{} \sqrt [3]{y}$,

$ \therefore S_5' \equal{} \int_{ \minus{} 2}^1\left(\frac {y \plus{} 2}{3} \minus{} \sqrt [3]{y}\right)\ dy$.

6.

The coordinates of $ x$ of two graphs : $ y \equal{} x^2 \minus{} 4x \plus{} 5: \equal{} f(x),\ y \equal{} \minus{} x \plus{} 5: \equal{} l(x)$ are $ x \equal{} 0,\ 3$.

$ \therefore V_6 \equal{} \pi \int_0^3 [\{l(x) \minus{} ( \minus{} 1)\}^2 \minus{} \{f(x) \minus{} ( \minus{} 1)\}^2]\ dx$.

7.

$ V_7 \equal{} 2*\int_0^2 2\pi x\cdot y\ dx \equal{} 2\int_0^2 2\pi x(2\sqrt {2}\sqrt {x})\ dx$.

10.

$ V_{10} \equal{} \pi \int_0^{\frac {\pi}{2}} \sin ^ 2 x\ dx$.

11.

$ V_a \equal{} \pi \int_{\frac {1}{2}}^1 \left(4 \minus{} \frac {1}{x^2}\right)\ dx \plus{} \pi \int_1^{\sqrt {2}} (4 \minus{} x^4)\ dx$.

Thank you for your help, kunny.
I am quite grateful.

I now only need to know if my solution to 7 is correct, and how to go about solving 8-11.

Regrettably, your solution to 7 is incorrect. I have just posted the solution.

[hide="#8 revised"]The volume of a sphere is $ \frac{4\pi r^3}{3}$. We take that, and subtract the cylinder of height six with a radius of $ \sqrt{r^2\minus{}9}$. That is to say we remove a cylindrical core with a volume totaling $ V_{cyl}\equal{}6\pi (r^2\minus{}9)$. In addition we removed two spherical caps that have a total height of $ r\minus{}3$. Using those as the basis for a disk integral on the generating circle for the sphere, we may obtain their volumes with \[ V_{caps}\equal{}2\pi\int^r_3 (r^2\minus{}x^2)\text{ }dx\equal{}\frac{4\pi r^3}{3}\plus{}2\pi(9\minus{}3r^2)\] Thus, the volume of the remaining solid is \[ V_T\equal{}V_s\minus{}V_{cyl}\minus{}V_{caps}\equal{}\frac{4\pi r^3}{3}\minus{}6\pi(r^2\minus{}9)\minus{}\frac{4\pi r^3}{3}\minus{}2\pi(9\minus{}3r^2)\] \[ \equal{}36\pi\][/hide] [hide="#9"]Use disk of the form $ A \equal{} \pi \int^{0}_{ \minus{} 3} [R(y)]^2 \minus{} [r(y)]^2\text{ }dy$, where the R/r denote the $ \pm$ representations of the parabola in $ y$. The big $ R$ is the one further from the axis of revolution.[/hide]

No shells involved.

What did I do wrong in seven? I can see that the actual answer is 2X mine, but I don't know why I'd get that.

So, now I just need to figure out 10/11.

Down to just 11.

I've no idea where to start on this one.

A picture is worth a thousand words. A well labeled, well diagrammed one is even better. On the AP test I believe they will not acknowledge a complimentary image that supports your solution unless it is well labeled (ex. Fig. 1).[hide="#7"]Shells is very natural for this problem. The radius is $ x$, the height is $ y=\sqrt{8x}$, and as always the width of each cylindrical shell is $ \lim_{\Delta x\to 0} \Delta x=dx$. The upper and lower bounds of integration $ (a,b)=(0,2)$. Now we slap it all together into an integral...
\begin{eqnarray*}V&=&4\pi\sqrt{2} \int^2_0 x^{3/2}\text{ }dx\\ &=&\frac{8\pi\sqrt{2}}{5}x^{5/2}\bigg|^2_0=\frac{64\pi}{5}\\\end{eqnarray*}[/hide] [hide="#11"]I can see why this one troubles you. Its definitely ugly. [hide="Disks"]Using disks you must define upper and lower bound functions of the region $ R$, $ f(x)$ and $ g(x)$ respectively over the interval $ (a,b)=I$. If [i]either[/i] one of these functions change at some point $ c\in(a,b)$, the you must define the subintervals $ I_1=(a,c),$ $ I_2=(c,b)$ upon which each has distinct upper bound and lower bound functions of $ R$, $ f_i(x)$ and $ g_i(x)$ respectively.

This being the case, we know there is some point $ c\in(0,1)$ such that $ f(c)=2=1/c \Rightarrow c=1/2$. It's quite clear that the lower-bound of $ R$ throughout $ (0,1)$ is $ g(x)=x^2$. However on the subintervals $ (0,0.5)\text{,  }(0.5, 1)$ the upper-bounds of $ R$ are $ f_1(x)=2$ and $ f_2(x)=1/x$ respectively. I'm sick of being meticulous now so here's the setup: \[ V=\pi\int^{1/2}_0 (4-x^4)\text{ }dx+\pi\int^1_{1/2} (x^{-2}-x^4)\text{ }dx\] which computes to $ 14\pi /5$.[/hide] [hide="Shells"][b]Much[/b] easier! Remember: the height of the cylindrical shells is parallel to the axis of revolution. Observation tells us the radius to each shell is $ y$, its width is $ dy$, and the height is $ x=y^{1/2}\text{ }\forall \text{ }y\in(0,1)$ and $ x=1/y\text{ }\forall \text{ }y\in(1,2)$. (Note: $ \forall$ is read 'for all', and $ \in$ is 'lies in', but can also denote 'belongs to'.) The setup is \[ V=2\pi\int^1_0 y^{3/2}\text{ }dy+2\pi\int^2_1 dy\] which also computes to $ 14\pi /5$.[/hide] [/hide]

Kunny, why is the method of shell integration not taught? Why are you fearful for posting it here?
It seems to be a very integral (no pun intended) part of calculus, and not teaching it to students seems unnecessarily silly.


Solve the inequation: $\frac{\log_{8}x}{\log_{2}(1+2x)}\le\frac{\log_{2}\sqrt[3]{1+2x}}{\log_{2}x}$

In my proof $x\in \left(0;\,\frac{1}{2}\right]\cup (1;\,+\infty)$


[quote="Elemennop"]How would you go about proving that inequality? I tried to work with the given one through homogenization but that didn't really help.[/quote]
Using the inequalilty that Argady posted and making the substotutions he showed we arrive at:
$2(a^{6}+b^{6}+c^{6})+3a^{2}b^{2}c^{2}\ge (a^{3}+b^{3}+c^{3})^{2}$
Expanding, get:
$2a^{6}+2b^{6}+2c^{6}+2a^{2}b^{2}c^{2}\ge a^{6}+b^{6}+c^{6}+2a^{3}b^{3}+2a^{3}c^{3}+2b^{3}c^{3}$
and then substracting from both sides $a^{6}+b^{6}+c^{6}$, we get the given inequality.
I couldn't find any other way to do this inequality...

Yeah, this inequality seems impossible (hard?) using Cauchy or other such elementary techniques.

Yep, I tried for half an hour to AMGM it...

The generalization is very hard.
But the original inequality enough easy.

[quote="Elemennop"](4,1,1) doesn't majorize (3,3,0).[/quote]

I'm sorry  :o

How easy can it be? (if you know Schur)

Schur on $(a^{2},b^{2},c^{2})$ and AM-GM gives $LHS\ge\sum_{sym}a^{4}b^{2}\ge RHS$

Just Schur for $x=a^{2},\ldots$ and AM-GM.